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# NCERT Class 5 Maths Area And Its Boundary Expected Questions
1. Find the perimeter of a triangle whose sides are 3 cm, 4 cm and 5 cm?
2. Find the side of a square whose perimeter is 40 m?
3. Find the cost of fencing a square of park of side 200 m at the rate of 20 rupees per meter.
4. A piece of string is 20 cm long. What will be the length of each side if the string is used to form a square?
5. Find the area of a square plot of side 6m.
6. Find the area of a rectangle whose length and breadth are 10 cm and 8 cm respectively.
7. The area of a rectangular garden is 42 square cm and its length is 7 cm. What is the width of the rectangular garden?
8. Find the area of a rectangle whose sides are 3 cm and 5 cm.
9. The area of a square carom board is 64 square cm. Find the length of each of side?
10. Find the length of a rectangular garden of area 330 square cm if its breadth is 15 cm?
1. Perimeter of a triangle = 3 + 4 + 5 = 12 cm
2. Perimeter of a square = 40 m (given)
Length of each side = 40 / 4 = 10m
3. Perimeter = 4 x side = 4 x 200 = 800 m
Cost of fencing = 800 x 20 = Rs 16000
4. Perimeter = 20 cm (given)
Length of each side of a square = 20 / 4 = 5 cm
5. Length of each side of a square = 6m (given)
Area of a square plot = side x side = 6 x 6 = 36 square m
6. Area of a rectangle = length x breadth
Length = 10 cm
Area = 10 x 8 = 80 square cm
7. Area of a rectangular garden = 42 square cm and length = 7 cm (given).
Area = l x b
Breadth = Area / length = 42 / 7 = 6 cm.
8. Area of a rectangle = l x b
Length =5 cm and breadth = 3 cm (given)
Area = 3 x 5 = 15 square cm
9. Area of a square carom board = 64 square cm
Length of each side = 64/4 = 16 cm
10. Area of a rectangular garden = 330 sq cm
Length = Area / Breadth = 330/15 = 22 cm.
### Expected questions
You can find more practice problems for this chapter below.
Area and Its Boundary Practice Worksheet 1
Area and Its Boundary Practice Worksheet 2
Area and Its Boundary Text book solutions
### 7 Responses
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nice quiz scored totally<|endoftext|>
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# Common Core Arithmetic
Arithmetic is the study of counting, comparison of quantities, number operations and properties of the rational number system. These studies include applications to daily and professional life.
##### Know number names and the count sequence.
1. Count to 100 by ones and by tens.
2. Count forward beginning from a given number within the known sequence (instead of having to begin at 1).
3. Write numbers from 0 to 20. Represent a number of objects with a written numeral 0–20 (with 0 representing a count of no objects).
##### Count to tell the number of objects.
4. Understand the relationship between numbers and quantities; connect counting to cardinality.
a. When counting objects, say the number names in the standard order, pairing each object with one and only one number name and each number name with one and only one object.
b. Understand that the last number name said tells the number of objects counted. The number of objects is the same regardless of their arrangement or the order in which they were counted.
c. Understand that each successive number name refers to a quantity that is one larger.
5. Count to answer “how many?” questions about as many as 20 things arranged in a line, a rectangular array, or a circle, or as many as 10 things in a scattered configuration; given a number from 1–20, count out that many objects.
##### Compare numbers.
6. Identify whether the number of objects in one group is greater than, less than, or equal to the number of objects in another group, e.g., by using matching and counting strategies.1
7. Compare two numbers between 1 and 10 presented as written numerals.
##### Extend the counting sequence.
1. Count to 120, starting at any number less than 120. In this range, read and write numerals and represent a number of objects with a written numeral.
##### Understand place value.
2. Understand that the two digits of a two-digit number represent amounts of tens and ones. Understand the following as special cases:
a. 10 can be thought of as a bundle of ten ones—called a “ten.”
b. The numbers from 11 to 19 are composed of a ten and one, two, three, four, five, six, seven, eight, or nine ones.
c. The numbers 10, 20, 30, 40, 50, 60, 70, 80, 90 refer to one, two, three, four, five, six, seven, eight, or nine tens (and 0 ones).
3. Compare two two-digit numbers based on meanings of the tens and ones digits, recording the results of comparisons with the symbols >, =, and <. Use place value understanding and properties of operations to add and subtract.
4. Add within 100, including adding a two-digit number and a one-digit number, and adding a two-digit number and a multiple of 10, using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a written method and explain the reasoning used. Understand that in adding two-digit numbers, one adds tens and tens, ones and ones; and sometimes it is necessary to compose a ten.
5. Given a two-digit number, mentally find 10 more or 10 less than the number, without having to count; explain the reasoning used.
6. Subtract multiples of 10 in the range 10–90 from multiples of 10 in the range 10–90 (positive or zero differences), using concrete models or drawings and strategies based on place value, properties of operations, and/or the between addition and subtraction; relate the strategy to a written method and explain the reasoning used.
##### Understand place value.
1. Understand that the three digits of a three-digit number represent amounts of hundreds, tens, and ones; e.g., 706 equals 7 hundreds, 0 tens, and 6 ones. Understand the following as special cases:
a. 100 can be thought of as a bundle of ten tens—called a “hundred.”
b. The numbers 100, 200, 300, 400, 500, 600, 700, 800, 900 refer to one, two, three, four, five, six, seven, eight, or nine hundreds (and 0 tens and 0 ones).
2. Count within 1000; skip-count by 2s, 5s, 10s, and 100s. CA
3. Read and write numbers to 1000 using base-ten numerals, number names, and expanded form.
4. Compare two three-digit numbers based on meanings of the hundreds, tens, and ones digits, using >, =, and < symbols to record the results of comparisons.
##### Use place value understanding and properties of operations to add and subtract.
5. Fluently add and subtract within 100 using strategies based on place value, properties of operations, and/or the relationship between addition and subtraction.
6. Add up to four two-digit numbers using strategies based on place value and properties of operations.
7. Add and subtract within 1000, using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a written method. Understand that in adding or subtracting three-digit numbers, one adds or subtracts hundreds and hundreds, tens and tens, ones and ones; and sometimes it is necessary to compose or decompose tens or hundreds.
7.1 Use estimation strategies to make reasonable estimates in problem solving. CA
8. Mentally add 10 or 100 to a given number 100–900, and mentally subtract 10 or 100 from a given number 100–900.
9. Explain why addition and subtraction strategies work, using place value and the properties of operations.
##### Use place value understanding and properties of operations to perform multi-digit arithmetic.
1. Use place value understanding to round whole numbers to the nearest 10 or 100.
2. Fluently add and subtract within 1000 using strategies and algorithms based on place value, properties of operations, and/or the relationship between addition and subtraction.
3. Multiply one-digit whole numbers by multiples of 10 in the range 10–90 (e.g., 9 × 80, 5 × 60) using strategies based on place value and properties of operations.
##### Develop understanding of fractions as numbers.
1. Understand a fraction 1/b as the quantity formed by 1 part when a whole is partitioned into b equal parts; understand a fraction a/b as the quantity formed by a parts of size 1/b.
2. Understand a fraction as a number on the number line; represent fractions on a number line diagram.
a. Represent a fraction 1/b on a number line diagram by defining the interval from 0 to 1 as the whole and partitioning it into b equal parts. Recognize that each part has size 1/b and that the endpoint of the part based at 0 locates the number 1/b on the number line.
b. Represent a fraction a/b on a number line diagram by marking off a lengths 1/b from 0. Recognize that the resulting interval has size a/b and that its endpoint locates the number a/b on the number line.
3. Explain equivalence of fractions in special cases, and compare fractions by reasoning about their size.
a. Understand two fractions as equivalent (equal) if they are the same size, or the same point on a number line.
b. Recognize and generate simple equivalent fractions, e.g., 1/2 = 2/4, 4/6 = 2/3). Explain why the fractions are equivalent, e.g., by using a visual fraction model.
c. Express whole numbers as fractions, and recognize fractions that are equivalent to whole numbers. Examples: Express 3 in the form 3 = 3/1; recognize that 6/1 = 6; locate 4/4 and 1 at the same point of a number line diagram.
d. Compare two fractions with the same numerator or the same denominator by reasoning about their size. Recognize that comparisons are valid only when the two fractions refer to the same whole. Record the results of comparisons with the symbols >, =, or <, and justify the conclusions, e.g., by using a visual fraction model.
##### Generalize place value understanding for multi-digit whole numbers.
1. Recognize that in a multi-digit whole number, a digit in one place represents ten times what it represents in the place to its right. For example, recognize that 700 ÷ 70 = 10 by applying concepts of place value and division.
2. Read and write multi-digit whole numbers using base-ten numerals, number names, and expanded form. Compare two multi-digit numbers based on meanings of the digits in each place, using >, =, and < symbols to record the results of comparisons.
3. Use place value understanding to round multi-digit whole numbers to any place.
Use place value understanding and properties of operations to perform multi-digit arithmetic.
4. Fluently add and subtract multi-digit whole numbers using the standard algorithm.
5. Multiply a whole number of up to four digits by a one-digit whole number, and multiply two two-digit numbers, using strategies based on place value and the properties of operations. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models.
6. Find whole-number quotients and remainders with up to four-digit dividends and one-digit divisors, using strategies based on place value, the properties of operations, and/or the relationship between multiplication and division. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models.2
##### Extend understanding of fraction equivalence and ordering.
1. Explain why a fraction a/b is equivalent to a fraction (n × a)/(n × b) by using visual fraction models, with attention to how the number and size of the parts differ even though the two fractions themselves are the same size. Use this principle to recognize and generate equivalent fractions.
2. Compare two fractions with different numerators and different denominators, e.g., by creating common denominators or numerators, or by comparing to a benchmark fraction such as 1/2. Recognize that comparisons are valid only when the two fractions refer to the same whole. Record the results of comparisons with symbols >, =, or <, and justify the conclusions, e.g., by using a visual fraction model. Build fractions from unit fractions by applying and extending previous understandings of operations on whole numbers.
3. Understand a fraction a/b with a > 1 as a sum of fractions 1/b.
a. Understand addition and subtraction of fractions as joining and separating parts referring to the same whole.
b. Decompose a fraction into a sum of fractions with the same denominator in more than one way, recording each decomposition by an equation. Justify decompositions, e.g., by using a visual fraction model. Examples:
3/8 = 1/8 + 1/8 + 1/8 ; 3/8 = 1/8 + 2/8 ; 2 1/8 = 1 + 1 + 1/8 = 8/8 + 8/8 + 1/8.
c. Add and subtract mixed numbers with like denominators, e.g., by replacing each mixed number with an equivalent fraction, and/or by using properties of operations and the relationship between addition and subtraction.
d. Solve word problems involving addition and subtraction of fractions referring to the same whole and having like denominators, e.g., by using visual fraction models and equations to represent the problem.
4. Apply and extend previous understandings of multiplication to multiply a fraction by a whole number.
a. Understand a fraction a/b as a multiple of 1/b. For example, use a visual fraction model to represent 5/4 as the product 5 × (1/4), recording the conclusion by the equation 5/4 = 5 × (1/4).
b. Understand a multiple of a/b as a multiple of 1/b, and use this understanding to multiply a fraction by a whole number. For example, use a visual fraction model to express 3 × (2/5) as 6 × (1/5), recognizing this product as 6/5 (In general, n × (a/b) = (n × a)/b.)
c. Solve word problems involving multiplication of a fraction by a whole number, e.g., by using visual fraction models and equations to represent the problem. For example, if each person at a party will eat 3/8 of a pound of roast beef, and there will be 5 people at the party, how many pounds of roast beef will be needed? Between what two whole numbers does your answer lie?
##### Understand decimal notation for fractions, and compare decimal fractions.
5. Express a fraction with denominator 10 as an equivalent fraction with denominator 100, and use this technique to add two fractions with respective denominators 10 and 100.4 For example, express 3/10 as 30/100, and add 3/10 + 4/100 = 34/100.2
6. Use decimal notation for fractions with denominators 10 or 100. For example, rewrite 0.62 as 62/100; describe a length as 0.62 meters; locate 0.62 on a number line diagram..
7. Compare two decimals to hundredths by reasoning about their size. Recognize that comparisons are valid only when the two decimals refer to the same whole. Record the results of comparisons with the symbols >, =, or <, and justify the conclusions, e.g., by using the number line or another visual model.
##### Understand the place value system.
1. Recognize that in a multi-digit number, a digit in one place represents 10 times as much as it represents in the place to its right and 1/10 of what it represents in the place to its left.
2. Explain patterns in the number of zeros of the product when multiplying a number by powers of 10, and explain patterns in the placement of the decimal point when a decimal is multiplied or divided by a power of 10. Use whole-number exponents to denote powers of 10.
3. Read, write, and compare decimals to thousandths.
a. Read and write decimals to thousandths using base-ten numerals, number names, and expanded form, e.g., 347.392 = 3 × 100 + 4 × 10 + 7 × 1 + 3 × (1/10) + 9 × (1/100) + 2 × (1/1000).
b. Compare two decimals to thousandths based on meanings of the digits in each place, using >, =, and < symbols to record the results of comparisons.
4. Use place value understanding to round decimals to any place. Perform operations with multi-digit whole numbers and with decimals to hundredths.
5. Fluently multiply multi-digit whole numbers using the standard algorithm.
6. Find whole-number quotients of whole numbers with up to four-digit dividends and two-digit divisors, using strategies based on place value, the properties of operations, and/or the relationship between multiplication and division. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models.
7. Add, subtract, multiply, and divide decimals to hundredths, using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a written method and explain the reasoning used.
##### Use equivalent fractions as a strategy to add and subtract fractions.
1. Add and subtract fractions with unlike denominators (including mixed numbers) by replacing given fractions with equivalent fractions in such a way as to produce an equivalent sum or difference of fractions with like denominators. For example, 2/3 + 5/4 = 8/12 + 15/12 = 23/12. (In general, a/b + c/d = (ad + bc)/bd.)
2. Solve word problems involving addition and subtraction of fractions referring to the same whole, including cases of unlike denominators, e.g., by using visual fraction models or equations to represent the problem. Use benchmark fractions and number sense of fractions to estimate mentally and assess the reasonableness of answers. For example, recognize an incorrect result 2/5 + 1/2 = 3/7, by observing that 3/7 < 1/2. Apply and extend previous understandings of multiplication and division to multiply and divide fractions.
3. Interpret a fraction as division of the numerator by the denominator (a/b = a ÷ b). Solve word problems involving division of whole numbers leading to answers in the form of fractions or mixed numbers, e.g., by using visual fraction models or equations to represent the problem. For example, interpret 3/4 as the result of dividing 3 by 4, noting that 3/4 multiplied by 4 equals 3, and that when 3 wholes are shared equally among 4 people each person has a share of size 3/4. If 9 people want to share a 50-pound sack of rice equally by weight, how many pounds of rice should each person get? Between what two whole numbers does your answer lie?
4. Apply and extend previous understandings of multiplication to multiply a fraction or whole number by a fraction.
a. Interpret the product (a/b) × q as a parts of a partition of q into b equal parts; equivalently, as the result of a sequence of operations a × q ÷ b. For example, use a visual fraction model to show (2/3) × 4 = 8/3, and create a story context for this equation. Do the same with (2/3) × (4/5) = 8/15. (In general, (a/b) × (c/d) = ac/bd.)
b. Find the area of a rectangle with fractional side lengths by tiling it with unit squares of the appropriate unit fraction side lengths, and show that the area is the same as would be found by multiplying the side lengths. Multiply fractional side lengths to find areas of rectangles, and represent fraction products as rectangular areas.
5. Interpret multiplication as scaling (resizing), by:
a. Comparing the size of a product to the size o one factor on the basis of the size of the other factor, without performing the indicated multiplication.
b. Explaining why multiplying a given number by a fraction greater than 1 results in a product greater than the given number (recognizing multiplication by whole numbers greater than 1 as a familiar case); explaining why multiplying a given number by a fraction less than 1 results in a product smaller than the given number; and relating the principle of fraction equivalence a/b = (n × a)/(n × b) to the effect of multiplying a/b by 1.
6. Solve real-world problems involving multiplication of fractions and mixed numbers, e.g., by using visual fraction models or equations to represent the problem.
7. Apply and extend previous understandings of division to divide unit fractions by whole numbers and whole numbers by unit fractions.1
a. Interpret division of a unit fraction by a non-zero whole number, and compute such quotients. For example, create a story context for (1/3) ÷ 4, and use a visual fraction model to show the quotient. Use the relationship between multiplication and division to explain that (1/3) ÷ 4 = 1/12 because (1/12) × 4 = 1/3.
b. Interpret division of a whole number by a unit fraction, and compute such quotients. For example, create a story context for 4 ÷ (1/5), and use a visual fraction model to show the quotient. Use the relationship between multiplication and division to explain that 4 ÷ (1/5) = 20 because 20 × (1/5) = 4.
c. Solve real-world problems involving division of unit fractions by non-zero whole numbers and division of whole numbers by unit fractions, e.g., by using visual fraction models and equations to represent the problem. For example, how much chocolate will each person get if 3 people share 1/2 lb of chocolate equally? How many 1/3-cup servings are in 2 cups of raisins?
##### Apply and extend previous understandings of multiplication and division to divide fractions by fractions.
1. Interpret and compute quotients of fractions, and solve word problems involving division of fractions by fractions, e.g., by using visual fraction models and equations to represent the problem. For example, create a story context for (2/3) ÷ (3/4) and use a visual fraction model to show the quotient; use the relationship between multiplication and division to explain that (2/3) ÷ (3/4) = 8/9 because 3/4 of 8/9 is 2/3. (In general, (a/b) ÷ (c/d) = ad/bc.) How much chocolate will each person get if 3 people share 1/2 lb of chocolate equally? How many 3/4-cup servings are in 2/3 of a cup of yogurt? How wide is a rectangular strip of land with length 3/4 mi and area 1/2 square mi?
##### Compute fluently with multi-digit numbers and find common factors and multiples.
2. Fluently divide multi-digit numbers using the standard algorithm.
3. Fluently add, subtract, multiply, and divide multi-digit decimals using the standard algorithm for each operation.
4. Find the greatest common factor of two whole numbers less than or equal to 100 and the least common multiple of two whole numbers less than or equal to 12. Use the distributive property to express a sum of two whole numbers 1–100 with a common factor as a multiple of a sum of two whole numbers with no common factor. For example, express 36 + 8 as 4 (9+ 2).
##### Apply and extend previous understandings of numbers to the system of rational numbers.
5. Understand that positive and negative numbers are used together to describe quantities having opposite directions or values (e.g., temperature above/below zero, elevation above/below sea level, credits/debits, positive/negative electric charge); use positive and negative numbers to represent quantities in real-world contexts, explaining the meaning of 0 in each situation.
6. Understand a rational number as a point on the number line. Extend number line diagrams and coordinate axes familiar from previous grades to represent points on the line and in the plane with negative number coordinates.
a. Recognize opposite signs of numbers as indicating locations on opposite sides of 0 on the number line; recognize that the opposite of the opposite of a number is the number itself, e.g., –(–3) = 3, and that 0 is its own opposite.
b. Understand signs of numbers in ordered pairs as indicating locations in quadrants of the coordinate plane; recognize red pairs differ only by signs, the locations of the points are related by reflections across one or both axes.
c. Find and position integers and other rational numbers on a horizontal or vertical number line diagram; find and position pairs of integers and other rational numbers on a coordinate plane.
7. Understand ordering and absolute value of rational numbers.
a. Interpret statements of inequality as statements about the relative position of two numbers on a number line diagram. For example, interpret –3 > –7 as a statement that –3 is located to the right of –7 on a number line oriented from left to right.
b. Write, interpret, and explain statements of order for rational numbers in real-world contexts. For example, write –3°C > –7°C to express the fact that –3°C is warmer than –7°C.
c. Understand the absolute value of a rational number as its distance from 0 on the number line; interpret absolute value as magnitude for a positive or negative quantity in a real-world situation. For example, for an account balance of –30 dollars, write |–30| = 30 to describe the size of the debt in dollars.
d. Distinguish comparisons of absolute value from statements about order. For example, recognize that an account balance less than –30 dollars represents a debt greater than 30 dollars.
8. Solve real-world and mathematical problems by graphing points in all four quadrants of the coordinate plane.
##### Understand ratio concepts and use ratio reasoning to solve problems.
1. Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, “The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every 2 wings there was 1 beak.” “For every vote candidate A received, candidate C received nearly three votes.”
2. Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0, and use rate language in the context of a ratio relationship. For example, “This recipe has a ratio of 3 cups of flour to 4 cups of sugar, so there is 3/4 cup of flour for each cup of sugar.” “We paid \$75 for 15 hamburgers, which is a rate of \$5 per hamburger.”1
3. Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.
a. Make tables of equivalent ratios relating quantities with whole number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios.
b. Solve unit rate problems including those involving unit pricing and constant speed. For example, if it took 7 hours to mow 4 lawns, then at that rate, how many lawns could be mowed in 35 hours? At what rate were lawns being mowed?
c. Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent.
d. Use ratio reasoning to convert measurement units; manipulate and transform units appropriately when multiplying or dividing quantities.
##### Apply and extend previous understandings of operations with fractions to add, subtract, multiply, and divide rational numbers.
1. Apply and extend previous understandings of addition and subtraction to add and subtract rational numbers; represent addition and subtraction on a horizontal or vertical number line diagram.
a. Describe situations in which opposite quantities combine to make 0. For example, a hydrogen atom has 0 charge because its two constituents are oppositely charged.
b. Understand p + q as the number located a distance |q| from p, in the positive or negative direction depending on whether q is positive or negative. Show that a number and its opposite have a sum of 0 (are additive inverses). Interpret sums of rational numbers by describing real-world contexts.
c. Understand subtraction of rational numbers as adding the additive inverse, p – q = p + (–q). Show that the distance between two rational numbers on the number line is the absolute value of their difference, and apply this principle in real-world contexts.
d. Apply properties of operations as strategies to add and subtract rational numbers.
2. Apply and extend previous understandings of multiplication and division and of fractions to multiply and divide rational numbers.
a. Understand that multiplication is extended from fractions to rational numbers by requiring that operations continue to satisfy the properties of operations, particularly the distributive property, leading to products such as (–1)(–1) = 1 and the rules for multiplying signed numbers. Interpret products of rational numbers by describing real-world contexts.
b. Understand that integers can be divided, provided that the divisor is not zero, and every quotient of integers (with non-zero divisor) is a rational number. If p and q are integers, then –(p/q) = (–p)/q = p/(–q). Interpret quotients of rational numbers by describing real-world contexts.
c. Apply properties of operations as strategies to multiply and divide rational numbers.
d. Convert a rational number to a decimal using long division; know that the decimal form of a rational number terminates in 0s or eventually repeats.
3. Solve real-world and mathematical problems involving the four operations with rational numbers.1
##### Analyze proportional relationships and use them to solve real-world and mathematical problems.
1. Compute unit rates associated with ratios of fractions, including ratios of lengths, areas and other quantities measured in like or different units. For example, if a person walks 1/2 mile in each 1/4 hour, compute the unit rate as the complex fraction ½/¼ miles per hour, equivalently 2 miles per hour.
2. Recognize and represent proportional relationships between quantities.
a. Decide whether two quantities are in a proportional relationship, e.g., by testing for equivalent ratios in a table or graphing on a coordinate plane and observing whether the graph is a straight line through the origin.
b. Identify the constant of proportionality (unit rate) in tables, graphs, equations, diagrams, and verbal descriptions of proportional relationships.
c. Represent proportional relationships by equations. For example, if total cost t is proportional to the number n of items purchased at a constant price p, the relationship between the total cost and the number of items can be expressed
as t = pn.
d. Explain what a point (x, y) on the graph of a proportional relationship means in terms of the situation, with specialattention to the points (0, 0) and (1, r) where r is the unit rate.
3. Use proportional relationships to solve multistep ratio and percent problems. Examples: simple interest, tax, markups and markdowns, gratuities and commissions, fees, percent increase and decrease, percent error.
##### Know that there are numbers that are not rational, and approximate them by rational numbers.
1. Know that numbers that are not rational are called irrational. Understand informally that every number has a decimal expansion; for rational numbers show that the decimal expansion repeats eventually, and convert a decimal expansion which repeats eventually into a rational number.
2. Use rational approximations of irrational numbers to compare the size of irrational numbers, locate them approximately on a number line diagram, and estimate the value of expressions (e.g.,π 2). For example, by truncating the decimal expansion of √2, show that √2 is between 1 and 2, then between 1.4 and 1.5, and explain how to continue on to get better approximations.<|endoftext|>
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# Mathematics – Class 8 – Chapter 5 – Data Handling- Exercise 5.2 – NCERT Exercise Solution
1. A survey was made to find the type of music that a certain group of young people liked in a city.
Adjoining pie chart shows the findings of this survey. From this pie chart, answer the following:
(i) If 20 people liked classical music, how many young people were surveyed?
(ii) Which type of music is liked by the maximum number of people?
(iii) If a cassette company were to make 1000 CD’s, how many of each type would they make?
Solution:
(i) Let 10% represents 100 people.
So, 20% represents = (100×20)/10 = 200
Hence, 200 people were surveyed.
(ii) Here 40% of the total people surveyed who liked light music and no other form of song. Hence, we can conclude that Light music is liked by the maximum number of people.
(iii)Total CD of classical music = (10 × 1000)/100 = 100
Total CD of semi-classical music = (20 × 1000)/100 = 200
Total CD of light music = (40 × 1000)/100 = 400
Total CD of folk music = (30 × 1000)/100 = 300
2. A group of 360 people were asked to vote for their favorite season from the three seasons rainy, winter and summer.
(i) Which season got the most votes?
(ii)Find the central angle of each sector.
(iii) Draw a pie chart to show this information
Solution:
(i) According to the given table, Winter season got the most votes.
(ii) According to question:
Central angle of summer season= (90×360)/360= 90o
Central angle of rainy season= (120×360)/360= 120o
Central angle of winter season= (150×360)/360= 150o
(iii)
3. Draw a pie chart showing the following information. The table shows the colours preferred by a group of people.
Solution:
Given, central angle = 360o
Total number of people = 36
Now,
4. The adjoining pie chart gives the marks scored in an examination by a student in Hindi, English, Mathematics, Social Science and Science. If the total marks obtained by the students were 540, answer the following questions.
(i) In which subject did the student score 105 marks?
(Hint: for 540 marks, the central angle = 360°. So, for 105 marks, what is the central angle?)
(ii) How many more marks were obtained by the student in Mathematics than in Hindi?
(iii) Examine whether the sum of the marks obtained in Social Science and Mathematics is more than that in Science and Hindi. (Hint: Just study the central angles).
Solution:
(i) The student scored 105 marks in Hindi.
(ii) Marks obtained in Mathematics by students = 135
Marks obtained in Hindi by students = 105
Difference = 135 – 105 = 30
Hence, Student obtained 30 more marks in Mathematics than in Hindi.
(iii) Marks in social science = 97.5
Marks in Mathematics = 225
Marks in Hindi = 105
Now,
The sum of marks in Social Science and Mathematics = 97.5 + 135
= 232.5
The sum of marks in Science and Hindi = 120 + 105 = 225
Hence, the sum of the marks in Social Science and Mathematics is more than that in Science and Hindi.
5. The number of students in a hostel, speaking different languages is given below. Display the data in a pie chart.
Solution:
👍👍👍
error:<|endoftext|>
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to built some excitement around the scientific concept of momentum. We learned that momentum is movement. We learned that the faster something is moving and the heavier it is, the more momentum it has. We also learned that momentum can transfer from one object to another (a rollerblading teacher to a chair on wheels, for example).
Bill Nye did a great job of showing us many examples of momentum, its transfer, and its conservation. After watching the video, we played “Explain that Momentum” with different examples of momentum. Ask your participants about the basketball and the tennis ball as well as the straw and the potato.
At the end of class we observed momentum in action with two types of rockets…a water rocket and a balloon rocket. As the birthday girl, Amanda got to send off the GIANT balloon rocket with the “Happy Birthday” message on it!<|endoftext|>
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Renaissance Art in Italy
Fifteenth-century Italy was unlike any other place in Europe. It was divided into independent city-states, each with a different form of government. Florence, where the Italian Renaissance began, was an independent republic. It was also a banking and commercial capital and, after London and Constantinople, the third-largest city in Europe. Wealthy Florentines flaunted their money and power by becoming patrons, or supporters, of artists and intellectuals. In this way, the city became the cultural center of Europe, and of the Renaissance.
Did You Know?
When Galileo died in 1642, he was still under house arrest. The Catholic Church did not pardon him until 1992.
The New Humanism: Cornerstone of the Renaissance
Thanks to the patronage of these wealthy elites, Renaissance-era writers and thinkers were able to spend their days doing just that. Instead of devoting themselves to ordinary jobs or to the asceticism of the monastery, they could enjoy worldly pleasures. They traveled around Italy, studying ancient ruins and rediscovering Greek and Roman texts.
To Renaissance scholars and philosophers, these classical sources held great wisdom. Their secularism, their appreciation of physical beauty and especially their emphasis on man’s own achievements and expression formed the governing intellectual principle of the Italian Renaissance. This philosophy is known as “humanism.”
Renaissance Science and Technology
Humanism encouraged people to be curious and to question received wisdom (particularly that of the medieval Church). It also encouraged people to use experimentation and observation to solve earthly problems. As a result, many Renaissance intellectuals focused on trying to define and understand the laws of nature and the physical world. For example, Renaissance artist Leonardo Da Vinci created detailed scientific “studies” of objects ranging from flying machines to submarines. He also created pioneering studies of human anatomy. Likewise, the scientist and mathematician Galileo Galilei investigated one natural law after another. By dropping different-sized cannonballs from the top of a building, for instance, he proved that all objects fall at the same rate of acceleration. He also built a powerful telescope and used it to show that the Earth and other planets revolved around the sun and not, as religious authorities argued, the other way around. (For this, Galileo was arrested for heresy and threatened with torture and death, but he refused to recant: “I do not believe that the same God who has endowed us with senses, reason and intellect has intended us to forgo their use, ” he said.)
However, perhaps the most important technological development of the Renaissance happened not in Italy but in Germany, where Johannes Gutenberg invented the mechanical movable-type printing press in the middle of the 15th century. For the first time, it was possible to make books–and, by extension, knowledge–widely available.
Renaissance Art and Architecture
During the Italian Renaissance, art was everywhere. Patrons such as Florence’s Medici family sponsored projects large and small, and successful artists became celebrities in their own right.
Renaissance artists and architects applied many humanist principles to their work. For example, the architect Filippo Brunelleschi applied the elements of classical Roman architecture–shapes, columns and especially proportion–to his own buildings. The magnificent eight-sided dome he built at the Santa Maria del Fiore cathedral in Florence was an engineering triumph–it was 144 feet across, weighed 37, 000 tons and had no buttresses to hold it up–as well as an aesthetic one.
Brunelleschi also devised a way to draw and paint using linear perspective. That is, he figured out how to paint from the perspective of the person looking at the painting, so that space would appear to recede into the frame. After the architect Leon Battista Alberti explained the principles behind linear perspective in his treatise Della Pittura (On Painting), it became one of the most noteworthy elements of almost all Renaissance painting. Later, many painters began to use a technique called chiaroscuro to create an illusion of three-dimensional space on a flat canvas.
The End of the Italian Renaissance
By the end of the 15th century, Italy was being torn apart by one war after another. The kings of England, France and Spain, along with the Pope and the Holy Roman Emperor, battled for control of the wealthy peninsula. At the same time, the Catholic Church, which was itself wracked with scandal and corruption, had begun a violent crackdown on dissenters. In 1545, the Council of Trent officially established the Roman Inquisition. In this climate, humanism was akin to heresy. The Italian Renaissance was over.<|endoftext|>
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A spectacular haul of stone tools discovered beneath a collapsed rock shelter in southern Arabia has forced a major rethink of the story of human migration out of Africa. The collection of hand axes and other tools shaped to cut, pierce and scrape bear the hallmarks of early human workmanship, but date from 125,000 years ago, around 55,000 years before our ancestors were thought to have left the continent.
The artefacts, uncovered in the United Arab Emirates, point to a much earlier dispersal of ancient humans, who probably cut across from the Horn of Africa to the Arabian peninsula via a shallow channel in the Red Sea that became passable at the end of an ice age. Once established, these early pioneers may have pushed on across the Persian Gulf, perhaps reaching as far as India, Indonesia and eventually Australia.
Michael Petraglia, an archaeologist at Oxford University who was not involved in the work, told the Science journal: "This is really quite spectacular. It breaks the back of the current consensus view."
Anatomically modern humans – those that resemble people alive today – evolved in Africa about 200,000 years ago. Until now, most archaeological evidence has supported an exodus from Africa, or several waves of migration, along the Mediterranean coast or the Arabian shoreline between 80,000 and 60,000 years ago.
A team led by Hans-Peter Uerpmann at the University of Tübingen in Germany uncovered the latest stone tools while excavating sediments at the base of a collapsed overhang set in a limestone mountain called Jebel Faya, about 35 miles (55km) from the Persian Gulf coast. Previous excavations at the site have found artefacts from the iron, bronze and neolithic periods, evidence that the rocky formation has provided millennia of natural shelter for humans.
The array of tools include small hand axes and two-sided blades that are remarkably similar to those fashioned by early humans in east Africa. The researchers tentatively ruled out the possibility of other hominins having made the tools, such as the Neanderthals that already occupied Europe and north Asia, as they were not in Arabia at the time.
The stones, a form of silica-rich rock called chert, were dated by Simon Armitage, a researcher at Royal Holloway, University of London, using a technique that measured how long sand grains around the artefacts had been buried. Another strand of the archaeologists' work, described in Science, focused on climate change records and historical sea levels in the area. They show that between 200,000 and 130,000 years ago, a global ice age caused sea levels to fall by up to 100 metres, while the Saharan and Arabian deserts expanded into vast, inhospitable wastelands.
But as the climate warmed at the end of the ice age, fresh rains fell on Arabia, opening up the region to human occupation. "The previously arid interior of Arabia would have been transformed into a landscape covered largely in savannah grasses, with extensive lakes and river systems," said Adrian Parker, a researcher at Oxford Brookes University and co-author of the paper.
The revival of Arabia coincided with record lows in sea level, which left only a shallow stretch of water about three miles wide at the Bab al Mandab Straits separating east Africa and the Arabian peninsula. Uerpmann said early humans may have walked or waded across, but added: "They could have used rafts or boats, which they certainly could make at that time."
The new arrivals would have found good hunting grounds at the end of their journey, with plentiful wild asses, gazelles and mountain ibex, Uerpmann said.
The discovery has sparked debate among archaeologists, some of whom say much stronger evidence is needed to back up the researchers' claims. "I'm totally unpersuaded," Paul Mellars, an archaeologist at Cambridge University, told Science. "There's not a scrap of evidence here that these were made by modern humans, nor that they came from Africa."
Chris Stringer, a palaeontologist at the Natural History Museum in London, said: "The region of Arabia has been terra incognita in trying to map the dispersal of modern humans from Africa during the last 120,000 years, leading to much theorising in the face of few data.
"Despite the confounding lack of diagnostic fossil evidence, this archaeological work provides important clues that early modern humans might have dispersed from Africa across Arabia, as far as the Straits of Hormuz, by 120,000 years ago."<|endoftext|>
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Measles under the microscope: facts & figures about the infectious disease
by Marisabelle Bonnici
Over the past few months, we have heard reports of measles outbreaks all over the world from the Philippines to France and the United States and even in Malta, and it seems to be a growing pandemic. So much so that several American states have declared a state of emergency. There are hundreds of cases recorded and the number keeps on growing every day.
What is measles?
Measles is an infection caused by a virus. It is a very contagious disease that can spread through contact with infected saliva droplets, which means that it can be present in the air for several hours if someone coughs, sneezes or even talks. So imagine an infectious person sneezing in the elevator at your place of work. Anyone riding that elevator for the next two hours could be exposed to this virus.
The measles virus can even live on surfaces for several hours. If you breathe the contaminated air or touch the infected surface then rub your eyes or mouth, you can quickly get infected. Measles is so contagious that if one person has it, up to 90% of the people close to that person who are not immune to it will in all likelihood become infected.
In effect, approximately 9 out of 10 individuals who have not been immunized against measles will get infected if exposed to the virus while before the measles vaccine was introduced in 1963 major epidemics occurred every 2–3 years with this virus causing an estimated 2.6 million deaths each year.
What are the symptoms of measles?
The symptoms of measles will usually appear approximately 14 days after you have been exposed to the virus. Symptoms include:
- red eyes
- muscle aches
- runny nose
- sore throat
- white spots inside the mouth
- light sensitivity
A widespread rash is one of the tell-tale signs of measles as many of the other symptoms can indicate a severe case of the common cold. Generally, the first symptom to appear will be a high fever, as well as red eyes and a runny nose. Then the rash, which usually consists of red, itchy bumps will start erupting. The outbreak will approximately occur around 14 days after exposure to the virus (within a range of 7 to 18 days and it is usually first noticed on the face and upper neck). The rash tends to spread over about 3 days, eventually reaching the hands and feet. The rash lasts for 5 to 6 days, after which it starts fading.
Infected people are considered contagious from about five days before the onset of the rash to four days afterwards, while a sick individual is considered to be most contagious when their cough is at its worst.
Who is most at risk of getting measles?
Measles primarily occurs in unvaccinated children but is not exclusive to them. The following are segments of the population considered at most risk of developing the disease:
- Babies who are unvaccinated because they are too young: The measles vaccine cannot be given to infants because their immune systems would not yet be able to provide the appropriate immunologic response to the vaccine.
- Being incompletely vaccinated: Those who haven't received a second booster dose of MMR do not have full immunity to the disease. This means that you could have given your child the first shot, but he or she comes in contact with the virus before the second shot is given. The first vaccine is around 93 percent effective, but the second one is 97 percent effective.
- People who are unvaccinated for medical reasons: Some people are unable to get vaccinated for several reasons including pregnancy, taking certain drugs like cancer chemotherapy or high doses of steroids, as well as those who suffer from conditions that can cause bleeding or bruising. Those patients who have had a life-threatening allergic reaction to gelatin, a medication called neomycin or a previous MMR vaccine should not get the vaccine.
- Being fully vaccinated but have not developed immunity: This is an infrequent occurrence that happens in approximately 3 percent of vaccinated people.
- Immunocompromised people: Even if you have previously been vaccinated, being immunocompromised due to conditions such as HIV/Aids/Cancer means that you are at a higher risk of getting the infection.
- Vitamin A deficiency: this is another risk factor. Children with too little vitamin A in their diets run a much higher risk of getting measles.
How is measles diagnosed?
If you're suspecting that you might have measles, you will need to see a doctor to properly diagnose you. The medical professional will typically examine your skin rash closely and check if you have any other symptoms that are characteristic of the disease, such as white spots in the mouth, fever, runny nose, muscle pains and sore throat.
If your doctor is unable to confirm a diagnosis based on observation, then blood tests can be ordered. Your doctor will guide you as to the proper management needed depending on the severity of your symptoms.
What is the treatment for measles?
There is no prescription medication to treat the infectious disease. The virus and symptoms will typically disappear within two to three weeks. However, your doctor may recommend:
- paracetamol to reduce fever and muscle aches
- drinking plenty of fluids (approximately two litres of water a day)
- electrolytes that will replace those lost due to the illness
- rest to help boost your immune system
- vitamin A supplements
- healthy food to boost your immunity
- immune system boosting herbs such as echinacea
- use of a humidifier to relieve cough and sore throat
- time off to rest your eyes - if you or your child find bright light bothersome keep the lights low or wear sunglasses. Also, avoid reading or watching television or spending too much time on the computer or tablet
- antibiotics could be prescribed in the case of complications such as bronchitis but will not have any effect on the measles virus itself
- probiotics can help if diarrhoea is a side effect
Are there any complications associated with measles?
Truth be told, measles can lead to several life-threatening complications, such as pneumonia and inflammation of the brain (encephalitis) amongst other more minor complications which may include:
- explosive diarrhoea
- bronchitis and other respiratory conditions
- miscarriage or preterm labour or even conditions such as blindness in the unborn child, depending on the stage the pregnancy has come to
- ear infection
Can you die if you get the measles?
The death rate in healthy children and adults is relatively low and the majority of people who do get infected make a full recovery. The risk of complications is higher in children and adults with a weak immune system as discussed above. However, measles can still cause a large number of deaths in children under the age of 5. In fact, the WHO reported that in 2017, there were 110, 000 measles deaths globally mostly among children under the age of five.
Every death due to a vaccine-preventable disease is a death too many and unfortunately, these numbers are on the rise in 2018 and 2019 as we are seeing many more cases.
How can measles be prevented?
The best way to completely prevent measles is by getting vaccinated. The MMR vaccine is a three-in-one vaccination that can protect you and your children from measles, mumps and rubella (German measles).
There is another vaccine also available, the four in one, which includes the chicken pox as well.
Children can receive their first MMR vaccination at 12 months or sooner if travelling internationally but this should be discussed with your child's paediatrician. The second dose is generally given between the ages of 3 and 4 and can be given up to 6 years of age. Adults who have never received an immunization can request the vaccine from their doctor and if you're considering getting pregnant but you're unvaccinated, it is highly recommended that you do get the vaccine beforehand.
If you or a family member contracts the measles virus, limit interaction with others. This includes staying home from school or work and avoiding social activities.
Proper hygiene is always recommended by doing the following:
- make sure you're washing your hands regularly with soap and water. If these are not available, use an alcohol rub to keep your hands free from microbes.
- do not pick your nose
- cover your mouth with your arm, sleeve or tissue instead of the palm of your hand to prevent spreading the virus
- do not share utensils or drink from someone else's glass or cup
- do not go to work or send your children to school if you think that you or they are developing contagious symptoms
Why are we seeing new cases now even though the vaccine is available?
In recent years, many people have lost their trust in the pharmaceutical and medical industry. This phenomenon is linked to the increasing influence of social media and celebrities over and above that of medical professionals. First of all, we are seeing people taking to Facebook groups, bloggers and Google itself for their medical questions, instead of consulting and discussing the issue with a qualified health care professional.
I am sure you have all come across anti-vaccination headlines on Facebook—like "HPV vaccine leaves another 17-year-old-girl paralyzed"— and "Mom researches vaccines, discovers vaccination horrors, goes vaccine-free", to name a few. These are all fake news articles available on the internet and it is important that you learn to distinguish between reputable sights and fake sights when it comes to issues like these. If you come across something along these lines, avoid sharing it on your social media and show it instead to a healthcare professional you trust who will be able to guide you and give you proper medical information.
In the past month alone I have come across several such posts on social media. I am sure that out of curiosity you would have thought of opening these articles and reading them. What's the harm, right? As a community pharmacist for the past decade and with a masters thesis in pharmacoepidemiology I have seen the damage that fraudulent science and unfounded claims can cause. I have spoken to highly educated mothers who were terrified of vaccinating their children following myths they would have read on the internet and to mothers of children with Leukemia who were putting them at further danger simply because they come in contact with individuals who have not been vaccinated for several conditions. For these children, even Influenza can be a huge threat and by vaccinating your healthy child you are not only protecting them from harm, but you are protecting other children as well.
The vaccine-autism myth is one overwhelming example of fraudulent science. February 28, 2018, marks the 20th anniversary of an article published in the medical journal, The Lancet, in which Andrew Wakefield, a former British doctor, falsely linked the MMR (measles, mumps and rubella) vaccine to autism. He did this for his gain with complete disregard to what this fake information would eventually cause. This study was ultimately retracted by the journal, and public apologies issued after millions of euros and thousands of cases were studied. Irrefutable proof was presented that Wakefield falsified the results and had also huge conflicts of interest with this study that would lead to monetary gains for him. Andrew Wakefield lost his medical licence for his deceit. Yet, 20 years later we are still dealing with the aftermath of this fraudulent article.
To prevent measles from spreading, we need to ensure that a minimum of 92 to 95 percent of the population gets immunized. So, we all have a joint responsibility at making sure that we get the right information out there. Do not share fake news on your social media profiles or public groups as this can cause confusion. Keep in mind that vaccines not only protect us from the virus but they protect others, especially infants and immunosuppressed people who cannot get the vaccine.
Does this mean that there are no side effects to the vaccine?
The MMR vaccine is very safe and it has been given safely to millions of children over the past forty years. It is highly likely that you have been vaccinated yourself when you were younger. The majority of side effects following vaccination are mild and short-lived and consist of the following:
Common side effects of the MMR vaccine
Fever and a mild rash as well some minor inflammation at the site of injection which usually goes down within 5 - 8 days are relatively common side effects. However, if you or your child are experiencing side effects that are concerning you, speak to your doctor.
Rare side effects of the MMR vaccine
In rare cases, a child may get a small, bruise-like spot and joint stiffness about two weeks after having the vaccine. There's a minimal chance of having a seizure (fit) around 6 to 11 days after getting vaccinated. I know that this may sound scary, however, this is rare and will happen in only about 1 in every 1,000 doses. MMR-related seizures are actually much less frequent than seizures that occur as a direct result of getting the measles infection.
Extremely rare side effects
In extremely rare cases, a child can have a severe allergic reaction (known medically as anaphylaxis) immediately after having the MMR vaccine. I understand that this may sound alarming, but this is why vaccines are administered by paediatricians who are trained in treating children quickly and effectively. Anaphylaxis can happen even with certain foods like nuts and shellfish if the child is predisposed to having such allergic reactions and that sort of allergic reaction can be more dangerous as it will typically not happen in a doctor's office.
The bottom line
The world is currently experiencing one of the worst measles outbreaks in recent history. In order for the highly contagious virus to be contained, more people must get vaccinated and limit exposure to the disease. If you will be travelling with a young unvaccinated child, please make sure you have a discussion with your doctor who will advise on what to do. Have a look at these paediatricians and general practitioners found in Malta & Gozo.
Keep on discovering local with Yellow!<|endoftext|>
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# Thread: Insect population growth problem
1. ## Insect population growth problem
An insect population is increasing exponentially. 4 days after the population was established there are 400 insects, and after 3 more days, there are 1600 insects. How many insects will there be 10 days after the population was established? Thank you in advance.
2. Originally Posted by lucky13
An insect population is increasing exponentially. 4 days after the population was established there are 400 insects, and after 3 more days, there are 1600 insects. How many insects will there be 10 days after the population was established? Thank you in advance.
Use the model $P(t) = P_0e^{rt}$ and find $P_0$ and $r$ via substitution of $(4,400)$ and $(7,1600)$
After this find $P(10)$ and you will have the answer.
3. Im not sure I understand, Where do I substitute those numbers to?
4. Originally Posted by lucky13
Im not sure I understand, Where do I substitute those numbers to?
In this case (4,400) = (t,P(t))
We get $400 = P_0e^{4r}$
Do the same for the other point.
What do you get?
5. 1600=Poe^7r?
6. Originally Posted by lucky13
1600=Poe^7r?
Yep, now solve your equation simultaneously with
$400 = P_0e^{4r}$
7. Solve it simultaneusly?
8. Originally Posted by lucky13
Solve it simultaneusly?
You have...
$P_0 e^{7r} = 1600$
$P_0 e^{4r} = 400$
Divide the top equation by the bottom one.
$\frac{P_0 e^{7r}}{P_0 e^{4r}} = \frac{1600}{400}$
=> $e^{(7-4)r} = 4$
=> $r = \frac{2}{3}\ln(2)$
Substitute that into $P_0 e^{4r} = 400$ to get...
$P_0 \cdot 4 \cdot 2^{2/3} = 400$
=> $P_0 = 50 \cdot 2^{1/3}$
You can simplify stuff if ya like, I've left it all in a kind of messy form.
Then solve...
$P_{10} = P_0 e^{10r}$
You have...
$P_0 e^{7r} = 1600$
$P_0 e^{4r} = 400$
Divide the top equation by the bottom one.
$\frac{P_0 e^{7r}}{P_0 e^{4r}} = \frac{1600}{400}$
=> $e^{(7-4)r} = 4$
=> $r = \frac{2}{3}\ln(2)$
Substitute that into $P_0 e^{4r} = 400$ to get...
$P_0 \cdot 4 \cdot 2^{2/3} = 400$
=> $P_0 = 50 \cdot 2^{1/3}$
You can simplify stuff if ya like, I've left it all in a kind of messy form.
Then solve...
$P_{10} = P_0 e^{10r}$
$P_{10} =50 \cdot 2^{1/3}e^{10(\frac{2}{3}\ln(2))}$
$=50 \cdot 2^{1/3}(e^{ln(2)})^{\frac{20}{3}}$
$=50 \cdot 2^{1/3}(2)^{\frac{20}{3}}$
$=50 \cdot 2^7$
$=6400$<|endoftext|>
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# Rational Numbers Between Two Unequal Rational Numbers
As we know that rational numbers are the numbers which are represented in the form of p/q where ‘p’ and ‘q’ are integers and ‘q’ is not equal to zero. So, we can call rational numbers as fractions too. So, in this topic we will get to know how to find rational numbers between two unequal rational numbers.
Let us suppose ‘x’ and ‘y’ to be two unequal rational numbers. Now, if we are told to find a rational number lying in the mid- way of ‘x’ and ’y’, we can easily find that rational number by using the below given formula:
$$\frac{1}{2}$$(x + y), where ‘x’ and ‘y’ are the two unequal rational numbers between which we need to find the rational number.
Rational numbers are ordered, i.e., given two rational numbers x, y either x > y, x < y or x = y.
Also, between two rational numbers there are infinite number of rational numbers.
Let x, y (x < y) be two rational numbers. Then
$$\frac{x + y}{2}$$ - x = $$\frac{y - x}{2}$$ > 0; Therefore, x < $$\frac{x + y}{2}$$
y - $$\frac{x + y}{2}$$ = $$\frac{y - x}{2}$$ = $$\frac{y - x}{2}$$ > 0; Therefore, $$\frac{x + y}{2}$$ < y.
Therefore, x < $$\frac{x + y}{2}$$ < y.
Thus, $$\frac{x + y}{2}$$ is a rational number between the rational numbers x and y.
For understanding it much better let us have a look at some of the below mentioned examples:
1. Find a rational number lying mid- way between $$\frac{-4}{3}$$ and $$\frac{-10}{3}$$.
Solution:
Let us assume x = $$\frac{-4}{3}$$
y = $$\frac{-10}{3}$$
If we try to solve the problem using formula mentioned above in the text, then it can be solved as:
$$\frac{1}{2}$${( $$\frac{-4}{3}$$)+ ($$\frac{-10}{3}$$)}
⟹ $$\frac{1}{2}$${( $$\frac{-14}{3}$$)}
⟹ $$\frac{-14}{6}$$
⟹ $$\frac{-7}{6}$$
Hence, ($$\frac{-7}{6}$$) or ($$\frac{-14}{3}$$) is the rational number lying mid- way between $$\frac{-4}{3}$$and $$\frac{-10}{3}$$.
2. Find a rational number in the mid- way of $$\frac{7}{8}$$ and $$\frac{-13}{8}$$
Solution:
Let us assume the given rational fractions as:
x = $$\frac{7}{8}$$,
y = $$\frac{-13}{8}$$
Now we see that the two given rational fractions are unequal and we have to find a rational number in the mid- way of these unequal rational fractional. So, by using above mentioned formula in the text we can find the required number. Hence,
From the given formula:
$$\frac{1}{2}$$(x + y) is the required mid- way number.
So, $$\frac{1}{2}$${ $$\frac{7}{8}$$+ ($$\frac{-13}{8}$$)}
⟹ $$\frac{1}{2}$$( $$\frac{-6}{8}$$)
⟹ $$\frac{-6}{16}$$
⟹ ($$\frac{-3}{8}$$)
Hence, ($$\frac{-3}{8}$$) or ($$\frac{-6}{16}$$) is the required number between the given unequal rational numbers.
In the above examples, we saw how to find the rational number lying mid- way between two unequal rational numbers. Now we would see how to find a given amount of unknown numbers between two unequal rational numbers.
The process can be better understood by having a look at following example:
1. Find 20 rational numbers in between ($$\frac{-2}{5}$$) and $$\frac{4}{5}$$.
Solution:
To find 20 rational numbers in between ($$\frac{-2}{5}$$) and $$\frac{4}{5}$$, following steps must be followed:
Step I: ($$\frac{-2}{5}$$) = $$\frac{(-2) × 5}{5 × 5}$$ = $$\frac{-10}{25}$$
Step II: $$\frac{4 × 5}{5 × 5}$$ = $$\frac{20}{25}$$
Step III: Since, -10 < -9 < -8 < -7 < -6 < -5 < -4 ...… < 16 < 17 < 18 < 19 < 20
Step IV: So, $$\frac{-10}{25}$$ < $$\frac{-9}{25}$$ < $$\frac{-8}{25}$$ < …… < $$\frac{16}{25}$$ < $$\frac{17}{25}$$ < $$\frac{18}{25}$$ < $$\frac{19}{25}$$.
Step V: Hence, 20 rational numbers between $$\frac{-2}{5}$$ and $$\frac{4}{5}$$ are:
$$\frac{-9}{25}$$, $$\frac{-8}{25}$$, $$\frac{-7}{25}$$, $$\frac{-6}{25}$$, $$\frac{-5}{25}$$, $$\frac{4}{25}$$ ……., $$\frac{2}{25}$$, $$\frac{3}{25}$$, $$\frac{4}{25}$$, $$\frac{5}{25}$$, $$\frac{6}{25}$$, $$\frac{7}{25}$$, $$\frac{8}{25}$$, $$\frac{9}{25}$$, $$\frac{10}{25}$$.
All the questions of this type can be solved using above steps.
Rational Numbers
Rational Numbers
Decimal Representation of Rational Numbers
Rational Numbers in Terminating and Non-Terminating Decimals
Recurring Decimals as Rational Numbers
Laws of Algebra for Rational Numbers
Comparison between Two Rational Numbers
Rational Numbers Between Two Unequal Rational Numbers
Representation of Rational Numbers on Number Line
Problems on Rational numbers as Decimal Numbers
Problems Based On Recurring Decimals as Rational Numbers
Problems on Comparison Between Rational Numbers
Problems on Representation of Rational Numbers on Number Line
Worksheet on Comparison between Rational Numbers
Worksheet on Representation of Rational Numbers on the Number Line<|endoftext|>
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The balancing bird is a toy that has its center of gravity located at the tip of the beak. The center of gravity (also called the center of mass) is a special point on an object. It's the point at which the mass of the body is perfectly balanced. For example, the center of gravity of a pencil is in the middle. So you can hold a pencil up using just one finger if you place it underneath the middle of the pencil. Similarly, you can hold the bird up using one finger as well, if you place it underneath the beak.
Check out a video of the bird:
It might seem strange that the bird is perfectly balanced at the beak location. But this is because the bird is designed this way. The wings are extended in front of the beak far enough, and made heavy enough, to balance the weight of the bird behind the beak.
In most objects it is not clear where the center of gravity is. If an object has a strange shape then you cannot know by looking at it where the center of gravity is. If an object has a uniform shape, like a pencil, or a square, then the center of mass is easy to determine. It's in the center of the object. But for objects with non-uniform shapes this is not known ahead of time. To find the center of gravity for these objects you need to use trial and error. You must try balancing the object on different points until you find the point where the object is perfectly balanced, and doesn't fall over. This point, once found, is the center of gravity.
Return to Science Toys page
Return to Real World Physics Problems home page<|endoftext|>
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On this page...
Ascidians belong to the subphylum Urochordata - one of the major groups of the phylum Chordata, which includes the vertebrates (fishes, amphibians, reptiles, birds and mammals). Although you don't look much like a sea squirt now, during development before you were born, you had the same characteristics present in all chordates at some stage of their life
These characteristics of chordates include:
- a nerve cord along the back of the body
- a 'notochord' or firm rod of cells beneath the nerve cord (this is your backbone)
- gill slits (they have disappeared now in some chordates, but were present during evolutionary development).
Ascidians are an evolutionary link between invertebrates and vertebrates. They have a primitive backbone at some stage of their life cycle, but in other aspects they resemble invertebrates.
Most ascidians are hermaphrodites (produce both eggs and sperm) and reproduce by external fertilisation (releasing eggs and sperm into the water). The free-swimming larva they produce are known as ascidian tadpoles. After a few hours, the 'tadpoles' secrete slime, and attach themselves to a rock surface head-first, and then absorb their tail. Adult ascidians are 'sessile' (unable to move around) and filter food particles from the water by pumping water in one siphon and out the other.
Their common name of sea squirt arises from their habit of squirting a jet of water when you stand on or near them when they are uncovered at low tide.<|endoftext|>
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The term ‚profit’ stands for the difference between revenue and costs. However, for one and the same activity, profit does not necessarily have to be the same number under different points of view. Different accounting standards or special regulations for taxation make organizations display different profits in financial statements for different purposes. On top of that, profit from the accountant’s point of view is not equal to profit from the economist’s point of view. This difference is based in fundamentally different understandings of costs and profits. This article describes the differences of profit in accounting and economics.
Approaches to Profit Maximization
The sciences of business management and economics have different approaches to profit maximization, due to their different subjects.
Managerial accounting is a discipline of business management. The subject of business management is the business activities and the development of economic procedures for achieving particular objectives in every single organization or business unit. Economics, on contrast, deals with the characteristics of the economic relationships between all business units that are linked in an economy; hence, it looks at the total economy.
Each individual business focuses its attention on factors like cost structures (e.g. ratio between costs of material, costs of labor, and overhead costs), the organization of production processes, working hours etc. The businesses management decisions influence these factors. In order to improve profitability, businesses can improve their individual cost basis, for instance by improving their production organization.
Contrary, economists consider these managerial costs as fixed data. They only look at the total level of these factors for the whole economy. They presume that all individual businesses operate at an optimal cost structure. From an economic view, production processes and the resulting cost structures cannot be changed (at least in the short term).
Economic data like pricing of factors, needs and demand structures, the impact of technological progress on useful lifes of assets, are fixed data for every single business unit from a managerial point of view. A single organization cannot influence these factors. It just has to consider them when planning and carrying out their activities.
Economists, however, analyze the impact of these economic factors on the whole economy.
This results in different approaches for profit maximization:
From the managerial view, increases in profits can be achieved by measures for reducing costs or increasing revenues. These could be rationalization programs (e.g. investment in equipment with higher efficiency, reorganization of process, standardization), measures for quality improvement or marketing activities for increasing sales.
In economics, total profit is maximized by increasing the volume of production and sales up to that point in which marginal costs of one additional unit produced and sold are equal to the marginal revenue generated with this additional unit. (Marginal costs / revenues are those costs / revenues are the additional costs / revenues that occur with every single additional unit.) Production systems and processes remain unchanged.
From an economical point of view, maximal profit does not result from an optimal and cost-effective design of all organizational processes, but from production and sales at the optimal level in the existing organizational situation.
The Scope of Costs
In both – accounting and economics – applies:
Profit = Total Revenue – Total Costs
However, the scope of the term ‚costs’ is wider in economics than in accounting. Both disciplines take into account all costs the organization actually incurs. (For ease of understanding, we will neglect neutral costs that are not directly related to the operations of the organization here.)
Accounting structures costs according to their cause. It distinguishes for instance costs of material, labor, depreciation, interests and other expenses. The economist distinguishes costs as fixed and variable costs by their dependency from production volume.
The term cost in a managerial sense includes all costs that are related with the production and selling of the organizations goods and services. In the simplest model, the company’s revenues less the costs that are incurred by producing and selling the goods and services sold equal profit (or loss).
