Datasets:
kaizen9
/
midtrain

Dataset card Data Studio Files
xet
Community
text
stringlengths
22
1.01M
# What does it mean when an angle is the included angle? ## What does it mean when an angle is the included angle? An included angle is the angle between two sides of a triangle. It can be any angle of the triangle, depending on its purpose. The included angle is used in proofs of geometric theorems dealing with congruent triangles. What is the difference between an included angle and a non included angle? The “included angle” in SAS is the angle formed by the two sides of the triangle being used. It is the side where the rays of the angles overlap. The “non-included” side in AAS can be either of the two sides that are not directly between the two angles being used. ### What is an example of an included angle? When two lines meet at a common point (vertex) the angle between them is called the included angle. The two lines define the angle. So for example in the figure above we could refer to the angle ∠ABC as the “included angle of BA and BC”. Or we could refer to “BA and BC and their included angle”. How do you work out the included angle? 5.7, a) the included angle = the difference of the two reduced bearings. ∠AOB = difference of bearings OA and OB. (b) If the lines are on the same side of the meridian and in the different quadrants (Fig. 5.7, b), the included angle = 180° – sum of the two reduced bearings. ## What does Included side mean? between two angles included side. • the side between two angles. What is not included angle? Students understand that two sides of a triangle and a 90° angle (or obtuse angle) not included between the. two sides determine a unique triangle. Lesson Notes. A triangle drawn under the condition of two sides and a non-included angle is often thought of as a condition that does not determine a unique triangle. ### What is included angle in SAS? Vocabulary Language: English ▼ English Term Definition Included Angle The included angle in a triangle is the angle between two known sides. SAS SAS means side, angle, side, and refers to the fact that two sides and the included angle of a triangle are known. What is the included angle? The included angle is the fixed SAI with the camber angle included. This is why it’s called the included angle. The larger of the two angles is the included angle and the smaller the SAI angle. ## What are the included angles of a triangle? An included angle is the angle between two line segments or rays. For any triangle, its three interior angles are each included between two sides. Here we have △ NAP, an equilateral triangle: To find the included angles, start with the sides: What does included mean in math? From Latin: includere “to shut in, enclose”. Definition: The angle made by two lines with a common vertex. When two lines meet at a common point (vertex) the angle between them is called the included angle. The two lines define the angle. ### What is included angle of BA and BC? When two lines meet at a common point ( vertex) the angle between them is called the included angle. The two lines define the angle. So for example in the figure above we could refer to the angle ∠ ABC as the “included angle of BA and BC”. Or we could refer to “BA and BC and their included angle”.
##### Praxis Core: 1,001 Practice Questions For Dummies Number line questions on the Praxis Core exam are usually pretty straightforward—they generally involve finding a missing number based on sequences or distances between points on the line. The first practice question is a simple problem (finding a labeled coordinate on a number line based on the surrounding numbers). The second question is a little trickier (finding the coordinate of a point on the line based on its distance from other points). ## Practice questions 1. The distance from one labeled coordinate to the next on the number line is the same in every case. What is the value of y? A. 17.5 B. 18.5 C. 17 D. 16 E. 18 2. For this number line, the distance from P to Q is half the distance from Q to R, and that distance is half the distance from R to S. The coordinate of P is 4, and the coordinate of R is 10. What is the coordinate of S? A. 18 B. 16 C. 20 D. 24 E. 14 1. The correct answer is Choice (E). The distance is the same from one labeled coordinate to the next in every case. Some of the labels indicate that each coordinate is 3 units away from the adjacent ones. The number that's 3 more than 15 and 3 less than 21 is 18. 2. The correct answer is Choice (A). You can figure out the distance from P to S if you determine the distance from P to Q, because the distance from P to S is that distance, plus twice that distance, plus twice that. The distance from P to R is 6 because 10 – 4 = 6. Q is in a position in which its distance from P is half its distance from R. The sum of those distances is 6. Therefore, the distance from P to Q is a number that can be added to twice itself to get 6. That number is 2. 2 + 2(2) = 6 You could use algebra to determine that, but it's not necessary. Because the distance from P to Q is 2, you know that the distance from Q to R is 4 and that the distance from R to S is 8. The coordinate of P is 4, so the coordinate of S is 4 + 2 + 4 + 8, or 18.
# How to Solve Word Problems Involving Money Welcome to this Mometrix lesson on word problems with money. Usually word problems with money will deal with purchasing something, taxing something, or maybe deducting something from a total price. When reading a word problem, it could get overwhelming, seeing as you’re expected to solve a paragraph of text; surely that’s not math, right? If this is your thought process, well, I’ve got some helpful tips for you! For starters, word problems are not all that scary and actually quite easy to maneuver once you know how, and second, yes, math problems can exist in a block of text! Today, we’re going to break down the scary assumptions that come with solving money word problems and learn how to solve them without hesitation! The best way to learn something is to try it for yourself, so let’s start by looking at a problem I’ve created: You and your friends want to order a large pizza for $14.99. You have a budget of:$20. Considering the tax of 4% and the tip at 30%, will you have enough money to buy the large pizza? When starting a word problem, a good rule of thumb is to identify information that is necessary and information that is unnecessary in the problem. Your budget of $20 is unnecessary information starting out, however it will become important once we solve the problem. At this point, we really only need to focus on the price of the pizza and finding out what the tax is. Then we can move on from there. First, we need to determine how much our tax will add onto the bill. To do this, we need to multiply 14.99 with 4%, or .04. (14.99)(.04)=.5996 Next, we need to add .5996, our tax, to 14.99, our total. To keep our number most accurate, we won’t round to the nearest cent until we have reached our final number. So, we have .5996+14.99=15.5896 Now that we know the sum of the original total and our tax, we need to factor in a 30% tip. We need to multiply 15.5896 with 30%, or .30. (15.5896)(.30)=4.67688 Lastly, we need to add 4.67688 with 15.5896 to reach our total number 15.5896+4.67688=20.266. After rounding up, we reach 20.27. The total for your order would be$20.27. You would be $.27 short of being able to order a pizza for you and your friends. I’d say it’s time to start checking the couches for spare change! When we break down word problems into manageable steps, they seem a lot less daunting, right? Next time you’re given a word problem with money to solve, be sure to start with what is known, what is necessary, and what is unnecessary information in the word problem. Identifying these things will help to clear up your understanding of what is being asked in the problem. After identifying the necessary information needed to solve the problem, perform the operations outlined in the question to reach your total(s) and compare your answers. Thanks for joining us for this video lesson on money word problems. If you enjoyed this lesson, be sure to “like” and “subscribe” to our channel for more videos like this one! ## Practice Questions Question #1: All of the items listed below, except for one, are helpful strategies to remember when solving word problems in math. Which item is NOT a helpful strategy? Identify the relevant information that is necessary to solve the problem. Perform the operations that are outlined in the question. Compare and check your final answer to see that it is reasonable. Always assume that every number provided in the text is relevant and useful for solving the problem. Answer: There will often be irrelevant information provided in math word problems. Make sure not to assume that all numbers or information provided in the text is crucial for solving the problem. Question #2: Determine the piece of information that is irrelevant to solving the problem. Jane is purchasing dog food at her local pet store. Jane notices that the store is having a 20% off sale on dog toys. Jane sees that Brand A of dog food costs$45 for a 20 pound bag, and Brand B of dog food costs $49 for a 22 pound bag. Which brand of dog food is less expensive when considering dollars per pound? Brand A costs$45 per pound Brand B costs $49 per pound The store is having a 20% off sale on dog toys All of the information provided is crucial for solving the problem. Answer: The price per pound for Brand A and Brand B is crucial for solving the problem. This information is what will allow us to determine which brand is the better deal. The fact that the store is having a 20% off sale on dog toys is irrelevant to the problem. Question #3: Dianne and Gary order sushi from their favorite restaurant for Gary’s birthday. The bill is$45.50 before the tip. Dianne offers to pay for the entire meal plus tip. She only brought $55 in cash, and she plans on leaving a 20% tip. Gary says that he has$44 in his wallet and he is willing to help with the bill. Does Dianne have enough to pay for the whole meal plus tip, or will Gary need to contribute to the bill? Yes, Dianne will have enough to pay for the entire bill plus tip. No, Dianne will not have enough to pay for the entire bill plus tip. Gary will need to contribute in order to cover the bill. Information that is irrelevant: Gary says that he has $44 in his wallet and he is willing to help with the bill. (This piece of information can be ignored because it is not crucial for solving the problem). The tip can be calculated by multiplying$45.40 by 20% or 0.2. $45.50 × 0.2 =$9.10 tip. This can be added to the bill in order to determine the total cost of the meal. $9.10 +$45.50 = $54.60 Dianne has$55, so she has enough to pay for the entire meal plus tip. Question #4: Eugene spends $24 on toy cars. He purchased 6 cars. His friend Julian spends$28 on toy cars. If each car is the same price, how much did Eugene spend on each car? $3 per car$4 per car $5 per car$6 per car Information that is irrelevant: His friend Julian spends $28 on toy cars. The important information in the text is that Eugene spends$24 on 6 cars. If each car is the same price, then we can divide $24 by 6 in order to determine the price per car.$24 divided by 6 is 4, so $4 per car. Question #5: Jason has a part-time lawn mowing job in the summers. This week Jason mows four lawns on Monday, five lawns on Tuesday, and three lawns on Thursday. He earns$35 per lawn, and he always saves 20% of the money he makes per lawn into his savings account. On Friday Jason spends $12 on ice cream with his friends. How much money did Jason place in his savings account this week?$84 is placed in his savings account. $420 is placed in his savings account.$100 is placed in his savings account. $35 is placed in his savings account. Answer: Information that is irrelevant: On Friday Jason spends$12 on ice cream with his friends. Jason earns $35 per lawn, and he mows 12 lawns this week.$35 times 12 is 420, so he earns $420 this week. 20% of this will be placed in his savings account. 0.2 times$420 is 84, so \$84 is placed in his savings account this week.
MATHS Class-5 NCERT Solutions,Chapter-10: Tenths And Hundredths The chapter Tenths and Hundredths focuses on introducing the idea of decimals to students. It helps the students to understand the concept of the tenth or the hundredth of any given number. It covers the topics: philosophy Measuring small values in millimetres philosophy Relative measurement units philosophyDifferences in sizes philosophyEstimating values NCERT Math-Magic questions are answered in a simple and engaging manner. We have also related 'Learning Concepts and interactive worksheets with the solutions. Our 'Learning Beyond' segment caters to all the probable questions that a child might think out of curiosity. Download Chapter 10 Tenths and Hundredths in PDF format for free here. Tenths and Hundredths Question 1 : a) What is the length of this pencil? __mm. b) What is the length of this pencil in centimetres? a) The length of the pencil is 6 mm. b) The length of the pencil is 0.6 cm. Question 2 : a) What was the length of the smallest pencil you have used? b) How long is this pencil? Guess___ cm c) Measure it using a scale. How good is your guess? a) Do it by yourself. Answers may vary. c) As shown on the ruler, the length of the pencil is 3.6 cm. Check how close your guess was. Frogs Question 1 : a) What does 0.9 cm mean? It is the same as __ millimetres. b) We can also say this is nine-tenths of a cm. Right? c) So, 30.5 cm is the same as __ cm and __ millimetre. d) About how many of the big frogs will fit on the 1 m scale? ___ e) If they sit in a straight line about how many of the small frogs will cover 1 m? ___ a) The relation between cm and mm is: 0.1 cm = 1 mm b) 1 mm is one-tenth of a cm. So, we can say 9 mm is nine-tenth of a cm. c) 30.5 cm is the same as 30 cm and 5 mm. d) The relation between m and cm is: 1 m = 100 cm Since 1 big frog is 30.5 cm long, about 3 big frogs will fit on the 1 m scale. e) Since 1 small frog is 0.9 cm long, about 111 small frogs will cover 1 m. Practice Time Question 1 : Length of the nail — 2 cm and __ mm or 2. __ cm. The length of the nail is 2 cm and 9 mm. Since 1 mm = 0.1 cm, we can write 2 cm and 9 mm also as 2.9 cm. Question 2 : The length of this lady’s finger (bhindi) is __ cm and __ mm. We can also write it as __ cm. The length of the lady’s finger is 8 cm and 4 mm. Since 1 mm = 0.1 cm, we can also write it as 8.4 cm. Question 3 : Guess the lengths to draw these things. Ask your friend to draw the same. After you make the drawing use a scale to measure the length. Whose drawing showed a better guess? Do it by yourself. Answers may vary. Question 4 : How many paise does a matchbox cost? The cost of a matchbox is Rs 0.50. 1 rupee = 100 paise ⇒ 0.50 rupee = 50 paise. Therefore, the cost of a match box is 50 paise. Question 5 : How many matchboxes can be got for Rs 2.50? Rs 0.50 is the cost of 1 matchbox. Therefore, Rs 2.50 is the cost of 2.50 ÷ 0.50 = 5 matchboxes. Thus, 5 matchboxes can be bought for Rs 2.50. Question 6 : How many rupees does the soap cost? The soap cost Rs 8.75. Question 7 : Arun wanted to buy a soap. He has a five-rupee coin, 2 one-rupee coins and 4 half-rupee coins. Write in rupees what money he will get back. Arun has: 1 five-rupee coin = Rs 5 2 one-rupee coins = Rs 2 4 half-rupee coins = Rs 2 Total amount = 5 + 2 + 2 = Rs 9 Since the cost of a soap is Rs 8.75, he will get back 9 – 8.75 = Rs 0.25. Question 8 : The price of two pens is Rs __. The price of one pen is Rs 6.50 ⇒ The price of two pens = 2 × 6.50 = Rs 13 Therefore, she can buy two pens for Rs 13. Question 9 : Match each yellow box with one green and one pink box. Match the rupee with equivalent paise, and the rupee written in decimals. The correct answer is: Question 10 : a) What part of this sheet is coloured blue?__/10 b) What part of the sheet is green? __ c) Which colour covers 0.2 of the sheet? The sheet can be divided into 10 long strips. So, each long strip is 1/10 or 0.1 of the sheet. a) 1 strip is coloured blue. Therefore, part of the sheet is coloured blue. b) Three strips are coloured green. Therefore, or 0.3 part of the sheet is green. c) There are 2 yellow strips. Therefore, the yellow colour covers or 0.2 of the sheet. Question 11 : a) Look at the second sheet. Each strip is divided into 10 equal boxes. How many boxes are there in all? b) Is each box 1/100 part of the sheet? How many blue boxes are there? __ Is blue equal to 10/100 of the sheet? We saw that blue is also equal to 1/10 of the sheet. We wrote it as 0.1 of the sheet. Can we say 10/100 = 1/10 = 0.10 = 0.1? c) Can we write ten paise as 0.1 of a rupee? d) How many boxes are red? What part of the sheet is this? 15/__ Can we also write it as 0.15 of the sheet? e) Now 3/100 of the sheet is black. We can say 0.__ sheet is black. f) How many white boxes are there in the sheet? What part of the second sheet is white? ___ a) Observe the given sheet. Each strip is divided into 10 equal boxes, and there are 10 strips in all. Therefore, the total number of boxes is: 10 × 10 = 100 So, there are 100 boxes in all. b) Since there are 100 equal boxes, each box is 1/100 of the sheet. There are 10 blue boxes. So, the blue portion is 10/100 of the sheet. 10/ 100 = 1/10 = 0.10 = 0.1 c) There are 100 paise in one rupee. Therefore, 10 paise = 10/100 rupee = 1/10 rupee = 0.10 rupee = 0.1 rupee d) There are 15 red boxes. Since the sheet is divided into 100 equal boxes, the red colour is 15/100 of the sheet. 15/100 = 0.15 So, the red colour is 0.15 of the sheet. e) 3/100 of the sheet is black. 3/100 = 0.30 Therefore, 0.30 of the sheet is black. f) There are 22 white boxes. Since there are 100 boxes in all, 22/100 part of the sheet is white. Question 12 : The school at Malappuram has its sports day. The first five children in the Long Jump are: Who is the winner in the long jump? ___ Write the names of the I, II and III winners on this stand. We can also write 1 cm as __m. Observe the given table and arrange the jumps in descending order. 4.50 > 4.05 > 3.50 > 3.35 > 3.05 Therefore, Rehana is the winner, and the names of the I, II and III winners are Rehana, Meena, and Teena, respectively. Since 100 cm = 1 m, 1 cm = 0.01 m Question 13 : What is Dinesh’s height in metres? __ m __ cm.
# What does simplify mean in order of operations? ## What does simplify mean in order of operations? Multiplication and Division — Simplify multiplication and division in the order that they appear from left to right. Addition and Subtraction — Simplify addition and subtraction in the order that they appear from left to right. ## What should be done first to simplify? According to the order of operations, simplify the terms with the exponents first, then multiply, then add. Multiply. ## How do you solve order of operations? First, we solve any operations inside of parentheses or brackets. Second, we solve any exponents. Third, we solve all multiplication and division from left to right. Fourth, we solve all addition and subtraction from left to right. ## How do you simplify order of operations? When simplifying mathematical expressions perform the operations in the following order: 1. Parentheses and other Grouping Symbols. Simplify all expressions inside the parentheses or other grouping symbols, working on the innermost parentheses first. 2. Exponents. 3. Multiplication and Division. ## Do you use Pemdas If there are no parentheses? Without parentheses, PEMDAS rules imply that you must do division first. With parentheses, the 3x now becomes a group. Multiplication technically must occur before division (but you can still do algebraic simplifications, like cancelling a common factor). ## What is the order of operations if there are no parentheses? There are no Parentheses. There are no Exponents. We start with the Multiplication and Division, working from left to right. NOTE: Even though Multiplication comes before Division in PEMDAS, the two are done in the same step, from left to right. ## Are brackets and parentheses the same? Brackets are tall punctuation marks that are used in pairs to separate texts within a sentence while parentheses are types of brackets. 2.In American usage, brackets refer to the box-type brackets while parentheses refer to a different type of punctuation mark. ## How do you simplify like terms? Like terms are combined in algebraic expression so that the result of the expression can be calculated with ease. For example, 7xy + 6y + 6xy is an algebraic equation whose terms are 7xy and 6xy. Therefore, this expression can be simplified by combining like terms as 7xy + 6xy + 6y = 13xy + y.
