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6,978	# NCERT/CBSE Class 6th Maths Chapter 2 Whole Numbers NCERT/CBSE Class 6th Maths Chapter 2 Whole Numbers: NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers are comprehensive study material for students preparing for the Class 6 Mathematics exam. Page Contents ## NCERT/CBSE Class 6th Maths Chapter 2 Whole Numbers • Counting numbers are called natural numbers. Natural numbers are denoted by N and given by N = {1, 2, 3, 4, …} • Natural numbers included with ‘0’ are called whole numbers. Whole numbers are represented by W and given by W = {0, 1, 2, 3, 4, …} • A number which comes after a given number is called its successor. Successor = given number + 1 • A number which comes before a given number is called its predecessor. Predecessor = given number – 1 • Every whole number has a successor. Every whole number except 0, has a predecessor. • The following number line represents a whole number line on which whole numbers are represented. The distance between two points is called unit distance. Number operations of addition, subtraction and multiplication can easily be performed on number line. • Closure Property for addition and multiplication: If we add or multiply two whole numbers the result is again a whole number. Closure property does not hold good for subtraction and division of whole numbers. • Commutativity of addition and multiplication: Whole number can be added or multiplied in any order. That is, for any two whole numbers ‘a’ and ‘b’, a + b = b + a and a × b = b × a Commutative property does not hold good for subtraction and division of whole numbers. • Associativity of addition and multiplication: Whole numbers can be grouped for the convenience of adding or multiplying. That is, for any three whole numbers ‘a’ , ‘b’ and ‘c’,a + (b + c) = (a + b) + c and a × (b × c) = (a × b) × c Associative property does not hold good for subtraction and division of whole numbers. • Distributivity of multiplication over addition: For any three whole numbers ‘a’ , ‘b’ and ‘c’, a × (b + c) = (a × b) + (a × c) • Identity for addition: 0 is the identity for addition; this means when 0 is added to any whole the result is the whole number itself. • Identity for multiplication: 1 is the identity for multiplication; this means when 1 is multiplied with any whole number the result is the whole number itself. • Division of a whole number by zero is not defined. Whole Numbers The numbers 1,2, 3, are called natural numbers or counting numbers. Let us add one more number i.e., zero (0), to the collection of natural numbers. Now the numbers are 0,1,2, … These numbers are called whole numbers We can say that whole nos. consist of zero and the natural numbers. Therefore, except zero all the whole nos. are natural numbers. Facts of Whole numbers • The smallest natural number is 1. • The number 0 is the first and the smallest whole nos. • There are infinitely many or uncountable number of whole-numbers. • All natural numbers are whole-numbers. • All whole-numbers are not natural numbers. For example, 0 is a whole-number but it is not a natural number. The first 50 whole nos. are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25,26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50 Number line A number line is a picture of a graduated straight and horizontal line in which numbers are written. A number written on the left-hand side of the number line is lesser and number written on the right-hand side of the number line is greater. Lets us look into some solved example problems. In the above figure, we can see a number line where integers are placed. Here, the positive and negative integers are placed on either side of the zero. Find 12 × 35 using distributivity. 12 × 35 = 12 × (30 + 5) = 12 × 30 +12 × 5 = 360 + 60 = 420. Calculate – (2 + 3) + 4 =? = 5+ 4 = 9. A number line is a pictorial representation of numbers on a straight line. It’s a reference for comparing and ordering numbers. It can be used to represent any real number that includes every whole number and natural number. Just to recollect, the whole number is a set of numbers that include all counting numbers (1, 2, 3,4,5,6 …….) and zero (0), whereas the natural number is the set of all counting numbers i.e. 1, 2, 3, 4, 5, 6…….. Writing numbers on a number line make it easier to compare the numbers. From the above figure, we can see that the integers on the left side are smaller than the integers on the right side. For example, 0 is less than 1, -1 is less than 0, -2 is less than -1, and so on. Numbers on a Number Line Arithmetic operations of numbers can be better explained on a number line. To begin with, one must know to locate numbers on a number line. Zero is the middle point of a number line. All (natural numbers) positive numbers occupy the right side of the zero whereas negative numbers occupy the left side of zero on the number line. As we move on to the left side value of a number decreases. For example, 1 is greater than -2. In a number line, integers, fractions, and decimals can also be represented easily. Check out the links given below to learn more. Number Line 1 to 100 Here are the number lines from 1 to 100 numbers. As we already we can extend the number line indefinitely, thus students can also try to draw the line beyond 100, to practice. Addition on Number Lines Adding Positive Numbers When we add two positive numbers, the result will always be a positive number. Hence, on adding positive numbers direction of movement will always be to the right side. For example, addition of 1 and 5 (1 + 5 = 6) Here the first number is 1 and the second number is 5; both are positive. First, locate 1 on the number line. Then move 5 places to the right will give 6. Adding Negative Numbers When we add two negative numbers, the result will always be a negative number. Hence, adding negative numbers direction of movement will always be to the left side. For example, the addition of -2 and -3 Here, the first number is -2 and the second number is -3; both are negative. Locate -2 on the number line. Then move 3 places to the left will give -5. Some other Important terms to remember Properties of Whole Numbers Closure Property Closure property on Addition for Whole Number 0 + 2 = 2 1 + 3 = 4 5 + 6 = 11 So Whole number are closed on Addition Closure property on Multiplication for Whole Number 0 × 2 = 0 1 × 4 = 4 5 × 1 = 5 So Whole number are closed on Multiplication Closure property on subtraction of Whole number 5 – 0 = 5 0 – 5 =? 1 – 3 =? 3 – 1 = 2 So Whole number are not closed on Subtraction Closure property on Division of Whole number 21=2 12=? 02=0 20=? (Division by Zero is undefined) So Whole Number are not closed on Division In short Closure Property If a and b are any two whole numbers, then a + b, a × b are also whole numbers. Commutative property Commutativity property on Addition for Whole Number So Whole number are Commutative on Addition 0 + 2 = 2 + 0 = 2 Commutativity property on Multiplication for Whole Number 0 × 2 = 0 or 2 × 0 = 0 So Whole number are Commutative on Multiplication Commutativity property on subtraction of Whole number 5 – 0 = 5 but 0 – 5 =? So Whole number are not Commutative on Subtraction Commutativity property on Division of Whole number 21=2 but 12=? So Whole Number are not Commutative on Division In short You can add two whole numbers in any order. You can multiply two whole numbers in any order. Commutative property: If a and b are any two whole numbers, then a + b = b + a and a × b = b × a Associative property Associativity property on Addition for Whole Number 0 + (2 + 3) = (0 + 2) + 3 = 50 + (2 + 3) = (0 + 2) + 3 = 5 1 + (2 + 3) = 6 = (1 + 2) + 31 + (2 + 3) = 6 = (1 + 2) + 3 So Whole number are Associative on Addition Associativity property on Multiplication for Whole Number 0 × (2 × 3) = 0 or (0 × 2) × 3 = 0 So Whole number are Associative on Multiplication Associativity property on subtraction of Whole number 10 – (2 – 1) = 9 but (10 – 2) – 1 = 7 So Whole number are not Associative on Subtraction Associativity property on Division of Whole number 16 ÷ (4 ÷ 2) = 8 but (16 ÷ 4) ÷ 2 = 2 So Whole Number are not Associative on Division So in Short If a, b and c are any two whole numbers, then (a + b) + c = a + (b + c) and (a × b) × c = a × (b × c). Distributive property If a, b and c are any two whole numbers, then a (b + c) = a × b + a × c Additive Identity If a is any whole number, then a + 0 = a = 0 + a Example 2 + 0 = 2 0 + 3 = 3 5 + 0 = 5 Multiplicative Identity If a is any whole number, then a × 1 = a = 1 × a Example 1 × 1 = 1 5 × 1 = 5 6 × 1 = 6 Multiplication by zero If a is any whole number, then a × 0 = 0 = 0 × a Example 1 × 0 = 0 5 × 0 = 0 0 × 0 = 0 Division by zero If a is any whole number, then a ÷ 0 is not defined Important Questions Multiple Choice Questions: 1. What is the successor of 2001? 1. 2003 2. 2001 3. 2002 4. 2000 5. The largest 5-digit number having three different digits is: 1. 98978 2. 99897 3. 99987 4. 98799 5. The difference between 85 and the number obtained by reversing the digits is: 1. 25 2. 26 3. 27 4. 72 5. The natural numbers along with zero form the collection of: 1. Whole numbers 2. Integers 3. Rational numbers 4. Real numbers 5. Find value of 297 × 17 + 297 × 3: 1. 5940 2. 5980 3. 5942 4. 5970 5. The smallest whole number is: 1. 1 2. 0 3. not defined 4. None of these 5. Find product 12 × 35: 1. 12600 2. 34840 3. 420 4. 400 5. The difference between the smallest 3 digit number and the largest two digit number is: 1. 0 2. 1 3. 2 4. None of these 5. What is the quotient of 64 ÷ 1? 1. 1 2. 0 3. 46 4. 64 5. What are the three consecutive predecessors of 70010? 1. 70009, 70008, 7007 2. 70009, 70008, 70010 3. 70009, 70008, 70007 4. 70009 5. 3 × 10000 + 7 × 1000 + 9 × 100 + 0 × 10 + 4 is the same as: 1. 3794 2. 37940 3. 37904 4. 379409 5. Study the pattern 1 × 8 + 1 = 9 12 × 8 + 2 = 98 Next step is- 1. 123 × 8 + 3 = 987 2. 1234 × 8 + 4 = 9876 3. 120 × 8 + 3 = 963 4. 13 × 8 + 3 = 987 5. Number of whole numbers between 38 and 68 is: 1. 31 2. 30 3. 28 4. 29 5. The value of 25 × 20 × 5 is: 1. 2505 2. 2503 3. 2500 4. None of these 5. Value of 0 ÷ 5 is: 1. 5 2. 0 3. 1 4. None of these Match The Following: Fill in the blanks: 1. Division by _____ is not defined. 2. A number remains unchanged when added to ______. 3. A number remains unchanged when multiplied to ______. 4. 13 × 100 × ________ = 1300000 True/ False: 1. All natural numbers are whole numbers. 2. All whole numbers are natural numbers. 3. The predecessor of a two digit number is never a single digit number. 4. 1 is the smallest whole number. Very Short Questions: 1. How many whole numbers are there between 32 and 53? 2. Are all whole numbers also natural numbers? 3. Complete pattern 1 1 = 1 11 11 = 121 111 111 = _______ 1111 1111 = 1234321 1. Write the next three consecutive whole numbers of the following numbers: 1. 39359 2. 8632157 1. A taxi driver filled his car petrol tank with 40 litres of petrol on Monday. The next day, he filled the tank with 50 litres of petrol. If the petrol costs Rs.44 per litre, how much did he spend in all on petrol? 2. Find the product by suitable rearrangement: 1. 8 × 391 × 125 2. 2 × 1234 × 50 1. If you’re on a diet and have a breakfast consisting of 150 calories, a lunch consisting of 350 calories, and a dinner consisting of 1000 calories, then find the sum of the calories consumed that day. 2. Write the smallest whole number. 3. What is the predecessor of whole number 0? 4. Which property do the following statements hold? (a) 6 + 4 = 4 + 6 (b) 3 + 2 = whole number 1. Add the following in three ways. Indicate the property used. (a) 25 + 36 + 15 (b) 30 + 18 + 22 1. Using distributive property, solve the following: (а) 360 × 102 (b) 35 × 98 1. Find the product of the greatest 3-digit number and the smallest 2-digit number. 2. Write the predecessor of the smallest 4-digit number. 3. For n = 5, verify the given statement 10 × n + 1 = n1 Short Questions: 1. Using the properties, find the values of each of the following: (a) 736 × 102 (b) 8165 × 169 – 8165 × 69 1. Observe the following patterns and extend them by two more terms. 1. Observe the following patterns and extend them by two more terms: 1. Using the properties of whole numbers, find the value of the following in suitable way: (a) 945 × 4 × 25 (b) 40 × 328 × 25 1. Represent the following on number line: (a) 3 + 4 (b) 6 – 2 (c) 2 × 4 1. Give one example for each of the following properties for whole numbers. (a) Closure property (b) Commutative property (c) Associative property (d) Distributive property 1. A dealer purchased 124 LED sets. If the cost of one set is ₹38,540, determine their total cost. 2. Find the product of the greatest 3-digit number and the greatest 2-digit number. 3. Write 10 such numbers which can be shown only as line. Long Questions: 1. 320 km distance is to be covered partially by bus and partially by train. Bus covers 180 km distance with a speed of 40 km/h and the rest of the distance is covered by the train at a speed of 70 km/h. Find the time taken by a passenger to cover the whole distance. 2. Solve the following and establish a pattern: (a) 84 × 9 (b) 84 × 99 (c) 84 × 999 (d) 84 × 9999 1. Solve the following with suitable and short-cut method: (a) 86 × 5 (b) 86 × 15 (c) 86 × 25 (d) 86 × 35 (e) 86 × 50 (f) 96 × 125 (g) 96 × 250 (h) 112 × 625 1. Ramesh buys 10 containers of juice from one shop and 18 containers of the same juice from another shop. If the capacity of each container is same and the cost of each of the container is ₹150, find the total money spend by Ramesh. 2. A housing complex built by DLF consists of 25 large buildings and 40 small buildings. Each large building has 15 floors with 4 apartments on each floor and each small building has 9 floors with 3 apartments on each floor. How many apartments are there in all? 3. A school principal places orders for 85 chairs and 25 tables with a dealer. Each chair cost ₹180 and each table cost ₹140. If the principal has given ₹2500 to the dealer as an advance money, then what amount to be given to the dealer now? Assertion Reason Questions: 1. Assertion (A) – 17 is the successor of 16 Reason (R) – any natural number, you can add 1 to that number and get the next number i.e. you get its successor. 1. Both A and R are true and R is the correct explanation of A 2. Both A and R are true but R is not the correct explanation of A 3. A is true but R is false 4. A is false but R is true 5. Assertion (A) –99 is the predecessor of 100 Reason (R) – any natural number, you can substract 1 to that number and get the next number i.e. you get its predecessor. 1. Both A and R are true and R is the correct explanation of A 2. Both A and R are true but R is not the correct explanation of A 3. A is true but R is false 4. A is false but R is true ANSWER KEY – Multiple Choice questions: 1. C. 2002 2. C. 99987 3. C. 27 4. A. Whole numbers 5. A. 5940 6. B. 0 7. C. 420 8. B. 1 9. D. 64 10. C. 70009, 70008, 70007 11. C. 37904 12. A. 123 × 8 + 3 = 987 13. D. 29 14. C. 2500 15. B. 0 Match The Following: Fill in the blanks: 1. Division by 0 is not defined. 2. A number remains unchanged when added to zero. 3. A number remains unchanged when multiplied to 1. 4. 13 × 100 × 1000 = 1300000 True /False: 1. True 2. False, every whole number except 0 is a natural number. 3. False, the predecessor of 10 is 9. 4. True Very Short Answer: 1. There are 20 whole numbers between 32 and 53. These are 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51 and 52. 2. No, all whole numbers are not natural numbers. (0 is a whole number but not a natural number.) 3. 12321 1. The next three consecutive whole numbers of 39359 are: 39360, 39361, 39362 2. The next three consecutive whole numbers of 8632157 are: 8632158, 8632159, 8632160. 3. Petrol filled on Monday = 40 liters. Petrol filled the next day = 50 liters. ∴ Total petrol filled on the two days = 40 liters + 50 liters = 90 liters. ∴ Cost of petrol per liter = Rs. 44 ∴ Cost of 90 liters petrol = Rs. 44 × 90 = Rs. 3960. 1. 8 × 391 × 125 = 391 × (125 × 8) = 391 × 1000 = 391,000 2. 2 × 1234 × 50 =1234 × (2 × 50) = 1234 × 100 = 123,400 3. Breakfast consisting of 150 calories. Lunch consisting of 350 calories. Dinner consisting of 1000 calories. The sum of the calories consumed that day is 150 + 350 + 1000 = 1500 calories. 1. 0 is the smallest whole number. 2. Whole number 0 has no predecessor. 3. (a) 6 + 4 = 4 + 6 holds commutative property of addition. (b) 3 + 2 = whole number holds closure property. 1. (a) 25 + 36 + 15 Way I: 25 + (36 + 15) = 25 + 51 = 76 Way II: (25 + 36) + 15 = 61 + 15 = 76 Way III: (25 + 15) + 36 = 40 + 36 = 76 Here, we have used associative property. (b) 30 + 18 + 22 Way I: 30 + (18 + 22) = 30 + 40 = 70 Way II: (30 + 18) + 22 = 48 + 22 = 70 Way III: (30 + 22) + 18 = 52 + 18 = 70 Here, we have used associative property. 1. (a) 36 × 102 = 36 × (100 + 2) = 36 × 100 + 36 × 2 = -36000 + 72 = 36072 (b) 35 × 98 = 35 × (100 – 2) = 35 × 100 – 35 × 2 = 3500 – 70 = 3430 1. The greatest 3-digit number = 999 The smallest 2-digit number = 10 ∴ Product = 999 x 10 = 9990 1. The smallest 4-digit number = 1000 ∴ The predecessor of 1000 = 1000 – 1 = 999 1. Given statement is 10 x n + 1 = n1 Put n = 5, 10 x 5 + 1 = 51 ⇒ 50 + 1 = 51 ⇒ 51 = 51. Hence, verified. Short Answer: 1. (a) 736 × 102 = 736 × (100 + 2) = 736 × 100 + 736 × 2 [Using distributive property] = 73600 + 1472 = 75072 (b) 8165 × 169 – 8165 × 69 = 8165 × (169 – 69) [Using distributive property] = 8165 × 100 = 816500 1. Next two terms are 1010101 × 1010101= 1020304030201 101010101 × 101010101=10203040504030201 1. Next two terms are 15873 × 7 × 3 = 333333 15873 × 7 × 4 = 444444 1. (a) 945 × 4 × 25 = 945 × (4 × 25) = 945 × 100 = 94500 (b) 40 × 328 × 25 = 328 × (40 × 25) = 328 × 1000 = 328000 (a) 3 + 4 (b) 6 – 2 (c) 2 × 4 (a) 3 + 4 = 7 (whole number) closure property (b) 4 + 5 = 5 + 4 Commutative property (c) 3 + (5 + 7) = (3 + 5) + 7 Associative property (d) 6 × (8 + 3) = 6 × 8 + 6 × 3 Distributive property. 1. Total cost of 124 LED sets = ₹ (38,540 × 124) = ₹ [38,540 × (100 + 20 + 4)] = ₹ [38,540 × 100 + 38,540 × 20 + 38,540 × 4] = ₹ [38,54,000 + 7,70,800 + 1,54,160] = ₹ 47,789,6 1. Greatest 3-digit number = 999 Greatest 2-digit number = 99 ∴ Product = 999 × 99 = 999 × (100 – 1) = 999 × 100 – 999 × 1 = 99900 – 999 = 98901 1. 2, 5, 7, 11, 13, 17, 19, 23, 29 and 31 are such numbers which can be shown only as line. 123 × 9 + 4 = 1111. Long Answer: 1. Total distance = 320 km Distance covered by the bus = 180 km Speed of the bus = 40 km/h ∴ Time Taken by the bus =DistanceSpeed =18040hours=92hours Distance covered by the train = 320 – 180 = 140 km. Speed of the train = 70 km/h ∴ Time taken by the train =DistanceSpeed=14070 hours = 2hours Hence, the total time taken by the passenger = 92 hours + 2 hours = 4 hours 30 min + 2 hours = 6 hours 30 min 1. (a) 84 × 9 = 84 × (10 – 1) = 84 × 10 – 84 × 1 = 840 – 84 = 756 (b) 84 × 99 = 84 × (100 – 1) = 84 × 100 – 84 × 1 = 8400 – 84 = 8316 (c) 84 × 999 = 84 × (1000 – 1) = 84 × 1000 – 84 × 1 = 84000 – 84 = 83916 (d) 84 × 9999 = 84 × (10000 – 1) = 84 x 10000 – 84 × 1 = 840000 – 84 = 839916 1. (a) 86×5=86×102=43×10=430 (b) 86×15=86×302=43×30=43×10×3=430×3=1290 (c) 86×25=86×1004=43×50=43×10×5=430×5=2150 (d) 86×35=86×702=43×70=43×10×7=430×7=3010 (e) 86×50=86×1002=43×100=4300 (f) 96×125=96×10008=12×1000=12000 (g) 96×250=96×10004=24×1000=24000 (h) 112×625=112×1000016=7×10000=70000 1. Ramesh buys 10 containers from one shop Cost of 1 container = ₹ 150 He buys 18 containers of the same capacity from another shop. Cost of 1 container = ₹150 ∴ Total money spent by Ramesh = ₹ [10 × 150 + 18 × 150] = ₹150 × (10 + 18) = ₹150 × 28 = ₹4200 1. Number of large buildings = 25 Number of floors = 15 Number of apartments on each floor = 4 ∴ Total number of apartments in large buildings = 25 × 15 × 4 Number of small building = 40 Number of floors = 9 Number of apartments on each floor = 3 ∴ Total number of apartments in small buildings = 40 × 9 × 3 Hence, the number of apartments in all = 25 × 15 × 4 + 40 × 9 × 3 = 1500 + 1080 = 2580. 1. Number of chairs = 85 Cost of one chair = ₹ 180 Cost of 85 chairs = ₹ (85 × 180) Number of tables = 25 Cost of one table = ₹ 140 Cost of 25 tables = ₹ (25 × 140) Total cost of all chairs and tables = ₹ (85 × 180 + 25 × 140) = ₹ (15300 + 3500) = ₹ 18800 Money given in advance = ₹ 2500 ∴ Balance money to be paid to the dealer = ₹ 18800 – ₹ 2500 = ₹ 16300 Assertion Reason Answers: 1) a) Both A and R are true and R is the correct explanation of A.2) a) Both A and R are true and R is the correct explanation of A. Sharing:<\|endoftext\|>	4.90625
691	# How do I find the derivative of y = arccos((x-3)^2)? $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left({\cos}^{-} 1 {\left(x - 3\right)}^{2}\right) = \frac{- 2 x + 6}{\sqrt{1 - {\left(x - 3\right)}^{4}}}$ #### Explanation: The formula to find the derivative of ${\cos}^{-} 1 u$ is $\frac{d}{\mathrm{dx}} \left({\cos}^{-} 1 u\right) = - \frac{1}{\sqrt{1 - {u}^{2}}} \cdot \frac{d}{\mathrm{dx}} \left(u\right)$ So from the given $y = {\cos}^{-} 1 {\left(x - 3\right)}^{2}$ Let $u = {\left(x - 3\right)}^{2}$ from the formula $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left({\cos}^{-} 1 u\right) = - \frac{1}{\sqrt{1 - {u}^{2}}} \cdot \frac{d}{\mathrm{dx}} \left(u\right)$ $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left({\cos}^{-} 1 {\left(x - 3\right)}^{2}\right) = - \frac{1}{\sqrt{1 - {\left({\left(x - 3\right)}^{2}\right)}^{2}}} \cdot \frac{d}{\mathrm{dx}} \left({\left(x - 3\right)}^{2}\right)$ $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left({\cos}^{-} 1 {\left(x - 3\right)}^{2}\right) =$ $- \frac{1}{\sqrt{1 - {\left(x - 3\right)}^{4}}} \cdot 2 \left(x - 3\right) \cdot \frac{d}{\mathrm{dx}} \left(x - 3\right)$ $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left({\cos}^{-} 1 {\left(x - 3\right)}^{2}\right) = \frac{- 2 \left(x - 3\right)}{\sqrt{1 - {\left(x - 3\right)}^{4}}} \cdot \left(1\right)$ $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left({\cos}^{-} 1 {\left(x - 3\right)}^{2}\right) = \frac{- 2 x + 6}{\sqrt{1 - {\left(x - 3\right)}^{4}}}$ have a nice day!<\|endoftext\|>	4.53125
722	# How to Solve Division Problems We hope that the ideas shared in this article will help you and your students to succeed in this potentially challenging topic. If you do find these tips helpful, check out what Happy Numbers offers its users by setting up your class and starting a free trial which is available only this week! Learn the parts of a division problem and how to solve them in a few easy steps. You have 20 cookies and 10 friends. How many cookies should you give each of your friends? This is a basic division problem. Division is one of the four basic operations: addition, subtraction, and multiplication are the other three. Division is a simple operation in which a number is divided. It’s easiest to think of it as a number of objects being divided among a certain number of people, such as in the above example. Of course, you always want to give the same amount to each person to be fair! That’s basically how division works, you divide numbers into equal groups of numbers. So, how can you solve a division problem? First, you have to know the parts of a division problem. ## Parts of a Division Problem There are three main parts to a division problem: the dividend, the divisor, and the quotient. The dividend is the number that will be divided. The divisor is the number of “people” that the number is being divided among. The quotient is the answer. ## How to Solve Division Problems Solving simple division problems is closely linked to multiplication. In fact, to check your work, you’ll have to multiply the quotient by the divisor to see if it equals the dividend. If it doesn’t, you’ve solved incorrectly. Let’s try solving one simple division problem. For example: 12 ÷ 2 = In this problem, you can see how Happy Numbers helps children visualize the problem. There are 12 oranges. There are 2 placed in each box. How many boxes are there? The answer is 6. You can check the answer by multiplying the quotient, 6, by the divisor, 2, (6 x 2) which gives us 12. So, the answer is correct. ## What Is a Remainder in Math? You may have heard of a remainder and wondered, what is a remainder in math? A remainder in math is used when a division problem doesn’t come out evenly. For example: 11 ÷ 4 = As you can see in the above example of tennis balls, first, the balls are divided into groups of 4. However, after making 2 groups of balls, 3 balls are left that can’t make a group of 4. These are the remainder. So, the quotient is 2 (2 groups of 4 can be made) and the remainder is 3. To check the work, multiply the quotient, 2, times the divisor, 4. The answer is 8. Then, add the remainder of 3. The answer is 11, which was the original dividend, so the answer is correct. Division can get more and more complicated as the numbers get bigger. Then, you must use strategies such as long division, estimation, and more to determine the answers. However, with these basic steps, you can solve just about any division problem. These are just a few examples of how Happy Numbers incorporates the ideas of small- group instruction into its curriculum. To find out more and try them out, register as a teacher on the website to start a free trial period which is available only this week!<\|endoftext\|>	4.625
2,212	Diabetes mellitus is a group of metabolic diseases in which the person has high blood glucose (blood sugar) level either due to inadequate insulin production or because the body’s cells do not respond properly to insulin or both. The term “Diabetes Mellitus” describes a metabolic disorder of multiple etiology characterized by chronic hyperglycemia with disturbances of carbohydrate, fat (dyslipidaemia) and protein metabolism resulting from defects in insulin secretion, insulin action, or both. The main symptoms are: – - Polyuria (frequent urination) - Polydipsia (increased thirst) - Polyphagia (increased hunger The main types of diabetes are: Type 1 diabetes: It is due to the body’s malfunction to produce insulin in the body, and requires the person to inject insulin. This form was previously referred to as “Insulin-Dependent Diabetes Mellitus” (IDDM) or “Juvenile Diabetes”. Type 2 diabetes: It is due to insulin resistance, a condition in which cells fail to use insulin properly, sometimes combined with an absolute insulin deficiency. This form was previously referred to as non insulin-dependent diabetes mellitus (NIDDM) or “adult-onset diabetes”. Type 2 diabetes can be prevented after following healthy life style such as healthy diet, proper exercise or maintaining healthy weight. The third main form, Gestational diabetes occurs when pregnant women without a previous diagnosis of diabetes develop a high blood glucose level. It may lead to type 2 DM. Other types of diabetes include those caused by: - Genetic defects of the beta cells, (the part of the pancreas that makes insulin) such as maturity-onset diabetes of the young (MODY) or neonatal diabetes mellitus (NDM) - Diseases of the pancreas or conditions that damage the pancreas, such as pancreatitis and cystic fibrosis - Excess amounts of certain hormones resulting from some medical conditions such as cortisol in Cushing’s syndrome that work against the action of insulin - Medications that reduce insulin action, such as glucocorticoids, or chemicals that destroy beta cells The main symptoms of diabetes are: - Polyuria: urinating frequently (particularly at night) - Polydipsia: feeling very thirsty - Polyphagia: feeling hungry frequently - Weight loss and loss of muscle bulk - Frequent episodes of thrush - Cuts or wounds that heal slowly - Blurred vision Type 1 diabetes can develop quickly, over weeks or even days. Many people have type 2 diabetes for years without realizing because early symptoms tend to be common. Type 1 Diabetes: The immune system of body attacks and destroys the cells that produce insulin. As no insulin is produced, glucose levels further increase, which can seriously damage the body’s organs. Type 1 diabetes is often known as insulin-dependent diabetes. It is also sometimes known as juvenile diabetes or early-onset diabetes because it usually develops before the age of 40, often during the teenage years. Type 1 diabetes is less common than type 2 diabetes. Type 2 Diabetes: Type 2 diabetes is where the body does not produce enough insulin or the body’s cells do not respond to insulin. This is known as insulin resistance. Type 2 diabetes, and is far more common than type 1 diabetes. Risk factors for type 2 diabetes: - Obesity or being overweight - Impaired glucose tolerance - High blood pressure - Dyslipidemia – Low levels of high-density lipoproteins (HDL) (“good”) cholesterol and high levels of triglycerides, high low-density lipoproteins (LDL) - Gestational diabetes - Sedentary lifestyle - Family history Gestational Diabetes: Some women tend to experience high levels of blood glucose as during pregnancy due to reduced sensitivity of insulin receptors. The clinical diagnosis of diabetics is often prompted by symptoms such as increased thirst and urination and recurrent infections. Blood Tests – Fasting plasma glucose, two-hour postprandial test and oral glucose tolerance test are done to know blood glucose levels. Glycated Haemoglobin (HbA1c) may be used to diagnose diabetes(if facilities are easily available). Diabetes can be diagnosed by blood glucose and HBA1c levels: \|Diabetes diagnostic criteria\| \|Condition\|\|2 hour* plasma glucose\|\|Fasting plasma glucose\|\|HbA1c\| \|Normal\|\|<7.8 (<140)\|\|<6.1 (<110)\|\|<6.0\| \|Impaired Fasting Glucose\|\|<7.8 (<140)\|\|≥ 6.1(≥110) & <7.0(<126)\|\|6.0–6.4\| \|Impaired Glucose Tolerance\|\|≥7.8 (≥140)\|\|<7.0 (<126)\|\|6.0–6.4\| \|Diabetes mellitus\|\|≥11.1 (≥200)\|\|≥7.0 (≥126)\|\|≥6.5\| - Fasting lipid profile, including total, LDL, and HDL cholesterol and triglycerides - Liver function tests - Kidney function tests - Thyroid stimulating hormone (TSH) in type 1 diabetes, dyslipidemia, or women over 50 years of age. Currently, six classes of oral antidiabetic drugs (OADs) are available: biguanides (e.g., metformin), sulfonylureas (e.g., glimepiride), meglitinides (e.g., repaglinide), thiazolidinediones (e.g., pioglitazone), dipeptidyl peptidase IV inhibitors (e.g., sitagliptin), and α-glucosidase inhibitors (e.g., acarbose). - Insulin: Type 1 diabetes is generally treated with combinations of regular and NPH (neutral protamine Hagedorn) insulin or synthetic insulin analogs. When insulin is used in type 2 diabetes, a long-acting formulation is usually added initially while continuing oral medications. Treatment of coexisting medical conditions (high blood pressure, dyslipidemia etc.) - Regular exercise - Proper diet - No smoking - No alcohol The complications of diabetes mellitus are less common and less severe in people who have well-controlled blood sugar levels. Its complications are: - Diabetic ketoacidosis (DKA): It is an intense and dangerous complication that can always result in a medical emergency. It is generally seen due to low insulin levels which may cause the liver to turn fatty acid to ketone for fuel as ketone bodies are intermediate substrates in that metabolic sequence. This is a normal condition if occurs periodically, but can become a serious problem if sustained. Elevated levels of ketone bodies in the blood decrease the blood’s pH leading to DKA. The patient with DKA is typically dehydrated and breathing rapidly and deeply. Abdominal pain is common and may be severe. - Hyperglycemia: Hyperglycemia is another acute complication. If a person has very high (usually considered to be above 300 mg/dl (16 mmol/L)) blood glucose levels, water is osmotically drawn out of cells into the blood and the kidneys eventually begin to dump glucose into the urine. This results in loss of water and an increase in blood osmolarity. If fluid is not replaced (by mouth or intravenously) the osmotic effect of high glucose levels combined with the loss of water will eventually lead to dehydration. The body’s cells become progressively dehydrated as water is taken from them and excreted. Electrolyte imbalances are also common and can be very dangerous. - Hypoglycemia: Hypoglycemia or abnormally low blood glucose is an acute complication of several diabetes treatments. It is rare otherwise, either in diabetic or non-diabetic patients. The patient may become agitated, sweaty, weak, and have many symptoms of sympathetic activation of the autonomic nervous system resulting in feelings akin to dread and immobilized panic. - Diabetic Coma: Diabetic coma is a medical emergency in which a person with diabetes mellitus is unconscious as of one of the acute complications of diabetes: - Severe diabetic hypoglycemia - Diabetic ketoacidosis advanced enough to result in unconsciousness from a combination of severe hyperglycemia, dehydration and shock and exhaustion - Hyperosmolar nonketotic coma in which extreme hyperglycemia and dehydration alone are sufficient to cause unconsciousness. (a)Micro vascular diseases (due to damage to small blood vessels): 1. Eye Diseases: - Retinopathy (nonproliferative/proliferative) - Macular oedema These conditions can lead to serious vision loss or blindness. - Sensory and motor(mono-and polyneuoropathy) neuropathy: sensory symptoms such as tingling , burning, stabbing pain or other abnormal sensations and motor symptoms such as sensory loss, weakness, numbness starting from feet and later on can involve finger and arm. - Autonomic neuropathy affects the digestive system, blood vessels, urinary system as well as sex organs. 3. Nephropathy: Damage to kidney can lead to chronic renal failure. (b) Macro vascular diseases: - Coronary artery disease - Peripheral arterial disease - Cerebrovascular disease (c) Other diseases: - High blood sugar level in blood can lead to cataract, glaucoma. - Dermatological complications- bacterial infections, fungal infections and some other skin conditions are more common in diabetic persons. - Infectious diseases are more frequent and/or serious in patients with diabetes mellitus. - Periodontal Disease-When diabetes is not controlled properly high glucose levels in mouth fluids may help germs to develop and cause periodontal diseases. - Cheiroarthropathy is a condition of limited joint mobility in patients with diabetes and is characterized by thickening of the skin resulting in contracture of the fingers. - Diabetic dyslipidemia: The characteristic features are a high plasma triglyceride concentration, low high density liprotein (HDL) concentration and increased concentration of low density lipoproteins (LDL).<\|endoftext\|>	3.671875
2,525	NCERT Class 10 Science Lab Manual – Introduction Hands-on experience promotes curiosity, provides opportunity for discussion and enhances enquiry skills. This helps the students to relate things scientifically and make sense of what they learn. The experiments should be conducted in a scientific method, i.e., carefully collect the evidences by observation and draw a conclusion from these evidences. - Experiments connect the theoretical world of textbooks to the real concrete world of experience. - It helps in creating interest for science among students. - It caters to the need of all types of learners, VAK, e., Visual, Auditory and Kinesthetic learners and enhances basic skills like enquiry skills, observation skills, manipulative skills, communication skills, skills of representing and interpreting data, etc. - Brief Up the Students: Students should have brief knowledge about the experiment that will be conducted in the laboratory for the practical period before they leave their classroom for laboratory. - Students must carry practical file/auxilliary copy, lab manual, stationery required (pencil, eraser, scale, protractor, compass, slide mounting material, pen, etc.) - Discipline and Punctuality: On their way to laboratory the students should maintain silence. In the lab, they should be assigned a desk/work station and the place should be fixed/permanent for the academic session. A student should remain present for the practical work and reach lab on time. A week for make-up of lost/absent experiments can be given to those students who were absent due to medical reason. Each student should keep its work station and surrounding area clean and tidy, similarly the practical files should also be kept neat and tidy. Knowledge of Apparatus - All the apparatus should be handled carefully and keep them at their respective place. - Each student should know how to select the right apparatus and correct way of using them. While performing the experiments, record the data in your copy (practical file/auxilliary note book), analyse the data and try to draw conclusions. Do not manipulate the readings. Reading: Take correct readings of the experiment performed by you. Recording: Record the readings with pencil, write units. Show the readings to the teacher. If it is wrong you may do the experiment again and record the new readings by erasing the previous one. Maintain your practical file neatly as shown below: - Never taste any chemical in the lab. - Never smell any gas released during a chemical reaction directly (sniff it or waft). - Never touch chemicals. - Never use cracked apparatus. - Never enter the lab in the absence of a teacher. - Always read the label on the chemical bottles before using it. - Use very little amount of chemicals. - You should know the main on/off connection of the gas (LPG/CNG). - Use a test tube holder and a hard glass apparatus for heating experiments. - Keep the mouth of the test tube away from your body. - If you smell the leakage of gas, immediately report to the teacher. - Stay away from concentrated acids. Use them only in the presence of a teacher. - You should know the location of fire-extinguisher, first-aid box and call for a teacher during an emergency. - Follow the general rules/instructions for the chemistry lab. - Do not try to dilute the acid by yourself. (For dilution add acid into water) - Do not open the tap completely as the water may come with pressure and splash over the work station area. - Do not wash the hot apparatus immediately after heating as it may crack/break. - Do not inhale gases or vapours directly. Always use your hand to waft fumes towards your nose. - Always use clean apparatus. - Wash the apparatus once the experiment is over. - Do not throw the chemicals in sink or dustbin, use separate container to throw the chemical waste - Keep the burners knob off when not in use. - Use tongs to hold hot objects. - Keep your work station clean. Recording the Observations - Enter all your observations in the practical file/auxilliary copy. - Record your readings with pencil and get them cross-checked by your teacher. - Maintain your practical file as instructed earlier. - Before you start with the next experiment get your previous experiment checked. Common Apparatus used in Chemistry Lab First Aid in Laboratory In Physics experiments we generally try to verify the laws, “establish the relationship”, “observe and compare things” A student must have a thorough knowledge of the theory/concepts for performing the experiment. He should know about the apparatus required for the experiment and how to handle it. - Record your readings neatly in the practical file with pencil. - Mention the units for each reading. - Repeat the observations and cross check the readings recorded. Do calculations to get the desired result. Draw neatly labelled and proportionate diagram for the experiment. Plot the graph wherever required. Label x-axis and y-axis properly. Correct way of measu ring the angle of incidence and angle of reflection. Discuss the doubts in the experiment and solve the MCQs all by yourself. Least count of an instrument: It is the minimum value that can be measured by an instrument. For example, if a measuring cylinder as 10 divisions between 10 mL to 20 mL then least count of measuring cylinder is (20 – 10)/10 = 1 mL. So, we can measure minimum 1 mL using this measuring cylinder. When the pointer of the measuring instrument don’t coincide with zero of the scale at rest, then the instrument has zero error. 1 kg = 1000 g 1 Newton = kg x 9.8 m/s2 (g = 9.8 m/s2) 1 kgf = 1 kg x 9.8 newton. 1 cc of water = 1 gf. Common Apparatus used in Physics Lab 1. Spring balance (a) It is used to measure the weight of an object. (b) It has a graduated scale and unit marked on it. (c) The unit may be kgf or N. The least count of spring balance should be studied and calculated before using it. Sources of error - The object/mass suspended should not oscillate/swing. - Do not handle/hold the spring balance on its side, hold it only at the top hook. - Parallax error — while recording the readings, the eye level and the reading should be parallel and coincide. 2. Measuring cylinder (a) It can be of glass/plastic. (b) Handle the glass measuring cylinder carefully. (c) Read the graduated scale and find the least count. Sources of error - It should be clean and placed on the plain table. - The scale marking should be readable. - While recording the readings, for liquid read the lower meniscus. - The eye-level and liquid level should be parallel. 3. Overflow can (a) It is used to verily Archimedes principle. It can be made of in glass, metal or plastic. (b) The outlet duct should not be blocked. (c) It should be filled above the level of duct and allow the extra water to flow into a beaker, to get the desired level of water. (d) It should be placed on plain table. 4. Tuning fork (a) It is used in experiments of sound, it is made up of metal alloys. The vibration of tuning fork produces sound. (b) The tuning fork is always held at the edge of the handle and banged on the rubber pad for vibrations. 5. Stop watch It is used to record the time in seconds. Follow the instructions given below for biology experiments. Before you enter a lab, be prepared for the experiment and read about the experiment that has to be performed. Read the experiments so that you get your thoughts organised about the experimental work and its reporting. Keep your desk top/work station area neat and clean. Always wash your hands when you are finished with practical. Be careful when you are working with blade, knife, cutter, scissors, etc. Diagram: Draw neat, proportionate and labelled diagrams wherever required. Read/observe the experimental work carefully i.e. specimens, slides, set-ups and record your observations in your practical file. Draw labelled diagram of each specimen in blank page on right hand page side. Common Equipments Used in Biology Lab - Microscopes: These are high powered, extremely expensive and sensitive pieces of equipment. Our naked eye can see 0.2 mm object i.e., 200 micrometer size. For objects smaller than 0.2 mm we use microscope. There are different types of microscopes with different magnifying power. (a) Hand lens: It is a biconvex lens mounted on a handle. It is based on the principle that when an object is placed between convex lens and its principal focus then an enlarged virtual image is formed. It is of different magnifying power 2X, 4X, 5X, etc. (b) Simple (dissecting) microscope: It has a single lens system. A single lens is mounted on a metallic frame and it can be moved up or down or sideways to get a magnified image. Its magnification ranges from 5X to 50X. (c) Compound microscope: Commonly used microscope in lab is compound microscope. It uses visible light to help us observe the specimens. Two lenses of different power are used to obtain the desired magnification. The magnification is achieved when light rays from any source of light fall on the concave mirror which then reflects the light into the condenser tube allowing it to pass through the slide and then through two lenses which magnify the image of slide taken as specimen. The lens near the eye is called eye lens and the lens near the object is called objective lens. To calculate the total magnification of a specimen multiply the power of objective lens with the eyepiece:Objective lens can have power of 10X (low power), 40X (big) and 100X (very big power) Eyepiece used is of order 10X E.g. for calculation of the power of magnification. - Slides: The permanent slides have a sample of blood, cell, muscles etc. with cover slips placed on it, these samples are fixed with a fixative on the slide. Temporary slides are those slides which a student make while performing an experiment, e.g. stomata, cheek cells, onion peel, blood, etc. - Petri dish: It is used to grow a culture or to carry out an experiment. - Watch glass: It is used to keep and stain the specimen. Forceps Used to hold tissues or pick up the structures for slide. - Needle: It is used to lift small cells or structures for slide. - Brush: It is used to transfer delicate specimen from one apparatus to another. - Dropper: Used to drop stain/water slowly during slide preparation. - Stain: These are the agents that colour the structures of cell and allow us to see them more clearly. - Stain: make invisible structures visible. Some common stains used are: (a) Safranin: is used to stain slides made of plant structures. (b) Iodine: Used to stain carbohydrates, flagella, nuclei etc. of the cells. (c) Methylene blue: used to stain cells, tissues, etc. (d) Eosin Y: used to stain cells, tissues, etc. - Preserved specimens: The animal specimens are preserved in a glass jar with formalin solution used as a preservative. While studying the specimens, find out the characteristics in it by observing carefully, draw diagram for your understanding.<\|endoftext\|>	3.859375
647	# Formulas for Area of a Triangle. There are a lot of formulas and techniques to find the area of a triangle. We can use many different formulas to calculate area of a triangle according to the given conditions. Here we shall derive some of the main formulas used to calculate area of a triangle. Formulas for Area of a Triangle: The area of a triangle is denoted by the symbol delta ( $\Delta$ ) We shall appeal to the formula: $\Delta = \frac{1}{2} bc \sin A = \frac{1}{2} ac \sin B = \frac{1}{2} ab \sin C$ And the half angle formula: $\sin \frac{1}{2} A = \sqrt{\dfrac{(s-b)(s-c)}{bc}} \, \, \, , \, \, \, \cos \frac{1}{2} A = \sqrt{\dfrac{s(s-a)}{bc}}$ etc. Where “s” is the semi circumference of the triangle or, $s = \frac{a+b+c}{2}$ We shall now derive different formula for the area of triangle: First formula for the area of a triangle: $\Delta = \frac{1}{2} bc \sin A = \frac{1}{2} bc . 2 \sin \frac{A}{2} \cos \frac{A}{2} \\ \\ \\ or , \Delta = bc . \sqrt{\dfrac{s(s-a)(s-b)(s-c)}{bc . bc}} \\ \\ \\ or , \Delta = \sqrt{s(s-a)(s-b)(s-c)}$ Second formula for the area of a triangle: $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$ Now , as 2s = (a+b+c) $\Delta = \frac{1}{4} \sqrt{(a+b-c)(b+c-a)(c+a-b)(a+b-c)} \\ \\ So , \Delta = \frac{1}{4} \sqrt{2b^2 c^2 + 2c^2 a^2 + 2a^2 b^2 - a^4 - b^4 - c^4}$ Third formula for the area of a triangle: $\Delta = \frac{1}{2} bc \sin A$ Now using sine law: $\Delta = \frac{1}{2} bc \frac{a}{2R} \\ \\ So , \Delta = \dfrac{abc}{4R}$ Related posts: 1. Half Angle formulas Half Angle formulas. Trigonometric half angles formula. Half angle formula... 2. Trigonometric transformation formulas Trigonometric transformation formulas. Trigonometric formulas used to transform an trigonometric... 3. Algebraic Formulas Algebra is one of the most basic part of mathematics... 4. Maths Formulas for Physics Maths Formulas for Physics. List of mathematical formulas used in... 5. Trigonometric multiple and sub-multiple angle formulas Trigonometric multiple and sub-multiple angle formulas. Trigonometric formulas for multiple...<\|endoftext\|>	4.8125
394	The first altars of the Aryans were those of the cult of fire. Agni, the fire, was celebrated in the Vedic hymns and kindled upon earth to carry prayers and offerings to the highest heavens. The central domestic rituals of the householder—daily rites, marriage rites, and finally funeral rites—took place at the fire altar. Laying and ritually kindling the fire altar was the job of brahmin priests. In ancient India, great rituals, called yajnas, were sponsored by kings and became increasingly elaborate, often involving several fires and many priestly specialists. In the world of the Aryans, as glimpsed in the Vedas, there were hymns addressed to Agni and other gods, but there were no temples, nor were there murtis of the gods. In the Vedic age, the gods were imaged only in words. Agni was described as the one with flaming hair and golden jaws, the messenger of the gods, through whom oblations were carried from earth to heaven. Even the most elaborately constructed fire altars were not permanent, but were constructed in a temporary ritual arena called a yajnashala. The Agnicayana, for example, involved the construction of an altar, built of layers of brick in the form of a large bird. When the rituals were completed, the site was burned and abandoned. Today, after three thousand years, the fire altar is still an important part of domestic ritual life. When a child is named, when his or her first hair-cutting is performed, when a boy receives his sacred thread, or when a couple is married, the fire altar is central to the rite. When a temple is consecrated in South India or in South Carolina, the fire altar is still central to the rites of consecration. The fire is the first flickering, portable, renewable icon and altar of the Divine.<\|endoftext\|>	3.71875
365	Crickets come in many varieties, from the familiar field cricket to tree and cave crickets. They go through incomplete or gradual metamorphosis, meaning the young insects resemble adults but don't have wings or reproductive organs. Crickets molt as they grow, shedding their skins anywhere from six to 18 times before reaching adulthood. Every adult female cricket has a conspicuous egg-laying tool called an ovipositor at the end of her abdomen; this enables you to tell males from females. Look for a mature cricket and examine the end of its abdomen. Locate the paired slender appendages protruding backward from the sides of the abdomen; these are called the cerci and function like a pair of backward antennae. Look between the cerci to see if an unpaired slender ovipositor projects backward from the end of the abdomen, resembling a spear or needle. (In camel crickets it is often half the length of the body.) Your cricket is female if the ovipositor is present and male if not. Sciencing Video Vault Examine a large immature cricket to see if the ovipositor has begun to form at the end of the abdomen, since ovipositors can begin to show before the cricket is an adult. Your young insect is female if you can see an ovipositor, even if it is short. Examine the wings of mature crickets as another means of separating males from females. Look at the base of the forewings for the thickened song-making structures called the file and scraper. Identify the insect as a male if these are present, since females can't sing. Look at the width of the wings in tree crickets: females have narrow wings and males have broad, paddle-shaped wings.<\|endoftext\|>	3.9375
2,011	Lymphocytes and macrophages are two types of white blood cells which serve various roles in our immune system. Lymphocytes are present in great numbers throughout the lymph and the lymphatic system and in smaller numbers in our blood. They serve a variety of purposes such as antibody makers, helper cells or killer cells. They are produced in the red bone marrow of certain bones and mature in organs such as the thymus and spleen. Reserves of lymphocytes are found in the lymph nodes, tonsils, adenoids, spleen, appendix and gastrointestinal tract. Macrophages digest everything that does not bear the mark of a healthy cell such as cell debris, viruses or bacteria, malignant cells and all sorts of foreign matter. A subtype of macrophage cells promotes regeneration processes and supports wound healing. Unlike lymphocytes which travel in the lymph, most macrophages remain in fixed points in the body where they intercept potential pathogens. What are white blood cells? White blood cells comprise several types and sybtypes of specialized cells involved in immunity. They are also called leucocytes (alternative spelling leukocytes). Despite performing several different functions within our body, all types of white blood cells serve the following purpose: protecting our body against disease, infection and any form of ill health caused by external pathogens such as bacteria, viruses, parasites or fungi. According to structure, white blood cells are divided into: 1) Granulocytes (white blood cells with a lobed nucleus and granules in their cytoplasm). 2) Agranulocytes (mononuclear leucocytes or white blood cells with one-lobed nucleus and no granules in their cytoplasm). White blood cells are further divided into 5 main types: 4) Lymphocytes (NK cells, helper T cells, cytotoxic T cells, B cells, gamma delta T cells). 5) Monocytes (macrophages, foam cells and dendritic cells). I. Lymphocytes. There are three main types of lymphocytes: 1) T cells. T cells are lymphocytes (a type of white blood cells) that are produced in the bone marrow but mature in the thymus and tonsils. There are two main types of T cells: T helper cells and cytotoxic T cells. Helper T cells release cytotoxins, while cytotoxic T cells destroy infected cells (cancer cells or cells infected with viruses or cells that are damaged in any way). T cells belong to the adaptive immune system, an immune system mechanism that does not involve the formation of antibodies. However, they require the help of macrophages to perform their immune role. 2) B cells. B cells are lymphocytes (a type of white blood cells) that are produced in the bone marrow. B cells belong to humoral immunity. They produce antibodies in response to viruses, bacteria and other threats to our immunity. B cells have special receptors along their cell membrane which allows them to bind to specific antigens (molecules that elicit an immune response) and produce an immune response to help protect the host from disease. 3) NK cells. NK or natural killer cells are lymphocytes that belong to our innate (or inborn) immune system response. These natural killer cells recognize infected cells and cancer cells from healthy cells and release cytotoxins to destroy them. Natural killer cells are quick to respond to infection (an estimated of 3 days) and also malignant cells. Like most white blood cells, NK cells have a large, one-lobed nucleus and little cytoplasm space with cytotoxic granules. II. Macrophages are a sybtype of monocytes (white blood cells) along with dendritic cells and foam cells. Macrophages basically eat anything that does not bear the mark of a healthy cell: damaged cells, debris, foreign matter, cells infected with viruses, parasites or bacteria and cancer cells. This process is called phagocytosis and belongs to both innate and adaptive immunity. While some macrophage digest every unhealthy cell, others repair damage within the body by repairing tissue. Macrophages may live up to several months. Where are lymphocytes and macrophages made? Lymphocytes and macrophages, and all other types of white blood cells, are made in the interior of some of our bones, in an area called red bone marrow. They are produced by a special type of stem cell along with red blood cells and platelets. Some types of lymphocytes such as B cells travel to our lymph nodes and spleen to mature, while T cells mature in our thymus gland, a gland situated behind the sternum and in between our lungs. Where are white blood cells such as lymphocytes and macrophages found? White blood cells are found in greater numbers in the lymph and lymphatic system and in lymph nodes and other key-organs such as the spleen, tonsils and special tissue in the gastrointestinal and respiratory tracts. They are found in smaller numbers in our blood. While some types of white blood cells migrate from the lymph to the blood and tissues, depending on where they are needed, other types of white blood cells become fixed in certain locations. For example, some types of macrophages remain fixed in the liver (and are known as Kupffer cells), others in the skin and mucosas that are in direct contact with the outside such as our lungs, stomach or nose mucosas (and are known as Langerhans cells) and so on. Lymphocytes are present in our circulatory system in the blood, lymph and lymphoid organs such as the lymph nodes, spleen etc. The difference between the two is that lymphocytes are generaly mobile, while some macrophages tend to hold a stabile position. What do lymphocytes and macrophages look like? Like all white blood cells, lymphocytes and macrophages are characterized by the presence of a large nucleus (red blood cells and platelets which do not have one). Macrophages have a one-lobed nucleus and are thus called mononuclear. Lymphocytes such as T cells, B cells and NK cells are also mononuclear. Generally, lymphocytes and monocytes (macrophages included) have a large nucleus about the size of an erythrocyte (or red blood cell) and very little cytoplasm. What can we do to help our immune system do its job? The immune system relies heavily on vitamins and minerals to work optimally. However, other nutrients from certain foods can be just as important in boosting immunity. 1) Vitamin C-rich foods. Vitamin C is the ultimate nutrient boasting not only potent regenerative and wound-healing properties, but also an excellent antibacterial, antiviral and anticancer action. Kiwifruit, red bell pepper, strawberries, papaya are only a few great alternatives to increase our vitamin C intake. Nonetheless, supplementation is needed if we want to increase our intake up to 2,000 mg a day, which is now the upper limit. 2) Vitamin A-rich foods. Most leafy green vegetables (spinach, kale, collard greens, mustard greens, turnip greens, beet greens, chard) and orange fruits and vegetables in general (sweet potato, carrots and squash in particular) are excellent sources. Vitamin A is required for maintaining healthy mucous membranes at the level of the nose, throat, lungs and digestive tract. These areas are directly exposed to the outside world and pathogens may breach unless they are as healthy as can be. 3) Garlic. While not anyone can eat it as they please, garlic has been found to possess wonderful antimicrobial and antiviral properties and some studies suggests it even exhibits a significant antibiotic action. 4) Dried fruits. Sun-dried fruits are excellent choices because they concentrate particularly high amounts of important nutrients such as dietary minerals of all sorts. This makes them a wonderful choice for anyone looking to keep healthy. 5) Mushrooms. Because of their selenium and beta-glucan intake, mushrooms are able to activate some types of white blood cells and thus actively contribute to good immunity. 6) Fresh salmon. We all know salmon is a posh food choices, but it’s actually healthier than it is fancy. This is because fresh salmon is rich in vitamin D (as well as healthy Omega-3 fatty acids). Vitamin D has been found to contribute to good immunity and even exert anticancer protection. 7) Herbal infusions. According to research at Harvard University, theanine in teas may help strengthen our immune system and contribute to better health. Green tea is a great choice in this respect, but there are other tea options to choose from. 8) Yogurt. Simple yogurt, also known as Greek yogurt, remains a healthy options for all of us. The probiotics in it help maintain gut health which is essential for the proper absorption of nutrients and good health. A just as good alternative to yogurt is Kefir or milk kefir. 9) Dark chocolate. While great amounts may overexcite one and cause sleepless nights or agitation, dark chocolate has also been found to improve immunity by enhancing T helper cells action. Read more about the health effects of cocoa beans here. 10) Zinc. While we can get all the zinc we need from food sources such as seafood, barley, peanuts or eggs, at times, our body needs higher amounts of the mineral to keep up with everything. In this case we either have to up our intake of zinc-rich foods or opt for some quality supplements to help meet our needs because zinc is essential for immune system health.<\|endoftext\|>	3.84375
648	# Substitution Method vs. Eliminations Methond Q: Solve by using the substitution and elimination method: (1) 3x – 4y = -1 (2) -10 + 8y = 6x A: Ok, this is 2 questions in 1. We need to solve it 2 different ways. Let’s first start with the substitution method: I am going to solve for x in equation (2). No reason that I picked equation (2) and no reason that I decided to solve for x (it could have been y). I just had to pick one…. So, solving for x in equation (2): -10 + 8y = 6x [divide everything by 6] -10/6 + 8/6y = 6x/6 [reduce the fractions] -5/3 + 4/3y = x OKAY, now look at equation (1): 3x – 4y = -1 We are going to put the blue equation that we solved for (from equation 2), into the x in equation (1) like so: 3(-5/3 + 4/3y) – 4y = -1 Now, multiply through: -15/3 + 12/3y -4y = -1 Reduce: -5 + 4y – 4y = -1 Simplify: -5 = -1 WHAT?!?! -5 = -1? Can’t be. We know that isn’t true. Therefore, this means that there are no solutions to this system of equations. OKAY, now let’s try solving using the elimination method. We should get the same answer, we just go about solving for it a different way: (1) 3x – 4y = -1 (2) -10 + 8y = 6x Re-arrange equation (2) so that the x’s and y’s are under those of equation (1)… I am going to subtract the 6x to the left side, and add the 10 to the right side to get: -6x + 8y = 10 So: (1) 3x – 4y = -1 (2) -6x + 8y = 10 Now, I am going to multiply equation (1) by 2 to each number: (1) 2[3x – 4y = -1] (2) -6x + 8y = 10 Simplify to get: (1) 6x – 8y = -2 (2) -6x + 8y = 10 Add the two equations together… notice the x’s and y’s cancel out on the left side? So you get: 0 = 8 What?? 0 does not equal 8. Therefore, again, we find the answer is no solutions!<\|endoftext\|>	4.84375