The owners of the company decide how to use this profit. Profits can be distributed to the owners, or they can be left in the organization to finance further investments. The distribution of profits to the owners provides them a return for the financial means they invested into the company. Hence, costs in a managerial sense do not include interests for the owners’ investments into the company. These interests have to be paid from profit.
The economical scope of costs exceeds this managerial view. Due to the concept of opportunity costs, costs in an economical sense include an adequate interest return on the capital invested by the owners. Opportunity costs are the costs or compensation for not exploited alternative opportunities. Thus, economics take into considerations that the owners could invest their money differently than into this particular organization (e.g. bonds, funds, property etc). As a compensation for the (secure) interest returns the owners would get from these alternative investments, the economist allocates an adequate interest return on the capital invested into a company to the owners. Hence, costs in an economical sense include all those costs that are necessary to keep the company in business – including a compensation for owners so that they do not invest their money in more lucrative investments.
Whereas in business management returns to owners have to be paid from profits, the economical scope of profit already includes an adequate interest on capital invested by the owners. What is adequate is determined by lost interests the owners would have earned from investing their money elsewhere.
Profit is calculated as total revenues less total costs in both – accounting and economics. Nevertheless, both sciences have a different definition for the term profit. These arise from different definitions and scopes of ‘costs’:
|Ways to increase profit||Optimize cost structures by optimizing organization and structure of production processes||Increase the volume of production up to that point in which marginal costs of one additional unit produced and sold are equal to the marginal revenue generated with this additional unit.|
|Factors that cannot be influenced||Supply and demand structures, factors determining cost structures at the level of total economy||Organization and structure of businesses production processes|
|Returns for capital invested by owners||Not considered a cost component, to be paid from profit||Adequate interest return is determined by opportunity costs and included in total costs|<|endoftext|>
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The Tonkawa are a Native American tribe indigenous to present-day Texas. They once spoke the now-extinct Tonkawa language, a language isolate. Today, many descendants are enrolled in the federally recognized Tonkawa Tribe of Indians of Oklahoma.
Seal of the Tonkawa Tribe of Oklahoma
|Regions with significant populations|
|United States ( Oklahoma)|
|English, Tonkawa language|
|Christianity, Native American Church, traditional tribal religions|
|Related ethnic groups|
|Wichita, Waco, Tawakoni, Kichai, Guichita|
In the 16th century, the Tonkawa tribe probably had around 1,879 members with their numbers diminishing to around 1,600 by the late 17th century due to fatalities from new infectious diseases and conflict with other tribes, most notably the Apache. By 1921, only 34 tribal members remained. Their numbers have since recovered to close to 700 in the early 21st century. Most live in Oklahoma.
The Tonkawa tribe operates a number of businesses which have an annual economic impact of over $10,860,657 (as of 2011). Along with several smoke shops, the tribe runs 3 different casinos: Tonkawa Indian Casino and Tonkawa Gasino located in Tonkawa, Oklahoma, and the Native Lights Casino in Newkirk, Oklahoma.
The annual Tonkawa Powwow is held on the last weekend in June to commemorate the end of the tribe's own Trail of Tears when the tribe was forcefully removed and relocated from its traditional lands to present-day Oklahoma.
Scholars once thought the Tonkawa originated in Central Texas. Recent research, however, has shown that the tribe inhabited northwestern Oklahoma in 1601. By 1700, the stronger and more aggressive Apache had pushed the Tonkawa south to the Red River which forms the border between current-day Oklahoma and Texas.
In 1824, the Tonkawa entered into a treaty with Stephen F. Austin to protect Anglo-American immigrants against the Comanche. At the time, Austin was an agent recruiting immigrants to settle in the Mexican state of Coahuila y Texas. In 1840 at the Battle of Plum Creek and again in 1858 at the Battle of Little Robe Creek, the Tonkawa fought alongside the Texas Rangers against the Comanche.
In 1859, the United States escorted the Tonkawa and a number of other Texas Indian tribes to a new home at the Wichita Agency in Indian Territory, and placed them under the protection of nearby Fort Cobb. When the American Civil War started, the troops at the fort received orders to march to Fort Leavenworth, Kansas, leaving the Indians at the Wichita Agency unprotected.
In response to years of animosity (in part regarding rumors that the Tonkawas engaged in cannibalism ), a number of pro-Union tribes, including the Delawares, Wichitas, and Penateka Comanches, attacked the Tonkawas as they tried to escape. The fight, known as the Tonkawa Massacre killed nearly half of the remaining Tonkawas, leaving them with little more than 100 people. The tribe returned to Fort Griffin, Texas where they remained for the rest of the Civil War. In October, 1884, the United States removed them, once again, to the new Oakland Agency in northern Indian Territory, where they remain to this day. This journey involved going to Cisco, Texas, where they boarded a railroad train that took them to Stroud in Indian Territory, where they spent the winter at the Sac and Fox Agency. The Tonkawas travelled 100 miles (160 km) to the Ponca Agency, and arrived at nearby Fort Oakland on June 30, 1885.[a]
The Tonkawa Tribe of Oklahoma incorporated in 1938.
The Tonkawa were actually made up of various groups, many of which are no longer known by name. These groups are generally counted as Tonkawa:
- 2011 Oklahoma Indian Nations Pocket Pictorial Directory. Archived 2012-04-24 at the Wayback Machine Oklahoma Indian Affairs Commission. 2011: 36. Retrieved 8 Feb 2012.
- International encyclopedia of linguistics. Frawley, William, 1953- (2nd ed.). New York, NY: Oxford University Press. 2003. ISBN 9780195307450. OCLC 66910002.CS1 maint: others (link)
- Hoijer, Harry (1933). Tonkawa, an Indian language of Texas. University of Pittsburgh Library System. New York : Columbia University Press.
- May, Jon D. Encyclopedia of Oklahoma History and Culture. "Tonkawa." Retrieved May 30, 2013."Archived copy". Archived from the original on 2012-02-21. Retrieved 2012-02-21.CS1 maint: Archived copy as title (link)
- Oklahoma Indian Casinos: Kay County. 50 Nations. (retrieved 8 Feb 2009)
- Tonkawa Tribal History. Archived 2009-03-04 at the Wayback Machine The Tonkawa Tribe. (retrieved 7 Feb 2009)
- May, Jon D. "Tonkawa" Archived 2012-02-21 at the Wayback Machine, Encyclopedia of Oklahoma History & Culture, Tulsa: Oklahoma Historical Society (retrieved 8 Feb 2009)
- Gary Clayton Anderson, The Indian Southwest, 1580-1830: Ethnogenesis and Reinvention (Norman: University of Oklahoma Press, 1999) p. 85
- Anderson, The Indian Southwest, p. 89
- Walker, Jeff (2007-11-16). "Chief returns » Local News » San Marcos Record, San Marcos, TX". Sanmarcosrecord.com. Archived from the original on 2011-09-27. Retrieved 2011-11-11.
- Gwynne, S. C. (2011). Empire of the Summer Moon: Quanah Parker and the Rise and Fall of the Comanches, the Most Powerful Indian Tribe in American History. Scribner. pp. 7, 211. ISBN 1-4165-9106-0.
- Barnes, Michael "With time, the story of the Tonkawa tribe evolves" The Washington Post (February 13, 2014)
- Jones, William K. 1969. “Notes on the History and Material Culture of the Tonkawa Indians.” Smithsonian Contributions to Anthropology. Vol. 2, No. 5.
- | Jerry Withers "The Tonkawan Indians of Texas" reprinted by SonsofDewittcolony.org)
- : John D. May "Tonkawa Massacre" OKHistory.org
- "Tonkawa Tribe of Indians of Oklahoma." Oklahoma State Department of Education. Oklahoma Indian Tribe Education Guide. July 2014. Accessed October 28, 2018.
- Deloria Jr., Vine J; DeMaille, Raymond J (1999). Documents of American Indian Diplomacy Treaties, Agreements, and Conventions, 1775-1979. University of Oklahoma Press. pp. 346–348. ISBN 978-0-8061-3118-4.
- Himmel, Kelly F. (1999). The conquest of the Karankawas and the Tonkawas, 1821-1859. College Station, Texas: Texas A&M University Press. ISBN 978-0-89096-867-3.
|Wikimedia Commons has media related to Tonkawa.|<|endoftext|>
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# 2.6 BOOLEAN FUNCTIONS
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1 2.6 BOOLEAN FUNCTIONS Binary variables have two values, either 0 or 1. A Boolean function is an expression formed with binary variables, the two binary operators AND and OR, one unary operator NOT, parentheses and equal sign. The value of a function may be 0 or 1, depending on the values of variables present in the Boolean function or expression. For example, if a Boolean function is expressed algebraically as: F = AB C then the value of F will be 1, when A = 1, B = 0, and C = 1. For other values of A, B, C the value of F is 0. Boolean functions can also be represented by truth tables. A truth table is the tabular form of the values of a Boolean function according to the all possible values of its variables. For an n number of variables, 2 n combinations of 1s and 0s are listed and one column represents function values according to the different combinations. For example, for three variables the Boolean function F = AB + C truth table can be written as below in Figure 2.7. A B C F Figure SIMPLIFICATION OF BOOLEAN EXPRESSIONS When a Boolean expression is implemented with logic gates, each literal in the function is designated as input to the gate. The literal may be a primed or unprimed variable. Minimization of the number of literals and the number of terms leads to less complex circuits as well as less number of gates, which should be a designer s aim. There are several methods to minimize the Boolean function. Here, simplification or minimization of complex algebraic expressions will be shown with the help of postulates and theorems of Boolean algebra. Example 2.1. Simplify the Boolean function F=AB+ BC + B C. Solution. F = AB + BC + B C = AB + C(B + B ) = AB + C 1
2 Example 2.2. Simplify the Boolean function F= A + A B. Solution. F = A+ A B = (A + A ) (A + B) = A + B Example 2.3. Simplify the Boolean function F= A B C + A BC + AB. Solution. F = A B C + A BC + AB = A C (B +B) + AB = A C + AB Example 2.4. Simplify the Boolean function F = AB + (AC) + AB C(AB + C). Solution. F = AB + (AC) + AB C(AB + C) = AB + A + C + AB C.AB + AB C.C = AB + A + C AB C (B.B = 0 and C.C = C) = ABC + ABC + A + C + AB C (AB = AB(C + C ) = ABC + ABC ) = AC(B + B ) + C (AB + 1) + A = AC + C +A (B + B = 1 and AB + 1 = 1) = AC + (AC) = CANONICAL AND STANDARD FORMS Logical functions are generally expressed in terms of different combinations of logical variables with their true forms as well as the complement forms. Binary logic values obtained by the logical functions and logic variables are in binary form. An arbitrary logic function can be expressed in the following forms. (i) Sum of the Products (SOP) (ii) Product of the Sums (POS) Product Term. In Boolean algebra, the logical product of several variables on which a function depends is considered to be a product term. In other words, the AND function is referred to as a product term or standard product. The variables in a product term can be either in true form or in complemented form. For example, ABC is a product term. Sum Term. An OR function is referred to as a sum term. The logical sum of several variables on which a function depends is considered to be a sum term. Variables in a sum term can also be either in true form or in complemented form. For example, A + B + C is a sum term. Sum of Products (SOP). The logical sum of two or more logical product terms is referred to as a sum of products expression. It is basically an OR operation on AND operated variables. For example, Y = AB + BC + AC or Y = A B + BC + AC are sum of products expressions. Product of Sums (POS). Similarly, the logical product of two or more logical sum terms is called a product of sums expression. It is an AND operation on OR operated variables. For example, Y = (A + B + C)(A + B + C)(A + B + C ) or Y = (A + B + C)(A + B + C ) are product of sums expressions. 2
3 Standard form. The standard form of the Boolean function is when it is expressed in sum of the products or product of the sums fashion. The examples stated above, like Y =AB + BC + AC or Y = (A + B + C)(A + B + C)(A + B + C ) are the standard forms. However, Boolean functions are also sometimes expressed in nonstandard forms like F = (AB + CD)(A B + C D ), which is neither a sum of products form nor a product of sums form. However, the same expression can be converted to a standard form with help of various Boolean properties, as: F = (AB + CD)(A B + C D ) = A B CD + ABC D Minterm A product term containing all n variables of the function in either true or complemented form is called the minterm. Each minterm is obtained by an AND operation of the variables in their true form or complemented form. For a two-variable function, four different combinations are possible, such as, A B, A B, AB, and AB. These product terms are called the fundamental products or standard products or minterms. In the minterm, a variable will possess the value 1 if it is in true or uncomplemented form, whereas, it contains the value 0 if it is in complemented form. For three variables function, eight minterms are possible as listed in the following table in Figure 2.9. Figure 2-9 So, if the number of variables is n, then the possible number of minterms is 2 n. The main property of a minterm is that it possesses the value of 1 for only one combination of n input variables and the rest of the 2 n 1 combinations have the logic value of 0. This means, for the above three variables example, if A = 0, B = 1, C = 1 i.e., for input combination of 011, there is only one combination A BC that has the value 1, the rest of the seven combinations have the value 0. Canonical Sum of Product Expression. When a Boolean function is expressed as the logical sum of all the minterms from the rows of a truth table, for which the value of the function is 1, it is referred to as the canonical sum of product expression. The same can be expressed in a compact form by listing the corresponding decimal- 3
4 equivalent codes of the minterms containing a function value of 1. For example, if the canonical sum of product form of a three-variable logic function F has the minterms A BC, AB C, and ABC, this can be expressed as the sum of the decimal codes corresponding to these minterms as below. F (A,B,C) = Σ (3,5,6) = m3 + m5 + m6 = A BC + AB C + ABC where Σ (3,5,6) represents the summation of minterms corresponding to decimal codes 3, 5, and 6. The canonical sum of products form of a logic function can be obtained by using the following procedure: 1. Check each term in the given logic function. Retain if it is a minterm, continue to examine the next term in the same manner. 2. Examine for the variables that are missing in each product which is not a minterm. If the missing variable in the minterm is X, multiply that minterm with (X+X ). 3. Multiply all the products and discard the redundant terms. Here are some examples to explain the above procedure. Example 2.6. Obtain the canonical sum of product form of the following function: F (A, B) = A + B Solution. The given function contains two variables A and B. The variable B is missing from the first term of the expression and the variable A is missing from the second term of the expression. Therefore, the first term is to be multiplied by (B + B ) and the second term is to be multiplied by (A + A ) as demonstrated below. F (A, B) = A + B = A.1 + B.1 = A (B + B ) + B (A + A ) = AB + AB + AB + A B = AB + AB + A B (as AB + AB = AB) Hence the canonical sum of the product expression of the given function is F (A, B) = AB + AB + A B. Example 2.7. Obtain the canonical sum of product form of the following function. F (A, B, C) = A + BC Solution. Here neither the first term nor the second term is minterm. The given function contains three variables A, B, and C. The variables B and C are missing from the first term of the expression and the variable A is missing from the second term of the expression. Therefore, the first term is to be multiplied by (B + B ) and (C + C ). The second term is to be multiplied by (A + A ). This is demonstrated below. F (A, B, C) = A + BC = A (B + B ) (C + C ) + BC (A + A ) = (AB + AB ) (C + C ) + ABC + A BC = ABC + AB C + ABC + AB C + ABC + A BC 4
5 = ABC + AB C + ABC + AB C + A BC (as ABC + ABC = ABC) Hence the canonical sum of the product expression of the given function is F (A, B,C) = ABC + AB C + ABC + AB C + A BC Maxterm A sum term containing all n variables of the function in either true or complemented form is called the maxterm. Each maxterm is obtained by an OR operation of the variables in their true form or complemented form. Four different combinations are possible for a two-variable function, such as, A + B, A + B, A + B, and A + B. These sum terms are called the standard sums or maxterms. Note that, in the maxterm, a variable will possess the value 0, if it is in true or uncomplemented form, whereas, it contains the value 1, if it is in complemented form. Like minterms, for a threevariable function, eight maxterms are also possible as listed in the following table in Figure Figure 2-10 So, if the number of variables is n, then the possible number of maxterms is 2 n. The main property of a maxterm is that it possesses the value of 0 for only one combination of n input variables and the rest of the 2 n 1 combinations have the logic value of 1. This means, for the above three variables example, if A = 1, B = 1, C = 0 i.e., for input combination of 110, there is only one combination A + B + C that has the value 0, the rest of the seven combinations have the value 1. Canonical Product of Sum Expression. When a Boolean function is expressed as the logical product of all the maxterms from the rows of a truth table, for which the value of the function is 0, it is referred to as the canonical product of sum expression. The same can be expressed in a compact form by listing the corresponding decimal equivalent codes of the maxterms containing a function value of 0. For example, if the canonical product of sums form of a three-variable logic function F has the maxterms A + B + C, A + B + C, and A + B + C, this can be expressed as the product of the decimal codes corresponding to these maxterms as below, 5
6 F (A,B,C) = Π (0,2,5) = M0 M2 M5 = (A + B + C) (A + B + C) (A + B + C ) where Π (0,2,5) represents the product of maxterms corresponding to decimal codes 0, 2, and 5. The canonical product of sums form of a logic function can be obtained by using the following procedure. 1. Check each term in the given logic function. Retain it if it is a maxterm, continue to examine the next term in the same manner. 2. Examine for the variables that are missing in each sum term that is not a maxterm. If the missing variable in the maxterm is X, add that maxterm with (X.X ). 3. Expand the expression using the properties and postulates as described earlier and discard the redundant terms. Some examples are given here to explain the above procedure. Example 2.8. Obtain the canonical product of the sum form of the following function. F (A, B, C) = (A + B ) (B + C) (A + C ) Solution. In the above three-variable expression, C is missing from the first term, A is missing from the second term, and B is missing from the third term. Therefore, CC is to be added with first term, AA is to be added with the second, and BB is to be added with the third term. This is shown below. F (A, B, C) = (A + B ) (B + C) (A + C ) = (A + B + 0) (B + C + 0) (A + C + 0) = (A + B + CC ) (B + C + AA ) (A + C + BB ) = (A + B + C) (A + B + C ) (A + B + C) (A + B + C) (A + B + C ) (A + B + C ) [using the distributive property, as X + YZ = (X + Y)(X + Z)] = (A + B + C) (A + B + C ) (A + B + C) (A + B + C) (A + B + C ) [as (A + B + C ) (A + B + C ) = A + B + C ] Hence the canonical product of the sum expression for the given function is F (A, B, C) = (A + B + C) (A + B + C ) (A + B + C) (A + B + C) (A + B + C ) Example 2.9. Obtain the canonical product of the sum form of the following function. F (A, B, C) = A + B C Solution. In the above three-variable expression, the function is given at sum of the product form. First, the function needs to be changed to product of the sum form by applying the distributive law as shown below. F (A, B, C) = A + B C = (A + B ) (A + C) Now, in the above expression, C is missing from the first term and B is missing from the second term. Hence CC is to be added with the first term and BB is to be added with the second term as shown below. F (A, B, C) = (A + B ) (A + C) 6
7 = (A + B + CC ) (A + C + BB ) = (A + B + C) (A + B + C ) (A + B + C) (A + B + C) [using the distributive property, as X + YZ = (X + Y) (X + Z)] = (A + B + C) (A + B + C ) (A + B + C) [as (A + B + C) (A + B + C) = A + B + C] Hence the canonical product of the sum expression for the given function is F (A, B, C) = (A + B + C) (A + B + C ) (A + B + C) Deriving a Sum of Products (SOP) Expression from a Truth Table The sum of products (SOP) expression of a Boolean function can be obtained from its truth table summing or performing OR operation of the product terms corresponding to the combinations containing a function value of 1. In the product terms the input variables appear either in true (uncomplemented) form if it contains the value 1, or in complemented form if it possesses the value 0. Now, consider the following truth table in Figure 2.11, for a three-input function Y. Here the output Y value is 1 for the input conditions of 010, 100, 101, and 110, and their corresponding product terms are A BC, AB C, AB C, and ABC respectively. Figure 2-11 The final sum of products expression (SOP) for the output Y is derived by summing or performing an OR operation of the four product terms as shown below. Y = A BC + AB C + AB C + ABC In general, the procedure of deriving the output expression in SOP form from a truth table can be summarized as below. 1. Form a product term for each input combination in the table, containing an output value of Each product term consists of its input variables in either true form or complemented form. If the input variable is 0, it appears in complemented form and if the input variable is 1, it appears in true form. 7
8 3. To obtain the final SOP expression of the output, all the product terms are OR operated Deriving a Product of Sums (POS) Expression from a Truth Table As explained above, the product of sums (POS) expression of a Boolean function can also be obtained from its truth table by a similar procedure. Here, an AND operation is performed on the sum terms corresponding to the combinations containing a function value of 0. In the sum terms the input variables appear either in true (uncomplemented) form if it contains the value 0, or in complemented form if it possesses the value 1. Now, consider the same truth table as shown in Figure 2.11, for a three-input function Y. Here the output Y value is 0 for the input conditions of 000, 001, 011, and 111, and their corresponding product terms are A + B + C, A + B + C, A + B + C, and A + B + C respectively. So now, the final product of sums expression (POS) for the output Y is derived by performing an AND operation of the four sum terms as shown below. Y = (A + B + C) (A + B + C ) (A + B + C ) (A + B + C ) In general, the procedure of deriving the output expression in POS form from a truth table can be summarized as below. 1. Form a sum term for each input combination in the table, containing an output value of Each product term consists of its input variables in either true form or complemented form. If the input variable is 1, it appears in complemented form and if the input variable is 0, it appears in true form. 3. To obtain the final POS expression of the output, all the sum terms are AND operated. 8
9 2.9.5 Conversion between Canonical Forms From the above example, it may be noted that the complement of a function expressed as the sum of products (SOP) equals to the sum of products or sum of the minterms which are missing from the original function. This is because the original function is expressed by those minterms that make the function equal to 1, while its complement is 1 for those minterms whose values are 0. According to the truth table given in Figure 2.11: F (A,B,C) = Σ ( 2,4,5,6) = m2 + m4 + m5 + m6 = A BC + AB C + AB C + ABC. This has the complement that can be expressed as F (A,B,C) = (0,1,3,7) = m0 + m1 + m3 + m7 Now, if we take complement of F by DeMorgan s theorem, we obtain F as F (A,B,C) = (m0 + m1 + m3 + m7) = m0 m1 m3 m 7 = M0M1M3M7 = Π(0,1,3,7) = (A + B + C)(A + B + C ) (A + B + C ) (A + B + C ). The last conversion follows from the definition of minterms and maxterms as shown in the tables in Figures 2.9 and It can be clearly noted that the following relation holds true m j = Mj. That is, the maxterm with subscript j is a complement of the minterm with the same subscript j, and vice versa. This example demonstrates the conversion between a function expressed in sum of products (SOP) and its equivalent in product of maxterms. A similar example can show the conversion between the product of sums (POS) and its equivalent sum of minterms. In general, to convert from one canonical form to other canonical form, it is required to interchange the symbols Σ and π, and list the numbers which are missing from the original form. Note that, to find the missing terms, the total 2 n number of minterms or maxterms must be realized, where n is the number of variables in the function. 9
### CHAPTER-2 STRUCTURE OF BOOLEAN FUNCTION USING GATES, K-Map and Quine-McCluskey
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### Combinational Logic Circuits
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### X Y Z F=X+Y+Z
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### Experiment 3: Logic Simplification
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1.) CLD2 problem 2.2 We are allowed to use AND gates, OR gates, and inverters. Note that all of the Boolean expression are already conveniently expressed in terms of AND's, OR's, and inversions. Thus,
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### BOOLEAN ALGEBRA. Logic circuit: 1. From logic circuit to Boolean expression. Derive the Boolean expression for the following circuits.
COURSE / CODE DIGITAL SYSTEMS FUNDAMENTAL (ECE 421) DIGITAL ELECTRONICS FUNDAMENTAL (ECE 422) BOOLEAN ALGEBRA Boolean Logic Boolean logic is a complete system for logical operations. It is used in countless<|endoftext|>
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is the process by which organisms, typically animals
. Terminology often uses either the
from Latin vorare
, meaning "to devour", or
, from Greek
, meaning "to eat".
The evolution of different feeding strategies is varied with some feeding strategies evolving several times in independent lineages. In terrestrial vertebrates, the earliest forms were large amphibious
400 million years ago. While amphibians continued to feed on fish and later insects, reptiles began exploring two new food types, other tetrapods (carnivory), and later, plants (herbivory). Carnivory was a natural transition from insectivory for medium and large tetrapods, requiring minimal adaptation (in contrast, a complex set of adaptations was necessary for feeding on highly fibrous plant materials).
The specialization of organisms towards specific food sources is one of the major causes of
of form and function, such as:
parts and teeth, such as in
mosquitos, predatory animals such as
and fishes, etc.
- distinct forms of
in birds, such as in
and other appendages, for apprehending or killing (including fingers in primates)
- changes in body colour for facilitating
camouflage, disguise, setting up traps for preys, etc.
- changes in the
system, such as the system of stomachs
of herbivores, commensalism
There are many modes of feeding that animals exhibit, including:
- Filter feeding: obtaining nutrients from particles suspended in water
- Deposit feeding: obtaining nutrients from particles suspended in soil
- Fluid feeding: obtaining nutrients by consuming other organisms' fluids
- Bulk feeding: obtaining nutrients by eating all of an organism
- Ram feeding
and suction feeding: ingesting prey via the fluids around it.
- Extra-cellular digestion: excreting digesting enzymes and then reabsorbing the products
- Myzocytosis: one cell pierces another using a feeding tube, and sucks out cytoplasm
- Phagocytosis: engulfing food matter into living cells, where it is digested
"Polyphagy" redirects here. For increased appetite as a medical symptom, see
Another classification refers to the specific food animals specialize in eating, such as:
The eating of non-living or decaying matter:
There are also several unusual feeding behaviours, either normal,
, or pathological, such as:
An opportunistic feeder sustains itself from a number of different food sources, because the species is behaviourally sufficiently flexible.
Some animals exhibit
behaviours in which they store or hide food for later use.
Alcohol—it is widely believed that some animals eat rotting fruit for this to ferment and make them drunk, however, this has been refuted in the case of at least elephants.
Sahney, S., Benton, M.J. & Falcon-Lang, H.J. (2010).
"Rainforest collapse triggered Pennsylvanian tetrapod diversification in Euramerica"
(12): 1079–1082. doi:10.1130/G31182.1.
Bakalar, N. (2005).
"Elephants drunk in the wild? Scientists put the myth to rest". Retrieved<|endoftext|>
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Sets: A Gentle Introduction
There are people who are fond of buying things in sets. For example, a pen collector may collect pens of different colors. Other people are obsessed with brands. For instance, Apple fans may buy the following an iPhone, iPod, iPad, and an iMac. If you are fond of reading Dan Brown, you probably have read or own some of the following books: The Lost SymbolDeception PointDigital Fortress, and The Da Vinci Code, and others.
In mathematics, a set is not very different. A set is can be described as a collection of objects with common characteristics. These objects are called elementsGiven a set and an object, it is clear if the object is an element of a set or not. If we let D be a set of Dan Brown novels, then D = { The Lost SymbolDeception PointDigital Fortress, The Da Vinci Code}. Also, if we let V be the set of vowel letters in the English alphabet, then V = {a, e, i, o, u}, and if we let R be the set of polygons whose number sides is less than 9, then the set would include the polygons in the figure below.
As we can observe, sets maybe described or listed. The vowel letters in the English alphabet is a description of the list {a, e, i, o, u} . It can also be noted when listing that the elements of the sets, by convention, they are enclosed with curly brackets and are separated by a commas as shown above.
As we have also mentioned, we can easily see if an object is a member of the set or not. We are sure that a is a member of V. We use the symbol $\in$ to denote membership in a set. Hence, we can write a is an element of V (or is a member of V) as $a \in V$. On the contrary, b is not an element of V and write $b \notin V$.
Sets can be finite or infinite. Finite sets are sets whose elements can be counted, otherwise infinite sets. The set of counting numbers N = {1, 2, 3, 4, … } is an infinite set. The … symbol denotes that the list continues indefinitely. The number of elements in a set is called its cardinality. Hence, the cardinality of V is 5 and the cardinality of D is 4.<|endoftext|>
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What to call a freshwater crustacean that resembles a small lobster? USDA Forest Service scientist Zanethia Barnett has a clever answer: “I study crayfish, but I eat crawfish.”
More than half of the nation’s 357 species of crayfish — also known as crawdads, mudbugs, or yabbies — can be found in the Southeast.
Crayfish break down plant materials, are an important food source for larger aquatic fauna, and can be an indicator of good water quality.
The vernal crayfish (Procambarus viaeviridis) spends part of the winter and early spring in shallow ponds, sloughs, or ephemeral streams. When these habitats begin to dry, they dig and retreat into burrows.
“These seasonal pools occur in bottomland hardwood forests and are valuable habitat for crayfish and other species: juvenile fish, tadpoles, and other amphibians,” says Barnett. “Because pools are more common at times when there are fewer visitors to the forests, most people don’t realize how important they are.”
In general, there are few chronicles of native crayfish. This information gap makes it difficult to develop effective management or conservation plans.
Barnett led a study of the vernal crayfish’s life history and habitat use near the southern extent of its range. She worked with SRS aquatic ecologist Susie Adams and Rebecca Rosamond, a wildlife biologist with the US Fish & Wildlife Service, to collect more than 3,500 vernal crayfish in the Dahomey National Wildlife Refuge in the Mississippi Delta.
Their research results were published in the Journal of Crustacean Biology.
The researchers collected samples from January to May 2012, November 2013 to June 2014, and February and March 2015. They found crayfish during every sample month.
They tallied the number of crayfish per trap, along with other crayfishes and fishes. Data about crayfish habitat on the refuge included the number and size of pools, amount of leaf litter, water quality, and air and water temperature.
Sample data also included species, sex, and reproductive condition – juvenile or adult, along with adult male form. Adult males molt between two forms over the course of a year. Form 1 males are reproductively active, while form 2 males are not.
Vernal crayfish in the study area had slightly different reproductive anatomy than other populations. In females, the annulus ventralis, or sperm receptacle, differed. In males, the gonopods differed. Gonopods are the first one or two pairs of swimmerets that are used for sperm transfer.
Such differences are often used to identify crayfish species and may indicate that the vernal crayfish population on the Dahomey are distinct.
Different numbers of juveniles and adults in each survey month gave the scientists clues about the timing of reproduction. Data pointed to a peak in breeding during May. Form 1 males were abundant, making up about 70 percent of the adults captured in May. Far fewer females were found during this month, suggesting that they began burrowing in April or May.
No females were found that had hatchlings, eggs, or glair – a substance the female uses to attach fertilized eggs that becomes visible at the onset of egg laying.
However, females kept in tanks for observation did have glair by May. By October, these same crayfish had more than a hundred ovarian (unfertilized) eggs each. Typically, about 40 percent of Procambarus eggs die. “That means each female could produce between 41 and 55 young. This is the first reported estimate of fecundity for this crayfish species,” says Barnett.
The scientists estimated that vernal crayfish in the study area live as long as two to three years.
The greatest numbers of vernal crayfish were caught in shallow pools that persisted for at least three months, when water temperatures exceeded 50 degrees F and other crayfish species were present. There was no significant difference in water quality across the different pools.
Adams and Barnett are conducting more vernal crayfish sampling across the Lower Mississippi Alluvial Valley and neighboring parts of Tennessee, Mississippi, and Alabama. Adams is leading a related taxonomic study to further understand morphologic and genetic differences among populations.
“Research on the status of the vernal crayfish is ongoing,” adds Barnett. “The Dahomey population may be a new species.”
In any case, more information about its specific habitat and potential management needs will be crucial.
For more information, email Zanethia Barnett at [email protected].<|endoftext|>
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Question Video: Expanding Algebraic Expressions Using Algebraic Identities | Nagwa Question Video: Expanding Algebraic Expressions Using Algebraic Identities | Nagwa
Question Video: Expanding Algebraic Expressions Using Algebraic Identities Mathematics • First Year of Preparatory School
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Expand (β8π₯ β 2π¦)Β² β (β8π₯ + 2π¦)Β².
05:07
Video Transcript
Expand negative eight π₯ minus two π¦ squared minus negative eight π₯ plus two π¦ squared.
In this question, we have two binomials, negative eight π₯ minus two π¦ and negative eight π₯ plus two π¦. Weβre squaring each of them and then finding the difference. Letβs deal with squaring each binomial separately. And we can use two different methods to do this. What we must remember though is that when weβre squaring a binomial, weβre multiplying that binomial by itself. So in the case of negative eight π₯ minus two π¦ all squared, weβre looking for the result of negative eight π₯ minus two π¦ multiplied by negative eight π₯ minus two π¦.
Weβll perform this expansion using the FOIL method. Remember, this is an acronym where each letter stands for a different pair of terms that we need to multiply together. F stands for firsts, so we multiply the first term in the first binomial by the first term in the second binomial. Thatβs negative eight π₯ multiplied by negative eight π₯. Negative eight multiplied by negative eight is 64. And π₯ multiplied by π₯ is π₯ squared. So we have 64π₯ squared. Next, O stands for outers or outside, so we multiply the terms on the outside of our binomials together. Thatβs the negative eight π₯ in the first binomial and the negative two π¦ in the second. Negative eight multiplied by negative two is positive 16 and π₯ multiplied by π¦ is π₯π¦. So we have positive 16π₯π¦.
Next, we have I, which stands for inners or inside. So we multiply together the terms on the inside of the expansion. Thatβs negative two π¦ by negative eight π₯. Again, this gives positive 16π₯π¦. Finally, L stands for last, so we multiply the last term in each binomial together. Thatβs the negative two π¦ in the first by the negative two π¦ in the second, giving positive four π¦ squared. So weβve completed our expansion. And we notice that, at this point, we have four terms and the middle two terms are identical. We simplify by collecting the like terms, giving 64π₯ squared plus 32π₯π¦ plus four π¦ squared for the result of expanding the first binomial.
For the second, negative eight π₯ plus two π¦ all squared, weβre looking for the result of multiplying negative eight π₯ plus two π¦ by itself. And this time, weβll use the distributive method. Weβll take one of our factors of negative eight π₯ plus two π¦ and distribute it over the other, giving negative eight π₯ multiplied by negative eight π₯ plus two π¦ plus two π¦ multiplied by negative eight π₯ plus two π¦. And now, we just have to expand or distribute a single set of brackets or parentheses.
For the first set, we have negative eight π₯ multiplied by negative eight π₯, giving 64π₯ squared, and then negative eight π₯ multiplied by positive two π¦, giving negative 16π₯π¦. And then, we have positive two π¦ multiplied by negative eight π₯, giving negative 16π₯π¦, and positive two π¦ multiplied by positive two π¦, giving positive four π¦ squared. As in our other expansion, we have four terms at this point, with two identical terms in the center. So we simplify to give 64π₯ squared minus 32π₯π¦ plus four π¦ squared.
So weβve squared each binomial, and now weβre ready to perform the subtraction. We must be really careful here because we must make sure weβre subtracting every term in our second expansion from the first. We have 64π₯ squared plus 32π₯π¦ plus four π¦ squared minus 64π₯ squared minus 32π₯π¦ plus four π¦ squared. Now, of course, these two algebraic expressions are very similar because the binomials we started off with were very similar. They just differed in the sign of the π¦ term. So some of the terms will cancel when we subtract. But we must be very careful.
We have 64π₯ squared minus 64π₯ squared, so those terms will cancel one another out. We also have four π¦ squared minus four π¦ squared. So those terms will also cancel each other out. What weβre left with is positive 32π₯π¦ minus negative 32π₯π¦. Now, because we are subtracting a negative, those two signs together will form a positive, giving us 32π₯π¦ plus 32π₯π¦, which is 64π₯π¦. So we have our answer to the problem. Negative eight π₯ minus two π¦ all squared minus negative eight π₯ plus two π¦ all squared is equal to 64π₯π¦. And weβve seen two different methods for squaring binomials, the FOIL method and the distributive method.
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Georgian alphabet is described as one of the most beautiful and unique scripts in the world.
There are three types of Georgian scripts known at this current moment. The first scripts are used in the writings of the Georgian Orthodox Church both in lower and upper case letters, called Khutsuri, which means priests’ alphabet.
Second type of Georgian scripts is . Asomtavruli that is only written in capitals.
Mostly widespread and used in modern Georgian language is Mkhedruli, the cursive script.
In addition to the national status of cultural heritage, the Georgian alphabet has been named among the top five most beautiful alphabets in the world by international travel website, Matador Network.
Georgians are proud of their unique writing system, which comprises of 33 characters. It is the only alphabet in the world that is pronounced exactly the same way it is written. In addition to that all letters are unicameral; they make no distinction between upper and lower case.
Some scholars believe that the Georgian alphabet was created in the 4th Century AD, or at the latest in the early 5th Century.<|endoftext|>
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Probability distributions when applied to grouped random variables gives rise to joint probability distributions. In this page we will focus only on two dimensional distributions though it can be applied to higher dimensions also.
Joint probability distributions describe situations where by both outcomes represented by random variables occur.
Joint probability distribution can be expressed in terms of joint probability density function or joint probability mass function and is defined as below:
$f(x, y)$ = $P(X = x, Y = y)$.
## Discrete and Continous Joint Probability Distributions
Discrete Probability Distributions
Discrete random variables when paired they give rise to discrete joint probability distributions.
The probability function, also known as the probability mass function for a joint probability distribution $f(x,y)$ is defined as
1) $f(x, y)$ $\geq$ 0 for all $(x, y)$
2) $\sum_{x}$$\sum_{y} f(x, y) = 1 3) f(x, y) = P(X = x, Y = y) 4) If x and y are independent then f(x, y) = f(x) \times f(y) Continuous Probability Distributions From a group of continuous random variables continuous joint probability distributions arise. They are characterized by the joint density function, The joint density function should have: 1) f(x, y) \geq 0 for all (x, y) 2) \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(x, y) dx dy = 1 3) For any region A lying in the xy plane P[(X, Y) \in A] = \int \int_{A} f(x, y) dx dy ## Joint Probability Distribution Function Let X and Y be two discrete random variables which are defined on a sample space. Let p(x, y) be a function such that p(x, y) = P(X = x ; Y = y) Then p(x, y) is called joint probability function of X and Y. Let p _{1}(x) be the probability function of X and let p _{2}(y) be the probability function of Y. Then p _{1}(x) and p _{2}(y) are called marginal probability functions of X and Y respectively. Here p _{1}(x) = P[X = x] = \sum_{y} P[X = x ; Y = y] = \sum_{y} p(x, y) p_{2}(y) = P[Y = y] = \sum_{x} P[X = x ; Y = y] = \sum_{x} p(x, y) Thus, p _{1}(x) = \sum_{y} p (x, y) and p _{2}(y) = \sum_{x} p(x, y) ## Examples Given below is an example on joint probability distribution. Example: For the following probability distribution, find E(3X + 5), E(2X^{2} - 5), Var(X), Var(2X), Var(4X - 3), Var(-X), Var(-5X + 2), SD(X) and SD(2X). x 10 12 p(x) 0.8 0.2 Solution : Given below is the table x p(x) x$$^{2}$ $x.p(x)$ $x$$^{2}$ $p(x)$ 10 0.8 100 8.0 80 12 0.2 144 2.4 28.8 $\sum$ = 1 $\sum$ = 10.4 $\sum$ = 108.8
$E(X)$ = $\sum$ $x$ $\times$ $p(x)$
= 10.4
$E(X^{2})$ = $\sum$ $x^{2} p(x)$
= 108.8
$E(3X + 5 )$ = 3 $E(X)$ + 5
= 3 $\times$ 10.4 + 5
= 36.2
$E(2X^{2} - 5)$ = 2$E(X^{2})$ - 5
= 2 $\times$ 108.8 - 5
= 212.6
$Var(X)$ = $E(X^{2})$ - $[E(X)]^{2}$
= 108.8 - (10.4)$^{2}$
= 0.64
$Var(2X)$ = 2$^{2}$ $\times Var(X)$
= 4 $\times$ 0.64
= 2.56
$Var(4X - 3)$ = 4$^{2}$ $\times Var(X)$
= 4$^{2}$ $\times$ 0.64
= 10.24
$Var(-X)$ = (-1)$^{2}$ $\times Var(X)$
= $Var(X)$
= 0.64
$Var(-5X + 2)$ = (-5)$^{2}$ $\times Var(X)$
= 25 $\times$ 0.64
= 16
$S.D (X)$ = $\sqrt{Var(X)}$
= $\sqrt{0.64}$
= 0.8
$S.D (2X)$ = $\sqrt{var(2X)}$
= $\sqrt{2.56}$
= 1.6<|endoftext|>
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Brouwer’s fixed point theorem, how to solve tricky mathematical problems, topology, and Brouwer
Brouwer’s fixed point theorem states that if h is a continuous function mapping a closed unit ball (or disc) into itself, then it must have at least one fixed point. That means: at least one point x where h(x)=x.
In other words, if the women above are sloshing their coffee about gently as they walk, then between any two moments of time at least one molecule in each drink must end up exactly back where it started.
“Continuous” means that the function has no jumps. Or: you could draw it without taking your pencil off the paper. Or: if x1 and x2 are close, then h(x1) and h(x2) are also close.
The “unit” ball (or disc) means a ball (or disc) of radius 1. “Closed” means that we include the surface (or circumference).
Here is a proof for the unit disc. It is a proof by contradiction, that is, it works by assuming that the theorem is wrong, and showing that that is impossible.
If the theorem is wrong, then h(x) is different from x for all x. So we can draw a ray from h(x) through x to the circumference, as shown in the diagram.
Call the point where the ray hits the circumference r(x). Then if h(x) is continuous, so is r(x). And since r(x)=x for x on the circumference, r(x) must take values all round the circumference.
There is a smooth, continuous process (called a “retraction”) which shrinks the whole unit disc down to one point: just make its radius smaller and smaller, smoothly, until it is down to zero.
When x is going through that process, r(x) (i.e., the whole circumference) must also shrink smoothly within itself down to a point.
But imagine a wire circle covered by an infinitely stretchy, infinitely squeezable sheath. Even though the covering is infinitely stretchy and infinitely squeezable, there is no way of stretching and squeezing it down to a point. You would have to break it or cut it to do that.
The conclusion that r(x) must shrink smoothly down to a point must be wrong; and so the initial assumption, that h(x) is different from x for all x, must be wrong. ▇
This proof is interesting for several reasons.
First
The modern French mathematician Alain Connes says that in mathematical research: “The main error to be avoided is trying to attack the problem head-on”.
If you want to solve a tricky mathematical problem, almost always you have to find some unexpected way of formulating the problem, some way of breaking it down into smaller steps, some zig-zag route.
For most A-level questions, going at them head-on is ok. But not for STEP, for example.
Here are three other neat examples of solving mathematical problems by an unexpected zig-zag route.
Morley’s miracle
Sum from n=1 to ∞ of 1n2 is π2/6
Second
Historically, Brouwer’s fixed point theorem was one of the first results in an area of maths called topology, which is concerned with the properties of shapes which stay the same when they are stretched or squeezed, but not when they are broken or cut.
Issues which are now reckoned to be part of topology had been studied by Leonhard Euler back in the 18th century, such as the Königsberg bridges problem and Euler characteristics, but topology as an established area of maths (alongside arithmetic, algebra, geometry, calculus, etc.) is usually reckoned to date only from an article by Henri Poincaré in 1895.
This fixed point theorem was proved by the Dutch mathematician L E J Brouwer in 1910.
Third
Although the proof above is generally reckoned by mathematicians to be sound, Brouwer himself came to think that it, and some other proofs of results in topology which had made him famous as a mathematician, were unsound.
He developed a philosophy which said that mathematical objects have no existence “out there”, and exist only as and when they are constructed by mathematicians. Since the proof does not tell us how to find the “fixed point” – it does not construct it – therefore it does not prove its existence.
In later years, according to another Dutch mathematician, Brouwer “never gave courses on topology, but always on – and only on – the foundations of [mathematics]. It seemed that he was no longer convinced of his results in topology because they were not correct from the point of view of [his philosophy], and he judged everything he had done before, his greatest output, false according to his philosophy. He was a very strange person, crazy in love with his philosophy”.
Brouwer’s philosophy is still discussed among mathematicians, though it is a minority view. I taught for a while at Marsden State High School, near Brisbane in Australia, when the head of the maths department was an advocate of Brouwer’s philosophy.
The American mathematician Errett Bishop, in the 1960s, reconstructed proofs of a lot of the standard results of mathematics to be satisfactory according to Brouwer’s philosophy.<|endoftext|>
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The Sustainable Development Goals are the blueprint to achieve a better and more sustainable future for all. They address the global challenges we face, including those related to poverty, inequality, climate, environmental degradation, prosperity, and peace and justice. The Goals interconnect and in order to leave no one behind, it ís important that we achieve each Goal and target by 2030.
Sustainable Development Goals (SDGs) (or Global Goals for Sustainable Development) are a collection of 17 global goals set by the United Nations General Assembly in 2015. The SDGs are part of Resolution 70/1 of the United Nations General Assembly: "Transforming our World: the 2030 Agenda for Sustainable Development." That has been shortened to "2030 Agenda." The goals are broad and interdependent, yet each has a separate list of targets to achieve. Achieving all 169 targets would signal accomplishing all 17 goals. The SDGs cover social and economic development issues including poverty, hunger, health, education, global warming, gender equality, water, sanitation, energy, urbanization, environment and social justice.
Learn more about 17 Sustainable Development Goals
Click in the one of the following buttons to learn more:<|endoftext|>
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In 1991, the Balkan nation of Yugoslavia was torn apart by civil war between opposing ethnic groups. Despite international intervention and attempts at peace negotiations, civil strife continued to dominate the country throughout the 1990s.
The war in Yugoslavia was directly linked to the circumstances surrounding the creation of the country. Following World War I, rival ethnic groups were yoked together into the Kingdom of Serbs, Croats, and Slovenes despite the demands of each group for separate homelands; the nation was renamed the Kingdom of Yugoslavia in 1929. Tensions between the ethnic groups were temporarily quelled during the Communist dictatorship of Marshal Tito, but after his death in 1980 and the fall of Communism in Eastern Europe in 1989, centuries-old tensions reemerged and erupted into civil war.
The conflict began in June 1991 after the republics of Slovenia and Croatia seceded from Yugoslavia and declared independence. Within days, Serbian leader Slobodan Milosevic—a strong nationalist who advocated the creation of a so-called Greater Serbia—ordered Serbian troops into Slovenia and Croatia under the pretext of protecting Serb populations in the territories. Within days, the Serbian army invaded the territories and forced Croatians and Slovenes from their homes. A cease-fire negotiated in January 1992 gave Serbia nearly a third of Croatia’s territory.
Within months, however, civil war broke out once again when Bosnia and Herzegovina seceded from Yugoslavia. The campaign launched by the Serbs against the breakaway republic was driven by ethnic and religious animosities. Supported by Milosevic, the Bosnian Serbs, who are primarily Eastern Orthodox, adopted a policy of ethnic cleansing—including systematic tortures, murders, and rapes—against non-Serb groups, particularly the Bosnian Muslims. By November 1995, when the presidents of Serbia, Croatia, and Bosnia and Herzegovina finally reached a peace agreement in Dayton, Ohio, nearly 2 million people had fled the region as refugees and more than 200,000 had been killed.
In the year following the peace agreement, the Bosnian capital of Sarajevo was reunified; the Muslims, Serbs, and Croats elected leaders to serve on a three-president committee; and Serbia and Bosnia restored diplomatic ties. Nevertheless, fighting between Muslims and Serbs broke out once again in November as Muslims attempted to return to their homes. The lingering conflicts were quieted by United Nations (UN) peacekeepers, who would remain in the region until December 1998.
The fragile peace in the Balkans, however, could not hold. In March 1998, less than two and a half years after the Dayton peace accords were signed, the region was again plunged into conflict when Milosevic, now the president of Yugoslavia, ordered the military to suppress pro-independence Albanians in the province of Kosovo in southern Yugoslavia. Although the government claimed that the crackdown targeted only members of the Kosovo Liberation Army (KLA), an organization dedicated to ending Serb authority in Kosovo through armed resistance, Albanians, as well as outside observers, reported that the attacks had also targeted civilians. Despite the imposition of economic sanctions by the United States and the European Union, Serbian attacks on the Albanians continued until military intervention by the North Atlantic Treaty Organization in 1999 forced Milosevic to withdraw Serb forces from Kosovo.<|endoftext|>
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2 Question 1The continuous movement of water from the oceans to the air and back to the oceans is called the water cycle. The Sun heats the surface of the oceans, which causes the water to evaporate.a) How does the rate of evaporation depend onThe wind speed (1)The temperature (1)The humidity? (1)b) Explain how evaporation causes a cooling effect?(3)
3 Question 1 AnswerThe higher the wind speed the greater the rate of evaporation.The greater the temperature the greater the rate of evaporation.The higher the humidity, the slower the rate of evaporation.b)The most energetic molecules leave the surface of the liquid, so theaverage energy of kinetic energy of the remaining molecules is less.The temperature depends on the average kinetic energy, so it isreduced.
4 Question 2Compare the similarities and differences between the process of conduction in metals and non-metals. (6) In this question you will assessed on using good English, organising information clearly using specialist language where appropriate.
5 Questions 2 Answer Similarities Involve particles Atoms vibrate causing neighbouring atoms to vibrate so energy is passed along.DifferencesMetals have free electronsThese collide with other free electrons and ions passing energy along.This process is much more effective, so metals are better conductors.
6 Question 3In a hot water system water is heated by burning gas in a boiler. The hot water is then stored in a tank. For every 111 J of energy released from the gas, 100 J of energy is absorbed by the water in the boiler,Calculate the percentage efficiency of the boiler. (4)The energy released from the gas but not absorbed by the ‘boiler’ is wasted. Explain why this energy is of little use for further energy transfers. (1)The tank in the hot water system is surrounded by a layer of insulation. Explain the effect of the insulation on the efficiency of the hot water system. (3)
7 Question 3 Answer Efficiency = 90%. The energy becomes too spread out to use.The insulation reduced the rate at which energy is lost from the tank to the surroundings. The water in the tank will stay hotter for longer. This increases the efficiency of the hot water system.
8 Question 4 A light bulb transfers electrical energy into useful light energy and wasted energy to the surroundings. For every100 J of energy supplied to the bulb, 5 J of energy istransferred to light.Draw and label a Sanky diagram for the light bulb. (3)
9 Question 4 AnswerThe width of the arrow should be scaled to the number ofJoules.
10 Question 5 A student uses some hair straighteners. The straighteners have a power of 90 W. What is meant by ‘a power of 90 W’. (2)Calculate how many kilowatt-hours of electricity are used when the straighteners are used for 15 minutes. (3)The electricity supplier is charging 14 p per KWh. Calculate how much it will cost to use the straighteners for 15 minutes a day for one year. (2)
11 Question 5 answer 90 Joules of energy are transferred every second. E= KWhCost = £1.15
12 Question 6 Fluorescent bulbs are being replaced by compact A 25 W compact bulb costs £12, a filament bulb costs50p.A 25 W compact bulb gives out as much light as a 100Wfilament bulb.A filament bulb lasts for about 1000 hours; a compact bulblasts for a about 8000 hours, although significantly shorter ifthe bulb is turned on and off very frequently.Compare the advantages and disadvantages of buying a compact fluorescentbulb ratherthan filament bulbs. (6)
13 Question 6 Answers Advantages More efficient/4 times more efficient. Cheaper to useLast longer/lasts 8 times as long.Better for the environment.DisadvantagesMore expensive to buy/cost 24 times as much to buy.Disposal a problem because of mercury vapour.Shortened lifespan if turned on and off very frequently.
14 Question 7 Explain why step-up transformers are used in the National grid (2)
15 Question 7 AnswerTo increase the voltage on the cables and reduce powerlosses.
16 Question 8 Electromagnetic waves travel at a speed of 300 000 000 m/s. BBC 4 is transmitted using a wavelength of 1500metres.Calculate the frequency of these waves. (3)Write down the equation you use. Show clearly how youwork out your answer and give the unit.
18 Question 9 Give one example of each of the following from everyday life.Reflection of light (1)Reflection of sound (1)Refraction of light (1)Diffraction of sound (1)We do not normally see the diffraction of light in everyday life. Suggest a reason for this (2)
19 Question 9 AnswerAny example using mirror/water or a shiny smooth surface.Any example of an echo.Any example using a lens e.g. spectacles, cameras.Any example of hearing sound around a corner.The wavelength of light is very small, so diffraction only occurs when light passes through a narrow gap.
20 Question 10 Red shift from distant galaxies provides evidence for the big bang theory.What is meant by red shift? (2)Explain how red-shift provides evidence for the Big Bangtheory. (6)
21 Question 10 AnswersThe wavelength of light from a galaxy is shifted towardsthe red end of the spectrum and the galaxy is movingaway from us.Light from (most) galaxies is red shifted.The further the galaxy the bigger the red shift.The furthest galaxies are moving fastest.This shows that the universe is expanding.If the universe has always been expandingIt must have once been very small.<|endoftext|>
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# 6 Fun Math Games to Use in Class or at Home!
in Activities
Games can be very entertaining. Math games provide opportunities for students to develop fluency, deepen mathematical thinking and strategies, and promote retention.
Of course, it may take some thought to create a game that develops math skills, but just treat it as a fun activity. Aside from online math games, you can try board games, card games, outdoor games, math puzzles, and even math tricks. Here are some enjoyable ones that develop different math skills.
## Computational Skills
### 1. 1089 Math Trick
You may have seen math tricks before. But one that never ceases to amaze is the 1089 trick! It uses addition and subtraction and will impress anyone.
Rules:
• First, choose a 3-digit number (The three digits used must be different): For example 251. Reverse the digits: 152
• Then, subtract the lesser number from the greater: 251 −152 = 099. (Always write the answer with 3 digits.) Reverse the difference: 990
• Finally, add those two numbers: 099 + 990 = 1089
No matter what number you choose in Step 1, your answer is always 1089! Use it to impress your friends! See if you can figure out why it always works.
### 2. Game of Codes
Codes have been used all over the world for hundreds of years to send secret messages. You can use a simple code like the one below to develop counting skills, or make it more complex by assigning numbers to letters using patterns or computation.
### 3. Dominoes: Five-Up
Domino games have been around for a very long time and are believed to have been invented in China. A similar game was played long ago in Egypt and was found in Tutankhamen’s tomb. There are many ways that dominoes can help develop computational skills and strategic thinking. Here’s a game that uses multiples of 5 and addition.
Rules
• Each of the 2 to 4 players takes 5 dominoes and leaves the rest (in the boneyard). The first player plays any domino.
• The next player plays a domino to match either end, and so on, to make a chain.
• The first double is placed crosswise in the chain, and up to four dominoes can be played against that double.
• If a player cannot play, they take dominoes from the boneyard until one can be played. The dominoes they could not play remain in that player’s hand.
Scoring
After each play, the ends are totaled. If the total is a multiple of 5, the player scores that total. The first player to play all their dominoes scores points from the dominoes remaining in the other players’ hands.
Let us know how much you liked that fun math game!
### 4. Magic Squares
Magic squares have fascinated people for thousands of years and in ancient times they were believed to be connected to magic. A magic square can have 3 rows and 3 columns, or 4 rows and 4 columns, etc. In a magic square, the sum of each row, column and diagonal is always the same; it’s the magic constant
To make this a fun math game, try creating a game board of squares. You can pre-fill some of the squares and have players fill in the remaining ones, or have players complete the entire board. Here is a way to create a magic square for any odd-number square board.
• Draw the square board. Then draw a pyramid of squares on each side, starting with 2 fewer squares than the side of the square board.
• Starting at 1, put the sequence of numbers in the diagonals.
• Then, put the numbers, from the squares that you added, into the original square board so that the number goes into the square at the opposite end of the row or column.
What is the magic number?
If you add or multiply the same number to every number in a magic square, it will remain magic!
Now, it’s your turn to play! Try this fun math game with your students or children.
## Geometry-Related Skills
### 5. draw a shape without lifting the pencil
Have you ever tried to draw a shape without lifting the pencil, or overlapping lines? Try to do this with the 5-pointed star. Then the 7-pointed star
What about other shapes? Try these…
In math, these problems are referred to as Euler paths or circuits. And yes, there is a mathematical approach to these math puzzles. Maybe you can figure it out!
## Strategic Thinking
### 6. Two Stones
Many of us know Tic-Tac-Toe or Chess as games of strategy. Here is another game that requires strategic thinking.
Two Stones
This game of strategy is called Ou-moul-ko-no in Korea and Pon Hau k’I in China.
Rules:
• First, one player places two game pieces at the top, and the other player places two at the bottom.
• Then, the players take turns moving a game piece to an empty space. Note that the first empty space is in the middle.
• The game ends when a player is blocked and cannot make a move.
Enjoy these fun math games and math puzzles with your students or children. You can find other math games, puzzles and tricks in online resources. Use a resource like Netmath to learn math skills and create fun math games.
Content Writer since 2010<|endoftext|>
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US withdrawal (1969 - 73)
President Richard Nixon, elected in November 1968, sought an exit strategy that would leave US credibility intact.
In June 1969 he announced a policy of "Vietnamization" – training and equipping the South Vietnamese military to enable the US to reduce troop numbers.
Over the following three years, more than 500,000 soldiers were withdrawn. This reduced already-low morale among troops, feeding high levels of desertion and drug abuse.
US public hostility continued, fuelled by several events. Two offensives against communist bases and supply routes in Cambodia, in 1970 and 1972, sparked waves of protest.
The June 1969 battle of "Hamburger Hill" raised fresh concerns about wasted US lives - 46 soldiers died fighting a successful but bloody battle for a site from which their comrades were withdrawn soon afterwards.
And evidence came to light of a 1968 massacre at My Lai, where US forces slaughtered more than 300 Vietnamese villagers during an assault on suspected Vietcong camps.
Ho Chi Minh died in 1969, but his successor Le Duan continued to fight. The communists launched another major offensive in 1972, but were turned back by massive US airpower.
Slow and convoluted talks were held in Paris from 1969. Punctuated in 1972 by an intense eight-day US bombing campaign targeting Hanoi, the negotiations eventually produced a peace deal in January 1973.
Under the agreement, US forces would leave and South Vietnam would have the right to determine its own future.<|endoftext|>
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Interstitial lung disease or ILD refers to a wide group of lung diseases affecting the interstitium (tissue that surrounds and separates the tiny air sacs/alveoale of the lungs). It is also known as diffuse parenchymal lung disease(DPLD).
In this disease the supportive tissues between the air sacs are inflamed rather than the air sacs themselves. Generally the inflammation spreads through out the lungs and is not confined to just one location. Interstitial lung diseases may also be referred to as interstitial pulmonary fibrosis or pulmonary fibrosis.
Alveolitis: inflammation that involves the alveoli (air sacs)
Bronchiolitis: inflammation that involves the bronchioles (small airways)
Vasculitis: inflammation that involves the small blood vessels (capillaries)
All the disorders grouped under ILD can cause inflammation and progressive scarring of the lung tissue. Breathing gets difficult and eventually affects oxygen flow to the bloodstream. This disease can put a person at risk of developing high blood pressure and serious heart problems. In simple terms ILD is a disease of the lung and affects it in three ways:
Each of these disorders has different causes, prognoses and treatment but they are clubbed together under ILD as they have similar clinical symptoms, and physiologic traits. ILD or interstitial lung disease develops gradually, but in few cases it does show up all of a sudden. Few cases of ILD have definite causes whereas a few do not. Lung scarring is generally irreversible.