# Mathematical Procedures Save this PDF as: Size: px Start display at page: ## Transcription 1 CHAPTER 6 Mathematical Procedures 2 168 CHAPTER 6 Mathematical Procedures The multidisciplinary approach to medicine has incorporated a wide variety of mathematical procedures from the fields of physics, chemistry, and engineering. The information presented in this chapter is designed as a self-teaching refresher course to be used as a review of basic mathematical procedures. Some of the more advanced mathematical concepts, including the section on descriptive statistics, should also help the practitioner to interpret data presented in medical journals and scientific articles. Fundamental Axioms Commutative Axiom a + b 5 b + a ab 5 ba When two or more numbers are added or multiplied together, their order does not affect the result. Associative Axiom (a + b) + c 5 a + (b + c) (ab)c 5 a(bc) When three or more numbers are added together, the way they are grouped or associated makes no difference in the result. The same holds true for multiplication. Distributive Axiom a(b + c) 5 ab + ac A coefficient (multiplier) of a sum may be distributed as a multiplier of each term. Order of Precedence A convention has been established for the order in which numerical operations are performed. This is to prevent confusion when evaluating expressions such as , which could be either 18 or 36. The following rules apply: 3 Fractions If the numerical expression does not contain fences (such as parentheses), then operations are carried out in the following order: a. Raising numbers to powers or extracting roots of numbers. b. Multiplication or division. c. Addition or subtraction If the numerical expression does contain fences, then follow the procedure in Rule 1, starting with the innermost set of parentheses. The sequence is round fences (parentheses), square fences [brackets], double fences {braces}. Once the fences have been eliminated, the expression can be evaluated following Rule {3 3 2 [ ( ) 20] + 12} {3 3 2 [ (6 4) 20] + 12} {3 3 2 [ ] + 12} {3 3 2 [ ] + 12} { } { } Fractions When a number is expressed as a fraction (e.g., 3 5), the number above the line (3) is called the numerator and the number below the line (5) the denominator. Multiplication Property of Fractions a 3 c b 3 c 5 a b 1c 2 02 The numerator and denominator of a fraction may be multiplied or divided by the same nonzero number to produce a fraction of equal value. 4 170 CHAPTER 6 Mathematical Procedures Simplify (reduce) the fraction 9 12 Solution 1. Find the largest integer that will evenly divide both the numerator and denominator. 2. Divide both the numerator and denominator by that number. The largest whole number is a9 3 b 4 a12 3 b Multiplication of Fractions a b 3 c d 5 ac bd ? Solution 1. Multiply the numerators Multiply the denominators Simplify the resulting fraction if possible Division of Fractions a b 4 c d 5 a b 3 d c 5 ad bc 5 Fractions 171 Find the quotient: Solution 1. Invert the divisor. Change 2 3 to Multiply the dividend by the inverted divisor Simplify if possible. Addition and Subtraction of Fractions with the Same Denominator a b 1 c 1a 1 c2 5 b b a b 2 c 1a 2 c2 5 b b Solution ? 1. Combine numerators Write the resultant fraction with the new numerator and the same denominator. 3. Simplify if possible 6 172 CHAPTER 6 Mathematical Procedures Addition and Subtraction of Fractions with Different Denominators To add or subtract fractions that do not have the same denominator, it is first necessary to express them as fractions having the same denominators. To find a common denominator, find an integer that is evenly divisible by each denominator. The smallest or least common denominator (LCD) is the most convenient. Solution 1. First find the LCD as follows: ? a. Express each denominator as the product of primes (integers greater than 1 that are evenly divisible by only themselves and 1). b. Note the greatest power to which an integer occurs in any denominator. c. The product of the integers noted in part b is the LCD. 2. Write fractions as equivalent fractions with denominators equal to the LCD. 3. Combine the numerators and use the LCD as the denominator. 4. Simplify if possible. a b. 2 2 is the greatest power of 2 in either denominator, 3 2 is the greatest power of 3 in either denominator. c 7 Ratios, Proportions, and Unit Conversion 173 Ratios, Proportions, and Unit Conversion The ratio of two numbers may be written as follows: a/b 5 a:b Two equivalent ratios form a proportion. a/b 5 c/d a:b 5 c:d a:b :: c:d Regardless of how the above proportions are expressed, they are read a is to b as c is to d. Ratios provide a convenient method for converting units. To change the units of a quantity, multiply by ratios whose values are equal to 1 (which does not change the value of the quantity). Select the dimensions of the ratios such that the unit to be changed occurs as a factor of the numerator or as a factor of the denominator. Thus, when the quantity is multiplied by the ratio, the unit is canceled and replaced by an equivalent unit and quantity. Convert 2 kilometers/hour to feet/second. Solution 1. Write an equation expressing the problem. 2 km/hr 5 x ft/s 2. Multiply the known quantity by ratios whose value is equal to 1, such that the desired unit remains after canceling pairs of equal dimensions that appear in both the numerator and the denominator. 1 km 5 1,000 m 1 m ft 1 hr 5 60 min 1 min 5 60 s 2 km hr 3 1,000 m 1 km 3.3 ft 3 1 m 3 1 hr 60 min 3 1 min 6,600 ft 1.8 ft s 3,600 s s 8 174 CHAPTER 6 Mathematical Procedures Exponents When a product is the result of multiplying a factor by itself several times, it is convenient to use a shorthand notation that shows the number used as a factor (a) and the number of factors (n) in the product. In general, such numbers are expressed in the form a n, where a is the base and n the exponent. The rules for these numbers are shown in Table 6 7. Table 6 7 Rules for Exponents Rule a n? a m 5 a n1m x 2? x 3 5 x 5 a n a m 5 a n2m z 7 z 5 5 z 2 (ab) n 5 a n b n (2wy) 2 5 4w 2 y 2 a 0 5 1, (a? 0) 9x a 1 5 a 3x 1 5 3x a 2n 5 1/a n x /x 2 a 1/n 5 2 n a y 1/ y a m/n 5 2 n a m z 2/ z 2 (a m ) n 5 a mn (w 2 ) 3 5 w 6 Scientific Notation A number expressed as a multiple of a power of 10, such as , is said to be written in scientific notation. Numbers written in this way have two parts: a number between 1 and 10 called the coefficient, multiplied by a power of 10 called the exponent. This notation has three distinct advantages: a. It simplifies the expression of very large or very small numbers that would otherwise require many zeros. For example, 681,000, and b. Scientific notation clarifies the number of significant figures in a large number. For example, if the radius of the earth is written as 6,378,000 m, it is not clear whether any of the zeros after the 8 is significant. However, when the same number is written as m, it is understood that only the first four digits are significant. c. Calculations that involve very large or very small numbers are greatly simplified using scientific notation. 9 Scientific Notation 175 Addition and Subtraction Solution Convert all numbers to the same power of 10 as the number with the highest exponent. 2. Add or subtract the coefficients and retain the same exponent in the answer. Multiplication and Division Solution ( )( ) Multiply (or divide) the coefficients. ( )( ) 5 8( ) 2. Combine the powers of 10 using the rules for exponents. Powers and Roots Solution ( ) 2 8( ) Raise the coefficient to the indicated power. ( ) 2 5 (4 2 )(10 5 ) (10 5 ) 2 2. Multiply the exponent by the indicated power. 16(10 5 ) ( ) 10 176 CHAPTER 6 Mathematical Procedures Significant Figures By convention, the number of digits used to express a measured number roughly indicates the error. For example, if a measurement is reported as 35.2 cm, one would assume that the true length was between and cm (i.e., the error is about 0.05 cm). The last digit (2) in the reported measurement is uncertain, although one can reliably state that it is either 1 or 2. The digit to the right of 2, however, can be any number (5, 6, 7, 8, 9, 0, 1, 2, 3, 4). If the measurement is reported as cm, it would indicate that the error is even less (0.005 cm). The number of reliably known digits in a measurement is the number of significant figures. Thus, the number 35.2 cm has three significant figures, and the number cm has four. The number of significant figures is independent of the decimal point. The numbers 35.2 cm and m are the same quantities, both having three significant figures and expressing the same degree of accuracy. The use of significant figures to indicate the accuracy of a result is not as precise as giving the actual error, but is sufficient for most purposes. Zeros as Significant Figures Final zeros to the right of the decimal point that are used to indicate accuracy are significant: significant figures significant figures significant figures For numbers less than one, zeros between the decimal point and the first digit are not significant: significant figure significant figures Zeros between digits are significant: significant figures significant figures significant figures 11 Significant Figures 177 If a number is written with no decimal point, the final zeros may or may not be significant. For instance, the distance between Earth and the sun might be written as 92,900,000 miles, although the accuracy may be only ±5000 miles. This would make only the first zero after the 9 significant. On the other hand, a value of 50 ml measured with a graduated cylinder would be expected to have two significant figures owing to the greater accuracy of the measurement. To avoid ambiguity, numbers are often written as powers of 10 (scientific notation), making all digits significant. Using this convention, 92,900,000 would be written , indicating that there are four significant figures. Calculations Using Significant Figures The least precise measurement used in a calculation determines the number of significant figures in the answer. Thus, rather than , since the least precise number (73.5) is accurate to only one decimal place. Similarly, , with only one significant figure. For multiplication or division, the rule of thumb is: The product or quotient has the same number of significant figures as the term with the fewest significant figures. As an example, in / , the term with the fewest significant figures is 4.6. Since this number has at most two significant figures, the result should be rounded off to 1.6. Rounding Off The results of mathematical computations are often rounded off to specific numbers of significant figures. This is done so that one does not infer an accuracy in the result that was not present in the measurements. The following rules are universally accepted and will ensure bias-free reporting of results (the number of significant figures desired should be determined first). 1. If the final digits of the number are 1, 2, 3, or 4, they are rounded down (dropped) and the preceding figure is retained unaltered. 2. If the final digits are 6, 7, 8, or 9, they are rounded up (i.e., they are dropped and the preceding figure is increased by one). 3. If the digit to be dropped is a 5, it is rounded down if the preceding figure is even and rounded up if the preceding figure is odd. Thus, 2.45 and 6.15 are rounded off to 2.4 and 6.2, respectively. 12 178 CHAPTER 6 Mathematical Procedures Functions A function is a particular type of relation between groups of numbers. The uniqueness of a function is that each member of one group is associated with exactly one member of another group. In general, let the variable x stand for the values of one group of numbers and the variable y stand for the values of another group. If each value of x is associated with a unique value of y, then this relation is a function. Specifically, y is said to be a function of x and is denoted y 5 f(x). With this notation, x is called the independent variable and y the dependent variable. A function may be represented graphically by using a two-dimensional coordinate (Cartesian) plane formed by two perpendicular axes intersecting each other at a point with coordinates designated as x 5 0, y 5 0 (Fig. 6 1). The vertical axis denotes values of y and the horizontal axis values of x. The function is plotted as a series of points whose coordinates are the values of x with their corresponding values of y as determined by the function. 10 Y axis y = f(x) X axis Figure 6 1 Graphic representation of the function y = ƒ(x), where ƒ(x)= 0.3x + 1. Linear Functions One of the simplest functions is expressed by the formula y 5 ax 13 Functions 179 where y and x are variables a is a constant The constant a is sometimes referred to as the constant of proportionality and y is said to be directly proportional to x (if y is expressed as y 5 a/x, y is said to be inversely proportional to x and the function is no longer linear). The graph of the equation y 5 ax is a straight line. The constant a is the slope of the line. General Linear Equation where y 5 ax + b y 5 dependent variable a 5 slope x 5 independent variable b 5 y-intercept (the value of y at which the graph of the equation crosses the y-axis) Solving Linear Equations To solve a linear equation, 1. Combine similar terms. 2. Use inverse operations to undo remaining additions and subtractions (i.e., add or subtract the same quantities to both sides of the equation). Get all terms with the unknown variable on one side of the equation. 3. If the equation involves fractions, multiply both sides by the least common denominator. 4. If there are multiplications or divisions indicated in the variable term, use inverse operations to find the value of the variable. 5. Check the result by substituting the value into the original equation. 14 180 CHAPTER 6 Mathematical Procedures s x x x 1 2x 3 10x x 1 2x x x 1 2x x 2 5x 2 2x x 5 5x 1 2x x 2 2x a5x 2 2x 3 b 5 3(39) 4. 13x x x 13 15x 2 2x x (9) (9) (9) Quadratic Equations A function of the form y 5 ax 2 is called a quadratic function. It is sometimes expressed in the more general form where a, b, and c are constants 3 y 5 ax 2 + bx + c 15 Quadratic Equations 181 The graph of this equation is a parabola. Frequently, it is of interest to know where the parabola intersects the x-axis. The value of y at any point on the x-axis is zero. Therefore, to find the values of x where the graph intersects the x-axis, the quadratic equation is expressed in standard form: ax 2 + bx + c 5 0 with a? 0. The solution of any quadratic equation expressed in standard form may be found using the quadratic formula: x 5 2b ; 2b2 2 4ac 2a where a, b, and c are the coefficients in the quadratic equation Solution 1. Write the equation in standard form. Solve 3x x 3x x Note the coefficients a, b, and c. a 5 3, b 5 210, c Substitute these values in the quadratic formula: 4. Simplify. x 5 2b ; 2b2 2 4ac 2a x ; x 5 10 ; ; or 1 16 182 CHAPTER 6 Mathematical Procedures Logarithms The logarithm of a number (N) is the exponent (x) to which the base (a) must be raised to produce N. Thus, if a x 5 N then log a N 5 x for a. 0 and a? 0. For example, log (read: the log to the base 2 equals 3) because Logarithms are written as numbers with two parts: an integer, called the characteristic, and a decimal, called the mantissa (e.g., log ). Common Logarithms Common logarithms are those that have the base 10. In this book, the base number will be omitted with the assumption that log means log 10. Table 6 1 shows the general rules of common logarithms. x log x log log log Table 6 1 Rules of Common Logarithms Rule 1. log ab 5 log a + log b x 5 (746)(384) (a. 0, b. 0) log x 5 log log 384 log log log x x 5 286, log 1/a 5 2log a x 5 1/273 (a. 0) log x 5 2log x 17 Logarithms 183 Table 6 1 Rules of Common Logarithms (continued) Rule 3. log a/b 5 log a 2 log b x 5 478/21 (a. 0, b. 0) log x 5 log log 21 log log log x x log a n 5 n log a x (a. 0, n is a real number) 5 (374) 1/3 log x 5 1/3 log 374 log log x 5 1/3(2.5729) x The Characteristic The integer or characteristic of the logarithm of a number is determined by the position of the decimal point in the number. The characteristic of a number can easily be found by expressing the number in scientific notation. Once in this form, the exponent is used as the characteristic. s Number Characteristic Note: The characteristics of logarithms of numbers less than 1 can be written in several ways. Thus, log log (not ) or Written as a negative number (as with handheld calculators) log 18 184 CHAPTER 6 Mathematical Procedures The Mantissa The mantissa is the decimal part of the logarithm of a number. The mantissa of a series of digits is the same regardless of the position of the decimal point. Thus, the logarithms of 1.7, 17, and 170 all have the same mantissa, which is Antilogarithms The number having a given logarithm is called the antilogarithm (antilog). The logarithm of 125 is approximately Therefore, the antilog of is , which is approximately 125. Antilogs of Negative Logarithms Using a calculator, negative logarithms can be solved simply by using the 10 x key. For example, the antilog of 2 is However, log and antilog tables in reference books are used with positive mantissas only. Therefore, a negative logarithm must be changed to a log with a positive mantissa to find its antilog. This change of form is accomplished by first adding and then subtracting 1, which does not alter the original value of the logarithm. Solution 1. Write the log as a negative characteristic minus the mantissa. 2. Subtract 1 from the characteristic and add 1 to the mantissa. Find antilog Express the result as a log having a negative characteristic and a positive mantissa Use the tables to find the antilog. antilog antilog 19 Logarithms 185 Natural Logarithms When a logarithmic function must be differentiated or integrated, it is convenient to rewrite the function with the number e as a base. The number e is approximately equal to Logarithms which have the base e are called natural logarithms and are denoted by ln (read: ell-en ). If e x 5 N, then log e N 5 ln N 5 x Any number of the form a x may be rewritten with e as the base: a x 5 e x ln a Note: e x is sometimes written as exp(x). The same rules apply to natural logarithms that apply to common logarithms. See Table 6 2. Table 6 2 Rules of Natural Logarithms 1. ln ab 5 ln a + ln b (a. 0, b. 0) 2. ln 1/a 5 2 ln a (a. 0) 3. ln a/b 5 ln a 2 ln b (a. 0, b. 0) 4. ln a x 5 x ln a (a. 0, x is a real number) 5. ln e ln e x 5 x 5 e ln x 7. a x 5 e x ln x (a. 0) 8. ln x 5 (ln 10)(log x) (log x) (x. 0) Change of Base Logarithms to one base can easily be changed to logarithms of another base using the following equation. log a x 5 log b x log b a 21 Trigonometry 187 Trigonometric Functions In trigonometry, an angle is considered positive if it is generated by a counterclockwise rotation from standard position, and negative if it is generated by a clockwise rotation (Fig. 6 2). The trigonometric functions of a positive acute angle u can be defined as ratios of the sides of a right triangle: sine of u 5 sin u 5 y/r cosine of u 5 cos u 5 x/r tangent of u 5 tan u 5 y/x cotangent of u 5 cot u 5 x/y secant of u 5 sec u 5 r/x cosecant of u 5 csc u 5 r/y These functions can also be expressed in terms of sine and cosine alone: tan u 5 sin u/cos u cot u 5 cos u/sin u sec u 5 1/cos u csc u 5 1/sin u Y X θ r x y Positive Negative Figure 6 2 An angle is generated by rotating a ray (or half-line) about the origin of a circle. The angle is positive if it is generated by a counterclockwise rotation from the x-axis and negative for a clockwise rotation. 22 188 CHAPTER 6 Mathematical Procedures Basic Trigonometric Identities sin 2 u + cos 2 u 5 1 sec 2 u tan 2 u Probability The probability of an event A is denoted p(a). It is defined as follows: If an event can occur in p number of ways and can fail to occur in q number of ways, then the probability of the event occurring is p/(p + q). The odds in favor of an event occurring are p to q. Addition Rule If A and B are any events, then p(a or B) 5 p(a) + p(b) p(a and B) The probability of drawing either a king or a black card from a deck of 52 playing cards is p(king or black card) 5 p(king) 1 p(black card) 2 p(king also black) 5 4/ /52 2 2/52 5 7/13 Note: If events A and B cannot occur at the same time, they are said to be mutually exclusive, and the addition rule can be simplified to p(a or B) 5 p(a) + p(b) Multiplication Rule If A and B are any events, then p(a and B) 5 p(a B) 3 p(b) where p(a B) 5 the probability of event A given that event B has occurred ### Algebra and Geometry Review (61 topics, no due date) Course Name: Math 112 Credit Exam LA Tech University Course Code: ALEKS Course: Trigonometry Instructor: Course Dates: Course Content: 159 topics Algebra and Geometry Review (61 topics, no due date) Properties ### Mathematics Placement Mathematics Placement The ACT COMPASS math test is a self-adaptive test, which potentially tests students within four different levels of math including pre-algebra, algebra, college algebra, and trigonometry. ### Copy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any. Algebra 2 - Chapter Prerequisites Vocabulary Copy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any. P1 p. 1 1. counting(natural) numbers - {1,2,3,4,...