4,117	# IBPS Explore popular questions from Geometry for IBPS. This collection covers Geometry previous year IBPS questions hand picked by experienced teachers. ## Hindi Language Geometry Correct Marks 1 Incorrectly Marks -0.25 Q 1. If (5, 1), (x, 7) and (3, -1) are 3 consecutive verticles of a square then x is equal to : -3 B -4 C 5 D 6 E None of these ##### Explanation For the verticles to form a square, we know that the length of each side of the square should be equal. Therefore, (x - 5){tex}^2{/tex} + (7 - 1){tex}^2{/tex} = (x - 3){tex}^2{/tex} + (7 + 1){tex}^2{/tex} [x{tex}^2{/tex} + 5{tex}^2{/tex} - 2 (x) (5) ] + [36] = [x{tex}^2{/tex} + 3{tex}^2{/tex} - 2 (x) (3) ] + [64] [5{tex}^2{/tex} + 36] - [9 + 64] = (10 - 6) x x = -{tex}\frac{12}{4}{/tex} = - 3 This gives the side of the square x = - 3. Correct Marks 1 Incorrectly Marks -0.25 Q 2. What is the area of an obtuse angled triangle whose two sides are 8 and 12 and the angle included between two sides is 150°? 24 sq units B 48 sq units C 24{tex}\sqrt 3{/tex} D 48{tex}\sqrt 3{/tex} E Such a triangle does not exist ##### Explanation If two sides of a triangle and the included angle 'y' is known, then the area of the triangle can be found using the formula {tex}\frac{1}{2}{/tex}* (product of sides) * sin y Substituting the values in the formula, we get {tex}\frac{1}{2}{/tex}* 8 * 12 * sin 150 = 24 sq units Correct Marks 1 Incorrectly Marks -0.25 Q 3. What is the measure of the radius of the circle that circumscribes a triangles whose sides measure 9, 40 and 41? A 6 B 4 C 24.5 20.5 E 12.5 ##### Explanation From the measure of the length of the sides of the triangle 9, 40 and 41 we can infer that the triangle is a right angled triangle. 9, 40, 41 is a Pythagorean triplet. In a right angled triangle, the radius of the circle that circumscribes the triangle is half the hypotenuse. In the given triangle, the hypotenuse = 41 Therefore, the radius of the circle that circumscribes the triangle ={tex}\frac{41}{2}{/tex} = 20.5 units Correct Marks 1 Incorrectly Marks -0.25 Q 4. Verticles of a quadrilateral ABCD are A (0, 0), B (4, 5), C (9, 9) and D (5, 4). What is the shape of the quadrilateral? A Square B Rectangle but not a square Rhombus D Parallelogram but not a rhombus E None of these ##### Explanation The lengths of the four sides AB, BC, CD and DA are all equal to {tex}\sqrt{41}{/tex}. Hence, the given quadrilateral is either a Rhombus or a Square. Now let us compute the lengths of the two diagonals AC and BD. The length of AC is {tex}\sqrt{162}{/tex} and the length of BD is {tex}\sqrt{2}{/tex} As the diagonals are not equal and the sides are equal, the given quadrilateral is a Rhombus. Correct Marks 1 Incorrectly Marks -0.25 Q 5. If the sum of the interior angles of a regular polygon measures upto 1440 degrees, how many sides does the polygon have? 10 sides B 8 sides C 12 sides D 9 sides E None of these ##### Explanation We know that the sum of an exterior angle and an interior angle of a polygon = 180° We also know that sum of all the exterior angles of a polygon = 360° The question states that the sum of all interior angles of the given polygon = 1440° Therefore, sum of all the interior and exterior angles of the polygon = 1440 + 360 = 1800 If there are ‘n' sides to this polygon, then the sum of all the exterior and interior angles = 180 x n = 10 Correct Marks 1 Incorrectly Marks -0.25 Q 6. What is the radius of the in circle of the triangle whose sides measure 5, 12 and 13 units? 2 units B 12 units C 6.5 units D 6 units E 7.5 units ##### Explanation The triangle given is a right angled triangle as its sides are 5, 12 and 13 which is one of the Pythagorean triplets. Note: In a right angled triangle, the radius of the incircle is given by the following relation {tex}\frac{sum\ of\ perpendicular\ sides - hypotenuse}{2}{/tex} As the given triangle is a right angled triangle, radius of its incircle = {tex}\frac{5+12-13}{2}{/tex}= 2 units Correct Marks 1 Incorrectly Marks -0.25 Q 7. How many diagonals does a 63 sided convex polygon have? A 3780 1890 C 3843 D 3906 E 1953 ##### Explanation The number of diagonals of an n-sided convex is {tex}\frac{n(n-3)}{2}{/tex} This polygon has 63 sides. Hence, {tex}n{/tex} = 63 Therefore, number of diagonals = {tex}\frac{63\times60}{2}{/tex} = 1890 Correct Marks 1 Incorrectly Marks -0.25 Q 8. If 10, 12 and ‘x' are sides of an acute angled triangle, how many integer values of ‘x' are possible? A 7 B 12 9 D 13 E 11 ##### Explanation For any triangle sum of any two sides must be greater than the third side. The sides are 10, 12 and ‘x'. From Rule 2, x can take the following values : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21 - A total of 19 values. When x = 3 or x = 4 or x = 5 or x = 6, the triangle is an OBTUSE angled triangle. The smallest value of x that satisfies both conditions is 7. (10{tex}^{2}{/tex} + 7{tex}^{2}{/tex} > 12{tex}^{2}{/tex}) The highest value of x that satisfies both conditions is 15. (10{tex}^{2}{/tex} + 12{tex}^{2}{/tex} + 15{tex}^{2}{/tex}) When x = 16 or x = 17 or x = 18 or x = 19 or x = 20 or x = 21, the triangle is an OBTUSE angled triangle. Hence, the values of x that satisfy both the rules are x = 7, 8, 9, 10, 11, 12, 13, 14, 15. A total of 9 values. Correct Marks 1 Incorrectly Marks -0.25 Q 9. Find the length of the hypotenuse of a right triangle if the lengths of the other two sides are 6 inches and 8 inches. 10 inches B 11 inches C 18 inches D 20 inches E None of these ##### Explanation Test the ratio of the lengths to see if it fits the 3n : 4n : 5n ratio. 6 : 8 : ?= 3 (2) : 4 (2) : ? Yes, it is a 3-4-5 triangle for n = Calculate the third side 5n = 5 x 2 = 10 The length of the hypotenuse is 10 inches. Correct Marks 1 Incorrectly Marks -0.25 Q 10. Find the length of one side of a right triangle if the length of the hypotenuse is 15 inches and the length of the other side is 12 inches. A 8 inches B 7 inches 9 inches D 13 inches E None of these ##### Explanation Test the ratio of the lengths to see if it fits the 3n : 4n : 5n ratio. ? : 12 : 15 = ? : 4 (3) : 5 (3) Yes, it is a 3-4-5 triangle for n = 3 Calculate the third side 3n = 3 x 3 = 9 The length of the side is 9 inches. Correct Marks 1 Incorrectly Marks -0.25 Q 11. Find the length of the hypotenuse of a right triangle if the lengths of the other two sides are both 3 inches. A {tex}5{/tex} inches B {tex}3\sqrt{4}{/tex} inches C {tex}6{/tex} inches {tex}3\sqrt{2}{/tex} inches E None of these ##### Explanation This is a right triangle with two equal sides so it must be a 45° - 45° - 90° triangle. You are given that the both the sides are 3. If the first and second value of the ratio n : n : n{tex}\sqrt{2}{/tex} is 3 then the length of the third side is 3{tex}\sqrt{2}{/tex} The length of the hypotenuse is 3{tex}\sqrt{2}{/tex} inches. Correct Marks 1 Incorrectly Marks -0.25 Q 12. Find the lengths of the other two sides of a right triangle if the length of the hypotenuse is 4{tex}\sqrt{2}{/tex} inches and one of the angles is 45°. 4 inches B 9 inches C 8 inches D 7 inches E None of these ##### Explanation This is a right triangle with a 45° so it must be a 45° - 45° - 90° triangle. You are given that the hypotenuse is 4{tex}\sqrt{2}{/tex} . If the third value of the ratio n : n : n{tex}\sqrt{2}{/tex} is 4{tex}\sqrt{2}{/tex} then the lengths of the other two sides must 4. The lengths of the two sides are both 4 inches. Correct Marks 1 Incorrectly Marks -0.25 Q 13. Find the length of the hypotenuse of a right triangle if the lengths of the other two sides are 4 inches and 4{tex}\sqrt{3}{/tex} inches. 8 inches B 9 inches C 10 inches D 11 inches E None of these ##### Explanation Test the ratio of the lengths to see if it fits the n : n{tex}\sqrt{3}{/tex} : 2n ratio. 4 : 4{tex}\sqrt{3}{/tex} : ? n : n{tex}\sqrt{3}{/tex} : 2n Yes, it is a 30° - 60° - 90° triangle for n = 4 Calculate the third side 2n = 2 x 4 = 8 The length of the hypotenuse is 8 inches. Correct Marks 1 Incorrectly Marks -0.25 Q 14. What is the area of the following square, if the length of BD is 2{tex}\sqrt{2}{/tex} ? A 1 B 2 C 3 4 E 5 ##### Explanation We need to find the length of the side of the square in order to get the area. The diagonal BD makes two 45° - 45° - 90° triangles with the sides of the square. Using the 45° - 45° - 90° special triangle ratio n: n : n{tex}\sqrt{2}{/tex}. If the hypotenuse is 2{tex}\sqrt{2}{/tex} then the legs must be 2. So, the length of the side of the square is 2. Area of square = 5{tex}^{2}{/tex} = 2{tex}^{2}{/tex} = 4 Correct Marks 1 Incorrectly Marks -0.25 Q 15. In the figure below, what is the value of y? A 40 B 50 60 D 100 E 120 ##### Explanation Vertical angles being equal allows us to fill in two angles in the triangle that y° belongs to. Sum of angles in a triangle = 180° So, y° + 40° + 80° = 180° y° + 120° = 180° y° = 60° Correct Marks 1 Incorrectly Marks -0.25 Q 16. Two circles both of radii 6 have exactly one point in common. If A is a point on one circle and B is a point on the other circle, what is the maximum possible length for the line segment AB? A 12 B 15 C 18 D 20 24 ##### Explanation Sketch the two circles touching at one point. The furthest that A and B can be would be at the two ends as shown in the above diagram. If the radius is 6 then the diameter is 2 x 6 = 12 and the distance from A to B would be 2 x 12 = 24 Correct Marks 1 Incorrectly Marks -0.25 Q 17. Note: Figures not drawn to scale. In the figures above, x = 60, How much more is the perimeter of triangle ABC compared with the triangle DEF. 0 B 2 C 4 D 6 E 8 ##### Explanation Note: Figures not drawn to scale Since x = 60°, triangle ABC is an equilateral triangle with sides all equal. The sides are all equal to 8. Perimeter of triangle ABC = 8 + 8 + 8 = 24 Triangle DEF has two angles equal, so it must be an isosceles triangle. The two equal sides will be opposite the equal angles So, the length of DF = length of DE = 10 Perimeter of triangle DEF = 10 + 10 + 4 = 24 Subtract the two perimeters. 24 - 24 = 0 Correct Marks 1 Incorrectly Marks -0.25 Q 18. An equilateral triangle has one side that measures 5 in. What is the size of the angle opposite that side? A 55° B 70° C 110° 60° E None of these ##### Explanation Since it is an equilateral triangle all its angles would be 60°. The size of the angle does not depend on the length of the side. The size of the angle is 60°. Correct Marks 1 Incorrectly Marks -0.25 Q 19. An isosceles triangle has one angle of 96°. What are the sizes of the other two angles? A 24° B 34° 42° D 96° E None of these ##### Explanation Since it is an isosceles triangle it will have two equal angles. The given 96° angle cannot be one of the equal pair because a triangle cannot have two obtuse angles. Let x be one of the two equal angles. The sum of all the angles in any triangle is 180°. x + x + 96° = 180° 2x = 84° x = 42° The sizes of the other two angles are 42° each. Correct Marks 1 Incorrectly Marks -0.25 Q 20. Find the circumference of the circle with a diameter of 8 inches? A 25 inches 25.163 inches C 29.45 inches D 35.62 inches E None of these ##### Explanation Formula C = {tex}\pi {/tex}d C = 8{tex}\pi {/tex} The circumference of the circle is 8{tex}\pi {/tex} = 25.163 inches Correct Marks 1 Incorrectly Marks -0.25 Q 21. Find the area of the circle with a diameter of 10 inches? A 55.78 sq. inches B 99.75 sq. inches C 92 inches 78.55 sq. inches E None of these ##### Explanation Formula A = {tex}\pi {/tex}r{tex}^{2}{/tex} Change diameter to radius r = {tex}\frac{1}{2}{/tex}d = {tex}\frac{1}{2}{/tex}x 10 = 5 Plug in the value: A = {tex}\pi {/tex}5{tex}^{2}{/tex} = 25 {tex}\pi {/tex} The area of the circle is 25{tex}\pi {/tex} 78.55 sq. inches Correct Marks 1 Incorrectly Marks -0.25 Q 22. Find the area of the circle with a radius of 10 inches? 314.2 sq. inches B 115 inches C 320.29 sq. inches D 56.12 sq. inches E None of these ##### Explanation Formula A = {tex}\pi {/tex}r2 Plug in the value A = {tex}\pi {/tex}102 = 100 {tex}\pi {/tex} The area of the circle is 100 {tex}\pi {/tex} 314.2 sq. inches<\|endoftext\|>	4.40625
4,263	<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> You are reading an older version of this FlexBook® textbook: CK-12 Algebra I - Second Edition Go to the latest version. # 8.1: Exponent Properties Involving Products Difficulty Level: At Grade Created by: CK-12 ## Learning Objectives • Use the product of a power property. • Use the power of a product property. • Simplify expressions involving product properties of exponents. ## Introduction In this chapter, we will discuss exponents and exponential functions. In Lessons 8.1, 8.2 and 8.3, we will be learning about the rules governing exponents. We will start with what the word exponent means. Consider the area of the square shown right. We know that the area is given by: But we also know that for any rectangle, $\text{Area} = (\text{width}) \cdot (\text{height})$, so we can see that: Similarly, the volume of the cube is given by: $\text{Volume} =\text{width} \cdot \text{depth} \cdot \text{height} = x \cdot x \cdot x$ But we also know that the volume of the cube is given by $\text{Volume} = x^3$ so clearly $x^3=x\cdot x \cdot x$ You probably know that the power (the small number to the top right of the $x$) tells you how many $x$'s to multiply together. In these examples the $x$ is called the base and the power (or exponent) tells us how many factors of the base there are in the full expression. $x^2 & = \underbrace{ x \cdot x }_{\text{2 factors of} \ x} && x^7 = \underbrace{ x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x }_{\text{7 factors of} \ x} \\x^3 & = \underbrace{ x \cdot x \cdot x }_{\text{3 factors of} \ x} && x^n = \underbrace{ x \cdot x \cdot \cdot \cdot \cdot \cdot \cdot \cdot x }_{\text{n factors of} \ x}$ Example 1 Write in exponential form. (a) $2\cdot 2$ (b) $(-3)(-3)(-3)$ (c) $y \cdot y \cdot y \cdot y \cdot y$ (d) $(3a)(3a)(3a)(3a)$ Solution (a) $2 \cdot 2 = 2^2$ because we have 2 factors of $2$ (b) $(-3)(-3)(-3) = (-3)^3$ because we have 3 factors of $(-3)$ (c) $y \cdot y \cdot y \cdot y \cdot y=y^5$ because we have 5 factors of $y$ (d) $(3a)(3a)(3a)(3a) = (3a)^4$ because we have 4 factors of $3a$ When we deal with numbers, we usually just simplify. We'd rather deal with 16 than with $2^4$. However, with variables, we need the exponents, because we'd rather deal with $x^7$ than with $x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x$. Let’s simplify Example 1 by evaluating the numbers. Example 2 Simplify. (a) $2\cdot 2$ (b) $(-3) (-3) (-3)$ (c) $y \cdot y \cdot y \cdot y \cdot y$ (d) $(3a) (3a)(3a)(3a)$ Solution (a) $2\cdot 2=2^2=4$ (b) $(-3) (-3) (-3)=(-3)^3=-27$ (c) $y \cdot y \cdot y \cdot y \cdot y=y^5$ (d) $(3a) (3a)(3a)(3a)=(3a)^4=3^4 \cdot a^4=81a^4$ Note: You must be careful when taking powers of negative numbers. Remember these rules. $& (\text{negative number}) \cdot (\text{positive number}) = \text{negative number}\\& (\text{negative number}) \cdot (\text{negative number}) = \text{positive number}$ For even powers of negative numbers, the answer is always positive. Since we have an even number of factors, we make pairs of negative numbers and all the negatives cancel out. $(-2)^6 = (-2)(-2)(-2)(-2)(-2)(-2) = \underbrace{ (-2)(-2) }_{+4 } \cdot \underbrace{ (-2)(-2) }_{+4} \cdot \underbrace{ (-2)(-2) }_{+4} = + 64$ For odd powers of negative numbers, the answer is always negative. Since we have an odd number of factors, we can make pairs of negative numbers to get positive numbers but there is always an unpaired negative factor, so the answer is negative: $\text{Ex:} \ (-2)^5 = (-2)(-2)(-2)(-2)(-2) = \underbrace{ (-2)(-2) }_{+4} \cdot \underbrace{ (-2)(-2) }_{+4} \cdot \underbrace{ (-2) }_{-2} = - 32$ ## Use the Product of Powers Property What happens when we multiply one power of $x$ by another? See what happens when we multiply $x$ to the power 5 by $x$ cubed. To illustrate better we will use the full factored form for each: $\underbrace{ (x \cdot x \cdot x \cdot x \cdot x) }_{x^5} \cdot \underbrace{ (x \cdot x \cdot x) }_{x^3} = \underbrace{ (x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x ) }_{x^8}$ So $x^5 \cdot x^3=x^8$. You may already see the pattern to multiplying powers, but let’s confirm it with another example. We will multiply $x$ squared by $x$ to the power 4: $\underbrace{ (x \cdot x)}_{x^2 } \cdot \underbrace{ (x \cdot x \cdot x \cdot x) }_{x^4 } = \underbrace{ (x \cdot x \cdot x \cdot x \cdot x \cdot x) }_{x^6}$ So $x^2 \cdot x^4=x^6$. Look carefully at the powers and how many factors there are in each calculation. 5 factors of $x$ times 3 factors of $x$ equals $(5 + 3) = 8$ factors of $x$. 2 factors of $x$ times 4 factors of $x$ equals $(2 + 4) = 6$ factors of $x$. You should see that when we take the product of two powers of $x$, the number of factors of $x$ in the answer is the sum of factors in the terms you are multiplying. In other words the exponent of $x$ in the answer is the sum of the exponents in the product. Product rule for exponents: $x^n \cdot x^m = x^{n+m}$ Example 3 Multiply $x^4 \cdot x^5$. Solution $x^4 \cdot x^5=x^{4+5}=x^9$ When multiplying exponents of the same base, it is a simple case of adding the exponents. It is important that when you use the product rule you avoid easy-to-make mistakes. Consider the following. Example 4 Multiply $2^2 \cdot 2^3$. Solution $2^2 \cdot 2^3 = 2^5=32$ Note that when you use the product rule you DO NOT MULTIPLY BASES. In other words, you must avoid the common error of writing $\xcancel{2^2 \cdot 2^3 = 4^5}$. Try it with your calculator and check which is right! Example 5 Multiply $2^2 \cdot 3^3$. Solution $2^2 \cdot 3^3 =4 \cdot 27=108$ In this case, the bases are different. The product rule for powers ONLY APPLIES TO TERMS THAT HAVE THE SAME BASE. Common mistakes with problems like this include $\xcancel{2^2 \cdot 3^3 = 6^5}$. ## Use the Power of a Product Property We will now look at what happens when we raise a whole expression to a power. Let’s take $x$ to the power $4$ and cube it. Again we will us the full factored form for each. $(x^4)^3 & = x^4 \cdot x^4 \cdot x^4 && 3 && \text{factors of} \ x \ \text{to the power}\ 4. \\\underbrace{ (x \cdot x \cdot x \cdot x) }_{x^4} \cdot \underbrace{ (x \cdot x \cdot x \cdot x) }_{x^4} \cdot \underbrace{ (x \cdot x \cdot x \cdot x) }_{x^4} & = \underbrace{ (x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x) }_{x^{12}}$ So $(x^4)^3=x^{12}$. It is clear that when we raise a power of $x$ to a new power, the powers multiply. When we take an expression and raise it to a power, we are multiplying the existing powers of $x$ by the power above the parenthesis. Power rule for exponents: $(x^n)^m=x^{n \cdot m}$ Power of a product If we have a product inside the parenthesis and a power on the parenthesis, then the power goes on each element inside. So that, for example, $(x^2y)^4=)(x^2)^4 \cdot (y)^4=x^8 y^4$. Watch how it works the long way. $\underbrace{ (x \cdot x \cdot y) }_{x^2y} \cdot \underbrace{ (x \cdot x \cdot y) }_{x^2y} \cdot \underbrace{ (x \cdot x \cdot y) }_{x^2y} \cdot \underbrace{ (x \cdot x \cdot y) }_{x^2y} = \underbrace{ (x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot y \cdot y \cdot y \cdot y ) }_{x^8 y^4}$ Power rule for exponents: $\left(x^n\right)^m = x^{nm}$ and $\left(x^ny^m\right)^p = x^{np}y^{mp}$ WATCH OUT! This does NOT work if you have a sum or difference inside the parenthesis. For example, $(x+y)^2 \neq x^2+y^2$. This is a commonly made mistake. It is easily avoidable if you remember what an exponent means $(x+y)^2=(x+y)(x+y)$. We will learn how to simplify this expression in a later chapter. Let’s apply the rules we just learned to a few examples. When we have numbers, we just evaluate and most of the time it is not really important to use the product rule and the power rule. Example 6 Simplify the following expressions. (a) $3^4 \cdot 3^7$ (b) $2^6 \cdot 2$ (c) $(4^2)^3$ Solution In each of the examples, we want to evaluate the numbers. (a) Use the product rule first: $3^5 \cdot 3^7=3^{12}$ Then evaluate the result: $3^{12}=531,441$ OR We can evaluate each part separately and then multiply them. $3^5 \cdot 3^7=243 \cdot 2,187 = 531,441$. (b) Use the product rule first. $2^6\cdot 2 = 2^7$ Then evaluate the result. $2^7 =128$ OR We can evaluate each part separately and then multiply them. $2^6\cdot 2=64 \cdot 2 =128$ (c) Use the power rule first. $(4^2)^3=4^6$ Then evaluate the result. $4^6= 4096$ OR We evaluate inside the parenthesis first. $(4^2)^3=(16)^3$ Then apply the power outside the parenthesis. $(16)^3= 4096$ When we have just one variable in the expression then we just apply the rules. Example 7 Simplify the following expressions. (a) $x^2\cdot x^7$ (b) $(y^3)^5$ Solution (a) Use the product rule. $x^2 \cdot x^7 = x^{2+7}=x^9$ (b) Use the power rule. $(y^3)^5=y^{3 \cdot 5}=y^{15}$ When we have a mix of numbers and variables, we apply the rules to the numbers or to each variable separately. Example 8 Simplify the following expressions. (a) $(3x^2 y^3)\cdot (4xy^2)$ (b) $(4 xyz) \cdot (x^2y^3) \cdot (2yz^4)$ (c) $(2a^3 b^3 )^2$ Solution (a) We group like terms together. $(3x^2y^3)\cdot (4xy^2) =(3\cdot 4 )\cdot (x^2 \cdot x) \cdot (y^3 \cdot y^2)$ We multiply the numbers and apply the product rule on each grouping. $12x^3 y^5$ (b) We groups like terms together. $(4xyz)\cdot (x^2 y^3) \cdot (2yz^4 )=(4 \cdot 2) \cdot (x \cdot x^2) \cdot (y \cdot y^3 \cdot y) \cdot (z \cdot z^4)$ We multiply the numbers and apply the product rule on each grouping. $8x^3 y^5 z^5$ (c) We apply the power rule for each separate term in the parenthesis. $(2a^3 b^3)^2=2^2 \cdot (a^3)^2 \cdot (b^3)^2$ We evaluate the numbers and apply the power rule for each term. $4a^6 b^6$ In problems that we need to apply the product and power rules together, we must keep in mind the order of operation. Exponent operations take precedence over multiplication. Example 9 Simplify the following expressions. (a) $(x^2)^2\cdot x^3$ (b) $(2x^2 y) \cdot (3x y^2)^3$ (c) $(4a^2 b^3 )^2 \cdot (2ab^4 )^3$ Solution (a) $(x^2)^2\cdot x^3$ We apply the power rule first on the first parenthesis. $(x^2)^2 \cdot x^3 = x^4 \cdot x^3$ Then apply the product rule to combine the two terms. $x^4 \cdot x^3 = x^7$ (b) $(2x^2 y ) \cdot (3xy^2)^3$ We must apply the power rule on the second parenthesis first. $(2x^2 y) \cdot (3xy^2)^3 = (2x^2y) \cdot (27x^3 y^6)$ Then we can apply the product rule to combine the two parentheses. $(2x^2 y) \cdot (27x^3 y^6) =54x^5 y^7$ (c) $(4a^2 b^3)^2 \cdot (2ab^4)^3$ We apply the power rule on each of the parentheses separately. $(4a^2 b^3)^2 \cdot (2ab^4)^3=(16a^4 b^6) \cdot (8a^3 b^{12})$ Then we can apply the product rule to combine the two parentheses. $(16a^4b^6) \cdot (8a^3 b^{12})=128a^7 b^{18}$ ## Review Questions Write in exponential notation. 1. $4 \cdot 4 \cdot 4 \cdot 4 \cdot 4$ 2. $3x \cdot 3x \cdot 3x$ 3. $(-2a)(-2a)(-2a)(-2a)$ 4. $6 \cdot 6 \cdot 6 \cdot x\cdot x \cdot y \cdot y \cdot y \cdot y$ Find each number: 1. $5^4$ 2. $(-2)^6$ 3. $(0.1)^5$ 4. $(-0.6)^3$ Multiply and simplify. 1. $6^3 \cdot 6^6$ 2. $2^2 \cdot 2^4 \cdot 2^6$ 3. $3^2 \cdot 4^3$ 4. $x^2 \cdot x^4$ 5. $(-2y^4) (-3y)$ 6. $(4a^2)(-3a)(-5a^4)$ Simplify. 1. $(a^3)^4$ 2. $(xy)^2$ 3. $(3a^2 b^3 )^4$ 4. $(-2xy^4 z^2)^5$ 5. $(-8x)^3(5x)^2$ 6. $(4a^2)(-2a^3)^4$ 7. $(12xy)(12xy)^2$ 8. $(2xy^2)(-x^2 y)^2 (3x^2 y^2)$ 1. $4^5$ 2. $(3x)^3$ 3. $(-2a)^4$ 4. $6^3 x^2 y^4$ 5. 625 6. 64 7. 0.00001 8. -0.216 9. 10077696 10. 4096 11. 576 12. $x^6$ 13. $6y^5$ 14. $60a^7$ 15. $a^{12}$ 16. $x^2 y^2$ 17. $81a^8 b^{12}$ 18. $-32x^5 y^{20} z^{10}$ 19. $12800x^5$ 20. $64a^{14}$ 21. $1728x^3 y^3$ 22. $6x^7 y^6$ Feb 23, 2012 Apr 02, 2015<\|endoftext\|>	4.96875
197	Communication and Relationships Common Sense Media ties communication online to relationship-building. As people connect more virtually, more often and in a variety of settings, it is important to understand that what may be clear in a face to face exchange can be complicated by the limitations of online exchanges. The ability to communicate in the virtual forums requires what is often referred to as “netiquette.” No communication is truly anonymous in the virtual world.Simple rules can keep communication productive and strengthen relationships: - Respect other participants and know that online communication is very personal and more easily misinterpreted. - Only say online what you would say standing three feet away from another person. - Reflect before hitting send - who are writing to and if forwarded would you feel confident in the message being received as you intended. Communication and relationships Common Sense Media family tips - Grades K-5: - Grades 6-8: - Grades 9-12:<\|endoftext\|>	3.796875
566	Bullying and Cyber-bullying Information • Harassment -repeatedly sending nasty, mean and insulting messages • Denigration -"dissing" someone online; sending or posting gossip or rumors about a person to damage his or her reputation or friendship • Impersonation -pretending to be someone else and sending or pasting material to get that person in trouble or danger or damage that person's reputation or friendships • Outing -sharing someone's secrets or embarrassing information or images online • Trickery - tricking someone into revealing secrets or embarrassing information, then sharing it online • Exclusion -intentionally and cruelly excluding someone form an online group • Cyberstalking -repeated, intense harassment and denigration that includes threats or creates significant fear • Almost 40% were disrespected, over 12% were threatened, and about 5% were scared for their safety. • Negative emotional responses included significant amounts of frustration, anger and sadness. • 28% of youth indicated that they had been bullied via email. • Over 40% of youth who were cyberbullied did not tell anyone about the incident. • Unexpectedly stops using the computer • Appears nervous or jumpy when an instant message or email appears • Appears uneasy about going to school or outside in general • Appears to be angry, depressed, or frustrated after using the computer • Avoids discussions about what he or she is doing on the computer • Becomes abnormally withdrawn from usual friends and family members A child may be cyberbullying others if he or she… • Quickly switches screens or closes programs when you walk by • Uses the computer at all hours of the night • Gets unusually upset if he or she cannot use the computer • Laughs excessively while using the computer • Is using multiple online accounts, or an account that is not his or her own • Parents must regularly monitor the online activities in which children are engaged. • Parents should also encourage an open dialogue with their children regarding issues of safety and responsible internet use. • Teachers must take care to supervise students as they use computers in the classroom and should consider incorporating discussions of issues related to cyber safety in their curriculum where appropriate. • School liaison officers and law enforcement officials must investigate all instances of harassment-including electronic bullying-and hold responsible parties accountable. Assume that EVERYONE has access to your profile and will use the information to cause your harm. Assume there are predators out there trying to find you based on the information you provide in your profile. "Popular internet social networking sites such as myspace.com, facebook.com, xanga.com, personal blogs and web pages are prime places for cyberbullying."<\|endoftext\|>	3.890625
588	# The theory of cryptography stems out of the term”Crypto” along with also the present day idea of cryptography is all about privacy, security and also the ability to transact securely. The majority of folks would concur that each of those things fall under the basic idea of cryptography, although of course there are additional applications for cryptography. Nicely, you may use numbers and paper writing math to help in the understanding of cryptography and this I shall explain to you the way exactly is mathematics. You will find several diverse means of working with numbers and the mathematics also you can use this when you want to greatly help in the comprehension of a general secret, or when you want to deliver a key word. If you would like to help provide the security for your data and defend it and you can also use it. First let us look at just how exactly is math. Let’s think about a secret. In cryptography we work with a secret key to decrypt it then to encrypt information. A secret key is made up of several facets, the range of secrets that’ll have to decrypt a selected bit of information. So that the primary my sources element is more than the issue it is likely to figure out how many there are required and just how many secrets that there are. It is possible todo the remainder by dividing by two, thereby multiplying the previous variable and multiplying with the range of keys required to decrypt the data, and to figure the square root of two. That is how is math employed in cryptography. Then there was another manner of applying is numbers and math to help in the comprehension of the way exactly is mathematics employed in cryptography. There is An essential factor needed to help in the calculation of a key. You have to select the square root of the variety of factors required to multiply the key, to receive the number if you require a public key to be multiplied with a few factors then. As there are other means to calculate it Utilizing the numbers alone to estimate that a key may possibly be insufficient. By way of example, you can calculate the square root of 2, split up the number of keys needed to decrypt the info by two, and then multiply by the number of keys required to encrypt the data, and then divide by 2 and multiply the result https://sreal.ucf.edu/ by the variety of keys needed to decrypt the information. That’s how is mathematics. You can use exactly the same way is numbers and mathematics to figure out just out is math employed in cryptography. You are able to utilize it in order to find the elements which are essential to multiply the people secret, to get the range. ### Author Este sitio web utiliza cookies para que usted tenga la mejor experiencia de usuario. Si continúa navegando está dando su consentimiento para la aceptación de las mencionadas cookies y la aceptación de nuestra política de cookies, pinche el enlace para mayor información. ACEPTAR<\|endoftext\|>	4.40625
570	As a way of helping commercial and hobbyist beekeepers track the many environmental stressors honey bees experience, Penn State’s Pollinator Research Center and the Huck Institutes of the Life Sciences, as well as several other collaborators, developed an app called Beescape, which can be used for monitoring environmental factors around hives, farms, and gardens where honey bees are flying around and pollinating. According to the release by Penn State announcing this new online tool, “Pennsylvania beekeepers lose nearly 50 percent of their honey bee colonies each winter,” and these insects are necessary for pollinating almost 90 percent of the flowering plants in that area. “Pollinators, particularly bees, play a vital role in supporting ecosystems in agricultural, urban, and natural landscapes,” says Christina Grozinger, who is a professor of entomology as well as Penn State’s director for their Pollinator Research Center. Some key stressors that are decreasing bee population numbers include reduced diversity and abundance of the plants bees need for food, insecticide exposure, and lost nesting habitats for wild bees. By using Beescape.org, beekeepers can choose a location—often where their hives are—and check surrounding regions as far away as five kilometers. This examination then reveals a score for the landscape’s quality and even lets beekeepers inspect local crops inside that radius. According to Maggie Douglas, who is an assistant professor with Dickinson College, “Beescape allows people to see the world as a bee, which will help them make decisions about where to place their colonies or steps they and their neighbors can take, such as planting pollinator gardens or reducing insecticide use, to make the landscape more friendly for bees.” Beescape.org, per Grozinger, acts as a bit of a go-between, partnering gardeners and beekeepers with researchers as a way of tracking a bee population’s collective health and working toward a better quality of life. Penn State also stated that wild bees like the bumblebee are threatened or endangered as well. “With data provided by beekeepers from agricultural, rural, and urban landscapes across multiple states, we will be able to develop high-quality predictive models that will be included in the website in the future,” says Melanie Kammerer Allen, who is a graduate ecology student with Penn State. Allen continued, “This will allow beekeepers to determine if they should provide the bees with supplementary food, for example, or for a grower to decide if they should add pollinator nesting habitats near their crops.” As this point, Illinois, Indiana, and Pennsylvania have been mapped out on Beescape, but there are plans to eventually include more U.S. states. Photo By halfpoint<\|endoftext\|>	3.671875
366	"Convention-al" Wisdom: Lessons Learned Before developing the workbooks, we sat down with a teacher from each grade and asked them two questions: 1) What skill-building practice do children need during the summer, and 2) What are some skills that you would like your child to experience before coming into your class? After listening and learning from our teachers, we created five (5) levels of workbooks: early preschool through second grade. Literacy Stories & Flip and Reads Basic math and number concepts, utilized in a preschool or kindergarten classroom, set the foundation for learning more advanced math concepts. Early exposure to math and number activities will promote your child’s comfort with these skills. As early exposure to letters and sight words help children feel confident that they are “readers”, early writing opportunities give children confidence that they are “writers”; and exposure to math skills help children feel like they are "mathematicians". We think numbers and counting are some of the easiest concepts to imbed naturally into almost any activity. Whether you are using our yearlong curriculum or not, we wanted to show you simple ways to integrate counting into your classroom or daily routine. Writing Lessons For Your Preschooler As we created The Reading Corner curriculum, we sat down as a team and decided which sight words children should know; in what sequence they should be written; and how often we should introduce new words. Literacy Stories follow the Sight Word scope embedded in our yearlong curriculum. The sentence strands get a little longer as the stories develop. Writing is the glue that cements the mastery of oral language and reading for the child. This is why it is so important that children not only learn to write, but learn to see themselves as writers.<\|endoftext\|>	3.6875
1,355	Combination of resistors in series and parallel connection Suppose you have two resistors of resistance 2 ohm each, but you need 1 ohm resistance to use or connect in a circuit. What will you do if there are no other sources to collect a resistor of resistance 1 ohm? Again, what will you do if you need 4 ohm resistance to use? Yes, the answer is that we need to use the combination of the available resistors. There are two types of combinations of resistors – Series combination and Parallel combination. In this article, we are going to explain the combination of resistors in series and parallel for class 10 and 12. Also, we will derive the equivalent resistance in series and parallel connection. 1. Series Combination of Resistors 2. Equivalent resistance of the combination of resistors in series 3. Parallel Combination of Resistors 4. Equivalent resistance for the combination of resistors in parallel 5. Numerical problems on the combinations of resistors Combination of Resistorsin series with circuit diagram The neat circuit diagram for series connection of resistors is shown above. In series combination of resistors we need to connect resistors one by one in a row. That means each resistor will be connected at ending point of previous resistor. The series combination of resistors has a significant role in electrical circuits. Using series combination of resistors one can increase the effective resistance to a higher value. The connection of a battery across this combination will provide the current flow through the resistors. Derive the formula for equivalent resistance for series combination of resistors Let, three resistors of resistances R1, R2 and R3 are connected in series and the battery across the series connection supplies the electric current through the circuit. Let, I be the electric current flowing through the circuit. Since, there is only one loop, then the same amount of current will flow through each resistance. Here, the battery supplies voltage across the resistors. If V1, V2 and V3 be the voltage drops across the resistors R1, R2 and R3, then V = ( V1 + V2 + V3 ) or, IReq = ( IR1 + IR2 + IR3 ) [ Using Ohm’s Law] or, Req = ( R1 + R2 + R3 ) ………………….(1) If we connect N number of resistors in series then the equivalent resistance of the series combination of resistors will be Req = ( R1 + R2 + R3 + ……. + RN ) ……………..(2) Equation-(2) gives the formula for equivalent resistance in series combination of resistors. This equation shows that in series combination, the effective resistance increases. Series equivalent resistance is greater than every individual resistance in that combination. Explain the Parallel combination of resistors with neat diagram The neat circuit diagram for parallel connection of resistors is shown above. In parallel combination of resistors we need to connect resistors in a column between same two points. The parallel combination of resistors also has a significant role in electric circuits. Using the parallel combination of larger resistors one can decrease the effective resistance to a lower value than individual resistors. The external battery will supply current through the resistors. Derive the formula of equivalent resistance for the parallel combination of resistors Let, three resistors of resistances R1, R2 and R3 are connected in parallel and the battery across the parallel connection supplies the electric current through the circuit. As the resistors are connected parallelly across the same two points, then the voltage difference across all resistances is same and is equal to the voltage of the external battery. Let, I be the electric current emitting from the battery. This current will divide into three parts to flow through three resistances. If I1, I2 and I3 be the current through the resistors R1, R2 and R3 then I = ( I1 + I2 + I3 ) or, $\small \frac{V}{R_{eq}}=\frac{V}{R_{1}}+\frac{V}{R_{2}}+\frac{V}{R_{3}}$ or, $\small \frac{1}{R_{eq}}=(\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}})$ ……………(3) If we connect N number of resistors in parallel combination then the formula for equivalent resistance in parallel combination is $\small \frac{1}{R_{eq}}=(\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+.........+\frac{1}{R_{N}})$ ………………(4) From the equation-(4), one can see that the equivalent resistance in parallel combination is smaller than individual resistances of the combination. So, in parallel combination, the effective resistance decreases. Facts related to the combination of resistors 1. Series connection of resistors increases the effective resistance and the parallel connection of resistors decreases the equivalent resistance. 2. Equivalent resistance of series combination is greater than the individual resistances used in the combination. Thus, if we connect more resistances in series in a circuit, the amount of current flow decreases if the external voltage remains same. Since, there is only one path in series connection, the amount current flow through each resistor will be same. But, the voltage drop across the resistors will be different. 3. The equivalent resistance of parallel combination is smaller than the individual resistances used in the combination. Thus, if we connect more resistances in parallel in a circuit, the battery will provide more current if the voltage remains same. This amount of current divides into the branches of resistors. That means the current will not be same through each resistor. As the resistors in parallel are connected between two same points, the voltage drop across all the resistors will be equal. Numeral problems on combination of resistances 1. Two equal resistors are connected once in series and once in parallel combination. Find the ratio of equivalent resistances in these cases. 2. How can you get 3 ohm resistance by using three 2 ohm resistances? 3. Find the minimum effective resistance from the combination of four 2 ohm resistors. This is all from the combination of resistors in series and parallel and their derivation. If you have any doubt on this topic you can ask me in the comment section. Thank you! Related posts:<\|endoftext\|>	4.625
1,983	Lesson 2: Image and Text \| Text and Image Photographs and written texts can complement one another in communicating concepts and ideas. In this lesson, students explore the relationship between image and text and how each can illustrate the other. First, the class discusses two more aspects of composing an image to aid them in their Photo Mission later in the lesson. In pairs, they then observe two photographs and select a text from Worksheet B that resonates with one of the images. They then take their own photographs that reflect the meaning of a chosen text. Outline (60 minutes) Introductory Composition Lesson (10 minutes) Meme Activity (10 minutes) Image and Text (15 minutes) Text and Image – Photo Mission (20 Minutes) Share Photographs (5 Minutes) - Laptop and projector - The photographs “La Chuppah” by Shayna Kling,”Summer Camp” by Zion Ozeri, and “Values” by Sofia Mijal Listovsky - The photographs “Freedom” by Max Yaffe, “Ray of Light” by Aaron Yadegar and “Between Light and Darkness” by Michael Sanders - Copies of Worksheet B: Image and Text / Text and Image - Cameras (phone cameras are acceptable) Set up projector to display the photographs. If laptop and projector are not available, make high-resolution photocopies of the photograph to distribute to students. Also, make enough copies of Worksheet B for the class, and be sure that there are enough cameras for each pair of students. - Strengthen students understanding of the relationship between image and text. - Show students how Judaism can be incorporated into art and photography with and without the aid of text. - Increase student’s knowledge of important visual composition principles. Introductory Composition Lesson (10 minutes) This section builds on the “Composing a Photograph” section from Lesson Plan 1: Reading a Photograph. Here the students will be encouraged to think more carefully about composing an image with an introduction to the ‘Rule of Thirds’ technique and a brief analysis of, and discussion about, the use of lighting in photography. The Rule of Thirds Show the students the following photograph: Zion Ozeri’s ‘Summer Camp.’ The rule of thirds is one of the fundamentals of photography, a technique designed to help photographers create a more interesting photograph. Show the class how with this technique, we divide a composition into nine squares of equal size (as we see in the image below), with two horizontal lines intersecting with two vertical lines. If we then align our subject with the guide lines and their intersection points, the subject isn’t centered but rather the focal point is slightly off to one side. Explain that this technique draws the viewer’s eye into the composition, instead of just glancing at the center. This is a method adopted by many photographers and can be used to enhance the student’s photography skills, however remember to mention that it is a compositional tool to be experimented with and is in no way a rule that must be adhered to. Tell the students that during their Photo Mission later in the lesson, they can play around with this technique by imagining the grid as they look through the viewfinder and compose their image. Light and Dark Lighting can make or break an image. To get your students to think carefully about light when they are using their cameras, use the following three images to discuss with the class the ways in which photographers use light and dark to create a particular effect. “The smallest glimmer of light stands out in the darkness, and no matter how dark things may seem, there is always light. This expresses the strength of the Jewish people. Others have tried to wipe out our people many times, but no one has succeeded. We are a minority. Although small, we always illuminate the darkness.” Michael has used candlelight in an unlit room to create a stark contrast here between light and dark. Our eyes are instantly drawn to the center of the image – to the candlelight, and the person holding it. The effect is dramatic and conjures up images of candle-lighting rituals, as well as a beacon of light – a light unto the nations. “Betty is a Holocaust survivor. The greatest lesson that I learned from Betty is that no matter what, if you have courage and strength in yourself, everything will turn out just fine.” Aaron has also used contrast between light and dark in his photograph – this time to highlight a person in a portrait. Here his subject, Betty, is lit up and the background is completely dark, so that there are no distractions. Generally, lighting should come from behind the camera in this way, students should avoid backlighting unless it is intended. Aaron has used artificial lighting, creating the effect of many different colors on Betty’s face and clothes – this colorfulness is perhaps intended to express something about her character. “This picture was taken in Kibbutz Hatzerim in the south of Israel and represents the freedom of life experienced in Israel and the fulfillment of the Zionist dream. Music and movement are expressions of Jewish creation and creativity, helping us face a new day that looks towards the future.” Here, Max has intentionally used backlighting – utilizing natural light – to create a silhouette effect. Experimenting with natural light and learning how to harness it is a great way for students to improve their images. Generally, natural light is at its best either at the beginning or at the end of the day. At midday, the light from the sun can be quite harsh, whereas shortly after the sun has risen in the morning and just before it sets at night (referred to by photographers as the ‘golden hour’), the light can change from being hard and dramatic (as you see in Max’s photo above) to softer and enhances colors. Encourage your students to experiment with natural light by taking notice of the quality and quantity of light depending on where they are, the weather conditions and time of day. Next time they are out taking photographs they can ask themselves: Does the light suit the subject? Would the light be better at a different time of day, or in different weather conditions? Meme Activity (10 Minutes) Ask the class: What is a meme? Display some memes to the class, such as the following: A meme is an idea, a behavior or a style that spreads from person to person within a culture. An internet meme spreads from person to person via social media. It can be made up of just an image, a word or phrase – in the memes above, we see the two combined. Here text and image work together to communicate a message more effectively than possible by one or the other alone. Today we are going to further analyzing the relationship between image and text. In the memes above, a funny message is relayed by combining text and image. We’re now going to attempt to pair text and image to portray other ideas and meaningful expressions of Jewish life. Show the students the following image and, in pairs or small groups, ask them to come up with a caption for it. Such as: Image and Text (15 minutes) Show the class the two photos: “Summer Camp” by Zion Ozeri and “Values” by Sofia Mijal Listovsky. Divide the class into pairs and give each a copy of Worksheet B. Each pair of students reads the texts on Worksheet B and chooses one text that they believe connects best with one of the above photographs. Ask a few pairs to share the texts they chose with explanation. Text and Image – Photo Mission (20 minutes) Have each pair go out into the building or the outdoors to take photographs of each other or of their surroundings that illustrate one of the texts on Worksheet B that “speaks” to them. The photographs can be literal or metaphorical, posed or candid, narrative or poetic. Remind students to plan their shots carefully. Before shooting, they should think about what they want to express and how they will do so. Encourage students to consider the compositional aspects from Lessons One (framing) and Two (Rule of thirds and lighting). Although the students should take many photographs to experiment, they will choose one photograph to show the rest of the class, to which they can make simple edits (cropping, rotation) on their phones if they desire. Share Photographs (5 minutes) Ask each pair to share one photograph. How do you think you emulated the quote you selected through your photo? Explain your group’s process. Was this different from past picture-taking experiences? How? Did your shots come out the way you intended? Why or why not? Ask students to upload their images to Instagram and tag the Beit Hatfutsot Jewish Lens Instagram page (www.instagram.com/jewishlens) Optional Take-Home Project For additional reflection, the questions above could form the basis of a journal-writing activity. Other option: Create Jewish meme upload to Instagram, and tag #JewishLens and share with class next week. Additionally, practice taking additional photos utilizing new composition knowledge that show your connection to your Judaism as practice for the Final Assignment. Then, upload a photo to Instagram and tag the Jewish Lens Instagram page.<\|endoftext\|>	3.671875
493	# How do you solve 5*11^(5a+10)=57? Oct 16, 2016 $a = - 1.797$ #### Explanation: $5 \cdot {11}^{5 a + 10} = 57$ $\frac{5 \cdot {11}^{5 a + 10}}{5} = \frac{57}{5} \textcolor{w h i t e}{a a a}$Divide both sides by 5 ${11}^{5 a + 10} = \frac{57}{5}$ $\log \left({11}^{5 a + 10}\right) = \log \left(\frac{57}{5}\right) \textcolor{w h i t e}{a a a}$Take the log of both sides $\left(5 a + 10\right) \log 11 = \log \left(\frac{57}{5}\right) \textcolor{w h i t e}{a a}$Use the log rule $\log {x}^{a} = a \log x$ $\frac{\left(5 a + 10\right) \log 11}{\log 11} = \frac{\log \left(\frac{57}{5}\right)}{\log 11} \textcolor{w h i t e}{a a}$Divide both sides by log11 $5 a + 10 = \frac{\log \left(\frac{57}{5}\right)}{\log 11}$ $\textcolor{w h i t e}{a a} - 10 \textcolor{w h i t e}{a a a} - 10 \textcolor{w h i t e}{a a a}$Subtract 10 from both sides $5 a = \frac{\log \left(\frac{57}{5}\right)}{\log 11} - 10$ $\frac{5 a}{5} = \frac{\frac{\log \left(\frac{57}{5}\right)}{\log 11} - 10}{5} \textcolor{w h i t e}{a a a}$Divide both sides by 5 $a = - 1.797 \textcolor{w h i t e}{a a a}$Use a calculator<\|endoftext\|>	4.75
739	Question # Number of roots of a quadratic equation are: A 1 B 2 C 3 D None Solution ## The correct option is A $$2$$Let us, consider the quadratic equation of the general form $$ax^{2}+bx+c=0$$ where, $$a≠0$$Now divide each term by $$a$$ (Since $$a≠0$$), we get $$\Rightarrow$$ $$x^{2}$$ $$+$$ $$\dfrac{b}{a}x$$$$+$$$$\dfrac{c}{a}=0$$$$\Rightarrow$$ $$x^{2}$$$$+$$$$\dfrac{2b}{2a}x$$ $$+$$ $$\Bigg(\dfrac{b}{2a}\Bigg)^{2}$$$$-$$$$\Bigg(\dfrac{b}{2a}\Bigg)^{2}$$$$+$$$$\dfrac{c}{a}$$ = $$0$$ After this we get$$\Rightarrow$$ $$\Bigg(x + \dfrac{b}{2a}\Bigg)^{2}$$ – $$\Bigg(\dfrac{\sqrt{b^{2} - 4ac}}{2a}\Bigg)^{2}=0$$$$\Rightarrow$$ $$\Bigg[x + \Bigg(\dfrac{b}{2a}\Bigg) + \Bigg(\dfrac{\sqrt{b^{2} - 4ac}}{2a}\Bigg)\Bigg]\Bigg[x +\Bigg(\dfrac{b}{2a}\Bigg) - \Bigg(\dfrac{\sqrt{b^{2} - 4ac}}{2a}\Bigg)\Bigg]=0$$$$\Rightarrow$$ $$\Bigg[x - \Bigg(\Bigg(\dfrac{-b - \sqrt{b^{2} - 4ac}}{2a}\Bigg)\Bigg)\Bigg]\Bigg[x - \Bigg(\Bigg(\dfrac{-b +\sqrt{b^{2} - 4ac}}{2a}\Bigg)\Bigg)\Bigg] = 0$$$$\Rightarrow$$ $$(x - α)(x - β) = 0,$$where $$α = \Bigg(\dfrac{- b -\sqrt{b^{2} - 4ac}}{2a}\Bigg)$$ and $$β = \Bigg(\dfrac{- b + \sqrt{b^{2} - 4ac}}{2a}\Bigg)$$Now we can clearly see that the equation $$ax^{2} + bx + c = 0$$ reduces to$$(x - α)(x - β) = 0$$ and the equation $$ax^{2} + bx + c = 0$$ is only satisfiedby the values $$x = α$$ and $$x = β$$.Except $$α$$ and $$β$$ no other values of x satisfies the equation $$ax^{2} + bx +c = 0$$Hence, we can say that the equation $$ax^{2} + bx + c = 0$$ has $$only$$ $$two$$ roots.Therefore, number of roots of Quadratic Equation are $$2$$Hence Answer is $$(B)$$ Mathematics Suggest Corrections 0 Similar questions View More People also searched for View More<\|endoftext\|>	4.4375
1,053	The average young person in the U.S. and in Vermont initiates sexual activity for the first time at age 17. To help young people choose safe and healthy sexual relationships, we must provide education and support regarding prevention of sexually transmitted infections (STIs). STIs are a common cause of disease in sexually active adolescents. Bacteria, viruses, protozoa, parasites, or fungi cause STIs. The most common viral STI is human papillomavirus (HPV). The most common bacterial STI is chlamydia. Other common infections in adolescents include gonorrhea, syphilis, trichomonas, and herpes simplex virus (HSV). Human immunodeficiency virus (HIV) is an STI, too. Some bacterial STIs, such as chlamydia, are curable with antibiotics. Others such as HSV persist in the body in dormant and active states, as there are no currently available cures. Assorted complications of STIs include: - Severe infections - Chronic pain - Ectopic pregnancy - Harmful effects in fetuses Every year, half of the 20 million new STIs reported in the US occur in youth between 15 and 24 years of age. This percentage is remarkable because 15 to 24 year-olds represent just 25 percent of the sexually experienced population in the United States. Because of a variety of factors, many adolescents’ sexual practices and behaviors put them at risk for acquiring STIs. These risk factors include: - Having increased biological susceptibility to infection - Failing to use barrier protection consistently and correctly - Having sequential sex partners of limited duration - Having multiple sex partners concurrently - Early age at sexual initiation - Experiencing multiple obstacles to accessing health care There are significant disparities in the STI rates for young people of different ages, gender, and sexual orientation. While the highest rates of STIs are in young adults and adolescents, detection of these infections depends greatly on the different screening recommendations for each STI, as many infections are asymptomatic. In terms of STI prevention a 2007 report by The National Campaign to Prevent Teen and Unplanned Pregnancy provided a detailed review of the important characteristics of effective curriculum-based sex education programs, and provided information about the primary factors that can reduce the probability of a young person contracting an STI: - Increase the correct and consistent use of condoms - Increase testing and treatment of STIs - Vaccinate against STIs for which vaccines are available - Increase abstinence (both delaying the initiation of sex and increasing the return to abstinence) - Reduce the number of sexual partners - Reduce the occurrence of concurrent partners - Increase the period of time between sexual partners - Decrease the frequency of sex - Circumcision (boys) Barrier protection is one of the key prevention factors in STI transmission and male condom use amongst young people has increased since 1991. Unfortunately, in 2015 only about 57 percent of males and females age 15-19 years old reported male condom use at last intercourse. Other barrier methods, dental dams, and female condoms are infrequently used. Although condom use has increased overall, consistent condom use is still a challenge. Unfortunately, our culture presents mixed messages related to condom use: irresponsibility for wanting to have sex, but responsibility for using condoms. Influencing consistent condom use requires messaging that supports personal responsibility, safety and trust in relationships. There are some additional key aspects to adolescent STI prevention: - Health Education: Engaged adolescents in sexual health discussions appropriate for their developmental level. Make health education messages as specific as possible, aimed at specific risk behaviors. Make these discussions thorough, genuine, nonjudgmental, and confidential. In addition, offer print or electronic educational materials, as patients may not absorb educational information while in clinic and/or receiving bad news or treatment. - Confidentiality: Confidentiality is an important element in providing quality care to adolescents. Discuss it with patients and parents/guardians. Most states have specific laws that guarantee minors’ rights to confidentiality for STI testing and treatment. - Expedited Partner Therapy (EPT): When providers dispense medication for partners to the patient, it is as good as or better than standard partner referral options. EPT is supported by numerous professional medical organizations. For more information on adolescent STIs: - CDC Sexually Transmitted Infections, Adolescents and Young Adults - Center for Young Women’s Health: Sexual Health - Young Men’s Health: Sexual Health - Vermont Department of Health: Adolescent Sexual Health - Vermont Department of Health: Sexually Transmitted Diseases Erica Gibson, MD, is a physician specializing in Adolescent Medicine at the University of Vermont Children’s Hospital. This article is excerpted from: Commonly Sexually Transmitted Infections in Adolescents; Primary Care: Clinics in Office Practice. Vol 41, Issue 3, pages 631-650. Elsevier, Sept 2014. Erica J. Gibson, David L. Bell & Sherine A. Powerful.<\|endoftext\|>	3.84375
646	Prior installments in this series (seen here, here, and here) examined the most famous Puritan grave symbol, the winged skull called a “death’s head.” Today’s post traces the evolution of the death’s head from a stern, menacing figure into softer imagery. Throughout the Puritan era, the winged skull remained the single most common gravestone decoration. Yet around 1690, an interesting change began. Fifty years later, the death’s head was transformed. We begin with a typical death’s head, from the 1680s, which stares insistently. This design was used by many engravers around Boston, without much variation: On the next stone (below), the death’s head wears finely curved eyebrows. This detail was popularized by Joseph Lamson (discussed here), and became common practice: In the next stone, representative of many graves from the early 1700s, we again see the eyebrows and also note a mustache. This gives the sense of a face superimposed upon the skull: Fifteen years later (below), the mustache resembles a small frown. The skull almost has two mouths, with massive teeth beneath and a small pout above. Again, we have the idea of a face and skull combined: Next, the teeth move downward and become smaller (below). The face appears more pronounced: As the teeth are made smaller (below), they come to resemble a neck, supporting the skull. Suddenly, there is more room for the frown, which now dominates the expression: By the mid-1700s, this pouting skull had almost entirely replaced the original death’s head. It expresses a softer emotion—sad but not menacing—and projects a more human form. It persisted as the single most prevalent gravestone design throughout New England until around 1800. And in some instances (below), the frown became a tiny smile. The fearsome death’s head was completely transformed: The reasons for this evolution are unclear. Puritan carvers began with a menacing death’s head, perched insistently upon the grave, and over fifty years, transitioned to a gently pouting (or even smiling) figure that looked slightly more like a face than a skull. The later graves expressed a different emotion, and it is tempting to think that this represented a changed approach to death. Certainly Puritan society became more secure as it entered its second century in America. The colonists gained familiarity with the land and seasons, they built an economic order, and their settlements expanded. Perhaps this societal maturation extended to the grave, as the confidence of cultural continuity might temper the loss of family and friends. They may have felt less need for fear and more room for sorrow. The gentler face of death may have been a reflection of this change. Tune in for the next essay, which will examine a curiosity of the Puritan calendar, captured forever in their slate gravestones. If you enjoy these posts, please consider subscribing; just scroll to the bottom and click ‘Follow this Blog.’ (All photographs taken by the author.)<\|endoftext\|>	3.984375
559	Quantum mechanics requires advanced mathematics to give numerical predictions for the outcome of measurements. However, one can understand many significant results of the theory from the basic properties of the probability waves. An important example is the behavior of electrons within atoms. Since such electrons are confined in some manner, we expect that they must be represented by standing waves that correspond to a set of allowed frequencies. Quantum mechanics states that for this new type of wave, its frequency is proportional to the energy associated with the microscopic particle. Thus, we reach the conclusion that electrons within atoms can only exist in certain states, each of which corresponds to only one possible amount of energy. The energy of an electron in an atom is an example of an observable which is quantized, that is it comes in certain allowed amounts, called quanta (like quantities). When an atom contains more than one electron, quantum mechanics predicts that two of the electrons both exist in the state with the lowest energy, called the ground state. The next eight electrons are in the state of the next highest energy, and so on following a specific relationship. This is the origin of the idea of electron "shells" or "orbits," although these are just convenient ways of talking about the states. The first shell is "filled" by two electrons, the second shell is filled by another eight, etc. This explains why some atoms try to combine with other atoms in chemical reactions. This idea of electron states also explains why different atoms emit different colors of light when they are heated. Heating an object gives extra energy to the atoms inside it and this can transform an electron within an atom from one state to another of higher energy. The atom eventually loses the energy when the electron transforms back to the lower-energy state. Usually the extra energy is carried away in the form of light which we say was produced by the electron making a transition, or a change of its state. The difference in energy between the two states of the electron (before and after the transition) is the same for all atoms of the same kind. Thus, those atoms will always give off a wavelength and frequency of light (i.e., color) that corresponds to that energy. Another element's atomic structure contains electron states with different energies (since the electron is confined differently) and so the differing energy levels produce light in other regions of the electromagnetic spectrum. Using this principle, scientists can determine which elements are present in stars by measuring the exact colors in the emitted light. Quantum mechanics theory has been extremely successful in explaining a wide range of phenomena, including a description of how electrons move in materials (e.g., through chips in a personal computer). Quantum mechanics is also used to understand superconductivity, the decay of nuclei, and how lasers work.<\|endoftext\|>	4.28125
1,298	# 2.1 Estimations, equations and variables (mental calculation test Page 1 / 1 4. Own attempts 5. (a) 202 (b) 485 (c) 8 (d) 8 (e)` 7 (f) 56 (g) 820 (h) 96 (i) 470 (j) 18 400 (k) 40 (l) 64 (m) 10 000 (n) 64 (o) 1 000 000 ## Activity: estimations, equations and variables (mental calculation test) [lo 1.7.2, lo 1.7.7, lo 1.8, lo 1.10] 4. Brainteaser! Can you form a square / rectangle / triangle with the following seven tangram pieces? Trace the shapes on a separate sheet of paper, cut them out and arrange the pieces. Then glue them in a frame.. 5. Let's first see how you cope with "ordinary" numbers! Complete the following mental calculation test as quickly and accurately as possible: a) 179 + 23 = _______________ b) 512 – 27 = _______________ c) 9 x _______________ = 72 d) _______________ x 6 = 48 e) 63 ÷ 9 = _______________ f) _______________ ÷ 8 = 7 g) 24 600 ÷ 30 = _______________ h) (60 ÷ 5) x (32 ÷ 4) = _______________ i) 3 x 47 + 7 x 47 = _______________ j) 25 x 184 x 4 = _______________ k) 158 – 130 + 12 = _______________ l) 82 = _______________ m) 1002 = _______________ n) 43 = _______________ o) 1003 = _______________....... 15 Complete by colouring in the appropriate block: I have performed ## Assessment Learning Outcome 1: The learner will be able to recognise, describe and represent numbers and their relationships, and to count, estimate, calculate and check with competence and confidence in solving problems. Assessment Standard 1.7: We know this when the learner estimates and calculates by selecting and using operations appropriate to solving problems that involve: 1.7.2: multiple operations with integers; 1.7.7: exponents. Assessment Standard 1.8: We know this when the learner performs mental calculations involving squares of natural numbers to at least 10 2 and cubes of natural numbers to at least 5 3 ; Assessment Standard 1.10: We know this when the learner uses a range of strategies to check solutions and judges the reasonableness of solutions. Introduction about quantum dots in nanotechnology what does nano mean? nano basically means 10^(-9). nanometer is a unit to measure length. Bharti do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment? absolutely yes Daniel how to know photocatalytic properties of tio2 nanoparticles...what to do now it is a goid question and i want to know the answer as well Maciej Abigail for teaching engĺish at school how nano technology help us Anassong Do somebody tell me a best nano engineering book for beginners? there is no specific books for beginners but there is book called principle of nanotechnology NANO what is fullerene does it is used to make bukky balls are you nano engineer ? s. fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball. Tarell what is the actual application of fullerenes nowadays? Damian That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes. Tarell what is the Synthesis, properties,and applications of carbon nano chemistry Mostly, they use nano carbon for electronics and for materials to be strengthened. Virgil is Bucky paper clear? CYNTHIA carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc NANO so some one know about replacing silicon atom with phosphorous in semiconductors device? Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure. Harper Do you know which machine is used to that process? s. how to fabricate graphene ink ? for screen printed electrodes ? SUYASH What is lattice structure? of graphene you mean? Ebrahim or in general Ebrahim in general s. Graphene has a hexagonal structure tahir On having this app for quite a bit time, Haven't realised there's a chat room in it. Cied what is biological synthesis of nanoparticles what's the easiest and fastest way to the synthesize AgNP? China Cied types of nano material I start with an easy one. carbon nanotubes woven into a long filament like a string Porter many many of nanotubes Porter what is the k.e before it land Yasmin what is the function of carbon nanotubes? Cesar I'm interested in nanotube Uday what is nanomaterials and their applications of sensors. what is nano technology what is system testing? how did you get the value of 2000N.What calculations are needed to arrive at it Privacy Information Security Software Version 1.1a Good Got questions? Join the online conversation and get instant answers!<\|endoftext\|>	4.40625