There is no cure for interstitial lung disease. Prescription medications can by and large slow down the damage. People with ILD never get to use their lungs completely. This disease affects men more than it does women and typically develops in people over age 50.
Interstitial lung disease types
Most of the interstitial lung diseases are diagnosed as pneumoconiosis (drug-induced disease) or hypersensitivity pneumonitis. The other types commonly reported are:
Primary symptoms of Interstitial lung disease
Breathlessness worsens with the progression of the disease. People complain of breathlessness while indulging in simple activities like talking over phone, eating, and dressing up, etc.
Interstitial lung disease causes
Environmental and work-related factors: Continuous exposure to pollutants and toxins like silica dust, asbestos dust, hard metal dust, chlorine or ammonia gases, chemical fumes, etc.
Exposure to organic dust like dust from sugar cane, grain, moldy hay, animals, birds, etc.
Radiation therapy: Continuous exposure to radiation therapy, particularly for breast and lung cancer. Severity of damage depends on the extent of exposure to radiation. Chemotherapy could also be a contributing factor.
Drugs: Few drugs are capable of damaging the tissues lining the lungs. List of drugs include chemotherapy drugs, few psychiatric medications, few antibiotics, medications used to treat heart arrhythmias, etc.
Connective tissue disease and other disorders: These conditions do not affect the lung directly but affect the tissues in the entire body. Conditions include rheumatoid arthritis, lupus, dermatomyositis, Sjogren's syndrome, etc. Few researchers point out that GERD (gastro esophageal reflux disease) could also contribute to the condition.
If the cause is not determined, then such disorders without clearly known cause are grouped under idiopathic pulmonary fibrosis or idiopathic interstitial lung disease.
Interstitial lung disease diagnosis
Identifying the cause can be difficult as ILD includes a broad range of disorders. The physician begins diagnosis by conducting a detailed investigation of the patient.
Interstitial lung disease treatment
Interstitial lung disease cannot be cured, but treatment can provide ease to the patient. Treatment for ILD will vary from person to person based on the following factors:<|endoftext|>
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Here are some games you can use for practising and revising a set of vocabulary with pictures. I have included a demonstration set but these games could be used with any lexical set with slight modifications. These games are motivating, memorable, easy to monitor and require very little preparation.
Board Race Variant
A ‘board race’ is fairly common idea; Divide the class into two or three teams and give each team a board pen. Give each team a section of the board and students write on the board, one answer per student, passing their team’s pen like in a relay.
You can play this game to see how much your students know within a lexical set (“Write as many animals on the board as you can, you have 3 minutes!”), grammar (“Complete this sentence: If I had a million dollars, … ), to check spelling or anything else really.
In this variant I have large picture cards stuck to the board (or projected) and students work in teams to write the vocab under the pictures. Each team uses a different coloured pen and at the end teams check each other’s answers, including spelling (two points for being first, one point for second).
There are a few ways to use the picture and word cards together. Divide students into pairs or groups of three and give each group a set of both cards. Students match the word and picture cards as a race. Play a few rounds to allow students plenty of practice (if one group keeps winning, you can give them a slight delay as a challenge).
You can add variety to the race above by putting the card sets on opposite sides of the classroom, where students take turns and can only move the cards one at a time.
There are also some classic games you could play, like pelmanism (the matching pairs memory game), go fish and snap.
Monikers (aka celebrities) is a a mix of mime, pictionary and descriptions you can play using the word cards. I’ve written about it before in this example here. Students play a few rounds in groups of three or four taking turns to draw a card and describe it to their group in a variety of different ways.
You can either set a different restriction for each round (e.g. “you can only use three words for your description”) or allow students choose from three options that you write on the board: (Draw, mime using only your hands or describe it silently by mouthing the words).
As a quick variant, students work in teams of 3 with a set of picture cards. They draw one at a time and say the answer together until they finish the set, racing against other teams. This is a quick way to revise and you can monitor for problems with pronunciation.
This game allows students to practise production of the vocabulary within a grammatical structure. Prepare sentence structures for some of your vocabulary e.g. “I usually get angry when people…” Introduce one for students to complete with as many answers as possible, working in pairs for 2 or 3 minutes. Monitor and give feedback, extending the time if necessary for students to come up with a couple of correct sentences.
Then, drill the sentence structure as a group chant with students taking turns clockwise around the room. Students have to complete the sentence with a unique ending on their turn or they’re out of the game (but they continue to chant the sentence starter with the group).
Move round the room until you have a winner or you’ve had three sentences from each student. Repeat with each prepared sentence. Finally, students mingle and try to remember as many sentences from the game as possible. This is a great way to integrate a language point into a vocabulary lesson or vice versa.
This is such a prevalent game in EFL that it feels a little redundant to include it, but there are a few different ways to use these cards and I still use them a lot myself. In the traditional game divide the class into groups of three or four and give each group a set of cards. They place one domino to start then take it in turns to place cards next to a match until all the cards are placed. This can be played competitively but I prefer students to work together.
You can also use them as a quick race. Working in groups of three or four students try to make a circle using the full set of dominoes as quickly as possible. When groups have finished they can check by looking at another group’s answers.
Similar to the matching races above, you can also make this more challenging by putting the dominoes across the room and only allowing students to take them one at a time. If the domino doesn’t match, then they have to take it back and try again. In this version students take turns getting cards, which can help to prevent more dominating students from taking over.
You can download the Powerpoint version here which includes a template that you can use to make your own domino sets. I highly encourage it, as they can be used to practise a wide variety of vocabulary sets, grammar rules and collocations (as I’m sure you’ve seen yourself in countless textbooks).
Thanks for reading!
Check out the front page, or use the search bar, to find dozens of games and activities on the site.<|endoftext|>
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A B CHomework
Solutions for Session 2, Part H
See solutions for Problems: H1 | H2 | H3 | H4 | H5 | H6| H7 | H8
Problem H1
a. It is possible. For example: b. It is possible. For example: c. It is possible. For example: d. It is possible. For example: e. It is possible. For example: f. This is impossible, because equal sides correspond to equal angles. This would mean that a triangle would have two consecutive obtuse angles. The sides extending from these two angles could not be connected for the same reason as in Problem A5, part (d). g. It is possible. In fact, all equilateral triangles are also isosceles triangles.
Problem H2
a. Constructions and answers will vary. But, if the diagonals have different lengths (such as 2 and 3 units), then the figure can never be a square or a rectangle. b. Constructions will vary, but the figure will always be a parallelogram. One possible example is the following: c. Constructions will vary, but the figure will either be a kite or a random quadrilateral. One possible example follows: d. In this case, there is only one possibility -- a rhombus:
Problem H3 You should find that the result must be a rectangle. Here's one explanation for that fact: The sum of the interior angles in any quadrilateral is 360°. Because the line segments marked "a" are all equal, the angles opposite to them inside the respective triangles are equal. Therefore, the sum of angles 4A + 4B = 360°; i.e., 4(A + B) = 360°; i.e., A + B = 90°. Hence, all of the interior angles are right angles, and the quadrilateral is indeed a rectangle.
Problem H4 The quadrilateral in question is a rectangle as described in the solution to Problem H3. In addition, two adjacent isosceles right triangles with hypotenuses a and b respectively are congruent since they have congruent legs, and the congruent (right) angles between them. So we must have a = b, and therefore the quadrilateral is a square.
Problem H5 In parts a-c, it is impossible to draw two different triangles. In other words, if we fix two sides and the angle between them, we uniquely determine a triangle.
Problem H6 This type of congruence can be called SAS (side-angle-side) congruence: If two triangles have two sides equal in length, and the angles between those sides are equal in their degree measure, then the two triangles are congruent.
Problem H7 In parts a-c, we can create two or more distinct triangles by keeping the angles fixed and changing the lengths of the sides. For example, we can build a second triangle where each side is twice as long as the original and the angles will remain the same size.
Problem H8 No. Problem H7 shows that two triangles can have the same size angles without being congruent. The triangles appear to have the same overall shape, but they might be larger or smaller than the original. They are not congruent, but they do have a relationship. They are called similar.<|endoftext|>
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by Justin Skycak on
Factoring is a method for solving quadratic equations.
This post is part of the book Justin Math: Algebra. Suggested citation: Skycak, J. (2018). Factoring Quadratic Equations. In Justin Math: Algebra. https://justinmath.com/factoring-quadratic-equations/
Factoring is a method for solving quadratic equations. It involves converting the quadratic equation to standard form, then factoring it into a product of two linear terms (called factors), and finally solving for the variable values that make either factor equal to $0$.
\begin{align*} \text{Original quadratic equation} \hspace{.5cm} &\Bigg| \hspace{.5cm} 2+x^2=-3x \\ \text{Convert to standard form} \hspace{.5cm} &\Bigg| \hspace{.5cm} x^2+3x+2=0 \\ \text{Factor} \hspace{.5cm} &\Bigg| \hspace{.5cm} (x+1)(x+2)=0 \\ \text{Set each factor to } 0 \hspace{.5cm} &\Bigg| \hspace{.5cm} x+1=0 \text{ or } x+2=0 \\ \text{Solve} \hspace{.5cm} &\Bigg| \hspace{.5cm} x=-1 \text{ or } x=-2 \end{align*}
When we factor, we are rearranging the equation to say that the product of two numbers is $0$. The equation is solved when either number is $0$, because any number multiplied by $0$ is $0$.
## How to Factor
Factoring is easiest in hindsight. Multiplying through, we see that the factored form is equivalent to the standard form:
\begin{align*} (x+1)(x+2) &= 0 \\ x(x+2) + 1(x+2) &= 0 \\ x^2 + 2x + 1x + 2 &= 0 \\ x^2 + 3x + 2 &= 0 \end{align*}
But how can we know this to begin with? In other words, if we want to factor an expression $x^2+bx+c$ into the form $(x+m)(x+n)$, how do we know what $m$ and $n$ are?
Here’s the trick: $m$ and $n$ need to multiply to $c$ and add to $b$.
To factor the expression $x^2+5x+4$, we need to find two numbers that multiply to $4$ and add to $5$. Although $2$ and $2$ multiply to $5$, they don’t add to $-2$. But $1$ and $-3$ multiply to $-3$ AND add to $-2$, so they work! The factored form is then $(x+1)(x-3)$.
Even with negatives, the method is still the same: to factor the expression $x^2-2x-3$, we need to find two numbers that multiply to $-3$ and add to $-2$. Although $-1$ and $3$ multiply to $-3$, they don’t add to $-2$. But $1$ and $-3$ multiply to $-3$ AND add to $-2$, so they work! The factored form is then $(x+1)(x-3)$.
## Case of Many Potential Factors
Factoring can become a little tricky when $c$ has a lot of factors. In such cases, it can be helpful to make a factor table.
For example, to factor $x^2+26x+144$, we can list out the factors of $144$ and find which pair adds to $26$. Since this pair is $8$ and $18$, the expression factors to $(x+8)(x+18)$.
\begin{align*} \textbf{Factor Pair} & \hspace{.5cm} \textbf{Sum} \\ \text{1 and 144} \hspace{.5cm} &\Bigg| \hspace{.5cm} \text{145} \\ \text{2 and 72} \hspace{.5cm} &\Bigg| \hspace{.5cm} \text{74} \\ \text{3 and 48} \hspace{.5cm} &\Bigg| \hspace{.5cm} \text{51} \\ \text{4 and 36} \hspace{.5cm} &\Bigg| \hspace{.5cm} \text{40} \\ \text{6 and 24} \hspace{.5cm} &\Bigg| \hspace{.5cm} \text{30} \\ \textbf{8}\text{ and }\textbf{18} \hspace{.5cm} &\Bigg| \hspace{.5cm} \textbf{26} \\ \text{9 and 16} \hspace{.5cm} &\Bigg| \hspace{.5cm} \text{25} \\ \text{12 and 12} \hspace{.5cm} &\Bigg| \hspace{.5cm} \text{24} \end{align*}
To speed up the process, notice that the sums are automatically ordered from biggest to smallest – so we don’t necessarily have to create the whole table.
We could have started with some intermediate pair, say $6$ and $24$, and realized that since the sum is too big, we need the first factor to be bigger than $6$.
Or, we could have noticed that sum of $12$ and $12$ is in the ballpark of $26$, and worked our way up from the bottom of the table.
## Case of Negative Terms
To deal with a negative value for $b$, we could use the same method as before, except that we would have to make both factors negative.
For example, since we know that $8$ and $18$ are factors of $144$ that add to $26$, we also know that $-8$ and $-18$ are factors of $144$ that add to $-26$, so the expression $x^2-26x+144$ factors to $(x-8)(x-18)$.
To deal with a negative value for $c$, we can think about the difference instead of the sum.
For example, to factor $x^2+32x-144$, we can find which factor pair of $144$ has a difference of $32$, and put a negative on the smaller factor to make the sum. Since this pair is $4$ and $36$, the expression factors to $(x-4)(x+36)$.
\begin{align*} \textbf{Factor Pair} & \hspace{.5cm} \textbf{Difference} \\ \text{1 and 144} \hspace{.5cm} &\Bigg| \hspace{.5cm} \text{143} \\ \text{2 and 72} \hspace{.5cm} &\Bigg| \hspace{.5cm} \text{70} \\ \text{3 and 48} \hspace{.5cm} &\Bigg| \hspace{.5cm} \text{45} \\ \textbf{4}\text{ and }\textbf{36} \hspace{.5cm} &\Bigg| \hspace{.5cm} \textbf{32} \\ \text{6 and 24} \hspace{.5cm} &\Bigg| \hspace{.5cm} \text{18} \\ \text{8 and 18} \hspace{.5cm} &\Bigg| \hspace{.5cm} \text{10} \\ \text{9 and 16} \hspace{.5cm} &\Bigg| \hspace{.5cm} \text{7} \\ \text{12 and 12} \hspace{.5cm} &\Bigg| \hspace{.5cm} \text{0} \end{align*}
If $b$ were negative as well – say, if we wanted to factor $x^2-32x-144$ – then we could use the same process but put the negative on the bigger factor to make the sum negative. That is, we would put the negative on the $36$ instead of the $4$, and the resulting factored form would then be $(x+4)(x-36)$.
## Case of a Common Factor
Sometimes, we can simplify quadratic expressions by factoring out something that ALL the terms have in common.
\begin{align*} \text{Original quadratic equation} \hspace{.5cm} &\Bigg| \hspace{.5cm} 3x^2-15+18=0 \\ \text{Factor a } 3 \text{ out of all terms} \hspace{.5cm} &\Bigg| \hspace{.5cm} 3(x^2-5x+6)=0 \\ \text{Factor the quadratic expression} \hspace{.5cm} &\Bigg| \hspace{.5cm} 3(x-2)(x-3)=0 \\ \text{Set each factor to } 0 \hspace{.5cm} &\Bigg| \hspace{.5cm} x-2=0 \text{ or } x-3=0 \\ \text{Solve} \hspace{.5cm} &\Bigg| \hspace{.5cm} x=2 \text{ or } x=3 \end{align*}
This makes it easy to factor quadratic expressions where $c$ is $0$ – just factor out the variable!
\begin{align*} \text{Original quadratic equation} \hspace{.5cm} &\Bigg| \hspace{.5cm} x^2+7x=0 \\ \text{Factor an } x \text{ out of all terms} \hspace{.5cm} &\Bigg| \hspace{.5cm} x(x+7)=0 \\ \text{Set each factor to } 0 \hspace{.5cm} &\Bigg| \hspace{.5cm} x=0 \text{ or } x+7=0 \\ \text{Solve} \hspace{.5cm} &\Bigg| \hspace{.5cm} x=0 \text{ or } x=-7 \\ \end{align*}
## Case when the Leading Coefficient is Not One
Factoring out the variable works even when $a$ is something other than $1$.
\begin{align*} \text{Original quadratic equation} \hspace{.5cm} &\Bigg| \hspace{.5cm} 2x^2-5x=0 \\ \text{Factor an } x \text{ out of all terms} \hspace{.5cm} &\Bigg| \hspace{.5cm} x(2x-5)=0 \\ \text{Set each factor to } 0 \hspace{.5cm} &\Bigg| \hspace{.5cm} x=0 \text{ or } 2x-5=0 \\ \text{Solve} \hspace{.5cm} &\Bigg| \hspace{.5cm} x=0 \text{ or } x=\frac{5}{2} \end{align*}
But what about when $a$ is something other than $1$, and $c$ is not zero?
There’s a little trick that lets us reduce this to a factoring problem with $a$ equal to $1$. We multiply $c$ by $a$, replace $a$ with $1$, factor the result, divide each constant in each factor by the original $a$, and move denominators onto our variables.
\begin{align*} \text{Original quadratic equation} \hspace{.5cm} &\Bigg| \hspace{.5cm} 6x^2+11x+3=0 \\ \begin{matrix} \text{Multiply } 3 \text{ by } 6 \text{, and} \\ \text{replace } 6 \text{ with } 1 \end{matrix} \hspace{.5cm} &\Bigg| \hspace{.5cm} x^2+11x+18=0 \\ \text{Factor normally} \hspace{.5cm} &\Bigg| \hspace{.5cm} (x+9)(x+2)=0 \\ \begin{matrix} \text{Divide each constant in} \\ \text{each factor by } 6 \end{matrix} \hspace{.5cm} &\Bigg| \hspace{.5cm} \left( x+\frac{9}{6} \right) \left( x+\frac{2}{6} \right)=0 \\ \text{Simplify} \hspace{.5cm} &\Bigg| \hspace{.5cm} \left( x+\frac{3}{2} \right) \left( x+\frac{1}{3} \right)=0 \\ \begin{matrix} \text{Move denominators onto} \\ \text{variables} \end{matrix} \hspace{.5cm} &\Bigg| \hspace{.5cm} (2x+3)(3x+1)=0 \\ \text{Set each factor to } 0 \hspace{.5cm} &\Bigg| \hspace{.5cm} 2x+3=0 \text{ or } 3x+1=0 \\ \text{Solve} \hspace{.5cm} &\Bigg| \hspace{.5cm} x=-\frac{3}{2} \text{ or } x=-\frac{1}{3} \end{align*}
We’ll talk about why this trick works in the next chapter, when we cover the quadratic formula.
## Case of No Middle Term
Lastly, what about when $b$ is $0$? Since the factors have to add to $b$, they must be negatives of each other. Since the factors have to multiply to $c$, and they are the same number (except one is negative), they must be the positive and negative square roots of $c$!
For example, $x^2-4$ factors to $(x+2)(x-2)$, and $x^2-9$ factors to $(x+3)(x-3)$.
This trick also works if $a$ is not equal to $1$ – we just have to factor $a$ out first.
\begin{align*} \text{Original quadratic equation} \hspace{.5cm} &\Bigg| \hspace{.5cm} 3x^2-48=0 \\ \text{Factor } 3 \text{ out of all terms} \hspace{.5cm} &\Bigg| \hspace{.5cm} 3(x^2-16)=0 \\ \text{Factor the quadratic} \hspace{.5cm} &\Bigg| \hspace{.5cm} 3(x+4)(x-4)=0 \\ \text{Solve} \hspace{.5cm} &\Bigg| \hspace{.5cm} x=-4 \text{ or } x=4 \end{align*}
## Exercises
Factor the following quadratic equations. Then, use the factored form to find the solutions. (You can view the solution by clicking on the problem.)
$1) \hspace{.5cm} x^2+7x+12=0$
Solution:
$(x+3)(x+4)=0$
$x=-3 \text{ or } x=-4$
$2) \hspace{.5cm} x^2+9x+14=0$
Solution:
$(x+7)(x+2)=0$
$x=-7 \text{ or } x=-2$
$3) \hspace{.5cm} x^2-7x=-10$
Solution:
$(x-2)(x-5)=0$
$x=2 \text{ or } x=5$
$4) \hspace{.5cm} x^2+18=9x$
Solution:
$(x-6)(x-3)=0$
$x=6 \text{ or } x=3$
$5) \hspace{.5cm} x^2+2x=8$
Solution:
$(x+4)(x-2)=0$
$x=-4 \text{ or } x=2$
$6) \hspace{.5cm} 21-4x=x^2$
Solution:
$(x+7)(x-3)=0$
$x=-7 \text{ or } x=3$
$7) \hspace{.5cm} 3x+10=x^2$
Solution:
$(x+2)(x-5)=0$
$x=-2 \text{ or } x=5$
$8) \hspace{.5cm} x^2-5x=36$
Solution:
$(x+4)(x-9)=0$
$x=-3 \text{ or } x=9$
$9) \hspace{.5cm} 4x^2=-52x$
Solution:
$4x(x+13)=0$
$x=0 \text{ or } x=-13$
$10) \hspace{.5cm} -8x^2+64x=0$
Solution:
$8x(x-8)=0$
$x=0 \text{ or } x=8$
$11) \hspace{.5cm} x^2-25=0$
Solution:
$(x+5)(x-5)=0$
$x=-5 \text{ or } x=5$
$12) \hspace{.5cm} x^2-144$
Solution:
$(x+12)(x-12)=0$
$x=-12 \text{ or } x=12$
$13) \hspace{.5cm} 12x^2+11x=-2$
Solution:
$(4x+1)(3x+2)=0$
$x=-\frac{1}{4} \text{ or } x=-\frac{2}{3}$
$14) \hspace{.5cm} 10x^2=27x-5$
Solution:
$(2x-5)(5x-1)=0$
$x=\frac{5}{2} \text{ or } x=\frac{1}{5}$
$15) \hspace{.5cm} 5x=4-6x^2$
Solution:
$(3x+4)(2x-1)=0$
$x=-\frac{4}{3} \text{ or } x=\frac{1}{2}$
$16) \hspace{.5cm} 21x^2-10=-29x$
Solution:
$(7x-2)(3x+5)=0$
$x=\frac{2}{7} \text{ or } x=-\frac{5}{3}$
This post is part of the book Justin Math: Algebra. Suggested citation: Skycak, J. (2018). Factoring Quadratic Equations. In Justin Math: Algebra. https://justinmath.com/factoring-quadratic-equations/
Tags:<|endoftext|>
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The Kilenge themselves are not sure of their origins: different legends variously ascribe their ancestors as coming from the north coast of New Guinea, the Siassi Islands, or the south coast of New Britain. Evidence suggests that their immediate forbears lived on the lower slopes of Mount Talave and slowly migrated down to the coast, arriving there about 150 years ago. The Germans began recruiting the Kilenge for labor around the turn of this century, establishing a pattern of wage-labor migration that persists today. Some depopulation resulted from a smallpox epidemic in the second decade of this century. World War II caused dislocation but few casualties. It also opened up new cultural and social horizons. Today, the Kilenge are marginally incorporated into the world economy. The Kilenge cultural repertoire, while related to those of other New Britain and Siassi Island groups, is unique in its particular configuration. The Kilenge are Primarily endogamous, and they distinguish themselves from other people, particularly their bush-dwelling Lolo neighbors, in terms of their particular combination of locality, language, marriage, and culture. In the past, the Kilenge participated in the overseas trade network organized and maintained by the Siassi Islanders, exchanging their pigs, coconuts, taro, and Talasea obsidian for carved bowls and clay pots needed in their bride-price payments. They also mediated the exchange between the Lolo and the Siassi and maintained ties with the Bariai, Kaliai, and Kove to the east.<|endoftext|>
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# How do you integrate int (1) / (x * ( x^2 - 1 )^2) using partial fractions?
Dec 25, 2017
The answer is $= - \frac{1}{2} \ln \left(| {x}^{2} - 1 |\right) + \ln \left(| x |\right) - \frac{1}{2 \left({x}^{2} - 1\right)} + C$
#### Explanation:
Reminder
${x}^{2} - 1 = \left(x + 1\right) \left(x - 1\right)$
Let's perform the decomposition into partial fractions
$\frac{1}{x {\left({x}^{2} - 1\right)}^{2}} = \frac{1}{x {\left(x + 1\right)}^{2} {\left(x - 1\right)}^{2}}$
$= \frac{A}{x} + \frac{B}{x + 1} ^ 2 + \frac{C}{x + 1} + \frac{D}{x - 1} ^ 2 + \frac{E}{x - 1}$
$= \frac{A {\left(x + 1\right)}^{2} {\left(x - 1\right)}^{2} + B \left(x\right) {\left(x - 1\right)}^{2} + C \left(x\right) {\left(x - 1\right)}^{2} \left(x + 1\right) + D \left(x\right) {\left(x + 1\right)}^{2} + E \left(x\right) {\left(x + 1\right)}^{2} \left(x - 1\right)}{x {\left({x}^{2} - 1\right)}^{2}}$
The denominators are the same, compare the numerators
$1 = A {\left(x + 1\right)}^{2} {\left(x - 1\right)}^{2} + B \left(x\right) {\left(x - 1\right)}^{2} + C \left(x\right) {\left(x - 1\right)}^{2} \left(x + 1\right) + D \left(x\right) {\left(x + 1\right)}^{2} + E \left(x\right) {\left(x + 1\right)}^{2} \left(x - 1\right)$
Let $x = 0$, $\implies$, $1 = A$
Let $x = 1$, $\implies$, $1 = 4 D$, $\implies$, $D = \frac{1}{4}$
Let $x = - 1$, $\implies$, $1 = - 4 B$, $\implies$, $B = - \frac{1}{4}$
Coefficients of ${x}^{4}$
$0 = A + C + E$, $\implies$, $C + E = - A = - 1$
Coefficients of ${x}^{3}$
$0 = B + D - C + E$, $\implies$, $B + D = C - E = 0$
$C = E = - \frac{1}{2}$
Therefore,
$\frac{1}{x {\left({x}^{2} - 1\right)}^{2}} = \frac{1}{x} + \frac{- \frac{1}{4}}{x + 1} ^ 2 + \frac{- \frac{1}{2}}{x + 1} + \frac{\frac{1}{4}}{x - 1} ^ 2 + \frac{- \frac{1}{2}}{x - 1}$
So,
$\int \frac{\mathrm{dx}}{x {\left({x}^{2} - 1\right)}^{2}} = \int \frac{\mathrm{dx}}{x} - \int \frac{\frac{1}{4} \mathrm{dx}}{x + 1} ^ 2 - \int \frac{\frac{1}{2} \mathrm{dx}}{x + 1} + \int \frac{\frac{1}{4} \mathrm{dx}}{x - 1} ^ 2 - \int \frac{\frac{1}{2} \mathrm{dx}}{x - 1}$
$= \ln \left(| x |\right) + \frac{1}{4 \left(x + 1\right)} - \frac{1}{2} \ln \left(| x + 1 |\right) - \frac{1}{4 \left(x - 1\right)} - \frac{1}{2} \ln \left(| x - 1 |\right) + C$<|endoftext|>
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# How do you solve this math problem?
## How do you solve this math problem?
Here are four steps to help solve any math problems easily:
1. Read carefully, understand, and identify the type of problem.
2. Draw and review your problem.
3. Develop the plan to solve it.
4. Solve the problem.
## What is the absolute value of 5?
The absolute value of 5 is 5, it is the distance from 0, 5 units.
How do I get math solutions on Google?
For this, all you have to do is type the math problem (the full problem) in the Search button. Or, you can use Google Lens and hover it over the math problem in your book or copy. Once Google captures the problem, you will be shown a series of steps that you can follow to solve the problem.
What is Bodmas rule in maths?
The Bodmas rule follows the order of the BODMAS acronym ie B – Brackets, O – Order of powers or roots, D – Division, M – Multiplication A – Addition, and S – Subtraction. The BODMAS rule states that mathematical expressions with multiple operators need to be solved from left to right in the order of BODMAS.
### What is the absolute value of negative 12?
It’s easy to understand that the absolute value of 12 and -12 are identical: they’re both 12, since each number is 12 units away from zero. Absolute value is always positive — to get the absolute value of a negative number, just take away the minus sign.
### What is the absolute value of (- 3?
3
For example, the absolute value of 3 is 3, and the absolute value of −3 is also 3. The absolute value of a number may be thought of as its distance from zero.
Does Google have a math solver?
Google Lens can solve simple equations such as “5+2” or more complex formulas such as “x2 – 3x + 2.” You can scan the problem from a real-world piece of paper or from a digital display. Open the “Google” app on your Android phone or tablet, iPhone, or iPad. Tap the “Lens” icon from the right side of the search bar.
How do you do homework on Google Lens?
3. At the bottom of the screen, choose “homework”.
## Is Pemdas or Bodmas correct?
To help students in the United States remember this order of operations, teachers drill the acronym PEMDAS into them: parentheses, exponents, multiplication, division, addition, subtraction. Other teachers use an equivalent acronym, BODMAS: brackets, orders, division and multiplication, and addition and subtraction.
## How to work with absolute value inequalities, step by step?
– 3x+2| < 0 | 3 x + 2 | < 0 – |x −9| ≤ 0 | x − 9 | ≤ 0 – |2x−4| ≥ 0 | 2 x − 4 | ≥ 0 – |3x−9| > 0 | 3 x − 9 | > 0
How can an absolute value inequality have no solution?
Isolate the absolute value by subtracting 9 from both sides of the inequality. The absolute value of a quantity can never be a negative number, so there is no solution to the inequality. Absolute inequalities can be solved by rewriting them using compound inequalities.
Which values are solutions to the inequality?
Eliminate fractions (multiplying all terms by the least common denominator of all fractions).
• Simplify (combine like terms on each side of the inequality).
• Add or subtract quantities (unknown on one side and the numbers on the other).
• ### How do you solve inequalities with two variables?
Add (or subtract) a number from both sides
• Multiply (or divide) both sides by a positive number
• Simplify a side<|endoftext|>
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# Tutoring math, you often get asked about sets of numbers. Let’s sort out what belongs where.
We’ll make this story as short as possible:
Naturals (N): {1,2,3,4…..} These might be referred to as counting numbers.
Wholes (W): {0,1,2,3,4…..} These include all the naturals, plus zero.
Integers (Z): {….-3,-2,-1,0,1,2,3….} These include all the whole numbers, plus the negatives of them.
Rationals (Q): You can’t list these numbers in order, since there is always another one between any two you name. However, you can define them as follows: a rational number consists of any integer divided by an integer other than zero.
In other words,
(1)
where integer1 can be zero, but integer2 cannot be zero. Therefore, rationals include the following examples:
Hence, we see that any integer, since it can be written as itself over 1, is rational.
It turns out that rationals also include repeating decimals as well as terminating ones. You can verify the facts on your calculator:
(2)
and of course
(3)
Up to and including the rationals, each set contains the previous one. That is, the rationals contain the integers, the integers contain the wholes and the wholes contain the naturals. However, the next set of numbers – called the Irrationals – is completely different from the rationals and separate from them. The irrationals contain non-repeating, non-terminating decimals. These numbers are written symbolically: examples are √11, as well as our friend pi (∏).
The Real Numbers (R) contain all the rationals, plus all the irrationals. There is yet another set: the Imaginary Numbers. We’ll save them for another post.
Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.
# Tutoring math, the FOIL method is used daily. We’ll explain it here.
In math, the distributive property is used constantly. An illustration:
(1)
To see what’s really happening here, look to the following:
We see that we take the outer number (in this case, 3) and multiply it by each term in the brackets, writing the result each time.
A binomial is a couple of terms that cannot be added. In the above example, 5x-7 is a binomial. When two binomials are multiplied together, it might look like this:
In order to multiply a binomial by a binomial, notice the steps below:
(The letters stand for First, Outer, Inner, Last: hence the acronym FOIL)
F: 2x times 4x gives 8x2
O: 2x times 9 gives 18x
I: -5 times 4x gives -20x
L: -5 times 9 gives -45
Writing the terms consecutively, we get
8x2 + 18x -20x -45
Now, we can combine the 18x with the -20x to finally obtain
8x2 -2x -45.
Indeed, (2x-5)(4x+9) = 8x2 -2x -45.
To verify the FOIL method, try it on (12)(10), written as
(10 +2)(9 + 1)
F: 10(9)=90
O: 10(1)=10
I: 2(9)=18
L: 2(1)=2
90 + 10 + 18 + 2 = 120, which of course we know is the answer to (12)(10).
Hope this helps.
Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC
# Tutoring high school physics, you get the privilege of retelling Newton’s three Laws. They can’t be discussed too often.
Newton’s First Law:
If no unbalanced force acts on an object, it either continues moving in a straight line at constant speed, or else remains in its state of rest.
Like his other laws, Newton’s First Law contains some surprises if you examine it closely. First of all, how could he have predicted it, when nothing on Earth does continue moving at a constant rate when left alone?
Back in high school, one of my textbooks (I can’t remember which one; it was over twenty years ago!) explained that Newton understood the tendency to retain constant velocity by comparing a ball to a brick. He noticed that when you throw a brick, it stops almost immediately when it hits the ground. However, a ball might continue moving for a long time after it contacts the ground.
Newton realized that the difference between the ball and the brick is that friction acts more emphatically on the brick. Given its shape, the brick catches the edges of the ground’s surface. The ground grips the brick, stopping it. The ball, with its round shape, makes smooth contact with the ground rather than rubbing along it. Friction is, of course, resistance to rubbing, so the brick’s movement across the ground is much more affected by friction than is the ball’s.
Newton realized that air resistance is just another case of friction. Therefore, everything moving on Earth – including the ball – is eventually arrested by friction. Friction is unbalanced unless you apply force to counteract it. Hence, a cyclist can continue at a constant speed if he or she is willing to pedal against friction. However, when that cyclist stops pedalling, the air gradually halts him/her.
In space, there is no air resistance, so objects can (and do) continue their straight line motion forever. Gravity can make things change direction, but gravity is then an unbalanced force. Unbalanced means that there is not an equivalent force opposing it.
Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.
# When you tutor high school chemistry, writing chemical formulas comes up every year. Having investigated ionic compounds, let’s sort out covalent ones.
If you look in the Chemistry category, you’ll see various posts about writing chemical formulas for ionic compounds. Those posts explain that, for ionic compounds, matching the combining capacities of the metals and the nonmetals is key to writing proper formulas. Another way to look at it is that the charges must be balanced in a viable ionic compound formula.
With covalent compounds, we needn’t worry about combining capacities; the name of the compound tells all. That’s because covalent names include prefixes that tell us the number of each type of atom involved:
prefix number mono 1 di 2 tri 3 tetra 4 penta 5 hexa 6 hepta 7 octa 8 nona 9 deca 10
Take, for example, dinitrogen pentoxide. The di before nitrogen tells us there are two nitrogens; the penta before oxide tells us there are five oxygens. Therefore, we have
dinitrogen pentoxide: N2O5
Going the other way, here is a point to mind: The prefix mono is rarely used. For example:
SF : sulphur hexafluoride.
Note that we don’t call it monosulphur hexafluoride, even though there is only one sulphur. However, we do call CO carbon monoxide. To my knowledge, mono is only used when another related compound is more common. In the case of carbon monoxide, we use mono to distinguish it from carbon dioxide, which of course is more often mentioned. Whatever the reason, the prefix mono is seldom used, even when there is only one of that atom. Other examples:
CCl4 : carbon tetrachloride
BF3 : boron trifluoride
Of course, one question that might evolve: “How can I tell if a compound is ionic or covalent?” The simple answer: If the compound is ionic, its name starts with a metal. If it’s covalent, it starts with a nonmetal. I explain how to tell a metal from a nonmetal here.
Remember: only covalent compounds use the numeric prefixes in the table above; ionic compounds don’t use them.
Source: Chemistry, Charles E. Mortimer, Sixth Edition, Wadsworth, Inc., 1986.
Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.
# When you tutor high school biology, cell biology is a mainstay. The particular role of the cell membrane sometimes gets neglected.
Membranes are very important in the construction of organisms. Every membrane has the basic function of keeping materials separate or else keeping them where they are. Every cell has a membrane surrounding (enveloping) it. Some people call it the cell membrane; others call it the plasma membrane.
The role of the cell membrane is sometimes described as protection – that it shelters the cell from its environment. To some extent you could argue that’s true, but really the cell membrane’s function is to control the movement of materials in and out of the cell.
You can think of the cell as a factory. It has valuable things inside that need to be there. Some of those valuables are destined to leave, but only under the correct circumstances. Until the appropriate time of departure, those valuables must remain safe inside.
The factory, however, can’t just be locked up until someone comes to take delivery of the merchandise. New supplies are arriving all the time, and need to be taken in. However, anything (and anyone) not meant to be there is stopped at the door.
A busy factory probably has its doors open all day, yet the movement of goods in and out of it is carefully controlled. Foremen watch the crews and the doors to make sure the right shipments leave at the right times. New shipments that arrive are confirmed before they are brought in. Securing the movement of the goods is a constant, involved process.
With a cell, the conduction of goods in and out is much the same. Water, oxygen and carbon dioxide can pass through the cell membrane freely, but many important molecules and ions need to be allowed in – or even brought in – by the membrane. Otherwise, they can’t enter. With exiting, it’s the same situation.
The cell “decides” what to let in at any given instant based on what it needs to maintain the composition of its cytoplasm. The cytoplasm is the nonspecific, jelly-like “body” of the cell. It’s a complex mix of water, ions, proteins, lipids, sugars, and other biological compounds. Cytoplasm’s precise mix comprises the “living condition” of the cell. If the mix goes wrong, the cell dies. Cytoplasm can only be found in a living cell.
Therefore, the cell membrane maintains the “living” mix of the cytoplasm by controlling what enters and leaves the cell. It’s a function that may not sound specific at first, but might be the most important of any the cell does. Just how the cell membrane manages its role will be discussed in future posts.
Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.
# One issue in English tutoring is apostrophes. We’ll sort out when to use them – and when not to.
In everyday English, you commonly see apostrophes misused. There are only two cases in which you do use them:
1) In contractions, such as can’t and don’t. The apostrophe means that letters have been left out to shorten the word. (For example, can’t means cannot.)
2) To show possession, such as John’s car or Heidi’s boots.
You don’t use apostrophes any other time.
Note that, when the possessor is plural, you put the apostrophe after the s:
That is my parents’ car.
Hope this saves you some marks :)
Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.
# Tutoring high school science, you might be asked about plate tectonics at any time. Here is the most condensed, pragmatic explanation you’ll likely find.
To believe in plate tectonics – which, by the way, is true beyond any doubt – you need to picture this: Earth’s surface consists of large islands of solid rock floating on a sea of magma. There are about 9 large plates, plus some smaller ones. Examples are the Indian Plate, the North American Plate, the South American Plate, the Eurasian Plate, the African Plate and the Pacific Plate.
The plates more or less cover the sea of magma beneath them. At the same time, because the plates float on the magma, they move around like toy boats in a bath tub. When they move apart, you have what’s called a divergent boundary. When they crash together, you have a convergent boundary. When they rub alongside each other, going in parallel but opposite directions (perhaps like people passing each other in a crowded hallway), you have a transform boundary. Essentially, the term fault can be substituted for boundary.
At a divergent boundary, a crack eventually forms between the plates, then magma leaks up between them, so you get volcanoes. The volcanoes can eventually form high ridges as they continue to erupt. Divergent boundaries are more common under the sea than on land.
At a convergent boundary, there are two possibilities. One is a head-on collision, in which case the plates buckle as they crash (like in a car crash). The plates deform upward, resulting in mountains. The Himalayas are such a mountain range, caused by the collision of the Indian Plate and the Eurasian Plate.
The second possible outcome at a convergent boundary is subduction, which is where one plate rides over top and the other slides below. If one plate is a continental one and the other is a sea plate, the sea plate slides under and the continental one rides up, forming a mountain chain. The Andes Mountains exemplify such a situation: there, the Nazca Plate is sliding under the South American Plate.
Although each type of boundary hosts earthquakes, a transform boundary results, especially, in earthquakes. As the plates try to shove past each other, they get stuck. The pressure builds, then they break loose, resulting in sudden movement – and an earthquake. The famous San Andreas Fault at San Francisco is a transform boundary: the Pacific Plate is running NNW, while the North American Plate is running SSE, along it.
Hope this gets you started :)
Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.
I pulled this article together from several high school textbooks:
Science Probe 10, Nelson Edition, Nelson Canada: 1996.
Earth Science, Spaulding and Namowitz, McDougal Littell: 2003.
The Changing Earth, McDougal Littell Science, McDougal Littell: 2005.
# When you tutor social studies or geography, you’ll likely have to explain the concepts of longitude and latitude. Now, we will.
The other day a kid came to me, embroiled in a conflict. One adult had told him lines of longitude lie north to south, while another one had told him “north and south is latitude.” Understandably, he was confused. What’s more, kids have a way of believing the adult they last talked to, rather than the one in front of them presently.
Luckily, I wasn’t either of the two adults he’d already talked to. Therefore, it was easy to explain to him that both those adults, in fact, had been right. As so often happens, he thought they’d been telling him opposing views, when really they’d been telling him the same truth but in different ways.
As the first adult said, lines of longitude do lie north to south. However, they don’t measure how far north or south you are. How far north or south you are is measured by your latitude, as the second adult pointed out. Of course, lines of latitude lie east to west.
“Think of a football field,” I told the kid. “The yard lines lie sideways across the field, but they don’t measure how far sideways you are. Rather, they measure how far forward you are. They lie sideways, but they measure your forward position. If you’re at 80 yards, it means you’ve crossed all the yard lines up to 80. Running forward, you cross them because they lie across your path rather than parallel to it.”
It’s the same with longitude. If you’re at 30° East, it means you’ve crossed the longitude lines from 1ºE through 29ºE, to land on the 30th one. Going East, you cross those lines of longitude because they lie North to South. If they ran East to West, then going East, you’d just stay on the same line forever.
In a similar way, lines of latitude lie East to West, but they measure your position north or south. Note, for example, the Equator: it’s at 0º latitude, yet obviously it runs East to West around the globe.
Although these ideas are obvious to anyone familiar with maps, they can be tough to grasp at first.
Well, the story has a happy ending. After explaining longitude and latitude to the kid, I told him to find the position of Moscow for me.
“Around 37E, 56N”, he reported five minutes later.
Good enough:)
Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.
# When you tutor math, decimal coefficients surface sometimes – especially in word problems. We’ll sort out a simple scenario.
Recently, someone asked me a question very similar to
(1)
Here’s what to do:
1) multiply the 206.5 by 0.7x:
(2)
2) combine the like terms on the left side:
(3)
3) of course, divide both sides by 256.55:
(4)
(5)
Hope this helps:)
Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.
# Whether tutoring French or learning it, these verb conjugations are critically important.
The French verb to be is être. Its present tense conjugation is as follows:
je suis (I am) nous sommes (we are) tu es (you [singular] are) vous êtes (you [plural] are) il est (he is) ils sont (they [masculine or mixed] are) elle est (she is) elles sont (they [females only] are)
The French verb to have is avoir. Its present tense conjugation:
j’ai (I have) nous avons (we have) tu as (you [singular] have) vous avez (you [plural] have) il a (he has) ils ont (they [masculine or mixed] have) elle a (she has) elles ont (they [females only] have)
Examples:
Nous sommes dans le salon. We are in the living room.
J’ai une gomme. I have an eraser.
Not only are these conjugations important for everyday communication, but also they comprise the auxiliary for the passé composé – the French past tense. We are headed in that direction – among others – for future posts.
Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC<|endoftext|>
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# 9.3: Simplification of Radical Expressions
Difficulty Level: At Grade Created by: CK-12
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Have you ever watched a baseball game? Have you ever seen a runner run so fast that he passed the intended base? Take a look at what Miguel saw.
Miguel was watching a game on Saturday night. During the fourth inning, the runner ran from second base to third base. He was running to fast that he ran past third base and had to come back to reach the base.
The runner ran 90+16\begin{align*}90 + \sqrt 16\end{align*}.
How far did the runner run?
This situation has been described using a radical expression. You will learn how to evaluate radical expressions in this Concept.
### Guidance
Sometimes, we can have an expression with a radical in it.
What is a radical?
A radical is the name of the sign that tells us that we are looking for a square root. We can call this “a radical.” Here is a radical symbol.
y\begin{align*}\sqrt{y}\end{align*}
Here we would be looking for the square root of y\begin{align*}y\end{align*}.
What is a radical expression?
Remember than an expressions is a number sentence that contains numbers, operations and now radicals. Just as we can have expressions without radicals, we can have expressions with them too.
Let's look at one.
24+7\begin{align*}2 \cdot \sqrt{4} + 7\end{align*}
Here we have two times the square root of four plus seven.
That’s a great question! We can evaluate this expression by using the order of operations.
P parentheses
E exponents (square roots too)
MD multiplication/division in order from left to right
AS addition/subtraction in order from left to right
According to the order of operations, we evaluate the square root of 4 first.
4=2\begin{align*}\sqrt{4} = 2\end{align*}
Next, we substitute that value into the expression.
22+7\begin{align*}2 \cdot 2 + 7\end{align*}
Next, we complete multiplication/division in order from left to right.
2×2=4\begin{align*}2 \times 2 = 4\end{align*}
Substitute that given value.
4+7=11\begin{align*}4 + 7 = 11\end{align*}
The answer is 11.
Let’s look at another one.
4163\begin{align*}\sqrt{4} \cdot \sqrt{16} - 3\end{align*}
Here we have two radicals in the expression. We can work the same way, by using the order of operations.
Evaluate the radicals first.
416=2=4\begin{align*}\sqrt{4} & = 2\\ \sqrt{16} & = 4\end{align*}
Substitute these values into the expression.
243\begin{align*}2 \cdot 4 - 3\end{align*}
Next, we complete multiplication/division in order from left to right.
2×4=8\begin{align*}2 \times 4 = 8\end{align*}
Finally, we complete the addition/subtraction in order from left to right.
83=5\begin{align*}8 - 3 = 5\end{align*}
The answer is 5.
Evaluate each radical expression.
#### Example A
6+949+5\begin{align*}6 + \sqrt{9} - \sqrt{49} + 5\end{align*}
Solution: 7\begin{align*}7\end{align*}
#### Example B
64÷4+13\begin{align*}\sqrt{64} \div \sqrt{4} + 13\end{align*}
Solution: 17\begin{align*}17\end{align*}
#### Example C
6(7)+1213\begin{align*}6(7) + \sqrt{121} - 3\end{align*}
Solution: 50\begin{align*}50\end{align*}
Here is the original problem once again.
Miguel was watching a game on Saturday night. During the fourth inning, the runner ran from second base to third base. He was running to fast that he ran past third base and had to come back to reach the base.
The runner ran 90+16\begin{align*}90 + \sqrt 16\end{align*}.
How far did the runner run?
To figure this out, we have to evaluate the radical expression.
The distance from one base to another is 90 feet.
The extra distance is the 16\begin{align*}\sqrt 16\end{align*}. Let's evaluate that part first.
16=4\begin{align*}\sqrt 16 = 4\end{align*}
Now let's add that into the original expression.
90+4=94\begin{align*}90 + 4 = 94\end{align*}
The runner ran 94 feet.
### Vocabulary
Here are the vocabulary words in this Concept.
Square Number
a number of units which makes a perfect square.
Square root
a number that when multiplied by itself equals the square of the number.
Perfect Square
square roots that are whole numbers.
Radical
the symbol that lets us know that we are looking for a square root.
Radical Expression
an expression with numbers, operations and radicals in it.
### Guided Practice
Here is one for you to try on your own.
Evaluate this radical expression.
8(7)+1449\begin{align*}8(7) + \sqrt{144} - 9\end{align*}
Answer
First, let's evaluate the square root.
144=12\begin{align*}\sqrt 144 = 12\end{align*}
Now we can substitute 12 back into the original expression and solve using the order of operations.
8(7)+129\begin{align*}8(7) + 12 - 9\end{align*}
56+129\begin{align*}56 + 12 - 9\end{align*}
689=59\begin{align*}68 - 9 = 59\end{align*}
Our answer is 59\begin{align*}59\end{align*}.
### Video Review
Here is a video for review.
### Practice
Directions: Evaluate each radical expression.
1. 2+9+152\begin{align*}2 + \sqrt{9} + 15-2\end{align*}
2. 34+169\begin{align*}3 \cdot 4 + \sqrt{169}\end{align*}
3. 1625+36\begin{align*}\sqrt{16} \cdot \sqrt{25} + \sqrt{36}\end{align*}
4. 8112+12\begin{align*}\sqrt{81} \cdot 12 + 12\end{align*}
5. 36+4716\begin{align*}\sqrt{36} + \sqrt{47} - \sqrt{16}\end{align*}
6. 6+36+252\begin{align*}6 + \sqrt{36} + 25-2\end{align*}
7. 4(5)+92\begin{align*}4(5) + \sqrt{9}-2\end{align*}
8. 15+16+5\begin{align*}15 + \sqrt{16} + 5\end{align*}
9. 3(2)+25+10\begin{align*}3(2) + \sqrt{25} + 10\end{align*}
10. 4(7)+4912\begin{align*}4(7) + \sqrt{49}-12\end{align*}
11. 2(4)+98\begin{align*}2(4) + \sqrt{9}-8\end{align*}
12. 3(7)+25+21\begin{align*}3(7) + \sqrt{25} + 21\end{align*}
13. 8(3)36+152\begin{align*}8(3) - \sqrt{36} + 15-2\end{align*}
14. 19+14422\begin{align*}19 + \sqrt{144}-22\end{align*}
15. 3(4)+64+25\begin{align*}3(4) + \sqrt{64} + \sqrt{25}\end{align*}
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# What is formula for volume of a cube?
## What is formula for volume of a cube?
Volume of a cube = side times side times side. Since each side of a square is the same, it can simply be the length of one side cubed. If a square has one side of 4 inches, the volume would be 4 inches times 4 inches times 4 inches, or 64 cubic inches.
## What is the volume of this rectangular pyramid?
Finding the Volume of a Rectangular Pyramid Rectangular pyramids have four-sided bases and four triangular sides coming together in an apex, or what we know more simply as the pointy tip. Your overall formula for finding the volume of these multi-faceted shapes is V = (l x w x h) / 3.
## What is the volume of rectangular?
To find the volume of a rectangular prism, multiply its 3 dimensions: length x width x height. The volume is expressed in cubic units.
## What is the formula for volume science?
Density offers a convenient means of obtaining the mass of a body from its volume or vice versa; the mass is equal to the volume multiplied by the density (M = Vd), while the volume is equal to the mass divided by the density (V = M/d).
## How do you calculate a cube?
What is a cubed number?
1. When you multiply a whole number (not a fraction) by itself, and then by itself again the result is a cube number.
2. The easiest way to do this calculation is to do the first multiplication (3×3) and then to multiply your answer by the same number you started with; 3 x 3 x 3 = 9 x 3 = 27.
## How do you cite in text APA 7?
When using APA format, follow the author-date method of in-text citation. This means that the author’s last name and the year of publication for the source should appear in the text, like, for example, (Jones, 1998). One complete reference for each source should appear in the reference list at the end of the paper.
## How do you find the length width and height if you have the volume?
Divide the volume by the product of the length and width to calculate the height of a rectangular object. For this example, the rectangular object has a length of 20, a width of 10 and a volume of 6,000. The product of 20 and 10 is 200, and 6,000 divided by 200 results in 30. The height of the object is 30.
## How do you cite Instagram in APA 7th edition?
Instagram profile Use the name of the profile page you want to cite in the title element of the reference (e.g., “Posts,” “IGTV,” “Tagged”). Include the description “[Instagram profile]” in square brackets. Provide a retrieval date because the content is designed to change over time and is not archived.
## How would you find the volume of a box?
To find the volume of a box, simply multiply length, width, and height – and you’re good to go! For example, if a box is 5x7x2 cm, then the volume of a box is 70 cubic centimeters.
## How do you find the surface area and volume of a cube?
The side length of the cube is 8 meters. Now use this value to determine the volume using the formula V = s3. The volume of the cube is 512 cubic meters. To determine the surface area of a cube, calculate the area of one of the square sides, then multiply by 6 because there are 6 sides.<|endoftext|>
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By Lynne Brehm, Statewide Coordinator, Rooted in Relationships
Flash cards and boot camps and Baby Einstein, oh my! When I mention the idea of school readiness, most parents envision training their kids up to be reading Tolstoy and building robots by the time they enroll at age 5. And it’s no wonder — parents are getting conflicting messages that can lead to high anxiety about expectations.
Relax. That is not at all what school readiness is.
School readiness has been achieved when the child has the social-emotional competency needed for success.
That’s right . . . it’s about social-emotional development. Social-emotional development is the developing capacity of the young child to:
- Experience, regulate and express emotions
- Form close and secure relationships
- Explore their environment and learn
Healthy social-emotional development occurs within a caregiving environment that includes family, community and cultural expectations.
A socially and emotionally prepared 5-year-old:
- Can follow developmentally appropriate directions
- Does not disrupt the class
- Can play cooperatively with others
- Knows how to communicate thoughts and needs
- Is curious and enthusiastic about new activities
- Regulates emotions under most circumstances
- Forms close and secure relationships
That’s the core of it. Of course, the child must be well fed, well rested and otherwise healthy as well to be ready to learn. These skills come along with those vital executive functions we’ve been talking about. If this all sounds like it’s too simple, just imagine a child who did not have these characteristics. What would learning look like for that child? How would the classroom environment be affected?
A child who is hungry or has had too little sleep can’t pay attention.
A child with little skill at self regulating cannot persist at a task, follow instructions, play well with others or suppress outbursts.
A child who has no interest in new experiences because he is distracted by issues at home or other stressors will find little inspiration in a classroom, and little energy to learn.
And a child who doesn’t know how to form and maintain relationships will be lonely and unhappy.
For many families, social-emotional school readiness is a challenge.
As many as 40% of children enter kindergarten lacking one or more of these critical social-emotional skills. About 10% of kindergartners exhibit challenging behaviors that disrupt the class, cause problems with other children or necessitate intensive interventions. And for 65% of the children identified as having emotional or behavior disorders will drop out of school.
We know that dropping out of school often leads to poor job outcomes, limited income, and patterns of failure that may persist into adulthood, including unplanned pregnancies and criminal activity. If the ability to form and sustain healthy relationships is not formed during those early years, it is much harder later and adults may struggle to attach with their partners and children….and the cycle continues.
This is one of those ounce-of-prevention moments.
From a family perspective, paying more attention to how our very young children are developing emotionally and socially than we pay to trying to make them the smartest kid on the block is the best course of action. From a systems perspective, focusing on the social and emotional development of those children most at risk will improve their outcomes, our classrooms and our communities.
Here’s what we need to do:
- Prevent Adverse Childhood Experiences – Infuse families and communities with the protective factors they need to ensure that the adverse childhood experiences that can derail social-emotional development don’t happen. And when ACEs do occur, we must provide immediate support to mitigate the toxic stress that can result. We can all do this. In schools, we can provide to support to families we see struggling. We can reach out to our neighbors and act as informal support to those who need help. We can support home visitation and early learning programs that are proven to get results.
- Teach parents to engage and attach – Home visitation programs like those in Sixpence communities work. They provide parents a powerful set of skills to engage with their children, understand serve-and-return interactions, and work for the better social-emotional development of their babies. Most parents want what’s best for their children . . . but many don’t know how to do it. Parenting is a skill like anything else, and skills are learned from good teachers.
- Focus on early childhood caregivers and environments – Preparing early childhood educators to build healthy relationships with young children and foster environments that are developmentally geared to social-emotional learning are fundamental practices that will help ensure that most young children reach their social-emotional milestones.
- Plan social emotional support and advanced interventions for pre-schoolers who need additional assistance – For those children who require more individual, intentional supports, early childhood educators should be well versed in the resources available so they can recommend next actions. Without a plan for interventions, we will continue to see very young children being expelled from preschool programs for behavioral issues.
The thing to remember is this: without these base social-emotional skills in place, all the flash cards and kindergarten boot camps in the world won’t matter. Let’s engage with our kids, build secure attachments and form the necessary relationships at home.. In our early childhood programs, let’s focus energy on that critical brain development. The result will be greater school success and more positive life outcomes.<|endoftext|>
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Continues 3 Biblical Models for Grounding Education. Read the series so far.
2. Paul’s Model for Transformative Learning
Paul recognizes that in order for us to understand how best to educate people, we must understand what a person actually is. These days he has competition, however, as five major contemporary theories of learning all make significant assumptions about what a person is and how they are best educated.
Behaviorism focuses on the learners’ response to stimuli, and postulates that if you can control the environment through operant conditioning, then you can create change in the behavior of the learner. B.F. Skinner was convinced that the person was essentially an active organism that was conditioned to behavior. Cognitivism focuses on “the representations and processes needed to give rise to activities ranging from pattern recognition, attention, categorization, memory, reasoning, decision making, problem solving, and language.”22 Humanity is essentially a computing device, processing and acting based on schemas. As the educator assesses where the student is in Piaget’s four stages of development (sensorimotor, pre-operational, concrete operational, or formal operational),23 the educator determines what information and tasks are age-appropriate for the computer to handle next.<|endoftext|>
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PP Section 1.5
# PP Section 1.5 - Graphs and Graphing Utilities Definition...
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Graphs and Graphing Utilities
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Definition and Example The graph of an equation in two variables x and y consists of the set of points in the xy-plane whose coordinates (x,y) satisfy the equation. Example: Since 5 = 2(2) + 1, the point (2, 5) is on the graph of the equation y = 2x + 1
Graph y x = - + 3 5 Example : x = 0 y = -3(0) + 5 = 5 x = 1 y = -3(1) + 5 = -3 + 5 = 2 x = 5 y = -3(5) + 5 = -15 + 5 = -10 x = -1 y = -3(-1) + 5 = 3 + 5 = 8 x y 0 5 1 2 5 -10 -1 8
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By plotting the points found in the previous step, the graph can be drawn: -12 -10 -8 -6 -4 -2 0 2 4 6 8 10 12 -10 -8 -6 -4 -2 0 2 4 6 8 10
The points, if any, at which the graph crosses or touches the coordinate axes are called the intercepts. Procedure for Finding Intercepts 1. To find the x -intercept(s), if any, of the graph of an equation, let y = 0 in the equation and solve for x. 2. To find the y -intercept(s), if any, of the graph of an equation, let x = 0 in the equation and solve for y .
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Example: Find the x-intercepts and y-intercepts
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One of the areas studied in mathematics is the inscribed angle. An angle using a circle is an inscribed angle, and by using the properties of the inscribed angle, we can calculate the angle.
There are properties of inscribed angles, which is called the inscribed angle theorem. If you do not understand the theorem, you will not be able to calculate the angle. Also, since there are advanced problems that use the theorem, you need to learn how to calculate angles in a circle, including special cases.
So, how can we use the inscribed angle theorem to solve calculation problems? Also, what is an inscribed angle?
In this article, we will explain the details of the inscribed angle theorem that we learn in mathematics.
## What Is an Inscribed Angle? Meaning of Arc and Central Angle
First of all, what is an inscribed angle? Before we understand the inscribed angle, we need to learn about the arc and the central angle. For a circle, the following parts are the arc and central angle.
When using the inscribed angle theorem, we will frequently use arcs and central angles. Therefore, it is important to understand which parts are arcs and central angles.
On the other hand, the inscribed angle refers to the following parts.
The arc AB is represented in mathematics as $\stackrel{\huge\frown}{AB}$. ∠APB corresponding to $\stackrel{\huge\frown}{AB}$ is called the inscribed angle. An inscribed angle always has an arc. Also, if there is an inscribed angle, there is always a central angle.
## There Are Two Theorems for Inscribed Angles
What are the theorems of inscribed angles? Inscribed angles have two properties. They are as follows.
• Half of the central angle is always an inscribed angle.
• If the arc lengths are the same, the inscribed angles are always equal.
Let’s check the details of the inscribed angle theorem.
### Half of the Central Angle Is Always an Inscribed Angle
The central angle is frequently used in the inscribed angle theorem. The reason for this is that there is a relationship between the inscribed angle and the central angle as follows.