} ### how to use dual base log log slide rules how to use dual base log log slide rules by Professor Maurice L. Hartung The University of Chicago Pickett The World s Most Accurate Slide Rules Pickett, Inc. Pickett Square Santa Barbara, California 93102 ### Understanding Basic Calculus Understanding Basic Calculus S.K. Chung Dedicated to all the people who have helped me in my life. i Preface This book is a revised and expanded version of the lecture notes for Basic Calculus and other ### 1 TRIGONOMETRY. 1.0 Introduction. 1.1 Sum and product formulae. Objectives TRIGONOMETRY Chapter Trigonometry Objectives After studying this chapter you should be able to handle with confidence a wide range of trigonometric identities; be able to express linear combinations of ### MATH 0110 Developmental Math Skills Review, 1 Credit, 3 hours lab MATH 0110 Developmental Math Skills Review, 1 Credit, 3 hours lab MATH 0110 is established to accommodate students desiring non-course based remediation in developmental mathematics. This structure will ### A Second Course in Mathematics Concepts for Elementary Teachers: Theory, Problems, and Solutions A Second Course in Mathematics Concepts for Elementary Teachers: Theory, Problems, and Solutions Marcel B. Finan Arkansas Tech University c All Rights Reserved First Draft February 8, 2006 1 Contents 25 ### 13. Write the decimal approximation of 9,000,001 9,000,000, rounded to three significant æ If 3 + 4 = x, then x = 2 gold bar is a rectangular solid measuring 2 3 4 It is melted down, and three equal cubes are constructed from this gold What is the length of a side of each cube? 3 What is the ### HIBBING COMMUNITY COLLEGE COURSE OUTLINE HIBBING COMMUNITY COLLEGE COURSE OUTLINE COURSE NUMBER & TITLE: - Beginning Algebra CREDITS: 4 (Lec 4 / Lab 0) PREREQUISITES: MATH 0920: Fundamental Mathematics with a grade of C or better, Placement Exam, ### Extra Credit Assignment Lesson plan. The following assignment is optional and can be completed to receive up to 5 points on a previously taken exam. Extra Credit Assignment Lesson plan The following assignment is optional and can be completed to receive up to 5 points on a previously taken exam. The extra credit assignment is to create a typed up lesson ### Exam 1 Sample Question SOLUTIONS. y = 2x Exam Sample Question SOLUTIONS. Eliminate the parameter to find a Cartesian equation for the curve: x e t, y e t. SOLUTION: You might look at the coordinates and notice that If you don t see it, we can ### PURSUITS IN MATHEMATICS often produce elementary functions as solutions that need to be Fast Approximation of the Tangent, Hyperbolic Tangent, Exponential and Logarithmic Functions 2007 Ron Doerfler http://www.myreckonings.com June 27, 2007 Abstract There are some of us who enjoy using our ### Mathematics Review for MS Finance Students Mathematics Review for MS Finance Students Anthony M. Marino Department of Finance and Business Economics Marshall School of Business Lecture 1: Introductory Material Sets The Real Number System Functions, ### Copyrighted Material. Chapter 1 DEGREE OF A CURVE Chapter 1 DEGREE OF A CURVE Road Map The idea of degree is a fundamental concept, which will take us several chapters to explore in depth. We begin by explaining what an algebraic curve is, and offer two ### 2010 Solutions. a + b. a + b 1. (a + b)2 + (b a) 2. (b2 + a 2 ) 2 (a 2 b 2 ) 2 00 Problem If a and b are nonzero real numbers such that a b, compute the value of the expression ( ) ( b a + a a + b b b a + b a ) ( + ) a b b a + b a +. b a a b Answer: 8. Solution: Let s simplify the ### The Australian Curriculum Mathematics The Australian Curriculum Mathematics Mathematics ACARA The Australian Curriculum Number Algebra Number place value Fractions decimals Real numbers Foundation Year Year 1 Year 2 Year 3 Year 4 Year 5 Year ### Connections Across Strands Provides a sampling of connections that can be made across strands, using the theme (integers) as an organizer. Overview Context Connections Positions integers in a larger context and shows connections to everyday situations, careers, and tasks. Identifies relevant manipulatives, technology, and web-based resources ### Section 1.1. Introduction to R n The Calculus of Functions of Several Variables Section. Introduction to R n Calculus is the study of functional relationships and how related quantities change with each other. In your first exposure to ### SPECIAL PRODUCTS AND FACTORS CHAPTER 442 11 CHAPTER TABLE OF CONTENTS 11-1 Factors and Factoring 11-2 Common Monomial Factors 11-3 The Square of a Monomial 11-4 Multiplying the Sum and the Difference of Two Terms 11-5 Factoring the ### AP Physics 1 and 2 Lab Investigations AP Physics 1 and 2 Lab Investigations Student Guide to Data Analysis New York, NY. College Board, Advanced Placement, Advanced Placement Program, AP, AP Central, and the acorn logo are registered trademarks ### Algebra 1 Course Information Course Information Course Description: Students will study patterns, relations, and functions, and focus on the use of mathematical models to understand and analyze quantitative relationships. Through ### Polynomial Operations and Factoring Algebra 1, Quarter 4, Unit 4.1 Polynomial Operations and Factoring Overview Number of instructional days: 15 (1 day = 45 60 minutes) Content to be learned Identify terms, coefficients, and degree of polynomials. ### DRAFT. New York State Testing Program Grade 8 Common Core Mathematics Test. Released Questions with Annotations DRAFT New York State Testing Program Grade 8 Common Core Mathematics Test Released Questions with Annotations August 2014 Developed and published under contract with the New York State Education Department ### Florida Math for College Readiness Core Florida Math for College Readiness Florida Math for College Readiness provides a fourth-year math curriculum focused on developing the mastery of skills identified as critical to postsecondary readiness ### 2009 Chicago Area All-Star Math Team Tryouts Solutions 1. 2009 Chicago Area All-Star Math Team Tryouts Solutions If a car sells for q 1000 and the salesman earns q% = q/100, he earns 10q 2. He earns an additional 100 per car, and he sells p cars, so his total ### SECTION 0.6: POLYNOMIAL, RATIONAL, AND ALGEBRAIC EXPRESSIONS (Section 0.6: Polynomial, Rational, and Algebraic Expressions) 0.6.1 SECTION 0.6: POLYNOMIAL, RATIONAL, AND ALGEBRAIC EXPRESSIONS LEARNING OBJECTIVES Be able to identify polynomial, rational, and algebraic ### Polynomials. Dr. philippe B. laval Kennesaw State University. April 3, 2005 Polynomials Dr. philippe B. laval Kennesaw State University April 3, 2005 Abstract Handout on polynomials. The following topics are covered: Polynomial Functions End behavior Extrema Polynomial Division ### Chapter 7 Outline Math 236 Spring 2001 Chapter 7 Outline Math 236 Spring 2001 Note 1: Be sure to read the Disclaimer on Chapter Outlines! I cannot be responsible for misfortunes that may happen to you if you do not. Note 2: Section 7.9 will ### Algebra I Credit Recovery Algebra I Credit Recovery COURSE DESCRIPTION: The purpose of this course is to allow the student to gain mastery in working with and evaluating mathematical expressions, equations, graphs, and other topics, ### 04 Mathematics CO-SG-FLD004-03. Program for Licensing Assessments for Colorado Educators 04 Mathematics CO-SG-FLD004-03 Program for Licensing Assessments for Colorado Educators Readers should be advised that this study guide, including many of the excerpts used herein, is protected by federal ### L 2 : x = s + 1, y = s, z = 4s + 4. 3. Suppose that C has coordinates (x, y, z). Then from the vector equality AC = BD, one has The line L through the points A and B is parallel to the vector AB = 3, 2, and has parametric equations x = 3t + 2, y = 2t +, z = t Therefore, the intersection point of the line with the plane should satisfy: ### Algebra Unpacked Content For the new Common Core standards that will be effective in all North Carolina schools in the 2012-13 school year. This document is designed to help North Carolina educators teach the Common Core (Standard Course of Study). NCDPI staff are continually updating and improving these tools to better serve teachers. Algebra ### ACCUPLACER. Testing & Study Guide. Prepared by the Admissions Office Staff and General Education Faculty Draft: January 2011 ACCUPLACER Testing & Study Guide Prepared by the Admissions Office Staff and General Education Faculty Draft: January 2011 Thank you to Johnston Community College staff for giving permission to revise ### Objectives After completing this section, you should be able to: Chapter 5 Section 1 Lesson Angle Measure Objectives After completing this section, you should be able to: Use the most common conventions to position and measure angles on the plane. Demonstrate an understanding ### Algebra II and Trigonometry Algebra II and Trigonometry Textbooks: Algebra 2: California Publisher: McDougal Li@ell/Houghton Mifflin (2006 EdiHon) ISBN- 13: 978-0618811816 Course descriphon: Algebra II complements and expands the ### Find the length of the arc on a circle of radius r intercepted by a central angle θ. Round to two decimal places. SECTION.1 Simplify. 1. 7π π. 5π 6 + π Find the measure of the angle in degrees between the hour hand and the minute hand of a clock at the time shown. Measure the angle in the clockwise direction.. 1:0. ### (1.) The air speed of an airplane is 380 km/hr at a bearing of. Find the ground speed of the airplane as well as its (1.) The air speed of an airplane is 380 km/hr at a bearing of 78 o. The speed of the wind is 20 km/hr heading due south. Find the ground speed of the airplane as well as its direction. Here is the diagram: ### SECTION 2.5: FINDING ZEROS OF POLYNOMIAL FUNCTIONS SECTION 2.5: FINDING ZEROS OF POLYNOMIAL FUNCTIONS Assume f ( x) is a nonconstant polynomial with real coefficients written in standard form. PART A: TECHNIQUES WE HAVE ALREADY SEEN Refer to: Notes 1.31 ### Chapter 111. Texas Essential Knowledge and Skills for Mathematics. Subchapter B. Middle School Middle School 111.B. Chapter 111. Texas Essential Knowledge and Skills for Mathematics Subchapter B. Middle School Statutory Authority: The provisions of this Subchapter B issued under the Texas Education ### DEFINITION 5.1.1 A complex number is a matrix of the form. x y. , y x Chapter 5 COMPLEX NUMBERS 5.1 Constructing the complex numbers One way of introducing the field C of complex numbers is via the arithmetic of matrices. DEFINITION 5.1.1 A complex number is a matrix of ### A Quick Algebra Review 1. Simplifying Epressions. Solving Equations 3. Problem Solving 4. Inequalities 5. Absolute Values 6. Linear Equations 7. Systems of Equations 8. Laws of Eponents 9. Quadratics 10. Rationals 11. Radicals ### Display Format To change the exponential display format, press the [MODE] key 3 times. Tools FX 300 MS Calculator Overhead OH 300 MS Handouts Other materials Applicable activities Activities for the Classroom FX-300 Scientific Calculator Quick Reference Guide (inside the calculator cover) ### Examples of Tasks from CCSS Edition Course 3, Unit 5 Examples of Tasks from CCSS Edition Course 3, Unit 5 Getting Started The tasks below are selected with the intent of presenting key ideas and skills. Not every answer is complete, so that teachers can ### Content. Chapter 4 Functions 61 4.1 Basic concepts on real functions 62. Credits 11 Content Credits 11 Chapter 1 Arithmetic Refresher 13 1.1 Algebra 14 Real Numbers 14 Real Polynomials 19 1.2 Equations in one variable 21 Linear Equations 21 Quadratic Equations 22 1.3 Exercises 28 Chapter ### 2.3. Finding polynomial functions. An Introduction: 2.3. Finding polynomial functions. An Introduction: As is usually the case when learning a new concept in mathematics, the new concept is the reverse of the previous one. Remember how you first learned ### Transition To College Mathematics Transition To College Mathematics In Support of Kentucky s College and Career Readiness Program Northern Kentucky University Kentucky Online Testing (KYOTE) Group Steve Newman Mike Waters Janis Broering ### FSCJ PERT. Florida State College at Jacksonville. assessment. and Certification Centers FSCJ Florida State College at Jacksonville Assessment and Certification Centers PERT Postsecondary Education Readiness Test Study Guide for Mathematics Note: Pages through are a basic review. Pages forward ### LAKE ELSINORE UNIFIED SCHOOL DISTRICT LAKE ELSINORE UNIFIED SCHOOL DISTRICT Title: PLATO Algebra 1-Semester 2 Grade Level: 10-12 Department: Mathematics Credit: 5 Prerequisite: Letter grade of F and/or N/C in Algebra 1, Semester 2 Course Description: ### FACTORING POLYNOMIALS 296 (5-40) Chapter 5 Exponents and Polynomials where a 2 is the area of the square base, b 2 is the area of the square top, and H is the distance from the base to the top. Find the volume of a truncated ### In this section, you will develop a method to change a quadratic equation written as a sum into its product form (also called its factored form). CHAPTER 8 In Chapter 4, you used a web to organize the connections you found between each of the different representations of lines. These connections enabled you to use any representation (such as a graph, ### Polynomial Degree and Finite Differences CONDENSED LESSON 7.1 Polynomial Degree and Finite Differences In this lesson you will learn the terminology associated with polynomials use the finite differences method to determine the degree of a polynomial ### Secondary Mathematics Syllabuses Secondary Mathematics Syllabuses Copyright 006 Curriculum Planning and Development Division. This publication is not for sale. All rights reserved. No part of this publication may be reproduced without ### Functional Math II. Information CourseTitle. Types of Instruction Functional Math II Course Outcome Summary Riverdale School District Information CourseTitle Functional Math II Credits 0 Contact Hours 135 Instructional Area Middle School Instructional Level 8th Grade ### MATD 0390 - Intermediate Algebra Review for Pretest MATD 090 - Intermediate Algebra Review for Pretest. Evaluate: a) - b) - c) (-) d) 0. Evaluate: [ - ( - )]. Evaluate: - -(-7) + (-8). Evaluate: - - + [6 - ( - 9)]. Simplify: [x - (x - )] 6. Solve: -(x + ### Course Outlines. 1. Name of the Course: Algebra I (Standard, College Prep, Honors) Course Description: ALGEBRA I STANDARD (1 Credit) Course Outlines 1. Name of the Course: Algebra I (Standard, College Prep, Honors) Course Description: ALGEBRA I STANDARD (1 Credit) This course will cover Algebra I concepts such as algebra as a language, ### Factoring and Applications Factoring and Applications What is a factor? The Greatest Common Factor (GCF) To factor a number means to write it as a product (multiplication). Therefore, in the problem 48 3, 4 and 8 are called the ### The Greatest Common Factor; Factoring by Grouping 296 CHAPTER 5 Factoring and Applications 5.1 The Greatest Common Factor; Factoring by Grouping OBJECTIVES 1 Find the greatest common factor of a list of terms. 2 Factor out the greatest common factor. ### 9 MATRICES AND TRANSFORMATIONS 9 MATRICES AND TRANSFORMATIONS Chapter 9 Matrices and Transformations Objectives After studying this chapter you should be able to handle matrix (and vector) algebra with confidence, and understand the ### averages simple arithmetic average (arithmetic mean) 28 29 weighted average (weighted arithmetic mean) 32 33 537 A accumulated value 298 future value of a constant-growth annuity future value of a deferred annuity 409 future value of a general annuity due 371 future value of an ordinary general annuity 360 future ### SECTION P.5 Factoring Polynomials BLITMCPB.QXP.0599_48-74 /0/0 0:4 AM Page 48 48 Chapter P Prerequisites: Fundamental Concepts of Algebra Technology Eercises Critical Thinking Eercises 98. The common cold is caused by a rhinovirus. The ### Licensed to: Printed in the United States of America 1 2 3 4 5 6 7 16 15 14 13 12 Licensed to: CengageBrain User This is an electronic version of the print textbook. Due to electronic rights restrictions, some third party content may be suppressed. Editorial review has deemed that any ### David Bressoud Macalester College, St. Paul, MN. NCTM Annual Mee,ng Washington, DC April 23, 2009 David Bressoud Macalester College, St. Paul, MN These slides are available at www.macalester.edu/~bressoud/talks NCTM Annual Mee,ng Washington, DC April 23, 2009 The task of the educator is to make the ### Blue Pelican Alg II First Semester Blue Pelican Alg II First Semester Teacher Version 1.01 Copyright 2009 by Charles E. Cook; Refugio, Tx (All rights reserved) Alg II Syllabus (First Semester) Unit 1: Solving linear equations and inequalities ### DOE FUNDAMENTALS HANDBOOK MATHEMATICS Volume 2 of 2 DOE-HDBK-1014/2-92 JUNE 1992 DOE FUNDAMENTALS HANDBOOK MATHEMATICS Volume 2 of 2 U.S. Department of Energy Washington, D.C. 20585 FSC-6910 Distribution Statement A. Approved for public release; distribution ### MATH 90 CHAPTER 1 Name:. MATH 90 CHAPTER 1 Name:. 1.1 Introduction to Algebra Need To Know What are Algebraic Expressions? Translating Expressions Equations What is Algebra? They say the only thing that stays the same is change. ### correct-choice plot f(x) and draw an approximate tangent line at x = a and use geometry to estimate its slope comment The choices were: Topic 1 2.1 mode MultipleSelection text How can we approximate the slope of the tangent line to f(x) at a point x = a? This is a Multiple selection question, so you need to check all of the answers that ### YOU CAN COUNT ON NUMBER LINES Key Idea 2 Number and Numeration: Students use number sense and numeration to develop an understanding of multiple uses of numbers in the real world, the use of numbers to communicate mathematically, and ### Algebra 1 If you are okay with that placement then you have no further action to take Algebra 1 Portion of the Math Placement Test Dear Parents, Based on the results of the High School Placement Test (HSPT), your child should forecast to take Algebra 1 this fall. If you are okay with that placement then you have no further action ### 6.1 The Greatest Common Factor; Factoring by Grouping 386 CHAPTER 6 Factoring and Applications 6.1 The Greatest Common Factor; Factoring by Grouping OBJECTIVES 1 Find the greatest common factor of a list of terms. 2 Factor out the greatest common factor. ### Determinants can be used to solve a linear system of equations using Cramer s Rule. 2.6.2 Cramer s Rule Determinants can be used to solve a linear system of equations using Cramer s Rule. Cramer s Rule for Two Equations in Two Variables Given the system This system has the unique solution ### SECTION 1-6 Quadratic Equations and Applications 58 Equations and Inequalities Supply the reasons in the proofs for the theorems stated in Problems 65 and 66. 