5,281	La NiñaA cooling of the waters The El Niño/La Niña connection La Niña and the jet stream Effects of La Niña The 1998–2000 La Niña Predictions for winter of 2006–2007 For More Information La Niña has traditionally received less attention than the more well-known El Niño. For example, during the years 1997 and 1998, El Niño (pronounced el NEE-nyo) was the weather phenomenon most in the news. However, from late 1998 through the first half of 2000 El Niño was succeeded by the less widely publicized La Niña (la NEE-nya) phenomenon. La Niña is the cold-water counterpart to El Niño. El Niño is a period of unusual warming in the Pacific near Peru and Ecuador, while La Niña is a period of abnormally cool water temperatures in the same region. La Niña sometimes follows in the wake of El Niño. La Niña's influence is typically felt in the form of colder-than-usual winters in the northern United States and warmer-than-usual winters in the southern United States. La Niña also promotes the formation of Atlantic Ocean hurricanes that can affect Caribbean islands and the U.S. East Coast. The term "la niña" is Spanish for "little girl." It was named for its relation to El Niño. Two other terms that are sometimes used to describe this cold-water phenomenon are El Viejo (el vee-AY-hoe), which means "the old man"; and "cold-phase ENSO event" (in contrast to El Niño, the warm phase of the El Niño/Southern Oscillation). According to guidelines set forth by the Japanese Meteorological Association, a La Niña year is one in which average sea temperatures along the equator in the Pacific are more than 1°F (0.5°C) colder than usual, and the water remains cold for at least six months. La Niña sometimes, but not always, follows El Niño. Since 1975 there have been half as many La Niñas as there have been El Niños. On average, La Niña is present one year in every four. Sometimes La Niñas occur as infrequently as once every ten years. La Niña conditions typically last from nine to twelve months, but may persist for up to two years. WORDS TO KNOW - air pressure: - the pressure exerted by the weight of air over a given area of Earth's surface. Also called atmospheric pressure or barometric pressure. - cold-phase ENSO (El Niño/Southern Oscillation): - another name for La Niña; colder-than-normal eastern Pacific waters. - convective zone: - the region of warm tropical water over which thunderstorms form; the ocean under the Intertropical Convergence Zone. - El Niño: - a term that means "the Christ child" or "little boy" in Spanish. A period of unusual warming of the Pacific Ocean off the coast of Peru and Ecuador. It usually starts around Christmas, which is how it got its name. - El Niño/Southern Oscillation (ENSO): - the simultaneous warming of the waters of the eastern Pacific and the accompanying shifts in air pressure over the eastern and western Pacific. - the inundation of water onto land that is normally dry. - gale-force wind: - any wind whose sustained speed is between 39 and 54 mph (63 and 87 kph). - Intertropical Convergence Zone: - a belt of warm, rising, unstable air formed from the inward-flowing trade winds from north and south of the equator. - jet stream: - the world's fastest upper-air winds. Jet streams travel in a west-to-east direction, at speeds of 80 to 190 miles (130 to 300 kilometers) per hour, around 30,000 feet (9,150 meters) above the ground. Jet streams occur where the largest differences in air temperature and air pressure exist. In North America, jet streams are typically found over southern Canada and the northern United States, as well as over the southern United States and Mexico. The northern jet stream is called the polar jet stream, and the southern jet stream is called the subtropical jet stream. - La Niña: - Spanish for "little girl," a period of cooler-than-normal water temperatures in the eastern Pacific near the coast of Peru and Ecuador. It often follows an El Niño. - a name for seasonal winds that result in a rainy season during the summer on tropical continents, when the land becomes warmer than the sea beside it. - mud slide: - a landslide of mostly mud mixed with debris, often caused by heavy rains on steep land with sparse vegetation. - polar jet stream: - a North American jet stream, typically found over southern Canada or the northern United States. - water in any form, such as rain, snow, ice pellets, or hail, that falls to Earth's surface. - pressure gradient: - the difference in air pressure between a high and low pressure area relative to the distance separating them. - research buoy: - a tethered or drifting buoy placed in the open ocean capable of recording atmospheric and ocean conditions and transmitting them to a satellite. - subtropical jet stream: - a North American jet stream, typically found over the southern United States or northern Mexico. - a storm resulting from strong rising air currents; characterized by heavy rain or hail along with thunder and lightning. - trade winds: - an area near the equator of prevailing winds that blow from the northeast north of the equator and the southeast south of the equator. - cold water flowing toward the surface, often caused by prevailing winds or pressure differences. - warm-phase ENSO (El Niño/Southern Oscillation): - another name for El Niño; warmer-than-normal eastern Pacific waters. The main characteristic of La Niña is a cooling of the waters in the tropical Pacific, from the coast of South America to the central equatorial region. Normally, the Pacific Ocean temperature off the coast of South America is around 68°F (20°C). During El Niño years the water temperature may be 10°F (6°C) warmer than normal. In contrast, the temperature of coastal waters falls as much as 15°F (8°C) below normal during La Niña years. La Niña's cold surface waters are produced by a strong upwelling—a churning up of cold water from below. The upwelling during La Niña is intensified by a strong easterly (from the east) trade wind, which is a prevailing wind near the equator. As the trade wind blows warm surface water westward across the ocean, cold water upwells to take its place. Like El Niño, La Niña is produced by shifts in water temperature and air pressure (also called atmospheric pressure) between the eastern and western portions of the southern tropical Pacific Ocean. While the shift in El Niño is toward warmer water and lower air pressure in the eastern Pacific, however, the shift in La Niña is toward colder water and higher air pressure in the eastern Pacific. La Niña and El Niño can be thought of as extremes in the range of conditions that may exist in the tropical Pacific Ocean, or as book-ends of the El Niño/Southern Oscillation (ENSO).Whereas El Niño is considered a warm-phase ENSO, La Niña is considered a cold-phase ENSO. In normal conditions, trade winds blow from the eastern Pacific (from the coast of South America), where pressure is higher, to the western Pacific, where pressure is lower. During El Niño, the pressure over the western Pacific rises and the pressure over the eastern Pacific drops, causing the trade winds to weaken or reverse course. During La Niña, the pressure gradient, or the difference between the high and low pressure areas related to the distance between them, tilts more steeply toward the west than usual. This is primarily because of the presence of very cold water in the eastern Pacific, which increases the air pressure there. This steepening of the pressure gradient causes trade winds to blow strongly toward the west. Meteorologists' understanding of La Niña is not as great as that of El Niño. Less effort has been spent researching La Niña, in part, because La Niña is not as powerful or potentially destructive a force as El Niño. Whereas El Niño provokes unusual weather patterns, La Niña merely intensifies typical weather patterns. La Niña, El Niño, or normal? Defining the condition of the tropical Pacific Ocean is tricky business. While some meteorologists assert that the ocean undergoes periods of "normalcy," others argue that El Niño or La Niña is always present to some degree. Weather expert Kevin Trenberth of the National Center for Atmospheric Research stated that in the tropical Pacific for the years 1950–1997: conditions were normal 46 percent of the time; El Niño was present 31 percent of the time; and La Niña was present 23 percent of the time. Scientists were driven to learn more about El Niño by the reduction of Peruvian fisheries in the 1970s. La Niña has no such obvious consequence. It was only in the 1980s, when the far-reaching impacts of La Niña (called teleconnections) were discovered, that research into La Niña began in earnest. During La Niña, the "obstacles" that divert the jet stream during El Niño are removed. That is to say, the towering thunderstorms (storms caused by rising air currents) that form off the coast of South America during El Niño years shift to the western-central Pacific and weaken, due to cooler waters. With the displacement of the convective zone (the region of warm water over which thunderstorms form) comes a weakening of the subtropical jet stream found over the southern United States and northern Mexico. Generally, during La Niña the subtropical jet stream occupies a region farther north than it does during El Niño. The area of strongest jet stream activity during La Niña is the central Rocky Mountains. It is there that the most severe weather can be expected. The jet stream also passes over the Great Lakes in the east. The polar jet stream, typically over the northern United States and southern Canada, is also weakened and pushed farther north, over Alaska and northern Canada. The effects of El Niño and La Niña are felt most intensely in the winter, when temperature contrasts between northern and southern states are greatest and the jet stream more directly influences the weather. The warming and cooling of the waters that characterize El Niño and La Niña also peak during the Northern Hemisphere winter. In many ways, La Niña's effects on the weather are the opposite of El Niño's. For instance, La Niña brings cold winters to the northern Great Plains states, the Pacific Northwest, the Great Lakes states, and Canada, and warmer-than-usual winters to the Southeast. El Niño, in contrast, often spells warmer-than-usual winters for the northern United States. and colder-than-usual winters for the southern United States. Whereas El Niño brings flooding (the inundation of normally dry land) to California, the Southwest, eastward across the states bordering the Gulf of Mexico, and Florida, La Niña brings drier-than-usual conditions to those locations. Also in contrast to El Niño, La Niña brings dry weather to the Central Plains states and the Southeast, and lots of precipitation to the Pacific Northwest. While El Niño brings unusually heavy rains to Peru and Ecuador, and drought to Southeast Asia and Australia, La Niña brings drought to the South American coast and flooding to the western Pacific region. Another contrast between the two phenomena is that while El Niño hinders development of Atlantic hurricanes, La Niña seems to promote hurricane formation in the region. El Niño's westerly winds (from west to east) blow in the opposite direction of low-altitude winds in the Atlantic Ocean, lopping the tops off thunderclouds before they can come together into hurricanes. La Niña's winds, in contrast, which blow strongly from east to west, encourage the development of Atlantic hurricanes. The effects of La Niña on weather patterns can be described as an exaggeration of normal conditions. For instance, while the northern United States typically has cold winters, La Niña makes them colder; and while coastal Peru and Ecuador are typically dry, La Niña makes them drier. Indonesia typically receives monsoon rains, caused by seasonal winds in the summer months, but La Niña makes the rains heavier. However, it is it is misleading to label a specific rainstorm, hurricane, or other weather occurrence a "La Niña event" or an "El Niño event." La Niña and El Niño influence the position and intensity of weather patterns, but they are only two among many other factors that influence local weather. In May 1998 a rapid cooling of the tropical Pacific waters began. This temperature drop signaled the end of El Niño and the start of La Niña. Over the next few months the cooling trend continued, after which temperatures varied. Ocean temperatures returned to normal at the end of June 1999, prompting some meteorologists to call a premature end to La Niña. The cooling resumed, however, leading to predictions that La Niña would continue through early 2000. The first clue to the arrival of La Niña came in the fall of 1997. Scientists detected a drop in temperature of a large pool of ocean water in the western Pacific, 420 feet (128 meters) below the surface. Temperature readings over the next several months, taken by research buoys (floating devices containing weather instruments) indicated that the cool water was moving eastward and upward, toward the coast of South America. Throughout May and June 1998, temperatures along a 5,000-mile (8,000 kilometer) strip of coastal water dropped more than 15°F (8°C), to about 65°F (18.4°C). Drought and wildfire La Niña, as expected, produced dry conditions in the summers of 1998 and 1999 for the Southwest, the Southeast, the central Plains states, and even the mid-Atlantic states. In addition to drought, some of these areas faced extended heat waves. In many places, non-essential uses of water (such as watering lawns and washing cars) were banned. Perhaps the greatest consequence of the hot, dry conditions was the outbreak of wildfires. Hundreds of thousands of acres burned during these two summers. A key reference to: The water cycle The water cycle (also called the hydrologic cycle) is the continuous movement of water between the atmosphere and Earth's surface (oceans and landmasses). On one side of the equation is precipitation and on the other side is evaporation—the process by which liquid water at the surface converts to a gas and enters the atmosphere. Eighty-five to ninety percent of the moisture that evaporates into the atmosphere comes from the oceans. The rest evaporates from the soil, lakes, and rivers that make up the continental land-masses. Plants lose water through tiny pores in the underside of their leaves in a process called transpiration. Some of the moist air above oceans is carried over land by the wind. Clouds form and drop rain and snow on the ground. When precipitation hits the ground, it either sinks in or runs off, depending on the surface composition. On soft ground, most of the water sinks into the soil to be absorbed by plant roots or to seep down into underground aquifers, which are underground layers of spongy rock, gravel, or sand in which water collects. Some of it runs off into streams and rivers. If the water strikes a hard surface, like rock or pavement, most of it runs directly into streams or man-made drains. Eventually this water also flows into rivers. The oceans lose water in this portion of the cycle. More water evaporates from them than returns as precipitation. The oceans get this water back when the rivers empty their water back into the oceans. Thus the global water budget—the volume of water coming and going from the oceans, atmosphere, and continental land-masses—is kept in balance. If the water cycle is kept in balance, that means that global precipitation levels remain fairly constant. So why do droughts occur? The answer is that rain and snow do not consistently fall in equal amounts in any given place. Moisture may evaporate from one place, travel through the atmosphere, and fall to the ground as rain in another. It is possible, then, for a given location to get lots of rainfall one year and almost no rainfall the next. In Florida, more than 500,000 acres were consumed by fire in the summer of 1998. In the first half of 1999, more than 35,000 acres burned. Several homes were lost to the blaze. A state of emergency for the entire state was declared in April 1999 and the National Guard was brought in to assist firefighters. In March 1999, fire blackened seventy-eight thousand acres of Nebraska prairie—an area the size of Rhode Island—and killed hundreds of cattle. That fire came one month after an unusual wintertime grass fire near North Platte, Nebraska, that burned fifteen thousand acres. For the first half of 1999, the mid-Atlantic region (New Jersey, New York, Pennsylvania, Delaware, Maryland, and Washington, D.C.) had precious few drops of precipitation. That spring wildfires raged in Georgia, eastern Tennessee, and western North Carolina, prompting the evacuation of hundreds of residents. The Southwest rebounded from its wet El Niño years with extraordinarily dry conditions in late 1998. The desert wildflowers that bloomed in abundance in the spring of 1998 were noticeably absent in 1999. Precipitation was so scarce in the mountains that Arizona ski resorts took the unusual step of closing for the 1998–1999 season. Even desert oases dried up in Arizona, prompting mountain lions and other desert-dwellers to enter populated areas and drink water out of chlorinated swimming pools. In some parts of Texas the drought began as early as August 1998. By April 1999, a state of emergency had been announced for 170 of the state's 254 counties. The wet zone As predicted for La Niña, the Pacific Northwest received above average precipitation during the winter of 1998–99. In Washington State, Mount Baker logged more than 90 feet (27 meters) of snow, rivaling Mount Rainier's record for snowfall in a single season. Lowland communities in the Pacific Northwest and in the Sierra Nevada Mountains in California faced threats of flooding. Around Puget Sound, precipitation fell for ninety-one straight days during the first quarter of 1999. In the early spring, the Midwest received heavy rains. The Midwest also experienced a series of tornadoes in early 1999; however, meteorologists are divided as to whether or not La Niña was responsible. As predicted, India, China, Southeast Asia, and Australia experienced unusually heavy monsoon rains and flooding. In the summer of 1998 flood waters stood chest-high in parts of eastern India and Bangladesh. River flooding in China claimed the lives of 4,150 people. In August 1999, several Asian countries were besieged by torrential downpours, and gale-force winds with speeds between 39 and 54 mph (63 and 87 kph). They also suffered from mud slides, landslides of mud caused by heavy rain, and flooding. Across South Korea, Thailand, the Philippines, and Vietnam tens of thousands of people were forced to evacuate their homes and at least fifty people perished in the floods. An active hurricane season A La Niña event began just in time for the 1998 hurricane season, which runs from June 1 through November 30. As predicted for a La Niña year, the Atlantic hurricane season was fast and furious. With a death toll of between twelve thousand and fifteen thousand, mostly in Central America and the Caribbean, 1998 was the deadliest Atlantic hurricane season 1780. The year 1998 saw fourteen tropical storms, ten of which became hurricanes and three of which became major hurricanes. Seven of the 1998 storms made landfall in the United States (the third highest number on record), causing twenty-one deaths and more than 3.2 billion dollars in damage. The real story of the 1998 hurricane season was a pair of monster hurricanes, Georges and Mitch, that brought death and destruction to Central America. A mild winter Predictions of an extra-cold winter of 1998–99 for the northern United States turned out to be wrong. The polar jet stream remained farther north than expected, trapping polar air north of the Canadian border. As a result, northerners enjoyed a relatively mild winter. The southern portion of the United States, south of the Rockies in the west and New York in the east, experienced spring-like conditions. Record high temperatures were set in many places. The winter of 1999–2000 In July 1999, meteorologists at the National Centers for Environmental Prediction, the National Oceanic and Atmospheric Administration, and other agencies predicted a continuing, but weakened, La Niña for the winter of 1999–2000. Forecasts included wet conditions for the Pacific Northwest and dry conditions for the South. The predictions were at least partly accurate. The winter proved to be one of the warmest and driest on record. It was driest in the Southwest and Southeast with Louisiana recording abnormally dry conditions. The Northwest was average to wet, with Montana and Wyoming experiencing above-average precipitation. Weather experts also predicted a lively hurricane season for autumn 1999, anticipating at least three major storms. This prediction proved accurate. There were twelve named storms, eight hurricanes, and five intense hurricanes during the season. Scientists continue to try and predict the effects of El Niño and La Niña. In the fall of 2006, the Climate Prediction Center/National Centers for Environmental Prediction (a branch of the National Weather Service) predicted El Niño conditions developing for the 2006–2007 winter season. These conditions included warmer-than-normal temperatures for western and central Canada, and the western and northern United States. Wetter-than-average conditions were predicted for portions of the U.S. Gulf Coast and Florida, while drier-than-average conditions were expected in the Ohio River Valley and the Pacific Northwest. As of February 2007 scientists saw a return to ENSO-neutral conditions with the possible development of La Niña conditions. D'Aleo, Joseph S. The Oryx Resource Guide to El Niño and La Niña. Phoenix, AZ: Oryx Press, 2002. Philander, S. George. El Niño, La Niña, and the Southern Oscillation. San Diego, CA: Academic Press, 1990. Rose, Sally. El Niño! And La Niña. New York: Simon and Schuster, 1999. Trenberth, Kevin. "Uncertainty in Hurricanes and Global Warming." Science. (17 Jun 2005): pp. 1753-1754. "La Niña." National Aeronautic and Space Administration: Earth Observatory. 〈http://earthobservatory.nasa.gov/Library/LaNina/〉 (accessed March 8, 2007). "La Niña." National Oceanic and Atmospheric Administration. 〈http://www.elnino.noaa.gov/lanino.html〉 (accessed March 9, 2007). "La Niña." UXL Encyclopedia of Weather and Natural Disasters. . Encyclopedia.com. (May 24, 2019). https://www.encyclopedia.com/environment/encyclopedias-almanacs-transcripts-and-maps/la-nina-0 "La Niña." UXL Encyclopedia of Weather and Natural Disasters. . Retrieved May 24, 2019 from Encyclopedia.com: https://www.encyclopedia.com/environment/encyclopedias-almanacs-transcripts-and-maps/la-nina-0 Encyclopedia.com gives you the ability to cite reference entries and articles according to common styles from the Modern Language Association (MLA), The Chicago Manual of Style, and the American Psychological Association (APA). Within the “Cite this article” tool, pick a style to see how all available information looks when formatted according to that style. Then, copy and paste the text into your bibliography or works cited list. 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1,482	> Reciprocal (Multiplicative Inverse) # Reciprocal (Multiplicative Inverse) The reciprocal meaning is something that is the opposite of the given value. It is a natural fact that the product of the number and its reciprocal will always be identity value i.e. 1. Reciprocal in math is a common computation. In this article, students will learn this concept in an easy manner with suitable examples. Let us learn it! ## Reciprocal (Multiplicative Inverse) In the number system theory of mathematics identity and inverse are the two essential computations. These two are related to each other. The multiplicative inverse of a given number n is the value represented as $$\frac{1}{n}$$. It is also termed as the reciprocal value of the number. Source: wikihow.com ### Reciprocal Definition: The multiplicative inverse or reciprocal of a number x is simply denoted as $$x^{-1}$$ and computed as $$\frac{1}{x}$$. Thus: Multiplicative inverse (x) = $$\frac{1}{x}$$ orÂ $$x^{-1}$$ Thus, xÂ Ã— $$x^{-1}$$ = 1 It is also popular as the reciprocal of the number given. It must be noted that 1 is known as the multiplicative identity of any real number. For example, the multiplicative inverse of 5 will be $$\frac{1}{5}$$. Also, the reciprocal of zero is termed as $$\frac{1}{0}$$ i.e. $$\infty$$. Thus the student can see that reciprocal mathematics is easy but important. ### Reciprocal Agreement Definition: In this regard, one important term is the reciprocal agreement. Reciprocal agreement definition says that it is a reciprocal action or agreement which involves two terms who do the same thing to each other or agree to help each other in a similar way. It means if x is reciprocal of y then, y will also be the reciprocal of x. We can see that, If x = $$\frac{1}{y}$$ then y = $$\frac{1}{x}$$ Thus x and y will be reciprocal of each other. ### How to find reciprocal? Reciprocal math is very easy for calculation. For many problems and applications, we can use this process. We can compute reciprocal as follows: 1. Reciprocal of a Number: If x is an integer number such as 1,2,3,4,â€¦ , then the multiplicative inverse of x will be $$\frac{1}{x}$$. For example, the multiplicative inverse of 9 is $$\frac{1}{9}$$. 1. Reciprocal of a Fraction: If x is any rational number like $$\frac{p}{q}$$, then its reciprocal will be $$\frac{q}{p}$$. For example, the multiplicative inverse of $$\frac{3}{4}$$ will be $$\frac{4}{3}$$. Consider some more examples. The multiplicative inverse of 3 will be $$\frac{1}{3}$$, of $$\frac{-1}{3}$$ is -3, of 8 is $$\frac{1}{8}$$ and of 4/7 will be $$\frac{-7}{4}$$. But the multiplicative inverse of zero will be infinite, as $$\frac{1}{0}$$ is infinity. Also, whereas the multiplicative inverse of 1 will be 1 itself. ## Solved Examples for You Q.1: Determine the reciprocals up to 30. Solution : As we know that reciprocal of x = $$\frac{1}{x}$$. Thus we have the solution as in the given table. Value of x Reciprocal = $$\frac{1}{x}$$ 1 $$\frac{1}{1}$$ 2 $$\frac{1}{2}$$ 3 $$\frac{1}{3}$$ 4 $$\frac{1}{4}$$ 5 $$\frac{1}{5}$$ 6 $$\frac{1}{6}$$ 7 $$\frac{1}{7}$$ 8 $$\frac{1}{8}$$ 9 $$\frac{1}{9}$$ 10 $$\frac{1}{10}$$ 11 $$\frac{1}{11}$$ 12 $$\frac{1}{12}$$ 13 $$\frac{1}{13}$$ 14 $$\frac{1}{14}$$ 15 $$\frac{1}{15}$$ 16 $$\frac{1}{16}$$ 17 $$\frac{1}{17}$$ 18 $$\frac{1}{18}$$ 19 $$\frac{1}{19}$$ 20 $$\frac{1}{20}$$ 21 $$\frac{1}{21}$$ 22 $$\frac{1}{22}$$ 23 $$\frac{1}{23}$$ 24 $$\frac{1}{24}$$ 25 $$\frac{1}{25}$$ 26 $$\frac{1}{26}$$ 27 $$\frac{1}{27}$$ 28 $$\frac{1}{28}$$ 29 $$\frac{1}{29}$$ 30 $$\frac{1}{30}$$ Q.2 : Find the reciprocal of $$\frac{5}{8}$$. Solution: let x = $$\frac{5}{8}$$. Since, $$\frac{5}{8}$$Â Ã— $$\frac{8}{5}$$ = 1 Therefore, $$x^{-1}$$ = $$\frac{8}{5}$$. Q.3: Find the reciprocal of $$a^{3}$$. Solution: The reciprocal of $$a^{3}$$ will be: $$\frac{1}{a^{3}}$$ , as $$\frac{1}{a^{3}}$$ Ã— $$a^{3}$$ = 1. Thus reciprocal of $$a^{3}$$ will be $$\frac{1}{a^{3}}$$ i.e. $$a^{-3}$$. Share with friends ## Customize your course in 30 seconds ##### Which class are you in? 5th 6th 7th 8th 9th 10th 11th 12th Get ready for all-new Live Classes! Now learn Live with India's best teachers. Join courses with the best schedule and enjoy fun and interactive classes. Ashhar Firdausi IIT Roorkee Biology Dr. Nazma Shaik VTU Chemistry Gaurav Tiwari APJAKTU Physics Get Started Subscribe Notify of ## Question Mark? Have a doubt at 3 am? Our experts are available 24x7. Connect with a tutor instantly and get your concepts cleared in less than 3 steps.<\|endoftext\|>	4.71875
984	# Find the values of x • Jul 1st 2009, 05:36 AM mark1950 Find the values of x such that $\displaystyle -2 < x^3 - 2x^2 + x - 2 < 0$. • Jul 1st 2009, 05:51 AM alexmahone $\displaystyle -2<x^3-2x^2+x-2<0$ $\displaystyle -2<(x-2)(x^2+1)<0$ $\displaystyle x^2+1$ is positive Dividing throughout by $\displaystyle x^2+1$, $\displaystyle -2<(x-2)<0$ $\displaystyle 0<x<2$ • Jul 1st 2009, 05:59 AM TheAbstractionist Quote: Originally Posted by mark1950 such that $\displaystyle -2 < x^3 - 2x^2 + x - 2 < 0$. Concentrate on one side of the inequality at a time. $\displaystyle -2 < x^3-2x^2+x-2$ $\displaystyle \implies\ 0<x^3-2x^2+x=x(x^2-2x+1)=x(x-1)^2$ Solive this, then go on to the next inequality. $\displaystyle x^3-2x^2+x-2<0$ Solve this (hint: $\displaystyle x-2$ is a factor of the LHS). Then combine the two solutions. Quote: Originally Posted by alexmahone $\displaystyle -2<(x-2)(x^2+1)<0$ $\displaystyle x^2+1$ is positive Dividing throughout by $\displaystyle x^2+1$, $\displaystyle -2<(x-2)<0$ $\displaystyle -2$ divided by $\displaystyle x^2+1$ is not $\displaystyle -2.$ (Shake) • Jul 1st 2009, 06:06 AM alexmahone -deleted- • Jul 1st 2009, 06:09 AM alexmahone Quote: Originally Posted by TheAbstractionist $\displaystyle -2$ divided by $\displaystyle x^2+1$ is not $\displaystyle -2.$ (Shake) I'm aware of the fact; nevertheless dividing by a positive term does not change the inequality. My solution is correct. • Jul 1st 2009, 06:11 AM TheAbstractionist I mean if you divide $\displaystyle -2<(x-2)(x^2+1)<0$ by $\displaystyle x^2+1$ you should get $\displaystyle \frac{-2}{x^2+1}<x-2<0$ not $\displaystyle -2<x-2<0.$ In any case, $\displaystyle x=1$ is not a solution. • Jul 1st 2009, 06:21 AM mark1950 Hmm...alex, the answer given is 0<x<1 or 1<x<2. Why is it so? • Jul 1st 2009, 06:30 AM alexmahone Quote: Originally Posted by mark1950 Hmm...alex, the answer given is 0<x<1 or 1<x<2. Why is it so? That's because x=1 is NOT a solution. • Jul 1st 2009, 06:31 AM pfarnall Quote: Originally Posted by alexmahone I'm aware of the fact; nevertheless dividing by a positive term does not change the inequality. My solution is correct. Think about this example and you'll see you're in error: $\displaystyle 0<2x<1$ We can all agree that 2 is positive so by your working: $\displaystyle 0<x<1$ Clearly this is not true and you must divide all parts of the inequality by 2 not just the middle term. • Jul 1st 2009, 06:33 AM alexmahone Quote: Originally Posted by pfarnall Think about this example and you'll see you're in error: $\displaystyle 0<2x<1$ We can all agree that 2 is positive so by your working: $\displaystyle 0<x<1$ Clearly this is not true and you must divide all parts of the inequality by 2 not just the middle term. Fair enough. Sorry.<\|endoftext\|>	4.40625
405	# 2016 AMC 8 Problems/Problem 23 ## Problem Two congruent circles centered at points $A$ and $B$ each pass through the other circle's center. The line containing both $A$ and $B$ is extended to intersect the circles at points $C$ and $D$. The circles intersect at two points, one of which is $E$. What is the degree measure of $\angle CED$? $\textbf{(A) }90\qquad\textbf{(B) }105\qquad\textbf{(C) }120\qquad\textbf{(D) }135\qquad \textbf{(E) }150$ ## Solutions ### Solution 1 Observe that $\triangle{EAB}$ is equilateral. Therefore, $m\angle{AEB}=m\angle{EAB}=m\angle{EBA} = 60^{\circ}$. Since $CD$ is a straight line, we conclude that $m\angle{EBD} = 180^{\circ}-60^{\circ}=120^{\circ}$. Since $BE=BD$ (both are radii of the same circle), $\triangle{BED}$ is isosceles, meaning that $m\angle{BED}=m\angle{BDE}=30^{\circ}$. Similarly, $m\angle{AEC}=m\angle{ACE}=30^{\circ}$. Now, $\angle{CED}=m\angle{AEC}+m\angle{AEB}+m\angle{BED} = 30^{\circ}+60^{\circ}+30^{\circ} = 120^{\circ}$. Therefore, the answer is $\boxed{\textbf{(C) }\ 120}$. ## Video Solution ~Education, the Study of Everything ~ pi_is_3.14 ~savannahsolver<\|endoftext\|>	4.40625
971	# What is meant by odd prime number? ## What is meant by odd prime number? n. (Mathematics) an integer that cannot be factorized into other integers but is only divisible by itself or 1, such as 2, 3, 5, 7, and 11. Sometimes shortened to: prime Compare composite number. ### How do you find odd prime numbers? In simple words, if a number is only divisible by 1 and itself, then it is a prime number. Every prime number is an odd number except number 2. What is the two odd prime? All the prime numbers are odd except 2. Prime numbers are those natural numbers which have no other factor than 1 and the number itself. Complete step-by-step answer: The odd prime numbers before 18 are 3, 5, 7, 11, 13, 17. What are the first 10 odd prime numbers? So, the first 10 odd prime numbers are as follows $3,\text{ 5, 7, 11, 13,17, 19, 23, 29, 31}$. ## Is 9 an odd prime number? Or a number having 1, 3, 5, 7 and 9 at its units place is called an odd number. Number s which have only two factors namely 1 and the number itself are called prime numbers. For example: 2, 3, 5, 7, 11, 19, 37 etc are prime numbers. ### Is an odd prime number? 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 (sequence A000040 in the OEIS). . Therefore, every prime number other than 2 is an odd number, and is called an odd prime. How do you express two odd primes? 36 – 17 = 19, is a prime number. So the 4th pair is (17,19). Therefore, The sum of two odd primes is (5,31),(7, 29),(13, 23),(17,19). Which is the smallest prime number? 2 2 is the smallest prime number. It also the only even prime number – all other even numbers can be divided by themselves, 1 and 2 at least, meaning they will have at least 3 factors. ## Which is the smallest even number? What Is the Smallest Even Number? 2 is the smallest even number. It is also the only even prime number. ### Is 1 a odd number? One is the first odd positive number but it does not leave a remainder 1. Some examples of odd numbers are 1, 3, 5, 7, 9, and 11. An integer that is not an odd number is an even number. If an even number is divided by two, the result is another integer. What are the even prime numbers? An even prime number is a number which is both even (i.e., divisible by 2) and prime (i.e., divisible by exactly 2 numbers, 1 and itself). This means that 2 is the only even number that is prime, as any other even number would be divisible by 2 and itself, which would make that number not prime. What are the first 100 prime numbers? The first prime numbers chart has the 25 prime numbers that are in the first 100 numbers (in sequential order: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97). Except for the number 1, the composite numbers are black and the prime numbers are light blue. ## What are all the even numbers? Even Numbers. Any integer that can be divided exactly by 2 is an even number. The last digit is 0, 2, 4, 6 or 8. Example: −24, 0, 6 and 38 are all even numbers. ### What are the uses of prime numbers? Prime numbers are also useful in generating random numbers. They helps us in avoid pattern and arrive at actual random series. Prime numbers are also used in designing gears.<\|endoftext\|>	4.5
718	Pan-European revolutions of 1830 manifested in different forms in different regions. In Netherlands and France they took a romantic hue, whereas in Poland and Switzerland the impact on the political establishment was less pronounced. In the United Kingdom of Netherlands and in France, the impact of the revolution was to establish constitutional monarchies (also called commonly as ‘popular monarchies’). This meant that the older aristocratic order was dismantled and republicanism was given a new thrust. For example, prior to the revolution, the king held dominion over his country through the mandate of God. His reference as the King of France testified this fact. But after the revolution, his title was changed to King of the French, indicating how his authority is derived from the collective will of the citizens. Likewise, in Belgium, King Leopold I took to the throne under the reconfigured political arrangement. At the same time in Congress Poland the revolt against the incumbent Czar of Russia ended as a failure. The reverberations of events of July 1830 were being felt elsewhere in Switzerland and Italy as well. In Switzerland, on the back of deep resentment from the peasantry, changes were made to the constitution. Greater representation was sought and implemented to the local legislatures and cantonments. In Italy, the 1830 revolution had the effect of unifying the erstwhile petty kingdoms and dukedoms. The role of the Duke of Modena (Francais IV) was instrumental to this unification project, partly owing to his own ambitions for power. The previously Papal Legations scattered across Italy revolted in unison and adopted the tricolor in the place of the Papal flag. A united Italian nation was thus born. The revolutions of 1830 were by and large politically progressive. They removed some of the ossified and regressive tendencies in prevailing monarchy systems. In their stead, the revolutions brought an element of republicanism, representation and overall reform. To this extent the 1830 revolutions will have to be considered as a success. The revolutions that occurred across Europe in 1848 are also known as Spring of Nations. Fitting to this metaphor, nations across Europe witnessed radical political upheavals. But unfortunately, the great promise that it held in its outbreak were soon quelled by reactionary forces. France was the epicenter for this great revolution, the influence of which was felt as far out as Latin America. The 1848 revolutions shared the same undercurrents as the 1830s event in terms of public resentment toward the functioning of monarchies. People everywhere felt great dissatisfaction over their lack of political franchise. This was more acutely felt by the working classes than the propertied classes. A renewed invocation of nationalist fervor was another crucial element in the outbreak of the 1848 revolutions. The outcomes of the 1848 revolutions were somewhat mixed. Since there was no central organization behind the revolts, their entirety did not produce synergy. Many brave people laid their lives for the cause, with fatalities running into tens of thousands. Feudalism was abolished in some nations as a result of the uprisings, just as absolute monarchy was made unviable after the events of 1848. The benign consequences of the revolutions had an enduring effect in Germany, France, Italy and Poland. But the rest of Europe remained tied to traditional systems of monarchy. All things considered, the 1848 revolutions were only partially successful in achieving its goals. Mark Kishlansky, Patrick Geary, Patricia O’Brien. Civilization in the West – Since 1555. Published by Penguin Academics.<\|endoftext\|>	3.859375
561	The Cancer Genome Atlas Project Scientists have mapped specific genes that turn normal healthy cells into cancerous cells through The Cancer Genome Atlas Project (TCGA). Through the TCGA, researchers have now developed a more reliable scientific method to identify these genes. As this research is shared around the globe, more accessible and effective cancer treatment options are being discovered. The ultimate goal of the TCGA is to create a catalogue of these “defective” genes, thereby offering increased detection, and ultimately, better prevention and treatment of cancer. As healthy cells grow and divide, they not only produce more healthy cells but also kill off cells that are no longer needed. A cancer cell is produced when an otherwise healthy cell grows abnormally or does not die off when it should. These cancerous cells then form a mass or tumor. Through the TCGA, researchers are trying to pinpoint why these specific genes change and how that can lead a cell to become cancerous. Once this information is gathered, cancer treatment will move to a more targeted form of treatment. Doctors will target only the abnormal gene, as opposed to today’s standard treatment of chemotherapy, which affects the whole body. Such a change in treatment will be beneficial because standard chemotherapy kills off cancerous cells but also some healthy cells in the process. If researchers can instead identify the “defective” genes and target them directly, healthy cells will stay intact and be less affected by the cancer treatment. For example, the National Institutes of Health recently announced that TCGA identified distinct subtypes of glioblastoma multiforme, the most common form of malignant brain cancer in adults. In the past, doctors treated this type of brain tumor as a single disease, whereas now they know that glioblastoma multiforme is, in fact, four distinct molecular subtypes. In addition, through the TCGA, researchers have also discovered that response to chemotherapy and radiation differed by gene subtype. Although the current standard of treatment will not change overnight, these new discoveries will help doctors to tailor a cancer treatment plan using genetic information. Originally a pilot project, TCGA has turned out to be quite a success in demonstrating the value of cancer research. In the fall of 2009, President Obama announced that the National Institutes of Health will spend $275 million to expand TCGA to other types of cancer, thus opening the door for more specific treatment options and hopefully more cures. To learn more about the cancer cells we have in our body, please read our January 27 blog post at: http://patientnavigator.com/blog/2010/01/27/we-all-carry-cancer-cells/ For further information on The Cancer Genome Atlas Project, please visit:<\|endoftext\|>	3.796875
2,635	The first Slovak orthography was proposed by Anton Bernolák (1762–1813) in his Dissertatio philologico-critica de litteris Slavorum, used in the six-volume Slovak-Czech-Latin-German-Hungarian Dictionary (1825–1927) and used primarily by Slovak Catholics. The standard orthography of the Slovak language is immediately based on the standard developed by Ľudovít Štúr in 1844 and reformed by Martin Hattala in 1851 with the agreement of Štúr. The then-current (1840s) form of the central Slovak dialect was chosen as the standard. It uses the Latin script with small modifications that include the four diacritics (ˇ, ´, ¨, ˆ) placed above certain letters. After Hattala's reform, the Slovak language remained mostly unchanged. The Slovak alphabet is an extension of the Latin alphabet used for writing the Slovak language. It has 46 letters which makes it the longest Slavic and European alphabet. The 46 letters of the Slovak alphabet are: \|Majuscule forms (also called uppercase or capital letters)\| \|Minuscule forms (also called lowercase or small letters)\| \|Á á\|\|dlhé á\| \|Ä ä\|\|prehlasované á;\| a s dvoma bodkami; \|É é\|\|dlhé é\| \|Í í\|\|dlhé í\| \|Ĺ ĺ\|\|dlhé el\| \|Ľ ľ\|\|mäkké el\| \|Ŕ ŕ\|\|dlhé er\| \|Ť ť\|\|mäkké té\| \|Ú ú\|\|dlhé ú\| \|W w1\|\|dvojité vé\| \|Ý ý\|\|dlhý ypsilon\| - The letters Q, W and X are only used in loanwords. \|ia\|\|/ɪ̯a/\|\|May be pronounced as two monophthongs /i.a/ in foreign words.\| \|ie\|\|/ɪ̯ɛ/\|\|May be pronounced as two monophthongs /i.ɛ/ in foreign words.\| \|iu\|\|/ɪ̯u/\|\|May be pronounced as two monophthongs /i.u/ in foreign words.\| \|l\|\|/l/\|\|Can be syllabic /l̩/.\| \|n\|\|/n/\|\|Becomes [ŋ] before /k/ and /ɡ/.\| \|q\|\|/kv/\|\|Only occurs in loanwords.\| \|r\|\|/r/\|\|Can be syllabic /r̩/.\| \|w\|\|/v/\|\|Only occurs in loanwords.\| \|x\|\|/ks/\|\|Only occurs in loanwords.\| Some additional notes includes the following (transcriptions in IPA unless otherwise stated): The primary principle of Slovak spelling is the phonemic principle. The secondary principle is the morphological principle: forms derived from the same stem are written in the same way even if they are pronounced differently. An example of this principle is the assimilation rule (see below). The tertiary principle is the etymological principle, which can be seen in the use of i after certain consonants and of y after other consonants, although both i and y are pronounced the same way. Finally, the rarely applied grammatical principle is present when, for example, the basic singular form and plural form of masculine adjectives are written differently with no difference in pronunciation (e.g. pekný = nice – singular versus pekní = nice – plural). Most foreign words receive Slovak spelling immediately or after some time. For example, "weekend" is spelled víkend, "software" - softvér, "gay" - gej (both not exclusively), and "quality" is spelled kvalita (possibly from Italian qualità). Personal and geographical names from other languages using Latin alphabets keep their original spelling unless a fully Slovak form of the name exists (e.g. Londýn for "London"). To accelerate writing, a rule has been introduced that the frequent character combinations ďe, ťe, ňe, ľe, ďi, ťi, ňi, ľi, ďí, ťí, ňí, ľí, ďie, ťie, ňie, ľie, ďia, ťia, ňia, ľia are written without a caron de, te, ne, le, di, ti, ni, li, dí, tí, ní, lí, die, tie, nie, lie, dia, tia, nia, lia. These combinations are usually pronounced as if a caron were found above the consonant. Some exceptions are as follows: - foreign words (e.g. telefón is pronounced with a hard t and a hard l) - the following words: ten (that), jeden (one), vtedy (then), teraz (now) - nominative masculine plural endings of pronouns and adjectives do not "soften" preceding n, d, t, l (e.g. tí odvážni mladí muži /tiː ɔdvaːʒni mladiː muʒi/, the/those brave young men) - short e in adjectival endings, which is derived from long é shortened by the "rhythmical rule" (see below), does not "soften" preceding n, d, t, l (e.g. krásne stromy /kraːsnɛ strɔmi/, beautiful trees, c.f. zelené stromy /zɛʎɛnɛː strɔmi/, green trees) The letter ľ is nowadays pronounced by many speakers, particularly from western Slovakia, as a non-palatalized l. When a voiced obstruent (b, d, ď, dz, dž, g, h, z, ž) is at the end of the word before a pause, it is pronounced as its voiceless counterpart (p, t, ť, c, č, k, ch, s, š, respectively). For example, pohyb is pronounced /pɔɦip/ and prípad is pronounced /priːpat/. When "v" is at the end of the syllable, it is pronounced as non-syllabic u [ʊ̯], with the exception of the position before "n" or "ň". For example, kov [kɔʊ̯] (metal), kravský [kraʊ̯skiː] (cow - adjective), but povstať [pɔfstac] (uprise), because the "v" is not at the end of the syllable (po-vstať) and hlavný [ɦlaʋniː] because "v" stands before the "n". The graphic group -ou (at the end of words) is pronounced [ɔʊ̯] but is not considered a diphthong. Its phonemic interpretation is /ɔv/. Consonant clusters containing both voiced and voiceless elements are entirely voiced if the last consonant is a voiced one, or voiceless if the last consonant is voiceless. For example, otázka is pronounced /ɔtaːska/ and vzchopiť sa is pronounced [fsxɔpit sa]. This rule applies also over the word boundary. One example is as follows: prísť domov [priːzɟ dɔmɔʊ̯] (to come home) and viac jahôd [ʋɪ̯adz jaɦʊ̯ɔt] (more strawberries). The voiced counterpart of "ch" /x/ is [ɣ], and the unvoiced counterpart of "h" /ɦ/ is [x]. A long syllable (that is, a syllable containing á, é, í, ý, ó, ú, ŕ, ĺ, ia, ie, iu, ô) cannot be followed by another long one in the next syllable within the same word. This rule has morphonemic implications for declension (e.g. žen-ám but tráv-am) and conjugation (e.g. nos-ím but súd-im). Several exceptions of this rule exist. It is typical of the literary Slovak language, and does not appear in Czech or in some Slovak dialects. One of the most important changes in Slovak orthography in the 20th century was in 1953 when s began to be written as z where pronounced [z] in prefixes (e.g. smluva into zmluva as well as sväz into zväz). The phonemic principle has been given priority over the etymological principle in this case. Slovak features some heterophonic homographs (words with identical spelling but different pronunciation and meaning), the most common examples being krásne /ˈkraːsnɛ/ (beautiful) versus krásne /ˈkraːsɲɛ/ (beautifully). The acute mark (in Slovak "dĺžeň", "prolongation mark" or "lengthener") indicates length (e.g. í = approximately [iː]). This mark may appear on any vowel except "ä" (wide "e", široké "e" in Slovak). It may also appear above the consonants "l" and "r", indicating the long [lː] and [rː] sounds. The umlaut ("prehláska", "dve bodky" = two dots) is only used above the letter "a". It indicates a raised vowel, almost an "e", similar to German ä. The caron (in Slovak "mäkčeň", "palatalization mark" or "softener") indicates a change of alveolar fricatives into either post-alveolar or palatal consonants, in informal Slovak linguistics often called just "palatalization". Eight consonants can bear a caron. Not all "normal" consonants have a "caroned" counterpart: - In printed texts, the caron is printed in two forms: (1) č, dž, š, ž, ň and (2) ľ, ď, ť (looking more like an apostrophe), but this is just a convention. In handwritten texts, it always appears in the first form. - Phonetically, two forms of "palatalization" exist: ľ, ň, ď, ť are palatal consonants, while č, dž, š, ž are postalveolar affricates and fricatives. The Slovak alphabet is available within the ISO/IEC 8859-2 “Latin-2” encoding, which generally supports Eastern European languages. All vowels, but none of the specific consonants (that is, no č, ď, ľ, ĺ, ň, ŕ, š, ť, ž) are available within the “Latin-1” encoding, which generally supports only Western European languages. - Kráľ, Ábel (1988). Pravidlá slovenskej výslovnosti. Bratislava: Slovenské pedagogické nakladateľstvo.<\|endoftext\|>	3.796875
792	‘Music is a universal language that embodies one of the highest forms of creativity. A high quality music education should engage and inspire pupils to develop a love of music and their talent as musicians, and so increase their self-confidence, creativity and sense of achievement. As pupils progress, they should develop a critical engagement with music, allowing them to compose, and to listen with discrimination to the best in the musical canon.’ (National Curriculum 2014) Early Years (Nursery) children will join in with dancing and ring games, learn a range of simple songs, tap out rhythms and explore sounds and how they can be changed. In addition, the Reception children will build a repertoire of songs and dances and explore the different sounds of instruments to make music. In Key stage 1 the children will use their voices expressively and creatively by singing songs and speaking chants and rhymes; play tuned and un-tuned instruments musically; listen with concentration and understanding to a range of music; and experiment with, create, select and combine sounds. The Key stage 2 children will sing and play musically with increasing confidence and control. They will work on developing an understanding of musical composition, organising and manipulating ideas within musical structures and reproducing sounds from aural memory. There will be opportunities to play and perform in solo and group contexts, using their voices and playing musical instruments with increasing accuracy, fluency, control and expression; improvise and compose music for a range of purposes; listen with attention to detail and recall sounds; use some musical notations; appreciate and understand a wide range of music drawn from different traditions and from great composers and musicians; and develop an understanding of the history of music. All pupils have the opportunity to participate in concerts during the year. These are performed to parents and invited guests and include aspects of music and dance. In years 3 and 4 the children will learn to play the recorder and in years 5 and 6 they will be able to learn to play the flute. In addition, there will be the opportunity for some children to learn to play the guitar and piano. Links with Other Curriculum Areas English – Music teaching helps to develop language and vocabulary and expression. It supports the teaching of rhyme and intonation and the development of breath control, essential for reading aloud. Maths – Rhythm in music can support the children’s understanding of pattern and time. Science – As children sing their bodies undergo particular changes related to their breathing and movement of their vocal chords. Instrumental work in music links with vibration and sound waves and knowledge of the senses. Art/DT – Children have opportunities to design and make simple percussion and stringed instruments using a range of materials. Looking at artworks can inspire children to create their own music, stimulated by the colours, shapes and items depicted and the moods and emotions they evoke. Humanities – By studying particular pieces of music and their composers and performers, the children will learn about the time and place from which the music originated. PSHE – By composing and performing as part of a group, the children learn how to work together and value the contributions of others. Talking about musical works helps children to learn more about themselves, their likes, dislikes and feelings. They will be encouraged to respond critically to pieces of music, providing reasons for their responses. From the autumn term 2016, individual piano lessons are being offered to pupils in years 3, 4, 5 and 6. They can be 15, 20 or 30 minutes as decided by parents and teachers. Pupils can be prepared for Grade exams and there will be opportunities for them to perform solos and/or duets in assemblies and at a pupils’ concert held annually at Wingfield Barns. Post Year 6, pupils will be able to continue their lessons with Dinah locally if they wish.<\|endoftext\|>	4.125
499	What is 0.56 as a Fraction? 0.56 as a fraction is 56/100. It can be reduced to 28/50. Checkout this video: What is a Fraction? In mathematics, a fraction is a number that represents a part of a whole. It is written with a numerator (a top number) over a denominator (a bottom number). For example, if we have 3 pieces of candy and we want to divide them equally among ourselves, each person would get 1 piece of candy. This can be represented using the fraction . The candy can be thought of as the whole and each person would be 1/3 of that whole. What is a Decimal? A decimal is a number that represents a fraction. The decimal point separates the whole number part from the fractional part of the number. In the number 0.56, the “0” is the whole number and the “56” is the fractional part. How to Convert 0.56 to a Fraction To convert a decimal like 0.56 to a fraction, we place the decimal point (or dot) in the middle, like this: 0.56 We then count how many digits are on each side of the dot. In this case, there are two digits on the left of the dot and two digits on the right. This means that our answer will be a mixed number (or whole number and a fraction). The next step is to make an equivalent fraction using 100 as our denominator (or bottom number). We do this by counting how many “tenths” there are in 1. In other words, we need to ask ourselves how many times can we divide 1 by 10 before we get to 0? The answer is 10: 1 ÷ 10 = 0.1 10 ÷ 10 = 1 As you can see from the division above, when we divide 1 by 10 we get 0.1 remaining. So, if we divide 100 by 10 we should get 10 remaining: 100 ÷ 10 = 10 (remainder) How to Convert 0.56 to a Decimal To convert 0.56 to a decimal, divide the numerator (56) by the denominator (100): 0.56 = 56 ÷ 100 = 0.56 Therefore, 0.56 as a fraction is 56/100.<\|endoftext\|>	4.65625
414	## Octagons and Squares By extending the sides of a regular polygon like this octagon, we can make a star. We call this star a stellation of the polygon. How do the side lengths of the octagon and the stellation compare? How do their perimeters compare? Keep reading to find out, or dive right into the challenge for a more extended question. Let's see if we can use the symmetry of both figures to help. All of the side lengths of the octagon are equal, and so are the side lengths of the stellation. That cuts down on our work since, by finding one side length of either shape, we've found every side length of the shapes. Let's label the length of one of the octagon's sides as $1.$ To which side (or sides) of the stellation should we relate this side? The octagon's side forms a right triangle with two sides of the stellation. Those other two sides have equal lengths, so our right triangle is isosceles. That makes it a $45^\circ\text{-}45^\circ\text{-}90^\circ$ right triangle. The hypotenuse of such a triangle is $\sqrt2$ times as long as either leg. This means the two legs of our triangle each have a length of $1 \div \sqrt2 = \frac{1}{\sqrt2}.$ The octagon's sides are $\mathit{\sqrt2}$ times as long as the stellation's sides. The stellation has twice as many sides, though, so its perimeter is \begin{aligned} 2 \times \frac{1}{\sqrt2} & = \frac{2}{\sqrt2} = \sqrt2 \end{aligned} times as long as the octagon's perimeter. Does either of these facts help with the challenge below? # Today's Challenge The figure below is made of regular octagons and squares. Which triangles have the greater total area?<\|endoftext\|>	4.5625
1,469	According to all four Gospel narratives, the people were hungry and a young boy surrendered his lunch of five loaves and two fishes. But what was that among a multitude of more than 5000 men, not counting the women and children? Still as Matthew Henry states: “Those who have but a little, yet when the necessity is urgent, must relieve others out of that little, and that is the way to make it more.” This paltry provision was brought in faith to Christ and was taken by the Lord, who then looked toward heaven and having blessed it, gave thanks. Afterward, so miraculous was the distribution of it that everyone did eat and was filled. So plentiful was the expansion of the meager measure there were twelve baskets full of left overs. America’s beginning was similar. When the Pilgrims arrived in the New World, escaping religious persecution, they became hungry. Food was scarce and supplies low. Nearly half of their band died in the first winter. However, Squanto, an English speaking Native American who was briefly a slave and had also been introduced to the Christian faith, offered to help. Like the Pilgrims, Squanto was also extremely impoverished. After having been kidnapped and stolen away from his kindred, upon his fateful return home he learned his entire tribe, the Patuxets, had been wiped out by a sudden plague. He wandered aimlessly for a time through the forests of his childhood grieving his loss. Having no where else to go, he settled with Massasoit, a chieftain of the Wampanoag. Later Sqaunto’s English speaking and interpreting skills would become instrumental in forging a forty year peace treaty between the Wampanoag and the Pilgrims. Before this, Squanto was in despair and thought he had no reason for living. But now with a renewed sense of purpose in life, he would adopt the Pilgrims as his own people and help them survive. He would give them what he had – his knowledge. He would teach them how to fish, plant and harvest corn. He would teach them to hunt, refine maple syrup from maple trees, harvest berries, and discern what plants were good for medicines. He would also introduce them to the trade of beaver pelts and prosper them economically. Peter Marshall, in his book, The Light and the Glory, says that by the fall of 1621: “The Pilgrims were brimming over with gratitude – not only to Squanto and the Wampanoags who had been so friendly, but to their God. In Him they had trusted, and He had honored their obedience beyond their dreams. So, Governor Bradford declared a public day of Thanksgiving…Massasoit was invited, and unexpectedly arrived early – with ninety Indians! Counting their numbers, the Pilgrims had to pray hard to keep from giving in to despair. To feed such a crowd would cut deeply into the food supply that was supposed to get them through the winter… But if they had learned one thing through their travails, it was to trust God implicitly. As it turned out, the Indians were not arriving empty-handed. Massasoit had commanded his braves to hunt for the occasion, and they arrived with no less than five dressed deer, and more than a dozen fat wild turkeys! And they helped with the preparations, teaching the pilgrim women to make hoecakes and a tasty pudding out of cornmeal and maple syrup. Finally, they showed them an Indian delicacy: how to roast corn kernels in an earthen pot until they popped, fluffy and white – popcorn! The Pilgrims in turn provided many vegetables from their household gardens: carrots, onions, turnips, parsnips, cucumbers, radishes, beets and cabbages. Also, using some of their precious flour, they took summer fruits which the Indians had dried and introduced them to the likes of blueberry, apple and cherry pie. A joyous occasion for all!....Things went so well that Thanksgiving Day was extended for three days.” America would do well to remember these humble beginnings. Although this is not meant as a sanction of illegal immigration, let us not forget there was a time when we were the immigrants in a foreign land, perceived as intruders and desperately in need of friends. We were a minority fleeing another land with hope for a better tomorrow. We were poor, ignorant of the ways of this land and unable to speak the language. But God was with us and countenanced us in our poverty, destroying those bent on our destruction and blessing those that made room for us. Need we in our own hour of testing forget these foundations and fail to find a way to make peace on the issue of illegal immigration? Dare we forget it is only when we are willing to share that we should be given more? And may God hasten the day when He would raise up someone like Squanto in our midst – someone who truly understands both cultures and would lead us to find a balance between compassion for the lowly and fairness for those who legally occupy and own the land. Moreover, let us remember whether it is a broken and impoverished life, or a broken and impoverished people, the miracle of the feeding of the five thousand remind us God specializes in transforming a little into an overflowing of abundance. The difference is whether we give our challenges to Christ. Though America has grown over the centuries to become one of the wealthiest and most powerful nations in human history, Dr. George Sweeting, former President and current Chancellor of Moody Bible Institute, rightly contends: “We have lost much of our basic trust in God. No longer do we see clearly His wisdom, power, and love. We have put our trust in men and they are failing. We have laid aside the Bible, not only in our schools, but in our homes and in our public life as well. Small wonder we have lost our concept of sin, our condemnation of wrongdoing. Bloodshed and violence fill our land. Again and again, we have affirmed the rights of men at the cost of God’s rights. We have sold our birthright for humanism’s pottage…We have made the fatal error of thinking we can be wise and good without God’s help, that we can be great and happy and still reject salvation on God’s terms. We must turn back.” This Thanksgiving season, let us remember from whence we came – God’s loving-kindness to us when we were weak and vulnerable – the source of our abundance – to make room and share with others – to trust God implicitly in matters both personal and public – to recognize our constant need of Him - to repent and be thankful! Dr. Creech is a regular columnist for Agape Press, the national news wire of the American Family Association. His columns have also appeared on numerous other sites across the web, including: The Christian Post, MichNews.com, The Intellectual Conservative, Capitol Hill Coffee House, The North Carolina Conservative, The Conservative Voice, Worldview Weekend Network, Renew America, as well as a number of others.<\|endoftext\|>	3.6875