• Inscribed angle × 2 = Central angle
For example, if the inscribed angle is 30°, the central angle will always be 60°; multiply the inscribed angle by two to get the central angle. On the other hand, if the central angle is 80°, the inscribed angle is always 40°. Half of the central angle is the inscribed angle.
### If the Arc Lengths Are the Same, the Inscribed Angles Are Equal
There is another theorem about the inscribed angle. For the same circle, if the arc lengths are the same, then the inscribed angles are equal. For example, the following inscribed angles are all the same.
By using the same arc, we can find other angles. For example, for the following figure, what is the angle of $a$?
Focusing on $\stackrel{\huge\frown}{BC}$, we see that ∠BDC is 20°. Therefore, we know that the angle of $a$ is 20°.
## If the Arc Is a Semicircle, the Inscribed Angle Is 90°
So far, we have discussed the most basic aspects of the inscribed angle theorem. However, there are more complicated calculation problems that use the inscribed angle theorem. Therefore, you need to be able to solve advanced problems. A typical example of this is the semicircle arc.
If the arc is a semicircle, we have the following.
If the arc is a semicircle, the central angle is 180°. Therefore, the inscribed angle will always be 90°. Any inscribed angle is 90°, as shown below.
If the central angle has a special shape, it is difficult to find the inscribed angle. In any case, remember that if the arc is a semicircle, the inscribed angle will always be 90°.
### How to Consider the Inscribed Angle, If the Arc Is Larger than a Semicircle?
On the other hand, how should we think about the inscribed angle when the arc is larger than a semicircle? In some cases, the arc length is small, while in other cases, the arc length is long, as shown below.
The idea is the same as what we have explained so far. If the arc is longer than a semicircle, the central angle will be greater than 180°. Therefore, the inscribed angle will always be greater than 90°.
The length of the arc is not necessarily shorter than the semicircle. There are cases where the arc length is longer than the semicircle, as in this case. Try to understand where the central angle is and where the inscribed angle is. For example, it is a mistake to think of the relationship between the central angle and the inscribed angle as follows.
In the figure above, the central angle is smaller than 180°. Nevertheless, the inscribed angle is larger than 90° in the figure. Therefore, the central angle and inscribed angle in this relationship are obviously wrong.
You need to find exactly where the central angle and inscribed angle are to be able to calculate the angle correctly. If the arc is longer than a semicircle, mistakes are more likely to occur. So when the central angle is greater than 180°, check to see if the inscribed angle is greater than 90°.
### If You Add the Opposite Angles of a Quadrilateral Inscribed in a Circle, You Get 180°
By learning what we have discussed so far, you will be able to understand the properties of angles of quadrilaterals inscribed in a circle. For quadrilaterals inscribed in a circle, adding the opposite angles always results in 180°.
Why do the two angles add up to 180°? Let’s use the inscribed angle theorem we have learned so far to find out why.
For each angle, the relationship between arc, central angle, and inscribed angle is as follows.
The important thing to remember is that the two central angles add up to 360°. Also, the inscribed angle is half of the central angle. Therefore, two inscribed angles (opposite angles) add up to 180°.
Remembering the property that the opposite angles of a quadrilateral inscribed in a circle add up to 180° will help you when calculating angles. Although the quadrilateral must be inscribed in a circle, it is important to remember that the opposite angles always add up to 180°.
### Arc Length and Inscribed Angle Are Proportional
For the same circle, the arc length and the inscribed angle are proportional. In other words, if the arc length is doubled, the inscribed angle will also be doubled.
It is easy to understand that the central angle is proportional to the arc length. As shown below, if the central angle increases from 30° to 60°, the arc length doubles.
The arc length can be calculated using the circumference and the central angle. For example, if the diameter of the circle is $b$ and the central angle is 30°, the arc length is as follows.
• $b×π×\displaystyle\frac{30°}{360°}$
On the other hand, if the central angle is 60°, the arc length is as follows.
• $b×π×\displaystyle\frac{60°}{360°}$
Thus, if the central angle is doubled, the arc length is doubled. In the same way, if the inscribed angle is doubled, the arc length is doubled. The central angle is proportional to the arc length. Also, the central angle and the inscribed angle are proportional. Therefore, the inscribed angle is proportional to the arc length.
Similarly, if the central angle or inscribed angle is tripled, the arc length will be tripled. If the arc length is quadrupled, the central angle and inscribed angle will be quadrupled. Understand that arc length and inscribed angle are proportional.
## Exercise: Using the Inscribed Angle Theorem to Calculate an Angle
Q1: Find the following angles.
In the problem of finding an angle, randomly write down the angles you can find. This way, you will be able to get the angle at some point.
Also, there is more than one way to find the angle. It doesn’t matter which method you use, as long as you can find the answer.
(a)
The arc corresponding to ∠ADC is $\stackrel{\huge\frown}{ABC}$. The central angle is twice the inscribed angle, so the central angle is 220°.
∠AOB is 180°. Therefore, $∠a$ is 40°.
• 220° – 180° = 40°
(b)
For complex problems, with the exception of a few geniuses, the answer cannot be given immediately. So, actively draw lines and write angles on the figure. In this way, you will eventually be able to get the answer.
If you draw a line from the center O, the length will always be the same. In other words, you can make an isosceles triangle. Since the base angles are the same in an isosceles triangle, we can find out the angles as follows.
Since it is an isosceles triangle, ∠BCO is 15°. Therefore, we know that ∠BOC is 150°. On the other hand, the arc for $∠b$ is $\stackrel{\huge\frown}{ABC}$. The central angle corresponding to $∠b$ is as follows.
• 70° + 150° = 220°
Since the central angle is 220°, $∠b$ is 110°.
To get this answer, we need to randomly draw lines on the figure and find angles as mentioned above. From there, find the angle you need to get the answer and do the calculation.
## Using the Property of Inscribed Angle to Calculate Unknown Angles
There are two theorems of inscribed angles to remember. The first is that half of the central angle is the inscribed angle. The other is that if the arc lengths are the same, the inscribed angles will always be the same.
However, although there are only two theorems for inscribed angles, there are several things you need to understand if you include advanced problems. It is important to understand beforehand what happens to the inscribed angle when the arc is a half-circle or longer than a semicircle.
Also, for a quadrilateral inscribed in a circle, the opposite angles always add up to 180°. Other than that, the arc length is proportional to the size of the inscribed angle.
Let’s use these facts to solve inscribed angle problems. Since inscribed angle problems can be complicated, it is best to find the answer by randomly writing down the angles you find.<|endoftext|>
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SII.MP.1.a: Explain the meaning of a problem and look for entry points to its solution. Analyze givens, constraints, relationships, and goals. Make conjectures about the form and meaning of the solution, plan a solution pathway, and continually monitor progress asking, “Does this make sense?” Consider analogous problems, make connections between multiple representations, identify the correspondence between different approaches, look for trends, and transform algebraic expressions to highlight meaningful mathematics. Check answers to problems using a different method.
SII.MP.3.a: Understand and use stated assumptions, definitions, and previously established results in constructing arguments. Make conjectures and build a logical progression of statements to explore the truth of their conjectures. Justify conclusions and communicate them to others. Respond to the arguments of others by listening, asking clarifying questions, and critiquing the reasoning of others.
SII.MP.4.a: Apply mathematics to solve problems arising in everyday life, society, and the workplace. Make assumptions and approximations, identifying important quantities to construct a mathematical model. Routinely interpret mathematical results in the context of the situation and reflect on whether the results make sense, possibly improving the model if it has not served its purpose.
SII.MP.7.a: Look closely at mathematical relationships to identify the underlying structure by recognizing a simple structure within a more complicated structure. See complicated things, such as some algebraic expressions, as single objects or as being composed of several objects.
SII.MP.8.a: Notice if reasoning is repeated, and look for both generalizations and shortcuts. Evaluate the reasonableness of intermediate results by maintaining oversight of the process while attending to the details.
(Framing Text): Extend the properties of exponents to rational exponents.
N.RN.1: Explain how the definition of the meaning of rational exponents follows from extending the properties of integer exponents to those values, allowing for a notation for radicals in terms of rational exponents.
(Framing Text): Use properties of rational and irrational numbers.
N.RN.3: Explain why sums and products of rational numbers are rational, that the sum of a rational number and an irrational number is irrational, and that the product of a nonzero rational number and an irrational number is irrational. Connect to physical situations (e.g., finding the perimeter of a square of area 2).
(Framing Text): Perform arithmetic operations with complex numbers.
N.CN.1: Know there is a complex number i such that i² = -1, and every complex number has the form a + bi with a and b real.
(Framing Text): Use complex numbers in polynomial identities and equations.
N.CN.7: Solve quadratic equations with real coefficients that have complex solutions.
(Framing Text): Interpret the structure of expressions.
A.SSE.1: Interpret quadratic and exponential expressions that represent a quantity in terms of its context.
A.SSE.1.a: Interpret parts of an expression, such as terms, factors, and coefficients.
A.SSE.1.b: Interpret increasingly more complex expressions by viewing one or more of their parts as a single entity. Exponents are extended from the integer exponents to rational exponents focusing on those that represent square or cube roots.
A.SSE.2: Use the structure of an expression to identify ways to rewrite it.
(Framing Text): Write expressions in equivalent forms to solve problems, balancing conceptual understanding and procedural fluency in work with equivalent expressions.
A.SSE.3: Choose and produce an equivalent form of an expression to reveal and explain properties of the quantity represented by the expression. For example, development of skill in factoring and completing the square goes hand in hand with understanding what different forms of a quadratic expression reveal.
A.SSE.3.a: Factor a quadratic expression to reveal the zeros of the function it defines.
A.SSE.3.c: Use the properties of exponents to transform expressions for exponential functions.
(Framing Text): Perform arithmetic operations on polynomials. Focus on polynomial expressions that simplify to forms that are linear or quadratic in a positive integer power of x.
A.APR.1: Understand that polynomials form a system analogous to the integers, namely, they are closed under the operations of addition, subtraction, and multiplication; add, subtract, and multiply polynomials.
(Framing Text): Create equations that describe numbers or relationships. Extend work on linear and exponential equations to quadratic equations.
A.CED.1: Create equations and inequalities in one variable and use them to solve problems. Include equations arising from linear and quadratic functions, and simple rational and exponential functions.
A.CED.2: Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales.
A.CED.4: Rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations; extend to formulas involving squared variables.
(Framing Text): Solve equations and inequalities in one variable.
A.REI.4: Solve quadratic equations in one variable.
A.REI.4.a: Use the method of completing the square to transform any quadratic equation in x into an equation of the form (x – p)² = q that has the same solutions. Derive the quadratic formula from this form.
A.REI.4.b: Solve quadratic equations by inspection (e.g., for x² = 49), taking square roots, completing the square, the quadratic formula and factoring, as appropriate to the initial form of the equation. Recognize when the quadratic formula gives complex solutions and write them as a ± bi for real numbers a and b.
(Framing Text): Interpret quadratic functions that arise in applications in terms of a context.
F.IF.5: Relate the domain of a function to its graph and, where applicable, to the quantitative relationship it describes. Focus on quadratic functions; compare with linear and exponential functions.
F.IF.6: Calculate and interpret the average rate of change of a function (presented symbolically or as a table) over a specified interval. Estimate the rate of change from a graph.
(Framing Text): Analyze functions using different representations.
F.IF.7: Graph functions expressed symbolically and show key features of the graph, by hand in simple cases and using technology for more complicated cases.
F.IF.7.a: Graph linear and quadratic functions and show intercepts, maxima, and minima.
F.IF.7.b: Graph piecewise-defined functions and absolute value functions. Compare and contrast absolute value and piecewise-defined functions with linear, quadratic, and exponential functions. Highlight issues of domain, range, and usefulness when examining piecewise-defined functions.
F.IF.8: Write a function defined by an expression in different but equivalent forms to reveal and explain different properties of the function.
F.IF.8.a: Use the process of factoring and completing the square in a quadratic function to show zeros, extreme values, and symmetry of the graph, and interpret these in terms of a context.
F.IF.8.b: Use the properties of exponents to interpret expressions for exponential functions.
F.IF.9: Compare properties of two functions, each represented in a different way (algebraically, graphically, numerically in tables, or by verbal descriptions). Extend work with quadratics to include the relationship between coefficients and roots, and that once roots are known, a quadratic equation can be factored.
(Framing Text): Build a function that models a relationship between two quantities.
F.BF.1: Write a quadratic or exponential function that describes a relationship between two quantities.
F.BF.1.a: Determine an explicit expression, a recursive process, or steps for calculation from a context.
F.BF.1.b: Combine standard function types using arithmetic operations. For example, build a function that models the temperature of a cooling body by adding a constant function to a decaying exponential, and relate these functions to the model.
(Framing Text): Build new functions from existing functions.
F.BF.3: Identify the effect on the graph of replacing f(x) by f(x) + k, k f(x), f(kx), and f(x + k) for specific values of k (both positive and negative); find the value of k given the graphs. Focus on quadratic functions and consider including absolute value functions. Experiment with cases and illustrate an explanation of the effects on the graph using technology. Include recognizing even and odd functions from their graphs and algebraic expressions for them.
(Framing Text): Construct and compare linear, quadratic, and exponential models and solve problems.
F.LE.3: Observe using graphs and tables that a quantity increasing exponentially eventually exceeds a quantity increasing linearly, quadratically, or (more generally) as a polynomial function. Compare linear and exponential growth to quadratic growth.
(Framing Text): Prove and apply trigonometric identities. Limit θ to angles between 0 and 90 degrees. Connect with the Pythagorean Theorem and the distance formula.
F.TF.8: Prove the Pythagorean identity sin²(θ) + cos²(θ) = 1 and use it to find sin (θ), cos (θ), or tan (θ), given sin (θ), cos (θ), or tan (θ), and the quadrant of the angle.
(Framing Text): Prove geometric theorems. Encourage multiple ways of writing proofs, such as narrative paragraphs, flow diagrams, two-column format, and diagrams without words. Focus on the validity of the underlying reasoning while exploring a variety of formats for expressing that reasoning.
G.CO.9: Prove theorems about lines and angles.
G.CO.10: Prove theorems about triangles.
G.CO.11: Prove theorems about parallelograms.
(Framing Text): Understand similarity in terms of similarity transformations.
G.SRT.1: Verify experimentally the properties of dilations given by a center and a scale factor.
G.SRT.1.b: The dilation of a line segment is longer or shorter in the ratio given by the scale factor.
G.SRT.2: Given two figures, use the definition of similarity in terms of similarity transformations to decide whether they are similar; explain using similarity transformations the meaning of similarity for triangles as the equality of all corresponding pairs of angles and the proportionality of all corresponding pairs of sides.
(Framing Text): Prove theorems involving similarity.
G.SRT.4: Prove theorems about triangles.
G.SRT.5: Use congruence and similarity criteria for triangles to solve problems and to prove relationships in geometric figures.
(Framing Text): Define trigonometric ratios and solve problems involving right triangles.
G.SRT.6: Understand that by similarity, side ratios in right triangles are properties of the angles in the triangle, leading to definitions of trigonometric ratios for acute angles.
G.SRT.8: Use trigonometric ratios and the Pythagorean Theorem to solve right triangles in applied problems.
(Framing Text): Understand and apply theorems about circles.
G.C.2: Identify and describe relationships among inscribed angles, radii, and chords.
(Framing Text): Translate between the geometric description and the equation for a conic section.
G.GPE.1: Derive the equation of a circle of given center and radius using the Pythagorean Theorem; complete the square to find the center and radius of a circle given by an equation.
(Framing Text): Explain volume formulas and use them to solve problems.
G.GMD.1: Give an informal argument for the formulas for the circumference of a circle, area of a circle, volume of a cylinder, pyramid, and cone. Informal arguments for area formulas can make use of the way in which area scale under similarity transformations: when one figure in the plane results from another by applying a similarity transformation with scale factor k, its area is k² times the area of the first. Use dissection arguments, Cavalieri’s principle, and informal limit arguments.
G.GMD.3: Use volume formulas for cylinders, pyramids, cones, and spheres to solve problems. Informal arguments for volume formulas can make use of the way in which volume scale under similarity transformations: when one figure results from another by applying a similarity transformation, volumes of solid figures scale by k³ under a similarity transformation with scale factor k.
(Framing Text): Summarize, represent, and interpret data on two categorical or quantitative variables.
S.ID.5: Summarize categorical data for two categories in two-way frequency tables. Interpret relative frequencies in the context of the data (including joint, marginal, and condition reltive frequencies). Recognize possible associations and trends in the date.
(Framing Text): Understand independence and conditional probability and use them to interpret data.
S.CP.1: Describe events as subsets of a sample space (the set of outcomes) using characteristics (or categories) of the outcomes, or as unions, intersections, or complements of other events (“or,” “and,” “not”).
(Framing Text): Use the rules of probability to compute probabilities of compound events in a uniform probability model.
S.CP.6: Find the conditional probability of A given B as the fraction of B’s outcomes that also belong to A, and interpret the answer in terms of the model.
Correlation last revised: 5/20/2019<|endoftext|>
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For centuries, farmers took their cues for planting times from observing what’s happening in nature—such as bird migration, the emergence of insects and amphibians (like peepers), and the flowering of native plants.
Nature’s Signs: Phenology
Phenology is the study of periodic plant and animal life cycle events and how these are influenced by seasonal and interannual variations in climate, as well as habitat factors. Watching for nature’s signs is called “phenology”, from the Greek for the” science of appearances”. Trees, shrubs and flowers are sensitive to temperature and day length and develop on a regular schedule based on local conditions.
It only makes sense to use them as indicators of when the weather is right for planting. Observations made over many years have led to some fairly reliable conclusions.
Following Nature’s Lead
Nature’s signs are different in every region, however, you should still relate to these examples:
- Blooming crocus are your cue to plant radishes, parsnips and spinach.
- Half-hardy vegetables including beets, carrots and chard can be planted when the daffodils blossom.
- When the forsythia is in bloom it is safe to plant peas, onion sets and lettuce.
- Look for dandelions to bloom before planting potatoes.
- Perennials can be planted when the maple trees begin to leaf out.
- When quince is blossoming, transplant cabbage and broccoli
- Wait for apple trees to bloom before planting bush beans.
- When the apple blossoms fall, plant pole beans and cucumbers.
- By the time the lilacs are in full bloom it will be safe to plant tender annual flowers and squashes.
- Transplant tomatoes when lily-of-the-valley is in full flower.
- Full-sized maple leaves signal time to plant morning glory seeds.
- Peppers and eggplant can be transplanted when the irises are blooming.
- When peonies blossom it is safe to plant heat-loving melons.
While not totally foolproof, following nature’s signs helps us tune in to the rhythm of life around us.<|endoftext|>
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According to investopedia, federal actions and guidelines that restrict or restrain international trade, often done with the intent of safeguarding local businesses and careers from international competition. Typical ways of protectionism are import tariffs, quotas, subsidies or tax slices to local businesses and immediate state intervention. Protectionism is the monetary plan of restraining trade between states through methods such as tariffs on imported goods, restrictive quotas, and a variety of other government legislation made to discourage imports preventing foreign take-over of local marketplaces and companies (source: Wikipedia).
Protectionism, coverage of protecting local industries against foreign competition by means of tariffs, subsidies, transfer quotas, or other constraints or handicaps positioned on imports of international challengers. (source: Britannica Encyclopedia)
PROTECTION OF LOCAL INDUSTRIES
Keeping money at home
Equalizing cost and price
Enhancing national security
Protecting child industry
Keeping Money at Home
Trade unions and protectionists often dispute that international trade will lead to an outflow of money, making foreigners richer and local people poorer. This argument is dependant on fallacy of regarding money as the only real indicators of wealth. Other resources, even products, can even be indicators of prosperity. Also, this protectionist discussion assumes that foreigners get money and never have to give something of value in return. Whether local consumers buy locally made products or foreign products, they have to spend money to cover such products.
It is a standard practice for trade unions and politicians to attack imports and international trade in name of job safety. The argument is based on the assumption that transfer reduction will create more demand for local products and eventually create more jobs.
Equalizing Cost and Price
Some protectionists attempt to justify their activities by invoking financial theory. They argue that overseas goods have lower prices because of lower development costs. Therefore, trade obstacles are had a need to make prices of brought in products less competitive and local items more competitive.
Enhancing Country wide Security
Protectionists often present themselves as patriots. They often claim that a country should be self-sufficient and even prepared to cover inefficiency in order to enhance countrywide security. Opponents of protectionism however dismiss attracts nationwide security. A region can never be completely self-sufficient because recycleables are not within the same percentage in all regions of the entire world.
Protecting toddler industry
The necessity to safeguard a child industry could very well be the most credible debate for protectionist actions. Some industries have to be shielded until they become practical. Here South Korea serves as an example. It has performed well by selectively safeguarding infant business for export purpose.
(Source: used from Sak Onkvisit, John J. Shaw, International Marketing: Analysis and Strategy)
Reasons for protectionism:
(source: implemented from econessays. com)
1. Baby industry debate: - small firms have to be protected so as to have a chance to expand and
gain economies of size to be able to be able to compete on an international basis later on.
However so far this has took place only in big companies like the material industry and it
gives a purpose for firms to stay lazy because they know they don't have to remain competitive on an
international level e. g. steel industry in america.
2. Dumping to prevent firms from offering goods at a loss to ruin the home industry. By
allowing free trade you can find assurance for low prices indefinitely because the moment one
firm becomes inefficient better ones will enter in the marketplace and take it away.
3. Raise earnings for the government through tariffs.
4. Prevent overspecialization and diseconomies of level in other words over development in a
country because of the need to export goods because this will also lead to misallocation of
resources which is exactly what we want to prevent by free trade.
5. To remove an equilibrium of repayments deficit without however tackling the challenge at its root
this is inefficiency.
Non-economic reason for protectionism:
1. Strategic hobbies: some companies including the security industry are easier to be kept
domestic. For example a country can't be based upon others for it weapons industry because in
the circumstance of war it would be remaining unarmed.
2. Politics reasons: lack of willingness to operate due to political differences. For instance China
and Japan don't operate due to politics disputes.
3. Avoidance of the transfer of demerit goods such as tobacco and liquor.
4. Way of life and maintenance of traditional approach to life.
5. Safety against low income economies: some countries gain comparative benefits by
offering lower wages. For example people are imposing trade restrictions on China because it
underpays its employees and therefore no other market has the capacity to contend with her.
Alternative for protectionism:
1. Offering subsidies to suppliers, which can be an unpopular alternative because the money will
have to be lifted through taxes.
2. Free trade area: free trade between member countries; members impose whatever tariffs they
wish towards non-member countries. Types of they are CAFTA, LAFTA, and NAFTA
3. Traditions union: free trade between member countries; members must charge a common
external tariff against non-member countries. The European union is the only existing such example.
Policies of Protectionism
Tariffs: Typically, tariffs (or fees) are enforced on brought in goods. Tariff rates usually differ according to the type of goods imported. Transfer tariffs will improve the cost to importers, and improve the price of brought in goods in the neighborhood markets, thus reducing the number of goods brought in. Tariffs can also be imposed on exports, and in an market with floating exchange rates, export tariffs have similar effects as import tariffs. However, since export tariffs tend to be perceived as 'hurting' local establishments, while import tariffs are regarded as 'assisting' local market sectors, export tariffs are rarely implemented.
Import quotas: To lessen the quantity and for that reason increase the market price of brought in goods. The economical effects of an transfer quota act like that of a tariff, except that the duty revenue gain from a tariff will instead be sent out to the people who receive transfer licenses. Economists often claim that import licenses be auctioned to the best bidder, or that import quotas be changed by an comparative tariff.
Administrative barriers: Countries are sometimes accused of utilizing their various administrative rules (e. g. regarding food safety, environmental standards, electronic security, etc. ) as a way to introduce obstacles to imports.
Anti-dumping legislation: Supporters of anti-dumping regulations argue that they prevent "dumping" of cheaper foreign goods that could cause local firms to close down. However, in practice, anti-dumping laws and regulations are usually used to impose trade tariffs on overseas exporters.
Direct subsidies: Authorities subsidies (by means of lump-sum repayments or cheap loans) are sometimes directed at local companies that cannot remain competitive well against overseas imports. These subsidies are purported to "protect" local careers, and to help local businesses adjust to the world markets.
Export subsidies: Export subsidies tend to be used by governments to increase exports. Export subsidies are the opposite of export tariffs, exporters are paid a percentage of the worthiness of the exports. Export subsidies improve the amount of trade, and in a country with floating exchange rates, have results similar to import subsidies.
Exchange rate manipulation: A authorities may intervene in the foreign exchange market to lower the value of its money by offering its money in the foreign exchange market. Doing this will raise the price of imports and lower the price tag on exports, leading to an improvement in its trade balance. However, such an insurance plan is only effective in the short run, as it'll most likely business lead to inflation in the country, which will subsequently raise the expense of exports, and decrease the relative price of imports.
International patent systems: There can be an argument for viewing countrywide patent systems as a cloak for protectionist trade procedures at a national level. Two strands of this argument are present: one when patents organised by one country form part of a system of exploitable relative advantage in trade negotiations against another another where adhering to a worldwide system of patents confers "good citizenship" status despite 'de facto protectionism'.
(Source: Protectionist Insurance policies, Wikipedia)
SOURCE: International online marketing strategy: analysis, development and implementationBy Isobel Doole, Robin Lowe
Non-tariff barriers to trade (NTBs) are trade barriers that restrict imports but are not in the most common form of a tariff. Some common examples of NTB's are anti-dumping measures and countervailing duties, which, although they are called "non-tariff" barriers, have the result of tariffs after they are enacted.
Their use has increased sharply following the WTO rules led to an extremely significant decrease in tariff use. Some non-tariff trade obstacles are expressly allowed in very limited circumstances, when they are deemed necessary to protect health, security, or sanitation, or to protect depletable natural resources. In other forms, they may be criticized as a way to evade free trade rules such as those of the World Trade Organization(WTO), the European Union (EU), or UNITED STATES Free Trade Arrangement (NAFTA) that restrict the use of tariffs.
Some of non-tariff barriers are not straight related to foreign economic regulations, but nevertheless they have a significant effect on foreign-economic activity and foreign trade between countries.
Trade between countries is referred to operate in goods, services and factors of production. Non-tariff obstacles to operate include import quotas, special licenses, unreasonable expectations for the grade of goods, bureaucratic delays at customs, export restrictions, limiting the actions of point out trading, export subsidies, countervailing tasks, technical obstacles to operate, sanitary and phyto-sanitary measures, rules of source, etc. Sometimes in this list they include macroeconomic actions impacting trade.
Six Types of Non-Tariff Barriers to Trade
Import Licensing requirements
Proportion limitations of overseas to domestic goods (local content requirements)
Minimum transfer price limits
Intergovernmental acceptances of examining methods and standards
Packaging, labeling, and marking
Government procurement policies
Domestic assistance programs
Prior import first deposit subsidies
Special supplementary duties
Import credit discriminations
Voluntary export restraints
Orderly marketing agreements
Examples of Non-Tariff Obstacles to Trade
General or product-specific quotas
Rules of Origin
Quality conditions imposed by the importing country on the exporting countries
Sanitary and phyto-sanitary conditions
Complex regulatory environment
Determination of eligibility of any exporting country by the importing country
Determination of eligibility of the exporting establishment(firm, company) by the importing country.
Additional trade documents like Qualification of Origin, Certificate of Authenticity
Occupational basic safety and health regulation
State subsidies, procurement, trading, condition ownership
Fixation of the very least import price
Foreign exchange market control buttons and multiplicity
"Buy nationwide" policy
Intellectual property laws (patents, copyrights)
Seasonal transfer regimes
Corrupt and/or long customs procedures
Types of Non-Tariff Barriers
There are several different variants of division of non-tariff barriers. Some scholars separate between internal taxes, administrative barriers, health insurance and sanitary legislation and federal government procurement procedures. Others split non-tariff barriers into more categories such as specific limits on trade, customs and administrative admittance procedures, standards, federal government involvement in trade, charges on transfer, and other categories. We choose traditional classification of non-tariff obstacles, according to that they are divided into 3 primary categories.
The first category includes methods to directly import constraints for safety of certain sectors of national market sectors: licensing and allocation of import quotas, antidumping and countervailing obligations, import debris, so-called voluntary export restraints, countervailing duties, the machine of minimum import prices, etc. Under second category follow methods that are not directly aimed at restricting international trade plus more related to the administrative bureaucracy, whose actions, however, restrict trade, for example: traditions procedures, technical requirements and norms, sanitary and veterinary expectations, requirements for labeling and product packaging, bottling, etc. The 3rd category involves methods that are not directly aimed at restricting the import or promoting the export, but the effects which often lead to this result.
The non-tariff barriers can include wide selection of restrictions to operate. Below are a few example of the "popular" NTBs.
The most typical instruments of immediate legislation of imports (and sometimes export) are licenses and quotas. Virtually all industrialized countries apply these non-tariff methods. The certificate system requires a express (through specially official office) issues permits for overseas trade deals of transfer and export commodities included in the lists of accredited merchandises. Product licensing may take many forms and procedures. The main types of licenses are general license that permits unrestricted importation or exportation of goods included in the lists for a certain period of time; and one-time permit for a certain product importer (exporter) to transfer (or export). One-time certificate indicates a quantity of goods, its cost, its country of origin (or vacation spot), and in some instances also traditions point by which import (or export) of goods should be completed. The use of licensing systems as an instrument for international trade regulation is based on lots of international level benchmarks agreements. Specifically, these agreements include some provisions of the overall Arrangement on Tariffs and Trade and the Contract on Transfer Licensing Types of procedures, concluded under the GATT (GATT).
Licensing of overseas trade is closely related to quantitative limitations - quotas - on imports and exports of certain goods. A quota is a restriction in value or in physical conditions, imposed on transfer and export of certain goods for a certain time frame. This category includes global quotas according to specific countries, seasonal quotas, and so-called "voluntary" export restraints. Quantitative adjustments on overseas trade transactions carried out through one-time certificate.
Quantitative limitation on imports and exports is a primary administrative form of authorities regulation of overseas trade. Licenses and quotas limit the freedom of businesses with a regard to entering overseas markets, narrowing the number of countries, which may be entered into exchange for certain commodities, regulate the number and range of goods permitted for transfer and export. However, the system of licensing and quota imports and exports, building stable control over international trade in certain goods, in many cases actually is more adaptable and effective than economical instruments of foreign trade regulation. This is explained by the actual fact, that licensing and quota systems are an important instrument of trade regulation of the vast majority of the planet.
Agreement on the "voluntary" export restraint
In the past decade, a wide-spread practice of concluding agreements on the "voluntary" export limitations and the establishment of import minimum prices imposed by leading American nations after weaker in inexpensive or politics sense exporters. The details of these kind of limitations is the establishment of unconventional techniques when the trade barriers of importing country, are created at the boundary of the exporting and not importing country. Thus, the contract on "voluntary" export restraints is imposed on the exporter under the risk of sanctions to limit the export of certain goods in the importing country. In the same way, the establishment of lowest transfer prices should be totally noticed by the exporting companies in agreements with the importers of the united states that has place such prices. Regarding reduction of export prices below the minimum level, the importing country imposes anti-dumping obligation that could lead to drawback from the marketplace. "Voluntary" export contracts affect trade in textiles, footwear, dairy products, consumer electronics, autos, machine tools, etc.
Problems come up when the quotas are sent out between countries, since it is necessary to ensure that products from one country aren't diverted in violation of quotas lay out in second country. Import quotas aren't necessarily made to protect domestic manufacturers. For example, Japan, retains quotas on many agricultural products it does not produce. Quotas on imports is a leverage when negotiating the sales of Japanese exports, as well as preventing excessive dependence on any other country in respect of necessary food, items of which may reduction in case of inclement weather or political conditions.
Export quotas can be set in order to provide local consumers with sufficient stocks and options of goods at low prices, to avoid the depletion of natural resources, as well concerning increase export prices by restricting supply to foreign marketplaces. Such constraints (through agreements on various types of goods) allow producing countries to utilize quotas for such commodities as espresso and essential oil; as the result, prices for these products increased in importing countries.
Embargo is a specific kind of quotas prohibiting the trade. Aswell as quotas, embargoes may be enforced on imports or exports of particular goods, regardless of destination, in respect of certain goods supplied to specific countries, or according of all goods transported to certain countries. However the embargo is usually introduced for politics purposes, the consequences, essentially, could be financial.
Standards take a special place among non-tariff barriers. Countries usually impose expectations on classification, labeling and assessment of products in order to be able to sell home products, but also to block sales of products of overseas manufacture. These expectations are sometimes came into under the pretext of safeguarding the security and health of local populations.
Administrative and bureaucratic delays at the entrance
Among the methods of non-tariff legislation should be described administrative and bureaucratic delays at the entry which increase uncertainty and the expense of preserving inventory.
Another example of foreign trade polices is import deposits. Import deposits is a kind of deposit, that your importer must pay the lender for a definite time period (non-interest bearing deposit) within an amount equal to all or part of the expense of imported goods.
At the countrywide level, administrative legislation of capital movements is carried out mainly in a framework of bilateral agreements, which include a specific definition of the legal program, the procedure for the entrance of investment funds and investors. It really is determined by method (reasonable and equitable, nationwide, most-favored-nation), order of nationalization and compensation, transfer revenue and capital repatriation and dispute quality.
Foreign exchange constraints and foreign exchange controls
Foreign exchange limitations and foreign exchange controls occupy a particular place among the non-tariff regulatory instruments of foreign financial activity. Forex limitations constitute the regulation of ventures of residents and nonresidents with money and other currency prices. Also an important part of the mechanism of control of foreign monetary activity is the establishment of the national currency against foreign currency.
The change from tariffs to non-tariff barriers
One of the reasons why industrialized countries have relocated from tariffs to NTBs is the fact that developed countries have resources of income apart from tariffs. Historically, in the formation of nation-states, governments was required to get funding. They received it through the release of tariffs. This explains the fact that a lot of growing countries still rely on tariffs as a way to funding their spending. Developed countries can afford not to depend on tariffs, at exactly the same time developing NTBs just as one way of international trade rules. The second reason for the move to NTBs is that these tariffs can be used to support weak industries or settlement of industries, which were affected adversely by the reduction of tariffs. The 3rd reason for the acceptance of NTBs is the power of interest groupings to influence the process in the absence of opportunities to acquire government support for the tariffs.
Non-tariff obstacles today
With the exception of export subsidies and quotas, NTBs are most like the tariffs. Tariffs for goods creation were reduced through the eight rounds of negotiations in the WTO and the General Arrangement on Tariffs and Trade (GATT). After bringing down of tariffs, the theory of protectionism demanded the introduction of new NTBs such as complex barriers to trade (TBT). Matching to claims made at US Meeting on Trade and Development (UNCTAD, 2005), the use of NTBs, predicated on the total amount and control of prices has decreased significantly from 45% in 1994 to 15% in 2004, while use of other NTBs increased from 55% in 1994 to 85% in 2004.
Increasing consumer demand for safe and environment friendly products also have had their effect on increasing popularity of TBT. Many NTBs are governed by WTO agreements, which originated in the Uruguay Circular (the TBT Contract, SPS Measures Arrangement, the Arrangement on Textiles and Clothing), as well as GATT articles. NTBs in the field of services have become as important as in the field of usual trade.
Most of the NTB can be defined as protectionist methods, unless they are simply related to complications in the market, such as externalities and information asymmetries information asymmetries between consumers and manufacturers of goods. A good example of this is protection requirements and labeling requirements.
The need to protect sensitive to transfer business, as well as a variety of trade restrictions, available to the government authorities of industrialized countries, forcing those to holiday resort to use the NTB, and putting serious obstacles to international trade and world financial expansion. Thus, NTBs can be known as a "new" of safety which has substituted tariffs as an "old" form of protection.
A case once and for all protectionism
Bharat Jhunjhunwala (source: The Hindu Business Line)
THE defeat of the NDA Federal government and the win of the Congress (I) backed by the Departed is yet another warning sign of the growing worldwide backlash against globalization. White- collar workers in professional countries are sacrificing their jobs to the cheap labor of India and China. Services, such as research, are now being outsourced because researchers in the expanding countries are cheaper.
On the other palm, personnel in the growing countries have found that their wages are stagnant while inequality is rising. The belief was that free trade brings about efficient development and also makes domestic government to lessen corruption. This provides alleviation to the people. Else businessmen would need to pay money to local thugs and politicians to avoid trouble.
Government officers would have to be bribed to run normal business. For instance, a boiler inspector can shut down a plant for 15 days on frivolous grounds. The money paid to politicians and officers by the entrepreneur adds to the expense of production and boosts the price tag on his produce say, towel to Rs 25 a metre instead of Rs 20. The expense of production of similar towel in other countries having good governance, however, remains low because they don't have to bribe politicians and officers.
The cost of other inputs, such as natural cotton, machines and chemicals, remains the same in every countries because of free trade. Material produced in another country can overcome Indian market segments if the expense of production for the reason that "clean" country is Rs 20 which is Rs 25 in "corrupt" India. Textile mills in India must down shutters.
Ultimately, politicians will have to reduce the money they extract from the businessmen failing which they will be killing the goose that lays fantastic eggs. The exact same pertains to inefficient businessmen.
Globalization will power the Indian entrepreneur to install latest looms to be able to endure. This provides good and cheap fabric to the Indian people. Globalization, indeed, begets clean governance and useful production.
The difficulty, however, is that free trade also works in the Labor market. Say, India and another country both have clean government authorities and the cost of production of material in both countries is Rs 20 a metre. The income rate in the other country is Rs 80 per day.
The Indian businessman will not be able to pay more than this rate to his employees as in any other case his cost of production will increase and he'll be priced out of the market. The country paying lowest income is victorious in free trade. Free trade causes equalization of salary rates to their global minimum levels. This decrease in wages nullifies the benefits from good governance and successful production. No marvel employees in the industrial countries are opposing free trade and outsourcing. Software developers are finding their wage rate declining as technology can help you transfer large sums of data at the click of the mouse.
The income rates in most expanding countries are also stagnant. Staff in East Asian countries are seeing their wage rates decline anticipated to competition from the less paid Chinese language personnel. Free trade works as a two-edged sword.
On the one hand, it leads to clean governance and efficient production but on the other it causes lowering of wage rates with their global minimum. What's the solution to the problem? How can the advantages of free trade be anchored while creating higher pay for the individuals?
Protectionism enables local prices to stay higher than the global prices. Such higher prices may be used to support corruption, inefficient production or higher wages. The answer originates from using protection not for corruption or inefficient development but also for higher wages. Suppose India were to impose an additional taxes of Rs 5 per metre on cloth imports. The price of fabric in the Indian market would become Rs 25 instead of Rs 20 prior.
This margin can be taken away by corrupt politicians and officials, or used to keep up inefficient development in outdated mills, or to raise income of the staff. The ability is based on preventing the first two uses and promoting the 3rd. If the federal government establishes, say, a system to snare corrupt politicians and officials, promotes local competition to avoid inefficient creation, and implements procedures that lead to raised pay, then this protectionism becomes pro-people.
Free trade is automatically anti-people because it brings about low pay even if it provides good governance and reliable production. Protectionism can possibly be pro-people if applied properly. What about exports, though? It is possible to prevent cheap imports by imposing tariffs. But how will exports be made if the domestic wage rates are high?
The solution is to use the receipts from import taxes to provide export subsidies to Labor-intensive products. The bigger cost scheduled to high income can be neutralised by the subsidies. It is clear that free trade won't lead to the welfare of folks anywhere in the world. Protectionism can help you secure people's welfare but only when applied properly. But bad protectionism that facilitates corruption is worse than free trade. The task is to embrace good protectionism.
FREE TRADE OR PROTECTIONISM?
The Circumstance Against Trade Restrictions
by Vincent H. Miller & James R. Elwood
(source: isil. org)
The Lure of Protectionism
The argument for so-called "protectionism" (called "good trade" by some) may at first sound appealing. Followers of "protectionist" regulations declare that keeping out overseas goods helps you to save jobs, offering ailing domestic business a chance to retrieve and prosper, and reduce the trade deficits. Are these cases valid?
Protectionism: What It Costs
LOST Careers: Protectionist laws and regulations raise taxes (tariffs) on brought in goods and/or impose boundaries (quotas) on the amount of goods governments permit to enter into a country. These are regulations that not only restrict the decision of consumer goods, but also contribute greatly both to the price of goods and the expense of conducting business. So under "protectionism" you conclude poorer, with less money for buying other activities you want and need. In addition, protectionist laws and regulations that reduce consumer spending vitality actually conclude destroying jobs. In the USA, for example, according to the US Office of Labor's own information, "protectionism" destroys eight careers in the overall economy for every one kept in a protected industry.
HIGHER PRICES: Japanese consumers pay five times the world price for rice because of import restrictions protecting Japanese farmers. Western consumers pay dearly for EC constraints on food imports and heavy fees for domestic farm subsidies. American consumers also have problems with the same dual burden, paying six times the world price for glucose because of trade restrictions (to give but one example). THE UNITED STATES Semiconductor Trade Pact, which pressured Japanese producers to lessen creation of computer memory chips, induced an severe worldwide shortage of the widely used parts. Prices quadrupled and companies using these components in the development of electric consumer goods, in a variety of countries round the world, were terribly hurt.
HIGHER TAXES: Protectionist regulations not only pressure someone to pay more fees on brought in goods, but also raise your general taxes as well. It is because governments invariably grow their Customs Section bureaucracies to force compliance using their new rounds of trade limitations (or regarding NAFTA, trade legislation). These bureaucrats must be paid. Addititionally there is the expense of more red tape and paperwork for trading companies and even more harassment of specific travelers passing through the borders.
THE DEBT Problems: Western Banks are owed a huge selection of billions of dollars by Eastern Western european and Third World countries. Trade limitations by Western government authorities, however, have cut off Western marketplaces for these countries, rendering it virtually impossible for them to earn the hard currencies necessary to repay their lending options. This increases the very real opportunity of your collapse of the world bank operating system.
Protectionism: Who Benefits?
In spite of evidence of destruction induced by trade constraints, pressure for much more "protectionist" regulations persists. Who is behind this, and just why?
Those who gain from "protectionist" laws and regulations are special-interest teams, such as some big firms, unions, and farmers' categories - all of whom would like to get away with charging higher prices and getting higher income than they could expect in a free current market. These special passions have the money and political clout for influencing politicians to go laws beneficial to them. Politicians in turn play on the doubts of uninformed voters to rally support for these laws.
THE LOSERS? YOU and all other common consumers. Your independence is being trampled in to the particles by these laws, and you are virtually being robbed, through taxes and higher prices, in order to sections the pockets of a few politically-privileged "fat felines. "
"Protectionism is a misnomer. The one people guarded by tariffs, quotas and trade limitations are those involved in uneconomic and wasteful activity. Free trade is the only real philosophy compatible with international serenity and success. "
Senior Economist, Fraser Institute (Canada)
Trade Wars: Both Factors Lose
When the government of Country "A" puts up trade barriers against the goods of Country "B", the government of Country "B" will in a natural way retaliate by erecting trade obstacles against the products of Country "A". The result? A trade battle in which both sides lose. But all too often a depressed overall economy is not the sole negative outcome of your trade war. . .
When Goods Don't Mix Borders,
Armies Often Do
Europe suffered with almost non-stop wars through the 17th and 18th hundreds of years, when restrictive trade coverage (mercantilism) was the rule; rival government authorities fought one another to develop their empires and to exploit captive market segments.
British tariffs provoked the American colonists to revolution, and later the Northern-dominated US federal imposed limitations on Southern cotton exports - a major factor resulting in the American Civil Conflict.
In the later 19th Century, following a half century of general free trade (which brought a half-century of tranquility), short-sighted politicians throughout European countries again started out erecting trade obstacles. Hostilities built up until they eventually exploded into World Battle I.
In 1930, facing only a gentle recession, US Chief executive Hoover ignored alert pleas in a petition by 1028 dominant economists and authorized the notorious Smoot-Hawley Work, which brought up some tariffs to 100% levels. Within the season, over 25 other governments acquired retaliated by passing similar laws. The effect? World trade arrived to a grinding halt, and the whole world was plunged into the "Great Major depression" for the rest of the decade. The depressive disorder in turn resulted in World War II.
The #1 Hazard To World Peace
The world enjoyed its biggest economic growth during the relatively free trade amount of 1945-1970, a period that also found no major wars. Yet we again see trade barriers being raised around the world by short-sighted politicians. Will the world again finish up in a firing war therefore of the economically-deranged insurance policies? Can we manage to allow this to occur in the nuclear get older?
"What generates war is the financial philosophy of nationalism: embargoes, trade and forex controls, monetary devaluation, etc. The idea of protectionism is a beliefs of warfare. "
Ludwig von Mises
The Solution: Free Trade
A century. 5 ago French economist and statesman Frederic Bastiat presented the practical case for free trade: "It is always beneficial, " he said, "for a nation to specialize in what it can produce best and then trade with others to acquire goods at costs less than it would try produce them at home. " In the 20th century, journalist Frank Chodorov made an identical observation: "Society thrives on trade due to the fact trade makes specialty area possible, and specialty area increases end result, and increased outcome reduces the cost in toil for the satisfactions men live by. That being so, the market place is a most humane establishment. "
What Can You Do?
Silence offers consent, and there should be no consent to the present waves of restrictive trade or capital control legislation being transferred. If you agree that free trade is an essential element in keeping world peacefulness, and that it is important to your future, we suggest that you notify the political market leaders in your country of your concern regarding their interference with free trade. Send them a backup of the pamphlet. We also suggest that you write letters to editors in the advertising and send this pamphlet to them. Discuss this matter with your friends and warn them of the threat of current "protectionist" developments. Check how the problem is being taught in the institutions. Widespread public understanding of this issue, followed by resident action, is the one solution. Free trade is too important an issue to leave in the hands of politicians.
"For thousands of years, the tireless effort of productive men and women has been put in trying to reduce the length between areas of the world by lowering the costs of commerce and trade.
"On the same course of record, the slothful and incompetent protectionist has endlessly sought to erect barriers to be able to prohibit competition - thus, effectively moving communities farther aside. When trade is cut off entirely, the true providers may as well be on different planets.
"The protectionist represents the most severe in mankind: concern with change, concern with obstacle, and the jealous envy of genius. The protectionist is not against the utilization of each kind of push, even warfare, to crush his competitor. If mankind is to endure, then these primeval anxieties must be defeated. "
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Frieden, J. , Lake, D. , International political market: perspectives on global ability and wealth, London: Routledge, 1995
Mansfield, E. , Busch, M. , The politics economy of Non-tariff barriers: a cross nationwide analysis; International Business, Vol. 49, No. 4, 1995
Oatley, T. , International political economy: hobbies and companies in the global market; Harlow: Longman, 2007
Roorbach, G. , Tariffs and Trade Obstacles in Relation to International Trade; Proceedings of the Academy of Political Technology, Vol. 15, No 2, 1993
Yu, Zhihao, A model of Substitution of Non-Tariff Obstacles for Tariffs; The Canadian Journal of Economics, Vol. 33, No. 4, 2000
World Trade Corporation Website, Non-tariff obstacles: red tape, etc; http://www. wto. org/english/thewto_e/whatis_e/tif_e/agrm9_e. htm
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## How to Find Proportional Ratios
### What are proportional ratios?
In mathematics, a ratio is defined as the comparison between two numbers. This is generally done to find out how big or small a number or a quantity is with respect to another. So, what method do we use to find these ratios? Well, we use the division method. In a ratio, two numbers are divided. The dividend part is known as the ‘antecedent’, whereas the divisor is known as the ‘consequent.’
So, let’s look at an example. Suppose, in a computer shop having $$25$$ computers, $$13$$ of them are desktops and the rest are laptops. So, the ratio of desktops to laptops would be $$13 \ : \ 12$$. Moreover, in mathematical terms, this is read as “$$13$$ is to $$12$$.”
So, now we move on to what exactly is a proportional ratio. It is a ratio between $$4$$ numbers, such that the simplified ratio between the first two is exactly equal to the simplified ratio between the other $$2$$ numbers. For example, $$3, \ 6, \ 9$$ and $$18$$ are in a proportion as $$3 \ : \ 6$$ is equal to $$9 \ : \ 18$$ as when we simplify both, we get $$1 \ : \ 2$$ as a result.
### How to Calculate Ratios?
To understand the exact process of calculating ratios, let’s take up the following problem.
Question: Suppose, $$22$$ elephants and $$19$$ hippos make up for an entire zoo. So, what is the ratio of elephants and hippos in that zoo?
• Firstly, identify the unique entities. In this case, $$22$$ elephants and $$19$$ hippos are unique entities.
• Next, write these in a fraction form. So, we write it as $$\frac{22}{19}$$
• Now, we need to check if this fraction can be further simplified or not. Here, it can’t be simplified further.
• So, the ratio of elephants and hippos in the zoo is $$22 \ : \ 19$$.
### Proportional Ratios
Another interesting type of ratios is the proportional ratios. If two distinct ratios can be simplified into the same fraction, then they are said to be proportional ratios. Suppose we have two ratios as $$1 \ : \ 2$$ and $$2 \ : \ 4$$. Now, we can clearly see that the second ratio can be reduced to $$1 \ : \ 2$$, and thus both of them are proportional. Now, let’s consider the following problem.
Q: Suppose two ratios $$1 \ : \ 3$$ and $$x \ : \ 9$$ is proportional. Then solve for $$x$$?
A: We know that proportional ratios are identical. So, $$\frac{1}{3} \ = \ \frac{x}{9}$$ , which solves as $$x=3$$.
### Exercises for Proportional Ratios
1) $$28 : 49 = x : 7$$$$\ \Rightarrow \$$
2) $$16 : 20 = x : 5$$$$\ \Rightarrow \$$
3) $$30 : 78 = x : 13$$$$\ \Rightarrow \$$
4) $$12 : 16 = x : 4$$$$\ \Rightarrow \$$
5) $$69 : 99 = 23 : x$$$$\ \Rightarrow \$$
6) $$10 : 35 = x : 7$$$$\ \Rightarrow \$$
7) $$18 : 69 = 6 : x$$$$\ \Rightarrow \$$
8) $$24 : 40 = 3 : x$$$$\ \Rightarrow \$$
9) $$65 : 80 = 13 : x$$$$\ \Rightarrow \$$
10) $$21 : 27 = 7 : x$$$$\ \Rightarrow \$$
1) $$28 : 49 = x : 7$$$$\ \Rightarrow \ \color{red}{49x = 28 \times 7}$$$$\ \Rightarrow \ \color{red}{ x \ = 4}$$
2) $$16 : 20 = x : 5$$$$\ \Rightarrow \ \color{red}{20x = 16 \times 5}$$$$\ \Rightarrow \ \color{red}{ x \ = 4}$$
3) $$30 : 78 = x : 13$$$$\ \Rightarrow \ \color{red}{78x = 30 \times 13}$$$$\ \Rightarrow \ \color{red}{ x \ = 5}$$
4) $$12 : 16 = x : 4$$$$\ \Rightarrow \ \color{red}{16x = 12 \times 4}$$$$\ \Rightarrow \ \color{red}{ x \ = 3}$$
5) $$69 : 99 = 23 : x$$$$\ \Rightarrow \ \color{red}{69x = 99 \times 23}$$$$\ \Rightarrow \ \color{red}{x \ = 33}$$
6) $$10 : 35 = x : 7$$$$\ \Rightarrow \ \color{red}{35x = 10 \times 7}$$$$\ \Rightarrow \ \color{red}{ x \ = 2}$$
7) $$18 : 69 = 6 : x$$$$\ \Rightarrow \ \color{red}{18x = 69 \times 6}$$$$\ \Rightarrow \ \color{red}{x \ = 23}$$
8) $$24 : 40 = 3 : x$$$$\ \Rightarrow \ \color{red}{24x = 40 \times 3}$$$$\ \Rightarrow \ \color{red}{x \ = 5}$$
9) $$65 : 80 = 13 : x$$$$\ \Rightarrow \ \color{red}{65x = 80 \times 13}$$$$\ \Rightarrow \ \color{red}{x \ = 16}$$
10) $$21 : 27 = 7 : x$$$$\ \Rightarrow \ \color{red}{21x = 27 \times 7}$$$$\ \Rightarrow \ \color{red}{x \ = 9}$$
## Proportional Ratios Quiz
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The Reading Workshop model is based off of Richard Allington's six T's of elementary literacy instruction. Time, texts, teaching, talk, tasks, and testing.
Research has shown that students need lots of time to read. It is also important that this time spent reading is done with texts that are "just right" for the reader. Explicit teaching of reading strategies and skills followed by meaningful tasks during their independent reading are the core of this model. Allowing students time to talk about or reflect on their book/skill ties it all together. Lastly, testing should be used to guide instruction to help the students grow as readers.
During the Reader's Workshop model, a 75 minute block of time is divided into 3 sections, the Mini-Lesson, Independent Reading and the Reflection Meeting/Sharing.
Mini lesson- Every Reading Workshop will begin with a 10-15 minute mini-lesson. Each mini-lesson, includes a read aloud and a specific teaching point in kid friendly language that typically focuses on a specific skill. The skill will be modeled and demonstrated, with and without purposeful mistakes, to set the readers up for success during their independent reading. Anchor charts will be created during this time for concrete visuals of the skill being taught.
Independent Reading- independent reading, or the Differentiated Workshop Core (DWC) is about 40 minutes long. During this time, students will be reading text at their "just right" level in their cozy reading nooks in the classroom. As the students are reading, they will be focusing on a specific task related to the teaching point. Students will be responding to reading in their reading journals.
During this time, there is one-on-one conferencing occurring, small group conferencing, and guided reading groups being led by the teacher.
Reflection/Sharing- The reflection meeting, approx. 10 minutes, is one more time for students to reflect on and deepen their understanding of the teaching point. Student(s) will be articulating their process to the class, referring to their text and reading notes based on the reflection question being asked. The whole class, or individual students may be asked to reflect on their process, and successes/mistakes.<|endoftext|>
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Conjunctivitis, informally called pink eye, is one of the most frequently encountered eye diseases, particularly in children. This condition can be caused by a virus, bacteria or sensitivities to pollen, chlorine in pools, and ingredients found in cosmetics, or other irritants, which penetrate your eyes. Certain forms of conjunctivitis are very transmittable and swiftly infect many people in close proximity such as at schools and at the home.
This kind of infection is seen when the conjunctiva, or thin transparent layer of tissue that protects the white part of the eye, gets inflamed. A sign that you have pink eye is if you notice eye itching, redness, discharge or inflamed eyelids and a crusty discharge surrounding the eyes early in the day. Conjunctivitis infections can be divided into three basic sub-types: viral, allergic and bacterial conjunctivitis.
Viral conjunctivitis is often caused by the same type of virus that produces the recognizable watery and red eyes, sore throat and runny nose of the common cold. Symptoms of the viral form of conjunctivitis will often last from seven to fourteen days and like other viruses cannot be treated with medication. You may however, be able to alleviate some of the symptoms by using soothing drops or compresses. The viral form of conjunctivitis is contagious until it is completely cleared up, so in the meantime wipe away discharge and avoid using communal pillowcases or towels. Children who have viral pink eye should be kept home for three days to a week until they are no longer contagious.
Bacterial conjunctivitis is caused by a common bacterial infection that enters the eye typically from something external entering the eye that is carrying the bacteria, such as a dirty finger. This type of pink eye is most commonly treated with antibiotic cream or drops. Usually one should notice the symptoms disappearing after three or four days of treatment, but make sure to take the entire course of antibiotics to stop pink eye from recurring.
Allergic conjunctivitis is not contagious or infectious. It usually occurs among people who already suffer from seasonal allergies or allergies to substances such as pets or dust. The allergic symptoms in the eyes may be just one aspect of their overall allergic reaction. The first step in relieving allergic conjunctivitis is to remove the irritant, if applicable. To ease discomfort, cool compresses and artificial tears may help. When the infection is more severe, non-steroidal anti-inflammatory medications and antihistamines might be prescribed. In cases of chronic allergic conjunctivitis, topical steroid eye drops might be tried.
Although conjunctivitis is typically a highly treatable eye infection, there is sometimes a chance it could worsen into a more serious issue. Any time you have signs of pink eye, be certain to visit your optometrist in order to decide what the best treatment will be.<|endoftext|>
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# Question #59ec4
##### 2 Answers
Apr 8, 2017
See the entire solution process below:
#### Explanation:
The sequence takes the difference between the two previous numbers, doubles it, and adds this to the last number in the sequence:
Looking at the first two terms in the sequence:
$9 - 4 = 5$; $5 \times 2 = 10$, add it to the last number, $9 + 10 = 19$
Now, look at $9$ and $19$:
$19 - 9 = 10$; $10 \times 2 = 20$, add it to the last number, $19 + 20 = 39$
Now, look at $19$ and $39$:
$39 - 19 = 20$; $20 \times 2 = 40$, add it to the last number, $39 + 40 = 79$
The missing term in the sequence is: $\textcolor{red}{39}$
Apr 8, 2017
$39$
#### Explanation:
Apparently the differnce between term doubles every time:
$4 + 5 = 9$
$9 + 10 = 19$
$19 + 20 = 39$
$39 + 40 = 79$
The next term will be:
$79 + 80 = 159$<|endoftext|>
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### Counting Factors
Is there an efficient way to work out how many factors a large number has?
### Summing Consecutive Numbers
Many numbers can be expressed as the sum of two or more consecutive integers. For example, 15=7+8 and 10=1+2+3+4. Can you say which numbers can be expressed in this way?
### Helen's Conjecture
Helen made the conjecture that "every multiple of six has more factors than the two numbers either side of it". Is this conjecture true?
# Magic Letters
##### Stage: 3 Challenge Level:
This was an interesting problem that was open to a lot of interpretation, so many of you came up with new and inventive ways of answering the question, or answered questions we hadn't even asked - it's great to see that you all enjoyed the challenge!
Many people gave ideas as to why Charlie might think that Alison's Magic V was the same as his. Shreya, from Claremont High School, wrote:
Charlie's V had 3, 4 and 1 down one side and 5, 2 and 1 down the other side, so each side added up to 8. Alison's V also had the numbers from 1 to 5 arranged in such positions so that one side of the V added up to 8. Therefore, she had the same magic V as Charlie had!
Ben, from Wilson's Grammar School, had the same thoughts. Ayngharran and Rohan, also from Wilson's, thought something a little more specific. Ayngharran said:
Charlie used the same numbers as Alison, with the same 'middle' number 1, and the same numbers on each side but just in a different order. So the pairs 3, 4 and 5, 2 are the same in both cases.
Zoya, from St. Hilda's C of E Primary School, said:
You will know if you have found all the Magic Vs adding up to a certain number, as each Magic V will always have eight ways of rearranging the arms to still add up to the same total.
Elliott, from Solihull School, found a different Magic V with the same numbers:
2 1 3 4 5
Ben also found a similar Magic V. Some people made interesting comments about what sorts of Magic Vs might exist with these numbers. Abopakr, from Globe Academy, said:
The base can only be an odd number if you are using numbers 1 - 5, because if you put an even number at the base there will be only one even number not at the base. So it has to be an odd number at the base. Whichever is the most common kind of number - even or odd - is the kind that goes at the base.
Michelle, also from Globe Academy, had the same idea, and so did Utkarsh, from Sancta Maria International School. Murat and Callum, from Globe Academy, said:
You need to put the lowest number on the same side as the highest number.
Adam, from Collis Primary School, made the following Magic V and commented:
4 5 3 2 1
All you do is start at the bottom, then zigzag up and right, then left, then up and left, then right.