65. Theorem: The complex numbers are commutative under addition. Proof: Let a bi and c di be ### Calculus. Contents. Paul Sutcliffe. Office: CM212a. Calculus Paul Sutcliffe Office: CM212a. www.maths.dur.ac.uk/~dma0pms/calc/calc.html Books One and several variables calculus, Salas, Hille & Etgen. Calculus, Spivak. Mathematical methods in the physical ### BX in ( u, v) basis in two ways. On the one hand, AN = u+ 1. Let f(x) = 1 x +1. Find f (6) () (the value of the sixth derivative of the function f(x) at zero). Answer: 7. We expand the given function into a Taylor series at the point x = : f(x) = 1 x + x 4 x ### An important theme in this book is to give constructive definitions of mathematical objects. Thus, for instance, if you needed to evaluate. Chapter 10 Series and Approximations An important theme in this book is to give constructive definitions of mathematical objects. Thus, for instance, if you needed to evaluate 1 0 e x2 dx, you could set ### POLYNOMIALS and FACTORING POLYNOMIALS and FACTORING Exponents ( days); 1. Evaluate exponential expressions. Use the product rule for exponents, 1. How do you remember the rules for exponents?. How do you decide which rule to use ### The following table lists metric prefixes that come up frequently in physics. Learning these prefixes will help you in the various exercises. Chapter 0 Solutions Circles Learning Goal: To calculate the circumference or area of a circle Every day, we see circles in compact disks, coins, and wheels, just to name a few examples Circles are also ### Polynomials and Factoring Lesson 2 Polynomials and Factoring A polynomial function is a power function or the sum of two or more power functions, each of which has a nonnegative integer power. Because polynomial functions are built ### Interpretation of Test Scores for the ACCUPLACER Tests Interpretation of Test Scores for the ACCUPLACER Tests ACCUPLACER is a trademark owned by the College Entrance Examination Board. Visit The College Board on the Web at: www.collegeboard.com/accuplacer ### Such As Statements, Kindergarten Grade 8 Such As Statements, Kindergarten Grade 8 This document contains the such as statements that were included in the review committees final recommendations for revisions to the mathematics Texas Essential ### 8th Grade Texas Mathematics: Unpacked Content 8th Grade Texas Mathematics: Unpacked Content What is the purpose of this document? To increase student achievement by ensuring educators understand specifically what the new standards mean a student must ### Algebra Isn t Hard or Demystifying Math: A Gentle Approach to Algebra. by Peter J. Hutnick Algebra Isn t Hard or Demystifying Math: A Gentle Approach to Algebra by Peter J. Hutnick 17th July 2004 2 Contents I Fundamentals of Algebra 17 1 Arithmetic 19 1.1 Vocabulary............................... ### Mathematics: Content Knowledge The Praxis Study Companion Mathematics: Content Knowledge 5161 www.ets.org/praxis Welcome to the Praxis Study Companion Welcome to the Praxis Study Companion Prepare to Show What You Know You have been ### Inversion. Chapter 7. 7.1 Constructing The Inverse of a Point: If P is inside the circle of inversion: (See Figure 7.1) Chapter 7 Inversion Goal: In this chapter we define inversion, give constructions for inverses of points both inside and outside the circle of inversion, and show how inversion could be done using Geometer ### December 4, 2013 MATH 171 BASIC LINEAR ALGEBRA B. KITCHENS December 4, 2013 MATH 171 BASIC LINEAR ALGEBRA B KITCHENS The equation 1 Lines in two-dimensional space (1) 2x y = 3 describes a line in two-dimensional space The coefficients of x and y in the equation ### College Algebra. Barnett, Raymond A., Michael R. Ziegler, and Karl E. Byleen. College Algebra, 8th edition, McGraw-Hill, 2008, ISBN: 978-0-07-286738-1 College Algebra Course Text Barnett, Raymond A., Michael R. Ziegler, and Karl E. Byleen. College Algebra, 8th edition, McGraw-Hill, 2008, ISBN: 978-0-07-286738-1 Course Description This course provides ### Microsoft Mathematics for Educators: Microsoft Mathematics for Educators: Familiarize yourself with the interface When you first open Microsoft Mathematics, you ll see the following elements displayed: 1. The Calculator Pad which includes ### Definitions 1. A factor of integer is an integer that will divide the given integer evenly (with no remainder). Math 50, Chapter 8 (Page 1 of 20) 8.1 Common Factors Definitions 1. A factor of integer is an integer that will divide the given integer evenly (with no remainder). Find all the factors of a. 44 b. 32 ### For example, estimate the population of the United States as 3 times 10⁸ and the CCSS: Mathematics The Number System CCSS: Grade 8 8.NS.A. Know that there are numbers that are not rational, and approximate them by rational numbers. 8.NS.A.1. Understand informally that every number Polynomials and Quadratics Want to be an environmental scientist? Better be ready to get your hands dirty!.1 Controlling the Population Adding and Subtracting Polynomials............703.2 They re Multiplying ### To define function and introduce operations on the set of functions. To investigate which of the field properties hold in the set of functions Chapter 7 Functions This unit defines and investigates functions as algebraic objects. First, we define functions and discuss various means of representing them. Then we introduce operations on functions ### 2After completing this chapter you should be able to After completing this chapter you should be able to solve problems involving motion in a straight line with constant acceleration model an object moving vertically under gravity understand distance time ### Factoring a Difference of Two Squares. Factoring a Difference of Two Squares 284 (6 8) Chapter 6 Factoring 87. Tomato soup. The amount of metal S (in square inches) that it takes to make a can for tomato soup is a function of the radius r and height h: S 2 r 2 2 rh a) Rewrite this ### Common Core Standards Practice Week 8 Common Core Standards Practice Week 8 Selected Response 1. Describe the end behavior of the polynomial f(x) 5 x 8 8x 1 6x. A down and down B down and up C up and down D up and up Constructed Response. ### CENTRAL TEXAS COLLEGE SYLLABUS FOR DSMA 0306 INTRODUCTORY ALGEBRA. Semester Hours Credit: 3 CENTRAL TEXAS COLLEGE SYLLABUS FOR DSMA 0306 INTRODUCTORY ALGEBRA Semester Hours Credit: 3 (This course is equivalent to DSMA 0301. The difference being that this course is offered only on those campuses ### Indiana University Purdue University Indianapolis. Marvin L. Bittinger. Indiana University Purdue University Indianapolis. Judith A. STUDENT S SOLUTIONS MANUAL JUDITH A. PENNA Indiana Universit Purdue Universit Indianapolis COLLEGE ALGEBRA: GRAPHS AND MODELS FIFTH EDITION Marvin L. Bittinger Indiana Universit Purdue Universit Indianapolis
# Factor Theory of Quadratic Equation Suppose when we assume that β be a root of the quadratic equation ax$$^{2}$$ + bx + c = 0, then we get (x - β) is a factor of the quadratic expression ax^2 + bx + c. Conversely, when we assume that (x - β) is a factor of the quadratic expression ax$$^{2}$$ + bx + c then β is a root of the quadratic equation ax$$^{2}$$ + bx + c = 0. Proof: The given quadratic equation ax$$^{2}$$ + bx + c = 0 According to the problem, β is a root of the quadratic equation ax$$^{2}$$ + bx + c = 0 Hence, aβ$$^{2}$$ + bβ + c = 0 Now, ax$$^{2}$$ + bx + c = ax$$^{2}$$ + bx + c - (aβ$$^{2}$$ + bβ + c), [Since, aβ$$^{2}$$ + bβ + c = 0] = a(x$$^{2}$$ - β$$^{2}$$) + b(x - β) = (x - β)[a(x + β) + b] Therefore, we clearly see that (x - β) is a factor of the quadratic expression ax$$^{2}$$ + bx + c. Conversely, when (x - α) is a factor of the quadratic expression ax$$^{2}$$ + bx + c then we can express, ax$$^{2}$$ + bx + c = (x - α)(mx + n), where m (≠ 0) and n are constants. Now, we need to substitute x = β on both sides of the identity ax$$^{2}$$ + bx + c = (x - β)(mx + n) then we get, aβ$$^{2}$$ + bβ + c = (β - β) × (mβ + n) = 0 aβ$$^{2}$$ + bβ + c = 0 × (mβ + n) = 0 It is evident that the equation ax$$^{2}$$ + bx + c = 0 is satisfied by x = β. Therefore, β is a root of the equation ax$$^{2}$$ + bx + c = 0. Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need. ## Recent Articles 1. ### Types of Fractions |Proper Fraction |Improper Fraction |Mixed Fraction Mar 02, 24 05:31 PM The three types of fractions are : Proper fraction, Improper fraction, Mixed fraction, Proper fraction: Fractions whose numerators are less than the denominators are called proper fractions. (Numerato… 2. ### Subtraction of Fractions having the Same Denominator | Like Fractions Mar 02, 24 04:36 PM To find the difference between like fractions we subtract the smaller numerator from the greater numerator. In subtraction of fractions having the same denominator, we just need to subtract the numera… 3. ### Addition of Like Fractions | Examples | Worksheet | Answer | Fractions Mar 02, 24 03:32 PM To add two or more like fractions we simplify add their numerators. The denominator remains same. Thus, to add the fractions with the same denominator, we simply add their numerators and write the com… 4. ### Comparison of Unlike Fractions | Compare Unlike Fractions | Examples Mar 01, 24 01:42 PM In comparison of unlike fractions, we change the unlike fractions to like fractions and then compare. To compare two fractions with different numerators and different denominators, we multiply by a nu…
# What is the difference between LCM and HCF and how to calculate these properly The highest common factor and least common multiple are such concepts that are important not only for the school-level but also for several exams. These exams might be MAT, CAT, and different recruitment exams. So, it is important to understand the basic concepts and properties of these two properly. In this article, we’re going to discuss what is LCM and HCF, their properties & differences, and how to calculate these properly with suitable examples. ## What are LCM and HCF? LCM is the smallest whole number that occurs in both of their time’s tables and it is the smallest integer that is a multiple of both numbers. It is also known as the lowest common multiple and shows the least number divisible by 2 or more numbers. For example, suppose two numbers 4 and 5. The common multiples of 4 are: 4, 8, 12, 16, 20, 24, 28 And of 5 are: 5, 10, 15, 20, 25. The LCM of two different numbers 4 & 5 is 20. Whereas, HCF is the highest common factor between different numbers. It is the largest number that divides two or more than two numbers. It is also known as the greatest common factor and can be found by several new methods. For example, suppose two numbers ‘30’ and ‘36’. The factors for 30 is (2 x 3 x 5) and for 36 (2 x 2 x 3 x 3). The HCF of these two numbers is 6 because it is the biggest number that divides each of these numbers. ## Difference between HCF and LCM LCM is the least common multiple and HCF is the highest common factor between different numbers. The main purpose of showing the difference between these two numbers is to show the difference between a factor and a multiple. A multiple of a whole number is an integer that occurs in the time table. For example, let’s have a look at the multiples of 3: 3, 6, 9, 12, 15, 18, and so on. Whereas, the factor of an integer is that number that divides the integer with no reminder at the end. For example, let’s have a look at the factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, and 36. And to find the relation between the LCM and HCF of two different numbers first, we need to find the highest common factor of 15 and 18 which is 3. And the exact LCM of 15 and 18 is 90. L.C.M x H.C.F = 90 X 3 = 270 Whereas the product of these two numbers is, 15 x 18 = 270 Thus, the product of L.C.M and H.C.F is equal to the product of these two numbers. ## Properties of LCM and HCF Some of the main properties of these two concepts are: • The highest common factor of two or more than two prime numbers is always 1. • The LCM of two or more than two numbers will be their products. • The HCF of the given numbers is never greater than the given numbers. • The least common multiple of the given numbers is never smaller than any of the given numbers • The product of two different numbers ‘a’ and ‘b’ is always equal to the product of their least common multiples and HCF. ## How to Calculate HCF and LCM properly? You can use the mentioned below ways to calculate the LCM or HCF of two numbers: ### ⦁ Prime Factorization Method for LCM and HCF To calculate the least common factor of random two numbers 45 and 60 simply follow: • List the prime factors of the given numbers as: 45 = 2 x 2 x 3 x 5 60 = 3 x 3 x 5 • Multiply each factor that occurs for the maximum times LCM = 2 x 2 x 3 x 3 x 5 = 180. And to calculate the HCF by the prime factorization method take two random numbers 104 and 144. The prime factors of these two numbers are: 104 = 23 x 13 144 = 2 x 2 x 2 x 2 x 3 x 3 The most common factors of these two numbers are 2 x 2 x 2. So, the HCF of the given numbers is: HCF = 2 x 2 x 2 = 8. ### ⦁ Division Method for LCM and HCF To calculate the least common multiple “LCM” of two different numbers 24 and 15, follow the mentioned below steps: • Divide the given number by the lowest prime number • Write number and quotient which are not divisible the least prime number • Further, divide these numbers with the other prime number • Keep the division number until the reminder becomes the prime number or 1. • Multiply the divisors and other prime numbers to get the LCM. And the LCM of the two numbers 24 and 15 is: LCM of these two numbers is: LCM = 2 x 2 x 2 x 3 x 5 = 120. To calculate HCF by division method, simply follow the mentioned below steps: • Divide the larger number by the smaller one • Further, divide the divisor of the above one by the reminder • Keep dividing the divisor of step 2 with the reminder until the remainder becomes zero. For example, suppose two different numbers 12 and 18. The greatest common factor of these numbers is: The HCF of these two numbers by the division method is 6. ### ⦁ Using Online Calculators Several useful online calculators are available to find the greatest common factor. These converters allow you to check the HCF, LCM, or GCD of different numbers just within a single click. These online calculators also allow you to select the desired method to find the LCM or HCF. Most of the tools provide the following methods to check the highest common factor and least common multiple: • Prime Factorization Method • Division Method • Euclidean Method • Binary stein’s Algorithm • And more 36% OFF X ## How Does It Work? ### 1. Find eBooks Locate the eBook you wish to purchase by searching for the test or title. ### 3. Checkout Complete the quick and easy checkout process. ## Why Buy eBook From Effortlessmath? Save up to 70% compared to print Help save the environment
# Vertical Subtraction Vertical subtraction of 1-digit number are done by arranging the numbers column wise i.e., one number under the other number. How to subtract 1-digit number vertically? 1. Here we subtracted 1 from 4? (Follow the steps along with the above picture) (i) First we will write the numbers column wise or vertically as we generally do subtraction. (ii) Beside the first number we will put four strokes since the first number is 4. (////) (iii) Since the second number is 1 we take off or cancel out one stroke from the four stokes. (iv) The remaining stokes are 3. 2. Subtract 2 from 5 vertically. 3. Subtract the following vertically. (i) Subtract 5 from 8. 8           -     5_        _______ ●●●●●●●● (ii) Subtract 3 from 6. 6           -     3_        _______ ●●●●●● (iii) Subtract 4 from 5. 5           -     4_        _______ ●●●●● (iv) Subtract 3 from 6. 6           -     2_        _______ ●●●●●● (v) Subtract 4 from 5. 5           -     3_        _______ ●●●●● (vi) Subtract 3 from 6. 7           -     5_        _______ ●●●●●●● 3. (i) 3 (ii) 3 (iii) 1 (iv) 4 (v) 2 (vi) 2 4. Fill in the blanks. (i) 4           -    3        _______ (ii) 5           -    4        _______ (iii) 6           -    5        _______ (iv) 7           -    6        _______ (v) 8           -    7        _______ (vi) 5           -    2        _______ (vii) 7           -    2        _______ (viii) 6           -    3        _______ (ix) 8           -    3        _______ (x) 4           -    1        _______ (xi) 8           -    4        _______ (xii) 9           -    7        _______ (xiii) 7           -    3        _______ (xiv) 6           -    4        _______ 4. (i) 1 (ii) 1 (iii) 1 (iv) 1 (v) 1 (vi) 3 (vii) 5 (viii) 3 (ix) 5 (x) 3 (xi) 4 (xii) 2 (xiii) 4 (xiv) 2 5. Five (5) birds were on the tree. One (1) flew away in search of food. How many are left? Solution: From 5 take away 1. 4 is left. 5 -    1 4 There are four (4) left on the tree. 6. Fill in the blanks. Do the subtraction word problem story sums vertically. One has been done for you. (i) Jennifer had 9 toy cars. She She gave 5 cars to her friend.She has __4__ toy cars now. 9       -  5            4 (ii) Daniel had 7 chocolates.He gave 4 chocolates to MatthewDaniel has _____ chocolates now. .....       -  .....            ..... (iii) There were 8 cups3 cups fell on the floor.There are ___ cups on the table now. .....       -  .....            ..... (iv) There were 6 in a van.4 persons got down.There are _____ persons left in the van. .....       -  .....            ..... (v) Tommy looked hungry. I had 8 biscuits.I gave 5 to Tommy.I now have _____ biscuits left. .....       -  .....            ..... (vi) David had 18 chocolates with him.He gave 5 chocolates to his friend.David _____ has chocolates left with him. .....       -  .....            ..... (vii) There were 16 birds sitting on a tree.3 birds flew away.There are _____ birds left on the tree. .....       -  .....            ..... (viii) There were 15 eggs in a nest.5 eggs fell down and broke.There are _____ eggs left in the nest. .....       -  .....            ..... (ix) There were 28 Coke bottles.7 were distributed among children._____ bottles were left. .....       -  .....            ..... (x) Jonathan has 36 comics.He has read 25 comics.Jonathan has not read _____ comics. .....       -  .....            ..... (xi) Elizabeth had 56 stamps.She gave 23 stamps to Jessica._____ stamps are left with Elizabeth. .....       -  .....            ..... (xii) Ronald had 69 biscuits.He gave away 16 biscuits to his friends._____ biscuits are left with him. .....       -  .....            ..... (xiii) There were 35 apples in the box.Mrs Bose used 21 to make jam.There are apples in the box now. .....       -  .....            ..... 5. (ii) 3 (iii) 5 (iv) 2 (v) 3 (vi) 13 (vii) 13 (viii) 10 (ix) 21 (x) 11 (xi) 33 (xii) 53 (xiii) 14 Vertical Subtraction Game: Play this game with your partner. You need 20 buttons each and a dice. Roll the dice and see the number. Give as many buttons to your partner. Your partner rolls the dice and sees the number. You get as many buttons back. Keep on playing. If at some point, a player does not have enough buttons to give to the partner, he or she waits for the next turn. The player whose buttons are finished first is the winner. ## You might like these • ### Worksheet on Tens and Ones | Math Place Value |Tens and Ones Questions In math place value the worksheet on tens and ones questions are given below so that students can do enough practice which will help the kids to learn further numbers. • ### Number Dot to Dot |Connect the Dots Worksheets|Joining the Dot Numbers These number dot to dot (or connect the dots worksheets) printable helps the kids to practice numbers while counting and joining the dot numbers to get a complete picture. These dot to dots worksheets • ### Greater or Less than and Equal to | Bigger Number | Smaller Number | > In math, greater or less than and equal to help the kids to understand how one number is different from the other number. Whether one number is bigger or smaller than the other number or both the numb • ### Ascending Order or Descending Order |Bigger to Smaller number| Smaller In math ascending order or descending order kids will learn to arrange the numbers from bigger to smaller number and also from smaller to bigger number. • ### Vertical Addition | How to Add 1-Digit Number Vertically? | Problems Now we will learn simple Vertical Addition of 1-digit number by arranging them one number under the other number. How to add 1-digit number vertically? • ### Subtraction Word Problems - 1-Digit Numbers |Basic Concept on Subtract Math basic concept on subtraction word problems - 1-digit numbers for the first grade. To understand the word problems with different digits combinations read the question carefully and follow • ### Addition Word Problems - 1-Digit Numbers | Basic Concept on Addition Math basic concept on addition word problems - 1-digit numbers for the first grade. To understand the word problems with different digits combinations read the question carefully and follow the • ### Adding Doubles | Adding Doubles is same as Multiplying by Two | Facts In math adding doubles plays an important role for the kids to know addition facts. This activity increases the math skills in adding the numbers. Adding a number with the same number makes the • ### 1st Grade Math Worksheets | Free Printable Colorful Worksheets |Answer Free printable 1st grade math worksheets for kids. These worksheets are very carefully planned which includes various topics. • ### Counting Numbers from 51 to 99 | Number Names | Write Numbers Let’s learn about counting numbers from fifty-one to ninety-nine in numerals and words. Counting from fifty one to ninety-nine with numbers or numerals: 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 73, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81 We add to put things together. When we count one forward from a number we get one more than that number. One more than number 3 is number 4. Counting forward means addition. The answer we get after adding numbers is called the sum. We will learn addition of tens. Observe the group of objects in each set. One group of objects represent 10. We put groups of 10 to find the sum. • ### Adding Tens and Ones | Find the Sum | Digits in Ones and Tens Column We will learn adding tens and ones. Write in expanded form and find the sum and arrange the numbers in columns. Add the digits in ones and tens column. In 1st grade word problems on addition students can practice the questions on word problems based on addition. This exercise sheet on addition can be practiced by the students to get more ideas to solve the worksheet on word problems on addition. Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need. ## Recent Articles 1. ### Types of Fractions |Proper Fraction |Improper Fraction |Mixed Fraction Jul 12, 24 03:08 PM The three types of fractions are : Proper fraction, Improper fraction, Mixed fraction, Proper fraction: Fractions whose numerators are less than the denominators are called proper fractions. (Numerato… 2. ### Worksheet on Fractions | Questions on Fractions | Representation | Ans Jul 12, 24 02:11 PM In worksheet on fractions, all grade students can practice the questions on fractions on a whole number and also on representation of a fraction. This exercise sheet on fractions can be practiced 3. ### Fraction in Lowest Terms |Reducing Fractions|Fraction in Simplest Form Jul 12, 24 03:21 AM There are two methods to reduce a given fraction to its simplest form, viz., H.C.F. Method and Prime Factorization Method. If numerator and denominator of a fraction have no common factor other than 1… 4. ### Conversion of Improper Fractions into Mixed Fractions |Solved Examples Jul 12, 24 12:59 AM To convert an improper fraction into a mixed number, divide the numerator of the given improper fraction by its denominator. The quotient will represent the whole number and the remainder so obtained…
# 14.1. Multiple Integration. Iterated Integrals and Area in the Plane. Objectives. Iterated Integrals. Iterated Integrals Save this PDF as: Size: px Start display at page: Download "14.1. Multiple Integration. Iterated Integrals and Area in the Plane. Objectives. Iterated Integrals. Iterated Integrals" ## Transcription 1 14 Multiple Integration 14.1 Iterated Integrals and Area in the Plane Copyright Cengage Learning. All rights reserved. Copyright Cengage Learning. All rights reserved. Objectives! Evaluate an iterated integral.! Use an iterated integral to find the area of a plane region. Iterated Integrals 3 4 Iterated Integrals Example 1 Integrating with Respect to y Evaluate Note that the variable of integration cannot appear in either limit of integration. Solution: Considering x to be constant and integrating with respect to y produces For instance, it makes no sense to write 5 6 2 Example 2 The Integral of an Integral Evaluate Solution: Using the result of Example 1, you have Iterated Integrals The integral in Example 2 is an iterated integral. The brackets used in Example 2 are normally not written. Instead, iterated integrals are usually written simply as The inside limits of integration can be variable with respect to the outer variable of integration. However, the outside limits of integration must be constant with respect to both variables of integration. 7 8 Iterated Integrals For instance, in Example 2, the outside limits indicate that x lies in the interval 1! x! 2 and the inside limits indicate that y lies in the interval 1! y! x. Together, these two intervals determine the region of integration R of the iterated integral, as shown in Figure Area of a Plane Region Figure Area of a Plane Region Consider the plane region R bounded by a! x! b and g 1 (x)! y! g 2 (x), as shown in Figure Area of a Plane Region Specifically, if you consider x to be fixed and let y vary from g 1 (x) to g 2 (x), you can write The area of R is given by the definite integral Combining these two integrals, you can write the area of the region R as an iterated integral Using the Fundamental Theorem of Calculus, you can rewrite the integrand g 2 (x) g 1 (x) as a definite integral. Figure 3 Area of a Plane Region Area of a Plane Region Placing a representative rectangle in the region R helps determine both the order and the limits of integration. A vertical rectangle implies the order dy dx, with the inside limits corresponding to the upper and lower bounds of the rectangle, as shown in Figure This type of region is called vertically simple, because the outside limits of integration represent the vertical lines x = a and x = b. Figure Similarly, a horizontal rectangle implies the order dx dy, with the inside limits determined by the left and right bounds of the rectangle, as shown in Figure This type of region is called horizontally simple, because the outside limits represent the horizontal lines y = c and y = d. Figure Area of a Plane Region Example 3 The Area of a Rectangular Region Use an iterated integral to represent the area of the rectangle shown in Figure Solution: The region shown in Figure 14.4 is both vertically simple and horizontally simple, so you can use either order of integration. By choosing the order dy dx, you obtain the following. Figure Example 3 Solution 14.2 Double Integrals and Volume 17 Copyright Cengage Learning. All rights reserved. 18 4 Objectives! Use a double integral to represent the volume of a solid region.! Use properties of double integrals.! Evaluate a double integral as an iterated integral. Double Integrals and Volume of a Solid Region! Find the average value of a function over a region Double Integrals and Volume of a Solid Region You know that a definite integral over an interval uses a limit process to assign measures to quantities such as area, volume, arc length, and mass. In this section, you will use a similar process to define the double integral of a function of two variables over a region in the plane. Double Integrals and Volume of a Solid Region Consider a continuous function f such that f(x, y) " 0 for all (x, y) in a region R in the xy-plane. The goal is to find the volume of the solid region lying between the surface given by z = f(x, y) Surface lying above the xy-plane and the xy-plane, as shown in Figure Figure Double Integrals and Volume of a Solid Region You can begin by superimposing a rectangular grid over the region, as shown in Figure Double Integrals and Volume of a Solid Region Next, choose a point (x i, y i ) in each rectangle and form the rectangular prism whose height is f(x i, y i ), as shown in Figure Figure 14.9 The rectangles lying entirely within R form an inner partition #, whose norm # is defined as the length of the longest diagonal of the n rectangles. 23 Because the area of the ith rectangle is #A i Area of ith rectangle it follows that the volume of the ith prism is f(x i, y i ) #A i Volume of ith prism Figure 5 Double Integrals and Volume of a Solid Region Example 1 Approximating the Volume of a Solid You can approximate the volume of the solid region by the Riemann sum of the volumes of all n prisms, Approximate the volume of the solid lying between the paraboloid as shown in Figure This approximation can be improved by tightening the mesh of the grid to form smaller and smaller rectangles. and the square region R given by 0! x! 1, 0! y! 1. Use a partition made up of squares whose sides have a length of Figure Example 1 Solution Example 1 Solution Begin by forming the specified partition of R. For this partition, it is convenient to choose the centers of the subregions as the points at which to evaluate f(x, y). Because the area of each square is approximate the volume by the sum you can This approximation is shown graphically in Figure The exact volume of the solid is. Figure Example 1 Solution You can obtain a better approximation by using a finer partition. For example, with a partition of squares with sides of length the approximation is Double Integrals and Volume of a Solid Region In Example 1, note that by using finer partitions, you obtain better approximations of the volume. This observation suggests that you could obtain the exact volume by taking a limit. That is, Volume 29 30 6 Double Integrals and Volume of a Solid Region The precise meaning of this limit is that the limit is equal to L if for every! > 0 there exists a " > 0 such that Double Integrals and Volume of a Solid Region Using the limit of a Riemann sum to define volume is a special case of using the limit to define a double integral. The general case, however, does not require that the function be positive or continuous. for all partitions # of the plane region R (that satisfy # < ") and for all possible choices of x i and y i in the ith region. 31 Having defined a double integral, you will see that a definite integral is occasionally referred to as a single integral. 32 Double Integrals and Volume of a Solid Region Sufficient conditions for the double integral of f on the region R to exist are that R can be written as a union of a finite number of nonoverlapping subregions (see Figure 14.14) that are vertically or horizontally simple and that f is continuous on the region R. Double Integrals and Volume of a Solid Region A double integral can be used to find the volume of a solid region that lies between the xy-plane and the surface given by z = f(x, y). Figure Properties of Double Integrals Double integrals share many properties of single integrals. Properties of Double Integrals 35 36 7 Evaluation of Double Integrals Consider the solid region bounded by the plane z = f(x, y) = 2 x 2y and the three coordinate planes, as shown in Figure Evaluation of Double Integrals 37 Figure Evaluation of Double Integrals Each vertical cross section taken parallel to the yz-plane is a triangular region whose base has a length of y = (2 x)/2 and whose height is z = 2 x. Evaluation of Double Integrals By the formula for the volume of a solid with known cross sections, the volume of the solid is This implies that for a fixed value of x, the area of the triangular cross section is 39 Figure This procedure works no matter how A(x) is obtained. In particular, you can find A(x) by integration, as shown in Figure Evaluation of Double Integrals That is, you consider x to be constant, and integrate z = 2 x 2y from 0 to (2 x)/2 to obtain Evaluation of Double Integrals To understand this procedure better, it helps to imagine the integration as two sweeping motions. For the inner integration, a vertical line sweeps out the area of a cross section. For the outer integration, the triangular cross section sweeps out the volume, as shown in Figure Combining these results, you have the iterated integral Figure 8 Evaluation of Double Integrals Example 2 Evaluating a Double Integral as an Iterated Integral Evaluate where R is the region given by 0! x! 1, 0! y! Solution: Because the region R is a square, it is both vertically and horizontally simple, and you can use either order of integration. Choose dy dx by placing a vertical representative rectangle in the region, as shown in Figure Figure Example 2 Solution This produces the following. Average Value of a Function Average Value of a Function For a function f in one variable, the average value of f on [a, b] is Given a function f in two variables, you can find the average value of f over the region R as shown in the following definition. Example 6 Finding the Average Value of a Function Find the average value of f(x, y) = over the region R, where R is a rectangle with vertices (0, 0), (4, 0), (4, 3), and (0, 3). Solution: The area of the rectangular region R is A = 12 (see Figure 14.23). 47 Figure 9 Example 6 Solution The average value is given by 14.3 Change of Variables: Polar Coordinates 49 Copyright Cengage Learning. All rights reserved. 50 Objective! Write and evaluate double integrals in polar coordinates. Double Integrals in Polar Coordinates Double Integrals in Polar Coordinates The polar coordinates (r,! ) of a point are related to the rectangular coordinates (x, y) of the point as follows. Example 1 Using Polar Coordinates to Describe a Region Use polar coordinates to describe each region shown in Figure Figure 10 Example 1 Solution a. The region R is a quarter circle of radius 2. It can be described in polar coordinates as Double Integrals in Polar Coordinates The regions in Example 1 are special cases of polar sectors as shown in Figure b. The region R consists of all points between concentric circles of radii 1 and 3. It can be described in polar coordinates as 55 Figure Double Integrals in Polar Coordinates To define a double integral of a continuous function z = f(x, y) in polar coordinates, consider a region R bounded by the graphs of r = g 1 (! ) and r = g 2 (! ) and the lines! = \$ and! = ". Instead of partitioning R into small rectangles, use a partition of small polar sectors. Double Integrals in Polar Coordinates The polar sectors R i lying entirely within R form an inner polar partition #, whose norm # is the length of the longest diagonal of the n polar sectors. Consider a specific polar sector R i, as shown in Figure On R, superimpose a polar grid made of rays and circular arcs, as shown in Figure Figure Figure Double Integrals in Polar Coordinates It can be shown that the area of R i is Double Integrals in Polar Coordinates The region R corresponds to a horizontally simple region S in the r! -plane, as shown in Figure where and. This implies that the volume of the solid of height above R i is approximately and you have The sum on the right can be interpreted as a Riemann sum for 59 Figure 11 Double Integrals in Polar Coordinates The polar sectors R i correspond to rectangles S i, and the area of S i is So, the right-hand side of the equation corresponds to the double integral Double Integrals in Polar Coordinates This suggests the following theorem 14.3 From this, you can write Double Integrals in Polar Coordinates The region R is restricted to two basic types, r-simple regions and! -simple regions, as shown in Figure Example 2 Evaluating a Double Polar Integral Let R be the annular region lying between the two circles and Evaluate the integral Solution: The polar boundaries are and as shown in Figure Figure Figure Example 2 Solution Example 2 Solution Furthermore, and So, you have 65 66 12 14.4 Center of Mass and Moments of Inertia Objectives! Find the mass of a planar lamina using a double integral.! Find the center of mass of a planar lamina using double integrals.! Find moments of inertia using double integrals. Copyright Cengage Learning. All rights reserved Mass If the lamina corresponding to the region R, as shown in Figure 14.34, has a constant density #, Mass Figure then the mass of the lamina is given by Mass A lamina is assumed to have a constant density. But now you will extend the definition of the term lamina to include thin plates of variable density. Double integrals can be used to find the mass of a lamina of variable density, where the density at (x, y) is given by the density function #. Example 1 Finding the Mass of a Planar Lamina Find the mass of the triangular lamina with vertices (0, 0), (0, 3), and (2, 3), given that the density at (x, y) is #(x, y) = 2x + y. Solution: As shown in Figure 14.35, region R has the boundaries x = 0, y = 3, and y = 3x/2 (or x = 2y/3). 71 Figure 13 Example 1 Solution Therefore, the mass of the lamina is Moments and Center of Mass Moments and Center of Mass For a lamina of variable density, moments of mass are defined in a manner similar to that used for the uniform density case. Moments and Center of Mass Assume that the mass of R i is concentrated at one of its interior points (x i, y i ). The moment of mass of R i with respect to the x-axis can be approximated by For a partition # of a lamina corresponding to a plane region R, consider the i th rectangle R i of one area #A i, as shown in Figure Similarly, the moment of mass with respect to the y-axis can be approximated by Figure Moments and Center of Mass By forming the Riemann sum of all such products and taking the limits as the norm of # approaches 0, you obtain the following definitions of moments of mass with respect to the x- and y-axes. Moments and Center of Mass For some planar laminas with a constant density #, you can determine the center of mass (or one of its coordinates) using symmetry rather than using integration 14 Moments and Center of Mass Example 3 Finding the Center of Mass For instance, consider the laminas of constant density shown in Figure Using symmetry, you can see that and for the second lamina. for the first lamina Find the center of mass of the lamina corresponding to the parabolic region 0 % y % 4 x 2 Parabolic region where the density at the point (x, y) is proportional to the distance between (x, y) and the x-axis, as shown in Figure Figure Figure Example 3 Solution Example 3 Solution Because the lamina is symmetric with respect to the y-axis and #(x, y) = ky the center of mass lies on the y-axis. So, To find, first find the mass of the lamina. Next, find the moment about the x-axis Example 3 Solution So, and the center of mass is Moments of Inertia 83 84 15 Moments of Inertia The moments of M x and M y used in determining the center of mass of a lamina are sometimes called the first moments about the x- and y-axes. In each case, the moment is the product of a mass times a distance. Moments of Inertia You will now look at another type of moment the second moment, or the moment of inertia of a lamina about a line. In the same way that mass is a measure of the tendency of matter to resist a change in straight-line motion, the moment of inertia about a line is a measure of the tendency of matter to resist a change in rotational motion. 85 For example, if a particle of mass m is a distance d from a fixed line, its moment of inertia about the line is defined as I = md 2 = (mass)(distance) Moments of Inertia As with moments of mass, you can generalize this concept to obtain the moments of inertia about the x- and y-axes of a lamina of variable density. These second moments are denoted by I x and I y, and in each case the moment is the product of a mass times the square of a distance. Example 4 Finding the Moment of Inertia Find the moment of inertia about the x-axis of the lamina in Example 3. Solution: From the definition of moment of inertia, you have The sum of the moments I x and I y is called the polar moment of inertia and is denoted by I Moments of Inertia The moment of inertia I of a revolving lamina can be used to measure its kinetic energy. For example, suppose a planar lamina is revolving about a line with an angular speed of \$ radians per second, as shown in Figure Moments of Inertia The kinetic energy E of the revolving lamina is On the other hand, the kinetic energy E of a mass m moving in a straight line at a velocity v is So, the kinetic energy of a mass moving in a straight line is proportional to its mass, but the kinetic energy of a mass revolving about an axis is proportional to its moment of inertia. Figure 16 Moments of Inertia The radius of gyration of a revolving mass m with moment of inertia I is defined as If the entire mass were located at a distance from its axis of revolution, it would have the same moment of inertia and, consequently, the same kinetic energy. For instance, the radius of gyration of the lamina in Example 4 about the x-axis is given by 14.5 Surface Area 91 Copyright Cengage Learning. All rights reserved. 92 Objective! Use a double integral to find the area of a surface. Surface Area Surface Area You know about the solid region lying between a surface and a closed and bounded region R in the xy-plane, as shown in Figure For example, you know how to find the extrema of f on R. Surface Area In this section, you will learn how to find the upper surface area of the solid. To begin, consider a surface S given by z = f(x, y) defined over a region R. Assume that R is closed and bounded and that f has continuous first partial derivatives. Figure 17 Surface Area To find the surface area, construct an inner partition of R consisting of n rectangles, where the area of the ith rectangle R i is #A i = #x i #y i, as shown in Figure In each R i let (x i, y i ) be the point that is closest to the origin. Surface Area The area of the portion of the tangent plane that lies directly above R i is approximately equal to the area of the surface lying directly above R i. That is, #T i \$ #S i. So, the surface area of S is given by At the point (x i, y i, z i ) = (x i, y i, f(x i, y i )) on the surface S, construct a tangent plane T i. Figure To find the area of the parallelogram #T i, note that its sides are given by the vectors and u = #x i i + f x (x i, y i ) #x i k v = #y i j + f y (x i, y i ) #y i k. 98 Surface Area Surface Area The area of #T i is given by, where So, the area of #T i is and Surface area of Surface Area As an aid to remembering the double integral for surface area, it is helpful to note its similarity to the integral for arc length. Example 1 The Surface Area of a Plane Region Find the surface area of the portion of the plane z = 2 x y that lies above the circle x 2 + y 2! 