4,639	# DAV Class 7 Maths Chapter 1 Brain Teasers Solutions The DAV Class 7 Maths Book Solutions Pdf and DAV Class 7 Maths Chapter 1 Brain Teasers Solutions of Rational Numbers offer comprehensive answers to textbook questions. ## DAV Class 7 Maths Ch 1 Brain Teasers Solutions Question 1. A. Tick (✓) the correct option: (i) The value ofx such that $$\frac{-3}{8}$$ and $$\frac{x}{-24}$$ are equivalent rational numbers is- (a) 64 (b) -64 (c) -9 (d) 9 $$\frac{-3}{8}, \frac{x}{-24}$$ will be equivalent if -3 × -24 = x × 8 ⇒ x = $$\frac{-3 \times-24}{8}$$ = -3 × -3 = 9 Therefore, x = 9 Hence, (d) is the correct option. (ii) Which of the following is a negative rational number? (a) $$\frac{-15}{-4}$$ (b) 0 (c) $$\frac{-5}{7}$$ (d) $$\frac{4}{9}$$ $$\frac{-5}{7}$$ is a negative rational number as the signs of the numerator and the denominator are different. Hence, (c) is the correct option. (iii) In the given number line, which of the following rational numbers does the point M represent? (a) $$\frac{2}{8}$$ (b) $$\frac{3}{5}$$ (c) $$\frac{2}{3}$$ (d) $$\frac{12}{5}$$ The point M is represented by fraction $$\frac{12}{20}$$. Now, HCF of 12 and 20 is 4. ∴ $$\frac{12 \div 4}{20 \div 4}=\frac{3}{5}$$ which is the standard form. Hence, (b) is the correct answer. (iv) Which is the greatest rational number out of $$\frac{5}{-11}, \frac{-5}{12}, \frac{5}{-17}$$? (a) $$\frac{5}{-11}$$ (b) $$\frac{-5}{12}$$ (c) $$\frac{5}{-17}$$ (d) cannot be compared First, write $$\frac{5}{-11}, \frac{-5}{12}, \frac{5}{-17}$$ in standard form, i.e., $$\frac{-5}{11}, \frac{-5}{12}, \frac{-5}{17}$$ Now, find the products The products are -5 × 12 = -60 and -5 × 11= -55. Since, -60 < -55, therefore, $$\frac{-5}{11}<\frac{-5}{12}$$ Now, find the products The products are -5 × 17 = -85 and -5 × 12 = -60. Since -85 < -60, therefore, $$\frac{-5}{12}<\frac{-5}{17}$$. Therefore, $$\frac{-5}{17}$$ is the greatest rational number. Hence, (c) is the correct option. (v) Which of the following rational numbers is the smalles? (a) $$\left\|\frac{7}{11}\right\|$$ (b) $$\left\|\frac{-8}{11}\right\|$$ (c) $$\left\|\frac{-2}{11}\right\|$$ (d) $$\left\|\frac{-9}{-11}\right\|$$ Absolute values of the given rational numbers are $$\left\|\frac{7}{11}\right\|=\frac{7}{11},\left\|\frac{-8}{11}\right\|=\frac{8}{11},\left\|\frac{-2}{11}\right\|=\frac{2}{11}$$ and $$\left\|\frac{-9}{-11}\right\|=\frac{9}{11}$$ Now, the rational numbers $$\frac{7}{11}, \frac{8}{11}, \frac{2}{11}$$ and $$\frac{9}{11}$$ have same denominator, therefore, smaller the numerator, smaller will be the rational number. Since 2 < 7 < 8 < 9, therefore, $$\frac{2}{11}<\frac{7}{11}<\frac{8}{11}<\frac{9}{11}$$ $$\left\|\frac{-2}{11}\right\|<\left\|\frac{7}{11}\right\|<\left\|\frac{-8}{11}\right\|<\left\|\frac{-9}{-11}\right\|$$ Therefore, $$\left\|\frac{-2}{11}\right\|$$ is the smallest rational number. Hence, (c) is the correct answer (i) Find the average of the rational numbers $$\frac{4}{5}, \frac{2}{3}, \frac{5}{6}$$. (ii) How will you write $$\frac{12}{-18}$$ in th e standard form? (iii) How many rational numbers are there between any two rational numbers ? (iv) On the number line, the rational number $$\frac{-5}{-7}$$ lies on which side of zero? (v) Express $$\frac{-7}{-8}$$ as a rational number with denominator 40. (i) We Know Average = $$\frac{\text { Sum of all observations }}{\text { Number of observations }}$$ Therefore, average of the rational numbers $$\frac{4}{5}, \frac{2}{3}, \frac{5}{6}$$ is given by Average = $$\frac{\left(\frac{4}{5}+\frac{2}{3}+\frac{5}{6}\right)}{3}$$ = $$\frac{\left(\frac{24+20+25}{30}\right)}{3}=\frac{\left(\frac{69}{30}\right)}{3}$$ = $$\frac{69}{30 \times 3}$$ = $$\frac{23}{30}$$ (ii) Step 1: Convert the denominator into positive by multiplying numerator and denominator by -1. i.e., $$\frac{(12) \times(-1)}{(-18) \times(-1)}=\frac{-12}{18}$$ Step 2: Find HCF of 12 and 18, which is 6 in this case, and divide numerator and denominator by it. i.e., $$\frac{-12 \div 6}{18 \div 6}=\frac{-2}{3}$$ which is the standard form. (iii) Infinitely many rational numbers lie between any two rational numbers. (iv) First, convert the denominator into positive by multiplying and denominator by -1. ∴ $$\frac{-5 \times(-1)}{-7 \times(-1)}=\frac{5}{7}$$ $$\frac{5}{7}$$ > 0, therefore, $$\frac{5}{7}$$ lies to the left of the zero. (v) To get denominator 40, we must multiply the given denominator -8 by -5. i.e., $$\frac{(-7) \times(-5)}{(-8) \times(-5)}=\frac{35}{40}$$ Hence, the required rational number is $$\frac{35}{40}$$. Question 2. State whether the following statements are true. If not, then give an example in support of your answer. (i) If $$\frac{p}{q}>\frac{r}{s}$$ then $$\left\|\frac{p}{q}\right\|>\left\|\frac{r}{s}\right\|$$ If $$\frac{p}{q}>\frac{r}{s}$$ is true if $$\frac{p}{q}$$ and $$\frac{r}{s}$$ are both positive rational numbers. But $$\left\|\frac{p}{q}\right\|>\left\|\frac{r}{s}\right\|$$ means that $$\frac{p}{q}>\frac{r}{s}$$ and $$\frac{-p}{q}>\frac{-r}{s}=\frac{p}{q}<\frac{r}{s}$$ $$\left\|\frac{p}{q}\right\|>\left\|\frac{r}{s}\right\|$$ is not true statements (ii) If \|x\| = \|y\| then x = y If \|x\| = \|y \| ∴ x = y and -x = -y So it is true. (iii) $$\frac{p}{q}$$ is a non-zero rational number in standard form. It is necessary that rational number $$\frac{p}{q}$$ will be in standard form. No it is not necessary Question 3. Represent 5$$\frac{1}{3}$$ and $$\frac{-29}{4}$$ on a number line. 5$$\frac{1}{3}$$ = $$\frac{16}{3}$$ Question 4. Arrange the following rational numbers in descending order $$\frac{-3}{10}, \frac{-7}{-5}, \frac{9}{-15}, \frac{18}{30}$$ Question 5. On a number line, what is the length of the line segment joining. (i) 3 and -3 3 – (-3) = 6 units (ii) $$\frac{1}{2}$$ and –$$\frac{1}{2}$$ $$\frac{1}{2}-\left(-\frac{1}{2}\right)=\frac{1}{2}+\frac{1}{2}$$ = 1 unit (iii) $$\frac{1}{2}$$ and 2$$\frac{1}{2}$$ $$\frac{1}{2}$$ and $$\frac{5}{2}=\frac{5}{2}-\frac{1}{2}$$ = 2 units (iv) –$$\frac{1}{2}$$, -2$$\frac{1}{2}$$ $$\frac{-1}{2}$$ and $$\frac{-5}{2}=\frac{-1}{2}-\left(\frac{-5}{2}\right)=\frac{-1}{2}+\frac{5}{2}$$ = 2 units Question 6. Find the value of x is each case. (i) $$\frac{23}{x}=\frac{2}{-8}$$ $$\frac{23}{x}=\frac{2}{-8}$$ ⇒ 2 × x = 23x – 8 ⇒ x = $$\frac{23 x-8}{2}$$ ∴ x = -92 (ii) $$\frac{x}{9}=\frac{19}{3}$$ $$\frac{x}{9}=\frac{19}{3}$$ ⇒ 3 × x = 19 × 9 ⇒ x = $$\frac{19 \times 9}{3}$$ = 57 ∴ x = 57 (iii) $$\frac{15}{-x}=\frac{1}{-7}$$ $$\frac{15}{-x}=\frac{1}{-7}$$ ⇒ -x = -105 ∴ x = 105 Question 7. (i) $$\frac{-5}{7}, \frac{9}{-13}$$ $$\frac{-5}{7}, \frac{9}{-13}$$ = -5 × -13, 7 × 9 = 65 > 63 ∴ $$\frac{-5}{7}>\frac{9}{-13}$$ (ii) $$\frac{-4}{9}, \frac{-3}{7}$$ $$\frac{-4}{9}, \frac{-3}{7}$$ = -4 × 7, -3 × 9 = -28 < – 27 (iii) $$\frac{-3}{-5}, \frac{12}{20}$$ $$\frac{-3}{-5}, \frac{12}{20}$$ = -3 × 20, 12 × -5 = -60 = -60 ∴ $$\frac{-3}{-5}=\frac{12}{20}$$ (iv) $$\left\|\frac{-4}{5}\right\|,\left\|\frac{-5}{4}\right\|$$ $$\left\|\frac{-4}{5}\right\|,\left\|\frac{-5}{4}\right\|$$ = = 4 × 4, 5 × 5 = 16 < 25 ∴ $$\left\|\frac{-4}{5}\right\|<\left\|\frac{-5}{4}\right\|$$ (v) $$\left\|\frac{5}{7}\right\|,\left\|\frac{-15}{21}\right\|$$ $$\left\|\frac{5}{7}\right\|,\left\|\frac{-15}{21}\right\|=\frac{5}{7}, \frac{15}{21},=\frac{5}{7}=\frac{5}{7}$$ ∴ $$\left\|\frac{5}{7}\right\|=\left\|\frac{-15}{21}\right\|$$ (vi) $$\left\|\frac{-8}{-9}\right\|,\left\|\frac{-3}{9}\right\|$$ $$\left\|\frac{-8}{-9}\right\|,\left\|\frac{-3}{9}\right\|=\frac{8}{9}, \frac{3}{9}$$ = 8 × 9 > 9 × 3 = 72 > 27 ∴ $$\left\|\frac{-8}{-9}\right\|>\left\|\frac{-3}{9}\right\|$$ Question 8. Fill in the following blanks. (i) $$\frac{3}{5}=\frac{138}{x}$$ ⇒ 3x = 5 × 138 ∴ x = $$\frac{5 \times 138}{3}$$ = 230 $$\frac{7}{9}=\frac{y}{108}$$ ⇒ 9 × y = 7 × 108 ∴ y = $$\frac{7 \times 108}{9}$$ = 84 $$\frac{z}{-15}=\frac{48}{90}$$ ⇒ 9 × z = -15 × 48 ∴ z = $$\frac{-15 \times 48}{90}$$ ∴ z = -8 $$\frac{121}{v}=\frac{-11}{12}$$ ⇒ -11v = 121 × 12 ∴ v = $$\frac{121 \times 12}{-11}$$ ∴ v = -132 ## DAV Class 7 Maths Chapter 1 HOTS Question 1. The points P, Q, R, S, T, U, A and B are on the number line representing integers such that TR = RS = SU and AP = PQ = QB Locate and write the rational numbers represented by points P, Q, R, and S. As AP = PQ = QB so it means that line-segment AB is divided into three equal parts. Now, point P represents 1$$\frac{1}{3}$$ i.e., $$\frac{4}{3}$$. and point Q represents 1$$\frac{2}{3}$$ i.e., $$\frac{5}{3}$$. Line segments UT is also divided into three equal parts such that TR = RS = SU. Point R represents -1$$\frac{1}{3}$$ i.e., $$\frac{-4}{3}$$. Point S represents -1$$\frac{2}{3}$$ i.e., $$\frac{-5}{3}$$. Question 1. Show that $$\frac{-2}{7}$$ and $$\frac{-12}{42}$$ are equivalent rational numbers. $$\frac{-2}{7}, \frac{-12}{42}$$ ⇒ -2 × 42 = -12 × 7 ⇒ – 84 = – 84 Hence they are equivalent rational numbers. Question 2. Write three rational numbers which are equivalent to $$\frac{-3}{5}$$. $$\frac{-3 \times 2}{5 \times 2}=\frac{-6}{10}, \frac{-3 \times 3}{5 \times 3}=\frac{-9}{15}$$ and $$\frac{-3 \times 4}{5 \times 4}=\frac{-12}{20}$$ Question 3. Express $$\frac{-2}{11}$$ as a rational number with denominator as (i) 33 $$\frac{-2}{11}=\frac{-2 \times 3}{11 \times 3}=\frac{-6}{33}$$ (ii) 165 $$\frac{-2}{11}=\frac{-2 \times 15}{11 \times 15}=\frac{-30}{165}$$ Question 4. Express $$\frac{-75}{81}$$ in standard form. $$\frac{-75}{81}=\frac{-75 \div 3}{81 \div 3}=\frac{-25}{27}$$ Question 5. Write the following rational numbers in ascending order $$\frac{2}{3}, \frac{3}{5}, \frac{4}{7}$$ $$\frac{2}{3}, \frac{3}{5}, \frac{4}{7}$$ L.C.M. of 3, 5 and 7 = 105 Now making the denominators same, we get Here 60 < 63 < 70 ∴ $$\frac{4}{7}<\frac{3}{5}<\frac{2}{3}$$ Question 6. Write the following rational numbers in descending order. $$\frac{1}{2}, \frac{1}{3}, \frac{1}{5}, \frac{1}{7}$$ $$\frac{1}{2}, \frac{1}{3}, \frac{1}{5}, \frac{1}{7}$$ L.C.M. of 2, 3, 5 and 7 = 210 Now making the denominators same, we get Here 105 > 70 > 42 > 30 ∴ $$\frac{1}{2}>\frac{1}{3}>\frac{1}{5}>\frac{1}{7}$$ Question 7. Compare the following rational numbers. (i) $$\frac{2}{7}$$ and $$\frac{5}{9}$$ latex]\frac{2}{7}[/latex] and $$\frac{5}{9}$$ = $$\frac{2 \times 9}{7 \times 9}$$ and $$\frac{5 \times 7}{9 \times 7}=\frac{18}{63}<\frac{35}{63}$$ ∴ $$\frac{2}{7}<\frac{5}{9}$$ (ii) $$\frac{3}{5}$$ and $$\frac{4}{9}$$ $$\frac{3}{5}$$ and $$\frac{4}{9}$$ = $$\frac{3 \times 9}{5 \times 9}$$ and $$\frac{4 \times 5}{9 \times 5}=\frac{27}{45}>\frac{20}{45}$$ ∴ $$\frac{3}{5}>\frac{4}{9}$$ Question 8. Fill in the blanks: Let us fill up the blanks with x and y as follows: $$\frac{x}{5}=\frac{-2}{3}=\frac{y}{4}$$ ⇒ $$\frac{x}{5}=\frac{-2}{3}$$ and $$\frac{-2}{3}=\frac{y}{4}$$ ⇒ 3x = -10 and 3y = -8 ⇒ x = $$\frac{-10}{3}$$ and y = $$\frac{-8}{3}$$ Question 9. Find the absolute values of the following rational numbers. $$\frac{-5}{7}, \frac{0}{-5}, \frac{-3}{-4}$$ Absolute value of $$\frac{-5}{7}=\left\|\frac{-5}{7}\right\|=\frac{5}{7}$$ Absolute value of $$\frac{0}{-5}=\left\|\frac{0}{-5}\right\|=\frac{0}{5}$$ = 0 Absolute value of $$\frac{-3}{-4}=\left\|\frac{-3}{-4}\right\|=\frac{3}{4}$$ (i) $$\frac{-3}{2}$$ (ii) $$\frac{3}{5}$$ (iii) $$\frac{-3}{5}$$ (iv) $$\frac{3}{2}$$ (v) $$\frac{-2}{3}$$<\|endoftext\|>	4.71875
688	We acquire language when we receptively and productively retrieve it. Growing up in school, some students don’t study. Maybe they drift through school, not knowing a purpose and reason for being there. Or maybe they find school too easy! Other students may study hard. They read their assigned homework. Before the test, they read the textbook again and again, and they review their notes, which they took in class. Maybe they pass the test, maybe not. But at least they can feel good about studying hard. We can praise the effort of students who studied hard. But actually, we cannot praise the way that they studied. If you re-read a text book and review notes, you might think you learned the material, but you really didn’t learn it! You only familiarized yourself with it. To really learn something, you need to show that you can retrieve it from memory without looking at the textbook and notes. When studying for a test, good learners practice retrieval. For example, they make flash cards based on the textbook and their notes. Then they quiz themselves on the contents. If they can retrieve the information from memory by saying it correctly in clear language, then they have learned it. If they fail to retrieve it, they see this failure as feedback. Now they know — that they did not know, so they go back and study. Then they practice retrieval again. With this kind of self-quizzing with flash cards, good learners practice an active or productive retrieval, which works better than just re-reading books and reviewing notes. How does retrieval work with language learning? For one, we can use productive retrieval for vocabulary study with word cards. Write the target word on one side of the card; write the translation in your native language on the other side. Since you want to learn the target language word, start with the native language side of the card. Since you already know the native language word, try to retrieve the target word because it’s the word you want to learn. Make lots of these word cards. Shuffle them. Quiz yourself with retrieval. Separate the cards into “I know” and “Not Yet” groups. Take breaks. (See the next Chapter on Spaced Repetition). And keep going till you learn the words. Besides productive retrieval, we also experience retrieval receptively by listening and reading. Let’s look at reading. As readers comprehend stories, they receptively retrieve the meanings of words and phrases. Instead of repeating with word cards, words and phrases naturally repeat themselves in stories and texts, and readers naturally acquire these words and phrases through receptive retrieval. Moreover, grammar patterns also repeat themselves in stories and texts, and readers acquire these grammar patterns also through receptive retrieval. When we read big, we meet many words, phrases, and grammar patterns thousands of times. When we review textbooks and notes, we only experience a few repetitions of contents. But with big reading, we experience the repetitions of words, phrases, and grammar patterns on a mass scale. Thus, retrieval works differently for big reading than it does for a short review of textbooks and notes. In the end, we study well and learn well when we do receptive and productive retrieval. But retrieval works best with spaced repetition and interleaving, which we will meet in the next chapters.<\|endoftext\|>	4.0625
1,130	Profiling the genome hundreds of variations at a time Geneticists have been using model organisms ranging from the house mouse to the single-cell bakers' yeast, Saccharomyces cerevisiae, to study basic biological processes that regulate human development and physiology, and that can be compromised in various diseases. This has been possible because many of the genes that control these processes in humans are also present with similar functions in those other species; and because genes in model organisms can be mutated and deleted in the laboratory at will. Thus far, however, even in easy-to-manipulate yeast, genes had to be deleted one-gene at a time, often with additional undesired sequence modifications left behind in their genome. A team at Harvard's Wyss Institute led by its Core Faculty member George Church now presents a CRISPR-Cas9-based strategy in Nature Biotechnology that solves both of these problems. Using baker's yeast, the researchers developed a high-throughput approach that allows researchers to precisely alter hundreds of different genes or features of a single gene at once in individual yeast cells with 80 to 100% efficiency, select cells from the population that show specific behaviors, and identify the gene alterations that either trigger or prevent them. "Our method not only offers a more efficient and precise way to perform high-throughput "functional genomics" in yeast than what was possible with previous methods. It will also allow us to model and test subtle human gene variations in yeast cells that have been loosely associated with certain traits or disorders, and find out which ones may actually be relevant," said Church, Ph.D., who also is Professor of Genetics at Harvard Medical School (HMS) and of Health Sciences and Technology at Harvard and the Massachusetts Institute of Technology (MIT). Variations in human genes normally do not occur as perfect deletions of their sequences from the genome, but rather consist of small point mutations—substitutions of single A, T, C, or G base units in the DNA code for one of the other ones—or the insertion or deletion of a few base units. To recreate such variations in the yeast genome in the absence of other potentially interfering variations, the team leveraged CRISPR-Cas9, which can be precisely targeted to pre-selected sequences in the DNA with the help of a small guide RNA (sgRNA). After the Cas9 enzyme has cut its target sequence, a process known as homology-directed recombination (HDR) can repair the gene by using information from an additionally supplied donor template sequence that carries a variation of interest. "We have developed a strategy that physically links the blueprints for sgRNA and donor template in one stable and heritable extra-chromosomal DNA molecule (guide+donor). This enabled us to construct large libraries of variants in one reaction, deliver multiple corresponding sgRNAs and donor templates en masse to yeast cells, and identify those that stimulate a certain cell behavior by next-generation sequencing," said Postdoctoral Fellow Xiaoge Guo, Ph.D., one of the study's first authors. In proof-of-concept studies, the team first focused on a single highly conserved gene encoding the DNA helicase and repair enzyme SGS1. They then broadly damaged the DNA of the yeast cell population carrying the guide+donor library with a toxic reagent, and sequenced the DNA of the surviving cells. This allowed them to uncover mutations compromising SGS1 features that are vital for the repair of damaged DNA and that guarantee the cells' continued survival. Next, the team applied their guide+donor strategy to delete 315 members of a poorly understood gene family, which encode so-called small open reading frames (smORFs) that are scattered throughout the genome, in one fell swoop. By analyzing how this affects yeast cells' survival in different environmental stress conditions, they could assign previously unknown, essential functions to specific smORFs, thereby opening a new gateway into their analysis. "Besides using the method to tease new functions out of genes and larger gene families, an intriguing potential also lies in the investigation of non-coding sequences in the genome to advance our understanding of gene regulation and chromosome biology," said first and co-corresponding author Alejandro Chavez, M.D., Ph.D., who as a Postdoctoral Fellow was co-mentored by Church and Wyss Institute Core Faculty member James Collins and is now Assistant Professor at Columbia University. Collins, Ph.D., who collaborated with the team on the study, is also the Termeer Professor of Medical Engineering & Science at MIT and a Professor of Biological Engineering at MIT. "We can also use the guide+donor method for synthetic biology applications that aim to engineer yeast cells with specific metabolic and industrially relevant abilities, or transfer it to pathological yeast strains for the discovery of genes and gene functions that affect their infectious properties," he said. "This newest application of the CRISPR-Cas9 technology that emerged through a dynamic collaboration between the Church and Collins labs opens yet another path towards discovery of previously hidden molecular mechanisms by which cells regulate their physiology, and when dysregulated, lead to infections as well as human disease," said Wyss Institute Founding Director Donald Ingber, M.D., Ph.D., who is also the Judah Folkman Professor of Vascular Biology at HMS and the Vascular Biology Program at Boston Children's Hospital, as well as Professor of Bioengineering at the Harvard John A. Paulson School of Engineering and Applied Sciences.<\|endoftext\|>	3.703125
322	This most commonly used central tendency is calculated by adding all the observations in the series and dividing it by the total number of observations. The formula for mean is: It represents the extent to which each value contributes to the total. Properties of Arithmetic Mean: The sum of deviations of the observations from their arithmetic mean is always Zero: Each value when subtracted from the mean then added gives us zero. This means in a way the A.M. (Arithmetic Mean) balances each value.E.g. the A.M. of: 3, 5, 7, 9, 11 will be: Only the arithmetic mean can be used for further mathematical treatment. E.g. we can add the arithmetic means of two different sets of data to find their combined arithmetic mean. Arithmetic mean is representative of each and every number in the series, therefore if even one number is missing or falls under open ended class interval, the arithmetic mean can’t be found. This means that the start point and the end point should be specified. E.g. we require to take out mean of 100 people having IQ>80 and IQ<135. 5 people don’t fall under this range therefore, we’ll take out the mean of the 95 which fall in that range and since there is no specific range for the rest of the 5, we ignore them. Due to this disadvantage and because that arithmetic mean can’t be used when there is a huge variation in the data of the series, Median is used.<\|endoftext\|>	4.0625
498	by Frederick H. Lowe NorthStar News Today.com readers Sharon Gale and General Parker were the only ones to participate in last week’s contest about the first southern state to abolish slavery. Gale said is was South Carolina. Later Gale wrote she had to do more research. Sharon, thanks for participating, but your answer was incorrect. South Carolina was the first southern state to secede from the Union to protect slavery. South Carolina seceded on December 20, 1860, a month after voters elected Abraham Lincoln president. General Parker, however, can go to the head of the class. Parker said it was West Virginia. West Virginia abolished slavery on July 4, 1863. Any African-American person born after that date in 1863 was free. When Virginia seceded from the Union in 1861, Unionists meeting in Wheeling repudiated Virginia’s decision to join the Confederacy. Voters elected Francis Pierpoint, a railroad attorney and a coal mine operator, the state’s legitimate governor. In 1863 West Virginia, which had been a part of Virginia, was admitted to the Union if it agreed to abolish slavery. West Virginia has played a historical role in the anti-slavery movement. White abolitionist John Brown and 21 men, including five black men, raided Harpers Ferry, Virginia (now West Virginia) on October 16 and 18, 1859. The raid on the United States armory mission was to spark a slave revolt. Brown planned his raid on Harpers Ferry in Chatham-Kent, Ontario, Canada, a town of 6,000. Half the residents either were escaped slaves or free black men. On May 10, 1859, Brown held a constitutional convention in Chatham-Kent with the help of Dr. Martin Delany, a Harvard- educated physician and the only black field officer to serve in the Civil War. Brown and his men were defeated and hanged at Harpers Ferry by U.S. Marines led by Robert E. Lee. Brown’s raid, however, was considered the meteor that sparked the Civil War. John Brown Days were celebrated in Chatham-Kent. It is not clear if the celebration is still being held. Next week’s question What was Reconstruction and how long did it last? Why did it end? 1). One year 2). Two years 3). Fourteen years<\|endoftext\|>	3.875
296	The Rhodesian Bush War (also known as the Second Chimurenga or the Zimbabwe War of Liberation) was a civil war that took place from July 1964 to December 1979 in the unrecognised country of Rhodesia. The conflict pitted three forces against one another: the Rhodesian government, under Ian Smith (later the Zimbabwe Rhodesian government of Bishop Abel Muzorewa); the Zimbabwe African National Liberation Army, the military wing of Robert Mugabe’s Zimbabwe African National Union; and the Zimbabwe People’s Revolutionary Army of Joshua Nkomo’s Zimbabwe African People’s Union. The war and its subsequent Internal Settlement, signed in 1978 by Smith and Muzorewa, led to the implementation of universal suffrage in June 1979 and the end of white minority rule in Rhodesia, which was renamed Zimbabwe Rhodesia under a black majority government. However, this new order failed to win international recognition and the war continued. Negotiations between the government of Zimbabwe Rhodesia, the British government and Mugabe and Nkomo’s united “Patriotic Front” took place at Lancaster House, London in December 1979, and the Lancaster House Agreement was signed. The country returned temporarily to British control and new elections were held under British and Commonwealth supervision in March 1980. ZANU won the election and Mugabe became the first Prime Minister of Zimbabwe on 18 April 1980, when the country achieved internationally recognised independence.<\|endoftext\|>	3.671875
1,572	Although many of us today think of summer as a time of rest and relaxation, for Ontario’s residents in the past, summer was a season of hard work and preparation for the winter months. Perhaps the most import task for ancestral Aboriginal and Euro-Canadian families was the preservation of foodstuffs for the long, cold months ahead. We’ve selected a few artifacts to highlight how the people in Ontario’s past preserved the tastes of summer. Strawberries, Blueberries and Raspberries Certainly some of the most important human advances have focused on the long-term preservation of food. Most preservation activities occurred in late spring, summer and early fall, coinciding with the availability of Ontario’s wide selection of wild berries. Preservation not only gave people the opportunity to enjoy summer flavours during the winter, but it also provided populations with much of the nutritional benefits of fresh berries, most importantly Vitamin C. As one would expect, archaeological evidence for the collection and consumption of berries exists at both Euro-Canadian homesteads and pre-contact indigenous settlements. Jams, jellies, and other conserves are often made from summer berries and fruits, and the historical containers used to store jams and jellies were often earthenware crocks and jars, as well as glass jars. In many of the ceramic preserve jars, a deep groove is present around the rim (see image below). To seal the jars, a circle of tissue paper brushed with egg white was added to the top, and a string (which rested in the groove) secured the paper to the jar, protecting the contents from exposure. Some people advocated the addition of brandy-soaked paper to the surface of the food to prevent rapid spoiling (Meschler 2013: 5). Evidence for the preservation of fruit is most common on mid-to-late nineteenth-century historical sites in Ontario. Glass preserve jars and lids were re-usable tools for preserving and storing food for long periods of time. Bill Lindsey (2013) explains the rise of such containers in the early nineteenth-century: As cities and relative affluence spread, the market and demand for bottled goods increased rapidly. At the same time, the expansion of the ever growing population in the farming regions created a need for methods and equipment to preserve foods. In The White House Cook Book (1887), author Mrs. F.L. Gillette references the earliest use of paraffin as an anti-microbial sealant, suggesting that: Mold can be prevented from forming on fruit jellies by pouring a little melted paraffin over the top. When cool, it will harden in a solid cake, which can be easily removed when the jelly is used, and saved to use over again another year. It is perfectly safe and harmless. To make preserves, fruits are boiled to reduce moisture and kill bacteria. Though necessary, the task of boiling, sugaring and preserving the various berries was hard work. In fact, it is referred to by the Journal of Antiques and Collectibles (2000) as “a hot job performed in hot weather.” Sugaring prevents further bacteria growth, while storage in an air-tight container keeps it safe from contamination, allowing the foods to remain edible through (at least) the winter (Bose 2013). While preservation methods have changed through the decades, the best recipes for preserves and jams remain the same. Click here for an old fashioned jam recipe dating from 1861! Sugar was very expensive until the mid-nineteenth-century when the development of alternative sweeteners and an increase in trade technology resulted in lower cost (Miller 1991). The sugar duty—a tax paid on sugar imports–was halved in Britain in 1870, and four years later the duty was abolished completely, making sugar much easier to purchase in the Commonwealth (Atkins 2013: 17). As a result, the preparation of jams, jellies, and other conserves became commonplace. These ceramic preserve jars were recovered from the Bishop’s Block site and the Toronto General Hospital site, both in downtown Toronto. They likely date from the mid- to late-nineteenth century, before glass jars became widely used. Similar preserve jars are common finds on urban sites. In 1858, New York metal worker John Landis Mason created and patented the “Mason Jar”—a glass canning jar with a specially-threaded neck which made preserving and canning safer and easier (Miller et al. 2000). This Mason-type jar liner is from the Woodbine site in Etobicoke and is embossed with a crown insignia. According to the Parks Canada Glass Glossary (1989: 160), glass liners were “part of a more complex closure and … therefore not a closure on its own.” This artifact would have had a metal screw band placed over it to seal it to the jar. Other lids, coupled with a rubber gasket, may have been held in place with a clamp-style (or “lightning”) closure. The lightning closure consists of a metal wire-toggle mechanism which works to clamp and seal the jar (Lindsey 2013). Evidence for the consumption of fruits is also found in abundance on many Iroquoian settlements in southern Ontario. At the large, mid-sixteenth-century Mantle Site in Stouffville, a great number of fruit seeds were found inside the longhouses, suggesting that perhaps the fruits were being dried and stored for winter use (Birch and Williamson 2013: 89). At the Joseph Picard site in Whitby, ASI staff recovered raspberry seeds from the remains of the mid-fifteenth-century village. Raspberries are in their peak season from June to September, and it is possible that the people at Joseph Picard prepared these fruits for storage during the summer months, although they no doubt consumed them fresh as well. Let’s turn our attention away from fruit and onto another summer food–maize (corn). The best time to harvest maize in Ontario is August and September, as the crop is cold-intolerant and must be planted in the spring. By the year AD 1300, maize made up approximately 50% of the diet of southern Ontario Iroquoians, as based on isotope research (Schwarcz et al. 1985). It was estimated that the people of Mantle were consuming roughly 1.3 pounds of maize a day, likely prepared in a soup or baked into a bread (Birch and Williamson 2013: 93). The dominance of corn in ancestral Aboriginal diets is supported not only by archaeological science but also by historical accounts. While living with the Huron, the Jesuits commented on the abundance of corn, writing “As to our food, I shall say this further, that God has shown his Providence very clearly to our eyes; we have obtained in eight days our provision of corn for a whole year, without making a single step beyond our Cabin” (Thwaites, Jesuit Relations 10: 70) We often find corn kernels and cobs on pre-contact sites, but perhaps some of the most interesting corn-related artifacts are the rare face effigy vessel fragments. The face depicted on the vessel below has eyebrows, closed eyes, a nose and an open mouth. Anthony Wonderley (2002, 2005) has suggested that these vessel fragments represent the mythical cornhusk people, based on their similarity to the St. Lawrence Iroquoian corn-ear motifs. It is also possible that the vessels were used to store corn, giving the artifacts both a literal and spiritual meaning.<\|endoftext\|>	3.671875
638	1). A class defined as abstract cannot be instantiated. Meaning, you cannot call just/only new AbstractClass().You need to override all the methods that are marked as abstract as well.(See AbstractMain) 2). Abstract keyword is non access modifier. 3). When a method is marked as abstract, there should not be any implementation for the method. 4). Abstract methods have to be overridden in the sub classes. 5). Abstract classes NEED NOT have abstract methods. Those methods which do not have abstract defined, should be implemented. Meaning there should be a implementation code for this method right in this abstract class method body. 6). If there is at–least one abstract method defined in the whole class, then that class should be declared as abstract. 7). Classes and Methods marked as private cannot be abstract 8). Classes and Methods marked as static cannot be abstract. 9). Classes and Methods marked as final cannot be abstract. 10). Abstract classes can have default and as well as overloaded constructors. When no constructor is defined – a default constructor will be generated by compiler. 11). Constructors of abstract classes are invoked when a subclass object is created. You can call the superclass constructor by calling super(). 12). Abstract classes hides the functionality of those methods that are defined as abstract. The actual functionality of the abstract methods are exposed by the implementation classes. 13). Abstract Method Overloading signatures are supported in abstract classes. 14). By making this class as abstract and implementing the methods as well, will not give any compile time error., but gives a run time error after Instantiating the AbstractSubClass. 15). @Override – Not necessary to provide this annotation, because, Compiler by default takes care of overriding, when it see’s extends keyword. 1). Interfaces does not have any method implementation. 2). All methods in an interface are public and/or abstract. 3). Any field that is declared, is public and static. 4). There is no multiple inheritance in Java. Which means, any Java class cannot extend more than one class at a time. Instead there is a Multiple Interface Inheritance. Which means, any Java can implement/extends more than one interface. 5). Any Java class that implements any interface, will be having a IS–A relationship, between the Java Class and the interface. 6). If a method deceleration in a interface is abstract, the whole interface need not be declared as abstract, this is unlike abstract classes. 7). Java does not support multiple inheritance, because of the Diamond of Death Problem. 8). Method Overriding is allowed in Interfaces. Each method signature in an interface is unique. Meaning, you can have multiple methods of same name with different arguments. 9). Multiple Interface Inheritance example can extend any number of interfaces. Ex: public interface MultipleInterfaceImplementation extends InterfaceExample,AnotherInterface * here InterfaceExample and AnotherInterface are both interfaces. Source code: https://github.com/chouhan/CoreJavaBasics<\|endoftext\|>	3.84375
516	Gravitational waves: What they are and why scientists are so excited about them You may have come across the term “gravitational waves” over the past week through social media. That’s because the scientific community is abuzz with a possible announcement today that the elusive signature in space finally may have been directly detected. Gravitational waves are a form of radiation, ripples in space-time that travel at the speed of light. Albert Einstein predicted the existence of gravitational waves in his theory of general relativity in 1916, but thus far no direct detection has been made. There has been indirect detection, however: In 1993, the Nobel Prize in physics was awarded to Russell A. Hulse and Joseph H. Taylor Jr. for the discovery of a new type of pulsar (a rapidly spinning star that sends out regular pulses of radio waves and electromagnetic radiation) which suggested the existence of gravitational waves. WATCH: Detection of gravitational waves being called a game-changer by scientific community Researchers using the Laser Interferometer Gravitational-Wave Observatory (LIGO) are searching for these waves using an interferometer. Two telescopes are set up kilometres apart — one in Hanford, Wash. and another in Livingston, La., splitting a single laser beam that travels perpendicularly. Those beams of light should return to the original source perfectly aligned — and we’re talking in very precise scales here. But a gravitational wave could change the distance that the beam travels compared to the other source. How slight a difference are we talking about? About 1/10,000th the width of an atom. Of course there’s a possibility that something else could interfere with the measurements, and that’s why LIGO must filter out all other sources of “noise.” On Monday, the LIGO Scientific Collaboration issued a press release announcing an update on the search for these waves. And now the scientific community is anticipating an announcement that the waves have been detected at last. So why does it even matter if we detect them or not? Gravitational waves could help astronomers listen in on cataclysmic processes, such as the merging of black holes or even distant supernovas. Basically, these waves could teach us much more about our universe. These waves will also help scientists to gain a better, more accurate understanding of our universe, its origins and workings as well as help further prove Einstein’s general theory of relativity. © 2016 Shaw Media<\|endoftext\|>	3.96875
1,111	# Factors of 180 \| Find the Factors of 180 by Factoring Calculator Factoring Calculator calculates the factors and factor pairs of positive integers. Factors of 180 can be calculated quickly with the help of Factoring Calculator i.e. 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90, 180 positive integers that divide 180 without a remainder. Factors of 180 are 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90, 180. There are 18 integers that are factors of 180. The biggest factor of 180 is 180. Factors of: ### Factor Tree of 180 to Calculate the Factors 180 2 90 2 45 3 15 3 5 Factors of 180 are 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90, 180. There are 18 integers that are factors of 180. The biggest factor of 180 is 180. Positive integers that divides 180 without a remainder are listed below. • 1 • 2 • 3 • 4 • 5 • 6 • 9 • 10 • 12 • 15 • 18 • 20 • 30 • 36 • 45 • 60 • 90 • 180 ### Factors of 180 in pairs • 1 × 180 = 180 • 2 × 90 = 180 • 3 × 60 = 180 • 4 × 45 = 180 • 5 × 36 = 180 • 6 × 30 = 180 • 9 × 20 = 180 • 10 × 18 = 180 • 12 × 15 = 180 • 15 × 12 = 180 • 18 × 10 = 180 • 20 × 9 = 180 • 30 × 6 = 180 • 36 × 5 = 180 • 45 × 4 = 180 • 60 × 3 = 180 • 90 × 2 = 180 • 180 × 1 = 180 ### Factors of 180 Table FactorFactor Number 1one 2two 3three 4four 5five 6six 9nine 10ten 12twelve 15fifteen 18eighteen 20twenty 30thirty 36thirty six 45forty five 60sixty 90ninety 180one hundred eighty ### How to find Factors of 180? As we know factors of 180 are all the numbers that can exactly divide the number 180 simply divide 180 by all the numbers up to 180 to see the ones that result in zero remainders. Numbers that divide without remainder are factors and in this case below are the factors 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90, 180 are the factors and all of them can exactly divide number 180. ### Frequently Asked Questions on Factors of 180 1. What are the factors of 180? Answer: Factors of 180 are the numbers that leave a remainder zero. The ones that can divide 180 exactly i.e. factors are 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90, 180. 2.What are Factor Pairs of 180? • 1 × 180 = 180 • 2 × 90 = 180 • 3 × 60 = 180 • 4 × 45 = 180 • 5 × 36 = 180 • 6 × 30 = 180 • 9 × 20 = 180 • 10 × 18 = 180 • 12 × 15 = 180 • 15 × 12 = 180 • 18 × 10 = 180 • 20 × 9 = 180 • 30 × 6 = 180 • 36 × 5 = 180 • 45 × 4 = 180 • 60 × 3 = 180 • 90 × 2 = 180 • 180 × 1 = 180 3. What is meant by Factor Pairs? Answer:Factor Pairs are numbers that when multiplied together will result in a given product.<\|endoftext\|>	4.40625
768	As a follow-up to the MDGs, the UN has developed 17 Sustainable Development Goals (SDGs), which will come into effect on January 1 2016. As with the MDGs, each SDG has specific targets, which can be found on the UN website, and a set of indicators to measure progress, which will be finalized by March 2016. This ambitious agenda has been proposed to shape the policies and actions of UN member states through to 2030: - End poverty in all its forms everywhere - End hunger, achieve food security and improved nutrition and promote sustainable agriculture - Ensure healthy lives and promote well-being for all - Ensure inclusive and equitable quality education and promote lifelong learning opportunities for all - Achieve gender equality and empower all women and girls - Ensure access to water and sanitation for all - Ensure access to affordable, reliable, sustainable and modern energy for all - Promote inclusive and sustainable economic growth, employment and decent work for all - Build resilient infrastructure, promote sustainable industrialization and foster innovation - Reduce inequality within and among countries - Make cities inclusive, safe, resilient and sustainable - Ensure sustainable consumption and production patterns - Take urgent action to combat climate change and its impacts - Conserve and sustainably use the oceans, seas and marine resources - Sustainably manage forests, combat desertification, halt and reverse land degradation, halt biodiversity loss - Promote just, peaceful and inclusive societies - Revitalize the global partnership for sustainable development What makes the SDGs different from the MDGs? The most obvious answer is that there are more than double the number of SDGs than there are MDGs, but there are more differences than just that! More SDGs translates to the UN being able to tackle a more broad range of issues related to development. The MDGs were grounded in anti-poverty, however, they often failed to address the root causes of poverty, opting instead to treat the symptoms. The SDGs aim to address the systemic issues that allow poverty to persist, including human rights, governance, peace and stability. The SDG targets also move beyond poverty, examining issues such as climate change and sustainable consumption, which will impact everyone, regardless of their socioeconomic status. The MDGs were primarily developed by member countries of the Organization for Economic Co-operation and Development and donor agencies focusing their efforts on low and middle-income countries. The process for determining the MDGs barely involved those who were most affected by their targets. The SDGs, however, have come out of international negotiations where low and middle-income countries were granted input. Furthermore, the SDGs are regarded as universal, applying to all countries and actors, rather than creating a divide between “rich donors” and “poor recipients.” More opportunities for partnership will (hopefully) be a product of, and a driver of, the SDGs. These include public-private partnerships, trade partnerships, global partnerships, local partnerships, and most importantly equitable partnerships. In order to meet the ambitious targets set out in the SDGs, a multidisciplinary and multi-stakeholder approach will be vital. The MDGs are about to expire, and though there has been considerable progress made on many of the targets, there is still much work to be done. About 1 billion people still live on less than $1.25 per day worldwide, gender inequality persists as women worldwide continue to face discrimination in their everyday lives and over 800 million people still suffer from hunger. The SDGs hope to finish the job of the MDGs, and ensure that this time around, no one is left behind. Over the next 15 years, we are excited to see how the SDGs will impact global health, as health is a key outcome of development.<\|endoftext\|>	3.78125
2,343	Bile Duct Cancer Bile duct cancer, also called cholangiocarcinoma, is a rare form of cancer that occurs in the duct that carries bile from the liver to the small intestine. Bile duct cancer is relatively slow growing. Its main symptom is jaundice (yellowing of skin and eyes). Surgery is the treatment of choice for bile duct cancer. Radiation and chemotherapy is commonly used before surgery to reduce the size of a tumor or as a follow-up treatment after surgery. The bile duct begins as many small channels in your liver that meet to form the hepatic duct. The hepatic duct is joined by the cystic duct from the gallbladder, and their union forms the common bile duct. The common bile duct continues to the duodenum, the first part of your small intestine. Your gallbladder works with your liver and pancreas to produce bile and digestive enzymes. Bile is a fluid that breaks down fat in food for digestion. Bile is produced in the liver and stored in the gallbladder until it is needed. When you eat high-fat or high-cholesterol foods, your gallbladder sends bile to your duodenum via the common bile duct. Bile duct cancer is a rare form of cancer. Most bile duct cancers are adenocarcinomas. The majority of cases are slow growing and late to metastasize. Cancer that has spread to other parts of the body is termed metastasized. However, by the time most bile duct cancers are diagnosed, they are too advanced for surgical removal. Researchers do not know the cause of most bile duct cancers. Cancer occurs when cells grow abnormally and out of control, instead of dividing in an orderly manner. Most bile duct cancer arises from the mucus glands that line the duct. It appears that chronic irritation of the bile duct, by inflammation or parasitic infection, is the top risk factor associated with bile duct cancer formation. Cancer can develop in any part of the bile duct. The cancer is classified based on its location. The majority of bile duct cancer develops in the hepatic duct at the site where the small channels in the liver join together. Cancers in this area are called perihilar cancers or Klatskin tumors. Distal bile duct cancers form in the common bile duct near the small intestine. A small percentage of bile duct cancers form in the channels within the liver and are called intrahepatic bile duct cancers. Symptoms occur when the bile ducts become blocked. Jaundice is the most common symptom of bile duct cancer. Jaundice is a condition caused by an excess of bilirubin. Symptoms of jaundice include yellowing of the eyes and skin, dark urine and pale-colored stools. You may also experience fever, nausea, vomiting, chills, and itching. You may lose your appetite and lose weight. You may feel pain in your right upper abdomen that may spread to your back. Imaging tests are used to identify the location and size of tumors and blockages. Common imaging tests include ultrasound, computed tomography (CT) scans, magnetic resonance imaging (MRI) scans, endoscopic retrograde cholangiopancreatography (ERCP), positron emission tomography (PET) scans, cholangiography, and angiography. Ultrasound uses sound waves to produce images of internal organs and detect abnormal tissues. An ultrasound device may be placed over the abdomen area, inserted through the mouth and into the stomach (endoscopic ultrasound), or through an incision in the side of the body (laparoscopic ultrasound). CT scans take cross-sectional images of the body. They may be used with a contrast agent or dye to take pictures of your bile duct and nearby organs. CT scans are useful for determining if cancer has metastasized. MRI scans produce even more detailed images and can outline the exact site of bile duct blockage. An ERCP uses an endoscope to view the biliary system. An endoscope is a thin tube with a light and viewing instrument at the end of it. After you are sedated, the thin tube is passed through your mouth and into your small intestine. An endoscope is used to take tissue samples with biliary brushing. It can administer dye to enhance views. A cholangiography can determine the exact location of bile duct cancer. It is helpful for determining if the cancer can be treated with surgery. For this procedure, contrast dye is injected into the bile duct before X-rays are taken. A laparoscopy is a procedure used to view the bile duct, gallbladder, liver, and other internal organs. It uses a thin-lighted instrument, a laparoscope, which is inserted through an incision in the abdomen. A laparoscope can take a biopsy. A biopsy is a tissue sample that is taken for evaluation of cancer cells. A CT scan is used to guide needle biopsies. If you have bile duct cancer, your doctor will assign your cancer a classification stage based on the results of all of your tests. Staging describes the cancer and how it has metastasized. Staging is helpful for treatment planning and recovery prediction. Surgery is the treatment of choice to remove bile duct cancer. Chemotherapy or radiation may be used if all of the cancer cannot be removed with surgery. The type of treatment that you receive depends on many factors, including the location and stage of your cancer. Intrahepatic surgery is used for bile duct cancer that originates in the liver. This procedure removes the part of the liver that contains cancer. Surgery for perihilar cancer usually includes removing the bile duct, gallbladder, and part of the pancreas, small intestine, and liver. The remaining bile ducts are connected to the small intestine. Part of the pancreas and small intestine is usually removed during surgical treatment of distal bile duct cancer. A Whipple procedure removes the bile ducts, part of the stomach, duodenum, pancreas, gallbladder, and lymph nodes. In select cases, a complete liver transplantation may be necessary to treat bile duct cancer. If all of the cancer cannot be removed, a bypass surgery is used to relieve symptoms of bile duct obstruction. Bypass surgery creates a new route from the bile duct to the small intestine. Chemotherapy, radiation therapy, or both may be used to reduce the size of a tumor before surgery or as a follow-up treatment after surgery. Radiation therapy uses high-energy rays to eliminate cancer cells. Chemotherapy uses cancer-fighting drugs to destroy cancer cells. There are different types of chemotherapy and radiation therapy. Even with treatment, some cases of bile duct cancer may return. This is termed “recurrent bile duct cancer.” The cancer may come back near the site of the original cancer or in other parts of the body. Your doctor can explain your risk for recurrent bile duct cancer and possible treatments if it does recur. The experience of bile duct cancer and cancer treatments can be an emotional process for people with cancer and their loved ones. It is important that you receive support from a positive source. Some people find comfort in their family, friends, counselors, co-workers, and faith. Cancer support groups are another good option. They can be a source of information and support from people who understand what you are experiencing. Ask your doctor for cancer support group locations in your area. PreventionYou may prevent bile duct cancer by reducing the risk factors that you can control. It may be helpful to avoid the hazardous chemicals associated with bile duct cancer. You can prevent hepatitis B with a vaccine and hepatitis C by avoiding blood-borne or sexually transmitted infections. Stopping alcohol abuse may prevent liver cirrhosis. When travelling in Asia, it is important to avoid contact with liver flukes. Am I at Risk Risk factors may increase your likelihood of developing bile duct cancer, although some people that experience this cancer may not have any risk factors. People with all of the risk factors may never develop bile duct cancer; however, the likelihood increases with the more risk factors you have. You should tell your doctor about your risk factors and discuss your concerns. Risk factors for bile duct cancer: _____ Long-term inflammation of the bile duct is associated with an increased risk of developing bile cancer. _____ Sclerosing cholangitis is a type of bile duct inflammation that leads to scar tissue formation and is associated with an increased risk of bile duct cancer. _____ Ulcerative colitis is an inflammation of the large intestine that can lead to sclerosing cholangitis. _____ Smoking increases the risk of bile duct cancer for people with sclerosing cholangitis. _____ Stones in the bile duct increase the risk for developing bile duct cancer. _____ Diseases of the liver and bile duct increase the risk for bile duct cancer. Such conditions include polycystic liver disease, choledochal cysts, congenital dilation of the intrahepatic bile ducts, and cirrhosis. _____ In Asian countries, parasites called liver flukes are a major cause of bile duct cancer. _____ Aging increases the risk for bile duct cancer. Bile duct cancer occurs most frequently in people over the age of 65. _____ Radioactive chemicals, including Thorotrast (thorium dioxide) which was used years ago during X-rays, are associated with an increased risk for bile duct cancer development. _____ Certain chemicals may be associated with bile duct cancer formation. These chemicals include dioxin, nitrosamines, and polychlorinated biphenyls (PCBs). _____ Viral hepatitis B or C is linked to intrahepatic bile duct cancer. The link is greater for hepatitis C. _____ An association with both diabetes and HIV have been suggested but not proven. Copyright © - iHealthSpot Interactive - www.iHealthSpot.com This information is intended for educational and informational purposes only. It should not be used in place of an individual consultation or examination or replace the advice of your health care professional and should not be relied upon to determine diagnosis or course of treatment. The iHealthSpot patient education library was written collaboratively by the iHealthSpot editorial team which includes Senior Medical Authors Dr. Mary Car-Blanchard, OTD/OTR/L and Valerie K. Clark, and the following editorial advisors: Steve Meadows, MD, Ernie F. Soto, DDS, Ronald J. Glatzer, MD, Jonathan Rosenberg, MD, Christopher M. Nolte, MD, David Applebaum, MD, Jonathan M. Tarrash, MD, and Paula Soto, RN/BSN. This content complies with the HONcode standard for trustworthy health information. The library commenced development on September 1, 2005 with the latest update/addition on April 13th, 2016. For information on iHealthSpot’s other services including medical website design, visit www.iHealthSpot.com.<\|endoftext\|>	3.859375
1,496	Common Core: 6th Grade Math : Understand Data is Collected to Answer a Question and has a Center, Spread, and Shape: CCSS.Math.Content.6.SP.A.2 Example Questions Example Question #1441 : Grade 6 Select the description that best describes the distribution of the data shown in the provided dot plot. Symmetric Right skewed Left skewed Left skewed Explanation: In order to answer this question correctly, we need to define our answer options: Symmetric: A symmetric distribution will have a middle, or center, and each side from the middle will look fairly similar. Many symmetrical distributions are bell shaped, with the middle being tall, and the two sides thinning out. Left skewed: A left skewed distribution will have most of the data on the right side of the number line. Right skewed: A right skewed distribution will have most of the data on the left side of the number line. The data shown in the number plot has most of the data on the right side; thus, left skewed is the correct answer. Example Question #12 : Statistics & Probability Select the description that best describes the distribution of the data shown in the provided dot plot. Left skewed Right skewed Symmetric Symmetric Explanation: In order to answer this question correctly, we need to define our answer options: Symmetric: A symmetric distribution will have a middle, or center, and each side from the middle will look fairly similar. Many symmetrical distributions are bell shaped, with the middle being tall, and the two sides thinning out. Left skewed: A left skewed distribution will have most of the data on the right side of the number line. Right skewed: A right skewed distribution will have most of the data on the left side of the number line. The data shown in the number plot provided has a bell shape look; thus, symmetric is the correct answer. Example Question #13 : Statistics & Probability Select the description that best describes the distribution of the data shown in the provided dot plot. Left skewed Right skewed Symmetric Right skewed Explanation: In order to answer this question correctly, we need to define our answer options: Symmetric: A symmetric distribution will have a middle, or center, and each side from the middle will look fairly similar. Many symmetrical distributions are bell shaped, with the middle being tall, and the two sides thinning out. Left skewed: A left skewed distribution will have most of the data on the right side of the number line. Right skewed: A right skewed distribution will have most of the data on the left side of the number line. The data shown in the number plot is mostly on the left side; thus, right skewed is the correct answer. Example Question #14 : Statistics & Probability Select the description that best describes the distribution of the data shown in the provided dot plot. Right skewed Left skewed Symmetric Symmetric Explanation: In order to answer this question correctly, we need to define our answer options: Symmetric: A symmetric distribution will have a middle, or center, and each side from the middle will look fairly similar. Many symmetrical distributions are bell shaped, with the middle being tall, and the two sides thinning out. Left skewed: A left skewed distribution will have most of the data on the right side of the number line. Right skewed: A right skewed distribution will have most of the data on the left side of the number line. The data shown in the number plot provided has a bell shape look; thus, symmetric is the correct answer. Example Question #15 : Statistics & Probability Select the description that best describes the distribution of the data shown in the provided dot plot. Symmetric Right skewed Left skewed Right skewed Explanation: In order to answer this question correctly, we need to define our answer options: Symmetric: A symmetric distribution will have a middle, or center, and each side from the middle will look fairly similar. Many symmetrical distributions are bell shaped, with the middle being tall, and the two sides thinning out. Left skewed: A left skewed distribution will have most of the data on the right side of the number line. Right skewed: A right skewed distribution will have most of the data on the left side of the number line. The data shown in the number plot is mostly on the left side; thus, right skewed is the correct answer. Example Question #1444 : Grade 6 Mrs. Frame's class counted the number pencils that each student had in his/her desk. The distribution of this data is show in the dot plot provided. What is the center of this distribution? Explanation: The data in the distribution shows the number of pencils each student found in his/her desk. Each student found pencils in their desks. Based on this, and by looking at the graph, we can see that the enter of the distribution is Example Question #1443 : Grade 6 Mrs. Frame's class counted the number pencils that each student had in his/her desk. The distribution of this data is show in the dot plot provided. What is the most number of pencils that her students found? Explanation: The dot plot is on a number line that is numbered from is the highest number on the number line, but there are no dots above that number, which means no one found pencils in their desk. is the next highest number, and there are two dots over the number on the number line; thus, the correct answer is Example Question #1445 : Grade 6 Mrs. Frame's class counted the number pencils that each student had in his/her desk. The distribution of this data is show in the dot plot provided. How many pencils did most of the students have? Explanation: If we look at the dot plot provided, the most number of dots is above the number on the number line. There are dots above the number , which means that students found pencils in their desks, which is the most students for the same number of pencils; thus, is the correct answer. Example Question #6 : Understand Data Is Collected To Answer A Question And Has A Center, Spread, And Shape: Ccss.Math.Content.6.Sp.A.2 A flower shop counted the number of flowers sold during a month period. The distribution of this data is show in the histogram provided. What is the center of this distribution? July March April May June May Explanation: The data in the distribution shows the number of flowers sold in each month. By looking at the graph, we can see that there are bars representing months, and the month of May is in the middles; thus, May is the center of this distribution. Example Question #1447 : Grade 6 A flower shop counted the number of flowers sold during a month period. The distribution of this data is show in the histogram provided. What is the most number of flowers the shop sold in a given month?<\|endoftext\|>	4.78125
1,002	Despite being a fictional account, the storylines featured in the groundbreaking drama Tenko reveal the many real horrors women endured at the hands of their enemies while imprisoned in concentration camps - facts that had previously not been common public knowledge. While treatment of POWs during wartime is undeniably harsh for all concerned, the fact is that women were and are treated differently from men. One of the most detailed account of women captured by the Japanese during World War II is based around the invasion of Corregidor. The Women of Santa Tomas Corregidor is an island in the Philippines' Manila Bay that has served as a vital defence structure to any naval attacks against the capital city of Manila. During World War II, the island had been the site of several battles but it finally fell to the Japanese on May 6th 1942. When the island fell a total of 79 women were taken prisoners and imprisoned in a concentration camp of Santa Tomas. All the captives were American and comprised of 66 Army Nurses, 1 civilian dietician, 1 civilian physiotherapist and 11 Navy nurses. According to the account of Lieutenant Colonel Madeline Ullom, Santa Tomas was the largest cohort of captured U.S. uniformed women on record. While incarcerated Ullom and fellow nurses created a hierarchy among the inmates. They organized shifts and began care for other prisoners who were captured, but despite the different roles their Japanese captors treated them equally badly. All these women had to constantly fight off starvation and disease, with an average weight loss being about 30% of their body weight. During that time all nurses suffered from diseases such as beriberi or scurvy. Fortunately all survived the ordeal but according to Ullom, "our atmosphere was one of a dusty pall, ever present, in which we moved, worked, tried to eat, tried to breathe in an endless nightmare." Women POWs in Europe During World War II POW camps for women were set up all over Europe. A particularly vivid account reserved for us comes from Janina Skrzynska, who was one of many women that formed part of the Polish Home Army and were imprisoned by the Nazis in the North West German region of Oberlangen, in a special penal facility just for women POWs. According to her account, the German authorities did not see the women as equal to the men and refused to give them "prisoner-of-war status", which afforded them certain internationally recognised conditions in which POWS had to be detained. While the male prisoners were sent to camps that were automatically under the care of the International Red Cross the women were kept in overcrowded barracks separated from the main POW camps by barbed wire. In Janina's account there were two hundred prisoners in each rotten wooden barrack, which comprised of three-tier bunks and only two cast-iron stoves burning damp peat "that produced more smoke than heat." "In these cramped conditions," Janina wrote "cold, frequently hungry, and lacking even the most basic sanitary facilities, these women had to endure the severe winter of 1944". In an attempt to survive the intense cold, the inmates made the most of the decrepit barracks. They ripped out planks from the bunks, floorboards and door and window frames for fuel. However, this was stopped when camp authorities started imposing penalties for destroying government property. Janina also describes the kind of meals they received. "In the mornings and evenings a tepid herbal tea, frequently mouldy bread, the occasional piece of margarine or a spoonful of beetroot marmalade. At midday we would receive soup from bitter cabbage or grubby peas with two or three jacket potatoes." Solidarity in Numbers In these grim surroundings it often falls on camaraderie and unity to survive. An excellent example of this can also be found in Janina's account from her time in Oberlangen. In January 1945, the first children were born in the camp, but the commandant ordered that the children would not be given clothes "because its mother has nothing." These words were enough!" Janina wrote. "Every woman who had anything to spare - a piece of bed linen, a handkerchief, a blouse or an undergarment - would undo the stitches, cut, sew and wash. So many bonnets, baby gowns and nappies were made for the first child that there was also enough for those who were born later. Cartons from Red Cross parcels were converted into cradles." Of course, women POWs is not something that is to be consigned to history. There are many women serving in the military who are imprisoned in military camps throughout the globe but because most of us still think of warfare as engaged largely by men, less consideration is given to those women who have to endure sometimes greater hardships than their male counterparts, including rape and abortion.<\|endoftext\|>	3.71875