That's a nice systematic way of constructing a Magic V - well spotted! (I wonder whether we can make bigger Magic Vs in this way?)
Kavi, from Wilson's, suggested:
To construct a Magic V with the numbers 2-6, just take a Magic V with the numbers 1-5 and raise each number by 1. Then each arm will sum to 3 more.
Maciej, from Wilson's, gave us the following example:
4 3 5 6 2
An anonymous student in the UK gave the following example:
7 9 8 6 5
Ben solved our 987-991 puzzle in the same way:
988 987 989 990 991
Some people came up with different types of Magic Vs. For example, Daniel, from Wilson's, solved our '1000' puzzle in this way:
2 1 3 4 5
Each arm had a total of ten. I then multiplied each number in the V by 100.
200 100 300 400 500
Interesting - this doesn't use consecutive numbers, but it's still a Magic V.
Shavindra, from Wilson's, gave us a formulaic way of finding a Magic V with each arm summing to N (as long as N is divisible by 3):
(N/3) + 1 (N/3) + 2 (N/3) - 1 (N/3) - 2 (N/3)
Ellis, from Westfield Middle School, gave us a Magic L, and a comment:
1 3 4 5 2 6
The number in the corner must be odd. After you've done one, you can jumble the numbers on each arm.
Finally, Rebecca and Angus came up with this observation:
If you take the middle number of the five consecutive numbers for a Magic V and multiply it by 3, it will give you the middle number of the range of totals you could get for that magic V.
Also, there are only 3 consecutive totals for each V depending on which number is on the bottom of the V.
For example, for numbers 12-16, the middle number is 14.
14 times 3 is 42, so the range of totals is 41 to 43.
Excellent! Thanks to everyone for all your contributions.<|endoftext|>
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|Computer network types|
by spatial scope
A backbone is a part of computer network that interconnects various pieces of network, providing a path for the exchange of information between different LANs or subnetworks. A backbone can tie together diverse networks in the same building, in different buildings in a campus environment, or over wide areas. Normally, the backbone's capacity is greater than the networks connected to it.
A large corporation that has many locations may have a backbone network that ties all of the locations together, for example, if a server cluster needs to be accessed by different departments of a company that are located at different geographical locations. The pieces of the network connections (for example: ethernet, wireless) that bring these departments together is often mentioned as network backbone. Network congestion is often taken into consideration while designing backbones.
The theory, design principles, and first instantiation of the backbone network came from the telephone core network, when traffic was purely voice. The core network was the central part of a telecommunications network that provided various services to customers who were connected by the access network. One of the main functions was to route telephone calls across the PSTN.
Typically the term referred to the high capacity communication facilities that connect primary nodes. A core network provided paths for the exchange of information between different sub-networks.
Core networks usually had a mesh topology that provided any-to-any connections among devices on the network. Many main service providers would have their own core/backbone networks that are interconnected. Some large enterprises have their own core/backbone network, which are typically connected to the public networks.
Core networks typically provided the following functionality:
- Aggregation: The highest level of aggregation in a service provider network. The next level in the hierarchy under the core nodes is the distribution networks and then the edge networks. Customer-premises equipment (CPE) do not normally connect to the core networks of a large service provider.
- Authentication: The function to decide whether the user requesting a service from the telecom network is authorized to do so within this network or not.
- Call Control/Switching: call control or switching functionality decides the future course of call based on the call signalling processing. E.g. switching functionality may decide based on the "called number" that the call be routed towards a subscriber within this operator's network or with number portability more prevalent to another operator's network.
- Charging: This functionality of the collation and processing of charging data generated by various network nodes. Two common types of charging mechanisms found in present-day networks are prepaid charging and postpaid charging. See Automatic Message Accounting
- Service Invocation: Core network performs the task of service invocation for its subscribers. Service invocation may happen based on some explicit action (e.g. call transfer) by user or implicitly (call waiting). It's important to note however that service "execution" may or may not be a core network functionality as third party network/nodes may take part in actual service execution.
- Gateways: Gateways shall be present in the core network to access other networks. Gateway functionality is dependent on the type of network it interfaces with.
Physically, one or more of these logical functionalities may simultaneously exist in a given core network node.
Besides above mentioned functionalities, the following also formed part of a telecommunications core network:
- O&M: Operations & Maintenance centre or Operations Support Systems to configure and provision the core network nodes. Number of subscribers, peak hour call rate, nature of services, geographical preferences are some of the factors which impact the configuration. Network statistics collection (Performance Management), alarm monitoring (Fault Management) and logging of various network nodes actions (Event Management) also happens in the O&M centre. These stats, alarms and traces form important tools for a network operator to monitor the network health and performance and improvise on the same.
- Subscriber Database: Core network also hosts the subscribers database (e.g. HLR in GSM systems). Subscriber database is accessed by core network nodes for functions like authentication, profiling, service invocation etc.
A distributed backbone is a backbone network that consists of a number of connectivity devices connected to a series of central connectivity devices, such as hubs, switches, or routers, in a hierarchy. This kind of topology allows for simple expansion and limited capital outlay for growth, because more layers of devices can be added to existing layers. In a distributed backbone network, all of the devices that access the backbone share the transmission media, as every device connected to this network is sent all transmissions placed on that network.
Distributed backbones, in all practicality, are in use by all large-scale networks. Applications in enterprise-wide scenarios confined to a single building are also practical, as certain connectivity devices can be assigned to certain floors or departments. Each floor or department possesses a LAN and a wiring closet with that workgroup's main hub or router connected to a bus-style network using backbone cabling . Another advantage of using a distributed backbone is the ability for network administrator to segregate workgroups for ease of management.
There is the possibility of single points of failure, referring to connectivity devices high in the series hierarchy. The distributed backbone must be designed to separate network traffic circulating on each individual LAN from the backbone network traffic by using access devices such as routers and bridges.
A collapsed backbone (inverted backbone, backbone-in-a-box) is a type of backbone network architecture. The traditional backbone network goes over the globe to provide interconnectivity to the remote hubs. In most cases, the backbones are the links while the switching or routing functions are done by the equipment at each hub. It is a distributed architecture.
In the case of a collapsed or inverted backbone, each hub provides a link back to a central location to be connected to a backbone-in-a-box. That box can be a switch or a router. The topology and architecture of a collapsed backbone is a star or a rooted tree.
The main advantages of the collapsed backbone approach are
- ease of management since the backbone is in a single location and in a single box, and
- since the backbone is essentially the back plane or internal switching matrix of the box, proprietary, high performance technology can be used.
However, the drawback of the collapsed backbone is that if the box housing the backbone is down or there are reachability problem to the central location, the entire network will crash. These problems can be minimized by having redundant backbone boxes as well as having secondary/backup backbone locations.
There are a few different types of backbones that are used for an enterprise-wide network. When organizations are looking for a very strong and trustworthy backbone they should choose a parallel backbone. This backbone is a variation of a collapsed backbone in that it uses a central node (connection point). Although, with a parallel backbone, it allows for duplicate connections when there is more than one router or switch. Each switch and router are connected by two cables. By having more than one cable connecting each device, it ensures network connectivity to any area of the enterprise-wide network.
Parallel backbones are more expensive than other backbone networks because they require more cabling than the other network topologies. Although this can be a major factor when deciding which enterprise-wide topology to use, the expense of it makes up for the efficiency it creates by adding increased performance and fault tolerance. Most organizations use parallel backbones when there are critical devices on the network. For example, if there is important data, such as payroll, that should be accessed at all times by multiple departments, then your organization should choose to implement a Parallel Backbone to make sure that the connectivity is never lost.
A serial backbone is the simplest kind of backbone network. Serial backbones consist of two or more internet working devices connected to each other by a single cable in a daisy-chain fashion. A daisy chain is a group of connectivity devices linked together in a serial fashion. Hubs are often connected in this way to extend a network. However, hubs are not the only device that can be connected in a serial backbone. Gateways, routers, switches and bridges more commonly form part of the backbone. The serial backbone topology could be used for enterprise-wide networks, though it is rarely implemented for that purpose.
- What is a Backbone?, Whatis.com, Accessed: June 25, 2007
- "Backbone Networks". Chapter 8. Angelfire. Retrieved 2 October 2013.
- Turner, Brough (12 September 2007). "Congestion in the Backbone: Telecom and Internet Solutions". CircleID. Retrieved 2 October 2013.
- Kashyap, Abhishek; Sun, Fangting; Shayman, Mark. "Relay Placement for Minimizing Congestion in Wireless Backbone Networks" (PDF). Department of Electrical and Computer Engineering, University of Maryland. Retrieved 2 October 2013.
- Howdie, Ben (28 January 2013). "The Backbone's connected to the…". KashFlow. Retrieved 2 October 2013.
- Tamara Dean. Network+ Guide to Networks. Course Technology, Cengage Learning, 2010, p. 202.
- BICSI Lan Design Manual - CD-ROM, Issue 1, Distributed backbone network, p.20 , 1996, accessed May 7, 2011.
- Dooley, Kevin. Designing Large-Scale Networks, p.23 , O'Reilly Online Catalog, January, 2002, accessed May 7, 2011.
- Distributed Backbone , accessed May 7, 2011.
- Boon & Kepekci (1996). BICSI Lan Design Manual. Tampa, FL. pp. 20–21.
- Dean, Tamara (2010). Network+ Guide to Networks 5th Edition. Boston, MA: Cengage Course Technology. pp. 203–204. ISBN 1-4239-0245-9.
- CompTIA Network+ In depth, Chapter 5 p. 169
- Dean, T. (2010) Network+ Guide to Networks, Fifth Edition
- , Backbone Networks<|endoftext|>
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A five digit number is formed by the digits $1,2,3,4,5$ with no digit being repeated. The probability that the number is divisible by $4$, is$A$. $\dfrac{1}{5}$$B. \dfrac{2}{5}$$C$. $\dfrac{3}{5}$$D$. $\dfrac{4}{5}$
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Hint: - The number of possibilities in each of these cases is $3!$ as the digits at one's and ten's places are fixed and the rest $3$ digits can be chosen in $3!$ ways. Because we know that rearranging the n numbers on n places is $n!$.
We know that the number is divisible by $4$ only when it ends with: when the last two digits are divisible by $4$.
So by the given numbers we can form number which is divisible by $4$ only when it ends with:
$12,\;,\;24,,\;32,\;\;52$
The number of possibilities in each of these cases is $3!$ as the digits at one's and ten's places are fixed and the rest $3$ digits can be chosen in $3!$ ways. Because we know that rearranging of n numbers on n places is $n!$.
Thus total possibilities for digit to be divisible by $4 = 4 \times 3!$ (because we made four conditions above for divisibility by 4).
The total number we can form by these five numbers are: - $5!$ , (arranging 5 numbers in 5 places).
Let $E = {\text{the number is divisible by}}\;4$
$P(E) = \frac{{{\text{Numbers divisible by}}4}}{{{\text{Total numbers}}}}$
$= \frac{{4 \times 3!}}{{5!}} = \frac{{4 \times 3 \times 2 \times 1}}{{5 \times 4 \times 3 \times 2 \times 1}} = \frac{1}{5}$
Hence, option A is the correct answer.
Note: - Whenever we face such a type of question first find out the favorable outcome. Here we can find out the number which is divisible by four by the divisibility rule by fixing the last two numbers which are divisible by four and rearrange the first three numbers to find a favorable outcome.<|endoftext|>
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• This is an assessment test.
• To draw maximum benefit, study the concepts for the topic concerned.
• Kindly take the tests in this series with a pre-defined schedule.
## Number System: Basics of Factors Test-1
Congratulations - you have completed Number System: Basics of Factors Test-1.You scored %%SCORE%% out of %%TOTAL%%.You correct answer percentage: %%PERCENTAGE%% .Your performance has been rated as %%RATING%%
Question 1
How many factors does 825 have
A 14 B 16 C 12 D 20
Question 1 Explanation:
Step 1: Prime factorisation, so N=825= 31 52 111
Power of 3 as 3, 31 ( 1+1=2)ways,
Power of 5 as 50 , 51, 52 ( 2+1=3)ways
Power of 11 as 110, 111 (1+1=2)ways
Step 2: Hence, the number of factors is (1+1)(2+1)(1+1)=2x3x2=12
Question 2
How many are number of factors N= 32537311213217 ?
A 700 B 456 C 864 D 900
Question 2 Explanation:
Step 1: Prime factorisation, so N=32537311213217 1
Power of 3 as 3, 31, 32
Power of 5 as 50 , 51, 52,53
Power of 7 as 70,71,72,73
Power of 11as 110111112
Power of 13 as 130131132
Power of 17 as 170171
Step 2:
Hence, the number of factors is(2+1)(3+1)(3+1)(2+1)(2+1)(1+1)=864
Question 3
How many even factors does the number 24 35 72 have?
A 76 B 72 C 36 D 128
Question 3 Explanation:
In this case we have to find number of even factors,an even factor is divisible by 2 or smallest power of 2 has to be 1 not 0.
Hence a factor must be 2(1 or 2 or 3 or 4)3(0 or 1,2,3,4,5)7(0 or 1 or 2)
Hence total number of factors= (4)(5 + 1 )(2 + 1) = 72
Question 4
How many odd factors does the number N= 24 35 72?
A 20 B 40 C 45 D 18
Question 4 Explanation:
To simply find odd factors, remove 2 from prime factors i.e from N remove \${{2}^{4}}\$
Then we are left with \$N={{3}^{5}}{{7}^{2}}\$ Since these are two odd factors, they will have no even factors.
Therefore, the total number of factors is= (5+1)(2+1)=18
Question 5
How many factors are there in N = 25 5272 which are divisible by 10?
A 45 B 35 C 30 D 55
Question 5 Explanation:
If a number is divisible by 10 then it must have minimum power of 2 and 5 as 1.
A factor divisible by 10 is 2(1 or 2 or 3or 4 or 5)5(1 or 2 )7(0 or 1 or 2 )
Hence, total number of factors divisible by 10 is = (5)(2)(2 + 1) = 30
Once you are finished, click the button below. Any items you have not completed will be marked incorrect.
There are 5 questions to complete.
← List →<|endoftext|>
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# With the next equation system, find possible values for $\frac{x}{y}$
Let $x,y,z,w$ $\in$ $\mathbb R^+$ such that:
$\left\{\begin{array}{l}x+y=z+w\\2xy=zw\end{array}\right.$
Find possible values of $\frac{x}{y}$
I can't see a way to start this problem, i played a lot with the equations, squaring the first equation and putting the second in it:
$x^2+y^2= z^2+zw+w^2$
But nothing more than that. Maybe the solution is to get values of $\frac{x}{y}$ directly and not try to get the possible values of $x,y$
Suppose $z = w$ this will maximize $zw$ (relative to $xy$)
$z = w = \frac {x+y}{2}$
$zw = \frac {x^2 + y^2 + 2xy}{4} = 2xy\\ x^2 + y^2 - 6xy = 0$
divide through by $y$
$(\frac {x}{y})^2 - 6\frac {x}{y} + 1 = 0$
$\frac {x}{y} = 3 \pm\sqrt {8}$
When we relax the constraint $z = w$ we force $\frac {x}{y}$ away from the moderate values.
$\frac {x}{y} \ge 3 + \sqrt {8}$ or $\frac {x}{y} \le 3 - \sqrt {8}$
Set $\frac{x}{y}=t$. We have $x=ty$, and then, by Vieta formulas, $z$ and $w$ are the roots of:
$$\xi^2-(t+1)y\xi+2ty^2=0$$
(as $z+w=x+y=(t+1)y, zw=2xy=2ty^2$). This has solutions if and only if the discriminant is $\ge 0$, i.e.
$$y^2((t+1)^2-8t)\ge 0$$
Cancelling (positive) $y^2$ and solving for positive $t$ yields:
$$t=\frac{x}{y}\in(0,3-2\sqrt 2]\cup[3+2\sqrt 2,\infty)$$
Yes indeed it is easier to find the values of $x/y$. Let $\tilde x=x/y,\tilde w=w/y,\tilde z=z/y$, Noe this is possible since $y\neq 0$. Then both equations become $$\tilde x+1=\tilde z+\tilde w,\quad 2\tilde x=\tilde z\tilde w$$
You may solve for $\tilde z=\tilde x+1-\tilde w$ in the first one and put it in the second one to get
$$\tilde x(2-\tilde w)=\tilde w(1-\tilde w).$$
As $\tilde w=2$ yields $0$ on the left hand side and $-2$ on the right hand side it is not a solution and we can divide by $2-\tilde w$. Then
$$\frac{x}{y}=\tilde x=\tilde w\frac{1-\tilde w}{2-\tilde w}$$
As $\tilde w$ runs from $0$ to $1$, $\tilde x$ goes from $0$ to a local maximum $M$ (that I am confident you can find, let me know if not) and back to $0$. From $\tilde w=1$ to $\tilde \omega=2$, $\tilde x$ is negative hence we don't want that solution. And from $\tilde w=2$ to $\tilde w\to \infty$, $\tilde x$ goes from $\infty$ to a local minimum $m$ and back to $\infty$.
Moreover $m>M$ then the posible values for $\tilde x$ are $(0,M)\cup[m,\infty)$<|endoftext|>
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# A Complement Union B Complement Venn Diagram
In set theory, the complement of a set A refers to elements not in A. When all sets under consideration are considered to be subsets of a given set U, the absolute complement of A is the set of elements in U but not in A. The relative complement of A with respect to a set B, also termed the If the set A is the suit of spades, then the complement of A is the union of the.
Note: A ∪ A = U, the union of a set with its complement gives the universal set. .
We begin by constructing a Venn diagram, we will use B for the Big Game and. Enter an expression like (A Union B) Intersect (Complement C) to describe a combination of two or three sets and get the notation and Venn diagram.
Here we are going to see how to draw a venn diagram for A union B whole complement. Venn diagram of (A U B)': To represent (A U B)' in venn diagram, we. Use Venn diagrams to illustrate data in a logical way which will enable you to see We can also find the union of A and B which is written as A ∪ B.
This means we can use the notation A' called the complement of A. In the example above.Here are some useful rules and definitions for working with sets.
The Venn diagram above illustrates the set notation and the logic of the answer. Since "union" means "everything in either of the sets", all of each circle is shaded in. (If you're not clear on the logic of the set notation, review set notation before proceeding further.).
The complement of a set A is everything that is not in A; it is represented by the magenta region in the Venn diagram below (hence the set A is represented by the white region). The union of A and B is everything which is in either A or B, as represented by the magenta shaded region in the following venn diagram.
The complement of a set using Venn diagram is a subset of U. Let U be the universal set and let A be a set such that A ⊂ U. Then, the complement of A with respect to U is denoted by A' or A\(^{C}\) or U – A or ~ A and is defined the set of all those elements of U which are not in A.
The complement of a set using Venn diagram is a subset of U. Let U be the universal set and let A be a set such that A ⊂ U. Then, the complement of A with respect to U is denoted by A' or A\(^{C}\) or U – A or ~ A and is defined the set of all those elements of U which are not in A.De Morgans Law - Proof with Examples - Set Theory - TeachooComplement of a Set using Venn Diagram | Example on Complement of a Set
## 6 thoughts on “A complement union b complement venn diagram”
1. I am sorry, this variant does not approach me. Perhaps there are still variants?<|endoftext|>
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# Lesson 10.2 - Comparing Two Population Means: Independent Samples
Unit Summary Sampling Distribution of the Differences Between the Two Sample Means for Independent Samples 2-Sample t-Procedures: Pooled Variances versus Non-Pooled Variances Performing the 2-Sample t-Procedure Using Minitab
An Introduction to Statistical Methods and Data Analysis, chapter 6.2.
### Sampling Distribution of the Differences Between the Two Sample Means for Independent Samples
The point estimate for - is: - .
In order to find a confidence interval for - and perform hypothesis testing, we need to find the sampling distribution of - .
We can show that when the sample sizes are large or the samples from each population are normal and the samples are taken independently, then - is normal with mean - and standard deviation .
In most cases, 1 and 2 are unknown and they have to be estimated. It seems natural to estimate 1 by s1 and 2 by s2. However, when the sample sizes are small, the estimates may not be that accurate and one may get a better estimate for the common standard deviation by pooling the data from both populations if the standard deviations for the two populations are not that different.
In view of this, there are two options for estimating the variances for the 2-sample t-test with independent samples:
1. 2-sample t-test using pooled variances
2. 2-sample t-test using separate variances
When to use which? When we are reasonably sure that the two populations have nearly equal variances, then we use the pooled variances test. Otherwise, we use the separate variances test.
### 2-Sample t-Procedures: Pooled Variances Versus Non-Pooled Variances
2-Sample (Independent Samples) t-Procedure Using Pooled Variances to Do Inferences for Two-Population Means (Standard Deviations are Assumed Equal)
When we have good reason to believe that the standard deviation for population 1 (also called sample) is about the same as that of population 2 (also called sample), we can estimate the common standard deviation by pooling information from samples from population 1 and population 2.
Let n1 be the sample size from population 1, s1 be the sample standard deviation of population 1.
Let n2 be the sample size from population 2, s2 be the sample standard deviation of population 2.
Then the common standard deviation can be estimated by the pooled standard deviation:
The test statistic is:
with degrees of freedom equal to df = n1 + n2 - 2.
In a packing plant, a machine packs cartons with jars. It is supposed that a new machine will pack faster on the average than the machine currently used. To test that hypothesis, the times it takes each machine to pack ten cartons are recorded. The results, in seconds, are shown in the following table.
New machine Old machine 42.1 41.3 42.4 43.2 41.8 42.7 43.8 42.5 43.1 44.0 41.0 41.8 42.8 42.3 42.7 43.6 43.3 43.5 41.7 44.1 = 42.14, s1 = 0.683 = 43.23, s2 = 0.750
Do the data provide sufficient evidence to conclude that, on the average, the new machine packs faster? Perform the required hypothesis test at the 5% level of significance.
It is given that:
= 42.14, s1 = 0.683
= 43.23, s2 = 0.750
Assumption 1: Are these independent samples? Yes, since the samples from the two machines are not related.
Assumption 2: Are these large samples or a normal population? We have n1 < 30, n2 < 30. We do not have large enough samples and thus we need to check the normality assumption from both populations.
From the normality plots, we conclude that both populations may come from normal distributions.
Assumption 3: Do the populations have equal variance? Yes, since s1 and s2 are not that different. We can thus proceed with the pooled t-test. (They are not that different as s1/s2=0.683/0.750=0.91 is quite close to 1. We will discuss this in more details and quantify what is "close" in Lesson 11.)
Let denote the mean for the new machine and denote the mean for the old machine.
Step 1. Ho: - = 0, Ha: - < 0
Step 2. Significance level: = 0.05.
Step 3. Compute the t-statistic:
Step 4. Critical value:
Left-tailed test
Critical value = - = - t0.05
Degrees of freedom = 10 + 10 - 2 = 18
- t0.05 = -1.734
Rejection region t < -1.734
Step 5. Check to see if the value of the test statistic falls in the rejection region and decide whether to reject Ho.
t* = -3.40 < -1.734
Reject Ho at = 0.05
Step 6. State the conclusion in words.
At 5% level of significance, the data provide sufficient evidence that the new machine packs faster than the old machine on average.
When one wants to estimate the difference between two population means from independent samples, then one will use a t-interval. If the sample variances are not very different, one can use the pooled 2-sample t-interval.
Step 1. Find with df = n1 + n2 - 2.
Step 2. The endpoints of the (1 - ) 100% confidence interval for - is:
the degrees of freedom of t is n1 + n2 - 2.
Continuing from the previous example, give a 99% confidence interval for the difference between the mean time it takes the new machine to pack ten cartons and the mean time it takes the present machine to pack ten cartons.
Step 1. = 0.01, = t0.005 = 2.878, where the degrees of freedom is 18.
Step 2.
The 99% confidence interval is (-2.01, -0.17).
Interpret the above result:
We are 99% confident that - is between -2.01 and -0.17.
Using Minitab to perform a pooled t-procedure:
1. Stat > Basic Statistics > 2-sample t. The following window will then be displayed.
Note: When entering values into the Samples in different columns input boxes, Minitab always subtracts the Second value (column entered second) from the First value (column entered first).
2. Select the Assume equal variances checkbox.
The Minitab output for the packing time example is as follows:
Two sample T for new machine vs present machine
N Mean StDev SE Mean new mach 10 42.140 0.683 0.22 present 10 43.230 0.750 0.24
99% CI for mu new mach - mu present: (-2.01, -0.17)
T-Test mu new mach = mu present (Vs <): T = -3.40
P = 0.0016 DF = 18
Both use Pooled StDev = 0.717
What to do if some of the assumptions are not satisfied:
A. What should we do if the assumption of independent samples is violated?
If the samples are not independent but paired,we can use the paired t-test.
B. What should we do if the sample sizes are not large and the populations are not normal?
We can use a nonparametric method to compare two samples such as the Mann-Whitney procedure.
C. What should we do if the assumption of equal variances is violated?
We can use the separate variances 2-sample t-test.
2-Sample (Independent Samples) t-Procedure Using Separate Variances to Do Inferences for Two-Population Means (Very Different Standard Deviations for the Two Samples)
Note: The formulas are provided in the following for your reference only. We can perform the separate variances test using Minitab.
with
(round down to nearest integer)
where
Using Minitab to perform a separate variance 2-sample t-procedure:
Stat > Basic Statistics > 2-sample t
For some examples, one can use both the pooled t-procedure and the separate variances (non-pooled) t-procedure with the results close to each other. However, when the sample standard deviations are very different from each other and the sample sizes are different, the separate variances 2-sample t-procedure is more reliable.
Independent random samples of 17 sophomores and 13 juniors attending a large university yield the following data on grade point averages:
Sophomores Juniors 3.04 2.92 2.86 2.56 3.47 2.65 1.71 3.60 3.49 2.77 3.26 3.00 3.30 2.28 3.11 2.70 3.20 3.39 2.88 2.82 2.13 3.00 3.19 2.58 2.11 3.03 3.27 2.98 2.60 3.13
At the 5% significance level, do the data provide sufficient evidence to conclude that the mean GPAs of sophomores and juniors at the university differ?
Check assumption 1: Are these independent samples?
Yes, the students selected from the sophomores are not related to the students selected from juniors.
Check assumption 2: Is this a normal population or large samples?
Since we don't have large samples from both populations, we need to check the normal probability plots of the two samples:
Now, we need to determine whether to use the pooled t-test or the non-pooled (separate variances) t-test.
We use the following Minitab commands:
Stat > Basic Statistics > Display Descriptive Statistics
To find the summary statistics for the two samples:
Descriptive Statistics
Variable N Mean Median TrMean StDev sophomor 17 2.840 2.920 2.865 0.520 juniors 13 2.9808 3.0000 2.9745 0.3093
Variable Minimum Maximum Q1 Q3 sophomor 1.710 3.600 2.440 3.200 juniors 2.5600 3.4700 2.6750 3.2300
Note: The standard deviations are 0.520 and 0.3093 respectively; both the sample sizes and the standard deviations are quite different from each other.
We, therefore, decide to use a non-pooled t-test.
Step 1. Set up the hypotheses:
Ho: - = 0
Ha: - 0
Step 2. Write down the significance level.
= 0.05
Step 3. Perform the 2-sample t-test in Minitab with the appropriate alternative hypothesis.
Note: The default for the 2-sample t-test in Minitab is the non-pooled one:
Two sample T for sophomores vs juniors
N Mean StDev SE Mean sophomor 17 2.840 0.520 0.13 juniors 13 2.981 0.309 0.086
95% CI for mu sophomor - mu juniors: ( -0.45, 0.173)
T-Test mu sophomor = mu juniors (Vs no =): T = -0.92
P = 0.36 DF = 26
Step 4. Find the p-value from the output.
p-value = 0.36
Step 5. Draw the conclusion using the p-value.
Since the p-value is larger than = 0.05, we cannot reject the null hypothesis.
Step 6. State the conclusion in words.
At 5% level of significance, the data does not provide sufficient evidence that the mean GPAs of sophomores and juniors at the university are different.
Continuing with the previous example, give a 95% confidence interval for the difference between the mean GPA of Sophomores and the mean GPA of Juniors.
Using Minitab:
95% CI for mu sophomor - mu juniors is:
( -0.45, 0.173)
Interpreting the above result:
We are 95% confident that the difference between the mean GPA of sophomores and juniors is between -0.45 and 0.173.
Remember: When entering values into the Samples in different columns input boxes, Minitab always subtracts the Second value (column entered second) from the First value (column entered first).<|endoftext|>
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RD Sharma Solutions Class 7 Chapter 4 Rational Numbers
Read RD Sharma Solutions Class 7 Chapter 4 Rational Numbers below, students should study RD Sharma class 7 Mathematics available on Studiestoday.com with solved questions and answers. These chapter wise answers for class 7 Mathematics have been prepared by teacher of Grade 7. These RD Sharma class 7 Solutions have been designed as per the latest NCERT syllabus for class 7 and if practiced thoroughly can help you to score good marks in standard 7 Mathematics class tests and examinations
Exercise 4.1
Question 1: Write down the numerator of each of the following rational numbers:
(i) ((-7)/5)
(ii) (15/(-4))
(iii) ((-17)/(-21))
(iv) (8/9)
(v) 5
Solution 1:
(i) ((-7)/5) = Numerator of ((-7)/5) is -7
(ii) (15/(-4)) = Numerator of (15/(-4)) is 15
(iii) ((-17)/(-21)) = Numerator of ((-17)/(-21)) is -17
(iv) (8/9) = Numerator of (8/9) is 8
(v) 5 = Numerator of 5 is 5
Question 2: Write down the denominator of each of the following rational numbers:
(i) ((-4)/5)
(ii) (11/(-34))
(iii) ((-15)/(-82))
(iv) 15
(v) 0
Solution 2:
(i) ((-4)/5) = Denominator of ((-4)/5) is 5
(ii) (11/(-34)) = Denominator of (11/(-34)) is -34
(iii) ((-15)/(-82))= Denominator of (15/(-82)) is -82
(iv) 15 = Denominator of 15 is 1
(v) 0 = Denominator of 0 is any non-zero integer
Question 3: Write down the rational number whose numerator is (-3) × 4, and whose denominator is (34 – 23) × (7 – 4).
Solution 3:
We found that Numerator is (-3) × 4 = -12
And Denominator = (34 – 23) × (7 – 4)
= 11 × 3 = 33
Hence, the rational number = ((-12)/33)
Question :4. Write down the rational numbers as integers: (7/1), ((-12)/1), (34/1), ((-73)/1), (95/1)
Solution 4:
Integers of (7/1), ((-12)/1), (34/1), ((-73)/1), (95/1) are 7, -12, 34, -73, 95
Question :5. Write the following integers as rational numbers: -15, 17, 85, -100
Solution 5:
The rational numbers of integers are ((-15)/1), (17/1), (85/1) and ((-100)/1)
Question :6. Write down the rational number whose numerator is the smallest three-digit number and denominator is the largest four-digit number.
Solution 6:
The smallest three-digit number are 100
The Largest four-digit number are 9999
Hence, the rational number of smallest three- digit number and the largest four digit number is = (100/9999)
Question :7. Separate positive and negative rational numbers from the following rational numbers:
((-5)/(-7)), (12/(-5)), (7/4), (13/(-9)), 0, ((-18)/(-7)), ((-95)/116), ((-1)/(-9))
Solution 7:
A rational number is positive if its numerator and denominator are either positive integers or negative integers. hence, positive rational numbers are: ((-5)/(-7)), , (7/4), ((-18)/(-7)), ((-1)/(-9)).
A rational number is negative integers if its numerator and denominator are such that one of them is positive integer and another one is a negative integer.
Hence, negative rational numbers are: (12/(-5)), (13/(-9)), ((-95)/116),
Question :8. Which of the following rational numbers are positive:
(i) ((-8)/7)
(ii) (9/8)
(iii) ((-19)/(-13))
(iv) ((-21)/13)
Solution 8:
A rational number is positive if its numerator and denominator are either positive integers or both negative integers.
Hence, the positive rational numbers are (9/8) and ((-19)/(-13))
Question :9. Which of the following rational numbers are negative:
(i) ((-3)/7)
(ii) ((-5)/(-8))
(iii) (9/(-83))
(iv) ((-115)/(-197))
Solution 9:
A rational number is negative integers if its numerator and denominator are such that one of them is positive integer and another one is a negative integer.
Hence, negative rational numbers are ((-3)/7) and (9/(-83))
Exercise 4.2
Question :1. Express each of the following as a rational number with positive denominator.
(i) ((-15)/(-28))
(ii) (6/(-9))
(iii) ((-28)/(-11))
(iv) (19/(-7))
Solution 1:
(i) ((-15)/(-28))
Multiplying the fraction’s numerator and denominator by -1:
= ((-15)/(-28)) × ((-1)/(-1)) = (15/28)
(ii) (6/(-9))
Multiplying the fraction’s numerator and denominator by -1:
= (6/(-9)) × ((-1)/(-1)) = ((-6)/9)
(iii) ((-28)/(-11))
Multiplying the fraction’s numerator and denominator by
= ((-28)/(-11)) × ((-1)/(-1)) = (28/11)
(iv) (19/(-7))
Multiplying the fraction’s numerator and denominator by
= (19/(-7)) × ((-1)/(-1)) = ((-19)/7)
Question 2: Express (3/5) as a rational number with numerator:
(i) 6
(ii) -15
(iii) 21
(iv) -27
Solution 2:
(i) (3/5)
To get 6 as a numerator of given value, we have to multiply both numerator and denominator by 2
Then, (3/5) × (2/2) = (6/10)
Hence, (3/5) as a rational number with numerator 6 is (6/10)
(ii) (3/5)
To get 15 as a numerator of the given value, we have to multiply both numerator and denominator by -5
Then, (3/5) × ((-5)/(-5)) = ((-15)/(-25))
Hence, (3/5) as a rational number with numerator -15 is ((-15)/(-25))
(iii) (3/5)
To get 21 as a numerator of the given value, we have to multiply both numerator and denominator by 7
Then, (3/5) × (7/7) = (21/35)
Hence, (3/5) as a rational number with numerator 21 is (21/35)
(iv) (3/5)
To get 27 as a numerator of the given value, we have to multiply both numerator and denominator by -9
Then, (3/5) × ((-9)/(-9)) = ((-27)/(-45))
Hence, (3/5) as a rational number with numerator -27 is ((-27)/(-45))
Question 3: Express (5/7) as a rational number with denominator:
(i) -14
(ii) 70
(iii) -28
(iv) -84
Solution 3:
(i) (5/7)
To get 14 as a denominator of the given value we have to multiply both numerator and denominator by -2
Then, (5/7) × ((-2)/(-2)) = ((-10)/(-14))
Hence, (5/7) as a rational number with denominator -14 is ((-10)/(-14))
(ii) (5/7)
To get 70 as a denominator of the given value we have to multiply both numerator and denominator by -2
Then, (5/7) × (10/10) = (50/70)
Hence, (5/7) as a rational number with denominator 70 is (50/70)
(iii) (5/7)
To get -28 as a denominator of the given value we have to multiply both numerator and denominator by -4
Then, (5/7) × ((-4)/(-4)) = ((-20)/(-28))
Hence, (5/7) as a rational number with denominator -28 is ((-28)/(-20))
(iv) (5/7)
To get-84 as a denominator of the given value we have to multiply both numerator and denominator by -12
Then, (5/7) × ((-12)/(-12)) = ((-60)/(-84))
Hence, (5/7) as a rational number with denominator -84 is ((-60)/(-84))
Question 4: Express (3/4) as a rational number with denominator:
(i) 20
(ii) 36
(iii) 44
(iv) -80
Solution 4:
(i) (3/4)
To get 20 as a denominator of the given value we have to multiply both by 5
Then, (3/4) × (5/5) = (15/20)
Hence (3/4) as a rational number with denominator 20 is (15/20)
(ii) (3/4)
To get 35as a denominator of the given value we have to multiply both by 9
Then, (3/4) × (9/9) = (27/36)
Hence (3/4) as a rational number with denominator 36 is (27/36)
(iii) (3/4)
To get 44 as a denominator of the given value we have to multiply both by 11
Then, (3/4) × (11/11) = (33/44)
Hence (3/4) as a rational number with denominator 44 is (33/44)
(iv) (3/4)
To get -80 as a denominator of the given value we have to multiply both by -20
Then, (3/4) × ((-20)/(-20)) = ((-60)/(-80))
Hence (3/4) as a rational number with denominator -80 is ((-60)/(-80))
Question :5. Express (2/5) as a rational number with numerator:
(i) -56
(ii) 154
(iii) -750
(iv) 500
Solution 5:
(i) (2/5)
To get-56 as a numerator of the given value we have to multiply both by -28
Then we get, (2/5) × ((-28)/(-28))
= ((-56)/(-140))
Hence (2/5) as a rational number with numerator -56 is ((-56)/(-140))
(ii) (2/5)
To get 154 as a numerator of the given value we have to multiply both by 77
Then we get, (2/5) × (77/77)
= (154/385)
Hence (2/5) as a rational number with numerator 154 is (154/385)
(iii) (2/5)
To get -750 as a numerator of the given value we have to multiply both by -375
Then we get, (2/5) × ((-375)/(-375)) = ((-750)/(-1875))
Hence (2/5) as a rational number with numerator -750 is ((-750)/(-1875))
(iv) (2/5)
To get 500 as a numerator of the given value we have to multiply both by 250
Then, (2/5) × (250/250) = (500/1250)
Hence (2/5) as a rational number with numerator 500 is (500/1250)
Question :6. Express ((-192)/108) as a rational number with numerator:
(i) 64
(ii) -16
(iii) 32
(iv) -48
Solution 6:
(i) ((-192)/108)
To get 64 as a numerator of the given value we have to divide both by -3
Then, ((-192)/108) ÷ ((-3)/(-3)) = (64/(-36))
Hence ((-192)/108) as a rational number with numerator 64 is (64/(-36))
(ii) ((-192)/108)
To get -16 as a numerator of the given value we have to divide both by 12
Then, ((-192)/108) ÷ (12/12) = ((-16)/9)
Hence ((-192)/108) as a rational number with numerator -16 is ((-16)/9)
(iii)((-192)/108)
To get 32 as a numerator of the given number we have to divide both by -6
The, ((-192)/108) ÷ ((-6)/(-6)) =(32/(-18))
Hence ((-192)/108) as a rational number with numerator 32 is (32/-18)
(iv) ((-192)/108)
To get -48 as a numerator of the given number we have to divide both by 4
Then, ((-192)/108) ÷ (4/4) = ((-48)/27)
Hence ((-192)/108) as a rational number with numerator -48 is ((-48)/27) (-48/27)
Question :7. Express (169/(-294)) as a rational number with denominator:
(i) 14
(ii) -7
(iii) -49
(iv) 1470
Solution 7:
(i) (169/(-294))
To get 14 as a denominator of the given number we have to divide both numerator and denominator by -21
Then, (169/(-294)) ÷ (21/(-21)) = ((-8)/14)
Hence (169/(-294)) as a rational number with denominator 14 is ((-8)/14)
(ii) (169/(-294))
To get -7 as a denominator of the given number we have to divide both numerator and denominator by 42
Then, (169/(-294)) ÷ (42/42) = (4/(-7))
Hence (169/(-294)) as a rational number with denominator -7 is (4/(-7))
(iii) (169/(-294))
To get -49 as a denominator of the given number we have to divide both numerator and denominator by 6
Then, (169/(-294)) ÷ (6/6) = (28/(-49))
Hence (169/(-294)) as a rational number with denominator -49 is (28/(-49))
(iv) (169/(-294))
To get 1407 as a denominator of the given number we have to multiply both numerator and denominator by -5
Then, (169/(-294)) × ((-5)/(-5)) = ((-840)/1470)
Hence (169/(-294)) as a rational number with denominator 1470 is ((-840)/1470)
Question :8. Write ((-14)/42) in a form so that the numerator is equal to:
(i) -2
(ii) 7
(iii) 42
(iv) -70
Solution 8:
(i) ((-14)/42)
To get -2 as a numerator of the given number we have to divide both numerator and denominator by 7
Then, ((-14)/42) ÷ (7/7) = ((-2)/6)
Hence ((-14)/42) as a rational number with numerator -2 is ((-2)/6)
(ii) ((-14)/42)
To get7 as a numerator of the given number we have to divide both numerator and denominator by -2
Then, ((-14)/42) ÷ ((-2)/(-2)) = (7/(-21))
Hence ((-14)/42) as a rational number with numerator -14 is (7/(-21))
(iii) ((-14)/42)
To get 42 as a numerator of the given number we have to multiply both numerator and denominator by -3
Then, ((-14)/42) × ((-3)/(-3)) = (42/(-126))
Hence ((-14)/42) as a rational number with numerator 42 is (42/(-126))
(iv) ((-14)/42)
To get -70 as a numerator of the given number we have to multiply both numerator and denominator by 5
Then, ((-14)/42) × (5/5) = ((-70)/210)
Hence ((-14)/42) as a rational number with numerator -70 is ((-70)/210)
Question 9: Select those rational numbers which can be written as a rational number with numerator 6: (1/22), (2/3), (3/4), (4/(-5)), ((-6)/7), ((-7)/8)
Solution 9:
Rational number with numerator 6 are:
=(1/22)
Multiplying by 6, (1/22) written as
((1 ×6)/(22 ×6)) = (6/132)
=(2/3)
Multiplying by 3, (2/3) written as
((2 ×3)/(3 ×3)) = (6/9)
=(3/4)
Multiplying by 2, (3/4) written as
((3×2)/(4×2) ) = (6/8)
= ((-6)/7)
Multiplying by -1, ((-6)/7) written as
((-6 ×(-1))/(7 ×(-1))) = (6/(-7))
Hence, rational numbers can be written as a rational number with numerator 6 are (1/22), (2/3), (3/4), and ((-6)/7)
Question :10. Select those rational numbers which can be written as rational number with denominator 4:(7/8), (64/16), (36/(-12)), ((-16)/17), (5/(-4)), (140/28)
Solution 10:
Rational numbers that can be written as rational number with denominator 4 are:
(7/8) ÷ (2/2)= (3.5/4)
On dividing both denominator and denominator by 2
(64/16) ÷ (4/4)= (16/4)
On dividing both denominator and numerator by 4
(36/(-12)) ÷ ((-3)/(-3))= ((-12)/4)
On dividing both denominator and numerator by -3
(5/(-4)) ÷ ((-1)/(-1))= ((-5)/4)
On multiplying both denominator and numerator by -1
(140/28) ÷ (7/7)= (20/4)
On dividing both numerator and denominator by 7
Question 11: In each of the following, find an equivalent form of the rational number having a common denominator:
(i) (3/4) and (5/12)
(ii) (2/3), (7/6) and (11/12)
(iii) (5/7), (3/8), (9/14) and (20/21)
Solution 11:
(i) (3/4) and (5/12) : multiply by 3
(3/4) = ((3 × 3))/((4 × 3)) = (9/12)
Equivalent forms are (9/12) and (5/12) having same denominators
(ii) (2/3), (7/6) and (11/12) : multiply by 4
(2/3) = ((2 × 4))/((3 × 4)) = (8/12)
(7/6) = ((7 × 2))/((6 × 2)) = (14/12)
Equivalent forms are (8/12), (14/12) and (11/12) having same denominators
(iii) (5/7), (3/8), (9/14) and (20/21)
(5/7) multiply by 24
= ((5 × 24))/((7 × 24)) = (120/168)
(3/8) multiply by 21
= ((3 × 21))/((8 × 21)) = (63/168)
(9/14) multiply by 12
= ((9 × 12))/((14 × 12)) = (160/168)
(20/21) multiply by 8
= ((20 × 8))/((21×8)) = (160/168)
Forms are (120/168), (63/168), (108/168) and (160/168) having same denominators.
Exercise 4.3
Question 1: Determine whether the following rational numbers are in the lowest form or not:
(i) (65/84)
(ii) ((-15)/32)
(iii) (24/128)
(iv) ((-56)/(-32))
Solution 1:
(i) (65/84) = 65 and 84 have no common factor their HCF is 1.
Thus, (65/84) is in its lowest form.
(ii) ((-15)/32) = -15 and 32 have no common factor i.e., their HCF is 1.
Thus, ((-15)/32) is in its lowest form.
(iii) (24/128) = HCF of 24 and 128 is not 1.
Thus, given rational number is not in its simplest form.
(iv) ((-56)/(-32)) = HCF of 56 and 32 is 8 and also not equal to 1.
Thus, given rational number is not in its simplest form.
Question 2: Express each of the following rational numbers to the lowest form:
(i) (4/22)
(ii) ((-36)/180)
(iii) (132/(-428))
(iv) ((-32)/(-56))
Solution 2:
(i) (4/22) HCF of 4 and 22 is 2
Divide the given number by its HCF
(4 ÷ 2/22 ÷ 2) = (2/11)
Hence, (2/11) is the simplest form of (4/22)
(ii) ((-36)/180) HCF of 36 and 180 is 36
Divide the given number by its HCF
(-36 ÷ 36/180 ÷ 36) = ((-1)/5)
Hence, ((-1)/5) is the simplest form of ((-36)/180)
(iii) (132/(-428)) HCF of 132 and 428 is 4
Divide the given number by its HCF
(132 ÷ (4/(-428)) ÷ 4) = (33/(-107))
Hence, (33/(-107)) is the simplest form of (132/(-428))
(iv) ((-32)/(-56)) HCF of 32 and 56 is 8
Divide the given number by its HCF
(-32 ÷ (8/(-56)) ÷ 8) = (4/7)
Hence, (4/7) is the simplest form of ((-32)/(-56))
Question :3. Fill in the blanks:
(i) ((-5)/7) = (…/35) = (…/49)
(ii) ((-4)/(-9)) = (…/18) = (12/…)
(iii) (6/(-13)) = ((-12)/…) = (24/…)
(iv) ((-6)/…) = (3/11) = (…/(-55))
Solution 3:
(i) ((-5)/7) = (…/35) = (…/49)= ((-5)/7) = ((-25)/35) = ((-35)/49)
Description:
((-5)/7) = (…/35) = (…/49)
On multiplying by 5:
((-5)/7) × (5/5) = ((-25)/35)
On multiplying by 7
((-5)/7) × (7/7) = ((-35)/49)
(ii) ((-4)/(-9)) = (…/18) = (12/…)= ((-4)/(-9)) = (8/18) = (12/27)
Description:
((-4)/(-9)) = (…/18) = (12/…)
On multiplying by -2
((-4)/(-9)) × ((-2)/(-2)) = (8/18)
on multiplying by -3
((-4)/(-9)) × ((-3)/(-3)) = (12/27)
(iii)(6/(-13)) = ((-12)/…) = (24/…)= (6/(-13)) = ((-12)/26) = (24/(-52))
Description:
(6/(-13)) = ((-12)/…) = (24/…)
On multiplying by -2
(6/(-13)) × ((-2)/(-2)) = ((-12)/26)
On multiplying by 4
(6/(-13)) × (4/4) = (24/(-52))
(iv) ((-6)/…) = (3/11) = (…/(-55))= ((-6)/(-22)) = (3/11) = ((-15)/(-55))
Description:
((-7)/…) = (3/11) = (…/(-55))
0n multiplying by -2
(3/11) × ((-2)/(-2)) = ((-6)/(-22))
On multiplying by -5
(3/11) × ((-5)/(-5)) = ((-15)/(-55))
Exercise 4.4
Question 1: Write each of the following rational numbers in the standard form:
(i) (2/10)
(ii) ((-8)/36)
(iii) (4/(-16))
(iv) ((-15)/(-35))
(v) (299/(-161))
(vi) ((-63)/(-210))
(vii) (68/(-119))
(viii) ((-195)/275)
Solution 1:
(i) (2/10) HCF of 2 and 10 is 2
Dividing the fraction by HCF i.e. 2:
(2/10) ÷ (2/2) = (1/5)
Hence, (1/5) is the standard form of (2/10).
(ii) ((-8)/36) HCF of 8 and 36 is 4
Dividing the fraction by HCF i.e. 4:
((-8)/36) ÷ (4/4) = ((-2)/9)
Hence ((-2)/9) is the standard form of ((-8)/36).
(iii) (4/(-16))
Denominator is negative here, so we multiply fraction by -1
(4/(-16)) × ((-1)/(-1)) = (4/(-16))
HCF of 4 and 16 is 4
Dividing the fraction by HCF i.e. 4:
(4/(-16)) ÷ (4/4) = ((-1)/4)
Hence ((-1)/4) is the standard form of (4/(-16)).
(iv) ((-15)/(-35))
Denominator is negative here, so we multiply fraction by -1
((-15)/(-35)) × ((-1)/(-1)) = (15/35)
HCF of 15 and 35 is 4
Dividing the numerator and denominator by HCF i.e. 5:
((-15)/(-35)) ÷ (5/5) = (3/7)
Hence (3/7) is the standard form of ((-15)/(-35))
(v) (299/(-161))
Denominator is negative here, so we multiply fraction by -1
(299/(-161)) × ((-1)/(-1)) = ((-299)/161)
The HCF of 299 and 161 is 23
Dividing the numerator and denominator by HCF i.e. 23:
(299/(-161)) ÷ (23/23) = ((-13)/7)
Hence ((-13)/7) is the standard form of (299/(-161))
(vi) ((-63)/(-210))
HCF of 63 and 210 is 21
Dividing the numerator and denominator by HCF i.e. 21:
((-63)/(-210)) ÷ (21/21) = ((-3)/(-10)) = (3/10)
Hence (3/10) is the standard form of ((-63)/(-210))
(vii) (68/(-119))
Here denominator is negative so we have multiply both fraction by -1
(68/(-119)) × ((-1)/(-1)) = ((-68)/119)
The HCF of 68 and 119 is 17
Dividing the fraction by HCF i.e. 17:
((-68)/119) ÷ (17/17) = ((-4)/7)
Hence, ((-4)/7) is the standard form of (68/(-119))
(viii) ((-195)/275)
The HCF of 195 and 257 is 5
Dividing the fraction by HCF i.e. 5:
((-195)/275) ÷ (5/5) = ((-39)/5)
Hence, ((-39)/5) is the standard form of ((-195)/275)
Exercise 4.5
Question :1. Which of the following rational numbers are equal?
(i) ((-9)/12) and (8/(-12))
(ii) ((-16)/20) and (20/(-25))
(iii) ((-7)/21) and (3/(-9))
(iv) ((-8)/(-14)) and (13/21)
Solution 1:
(i) ((-9)/12) and (8/(-12))
By dividing the fraction by their HCF i.e. 3
Then the standard form of ((-9)/12) is ((-3)/4)
Dividing the fraction of given number by their HCF i.e. by 4
The standard form of (8/(-12)) = ((-2)/3)
Since, the standard forms of two rational numbers are not same. Hence, they are not equal.
(ii) ((-16)/20) and (20/(-25))
Multiplying fraction of ((-16)/20) by the denominator of (20/(-25)) i.e. -25.
((-16)/20) × ((-25)/(-25)) = (400/(-500))
Multiplying the fraction of (20/(-25)) by the denominator of
((-16)/20) i.e. 20
(20/(-25)) × (20/20) = (400/(-500))
Clearly, the numerators of the above obtained rational numbers are equal.
Hence, the given rational numbers are equal
(iii) ((-7)/21) and (3/(-9))
Multiplying fraction of ((-7)/21) by the denominator of (3/(-9))
i.e. -9.
((-7)/21) × ((-9)/(-9)) = (63/(-189))
Now multiply the fraction of (3/(-9)) by the denominator of
((-7)/21) i.e. 21
(3/(-9)) × (21/21) = (63/(-189))
Clearly, the numerators of the above obtained rational numbers are equal.
Hence, the given rational numbers are equal
(iv) ((-8)/(-14)) and (13/21)
Multiplying fraction of ((-8)/(-14)) by the denominator of (13/21)
i.e. 21
((-8)/(-14)) × (21/21) = ((-168)/(-294))
Now multiply the fraction of (13/21) by the denominator of
((-8)/(-14)) i.e. -14
(13/21) × ((-14)/(-14)) = ((-182)/(-294))
Clearly, the numerators of the above obtained rational numbers are not equal.
Hence, the given rational numbers are also not equal
Question 2: In each of the following pairs represent a pair of equivalent rational numbers, find the values of x.
(i) (2/3) and (5/x)
(ii) ((-3)/7) and (x/4)
(iii) (3/5) and (x/(-25))
(iv) (13/6) and ((-65)/x)
Solution 2:
(i) (2/3) and (5/x)
They are equivalent rational number so (2/3) = (5/x)
x = ((5 × 3))/2
x = (15/2)
(ii) ((-3)/7) and (x/4)
they are equivalent rational number so ((-3)/7) = (x/4)
x = ((-3 × 4))/7
x = ((-12)/7)
(iii) Given (3/5) and (x/(-25))
They are equivalent rational number so (3/5) = (x/(-25))
x = ((3 × -25))/5
x = ((-75)/5)
x = -15
(iv) (13/6) and ((-65)/x)
They are equivalent rational number so (13/6) = ((-65)/x)
x = 6/13 x (- 65)
x = 6 x (-5)
x = -30
Question 3: In each of the following, fill in the blanks so as to make the statement true:
(i) A number which can be expressed in the form p/q, where p and q are integers and q is not equal to zero, is called a _______.
(ii) If the integers p and q have no common divisor other than 1 and q is positive, then the rational number (p/q) is said to be in the _______.
(iii) Two rational numbers are said to be equal, if they have the same _______ form
(iv) If m is a common divisor of a and b, then (a/b) = (a ÷ m)/ _______.
(v) If p and q are positive Integers, then p/q is a _______ rational number and (p/(-q)) is a ______ rational number.
(vi) The standard form of -1 is _______.
(vii) If (p/q) is a rational number, then q cannot _______.
(viii) Two rational numbers with different numerators are equal, if their numerators are in the same _______ as their denominators.
Solution 3:
(i) A number which can be expressed in the form p/q, where p and q are integers and q is not equal to zero, is called a Rational number.
(ii) If the integers p and q have no common divisor other than 1 and q is positive, then the rational number (p/q) is said to be in the Standard form.
(iii) Two rational numbers are said to be equal, if they have the same Standard form
(iv) If m is a common divisor of a and b, then (a/b) = (a ÷ m)/ b ÷ m.
(v) If p and q are positive Integers, then p/q is a Positive rational number and (p/(-q)) is a negative rational number.
(vi) The standard form of -1 is((-1)/1) .
(vii) If (p/q) is a rational number, then q cannot Zero .
(viii) Two rational numbers with different numerators are equal, if their numerators are in the same Ratio as their denominators.
Question 4: In each of the following state if the statement is true (T) or false (F):
(i) The quotient of two integers is always an integer.
(ii) Every integer is a rational number.
(iii) Every rational number is an integer.
(iv) Every traction is a rational number.
(v) Every rational number is a fraction.
(vi) If (a/b) is a rational number and m any integer, then (a/b) = ((a x m))/((b x m))
(vii) Two rational numbers with different numerators cannot be equal.
(viii) 8 can be written as a rational number with any integer as denominator.
(ix) 8 can be written as a rational number with any integer as numerator.
(x) (2/3) is equal to (4/6).
Solution 4:
(i) The quotient of two integers is always an integer.
(ii) Every integer is a rational number.
(iii) Every rational number is an integer.
(iv) Every traction is a rational number.
(v) Every rational number is a fraction.
(vi) If (a/b) is a rational number and m any integer, then (a/b) = ((a x m))/((b x m))
(vii) Two rational numbers with different numerators cannot be equal.
(viii) 8 can be written as a rational number with any integer as denominator.
(ix) 8 can be written as a rational number with any integer as numerator.
(x) (2/3) is equal to (4/6).
(i) False
Description:
The quotient of two integers is not necessary to be an integer
(ii) True
Description:
Every integer can be expressed in the form of p/q, where q is not zero.
(iii) False
Description:
Every rational number is not necessary to be an integer
(iv) True
Description:
According to definition of rational number i.e. every integer can be expressed in the form of p/q, where q is not zero.
(v) False
Description:
It is not necessary that every rational number is a fraction.
(vi) True
Description:
If (a/b) is a rational number and m any integer, then (a/b) = ((a x m))/((b x m)) is one of the rule of
rational numbers
(vii) False
Description:
They can be equal, when simplified further.
(viii) False
Description:
8 can be written as a rational number but we can’t write 8 with any integer as denominator.
(ix) False
Description:
8 can be written as a rational number but we can’t with any integer as numerator.
(x) True
Description:
When convert it into standard form they are equal
Exercise 4.6
Question :1. Draw the number line and represent following rational number on it:
(i) (2/3)
(ii) (3/4)
(iii) (3/8)
(iv) ((-5)/8)
(v) ((-3)/16)
(vi) ((-7)/3)
(vii) (22/(-7))
(viii) ((-31)/3) (-31/3)
Solution 1:
(i) We know that (2/3) is greater than 2 and less than 3.
∴ it lies between 2 and 3. It can be represented on number line as,
Question :2. Which of the two rational numbers in each of the following pairs of rational number is greater?
(i) ((-3)/8), 0
(ii) (5/2), 0
(iii) ((-4)/11), (3/11)
(iv) ((-7)/12), (5/(-8))
(v) (4/(-9)), ((-3)/(-7))
(vi) ((-5)/8), (3/(-4))
(vii) (5/9), ((-3)/(-8))
(viii) (5/(-8)), ((-7)/12)
Solution 2:
(i) Given ((-3)/8), 0
We know that every positive rational number is greater than zero and every negative
rational number is smaller than zero. Thus, – ((-3)/8) > 0
(ii) Given (5/2), 0
We know that every positive rational number is greater than zero and every negative rational number is smaller than zero. Thus, (5/2) > 0
(iii) Given ((-4)/11), (3/11)
We know that every positive rational number is greater than zero and every negative rational number is smaller than zero, also the denominator is same in given question now we have to compare the numerator, thus((-4)/11) < (3/11).
Question :3. Which of the two rational numbers in each of the following pairs of rational numbers is smaller?
(i) ((-6)/(-13)), (7/13)
(ii) (16/(-5)), 3
(iii) ((-4)/3), (8/(-7))
(iv) ((-12)/5), (-3)
Solution 3:
(i) Given ((-6)/(-13)), (7/13)
Here denominator is same Hence compare the numerator,
Thus ((-6)/(-13)) < (7/13)
(ii) Given (16/(-5)), 3
We know that 3 is a whole number with positive sign
Hence (16/(-5)) < 3
(iii) Given ((-4)/3), (8/(-7))
Consider ((-4)/3)
Multiply both numerator and denominator by 7 then we get
((-4)/3) × (7/7) = ((-28)/21)…… (1)
Now consider (8/(-7))
Multiply both numerator and denominator by 3 we get
(8/(-7)) × (3/3) = ((-24)/21)…… (2)
The denominator is same in equation (1) and (2) now we have to compare the numerator, thus ((-4)/3) < (8/(-7))
(iv) Given ((-12)/5), (-3)
Now consider ((-3)/1)
Multiply both numerator and denominator by 5 we get
((-3)/1) × (5/5) = ((-15)/5)
The denominator is same in above equation, now we have to compare the numerator, thus ((-12)/5) > (-3)
Question :4. Fill in the blanks by the correct symbol out of >, =, or <:
(i) ((-6)/7) …. (7/13)
(ii) ((-3)/5) …. ((-5)/6)
(iii) ((-2)/3) …. (5/(-8))
(iv) 0 …. ((-2)/5)
Solution 4:
(i) ((-6)/7) < (7/13)
Description:
Because every positive number is greater than a negative number.