1 in the first quadrant, as shown in Figure Figure 18 Example 1 Solution Example 1 Solution Because f x (x, y) = %1 and f y (x, y) = %1, the surface area is given by Note that the last integral is simply region R. times the area of the R is a quarter circle of radius 1, with an area of or. So, the area of S is Triple Integrals and Applications Objectives! Use a triple integral to find the volume of a solid region.! Find the center of mass and moments of inertia of a solid region. Copyright Cengage Learning. All rights reserved Triple Integrals The procedure used to define a triple integral follows that used for double integrals. Triple Integrals Consider a function f of three variables that is continuous over a bounded solid region Q. 107 Then, encompass Q with a network of boxes and form the inner partition consisting of all boxes lying entirely within Q, as shown in Figure Figure 19 Triple Integrals Triple Integrals The volume of the ith box is Taking the limit as leads to the following definition. The norm of the partition is the length of the longest diagonal of the n boxes in the partition. Choose a point (x i, y i, z i ) in each box and form the Riemann sum Triple Integrals Triple Integrals Some of the properties of double integrals can be restated in terms of triple integrals Example 1 Evaluating a Triple Iterated Integral Evaluate the triple iterated integral Example 1 Solution For the second integration, hold x constant and integrate with respect to y. Solution: For the first integration, hold x and y constant and integrate with respect to z. Finally, integrate with respect to x 20 Triple Integrals To find the limits for a particular order of integration, it is generally advisable first to determine the innermost limits, which may be functions of the outer two variables. Triple Integrals For instance, to evaluate Then, by projecting the solid Q onto the coordinate plane of the outer two variables, you can determine their limits of integration by the methods used for double integrals. first determine the limits for z, and then the integral has the form Triple Integrals By projecting the solid Q onto the xy-plane, you can determine the limits for x and y as you did for double integrals, as shown in Figure Center of Mass and Moments of Inertia Figure Center of Mass and Moments of Inertia Center of Mass and Moments of Inertia Consider a solid region Q whose density is given by the density function!. The center of mass of a solid region Q of mass m is given by, where and The quantities M yz, M xz, and M xy are called the first moments of the region Q about the yz-, xz-, and xy-planes, respectively. 119 The first moments for solid regions are taken about a plane, whereas the second moments for solids are taken about a line. 120 21 Center of Mass and Moments of Inertia The second moments (or moments of inertia) about the and x-, y-, and z-axes are as follows. Center of Mass and Moments of Inertia For problems requiring the calculation of all three moments, considerable effort can be saved by applying the additive property of triple integrals and writing where I xy, I xz, and I yz are as follows Example 5 Finding the Center of Mass of a Solid Region Find the center of mass of the unit cube shown in Figure 14.61, given that the density at the point (x, y, z) is proportional to the square of its distance from the origin. Example 5 Solution Because the density at (x, y, z) is proportional to the square of the distance between (0, 0, 0) and (x, y, z), you have You can use this density function to find the mass of the cube. Because of the symmetry of the region, any order of integration will produce an integral of comparable difficulty. Figure Example 5 Solution Example 5 Solution The first moment about the yz-plane is Note that x can be factored out of the two inner integrals, because it is constant with respect to y and z. After factoring, the two inner integrals are the same as for the mass m 22 Example 5 Solution Therefore, you have 14.7 Triple Integrals in Cylindrical and Spherical Coordinates So, Finally, from the nature of! and the symmetry of x, y, and z in this solid region, you have, and the center of mass is. 127 Copyright Cengage Learning. All rights reserved. 128 Objectives! Write and evaluate a triple integral in cylindrical coordinates.! Write and evaluate a triple integral in spherical coordinates. Triple Integrals in Cylindrical Coordinates Triple Integrals in Cylindrical Coordinates Triple Integrals in Cylindrical Coordinates The rectangular conversion equations for cylindrical coordinates are x = r cos " y = r sin " z = z. In this coordinate system, the simplest solid region is a cylindrical block determined by r 1! r! r 2, " 1! "! " 2, z 1! z! z 2 as shown in Figure To obtain the cylindrical coordinate form of a triple integral, suppose that Q is a solid region whose projection R onto the xy-plane can be described in polar coordinates. That is, and Q = {(x, y, z): (x, y) is in R, h 1 (x, y)! z! h 2 (x, y)} R = {(r, "): " 1! "! " 2, g 1 (")! r! g 2 (")}. Figure 23 Triple Integrals in Cylindrical Coordinates If f is a continuous function on the solid Q, you can write the triple integral of f over Q as where the double integral over R is evaluated in polar coordinates. That is, R is a plane region that is either r-simple or "-simple. If R is r-simple, the iterated form of the triple integral in cylindrical form is Triple Integrals in Cylindrical Coordinates To visualize a particular order of integration, it helps to view the iterated integral in terms of three sweeping motions each adding another dimension to the solid. For instance, in the order dr d" dz, the first integration occurs in the r-direction as a point sweeps out a ray. Then, as " increases, the line sweeps out a sector Triple Integrals in Cylindrical Coordinates Finally, as z increases, the sector sweeps out a solid wedge, as shown in Figure Example 1 Finding Volume in Cylindrical Coordinates Find the volume of the solid region Q cut from the sphere x 2 + y 2 + z 2 = 4 by the cylinder r = 2 sin ", as shown in Figure Figure Figure Example 1 Solution Because x 2 + y 2 + z 2 = r 2 + z 2 = 4, the bounds on z are Example 1 Solution Let R be the circular projection of the solid onto the r"-plane. Then the bounds on R are 0! r! 2 sin " and 0! "! #. So, the volume of Q is 24 Triple Integrals in Spherical Coordinates Triple Integrals in Spherical Coordinates 139 The rectangular conversion equations for spherical coordinates are x =! sin % cos " y =! sin % sin " z =! cos %. In this coordinate system, the simplest region is a spherical block determined by {(!, ", %):! 1!!!! 2, " 1! "! " 2, % 1! %! % 2 } where! 1 " 0, " 2 " 1! 2#, and 0! % 1! % 2! #, as shown in Figure Figure Triple Integrals in Spherical Coordinates If (!, ", %) is a point in the interior of such a block, then the volume of the block can be approximated by #V \$! 2 sin % #! #% #" Using the usual process involving an inner partition, summation, and a limit, you can develop the following version of a triple integral in spherical coordinates for a continuous function f defined on the solid region Q. Triple Integrals in Spherical Coordinates Triple integrals in spherical coordinates are evaluated with iterated integrals. You can visualize a particular order of integration by viewing the iterated integral in terms of three sweeping motions each adding another dimension to the solid Triple Integrals in Spherical Coordinates For instance, the iterated integral is illustrated in Figure Example 4 Finding Volume in Spherical Coordinates Find the volume of the solid region Q bounded below by the upper nappe of the cone z 2 = x 2 + y 2 and above by the sphere x 2 + y 2 + z 2 = 9, as shown in Figure Figure Figure 14.70 25 Example 4 Solution In spherical coordinates, the equation of the sphere is! 2 = x 2 + y 2 + z 2 = 9 Furthermore, the sphere and cone intersect when (x 2 + y 2 ) + z 2 = (z 2 ) + z 2 = 9 and, because z =! cos %, it follows that Example 4 Solution Consequently, you can use the integration order d! d% d", where 0!!! 3, 0! %! #/4, and 0! "! 2#. The volume is Change of Variables: Jacobians Objectives! Understand the concept of a Jacobian.! Use a Jacobian to change variables in a double integral. Copyright Cengage Learning. All rights reserved Jacobians For the single integral Jacobians you can change variables by letting x = g(u), so that dx = g\$(u) du, and obtain where a = g(c) and b = g(d). 149 The change of variables process introduces an additional factor g\$(u) into the integrand. 150 26 Jacobians This also occurs in the case of double integrals Jacobians In defining the Jacobian, it is convenient to use the following determinant notation. where the change of variables x = g(u, v) and y = h(u, v) introduces a factor called the Jacobian of x and y with respect to u and v Example 1 The Jacobian for Rectangular-to-Polar Conversion Find the Jacobian for the change of variables defined by x = r cos " and y = r sin ". Solution: From the definition of the Jacobian, you obtain Jacobians Example 1 points out that the change of variables from rectangular to polar coordinates for a double integral can be written as 153 where S is the region in the r"-plane that corresponds to the region R in the xy-plane, as shown in Figure Figure Jacobians In general, a change of variables is given by a one-to-one transformation T from a region S in the uv-plane to a region R in the xy-plane, to be given by T(u, v) = (x, y) = (g(u, v), h(u, v)) where g and h have continuous first partial derivatives in the region S. Note that the point (u, v) lies in S and the point (x, y) lies in R. Change of Variables for Double Integrals 27 Change of Variables for Double Integrals Example 3 Using a Change of Variables to Simplify a Region Let R be the region bounded by the lines x 2y = 0, x 2y = 4, x + y = 4, and x + y = 1 as shown in Figure Evaluate the double integral Figure Example 3 Solution You can use the following change of variables. and The partial derivatives of x and y are Example 3 Solution which implies that the Jacobian is So, by Theorem 14.5, you obtain Example 3 Solution 161 ### Solutions to Homework 10 Solutions to Homework 1 Section 7., exercise # 1 (b,d): (b) Compute the value of R f dv, where f(x, y) = y/x and R = [1, 3] [, 4]. Solution: Since f is continuous over R, f is integrable over R. Let x ### 11.1. Objectives. Component Form of a Vector. Component Form of a Vector. Component Form of a Vector. Vectors and the Geometry of Space 11 Vectors and the Geometry of Space 11.1 Vectors in the Plane Copyright Cengage Learning. All rights reserved. Copyright Cengage Learning. All rights reserved. 2 Objectives! Write the component form of ### MULTIPLE INTEGRALS. h 2 (y) are continuous functions on [c, d] and let f(x, y) be a function defined on R. Then MULTIPLE INTEGALS 1. ouble Integrals Let be a simple region defined by a x b and g 1 (x) y g 2 (x), where g 1 (x) and g 2 (x) are continuous functions on [a, b] and let f(x, y) be a function defined on. ### This function is symmetric with respect to the y-axis, so I will let - /2 /2 and multiply the area by 2. INTEGRATION IN POLAR COORDINATES One of the main reasons why we study polar coordinates is to help us to find the area of a region that cannot easily be integrated in terms of x. In this set of notes, ### 3. Double Integrals 3A. Double Integrals in Rectangular Coordinates 3. Double Integrals 3A. Double Integrals in ectangular Coordinates 3A-1 Evaluate each of the following iterated integrals: c) 2 1 1 1 x 2 (6x 2 +2y)dydx b) x 2x 2 ydydx d) π/2 π 1 u (usint+tcosu)dtdu u2 ### 15.1. Vector Analysis. Vector Fields. Objectives. Vector Fields. Vector Fields. Vector Fields. ! Understand the concept of a vector field. 15 Vector Analysis 15.1 Vector Fields Copyright Cengage Learning. All rights reserved. Copyright Cengage Learning. All rights reserved. Objectives! Understand the concept of a vector field.! Determine ### Module 1 : A Crash Course in Vectors Lecture 2 : Coordinate Systems Module 1 : A Crash Course in Vectors Lecture 2 : Coordinate Systems Objectives In this lecture you will learn the following Define different coordinate systems like spherical polar and cylindrical coordinates ### Calculus C/Multivariate Calculus Advanced Placement G/T Essential Curriculum Calculus C/Multivariate Calculus Advanced Placement G/T Essential Curriculum UNIT I: The Hyperbolic Functions basic calculus concepts, including techniques for curve sketching, exponential and logarithmic ### 7.3 Volumes Calculus 7. VOLUMES Just like in the last section where we found the area of one arbitrary rectangular strip and used an integral to add up the areas of an infinite number of infinitely thin rectangles, we are ### MULTIPLE INTEGRATION NOTES AND EXERCISES. Antony Foster MULTIPL INTGRATION NOTS AN XRCISS by Antony Foster Overview In this discussion we consider the integral of a function of two variables z f(x, y) over a region in the xy-plane (-space) and the integral ### 1.7 Cylindrical and Spherical Coordinates 56 CHAPTER 1. VECTORS AND THE GEOMETRY OF SPACE 1.7 Cylindrical and Spherical Coordinates 1.7.1 Review: Polar Coordinates The polar coordinate system is a two-dimensional coordinate system in which the ### Solutions to Vector Calculus Practice Problems olutions to Vector alculus Practice Problems 1. Let be the region in determined by the inequalities x + y 4 and y x. Evaluate the following integral. sinx + y ) da Answer: The region looks like y y x x ### Double integrals. Notice: this material must not be used as a substitute for attending the lectures ouble integrals Notice: this material must not be used as a substitute for attending the lectures . What is a double integral? Recall that a single integral is something of the form b a f(x) A double integral ### 27.3. Introduction. Prerequisites. Learning Outcomes olume Integrals 27. Introduction In the previous two sections, surface integrals (or double integrals) were introduced i.e. functions were integrated with respect to one variable and then with respect ### Figure 1: u2 (x,y) D u 1 (x,y) yρ(x, y, z)dv. xρ(x, y, z)dv. M yz = E Contiune on.7 Triple Integrals Figure 1: [ ] u2 (x,y) f(x, y, z)dv = f(x, y, z)dz da u 1 (x,y) Applications of Triple Integrals Let be a solid region with a density function ρ(x, y, z). Volume: V () = ### Practice Final Math 122 Spring 12 Instructor: Jeff Lang Practice Final Math Spring Instructor: Jeff Lang. Find the limit of the sequence a n = ln (n 5) ln (3n + 8). A) ln ( ) 3 B) ln C) ln ( ) 3 D) does not exist. Find the limit of the sequence a n = (ln n)6 ### AP CALCULUS AB 2008 SCORING GUIDELINES AP CALCULUS AB 2008 SCORING GUIDELINES Question 1 Let R be the region bounded by the graphs of y = sin( π x) and y = x 4 x, as shown in the figure above. (a) Find the area of R. (b) The horizontal line ### PROBLEM SET. Practice Problems for Exam #1. Math 1352, Fall 2004. Oct. 1, 2004 ANSWERS PROBLEM SET Practice Problems for Exam # Math 352, Fall 24 Oct., 24 ANSWERS i Problem. vlet R be the region bounded by the curves x = y 2 and y = x. A. Find the volume of the solid generated by revolving ### Engineering Geometry Engineering Geometry Objectives Describe the importance of engineering geometry in design process. Describe coordinate geometry and coordinate systems and apply them to CAD. Review the right-hand rule. ### MATH SOLUTIONS TO PRACTICE FINAL EXAM. (x 2)(x + 2) (x 2)(x 3) = x + 2. x 2 x 2 5x + 6 = = 4. MATH 55 SOLUTIONS TO PRACTICE FINAL EXAM x 2 4.Compute x 2 x 2 5x + 6. When x 2, So x 2 4 x 2 5x + 6 = (x 2)(x + 2) (x 2)(x 3) = x + 2 x 3. x 2 4 x 2 x 2 5x + 6 = 2 + 2 2 3 = 4. x 2 9 2. Compute x + sin ### Week #15 - Word Problems & Differential Equations Section 8.1 Week #15 - Word Problems & Differential Equations Section 8.1 From Calculus, Single Variable by Hughes-Hallett, Gleason, McCallum et. al. Copyright 25 by John Wiley & Sons, Inc. This material is used by ### Chapter 5 Polar Coordinates; Vectors 5.1 Polar coordinates 1. Pole and polar axis Chapter 5 Polar Coordinates; Vectors 5.1 Polar coordinates 1. Pole and polar axis 2. Polar coordinates A point P in a polar coordinate system is represented by an ordered pair of numbers (r, θ). If r > ### Copyright 2011 Casa Software Ltd. www.casaxps.com. Centre of Mass Centre of Mass A central theme in mathematical modelling is that of reducing complex problems to simpler, and hopefully, equivalent problems for which mathematical analysis is possible. The concept of ### Two vectors are equal if they have the same length and direction. They do not Vectors define vectors Some physical quantities, such as temperature, length, and mass, can be specified by a single number called a scalar. Other physical quantities, such as force and velocity, must ### Vectors, Gradient, Divergence and Curl. Vectors, Gradient, Divergence and Curl. 1 Introduction A vector is determined by its length and direction. They are usually denoted with letters with arrows on the top a or in bold letter a. We will use ### If Σ is an oriented surface bounded by a curve C, then the orientation of Σ induces an orientation for C, based on the Right-Hand-Rule. Oriented Surfaces and Flux Integrals Let be a surface that has a tangent plane at each of its nonboundary points. At such a point on the surface two unit normal vectors exist, and they have opposite directions. ### m i: is the mass of each particle Center of Mass (CM): The center of mass is a point which locates the resultant mass of a system of particles or body. It can be within the object (like a human standing straight) or outside the object ### Section 16: Neutral Axis and Parallel Axis Theorem 16-1 Section 16: Neutral Axis and Parallel Axis Theorem 16-1 Geometry of deformation We will consider the deformation of an ideal, isotropic prismatic beam the cross section is symmetric about y-axis All parts ### AB2.5: Surfaces and Surface Integrals. Divergence Theorem of Gauss AB2.5: urfaces and urface Integrals. Divergence heorem of Gauss epresentations of surfaces or epresentation of a surface as projections on the xy- and xz-planes, etc. are For example, z = f(x, y), x = ### Centroid: The point of intersection of the three medians of a triangle. Centroid Vocabulary Words Acute Triangles: A triangle with all acute angles. Examples 80 50 50 Angle: A figure formed by two noncollinear rays that have a common endpoint and are not opposite rays. Angle Bisector: ### 15.3. Calculating centres of mass. Introduction. Prerequisites. Learning Outcomes. Learning Style Calculating centres of mass 15.3 Introduction In this block we show how the idea of integration as the limit of a sum can be used to find the centre of mass of an object such as a thin plate, like a sheet ### Double Integrals in Polar Coordinates Double Integrals in Polar Coordinates. A flat plate is in the shape of the region in the first quadrant ling between the circles + and +. The densit of the plate at point, is + kilograms per square meter ### Example 1. Example 1 Plot the points whose polar coordinates are given by Polar Co-ordinates A polar coordinate system, gives the co-ordinates of a point with reference to a point O and a half line or ray starting at the point O. We will look at polar coordinates for points ### Physics 210 Q ( PHYSICS210BRIDGE ) My Courses Course Settings 1 of 16 9/7/2012 1:10 PM Logged in as Julie Alexander, Instructor Help Log Out Physics 210 Q1 2012 ( PHYSICS210BRIDGE ) My Courses Course Settings Course Home Assignments Roster Gradebook Item Library ### Differential Relations for Fluid Flow. Acceleration field of a fluid. The differential equation of mass conservation Differential Relations for Fluid Flow In this approach, we apply our four basic conservation laws to an infinitesimally small control volume. The differential approach provides point by point details of ### 4 More Applications of Definite Integrals: Volumes, arclength and other matters 4 More Applications of Definite Integrals: Volumes, arclength and other matters Volumes of surfaces of revolution 4. Find the volume of a cone whose height h is equal to its base radius r, by using the ### Section 2.6 Cylindrical and Spherical Coordinates A) Review on the Polar Coordinates Section.6 Cylindrical and Spherical Coordinates A) Review on the Polar Coordinates The polar coordinate system consists of the origin O,the rotating ray or half line from O with unit tick. A point P in ### Math 1B, lecture 5: area and volume Math B, lecture 5: area and volume Nathan Pflueger 6 September 2 Introduction This lecture and the next will be concerned with the computation of areas of regions in the plane, and volumes of regions in ### Lecture 8 : Coordinate Geometry. The coordinate plane The points on a line can be referenced if we choose an origin and a unit of 20 Lecture 8 : Coordinate Geometry The coordinate plane The points on a line can be referenced if we choose an origin and a unit of 0 distance on the axis and give each point an identity on the corresponding ### Coefficient of Potential and Capacitance Coefficient of Potential and Capacitance Lecture 12: Electromagnetic Theory Professor D. K. Ghosh, Physics Department, I.I.T., Bombay We know that inside a conductor there is no electric field and that ### Chapter 3. Gauss s Law 3 3 3-0 Chapter 3 Gauss s Law 3.1 Electric Flux... 3-2 3.2 Gauss s Law (see also Gauss s Law Simulation in Section 3.10)... 