2,487	Government Subsidies (Farm, Oil, Export, Etc) What Are the Major Federal Government Subsidies? Each year, the U.S. federal government subsidizes a wide range of economic activities that it wants to promote. What exactly are subsidies? The definition may be broader than you think. Find out about the most well-known subsidies, the history of these subsidies, and some of their costs. What Is a Subsidy? Most subsidies are cash grants or loans that the government gives to businesses. It encourages activities the government wishes to promote. The subsidy depends on the amount of the goods or services provided. One level of government can also give subsidies to another. This includes federal grants given to state or local governments and state grants given to municipal governments. The World Trade Organization has a broader definition of subsidies. It says a subsidy is any financial benefit provided by a government which gives an unfair advantage to a specific industry, business, or even individual. The WTO mentions five types of subsidies: - Cash subsidies, such as the grants mentioned above. - Tax concessions, such as exemptions, credits, or deferrals. - Assumption of risk, such as loan guarantees. - Government procurement policies that pay more than the free-market price. - Stock purchases that keep a company's stock price higher than market levels. These are all considered subsidies because they reduce the cost of doing business. Many experts argue that U.S. farms don't even need subsidies. After all, they are located in one of the world's most favorable geographic regions. It has rich soil, abundant rainfall, and access to rivers for irrigation when rainfall fails. Today's farms have all the advantages of modern business. They have highly trained labor, computerized equipment, and cutting-edge chemical research in fertilizers and seeds. But America's food supply must also be protected from droughts, tornadoes, and recessions. In fact, agricultural subsidies were originally created to help farmers ravaged by the Dust Bowl and the Great Depression of 1929. This price support system lasted until the 1990s. The federal government guaranteed farmers a high enough price to remain profitable. How did it do this? It paid farmers to make sure supply did not exceed demand. The government subsidized farmers to keep croplands idle in order to prevent overproduction. It also bought excess crops. It then either stored them or gave them away to feed low-income people throughout the world. Most subsidies went to farmers of grains, such as corn, wheat, and rice. It’s because grains provide 80 percent of the world's caloric needs. By 1999, farm subsidies had reached a record $22 million. Between 2001 and 2006, farm subsidies tapered off a bit, averaging $19 billion a year. Of this, about 15 percent was wasteful, unnecessary, or redundant. Between 1995 and 2010, farm subsidies had ballooned to $52 billion a year on average. Of this, more than 6 percent went toward four "junk food" components: corn syrup, high-fructose corn syrup, corn starch, and soy oils. Many people wondered why the federal government was subsidizing food that contributed to America's obesity problem. During the recession, as lawmakers looked for ways to cut the budget, many asked, "Do corn growers need subsidies?" In 2011, a record 12.4 billion bushels of corn were produced. In 2012, 94 million acres of corn were scheduled to be planted. This was more than in any year since World War II. By 2017, large farms dominated the industry. Farms generating $1 million or more in sales produced two-thirds of the nation's agricultural output. Only 4 percent of farms were that large. Big farms gobbled up small ones that couldn't compete. They relied on economies of scale to produce more food at a cheaper price. That sent prices down even more, putting more small farmers out of business. The 2012 budget proposed a 22 percent cut to farm subsidies, including the $5 billion direct payment program. Half of farmers receiving subsidies made more than $100,000 a year. Between 1995 and 2016, the top 10 percent of farmers received 77 percent of subsidies. The top 1 percent received 26 percent or $1.7 million per recipient. The top recipient was Deline Farms Partnership, which received $4 million in 2016. The House budget also proposed $180 billion in cuts to the farm subsidy program. But $133 billion of the cuts were to the food stamp program, affecting 8 million consumers, not farmers. In March 2012, President Obama called for an end to the $4 billion in oil industry subsidies. Some estimates indicated that the real level of oil industry subsidies is higher, between $10 and $40 billion. At the same time, oil company profits benefited when oil prices reached a record of $145 a barrel in 2008. The oil industry subsidies have a long history in the United States. As early as World War I, the government stimulated oil and gas production in order to ensure a domestic supply. In 1995, Congress established the Deep Water Royalty Relief Act. It allowed oil companies to drill on federal property without paying royalties. This encouraged the expensive form of extraction since oil was only $18 a barrel. The Treasury Department reported that the federal government has missed $50 billion in foregone revenue over the program's lifetime. It argued that this may no longer be needed now that deepwater extraction has become profitable. Here is a summary of the 2011 oil industry subsidies compiled by Taxpayers for Common Sense in its report, "Subsidy Gusher." - Volumetric Ethanol Excise Tax Credit - $31 billion. - Intangible Drilling Costs - $8.9 billion. - Oil and Gas Royalty Relief - $6.9 billion. - Percentage Depletion Allowance - $4.327 billion. - Refinery Equipment Deductions - $2.3 billion. - Geological and Geophysical Costs Tax Credit - $698 million. - Natural Gas Distribution Lines - $500 million. - Ultradeepwater and Unconventional Natural Gas and other Petroleum Resources R&D - $230 million. - Passive Loss Exemption - $105 million. - Unconventional Fossil Technology Program - $100 million. - Other subsidies - $161 million. Greenpeace argues that the oil industry subsidies should also include the following activities: - The Strategic Petroleum Reserve. - Defense spending that involves military action in oil-rich countries in the Persian Gulf. - The construction of the U.S. federal highway system which encourages reliance on gas-driven cars. The BEA argues that these federal government activities were primarily done to protect national security and not promote specific activities within the oil industry. Even though the intent was not to directly subsidize it, they may have benefited the industry indirectly. Between 1979 and 2010, the corn industry received $20 billion in federal subsidies. Congress wanted to divert production into ethanol, a component of gasoline. The subsidies were meant to help producers meet a 2005 federal law that required 7.5 billion gallons of renewable fuel to be produced by 2012. In 2007, a revision increased the goal to 36 billion gallons by 2022. Only 6.25 billion gallons were produced in 2011. The corn subsidy, a tax credit of $0.46 a gallon, ended in January 2012. Ethanol producers would have liked to see a larger credit of $1.10 per gallon remain. The credit was to research cost-effective ways to convert other bio-fuels, like switchgrass, wood chips, and nonfood corn byproducts. When the corn subsidy ended in 2012, ethanol producers were left in a bit of a glut. But that was because gasoline refiners stocked up on subsidized ethanol before prices went up. The glut was absorbed over time. Demand increased during the U.S. summer driving season. Growing markets, such as Brazil, couldn’t keep up with their own need for ethanol. They began importing it from the United States. Converting corn for fuel became controversial when it helped drive food prices higher in 2008. That created food riots throughout the world. That was just one reason for the high price for corn and other commodities. Also, investors fled to the commodities markets in response to the global financial crisis of 2008. Many experts argue that using corn for fuel is a poor allocation of natural resources when 60 percent of the world's population is malnourished. Furthermore, corn is not an efficient fuel source. Even if all the corn in the United States were converted to ethanol, it would only meet 4 percent of America's fuel consumption needs. (Source: “Ethanol Subsidy Dies But Wait There's More,” MSNBC.com, December 29, 2011.) The WTO bans export subsidies. But it allows two U.S. federal government export subsidy programs. They help U.S. farmers compete with other countries' subsidized exports. The U.S. Department of Agriculture promotes: - The Export Credit Guarantee Program, which finances U.S. farm exports. The USDA guarantees the buyers' credit when they can't get credit approval locally. - The Dairy Export Incentive Program, which pays cash subsidies to dairy exporters. It helps them meet the subsidized prices of foreign dairy producers. Housing subsidies promote homeownership and support the construction industry. They total about $15 billion a year. Housing subsidies come in two forms: interest rate subsidies and down-payment assistance. The biggest interest rate subsidy is the mortgage interest deduction on the federal income tax. There are also some smaller interest subsidies that reduce mortgage costs for low-income families. The federal government also matches the amount low-income families save for a down-payment. This came to $10.9 million in 2008. (Source: “Homeowner Subsidies,” Federal Reserve Bank of Cleveland, February 23, 2011.) These direct homeowner subsidies paled in comparison to what the federal government spent to support its Federal Housing Authority mortgage loan guarantee program. The real trouble started when it created two government-sponsored enterprises. Fannie Mae and Freddie Mac provided a secondary market to buy these mortgages from banks. But they bought too many. That forced the government to spend up to $100 billion to bail out Fannie and Freddie. Even this wasn't enough, and the government nationalized them. Was the bailout a subsidy? Yes, in a sense. That’s because without it, there would have been no housing activity whatsoever after the subprime mortgage crisis. Fannie, Freddie and the Federal Home Loan Guaranty Corporation were behind 90 percent of all home loans. The agencies replaced the private sector's role in the home mortgage market in the United States. The U.S. federal government offers many more subsidies that it thinks will improve the economy. For example, the 2009 Cash for Clunkers program was a subsidy to auto dealers, according to the BEA. In the program, dealers received a subsidy of up to $4,500 from the federal government after discounting a new vehicle to a consumer who traded in an old car. The goal was to jump-start the economy after the recession. It also aimed to encourage people to buy more fuel-efficient vehicles and lessen U.S. reliance on foreign oil. More than half of the Obamacare subsidies are designed to go to middle-income families. These are hard-working parents. They hold jobs as food service workers, administrative personnel, and health aides. These are also jobs that don't provide health insurance. Although 10.6 million Americans were eligible for subsidies as of February 2018, most didn't get them. Why? It’s because they didn't sign up for insurance on the exchanges. Obamacare is budgeted to spend $1.039 trillion on subsidies for these middle-class working families between 2015 and 2024. It only expects to spend $792 billion on expanded Medicaid and Childrens’ Health Insurance Program for the poor.<\|endoftext\|>	3.71875
212	# How do you write an equation of a line given slope -3 and passes through (5,2)? Mar 3, 2017 $y = - 3 x + 17$ #### Explanation: The general formula for a line is $y = m x + c$ where $m =$ gradient and $c =$ the $y$ intercept. Using the numbers you have been given in the question, you can put these numbers into the general formula to calculate $c$. From the coordinates given, the $x$ is 5 and $y$ is 2. $2 = - 3 \cdot 5 + c$ $2 = - 15 + c$ $c = 17$ Now you have worked out the value of $c$, (the $y$ intercept), you can give the equation of the line, by inserting $- 3$ in place of the $m$ and $17$ in place of the $c$: $y = - 3 x + 17$<\|endoftext\|>	4.4375
1,023	Anatomy Knee Joint To understand conditions and injuries of the knee, it is essential to look at the joint’s structure, meaning the anatomy of the knee and the function of the important structures such as the meniscus, the cruciate ligament, and the cartilage. The knee joint is the articulation between the femur, tibia and patella. That the structure is somewhat complicated shows a look inside the joint: The contact points between the round end of the femur and the tibial plateau only consists two rounded condyles at the end of the femur, which, however, does not provide sufficient support. A complex apparatus of ligaments, tendons and buffers is needed to ensure a good mobility. The knee ensures stability supporting the body weight; it allows for rotations and absorbs movements. The result is a complex construct, which is on the one hand quite capable of tolerating stress and carrying the body’s weight, but on the other hand is also vulnerable to injuries and wears and tears throughout life. The main structures of the knee joint The ends of the upper and lower leg bones are coated by a three to five millimeter thick layer of hyaline cartilage acting as a buffer. In conjunction with the synovial fluid, this layer permits a friction-free movement. The cartilage of adults has no blood supply; consequently, injuries do not heal by themselves. On the contrary, when an injury occurs, the cartilage damage increases and threatens to turn into osteoarthritis. The internal (or medial) meniscus and the outside of the lateral meniscus function as buffers and are located between the upper and lower leg bones. The menisci are made of fibrocartilage. In addition to the buffer function, the meniscus is responsible for an enlargement of the contact surface creating a better pressure distribution between upper and lower leg bones. A meniscus tear can be caused by an accident, but also normal strain and tear. Ligaments - cruciate ligament, lateral ligament The knee joint is secured by four complex ligament structures. These include the two collateral ligaments, the anterior, and the posterior cruciate ligaments. The collateral ligaments (medial collateral ligament (MCL) and Lateral also called figural collateral ligament (LCL)) are used to stabilize the knee joint to toward the side. According to their names, the medial collateral ligament is located on the inner and the lateral collateral ligament on the outside of the knee. The cruciate ligaments extend into the depth of the knee joint and stabilize the lower leg forward or backward against the thigh and in rotation movements. A serious accident may cause a ligament injury. Depending on the severity of the injury, we differentiate between overstretching, partial tear or complete cruciate ligament tear (Torn ACL)) Muscles and tendons They are essential for the active movement of the knee. The muscles end in tendons and are additional connected with the bones. With tension, the muscles will shorten and the bones are moved in direction of the tension. Simultaneously, the muscle pull secures the joint in its position. As a result, chronic overuse can lead to tendon disorders, while severe stress or sudden movements can cause muscle injuries (patellar tendinitis / iliotibial band syndrome (ITBS) / runner's knee). The patella is the bony part of the patellar tendon (patella tendon), which connects the large thigh muscle (M. quadriceps) to the tibia. With every more, the tibia slides over the fermur. On the back, it is coated with a thick layer of cartilage. If cartilage surface is damaged, the damage becomes especially noticeable through pain when climbing stairs. Advanced cartilage damage is known as osteoarthritis. Joint capsule, synovial membrane, synovial fluid Just like other joints, the knee joint is surrounded by a strong joint capsule. It is lined with mucous membrane. The membrane produces synovial fluid to provide nutrition to the cartilage and lubrication of the joint. If an inflammation of the synovial membrane occurs, for example, caused by cartilage damage, the bursa produces more fluid, similar to the nasal mucosa in rhinitis. As a result, the knee swells and is painful. In addition to an anti-inflammatory therapy, if necessary, the condition should be treated with a medicinal cartilage therapy such as hyaluronic acid. The knee joint is surrounded by numerous bursae (fluid sacs). They serve as buffers to minimize friction and facilitate movement between different sections of the joint. If an inflammation is present, the bursae increase, i.e. they swell, which can cause the appearance of the leg surface to change.<\|endoftext\|>	4
11,644	Malaria is a mosquito-borne infectious disease caused by a eukaryotic protist of the genus Plasmodium. It is widespread in tropical and subtropical regions, including parts of the Americas, Asia, and Africa. Each year, there are approximately 350–500 million cases of malaria, killing between one and three million people, the majority of whom are young children in sub-Saharan Africa. Ninety percent of malaria-related deaths occur in sub-Saharan Africa. Malaria is commonly associated with poverty, but is also a cause of poverty and a major hindrance to economic development. Five species of the plasmodium parasite can infect humans; the most serious forms of the disease are caused by Plasmodium falciparum. Malaria caused by Plasmodium vivax, Plasmodium ovale and Plasmodium malariae causes milder disease in humans that is not generally fatal. A fifth species, Plasmodium knowlesi, is a zoonosis that causes malaria in macaques but can also infect humans. Malaria is naturally transmitted by the bite of a female Anopheles mosquito. When a mosquito bites an infected person, a small amount of blood is taken, which contains malaria parasites. These develop within the mosquito, and about one week later, when the mosquito takes its next blood meal, the parasites are injected with the mosquito’s saliva into the person being bitten. After a period of between two weeks and several months (occasionally years) spent in the liver, the malaria parasites start to multiply within red blood cells, causing symptoms that include fever and headache. In severe cases, the disease worsens, leading to coma and death. A wide variety of antimalarial drugs are available to treat malaria. In the last 5 years, treatment of P. falciparum infections in endemic countries has been transformed by the use of combinations of drugs containing an artemisinin derivative. Severe malaria is treated with intravenous or intramuscular quinine or, increasingly, the artemisinin derivative artesunate. Several drugs are also available to prevent malaria in travellers to malaria-endemic countries (prophylaxis). Resistance has developed to several antimalarial drugs, most notably chloroquine. Malaria transmission can be reduced by preventing mosquito bites by distribution of inexpensive mosquito nets and insect repellents, or by mosquito-control measures such as spraying insecticides inside houses and draining standing water where mosquitoes lay their eggs. Signs and symptoms Main symptoms of malaria. Symptoms of malaria include fever, shivering, arthralgia (joint pain), vomiting, anemia (caused by hemolysis), hemoglobinuria, retinal damage and convulsions. The classic symptom of malaria is cyclical occurrence of sudden coldness followed by rigor and then fever and sweating lasting four to six hours, occurring every two days in P. vivax and P. ovale infections, while every three for P. malariae. P. falciparum can have recurrent fever every 36–48 hours or a less pronounced and almost continuous fever. For reasons that are poorly understood, but that may be related to high intracranial pressure, children with malaria frequently exhibit abnormal posturing, a sign indicating severe brain damage. Malaria has been found to cause cognitive impairments, especially in children. It causes widespread anemia during a period of rapid brain development and also direct brain damage. This neurologic damage results from cerebral malaria to which children are more vulnerable. Cerebral malaria is associated with retinal whitening, which may be a useful clinical sign in distinguishing malaria from other causes of fever. Chronic malaria is seen in both P. vivax and P. ovale, but not in P. falciparum. Here, the disease can relapse months or years after exposure, due to the presence of latent parasites in the liver. Describing a case of malaria as cured by observing the disappearance of parasites from the bloodstream can, therefore, be deceptive. The longest incubation period reported for a P. vivax infection is 30 years. Approximately one in five of P. vivax malaria cases in temperate areas involve overwintering by hypnozoites (i.e., relapses begin the year after the mosquito bite). A Plasmodium sporozoite traverses the cytoplasm of a mosquito midgut epithelial cell in this false-color electron micrograph.Malaria parasites Malaria parasites are members of the genus Plasmodium (phylum Apicomplexa). In humans malaria is caused by P. falciparum, P. malariae, P. ovale, P. vivax and P. knowlesi. P. falciparum is the most common cause of infection and is responsible for about 80% of all malaria cases, and is also responsible for about 90% of the deaths from malaria. Parasitic Plasmodium species also infect birds, reptiles, monkeys, chimpanzees and rodents. There have been documented human infections with several simian species of malaria, namely P. knowlesi, P. inui, P. cynomolgi, P. simiovale, P. brazilianum, P. schwetzi and P. simium; however, with the exception of P. knowlesi, these are mostly of limited public health importance. Mosquito vectors and the Plasmodium life cycle The parasite’s primary (definitive) hosts and transmission vectors are female mosquitoes of the Anopheles genus, while humans and other vertebrates are secondary hosts. Young mosquitoes first ingest the malaria parasite by feeding on an infected human carrier and the infected Anopheles mosquitoes carry Plasmodium sporozoites in their salivary glands. A mosquito becomes infected when it takes a blood meal from an infected human. Once ingested, the parasite gametocytes taken up in the blood will further differentiate into male or female gametes and then fuse in the mosquito gut. This produces an ookinete that penetrates the gut lining and produces an oocyst in the gut wall. When the oocyst ruptures, it releases sporozoites that migrate through the mosquito’s body to the salivary glands, where they are then ready to infect a new human host. This type of transmission is occasionally referred to as anterior station transfer. The sporozoites are injected into the skin, alongside saliva, when the mosquito takes a subsequent blood meal. Only female mosquitoes feed on blood, thus males do not transmit the disease. The females of the Anopheles genus of mosquito prefer to feed at night. They usually start searching for a meal at dusk, and will continue throughout the night until taking a meal. Malaria parasites can also be transmitted by blood transfusions, although this is rare.[ The life cycle of malaria parasites in the human body. A mosquito infects a person by taking a blood meal. First, sporozoites enter the bloodstream, and migrate to the liver. They infect liver cells (hepatocytes), where they multiply into merozoites, rupture the liver cells, and escape back into the bloodstream. Then, the merozoites infect red blood cells, where they develop into ring forms, then trophozoites (a feeding stage), then schizonts (a reproduction stage), then back into merozoites. Sexual forms called gametocytes are also produced, which, if taken up by a mosquito, will infect the insect and continue the life cycle.Malaria in humans develops via two phases: an exoerythrocytic and an erythrocytic phase. The exoerythrocytic phase involves infection of the hepatic system, or liver, whereas the erythrocytic phase involves infection of the erythrocytes, or red blood cells. When an infected mosquito pierces a person’s skin to take a blood meal, sporozoites in the mosquito’s saliva enter the bloodstream and migrate to the liver. Within 30 minutes of being introduced into the human host, the sporozoites infect hepatocytes, multiplying asexually and asymptomatically for a period of 6–15 days. Once in the liver, these organisms differentiate to yield thousands of merozoites, which, following rupture of their host cells, escape into the blood and infect red blood cells, thus beginning the erythrocytic stage of the life cycle. The parasite escapes from the liver undetected by wrapping itself in the cell membrane of the infected host liver cell. Within the red blood cells, the parasites multiply further, again asexually, periodically breaking out of their hosts to invade fresh red blood cells. Several such amplification cycles occur. Thus, classical descriptions of waves of fever arise from simultaneous waves of merozoites escaping and infecting red blood cells. Some P. vivax and P. ovale sporozoites do not immediately develop into exoerythrocytic-phase merozoites, but instead produce hypnozoites that remain dormant for periods ranging from several months (6–12 months is typical) to as long as three years. After a period of dormancy, they reactivate and produce merozoites. Hypnozoites are responsible for long incubation and late relapses in these two species of malaria. The parasite is relatively protected from attack by the body’s immune system because for most of its human life cycle it resides within the liver and blood cells and is relatively invisible to immune surveillance. However, circulating infected blood cells are destroyed in the spleen. To avoid this fate, the P. falciparum parasite displays adhesive proteins on the surface of the infected blood cells, causing the blood cells to stick to the walls of small blood vessels, thereby sequestering the parasite from passage through the general circulation and the spleen. This “stickiness” is the main factor giving rise to hemorrhagic complications of malaria. High endothelial venules (the smallest branches of the circulatory system) can be blocked by the attachment of masses of these infected red blood cells. The blockage of these vessels causes symptoms such as in placental and cerebral malaria. In cerebral malaria the sequestrated red blood cells can breach the blood brain barrier possibly leading to coma. Although the red blood cell surface adhesive proteins (called PfEMP1, for Plasmodium falciparum erythrocyte membrane protein 1) are exposed to the immune system, they do not serve as good immune targets, because of their extreme diversity; there are at least 60 variations of the protein within a single parasite and effectively limitless versions within parasite populations. The parasite switches between a broad repertoire of PfEMP1 surface proteins, thus staying one step ahead of the pursuing immune system. Some merozoites turn into male and female gametocytes. If a mosquito pierces the skin of an infected person, it potentially picks up gametocytes within the blood. Fertilization and sexual recombination of the parasite occurs in the mosquito’s gut, thereby defining the mosquito as the definitive host of the disease. New sporozoites develop and travel to the mosquito’s salivary gland, completing the cycle. Pregnant women are especially attractive to the mosquitoes, and malaria in pregnant women is an important cause of stillbirths, infant mortality and low birth weight, particularly in P. falciparum infection, but also in other species infection, such as P. vivax. Further information: Romanowsky stain, Malaria antigen detection tests Blood smear from a P. falciparum culture (K1 strain). Several red blood cells have ring stages inside them. Close to the center there is a schizont and on the left a trophozoite.Since Charles Laveran first visualised the malaria parasite in blood in 1880,the mainstay of malaria diagnosis has been the microscopic examination of blood. Fever and septic shock are commonly misdiagnosed as severe malaria in Africa, leading to a failure to treat other life-threatening illnesses. In malaria-endemic areas, parasitemia does not ensure a diagnosis of severe malaria, because parasitemia can be incidental to other concurrent disease. Recent investigations suggest that malarial retinopathy is better (collective sensitivity of 95% and specificity of 90%) than any other clinical or laboratory feature in distinguishing malarial from non-malarial coma. Although blood is the sample most frequently used to make a diagnosis, both saliva and urine have been investigated as alternative, less invasive specimens. Areas that cannot afford even simple laboratory diagnostic tests often use only a history of subjective fever as the indication to treat for malaria. Using Giemsa-stained blood smears from children in Malawi, one study showed that when clinical predictors (rectal temperature, nailbed pallor, and splenomegaly) were used as treatment indications, rather than using only a history of subjective fevers, a correct diagnosis increased from 21% to 41% of cases, and unnecessary treatment for malaria was significantly decreased. Microscopic examination of blood films For more details on individual parasites, see P. falciparum, P. vivax, P. ovale, P. malariae, P. knowlesi. The most economic, preferred, and reliable diagnosis of malaria is microscopic examination of blood films because each of the four major parasite species has distinguishing characteristics. Two sorts of blood film are traditionally used. Thin films are similar to usual blood films and allow species identification because the parasite’s appearance is best preserved in this preparation. Thick films allow the microscopist to screen a larger volume of blood and are about eleven times more sensitive than the thin film, so picking up low levels of infection is easier on the thick film, but the appearance of the parasite is much more distorted and therefore distinguishing between the different species can be much more difficult. With the pros and cons of both thick and thin smears taken into consideration, it is imperative to utilize both smears while attempting to make a definitive diagnosis. From the thick film, an experienced microscopist can detect parasite levels (or parasitemia) down to as low as 0.0000001% of red blood cells. Diagnosis of species can be difficult because the early trophozoites (“ring form”) of all four species look identical and it is never possible to diagnose species on the basis of a single ring form; species identification is always based on several trophozoites. One important thing to note is that P. malariae and P. knowlesi (which is the most common cause of malaria in South-east Asia) look very similar under the microscope. However, P. knowlesi parasitemia increases very fast and causes more severe disease than P. malariae, so it is important to identify and treat infections quickly. Therefore modern methods such as PCR (see “Molecular methods” below) or monoclonal antibody panels that can distinguish between the two should be used in this part of the world. Further information: Malaria antigen detection tests For areas where microscopy is not available, or where laboratory staff are not experienced at malaria diagnosis, there are commercial antigen detection tests that require only a drop of blood. Immunochromatographic tests (also called: Malaria Rapid Diagnostic Tests, Antigen-Capture Assay or “Dipsticks”) been developed, distributed and fieldtested. These tests use finger-stick or venous blood, the completed test takes a total of 15–20 minutes, and the results are read visually as the presence or absence of colored stripes on the dipstick, so they are suitable for use in the field. The threshold of detection by these rapid diagnostic tests is in the range of 100 parasites/µl of blood (commercial kits can range from about 0.002% to 0.1% parasitemia) compared to 5 by thick film microscopy. One disadvantage is that dipstick tests are qualitative but not quantitative – they can determine if parasites are present in the blood, but not how many. The first rapid diagnostic tests were using P. falciparum glutamate dehydrogenase as antigen. PGluDH was soon replaced by P.falciparum lactate dehydrogenase, a 33 kDa oxidoreductase [EC 18.104.22.168]. It is the last enzyme of the glycolytic pathway, essential for ATP generation and one of the most abundant enzymes expressed by P.falciparum. PLDH does not persist in the blood but clears about the same time as the parasites following successful treatment. The lack of antigen persistence after treatment makes the pLDH test useful in predicting treatment failure. In this respect, pLDH is similar to pGluDH. Depending on which monoclonal antibodies are used, this type of assay can distinguish between all five different species of human malaria parasites, because of antigenic differences between their pLDH isoenzymes. Molecular methods are available in some clinical laboratories and rapid real-time assays (for example, QT-NASBA based on the polymerase chain reaction) are being developed with the hope of being able to deploy them in endemic areas. PCR (and other molecular methods) is more accurate than microscopy. However, it is expensive, and requires a specialized laboratory. Moreover, levels of parasitemia are not necessarily correlative with the progression of disease, particularly when the parasite is able to adhere to blood vessel walls. Therefore more sensitive, low-tech diagnosis tools need to be developed in order to detect low levels of parasitemia in the field. Anopheles albimanus mosquito feeding on a human arm. This mosquito is a vector of malaria and mosquito control is a very effective way of reducing the incidence of malaria.Methods used to prevent the spread of disease, or to protect individuals in areas where malaria is endemic, include prophylactic drugs, mosquito eradication, and the prevention of mosquito bites. The continued existence of malaria in an area requires a combination of high human population density, high mosquito population density, and high rates of transmission from humans to mosquitoes and from mosquitoes to humans. If any of these is lowered sufficiently, the parasite will sooner or later disappear from that area, as happened in North America, Europe and much of Middle East. However, unless the parasite is eliminated from the whole world, it could become re-established if conditions revert to a combination that favors the parasite’s reproduction. Many countries are seeing an increasing number of imported malaria cases due to extensive travel and migration. Many researchers argue that prevention of malaria may be more cost-effective than treatment of the disease in the long run, but the capital costs required are out of reach of many of the world’s poorest people. Economic adviser Jeffrey Sachs estimates that malaria can be controlled for US$3 billion in aid per year. The distribution of funding varies among countries. Countries with large populations do not receive the same amount of support. The 34 countries that received a per capita annual support of less than $1 included some of the poorest countries in Africa. Brazil, Eritrea, India, and Vietnam have, unlike many other developing nations, successfully reduced the malaria burden. Common success factors included conducive country conditions, a targeted technical approach using a package of effective tools, data-driven decision-making, active leadership at all levels of government, involvement of communities, decentralized implementation and control of finances, skilled technical and managerial capacity at national and sub-national levels, hands-on technical and programmatic support from partner agencies, and sufficient and flexible financing. Further information: Mosquito control Efforts to eradicate malaria by eliminating mosquitoes have been successful in some areas. Malaria was once common in the United States and southern Europe, but vector control programs, in conjunction with the monitoring and treatment of infected humans, eliminated it from those regions. In some areas, the draining of wetland breeding grounds and better sanitation were adequate. Malaria was eliminated from the northern parts of the USA in the early 20th century by such methods, and the use of the pesticide DDT eliminated it from the South by 1951. In 2002, there were 1,059 cases of malaria reported in the US, including eight deaths, but in only five of those cases was the disease contracted in the United States. Before DDT, malaria was successfully eradicated or controlled also in several tropical areas by removing or poisoning the breeding grounds of the mosquitoes or the aquatic habitats of the larva stages, for example by filling or applying oil to places with standing water. These methods have seen little application in Africa for more than half a century. In the 1950s and 1960s, there was a major public health effort to eradicate malaria worldwide by selectively targeting mosquitoes in areas where malaria was rampant. However, these efforts have so far failed to eradicate malaria in many parts of the developing world—the problem is most prevalent in Africa. Sterile insect technique is emerging as a potential mosquito control method. Progress towards transgenic, or genetically modified, insects suggest that wild mosquito populations could be made malaria-resistant. Researchers at Imperial College London created the world’s first transgenic malaria mosquito, with the first plasmodium-resistant species announced by a team at Case Western Reserve University in Ohio in 2002. Successful replacement of current populations with a new genetically modified population, relies upon a drive mechanism, such as transposable elements to allow for non-Mendelian inheritance of the gene of interest. However, this approach contains many difficulties and success is a distant prospect. An even more futuristic method of vector control is the idea that lasers could be used to kill flying mosquitoes. Main article: Malaria prophylaxis Several drugs, most of which are also used for treatment of malaria, can be taken preventively. Generally, these drugs are taken daily or weekly, at a lower dose than would be used for treatment of a person who had actually contracted the disease. Use of prophylactic drugs is seldom practical for full-time residents of malaria-endemic areas, and their use is usually restricted to short-term visitors and travelers to malarial regions. This is due to the cost of purchasing the drugs, negative side effects from long-term use, and because some effective anti-malarial drugs are difficult to obtain outside of wealthy nations. Quinine was used starting in the 17th century as a prophylactic against malaria. The development of more effective alternatives such as quinacrine, chloroquine, and primaquine in the 20th century reduced the reliance on quinine. Today, quinine is still used to treat chloroquine resistant Plasmodium falciparum, as well as severe and cerebral stages of malaria, but is not generally used for prophylaxis. Modern drugs used preventively include mefloquine (Lariam), doxycycline (available generically), and the combination of atovaquone and proguanil hydrochloride (Malarone). The choice of which drug to use depends on which drugs the parasites in the area are resistant to, as well as side-effects and other considerations. The prophylactic effect does not begin immediately upon starting taking the drugs, so people temporarily visiting malaria-endemic areas usually begin taking the drugs one to two weeks before arriving and must continue taking them for 4 weeks after leaving (with the exception of atovaquone proguanil that only needs be started 2 days prior and continued for 7 days afterwards). The use of prophylactic drugs where malaria-bearing mosquitoes are present may encourage the development of partial immunity. Indoor residual spraying Main articles: Indoor residual spraying and DDT use against malaria Indoor residual spraying (IRS) is the practice of spraying insecticides on the interior walls of homes in malaria affected areas. After feeding, many mosquito species rest on a nearby surface while digesting the bloodmeal, so if the walls of dwellings have been coated with insecticides, the resting mosquitos will be killed before they can bite another victim, transferring the malaria parasite. The first pesticide used for IRS was DDT.Although it was initially used exclusively to combat malaria, its use quickly spread to agriculture. In time, pest-control, rather than disease-control, came to dominate DDT use, and this large-scale agricultural use led to the evolution of resistant mosquitoes in many regions. The DDT resistance shown by Anopheles mosquitoes can be compared to antibiotic resistance shown by bacteria. The overuse of anti-bacterial soaps and antibiotics led to antibiotic resistance in bacteria, similar to how overspraying of DDT on crops led to DDT resistance in Anopheles mosquitoes. During the 1960s, awareness of the negative consequences of its indiscriminate use increased, ultimately leading to bans on agricultural applications of DDT in many countries in the 1970s. Since the use of DDT has been limited or banned for agricultural use for some time, DDT may now be more effective as a method of disease-control. Although DDT has never been banned for use in malaria control and there are several other insecticides suitable for IRS, some advocates have claimed that bans are responsible for tens of millions of deaths in tropical countries where DDT had once been effective in controlling malaria. Furthermore, most of the problems associated with DDT use stem specifically from its industrial-scale application in agriculture, rather than its use in public health. The World Health Organization (WHO) currently advises the use of 12 different insecticides in IRS operations. These include DDT and a series of alternative insecticides (such as the pyrethroids permethrin and deltamethrin), to combat malaria in areas where mosquitoes are DDT-resistant and to slow the evolution of resistance. This public health use of small amounts of DDT is permitted under the Stockholm Convention on Persistent Organic Pollutants (POPs), which prohibits the agricultural use of DDT However, because of its legacy, many developed countries discourage DDT use even in small quantities. One problem with all forms of Indoor Residual Spraying is insecticide resistance via evolution of mosquitos. According to a study published on Mosquito Behavior and Vector Control, mosquito species that are affected by IRS are endophilic species (species that tend to rest and live indoors), and due to the irritation caused by spraying, their evolutionary descendants are trending towards becoming exophilic (species that tend to rest and live out of doors), meaning that they are not as affected—if affected at all—by the IRS, rendering it somewhat useless as a defense mechanism. Mosquito nets and bedclothes Main article: Mosquito net Mosquito nets help keep mosquitoes away from people and greatly reduce the infection and transmission of malaria. The nets are not a perfect barrier and they are often treated with an insecticide designed to kill the mosquito before it has time to search for a way past the net. Insecticide-treated nets (ITN) are estimated to be twice as effective as untreated nets and offer greater than 70% protection compared with no net. Although ITN are proven to be very effective against malaria, less than 2% of children in urban areas in Sub-Saharan Africa are protected by ITNs. Since the Anopheles mosquitoes feed at night, the preferred method is to hang a large “bed net” above the center of a bed such that it drapes down and covers the bed completely. The distribution of mosquito nets impregnated with insecticides such as permethrin or deltamethrin has been shown to be an extremely effective method of malaria prevention, and it is also one of the most cost-effective methods of prevention. These nets can often be obtained for around $2.50–$3.50 (2–3 euros) from the United Nations, the World Health Organization (WHO), and others. ITNs have been shown to be the most cost-effective prevention method against malaria and are part of WHO’s Millennium Development Goals (MDGs). While some experts argue that international organizations should distribute ITNs and LLINs to people for free in order to maximize coverage (since such a policy would reduces price barriers), others insist that cost-sharing between the international organization and recipients would led to greater usage of the net (arguing that people will value a good more if they pay for it). Additionally, proponents of cost-sharing argue that such a policy ensures that nets are efficiently allocated to those people who most need them (or are most vulnerable to infection). Through a “selection effect”, they argue, those people who most need the bed nets will choose to purchase them, while those less in need will opt out. However, a randomized controlled trial study of ITNs uptake among pregnant women in Kenya, conducted by economists Pascaline Dupas and Jessica Cohen, found that cost-sharing does not necessarily increase the usage intensity of ITNs nor does it induce uptake by those most vulnerable to infection, as compared to a policy of free distribution. In some cases, cost-sharing can actually decrease demand for mosquito nets by erecting a price barrier. Dupas and Cohen’s findings support the argument that free distribution of ITNs can be more effective than cost-sharing in both increasing coverage and saving lives. In a cost-effectiveness analysis, Dupas and Cohen note that “cost-sharing is at best marginally more cost-effective than free distribution, but free distribution leads to many more lives saved.” The researchers base their conclusions about the cost-effectiveness of free distribution on the proven spillover benefits of increased ITN usage. When a large number of nets are distributed in one residential area, their chemical additives help reduce the number of mosquitoes in the environment. With fewer mosquitoes in the environment, the chances of malaria infection for recipients and non-recipients are significantly reduced. (In other words, the importance of the physical barrier effect of ITNs decreases relative to the positive externality effect of the nets in creating a mosquito-free environment when ITNs are highly concentrated in one residential cluster or community.) For maximum effectiveness, the nets should be re-impregnated with insecticide every six months. This process poses a significant logistical problem in rural areas. New technologies like Olyset or DawaPlus allow for production of long-lasting insecticidal mosquito nets (LLINs), which release insecticide for approximately 5 years, and cost about US$5.50. ITNs protect people sleeping under the net and simultaneously kill mosquitoes that contact the net. Some protection is also provided to others by this method, including people sleeping in the same room but not under the net. While distributing mosquito nets is a major component of malaria prevention, community education and awareness on the dangers of malaria are associated with distribution campaigns to make sure people who receive a net know how to use it. “Hang Up” campaigns such as the ones conducted by volunteers of the International Red Cross and Red Crescent Movement consist of visiting households that received a net at the end of the campaign or just before the rainy season, ensuring that the net is being used properly and that the people most vulnerable to malaria, such as young children and the elderly, sleep under it. A study conducted by the CDC in Sierra Leone showed a 22 percent increase in net utilization following a personal visit from a volunteer living in the same community promoting net usage. A study in Togo showed similar improvements. Mosquito nets are often unaffordable to people in developing countries, especially for those most at risk. Only 1 out of 20 people in Africa own a bed net. Nets are also often distributed though vaccine campaigns using voucher subsidies, such as the measles campaign for children. A study among Afghan refugees in Pakistan found that treating top-sheets and chaddars (head coverings) with permethrin has similar effectiveness to using a treated net, but is much cheaper. Another alternative approach uses spores of the fungus Beauveria bassiana, sprayed on walls and bed nets, to kill mosquitoes. While some mosquitoes have developed resistance to chemicals, they have not been found to develop a resistance to fungal infections. Further information: Malaria vaccine Immunity (or, more accurately, tolerance) does occur naturally, but only in response to repeated infection with multiple strains of malaria. Vaccines for malaria are under development, with no completely effective vaccine yet available. The first promising studies demonstrating the potential for a malaria vaccine were performed in 1967 by immunizing mice with live, radiation-attenuated sporozoites, providing protection to about 60% of the mice upon subsequent injection with normal, viable sporozoites. Since the 1970s, there has been a considerable effort to develop similar vaccination strategies within humans. It was determined that an individual can be protected from a P. falciparum infection if they receive over 1,000 bites from infected, irradiated mosquitoes. It has been generally accepted that it is impractical to provide at-risk individuals with this vaccination strategy, but that has been recently challenged with work being done by Dr. Stephen Hoffman, one of the key researchers who originally sequenced the genome of Plasmodium falciparum. His work most recently has revolved around solving the logistical problem of isolating and preparing the parasites equivalent to 1000 irradiated mosquitoes for mass storage and inoculation of human beings. The company has recently received several multi-million dollar grants from the Bill & Melinda Gates Foundation and the U.S. government to begin early clinical studies in 2007 and 2008. The Seattle Biomedical Research Institute (SBRI), funded by the Malaria Vaccine Initiative, assures potential volunteers that “the clinical trials won’t be a life-threatening experience. While many volunteers [in Seattle] will actually contract malaria, the cloned strain used in the experiments can be quickly cured, and does not cause a recurring form of the disease.” “Some participants will get experimental drugs or vaccines, while others will get placebo.” Instead, much work has been performed to try and understand the immunological processes that provide protection after immunization with irradiated sporozoites. After the mouse vaccination study in 1967, it was hypothesized that the injected sporozoites themselves were being recognized by the immune system, which was in turn creating antibodies against the parasite. It was determined that the immune system was creating antibodies against the circumsporozoite protein (CSP) which coated the sporozoite. Moreover, antibodies against CSP prevented the sporozoite from invading hepatocytes. CSP was therefore chosen as the most promising protein on which to develop a vaccine against the malaria sporozoite. It is for these historical reasons that vaccines based on CSP are the most numerous of all malaria vaccines. Presently, there is a huge variety of vaccine candidates on the table. Pre-erythrocytic vaccines (vaccines that target the parasite before it reaches the blood), in particular vaccines based on CSP, make up the largest group of research for the malaria vaccine. Other vaccine candidates include: those that seek to induce immunity to the blood stages of the infection; those that seek to avoid more severe pathologies of malaria by preventing adherence of the parasite to blood venules and placenta; and transmission-blocking vaccines that would stop the development of the parasite in the mosquito right after the mosquito has taken a bloodmeal from an infected person. It is hoped that the knowledge of the P. falciparum genome, the sequencing of which was completed in 2002, will provide targets for new drugs or vaccines. The first vaccine developed that has undergone field trials, is the SPf66, developed by Manuel Elkin Patarroyo in 1987. It presents a combination of antigens from the sporozoite (using CS repeats) and merozoite parasites. During phase I trials a 75% efficacy rate was demonstrated and the vaccine appeared to be well tolerated by subjects and immunogenic. The phase IIb and III trials were less promising, with the efficacy falling to between 38.8% and 60.2%. A trial was carried out in Tanzania in 1993 demonstrating the efficacy to be 31% after a years follow up, however the most recent (though controversial) study in The Gambia did not show any effect. Despite the relatively long trial periods and the number of studies carried out, it is still not known how the SPf66 vaccine confers immunity; it therefore remains an unlikely solution to malaria. The CSP was the next vaccine developed that initially appeared promising enough to undergo trials. It is also based on the circumsporoziote protein, but additionally has the recombinant (Asn-Ala-Pro15Asn-Val-Asp-Pro)2-Leu-Arg(R32LR) protein covalently bound to a purified Pseudomonas aeruginosa toxin (A9). However at an early stage a complete lack of protective immunity was demonstrated in those inoculated. The study group used in Kenya had an 82% incidence of parasitaemia whilst the control group only had an 89% incidence. The vaccine intended to cause an increased T-lymphocyte response in those exposed, this was also not observed. The efficacy of Patarroyo’s vaccine has been disputed with some US scientists concluding in The Lancet (1997) that “the vaccine was not effective and should be dropped” while the Colombian accused them of “arrogance” putting down their assertions to the fact that he came from a developing country. The RTS,S/AS02A vaccine is the candidate furthest along in vaccine trials. It is being developed by a partnership between the PATH Malaria Vaccine Initiative (a grantee of the Gates Foundation), the pharmaceutical company, GlaxoSmithKline, and the Walter Reed Army Institute of Research. In the vaccine, a portion of CSP has been fused to the immunogenic “S antigen” of the hepatitis B virus; this recombinant protein is injected alongside the potent AS02A adjuvant. In October 2004, the RTS,S/AS02A researchers announced results of a Phase IIb trial, indicating the vaccine reduced infection risk by approximately 30% and severity of infection by over 50%. The study looked at over 2,000 Mozambican children. More recent testing of the RTS,S/AS02A vaccine has focused on the safety and efficacy of administering it earlier in infancy: In October 2007, the researchers announced results of a phase I/IIb trial conducted on 214 Mozambican infants between the ages of 10 and 18 months in which the full three-dose course of the vaccine led to a 62% reduction of infection with no serious side-effects save some pain at the point of injection. Further research will delay this vaccine from commercial release until around 2011. Education in recognizing the symptoms of malaria has reduced the number of cases in some areas of the developing world by as much as 20%. Recognizing the disease in the early stages can also stop the disease from becoming a killer. Education can also inform people to cover over areas of stagnant, still water e.g. Water Tanks which are ideal breeding grounds for the parasite and mosquito, thus cutting down the risk of the transmission between people. This is most put in practice in urban areas where there are large centers of population in a confined space and transmission would be most likely in these areas. The Malaria Control Project is currently using downtime computing power donated by individual volunteers around the world (see Volunteer computing and BOINC) to simulate models of the health effects and transmission dynamics in order to find the best method or combination of methods for malaria control. This modeling is extremely computer intensive due to the simulations of large human populations with a vast range of parameters related to biological and social factors that influence the spread of the disease. It is expected to take a few months using volunteered computing power compared to the 40 years it would have taken with the current resources available to the scientists who developed the program. An example of the importance of computer modeling in planning malaria eradication programs is shown in the paper by Águas and others. They showed that eradication of malaria is crucially dependent on finding and treating the large number of people in endemic areas with asymptomatic malaria, who act as a reservoir for infection. The malaria parasites do not affect animal species and therefore eradication of the disease from the human population would be expected to be effective. Other interventions for the control of malaria include mass drug administrations and intermittent preventive therapy. Further information: Antimalarial drug Active malaria infection with P. falciparum is a medical emergency requiring hospitalization. Infection with P. vivax, P. ovale or P. malariae can often be treated on an outpatient basis. Treatment of malaria involves supportive measures as well as specific antimalarial drugs. Most antimalarial drugs are produced industrially and are sold at pharmacies. However, as the cost of such medicines are often too high for most people in the developing world, some herbal remedies (such as Artemisia annua tea) have also been developed, and have gained support from international organisations such as Médicins Sans Frontières. When properly treated, someone with malaria can expect a complete recovery. Sophisticated counterfeits have been found in several Asian countries such as Cambodia, China, Indonesia, Laos, Thailand, Vietnam and are an important cause of avoidable death in those countries. WHO have said that studies indicate that up to 40% of artesunate based malaria medications are counterfeit, especially in the Greater Mekong region and have established a rapid alert system to enable information about counterfeit drugs to be rapidly reported to the relevant authorities in participating countries. There is no reliable way for doctors or lay people to detect counterfeit drugs without help from a laboratory. Companies are attempting to combat the persistence of counterfeit drugs by using new technology to provide security from source to distribution. Further information: Diseases of poverty, Tropical disease Countries which have regions where malaria is endemic as of 2003 (coloured yellow). Countries in green are free of indigenous cases of malaria in all areas. Disability-adjusted life year for malaria per 100,000 inhabitants in 2002. Malaria causes about 250 million cases of fever and approximately one million deaths annually. The vast majority of cases occur in children under 5 years old; pregnant women are also especially vulnerable. Despite efforts to reduce transmission and increase treatment, there has been little change in which areas are at risk of this disease since 1992. Indeed, if the prevalence of malaria stays on its present upwards course, the death rate could double in the next twenty years Precise statistics are unknown because many cases occur in rural areas where people do not have access to hospitals or the means to afford health care. As a consequence, the majority of cases are undocumented. Although co-infection with HIV and malaria does cause increased mortality, this is less of a problem than with HIV/tuberculosis co-infection, due to the two diseases usually attacking different age-ranges, with malaria being most common in the young and active tuberculosis most common in the old. Although HIV/malaria co-infection produces less severe symptoms than the interaction between HIV and TB, HIV and malaria do contribute to each other’s spread. This effect comes from malaria increasing viral load and HIV infection increasing a person’s susceptibility to malaria infection. Malaria is presently endemic in a broad band around the equator, in areas of the Americas, many parts of Asia, and much of Africa; however, it is in sub-Saharan Africa where 85– 90% of malaria fatalities occur. The geographic distribution of malaria within large regions is complex, and malaria-afflicted and malaria-free areas are often found close to each other. In drier areas, outbreaks of malaria can be predicted with reasonable accuracy by mapping rainfall. Malaria is more common in rural areas than in cities; this is in contrast to dengue fever where urban areas present the greater risk. For example, the cities of Vietnam, Laos and Cambodia are essentially malaria-free, but the disease is present in many rural regions. By contrast, in Africa malaria is present in both rural and urban areas, though the risk is lower in the larger cities. The global endemic levels of malaria have not been mapped since the 1960s. However, the Wellcome Trust, UK, has funded the Malaria Atlas Project to rectify this, providing a more contemporary and robust means with which to assess current and future malaria disease burden. Further information: History of malaria Malaria has infected humans for over 50,000 years, and Plasmodium may have been a human pathogen for the entire history of the species. Close relatives of the human malaria parasites remain common in chimpanzees. References to the unique periodic fevers of malaria are found throughout recorded history, beginning in 2700 BC in China. The term malaria originates from Medieval Italian: mala aria—”bad air”; and the disease was formerly called ague or marsh fever due to its association with swamps and marshland. Malaria was once common in most of Europe and North America, where it is no longer endemic, though imported cases do occur. Scientific studies on malaria made their first significant advance in 1880, when a French army doctor working in the military hospital of Constantine in Algeria named Charles Louis Alphonse Laveran observed parasites for the first time, inside the red blood cells of people suffering from malaria. He, therefore, proposed that malaria is caused by this organism, the first time a protist was identified as causing disease. For this and later discoveries, he was awarded the 1907 Nobel Prize for Physiology or Medicine. The malarial parasite was called Plasmodium by the Italian scientists Ettore Marchiafava and Angelo Celli. A year later, Carlos Finlay, a Cuban doctor treating patients with yellow fever in Havana, provided strong evidence that mosquitoes were transmitting disease to and from humans. This work followed earlier suggestions by Josiah C. Nott, and work by Patrick Manson on the transmission of filariasis. However, it was Britain’s Sir Ronald Ross working in the Presidency General Hospital in Calcutta who finally proved in 1898 that malaria is transmitted by mosquitoes. He did this by showing that certain mosquito species transmit malaria to birds and isolating malaria parasites from the salivary glands of mosquitoes that had fed on infected birds. For this work Ross received the 1902 Nobel Prize in Medicine. After resigning from the Indian Medical Service, Ross worked at the newly-established Liverpool School of Tropical Medicine and directed malaria-control efforts in Egypt, Panama, Greece and Mauritius. The findings of Finlay and Ross were later confirmed by a medical board headed by Walter Reed in 1900, and its recommendations implemented by William C. Gorgas in the health measures undertaken during construction of the Panama Canal. This public-health work saved the lives of thousands of workers and helped develop the methods used in future public-health campaigns against this disease. The first effective treatment for malaria came from the bark of cinchona tree, which contains quinine. This tree grows on the slopes of the Andes, mainly in Peru. A tincture made of this natural product was used by the inhabitants of Peru to control malaria, and the Jesuits introduced this practice to Europe during the 1640s, where it was rapidly accepted. However, it was not until 1820 that the active ingredient, quinine, was extracted from the bark, isolated and named by the French chemists Pierre Joseph Pelletier and Joseph Bienaimé Caventou. In the early 20th century, before antibiotics became available, Julius Wagner-Jauregg discovered that patients with syphilis could be treated by intentionally infecting them with malaria; the resulting fever would kill the syphilis spirochetes, and quinine would then be administered to control the malaria. Although some patients died from malaria, this was considered preferable to the almost-certain death from syphilis. A continuous P. falciparum culture was established in 1976.The first successful continuous malaria culture was established in 1976 by William Trager and James B. Jensen, which facilitated research into the molecular biology of the parasite and the development of new drugs substantially. Although the blood stage and mosquito stages of the malaria life cycle were identified in the 19th and early 20th centuries, it was not until the 1980s that the latent liver form of the parasite was observed. The discovery of this latent form of the parasite finally explained why people could appear to be cured of malaria but still relapse years after the parasite had disappeared from their bloodstreams. Evolutionary pressure of malaria on human genes Further information: Genetic resistance to malaria Malaria is thought to have been the greatest selective pressure on the human genome in recent history. This is due to the high levels of mortality and morbidity caused by malaria, especially the P. falciparum species. Frequency and origin of malaria cases in 1996. The most-studied influence of the malaria parasite upon the human genome is a hereditary blood disease, sickle-cell disease. The sickle-cell trait causes disease, but even those only partially affected by sickle-cell have substantial protection against malaria. In sickle-cell disease, there is a mutation in the HBB gene, which encodes the beta-globin subunit of haemoglobin. The normal allele encodes a glutamate at position six of the beta-globin protein, whereas the sickle-cell allele encodes a valine. This change from a hydrophilic to a hydrophobic amino acid encourages binding between haemoglobin molecules, with polymerization of haemoglobin deforming red blood cells into a “sickle” shape. Such deformed cells are cleared rapidly from the blood, mainly in the spleen, for destruction and recycling. In the merozoite stage of its life cycle, the malaria parasite lives inside red blood cells, and its metabolism changes the internal chemistry of the red blood cell. Infected cells normally survive until the parasite reproduces, but, if the red cell contains a mixture of sickle and normal haemoglobin, it is likely to become deformed and be destroyed before the daughter parasites emerge. Thus, individuals heterozygous for the mutated allele, known as sickle-cell trait, may have a low and usually-unimportant level of anaemia, but also have a greatly reduced chance of serious malaria infection. This is a classic example of heterozygote advantage. Individuals homozygous for the mutation have full sickle-cell disease and in traditional societies rarely live beyond adolescence. However, in populations where malaria is endemic, the frequency of sickle-cell genes is around 10%. The existence of four haplotypes of sickle-type hemoglobin suggests that this mutation has emerged independently at least four times in malaria-endemic areas, further demonstrating its evolutionary advantage in such affected regions. There are also other mutations of the HBB gene that produce haemoglobin molecules capable of conferring similar resistance to malaria infection. These mutations produce haemoglobin types HbE and HbC, which are common in Southeast Asia and Western Africa, respectively. Another well-documented set of mutations found in the human genome associated with malaria are those involved in causing blood disorders known as thalassaemias. Studies in Sardinia and Papua New Guinea have found that the gene frequency of β-thalassaemias is related to the level of malarial endemicity in a given population. A study on more than 500 children in Liberia found that those with β-thalassaemia had a 50% decreased chance of getting clinical malaria. Similar studies have found links between gene frequency and malaria endemicity in the α+ form of α-thalassaemia. Presumably these genes have also been selected in the course of human evolution. The Duffy antigens are antigens expressed on red blood cells and other cells in the body acting as a chemokine receptor. The expression of Duffy antigens on blood cells is encoded by Fy genes (Fya, Fyb, Fyc etc.). Plasmodium vivax malaria uses the Duffy antigen to enter blood cells. However, it is possible to express no Duffy antigen on red blood cells (Fy-/Fy-). This genotype confers complete resistance to P. vivax infection. The genotype is very rare in European, Asian and American populations, but is found in almost all of the indigenous population of West and Central Africa. This is thought to be due to very high exposure to P. vivax in Africa in the last few thousand years. Glucose-6-phosphate dehydrogenase (G6PD) is an enzyme that normally protects from the effects of oxidative stress in red blood cells. However, a genetic deficiency in this enzyme results in increased protection against severe malaria. HLA and interleukin-4 HLA-B53 is associated with low risk of severe malaria. This MHC class I molecule presents liver stage and sporozoite antigens to T-Cells. Interleukin-4, encoded by IL4, is produced by activated T cells and promotes proliferation and differentiation of antibody-producing B cells. A study of the Fulani of Burkina Faso, who have both fewer malaria attacks and higher levels of antimalarial antibodies than do neighboring ethnic groups, found that the IL4-524 T allele was associated with elevated antibody levels against malaria antigens, which raises the possibility that this might be a factor in increased resistance to malaria. Resistance in South Asia The lowest Himalayan Foothills and Inner Terai or Doon Valleys of Nepal and India are highly malarial due to a warm climate and marshes sustained during the dry season by groundwater percolating down from the higher hills. Malarial forests were intentionally maintained by the rulers of Nepal as a defensive measure. Humans attempting to live in this zone suffered much higher mortality than at higher elevations or below on the drier Gangetic Plain, however the Tharu people had lived in this zone long enough to evolve resistance via multiple genes. Endogamy along caste and ethnic lines appear to have confined these to the Tharu community. Otherwise these genes probably would have become nearly universal in South Asia and beyond because of their considerable survival value and the apparent lack of negative effects comparable to Sickle Cell Anemia. Society and culture Malaria is not just a disease commonly associated with poverty but also a cause of poverty and a major hindrance to economic development. Tropical regions are affected most, however malaria’s furthest extent reaches into some temperate zones with extreme seasonal changes. The disease has been associated with major negative economic effects on regions where it is widespread. During the late 19th and early 20th centuries, it was a major factor in the slow economic development of the American southern states. A comparison of average per capita GDP in 1995, adjusted for parity of purchasing power, between countries with malaria and countries without malaria gives a fivefold difference ($1,526 USD versus $8,268 USD). In countries where malaria is common, average per capita GDP has risen (between 1965 and 1990) only 0.4% per year, compared to 2.4% per year in other countries. Poverty is both cause and effect, however, since the poor do not have the financial capacities to prevent or treat the disease. The lowest income group in Malawi carries (1994) the burden of having 32% of their annual income used on this disease compared with the 4% of household incomes from low-to-high groups. In its entirety, the economic impact of malaria has been estimated to cost Africa $12 billion USD every year. The economic impact includes costs of health care, working days lost due to sickness, days lost in education, decreased productivity due to brain damage from cerebral malaria, and loss of investment and tourism. In some countries with a heavy malaria burden, the disease may account for as much as 40% of public health expenditure, 30-50% of inpatient admissions, and up to 50% of outpatient visits.<\|endoftext\|>	4.09375