(ii) ((-3)/5) > ((-5)/6)
Description:
Consider ((-3)/5)
Multiply both numerator and denominator by 6 then we get
((-3)/5) × (6/6) = ((-18)/30)…… (1)
Now consider (-5/6)
Multiply both numerator and denominator by 5 we get
((-5)/6) × (5/5) = ((-25)/30)…… (2)
The denominator is same in equation (1) and (2) now we have to compare the numerator, thus ((-3)/5) >((-5)/6)
(iii) ((-2)/3) < (5/(-8))
Description:
Consider ((-2)/3)
Multiply both numerator and denominator by 8 then we get
((-2)/3) × (8/8) = ((-16)/24)…… (1)
Now consider (5/(-8))
Multiply both numerator and denominator by 3 we get
(5/(-8)) × (3/3) = (15/(-24))…… (2)
The denominator is same in equation (1) and (2) now we have to compare the numerator, thus ((-2)/3) < (5/(-8))
(iv) 0 > ((-2)/5)
Description:
Because every positive number is greater than a negative number
Question :5. Arrange the following rational numbers in ascending order:
(i) (3/5), ((-17)/(-30)), (8/(-15)), ((-7)/10)
(ii) ((-4)/9), (5/(-12)), (7/(-18)), (2/(-3))
Solution 5:
(i) Given (3/5), ((-17)/(-30)), (8/(-15)), ((-7)/10)
The LCM of 5, 30, 15 and 10 is 30
Multiplying the numerators and denominators to get the denominator equal to the LCM i.e. 30
Consider (3/5)
Multiply both numerator and denominator by 6, then we get
(3/5) × (6/6) = (18/30) ….. (1)
Consider (8/(-15))
Multiply both numerator and denominator by 2, then we get
(8/(-15)) × (2/2) = (16/(-30)) ….. (2)
Consider ((-7)/10)
Multiply both numerator and denominator by 3, then we get
((-7)/10) × (3/3) = ((-21)/30) ….. (3)
In the above equation, denominators are same
Now on comparing the ascending order is:
((-7)/10) < (8/(-15)) < ((-17)/(-30)) < (3/5)
(ii) Given ((-4)/9), (5/(-12)), (7/(-18)), (2/(-3))
The LCM of 9, 12, 18 and 3 is 36
Multiplying the numerators and denominators to get the denominator equal to the LCM i.e. 36
Consider ((-4)/9)
Multiply both numerator and denominator by 4, then we get
((-4)/9) × (4/4) = ((-16)/36) ….. (1)
Consider (5/(-12))
Multiply both numerator and denominator by 3, then we get
(5/(-12)) × (3/3) = (15/(-36)) ….. (2)
Consider (7/(-18))
Multiply both numerator and denominator by 2, then we get
(7/(-18)) × (2/2) = (14/(-36)) ….. (3)
Consider (2/(-3))
Multiply both numerator and denominator by 12, then we get
(2/(-3)) × (12/12) = (24/(-36)) ….. (4)
In the above equation, denominators are same
Now on comparing the ascending order is:
(2/(-3)) < ((-4)/9) < (5/(-12)) < (7/(-18))
Question :6. Arrange the following rational numbers in descending order:
(i) (7/8), (64/16), (39/(-12)), (5/(-4)), (140/28)
(ii) ((-3)/10),(17/(-30)),(7/(-15)), ((-11)/20)
Solution 6:
(i) Given (7/8), (64/16), (39/(-12)), (5/(-4)), (140/28)
The LCM of 8, 16, 12, 4 and 28 is 336
Multiplying the numerators and denominators to get the denominator equal to the LCM i.e. 336
Consider (7/8)
Multiply both numerator and denominator by 42, then we get
(7/8) × (42/42) = (294/336) ….. (1)
Consider (64/16)
Multiply both numerator and denominator by 21, then we get
(64/16) × (21/21) = (1344/336) ….. (2)
Consider (39/(-12))
Multiply both numerator and denominator by 28, then we get
(39/(-12)) × (28/28) = ((-1008)/336) (-1008/336) ….. (3)
Consider (5/(-4))
Multiply both numerator and denominator by 84, then we get
(5/(-4)) × (84/84) = ((-420)/336) ….. (4)
In the above equation, denominators are same
Now on comparing the descending order is:
(140/28) > (64/16) > (7/8) > (5/(-4)) > (39/(-12))
(ii) Given ((-3)/10),(17/(-30)),(7/(-15)), ((-11)/20)
The LCM of 10, 30, 15 and 20 is 60
Multiplying the numerators and denominators to get the denominator equal to the LCM i.e. 60
Consider ((-3)/10)
Multiply both numerator and denominator by 6, then we get
((-3)/10) × (6/6) = ((-18)/60) ….. (1)
Consider (17/(-30))
Multiply both numerator and denominator by 2, then we get
(17/(-30)) × (2/2) = (34/(-60)) ….. (2)
Consider (7/(-15))
Multiply both numerator and denominator by 4, then we get
(7/(-15)) × (4/4) = (28/(-60)) ….. (3)
In the above equation, denominators are same
Now on comparing the descending order is:
((-3)/10) > (7/(-15)) > ((-11)/20) > (17/(-30))
Question :7. Which of the following statements are true:
(i) The rational number (29/23) lies to the left of zero on the number line.
(ii) The rational number ((-12)/(-17)) lies to the left of zero on the number line.
(iii) The rational number (3/4) lies to the right of zero on the number line.
(iv) The rational number ((-12)/(-5)) and ((-7)/17) are on the opposite side of zero on the number line.
(v) The rational number ((-2)/15) and (7/(-31)) are on the opposite side of zero on the number line.
(vi) The rational number ((-3)/(-5)) is on the right of ((-4)/7) on the number line.
Solution 7:
(i) False
Description:
It lies to the right of zero because it is a positive number.
(ii) False
Description:
It lies to the right of zero because it is a positive number.
(iii) True
Description:
Always positive number lie on the right of zero
(iv) True
Description:
Because they are of opposite sign
(v) False
Description:
Because they both are of same sign
(vi) True
Description:
They both are of opposite signs and positive number is greater than the negative number. Thus, it is on the right of the negative number.<|endoftext|>
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# How do you solve x^2-1>=4x using a sign chart?
Dec 14, 2016
The answer is x in ] -oo,(2-sqrt5) ] uu [(2+sqrt5), +oo[
#### Explanation:
Let's rewrite the equation
${x}^{2} - 4 x - 1 \ge 0$
Let $f \left(x\right) = {x}^{2} - 4 x - 1$
We need the values of $x$, when $f \left(x\right) = 0$
that is, ${x}^{2} - 4 x - 1 = 0$
Let's calculate $\Delta = 16 - 4 \cdot 1 \cdot - 1 = 20$
$\Delta > 0$. there are 2 real solutions
${x}_{1} = \frac{4 + \sqrt{20}}{2} = \frac{4 + 2 \sqrt{5}}{2} = 2 + \sqrt{5}$
${x}_{2} = \frac{4 - \sqrt{20}}{2} = \frac{4 - 2 \sqrt{5}}{2} = 2 - \sqrt{5}$
Now, we can do our sign chart
$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a a a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$\left(2 - \sqrt{5}\right)$$\textcolor{w h i t e}{a a a a}$$\left(2 + \sqrt{5}\right)$$\textcolor{w h i t e}{a a a a}$$+ \infty$
$\textcolor{w h i t e}{a a a a}$$x - 2 + \sqrt{5}$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a a a a a a a}$$+$$\textcolor{w h i t e}{a a a a a a a a a}$$+$
$\textcolor{w h i t e}{a a a a}$$x - 2 - \sqrt{5}$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a a a a a a}$$+$
$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a a a a a a}$$+$$\textcolor{w h i t e}{a a a a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a a a a a a}$$+$
Therefore,
$f \left(x \ge 0\right)$ when x in ] -oo,(2-sqrt5) ] uu [(2+sqrt5), +oo[
graph{x^2-4x-1 [-11.25, 11.25, -5.625, 5.625]}<|endoftext|>
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You should complete the VLSI CAD Part I: Logic course before beginning this course.
A modern VLSI chip is a remarkably complex beast: billions of transistors, millions of logic gates deployed for computation and control, big blocks of memory, embedded blocks of pre-designed functions designed by third parties (called “intellectual property” or IP blocks). How do people manage to design these complicated chips? Answer: a sequence of computer aided design (CAD) tools takes an abstract description of the chip, and refines it step-wise to a final design. This class focuses on the major design tools used in the creation of an Application Specific Integrated Circuit (ASIC) or System on Chip (SoC) design. Our focus in this part of the course is on the key logical and geometric representations that make it possible to map from logic to layout, and in particular, to place, route, and evaluate the timing of large logic networks. Our goal is for students to understand how the tools themselves work, at the level of their fundamental algorithms and data structures. Topics covered will include: technology mapping, timing analysis, and ASIC placement and routing.
Programming experience (C, C++, Java, Python, etc.) and basic knowledge of data structures and algorithms (especially recursive algorithms). An understanding of basic digital design: Boolean algebra, Kmaps, gates and flip flops, finite state machine design. Linear algebra and calculus at the level of a junior or senior in engineering. Elementary knowledge of RC linear circuits (at the level of an introductory physics class).
Orientation In this module you will become familiar with the course and our learning environment. The orientation will also help you obtain the technical skills required for the course.
ASIC Placement In this second part of our course, we will talk about geometry. We will begin with an overview of the ASIC layout process, and discuss the role of technology libraries, tech mapping (a topic we delay until the following week, to let those who want to do the Placer programming assignment have more time), and placement and routing. In this set of lectures, we focus on the placement process itself: you have a million gates from the result of synthesis and map, so, where do they go? This process is called “placement”, and we describe an iterative method, and a mathematical optimization method, that can each do very large placement tasks.
Technology Mapping Technology Mapping! We omitted one critical step between logic and layout, the process of translating the output of synthesis -- which is NOT real gates in your technology library -- into real logic gates. The Tech Mapper performs this important step, and it is a surprisingly elegant algorithm involving recursive covering of a tree. Another place where knowing some practical computer science comes to the rescue in VLSI CAD.
ASIC Routing Routing! You put a few million gates on the surface of the chip in some sensible way. What's next? Create the wires to connect them. We focus on Maze Routing, which is a classical and powerful technique with the virtue that one can "add" much sophisticated functionality on top of a rather simple core algorithm. This is also the topic for final (optional) programming assignment. Yes, if you choose, you get to route pieces of the industrial benchmarks we had you place in the placer software assignment.
Timing Analysis You synthesized it. You mapped it. You placed it. You routed it. Now what? HOW FAST DOES IT GO? Oh, we need some new models, to talk about how TIMING works. Delay through logic gates and big networks of gates. New numbers to understand: ATs, RATs, SLACKS, etc. And some electrical details (minimal) to figure out how delays happen through the physical geometry of physical routed wires. All together this is the stuff of Static Timing Analysis (STA), which is a huge and important final "sign off" step in real ASIC design.
Final Exam There is no new content this week. Instead, you should focus on finishing the last problem set and completing the Final Exam.<|endoftext|>
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Ultraviolet (UV) radiation is the part of the invisible radiation spectrum created by the sun. UV radiation is made up of elementary particles known as photons, which are categorised into wavelengths. The wavelength varies inversely in accordance to its energy – the shorter the wavelength, the greater its energy. The skin reflects some UV radiation naturally, but care must be taken since that which penetrates the skin tends to damage the cells permanently.
There are three types of UV rays: UVA, UVB and UVC. The UV radiation wavelength range is 100-400nm, often referred to as the Broad Spectrum. These levels are not constant and are subject to change from seasons, atmosphere and even the time of day.
UVA wavelength range: 320 – 400nm
UVA accounts for the largest amount of UV radiation to reach the Earth’s surface and accounts for 10% of all solar rays – 90% of UV energy received by unprotected skin comes from teh UVA range (Osterwalder et al).
Although lower in energy than UVB, it is believed that UVA causes damage to the connective tissue and increases the chances of skin cancer. 80% of UVA rays penetrate to the dermal layer of the skin and a further 20% penetrate even further.
UVA radiation can travel through clear glass and is mainly responsible for the ageing effect, causing wrinkling of the skin.<|endoftext|>
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Get first steps for creating a respectful yet vibrant environment for students to explore diverse ideas on controversial topics, from politics to profanity, religion to racism. Four guidelines and a debate leader checklist provide a foundation for those seeking to steer productive conversations about controversial subjects.
- Lesson Plans, Modules (Teaching Unit)
- Subject: Foundations of Democracy
- Grades: 8, 9, 10, 11, 12<|endoftext|>
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# Lesson 15
Mayor que, menor que
## Warm-up: Conversación numérica: Sumemos o restemos 10 (10 minutes)
### Narrative
The purpose of this Number Talk is to elicit strategies and understandings students have for adding and subtracting 10 from a two-digit number.
When students notice how the tens place changes while the ones place doesn’t, they are making sense of the base-ten structure of numbers (MP7).
### Launch
• Display one expression.
• “Hagan una señal cuando tengan una respuesta y puedan explicar cómo la obtuvieron” // “Give me a signal when you have an answer and can explain how you got it.”
• 1 minute: quiet think time
### Activity
• Keep expressions and work displayed.
• Repeat with each expression.
### Student Facing
Encuentra mentalmente el valor de cada expresión.
• $$35 + 10$$
• $$52 + 10$$
• $$52 -10$$
• $$83-10$$
### Activity Synthesis
• “¿Cuál expresión tiene un valor mayor, $$52 + 10$$ o $$52 - 10$$? ¿Pueden compararlas sin encontrar el valor de las expresiones?” // “Which expression has a greater value $$52 + 10$$ or $$52 - 10$$? Can you compare without finding the value of the expressions?” ($$52 + 10$$ will be greater than $$52 - 10$$ because it is getting larger rather than having something taken away.)
## Activity 1: ¿Cuál es mayor?, ¿cuál es menor? (20 minutes)
### Narrative
The purpose of this activity is for students to interpret comparison symbols and compare two-digit numbers based on the value of the digits using drawings, numbers, or words. During the launch, students notice and wonder about two related comparison statements that use symbols rather than words. The teacher creates a chart with the comparison statements and what the symbols mean in words for students to refer to during the activity. Students may use connecting cubes to build each number, use the value of each number’s tens or ones place, or use expressions that show the value of tens and ones to justify their reasoning. Students then circle the true comparison statement.
MLR7 Compare and Connect. Synthesis: After both examples have been presented and discussed, lead a discussion comparing, contrasting, and connecting the different approaches. Ask, “¿En qué se parecían sus estrategias al comparar 21 y 12 y al comparar 74 y 78? ¿En qué eran diferentes?” // “How were your approaches similar when comparing 21 and 12 and when comparing 74 and 78? How were they different?”
Engagement: Develop Effort and Persistence. Chunk this task into more manageable parts. Check in with students to provide feedback and encouragement after each chunk.
Supports accessibility for: Attention, Social-Emotional Functioning
### Required Preparation
• Write $$78 > 45$$ and $$45 < 78$$ on a piece of chart paper.
### Launch
• Groups of 2
• Give students access to connecting cubes in towers of 10 and singles.
• Display $$78 > 45$$ and $$45 < 78$$.
• “¿Qué observan? ¿Qué se preguntan?” // “What do you notice? What do you wonder?” (The wider part of the symbol points toward the larger number. The point is toward the smaller number. Is this always true? Does the order you write the comparison matter? Does the order of the numbers matter?)
• 1 minute: quiet think time
• 2 minutes: partner discussion
• Record responses.
• “Estos son símbolos de comparación. Los usamos para mostrar que un valor es mayor que o menor que otro sin escribir las palabras. El lado abierto, o el lado del símbolo con la mayor cantidad de espacio entre la parte superior y la inferior, siempre mira hacia el número mayor” // “These are comparison symbols. We can use them to show that one value is greater than or less than another without writing the words. The open side, or the side of the symbol with the greater amount of space between the top and bottom, always faces the greater number.”
• Record “78 es mayor que 45” // “78 is greater than 45” under $$78 > 45$$. Consider writing “greater than” in a different color.
• “El lado puntiagudo, o el lado del símbolo con menos espacio entre las líneas, siempre mira hacia el número menor” // “The pointy side, or side of the symbol with less space between the lines, always faces the lesser number.”
• Record “45 es menor que 78” // “45 is less than 78” under $$45 < 78$$. Consider writing “less than” in a different color.
### Activity
• 10 minutes: partner work time
• Monitor for students who:
• verbally describe the relationship between two numbers using “greater than” and “less than”
• compare numbers using their place value understanding
### Student Facing
Prepárate para explicar cómo lo sabes de una forma que los demás entiendan.
$$27 < 17$$
$$17<27$$
$$34 < 36$$
$$36 < 34$$
$$25 < 52$$
$$52 < 25$$
$$24 > 54$$
$$54 > 24$$
$$21 > 29$$
$$29 > 21$$
$$85 > 58$$
$$58 > 85$$
$$45 < 54$$
$$45 > 54$$
$$74 < 78$$
$$74 > 78$$
$$21 < 12$$
$$21>12$$
### Activity Synthesis
• Display $$21 < 12$$ and $$21 > 12$$.
• Invite previously identified students to share.
• “Sabemos que 21 es mayor que 12. ¿Cómo recuerdan qué símbolo representa mayor que?” // “We know that 21 is greater than 12. How do you remember which symbol represents greater than?” (I think about the part of the symbol with a greater amount of space being next to the greater number.)
• Display $$74 < 78$$ and $$74 > 78$$.
• Invite students to share their thinking.
• “74 es menor que 78. ¿Cómo saben qué símbolos muestran esto?” // “74 is less than 78. How do you know which symbols show this?” (The side of the symbols with less space between the lines needs to be closer to 74.)
## Activity 2: Comparaciones verdaderas o falsas (15 minutes)
### Narrative
The purpose of this activity is for students to determine if comparison statements are true or false and explain why. In the previous activity, students focused on using their understanding of place value to determine if the symbols were facing the appropriate numbers. In this activity, students are encouraged to read the statements from left to right before determining whether the statement is true or false. Encourage students to use the display created in the previous activity to help them interpret the symbols and read the statements.
### Launch
• Groups of 2
• Give students access to connecting cubes in towers of 10 and singles.
• Display $$45 < 54$$.
• “En la actividad anterior, decidimos que esta afirmación es verdadera. ¿Cómo se lee esta afirmación?” // “In the last activity, we decided that this statement is true. How would we read this statement?” (45 is less than 54.)
• 30 seconds: quiet think time
• 1 minute: partner discussion
• Share responses.
• Display $$21 > 12$$.
• “También decidimos que esta afirmación era verdadera. ¿Cómo se lee esta afirmación?” // “We also decided this statement was true. How would we read this statement?” (21 is greater than 12.)
• 30 seconds: quiet think time
• 1 minute: partner discussion
• Share responses.
### Activity
• “Van a trabajar con su pareja en esta actividad. Asegúrense de que cada uno tenga tiempo para pensar en el problema y darle sentido antes de compartir lo que pensaron” // “You are going to work with your partner on this activity. Make sure that each partner has time to think on their own and make sense of the problem before sharing your thinking.”
• 10 minutes: partner work time
• Monitor for students who determined $$58 = 53$$ is false using the values of the tens or ones place to share during the synthesis.
### Student Facing
Prepárate para explicar cómo lo sabes de una forma que los demás entiendan.
1. $$17 < 47$$
2. $$58 = 53$$
3. $$45 > 63$$
4. $$39 < 93$$
5. $$4 = 46$$
Si te queda tiempo, escribe cada afirmación falsa de otra forma para que sea verdadera.
### Student Response
If students discuss whether each statement is true or false, but do not read the statements, consider asking:
• “¿Cómo puedes usar la presentación que hicimos como ayuda para leer la afirmación?” // “How could you use the display we made to help you read the statement?”
• “Lee la afirmación. ¿Qué observaste sobre el símbolo que podría ayudarte a recordar cómo se dice la próxima vez que lo leas?” // “Read the statement. What did you notice about the symbol that could help you remember how to say it the next time you read it?”
### Activity Synthesis
• Display answers for students to check their work.
• Invite selected students to share their explanations for $$58 = 53$$.
• “¿Cómo podemos cambiar esta afirmación para que sea verdadera?” // “How can we change this statement so it is true?” ($$58 > 53$$)
• Read the new comparison statement.
## Lesson Synthesis
### Lesson Synthesis
Display 43 and 48.
“Hoy conocimos los símbolos que significan ‘mayor que’ y ‘menor que’, y comparamos más números de dos dígitos. Usen las palabras ‘mayor que’ para comparar los números” // “Today we learned symbols that mean 'greater than' and 'less than' and compared more two-digit numbers. Use the words ‘greater than’ to compare the numbers.” (48 is greater than 43.)
“¿Cómo podría escribir esto usando el símbolo de mayor que?” // “How might I write this using the greater than symbol?” ($$48 > 43$$)
Display 85 and 65.
“Usen las palabras ‘menor que’ para comparar los números” // “Use the words ‘less than’ to compare the numbers.” (65 is less than 85.)
“¿Cómo podría escribir esto usando el símbolo de menor que?” // “How might I write this using the less than symbol?” ($$65 < 85$$)<|endoftext|>
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Copper Element in Periodic Table | Atomic Number Atomic Mass
Copper is a chemical element and it is a member of group 11 and period 4 in the periodic table. The atomic number of copper is 29 and it is indicated by the symbol Cu. Cu is a soft, malleable and ductile metal with very high thermal and electrical conductivity. It is one of a few metallic elements with natural Color other than gray or silver. 63Cu and 65Cu are the stable isotopes of copper.
Uses of Copper Element:
- Cu is used to make coins, along with silver and gold.
- Cu sulphate is used as an agricultural poison and as an algicide in water purification.
- Cu compounds are used in chemical tests for sugar detection.
- It is also used in guttering, roofing and as rainspouts on buildings.
- Copper is used in plumbing and in cookware and cooking utensils.
Copper Element Information:
Discovery year:9000 BC
Discovered by:Middle East
Atomic number: 29
Relative atomic mass: 63.546
Electron configuration: [Ar] 3d10 4s1
Other elements in the same block:
Scandium, Titanium, Vanadium, Chromium, Manganese, Iron, Cobalt, Nickel, Zinc, Yttrium, Zirconium, Niobium, Molybdenum, Technetium, Ruthenium, Rhodium, Palladium, Silver, Cadmium, Lanthanum, Hafnium, Tantalum, Tungsten, Rhenium, Osmium, Iridium, Platinum, Gold, Mercury, Actinium, Rutherfordium, Dubnium, Seaborgium, Bohrium, Hassium, Meitnerium, Darmstadtium, Roentgenium and Copernicium are the elements in the same block.
Period: period 4
Other elements in the same period:
Group: group 11
Other elements in the same group:
Other elements in the same orbital:
Key isotopes: 63Cu
Melting point: 1084.62°C, 1984.32°F, 1357.77 K
Boiling point: 2560°C, 4640°F, 2833 K
Element category: transition metal
Density (g cm−3): 8.96
CAS number: 7440-50-8
Color: red-orange metallic luster
You can know detailed information for each element,
Hydrogen, Helium, LithiumBeryllium, Boron, Carbon, Nitrogen, Oxygen, Fluorine, Neon , Sodium, Magnesium, Aluminium, Silicon, Phosphorus, Sulfur, Chlorine, Argon, Potassium, Calcium, Scandium, Titanium, Vanadium, Chromium, Manganese, Iron, Cobalt, Nickel, Copper, Zinc, Gallium, Germanium, Arsenic, Selenium, Bromine, Krypton, Rubidium, Strontium, Yttrium, Zirconium, Niobium, Molybdenum, Technetium, Ruthenium, Rhodium, Palladium, Silver, Cadmium, Indium, Tin, Antimony, Tellurium, Iodine, Xenon, Caesium, Barium, Lanthanum, Cerium, Praseodymium, Neodymium, Promethium, Samarium, Europium, Gadolinium, Terbium, Dysprosium, Holmium, Erbium, Thulium, Ytterbium, Lutetium, Hafnium, Tantalum, Tungsten, Rhenium, Osmium, Iridium, Platinum, Gold, Mercury, Thallium, Lead, Bismuth, Polonium, Astatine, Radon, Francium, Radium, Actinium, Thorium, Protactinium, Uranium, Neptunium, Plutonium, Americium, Curium, Berkelium, Californium, Einsteinium, Fermium, Mendelevium, Nobelium, Lawrencium, Rutherfordium, Dubnium, Seaborgium, Bohrium, Hassium, Meitnerium, Darmstadtium, Roentgenium, Copernicium, Nihonium, Flerovium, Moscovium, Livermorium, Tennessine, Oganesson.<|endoftext|>
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An ORNL study found that complex oxide materials can self-organize into electrical circuits, which creates the possibility for new types of computer chips. (CREDIT: Oak Ridge National Laboratory)
Researchers at the Department of Energy’s Oak Ridge National Laboratory are currently analyzing the behavior of nanoscale materials. The outstanding behavior identified by them could help improve microprocessors beyond the existing silicon-based chips.
This research featuring on the cover of Advanced Electronic Materials demonstrates that a single crystal complex oxide material tends to behave like a multi-component electrical circuit when limited to micro- and nanoscales. This behavior originates from a unique feature of specific complex oxides known as phase separation, in which small regions in the material display extremely different magnetic and electronic properties.
This explains the fact that separate nanoscale regions in complex oxide materials are capable of acting as self-organized circuit elements, which may support the latest multifunctional types of computing architectures.
Within a single piece of material, there are coexisting pockets of different magnetic and/or electronic behaviors. What was interesting in this study was that we found we can use those phases to act like circuit elements. The fact that it is possible to also move these elements around offers the intriguing opportunity of creating rewritable circuitry in the material.
Zac Ward, ORNL
It is possible to control the material in different ways as the phases respond to both electrical and magnetic fields, resulting in the likelihood for new varieties of computer chips.
“It’s a new way of thinking about electronics, where you don’t just have electrical fields switching off and on for your bits,” Ward said. “This is not going for raw power. It’s looking to explore completely different approaches towards multifunctional architectures where integration of multiple outside stimuli can be done in a single material.”
With the computer industry looking to surpass the limitations of silicon-based chips, the ORNL proof-of-principle experiment highlights that phase separated materials can be considered as a way ahead of the “one-chip-fits-all” approach. A multifunctional chip, unlike a chip capable of performing only one role, could manage a number of outputs and inputs that are customized to meet the requirements of a particular application.
Typically you would need to link several different components together on a computer board if you wanted access to multiple outside senses. One big difference in our work is that we show certain complex materials already have these components built in, which may cut down on size and power requirements.
Zac Ward, ORNL
The researchers used a material known as LPCMO to demonstrate their approach. However, Ward highlights that other phase-separated materials comprise varied properties that engineers could tap into.
“The new approach aims to increase performance by developing hardware around intended applications,” he said. “This means that materials and architectures driving supercomputers, desktops, and smart phones, which each have very different needs, would no longer be forced to follow a one-chip-fits-all approach.”
The research has been published under the title “Multimodal Responses of Self-Organized Circuitry in Electronically Phase Separated Materials.” Coauthors are Andreas Herklotz, Hangwen Guo, Anthony Wong, Ho Nyung Lee, Philip Rack and Thomas (Zac) Ward.
The work was supported by DOE’s Office of Science and used resources at the ORNL’s Center for Nanophase Materials Sciences, which is a DOE Office of Science User Facility.<|endoftext|>
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# Question Video: Understanding the Equation That Links Force- Momentum and Time Physics • 9th Grade
Which of the following formulas correctly relates the force acting on an object, the time for which that force acts, and the change in the momentum ๐ of the object? [A] ๐น = ฮ๐ฮ๐ก [B] ๐น = ฮ๐ก/ฮ๐ [C] ๐น = 2ฮ๐ฮ๐ก [D] ๐น = ฮ๐/ฮ๐ก [E] ๐น = (1/2)ฮ๐ฮ๐ก
02:19
### Video Transcript
Which of the following formulas correctly relates the force acting on an object, the time for which that force acts, and the change in the momentum ๐ of the object? (A) ๐น is equal to ฮ๐ times ฮ๐ก. (B) ๐น is equal to ฮ๐ก divided by ฮ๐. (C) ๐น is equal to two times ฮ๐ times ฮ๐ก. (D) ๐น is equal to ฮ๐ divided by ฮ๐ก. (E) ๐น is equal to a half times ฮ๐ times ฮ๐ก.
Okay, so in this question, weโre given a bunch of different potential formulas which could relate the force acting on an object, the time that force acts for, and the change in the objectโs momentum. In these formulas, the force is denoted as ๐น, the time interval is ฮ๐ก, and the change in momentum is ฮ๐. We are asked to work out which of the formulas is the correct one.
We can recall that whenever a force acts on an object, it causes the momentum of the object to change. Specifically, the force acting on the object is equal to the rate of change of the objectโs momentum over time. Another way of writing this statement is to say that the force is equal to the change in the momentum of the object divided by the change in time over which this momentum change occurs. We have seen that in the formulas that weโve been given, the force is denoted as ๐น. The change in momentum of the object is denoted as ฮ๐. And the change in time over which the momentum change occurs is denoted as ฮ๐ก. And so, if we write out this formula in symbols instead of words, that formula becomes ๐น, so thatโs the force, is equal to ฮ๐, thatโs the change in momentum, divided by ฮ๐ก, the change in time.
Looking at this formula, we see that it matches the one given in option (D), and so this option must be the correct answer. That is, the formula which correctly relates the force acting on an object, the time for which that force acts, and the change in the objectโs momentum is ๐น is equal to ฮ๐ divided by ฮ๐ก.<|endoftext|>
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# A motor boat covers the distance between two spots on the river banks in ${t_1} = 8h$ and ${t_2} = 12h$ downstream and upstream respectively. The time required for the boat to cover this distance in still water will be?
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Hint: In order to answer this problem let us first get some idea about Relative Velocity. The relative velocity is the speed at which body $A$ appears to an observer on body B, and vice versa. The vector difference between the velocities of two bodies is the relative velocity in mathematics. ${{\text{V}}_{AB}}{\text{ }} = {\text{ }}{{\text{v}}_A} - {v_B}$ is the equation.
Let us get some ideas about rivers and boats to solve this problem. The wind's effect on the plane is comparable to the river current's effect on the motorboat. A motorboat would not meet the shore directly across from its starting point if it were to head straight across a river (that is, if the boat were to point its bow straight towards the other side). The boat's motion is influenced by the river current, which takes it downstream.
To solve river boat problems, we need to understand two concepts:
-A boat's speed in relation to the sea is the same as a boat's speed in still water.
-The velocity of a boat relative to the water (${v_{b/w}}$) is equal to the difference in velocity of the boat relative to the ground (${v_b}$) and velocity of water relative to the ground (${v_w}$) i.e.
${v_{b/w}} = {v_b} - {v_w}$
Let apply this concept to our problem: let the time needed in still water be $t$ $hr$ and the boat's speed be $v$ and ${v^1}$.For still water
$d = vt$..................$[equation - 1]$
where, $d$=distance, $v$=velocity and $t$=time.
For downstream
$d = (v + {v^1}) \times 8$...................$[equation - 2]$
For upstream
$d = (v - {v^1}) \times 12$....................$[equation - 3]$
Solving $2$ & $3$ we get
$3 \times (v - {v^1}) = 2 \times (v + {v^1})$
$\Rightarrow v = 5{v^1}$.....................$[equation - 4]$
Putting value of $v$ in $equ - 2$ and also value of $d$ from $equ - 1$
$vt = (v + \dfrac{v}{5}) \times 8$
$\Rightarrow t = \dfrac{{6 \times 8}}{5} \\ \Rightarrow t= \dfrac{{48}}{5} \\ \therefore t= 9.6\,hr$
Note:To solve problems based on boat and stream above two concepts are necessary to keep in mind. If you are going to remember the concept then it will be very easy for you to solve the problem.<|endoftext|>
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Climate Justice is a global movement to redress the material inequities created by climate change. A form of environmental justice, climate justice demands the fair treatment of all people and freedom from discrimination, and requires the creation of policies and projects that address the systems which cause climate change and perpetuate discrimination.*
It turns out that only a handful of countries are responsible for generating the greenhouse gas pollution responsible for global warming. Those countries are the United States, Japan, Australasia, and the European nations. The Climate Justice movement asserts that these nations are ethically obliged to underwrite the efforts of poor countries, mainly in the Global South, to mitigate the effects of global warming. Those who are least to blame for causing global warming—Latin America, Africa, South and Southeast Asia—not only stand to suffer the worst from climate change, but also have the least resources to invest in preparing their countries for the excruciatingly difficult decades ahead.
Only solutions that privilege the needs and concerns of Indigenous peoples and people of color, who are impacted first and worst by climate change, make sense in a climate justice context.
To learn more about climate justice, explore the following links:<|endoftext|>
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The light from distant stars and distant galaxies has distinct spectral features of atoms. When these spectra are examined, they are found to be shifted towards the red end of the spectrum. This spectrum shift is called red shift . In fact, red shift is apparently a Doppler shift and indicates that essentially all of the galaxies are moving away from the nearer ones. It is measured by the quantity
where λ’ is observed wavelength of the same spectral line in the spectrum of galaxy.
λ is real wavelength of the same spectral line in the spectrum of galaxy if they were stationary
If λ’> λ then Z is positive which means red shift exists.
If the galaxy is away with speed v then red shift according to doppler’s effect can be given by,
Thus the measurement of the red shift for a particular galaxy helps us to estimate the recession velocity of the galaxy. This study shows that the galaxies are rushing away from us with enormous speed and the universe built up by the galaxies is not static but ever expanding.
Have a question? Ask us in our discussion forum.<|endoftext|>
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Theory:
Negative numbers are numbers that are less than zero.
For example, The negative of $$2$$ is $$-2$$ because $$2 + (-2) = (-2) + 2 = 0$$.
Important!
Negative number symbol: Negative numbers are indicated by placing a dash or minus ($$-$$) sign in front, such as $$–5$$, $$–12.77$$. A negative number, such as $$-1$$, is termed as 'negative one'.
An integer $$a$$, we have, $$a + (-a) = (-a) + a = 0$$; So, $$a$$ is the negative of $$-a$$, and $$-a$$ is the negative of $$a$$.
For example, $$2 + (-2) = (-2) + 2 = 0$$, So we say $$2$$ is the negative or additive inverse of $$-2$$ and vice-versa.
Rational Number: A number can be made by dividing two integers. (Note: An integer is a number with no fractional part.)
For example, 1.5 is a rational number because 1.5 $$=$$ $\frac{3}{2}$ (3 and 2 are both integers.)
Negative numbers: A rational number is supposed to be negative if its numerator and denominator are of opposite signs such that, one of them is a positive integer, and another one is a negative integer. In additional words, a rational number is negative, if its numerator and denominator are of the opposite signs.
Example:
Each of the rational numbers $$2/-8$$, $$-30/16$$, $$13/-18$$, $$-15/24$$ are negative rationals, but $$-11/-17$$, $$2/9$$, $$-3/-7$$, $$1/7$$ are not negative rationals.
For example, The rational number $\frac{2}{3}$.
$\frac{2}{3}$ $$+$$ $-\frac{2}{3}$ $$=$$ $\frac{\left(2+\left(-2\right)\right)}{3}$ $$=$$ 0.
Also, $-\frac{2}{3}$ $$+$$ $\frac{2}{3}$ $$=$$ $$0$$.
Similarly, $\frac{-8}{9}$ $$+$$ $\frac{8}{9}$ $$=$$ $\frac{8}{9}$ $$+$$$\frac{-8}{9}$ $$=$$ $$0$$.
$\frac{11}{6}$ $$+$$ $-\frac{11}{6}$ $$=$$ $-\frac{11}{6}$ $$+$$ $\frac{11}{6}$ $$=$$ $$0$$.
In general, for a rational number $\frac{a}{b}$, we have, $\frac{a}{b}$ $$+$$ $\frac{-a}{b}$ $$=$$ $\frac{-a}{b}$ $$+$$ $\frac{a}{b}$ $$=$$ $$0$$.
We say that $\frac{-a}{b}$ is the additive inverse of $\frac{a}{b}$ and $\frac{a}{b}$ is the additive inverse of $\frac{-a}{b}$.<|endoftext|>
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In Cisco, Utah, a discovery of a new mammal similar to the reptile has changed the way of Scientists looking at the history. Paleontologists in 2006 had found an undamaged skull, but it is recently, that the investigators are done with the identification of the species and its migration.
The mammal got its name derived from the paleontologist, Richard Cifelli, who found it, and named “Cifellidon Wahkarmoosuch.” The skull was 130M years old, it shows it belongs to Mesozoic Era and based on the size of the skull, it is evaluated that the animal was very big compared to others in its species. The mammal also has legs, which means it can burrow.
James Kirkland, Paleontologist, Utah, said, “A lot of species of that era were able to burrow to safeguard them from the steps of big dinosaurs.” The discovered mammal is of a size of a rabbit, which is like a Godzilla for a Mesozoic mammal.”
Kirkland further added, however, the skull was discovered in Cisco, Utah, the teeth of the same animal were discovered in North Africa. This indicated that before the continental split which happened millions of years ago, the mammal migrated from one place to another.
The recent finding of the study, which was executed on the same animal’s fossils, which were found on two different continents, also challenges the phenomenon of geologists’ history of Pangaea.
Kirkland further said, “These mammals were not possibly swimming, these animals at least were hopping around the island between these continents and indicate the connection. This finding is really exciting, as it brought us back in ages at least 15M years ago that we were considering there would be any sort of connections in the southern hemisphere. In fact, we have the proofs that the connection was there a little later than ever known to us.”
The new mammal was available for display in “Past Worlds Gallery”, at Natural History Museum, Utah.<|endoftext|>
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This section sets out the state of our atmosphere and climate. It presents information on greenhouse gases, ozone and ultraviolet intensity, temperature, rainfall, and extreme weather.
Greenhouse gases, ozone, and ultraviolet intensity
This section presents information on greenhouse gases, ozone concentrations, and ultraviolet (UV) intensity in our atmosphere.
Greenhouse gases are increasing
Greenhouse gases in the atmosphere absorb heat radiating from the Earth’s surface causing warmer temperatures. The main greenhouse gases generated by human activities are carbon dioxide, methane, nitrous oxide, and carbon monoxide. Concentrations of these gases are measured by NIWA at Baring Head near Wellington. Carbon dioxide has the greatest impact over the long term because it persists in the atmosphere longer than the other greenhouse gases. However, the other greenhouse gases have a significant impact in the short term, because they absorb much more heat per kilogram than carbon dioxide.
Carbon dioxide concentrations measured over New Zealand increased 21 percent since measurements began in 1972 (see figure 15). This is an increase of about 1.6 parts per million a year (0.5 percent).
Note: Baring Head is near Wellington. It is a site that is less likely to be influenced by local sources of pollution. These observations are made only when the wind is blowing from the south and away from any likely local sources of gas emissions. This gives a measure representative of the concentrations over the Southern Ocean. Data are unavailable for some periods.
This graph shows the carbon dioxide concentrations at Baring Head between 1972 and 2013. Visit the MfE data service for the full breakdown of the data.
Methane and nitrous oxide concentrations also increased significantly since they were first measured in 1989 and 1996, respectively. On the other hand, carbon monoxide concentrations decreased significantly since first measured in 2000. Globally, technological improvements and stricter controls on vehicle and industrial emissions are leading to lower emissions of carbon monoxide (Mikaloff-Fletcher & Nichol, 2014).
The measurements taken in New Zealand are consistent with other observations around the globe. This is expected, because these gases get mixed around the globe by weather systems (Mikaloff-Fletcher & Nichol, 2014).
For more detail see Environmental indicators Te taiao Aotearoa: Greenhouse gas concentrations.
Ozone concentrations over New Zealand are comparable to countries of similar latitudes
The thickness of the ozone layer is significant because ozone absorbs UV light from the sun and acts as a filter against sunburn and other damage (eg to plastic and fabric). Changes in ozone concentrations can also influence the risk of skin cancer.
The amount of ozone in an atmospheric column is measured in Dobson units, where 1 Dobson unit corresponds to the amount of ozone required to form a 0.01 millimetre layer of pure ozone. The global average ozone is 300 Dobson units (Liley & McKenzie, 2007), slightly lower than the average ozone over New Zealand (307 Dobson units). Over New Zealand, the ozone layer ranges from about 275 Dobson units in autumn to 345 Dobson units in spring (see figure 16). Ozone varies by around 15 Dobson units from day to day (around 5 percent). Ozone concentrations over New Zealand are similar to other places around the world at similar latitudes.
Note: Maximum, minimum, and average ozone concentrations are shown.
This graph shows the daily maximum, average, and minimum column ozone by the day of the year over the period 1979 to 2013. Visit the MfE data service for the full breakdown of the data.
A small but discernible downward trend is evident in ozone concentrations over New Zealand. However, this may only have had a small impact on our UV levels.
For more detail see Environmental indicators Te taiao Aotearoa: Ozone concentrations.
The ozone hole only affects New Zealand’s ozone and UV levels when it breaks up in spring – when it can send plumes of air with lower ozone concentrations over the country. Even then, the ozone above New Zealand drops only about 5 percent, which is within the normal daily variation.
For more detail see Environmental indicators Te taiao Aotearoa: Ozone hole.
New Zealand’s UV intensity is relatively high
Peak UV intensities in New Zealand are about 40 percent greater than at comparable latitudes in the Northern Hemisphere. The strength of UV light is measured using a solar UV index (UVI) or sun index. A UVI above 11 is extreme.
In New Zealand, UV indexes vary day to day, and showed no consistent trend across the five monitored sites between 1981 and 2014. In summer, the UVI often exceeds 11. For example, the northernmost site, Leigh, had an average of 73 days with a UVI greater than 11. In winter, the UVI can be as low as 1 even on clear days, mainly because the sun is low in the sky.
For more detail see Environmental indicators Te taiao Aotearoa: UV intensity.
Temperature, rainfall, and extreme weather
This section presents information on average temperatures, rainfall, and extreme weather. We also collect data on sunshine hours, but that is not discussed here (for information on sunshine hours see Environmental indicators Te taiao Aotearoa: Sunshine hours).
Average temperatures have risen
The average annual temperature in New Zealand increased by about 0.9 degrees Celsius over the last century (a statistically significant trend), slightly lower than the global average land-based temperature increase of around 1–1.2 degrees Celsius (NIWA, 2015) (see figure 17).
The Intergovernmental Panel on Climate Change concluded that global warming is beyond doubt, and it is extremely likely that human-caused pressures such as the burning of fossil fuels has been the primary cause since 1950 (IPCC, 2013).
Note: Data from seven sites around the country, adjusted for changes in site location. The plot shows the difference from the 1981–2010 average (known as ‘normal’). Climatologists use the term ‘normal’ to refer to an average of the weather across a 30-year period.
This graph shows the national average temperature as the difference from 'normal' between 1909 and 2013. Visit the MfE data service for the full breakdown of the data.
New Zealand’s average temperature, as estimated from seven monitored sites, fluctuates by about 0.4 degrees Celsius from year to year. This is a natural variation not directly linked to human-caused pressures. These year-to-year changes can be linked in part to climate oscillations such as the El Niño Southern Oscillation.
The increase in temperature is also reflected in ‘growing degree days’. Growing degree days is the total days over a year in which the mean daily temperature is higher than a base value (10 degrees Celsius).
Of the 29 sites assessed, 18 showed upward trends, with a median increase across all sites of 0.22 growing degree days per year.
Rainfall is highly variable with no clear trend
Rainfall is highly variable both geographically and from year to year, so we have been unable to determine any long-term trends. However, it is expected that climate change will influence rainfall (and other forms of precipitation) over the longer term.
Geographically, annual average rainfall varies from less than 500 millimetres in dry years in places like Central Otago, to over 4,000 millimetres in mountainous areas such as the Tararua Range, and even more in the Southern Alps (see figure 18).
For more detail see Environmental indicators Te taiao Aotearoa: Annual rainfall.
This map illustrates the average annual rainfall across New Zealand between 1972 and 2013. Visit the MfE data service for the full breakdown of the data.
Fewer days with gale-force winds but no clear trends in other extreme weather
This section presents information about three common elements of severe weather events: annual three-day rainfall maximums, the number of days with wind gusts above gale force, and lightning frequency.
Three-day rainfall data capture rainfall events likely to cause widespread flooding in affected areas. These data show the greatest rainfall total over any three-day period for each year, at each site. The data vary from year to year, so we cannot identify any statistically significant trends. While these data (see figure 19) show the maximum three-day rain totals for each year, they do not show the frequency of heavy rainfall events in a year.
Note: Data was unavailable for Wellington in 1993.
This graph shows the annual maximum three-day rain totals for Christchurch, Wellington, and Auckland, between 1950 and 2013. Visit the MfE data service for the full breakdown of the data.
We estimate damaging wind by the number of days a year wind gusts exceed gale force (61 kilometres an hour). The data show that Auckland, Wellington, and Christchurch regularly experience gale-force gusts, but Wellington is particularly prone to gale-force winds. Since 1975, the occurrence of potentially damaging wind has decreased in Wellington. Monthly (but incomplete) data for Christchurch and Auckland also suggest a smaller decrease (see figure 20). It is not clear whether this decrease is part of a natural cycle or is influenced by climate change. Studies predict global climate change may increase the frequency of damaging wind events in winter in almost all areas in New Zealand, and decrease the frequency in summer (Mullan et al, 2011).
Note: Two of the three selected sites have missing data for some years. Gusts above gale force – above 33 knots, approximately 61 km/h.
This graph shows the number of days with wind gusts above gale force in Wellington, Auckland, and Christchurch, between 1975 and 2013. Visit the MfE data service for the full breakdown of the data.
Annually, there are about 190,000 lightning strikes on the New Zealand land mass and over coastal seas. This is not frequent by international standards. Lightning strikes are most frequent on the West Coast of the South Island. The data show a high degree of variability over the 14-year period from September 2000 to December 2014. However, thunderstorms, and therefore lightning, are expected to increase in frequency and intensity as a result of climate change (Mullan et al, 2011).<|endoftext|>
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Solar energy is one more science fiction fantasy come true, and it joins ranks with the moon landing, orbiting satellites, nanotechnology and many other things we accept as commonplace.
We investigated how solar energy works and how and why it would interesting to install. First of all, what is solar-generated electricity, and where does it comes from? It comes from the energy of the light emanating from the sun, under the form of photons, which is captured by solar or photovoltaic panels which generate electricity using the photovoltaic effect.
Sunlight Vs Daylight
It is a common misconception that only direct sunlight can activate solar panels, when in fact daylight is sufficient. This means that solar panels work effectively even under a cloud cover, and that the crucial feature in assessing whether to install a solar panel system is exposure to daylight, rather than just direct sunlight. This is also because it is not the whole sunlight spectrum that can be converted into energy. Ultraviolet and infrared light for example are not useful, in fact a photovoltaic system typically converts between 15 and 17% of sunlight into actual energy.
In a solar panel system, photons are captured in solar panels, relatively thin “wafers” made of crystalline silicon or cadmium telluride and silicon. They are assembled in an array of several panels, usually mounted on a flat, glass-covered aluminium frame, which is then connected to an inverter that transforms the direct current produced into the alternate type useful for household and industrial purposes.
Where To Install A Solar System
Where is it feasible to install a solar energy system? Anywhere flat, with good exposure to sunlight and good weather generally (hot weather is not a requirement, in fact solar panels become less energy-efficient in the heat). The sunny side of a hill, a flat agricultural field with no houses or trees in the vicinity. The top of any residential building, especially if flat and no taller buildings overshadow it. The roof of a detached house is one of the most common locations, though also one of the most difficult to use, for several reasons. Roofs tend to be steep, so installing anything requires craftsmanship that’s sometimes expensive. A roof needs to be solid and tough, to take the weight of the system without structural damage risks. And finally the roof needs to have the proper orientation, East and South-East are obviously the best, North the worst, and to be clear of tall trees ad higher buildings.
The Benefits Of Going Solar
So why install a solar system if it is so difficult? In certain locations, think mountain huts for example, electricity may not be available through the conventional grid system. But if you have a “normal” house in an ordinary town or city neighbourhood, with plenty of power lines within easy reach, why go solar? We asked several people who have installed a solar panel system why they did it, and replies came from two main directions. The first is more philosophical: because by turning solar people want to reduce dependence from fossil fuel energy-generation systems, and do their own bit to preserve our overused planet.
The second direction is more prosaic: because solar energy is cheap, and because there are incentives to use it. The word “cheap” must be qualified. To install a single-home solar panel system in an average detached house in the UK costs between £ 7.000 and £ 10.000. The key to it is the fact that household energy bills, if the system works properly – and most of them do these days – can be reduced up to 40%. An average detached house where the energy bill (electricity plus hot water) is around £ 1000 a year on national grid, can save up to £ 400 pounds a year by going solar. Given that the government provides incentives which can reduce the installation cost, it is clear that the household in question can expect to amortize the initial cost in between 10 and 15 years, not an unreasonable time period. All the while, saving on energy bills and creating greater liquidity.
So it is time to look up to the sun and start converting its huge energy into energy we can use.
This post is written by Mark Jenkins and he works at CouponAudit as a writer, where thousands of valid and working online coupons are available for different stores. For example you can use soap.com coupon, petcarerx coupon etc. to get discount on your order while shopping on that store.<|endoftext|>
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# Probability
Announcements
#1
Sheila either walks or cycles to school
The probability that she walks to school is 0.65
If she walks,the probability that she is late is 0.4
If she cycles, the probability that she is late is 0.1
Work out the probability that on any one day sheila will not be late for school
0
3 years ago
#2
(Original post by XxevexX)
Sheila either walks or cycles to school
The probability that she walks to school is 0.65
If she walks,the probability that she is late is 0.4
If she cycles, the probability that she is late is 0.1
Work out the probability that on any one day sheila will not be late for school
Bet Sheila gets a lot of detentions
0
3 years ago
#3
find out all the probabilities of them being late and then add them all together and take away from 1 to find the answer. (Either use a tree diagram or figure it out with just the numbers.)
0
#4
So if she walks the probability that she is late is 0.4/0.65 ??
(Original post by monkeyman0121)
find out all the probabilities of them being late and then add them all together and take away from 1 to find the answer. (Either use a tree diagram or figure it out with just the numbers.)
0
3 years ago
#5
(Original post by XxevexX)
So if she walks the probability that she is late is 0.4/0.65 ??
I don't think you divide them, I think you have to multiply them to get the probability she will be late if she walks. After that do that with the probability she is late whilst cycling, add them together and take from 1.
0
3 years ago
#6
Probability that sheila:
walks=0.65 and is late=0.4;is not late=0.6
cycles=0.35 and is late=0.1;is not late=0.9
so if you need to find the probability of sheila walking and not being late:
0.65*0.6=0.39
AND probability of being not late and cycling:
0.35*0.9=0.315
therefore,the probability of sheila walking or cycling and still not being late:
0.39+0.315=0.705
0
#7
Thank you 😊
(Original post by smiley_21)
Probability that sheila:
walks=0.65 and is late=0.4;is not late=0.6
cycles=0.35 and is late=0.1;is not late=0.9
so if you need to find the probability of sheila walking and not being late:
0.65*0.6=0.39
AND probability of being not late and cycling:
0.35*0.9=0.315
therefore,the probability of sheila walking or cycling and still not being late:
0.39+0.315=0.705
0
3 years ago
#8
(Original post by XxevexX)
Thank you 😊
That's okay😄
0
X
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• Study Resource
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Transcript
5.2 The Standard Normal
Distribution
Statistics
Mrs. Spitz
Fall 2008
Objectives/Assignment
• How to find and interpret standard zscores and how to find the value of a
variable when its standard score is given
• How to find areas under the standard
normal curve and how to find areas under
any normal curve using a table.
Assignment: pp 209-211 #1-52
Intro
• In 5.1, you learned to calculate areas
under a normal curve when values of the
random variable, x corresponded to -3, -2,
-1, 0, 1, 2, or 3 standard deviations from
the mean. In this section, you will learn to
calculate areas corresponding to other xvalues. To do that, you will use the
standard score.
Definition
• The standard score, or z-score represents the number of
standard deviations a random variable, x, falls from the
mean, . To transform the random variable to a z-score,
use the following formula
value mean x
z
St.Dev
Ex. 1: Finding z-scores
• The mean speed of vehicles along a stretch of highway is 56 mph
with a standard deviation of 4 mph. You measure the speed of three
cars traveling this stretch of highway as 62mph, 47 mph, and 56
mph. Find the z-score that corresponds to each speed. What can
you conclude?
z
x
62 56
z
4
6
z 1.5
4
z
x
47 56
z
4
9
z
2.25
4
z
x
56 56
z
4
0
z 0
4
Ex. 1: Finding z-scores
• From the z-scores, you can conclude that a speed of 62
mph is 1.5 standard deviations above the mean, a speed
of 47 is 2.25 standard deviations below the mean, and a
speed of 56 is equal to the mean.
z
x
62 56
z
4
6
z 1.5
4
z
x
47 56
z
4
9
z
2.25
4
z
x
56 56
z
4
0
z 0
4
In terms of z
• The formula prior
gives z in terms of x.
If you solve this
formula for x, you get
a new formula that
gives x in terms of z.
z
x
z x
z x
x z
Transforming a z-score to an x-value
• To transform a standard z-score to a data
value x in a given population, use the
formula
x z
Ex. 2: Finding an x-value
• The speeds of vehicles along a stretch of
highway have a mean of 56 mph and a standard
deviation of 4 mph. Find the speeds x
corresponding to z-scores of 1.96, -2.33, and 0.
x z
x z
x 56 1.96(4) x 56 (2.33)(4)
x 63.84mph
x 46.68mph
x z
x 56 0(4)
x 56mph
You can see that 63.84 mph is above the mean,
46.68 is below the mean and 56 is equal to the
mean.
The Standard Normal Distribution
• There are infinitely many normal distributions, each with
its own mean and standard deviation. The normal
distribution with a mean of 0 and a standard deviation of
1 is called the standard normal deviation.
• If each data value of a normally distributed random
variable, x is transformed into a standard z-score, the
result will be the standard normal distribution. When this
transformation takes place, the area that falls in the
interval of the nonstandard normal curve is the SAME as
that under the standard normal curve within the
corresponding z-boundaries.
Definition
• The standard normal distribution is a
normal distribution with a mean of 0 and a
standard deviation of 1.
Nonstandard v. standard
• It is important that you know the difference
between x and z. The random variable x,
is sometimes called a raw score and
represents values in a nonstandard
normal distribution, while z represents
values in the standard normal distribution.
Standard Normal Table—Appendix B
• Because every normal distribution can be
transformed to the standard normal
distribution, you can use z-scores and the
standard normal curve to find areas under
any normal curve. The Standard Normal
Table in Appendix B lists the cumulative
area under the standard normal curve to
the left of z for z-scores from -3.49 to 3.49.
As you examine the table, notice the
following:
Properties of the Standard Normal
Distribution
Ex. 3: Using the Standard Normal Table
1. Find the cumulative area
that corresponds to a zscore of 1.15.
2. Find the z-scores that
corresponds to a
cumulative area of 0.4052
Solution: Find the area that
corresponds to z=1.15 by
finding 1.1 in the left
column and then moving
across the row to the
column under 0.05. The
number in that row and
column is:
Ex. 3: Using the Standard Normal Table
1. Find the cumulative area that corresponds to a
z-score of 1.15.
0.8749
Ex. 3: Using the Standard Normal Table
2. Find the z-scores that
corresponds to a cumulative
area of 0.4052
Solution: Find the z-score that
corresponds to an area of
0.4052 by locating 0.4052 in
the table. The values at the
beginning of the corresponding
row and at the top of the
column give the z-score. For
an area of 0.4052, the row
value is -0.2 and the column
value is 0.04, so the z-scores
is 0.2 – 0.04 = -.24
Ex. 3: Using the Standard Normal Table
2. Find the z-scores that corresponds to a cumulative area
of 0.4052
For an area of 0.4052, the row value is -0.2 and the column
value is 0.04, so the z-scores is 0.2 – 0.04 = -.24
You can use the following guidelines to find various types of
areas under the standard normal curve.
You can use the following guidelines to find various types of
areas under the standard normal curve.
You can use the following guidelines to find various types of
areas under the standard normal curve.
Ex. 4: Finding Area under the Standard
Normal Curve
• Find the area under the standard normal curve to the left
of z = -0.99
• SOLUTION: The area under the standard normal curve
to the left of z = -0.99 is shown:
Ex. 5: Finding Area under the Standard
Normal Curve
• Find the area under the standard normal curve to the
right of z = 1.06
• SOLUTION: The area under the standard normal curve
to the left of z = 1.06 is shown:
From the Standard Normal Table, the area to the left of z =
1.06 is 0.8554. Since the total area under the curve is 1, the
area to the right of z = 1.06 is 1 - .8554 = .1446
Ex. 5: Finding Area under the Standard
Normal Curve
• Find the area under the standard normal curve to the
right of z = 1.06
• SOLUTION: The area under the standard normal curve to the left of
z = 1.06 is shown:
1 - .8554 = .1446
Ex. 6: Finding Area under the Standard
Normal Curve
• Find the area under the standard normal curve between
z = -1.5 and z = 1.25
• The area under the standard normal curve between z = 1.5 and z = 1.25 is shown below
Ex. 6: Finding Area under the Standard
Normal Curve
• From the Standard Normal Table, the area
to the left of z = 1.25 is 0.8944 and the
area to the left of z = -1.5 is 0.0668. So
the area between z = -1.5 and z = 1.25 is:
Area = .8944 - .0668 = .8276
So 82.76% of the area under the curve falls
between z = -1.5 and z = 1.25
Related documents<|endoftext|>
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# decimal fraction
A rational number $d$ is called a decimal fraction if $10^{k}d$ is an integer for some non-negative integer $k$. For example, any integer, as well as rationals such as
$0.23123,\qquad\frac{3}{4},\qquad\frac{236}{125}$
are all decimal fractions. Rational numbers such as
$\frac{1}{3},\qquad-\frac{227}{12},\qquad 2.\overline{312}$
are not.
There are two other ways of characterizing a decimal fraction: for a rational number $d$,
1. 1.
$d$ is as in the above definition;
2. 2.
$d$ can be written as a fraction $\displaystyle{\frac{p}{q}}$, where $p$ and $q$ are integers, and $q=2^{m}5^{n}$ for some non-negative integers $m$ and $n$;
3. 3.
$d$ has a terminating decimal expansion, meaning that it has a decimal representation
$a.d_{1}d_{2}\cdots d_{n}000\cdots$
where $a$ is a nonnegative integer and $d_{i}$ is any one of the digits $0,\ldots,9$.
A decimal fraction is sometimes called a decimal number, although a decimal number in the most general sense may have non-terminating decimal expansions.
Remarks. Let $D\subset\mathbb{Q}$ be the set of all decimal fractions.
• If $a,b\in D$, then $a\cdot b$ and $a+b\in D$ as well. Also, $-a\in D$ whenever $a\in D$. In other words, $D$ is a subring of $\mathbb{Q}$. Furthermore, as an abelian group, $D$ is $2$-divisible and $5$-divisible. However, unlike $\mathbb{Q}$, $D$ is not divisible (http://planetmath.org/DivisibleGroup).
• As inherited from $\mathbb{Q}$, $D$ has a total order structure. It is easy to see that $D$ is dense (http://planetmath.org/DenseTotalOrder): for any $a,b\in D$ with $a, there is $c\in D$ such that $a. Simply take $c=\displaystyle{\frac{a+b}{2}}$.