3-4 Example 3.1: Infinitely Long Rod of Uniform Charge Density... 3-9 Example ### Solutions - Homework sections 17.7-17.9 olutions - Homework sections 7.7-7.9 7.7 6. valuate xy d, where is the triangle with vertices (,, ), (,, ), and (,, ). The three points - and therefore the triangle between them - are on the plane x + ### Section 16.7 Triple Integrals in Cylindrical Coordinates Section 6.7 Triple Integrals in Cylindrical Coordinates Integrating Functions in Different Coordinate Systems In Section 6.4, we used the polar coordinate system to help integrate functions over circular ### Solutions for Review Problems olutions for Review Problems 1. Let be the triangle with vertices A (,, ), B (4,, 1) and C (,, 1). (a) Find the cosine of the angle BAC at vertex A. (b) Find the area of the triangle ABC. (c) Find a vector ### Fundamental Theorems of Vector Calculus Fundamental Theorems of Vector Calculus We have studied the techniques for evaluating integrals over curves and surfaces. In the case of integrating over an interval on the real line, we were able to use ### L 2 : x = s + 1, y = s, z = 4s + 4. 3. Suppose that C has coordinates (x, y, z). Then from the vector equality AC = BD, one has The line L through the points A and B is parallel to the vector AB = 3, 2, and has parametric equations x = 3t + 2, y = 2t +, z = t Therefore, the intersection point of the line with the plane should satisfy: ### Volumes of Revolution Mathematics Volumes of Revolution About this Lesson This lesson provides students with a physical method to visualize -dimensional solids and a specific procedure to sketch a solid of revolution. Students ### Engineering Math II Spring 2015 Solutions for Class Activity #2 Engineering Math II Spring 15 Solutions for Class Activity # Problem 1. Find the area of the region bounded by the parabola y = x, the tangent line to this parabola at 1, 1), and the x-axis. Then find ### Solutions to old Exam 1 problems Solutions to old Exam 1 problems Hi students! I am putting this old version of my review for the first midterm review, place and time to be announced. Check for updates on the web site as to which sections ### Number Sense and Operations Number Sense and Operations representing as they: 6.N.1 6.N.2 6.N.3 6.N.4 6.N.5 6.N.6 6.N.7 6.N.8 6.N.9 6.N.10 6.N.11 6.N.12 6.N.13. 6.N.14 6.N.15 Demonstrate an understanding of positive integer exponents ### Chapter 10 Rotational Motion. Copyright 2009 Pearson Education, Inc. Chapter 10 Rotational Motion Angular Quantities Units of Chapter 10 Vector Nature of Angular Quantities Constant Angular Acceleration Torque Rotational Dynamics; Torque and Rotational Inertia Solving Problems ### New York State Student Learning Objective: Regents Geometry New York State Student Learning Objective: Regents Geometry All SLOs MUST include the following basic components: Population These are the students assigned to the course section(s) in this SLO all students ### AP Calculus AB 2010 Free-Response Questions Form B AP Calculus AB 2010 Free-Response Questions Form B The College Board The College Board is a not-for-profit membership association whose mission is to connect students to college success and opportunity. ### Right Prisms Let s find the surface area of the right prism given in Figure 44.1. Figure 44.1 44 Surface Area The surface area of a space figure is the total area of all the faces of the figure. In this section, we discuss the surface areas of some of the space figures introduced in Section 41. ### SURFACE AREA AND VOLUME SURFACE AREA AND VOLUME In this unit, we will learn to find the surface area and volume of the following threedimensional solids:. Prisms. Pyramids 3. Cylinders 4. Cones It is assumed that the reader has ### 10.4 Surface Area of Prisms, Cylinders, Pyramids, Cones, and Spheres. 10.4 Day 1 Warm-up 10.4 Surface Area of Prisms, Cylinders, Pyramids, Cones, and Spheres 10.4 Day 1 Warm-up 1. Which identifies the figure? A rectangular pyramid B rectangular prism C cube D square pyramid 3. A polyhedron ### www.mathsbox.org.uk Displacement (x) Velocity (v) Acceleration (a) x = f(t) differentiate v = dx Acceleration Velocity (v) Displacement x Mechanics 2 : Revision Notes 1. Kinematics and variable acceleration Displacement (x) Velocity (v) Acceleration (a) x = f(t) differentiate v = dx differentiate a = dv = d2 x dt dt dt 2 Acceleration Velocity ### Section 11.1: Vectors in the Plane. Suggested Problems: 1, 5, 9, 17, 23, 25-37, 40, 42, 44, 45, 47, 50 Section 11.1: Vectors in the Plane Page 779 Suggested Problems: 1, 5, 9, 17, 3, 5-37, 40, 4, 44, 45, 47, 50 Determine whether the following vectors a and b are perpendicular. 5) a = 6, 0, b = 0, 7 Recall ### Differentiation of vectors Chapter 4 Differentiation of vectors 4.1 Vector-valued functions In the previous chapters we have considered real functions of several (usually two) variables f : D R, where D is a subset of R n, where 1.2 GRAPHS OF EQUATIONS Copyright Cengage Learning. All rights reserved. What You Should Learn Sketch graphs of equations. Find x- and y-intercepts of graphs of equations. Use symmetry to sketch graphs ### The Fourth International DERIVE-TI92/89 Conference Liverpool, U.K., 12-15 July 2000. Derive 5: The Easiest... Just Got Better! The Fourth International DERIVE-TI9/89 Conference Liverpool, U.K., -5 July 000 Derive 5: The Easiest... Just Got Better! Michel Beaudin École de technologie supérieure 00, rue Notre-Dame Ouest Montréal ### Adding vectors We can do arithmetic with vectors. We ll start with vector addition and related operations. Suppose you have two vectors 1 Chapter 13. VECTORS IN THREE DIMENSIONAL SPACE Let s begin with some names and notation for things: R is the set (collection) of real numbers. We write x R to mean that x is a real number. A real number ### GEOMETRY COMMON CORE STANDARDS 1st Nine Weeks Experiment with transformations in the plane G-CO.1 Know precise definitions of angle, circle, perpendicular line, parallel line, and line segment, based on the undefined notions of point, ### Additional Topics in Math Chapter Additional Topics in Math In addition to the questions in Heart of Algebra, Problem Solving and Data Analysis, and Passport to Advanced Math, the SAT Math Test includes several questions that are ### Introduction. The Aims & Objectives of the Mathematical Portion of the IBA Entry Test Introduction The career world is competitive. The competition and the opportunities in the career world become a serious problem for students if they do not do well in Mathematics, because then they are ### (a) We have x = 3 + 2t, y = 2 t, z = 6 so solving for t we get the symmetric equations. x 3 2. = 2 y, z = 6. t 2 2t + 1 = 0, Name: Solutions to Practice Final. Consider the line r(t) = 3 + t, t, 6. (a) Find symmetric equations for this line. (b) Find the point where the first line r(t) intersects the surface z = x + y. (a) We ### SAT Subject Test Practice Test II: Math Level II Time 60 minutes, 50 Questions SAT Subject Test Practice Test II: Math Level II Time 60 minutes, 50 Questions All questions in the Math Level 1 and Math Level Tests are multiple-choice questions in which you are asked to choose the ### Angles and Quadrants. Angle Relationships and Degree Measurement. Chapter 7: Trigonometry Chapter 7: Trigonometry Trigonometry is the study of angles and how they can be used as a means of indirect measurement, that is, the measurement of a distance where it is not practical or even possible ### Applications of the definite integral to calculating volume and length Chapter 5 Applications of the definite integral to calculating volume and length In this chapter we consider applications of the definite integral to calculating geometric quantities such as volumes. The ### Chapter 1: Essentials of Geometry Section Section Title 1.1 Identify Points, Lines, and Planes 1.2 Use Segments and Congruence 1.3 Use Midpoint and Distance Formulas Chapter 1: Essentials of Geometry Learning Targets I Can 1. Identify, ### Change of Variables in Double Integrals Change of Variables in Double Integrals Part : Area of the Image of a egion It is often advantageous to evaluate (x; y) da in a coordinate system other than the xy-coordinate system. In this section, we ### Figure 1.1 Vector A and Vector F CHAPTER I VECTOR QUANTITIES Quantities are anything which can be measured, and stated with number. Quantities in physics are divided into two types; scalar and vector quantities. Scalar quantities have ### Review Sheet for Test 1 Review Sheet for Test 1 Math 261-00 2 6 2004 These problems are provided to help you study. The presence of a problem on this handout does not imply that there will be a similar problem on the test. And ### Geometry Honors: Extending 2 Dimensions into 3 Dimensions. Unit Overview. Student Focus. Semester 2, Unit 5: Activity 30. Resources: Online Resources: Geometry Honors: Extending 2 Dimensions into 3 Dimensions Semester 2, Unit 5: Activity 30 Resources: SpringBoard- Geometry Online Resources: Geometry Springboard Text Unit Overview In this unit students ### Biggar High School Mathematics Department. National 5 Learning Intentions & Success Criteria: Assessing My Progress Biggar High School Mathematics Department National 5 Learning Intentions & Success Criteria: Assessing My Progress Expressions & Formulae Topic Learning Intention Success Criteria I understand this Approximation ### (b)using the left hand end points of the subintervals ( lower sums ) we get the aprroximation (1) Consider the function y = f(x) =e x on the interval [, 1]. (a) Find the area under the graph of this function over this interval using the Fundamental Theorem of Calculus. (b) Subdivide the interval ### AP Calculus BC 2003 Free-Response Questions AP Calculus BC 2003 Free-Response Questions The materials included in these files are intended for use by AP teachers for course and exam preparation; permission for any other use must be sought from the ### ( ) ( ) 1. Let F = ( 1yz)i + ( 3xz) j + ( 9xy)k. Compute the following: A. div F. F = 1yz x. B. curl F. i j k. = 6xi 8yj+ 2zk F = z 1yz 3xz 9xy . Let F = ( yz)i + ( 3xz) j + ( 9xy)k. Compute the following: A. div F F = yz x B. curl F + ( 3xz) y + ( 9xy) = 0 + 0 + 0 = 0 z F = i j k x y z yz 3xz 9xy = 6xi 8yj+ 2zk C. div curl F F = 6x x + ( 8y) ### 11. Rotation Translational Motion: Rotational Motion: 11. Rotation Translational Motion: Motion of the center of mass of an object from one position to another. All the motion discussed so far belongs to this category, except uniform circular motion. Rotational ### 52. The Del Operator: Divergence and Curl 52. The Del Operator: Divergence and Curl Let F(x, y, z) = M(x, y, z), N(x, y, z), P(x, y, z) be a vector field in R 3. The del operator is represented by the symbol, and is written = x, y, z, or = x, ### Stress and Deformation Analysis. Representing Stresses on a Stress Element. Representing Stresses on a Stress Element con t Stress and Deformation Analysis Material in this lecture was taken from chapter 3 of Representing Stresses on a Stress Element One main goals of stress analysis is to determine the point within a load-carrying ### Geometry and Measurement The student will be able to: Geometry and Measurement 1. Demonstrate an understanding of the principles of geometry and measurement and operations using measurements Use the US system of measurement for ### Section 1.1. Introduction to R n The Calculus of Functions of Several Variables Section. Introduction to R n Calculus is the study of functional relationships and how related quantities change with each other. In your first exposure to ### HOMEWORK 4 SOLUTIONS. All questions are from Vector Calculus, by Marsden and Tromba HOMEWORK SOLUTIONS All questions are from Vector Calculus, by Marsden and Tromba Question :..6 Let w = f(x, y) be a function of two variables, and let x = u + v, y = u v. Show that Solution. By the chain ### Derive 5: The Easiest... Just Got Better! Liverpool John Moores University, 1-15 July 000 Derive 5: The Easiest... Just Got Better! Michel Beaudin École de Technologie Supérieure, Canada Email; mbeaudin@seg.etsmtl.ca 1. Introduction Engineering ### Mathematical Procedures CHAPTER 6 Mathematical Procedures 168 CHAPTER 6 Mathematical Procedures The multidisciplinary approach to medicine has incorporated a wide variety of mathematical procedures from the fields of physics, ### Maximum / Minimum Problems 171 CHAPTER 6 Maximum / Minimum Problems Methods for solving practical maximum or minimum problems will be examined by examples. Example Question: The material for the square base of a rectangular box ### Chapter 28 Fluid Dynamics Chapter 28 Fluid Dynamics 28.1 Ideal Fluids... 1 28.2 Velocity Vector Field... 1 28.3 Mass Continuity Equation... 3 28.4 Bernoulli s Principle... 4 28.5 Worked Examples: Bernoulli s Equation... 7 Example ### MATHEMATICS FOR ENGINEERING INTEGRATION TUTORIAL 1 BASIC INTEGRATION MATHEMATICS FOR ENGINEERING INTEGRATION TUTORIAL 1 ASIC INTEGRATION This tutorial is essential pre-requisite material for anyone studying mechanical engineering. This tutorial uses the principle of learning ### Rotation. Moment of inertia of a rotating body: w I = r 2 dm Rotation Moment of inertia of a rotating body: w I = r 2 dm Usually reasonably easy to calculate when Body has symmetries Rotation axis goes through Center of mass Exams: All moment of inertia will be ### 4B. Line Integrals in the Plane 4. Line Integrals in the Plane 4A. Plane Vector Fields 4A-1 Describe geometrically how the vector fields determined by each of the following vector functions looks. Tell for each what the largest region ### Working with Wireframe and Surface Design Chapter 9 Working with Wireframe and Surface Design Learning Objectives After completing this chapter you will be able to: Create wireframe geometry. Create extruded surfaces. Create revolved surfaces. ### Geometry Enduring Understandings Students will understand 1. that all circles are similar. High School - Circles Essential Questions: 1. Why are geometry and geometric figures relevant and important? 2. How can geometric ideas be communicated using a variety of representations? ******(i.e maps, ### 5. Surface Area Practice Chapter Test ID: A Date: / / Name: Block ID: 5. Surface Area Practice Chapter Test Multiple Choice Identify the choice that best completes the statement or answers the question. Choose the best answer. 1. Which combination ### Algebra Geometry Glossary. 90 angle lgebra Geometry Glossary 1) acute angle an angle less than 90 acute angle 90 angle 2) acute triangle a triangle where all angles are less than 90 3) adjacent angles angles that share a common leg Example: ### 13.4 THE CROSS PRODUCT 710 Chapter Thirteen A FUNDAMENTAL TOOL: VECTORS 62. Use the following steps and the results of Problems 59 60 to show (without trigonometry) that the geometric and algebraic definitions of the dot product ### Chapter 8 Geometry We will discuss following concepts in this chapter. Mat College Mathematics Updated on Nov 5, 009 Chapter 8 Geometry We will discuss following concepts in this chapter. Two Dimensional Geometry: Straight lines (parallel and perpendicular), Rays, Angles ### Area and Arc Length in Polar Coordinates Area and Arc Length in Polar Coordinates The Cartesian Coordinate System (rectangular coordinates) is not always the most convenient way to describe points, or relations in the plane. There are certainly
# GMAT Math : Calculating ratio and proportion ## Example Questions ### Example Question #11 : Calculating Ratio And Proportion A fathom is a unit of underwater depth equal to six feet. Express two miles in fathoms. Explanation: Multiply 2 miles by 5,280 feet per mile, then divide by 6 feet per fathom: ### Example Question #12 : Calculating Ratio And Proportion A furlong is a measure of length used in horse racing. Eight furlongs are equal to one mile. Express 20 furlongs in feet. Explanation: Divide 20 furlongs by 8 furlongs per mile, then multiply by 5,280 feet per mile. ### Example Question #13 : Calculating Ratio And Proportion 127 people signed up for a subscription service. 45 signed up for weekly deliveries, 34 for monthly deliveries, and the rest for annual deliveries. What is the ratio of weekly subscribers to annual subscribers? Explanation: Let's first find the number of annual subscribers: The ratio of weekly subscribers to annual subscribers is therefore . ### Example Question #14 : Calculating Ratio And Proportion A clothing store is having a sales for the holiday season. They mark down all items by . What is the sales price of a suit whose original price is ? Explanation: We start by finding the value of the mark down: Therefore, the mark down is 43.8% or 0.438, which means that the new price of the suit is: ### Example Question #15 : Calculating Ratio And Proportion Noah and Lilly both work for Company ABC. Lilly earns  more than Noah, and each year every employee gets a  salary increase. In 2012, Noah's salary was . What was Lilly's salary in 2013? Explanation: We start by finding Lilly's salary in 2012: $60,000 x 1.20 =$60,000 + $12,000 =$72,000 Then we account for the annual 10% increase to find Lilly's salary in 2013: $72,000 x 1.10 =$72,000 + $7,200 =$79,200 So Lilly's salary in 2013 was $79,200. ### Example Question #16 : Calculating Ratio And Proportion Company sells liters of soda for , and Company sells liters of soda for . If the amount of soda sold per dollar is the same for each company, what is ? Possible Answers: Correct answer: Explanation: Since the amount of soda per dollar is the same for each company, we can equate the 2 ratios And solve (Cross multiply) ### Example Question #17 : Calculating Ratio And Proportion Ms. Lopez earned$236.25 after working 15 hours last week. At this rate, how much will she have earned if she works 25 hours? Explanation: The first thing we must do is set up our equation. We do this by writing our earnings to hours worked ratio: where  represents her earnings after working 25 hours. We can then solve this equation for . Then, ### Example Question #18 : Calculating Ratio And Proportion George just finished the first week of his new job. He calculated that if he works 50 hours in two weeks, his biweekly paycheck will be $812.50. If he has worked a total of 27 hours so far, how much additional money must he earn in order to reach$812.50? $432.75$621.00 $373.75$438.75 $373.75 Explanation: He has worked 27 hours out of the 50 hours calculated. As such, we are asked how much he will earn after working the remaining 23 hours. We can first find his hourly pay and then calculate his earnings for 23 hours. where represents his hourly earnings He earns$16.25 an hour We can now calculate how much he earns for the remaining 23 hours: So, George needs \$373.75. ### Example Question #11 : Proportion / Ratio / Rate A small company's workforce consists of store employees, store managers, and corporate managers in the ratio 10:3:1. How many employees are either corporate managers or store managers if the company has a total of  employees? Explanation: Let  be the number of store employees,  the number of store managers, and  the number of corporate managers. , so the number of store employees is . , so the number of store managers is . , so the number of corporate managers is . Therefore, the number of employees who are either store managers or corporate managers is . ### Example Question #19 : Calculating Ratio And Proportion A scale model of the newest Mars rover has a wheel radius of  and its antennae are each  long. If the radius of the wheel on the actual rover is , what is the length of the antennae on the actual rover?