825	GMAT Question of the Day: Daily via email \| Daily via Instagram New to GMAT Club? Watch this Video It is currently 28 Feb 2020, 01:32 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # Three years back, a father was 24 years older than his son. At present Author Message TAGS: ### Hide Tags Math Expert Joined: 02 Sep 2009 Posts: 61540 Three years back, a father was 24 years older than his son. At present [#permalink] ### Show Tags 14 Feb 2016, 04:30 00:00 Difficulty: 35% (medium) Question Stats: 79% (01:43) correct 21% (01:27) wrong based on 20 sessions ### HideShow timer Statistics Three years back, a father was 24 years older than his son. At present the father is 5 times as old as the son. How old will the son be three years from now? A. 12 years B. 6 years C. 3 years D. 9 years E. 27 years _________________ Intern Joined: 23 Dec 2015 Posts: 3 Three years back, a father was 24 years older than his son. At present [#permalink] ### Show Tags 14 Feb 2016, 07:42 According to the statement, 3 years back the father was 24 years older than his son. Mathematically we can write it like that : (Y-3)-(X-3) = 24 Now the faher is 5 time older than his son, again we can write it like that : 5x = Y thats give us a system of linear equation (Y-3) - ( X - 3) = 24 5X = Y Where Y is the age of the father and X the age of the son nowadays. According to this we can easily solve the system by remplacing the Y in the first equation by the 5X in the second equation : 5X -3 - (x - 3) = 24 X= 6 According to X equal 6 we can conclude that X + 3 is the age of the son 3 years from now. Intern Joined: 14 Dec 2014 Posts: 18 Concentration: Technology Re: Three years back, a father was 24 years older than his son. At present [#permalink] ### Show Tags 14 Feb 2016, 09:12 Using linear equations to convert the word problem: ** F = fathers current age & S = sons current age #1 Three years back, a father was 24 years older than his son: F-3 = 24 + (S - 3) #2 At present the father is 5 times as old as the son: F = 5S How old will the son be three years from now?: S + 3 = ? With two variables and two linear equations we are able to solve the problem: (S5) - 3 = 24 +S -3 4S=24 S = 6 S + 3 = 6 + 3 = 9 Re: Three years back, a father was 24 years older than his son. At present [#permalink] 14 Feb 2016, 09:12 Display posts from previous: Sort by<\|endoftext\|>	4.40625
1,862	12.17: Exponential Growth and Decay Difficulty Level: At Grade Created by: CK-12 Estimated6 minsto complete % Progress Practice Exponential Growth and Decay MEMORY METER This indicates how strong in your memory this concept is Progress Estimated6 minsto complete % Estimated6 minsto complete % MEMORY METER This indicates how strong in your memory this concept is Have you ever been to a condominium? Take a look at this dilemma. A condominium complex charges $185 per month for the homeowners’ association fee. The rates can rise every year because of inflation but they promise not to raise the rates more than 10% each year. Keep in mind, though, that if they raise the rate by 10% the first year, the second year is now more expensive. If they raise the maximum again, they are increasing the original$185 plus the first year’s adjustment by 10%. Graph the situation for 10 years. How much could the homeowners’ fee be in ten years? Use the function \begin{align}f=185 \cdot 1.1^t\end{align} where \begin{align}f\end{align} is the fee after \begin{align}t\end{align} years. Understanding exponential growth and decay will help you with this dilemma. Pay attention and you will see it again at the end of the Concept. Guidance Do you know how to tell if a function is an exponential function? If your function can be written in the form \begin{align}y=ab^x\end{align}, where \begin{align}a\end{align} and \begin{align}b\end{align} are constants, \begin{align}a \ne 0, b>0,\end{align} and \begin{align}b \ne 1\end{align}, then it must be exponential. In quadratic equations, your functions were always to the \begin{align}2^{nd}\end{align} power. In exponential functions, the exponent is a variable. Their graphs will have a characteristic curve either upward or downward. Exponential Functions 1. \begin{align}y=2^x\end{align} 2. \begin{align}c=4 \cdot 10^d\end{align} 3. \begin{align}y=2 \cdot \left(\frac{2}{3}\right)^x\end{align} 4. \begin{align}t=4 \cdot 10^u\end{align} Not Exponential Functions \begin{align} & 1.\ y=3 \cdot 1^x && 2.\ n=0 \cdot 3^p && 3. \ y=(-4)^x && 4. \ y=-6 \cdot 0^x\\ \text{because} & \quad \ b=1 && \quad \ a=0 && \quad \ b<0 && \quad \ b \le 1\end{align} In some cases with exponential functions, as the \begin{align}x\end{align} value increased, the \begin{align}y\end{align} value increased, too. This was a direct relationship known as exponential growth. As the \begin{align}x\end{align} value increases, the \begin{align}y\end{align} value grows at a very fast rate! The other graph you saw showed the opposite—as the \begin{align}x\end{align} value increased, the \begin{align}y\end{align} value decreased. This relationship is an inverse relationship known as decay. The graphs of these functions are opposites, reflected on the \begin{align}y\end{align}-axis. We can also analyze growth and decay functions in real – life situations. A famous story tells about a courtier who presented a Persian king with a beautiful handmade chessboard. The king asked him what he would like in return for his gift and the courtier surprised the king by asking him for one grain of rice on the first square of the chessboard, two grains of rice on the second, four grains on the third, etc. The king agreed and ordered for the rice to be brought. By the \begin{align}21^{st}\end{align} square, over a million grains of rice were required and, by the \begin{align}41^{st}\end{align} square, over a quadrillion grains of rice were needed. There was simply not enough rice in all the world for the final squares. This story reminds us of the drastic increases that we can see in exponential functions. Although this story is a fable, there are many instances in the real-world where exponential growth can be seen. Graph each function and tell whether it will represent exponential growth or decay. Example A \begin{align}y=\frac{1}{2}^x\end{align} Solution: Exponential decay Example B \begin{align}y=4^x\end{align} Solution: Exponential growth Example C \begin{align}y=5^x\end{align} Solution: Exponential growth Now let's go back to the dilemma from the beginning of the Concept. First, make a table of values. \begin{align}t\end{align} \begin{align}f\end{align} 0 185 1 203.5 2 223.85 3 246.24 4 270.86 5 297.94 6 327.74 7 360.51 8 396.56 9 436.22 10 479.84 Now we graph the function. Vocabulary Exponential Functions results that expand exponentially. The graph curves upward or downward. Exponential Growth Graph a direct relationship graph each variable increases. Decay Graph an indirect relationship graph, one variable increases as the other one decreases. Guided Practice Here is one for you to try on your own. Does the following function represent an exponential function? \begin{align}y=3 \times 1^x\end{align} Solution No, it does not represent an exponential function because the \begin{align}b\end{align} value is 1. Practice Directions:Graph each function. Then say whether it represents economic growth or decay. There will be two answers for each problem. 1. \begin{align}y = 4^x\end{align} 2. \begin{align}y =\frac{1}{2}^x\end{align} 3. \begin{align}y =\frac{1}{3}^x\end{align} 4. \begin{align}y =7^x\end{align} 5. \begin{align}y =5^x\end{align} 6. \begin{align}y =2^x\end{align} 7. \begin{align}y =\frac{1}{4}^x\end{align} 8. \begin{align}y =\frac{3}{4}^x\end{align} 9. \begin{align}y =6^x\end{align} 10. \begin{align}y =11^x\end{align} 11. \begin{align}y =9^x\end{align} 12. \begin{align}y =\frac{1}{8}^x\end{align} 13. \begin{align}y = 12^x\end{align} 14. \begin{align}y =\frac{2}{5}^x\end{align} 15. \begin{align}y =13^x\end{align} Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Vocabulary Language: English TermDefinition Exponential decay Exponential decay occurs when a quantity decreases by the same proportion in each given time period. Exponential Function An exponential function is a function whose variable is in the exponent. The general form is $y=a \cdot b^{x-h}+k$. Exponential growth Exponential growth occurs when a quantity increases by the same proportion in each given time period. Show Hide Details Description Difficulty Level: Authors: Tags: Subjects:<\|endoftext\|>	4.75
1,323	why were japanese americans placed in internment camps Jane Yanagi Diamond taught American History at a California high school, but I couldn t talk about the internment, she says. My voice would get all strange. Born in Hayward, California, in 1939, she spent most of World War II interned with her family at a camp in Utah. Seventy-five years after the fact, the federal government s incarceration of some 120,000 Americans of Japanese descent during that war is seen as a shameful aberration in the U. S. victory over militarism and totalitarian regimes. Though President Ford issued a formal apology to the internees in 1976, saying their incarceration was a setback to fundamental American principles, and Congress authorized the payment of reparations in 1988, the episode remains, for many, a living memory. Now, with immigration-reform proposals targeting entire groups as suspect, it resonates as a painful historical lesson. The roundups began quietly within 48 hours after the Japanese attacked Pearl Harbor, on December 7, 1941. The announced purpose was to protect the West Coast. Significantly, the incarceration program got underway despite a warning; in January 1942, a naval intelligence officer in Los Angeles reported that Japanese-Americans were being perceived as a threat almost entirely because of the physical characteristics of the people. Fewer than 3 percent of them might be inclined toward sabotage or spying, he wrote, and the Navy and the FBI already knew who most of those individuals were. Still, the government took the position summed up by John DeWitt, the Army general in command of the coast: A Jap s a Jap. They are a dangerous element, whether loyal or not. That February, President Franklin D. Roosevelt signed Executive Order 9066, empowering DeWitt to issue orders emptying parts of California, Oregon, Washington and Arizona of issei immigrants from Japan, who were precluded from U. S. citizenship by law and nisei, their children, who were U. S. citizens by birth. Photographers for the War Relocation Authority were on hand as they were forced to leave their houses, shops, farms, fishing boats. For months they stayed at assembly centers, living in racetrack barns or on fairgrounds. Then they were shipped to ten relocation centers, primitive camps built in the remote landscapes of the interior West and Arkansas. The regime was penal: armed guards, barbed wire, roll call. Years later, internees would recollect the cold, the heat, the wind, the dust and the isolation. There was no wholesale incarceration of U. S. residents who traced their ancestry to Germany or Italy, America s other enemies. The exclusion orders were rescinded in December 1944, after the tides of battle had turned in the Allies favor and just as the Supreme Court ruled that such orders were permissible in wartime (with three justices dissenting, bitterly). By then the Army was enlisting nisei soldiers to fight in Africa and Europe. After the war, President Harry Truman told the much-decorated, all-nisei 442nd Regimental Combat Team: You fought not only the enemy, but you fought prejudice and you have won. If only: Japanese-Americans met waves of hostility as they tried to resume their former lives. Many found that their properties had been seized for nonpayment of taxes or otherwise appropriated. As they started over, they covered their sense of loss and betrayal with the Japanese phrase Shikata ga nai It can t be helped. It was decades before nisei parents could talk to their postwar children about the camps. Paul Kitagaki Jr. , a photojournalist who is the son and grandson of internees, has been working through that reticence since 2005. At the National Archives in Washington, D. C. , he has pored over more than 900 pictures taken by War Relocation Authority photographers and others including one of his father s family at a relocation center in Oakland, California, by one of his professional heroes, Dorothea Lange. From fragmentary captions he has identified more than 50 of the subjects and persuaded them and their descendants to sit for his camera in settings related to their internment. His pictures here, published for the first time, read as portraits of resilience. Jane Yanagi Diamond, now 77 and retired in Carmel, California, is living proof. I think I m able to talk better about it now, she told Kitagaki. I learned this as a kid you just can t keep yourself in gloom and doom and feel sorry for yourself. You ve just got to get up and move along. I think that s what the war taught me. Subject interviews conducted by During World War II, the United States set up internment camps for Japanese-Americans. There were reasons for doing this, although we later regretted our actions and formally apologized to them. We also made restitution to those Japanese-Americans who were still surviving. One reason for setting up these camps was a fear that Japanese-Americans would aid the Japanese during World War II. Since we were fighting Japan, people worried that the loyalty of the Japanese-Americans would be with Japan and not with the United States. It turned out that this fear was unfounded because no Japanese-Americans were convicted of aiding Japan during World War II. Another reason for setting up these camps was that many Americans resented the economic success of the Japanese-Americans. There were fears that the Japanese-Americans were taking jobs and economic opportunities away from the American people. World War II gave people an opportunity to remove the Japanese-Americans from the economic picture. This would create more opportunities for Americans. Americans could take over the jobs the Japanese-Americans were doing and run the businesses the Japanese-Americans were operating. Many times this economic fear was used to cover up the anti-immigrant feelings many Americans had toward the Japanese, especially for those Americans who lived near the west coast. The United States government formally apologized to the Japanese-Americans in 1988. Each surviving Japanese-American was offered $20,000 as a form of restitution for our government's actions during World War II. - Views: 45 why does google not have an easter logo why do we have pearl harbor day why do the chinese eat fortune cookies why do we have pearl harbor day why were thousands of japanese americans interned in relocation camps why was the attack on pearl harbor important why was the articles of confederation replaced<\|endoftext\|>	3.65625
520	What’s a ramp and what can it do? - a block - small ball - two pieces of identical cardboard - two toilet paper tubes Key Science Concepts - A ramp is a surface with one end higher than the other. - An object placed on a ramp will roll, slide, or stay put. Encourage children to use words such as ramp, slanted, up, down, roll, and slide. - To introduce ramps, place a small block on a piece of cardboard or wood in the center of the circle. Ask, How can we get the block to move? Let children demonstrate their ideas and ask the others to describe what they did: Sam pushed the block. Aisha shoved the block with her foot. Now ask: Without touching the block, how can we get it to move? If a child doesn’t introduce the idea of slanting the piece of cardboard so the block slides off, demonstrate this idea yourself. Ask children to describe the motion of the block. When does the block start sliding? Does it slide right away? Slant the cardboard slowly to show how far you need to slant it in order to make the block slide. Repeat the activity using a ball instead of a block. Ask children to suggest and demonstrate some different ways to get the ball to move. When a child uses the cardboard as a ramp, ask the group to describe what happened. Ask, What was different about the way the block and the ball moved? (This is a good time to introduce or review the concepts slide and roll.) - Hold up two identical pieces of cardboard and say, Let’s make these pieces of cardboard into ramps. We’ll need something to hold up one end. What can we use? Let children suggest different things to support the ramps (books, blocks, chairs, etc.). Use the pieces of cardboard to make two ramps, side by side, each supported by a different object. Ask children to compare the two ramps—how are they different? How are they the same? Roll toilet paper tubes simultaneously down the two ramps, and talk about what happens. Does one toilet paper tube roll farther than the other? What do you think made it go farther? Try “racing” the tubes again. Does the same thing happen? Does one tube go faster than the other or is the speed about the same? Explain that over the next few weeks, you’ll be exploring ramps and things that slide and roll down ramps.<\|endoftext\|>	4.53125
184	Student Model Print First-grader Traci begins and ends her report with a main idea about parrots: they are pretty, colorful birds. Parrots are colorful. They have different colors like blue, red, black, orange, green, yellow, and white. Parrots live in cactuses, nests, and underground holes. Some parrots eat roots, and some eat nectar and seeds. They have different bills. Macaws (a kind of parrot) crack things open. Other parrots dig up roots and bulbs from the ground. There are different kinds of parrots, but they all lay eggs. Parrots are pretty birds. Parrots by Thoughtful Learning is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. Based on a work at k12.thoughtfullearning.com/studentmodels/parrots.<\|endoftext\|>	4.0625
1,420	The reported discovery of the De Loys’ Ape by Swiss Zoologist Francois De Loys in 1920 is just one of many unknown primate claims made over the years. South American forest natives often claim encounters with large monkeys. These elusive creatures have been known by many names for many centuries . Natives commonly referred to a “mono grande”, or king monkey, also known as “mono rey.” Pedro Cieza, a Spanish explorer claimed to have seen large apes referred to as “marimondas” in 1533; Dr. Edward Bancroft witnessed 5 ft apes referred to as “didi” in 1769; and Alexander von Humbolt recorded stories of “salvaje” or hairy ape men between 1799 and 1804. These anecdotal sightings along with one lone picture have kept the controversy surrounding De Loys’ Ape alive for nearly 100 years. Sighting of the De Loys’ Ape Swiss geologist Francois De Loys and his crew were among the numerous Europeans taken to Venezuela by oil companies to prospect the vast oil reserves. During the expedition, his group was ambushed by two threatening creatures that resembled apes. De Loys’ men shot one of the creatures while the other one escaped. Marveling at the unique appearance of the primate, they decided to take a photo of it. The ape corpse was sat on a crate with its head propped using a stick placed under its chin. After the photo was taken, the creature was skinned in order to prepare the skeleton for transportation. Unfortunately, the hardships of de Loys’ expedition resulted in the loss of all his equipment and specimen, including the creature’s remains. Only the photograph was saved, though it was enough to cause a sensation. Description of De Loys’ Ape The apelike South American primate was given the scientific name “Ameranthropoides Loysi” by George Montandon in 1929, while Arthur Keith proposed that it be called “Ateles Loysi”. Scientists claim that it is an unknown species of primate that walks bipedally in the jungles of South America. Its physical appearance resembles that of a monkey but lacks a tail like the apes. Based on the single photograph of the creature that was saved during the expedition, it was determined that the primate was around 5 feet tall. This was established from the fact that the creature was set on a Standard Oil Company packing crate that was known to have a standard size. In addition, the creature was described as having a thick coat comprising long grayish-brown fur and an oval face with a developed forehead. There was also an indication of a triangular pale pigment patch on the forehead, and round ridges around the eye sockets. It had a flat nose with flared nostrils and a strong jaw with thirty two teeth, which contradicts the typical thirty six in platyrrhine monkeys in South America. An examination of the creature revealed that it did not have a tail, which is an attribute of great apes as opposed to monkeys. Other distinct features include its broad shoulders, flat chest, sturdy arms, monkey like hands with long fingers, vestigial thumbs and long toes. There were strong arguments against de Loys’ discovery, with individuals such as Sir Arthur Keith suggesting that the entire saga had been fabricated. While George Montandon, an anthropologist friend to de Loys thought that his friend had discovered an unknown South American species, Keith claimed that the primate was just a large Black spider monkey whose tail had been cut off. Montandon promoted the existence of the De Loys’ Ape since he found it quite convenient in filling the missing pieces in his evolution theory called “Polygenism”. This evolution theory claimed that different human races had evolved autonomously from various species of ape. The existence of an American anthropoid allowed Montandon to justify his theory by claiming that it could have evolved into the Native American race. Keith, on the other hand, suspected De Loys of manipulating the creature for the camera. However, this explanation is doubtful since spider monkeys are significantly shorter with a height of about 3 feet, six inches when standing upright on the hind legs. The absence of a tail, less body hair, and a massive body also disagree with the features of these monkeys. However, it is highly possible that the creature was a white-bellied spider monkey, scientifically known as “ Ateles belzebuth “. This assumption is based on Montandon’s estimation of the size of the crate on which the creature was seated: measures 16 inches instead of the usual 18-20 inches. With just a photo to support the newly found species, De Loys’ Ape was never included in any analysis of primate evolution. There were other efforts to find the creature along Guyana’s Maruzuni River, where De Loys was attacked by the ferocious creatures. However, Italian explorers in 1931 never found it though they claimed that there were eyewitnesses supporting its existence. American scientist, Phillip Herschkowitz, also retraced the route taken by De Loys and his crew and came to the conclusion that the mono grande was a hoax and a mythical exaggeration of the spider monkeys. In 1951, a Frenchman called Roger Courteville revealed another photo of a mysterious ape from Rio Tarra in South America, but he was accused by skeptics of modifying the original photo to uphold the hoax. Keith’s deductions that the discovery was a misidentification were also supported by a variety of other anthropologists and cryptozoologists such as Ivan Sanderson in the 1960s, after re-examination of the photo. The primary question regarding this mysterious creature is whether De Loys’ Ape is indeed, a new species. Based on the lack of supporting evidence besides the single photo, the other two possibilities are that it was a deliberate hoax, or that it was an honest misidentification supported by trumped-up reporting. The idea of a hoax could be true especially when argued from Montandon’s point of view, since he wanted to support his evolution theory to the scientific world. The integrity of De Loys can also be put to question, based on a letter written by Dr. Enrique Tejera in reaction to a story published about the ape. In his letter, Dr. Tejera claimed to have been on the expeditions with de Loys, and that De Loys was a practical joker, and he had been given a white-bellied spider monkey, whose tail had been cut off. The idea to photograph it came to him after it died in 1919. The truth of the matter is not known, but to this day, scientific interest in mono grande persists, though with a lot of skepticism.<\|endoftext\|>	3.828125
808	BAMBERG, city in Bavaria, Germany. There were Jews living in Bamberg before the First Crusade (1096), when they were forcibly baptized but later allowed to return to Judaism. Establishments in the medieval "Jewish Lane" (today Pfahlplaetzchen) included a dance hall for weddings, a hostel (hekdesh) for the needy sick and transients, a mikveh, and a synagogue. In 1298 during Rindfleisch massacres 135 Jews were martyred in Bamberg. During the persecution following the outbreak of the Black Death in 1348 the Jews there set fire to their homes and perished in flames. Between the 14th and 17th centuries Jews repeatedly attempted to settle in Bamberg, paying high "protection" taxes, only to be later attacked and expelled. In 1633 they numbered ten families, whose right of residence was recognized in 1644. An annual "plum fast" (Zwetschgen Taanit) was observed by the Bamberg community, to commemorate the preservation of the Jews there during the riots of 1699 by one of their number who averted greater damage by pouring plums over the mob. The community increased from 287 in 1810 to 1,270 in 1880 (4.3% of the total population), subsequently declining to 812 in 1933 (1.6%) and 418 in May 1939. Prominent members of the community included the talmudist and paytan Samuel b. Baruch Bamberg (13th century). Notable rabbis were Moses Mintz who served there from c. 1469 to 1474; Samuel Meseritz (c. 1661–65), author of Naḥalat Shivah; and Joseph Kobak (1862–82), editor of Jeschurun. A. Eckstein, rabbi of Bamberg (1888–1935), wrote a number of studies on the history of the Jewish communities in Bavaria. During the Nazi regime, the synagogue was burned down on Nov. 10, 1938, and 30 to 40 Torah scrolls were destroyed. In 1933–41, 443 Bamberg Jews left Germany and another 66 fled to other German cities. The 300 who remained at the end of 1941 were deported to Riga, Izbica/Lublin, Theresienstadt, and Auschwitz. After the war many displaced persons assembled in Bamberg (14,000 in 1947), but only 17 of the former Jewish residents were among them. In 1965 the cemetery was desecrated. The community then numbered 70. In 1989, there were 106 community members; their number rose to 893 in 2003 as a result of the immigration of Jews from the former Soviet Union. PK; Germ Jud, S.V.; H.F. Brettinger, Juden in Bamberg (1963); A. Eckstein, Geschichte der Juden im ehemaligen Fuerstbistum Bamberg (1898); idem, in: Festschrift zur Einweihung der neuen Synagoge in Bamberg (1910); idem, Die israelitische Kultusgemeinde Bamberg, 1803–53 (1910); Bilder aus der Vergangenheit der israelitischen Gemeinde Bamberg (1933); R.M. Kloos, in: Bericht des historischen Vereins Bamberg, 103 (1967), 341–86. ADD. BIBLIOGRAPHY: N. Haas, Juden in Bamberg 1868–1906; H. Loebl, Juden in Bamberg. Die Jahrzehnte vor dem Holocaust (1999). [Ze'ev Wilhem Falk] Source: Encyclopaedia Judaica. © 2008 The Gale Group. All Rights Reserved.<\|endoftext\|>	3.765625
190	# In a certain area, there are 35 houses to 55 businesses. How do you write the ratio of houses to businesses as a fraction in simplest form. Then explain its meaning? Nov 4, 2016 There are $\frac{7}{11}$ times as many houses as businesses. #### Explanation: $\text{houses":"businesses} = 35 : 55$ This ratio can also be written in fractional form as: $\textcolor{w h i t e}{\text{XXX}} \frac{35}{55}$ which can be simplified (by dividing each of the numerator and denominator by $5$) $\textcolor{w h i t e}{\text{XXX}} = \frac{7}{11}$ If you divided the businesses up into (5 groups of) $11$ and then evenly assigned houses among those groups, each group would have $7$ houses.<\|endoftext\|>	4.53125
415	Tangent to the Graph of a Circle # Tangent to the Graph of a Circle GCSE(H), If a circle has a radius with centre at the origin, then a point on the circle can be described as being at a point (x, y). The gradient of the line to that point is frac(y)(x). A line perpendicular to that radius is given by frac(-1)(text(gradient)) (see perpendicular). If the original gradient is frac(y)(x), then the gradient of the perpendicular line is frac(-x)(y). ## Examples 1. A circular function, centred on the origin, has a radius of 5. What is the gradient of the tangent to the circle when x = 4? Answer: -frac(4)(3) The y-coordinate can be found by solving r^2 = x^2 + y^2; substituting 5^2 = 4^2 + y^2 25 = 16 + y^2, giving y = 3 The gradient of the radius at (4, 3) is frac(3)(4). Use the inverse of gradient, and multiply by -1: the tangent is -frac(4)(3). 2. The gradient of a tangent to a circle is -frac(4)(7). The circle is centred on the origin. If the radius of the circle is sqrt(585), what are the coordinates of the tangent? Gradient of the radius to the tangent is the negative inverse of -frac(7)(4) = frac(4)(7). The radius of a circle is given by r^2 = x^2 + y^2: in this instance the radius is a multiple: (mx)^2 + (my)^2 = (mr)^2 (7m)^2 + (4m)^2 = (sqrt(585))^2 49m^2 + 16m^2 = 585 65m^2 = 585, therefore m = 3<\|endoftext\|>	4.46875
1,288	# Question Video: Differentiating Trigonometric Functions Involving Trigonometric Ratios Using the Product Rule Mathematics • Higher Education Consider the series π(π₯) = 1/(1 + π₯) = β_(π = 0)^(β) (β1)^(π) π₯^(π). Differentiate the given series expansion of π term by term to find the corresponding series expansion for the derivative of π. Use the result of the first part to evaluate the sum of the series β_(π = 0)^(β) ((β1)^(π + 1) (π + 1))/3^(π). 03:17 ### Video Transcript Consider the series π of π₯ equals one over one plus π₯, which equals the sum of negative one to the πth power times π₯ to the πth power, for values of π between zero and β. Differentiate the given series expansion of π term by term to find the corresponding series expansion for the derivative of π. Then use the result of the first part to evaluate the sum of the series the sum of negative one to the π plus one times π plus one over three to the πth power for values of π between zero and β. Weβve been given the series expansion of π. So letβs work out the first few terms. The first term is when π is equal to zero. Itβs negative one to the power of zero times π₯ to the power of zero, which is equal to one. The second term is when π is equal to one. So itβs negative one to the power of one times π₯ to the power of one, which is negative π₯. We then add negative one squared times π₯ squared, which is π₯ squared. And we continue in this manner. And we see that π of π₯ is equal to one minus π₯ plus π₯ squared minus π₯ cubed plus π₯ to the fourth power minus π₯ to the fifth power, and so on. The question tells us to differentiate the given series expansion of π that will give us π prime. And of course we can simply do this term by term. The derivative of one is zero, and the derivative of negative π₯ is negative one. The derivative of π₯ squared is two π₯. And then our next term is negative three π₯ squared. We continue in this way, multiplying each term by its exponent and then reducing that exponent by one. Without writing the zero, we see that we can write π prime of π₯ as negative π₯ to the power of zero β remember, thatβs just negative one β plus two π₯ to the power of one minus three π₯ squared plus four π₯ cubed, and so on. Letβs think about how we can write this as a sum. We know that we start with a negative term, and then the sign alternates. To achieve this, we need negative one to the power of π plus one. This will work for values of π between zero and β. We then multiply each term by π plus one. So the first term is when π is equal to zero. And weβre multiplying that by one. The second term is when π is equal to one. And weβre multiplying that by two, and so on. And this is all times π₯ to the πth power. And so we found the series expansion for the derivative of π. Itβs the sum of negative one to the power of π plus one times π plus one times π₯ to the πth power for values of π between zero and β. Part two says to use the result of the first part to evaluate the sum of the series the sum of negative one to the power of π plus one times π plus one over three to the πth power for values of π between zero and β. Letβs compare this sum to the one we generated in the first part of this question. If we rewrite this as the sum of negative one to the power of π plus one times π plus one times a third to the πth power, then we see itβs of the same form. Just π₯ equals a third. So what weβre going to do is weβre going to differentiate our original function one over one plus π₯ and evaluate that when π₯ is equal to one-third. Letβs write π of π₯ as one plus π₯ to the power of negative one. And then we use the chain rule. When we do, we find its derivative to be equal to negative one over one plus π₯ squared. The sum of our series will be the value of the derivative when π₯ is equal to one-third. So thatβs negative one over one plus one-third squared. Thatβs negative one over 16 over nine, which is simply negative nine sixteenths.<\|endoftext\|>	4.75
186	It is well known that today, anyone with an Internet connection has instant access to other languages. That is why teachers should take advantage of using the Internet and other modern technology tools while teaching foreign languages. The use of multimedia helps exposing students to these authentic speech exchanges, increases time on task and the effectiveness of the study time. Incorporating computers and technology into class discussions can make difficult courses much easier for students to grasp. Using technology as both a learning tool and a subject matter discipline provides teachers with a number of new ways to inspire children to learn. The present paper deals with different ways of using education technology to promote enhanced classroom learning, such as: using connected learning, sharing content online, having fun with Twitter, using video chats, creating a class blog, creating interactive maps. We will develop each of these in turn, focusing on advantages and disadvantages. Key words: computers, foreign languages, learning, interaction, technology<\|endoftext\|>	3.890625
3,068	\|Humidity and hygrometry\| \|Measures and Instruments\| The dew point is the temperature to which air must be cooled to become saturated with water vapor. When further cooled, the airborne water vapor will condense to form liquid water (dew). When air cools to its dew point through contact with a surface that is colder than the air, water will condense on the surface. When the temperature is below the freezing point of water, the dew point is called the frost point, as frost is formed rather than dew. The measurement of the dew point is related to humidity. A higher dew point means there will be more moisture in the air. If all the other factors influencing humidity remain constant, at ground level the relative humidity rises as the temperature falls. This is because less vapor is needed to saturate the air, so vapor condenses as the temperature falls. In normal conditions, the dew point temperature will not be greater than the air temperature because relative humidity cannot exceed 100%. In technical terms, the dew point is the temperature at which the water vapor in a sample of air at constant barometric pressure condenses into liquid water at the same rate at which it evaporates. At temperatures below the dew point, the rate of condensation will be greater than that of evaporation, forming more liquid water. The condensed water is called dew when it forms on a solid surface, or frost if it freezes. The condensed water is called either fog or a cloud, depending on its altitude when it forms in the air. If the temperature is below the dew point, the vapor is called supersaturated. This can happen if there are not enough particles in the air to act as condensation nuclei. A high relative humidity implies that the dew point is closer to the current air temperature. A relative humidity of 100% indicates the dew point is equal to the current temperature and that the air is maximally saturated with water. When the moisture content remains constant and temperature increases, relative humidity decreases, but the dew point remains constant. Increasing the barometric pressure increases the dew point. This means that, if the pressure increases, the mass of water vapour in the air must be reduced in order to maintain the same dew point. For example, consider New York (33 ft or 10 m elevation) and Denver (5,280 ft or 1,610 m elevation). Because Denver is at a higher elevation than New York, it will tend to have a lower barometric pressure. This means that if the dew point and temperature in both cities are the same, the amount of water vapor in the air will be greater in Denver. Relationship to human comfort This section needs additional citations for verification. (October 2016) (Learn how and when to remove this template message) When the air temperature is high, the human body uses the evaporation of sweat to cool down, with the cooling effect directly related to how fast the perspiration evaporates. The rate at which perspiration can evaporate depends on how much moisture is in the air and how much moisture the air can hold. If the air is already saturated with moisture, perspiration will not evaporate. The body's thermoregulation will produce perspiration in an effort to keep the body at its normal temperature even when the rate it is producing sweat exceeds the evaporation rate, so one can become coated with sweat on humid days even without generating additional body heat (such as by exercising). As the air surrounding one's body is warmed by body heat, it will rise and be replaced with other air. If air is moved away from one's body with a natural breeze or a fan, sweat will evaporate faster, making perspiration more effective at cooling the body. The more unevaporated perspiration, the greater the discomfort. Discomfort also exists when the dew point is low (below around −30 °C or −22 °F). The drier air can cause skin to crack and become irritated more easily. It will also dry out the airways. The US Occupational Safety and Health Administration recommends indoor air be maintained at 20–24.5 °C (68–76 °F) with a 20–60% relative humidity, equivalent to a dew point of −4.5 to 15.5 °C (24 to 60 °F). Lower dew points, less than 10 °C (50 °F), correlate with lower ambient temperatures and the body requires less cooling. A lower dew point can go along with a high temperature only at extremely low relative humidity, allowing for relatively effective cooling. People inhabiting tropical and subtropical climates acclimatize somewhat to higher dew points. Thus, a resident of Singapore or Miami, for example, might have a higher threshold for discomfort than a resident of a temperate climate like London or Chicago. People accustomed to temperate climates often begin to feel uncomfortable when the dew point gets above 15 °C (59 °F), while others might find dew points up to 18 °C (64 °F) comfortable. Most inhabitants of temperate areas will consider dew points above 21 °C (70 °F) oppressive and tropical-like, while inhabitants of hot and humid areas may not find this uncomfortable. Thermal comfort depends not just on physical environmental factors, but also on psychological factors. Dew point Relative humidity at 32 °C (90 °F) Over 26 °C Over 80 °F 73% and higher 24–26 °C 75–80 °F 62–72% 21–24 °C 70–74 °F 52–61% 18–21 °C 65–69 °F 44–51% 16–18 °C 60–64 °F 37–43% 13–16 °C 55–59 °F 31–36% 10–12 °C 50–54 °F 26–30% Under 10 °C Under 50 °F 25% and lower Devices called hygrometers are used to measure dew point over a wide range of temperatures. These devices consist of a polished metal mirror which is cooled as air is passed over it. The temperature at which dew forms is, by definition, the dew point. Manual devices of this sort can be used to calibrate other types of humidity sensors, and automatic sensors may be used in a control loop with a humidifier or dehumidifier to control the dew point of the air in a building or in a smaller space for a manufacturing process. A dew point of 33 °C (91 °F) was observed at 14:00 EDT on July 12, 1987, in Melbourne, Florida. A dew point of 32 °C (90 °F) has been observed in the United States on at least two other occasions: Appleton, Wisconsin, at 17:00 CDT on July 13, 1995, and New Orleans Naval Air Station at 17:00 CDT on July 30, 1987. A dew point of 35 °C (95 °F) was observed at Dhahran, Saudi Arabia, at 15:00 AST on July 8, 2003, which caused the heat index to reach 81 °C (178 °F), the highest value recorded. Calculating the dew point A well-known approximation used to calculate the dew point, Tdp, given just the actual ("dry bulb") air temperature, T (in degrees Celsius) and relative humidity (in percent), RH, is the Magnus formula: The more complete formulation and origin of this approximation involves the interrelated saturated water vapor pressure (in units of millibars, also called hectopascals) at T, Ps(T), and the actual vapor pressure (also in units of millibars), Pa(T), which can be either found with RH or approximated with the barometric pressure (in millibars), BPmb, and "wet-bulb" temperature, Tw is (unless declared otherwise, all temperatures are expressed in degrees Celsius): For greater accuracy, Ps(T) (and therefore γ(T, RH)) can be enhanced, using part of the Bögel modification, also known as the Arden Buck equation, which adds a fourth constant d: - a = 6.1121 mb, b = 18.678, c = 257.14 °C, d = 234.5 °C. - a = 6.112 mb, b = 17.67, c = 243.5 °C. These valuations provide a maximum error of 0.1%, for −30 °C ≤ T ≤ 35°C and 1% < RH < 100%. Also noteworthy is the Sonntag1990, - a = 6.112 mb, b = 17.62, c = 243.12 °C; for −45 °C ≤ T ≤ 60 °C (error ±0.35 °C). Another common set of values originates from the 1974 Psychrometry and Psychrometric Charts, as presented by Paroscientific, - a = 6.105 mb, b = 17.27, c = 237.7 °C; for 0 °C ≤ T ≤ 60 °C (error ±0.4 °C). Also, in the Journal of Applied Meteorology and Climatology, Arden Buck presents several different valuation sets, with different minimum accuracies for different temperature ranges. Two particular sets provide a range of −40 °C to +50 °C between the two, with even greater minimum accuracy than all of the other, above sets (maximum error at extremes of temperature range): - a = 6.1121 mb, b = 17.368, c = 238.88 °C; for 0 °C ≤ T ≤ 50 °C (error ≤ 0.05%). - a = 6.1121 mb, b = 17.966, c = 247.15 °C; for −40 °C ≤ T ≤ 0 °C (error ≤ 0.06%). There is also a very simple approximation that allows conversion between the dew point, temperature, and relative humidity. This approach is accurate to within about ±1 °C as long as the relative humidity is above 50%: This can be expressed as a simple rule of thumb: For every 1 °C difference in the dew point and dry bulb temperatures, the relative humidity decreases by 5%, starting with RH = 100% when the dew point equals the dry bulb temperature. The derivation of this approach, a discussion of its accuracy, comparisons to other approximations, and more information on the history and applications of the dew point are given in the Bulletin of the American Meteorological Society. For temperatures in degrees Fahrenheit, these approximations work out to For example, a relative humidity of 100% means dew point is the same as air temp. For 90% RH, dew point is 3 °F lower than air temperature. For every 10 percent lower, dew point drops 3 °F. The frost point is similar to the dew point in that it is the temperature to which a given parcel of humid air must be cooled, at constant atmospheric pressure, for water vapor to be deposited on a surface as ice crystals without undergoing the liquid phase (compare with sublimation). The frost point for a given parcel of air is always higher than the dew point, as the stronger bonding between water molecules on the surface of ice requires higher temperature to break. - Glossary – NOAA's National Weather Service - John M. Wallace; Peter V. Hobbs (24 March 2006). Atmospheric Science: An Introductory Survey. Academic Press. pp. 83–. ISBN 978-0-08-049953-6. - Glossary – NOAA's National Weather Service - "Observed Dew Point Temperature". Department of Atmospheric Sciences (DAS) at the University of Illinois at Urbana-Champaign. Retrieved 15 February 2018. - Dew Point \| Definition of dew point by Merriam-Webster - Skilling, Tom (20 July 2011). "Ask Tom why: Is it possible for relative humidity to exceed 100 percent?". Chicago Tribune. Retrieved 24 January 2018. - Horstmeyer, Steve (2006-08-15). "Relative Humidity....Relative to What? The Dew Point Temperature...a better approach". Steve Horstmeyer, Meteorologist, WKRC TV, Cincinnati, OH. Retrieved 2009-08-20. - "Dew Point in Compressed Air – Frequently Asked Questions" (PDF). Vaisala. Retrieved 15 February 2018. - "Denver Facts Guide – Today". The City and County of Denver. Archived from the original on February 3, 2007. Retrieved March 19, 2007. - Lin, Tzu-Ping (10 February 2009). "Thermal perception, adaptation and attendance in a public square in hot and humid regions" (PDF). Building and Environment. 44 (10): 2017–2026. Retrieved 23 January 2018. - Relative Humidity and Dewpoint Temperature from Temperature and Wet-Bulb Temperature - Bolton, David (July 1980). "The Computation of Equivalent Potential Temperature" (PDF). Monthly Weather Review. 108 (7): 1046–1053. Bibcode:1980MWRv..108.1046B. doi:10.1175/1520-0493(1980)108<1046:TCOEPT>2.0.CO;2. - SHTxx Application Note Dew-point Calculation - "MET4 and MET4A Calculation of Dew Point". Archived from the original on May 26, 2012. Retrieved 7 October 2014. - Buck, Arden L. (December 1981). "New Equations for Computing Vapor Pressure and Enhancement Factor" (PDF). Journal of Applied Meteorology. 20 (12): 1527–1532. Bibcode:1981JApMe..20.1527B. doi:10.1175/1520-0450(1981)020<1527:NEFCVP>2.0.CO;2. - Lawrence, Mark G. (February 2005). "The Relationship between Relative Humidity and the Dewpoint Temperature in Moist Air: A Simple Conversion and Applications". Bulletin of the American Meteorological Society. 86 (2): 225–233. Bibcode:2005BAMS...86..225L. doi:10.1175/BAMS-86-2-225. - Haby, Jeff. "Frost point and dew point". Retrieved September 30, 2011. - Often Needed Answers about Temp, Humidity & Dew Point from the sci.geo.meteorology<\|endoftext\|>	4.4375
142	MS Excel adheres to the standard order of calculation and will calculate percentages, exponenets, multiplications, and division in order. For an example, follow these steps: - First, open a spreadsheet or create a new file. - Enter =7+53 and the result will be 22, not 36. The multiplication is done first, then the addition. - Now, use parenthesis to change the order of the calculation. - Enter =(7+5)3 = 36. Parenthesis make the difference. Tip comes from using Office. Just click on the button for the Tiproom's Home Page. Date of last revision: 20 January 2006.<\|endoftext\|>	3.96875
352	Math Lesson: Direct Variation By Guido Feliz, Jr When two variable quantities have a constant ratio, their relationship is called a direct variation. It is said that one variable "varies directly" as the other variable. The constant ratio is called the constant of variation. The formula normally used for direct variation is y = kx, where k is the constant of variation. The equation above is read: "y varies directly as x" To solve for k, divide both sides of the equation by x. Doing so, we get k = (y/x), where y = numerator; x = denominator. In a direct variation problem, the two variables change at the same time. In other words, if one increases, so does the other. The weekly salary a man earns, S, varies directly as the number of hours, h, which he works. Express this relation as a formula. The formula for direct variation is y = kx, where k is the constant of variation. The equation is read: "y varies directly as x." Since S varies directly as h in the question, let y = S and x = h in the equation y = kx. In other words, replace y with S and x with h. Leave k where it is. Doing so, we can then write the relation between S and h as a formula S = kh. Final answer: S = kh NOTE: To find k in the equation S = kh, simply isolate k (place k alone on one side of the equation) by dividing both sides by h. It then looks like this: S/h = k.<\|endoftext\|>	4.1875
931	This piece can be used off and on throughout the elementary school years. It has a “perfect rhythm.” Kindergarten and 1st Grade, beginning of the year: 4 Voices: Whisper Shout Talk Sing – Create a story about a king and queen. 1) All the people in the castle were whispering about the queen’s hair (whisper the rhyme). 2) The prince rode into the town and shouted to the town (shout the rhyme). 3) The townspeople couldn’t believe it! They all started whispering (whisper the rhyme). 4) They had a parade. The king and queen stopped and shouted about it to the town (shout the rhyme). After students know it, then say, “Can you perform it with your talking voice?” After that, teach melody with two notes, so and mi. Then, “Can you perform it with your singing voice?” Then, using a chart with pictures and names of the four voices, “Can you change your voice when I change the picture that I’m pointing to?!” Start by alternating only two voices (whisper/shout … talk/sing), then a few weeks later challenge class to do all four. 1st Grade, mid-year (spring): Introduce “Rhythm, the way the words go!” “Pull out your keys, open your memory banks.” “I have a story for you, but I’m not going to tell you with words … the drum is going to tell you with the RHYTHM!” Play rhythm Queen Caroline on hand drum. Students identify the rhyme. “You know how to play the heartbeat. Today you will learn about RHYTHM and play it on instruments!” (WHEN INTRODUCING THE WORD “RHYTHM,” IT IS HELPFUL TO SPEAK SLOWLY, CLAP EVERY SYLLABLE YOU SPEAK, THEN STUDENTS ECHO.) SAY: “To-day . . . we play `RHY-THM’ . . . ev’-ry part, ev’-ry word . . . the way, the words, go . . . that’s the `RHY-THM!’ . . . NICE JOB!” Echo clap rhythm while speaking words of Queen Caroline 4 beats at a time. “There’s another way we can say this rhyme, using the rhythm words.” (Use your favorite rhythm syllable system for this.) Say each 4 beat section of the rhyme and immediately speak the rhythm syllables – students echo speak the rhythm syllables. Write definitions on board: BEAT – THE WAY YOUR HEART GOES. RHYTHM – THE WAY THE WORDS GO. Perform rhythm on hand drums, half the class at a time. The half that is not performing is the audience. “Look for someone who is relaxing while they play the drum!” Next Class Period: Clap rhythm of the rhyme, C identify. Review rhythm words. FOUR CHAIRS: each chair is a beat; if one student sits in a chair, it is a “tah,” if two students sit in a chair, it is a “ta-te” (or ti-ti, depending on your rhythm syllable system). If no one sits in a chair, it is a rest – use the rhyme Mama Caught A Flea for this. Notate each four beats first in chairs. then immediately put the rhythm on the board with stem notation. Use only stem notation for the students’ first reading experiences. Can add note heads shortly thereafter. 2nd or 3rd Grade: Create melody on Orff instruments using Queen Caroline as the basic rhythm. First day of recorder: create melody on recorder using C2 and A using Queen Caroline as the basic rhythm. (When beginning recorder, I prefer to start with C2 and A, then add D2, then go to B A G.) Click here to download this lesson from Jim Solomon. To learn more about Jim Solomon’s workshops or to purchase one of his publications, please visit CongaTown.com<\|endoftext\|>	3.953125
641	Plate Tectonics Lesson Plans - 2. Explore: (15 mins) 1. Anyone goes to be paired with a associate and given an orange. Every orange goes to symbolize the earth. Now, look ahead to the relaxation of my education earlier than you begin so that you understand what to do. [Pass out oranges, one per group] ask students while passing out oranges, “how is the orange much like the earth??? [wait for students’ response] 2. ??at the same time as seeking to maintain the peal of the orange as entire as possible, the companion who is taller can also start to peel the orange. Because the piece of the orange peel wreck and come off attempt to location them inside the particular order they got here off in.?? [teacher walks around to assist if the students need help] 3. ??as soon as the orange is peeled, the partner that did now not peel the orange can also begin setting the orange peel back on the orange by way of using the toothpicks.?? [demonstrate what this may look like for the students while they do it] 4. ??what makes the orange and the earth similar??? [they are both round] “correct, this is what we name a sphere! Due to the fact it's miles a three-demontional object meaning that the earth is not flat” five. ??did you know that the earth’s crust is the tough-rocky surface that we walk on and spot??? [i knew that, or no] 6. ??how does the peel of an orange evaluate to the crust of the earth??? [they are both the outer most layer] [*note: the earth is broken into 9 plates] 3. Provide an explanation for: (13 mins) 7. ??just like we had been able to positioned the peel of the orange back on, or the crust of our earth. We were showing something referred to as plate tectonics. The earth’s crust is without a doubt broken down into extraordinary plates. [Show map of the earth’s plates]. Reflect onconsideration on the earth’s crust, is it changing over time? [Yes]. [Show students map of plate tectonics] eight. Additionally display college students a map of the continents and give an explanation for the differences among these things. 9. ??sure, so the distinct parts of earth referred to as the plate tectonics are the purpose that mountains shape and earthquakes appear!?? this happens while the plates are pushed collectively! Or pushed faraway from every different. 10. ??from mastering approximately the plates, do you observed that the plate remains nonetheless or pass? [Move] why now not? [Because the earth’s crust is constantly changing or moving]. 11. Plate tectonics is the motive that we've got distinctive continents. There are nine predominant plates in the international! 12. Watch video of bill nye explaining plate tectonics to review what we learned: ?V=1pvms2nsdmc (2 minutes in duration) 13. Have a small dialogue on.<\|endoftext\|>	4.0625
550	# What is a Median? Definition and Examples What is a median? A median is the number in the middle for a set of data when the data are arranged in order form least to greatest or from greatest to least. For example, the median of the data set below is 56 because 56 is the value right in the middle. 45 48 56 60 63 What is the median though if you remove 63 from the data set? 45 48 56 60 The value in the middle is not there this time because the number of values is even. Just find the value between 48 and 56 mathematically by taking the average of 48 and 56. (48 + 56) / 2 = 104 / 2 = 52 The median is 52 How to find the median: If the number of values in the set is odd, just locate the number in the middle. If the number of values in the set is even, just take the average of the two numbers in the middle of the data set. ## Real life examples showing what the median is and how to find it Example #1 The price for 1 scoop of ice cream on a cone at 10 different ice cream shops is shown below. What is the median price? 2.50 3.75 2.25 4 1.75 3 1 3.25 2.75 1.50 Order the data set from greatest to least 4 3.75 3.25 3 2.75 2.50 2.25 1.75 1.50 1 Since the number of values in the set is even, the median is between 2.75 and 2.50. (2.75 + 2.50) / 2 = 5.25 / 2 = 2.625 The median is 2.625 Example #2 The yearly salary for 5 people are shown below in USD . What is the median salary? 70,000 55,000 30,000 73,000 60,000 Order the set from greatest to least 73,000 70,000 60,000 55,000 30,000 Since the number of values in the set is odd, just locate the number in the middle. The number in the middle is 60,000, so the median salary is 60,000<\|endoftext\|>	4.40625