• From a topological point of view, $D$, as a subset of $\mathbb{R}$, is dense in $\mathbb{R}$. This is essentially the fact that every real number has a decimal expansion, so that every real number can be “approximated” by a decimal fraction to any degree of accuracy.
• We can associate each decimal fraction $d$ with the least non-negative integer $k(d)$ such that $10^{k(d)}d$ is an integer. This integer is uniquely determined by $d$. In fact, $k(d)$ is the last decimal place where its corresponding digit is non-zero in its decimal representation. For example, $k(1.41243)=5$ and $k(7/25)=2$. It is not hard to see that if we write $d=\displaystyle{\frac{p}{2^{m}5^{n}}}$, where $p$ and $2^{m}5^{n}$ are coprime, then $k(d)=\max(m,n)$.
• For each non-negative integer $i$, let $D(i)$ be the set of all $d\in D$ such that $k(d)=i$. Then $D$ can be partitioned into sets
$D=D(0)\cup D(1)\cup\cdots\cup D(n)\cup\cdots.$
Note that $D(0)=\mathbb{Z}$. Another basic property is that if $a\in D(i)$ and $b\in D(j)$ with $i, then $a+b\in D(j)$.
Title decimal fraction DecimalFraction 2013-03-22 17:27:15 2013-03-22 17:27:15 CWoo (3771) CWoo (3771) 10 CWoo (3771) Definition msc 11-01 RationalNumber decimal number<|endoftext|>
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By Olivia Solon, Wired UK
A team of roboticists at Cornell University have created tiny flying robotic insects using 3-D printing.
The flapping wings of the hovering robotic insects (known as ornithopters) are very thin, lightweight and yet strong. Traditionally, the manufacturing process for these wings is time-consuming and a case of trial and error. However, advances in rapid prototyping have greatly expanded the possibilities for wing design, allowing wing shapes to replicate those of real insects or virtually any other shape. Furthermore, this can be done in minutes.
Researchers have managed to create a ornithopter with 3-D-printed wings weighing just 3.89 grams that can hover untethered for 85 seconds.
In order to create the ornithopters, the Cornell roboticists used an Objet EDEN260V 3-D printer. The wings are made of a polythene film stretched over a carbon fiber frame. The "fuselage" of the robotic insect was designed to hold a small GM14 motor, crank and wing hinge. The wings are driven by a crankshaft that is connected to a gearbox. Having initially connected the ornithopter using a DC power source, they realized that the device could lift 1.5 grams of payload, which was roughly the mass of the batteries required for flight.
When they experimented with a free-flying ornithopter, they needed to introduce lightweight sails above and below the winged robot in order to maintain stability (see second part of the video above).
The work at Cornell is testing hypotheses of insect propulsion and control. Researchers will test how different wing angles affect flight. If successful, these principles could form the basis of hovering ornithopter control. In other words, we could be seeing some seriously cool navigable robot insects.
Engineers have only been able to replicate flapping wing flight in the last decade. Major challenges include the lack of any established body of theoretical and experimental work on the unsteady aerodynamics of flapping wing flight for the purpose of wing design (most aircraft are designed for smooth flight). Researchers also need a sophisticated solution to make sure the robot remains upright. And they must solve problems with the energy density of batteries.<|endoftext|>
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# MODULE 3
LINEAR LAW
## CHAPTER 2 : LINEAR LAW Page Contents
3.0 CONCEPT MAP 2
3.1
UNDERSTAND AND USE THE CONCEPT OF LINES OF BEST FIT Draw lines of best fit by inspection of given data Exercises 1 Write equations for lines of best fit Exercises 2 Determine values of variables from a) lines of best fit b) equations of lines of best fit Exercises 3
3.1.1
3 4
3.1.2
3.1.3
3.2 3.2.1
APPLY LINEAR LAW TO NON- LINEAR RELATIONS Reduce non-linear relations to linear form Exercises 4 Determine values of constants of non-lineas relations given Exercises 5
7 7
3.2.2
3.2.3
Exercises 6
10
17 27 30
LINEAR LAW
Linear Relation
## Determining values of constants Formation of equation
Obtaining information
## UNDERSTAND AND USE THE CONCEPT OF LINES OF BEST FIT
THE CONCEPT OF LINES OF BEST FIT.
The properties of the line of best fit. The straight line is drawn such away that it passes through as many points as possible. The number of points that do not lie on the straight line drawn should be more or less the same both sides of the straight line.
Example. Draw the line of best fit for the given graph below. y
x EXERCISE 1 1. Draw the line of best fit for the following diagrams. x
Exercises 1
(a) y x (b) y x
Exercises 2
Example 1 Example 2 y
y x (3 , 8) 2 0 1 x (1, 3)
x (3, 7)
## The equation that can be formed is: Y = mX + c 82 =2Gradient , m = 30 Y intercept = 2 Therefore, y = 2x + 2
Linear form
0x The equation that can be formed is: Y = mX + c 73 =2Gradient , m = 3 1 The straight line also passes through the Point (1, 3) xY Another equation can be formed is, 3 = 2(1) + c .c = 1 Therefore, y = 2x + 1 Linear form
y a) 9 .x (8, 1) 0 x b)
y x (3, 6)
y c) x ( 5 , 2) d) x (8, 8)
x ( 6,4) 0 7 x 0 x
a)
y X (7, k)
## b) Equations of lines of best fit
Example 1
Example 1
y x (3, k) y = 3x + 4
## x (2, 4) 2 42 Gradient = =1 20 k4 Therefore,=1 72 .k 2 = 5 .k = 7 x 0 4
The straight line also passes through the point (3, k) Another equation that can be formed is .k = 3(3) + 4 .k = 9 + 4 = 13
i) y X (6, k)
## b) Equations of lines of best fit
i) y X
x (3, 3) 1 0 x (k, 3) x 0
.y = 2x - 3
ii)
y X (k, 6)
ii)
y y= 6 - x
x (3, k) x (5, 2) 0 0 7 x x
iii)
y X (8, 8)
iii) y X (4, k)
x (6, 4) .y = x + 2 x 0 (k, 2) 0 x x
## APPLY LINEAR LAW TO NON- LINEAR RELATIONS
3.2.1 Reduce non- linear relations to linear form Example 1
2
Y = mX + c
Example 1
x y = ax + b
Non- linear
Divided by x
Linear form
## log10 y = log10 p + qlog10 x Y = c mX
Linear form
Exercises 4
Equation a) y2 = ax + b Linear form Y X m c
b) y = ax2 + bx
c)
ab =+ 1 yx
d) y2 = 5x2 + 3x
e) y = 3 x +
5 x
f) y = abx
g) y =
4 (x + b)2 a2
Exercises 5
xy Example 1 X (11, 6)
a)
y x 10
x (1, 2) x2 The above figure shows part of a straight line graph drawn to represent the equation . xy = ax2 + b Find the value of a and b
.x (4, 2) 0x The above figure shows part of a straight line graph drawn to represent the equation . y = ax2 + bx Find the value of a and b,
62 Gradient , a = 11 1 2 A= 5 22 Therefore, xy = x +b 5 Another equation that can be formed is 2 .(1)(2) = (1) 2 + b 5 28 b, = 2 -= 55 2 8 Hence a = 5 , b= 5
y b) x (1, 9) c) x (3, 6)
1 .lg x 0 The above figure shows part of a straight line graph drawn to represent the equation .of y = axb Find the value of a and b,
1 x The above figure shows part of a straight line graph drawn to represent the equation 0 1 .of xy = a + bx Find the value of a and b,
## STEPS TO PLOT A STRAIGHT LINE 3.3.1. Using a graph paper.
Change the non-linear function with variables x and y to a linear form Y = mX + c
Construct a table
## Determine : gradient m Y-intercept c
QUESTIONS X Y 2 2 3 9 4 20 5 35 6 54
Table
The above table shows the experimental values of two variables, x and y. It is know that x and y are related by the equation y = px2 + qx
Non- linear
a) Draw the line of best fit for b) From your graph, find, i)the initial velocity ii)the acceleration
y against x x
10
## STEP 1 y = px2 + qx ypx = xx y = px + q x Y = mX + c
2
Reduce the non-linear To the linear form The equation is divided throughout dy x To create a constant that is free from x On the right-hand side i.e, q
qx + x
Linear form Y = mX + c
STEP 2
construct table
.x
.y y x
20
35
54
STEP 3
Using graph paper, - Choose a suitable scale so that the graph drawn is as big as possible. - Label both axis - Plot the graph of Y against X and draw the line of best fit
11
12
10 x 8 x 6 x 4 x 2 x
1
-2
-4
STEP 4
## From the graph, find m and c
9 1 Gradient , m ==2 62
Construct a right-angled triangle, So that two vertices are on the line of best fit, calculate the gradient, m
Y- intercept = c.= -3
## Determine the Y-intercept, c .from the straight line graph
12
1. The table below shows some experimental data of two related variable x and y It is know that x and y are related by an equation in the form y .x = ax + bx2 ,1where. a and b are constants 2 3 4 .y 7 16 24 24 . y a) Draw the straight line graphagainst x of x b) Hence, use the graph to find the values of a and b 5 16 6 0 7 -24
13
14
.x y
0 1.67
2 1.9
4 2.21
6 2.41
8 2.65
10 2.79
It is known that x and y are related by an equation in the form axb .y =+, where a and b are constants. yy a) Draw the straight line graph y2 against x b) Hence, use the graph to find the values of a and b
15
16
1. SPM 2003( paper 1, question no 10) x and y are related by equation y = px2 + qx , where p and q are constants. A straight y against x, as show in Diagram belowline is obtained by plotting x y x x (2, 9)
17
y against x x
x (2, k)
## x (h, 3) a. x Gigen that y = 6x x2 , calculate the value of k and of h
18
SPM 2005( paper 1, question no 13) The variable x and y are related by the equation y = kx4 where k is a constant. a) Convert the equation y = kx4 to linear form b) Diagram below shows the straight line obtained by plotting log10 y against log10 x log10 y
## . x (2, h) (0, 3) x log10 y 0 Find the value of i) log10 k ii) h.
19
SPM 2003( paper 2, question no 7) Table 1 shows the values of two variables, x and y, obtained from on experiment. It is x 2 , where p and k are constants.known that x and y are related by the equation y = pk x y 1.5 1.59 2.0 1.86 2.5 2.40 3.0 3.17 3.5 4.36 4.0 6.76
(a) Plot log y against x2. Hence, draw the line of best fit (b) Use the graph in (a) to find value of (i )p ( ii ) k
20
21
Table 1 shows the values of two variables, x and y, obtained from on experiment. Variables x and y are related by the equation y = pkx , where p and k are constants. x y 2 3.16 4 5.50 6 9.12 8 16.22 10 28.84 12 46.77
(a) Plot log10 y against x by using scala of 2 cm to 2 units on the x-axis and 2 cm to 0.2 unit on the log 10-axis. Hence, draw the line of best fit (b) Use the graph in (a) to find value of (i )p ( ii ) k
22
23
Table 1 shows the values of two variables, x and y, obtained from on experiment. r ,The variables x and y are related by the equation y = px + x 1.0 y 5.5 2.0 4.7 px 3.0 5.0 4.0 6.5 5.0 7.7 5.5 8.4
(b)where p against x2constants.scala of 2 cm to 5 units on both axes. Plot xy and r are by using Hence, draw the line of best fit (b) Use the graph in (a) to find value of (i )p ( ii ) r
24
25
1.
xy
(0, h)
.x 0
(8, k) x
The above figure shows part of a straight-line graph drawn to represent the equation 61 -Find the value of h and of ky = x2
26
.x Y
1 - 1.5
2 9
3 43.5
4 114
5 232.5
It is know that x and y are related by an equation in the from .y = ax3 - bx , where a and b are constants. a) Change the equation to the linear form and hence draw the straight line graph for values of x and y b) From your graph, I).determine the values of a and b II).find the value of y when x = 3.5
27
28
## Exercises 3 i) k = 5 ii) k = 1 iii) k = 5 b) i) k = 3 ii) k = 3 iii) k = 6
Exercises 4
Exercises 5
Y
a) b) c) d) .y
2
X
x x
m
a a
c
b b
a) a = 1, b = 2 b) a = 11, b = 8 c) a = 3 , b = 3
y x 1 y y2 x
1 x
x x x
b a
5 3 lg b
1 a
3 5 lg a
Exercises 5
1) a = - 1 , b = 10 2) a = 0.5 , b = 2.8
e) y x f) lg y g)
2 a
2b a
SPM Questions 1) p = -2 , q = 13 2) h = 3 , k = 4 3) a) lg y = 4lg x + lg k b) k = 100, h = 11 4) p = 1.259 , k = 1.109 5) p = 1.82, k = 1.307 6) P = 1.37, r = 5.48
Assessment test
1) h = 6 ,k=2
2)
a)
29<|endoftext|>
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# How to Find Surface Area
How to Find Surface Area
Go back to 'Geometry'
Let's say there is a rectangular box or a cube that needs to be painted. For doing so, you need to calculate the amount of paint that will be needed to cover it. How are you going to do that?
The first step will be finding the area of the cube. Only then can you estimate the amount of paint required. This is exactly what we are going to learn on this page.
Once you know the basic concepts and the various surface area formulae, you can calcuate the area of 3D shapes without a surface area calculator.
Check out the interactive simulations to know more about the lesson and try your hand at solving a few interesting practice questions at the end of the page.
## What is Surface Area?
Surface Area is the measure of the total area that the surface of a particular object occupies.
Real-life examples-
Sphere - Ball
Rectangular prism - Book
Cube - Rubik's cube
Cone - Ice cream cone
Pyramid - Pyramids in Egypt
Cylinder - Bucket
Tips and Tricks
1. Do you know a cube is also known as hexahedron because of its $$6$$ identical square faces. So, whenever you are asked to find the surface area of a cube, just multiply the square of a side with $$6$$.
2. Three-dimensional objects have $$3$$ dimensions, namely length, breadth, and height.
3. $$3$$ D shapes have faces, edges, and vertices.
4. Surface area is classified into:
1. Curved surface area
2. Lateral surface area
3. Total surface area
## How to Find Surface Area
In simple words, the area of the whole of anything, whether it's an object or a surface, is the sum of the area of its constituent parts.
The surface area of a three-dimensional object is the total area of all of its surfaces.
The surface area is important to know in situations where we want to wrap something, paint something, or build things to get the best possible design.
## What is the General Formula of Surface Area of Sphere, Cylinder, Cone, Cube, Pyramid, Rectangular Prism, Triangular Prism.
### General Formula of Surface Area of Sphere-
The surface area of the sphere is $$4\pi r^{2}=\pi d^{2}$$
Where, $$r$$ = radius of the sphere.
$$d$$ = diameter
### Formula of Surface Area of Cylinder-
The surface area of cylinder is $$2\pi r^{2}+2\pi rh = 2\pi r\left ( r+h \right )$$
Where, $$r$$ = radius of the cylinder.
$$h$$ = height of the cylinder
### Formula of Lateral Surface Area of Cone-
$$\pi r\left ( \sqrt{r^{2}+h^{2}} \right )=\pi rs$$
Where,$$r$$ = radius of the circular base.
$$h$$ = height of the cone
and $$s$$ = slant height of cone
### Formula of Total Surface Area of Cone-
$$\pi r\left ( r+\sqrt{r^{2}+h^{2}} \right )=\pi r\left ( r+s \right )$$
Where, $$r$$ = radius of the circular base
$$h$$ = height of the cone
and $$s$$ = slant height of cone
### Formula of Surface Area of Pyramid-
$$B+\left ( \frac{PL}{2} \right )$$
Where, $$B$$ = area of base.
$$P$$ = perimeter of base
and $$L$$ = slant height
### Formula of Surface Area of Cube-
$$6s^{2}$$
Where, $$s$$ = side length
### Formula of Surface Area of Triangular Prism-
$$bh+l\left ( a+b+c \right )$$
Where, $$b$$ = base length of triangle
$$h$$ = height of triangle
$$l$$ = distance between triangular bases
$$a,b,c$$ = sides of triangle
### Formula of Surface Area of Rectangular Prism-
$$A=2wl+2lh+2hw$$ or $$A=2\left ( wl+lh+hw \right )$$
Where, $$w$$ = width of the prism
$$h$$ = height of the prism
$$l$$ = length of the prism
Challenging Questions
1. For her birthday, Emily wanted to make birthday caps in the form of right circular cones for all her friends.
She made 10 cones of base radius $$7$$ inches and height $$24$$ inches each.
Find the area of the sheet required to make $$10$$ such caps for her friends.
## The Formula of Different Surface Areas with Sample Problems
### The surface area of a cone with a sample problem
Surface area of cone is $$\pi r\left ( r+\sqrt{r^{2}+h^{2}} \right )=\pi r\left ( r+s \right )$$
Where, $$r$$ = radius of the circular base.
$$h$$ = height of the cone
and $$s$$ = slant height of cone
Example 1
Dylan observed the cone of the ice cream he was eating. To understand the concept and shape of a cone,
his father suggested that he would help him find the total surface area of the cone. Given that the radius = $$4$$ inches and height = $$7$$ inches.
what would be the surface area of the ice cream cone?
Solution
Given:
Radius = $$4$$ inches and height =$$7$$ inches
Surface area of cone
\begin{align} &= \pi rh+\pi r^2\\&=3.14\times 4\times 7+3.14\times 4^2\\&= 87.92 + 50.24\\&= 138.16 \text{ inches}^2\end{align}
$$\therefore$$The surface area will be $$138.16$$ $$inches^{2}$$
### The surface area of a cylinder with a sample problem
The surface area of the cylinder is $$2\pi r^{2}+2\pi rh = 2\pi r\left ( r+h \right )$$
Where, $$r$$ = radius of the cylinder.
$$h$$ = height of the cylinder
Example 1
While painting a cylindrical tank of radius $$3.5$$ yd and height $$6$$ yd, Ryan thought of painting the outer surface.
If the cost of the painting is \$ $$5$$ per $$yd^{2}$$, what will be the total cost of painting?
Solution
Total Surface Area = Curved Surface Area + area of top and bottom faces
\begin{align} &=2\pi rh + 2\pi r^2 = 2πr (r + h)\\ &=2 × 22/7 × 3.5 × 3.5 + 6\\ &=22 × 9.5\\&= 209 \text{yd}^2\end{align}
Cost of painting at \begin{align} 5\text{ per yd}^2=209\times 5=1045\end{align}
$$\therefore$$Cost of painting =$$1045$$
### The surface area of a cube with a sample problem
The surface area of the cube is $$6s^{2}$$
Where, $$s$$ = side length
Example 1
While playing with a Rubik's cube of side length $$3 inches$$, Antonia closely observed the shape that the shape resembled a cube that she learn about while learning shapes in school,
she thought of calculating its surface area to get a practical idea. What would be the surface area of the cube?
Solution
Given,
Side length \begin{align}&=3\text{inches}\\&=6s^{2}\\&=6\times 3^2\\&=6\times 9\\&=54\text{inches}^2\end{align}
$$\therefore$$The surface area of cube will be $$54 inches^{2}$$
### The surface area of cuboid with a sample problem
The surface area of a cuboid $$2×(lb+bh+lh)$$
Example 2
Ricky ordered a toy for his son from an online store. When the package came, he noticed that the box was a bit larger than the regular size for such a toy package.
If the length, width, and height of the box are $$7$$ inches, $$8$$ inches, and $$6$$ inches respectively, find the surface area of the cuboid-shaped box.
Solution
Given that,
Length of the cuboid = $$7$$ inches
Breadth of the cuboid = $$8$$ inches
Height of the cuboid = $$6$$ inches
Surface area of the cuboid is
\begin{align}&= 2×(lb+bh+lh)\\&=2\left ( 7\times 8+8\times 6+6\times 7 \right)\\&=2(56+48+42)\\ &= 2(146)\\&=292\text{ inches}^2\end{align}
$$\therefore$$The surface area will be $$292 inches^{2}$$
### The surface area of a triangular prism with a sample problem
$$bh+l\left ( a+b+c \right )$$
Where, $$b$$ = base length of triangle
$$h$$ = height of triangle
$$l$$ = distance between triangular bases
$$a,b,c$$ = sides of triangle
Example 1
Ricky was curious to know the surface area of his kaleidoscope, so he asked his teacher who explained it to him.
The teacher gave an example of the surface area of a triangular prism with base $$4$$ ft, height $$3$$ ft, the distance between triangular bases $$10$$ ft, and sides a $$8$$ ft, b $$4$$ ft, c $$6$$ ft.
What would be the surface area of the kaleidoscope?
Solution
Given,
$$b$$ = $$4$$ ft
$$h$$ = $$3$$ ft
$$l$$ = $$10$$ ft
$$a$$ = $$8$$ ft
$$b$$ = $$4$$ ft
$$c$$ = $$6$$ ft
Surface area of the triangular prism:
\begin{align}&=bh+l\left ( a+b+c \right )\\&= 4\times 3+10\left ( 8+4+6 \right) \\&= 12+40+60+80\\&= 192\text{ ft}^2\end{align}
$$\therefore$$The surface area of a triangular prism will be $$192 ft^{2}$$.
## Solved Examples of Surface Area
Example 1
Stuart has a Rubik's cube with each side length = 5 inches. Can you help Stuart understand how to find the surface area of a cube by solving the question?
Solution
Given,
Length of side of the cube = a = $$5$$ inch
Surface area of cube
\begin{align}&= 6a^2\\ &= 6\times 5^2\text{inch}^2\\ &= 6\times 25 \text{inch}^2\\&= 150 \text{inch}^2\end{align}
$$\therefore$$The surface area of the cube will be $$150$$ $$inch^{2}$$
Example 2
Wilson gave a sphere-shaped balloon to his son. The balloon had some chocolates inside it. To test his son’s knowledge, Wilson asked his son to calculate the total surface area of the balloon (given that the radius is $$12 inches$$).
Solution
Given,
Radius of the sphere = $$12 inch$$
Surface area of the sphere
\begin{align}&=4\pi r^2\\&= 4\pi \left (12^2 \right)\\&= 4\pi \left ( 144 \right)\\&= 576\pi\text{ or}576\times 3.14\text{ inch}^2\end{align}
$$\therefore$$The surface area will be $$576\pi$$ or $$576\times 3.14 inches^{2}$$.
Example 3
Help Kaley understand how to find the surface area of a cylinder with diameter $$12$$ inch and height $$8$$ inch.
Solution
Given,
Diameter = $$12$$ inch so radius will be,
$$r=\frac{d}{2}$$ i.e. $$r=\frac{12}{2} = 6 inch$$
Surface area of cylinder
\begin{align}&= 2\pi r\left ( r+h \right)\\&= 2\times \dfrac{22}{7}\times 6\left ( 6+8 \right)\\&= 2\times \dfrac{22}{7}\times 6\left ( 14 \right)\\&= 2\times \dfrac{22}{7}\times 84\\&= 2\times 22\times 12\\&= 528 \text{inch}^2\end{align}
$$\therefore$$The surface area of the cylinder will be $$528$$ $$inch^{2}$$.
Example 4
Find the surface area of a pyramid with a square base of edge $$16$$ inch, slant height $$17$$ inch, and altitude $$15$$ inch.
Solution
Given,
base length = $$12$$ inch.
So perimeter of base will be
\begin{align}P& = 4s\\P &= 4\times 12=48\text {inches}\end{align}
Area of base is $$s^{2}$$
\begin{align}B& = 12^2\\&= 144\text{ inches}^2\end{align}
Total surface area
\begin{align}&= B+\frac{PL}{2}\\&= 144+\frac{48\times 17}{2}\\&= 144+\frac{816}{2}\\&= 144+408\\ &= 552 \text{ inches}^2\end{align}
$$\therefore$$The surface area will be $$552$$ $$inches^{2}$$
Example 5
Can you help Sophia find the surface area of a rectangular prism whose width is $$30$$ inch, the length is $$15$$ inch, and height is $$20$$ inch
Solution
Given,
$$W$$ = $$30 inch$$
$$L$$ = $$15 inch$$
$$H$$ = $$20 inch$$
\begin{align}A &= 2\left ( WL+LH+HW \right )\\A &= 2\left ( 450+300+600 \right )\\ &= 2\left ( 1350 \right )\\&= 2700\text{ inch}^2\end{align}
$$\therefore$$The surface area of the rectangular prism will be $$2700$$ $$inch^{2}$$
Surface Area Calculator
Talking about the surface area calculator, it helps you find the surface area of several shapes.
It simply works like any other calculator in which you put the input in the form of dimensions and
it will put it in the place of the formula and you will get the answer
## Let's Summarize
We hope you enjoyed learning about how to find the surface area with the simulations and practice questions. Now you will be able to easily solve problems on the surface area, surface area formula, surface area calculator, how to find the surface area of a rectangular prism, how to find the surface area of the cylinder
At Cuemath, our team of math experts is dedicated to making learning fun for our favorite readers, the students!
Through an interactive and engaging learning-teaching-learning approach, the teachers explore all angles of a topic.
Be it worksheets, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we, at Cuemath, believe in.
## 1. Is surface area the same as area?
The main difference is that surface area is the area of all the constituent parts of 3D shapes such as a sphere, cylinder, etc. whereas area is the measurement of the size of a plane surface, i.e. a 2D shape such as triangle, square, etc.
## 2. How do you find the surface area of a solid?
Let us take an example of a cuboid-shaped solid of length $$8$$ inches, breadth(width) $$6$$ inches, and height $$5$$ inches.
What will be the surface area of the cuboid?
Given, a = $$8$$ inches, b = $$6$$ inches, h = $$5$$ inches
The total surface area \begin{align}&= 2(ab + ah + bh)\\ & = 2 × [(8 × 6) + (5 × 6) + (8 × 5)]\\ &= 236\text { sq.inches}\end{align}
## 3. What is the surface area of a circle?
The surface area of a circle is $$\pi r^{2}$$
More Important Topics
Numbers
Algebra
Geometry
Measurement
Money
Data
Trigonometry
Calculus<|endoftext|>
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## Linear Algebra and Its Applications, Exercise 3.4.1
Exercise 3.4.1. a) Given the following four data points:
$y = -4 \quad\textrm{at}\quad t = -2 \quad\textrm{and}\quad y = -3 \quad\textrm{at}\quad t = -1$
$y = -1 \quad\textrm{at}\quad t = 1 \quad\textrm{and}\quad y = 0 \quad\textrm{at}\quad t = 2$
write down the four equations for fitting $C + Dt$ to the data.
b) Find the line fit by least squares and calculate the error $E^2$.
c) Given the value of $E^2$ what is $b$ in relation to the column space? What is the projection $p$ of $b$ on the column space?
Answer: a) This corresponds to a system of the form $Ax = b$ as follows:
$\begin{bmatrix} 1&-2 \\ 1&-1 \\ 1&1 \\ 1&2 \end{bmatrix} \begin{bmatrix} C \\ D \end{bmatrix} = \begin{bmatrix} -4 \\ -3 \\ -1 \\ 0 \end{bmatrix}$
The equivalent system of equations is:
$C - 2D = -4$
$C - D = -3$
$C + D = -1$
$C + 2D = 0$
b) To find the least squares solution we multiply both sides by $A^T$ to create a system of the form $A^TA\bar{x} = A^Tb$. We have
$A^TA = \begin{bmatrix} 1&1&1&1 \\ -2&-1&1&2 \end{bmatrix} \begin{bmatrix} 1&-2 \\ 1&-1 \\ 1&1 \\ 1&2 \end{bmatrix} = \begin{bmatrix} 4&0 \\ 0&10 \end{bmatrix}$
and
$A^Tb = \begin{bmatrix} 1&1&1&1 \\ -2&-1&1&2 \end{bmatrix} \begin{bmatrix} -4 \\ -3 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -8 \\ 10 \end{bmatrix}$
so that the new system is
$\begin{bmatrix} 4&0 \\ 0&10 \end{bmatrix} \begin{bmatrix} \bar{C} \\ \bar{D} \end{bmatrix} = \begin{bmatrix} -8 \\ 10 \end{bmatrix}$
From the second equation we have $\bar{D} = \frac{10}{10} = 1$ and from the first equation we have $\bar{C} = \frac{-8}{4} = -2$.
The resulting graph is a line $-2 + t$ with slope of 1 and $y$-intercept of -2. For the values of $t$ of -2, -1, 1, and 2 the values of $-2 + t$ are -4, -3, -1, and 0 respectively. But these are exactly the same as the values of $y$ in the given data points. Therefore we have $E^2 = 0$.
c) Since $E^2 = 0$ the vector $b$ must be in the column space of $A$, and the projection $p$ of $b$ onto the column space is simply $b$ itself.
NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.
If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fifth Edition and the accompanying free online course, and Dr Strang’s other books.
This entry was posted in linear algebra and tagged , , . Bookmark the permalink.<|endoftext|>
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You might have heard a lot about measles, but separating fact from fiction can be difficult in the age of social media.
Here’s what you should know about the disease.
What is measles?
Measles, also called rubeola, is an illness caused by a virus. It’s usually marked by a rash and fever that clears up within a few days, but measles can cause serious health complications, especially in very young children.
“When you first start to show symptoms, they can be very mild – a mild fever, runny nose, red eyes, cough and little white spots in your mouth, alongside your cheek. You can then develop a very high fever and a rash that starts on the head and face then spreads down the body. The rash is the classic symptom everyone is familiar with,” said Lori Grooms, OSF HealthCare director of infection control and prevention.
“But measles can be very serious, especially for children under 5 years of age and adults over 20 years old.”
According to the Centers for Disease Control and Prevention:
- About 1 in 4 people in the U.S. who get measles will require hospitalization
- About 1 in every 1,000 will develop brain swelling, which could cause permanent brain damage
- 1 to 2 out of 1,000 people with measles will die, even after receiving appropriate care
How does measles spread?
Measles spreads through the air when an infected person coughs or sneezes. Measles is considered highly contagious, and an unprotected person has about a 90 percent chance of contracting the disease if they are exposed to it.
Because the virus can live in the air for up to two hours, an infection can spread from person to person even if they aren’t in a room together at the same time. Additionally, an infected person can be contagious for several days before they show any symptoms of disease. For these reasons, measles spreads very quickly among unprotected populations.
How can I protect myself and my child from measles?
Measles can be prevented with the MMR vaccine, which protects against measles, mumps and rubella.
The CDC recommends children get two doses of MMR vaccine: the first between the ages of 12 to 15 months and the second between ages 4 and 6 years.
Despite concerns, research shows that the vaccine is safe and up to 97 percent effective at preventing the spread of disease.
“Since the discovery of vaccinations, many diseases once considered to be very dangerous have decreased in recent years. Many physicians today have never seen cases of measles, mumps, polio or other communicable diseases,” Grooms said. “Because of this many parents today grew up in an era when these diseases were unseen and are unaware of the complications they can cause.”
As a result, more parents have chosen not to vaccinate their children. With this decline in vaccination rates, we have seen an increase in the number of cases of these diseases over the past several years, which could have been prevented through vaccination.
“Vaccination really is the safest way to protect both your children and the public at large. Many of the vaccine preventable diseases can cause lifelong complications or even death. The safety and efficacy of vaccines has improved over the years and the benefits of vaccines outweighs the risks associated,” Grooms said.<|endoftext|>
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A botnet is a network of malicious computers infected with bot malware and remotely controlled by the cyber attackers. Botnets are used by hackers for different attack purposes such as to send spam/phishing emails, launch Distributed Denial of Service Attacks or in some scenarios, botnet authors rent them out to other hackers to use or launch an attack.
Hackers expand botnets by passing on the malware infection to clean internet-connected devices. The bot-masters control botnets by a command and control server. Once one device on a network is compromised, the other devices connected to the same network become vulnerable and will be easily infected. Some of the notable botnet attacks which had taken the cyberworld by storm are Zeus, Gameover Zues, Srizbi, Methbot and Mirai.
There are different symptoms to detect if the system or the complete network is attacked by botnet. If the machine or the network is compromised by a botnet,
- The hacker connects the infected system with a command & control server to instruct and control the infected system
- Establishes Internet Relay Chat traffic to facilitate communication based on a set of rules.
- Creates similar DNS requests
- Generates SMTP (Simple Mail Transfer Protocol) which is a communication protocol that moves your email on and across networks.
How Botnet Works?
The word botnet is derived from two words robot and network. Botnet malware infects vulnerable devices connected to the internet. Botnets aim to attack multiple device connected to across a network; They then exploit the system resources and power of the infected devices to generate automated tasks concealed from the users of the devices.
The typical botnet architecture is built in such a way that the infection is carried by Trojan horses. It scans the target system for vulnerabilities, outdated security applications to possibly pass on the infection. Once a successful number of infections are carried by, attackers take control over the bots through two different methods
Client/Server approach implements the use of Command-and-Control Server to send commands instantly to infect target devices via Internet Relay Chat.
The other method involves the use of peer- to-peer network to take control of the bots. The infected devices are programmed to check for malicious websites or for any other malware infected devices within the same network. This will enable the bots to share the latest commands or versions of the botnet malware.
Botnet Detection at the Endpoint
Host-based detection on an endpoint includes rootkit installations, annoying pops, impromptu changes to Windows host files to limit the outbound server access attempts.
Botnet Detection on the Network
This is more complicated which involves the detection process by monitoring IRC traffic (Internet Relay Chat), which must be denied on a company’s network. The IRC traffic is unencrypted, which can be therefore accessed by the packet sniffer. 6667 is the default IRC port number, but the bots use the complete port range from 6000-6669 and 7000.
How to Prevent Botnet Attacks?
With an alarming rise in botnet attacks, it is important to prevail with effective preventive measures.
Installation of an effective and comprehensive antivirus solution would enable computers and the networks with well-informed virus protection techniques. Comodo offers free antivirus software for Windows devices with best-in-class features to outplay the botnet malware attacks. To get more insights on Comodo’s Free Antivirus, visit our official page!<|endoftext|>
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Question
# Describe the key characteristics of the graphs of rational functions of the form f(x)=(ax+b)/(cx+d). Explain how you can determine these characteristi
Rational functions
Describe the key characteristics of the graphs of rational functions of the form $$\displaystyle{f{{\left({x}\right)}}}=\frac{{{a}{x}+{b}}}{{{c}{x}+{d}}}$$. Explain how you can determine these characteristics using the equations of the functions. In what ways are the graphs of all the functions in this family alike? In what ways are they different? Use examples in your comparison.
This kind of function will have vertical asymptote at $$\displaystyle{x}=-{\left(\frac{{d}}{{c}}\right)}$$, horizontal asymptote at $$\displaystyle{y}=\frac{{a}}{{c}}$$. Its points of untersects will be: $$\displaystyle{\left({0},\frac{{b}}{{d}}\right)}$$ with y-axis $$\displaystyle{\left(-{\left(\frac{{b}}{{d}}\right)},{0}\right)}$$ with x-axis
For example, let`s take following function for values $$a=2, b=4, c=1. d=3$$: $$\displaystyle{y}=\frac{{{2}{x}+{4}}}{{{x}+{3}}}$$
This function will have: vertical asymptote" $$\displaystyle{x}=-{\left(\frac{{d}}{{c}}\right)}=-{\left(\frac{{3}}{{1}}\right)}=-{3}$$ horizontal asymptote: $$\displaystyle{y}=\frac{{a}}{{c}}=\frac{{2}}{{1}}={2}$$ y-intercept: $$\displaystyle{y}=\frac{{b}}{{d}}=\frac{{4}}{{3}}$$ x-intercept: $$\displaystyle{x}=-{\left(\frac{{b}}{{a}}\right)}=-{\left(\frac{{4}}{{2}}\right)}=-{2}$$<|endoftext|>
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You know base ten cards? They are little cards with pictures of cubes on them. Some have just one cube, others ten, others 100, and others 1000. Using the cards helped my oldest son learn adding and subtracting from four digit numbers.
The other day I made binary cards. The first one had one square, the second two, the third 4, the fourth 8 etc, etc. Then I could show my children that just as they could use base ten cards to make up any number, they could use binary cards. We talked about how we only need one set of each number to be able to make up any other number, but that having two sets would allow us to add and subtract numbers. Using the cards gives us a physical way of seeing that every number can be represented by the sum of powers of two.
Using binary cards gives an easy way to start writing numbers in binary. Lay out all the cards with the one with the most squares on the left and then with numbers going progressively lower. Choose any number under 63 (or you have to make more cards than the ones pictured above). Pick the cards needed to make up that number and flip the rest over. Now think of the flipped up cards as representing 1s and the flipped down cards as representing 0s.
The ancient Egyptians used this fact in how they did multiplication. A video about it is found at: http://bestofyoutube.com/egyptian-maths and there is a description of the method in The secret life of math : discover how (and why) numbers have survived from the cave dwellers to us! by Ann McCallum. A hands-on way to do this is with sets of binary cards. Try printing out the cards illustrated. Take the first set of six cards and use them to multiply by three. Try 3 x 9. Thinking in terms of binary and powers of two, nine equals one group of eight plus a one, so take the card with the eight squares and the card with the single square, add the numbers on them together and you get: 3 x 9 = 27. Try it with a higher number. What is 3 x 19? 19 = 16 + 2 + 1, so 3 x 19 = 48 + 6 + 3 or 57.
Mathemagic! : number tricks by Lynda Colgan has a “magic” trick that relies on binary. In the trick you ask your volunteer to choose a number up to 63, and you give the volunteer six special cards. The cards all contain a list of numbers and the volunteer has to give you back the cards that contain his or her special number. The first number on each card is a power of two and the rest of the numbers on the card are numbers that require that power of two as one of the sums, if the number is going to be made as a sum of the powers of two. By quickly adding up the first number in each card, the person can easily name the secret number.
Although no knowledge of binary is necessary for the trick it can be thought of as having the person identify in which column the number 1 would be and in which a 0 would be. So the number 14, written as 1110 in binary, would require the card starting with 8, the card starting with 4 and the card starting with 2. Using Ms. Colgan’s cards, from the book, gives a chance to explore binary in a different way.
Of course another way to explore binary is with Vi Hart’s wonderful hand dance.
• ### Carolyn Wilhelm
I love how you did this! Your visuals are wonderful and show so much. Your children (again) are going to be so smart! Thanks for this excellent lesson. Carolyn
• ### Michelle
Thanks for this post! I learned a lot and was able to share this with my three kids today.
• ### Sue VanHattum
I do the magic trick with 5 cards (1 to 31), and say if they show me which cards their birthday is on, I can tell them what it is. Kids have loved it. And it’s pretty quick to make the cards.
This site uses Akismet to reduce spam. Learn how your comment data is processed.<|endoftext|>
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# AP Statistics Curriculum 2007 Normal Prob
(Difference between revisions)
Revision as of 04:51, 26 October 2009 (view source)IvoDinov (Talk | contribs) (added a link to the Problems set)← Older edit Revision as of 16:26, 4 February 2011 (view source)IvoDinov (Talk | contribs) m (→General Normal Distribution: typo)Newer edit → Line 8: Line 8: * The area: $\mu -2\sigma < x < \mu+2\sigma = 0.9772 - 0.0228 = 0.9544$ * The area: $\mu -2\sigma < x < \mu+2\sigma = 0.9772 - 0.0228 = 0.9544$ * The area: $\mu -3\sigma < x < \mu +3\sigma= 0.9987 - 0.0013 = 0.9974$ * The area: $\mu -3\sigma < x < \mu +3\sigma= 0.9987 - 0.0013 = 0.9974$ - * Note that the [http://en.wikipedia.org/wiki/Inflection_point inflection points] ($f ''(x)=0$)of the (general) Normal density function are $\pm \sigma$. + * Note that the [http://en.wikipedia.org/wiki/Inflection_point inflection points] ($f ''(x)=0$) of the (general) Normal density function are $\pm \sigma$.
[[Image:SOCR_EBook_Dinov_RV_Normal_013108_Fig6.jpg|500px]]
[[Image:SOCR_EBook_Dinov_RV_Normal_013108_Fig6.jpg|500px]]
## General Advance-Placement (AP) Statistics Curriculum - Non-Standard Normal Distribution and Experiments: Finding Probabilities
Due to the Central Limit Theorem, the Normal Distribution is perhaps the most important model for studying various quantitative phenomena. Many numerical measurements (e.g., weight, time, etc.) can be well approximated by the normal distribution. While the mechanisms underlying natural processes may often be unknown, the use of the normal model can be theoretically justified by assuming that many small, independent effects are additively contributing to each observation.
### General Normal Distribution
The (general) Normal Distribution, N(μ,σ2), is a continuous distribution that has similar exact areas, bound in terms of its mean, like the Standard Normal Distribution and the x-axis on the symmetric intervals around the origin:
• The area: μ − σ < x < μ + σ = 0.8413 − 0.1587 = 0.6826
• The area: μ − 2σ < x < μ + 2σ = 0.9772 − 0.0228 = 0.9544
• The area: μ − 3σ < x < μ + 3σ = 0.9987 − 0.0013 = 0.9974
• Note that the inflection points (f''(x) = 0) of the (general) Normal density function are $\pm \sigma$.
• General Normal density function $f(x)= {e^{{-(x-\mu)^2} \over 2\sigma^2} \over \sqrt{2 \pi\sigma^2}}.$
• General Normal cumulative distribution function $\Phi(y)= \int_{-\infty}^{y}{{e^{{-(x-\mu)^2} \over 2\sigma^2} \over \sqrt{2 \pi\sigma^2}} dx}.$
• The relation between the Standard and the General Normal Distribution is provided by these simple linear transformations (Suppose X denotes General and Z denotes Standard Normal Random Variables):
$Z = {X-\mu \over \sigma}$ converts general normal scores to standard (Z) values.
$X = \mu +Z\times\sigma$ converts standard scores to general normal values.
### Examples
#### Systolic Arterial Pressure Example
Suppose that the average systolic blood pressure (SBP) for a Los Angeles freeway commuter follows a Normal distribution with mean 130 mmHg and standard deviation 20 mmHg. Denote X to be the random variable representing the SBP measure for a randomly chosen commuter. Then $X\sim N(\mu=130, \sigma^2 =20^2)$.
• Find the percentage of LA freeway commuters that have a SBP less than 100. That is compute the following probability: p=P(X<100)=? (p=0.066776)
• If normal SBP is defined by the range [110 ; 140], and we take a random sample of 1,000 commuters and measure their SBP, how many would be expected to have normal SBP? (Number = 1,000P(110<X<140)= 1,000*0.532807=532.807).
• What is the 90th percentile for the SBP? That is what is xo, so that P(X < xo) = 0.9?
• What is the range of SBP values that contain the central 80% of the SBPs for all commuters? That is what are xo,x1, so that P(x0 < X < x1) = 0.8 and ${x_o+x_1\over2}=\mu=130$ (i.e., they are symmetric around the mean)? (xo = 104,x1 = 156)
### Assessing Normality
How can we tell if data collected from a process or experiment we observe is normally distributed? There are several methods for checking normality:
• Why do we care if the data is normally distributed? Having evidence that the data we are analyzing is normally distributed allows us to use the (General) Normal distribution as a model to calculate the probabilities of various events and assess significant observations.
• Example: Suppose we are given the heights for 11 women.
• First we need to show that there is no evidence suggesting that the Normal and Data distributions are significantly distinct.
• Then, we want to use the normal distribution to make inference on women heights. If the height of a randomly chosen woman is measured, how likely is that she'll be taller than 60 inches? 70 inches? Between 55 and 65 inches?
Height (in.) 61 62.5 63 64 64.5 65 66.5 67 68 68.5 70.5<|endoftext|>
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# Slope Intercept Form – Equation of a Line Example
Slope intercept form is a function of a straight line with the form
y = mx + b
where
m is the slope of the line and
b is the y-intercept.
Slope refers to how steep the line changes as the values of x change where the y-intercept is where the line crosses the y-axis.
Here are a pair of worked examples to show how to find the equation of a line from different initial conditions.
### Slope Intercept Form Example 1
This example shows how to find the equation of the line when given the slope and one point on the line.
Q: What is the Slope intercept form of the line with slope 23 and passes through the point (-3, -4)
Here’s the graph of that line. A slope of 23 means the line will rise 2 points for every 3 points of x to the right. If we start at the point (-3, -4) we can count 2 up and three over to see the way the line will progress.
The slope intercept form is
y = mx + b
To solve this, we need to know both the slope and the y-intercept. We know the slope, so we need to find the value of b. We know from the given:
m = 23
x = -3
y = -4
Plug these into the equation
-4 = 23 (-3) + b
-4 = -2 + b
-2 = b
Now we have all we need for the equation of the line.
y = 23 x + (-2)
y = 23 x – 2
We can see from the graph the line crosses the y-axis at -2 but it nice to have a second opinion.
### Slope Intercept Form Example 1 – Alternate Method
There is an alternate method to solve this problem where you can memorize an equation to eliminate the first couple steps. That formula is
(y – y1) = m(x – x1)
where x1 and yare the coordinates of the given point.
Let’s plug in the point (-3, -4) from above.
(y – y1) = m(x – x1)
(y – (-4)) = 23(x – (-3))
y + 4 = 23(x + 3)
y + 4 = 23x + 23(3)
y + 4 = 23x + 2
y = 23x + 2 – 4
y = 23x – 2
### Slope Intercept Form Example 2
The second example, two points of the line are given in the initial conditions.
Let’s find the same line as before. The red line passes through points (-6, -6) and (9, 4). Find the equation of the line.
First we need to find the slope.
The formula to find slope between two points (x1, y1) and (x2, y2) is
For our two points:
x1 = -6
y1 = -6
x2 = 9
y2 = 4
Plug these into the formula
m = 1015
reduce the fraction by factoring out a 5
m = 23
Now we can find the y-intercept in the same way we did above. Choose one of the points for x and y. Let’s try the (-6, -6) point.
y = mx + b
-6 = 23(-6) + b
-6 = –123 + b
-6 = -4 + b
-2 = b
Just to show it doesn’t matter which point you choose, let’s use the (9, 4) point.
y = mx + b
4 = 2(9) + b
4 = 183 + b
4 = 6 + b
-2 = b
Plug into the slope intercept formula
y = 23 x + (-2)
y = 23 x – 2
Which matches what we expected to see from the first example.
The key to this type of problem is to find the slope of the line. Once you have the slope, finding the y-intercept is easy. For more examples of finding the slope, check out the What is Slope? page and examples
# SOHCAHTOA Example Problem – Trigonometry Help
SOHCAHTOA is the mnemonic used to remember which sides of a right triangle are used to find the ratios needed to determine the sine, cosine or tangent of an angle. Here are a pair of SOHCAHTOA example problems to help show how to use these relationships. If you have no idea what SOHCAHTOA means, check out this introduction to SOHCAHTOA.
### SOHCAHTOA Example Problem 1
A mathematically inclined squirrel sits atop a 14-foot tall tree. He spies a nut on the ground some distance away. After careful measurements, he determines the nut is 74º from the base of his tree.
How far away is the nut from:
A) The base of the tree?
B) The math squirrel?
Here is a layout of the problem.
Part A: How far is the nut from the base of the tree?
Looking at our triangle, we see we know the angle and the length of the adjacent side. We want to know the length of the opposite side of the triangle. The part of SOHCAHTOA with these three parts is TOA, or tan θ = opposite / adjacent.
tan θ = opposite / adjacent
tan ( 74º ) = opposite / 14 ft
solve for ‘opposite’
opposite = 14 ft ⋅ tan ( 74º )
opposite = 14 ft ⋅ 3.487
opposite = 48.824 ft
The nut is 48.824 ft from the base of the tree.
Part B: How far is the nut from the math squirrel?
This time, the needed distance is the hypotenuse of the triangle. The part of SOHCAHTOA with both adjacent and hypotenuse is CAH, or cos θ = adjacent / hypotenuse.
cos θ = adjacent / hypotenuse
cos ( 74º ) = 14 ft / hypotenuse
hypotenuse = 14 ft / cos ( 74º )
hypotenuse = 14 ft / 0.276
hypotenuse = 50.791 ft
We can use the Pythagorean Theorem to check our work.
a2 + b2 = c2
(14 ft)2 + (48.824 ft)2 = c2
196 ft2+ 2383.783 ft2= c2
2579.783 ft2 = c2
50.792 ft = c
The two numbers are close enough to account for rounding errors.
### SOHCAHTOA Example Problem 2
A prince has arrived to rescue a princess standing in a tower balcony 15 meters up across a moat 12 meters wide. He knew the princess was high up and there was a moat, so he brought the longest ladder in the kingdom, a 20-meter monster.
A) Is the ladder long enough?
B) If the top of the 20 m ladder touches the balcony, how far away from the tower wall is the bottom of the ladder?
C) What is the angle between the ground and the ladder?
Here is an illustration of our situation.
Part A: Is the ladder long enough? To know if the ladder is long enough, we need to know how long it needs to be. At the very least, the prince’s ladder must reach the 15-meter balcony from the edge of the 12-meter moat. Use the Pythagorean theorem to find out how far up the tower wall a 20-meter ladder will reach.
a2 + b2 = c2
height2 = (20 m)2 – (12 m)2
height2 = 400 m2 – 144 m2
height2 = 256 m2
height = 16 m
A 20-meter ladder can reach 16 meters up the side of the tower. This is a meter longer than the 15 meters needed to reach the princess balcony.
Part B: When the ladder touches the edge of the balcony, how far away from the tower wall is the bottom of the ladder?
The first part measured the height the 20-meter ladder reached when the ladder was placed at the edge of the moat. We found we had more ladder than needed. If the ladder touches the balcony, we know the height it reaches is 15 meters. The ladder is still 20 meters long. We just need to find out how far from the tower to stick the bottom of the ladder. Use the Pythagorean theorem again.
a2 + b2 = c2
(balcony height)2 + (ground distance)2 = ladder2
(ground distance)2 = ladder2 – (balcony height)2
(ground distance)2 = (20 m)2 – (15 m)2
(ground distance)2 = 400 m2 – 225 m2
(ground distance)2 = 175 m2
ground distance = 13.23 m
The ladder’s base is 13.23 m away from the tower.
Part C: What is the angle between the ground and the ladder?
We were given the height of the balcony, which in this case is the ‘opposite’ side of the triangle from the needed angle. We also know the length of the ladder that forms the hypotenuse of the triangle. The part of SOHCAHTOA that has both of these parts is SOH, or sin θ = opposite/hypotenuse. Use this to solve for the angle.
sin θ = opposite/hypotenuse
sin θ = 15 m / 20 m
sin θ = 0.75
θ = sin-1(0.75)
θ = 48.59°
You can check your work in Part B if you use the answer we got as the ‘adjacent’ side of the triangle.
If you use the ladder as the hypotenuse, use CAH, or cos θ = adjacent / hypotenuse.
cos θ = adjacent / hypotenuse
cos θ = 13.23 m / 20 m
cos θ = 0.662
θ = cos-1(0.662)
θ = 48.59°
The angle matches the value above.
If you choose the balcony height as the opposite side, use TOA, or tan θ = opposite / adjacent
tan θ = opposite / adjacent
tan θ = 15 m / 13.23 m
tan θ = 1.134
θ = tan-1(1.134)
θ = 48.59°
Pretty neat how it doesn’t really matter which sides of the right triangle you use, as long as you use the correct part of SOHCAHTOA.
# Vertical Motion Example Problem – Coin Toss Equations of Motion
This equations of motion under constant acceleration example problem shows how to determine the maximum height, velocity and time of flight for a coin flipped into a well. This problem could be modified to solve any object tossed vertically or dropped off a tall building or any height. This type of problem is a common equations of motion homework problem.
Problem:
A girl flips a coin into a 50 m deep wishing well. If she flips the coin upwards with an initial velocity of 5 m/s:
a) How high does the coin rise?
b) How long does it take to get to this point?
c) How long does it take for the coin to reach the bottom of the well?
d) What is the velocity when the coin hits the bottom of the well?
Solution:
I have chosen the coordinate system to begin at the launch point. The maximum height will be at point +y and the bottom of the well is at -50 m. The initial velocity at launch is +5 m/s and the acceleration due to gravity is equal to -9.8 m/s2.
The equations we need for this problem are:
1) y = y0 + v0t + ½at2
2) v = v0 + at
3) v2 = v02 + 2a(y – y0)
Part a) How high does the coin rise?
At the top of the coin’s flight, the velocity will equal zero. With this information, we have enough to use equation 3 from above to find the position at the top.
v2 = v02 – 2a(y – y0)
0 = (5 m/s)2 + 2(-9.8 m/s2)(y – 0)
0 = 25 m2/s2 – (19.6 m/s2)y
(19.6 m/s2)y = 25 m2/s2
y = 1.28 m
Part b) How long does it take to reach the top?
Equation 2 is the useful equation for this part.
v = v0 + at
0 = 5 m/s + (-9.8 m/s2)t
(9.8 m/s2)t = 5 m/s
t = 0.51 s
Part c) How long does it take to reach the bottom of the well?
Equation 1 is the one to use for this part. Set y = -50 m.
y = y0 + v0t + ½at2
-50 m = 0 + (5 m/s)t + ½(-9.8 m/s2)t2
0 = (-4.9 m/s2)t2 + (5 m/s)t + 50 m
This equation has two solutions. Use the quadratic equation to find them.
where
a = -4.9
b = 5
c = 50
t = 3.7 s or t = -2.7 s
The negative time implies a solution before the coin was tossed. The time that fits the situation is the positive value. The time to the bottom of the well was 3.7 seconds after being tossed.
Part d) What was the velocity of the coin at the bottom of the well?
Equation 2 will help here since we know the time it took to get there.
v = v0 + at
v = 5 m/s + (-9.8 m/s2)(3.7 s)
v = 5 m/s – 36.3 m/s
v = -31.3 m/s
The velocity of the coin at the bottom of the well was 31.3 m/s. The negative sign means the direction was downward.
If you need more worked examples like this one, check out these other constant acceleration example problems.
Equations of Motion – Constant Acceleration Example Problem
Equations of Motion – Interception Example Problem
Projectile Motion Example Problem
# Friction Example Problem – Physics Homework Help
Friction is the resistance force generated between two bodies as they move across each other. It is proportional to the force that presses the two bodies together. This diagram shows the forces acting on a block sitting on a surface.
The block is pulled down onto the surface by the force of gravity while the surface pushes back in an equal and opposite force known as the normal force: N. Note there are no horizontal forces. If a horizontal force is applied such as pushing the block to the right, the block will begin to accelerate. Experience tells us this does not always happen. If you try to push something heavy, it doesn’t always move until you’ve pushed HARD enough. There must be a force working in the opposite direction of the push to resist the motion. This force is the force of friction, Ff.
Experiments have shown the magnitude of this force is dependent on the normal force. The magnitude of the friction force is directly proportional to the magnitude of the normal force. The proportionality constant between them is called the coefficient of friction, μf. The f subscript is commonly left off, and is not unusual to see just the μ listed.
The coefficient of friction depends on two factors.
The first depends on the materials the two objects are made of. It is generally easier to move 50 kg of ice on a glass surface than a 50 kg stone on sand. Each two materials have their own coefficients of friction.
The second factor is whether or not the block is moving. You may have noticed it is usually easier to move a heavy object once it is already moving. This means there are two different coefficients of friction. One for when the block is stationary, μs (static) and one for when the block begins to move, μk (kinetic).
The static coefficient is used whenever the block is stationary. As the force pushing the block increases, it will eventually reach a point where the block is on the verge of moving. The coefficient of static friction, μs is experimentally determined by carefully measuring the force at this point. The frictional force required to reach this point is F = -μsN. The minus sign indicates the direction of the force. Frictional forces oppose the force trying to move the object and act in the opposite direction. Any magnitude of force less than musN, the block will not move.
When the block is moving, the coefficient of kinetic friction is used. This value is experimentally calculated by measuring the force necessary to keep the block moving at a constant velocity. This force will equal -μkN.
Now, let’s try a friction example problem.
Example Problem:
A block weighing 200 N is pushed along a surface. If it takes 80 N to get the block moving and 40 N to keep the block moving at a constant velocity, what are the coefficients of friction μs and μk?
Solution:
For the coefficient of static friction, we need the force needed to get the block moving. In this case, 80 N.
From the description above:
Ff = μsN
N is equal to the weight of the block, so N = 200 N. Put these values into the formula.
80 N = μs·200 N
or
μs = 0.4
For the coefficient of kinetic friction, the force needed to maintain a constant velocity was 40 N. Use the formula:
Ff = μkN
40 N = μk·200 N
μk = 0.2
The two coefficients of friction for this system are μs = 0.4 and μk = 0.2.
There are two important things to remember in friction homework problems. The first is the normal force N is always perpendicular to the surface. The normal force is not always ‘up’. The second is the friction force works opposite in direction to the motion of the block. Friction is a resistive force.
# Calculate Percent Error
Percent error is the percent difference between a measured and expected value. (image: Sherman Geronimo-Tan)
Percent Error Definition
Percent error, sometimes referred to as percentage error, is an expression of the difference between a measured value and the known or accepted value. It is often used in science to report the difference between experimental values and expected values.
The formula for calculating percent error is:
Note: occasionally, it is useful to know if the error is positive or negative. If you need to know the positive or negative error, this is done by dropping the absolute value brackets in the formula. In most cases, absolute error is fine. For example, in experiments involving yields in chemical reactions, it is unlikely you will obtain more product than theoretically possible.
### Steps to Calculate the Percent Error
1. Subtract the accepted value from the experimental value.
2. Take the absolute value of step 1
3. Divide that answer by the accepted value.
4. Multiply that answer by 100 and add the % symbol to express the answer as a percentage.
Now let’s try an example problem.
You are given a cube of pure copper. You measure the sides of the cube to find the volume and weigh it to find its mass. When you calculate the density using your measurements, you get 8.78 grams/cm3. Copper’s accepted density is 8.96 g/cm3. What is your percent error?
Solution:
experimental value = 8.78 g/cm3
accepted value = 8.96 g/cm3
Step 1: Subtract the accepted value from the experimental value.
8.96 g/cm3 – 8.78 g/cm3 = -0.18 g/cm3
Step 2: Take the absolute value of step 1
|-0.18 g/cm3| = 0.18 g/cm3
Step 3: Divide that answer by the accepted value.
Step 4: Multiply that answer by 100 and add the % symbol to express the answer as a percentage.
0.02 x 100 = 2
2%
The percent error of your density calculation was 2%.
# Guidelines for Balancing Chemical Equations
One of the basic skills you will develop as you study chemistry is the ability to balance a chemical equation. One you have written a balanced equation, you will be in a good position to perform all manner of calculations! Balancing a chemical equation refers to establishing the mathematical relationship between the quantity of reactants and products. The quantities are expressed as grams or moles.
It takes practice to be able to write balanced equation. There are essentially three steps to the process:
• Write the unbalanced equation.
• Chemical formulas of reactants are listed on the lefthand side of the equation.
• Products are listed on the righthand side of the equation.
• Reactants and products are separated by putting an arrow between them to show the direction of the reaction. Reactions at equilibrium will have arrows facing both directions.
• Balance the equation.
• Apply the Law of Conservation of Mass to get the same number of atoms of every element on each side of the equation. Tip: Start by balancing an element that appears in only one reactant and product.
• Once one element is balanced, proceed to balance another, and another, until all elements are balanced.
• Balance chemical formulas by placing coefficients in front of them. Do not add subscripts, because this will change the formulas.
• Indicate the states of matter of the reactants and products.
• Use (g) for gaseous substances.
• Use (s) for solids.
• Use (l) for liquids.
• Use (aq) for species in solution in water.
• Write the state of matter immediately following the formula of the substance it describes.
### Worked Example Problem
Tin oxide is heated with hydrogen gas to form tin metal and water vapor. Write the balanced equation that describes this reaction.
1. Write the unbalanced equation.SnO2 + H2 → Sn + H2O
2. Balance the equation.Look at the equation and see which elements are not balanced. In this case, there are two oxygen atoms on the lefthand side of the equation and only one on the righthand side. Correct this by putting a coefficient of 2 in front of water:SnO2 + H2 → Sn + 2 H2O
This puts the hydrogen atoms out of balance. Now there are two hydrogen atoms on the left and four hydrogen atoms on the right. To get four hydrogen atoms on the right, add a coefficient of 2 for the hydrogen gas. Remember, coefficients are multipliers, so if we write 2 H2O it denotes 2×2=4 hydrogen atoms and 2×1=2 oxygen atoms.
SnO2 + 2 H2 → Sn + 2 H2O
The equation is now balanced. Be sure to double-check your math! Each side of the equation has 1 atom of Sn, 2 atoms of O, and 4 atoms of H.
3. Indicate the physical states of the reactants and products.To do this, you need to be familiar with the properties of various compounds or you need to be told what the phases are for the chemicals in the reaction. Oxides are solids, hydrogen forms a diatomic gas, tin is a solid, and the term ‘water vapor’ indicates that water is in the gas phase:SnO2(s) + 2 H2(g) → Sn(s) + 2 H2O(g)
There you go! This is the balanced equation for the reaction. Remember, no elements appear on one side of the reaction and not on the other. It’s always a good idea to check by counting up the number of atoms of each element just to be sure you have balanced your equation correctly. This was a straightforward example, using conservation of mass. For reactions involving ions, conservation of charge will also come into play.<|endoftext|>
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A brand-new analysis of fossils recovered in the 1990’s in the village of Nikiti, northern Greece, supports the controversial idea that the apes which gave rise to humans evolved in south-eastern Europe instead of Africa.
The 8 or 9-million-year-old fossils had first been linked to the extinct ape called Ouranopithecus. However, a team led by David Begun from the University of Toronto’s Department of Anthropology has recently analyzed the remains and has determined that they likely belonged to a male animal from a potentially new species.
Charles Darwin proposed in 1871 that all hominins, including both modern and extinct humans, descended from a group in Africa, and this is the most widely accepted theory today.
On the other hand, Darwin also speculated that hominins could also have originated in Europe, where fossils of large apes had already been discovered, and the new analysis supports this theory.
While Begun does not believe the Nikiti ape was a hominin, he speculates that it could represent the group from which hominins directly evolved.
The research team led by Begun had determined in 2017 that a 7.2-million-year-old ape called Graecopithecus, which also lived in what is now Greece, could possibly be a hominin. In this case, the 8 to 9-million-year-old Nikiti ape would have directly preceded the first hominin, Graecopithecus, before hominins migrated to Africa 7 million years ago.
According to a report in the journal New Scientist, Begun foresees that this new concept will be rejected by many experts who believe in African hominin origins, but he hopes that the new scenario will at least be considered.
Begun presented the research last month at a conference of the American Association of Physical Anthropologists.
Sources: New Scientist, bigthing.com<|endoftext|>
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# If the pth and qth terms of a G.P. are q and p, respectively,
Question:
If the $p$ th and qth terms of a G.P. are $q$ and $p$, respectively, then show that $(p+q)$ th term is $\left(\frac{q^{p}}{p^{q}}\right)^{\frac{1}{p-q}}$.
Solution:
As, $a_{p}=q$
$\Rightarrow a r^{(p-1)}=q \quad \ldots$ (i)
Also, $a_{q}=p$
$\Rightarrow a r^{(q-1)}=p \quad \ldots \ldots$ (ii)
Dividing (i) by (ii), we get
$\frac{a r^{(p-1)}}{a r^{(q-1)}}=\frac{q}{p}$
$\Rightarrow r^{(p-1-q+1)}=\frac{q}{p}$
$\Rightarrow r^{(p-q)}=\frac{q}{p}$
$\Rightarrow r=\left(\frac{q}{p}\right)^{\frac{1}{(p-q)}}$
Substituting the value of $r$ in (ii), we get
$a\left[\left(\frac{q}{p}\right)^{\frac{1}{(p-q)}}\right]^{(q-1)}=p$
$\Rightarrow a\left[\left(\frac{q}{p}\right)^{\frac{(q-1)}{(p-q)}}\right]=p$
$\Rightarrow a=p \times\left(\frac{p}{q}\right)^{\frac{(q-1)}{(p-q)}}$
$\Rightarrow a=p\left(\frac{p}{q}\right)^{\frac{(q-1)}{(p-q)}}$
Now,
$a_{(p+q)}=a r^{(p+q-1)}$
$=p\left(\frac{p}{q}\right)^{\frac{(q-1)}{(p-q)}} \times\left[\left(\frac{q}{p}\right)^{\frac{1}{(p-q)}}\right]^{(p+q-1)}$
$=p\left(\frac{p}{q}\right)^{\frac{(q-1)}{(p-q)}} \times\left(\frac{q}{p}\right)^{\frac{(p+q-1)}{(p-q)}}$$=p \times\left(\frac{q}{p}\right)^{\frac{-(q-1)}{(p-q)}+\frac{(p+q-1)}{(p-q)}}$
$=p\left(\frac{q}{p}\right)^{\frac{-(q-1)}{(p-q)}} \times\left(\frac{q}{p}\right)^{\frac{(p+q-1)}{(p-q)}}$
$=p \times\left(\frac{q}{p}\right)^{\frac{-(q-1)}{(p-q)}+\frac{(p+q-1)}{(p-q)}}$
$=p \times\left(\frac{q}{p}\right)^{\frac{-q+1+p+q-1}{(p-q)}}$
$=p \times\left(\frac{q}{p}\right)^{\frac{p}{(p-q)}}$
$=\frac{p \times q^{\frac{p}{(p-q)}}}{p \frac{p}{(p-q)}}$
$=\frac{q^{\frac{p}{(p-q)}}}{p \frac{p}{(p-q)}-1}$
$=\frac{q \frac{p}{(p-q)}}{p^{\frac{p-p+q}{(p-q)}}}$
$=\frac{q \frac{p}{(p-q)}}{p^{\frac{q}{(p-q)}}}$
$=\frac{q^{p \times \frac{1}{(p-q)}}}{p^{q \times \frac{1}{(p-q)}}}$
$=\left(\frac{q^{p}}{p^{q}}\right)^{\frac{1}{p-q}}$<|endoftext|>
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What does ‘family’ mean in Aboriginal and Torres Strait Islander culture?
Family is the cornerstone of Aboriginal and Torres Strait Islander culture, spirituality and identity.
Family is often more broadly defined within Aboriginal and Torres Strait Islander culture than within white culture. Those involved in children’s lives, and helping to raise them, commonly include grandparents, aunts, uncles, cousins, nieces and nephews, and members of the community who are considered to be family.
Aboriginal and Torres Strait Islander people have a complex system of family relations. Extended family relationships are the core of Aboriginal and Torres Strait Islander kinship system that are central to the way culture is passed on and society is organized.
The kinship system is a complex system that determines how people relate to each other and their role, responsibilities and obligations in relation to one another, to ceremonial business and to land. These systems vary across Australia.
Why is it important for a child to be in contact with their family?
How can I help children to keep a connection with family and community?
Part of your role as a carer is helping children to stay connected to their family.
Tips for supporting ongoing family connection
- Keep parents informed about events/achievements in the child’s life. This could include school sports day, concerts or sporting events.
- Keep a Life Story Book for the child.
- It is important for the child to know you don’t view their family negatively.
- Avoid blaming or criticizing the child’s family.
- Talk to the kids about how they feel about seeing their family. Be prepared for contact to stir-up some mixed feelings – anger, sadness, confusion.
- Plan ahead. Talk about travel arrangements in care plan meetings. Be sure to let the caseworker know early if you are unable to take the child to visits.
- Prepare children for visits. Be sure the child is dressed neatly and appropriately for the occasion.
- Don’t forget special occasions like Mother’s Day, Father’s Day, birthdays or anniversaries.
- Encourage the child to take items to share with their family. For example a school art project, photographs, homework, or their Life Story Book to share with family.
- Be there for the child after the visit – ask questions and listen to the child if they are ready to talk about it.
- Provide various options for different kinds of contact, for example, phone, e-mail, Skype, writing letters, or sending photographs or art works.
- Speak with respect to family members and when speaking about them to others.
- Ask the family about cultural events and activities that they feel are important for the child to attend and be involved with.
- Share stories with parents about activities and events you have taken the children to.
- Learn more about Aboriginal or Torres Strait Islander culture.
“It’s a team effort – you as the carer family, the worker and the parents – we’re all important to the kids.”
“I don’t really like the parents but I never put them down. I try to be positive and say to my child you have two sets of parents who all care about you.”
“I took my child to Kempsey from Sydney to visit her birth Mum. I think they both really appreciated that.”
What can I do when contact with family is difficult?
Yes, carers will at times find family contact challenging.
Challenges may include:
- Mixed emotions toward the child’s parents relating to issues of abuse or neglect.
- The child’s challenging behaviour before and after a visit.
- Resentment and anger toward you from birth parents.
- Parents trying to see children outside arranged visits.
A guide to managing contact that is difficult:<|endoftext|>
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Solar Photovoltaic Systems
"Solar photovoltaic systems" refers to a wide variety of solar electricity systems. Solar photovoltaic systems use solar panels made of silicon to convert sunlight into electricity.
Solar photovoltaics are used in a number of ways, primarily to power homes that are inter-tied or interconnected with the grid.
In a grid-interconnected solar power system, your photovoltaic panel or array of panels is added onto a regular source of electricity (provided by your utility company). The use of solar energy can reduce your reliance on other forms of energy to any degree you choose, depending on how big a system you purchase. Learn more about grid inter-tied solar.
You may also want to know about the following:
- tips on selecting a solar supplier or service company
- state and federal incentives for renewable energy
- net metering
- tips on solar panels and grid-inter-tied inverters
Some homes and businesses require a guaranteed source of power - and this can be provided by a grid-interconnected system with a battery backup.
Solar Photovoltaics are also used:
- To independently power homes and other buildings
- In portable photovoltaic systems
- In small remote solar systems that power electric fences and traffic signs.
- In solar powered products - everything from calculators to clothing
With enough panels and a high enough rate of sunlight, a large solar photovoltaic system will be able to power a home. You can learn more about off grid solar.
We also have a separate page and section on portable solar power.<|endoftext|>
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the streets of mea shearim
During the 1870s the city within the walls of Jerusalem were undergoing a serious crisis. An increase in population, especially in the Jewish quarter, resulted in high housing prices and poor sanitation.The Ottoman government failed to remove garbage dumps and eventually the pollution seeped into the water pits, causing a rise in disease and mortality rates among the population within the walls. This drove the Jewish community to establish neighborhoods outside the walls, and by 1873 four such neighborhoods were built - “Mishkenot Sha’ananim” (1880), “Mahane Israel” (1886), “Nahalat Shiva” (1869) and “Beit David” (1873).
A small group of about one-hundred young Ashkenazi Jews who believed that moving outside the walls would help them improve their standard of living, decided in 1874 to combine their resources. They were able to purchase a tract of land outside the walls for a new settlement. It would have one-hundred houses and would serve as the fifth neighborhood outside the city walls. The name which they chose for that piece of land, Mea Shearim, was derived from a verse in the Torah portion that was read in the week the neighborhood was founded: “Isaac sowed in that land, and in that year he reaped a hundredfold (מאה שערים , Mea Shearim); God had blessed him” (Genesis 26:12).
Construction began around April 1874, by both Jewish and non-Jewish workers. Contractors, builders and plasterers were Christian Arabs from Bethlehem, and Jewish craftsmen also contributed. By December 1874, the first ten houses were standing. At first Mea Shearim was a courtyard neighborhood, surrounded by four walls with gates that were locked every evening. By October 1880, 100 apartments were ready for occupancy and a lottery was held to assign them to families. Between the years 1881 and 1917, more houses and neighborhoods were built. New neighborhoods surrounded Mea Shearim and helped establish a large Jewish presence outside the walls. By the turn of the century there were 300 houses, a flour mill, and a bakery.
Mandatory Palestine under British administration had been carved out of Ottoman southern Syria after World War I. The British civil administration in Palestine operated from 1920 until 1948. During its existence the country was known simply as Palestine. The British regime was welcomed by the residents of Mea Shearim, who maintained good relations with the authorities for the good of the neighborhood. As a result, access roads to the area were improved, the neighborhood markets prospered, old shops were renovated, and new shops opened. Mea Shearim continued to grow, and by 1931 it was the third largest neighborhood in Jerusalem. This growth enhanced the neighborhood’s status and importance, but daily life became more difficult, as many of the houses were populated with a large number of people resulting in sanitary conditions that endangered their health. The neglect of the Ottoman regime continued to set the tone, and lack of proper drainage caused rain to flood the streets and even people homes. There was a rise in poverty, resulting not only in a deterioration of the houses outer appearance but also in a spread of diseases. The neighborhood’s uniform appearance also began to change, as different kinds of constructions materials came into use, resulting in non-uniform façades. Cheap tin became an alternative to the Jerusalem stone commonly used for construction.
In 1948 the Arab–Israeli war broke out and Jerusalem was divided between two countries - Israel and Jordan. The border was very close to Mea Shearim and the neighborhood suffered from military attacks and damage to buildings. Within the next 20 years ,the neighborhood would suffer from decreasing population as the children of the second founding generation moved to orthodox neighborhoods nearby, leaving as few as 170 houses occupied out of a total of 304. In later years the residents returned and the population grew once again. The population remained isolated and segregated, because it refused to cooperate with the government of Israel. Street posters (Pashkvilim) began to appear on a public walls calling on residents not to serve in the Israeli army, not to vote or be elected to the Israeli parliament, and not to participate in Israel’s Independence Day celebrations.
Today, Mea Shearim remains loyal to its old customs and preserves its isolation in the heart of Jerusalem while trying to stave off the modern world; it is, in a way, frozen in time. The numerous renovations of houses at the end of the 20th century hardly affected the appearance of the neighborhood. They are still common today but fewer in numbers. Houses that were built over one- hundred years ago stand alongside a few new ones. The life of the Hassidic community still revolves around strict adherence to Jewish law, prayer, and the study of Jewish religious texts. The large majority of the people are Ashkenazim; there are hardly Sephardic Jews in the neighborhood. In addition to some well-to-do family there are also many needy ones, which are helped by local charity institutions. The traditional dress code remains in effect here; for men and boys it includes black frock coats and black hats. Long, black beards cover their faces and many of them grow side curls called “payots”.Women and girls are urged to wear what is considered to be modest dress – knee-length or longer skirts, no plunging necklines or midriff tops, no sleeveless blouses or bare shoulders. Some women wear thick black stockings all year long, and married women wear a variety of hair coverings, from hats to wigs and headscarves. The common language of daily communication in Mea Shearim is Yiddish, in contrast to the Hebrew spoken by the majority of Israel’s Jewish population. Hebrew is used by the residents only for prayer and religious study, as they believe that Hebrew is a sacred language to be used only for religious purposes.
This is the story of the ongoing battle between the old and the new, the past versus the present, this is the everyday life of a city within a city.
My grandmother and I had a special bond.
We developed a habit that once a week, usually on Mondays, we cleared our schedule and sat down to discuss the photographs I took. We talked the stories behind the photos, the people, even how the weather affected the light in the pictures. At first, photography was something foreign for both of us and with time, we developed a passion for it. We loved our gatherings and anticipated them every week. In early 2014 things changed, we had fewer opportunities for our weekly routine as her health had begun to deteriorate. She received treatments on a weekly basis and eventually had to be under medical supervision and hospitalized. On one of the visits as I sat by her bed, I wanted to ease her mind from the treatments she received and asked if she would like to see a photograph I took the day before. She immediately said yes and was very enthused when I showed her the photograph. We ended up taking and analyzing the photo as we used to, freeing our minds from the hospital room we were in. Neither of us knew that it would be our last time together.
After her death, I decided to do a project based on the last photograph she saw. This one photo has led me on a three-year journey, photographing the streets of Mea Shearim.<|endoftext|>
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# What Is The Cube Root Of 18?
• Cube root ∛18 cannot be reduced, because it already is in its simplest form.
• All radicals are now simplified. The radicand no longer has any cubed factors.
## Determine The Cubed Root Of 18?
• The cubed root of eighteen ∛18 = 2.6207413942089
## How To Calculate Cube Roots
• The process of cubing is similar to squaring, only that the number is multiplied three times instead of two. The exponent used for cubes is 3, which is also denoted by the superscript³. Examples are 4³ = 4*4*4 = 64 or 8³ = 8*8*8 = 512.
• The cubic function is a one-to-one function. Why is this so? This is because cubing a negative number results in an answer different to that of cubing it's positive counterpart. This is because when three negative numbers are multiplied together, two of the negatives are cancelled but one remains, so the result is also negative. 7³ = 7*7*7 = 343 and (-7)³ = (-7)*(-7)*(-7) = -343. In the same way as a perfect square, a perfect cube or cube number is an integer that results from cubing another integer. 343 and -343 are examples of perfect cubes.
## Mathematical Information About Numbers 1 8
• About Number 1. The number 1 is not a prime number, but a divider for every natural number. It is often taken as the smallest natural number (however, some authors include the natural numbers from zero). Your prime factorization is the empty product with 0 factors, which is defined as having a value of 1. The one is often referred to as one of the five most important constants of analysis (besides 0, p, e, and i). Number one is also used in other meanings in mathematics, such as a neutral element for multiplication in a ring, called the identity element. In these systems, other rules can apply, so does 1 + 1 different meanings and can give different results. With 1 are in linear algebra and vectors and one Einsmatrizen whose elements are all equal to the identity element, and refers to the identity map.
• 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 is a cube root?
In arithmetic and algebra, the cube of a number n is its third power: the result of the number multiplied by itself twice: n³ = n * n * n. It is also the number multiplied by its square: n³ = n * n².
This is also the volume formula for a geometric cube with sides of length n, giving rise to the name. The inverse operation of finding a number whose cube is n is called extracting the cube root of n. It determines the side of the cube of a given volume. It is also n raised to the one-third power.
Both cube and cube root are odd functions: (-n)³ = -(n³). The cube of a number or any other mathematical expression is denoted by a superscript 3, for example 2³ = 8 or (x + 1)³.
© Mathspage.com | Privacy | Contact | info [at] Mathspage [dot] com<|endoftext|>
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For some children, despite having no known physical or mental disability, learning to read, write, spell, do maths, dress, throw and catch a ball, or organise themselves presents significant challenges. When childhood milestones involving speech and movement are slow to develop in young children, should parents be concerned that their child might have a learning disability?
Below we outline three learning disabilities and what to look out for.
One of the most common specific learning disabilities is dyslexia. Dyslexia is described as a difference in how children process information. It is often diagnosed shortly after starting school, when formal reading instruction begins. Some early signs of dyslexia are also observed during pre-school years.
Early indicators of dyslexia include poor phonological skills (the ability to manipulate sounds and words), difficulties with rhyme and rhythm, problems with remembering more than two pieces of information, and general forgetfulness. Physical co-ordination problems are often noticed in pre-schoolers who are later diagnosed with dyslexia. It is estimated that between 3% and 10% of children may have dyslexia.
The most common symptoms of dyslexia in school-aged children include difficulties with reading (reading accurately), reading comprehension, spelling and writing. Children with dyslexia may also have difficulties with following instructions, memory and organisational skills (organising school equipment, time management). Dyslexia also affects children’s self-esteem, which can be further lowered by bullying or teasing from classmates.
A child with dyslexia can be supported in a number of ways. Teachers need to be acquainted with appropriate reading strategies, such as decoding words (sounding out letters to form words – cat, c-a-t) and word attack skills (sounding out nonsense words – bem, b-e-m).
Parents can assist by reading aloud and having their children follow along in the book. Involving children in conversations over dinner, or asking them to retell a story they heard, can be of great help as well. Using tablets or smartphones, in addition to traditional teaching strategies, can assist children with dyslexia with their reading, writing and organisation skills.
Developmental dyscalculia is a specific learning disability involving mathematics. Children can experience problems understanding numbers, learning how to manipulate numbers and learning basic maths facts.
Children are born with dyscalculia. It occurs in those with average to above average mental ability. It is estimated that between 5-15% of school-aged children and young adults have dyscalculia. One-quarter of students with dyscalculia are also diagnosed as having attention deficit hyperactivity disorder (ADHD).
The first signs of developmental dyscalculia are problems with counting, or being able to tell how many objects are in a group, even when there are just a few objects. Other signs of dyscalculia include: difficulty telling time or understanding time-related concepts, inability to discern which of two numbers is the larger, difficulty with memorising basic maths facts (like multiplication tables) and doing mental arithmetic, and difficulty with understanding mathematical formulae and concepts. Additional problems may include telling left from right, reading music and map reading.
Children with dyscalculia are best assisted by providing individualised instruction. Recent research has also shown promising results from computer-based instruction, where repeated opportunities to practice are provided in an engaging way.
When planning interventions, educators should consider the broad scope of mathematics skills to be learnt. Careful assessment is needed to determine what areas require attention. Skills should then be taught step-by-step. Other useful support strategies include the use of number lines, multiplication charts, an abacus and math games to support and reinforce student learning.
Developmental dyspraxia is also know as developmental co-ordination disorder (DCD). It is a condition where the co-ordination of arms, legs, fingers, mouth or eyes is less smooth, fluent or precise than seen in age-matched peers. Dyspraxia is thought to affect around 6% of children. Boys are four times more likely to present with symptoms than girls.
Dyspraxia can affect academic achievement (handwriting, speech) and everyday activities (use of cutlery, skipping). Problems with speech co-ordination are known as verbal dyspraxia. A high number of children diagnosed with an autism spectrum disorder, attention deficit hyperactivity disorder, language disorders, or born prematurely have dyspraxia.
The symptoms of dyspraxia vary from child to child, and from one developmental stage to the next. Speech co-ordination problems can include forming sounds, slower speech, limited vocabulary and perseveration (getting stuck on a word and repeating it). Poor oral co-ordination problems can also affect the ability to blow or suck.
Physical co-ordination problems may appear as clumsiness, poor balance, avoidance of toys that require good finger co-ordination (jigsaw puzzles), problems with pencil grip, or using scissors. There is a wide range of treatment options for dyspraxia. These include occupational therapy to improve motor co-ordination fluency and perceptual motor therapy, but research indicates some treatments are more effective than others.
It is important to keep in mind that regardless of the learning disability, the use of evidence-based interventions from a young age and throughout the school years can greatly improve a child’s academic, physical and social-emotional skills.
Having a learning disability is not a life sentence: just look at what Richard Branson and Steve Jobs have achieved.<|endoftext|>
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# Algebra II : Quadratic Inequalities
## Example Questions
### Example Question #1 : Quadratic Inequalities
Explanation:
The problem is already in standard form, so all we have to at first do is set the quadratic expression = 0 and factor as normal.
Negative x^2's are hard to work with, so we multiply through by -1.
Now we can factor easily.
By the zero product property, each of these factors will be equal to 0.
Since -9 and 1 are our zeros, we just have to test one point in the region between them to find out which region our answer set goes in. Let's test x = 0 in the original inequality.
Since this statement is false, the region between -9 and 1 is not correct. So it must be the region on either side of those points. Since the original inequality was less than or equal to, the boundary points are included. So all values from -infinity to -9 inclusive, and from 1 inclusive to infinity, are solutions. In interval notation we write this as:
### Example Question #352 : Quadratic Equations And Inequalities
Explanation:
First we want to rewrite the quadratic in standard form:
Now we want to set it = 0 and factor and solve like normal.
Using the zero product property, both factors produce a zero:
So the two zeros are -2 and 3, and will mark the boundaries of our answer interval. To find out if the interval is between -2 and 3, or on either side, we simply take a test point between -2 and 3 (for instance, x = 0) and evaluate the original inequality.
Since the above is a true statement, we know that the solution interval is between -2 and 3, the same region where we picked our test point. Since the original inequality was less than or equal, we include the endpoints.
Ergo, .
### Example Question #353 : Quadratic Equations And Inequalities
What is the discriminant of the following quadratic equation:
Explanation:
The discriminant of a quadratic equation in form is equal to . The given equation is not in that form however, so we must first multiply it out to get it into that form. We therefore obtain:
We therefore have , and . Our discriminant is therefore:
### Example Question #4 : Quadratic Inequalities
and
Explanation:
1. Rewrite the equation in standard form.
2. Set the equation equal to and solve by factoring.
So, and are our zeroes.
3. Test a point between your zeroes to find out if the solution interval is between them or on either side of them. (Try testing by plugging it into your original inequality.)
Because the above statement is true, the solution is the interval between and .
### Example Question #5 : Quadratic Inequalities
Solve this inequality.
Explanation:
Combine like terms first.
Factor
The zeroes are 3 and 8 so a number line can be divided into 3 sections.
X<3 works, 3<x<8 does not work, and x>8 works
### Example Question #6 : Quadratic Inequalities
Solve:
Explanation:
Start by setting the inequality to zero and by solving for .
Now, plot these two points on to a number line.
Notice that these two numbers effectively divide up the number line into three regions:
, and
Now, choose a number in each of these regions and put it back in the factored inequality to see which cases are true.
For , let
Since this is not less than , the solution to this inequality cannot lie in this region.
For , let .
Since this will make the inequality true, the solution can lie in this region.
Finally, for , let
Since this number is not less than zero, the solution cannot lie in this region.
Thus, the solution to this inequality is
### Example Question #7 : Quadratic Inequalities
Solve:
The solution cannot be determined with the information given.
Explanation:
First, set the inequality to zero and solve for .
Now, plot these two numbers on to a number line.
Notice how these numbers divide the number line into three regions:
Now, you will choose a number from each of these regions to test to plug back into the inequality to see if the inequality holds true.
For , let
Since this is not less than zero, the solution to the inequality cannot be found in this region.
For , let
Since this is less than zero, the solution is found in this region.
For , let
Since this is not less than zero, the solution is not found in this region.
Then, the solution for this inequality is
### Example Question #354 : Quadratic Equations And Inequalities
Solve:
Explanation:
Start by changing the less than sign to an equal sign and solve for .
Now, plot these two numbers on a number line.
Notice how the number line is divided into three regions:
Now, choose a number fromeach of these regions to plug back into the inequality to test if the inequality holds.
For , let
Since this number is not less than zero, the solution cannot be found in this region.
For , let
Since this number is less than zero, the solution can be found in this region.
For let .
Since this number is not less than zero, the solution cannot be found in this region.
Because the solution is only negative in the interval , that must be the solution.
### Example Question #9 : Quadratic Inequalities
Solve:
Explanation:
First, set the inequality to zero and solve for .
Now, plot these two numbers on to a number line.
Notice how these numbers divide the number line into three regions:
Now, you will choose a number from each of these regions to test to plug back into the inequality to see if the inequality holds true.
For , let
Since this solution is greater than or equal to , the solution can be found in this region.
For , let
Since this is less than or equal to , the solution cannot be found in this region.
For , let
Since this is greater than or equal to , the solution can be found in this region.
Because the solution can be found in every single region, the answer to this inequality is
### Example Question #1 : Quadratic Inequalities
Which value for would satisfy the inequality ?
Not enough information to solve
Explanation:
First, we can factor the quadratic to give us a better understanding of its graph. Factoring gives us: . Now we know that the quadratic has zeros at and . Furthermore this information reveals that the quadratic is positive. Using this information, we can sketch a graph like this:
We can see that the parabola is below the x-axis (in other words, less than ) between these two zeros and .
The only x-value satisfying the inequality is .
The value of would work if the inequality were inclusive, but since it is strictly less than instead of less than or equal to , that value will not work.<|endoftext|>
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Meniere’s (Disease, Disorder, or Syndrome)
(Keywords: Dizziness/Imbalance, hearing loss or distortion, abnormal rushing water or rumbling sounds, ear pressure, nausea, vomiting.)
Audiological tests required: Audiogram (baseline and follow-up/monitor audiograms), distortion product otoacoustic emissions (DPOAE), auditory brainstem response, auditory brainstem response (ABR or stacked ABR), electrocochleography (eCocHG).
Treatment: This will vary depending on the patient’s circumstances and the physician’s specific approach. Dietary changes, medication, and other therapy may be implemented to help treat Meniere’s.
Description: Typically a person with Meniere’s will have bouts of dizziness/imbalance, fluctuating hearing loss or distorted hearing, ear pressure, and rushing type tinnitus (roaring or water running like sound). The dizziness related to Meniere’s will typically last a few minutes to hours. During that time nausea, vomiting, and imbalance may occur. The symptoms may or may not all occur at the same time. These bouts may occur regularly, a few times a year, or intermittently over a course of years. People suffering from Meniere’s may develop permanent hearing loss. It is important to distinguish Meniere’s from Migraine Disorder for proper treatment of the symptoms.
Meniere’s is believed to be caused by an excess of inner ear (cochlear) fluid. The picture depicts an inner ear without the bout of Meniere’s and an inner ear during a bout of Meniere’s.<|endoftext|>
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# Convert g units to mile / square minute
Learn how to convert 1 g units to mile / square minute step by step.
## Calculation Breakdown
Set up the equation
$$1.0\left(g \text{ } units\right)={\color{rgb(20,165,174)} x}\left(\dfrac{mile}{square \text{ } minute}\right)$$
Define the base values of the selected units in relation to the SI unit $$\left(\dfrac{meter}{square \text{ } second}\right)$$
$$\text{Left side: 1.0 } \left(g \text{ } units\right) = {\color{rgb(89,182,91)} 9.80665\left(\dfrac{meter}{square \text{ } second}\right)} = {\color{rgb(89,182,91)} 9.80665\left(\dfrac{m}{s^{2}}\right)}$$
$$\text{Right side: 1.0 } \left(\dfrac{mile}{square \text{ } minute}\right) = {\color{rgb(125,164,120)} \dfrac{1609.344}{3.6 \times 10^{3}}\left(\dfrac{meter}{square \text{ } second}\right)} = {\color{rgb(125,164,120)} \dfrac{1609.344}{3.6 \times 10^{3}}\left(\dfrac{m}{s^{2}}\right)}$$
Insert known values into the conversion equation to determine $${\color{rgb(20,165,174)} x}$$
$$1.0\left(g \text{ } units\right)={\color{rgb(20,165,174)} x}\left(\dfrac{mile}{square \text{ } minute}\right)$$
$$\text{Insert known values } =>$$
$$1.0 \times {\color{rgb(89,182,91)} 9.80665} \times {\color{rgb(89,182,91)} \left(\dfrac{meter}{square \text{ } second}\right)} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} {\color{rgb(125,164,120)} \dfrac{1609.344}{3.6 \times 10^{3}}}} \times {\color{rgb(125,164,120)} \left(\dfrac{meter}{square \text{ } second}\right)}$$
$$\text{Or}$$
$$1.0 \cdot {\color{rgb(89,182,91)} 9.80665} \cdot {\color{rgb(89,182,91)} \left(\dfrac{m}{s^{2}}\right)} = {\color{rgb(20,165,174)} x} \cdot {\color{rgb(125,164,120)} \dfrac{1609.344}{3.6 \times 10^{3}}} \cdot {\color{rgb(125,164,120)} \left(\dfrac{m}{s^{2}}\right)}$$
$$\text{Cancel SI units}$$
$$1.0 \times {\color{rgb(89,182,91)} 9.80665} \cdot {\color{rgb(89,182,91)} \cancel{\left(\dfrac{m}{s^{2}}\right)}} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} \dfrac{1609.344}{3.6 \times 10^{3}}} \times {\color{rgb(125,164,120)} \cancel{\left(\dfrac{m}{s^{2}}\right)}}$$
$$\text{Conversion Equation}$$
$$9.80665 = {\color{rgb(20,165,174)} x} \times \dfrac{1609.344}{3.6 \times 10^{3}}$$
Switch sides
$${\color{rgb(20,165,174)} x} \times \dfrac{1609.344}{3.6 \times 10^{3}} = 9.80665$$
Isolate $${\color{rgb(20,165,174)} x}$$
Multiply both sides by $$\left(\dfrac{3.6 \times 10^{3}}{1609.344}\right)$$
$${\color{rgb(20,165,174)} x} \times \dfrac{1609.344}{3.6 \times 10^{3}} \times \dfrac{3.6 \times 10^{3}}{1609.344} = 9.80665 \times \dfrac{3.6 \times 10^{3}}{1609.344}$$
$$\text{Cancel}$$
$${\color{rgb(20,165,174)} x} \times \dfrac{{\color{rgb(255,204,153)} \cancel{1609.344}} \times {\color{rgb(99,194,222)} \cancel{3.6}} \times {\color{rgb(166,218,227)} \cancel{10^{3}}}}{{\color{rgb(99,194,222)} \cancel{3.6}} \times {\color{rgb(166,218,227)} \cancel{10^{3}}} \times {\color{rgb(255,204,153)} \cancel{1609.344}}} = 9.80665 \times \dfrac{3.6 \times 10^{3}}{1609.344}$$
$$\text{Simplify}$$
$${\color{rgb(20,165,174)} x} = \dfrac{9.80665 \times 3.6 \times 10^{3}}{1609.344}$$
Solve $${\color{rgb(20,165,174)} x}$$
$${\color{rgb(20,165,174)} x}\approx21.936851288\approx21.9369$$
$$\text{Conversion Equation}$$
$$1.0\left(g \text{ } units\right)\approx{\color{rgb(20,165,174)} 21.9369}\left(\dfrac{mile}{square \text{ } minute}\right)$$<|endoftext|>
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CBSE Class 7 Maths worksheet for chapter- Congruence of Triangles
Sep 29, 2022, 16:45 IST
Worksheet For CBSE Class 7 Maths Chapter - Congruence of Triangles
This page consists of CBSE Class 7 Maths Worksheet for chapter - Congruence of Triangle prepared by Physics Wallah. It states that the two triangles are said to be congruent when they are related and that they are covered exactly. Class 7th chapter Congruence of triangles helps students learn some of the axiom rules that every student needs to continue their higher education.
All questions of class 7 maths chapter - Congruence of Triangles are with detail solutions for reference. Find out all worksheet for class 7 Maths. Do follow NCERT Solutions for class 7 Maths prepared by Physics Wallah.
Solved Examples
Q1. Two line segments are congruent if …………
1. They have equal length.
2. They have unequal length.
3. First is equal to half of second.
4. None of them.
Ans. Correct option is 1
We know that “Two line segments are congruent if they have equal length”.
Q2. If △ABC and △PQR are congruent under the correspondence: ABC⇔ RQP then, P = …….
1. ∠A
2. ∠B
3. ∠C
4. ∠D
Ans. Correct option is 3
Q3. If triangle ABC congruence to triangle CBD, then ∠ABC = ………
1. ∠CDB
2. ∠CBD
3. ∠DCB
4. none of them
Ans. Correct option is 2
Since,△ABC ≅ △CBD
Then, ∠ABC = ∠CBD ( by corresponding part of congruence triangle )
Q4. Among two congruent angles, one has a measure of 600; the measure of the other angle is …….
1. 500
2. 700
3. 600
4. 800
Ans. Correct option is 2
When two angles are congruent, then measure of each angle is equal.
Q5. If you want to show two triangles are congruent, using the ASA rule then you need to show ………….
1. All angles are equal.
2. All sides are equal.
3. Two angle and 1 sides are equal.
4. One angle and two sides are equal.
Ans. Correct option is 3
Two angle and 1 sides are equal.
Q6. What is the side included the angles M and N of MNP?
1. MP
2. NP
3. MN
4. none of them
Ans. Correct option is 3
Q7. If you want to show two triangles are congruent, using the SAS rule then you need to show ………….
1. All angles are equal.
2. All sides are equal.
3. Two angle and 1 sides are equal.
4. One angle and two sides are equal.
Ans. Correct option is 4
one angle and two sides are equal.
For additional information related to the subject you can check the Maths Formula section. Students can also access the Congruence of Triangles notes from here.
Also Check
Find Below pdf of CBSE Class 7 Maths Worksheet of Chapter - Congruence of Triangles<|endoftext|>
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## Indices with brackets and powers
Note:The words powers, exponents and indices all mean the same thing. We will now examine what happens when we raise numbers in brackets to a power.
The a represents the number in the bracket while the m and n represent the two powers (one inside and one outside of the bracket). Here is an example in which this rule is applied. Example: In this example, the powers were multiplied together to give the answer which is 3 to the power of 6. How to do brackets to a power. Indices and brackets. Otherwise known as parentheses. Multiplying powers. If you learnt something new and are feeling generous Expanding brackets with powers Powers or index numbers are the floating numbers next to terms that show how many times a letter or number has been multiplied by itself. For example, and. Using In general: This formula tells us that when a power of a number is raised to another power, multiply the indices. This is the fourth index law and is known as the Index Law for Powers. In general, we have for any base a and indices m and n: (a m) n = a mn. Raising a Product to a Power. Number example: (5 × 2) 3 = 5 3 × 2 3. In this case, with numbers, it would be better to perform the multiplication in brackets first and then raise our answer to the power 3. This feature is not available right now. Please try again later. Negative indices are all exponents or powers that have a minus sign in front of them and are as result negative. They are quite easy to deal with as there is only one thing that you have to do. They are quite easy to deal with as there is only one thing that you have to do.
## 4 Jun 2019 Product of powers: Add powers together when multiplying like bases First, redistribute the power to the inside of the brackets, following the
The a represents the number in the bracket while the m and n represent the two powers (one inside and one outside of the bracket). Here is an example in which this rule is applied. Example: In this example, the powers were multiplied together to give the answer which is 3 to the power of 6. How to do brackets to a power. Indices and brackets. Otherwise known as parentheses. Multiplying powers. If you learnt something new and are feeling generous Expanding brackets with powers Powers or index numbers are the floating numbers next to terms that show how many times a letter or number has been multiplied by itself. For example, and. Using In general: This formula tells us that when a power of a number is raised to another power, multiply the indices. This is the fourth index law and is known as the Index Law for Powers. In general, we have for any base a and indices m and n: (a m) n = a mn. Raising a Product to a Power. Number example: (5 × 2) 3 = 5 3 × 2 3. In this case, with numbers, it would be better to perform the multiplication in brackets first and then raise our answer to the power 3. This feature is not available right now. Please try again later.
### Always remove brackets first. Example 11. Simplify each of the following: Solution : Key Terms. index law for powers
Powers of brackets can be expanded using the distributive law. This property is shown below: \begin{align*} (x \times y)^n = x^n \times y^n \end{align*}. Indices: powers/Surds. Multiplication. Division. Addition. Subtraction. Order of calculations: 1st: Solve any brackets. 2nd: Solve any power/indices. 3rd: Solve any
### Negative indices are all exponents or powers that have a minus sign in front of them and are as result negative. They are quite easy to deal with as there is only one thing that you have to do. They are quite easy to deal with as there is only one thing that you have to do.
order of operations > when to work out powers > brackets and powers. worksheets. bidmas This section covers Indices and the uses of Indices in algebra. After studying this section, you will be able to: divide and multiply algebraic expressions using indices; find roots using indices. This video shows a guide to indices and powers. Multiplying and dividing indices, raising indices to a power and using standard form are explained. Indices or Powers mc-TY-indicespowers-2009-1 A knowledge of powers, or indices as they are often called, is essential for an understanding of most algebraic processes. In this section of text you will learn about powers and rules for manipulating them through a number of worked examples. 3D shapes Adding algebraic fractions Adding and subtracting vectors Adding decimals Adding fractions Adding negative numbers Adding surds Algebraic fractions Algebraic indices Algebraic notation Algebraic proof Alternate angles Alternate segment theorem Angle at the centre Angle in a semi-circle Angles Angles at a point Angles in a polygon RULES FOR INDICES; BRACKETS AND FACTORISING; SOLVING EQUATIONS; REARRANGING FORMULAE; ALGEBRAIC PROOF; INEQUALITIES; Brackets with all Four Operations [First Steps] BiDMAS. Squares and Cubes [First Steps] Powers of Two and Ten. NEW [First Steps] Squares and Cubes. NEW [Strengthen] Powers of Two and Ten. NEW Exponents are also called Powers or Indices. Let us first look at what an "exponent" is: The exponent of a number says how many times to use the number in a multiplication. In this example: 8 2 = 8 × 8 = 64. In words: 8 2 can be called "8 to the second power", "8 to the power 2" To add or subtract with powers, both the variables and the exponents of the variables must be the same. You perform the required operations on the coefficients, leaving the variable and exponent as they are. When adding or subtracting with powers, the terms that combine always have exactly the same variables with exactly the same […]
## Note:The words powers, exponents and indices all mean the same thing. We will now examine what happens when we raise numbers in brackets to a power.
Always remove brackets first. Example 11. Simplify each of the following: Solution : Key Terms. index law for powers Expanding brackets with powers. Powers or index numbers are the floating numbers next to terms that show how many times a letter or number has been The third law: brackets. find roots using indices. This video shows a guide to indices and powers. Multiplying and dividing indices, raising indices to a power and using standard form are When a number written in exponential notation is raised to a power, it is called a “ power of a power.” In this expression, the base is 52 and the exponent is 4: 52 is
Negative indices are all exponents or powers that have a minus sign in front of them and are as result negative. They are quite easy to deal with as there is only one thing that you have to do. They are quite easy to deal with as there is only one thing that you have to do. This formula tells us that when a power of a number is raised to another power, multiply the indices. This is the fourth index law and is known as the Index Law for Powers. Example 10 Solution: Note: Always remove brackets first. Example 11. Simplify each of the following: Solution: Key Terms. index law for powers Indices (Powers) & Roots The superscript numbers (2, 3 & 4 above) are known as indices or powers. When the power is 2 we say “squared”, when the power is 3 we say “cubed” and for all other powers power outside of a bracket you multiply the powers. (45)2=410<|endoftext|>
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# NCERT solutions for Class 10 Maths: Chapter 13 Surface Area and Volume Exercise 13.2
In this page we have NCERT solutions for Class 10 Maths: Chapter 13 Surface Area and Volume for Exercise 13.3 on pages 251 and 252. Hope you like them and do not forget to like , social_share and comment at the end of the page.
This exercise is based on the volume of an object formed by combining any two of the basic solids, namely, cuboid, cone, cylinder, sphere and hemisphere
Formula to used
The total Volume of the any new solid is the sum of the Volume of each of the individual parts.
Volume of new solid = Volume of One part + Volume of Second part + Volume of Third part
Volume of Common Shapes
Volume of cone
$=\frac {1}{3}\pi r^2h$
Volume of Sphere
$=\frac {4}{3} \pi r^3$
Volume of Cylinder
$=\pi r^2 h$
Unless stated otherwise, take $\pi = \frac {22}{7}$.
## Surface Area and Volume Exercise 13.3
Question 1
A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.
Solution
Let h be the height then
Then , volume of cylinder =Volumne of Sphere
$\pi (6)^2 h =\frac {4}{3} \pi (4.2)^3$
then h =2.74 cm
Question 2
Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.
Solution
Now
Volume of resulting sphere = Volume of 6 cm sphere + Volume of 8 cm sphere + Volume of 10 cm sphere
$\frac {4}{3} \pi R^3 = \frac {4}{3}\pi 6^3 + \frac {4}{3}\pi 8^3 + \frac {4}{3}\pi 10^3$
or
$R^3= 6^3 + 8^3 + 10^3$
$R^3=1728$
R=12 cm
Question 3
A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.
Solution
Let h be the height of platform
Now
Volume Of well = Volume of the platform formed
$\pi (3.5)^2 20 = 22 \times 14 \times h$
h=2.5 m
Question 4
A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.
Solution
Let h be the height of the embankment.
It is aparant from figue the embankment is a cylinderical right with inner radii= 1.5 m and outer radii = 5.5 m
Now
Volume Of well = Volume of the embankment formed
$\pi \times (1.5)^2 \times 14 = \pi \times h \times (5.5^2 - 1.5^2)$
h=1.125 m
Question 5
A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.
Solution
Radius of Cone(r) = 6/2 = 3 cm
Height of cone(h)= 12 cm
Radius of hemispherical shape on the top(r) = 3cm
Volume of One cone icecream = $\frac {1}{3} \pi r^2 h + \frac {2}{3} \pi r^3 = \frac {1}{3} \pi ( 108 + 54)$
Let n be the number of icecream cones filled
Then
Volume of Cylinder = n × Volume of One cone icecream
$\pi (6)^2 15 = \frac {n}{3} \pi ( 108 + 54)$
or n=10
Question 6
How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm x 10 cm x 3.5 cm?
Solution
Let n be the silver coins needed
n × Volume of one silver coin= volume of cuboid
$n \times \pi \times (.875)^2 \times .2 = 5.5 \times 10 \times 3.5$
n=400
Question 7
A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
Solution
Let r be the radius of the conical heap
Now Volume of cylinderical bucket = Volume of conical heap
$\pi \times (18)^2 \times 32 = \frac {1}{3} r^2 \times 24$
r=36 cm
Now slant height = $\sqrt {r^2 + h^2} = \sqrt {36^2 + 24^2} = 12 \sqrt {13}$ cm
Question 8
Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?
Solution
Cross-section of the Canal = $6 \times 1.5 = 9 m^2$
Now Water speed is 10 km/hr = 1000/6 m/min
So volume of Water flowed in 1 min = $9 \times \frac {1000}{6}$
Therefore Volume of water in 30 min = $9 \times \frac {1000}{6} \times 30=45000 m^3$
Let A be the Area irrigated then
Then
$A \times .08 = 45000$
A=562500 m2
Question 9
A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?
Solution
Cross-section of the pipe = $\pi \times (.1)^2 = .01 \pi m^2$
Speed of water = 3km/h = 3000/60 m/min = 50 m/min
Let t be the time time taken in min, then Volume of water out of pipe
$= .01 \pi \times 50 \times t = .5 t \pi m^2$
Now Volume of cylindrical tank= $\pi (5)^2 2= 50 \pi m^2$
Now Volume of water out of pipe=Volume of cylindrical tank
$.5 t \pi=50 \pi$
t= 100 min
## Summary
1. Class 10 Maths NCERT solutions for Surface Area and Volume Exercise 13.3 has been prepared by Expert with utmost care. If you find any mistake.Please do provide feedback on mail.You can download this as pdf
2. This chapter 13 has total 5 Exercise 13.1 ,13.2,13.3 ,13.4 and 13.5. This is the Third exercise in the chapter.You can explore previous exercise of this chapter by clicking the link below
• Notes
• Assignments
• NCERT Solutions
• NCERT Solution Surface Area and Volume Class10 Exercise 13.1
• NCERT Solution Surface Area and Volume Class10 Exercise 13.2
• NCERT Solution Surface Area and Volume Class10 Exercise 13.3
Go back to Class 10 Main Page using below links
### Practice Question
Question 1 What is $1 - \sqrt {3}$ ?
A) Non terminating repeating
B) Non terminating non repeating
C) Terminating
D) None of the above
Question 2 The volume of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is?
A) 19.4 cm3
B) 12 cm3
C) 78.6 cm3
D) 58.2 cm3
Question 3 The sum of the first three terms of an AP is 33. If the product of the first and the third term exceeds the second term by 29, the AP is ?
A) 2 ,21,11
B) 1,10,19
C) -1 ,8,17
D) 2 ,11,20<|endoftext|>
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# Complex number
(Redirected from Complex numbers)
A complex number is a number, but is different from common numbers in many ways. A complex number is made up using two numbers combined together. The first part is a real number, and the second part is an imaginary number. The most important imaginary number is called ${\displaystyle i}$, defined as a number that will be -1 when squared ("squared" means "multiplied by itself"): ${\displaystyle i^{2}=i\times i=-1\ }$. All the other imaginary numbers are ${\displaystyle i}$ multiplied by a real number, in the same way that all real numbers can be thought of as 1 multiplied by another number. Arithmetic functions such as addition, subtraction, multiplication, and division can be used with complex numbers. They also follow commutative, associative and distributive properties, just like real numbers. The set of complex numbers is often represented using the symbol ${\displaystyle \mathbb {C} }$.[1][2]
Complex numbers were discovered while attempting to solve special equations that have exponents in them. These began to pose real problems for mathematicians. As a comparison, using negative numbers, it is possible to find the x in the equation ${\displaystyle a+x=b}$ for all real values of a and b, but if only positive numbers are allowed for x, it is sometimes impossible to find a positive x, as in the equation 3 + x = 1.
With exponentiation, there is a difficulty to be overcome.[3] There is no real number that gives −1 when it is squared. In other words, −1 (or any other negative number) has no real square root. For example, there is no real number ${\displaystyle x}$ that solves the equation ${\displaystyle (x+1)^{2}=-9}$. To solve this problem, mathematicians introduced a symbol i and called it the imaginary unit.[1] This is the imaginary number that will give −1 when it is squared.
The first mathematicians to have thought of this were probably Gerolamo Cardano and Raffaele Bombelli. They lived in the 16th century.[2] It was probably Leonhard Euler who introduced writing ${\displaystyle \mathrm {i} }$ for that number.
All complex numbers can be written as ${\displaystyle a+bi}$[3] (or ${\displaystyle a+b\cdot i}$), where a is called the real part of the number, and b is called the imaginary part. We write ${\displaystyle \Re (z)}$ or ${\displaystyle \operatorname {Re} (z)}$ for the real part of a complex number ${\displaystyle z}$. So, if ${\displaystyle z=a+bi}$, we write ${\displaystyle a=\Re (z)=\operatorname {Re} (z)}$. Similarly, we write ${\displaystyle \Im (z)}$ or ${\displaystyle \operatorname {Im} (z)}$ for the imaginary part of a complex number ${\displaystyle z}$; ${\displaystyle b=\Im (z)=\operatorname {Im} (z)}$, for the same z.[1] Every real number is also a complex number; it is a complex number z with ${\displaystyle \Im (z)=0}$.
A complex number can also be written as an ordered pair (a, b), where both a and b are real numbers. Any real number can simply be written as ${\displaystyle a+0\cdot i}$, or as the pair (a, 0).[3]
Sometimes, ${\displaystyle j}$ is written instead of ${\displaystyle i}$. In electrical engineering for instance, ${\displaystyle i}$ means electric current, so writing ${\displaystyle i}$ can cause a lot of problems because some numbers in electrical engineering are complex numbers.
The set of all complex numbers is usually written as ${\displaystyle \mathbb {C} }$.[1]
## Operations over complex numbers
Addition, subtraction, multiplication and exponentiation (raising numbers to exponents) are all possible with complex numbers. Division is also possible with complex numbers—as long as the divisor is not zero, Some other calculations are also possible with complex numbers.
The rule for addition and subtraction of complex numbers is pretty simple:
Let ${\displaystyle z=(a+bi),w=(c+di)}$, then ${\displaystyle z+w=(a+bi)+(c+di)=(a+c)+(b+d)i}$, and ${\displaystyle z-w=(a+bi)-(c+di)=(a-c)+(b-d)i}$.
Multiplication is a bit different:
${\displaystyle z\cdot w=(a+bi)(c+di)=ac+bci+adi+bdi^{2}=(ac-bd)+(bc+ad)i.}$
Another notable operation for complex numbers is conjugation. A complex conjugate ${\displaystyle {\overline {z}}}$ to ${\displaystyle z=a+bi}$ is ${\displaystyle a-bi}$. It is pretty simple, but is important for calculations, because ${\displaystyle z\times {\overline {z}}}$ is actually a real number for all complex ${\displaystyle z}$:
${\displaystyle z{\bar {z}}=(a+bi)(a-bi)=(a^{2}+b^{2})+(ab-ab)i=a^{2}+b^{2}}$.
Because of that, we can use it to do division:
${\displaystyle {\frac {1}{z}}={\frac {\bar {z}}{z{\bar {z}}}}={\frac {a-bi}{a^{2}+b^{2}}}={\frac {a}{a^{2}+b^{2}}}-{\frac {b}{a^{2}+b^{2}}}i}$
${\displaystyle {\frac {w}{z}}=w({\frac {1}{z}})=(c+di)\cdot \left({\frac {a}{a^{2}+b^{2}}}-{\frac {b}{a^{2}+b^{2}}}i\right)={\frac {1}{a^{2}+b^{2}}}\left((cx+dy)+(dx-cy)i\right).}$
## Other forms of describing complex numbers
Complex numbers can be shown on a so-called complex plane. If you have a number ${\displaystyle z=a+bi}$, you can go to point a on the real axis and point b on the imaginary axis, and draw a vector from ${\displaystyle (0,0)}$ to ${\displaystyle (a,b)}$. The length of this vector can be calculated using the Pythagorean theorem, and the angle of this vector is simply the angle between the positive real axis and this vector—going counterclockwise. The length of a vector for a number ${\displaystyle z}$ is called its modulus or absolute value (written as ${\displaystyle |z|}$), and the angle is called its argument (${\displaystyle \arg z}$).[1]
A complex number can be visually shown as two numbers which form a vector on an Argand diagram, representing the complex plane.
This leads to the trigonometrical form of describing complex numbers: by the definitions of sine and cosine, it follows that for all ${\displaystyle z}$:
${\displaystyle z=|z|(\cos \arg z+i\sin \arg z).}$
This is closely connected to De Moivre's formula.
There exists even another form, called exponential form.
## Conclusion
With the introduction of complex numbers to math, every polynomial with complex coefficients has roots in complex numbers. This introduction also helped to open a path to the creation of another kind of numbers, which could help resolve and explain many different problems. These include the hypercomplex numbers, sedenion, hyperreal numbers, surreal numbers and many others. For more, see types of numbers.
## References
1. "Comprehensive List of Algebra Symbols". Math Vault. 2020-03-25. Retrieved 2020-08-12.
2. "Complex Numbers | Brilliant Math & Science Wiki". brilliant.org. Retrieved 2020-08-12.
3. Weisstein, Eric. "Complex numbers". Wolfram MathWorld. Retrieved August 11, 2020.<|endoftext|>
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Sunlight streaks through the leafy canopy, painting the Elwha River and surrounding forest in dappled light. Two men in chest waders splash through the swift-moving current, kicking up mushroom clouds of silt. Their eyes scan the shallows as they make their way downstream.
John McMillan, a fish biologist with the National Oceanic and Atmospheric Administration, and Jeff Duda, a research ecologist with the U.S. Geological Survey, are here looking for steelhead and salmon nests, or redds, where fish have laid eggs. The signs are subtle but unmistakable to their trained eyes — a depression in the river bottom and disturbed gravel.
McMillan and Duda are just two of the many scientists from federal and state agencies and the local Native American tribe, who are studying the dramatic changes taking place on this river. Two massive hydroelectric dams are in the process of being removed. And these scientists are trying to understand how the largest dam removal project in the nation’s history is impacting fish.
“We’re trying to see how fast these fish recolonize after the dams have been removed,” McMillan said.
The Elwha River in western Washington State was once home to one of the biggest salmon runs in the continental United States. But in the early 1900s two dams were built on this pristine river, which today lies mostly within the protected confines of Olympic National Park. The dams provided power to the nearby town of Port Angeles and helped the frontier community grow and thrive.
But the dams were built without fish ladders and had devastating consequences for fish; salmon returning from the ocean to spawn were cut off from their natal waters. Only the lower five miles of the 45-mile river were accessible to them. Over time the dams starved those five miles of the sediment salmon need for building their nests. The diminished and degraded habitat decimated fish populations in a river once known for producing 100-pound salmon.
The Lower Elwha Klallam tribe, which has relied on the river for sustenance and its spiritual traditions for thousands of years, has been fighting for nearly 100 years to have the dams removed. In the early 1990s, Congress finally authorized dam removal as a way to help restore the river and its salmon runs. But it would take another two decades to secure funding for the $300 million project.
Dam removal began in 2011 and will connect the headwaters of the Elwha with the mouth of the river. The project is expected to be completed by 2014. One of the major challenges in removing the dams was figuring out how to deal with all the sediment stored behind the dams – enough sediment to fill 11 NFL football stadiums. To deal with the sediment, construction crews are taking the dams down gradually, piece by piece, rather than dynamiting them. Even so, the river is swollen with sediment. And the physical shape of the river is changing as the river deposits sediment along its banks.
“This place is changing every day,” Duda said. “Every time I come out there’s something new, something different.”
In five to 10 years, the sediment is expected to make its way downstream and the river will run clear once again. But for now, salmon returning to the river face some harsh conditions. Sediment can clog their gills and make it difficult to find food.
“It’s almost like you’re tearing off a Band-Aid,” Duda said. “You have to go through a little bit of pain in order to get to that final state of healing.”
On this day, McMillan and Duda walked the main channel of the Elwha River between the lower dam and the upper dam. The swift-flowing river, suffused with fine sediment from dam removal, ran a milky gray. McMillan fixed a keen eye on the clearer water near shore.
“They dug here! These are digs!”
“So you think this [is a nest] right here?” Duda said.
“This is a redd clearly,” McMillan replied. “It’s the first redd we’ve found by a steelhead in the mainstem Elwha River.”
There’s still a long way to go, but McMillan and Duda are hopeful that salmon and steelhead will thrive here once again.<|endoftext|>
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# How do you solve $\log x + \log (x - 3) = 1$ ?
Last updated date: 21st Feb 2024
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Hint: In this particular question we need to use basic logarithmic properties to simplify the equation. Then we need to further solve the equation and get the desired answer.
Complete step by step solution:
In the above question, it is given that,
$\log x + \log (x - 3) = 1$
(Since $\log a + \log b = \log (ab)$ )
Using the above stated property we get,
$\Rightarrow \log (x(x - 3)) = 1$
Taking antilog on both sides of the equation we get,
$\Rightarrow anti\log (\log (x(x - 3))) = anti\log 1$
$\Rightarrow x(x - 3) = {10^1}$
On solving the above equation we get a quadratic equation
$\Rightarrow {x^2} - 3x = 10$
Subtracting 10 from both sides of the equation
$\Rightarrow {x^2} - 3x - 10 = 0$
Now solve the above quadratic equation for x
$\Rightarrow {x^2} - 5x + 2x - 10 = 0$
$\Rightarrow x(x - 5) + 2(x - 5) = 0$
$\Rightarrow (x + 2)(x - 5) = 0$
This implies that either $x = - 2$or $x = 5$
As log can not be a negative value therefore the $x = - 2$ is rejected and hence $x = 5$ is the required solution to the above logarithmic equation.
Note:
Remember to recall the basic logarithmic properties to solve the above question. Note that
$\log x = 1 \\ \Rightarrow x = {10^1} \\ \Rightarrow x = 10 \\$
The basic logarithmic algebra includes the following properties:
$\log a + \log b = \log (ab) \\ \log a - \log b = \log \left( {\dfrac{a}{b}} \right) \\ \log {a^b} = b\log a \\ {\log _a}a = 1 \\$<|endoftext|>
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Megalodon (/ˈmɛɡəloʊˌdɒn/ meg-ə-lə-don, /ˈmeɪɡə-/ may-gə-; meaning “big tooth”, from Ancient Greek: μέγας (megas) “big, mighty” and ὀδoύς (odoús), “tooth”—whose stem is odont-, as seen in the genitive case form ὀδόντος, odóntos) is an extinct species of shark that lived approximately 23 to 2.6 million years ago, during the Cenozoic Era (early Miocene to end of Pliocene).
The taxonomic assignment of C. megalodon has been debated for nearly a century, and is still under dispute.
- Extra: There may be more details in Wikipedia
- Scientific name: Carcharodon megalodon
- Class: Chondrichthyes
- Family: Lamnidae or †Otodontidae
- Rank: Species
- Phylum: Chordata
- Did you know: Megalodon hunted large and medium-sized whales, attacking the bony areas, such as chest, fins, or tail.<|endoftext|>
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