5,302	OBJECTIVE: The medical home concept encompasses the elements of pediatric care considered essential for all children. We describe here the characteristics of children with medical homes and the relationship between presence of a medical home and selected health care outcomes by using new data from the 2007 National Survey of Children's Health (NSCH). METHODS: We used a medical home measure comprising 5 components: having a usual source of care; having a personal physician or nurse; receiving all needed referrals for specialty care; receiving help as needed in coordinating health and health-related care; and receiving family-centered care. A total of 83 448 children aged 1 to 17 years had valid data for all applicable medical home components. The NSCH is a random-digit-dial population-based telephone survey. RESULTS: In 2007, 56.9% of US children aged 1 to 17 years received care in medical homes. Younger children were more likely to have a medical home than their older counterparts. Substantial racial/ethnic, socioeconomic, and health-related disparities were present. Children who received care in medical homes were less likely to have unmet medical and dental needs and were more likely to have annual preventive medical visits. CONCLUSIONS: Approximately half of the children in the United States have access to all components of a pediatric medical home. Because the medical home is increasingly promoted as the standard for provision of high-quality comprehensive health care, these findings reinforce the need to continue and expand federal, state, and community efforts to ensure that all children have access to this model of care. - community pediatrics - health care delivery/access - quality of care - health outcomes - health policy - medical home WHAT'S KNOWN ON THIS SUBJECT: The medical home is recognized as a mechanism for ensuring quality health care for children with special health care needs and adults with chronic conditions. Few studies address the extent to which all children have a medical home. WHAT THIS STUDY ADDS: This article provides a comprehensive assessment of the proportion of children who have a medical home, the health and social correlates of having a medical home, and its impact on receipt of preventive care and unmet need. The medical home concept encompasses the characteristics of pediatric care considered essential for all children.1,2 The American Academy of Pediatrics (AAP) developed and has championed the medical home concept for decades,3 and currently defines the medical home as a model of primary care that is accessible, continuous, comprehensive, family centered, coordinated, compassionate, and culturally effective.1 The medical home has received widespread national attention as a mechanism for ensuring quality in health care for children with special health care needs (CSHCN)4,–,10 and more recently for adults with chronic conditions.11,–,13 Families, child health professionals, policy makers, and insurers endorse this model as a standard of care1,2,14,–,17 and it now serves as a centerpiece for national quality assurance measures.18 Although existing research on the pediatric population supports a positive relationship between some components of the medical home and desired child and family health-related outcomes, few studies have incorporated a medical home definition reflecting the comprehensive elements articulated by the AAP19 or studied the extent to which the medical home is available to the pediatric population as a whole.20 Furthermore, to our knowledge, none have studied the association between having a medical home and receipt of other important components of health such as dental care. The purpose of this article was to provide an up-to-date, population-based assessment of medical home access for all children using a comprehensive definition and to describe the relationship between presence of a medical home and receipt of preventive medical and dental care, and unmet medical and dental needs. Dental care is included because existing policy guidelines and experts promote the integration of oral health services in the medical home.21,–,27 The 2007 National Survey of Children's Health (NSCH) is a random-digit-dial population-based telephone survey designed and directed by the Health Resources and Services Administration's Maternal and Child Health Bureau and conducted by the Centers for Disease Control and Prevention's National Center for Health Statistics, using the State and Local Area Integrated Telephone Survey mechanism.28 Interviews were completed in 66.0% of identified households with children. A total of 91 642 interviews were conducted in households with children ages birth through 17 years between April 2007 and July 2008. The survey was administered for 1 randomly selected child in each household with an age-eligible child. The parent or guardian who knew the most about the health and health care of the selected child served as the respondent for the interview. Because many of the survey items used in this analysis encompassed a 1-year recall period, we restricted our analysis to children aged 1 to 17 years rather than 0 to 17 years. Medical Home Measurement The medical home measure used here was designed to approximate the components of the AAP-defined medical home concept and is the most robust, comprehensive measure used in a national survey.19,29 With the exception of the element of “continuity,” all elements of the AAP medical home measure are addressed through the NSCH medical home measure. The cross-sectional nature of the NSCH creates methodologic barriers to measuring continuity of care over time. The NSCH medical home measure is a composite of 5 components: having a usual source of care, having a personal physician or nurse, receiving all needed referrals for specialty care, receiving needed help coordinating health and health-related care, and receiving family-centered care. Each component was operationalized using ≥1 survey items. For example, the family-centered care component was measured by using 6 items: (1) whether the family reports that the child's physicians spend enough time with the child; (2) whether physicians listen carefully to family concerns; (3) whether physicians are sensitive to family values and customs; (4) whether physicians provide needed information; (5) whether physicians make the family feel like a partner in the child's care; and (6) whether interpretation services are available, if needed. If the respondent answered “usually” or “always” to each item, the child was considered to have received family-centered care. A similar process was used to operationalize the other 4 components of the medical home. The components and subcomponents of the medical home measure are included in Table 1. It should be noted that the components do not apply universally to all children in the sample. Specifically, the component on receiving referrals applies only to children who were reported to need referrals (n = 14 349). Similarly, the component on receiving effective care coordination applies only to children reported to need care coordination (n = 36 889) and the component on receipt of family-centered care applies only to children with at least 1 physician visit in the past year (n = 82 354). To qualify as successfully attaining the medical home, all applicable components must be met. A success rate was calculated by dividing the number of children whose providers delivered all applicable components by the total number of children with valid data. When questions on the referrals, care coordination, and family-centered care components were legitimately skipped, then the child was included in the analysis and classified as having a medical home on the basis of responses to the remaining components. Analyses excluded children with missing data. A total of 83 448 children had valid data for all applicable components of the medical home. The impact of having a medical home was measured by using 4 variables meant to capture the spectrum of health care experiences associated with difficulty accessing medical and dental care, and timely receipt of routine preventive medical and dental care. The variables used to measure these concepts were having an unmet medical need, not receiving a preventive medical care visit, having an unmet dental need, and not receiving a preventive dental care visit. Presence of a medical home was computed for children according to the demographic, social, and health status variables described here. The χ2 statistic was used to test bivariate associations between each covariate and prevalence of a medical home. The independent effects of the medical home on the outcome variables were ascertained using logistic regression analyses that controlled for confounders. Confounding variables were selected on the basis of Andersen's health behavior model.30 Predisposing variables included age, gender, race/ethnicity, primary language spoken at home, mother's educational attainment, perception of neighborhood safety, region, and urban/rural residence. Need variables included perceived health status and perceived oral health (dental outcomes only). Enabling variables included household poverty status and the child's health insurance coverage status at the time of the interview. SUDAAN software (Research Triangle Institute, Research Triangle Park, NC) was used to conduct all analyses with the weighted survey data, adjusted for the complex, multistage sample design. Unless otherwise indicated, all differences described in the text are significant at the ≤.05 level. Correlates of Success in Medical Home Attainment Table 1 lists the 5 components and 9 subcomponents of the medical home measure and the number of cases with valid data for each component among children aged 1 to 17 years. Table 2 lists the proportion of children with a medical home and each of the 5 components of the medical home. Nationally, 56.9% of children aged 1 to 17 years had a medical home in 2007. Much higher proportions of children met the individual components: 93.1% of children had a usual source of care, 92.1% had a personal physician or nurse, 81.9% of children had no problems in obtaining referrals when needed, 68.8% received effective care coordination when needed, and 66.7% of children received family-centered care. Substantial differences in medical home attainment rates are apparent across demographic, socioeconomic, and health characteristics. Younger children were more likely to have medical homes than their older counterparts, whereas no significant differences were found for gender. There are large racial and ethnic disparities; non-Hispanic white children had the highest attainment rate, and Hispanic children had the lowest. Non-Hispanic black children fared only modestly better than Hispanic children. Prevalence of medical homes was twofold higher for children in families in which English was the primary language compared with children in families in which other languages were primarily spoken. Where a child lives was also related to attainment of a medical home. Attainment rates were highest in the Midwest and lowest in the West. Although only a modest difference was found in attainment rates for children living in urban and rural areas, children who lived in neighborhoods considered “safe” by their parents were much more likely to have medical homes than children whose parents did not consider their neighborhood to be safe. Strong gradients are apparent for the 2 measures of socioeconomic status in Table 2. Maternal educational attainment beyond high school conferred a twofold advantage in the likelihood of having a medical home compared with mothers with less than a high school education. Children in families with incomes at <100% of the federal poverty level were only about half as likely to meet the criteria for having a medical home as children in families with incomes at ≥400% of the federal poverty threshold. In addition, uninsured children were about half as likely as insured children to have a medical home. Finally, children who were reported to be in excellent or very good overall health, as perceived by their parents, were more than twice as likely to have medical homes as their counterparts in fair or poor health. A similar, but less steep, gradient exists for parent-reported oral health. Recognizing that many of the demographic, socioeconomic, and health characteristics listed in Table 2 are correlated, we also examined whether the bivariate results retained their significance after multivariable analysis. Comparison of the unadjusted and adjusted odds ratios in Table 3 reveals some attenuation of effect sizes for most of the associated factors. However, the adjusted results, which show the independent effect of each covariate on the likelihood of having a medical home, remain significant with the exception of region and residential location. Impact on Medical Care and Dental Care Table 4 reveals the impact of having a medical home on access and use of medical and dental care. Unmet medical care needs were reported for 3.7% of children. A significantly greater percentage of children without a medical home (6.4%) were reported as having an unmet health care need than children with a medical home (1.6%). The adjusted analysis shows that children without a medical home had almost 4 times the odds of having unmet health care needs as children who have a medical home. Overall, 11.7% of children did not receive a preventive care visit in the past year. Children without medical homes were more likely than children with a medical home to have gone without a visit (14.0% vs 9.9%). This difference remained significant in the adjusted analysis, as shown in the lower half of the table. The association between presence of a medical home and dental care is shown in the last 2 columns of Table 4. The prevalence of unmet dental care needs was 2.9% for all children. Those without a medical home were 3 times more likely to have unmet dental needs than those with medical homes (4.8% vs 1.5%). After adjusting for potential confounders, absence of a medical home was associated with nearly threefold higher odds of having an unmet dental care need. Overall, 17.4% of children did not have a preventive dental visit in the past year. On an unadjusted basis, children without medical homes were slightly more likely than those with medical homes to go without preventive dental care (16.6% vs 18.4%); however, this relationship reversed in the adjusted analysis. The medical home is increasingly accepted as the standard for provision of high-quality comprehensive health care. The definition used here requires that children have not only a usual source of care and a personal physician or nurse, but also care that is family centered and provides ready access to referrals and care coordination when needed. These same principles are at the core of the AAP's definition of medical home. Although many articles have assessed the prevalence and impact of medical homes for CSHCN,6,31,–,33 few have done so for the pediatric population as a whole. The results presented here provide the most recent, comprehensive assessment of the proportion of US children who receive their care in medical homes, the health and social correlates of having a medical home, and the impact of medical homes on receipt of preventive care and presence of unmet health needs. We found that most children had 1 or more of the 5 medical home components but only about half of children (56.9% of children aged 1–17 years, or ∼38 million, nationally) had a medical home in 2007. Aspects of the patient-provider relationship (including access to needed referrals, care coordination, and receipt of family-centered care) remain problematic for many children. Strategies to address these bottlenecks include education and technical assistance for practice transformation, including improved care coordination, use of electronic medical records to monitor referrals and follow-up care, and shifting financial incentives to create greater parity in reimbursement of cognitive and procedure-oriented care. Similar to findings from the 2003 NSCH,20 we found significant disparities in receipt of care in medical homes by race and ethnicity and poverty. Among racial and ethnic groups, Hispanic children fared worse, followed closely by blacks. A strong gradient across poverty categories was also documented in our study. Although these racial/ethnic and income-related disparities attenuated when confounding variables were considered, they remained significant. Notably, our analysis also revealed large disparities in access to medical homes across health status. Children who could conceivably benefit most (those reported in fair or poor health) were only half as likely as those rated in excellent or very good health to have a medical home. Together, these health and social disparities indicate a need to target interventions toward the most vulnerable children. There may be additional benefits to improving access to medical homes for vulnerable populations, especially minority racial and ethnic groups. An analysis of the Commonwealth Fund's 2006 Health Care Quality Survey reported that health care settings with features of a medical home (including a regular source of care, enhanced access to physicians, and timely, well-organized care) can eliminate racial and ethnic disparities in access to quality care.11 Although that study used a narrow definition of medical home and included only adults, the findings suggest that expanding access to medical homes could improve quality and increase equity among children. Insurance provides an important tool for reducing disparities by increasing access to medical homes. Our analysis revealed that insured children were almost twice as likely to have medical homes as uninsured children. In this regard, the new health care reform law34 is particularly salient, containing several provisions that should increase access to medical homes. First, by 2014, all children will be required to have health insurance coverage. Given our finding that health insurance is highly correlated with medical home access, this provision alone should have a large impact on the proportion of children with medical homes. In addition, the new law provides for raising Medicaid reimbursement rates for primary care to current Medicare levels. That substantial boost in payment rates should increase access to primary care and, by extension, medical homes for Medicaid-enrolled children. Other health care reform provisions, including requiring private insurance plans to provide preventive care according to the Bright Futures guidelines23 at no out-of-pocket expense to enrollees, will also support the medical home movement. Our study found strong associations between presence of a medical home and unmet health care needs. Even after adjusting for confounding variables, lacking a medical home was associated with a three- to fourfold increased risk of having an unmet need for medical or dental care. These results add to the growing body of evidence supporting the medical home as a model of comprehensive health care for children. However, although we found a small salutatory effect of medical home on receipt of preventive medical care, the association was not as strong as expected. The modest effect size suggests that medical home providers may need to use new strategies to ensure that children receive routine preventive care at recommended intervals. Surprisingly, presence of a medical home was inversely related to receipt of preventive dental care after adjusting for other variables, albeit the effect was small and only marginally significant. This finding may reflect the significant shortage of dentists providing preventive oral health services to young children25 as well as the fact that incorporating preventive oral health care in the medical home, although recommended, is not yet a common practice for most physicians.25,–,27,35 Although the medical home may play a significant role in assuring that children receive appropriate referral and follow-up for dental problems, it does not seem to be influential in assuring receipt of recommended preventive oral health care. Parents may be unaware of professional guidelines for preventive oral health care and thus may neither seek nor expect these services from the medical home. Compared with previous studies of CSHCN, we found similar disparity patterns in access to a medical home as well as benefits in the form of lower rates of delayed care and unmet needs associated with care in a medical home setting.6 The main difference in findings relates to the prevalence of medical homes. Perhaps because they place more demands on the health care system, CSHCN are less likely than children in general to have care that meets all of the components of the medical home. There are limitations to this study. First, the 5 components used here to define a medical home, although designed to align with the main components of the AAP definition, are not identical; they are operational approximations.19,29 In particular, the element of continuity included in the AAP definition of medical home cannot be assessed because of the methodologic difficulties of measuring continuity of care over time in a reliable way using cross-sectional data.29 Second, the NSCH is based on parent report, which is both a limitation and strength. Although the estimates provided are limited by the knowledge and recollection of the parent, these data represent a consumer-based national measurement of the medical home concept. Third, some children are underrepresented or not represented in the survey, including those in institutional settings, homeless, or in migrant families, and those without landline telephones. Adjustments in the sample weights are made to account for these differences. Finally, because of the cross-sectional nature of the survey data set, we are limited in drawing causal inferences from the data. Many of these limitations could be addressed through thoughtfully designed longitudinal comparison studies of children receiving care in medical homes and traditional practice settings. Overall, slightly more than half of US children receive their care in medical homes. Receipt of care in medical homes is shown here to be associated with reduced access problems for medical and dental care. These findings reinforce the need to continue and expand federal, state, and community efforts to ensure that all children have access to a medical home. Given the presence of socioeconomic, racial/ethnic, and health disparities in receipt of care in medical homes, targeted initiatives addressing disadvantaged segments of the child population are needed. - Accepted December 16, 2010. - Address correspondence to Bonnie B. Strickland, PhD, 5600 Fishers Lane, Room 18A27, Parklawn Building, Rockville, MD 20857. E-mail: FINANCIAL DISCLOSURE: The authors have indicated they have no financial relationships relevant to this article to disclose. - AAP = - American Academy of Pediatrics • - CSHCN = - children with special health care needs • - NSCH = - National Survey of Children's Health - 1.↵American Academy of Pediatrics, Medical Home Initiatives for Children With Special Needs Project Advisory Committee. The medical home. Pediatrics. 2002;110(1 pt 1):184–186 - Sia C, - Tonniges TF, - Osterhus E, - Taba S - Cooley WC, - McAllister JW, - Sherrieb K, - Kuhlthau K - Homer CJ, - Klatka K, - Romm D, - et al - Strickland B, - McPherson M, - Weissman G, - van Dyck P, - Huang ZJ, - Newacheck P - Kogan MD, - Strickland BB, - Blumberg SJ, - Singh GK, - Perrin JM, - van Dyck PC - Palfrey JS, - Sofis LA, - Davidson EJ, - Liu J, - Freeman L, - Ganz ML - Beal AC, - Doty MM, - Hernandez SE, - et al - Rosenthal TC - Fields D, - Leshen E, - Patel K - 14.↵Institute of Medicine, Committee on Quality of Health Care in America. Crossing the Quality Chasm: A New Health System for the 21st Century. Washington, DC: National Academy Press; 2000 - 15.↵Patient-Centered Primary Care Collaborative. Available at: www.pcpcc.net. - 16.↵American Academy of Family Physicians; American Academy of Pediatrics; American College of Physicians; American Osteopathic Association. Joint principles of the patient-centered medical home. Available at: www.pcpcc.net/content/joint-principles-patient-centered-medical home. Accessed May 17, 2010 - Albuquerque NM - 18.↵National Committee for Quality Assurance. Patient-centered medical home. Available at: www.ncqa.org/tabid/631/Default.aspx. Accessed October 21, 2009 - Bethell CD, - Read D, - Brockwood K - Hale KJ - 22.↵US Department of Health and Human Services. Oral Health in America: A Report of the Surgeon General. Rockville, MD: US Department of Health and Human Services, National Institute of Dental and Craniofacial Research, National Institutes of Health; 2000 - 23.↵American Academy of Pediatrics. Bright Futures Guidelines for Health Supervision of Infants, Children, and Adolescents. 3rd ed.Elk Grove Village, IL: American Academy of Pediatrics; 2008 - 24.↵American Academy of Pediatric Dentistry. Policy on the dental home. Available at www.aapd.org/media/Policies_Guidelines/P_DentalHome.pdf. Accessed May 17, 2010 - Kenney MK - Blumberg SJ, - Foster EB, - Frasier AM, - et al - 29.↵Child and Adolescent Health Measurement Initiative. Measuring Medical Home for Children and Youth: Methods and Findings From the National Survey of Children With Special Health Care Needs and the National Survey of Children's Health. Portland, OR. Data Resource Center for Child and Adolescent Health; 2009. Available at: http://ftp.cdc.gov/pub/health_statistics/nchs/slaits/nsch07/2_Methodology_Report/NSCH_Design_and_Operations_052109.pdf. Accessed February 19, 2011 - Mulvihill BA, - Altarac M, - Swaminathan S, - Kirby RS, - Kulczycki A, - Ellis DE - Singh GK, - Strickland BB, - Ghandour RM, - van Dyck PC - 34.↵Patient Protection and Affordable Care Act (Pub L No. 111-148). - Copyright © 2011 by the American Academy of Pediatrics<\|endoftext\|>	3.6875
1,125	# If n + 1 n what is n A Sequence of numbers is with an designated and their Sequential members obey that Education Act the sequence of numbers. Natural numbers (sometimes also with 0) are used for n. For example: an = n + 2 (n + 2 is the law of formation; values for n = ... are elements of the sequence) or: an = (½)n or also: an = (-1)n (n / 2) + 1 While in the second example the sequence values for larger n are getting smaller and smaller, in the third example you can see that the values for even n are getting larger and smaller for odd n. A sequence of numbers can also recursive can be defined as the following example shows: The first seven terms of a sequence are shown here, in which an ever higher power of -1/2 is added and subtracted alternately. The consequence obeys the following law: an + 1 = an + (-½) n and a0 = 0 This is called a recursively (going backwards) defined sequence, since a sequence term can only be calculated if you know its predecessor. In the figure below, the graph is the sequence an= 1 (1 / n) shown: This consequence is increasing monotonouslybecause each sequence value is greater than its predecessor. This can be shown by proving: an + 1 an> 0. There is also analog monotonically falling Follow like an = 1 + (1 / n). (Proof by: an + 1 an< 0.) If you look at the above illustration, you will notice that the values are getting closer and closer to 1. For example, a= 1 (1/4) = 3/4. a1000 = 1 (1/1000) = 999/1000 is already much closer to 1. Now you can ask yourself what happens when you look at n larger and larger. Since the sequence increases monotonically, one gets as close as desired to 1 with ever larger n, but never reaches this, since 1 / n would have to become 0 for this. Here the sequence an= 1 (1 / n) is no longer shown in the Cartesian coordinate system, but only its individual terms on the number line. To the (presumed) limit a stripe is placed at a distance epsilon (a very small positive number) and the elements in the sequence that are not in it are counted. In every little one Epsilon strips almost all of the elements of the sequence must be found. (almost all = all but a finite number). This is the definition of the limit. In our example above, the limit value would be written as follows: If a sequence of numbers has a limit, it is called convergent (converging), the sequence converges towards the limit value. If it has no limit, it is called divergent (diverging), the sequence diverges. (see example 3 above) With this so-called Square plant the area that is added by the small squares becomes smaller and smaller; so where does the area of the outer squares strive? Here is the calculation with which one comes to the conclusion from above (see picture): (a is the edge length of the original square) The area of a square of the kth generation is (a / 3k)2. Of a certain generation k there are 3k Squares, so that the area of all squares of a generation a2/3k is. For the area of all squares one must have the total form, and let n go towards infinity: Consequences appear in many mathematical processes. An example are Approximation method. The Greek mathematician Archimedes (around 287 to 212 BC) tried to determine the relationship between the diameter and the circumference of a circle by means of complex approximations. He calculated the circumference of n-vertices, which he inscribed and circumscribed the circle, and continued his investigations up to the value n = 96! We can just surpass this performance. The picture below shows how a 9-corner in as well as another around a circle was laid. The associated construction with The Geometer's Sketchpad allows you to drag the blue circle (above) to turn the 9-corner into a triangle, square, ... up to a 100-corner. In addition to the circumference, the area of the circle is also specified (as is well known, the ratio between the area and the radius of each circle). The correct mathematical description for this approximation process: We define two sequences, one each for the circumference of the inner and the outer n-gon. (Because we're interested in, we cut each circumference in half.) • Insiden = [half the circumference of the inscribed n-gon] • • Outsiden = [half the circumference of the circumscribed n-gon] Without calculation, purely from the geometric context, it is obvious: lim Insiden = lim Outsiden = In addition, it is easy to see that the two sequences "nest" the limit value with increasingly good approximation, that is: Insiden n + 1 <n + 1 n for all n. There is little point in proving this mathematically. Conversely, however, we could try to use the convergence of the sequences to calculate approximate values for.<\|endoftext\|>	4.65625
649	The start of summer vacation doesn’t have to mean the end of science class! Long summer days are perfect for creative, hands on science experiments and projects. Whether you’re getting colorful with chemistry in the kitchen or innovating an engineering project, the following five ideas are ways to teach children science even when school is out of session. Grow your own rock candy Kids of all ages will enjoy this fun - and sweet - science experiment. With materials that can be easily found around the house such as sugar, string, and wax paper, plus a little bit of patience - and adult supervision for the boiling sugar solution - you can test whether or not seed crystals affect the growth rate of rock candy. Find more instructions for this experiment try this site [http://www.sciencebuddies.org/science-fair-projects/project_ideas/FoodSci_p005.shtml?from=Blog#materials]. Learn about absorption with sponges Not all science experiments need elaborate setups; in fact, ones like this can be put together in a matter of minutes! Using a sponge, washcloth, plastic blocks or toys, and a bowl or tub of water have children predict which objects with absorb the most water. Then experiment by placing each object into the bowl; which objects absorb the most water? The least? What can you do to increase or decrease absorption? Kids of all ages will enjoy cooling down on a hot day with this hands-on experiment. Plant a vegetable garden Whether you have a large garden or a few pots on a balcony, growing a vegetable garden during the summer can provide you with fresh produce while helping kids practice their science skills. Kids will enjoy predicting which plants will grow best in their climate, which seedlings will sprout first, and how different amounts of sunlight and water affect plant growth. If you don’t want to be digging and weeding in the dirt, try growing plants without soil with a hydroponic terrarium [http://www.sciencebuddies.org/science-fair-projects/project_ideas/PlantBio_p045.shtml?from=Blog]. Cool down with homemade ice cream Nothing beats the heat during the heat of summer quite like a bowl of cold ice cream. Kids can learn about lowering the freezing point of water while also making their own delicious treats. With cream, sugar, ice, rock salt, plastic bags, and the flavorings of their choice, kids can create their own ice cream concoctions. Learn more about making your own ice cream here [http://www.stevespanglerscience.com/lab/experiments/homemade-ice-cream-sick-science/]. Build an egg parachute As long as you don’t mind cracking a dozen - or more - eggs in the process, kids will enjoy the challenge of building an egg parachute and carton that can withstand being dropped from a ladder or balcony. This engineering-based activity encourages kids to think creatively and problem solve while using recycled materials from around the house. This science project can keep kids occupied for days at a time!<\|endoftext\|>	3.96875
675	Better monitoring is needed to safeguard the Antarctic against threats posed by invasive alien species, according to a new study. The authors developed 'the Antarctic Biological Invasions Indicator' (ABII) to help generate data for tracking trends in alien invasions and the measures taken to prevent them. Invasive alien species are a leading threat to biodiversity. Currently, there are no systems for tracking invasion trends in the Antarctic, or for documenting the impacts of invasions on other wildlife or the management responses taken. Under the Antarctic Treaty System, Parties of the Treaty have committed to preventing accidental introductions of alien species, but there is no requirement to report on new species, their impacts or eradications. A global indicator of invasion developed under the Convention on Biological Diversity (CBD) is not applied in Antarctica. In an attempt to develop a framework for monitoring invasions in the region, the authors adapted the global CBD invasion indicator and applied it to generate baseline data for future monitoring efforts. They collated data from a variety of different sources to meet the requirements of the new indicator. The original, global indicator system combines measures of the numbers of alien species; the conservation status of native species; and the responses taken. Responses include implementation of international and national-level policy instruments such as the Cartagena Protocol on Biosafety (international) and Invasive Non-Native Species Framework Strategy (UK). The new indicator — the ABII — also includes a fourth measure for drivers of invasion, which is intended to keep track of growing levels of human activity across the region, including trade, tourism and construction. Other important differences are that the new ABII includes all alien species rather than only those with documented impacts on biodiversity. It also defines reporting regions based on biological similarities, rather than country boundaries. According to the results, the Southern Ocean Islands have the highest concentrations of alien species in Antarctic. However, there are many areas that are rarely visited and that it was difficult to find data for. Only for seabirds were there enough data to examine the impacts of invasive species over time, and the results from over a hundred species suggest that, overall, they are at increasing risk of extinction. More than half of the alien species recorded — a higher proportion than for other regions of the world — are already known to have negative impacts on biodiversity, meaning that there is evidence that they are invasive. Given the apparent vulnerability of the region, the researchers suggest adopting more stringent targets than in the rest of the world, aiming for zero new alien introductions as well as the eradication of all existing alien species, for example, as opposed to focusing on priority species. The authors highlight the lack of information about management responses to alien species invasions. They say that while awareness about invasions is rising, more practical management actions need to be taken to tackle them. Monitoring and reporting of such actions could help demonstrate success and support ongoing policy development. Source: McGeoch, M. A., Shaw, J. D., Terauds, A., Lee, J. E. & Chown, S. L. (2015). Monitoring biological invasion across the broader Antarctic: A baseline and indicator framework. Global Environmental Change 32, 108–125. DOI:10.1016/j.gloenvcha.2014.12.012.<\|endoftext\|>	3.703125
4,720	# Into Math Grade 6 Module 10 Lesson 3 Answer Key Write Equations from Tables and Graphs We included HMH Into Math Grade 6 Answer Key PDF Module 10 Lesson 3 Write Equations from Tables and Graphs to make students experts in learning maths. ## HMH Into Math Grade 6 Module 10 Lesson 3 Answer Key Write Equations from Tables and Graphs I Can identify patterns in tables and graphs and use the patterns to write equations to represent real-world situations. Step It Out Question 1. Use the table to complete each statement. A. Look at the table for patterns. Describe the patterns found in the table. As x increases by ___________, y increases by ____________. Each value of y is ___________ units ____________ than the corresponding value of x. As x increases by 1.25, y increases by 1.25. Each value of y is 8 units greater than the corresponding value of x. B. Write a verbal model describing an equation in the form y = x + p to represent the relationship between the values in the table. The equation that represents the table is y = x + 8 The equation in the verbal model is some numbers increased by 8. C. Write an equation using the verbal model. Some number increased by 8 in the verbal model Consider some number = x Then the equation is y = x + 8 Turn and Talk What would be the value of x in the table if the corresponding value of y were 25.2? Explain. Given that, The value pf x is 25.2 Then the value of y is 25.2 + 8 = 33.2 Question 2. Mara is making bracelets. For every red bead, she uses 4 green beads. The graph represents this situation. Use the graph to complete each part. A. Read the ordered pairs from the graph. Use them to complete the table. The given coordinates (x, y) in the graph are (1, 4), (2, 8), (3, 12), (4, 16). B. Look at the table for patterns. Describe the patterns found in the table. As x increases by ___________, y increases by ___________. Each value of y is ______________ the corresponding value of x. From the table the equation is y = 4x. As x increases by 1, y increases by 4. Each value of y is 4 the corresponding value of x. C. Write a verbal model of the form y = px to represent the relationship between the values in the table From the table the equation is y = 4x. The verbal model of the equation is some number is multiplied by the 4. D. Write an equation using the verbal model. Some number is multiplied by 4 is the verbal model Consider some number = x The equation using the verbal model is y = 4x. Turn and Talk In Task 2, should the dots be connected in the graph? Explain why or why not. You can model real-world situations using an equation. Question 3. The graph shows the distance d, in miles, a car traveled over time t, in hours. A. Write the coordinates of the points on the graph as ordered pairs. Answer: (1, 50) (2, 100) (3, 150) and (4, 200) B. Complete the sentence to describe the pattern shown in the graph. As the first coordinate of each of the points increases by ___________, the second coordinate of each of the points increases by ____________. Answer: As the first coordinate of each of the points increase by 1, the second coordinate of each of the points increases by 50. C. The value of the distance d, in miles, is always _____________ the value of the time t, in hours. So the equation that models the relationship is ___________. Answer: The value of the distance d, in miles, is always 50 times the value of the time t, in hours. So, the equation that models the relationship is y = 50x Turn and Talk Why does it make sense to have the line connecting the points on the graph in Task 3 but not in Task 2. Explain. Answer: The line connecting the points on the graph is based on the x and y coordinates. Check Understanding Write an equation representing each table or graph. Question 1. The equation that represents the table is y = x + 12. If x = 0 then y = 0 + 12 = 12 x = 1 then y = 1 + 12 = 13 x = 2 then y = 2 + 12 = 14 x = 3 then y = 3 + 12 = 15 x = 4 then y = 4 + 12 = 16 Question 2. From the graph the coordinates points are (1,3), (2,6), (3,9), (4,12). The equation is y = 3x. If x = 1 then y = 3(1) = 3 x = 2 then y = 3(2) = 6 x = 3 then y = 3(3) = 9 x = 4 then y = 3(4) = 12. Question 3. Construct Arguments The table represents the distance driven using various amounts of gas. Write an equation that models the distance d, in miles, with respect to gas g, in gallons. Explain your answer by describing patterns in the table. The equation that models the distance d, in miles, with respect to gas g, in gallons is d = 24g If g = 5 then d = 24(5) = 120 g = 6 then d = 24(6) = 144 g = 7 then d = 24(7) = 168 g = 8 then d = 24(8) = 192 g = 9 then d = 24(9) = 216 Question 4. The graph represents the relationship between the total cost, in dollars, of a city taxi and the number of miles driven. A. Write an equation that relates C, the cost in dollars, to n, the miles driven. The coordinates points are (1, 1.5), (2,3), (3,4.5). The equation that relates to the C, the cost in dollars to n is y = 1.5x If x = 1.5 then y = 1.5(1) = 1.5 x = 2 then y= 1.5(2) = 3. x = 3 then y = 1.5(3) = 4.5 B. Reason Suppose the points in the graph are all increased by three dollars. Write an equation that relates C, the new cost in dollars, to n, the miles driven. If the number of miles driven is 6, what would the cost be? Explain how you found the new equation. The equation that relates to the C, the cost in dollars to n is y = 1.5x x = 6 then y = 1.5(6) = 9 If the number of miles driven is 6 then the cost is 9. Question 5. Model with Mathematics The table represents the cost C, in dollars, of a cell phone data plan for n months. Write an equation that represents the data in the table. The equation that represents the cost C, in dollars, of a cell phone data plan for n months is C = 45.25n If n = 1 then C = 45.25(1) = 45.25 n = 2 then C = 45.25(2) = 90.50 n = 3 then C = 45.25(3) = 135.75 n = 4 then C = 45.25(4) = 181 Question 6. Model with Mathematics Write an equation representing each table or graph. A. The equation representing the table is y = 14 + x = 0 If x = 0 then y = 14 + 0 = 14 x = 1 then y = 14 + 1 = 15 x = 2 then y = 14 + 2 = 16. x = 3 then y = 14 + 3 = 17. x = 4 then y = 14 + 4 = 18. B. The equation that represents the table is y = 12x + x If x = 0 then y = 12x + x = 12(0) + 0 = 0 x = 1 then y = 12(1) + 1 = 13 x = 2 then y = 12(2) + 2 = 26 x = 3 then y = 12(3) + 3 = 39 x = 4 then y = 12(4) + 4 = 52 C. The given coordinates are (0,0), (1, 0.5), (2,1), (3,1.5), (4,2), (5,2.5), (6,3). The equation that represents the table is y = 0.5x If x = 0 then y = 0.5(0) = 0 If x = 1 then y = 0.5(1) = 0.5 x = 2 then y = 0.5(2) = 1 x = 3 then y = 0.5(3) = 1.5 x = 4 then y = 0.5(4) = 2. x = 5 then y = 0.5(5) = 2.5 x = 6 then y = 0.5(6) = 3 D. The given coordinates are (0,1), (1,2), (2,3), (3,4), (4,5), (5,6), (6,7), (7,8). The equation that represents the table is y = x + 1 If x = 0 the y = 0 + 1 = 1 x = 1 then y = 1 + 1 = 2 x = 2 then y = 2 + 1 = 3 x = 3 then y = 3 + 1 = 4 x = 4 then y = 4 + 1 = 5 x = 5 then y = 5 + 1 = 6 x = 6 then y = 6 + 1 = 7 x = 7 then y = 7 + 1 = 8. E. The equation that represents the table is y = 16 + x. If x = 0 then y = 16 + 0 = 16. x = 1 then y= 16 + 1 = 17. x = 2 then y = 16 + 2 = 18. x = 3 then y = 16 + 3 = 19 x = 4 then y = 16 + 4 = 20 F. The equation that represents the table is y = 14x If x = 0 then y =14(0) = 0 x = 1 then y = 14(1) = 14 x = 2 then y = 14(2) = 28 x = 3 then y = 14(3) = 42 x = 4 then y = 14(4) = 56. G. Look for Repeated Reasoning Examine the tables and graphs. What do you notice about the equations for the tables and graphs that include the point (0, 0)? What do you notice about the equations for the tables and graphs that do not include the point (0, 0)? Answer: We noticed that the equations for the tables and the graphs are included the points (0,0). Question 7. The table shows the monthly costs for a smartphone plan with Company A. The advertisement shows the cost of Company B’s plan. A. Use Structure Write an equation that models the total cost C, in dollars, for n months of Company B’s plan. Explain. B. Complete the table for Company B’s plan. C. After how many months will Company B’s plan cost less than Company A’s plan? Explain how you arrived at your answer. Model with Mathematics For Problems 8-9, write an equation representing each table or graph. Question 8. The equation that represents the table is y = 6x If x = 0 then y = 6(0) = 0 x = 2 then y = 6(2) = 12 x = 5 then y = 6(5) = 30. x = 8 then y = 6(8) = 48 x = 12 then y = 6(12) = 72 Question 9. The coordinates in the graph are (2,0), (3,1), (4,2), (5,3), (6,4), (7,5). The equation that represents the graph is y = x – 2 If x = 2 then y = 2 – 2 = 0 x = 3 then y = 3 – 2 = 1 x = 4 then y = 4 – 2 = 2 x = 5 then y = 5 – 2 = 3 x = 6 then y = 6 – 2 = 4 x = 7 then y = 7 – 2 = 5 Lesson 10.3 More Practice/Homework Question 1. The graph shows the total sales of tickets to a school performance. A. Write an equation that relates the total sales to the number of tickets sold. Use T for the total sales and n for the number of tickets. The coordinates are (0,0), (1,4), (2,8), (3,12), (4,16), (5,20), (6,24), (7,28), (8,32). The equation that relates the total sales to the number of tickets sold is T = 4n If n = 0 then T = 4(0) = 0 n = 1 then T = 4(1) = 4 n = 2 then T = 4(2) = 8 n = 3 then T = 4(3) = 12 n = 4 then T = 4(4) = 16 n = 5 then T = 4(5) = 20 n = 6 then T = 4(6) = 24 n = 7 then t = 4(7) = 28 n = 8 then T = 4(8) = 32 B. If the number of tickets sold is 125, what will be the total sales? Given that the number of tickets sold is 125. The equation that relates the total sales to the number of tickets sold is T = 4n If n = 125 then T = 4(125) = 500. Therefore the total sales are 500. Question 2. STEM The table shows the relationship between the mass m and the force w of weights on the ground. Write an equation representing this relationship. The equation that representing the relationship between the mass m and the force w of weights on the ground is f = 9.8m If m = 1 then f = 9.8(1) = 9.8 m = 2 then f = 9.8(2) = 19.6 m = 3 then f = 9.8(3) = 29.4 m = 4 then f = 9.8(4) = 39.2 Model with Mathematics For Problems 3-6, write an equation representing each graph or table. Question 3. The (x, y) coordinates in the graph are (0,0), (1,7.5), (2,15), (3,22.5), (4,30), (5,37.5), (6,45). The equation that represents the table is y = 7.5x If x= 0 then y = 7.5(0) = 0 x = 1 then y = 7.5(1) = 7.5 x = 2 then y = 7.5(2) = 15 x = 3 then y = 7.5(3) = 22.5 x = 4 then y = 7.5(4) = 30 x = 5 then y = 7.5(5) = 37.5 x = 6 then y = 7.5(6) = 45 Question 4. The (x, y) coordinates in the graph is (0,6), (1,7), (2,8), (3,9). The equation that represents the graph is y = 6 + x If x = 0 then y = 6 + 0 = 6 x = 1 then y = 6 + 1 = 7 x = 2 then y = 6 + 2 = 8 x = 3 then y = 6 + 3 = 9 Question 5. The equation that represents the table is y = x + 11.2 If x = 0 then y = 11.2 x = 3 then y = 3 + 11.2 = 14.2 x = 6 then y = 6 + 11.2 = 17.2 x = 9 then y = 9 + 11.2 = 20.2 x = 12 then y = 12 + 11.2 = 23.2 x = 15 then y = 15 + 11.2 = 26.2 Question 6. The equation that represents the table is y = 8.5x If x = 0 then y = 8.5(0) = 0 x = 2 then y = 8.5(2) = 17 x= 4 then y = 8.5(4) = 34 x = 6 then y = 8.5(6) = 51. x = 8 then y = 8.5(8) = 68. x = 10 then y = 8.5(10) = 85 Test Prep Question 7. The partially completed table contains some values for x and y. A. Use patterns to complete the table. The equation relationship between the x and y is y = 7x If x = 1 then y = 7(1) = 7 x = 3 then y = 7(3) = 21 x = 5 then y = 7(5) = 35 x = 7 then y = 7(7) = 49 x = 9 then y = 7(9) = 63 B. Write an equation that models the relationship between x and y. The equation relationship between the x and y is y = 7x If x = 1 then y = 7(1) = 7 x = 3 then y = 7(3) = 21 x = 5 then y = 7(5) = 35 x = 7 then y = 7(7) = 49 x = 9 then y = 7(9) = 63 Question 8. Brenda sells balloon bouquets. She charges the same price for each balloon in a bouquet. The costs for several bouquets are shown in the table. Write an equation that relates the cost of a bouquet to the number of balloons in the bouquet. The equation that relates the cost of a bouquet to the number of balloons in the bouquet is y = x/2 If x = 6 then y = 6/2 = 3 x = 9 then y = 9/2 = 4.50 x = 12 then y = 12/2 = 6 Question 9. Which equation represents the relationship between x and y shown in the graph? (A) y = 0.25x (B) y = 0.5x (C) y = 2x (D) y = 4x The equation y = 0.25x represents the relationship between x and y. The (x, y) coordinates in the graph are (4,1), (8,2), (12,3), (16,4) If x = 4 then y = 0.25(4) = 1 x = 8 then y = 0.25(8) = 2 x = 12 then y = 0.25(12) = 3 x = 16 then y = 0.25(16) = 4 Option A is the correct answer. Spiral Review Question 10. Evaluate the expression for a = 3 and b = $$\frac{1}{4}$$. 5a – 16b + 7 5a – 16b + 7 a = 3 and b = $$\frac{1}{4}$$ 5(3) – 16($$\frac{1}{4}$$) + 7 15 – 4 + 7 11 + 7 = 18 Question 11. What is the opposite of -3.5?<\|endoftext\|>	4.75
634	Mathematical thinking comes more naturally to some people than to others, but everyone can succeed in math. Math is used on the job and in everyday life, so it’s important for students of all ages to grasp mathematical concepts. Students in many locales must maintain at least a C average to avoid repeating math class or attending summer school. Whether you’re in fifth grade or college, if you are proactive you can earn a better grade in math. Develop a Positive Attitude A student might feel defeated before he even opens a math book, and this stands in the way of achievement. For example, when a parent says, “You’re bad in math, just like I was,” a child might think there’s no point in trying for a better grade. Since math success is based more on effort than ability, thinking you can do it is half the battle. Shedding preconceived notions is the first step toward reaching goals in math class. Teach Someone Else Teaching a math concept to another person increases understanding, so work with a study partner. A person learns 95 percent of what he teaches someone else, according to renowned psychiatrist, William Glasser. Strategies such as modeling and drawing diagrams should be used in explanations. It’s best to teach concepts in small chunks so you can revisit parts that are unclear. Parents should encourage their children to teach them the lessons they learned in math class. Their confidence will increase, resulting in higher homework and test scores. Tap Available Resources A student who seeks help typically achieves a better grade in math. Ask questions in class or request that the teacher explain a concept in a different way. Most schools make after-school tutoring sessions available, so struggling students should attend. Additionally, a variety of math tutorials are available online, and they offer visual and auditory explanations as well as sample problems and solutions. Working with a peer who has a command of the concepts is another alternative. Get Plenty of Practice Many new concepts are introduced each week in math class, and it takes practice to fully understand them and remember them for tests. You can often find answers to odd-numbered problems in the back of a math book, so teachers using these textbooks will assign even numbers for homework. Before you begin the homework, do several of the odd numbers and check your work to determine whether you’re getting correct answers. Once you’re confident, advance to the actual homework problems. Math is a series of building blocks, and when you don’t understand a concept and try to move to the next step, it can cause frustration. For example, in order to do algebra problems, you must be able to add and subtract positive and negative numbers. When you receive a low grade on a homework assignment, figure out what you did wrong and then redo the assignment. Students who are absent should make up missed work in a timely manner and obtain any notes that were taken in class. Some teachers allow students to retake tests in order to improve their grades and their understanding. - Digital Vision./Digital Vision/Getty Images<\|endoftext\|>	3.984375
815	The malaria life cycle requires two hosts, as the malarial parasite, Plasmodium, must shift from mosquito to human, and vice versa. The mosquito is the intermediary invertebrate host, whereas the human is the perfect vertebrate host. A Look At The Malaria Life Cycle Here’s a detailed look at the malaria life cycle, starting with the causes of malaria. When a person is bitten by a female Anopheles mosquito, sporozoites are introduced into his blood. Within 40 minutes, these sporozoites travel through the person’s blood stream into his liver cells, where they mature and develop into the next stage, known as schizonts. These schizonts then rupture to release thousands of tiny merozoites in the blood after a period of about 16 days. The Plasmodium vivax and Plasmodium ovale strains of the parasite can continue to remain in the person’s liver in an inactive state for a long time. They are called hypnozoites. On re-activation of these parasites, the patient can suffer a relapse after several weeks, months, or years following the malarial infection. This preliminary reproduction of the malarial parasite in the liver is known as exo-erythrocytic schizogony. In this phase, the parasites undergo a phase of asexual multiplication in the red blood cells. This phase is called erythrocytic schizogony and it releases merozoites into the patient’s blood. The other red blood cells are invaded by merozoites and they develop into trophozoites. During their developmental process, the trophozoites take in the hemoglobin present in the red cells, leaving behind a pigment known as hemozoin, which is a mixture of hematin and protein. This pigment contains iron and it is visualized as dark granules in the body of the parasite. These granules are more distinct in the later stages of the development of the parasite. The Development Of Parasitemia Following the division of the mature trophozoites in the red blood cells, the formation of separate merozoites occurs, causing a schizont. Following complete development, the schizont bursts out of the red blood cell containing it, releasing the merozoites into the bloodstream. Then new red blood cells are infected by these merozoites and the procedure of asexual division in the blood resumes. Some of the merozoites that enter the red blood cells do not develop into trophozoites, instead they develop into gametocytes. This process occurs in the capillaries present in deep tissues. Female/macro-gametocytes or male/micro-gametocytes enter the mosquito as an ookinete, which penetrates the wall of the stomach. The development of an oocyst then takes place. The oocyst ruptures and releases sporozoites that break into the salivary glands. This cycle of erythrocytic schizogony is repeated time and again in due course of the infection, resulting in an aggravating a condition called parasitemia. The Length Of The Malaria Life Cycle Certain factors, like the species of Plasmodium, the ambient temperature, and the mosquito host, all play an important role in determining the duration of the developmental stage in the mosquito. This may last for about 8 days, as in Plasmodium vivax, or even for a month, as seen in Plasmodium malariae. The malaria life cycle is essentially very similar in all strains of the Plasmodium parasite. Malaria affects millions of people worldwide, especially those hailing from tropical and sub-tropical regions, so it very important to be have up-to-date information about this disease. You can begin by studying the life cycle of the disease, along with other common malarial symptoms.<\|endoftext\|>	3.84375
104	a way of selecting items from a collection, such that (unlike permutations) the order of selection does not matter. In smaller cases, it is possible to count the number of combinations. For example, given three fruits, say an apple, an orange, and a pear, there are three combinations of two that can be drawn from this set: an apple and a pear; an apple and an orange; or a pear and an orange. - Explain in your own words - Provide 1 - 2 examples<\|endoftext\|>	4.1875
269	To truncate something is to shorten it, or cut part of it off. In computer science, the term is often used in reference to data types or variables, such as floating point numbers and strings. For example, a function may truncate the decimal portion of a floating point number to make it an integer. If the number 3.875 is truncated, it becomes 3. Note that this is different than if the number had been rounded to the nearest integer, which would be 4. Strings may also be truncated, which can be useful if a string exceeds the maximum character limit for a certain application. Several programming languages use the function trunc() to truncate a variable. PHP uses strlen() to truncate a string to a set limit of characters. TechTerms - The Tech Terms Computer Dictionary This page contains a technical definition of Truncate. It explains in computing terminology what Truncate means and is one of many technical terms in the TechTerms dictionary. All definitions on the TechTerms website are written to be technically accurate but also easy to understand. If you find this Truncate definition to be helpful, you can reference it using the citation links above. If you think a term should be updated or added to the TechTerms dictionary, please email TechTerms!<\|endoftext\|>	4.1875
3,825	# Logic for Computer Scientists/Propositional Logic/Resolution ## Resolution In this subsection we will develop a calculus for propositional logic. Until now we have a language, i.e. a set of formulae and we have investigated into semantics and some properties of formulae or sets of clauses. Now we will introduce an inference rule, namely the resolution rule, which allows to derive new clauses from given ones. ## Definition 10 A clause ${\displaystyle R}$ is a resolvent of clauses ${\displaystyle C_{1}}$ and ${\displaystyle C_{2}}$ , if there is a literal ${\displaystyle L\in C_{1}}$ and ${\displaystyle {\overline {L}}\in C_{2}}$ and ${\displaystyle R=(C_{1}-\{L\})\cup (C_{2}-\{{\overline {L}}\})}$ where ${\displaystyle {\overline {L}}={\begin{cases}\;\;\,\lnot A\;{\text{ if }}L=A\\\;\;\,A\;{\text{ if }}L=\lnot A\end{cases}}}$ Note that there is a special case, if we construct the resolvent out of two literals, i.e. ${\displaystyle L}$ and ${\displaystyle {\overline {L}}}$ can be resolved upon and yield the empty set. This empty resolvent is depicted by the special symbol ${\displaystyle \square }$ . ${\displaystyle \square }$ denotes an unsatisfiable formula. We define a clause set, which contains this empty clause to be unsatisfiable, In the following we investigate in properties of the resolution rule and the entire calculus. The following Lemma is stating the correctness of one single application of the resolution rule. ## Theorem 7 If ${\displaystyle S}$ is a set of clauses and ${\displaystyle R}$ a resolvent of ${\displaystyle C_{1},C_{2}\in S}$ , then ${\displaystyle S\equiv S\cup \{R\}}$ Proof: Let ${\displaystyle {\mathcal {A}}}$ be an assignment for ${\displaystyle S}$ ; hence it is an assignment for ${\displaystyle S\equiv S\cup \{R\}}$ as well. Assume ${\displaystyle {\mathcal {A}}\models S}$ : hence for all clauses ${\displaystyle C}$ from ${\displaystyle S}$ , we have that ${\displaystyle {\mathcal {A}}\models C}$ . The resolvent ${\displaystyle R}$ of ${\displaystyle C_{1}}$ and ${\displaystyle C_{2}}$ looks like: ${\displaystyle R=(C_{1}-\{L\})\cup (C_{2}-\{{\overline {L}}\})}$ where ${\displaystyle L\in C_{1}}$ and ${\displaystyle {\overline {L}}\in C_{2}}$ . Now there are two cases: • ${\displaystyle {\mathcal {A}}\models L}$ : From ${\displaystyle {\mathcal {A}}\models C_{2}}$ and ${\displaystyle {\mathcal {A}}\not \models {\overline {L}}}$ , we conclude ${\displaystyle {\mathcal {A}}\models (C_{2}-\{{\overline {L}}\})}$ and hence ${\displaystyle {\mathcal {A}}\models R}$ . • ${\displaystyle {\mathcal {A}}\not \models L}$ : From ${\displaystyle {\mathcal {A}}\models C_{1}}$ we conclude ${\displaystyle {\mathcal {A}}\models (C_{1}-\{L\})}$ and hence ${\displaystyle {\mathcal {A}}\models R}$ . The opposite direction of the lemma is obvious. ## Definition 11 Let ${\displaystyle S}$ be s set of clauses and ${\displaystyle Res(S)=S\cup \{R\mid R{\text{ is a resolvent of two clauses in }}S\}}$ then ${\displaystyle Res^{0}(S)=S}$ ${\displaystyle Res^{n+1}(S)=Res(Res^{n}(S)),n\geq 0}$ ${\displaystyle Res^{}(S)=\bigcup _{n\geq 0}Res^{n}(S)}$ If we understand the process of iterating the Res-operator as a procedure for deriving new clauses from a given set, and in particular to derive possibly the empty clause, we have to ask, under which circumstances we get the empty clause, and vice versa, what does it mean if we get it. These properties are investigates in the following two Theorems. ## Theorem 8 (Correctness) Let ${\displaystyle S}$ be a set of clauses. If ${\displaystyle \square \in Res^{}(S)}$ then ${\displaystyle S}$ is unsatisfiable. Proof: From ${\displaystyle \square \in Res^{}(S)}$ we conclude, that ${\displaystyle \square }$ is obtained by resolution from two clauses ${\displaystyle C_{1}=\{L\}}$ and ${\displaystyle C_{2}=\{{\overline {L}}\}}$ . Hence there is a ${\displaystyle \exists n\geq 0}$ such that ${\displaystyle \square \in Res^{n}(S)}$ and ${\displaystyle C_{1},C_{2}\in Res^{n}(S)}$ and therefore ${\displaystyle Res^{n}(S)}$ is unsatisfiable. From Theorem (resolution-lemma) we conclude that ${\displaystyle Res^{n}(S)\equiv S}$ and hence ${\displaystyle S}$ is unsatisfiable. ## Theorem 9 (Completeness) Let ${\displaystyle S}$ be a finite set of clauses. If ${\displaystyle S}$ is unsatisfiable then ${\displaystyle \square \in Res^{}(S)}$ . Proof: Induction over the number ${\displaystyle n}$ of atomic formulae in ${\displaystyle S}$ . With ${\displaystyle n=0}$ we have ${\displaystyle S=\{\square \}}$ and hence ${\displaystyle \square \in S\subseteq Res^{}(S)}$ . Assume ${\displaystyle n}$ fixed and for every unsatisfiable set of clauses ${\displaystyle S}$ with ${\displaystyle n}$ atomic formulae ${\displaystyle A_{1},\cdots ,A_{n}}$ it holds that ${\displaystyle \square \in Res^{}(S)}$ . Assume a set of clauses ${\displaystyle S}$ with atomic formulae ${\displaystyle A_{1},\cdots ,A_{n},A_{n+1}}$ . In the following we construct two clause sets ${\displaystyle S_{f}}$ and ${\displaystyle S_{t}}$ : • ${\displaystyle S_{f}}$ is received from ${\displaystyle S}$ by deleting every occurrence of ${\displaystyle A_{n+1}}$ in a clause and by deleting every clause which contains an occurrence of ${\displaystyle \lnot A_{n+1}}$ . This transformation obviously corresponds to interpreting the atom ${\displaystyle A_{n+1}}$ with ${\displaystyle false}$ , • ${\displaystyle S_{t}}$ results from a similar transformation, where occurrences of ${\displaystyle \lnot A_{n+1}}$ and clauses containing ${\displaystyle A_{n+1}}$ are deleted, hence ${\displaystyle A_{n+1}}$ is interpreted with ${\displaystyle true}$ . Let us show, that both ${\displaystyle S_{f}}$ and ${\displaystyle S_{t}}$ are unsatisfiable: Assume an assignment ${\displaystyle {\mathcal {A}}}$ for the atomic formale ${\displaystyle \{A_{1},\cdots ,A_{n}\}}$ which is a model for ${\displaystyle S_{f}}$ . Hence the assignment ${\displaystyle {\mathcal {A}}'(B)={\begin{cases}\;\;\,{\mathcal {A}}(B)\;{\text{ if }}B\in \{A_{1},\cdots ,A_{n}\}\\\;\;\,false{\text{ if }}B=A_{n+1}\end{cases}}}$ is a model for ${\displaystyle S}$ , which leads to a contradiction. A similar construction shows that ${\displaystyle S_{t}}$ is unsatisfiable. Hence then we can use the induction assumption to conclude that ${\displaystyle \square \in Res^{}(S_{t})}$ and ${\displaystyle \square \in Res^{}(S_{f})}$ . Hence there is a sequence of clauses ${\displaystyle C_{1},\cdots ,C_{m}=\square }$ such that ${\displaystyle \forall 1\leq i\leq m}$ it holds ${\displaystyle C_{i}\in S_{f}}$ or ${\displaystyle C_{i}}$ is a resolvent of two clauses ${\displaystyle C_{a}}$ and ${\displaystyle C_{b}}$ with ${\displaystyle a,b .There is an analog sequence for ${\displaystyle S_{t}}$ : ${\displaystyle C_{1}',\cdots ,C_{t}'=\square }$ Now we are going to reintroduce the previously deleted literals ${\displaystyle A_{n+1}}$ and ${\displaystyle \lnot A_{n+1}}$ in the two sequences: • Clause ${\displaystyle C_{i}}$ which has been the result of deleting ${\displaystyle A_{n+1}}$ from the original clause in ${\displaystyle S}$ are again modified to ${\displaystyle C_{i}\cup \{A_{n}+1\}}$ . This results in a sequence ${\displaystyle {\overline {C_{1}}},\cdots ,{\overline {C_{m}}}}$ where ${\displaystyle {\overline {C_{m}}}}$ is either ${\displaystyle \square }$ or ${\displaystyle A_{n+1}}$ . • Analogous we introduce ${\displaystyle \lnot A_{n+1}}$ in the second sequence, such that ${\displaystyle {\overline {C_{t}'}}}$ is either ${\displaystyle \square }$ or ${\displaystyle \lnot A_{n+1}}$ In any of the above cases we get ${\displaystyle \square \in Res^{}(S)}$ latest after one resolution step with ${\displaystyle {\overline {C_{m}}}}$ and ${\displaystyle {\overline {C_{t}'}}}$ . Based on the theorems for correctness and completeness, we give a procedure for deciding the satisfiability of propositional formulae. Deciding Satisfiability of Propositional Formulae Given a propositional formula ${\displaystyle F}$ . • Transform ${\displaystyle F}$ into an equivalent CNF ${\displaystyle S}$ . • Compute ${\displaystyle Res^{n}(S)}$ for ${\displaystyle n=0,1,2,\cdots }$ • If ${\displaystyle \square \in Res^{n}(S)}$ then Stop: unsatisfiable . • if ${\displaystyle Res^{n}(S)=Res^{n+1}(S)}$ then Stop: satisfiable . ## Theorem 10 If ${\displaystyle S}$ is a finite set of clauses, then there exists a ${\displaystyle k\geq 0}$ such that ${\displaystyle Res^{k}(S)=Res^{k+1}(S)}$ Until now, we have been dealing with sets of clauses. In the following it will turn out, that it is helpful to talk about sequences of applications of the resolution rule. ## Definition 12 A deduction of a clause ${\displaystyle C}$ from a set of clauses ${\displaystyle S}$ is a sequence ${\displaystyle C_{1},\cdots ,C_{n}}$ , such that • ${\displaystyle C_{n}=C}$ and • ${\displaystyle \forall 1\leq i\leq n:(C_{i}\in S{\text{ or }}\exists l,r A deduction of the empty clause ${\displaystyle \square }$ from ${\displaystyle S}$ is called a refutation of ${\displaystyle S}$ . Example We want to show, that the formula ${\displaystyle K=((B\land \lnot A)\lor C)}$ is a logical consequence of ${\displaystyle F=((A\lor (B\lor C))\land (C\lor \lnot A))}$ . For this negate ${\displaystyle K}$ and prove the unsatisfiability of ${\displaystyle F\land \lnot K}$ For this you can use the interaction in this book in various forms: • Use the interaction Truth Tables for proving the unsatisfiability, or • use the interaction CNF Transformation for transforming the formula into CNF, and then • use the following interaction Resolution. ## Problems #### Problem 23 (Propositional) Compute ${\displaystyle Res^{n}(M)}$ for all ${\displaystyle n\geq 0}$ and ${\displaystyle Res^{}(M)}$ for the following set of clauses: 1. ${\displaystyle M=\{\{A\},\{B\},\{\lnot A,C\},\{B,\lnot C,\lnot D\},\{\lnot C,D\},\{\lnot D\}}$ 2. ${\displaystyle M=\{\{A,\lnot B\},\{A,B\},\{\lnot A\}\}}$ 3. ${\displaystyle M=\{\{A,B,C\},\{\lnot B,\lnot C\},\{\lnot A,C\}\}}$ 4. ${\displaystyle M=\{\{\lnot A,\lnot B\},\{B,C\},\{\lnot C,A\}\}}$ Which formula is satisfiable or which is unsatisfiable? ${\displaystyle \Box }$ #### Problem 24 (Propositional) Indicate all resolvent of the clauses in S, where ${\displaystyle S=\{\{A,\lnot B,C\},\{A,B,E\},\{\lnot A,C,\lnot D\},\{A,\lnot E\}\}}$ ${\displaystyle \Box }$ #### Problem 25 (Propositional) Prove: A resolvent ${\displaystyle R}$ of two clauses ${\displaystyle C_{1}}$ and ${\displaystyle C_{2}}$ is a logical consequence from ${\displaystyle C_{1}}$ and ${\displaystyle C_{2}}$ . Note: Use the definition of "consequence". ${\displaystyle \Box }$ #### Problem 26 (Propositional) Let ${\displaystyle M}$ be a set of formulae and ${\displaystyle F}$ a formula. Prove: ${\displaystyle M\models F}$ iff ${\displaystyle M\cup \{\lnot F\}}$ is unsatisfiable. ${\displaystyle \Box }$ #### Problem 27 (Propositional) Compute ${\displaystyle Res^{n}(S)}$ with ${\displaystyle n=0,1,2}$ and ${\displaystyle S=\{\{A,\lnot B,C\},\{B,C\},A\{\lnot ,C\},\{B,\lnot C\},\{\lnot C\}\}}$ ${\displaystyle \Box }$ #### Problem 28 (Propositional) Show that the following set ${\displaystyle S}$ of formulae is unsatisfiable, by giving a refutation. ${\displaystyle S=(B\lor C\lor D)\land (\lnot C)\land (\lnot B\lor C)\land (B\lor \lnot D)}$ ${\displaystyle \Box }$ #### Problem 29 (Propositional) Show by using the resolution rule, that ${\displaystyle \lnot A\land \lnot B\land C}$ is an inference from the set of clauses ${\displaystyle F=\{\{A,C\},\{\lnot B,\lnot C\},\{\lnot A\}\}}$ . ${\displaystyle \Box }$ #### Problem 30 (Propositional) Show by using the resolution rule, that ${\displaystyle ((P\to Q{\text{ is }})\land P)\to Q}$ is a tautology. ${\displaystyle \Box }$<\|endoftext\|>	4.5625
946	# Mathematical Formulation Of Second Law Of Motion We often observe, that if the same magnitude of force is used to push two blocks of wood, where one of the blocks is heavier than the other, the rate of change of position of the lighter block will be more than the heavier one. Similarly, when two same forces are applied to push a car and a bus, the car will have more acceleration compared to the bus. From these examples, it is clear that the acceleration gained by an object when subjected to the same magnitude of the force is a factor of the mass of the object. When a car is given a momentary jerk, it may not move from its initial position whereas when an extended and continuous force of the same magnitude is applied to the car, it experiences a displacement. With these examples, we can conclude that the impact produced by an object depends on its mass and velocity i.e., its momentum and the time rate at which the change in momentum is occurring. The second law of motion is used to validate this phenomenon. In this section, we shall learn about the formulation of the second law of motion. ## Newton’s Second Law of Motion Newton’s second law of motion can be formally stated as follows: The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object. ## Second Law of Motion Formula Let us consider an object of mass m, moving along a straight line with an initial velocity of u. Let us say, after a certain time t, with a constant acceleration, the final velocity becomes v. Here we see that, the initial momentum is: $$\begin{array}{l}(p_1)=m \times u\end{array}$$ The final momentum, $$\begin{array}{l}(p_2)=m \times v\end{array}$$ The change in momentum can be written as, $$\begin{array}{l}(p_2) – (p_1) =(m \times v)- (m \times u)=m\times(v-u)\end{array}$$ As we know, the rate of change of momentum with respect to time is proportional to the applied force. The applied force, $$\begin{array}{l}F \propto \frac{(m\times(v-u))}{t}\end{array}$$ Or, $$\begin{array}{l}F\propto m \times a \end{array}$$ as acceleration (a) = rate of change of velocity with respect to time. $$\begin{array}{l}F= k\times m \times a\end{array}$$ Above is the second law of motion formula. ### Notations Used In The Formula • F is the force • k is the constant of proportionality • a is the acceleration The SI units of mass and acceleration are kg and m.s-2 respectively. So, $$\begin{array}{l} Unit\,of\,Force = k \times ( 1\,kg) \times (1\, ms^{-2})\end{array}$$ The second law of motion gives us a method to measure the force acting on an object as a product of the mass of the object and the acceleration of the object which is the change in velocity with respect to time. ## Frequently Asked Questions – FAQs Q1 ### What are Newton’s laws of motion? Newton’s laws of motion are three laws of classical physics that describe the connection between the motion of bodies and the forces acting on them. Q2 ### What are the three laws of motion in classical mechanics? The law of inertia (Newton’s first law), the law of force and acceleration (Newton’s second law), and the law of action and reaction (Newton’s third law). Q3 ### Define Newton’s first law of motion. Newton’s first law of motion states that an object at rest or uniform motion will continue to be at rest or uniform motion until and unless an external force acts on it. Q4 ### Define Newton’s second law of motion. Newton’s second law of motion states that the acceleration of a body as generated by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the body’s mass. Q5 ### Define Newton’s third law of motion. Newton’s third law of motion states that there is an equal and opposite reaction for every action.<\|endoftext\|>	4.875
1,173	489 views ### Show that substituting y = xv, where v is a function of x, in the differential equation "xy(dy/dx) + y^2 − 2x^2 = 0" (with x is not equal to 0) leads to the differential equation "xv(dv/dx) + 2v^2 − 2 = 0" This is the first part of a Step 1 question (2012 question 8), and is fairly typical in that it requires A Level Maths understanding to be applied in a number of different ways, out of the usual context. We begin by finding a new expression for (dy/dx), in terms of v. We do this by differentiating "y = xv", and getting: dy/dx = x(dv/dx) + v By substituting this and "y = xv" into the original differential equation, we get: (x)(xv)(x(dv/dx) + v) + (xv)^2 - 2(x^2) = 0 (x^3)v(dv/dx) + 2(xv)^2 - 2(x^2) = 0 We can divide by x^2, since we know that x does not equal zero. This gives the required result: xv(dv/dx) + 2(v^2) - 2 = 0 The rest of the question first asks you to find a solution to the original equation. The new equation is now separable, and can be solved to find v. This can be substituted back into "y = xv", giving a solution for y. Try it if you like. It should look something like: (x^2)(y^2 - x^2) = C (where C is a constant) The question then asks you to solve (again with x is not equal to zero): y(dy/dx) + 6x + 5y = 0 This can be done with the same substitution, which again makes it separable. Using partial fractions, integrating, finding v in terms of x (or vice versa), and substituting allows a solution to be found. This should (although it is still pretty ugly, particularly on a computer screen) look like: ((y + 3x)^3)/((y + 2x)^2) = B (where B is a constant) 10 months ago ## Still stuck? Get one-to-one help from a personally interviewed subject specialist #### 31 SUBJECT SPECIALISTS £36 /hr Degree: Mathematics G100 (Bachelors) - Bath University Subjects offered:.STEP., Maths+ 3 more .STEP. Maths Further Mathematics .MAT. -Personal Statements- “Hi! I'm Joe, a friendly, experienced and patient tutor with in-depth knowledge of both the old and new A Level Maths & Further Maths specifications.” £28 /hr Degree: Aerospace Engineering PhD Spacecraft Control (Doctorate) - Bristol University Subjects offered:.STEP., Physics+ 4 more .STEP. Physics Maths Further Mathematics Extended Project Qualification .PAT. “Aerospace Engineering PhD candidate in spacecraft control with 7 years of experience in tutoring.” £25 /hr Degree: Mathematics (Bachelors) - Cambridge University Subjects offered:.STEP., Maths+ 3 more .STEP. Maths Further Mathematics .MAT. -Oxbridge Preparation- “Friendly first-year Mathematics student studying at Trinity College, Cambridge. Two years tutoring experience.” MyTutor guarantee \| 7 completed tutorials £30 /hr Degree: Mathematics (Bachelors) - Durham University Subjects offered:.STEP., Physics+ 2 more .STEP. Physics Maths Further Mathematics “Hi, I'm Will. I study Maths and Physics and really enjoy talking about them. I am patient and caring, so will take the time to refine a student’s understanding.” ### You may also like... #### Posts by Will How do you calculate the derivative of cos inverse x? How do you integrate (sinx)^2? How much work must be done on a 4.0kg frictionless trolley, to accelerate it from rest to a velocity of 5.0m/s? Show that substituting y = xv, where v is a function of x, in the differential equation "xy(dy/dx) + y^2 − 2x^2 = 0" (with x is not equal to 0) leads to the differential equation "xv(dv/dx) + 2v^2 − 2 = 0" #### Other Uni Admissions Test .STEP. questions Prove that any number of the form pq, where p and q are prime numbers greater than 2, can be written as the difference of two squares in exactly two distinct ways. Show that substituting y = xv, where v is a function of x, in the differential equation "xy(dy/dx) + y^2 − 2x^2 = 0" (with x is not equal to 0) leads to the differential equation "xv(dv/dx) + 2v^2 − 2 = 0" Show that if a polynomial with integer coefficients has a rational root, then the rational root must be an integer. Hence, show that x^n-5x+7=0 has no rational roots. How can I integrate e^x sin(x)? We use cookies to improve your site experience. By continuing to use this website, we'll assume that you're OK with this.<\|endoftext\|>	4.5625
481	The following was provided by the National Institute of Neurological Disorders and Stroke, part of the National Institutes of Health. What Causes ALS? The cause of ALS is not known, and scientists do not yet know why ALS strikes some people and not others. An important step toward answering that question came in 1993 when scientists supported by the National Institute of Neurological Disorders and Stroke (NINDS) discovered that mutations in the gene that produces the SOD1 enzyme were associated with some cases of familial ALS. This enzyme is a powerful antioxidant that protects the body from damage caused by free radicals. Free radicals are highly unstable molecules produced by cells during normal metabolism. If not neutralized, free radicals can accumulate and cause random damage to the DNA and proteins within cells. Although it is not yet clear how the SOD1 gene mutation leads to motor neuron degeneration, researchers have theorized that an accumulation of free radicals may result from the faulty functioning of this gene. In support of this, animal studies have shown that motor neuron degeneration and deficits in motor function accompany the presence of the SOD1 mutation. Studies also have focused on the role of glutamate in motor neuron degeneration. Glutamate is one of the chemical messengers or neurotransmitters in the brain. Scientists have found that, compared to healthy people, ALS patients have higher levels of glutamate in the serum and spinal fluid. Laboratory studies have demonstrated that neurons begin to die off when they are exposed over long periods to excessive amounts of glutamate. Now, scientists are trying to understand what mechanisms lead to a buildup of unneeded glutamate in the spinal fluid and how this imbalance could contribute to the development of ALS. Autoimmune responses, which occur when the body’s immune system attacks normal cells, have been suggested as one possible cause for motor neuron degeneration in ALS. Some scientists theorize that antibodies may directly or indirectly impair the function of motor neurons, interfering with the transmission of signals between the brain and muscles. In searching for the cause of ALS, researchers have also studied environmental factors such as exposure to toxic or infectious agents. Other research has examined the possible role of dietary deficiency or trauma. However, as of yet, there is insufficient evidence to implicate these factors as causes of ALS. Future research may show that many factors, including a genetic predisposition, are involved in the development of ALS.<\|endoftext\|>	3.671875
3,101	Notes On Conditional Probability - CBSE Class 12 Maths The set of all the possible outcomes of a random experiment is called sample space. It is denoted by S. Each outcome in a sample space is called a sample point. In the sample space for the experiment when three fair coins are tossed. S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} Here, each sample point is equally likely. Probability of each sample point = 1/8 Let A be an event that at least two tails appear. That is, two or more than two tails appear. A = {HTT, THT, TTH, TTT} P (A) = P ({HTT}) + P ({THT}) + P ({TTH}) + P ({TTT}) = 1/8 + 1/8 + 1/8 + 1/8 = 1/2 Let B is event first coin shows head. B = {HTT, HHT, HTH, HHH} P (B) = P ({HTT}) + P ({HHT}) + P ({HTH}) + P ({HHH}) = 1/8 + 1/8 + 1/8 + 1/8 = 1/2 S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} A = {HTT, THT, TTH, TTT} B = {HTT, HHT, HTH, HHH} A âˆ© B = {HTT} P (A âˆ© B) = P ({HTT}) = 1/8 Let's find the probability of A (at least two tails), if event B has already occurred. S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} B = {HTT, HHT, HTH, HHH} Now, sample space for event A is the set of all favourable occurrences of event B. Sample space for event A = {HTT, HHT, HTH, HHH} Event A: At least two tails The sample point of B favourable to the occurrence of event A is HTT. A = {HTT} The sample points in B are equally likely. Probability of each sample point in B = 1/4 P (A) = P ({HTT}) = 1/4 Conditional probability The probability of event A is called the conditional probability of A given that event B has already occurred. This is denoted by P (A\|B) and read as probability of A given B. S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} A = {HTT, THT, TTH, TTT} B = {HTT, HHT, HTH, HHH} {HTT} is common to events A and B. The outcomes of event A = The sample points of event "A âˆ© B" P (A\|B) = Number of elementary events favourable to A âˆ© B / Number of elementary events favourable to B = n(A âˆ© B)/n(B) = (n(A âˆ© B)/n(S))/(n(B)/n(S)) Probability of A, given that B has already occurred = P (A\|B) = P(A âˆ© B)/P(B) If A and B are two events of same sample space S, then the conditional probability of event A given that event B has occurred, is given by P (A\|B) = P(A âˆ© B)/P(B), P (B) â‰ 0 S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} A = {HTT, THT, TTH, TTT} B = {HTT, HHT, HTH, HHH} Let's find the probability of A, if event B has already occurred. A âˆ© B = {HTT} P (A âˆ© B) = 1/8 P (B) = 1/2 Probability of A, given that B has already occurred = P (A\|B) = (1/8)/(1/2) = 1/4 The definition of conditional probability can also be applied in a general case, where the elementary events are not equally likely. Properties of Conditional Property Property: If A is an event associated with sample space S, then P (S\|A) = P (A\|A) = 1. P (S\|A) = P(S âˆ© A)/P(A) = P(A)/P(B) = 1 P (A\|A) = P(A âˆ© A)/P(A) = P(A)/P(A) = 1 Hence, P (S\|A) = P (A\|A) = 1 Property: If A and B are two events of sample space S, and E is an event of S such that P(E) â‰ 0, then P((A âˆª B)\|E) = P(A\|E) + P(B\|E) - P((A âˆ© B)\|E). LHS = P ((A âˆª B) \|E) = P[(A âˆª B) âˆ© E]/P(E) = P[(A âˆ© E) âˆª (B âˆ© E)]/P(E) (by distributive law of union of sets over intersection) = (P(A âˆ© E) + P(B âˆ© E) - P(A âˆ© B âˆ© E))/P(E) [âˆµ P(X âˆª Y) = P(X) + P(Y) - P(X âˆ© Y)] = P(A âˆ© E)/P(E) + P(B âˆ© E)/P(E) - (P(A âˆ© B) âˆ© E)/P(E) = P(A\|E) + P(B\|E) - P((A âˆ© B)\|E) = RHS. Hence, P((A âˆª B)\|E) = P(A\|E) + P(B\|E) - P((A âˆ© B)\|E) For two disjoint events A and B, A âˆ© B = âˆ… â‡’ P((A âˆ© B)\|E) = 0 P((A âˆª B)\|E) = P(A\|E) + P(B\|E), if A and B are disjoint sets. Property: If A and B are two events of sample space S, then P(A'\|B) = 1 - P(A\|B). We know that P(S\|B) = 1 (From property I) â‡’ P(A âˆª A'\|B) = 1 (âˆµ A âˆª A' = S) â‡’ P(A\|B) + P (Aâ€²\|B) = 1 (âˆµ A and A' are disjoint events) â‡’ P (Aâ€²\|B) = 1 - P (A\|B). #### Summary The set of all the possible outcomes of a random experiment is called sample space. It is denoted by S. Each outcome in a sample space is called a sample point. In the sample space for the experiment when three fair coins are tossed. S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} Here, each sample point is equally likely. Probability of each sample point = 1/8 Let A be an event that at least two tails appear. That is, two or more than two tails appear. A = {HTT, THT, TTH, TTT} P (A) = P ({HTT}) + P ({THT}) + P ({TTH}) + P ({TTT}) = 1/8 + 1/8 + 1/8 + 1/8 = 1/2 Let B is event first coin shows head. B = {HTT, HHT, HTH, HHH} P (B) = P ({HTT}) + P ({HHT}) + P ({HTH}) + P ({HHH}) = 1/8 + 1/8 + 1/8 + 1/8 = 1/2 S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} A = {HTT, THT, TTH, TTT} B = {HTT, HHT, HTH, HHH} A âˆ© B = {HTT} P (A âˆ© B) = P ({HTT}) = 1/8 Let's find the probability of A (at least two tails), if event B has already occurred. S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} B = {HTT, HHT, HTH, HHH} Now, sample space for event A is the set of all favourable occurrences of event B. Sample space for event A = {HTT, HHT, HTH, HHH} Event A: At least two tails The sample point of B favourable to the occurrence of event A is HTT. A = {HTT} The sample points in B are equally likely. Probability of each sample point in B = 1/4 P (A) = P ({HTT}) = 1/4 Conditional probability The probability of event A is called the conditional probability of A given that event B has already occurred. This is denoted by P (A\|B) and read as probability of A given B. S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} A = {HTT, THT, TTH, TTT} B = {HTT, HHT, HTH, HHH} {HTT} is common to events A and B. The outcomes of event A = The sample points of event "A âˆ© B" P (A\|B) = Number of elementary events favourable to A âˆ© B / Number of elementary events favourable to B = n(A âˆ© B)/n(B) = (n(A âˆ© B)/n(S))/(n(B)/n(S)) Probability of A, given that B has already occurred = P (A\|B) = P(A âˆ© B)/P(B) If A and B are two events of same sample space S, then the conditional probability of event A given that event B has occurred, is given by P (A\|B) = P(A âˆ© B)/P(B), P (B) â‰ 0 S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} A = {HTT, THT, TTH, TTT} B = {HTT, HHT, HTH, HHH} Let's find the probability of A, if event B has already occurred. A âˆ© B = {HTT} P (A âˆ© B) = 1/8 P (B) = 1/2 Probability of A, given that B has already occurred = P (A\|B) = (1/8)/(1/2) = 1/4 The definition of conditional probability can also be applied in a general case, where the elementary events are not equally likely. Properties of Conditional Property Property: If A is an event associated with sample space S, then P (S\|A) = P (A\|A) = 1. P (S\|A) = P(S âˆ© A)/P(A) = P(A)/P(B) = 1 P (A\|A) = P(A âˆ© A)/P(A) = P(A)/P(A) = 1 Hence, P (S\|A) = P (A\|A) = 1 Property: If A and B are two events of sample space S, and E is an event of S such that P(E) â‰ 0, then P((A âˆª B)\|E) = P(A\|E) + P(B\|E) - P((A âˆ© B)\|E). LHS = P ((A âˆª B) \|E) = P[(A âˆª B) âˆ© E]/P(E) = P[(A âˆ© E) âˆª (B âˆ© E)]/P(E) (by distributive law of union of sets over intersection) = (P(A âˆ© E) + P(B âˆ© E) - P(A âˆ© B âˆ© E))/P(E) [âˆµ P(X âˆª Y) = P(X) + P(Y) - P(X âˆ© Y)] = P(A âˆ© E)/P(E) + P(B âˆ© E)/P(E) - (P(A âˆ© B) âˆ© E)/P(E) = P(A\|E) + P(B\|E) - P((A âˆ© B)\|E) = RHS. Hence, P((A âˆª B)\|E) = P(A\|E) + P(B\|E) - P((A âˆ© B)\|E) For two disjoint events A and B, A âˆ© B = âˆ… â‡’ P((A âˆ© B)\|E) = 0 P((A âˆª B)\|E) = P(A\|E) + P(B\|E), if A and B are disjoint sets. Property: If A and B are two events of sample space S, then P(A'\|B) = 1 - P(A\|B). We know that P(S\|B) = 1 (From property I) â‡’ P(A âˆª A'\|B) = 1 (âˆµ A âˆª A' = S) â‡’ P(A\|B) + P (Aâ€²\|B) = 1 (âˆµ A and A' are disjoint events) â‡’ P (Aâ€²\|B) = 1 - P (A\|B). Previous Next âž¤<\|endoftext\|>	4.46875
1,750	Silicosis, (previously miner’s phthisis, grinder’s asthma, potter’s rot and other occupation-related names) is a form of occupational lung disease caused by inhalation of crystalline silica dust, and is marked by inflammation and scarring in the form of nodular lesions in the upper lobes of the lungs. It is a type of pneumoconiosis. Silicosis (particularly the acute form) is characterized by shortness of breath, cough, fever, and cyanosis (bluish skin). It may often be misdiagnosed as pulmonary edema (fluid in the lungs), pneumonia, or tuberculosis. The recognition of respiratory problems from breathing in dust dates to ancient Greeks and Romans. Agricola, in the mid-16th century, wrote about lung problems from dust inhalation in miners. In 1713, Bernardino Ramazzini noted asthmatic symptoms and sand-like substances in the lungs of stone cutters. With industrialization, as opposed to hand tools, came increased production of dust. The pneumatic hammer drill was introduced in 1897 and sandblasting was introduced in about 1904, both significantly contributing to the increased prevalence of silicosis. Classification of silicosis is made according to the disease’s severity (including radiographic pattern), onset, and rapidity of progression. These include: Chronic simple silicosis Usually resulting from long-term exposure (10 years or more) to relatively low concentrations of silica dust and usually appearing 10–30 years after first exposure. This is the most common type of silicosis. Patients with this type of silicosis, especially early on, may not have obvious signs or symptoms of disease, but abnormalities may be detected by x-ray. Chronic cough and exertional dyspnea are common findings. Radiographically, chronic simple silicosis reveals a profusion of small (<10 mm in diameter) opacities, typically rounded, and predominating in the upper lung zones. Silicosis that develops 5–10 years after first exposure to higher concentrations of silica dust. Symptoms and x-ray findings are similar to chronic simple silicosis, but occur earlier and tend to progress more rapidly. Patients with accelerated silicosis are at greater risk for complicated disease, including progressive massive fibrosis (PMF). Silicosis can become "complicated" by the development of severe scarring (progressive massive fibrosis, or also known as conglomerate silicosis), where the small nodules gradually become confluent, reaching a size of 1 cm or greater. PMF is associated with more severe symptoms and respiratory impairment than simple disease. Silicosis can also be complicated by other lung disease, such as tuberculosis, non-tuberculous mycobacterial infection, and fungal infection, certain autoimmune diseases, and lung cancer. Complicated silicosis is more common with accelerated silicosis than with the chronic variety. Silicosis that develops a few weeks to 5 years after exposure to high concentrations of respirable silica dust. This is also known as silicoproteinosis. Symptoms of acute silicosis include more rapid onset of severe disabling shortness of breath, cough, weakness, and weight loss, often leading to death. The x-ray usually reveals a diffuse alveolar filling with air bronchograms, described as a ground-glass appearance, and similar to pneumonia, pulmonary edema, alveolar hemorrhage, and alveolar cell lung cancer. Because chronic silicosis is slow to develop, signs and symptoms may not appear until years after exposure. Signs and symptoms include: Dyspnea (shortness of breath) exacerbated by exertion Cough, often persistent and sometimes severe Tachypnea (rapid breathing) which is often labored Loss of appetite and weight loss Gradual dark shallow rifts in nails eventually leading to cracks as protein fibers within nail beds are destroyed. In advanced cases, the following may also occur: Cyanosis (blue skin) Cor pulmonale (right ventricle heart disease) Patients with silicosis are particularly susceptible to tuberculosis (TB) infection—known as silicotuberculosis. The reason for the increased risk—3 fold increased incidence—is not well understood. It is thought that silica damages pulmonary macrophages, inhibiting their ability to kill mycobacteria. Even workers with prolonged silica exposure, but without silicosis, are at a similarly increased risk for TB. Pulmonary complications of silicosis also include Chronic Bronchitis and airflow limitation (indistinguishable from that caused by smoking), non-tuberculous Mycobacterium infection, fungal lung infection, compensatory emphysema, and pneumothorax. There are some data revealing an association between silicosis and certain autoimmune diseases, including nephritis, Scleroderma, and Systemic Lupus Erythematosus, especially in acute or accelerated silicosis. When small silica dust particles are inhaled, they can embed themselves deeply into the tiny alveolar sacs and ducts in the lungs, where oxygen and carbon dioxide gases are exchanged. There, the lungs cannot clear out the dust by mucous or coughing. Silicon (Si) is the second most common element in the Earth’s crust (oxygen is the most common). The compound silica, also known as silicon dioxide (SiO2), is formed from silicon and oxygen atoms. Since oxygen and silicon make up about 75% of the Earth, the compound silica is quite common. It is found in many rocks, such as marble, sandstone, flint and slate, and in some metallic ores. Silica can be a main component of sand. It can also be in soil, mortar, plaster, and shingles. The cutting, breaking, crushing, drilling, grinding, or abrasive blasting of these materials may produce fine silica dust. Silica occurs in 3 forms: crystalline, microcrystalline (or cryptocrystalline) and amorphous (non-crystalline). “Free” silica is composed of pure silicon dioxide, not combined with other elements, whereas silicates (e.g. talc, asbestos, and mica) are SiO2 combined with an appreciable portion of cations. Microcrystalline silica consists of minute quartz crystals bonded together with amorphous silica. Examples include flint and chert. Amorphous silica consists of kieselgur (diatomite), from the skeletons of diatoms, and vitreous silica, produced by heating and then rapid cooling of crystalline silica. Amorphous silica is less toxic than crystalline, but not biologically inert, and diatomite, when heated, can convert to tridymite or cristobalite. Silica flour is nearly pure SiO2 finely ground. Silica flour has been used as a polisher or buffer, as well as paint extender, abrasive, and filler for cosmetics. Silica flour has been associated with all types of silicosis, including acute silicosis. Silicosis is due to deposition of fine respirable dust (less than 10 micrometers in diameter) containing crystalline silicon dioxide in the form of alpha-quartz, cristobalite, or tridymite. Silicosis is an irreversible condition with no cure. Treatment options currently focus on alleviating the symptoms and preventing complications. These include: Stopping further exposure to silica and other lung irritants, including tobacco smoking. Antibiotics for bacterial lung infection. TB prophylaxis for those with positive tuberculin skin test or IGRA blood test. Prolonged anti-tuberculosis (multi-drug regimen) for those with active TB. Chest physiotherapy to help the bronchial drainage of mucus. Oxygen administration to treat hypoxemia, if present. Bronchodilators to facilitate breathing. Lung transplantation to replace the damaged lung tissue is the most effective treatment, but is associated with severe risks of its own. For acute silicosis, bronchoalveolar lavage may alleviate symptoms, but does not decrease overall mortality.<\|endoftext\|>	3.8125
410	#### Need solution for R.D.Sharma maths class 12 chapter Maxima and Minima exercise 17.3 Question 1 sub Question 6. Point of local minima value is 2 and it’s local minimum value is 2. Hint: First find critical values of f(x) by solving f'(x) =0 then find f''(x). If $f^{\prime \prime}\left(c_{1}\right)>0$ then $\mathrm{c}_{1}$ is point of local minima. If $f^{\prime \prime}\left(c_{2}\right)>0$ then $c_2$ is point of local maxima . where $c_1$ &$c_2$ are critical points. Put $c_1$ and $c_2$ in f(x) to get minimum value & maximum value. Given: $\frac{x}{2}+\frac{2}{x}, \underline{x}>0$ Explanation: We have, \begin{aligned} &f(x)=\frac{x}{2}+\frac{2}{x} \quad, x>0 \\ &f^{\prime}(x)=\frac{1}{2}-\frac{2}{x^{2}} \ \\ &f^{\prime \prime}(x)=\frac{4}{x^{3}} \end{aligned} For local maxima or minima, we have $f^{\prime}(x)=0$ $\begin{gathered} \frac{1}{2}-\frac{2}{x^{2}}=0 \\ x^{2}=4 \\ x=2 \text { or } x=-2 \end{gathered}$ since, $x>0$ $x = 2$ Now, at x=2 , $f(2)=\frac{2}{2}+\frac{2}{2}=2$ Thus, its min. value at 2 is 2.<\|endoftext\|>	4.5
654	The above button for the current speaker of the house Paul Ryan’s 2016 campaign button. The Speaker of the House works as the presiding officer of the United States of Representatives. The position was established by the virtue of Article I, Section 2 of the United States Constitution in 1789. Next to the vice-president, the Speaker falls second in the United States presidential line of succession. In the United States, the Speaker is primarily concerned with the majority party’s legislative goals and agenda. He more specifically assigns tasks to various members of the House from the majority party. During debate, he does not take sides and seldom votes when it comes to certain political matters. In addition, the Speaker of the House is also assigned with administrative and procedural responsibilities and representing their Congressional districts. HOW THE SPEAKER OF THE HOUSE CAME TO BE The first person given the task of being the Speaker was Frederick Muhlenberg. However, the position started to have pertinent influence on politics during the tenure of Henry Clay. Clay became popular for being highly involved in debates and using his influence to achieve things that were in his favor. After Clay’s service ended, the prowess possessed by the Speaker weakened for quite some time. In the 19th century, the Speaker started to gain political power again. This could be exemplified by the period when the Speaker also served as the Chairman of the Committee on Rules, which eventually became one of the most influential committees of the House. In addition, some Speakers eventually became prominent figures in their political parties just like, Samuel J. Randall, John Griffin Carlisle, James G. Blaine, and Joseph Gurney Cannon. The person who greatly played a role in increasing the power held by the Speaker was Thomas Beckett Reed. He usually opposed the bills proposed by the minorities and used the “disappearing quorum” to thwart them. The pinnacle of the Speaker of the House was reached during the service of Joseph Gurney Cannon. During his term, he posed certain control over judicial matters. This extraordinary political powers made a number of Republicans and Democrats work against him and strip him of his power. Another powerful Speaker was Sam Rayburn who held the position the longest. Some of his works included passing a number of bills and implementation of some domestic and foreign assistance programs. In 1975, the Speaker was given the power to appoint members of the Rules Committee. ROLE OF THE SPEAKER Aside from serving a partisan role in government, the Speaker also contributes in passing the legislation’s presented by House’s majority parties. Sometimes, the role played by the Speaker seems to be more important than that of the President especially during the instances that the people holding those positions belong to the same political parties. OTHER CHIEF FUNCTIONS OF THE SPEAKER The Speaker poses certain influence on some matters that concern the House and the government. For example, he can assign another member of the House to be his replacement during his absence or just let them experience the power exercised by one who works as the Speaker. The Speaker also makes sure that each member of the House are in their best behavior during House meetings.<\|endoftext\|>	3.9375
698	Linear Scale Given a scale diagram we can make drawing measurements. Similarly, real (actual) measurements can be made on a real object. Linear scale shows, the relationship between drawing (scale) length and actual length. Linear scale of a diagram is given in statement or ratio form. When the ratio form is used the unit of measurement for the drawing length and actual length is the same. It is not always possible to draw on paper the actual size of real-life objects such as the real size of a car, an airplane, house we need scale drawings to represent the size like the one you have seen. #### Example 1 In a scale diagram 1cm represents 50 m (a) Write the scale in ratio form. (b) Find the drawing length for 1.25 km. (c) Find the actual length in kilometers corresponding to a length of 10.5cm on the diagram. #### Example 2 In a diagram of a plan of a factory a length of 2.5 cm represents an actual length of 12.5 m (a) Work out the linear scale in ratio form. (b) Find the distance on the plan between two buildings which are 35 m apart. ## Conversions to recall • 1 km =1000 m • 1 km = 100,000 cm • 1 m = 100 cm • 1 ha = 10,000 m2 ## Scale in statement form 1cm represents 3m can be written as: 1cm represents 300cm. It can also be written as a ratio 1:300 Read as; one to three hundred. For every unit of drawing there are three hundred similar units of real length. Drawing length: real length #### Write each of the following scales as a ratio (i)1cm represents 4km (ii)1cm represents 5m (iii)1cm represents 200m #### Complete the following statements (i)The scale 1:10000 can be written as 1cm represents____metres. (ii)The scale 1:4000000 can be written as 1cm represents ____km. #### Solution (i)1:400000 (ii)1:500 (iii)1:20000 (iv)100m (v)40km ### Why are scale drawings important? Careers that involve construction, architecture, city planning, design and map-reading all require knowledge of scale drawings You also need to able to use scale-drawings when you are travelling (maps). #### Example use the map of Maine to estimate the distance of China and New Sweden. ### Sources • liner-scale_img1 by elimu used under CC_BY-SA • 93bc2d6c-012a-4fc7-a5a0-0bb031654c97 by elimu used under CC_BY-SA • liner-scale_img3 by elimu used under CC_BY-SA • liner-scale_img3 by elimu used under CC_BY-SA • liner-scale_img4 by elimu used under CC_BY-SA • liner-scale_img4 by elimu used under CC_BY-SA • liner-scale_img5 by elimu used under CC_BY-SA • SCALE_11 by Unknown used under Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License<\|endoftext\|>	4.75
1,178	Summary of hand graphing techniques The knowledge that you carry in the unit circle can be used to graph the sine, cosine, and tangent functions. Of course, you can graph these quickly on a graphing calculator or computer grapher, but sometimes it is very useful to quickly analyze a function by sketching it on paper. Therefore, I expect you to be able to draw the functions and label the critical points. Section 4.5 of the book has a very complete and very good step by step description of how to do this. But, here is a quick and dirty summary! Here is a video clip that demonstrates how the unit circle can be "unwrapped" into the sine and cosine functions. Although this clip emphasizes that it is "unwrapping" sin(θ) rather than sin(x), it works just fine for you to think of the horizontal axis as x rather than θ. The point that I want to stress is that you already have all the important values you need to graph these functions at your fingertips in the unit circle. The period of sinx is because that is the distance around the unit circle. The amplitude of sin x is 1 because that is the radius of the unit circle. The phase shift of sin x is 0 because the unit circle starts at 0. Your book identifies five key points for graphing y = sin x (page 518). They are the three x-intercepts, the maximum point, and the minimum point. All of these are on your unit circle. The values of sin x correspond to the y-values, so those key points are (angle, y-value) or (0,0), (π/2, 1), (π, 0), (3π/2, -1), (2π, 0). The values of cos x correspond to the x-values, so those key points are (angle, x-value) or (0,1), (π/2, 0), (π, -1), (3π/2, 0), (2π, 1). To sketch one period of the graph, you simple need to move these five key points; For example, to sketch the much more complicated function , you must first rewrite it in a form that shows the phase shift Now you can tell the amplitude is 3 the period is the phase shift is to the right. To graph this you need to take each original key point and find the transformation. 1. Find the new period values The new period is 4π so all fo the x-values will double: Instead of 0, π/2, π, 3π/2, 2π we will double the values and use 0, π, 2π, 3π, 4π 2. Apply the phase shift Now apply the phase shift: Doing the addition, the new x-values will be or The original y-values for sin x are 0, 1, 0, -1, and 0. 3. Apply the amplitude Applying the amplitude of 3, the new y-values are 0, 3, 0, -3, and 0. The transformed points are Now you can choose an appropriate scale on your graph paper Then graph and label your five key points:: Now sketch the curve: This same procedure will work for tangent, but tangent is defined by 3 points and two asymptotes. rather than five points. The period of tangent is π, so use quadrants I and IV. (angle, y-value/x-value) or (π/2, 1/0), (π/4, 1),(0,0/1), (-π/4, -1), (-π/2, -1/0) Since division by 0 is undefined, this gives three points (π/4,1), (0,0) and (-π/4,-1) and two vertical asymptotes, x=π/2 and x=-π/2 .Remember that tangent does not have an amplitude (although it can have a stretch which is why we included the points at π/4.) The period of tangent is π. The phaseshift is 0. Let's look at an example of a tangent function that has a stretch, an altered period, and a phase shift, We need to isolate the phase shift so we will factor a 2 out: and write it as a subtraction Now we can see that the stretch factor is 3 the period is π/2 the phase shift is -3π/2 or 3π/2 to the left To graph this you need to take each original key point and find the transformation. 1. Find the new period values The new period is π /2so all fo the x-values will be halved: Instead of π/2, π/4, 0, -π/4, -π/2 we will halve the values and use π/4, π/8, 0, -π/8, -π/4 2. Apply the phase shift Now apply the phase shift: Doing the addition, the new x-values will be or The original y-values for tan x are undefined, 1, 0, -1, and undefined. 3. Apply the stretch Applying the stretch of 3, the new y-values are undefined, 3, 0, -3, and undefined. The transformed key values are Now you can choose an appropriate scale on your graph paper: If you only need to graph one period, you can draw in your tangent. You can also use this pattern to fill in your other periods:<\|endoftext\|>	4.5
680	# Problem Solving Using Inequalities The number two is shaded on both the first and second graphs. This is how we will show our solution in the next examples. Just as the United States is the union of all of the 50 states, the solution will be the union of all the numbers that make either inequality true.There are no numbers that make both inequalities true. There are no numbers that make both inequalities true. There are no numbers that make both inequalities true. To find the solution of the compound inequality, we look at the graphs of each inequality, find the numbers that belong to either graph and put all those numbers together.We will use the same problem solving strategy that we used to solve linear equation and inequality applications. Tags: Ocr Pe CourseworkFifth Grade Problem SolvingDisability In The Media EssayJane Schaffer Essay FormatCompare Contrast 2 Articles Essay10 Writing Tools And Essay WritingResearch Paper Topic OutlineRandom Assignment Experiment The bill for Normal Usage would be between or equal to .06 and 1.02. For the compound inequality $$x−3$$ and $$x\leq 2$$, we graph each inequality. The numbers that are shaded on both graphs, will be shaded on the graph of the solution of the compound inequality. See We can see that the numbers between $$−3$$ and $$2$$ are shaded on both of the first two graphs. The number $$−3$$ is not shaded on the first graph and so since it is not shaded on both graphs, it is not included on the solution graph. $\begin & & \ & & \ \end \nonumber$ To solve a compound inequality means to find all values of the variable that make the compound inequality a true statement. We solve compound inequalities using the same techniques we used to solve linear inequalities. ## Comments Problem Solving Using Inequalities • ###### Inequalities Calculator - Symbolab Free inequality calculator - solve linear, quadratic and absolute value inequalities step-by-step.… • ###### Solve Compound Inequalities - Mathematics LibreTexts Days ago. If you missed this problem, review link. We solve compound inequalities using the same techniques we used to solve linear inequalities.… • ###### Solving Linear Inequalities - Interactive Mathematics In this section we see how to solve inequalities in one variable and includes. Following are several examples of solving equations involving inequalities. textbooks "don't immediately tell you how to solve math problems.… • ###### Inequalities Word Problems Worksheet - Helping With Math A 2-page worksheet with real-world problems for which inequalities must be. Note You will find guidance on using inequalities to represent real-world problems. problem, and construct simple equations and inequalities to solve problems.… • ###### Word Problems with Inequalities Word Problems with Inequalities. Don't panic! Remember to Read the questions carefully. Define the variable. Write an inequality. Solve the inequality.… • ###### Writing inequalities from worded problems by KatieRoseW. I struggled to find much on this, so I've put several resources/questions together to create a PowerPoint to suit and thought I'd share it!… • ###### Algebra - Solving Equations and Inequalities Practice. Here is a set of practice problems to accompany the Solving Equations and Inequalities chapter of the notes for Paul Dawkins Algebra course.…<\|endoftext\|>	4.59375
419	16. Verify xx (y + 2) = x xy + x x z for:x=3/5,y=-2/5,z=-7/5 Question 16. Verify xx (y + 2) = x xy + x x z for: x=3/5,y=-2/5,z=-7/5 in progress 0 5 months 2021-06-23T04:27:57+00:00 1 Answers 0 views 0 1. ★ How to do :- Here, we are given with some of the fractions on LHS of the statement and the same three fractions in the RHS of the statement. On LHS, the last two fractions are grouped by brackets and on the RHS, the first two fractions are grouped in the brackets. If we are given with the fractions by grouping in this format, then this property can be classified as the distributive property. This property can only be done with the multiplication and addition of fractions. It cannot be done in the subtraction and division of fractions or integers. As in this question we are given with multiplication and addition, we can verify it. We should see whether the results of LHS and RHS are equal or not. So, let’s solve!! ➤ Solution :- Substitute the values of x, y and z. Let’s solve the LHS and RHS separately. LHS :– Write both numerators in bracket with a common denominator. Write the second number in numerator with one sign. Subtract the numerators in bracket. Write both numerators and denominators with a common fraction. Multiply the numbers. RHS :– Write both numerators and denominators with common fraction of both brackets. Multiply the numerators and denominators of both brackets. Write both numerators with a common denominator. Write the second number in numerator with one sign. Subtract the numbers. Now, let’s compare the results of both LHS and RHS. Comparison :– As we can see that those both are equal. So, So,<\|endoftext\|>	4.5625
1,965	# How to Calculate Acceleration – 3 Formulas You Must Know March 30, 2024 Every day you can experience differences in rates of acceleration whether you are aware of it or not. Take driving, for instance: when you hear a sports car rushing past you or a plane soaring past another one above you, you’re witnessing how acceleration changes in real-time. It’s a common mistake to confuse acceleration for speed or velocity, but the acceleration formula involves calculating the rate of change in the velocity of something as it moves over time. Therefore, in order to know how to calculate acceleration, you’ll need to know the velocity and set period of time. Acceleration differs from velocity because velocity is the speed of an object in a given direction. The difference between the two is that to calculate acceleration, you need not just velocity but also time. Acceleration measures how much something is becoming faster or slower, depending on these factors. To calculate acceleration, you’ll have to familiarize yourself with challenging math formulas that are prevalent in physics. Here are 3 acceleration formulas you must know: ## 1) Calculating the average acceleration from two velocities and two points in time The arguably most widely used acceleration formula is as follows: a = △v / △t Here, △v is the change in velocity and △t is the change in time. Looking at it more closely, the equation can also be written as: a = [v(f) – v(i)] / [t(f) – t(i)] In this acceleration formula, you are calculating the difference in final and initial velocity with final and initial time, where v(f) is the final velocity, v(i) is the initial velocity, t(f) is the final time and t(i) is the initial time. It is crucial to always subtract the initial velocity or time from the final velocity or time, and not the other way around. Switching these can result in an incorrect direction of your acceleration. Bear in mind that it’s still okay if your final time (t(f)) is smaller than your initial time (t(i)) because this just means that you will have a negative acceleration, which translates into “you’re slowing down.” And if, for some reason, you do not have a starting time (initial time), then you can plug in “0” as t(i). ### How to Calculate Acceleration Formulas (Continued) This acceleration formula of change in velocity over change in time can also be applied in the angular acceleration formula. The angular acceleration formula calculates the rate the angular acceleration of a rotating object changes with respect to time. Similar to the average acceleration formula we saw above, this formula is written as: a = (change in angular velocity) / (change in time) ## 2) How to calculate acceleration using Newton’s second law of motion Another way to figure out how to calculate acceleration is through Newton’s second law of motion, which states that an object will accelerate when the forces acting upon it are unbalanced. Acceleration here is contingent on the mass of an object and the net forces acting upon it. In a nutshell, Newton’s law points out that acceleration can be calculated if you know the force that is acting on the object. The acceleration formula is as follows: F(net) = m x a ### How to Calculate Acceleration Formulas (Continued) Here, F(net) represents the total force that is acting on the object, while m represents the mass of the object and a represents the acceleration of the object. To know how to calculate acceleration for Newton’s second law of motion, it’s required to use units in the metric system, which means that you must use newtons (N) for force, kilograms (kg) for mass and meters per second squared (m/s2) for acceleration. 1) To find the mass of the object, you are going to weigh the object in order to identify its mass in grams. Utilizing a scale or a balance is the ideal method, and the larger object you use, the greater the chances are that the mass will be measured in kilograms. To convert the mass of an object into kilograms from grams, you just need to divide the mass by 1000. ### How to Calculate Acceleration Formulas (Continued) 2) To find the net force acting on an object, you will have to look at the unbalanced forces at play. If one force is larger than the other, then the resulting net force will go in the direction of the larger force. In this example, because acceleration occurs when the unbalanced force impacts an object, the object will go in the direction that is more powerful. The units used when calculating the net force is by newton (N). So it’s important to know that 1 newton (N) is equivalent to 1 kilogram (kg) x meter/second squared (m/s2). For example, if a larger dog and a smaller dog are both pulling at the same toy, but the larger dog is pulling with a force of 6 newtons and the smaller dog is pulling with a force of 4 newtons, then the net force will be 2 newtons going in the direction of the larger dog, from where the larger dog has been pulling. 3) Find the acceleration by reordering Newton’s second law of motion. With the acceleration formula of F(net) = m x a, you can calculate for acceleration by dividing the two sides by mass (m). This essentially means that you’ll have to divide the net force by the mass of the object in order to calculate the acceleration, which results in the following equation: a = F(net) / m Some things to keep in mind here are that force is directly proportional to acceleration, while mass does the exact opposite in how it is inversely proportional. This means that with a greater force, you can expect a greater acceleration. And with a greater mass, you can expect a lower (decreasing) acceleration. ## 3) Utilizing distance traveled during acceleration Take a look at the following acceleration formula that uses distance: a = 2 × (Δd − v(i) × Δt) / Δt² Here, like in the aforementioned formulas above, a represents acceleration, v(i) represents the initial velocity, and Δt represents acceleration time and Δd represents the distance traveled during acceleration time. ## Further examples of acceleration How do you calculate acceleration when an object is moving inwards towards the center of a circle, like Earth? Centripetal acceleration. It can change the direction of the velocity, and as a result, the shape of its course, but it does not affect the value of the velocity. Due to the gravity of the sun, the Earth has centripetal acceleration and moves at a constant value, although the speed does change here and there throughout the year. The centripetal acceleration formula calculates the rate of the object’s motion that moves in such a manner. The centripetal acceleration equation is as follows: a(c) = v2/ r In this acceleration formula, you have a(c) which represents acceleration (centripetal), v which represents velocity and r which represents radius. Other examples of acceleration include angular acceleration, mentioned earlier above, and gravitational acceleration, which is the acceleration of an object in free fall within a vacuum. ## Comprehending the key elements of the acceleration formula It’s one thing to memorize these three formulas, and another thing to actually understand them. Some of the most interesting research topics involve understanding acceleration. When you’re learning how to calculate acceleration, take the time you need to understand the following elements: A. The direction of force A force can only cause acceleration in the direction the force is going in. If, for example, you are solving a problem that tells you that a miniature plane with a mass of 20 kg is accelerating south at 4 m/s2 and there is a wind blowing east at 100 Newtons, what is the car’s acceleration? Since the force of the wind blowing against the car is perpendicular to its southbound direction, the miniature plane will continue accelerating south at 4 m/s2. ### How to Calculate Acceleration Formulas (Continued) 1) Net force Always make sure to check if there is more than one force acting on the object, and if there is, then combine the forces into a net force. Forces could be going in the same direction or opposite directions, so you’ll need to calculate the net force to end up with the accurate acceleration. 2) The direction of acceleration It’s important to remember that the direction of acceleration might not correspond to our usual understanding of it in our daily lives. Take for instance the example of driving a car. The direction of acceleration is usually positive if it is moving up or right, and negative if moving down or left. A driver moving right and pressing on the gas pedal will result in a positive direction of acceleration; a driver moving right and pressing on the brakes will result in a less positive velocity, therefore creating a negative direction of acceleration. ## How to Calculate Acceleration Formulas – Tip of the iceberg The more you learn about acceleration, the more you’ll come across formulaic rules that might remind you of certain formulas on SAT/ACT math sections or natural log rules. That’s because these mathematical formulas are foundational to understanding the mechanics of such matters in physics. When you learn how to calculate acceleration, you’ll be stepping into a world, if not worlds, of equations that seek to explain the nature of motion. You may also find the following calculators to be of interest: bookmark bookmark Filter by Category<\|endoftext\|>	4.40625
510	When ancient Britons went for a pig roast, they went big. In a paper published today in , a team of Cardiff University researchers outlined the results of a study of pig bones left near Stonehenge and other similar formations that led them to an important discovery: It seems these late Stone Age pig roasts were big enough to draw people from across the British mainland. The proof was in the isotopes. The researchers looked at stronium, oxygen, nitrogen, carbon, and sulfur isotopes from 131 examples of pig remains found at four sites near Stonehenge. They found the pigs came from all over Britain. Given that pigs aren't exactly easy to herd, this also means that there probably was some ritual aspect to the pig feasts, important enough to drive early Welsh and Scottish settlers back to the south of modern-day England. "They are an absolute gold mine in terms of where the people came from," lead author Richard Madgwick of Cardiff says of the pig bones, which were seemingly "unspectacular" on the outside, according to Madgwick. "They provide a perfect proxy for humans and where humans are moving." Madgwick says that the event probably happened just after the fall harvest, and was a "massive cultural event." Some of the pigs were embedded with flint as if shot by bows and arrows, meaning they "may have been shot in some kind of pre-feasting ritual," he says. These yearly mid-winter feasts were "the way they interacted with each other, and this is the time when they came and worked and played together." The study also points to the importance of centers outside of Stonehenge, showing they may have had as much as an importance in ancient British life. In addition, Madgwick says, Stonehenge itself may have been overemphasized in importance. Madgwick would like to examine sites from across Britain to see if they, too, had a similar cultural import outside of southern England. They may point to a greater-than-believed centralization of power among certain leaders able to mobilize labor for feasts and formation building. The actual purposes of the feast may be hard to disentangle, at least beyond the potential camaraderie. One thing it does prove, though, is that we've had pork belly and ribs on the menu for a long, long time—and people would bring their prized pig from across the country as part of the pan-British ancient barbecue feast.<\|endoftext\|>	3.75
267	# How do you write the nth term rule for the arithmetic sequence with d=5/3 and a_8=24? Jul 22, 2016 ${a}_{n} = \frac{1}{3} \left(32 + 5 n\right)$ #### Explanation: The ${n}^{t h}$ term of an arithmetic sequence in terms of the ${m}^{t h}$ term and the common difference $\left(d\right)$ is given by: ${a}_{n} = {a}_{m} + \left(n - m\right) d$ In this example; $m = 8 , d = \frac{5}{3}$ and ${a}_{8} = 24$ Replacing these values into the general formula above$\to$ ${a}_{n} = {a}_{8} + \left(n - 8\right) \cdot \frac{5}{3}$ ${a}_{n} = 24 + \left(n - 8\right) \cdot \frac{5}{3}$ $= \frac{1}{3} \left(72 + 5 n - 40\right)$ $= \frac{1}{3} \left(32 + 5 n\right)$<\|endoftext\|>	4.53125
695	Counting is the foundation of math. When tested for counting, Chinese children outperform American children by age three. Studies have identified language as a key factor . Arabic numerals are organized in a base-10 structure. To count correctly, it is crucial to recognize and follow the recurring pattern. For example, 2 recurs in every tens, 2,12, 22, … , 92. Children learn counting through words. English words for numbers 11-19 are irregular and disrupt the recurring pattern. To illustrate the disruption, they are compared with Chinese in the Table. For example, in the sequence of two (2), twelve (12), twenty-two (22),…, Ninety-two (92), all words but twelve fit the recurring pattern. The Chinese word for 12 is literally translated as “ten-two;” this aligns with the recurring pattern. The English word twelve disrupts the recurring pattern because it bears no relation to ten and two. The irregularity of English words masks the base-10 structure. Numbers 11 to 19, in particular, form a stumbling block for American children. Besides the base-10 structure, English words lack transparency in place value. For example, in the number 11, each 1 represents a different value depending on its place. The first 1 represents ten and the second 1 represents one. The Chinese language distinguishes the two 1s with the word ten-one, which reveals two distinct values in relation to their places. Eleven in English conveys no place value at all. Place value is critical to counting and arithmetic beyond 10. Most children learn addition through fingers. For example, to add 8 and 4, children often start with 8 fingers and then count “nine, ten, eleven, twelve” continuously. In China, children count “nine, ten, ten-one, ten-two.” The place value breaks a big number beyond 10 into small numbers within 10. There is no cross-cultural difference in arithmetic within 10. Beyond 10, Chinese children gain linguistic advantage in “make-a-ten” strategy, a fundamental skill in arithmetic. For example, 9 + 3 = 9 + (1+2) = (9+1) + 2 = 10 (ten) + 2 (two) = 12 (ten-two) Chinese children connect ten and two to ten-two in words intuitively. American children struggle to make the same connection because twelve does not relate to ten and two in word. The irregularity of those words blocks their intuition and delays their math development for two years . What is the solution? First, teach the numbers 11-19 in words ten-one through ten-nine as shown in the Table. These words ought to be taught at homes, daycare centers, and preschools because children develop number sense by age two. This would free a child's intuition because those regular words restore the recurring pattern and place value. Then, teach words eleven through nineteen as nicknames. This remedy would fit seamlessly into existing school teaching curricula. Spread the words literally, Ten-one (11), Ten-two (12), ... , Ten-nine (19). Kevin F. Miller, Melissa Kelly, Xiaobin Zhou, Learning Mathematics in China and the United States. In J.I.D. Campbell (Ed.) Handbook of Mathematical Cognition.<\|endoftext\|>	4.03125
503	Not enough sediments in the ocean for an old earth (Talk.Origins) At current rates of erosion, only thirty million years are needed to account for all the sediments in the ocean. If the earth were as ancient as is claimed, there should be more sediments. The wording for this claim is not consistent with what Morris says in his book. Among other things he estimates a maximum age of 75 million years not 30 million. (Talk.Origins quotes in blue) 1. The thickness of sediment in the oceans varies, and it is consistent with the age of the ocean floor. The thickness is zero at the mid-Atlantic Ridge, where new ocean crust is forming, and there is about 150 million years' worth of sediment at the continental margins. This pattern of thickness does not necessarily indicate differences in the age of the ocean. The same pattern would be a natural result of the run off from the Global flood. Sediment coming off of the continents in this manner would tend to dump more near the continents and with it tapering off as one got further out into the ocean. The claim that there is "150 million years' worth of sediment at the continental margins," is an estimate of age not depth. This estimation is based on Uniformitarian estimates of the age of the seafloor and as such it is not an independent date. However what Morris shows is that this sediment would have accumulated in only 75 million years at current erosion rates. Simply put, if the Atlantic ocean were really 150 million years old as Talk Origins claims then there should be about twice as much sediment as there actually is. The average age of the ocean floor is younger than the earth due to subduction at some plate margins and formation of new crust at others. Fine, because at issue is the maximum age not the average age. 2. The age of the ocean floor can be determined in various ways -- measured via radiometric dating, estimated from the measured rate of seafloor spreading as a result of plate tectonics, and estimated from the ocean depth that predicted from the sea floor sinking as it cools. All these measurements are consistent, and all fit with sediment thickness. It's no surprise that methods are consistent when they are based on and / or calibrated to one another. For example the above "150 million years' worth of sediment" is clearly not based on erosion rates, but is clearly based on other age estimates of the sea floor.<\|endoftext\|>	4.09375
1,776	<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # 8.1: Exponential Properties Involving Products Difficulty Level: At Grade Created by: CK-12 Estimated9 minsto complete % Progress Practice Exponential Properties Involving Products MEMORY METER This indicates how strong in your memory this concept is Progress Estimated9 minsto complete % Estimated9 minsto complete % MEMORY METER This indicates how strong in your memory this concept is What if you wanted to simplify a mathematical expression containing exponents, like 4342? How would you do so? After completing this Concept, you'll be able to use the product of powers property to simplify exponential expressions like this one. ### Guidance Back in chapter 1, we briefly covered expressions involving exponents, like 35 or x3. In these expressions, the number on the bottom is called the base and the number on top is the power or exponent. The whole expression is equal to the base multiplied by itself a number of times equal to the exponent; in other words, the exponent tells us how many copies of the base number to multiply together. #### Example A Write in exponential form. a) 22 b) (3)(3)(3) c) yyyyy d) (3a)(3a)(3a)(3a) Solution a) 22=22 because we have 2 factors of 2 b) (3)(3)(3)=(3)3 because we have 3 factors of (-3) c) yyyyy=y5 because we have 5 factors of y d) (3a)(3a)(3a)(3a)=(3a)4 because we have 4 factors of 3a When the base is a variable, it’s convenient to leave the expression in exponential form; if we didn’t write x7, we’d have to write xxxxxxx instead. But when the base is a number, we can simplify the expression further than that; for example, 27 equals 2222222, but we can multiply all those 2’s to get 128. Let’s simplify the expressions from Example 1. #### Example B Simplify. a) 22 b) (3)3 c) y5 d) (3a)4 Solution a) 22=22=4 b) (3)3=(3)(3)(3)=27 d) (3a)4=(3a)(3a)(3a)(3a)=3333aaaa=81a4 Be careful when taking powers of negative numbers. Remember these rules: (negative number)(positive number)=negative number(negative number)(negative number)=positive number So even powers of negative numbers are always positive. Since there are an even number of factors, we pair up the negative numbers and all the negatives cancel out. (2)6=(2)(2)(2)(2)(2)(2)=(2)(2)+4(2)(2)+4(2)(2)+4=+64 And odd powers of negative numbers are always negative. Since there are an odd number of factors, we can still pair up negative numbers to get positive numbers, but there will always be one negative factor left over, so the answer is negative: (2)5=(2)(2)(2)(2)(2)=(2)(2)+4(2)(2)+4(2)2=32 Use the Product of Powers Property So what happens when we multiply one power of x by another? Let’s see what happens when we multiply x to the power of 5 by x cubed. To illustrate better, we’ll use the full factored form for each: (xxxxx)x5(xxx)x3=(xxxxxxxx)x8 So x5×x3=x8. You may already see the pattern to multiplying powers, but let’s confirm it with another example. We’ll multiply x squared by x to the power of 4: (xx)x2(xxxx)x4=(xxxxxx)x6 So x2×x4=x6. Look carefully at the powers and how many factors there are in each calculation. 5 x’s times 3 x’s equals (5+3)=8 x’s. 2 x’s times 4 x’s equals (2+4)=6 x’s. You should see that when we take the product of two powers of x, the number of x’s in the answer is the total number of x’s in all the terms you are multiplying. In other words, the exponent in the answer is the sum of the exponents in the product. Product Rule for Exponents: xnxm=x(n+m) There are some easy mistakes you can make with this rule, however. Let’s see how to avoid them. #### Example C Multiply 2223. Solution 2223=25=32 Note that when you use the product rule you don’t multiply the bases. In other words, you must avoid the common error of writing 2223=45. You can see this is true if you multiply out each expression: 4 times 8 is definitely 32, not 1024. #### Example D Multiply 2233. Solution 2233=427=108 In this case, we can’t actually use the product rule at all, because it only applies to terms that have the same base. In a case like this, where the bases are different, we just have to multiply out the numbers by hand—the answer is not 25 or 35 or 65 or anything simple like that. Watch this video for help with the Examples above. ### Vocabulary • An exponent is a power of a number that shows how many times that number is multiplied by itself. An example would be 23. You would multiply 2 by itself 3 times: 2×2×2. The number 2 is the base and the number 3 is the exponent. The value 23 is called the power. • Product Rule for Exponents: xnxm=x(n+m) ### Guided Practice Simplify the following exponents: a. (2)5 b. (10x)2 Solutions: a. (2)5=(2)(2)(2)(2)(2)=32 b. (10x)2=102x2=100x2 ### Practice Write in exponential notation: 1. 44444 2. 3x3x3x 3. (2a)(2a)(2a)(2a) 4. 666xxyyyy 5. 2xy22yx Find each number. 1. 54 2. (2)6 3. (0.1)5 4. (0.6)3 5. (1.2)2+53 6. 32(0.2)3 Multiply and simplify: 1. 6366 2. 222426 3. 3243 4. x2x4 5. (2y4)(3y) 6. (4a2)(3a)(5a4) ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes ### Vocabulary Language: English Base When a value is raised to a power, the value is referred to as the base, and the power is called the exponent. In the expression $32^4$, 32 is the base, and 4 is the exponent. Exponent Exponents are used to describe the number of times that a term is multiplied by itself. Power The "power" refers to the value of the exponent. For example, $3^4$ is "three to the fourth power". 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1,044	Mammals first appeared 215 million years ago during the Triassic period, according to a new study in the journal Nature. Wait, no, they first showed up 175 million years ago during the Jurassic period, said another study in the same issue of the journal. Who’s right? The answer may rest in the middle of your ear. A bit of background: The origin of our mammalian ancestors has long been shrouded by an incomplete fossil record. The first breakthrough occurred during World War II, when German paleontologist Walter Georg Kühne was imprisoned in Britain and filled his hours by examining fossils. He described Morganucodon, an unremarkable shrewlike creature thought to be among the earliest mammals, if not the earliest mammal. Over the ensuing decades, paleontologists discovered a smattering of other proto- and early-mammal fossils that looked and acted a lot like Morganucodon, “omnivore-kind-of-things running around the feet of dinosaurs,” said University of Oregon paleontologist Samantha Hopkins, who was not involved with either Nature study. But one group of fossils baffled the paleo crew — those belonging to the order Haramiyida, a furry band of pseudo-rodents that in some ways seemed like modern mammals, and in others like mammal ancestors. Everything the scientists knew about Haramiyida came from examining scattered teeth and a piece of jawbone. The scattered teeth were crucial, because teeth are the single most important body part for identifying mammal fossils, reflecting much about a species’ diet and lifestyle. “Give me one tooth, and I can probably identify the species,” said Jin Meng, curator of paleontology at the American Museum of Natural History and the senior author on one of the studies. But “if you give me the whole skeleton without the teeth, I will hesitate to say what it is.” The Nature papers deal with two separate fossils from China, dated to 160 to 165 million years ago, that shed light on Haramiyida’s true identity. One fossil included a complete skeleton, and the other was mostly complete, representing a significant upgrade over the handful of teeth previously known. The fossils were discovered in the same rock formation by two independent teams of paleontologists. The group led by Zhe-Xi Luo, senior author of the second study and a paleontologist at the University of Chicago, acquired its fossil — a complete skeleton embedded in shale, albeit crushed and flattened — from the Paleontological Museum of Liaoning in China, which bought it from the community of fossil prospectors who scour remote regions of the country. Naming the creature Megaconus, Luo and his team used CT scans and advanced microscopy to examine as many parts of its anatomy as possible, paying special attention to the teeth and ear bones, which are also vital for identification. They found that the beast was quite the Frankenmammal — it was the size of a large squirrel, walked like an armadillo and had fur, except on its stomach. When their analysis was complete and they had built a family tree of Megaconus and several of its relatives (both fossilized and modern), they determined that the animal belonged to the Haramiyida, and that the enigmatic group preceded the rise of the mammals, believed to have occurred 175 million years ago. The other team, led by Jin Meng, came to a different conclusion: that their creature, dubbed Arboroharamiya because of its tree-dwelling habits, was also a Haramiyidan but that Haramiyidans were, in fact, mammals. Because past research indicates that Haramiyidans arose 215 million years ago, they reasoned further, so too did the mammals — about 40 million years earlier than Luo’s team believes. The discrepancies between the team’s conclusions hang on the anatomy of the middle ear, said the University of Oregon’s Hopkins. Virtually all mammals have a middle ear bone that enables them to amplify sound. In fish and in reptiles, the bone is attached to the jaw, but as those animals evolved and mammals arose, the bone migrated to the middle ear, she said. Both fossil teams were limited by imperfect material (Arboroharamiya was missing its skull), but felt they had enough evidence to know what the ear bones looked like. Luo’s team said that the middle ear bone of Megaconus was still attached to the jaw, showing it wasn't a mammal. Meng’s team said Arboroharamiya’s middle ear bone was where it should be in mammals. Both teams said they were not aware of each other’s research until shortly before the studies went to press. Now that their work is out there, perhaps it’s time for a collaboration? “Put those specimens in a box and drive on over there!” Hopkins implored the groups. Return to Science Now. Follow on Twitter: @BradBalukjian<\|endoftext\|>	4.03125

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