## College Algebra (6th Edition) Point-slope form: $y-3=\dfrac{3}{2}(x+1)$ General form: $3x-2y+9=0$ Passing through $(-1,3)$ and parallel to the line whose equation is $3x-2y-5=0$ Rewrite the equation given in slope-intercept form by solving it for $y$: $3x-2y-5=0$ $3x-5=2y$ $2y=3x-5$ $y=\dfrac{3}{2}x-\dfrac{5}{2}$ Since the slope-intercept form of the equation of a line is $y=mx+b$, the slope of the given equation can be identified as $m=\dfrac{3}{2}$ Since parallel lines have the same slope, the slope of the line whose equation must be found is also $m=\dfrac{3}{2}$ The slope of the line whose equation must be found and a point through which it passes are now known. Substitute them into the point-slope form of the equation of a line formula, which is $y-y_{1}=m(x-x_{1})$ $y-3=\dfrac{3}{2}[x-(-1)]$ $y-3=\dfrac{3}{2}(x+1)$ This is the point-slope form of the equation of the line. Take all terms to the right side and simplify: $0=\dfrac{3}{2}(x+1)-y+3$ $0=\dfrac{3}{2}x+\dfrac{3}{2}-y+3$ $0=\dfrac{3}{2}x-y+\dfrac{9}{2}$ Rearrange: $\dfrac{3}{2}x-y+\dfrac{9}{2}=0$ Multiply the whole equation by $2$: $2\Big(\dfrac{3}{2}x-y+\dfrac{9}{2}=0\Big)$ $3x-2y+9=0$ This is the general form of the equation of the line
# Cramer’s Rule ## What is Cramer’s Rule and how to solve linear equation problems using Cramer’s rule? In mathematics, to solve linear equations, we use matrix form. Matrix is the rectangular array of elements enclosed in brackets “[]”. There are some methods and rules of matrix used to solve the linear equations. Creamer’s rule is one of them. Determinants are also used in this rule. In this post, we will learn the basic concept of this rule and use some examples to understand how to solve the problems of linear equations by using Cramer’s rule. ## What is Cramer’s Rule? In matrices, Cramer’s rule is used to find the values of the unknown variables present in the linear equations. Cramer’s rule is used to find the system of the equations and the equations must be linear. In this method, all the calculations are based on determinants so this method is also named the determinant method. The rectangular array of matrices is not used in Cramer’s rule, so we use the square matrix form. In the square matrix, there is an equal number of rows and columns are present. Since the definition of the matrix states that the rectangular array of the matrix then the question must arise in the mind why do we use the square matrix for Cramer’s rule? The answer is that all the square matrices are rectangular matrices in general. So, we always conclude that the square matrix is rectangular. ## The formula of Cramer’s rule Cramer’s rule is following the equation such, Ax = b In this equation, A is the square matrix having an equal number of rows and columns, b is the right-hand side of the linear equation, and X is the unknown variables to be found. The equations are taken in general in the form of. A1x + B1y = C1 A2x + B2y = C2 The matrix form of this general linear equation, along with x-matrix and y-matrix is given by The first step of Cramer’s rule is to convert the matrix into matrix form. And then find the determinant of all the square matrices that we made. In Cramer’s rule, if the determinant is zero of a square matrix, then this rule is not applicable. The following forms are used to calculate the unknown variables x and y of the linear equations. To determine the value of x X = det[x]/det To determine the value of y y = det[y]/det ## How to solve linear equation problems by using Cramer’s Rule? To solve the linear equation problems, you must be familiar with the basics formula and working of Cramer’s rule. In Cramer’s rule, determinants are used, determinants are to be calculated by applying “||” for matrices instead of [] brackets. There is an online tool Cramer’s rule calculator used for such calculations to reduce the difficulty of large calculations. The steps to determine the determinant for 2x2 matrices are very simple just follow the given method. abcd = (a x d) – (b x c) The steps to determine the determinant for 3x3 matrices are pretty difficult and lengthy just follow the given method. ### Example 1 Find the unknowns of the linear equation by using Cramer’s rule. 5x + 3y = 6 x + 3y = 2 Solution Step 1: First write in AX = B form. 5313x y = 6 2 Step 2: Find the determinant of square matrix A. A = 5313 Step 3: Write the formula to determine the determinant of square matrix A. abcd = (a x d) – (b x c) Step 4: Place the values of each term in the above formula. 5313 = (5 x 3) – (3 x 1) 5313 = 15 – 3 5313 = 12 Det A = 12 Step 5: Now take the values of b in the first column of matrix A and give the name of x value. X values = 6323 Step 6: Find the determinant of the above matrix. 6323 = (6 x 3) – (3 x 2) 6323 = 18 – 6 6323 = 12 detx = 12 Step 7: Now take the values of b in the second column of matrix A and give the name of y value. X values = 5612 Step 8: Find the determinant of the above matrix. 5612 = (5 x 2) – (6 x 1) 5612 = 10 – 6 5612 = 4 dety = 4 Step 9: By the formula of Cramer’s rule to find the unknown variables, we have. X = detx/det Y = dety/det Step 10: Place the values of the calculated determinants in the above formulas. X = 12/12 = 1 Y = 4/12 = 1/3 = 0.3333 ### Example 2 Find the unknowns of the linear equation by using Cramer’s rule. 5x – 4y = 6 3x + 9y = 2 Solution Step 1: First write in AX = B form. 5-439x y = 6 2 Step 2: Find the determinant of square matrix A. A = 5-439 Step 3: Write the formula to determine the determinant of square matrix A. abcd = (a x d) – (b x c) Step 4: Place the values of each term in the above formula. 5-439 = (5 x 9) – (3 x (-4)) 5-439 = 45 – (-12) 5-439 = 45 + 12 5-439 = 57 Det A = 57 Step 5: Now take the values of b in the first column of matrix A and give the name of x value. X values = 6-429 Step 6: Find the determinant of the above matrix. 6-429 = (6 x 9) – (-4 x 2) 6-429 = 54 – (-8) 6-429 = 54 + 8 6-429 = 62 detx = 62 Step 7: Now take the values of b in the second column of matrix A and give the name of y value. X values = 5632 Step 8: Find the determinant of the above matrix. 5632 = (5 x 2) – (6 x 3) 5632 = 10 – 18 5632 = -8 dety = -8 Step 9: By the formula of Cramer’s rule to find the unknown variables, we have. X = detx/det Y = dety/det Step 10: Place the values of the calculated determinants in the above formulas. X = 62/57 = 1.0877 Y = -8/57 = -0.1404 ## Summary Now you grab all the basis of Cramer’s rule. The working of this rule is very simple, just some practice is required, this rule is also defined in the vedic maths. From these given examples, you can also solve Cramer’s rule for 3x3 matrices by following the same steps.
# Lesson: Multiplying and dividing integers 3783 Views 2 Favorites ### Lesson Objective We will be able to multiply and divide integers. ### Lesson Plan Do Now: (5 minutes) 1) -18 - - 7 2) Identify the coefficient in the following equation: 3x + 2 = 9 3) 12 – 17 Direct Instruction:(15 minutes) Yesterday we learned that when adding and subtracting integers, it’s important to remember “same signs add, different signs subtract”. Today we are going to take our knowledge of integer operations one step further and learn how to multiply and divide integers. When we are multiplying integers, we are multiplying both the numbers and the signs. For example, -x time – y gives us –x-y. We remember from last week that same signs means add and different signs means subtract. So since we have two negatives, that really means that it is a positive. If I multiply –x times y, I get – xy. Different signs, subtract, so I get –xy as my final answer. We can also think about multiplication as repeated addition.  If I have -4 x 2, it means I have -4 2 times, which is -4 + -4.  If -4 + -4 = 8 then -4 x 2 = - 8 Students will write the following rule and examples in their notes: Rule:  When multiplying integers, two negatives become a positive product. A negative and a positive make a negative product. Ex. – 5 x 3                 Ex. -4 x -5                       Ex. – 8 x 3 We can use what we know about fact families to help us with integer division. Dividing a negative and a positive gives us a negative. Dividing two negatives gives us a positive. Have students write the following rule and examples in their notes: Rule: A negative divided by a negative is a positive.  A positive divided by a negative is a negative. Ex. -18 ÷ -3 = 6                         Ex. 24 ÷ -3 = -8 Ex. -12 ÷ 3 = -4 Guided Practice: (5 minutes) Complete the following problems as a class: -3 x 6 12 ÷ -2 -12 x – 9 -72 ÷ -6 Additional reinforcement of the skill can come from flashcards with multiplying and dividing integers. Assessment: (5 minutes) Multiplying and dividing integers work-out ### Lesson Resources multiplying and dividing integers work out   Assessment 1,786
Search IntMath Close 450+ Math Lessons written by Math Professors and Teachers 5 Million+ Students Helped Each Year 1200+ Articles Written by Math Educators and Enthusiasts Simplifying and Teaching Math for Over 23 Years # Triangle Congruence Theorem: All You Need to Know Geometry can be a difficult subject for many students. However, understanding the triangle congruence theorem is critical for success in the class. This blog post will explain everything you need to know about the triangle congruence theorem so that you can ace your next test! ## What is the Triangle Congruence Theorem? In geometry, the triangle congruence theorem states that if two triangles have all three sides equal, then the triangles arecongruent. This theorem is also sometimes referred to as the SSS congruence criterion. ## How to Prove the Triangle Congruence Theorem? There are a few different ways that you can prove the triangle congruence theorem. One way is using a technique called proof by contradiction. Proof by contradiction works by assuming that the opposite of what you're trying to prove is true and then showing that this leads to a contradiction. For example, let's say you're trying to prove that all cats are animals. To do this using proof by contradiction, you would first assume that there exists a cat that is not an animal. But since we know that all cats are animals, this assumption must be false and therefore our original statement must be true! We can use a similar approach to prove the triangle congruence theorem. Assume that two triangles are not congruent even though they have all three sides equal. This means that the triangles must be different shapes, which contradicts our initial assumption that the triangles were congruent! Therefore, we can conclude that if two triangles have all three sides equal, then they are indeed congruent. Another way to prove the triangle congruence theorem is by using a technique called induction. Induction works by proving a statement true for a base case (usually n = 1 or n = 0) and then showing that if it's true for one case, it must be true for the next case. For example, let's say we want to prove that 1 + 2 + 3 + ... + n = n(n+1)/2 for all positive integers n. We can start with our base case of n = 1: 1 = 1(1+1)/2 which is indeed true! Now let's assume that it's true for some arbitrary case of k: 1 + 2 + 3 + ... + k = k(k+1)/2 . We want to show that it must also be true for k+1: 1 + 2 + 3 + ... + k+(k+1) = (k+1)(k+2)/2 . We can do this by simply adding k+1 to both sides of our equation: 1 + 2 + 3 + ... + k+(k+1) = k(k+1)/2+(k+1) = (k+1)(k+(k+1))/2 , which is exactly what we wanted to show! Therefore, by induction, we can conclude that 1 + 2 + 3 + ... n = n(n+1)/2 for all positive integers as desired. ## Conclusion The triangle congruence theorem is a critical concept in geometry. By understanding how to proves this theorem using different methods, you'll be well on your way to getting an A in your geometry class! ## FAQ ### What is the Triangle Congruence Theorem? The Triangle Congruence Theorem states that two triangles are congruent if all three sides of one triangle are equal to the corresponding three sides of the other triangle. This theorem is a fundamental concept in geometry and can be used to compare and contrast the properties of different triangles. ### How is the Triangle Congruence Theorem used? The Triangle Congruence Theorem can be used to solve various geometric problems, such as finding the area or perimeter of a triangle. Additionally, this theorem can also be used to prove whether two sides are equal, which can help determine the congruence of two triangles. ### What is the importance of the Triangle Congruence Theorem? The Triangle Congruence Theorem is an important tool for analyzing geometry and understanding the properties of different shapes. It allows us to make deductions about a triangle based on its sides, which can be useful when studying various topics in geometry. Additionally, this theorem can be used to prove theorems and problems related to angles, triangles, and other shapes in geometry. By understanding the Triangle Congruence Theorem, we can gain a better understanding of geometry overall. ### Are there any special cases for the Triangle Congruence Theorem? Yes, there are special cases for the Triangle Congruence Theorem. For instance, if two triangles have only two sides equal to each other, then they can still be considered congruent according to this theorem. Additionally, if two angles and a side of one triangle are equal to their respective counterparts in another triangle, then they can also be considered congruent. These special cases can help us to better understand the properties of triangles and other shapes in geometry.
# Systems of Trigonometric Equations A system of trigonometric equations is a set of equations composed either only from trigonometric equations or from trigonometric and algebraic equations. In this section, we consider several types of systems of trigonometric equations with two variables $$x$$ and $$y$$ and describe possible ways to solve them. ## Simplest Systems of Equations It may happen that one of the equations of the system contains trigonometric functions in the unknowns x and y, and the other equation is linear in x and y. In such case, we act in an obvious way: one of the unknowns is expressed from the linear equation and substitute into another equation of the system. ### Example Solve the system: $\left\{ \begin{array}{l} x + y = \frac{\pi}{2}\\ \sin x + \sin y = 1 \end{array} \right..$ Solution. From the first equation, we express $$y$$ in terms of $$x:$$ $y = \frac{\pi}{2} - x,$ and substitute into the second equation: $\sin x + \sin y = 1, \Rightarrow \sin x + \sin \left({\frac{\pi}{2} - x}\right) = 1, \Rightarrow \sin x + \cos x = 1.$ We got a linear trigonometric equation. Let's solve it using the R method. Multiply both sides of the equation by $$\frac{\sqrt{2}}{2}$$ and convert the expression in the left-hand side into one trig function: $\frac{\sqrt{2}}{2}\sin x + \frac{\sqrt{2}}{2}\cos x = \frac{\sqrt{2}}{2}, \Rightarrow \cos\frac{\pi}{4}\sin x + \sin\frac{\pi}{4}\cos x = \frac{\sqrt{2}}{2}, \Rightarrow \sin\left({x + \frac{\pi}{4}}\right) = \frac{\sqrt{2}}{2}.$ The solution of this equation is written in the form of two branches: $x + \frac{\pi}{4} = \frac{\pi}{4} + 2\pi n, \Rightarrow x_1 = 2\pi n,$ $x + \frac{\pi}{4} = \frac{3\pi}{4} + 2\pi n, \Rightarrow x_2 = \frac{\pi}{2} + 2\pi n,$ where $$n \in \mathbb{Z}.$$ Find now the corresponding values of $$y:$$ $y_1 = \frac{\pi}{2} - x_1 = \frac{\pi}{2} - 2\pi n,$ $y_2 = \frac{\pi}{2} - x_2 = \cancel{\frac{\pi}{2}} - \cancel{\frac{\pi}{2}} - 2\pi n = - 2\pi n, \;n \in \mathbb{Z}.$ The final answer is written as $$\left({x,y}\right)$$ pairs: $\left({2\pi n,\frac{\pi}{2} - 2\pi n}\right), \left({\frac{\pi}{2} + 2\pi n,- 2\pi n}\right),\;n\in\mathbb{Z}.$ ## Reduction of a Trigonometric System to an Algebraic One In a number of cases, the trigonometric system can be reduced to a system of algebraic equations by a suitable change of variables. ### Example Solve the system: $\left\{ \begin{array}{l} \sin x + \cos y = 1\\ \cos2x - \cos 2y = 1 \end{array} \right..$ Solution. Remember that $\cos2x = 1 - 2\sin^2x,\;\cos2y = 2\cos^2y - 1.$ Then we can make the substitution $$u = \sin x,$$ $$v = \cos y$$ which leads to an algebraic system with respect to $$u$$ and $$v:$$ $\left\{ \begin{array}{l} u + v = 1\\ 1 - 2u^2 - \left({2v^2 - 1}\right) = 1 \end{array} \right., \Rightarrow \left\{ \begin{array}{l} u + v = 1\\ 1 - 2u^2 - 2v^2 + 1 = 1 \end{array} \right., \Rightarrow \left\{ \begin{array}{l} u + v = 1\\ u^2 + v^2 = \frac{1}{2} \end{array} \right..$ Express $$v$$ from the first equation: $$v = 1 - u.$$ Substituting $$v$$ into the second equation, we get $\left\{ \begin{array}{l} v = 1 - u\\ u^2 + \left({1 - u}\right)^2 = \frac{1}{2} \end{array} \right., \Rightarrow \left\{ \begin{array}{l} v = 1 - u\\ u^2 + 1 - 2u + u^2 = \frac{1}{2} \end{array} \right., \Rightarrow \left\{ \begin{array}{l} v = 1 - u\\ 4u^2 - 4u + 1 = 0 \end{array} \right., \Rightarrow \left\{ \begin{array}{l} v = 1 - u\\ \left({2u - 1}\right)^2 = 0 \end{array} \right., \Rightarrow \left\{ \begin{array}{l} v = \frac{1}{2}\\ u = \frac{1}{2} \end{array} \right..$ From here we find $$x$$ and $$y:$$ $\sin x = u = \frac{1}{2}, \Rightarrow x = \left({-1}\right)^n\frac{\pi}{6} + \pi n,\;n\in\mathbb{Z};$ $\cos y = v = \frac{1}{2}, \Rightarrow y = \pm\frac{\pi}{3} + 2\pi k,\;k\in\mathbb{Z}.$ $\left({\left({-1}\right)^n\frac{\pi}{6} + \pi n,\,\pm\frac{\pi}{3} + 2\pi k}\right), \;n,k\in\mathbb{Z}.$ ## Addition and Subtraction of Equations Sometimes a trigonometric system can be simplified by adding or subtracting equations. ### Example Solve the system: $\left\{ \begin{array}{l} \sin x \cos y = \frac{1}{4}\\ \cos x \sin y = \frac{3}{4} \end{array} \right..$ Solution. $\sin x \cos y + \cos x \sin y = \frac{1}{4} + \frac{3}{4},$ $\Rightarrow \sin \left({x + y}\right) = 1.$ Similarly, subtracting the second equation from the first one and using the sine subtraction identity, we have $\sin x \cos y - \cos x \sin y = \frac{1}{4} - \frac{3}{4},$ $\Rightarrow \sin \left({x - y}\right) = -\frac{1}{2}.$ As a result, we obtain the following equivalent system of equations: $\left\{ \begin{array}{l} \sin \left({x + y}\right) = 1\\ \sin \left({x - y}\right) = -\frac{1}{2} \end{array} \right..$ The solution of the first equation is written in the form $x + y = \frac{\pi}{2} + 2\pi n,\;n\in\mathbb{Z}.$ The second equation has two families of solutions: $x - y = -\frac{\pi}{6} + 2\pi k,$ $x - y = -\frac{5\pi}{6} + 2\pi k,\;k\in\mathbb{Z}.$ Accordingly, we get two solutions: $\left[ \begin{array}{l} \left\{ \begin{array}{l} x + y = \frac{\pi}{2} + 2\pi n\\ x - y = -\frac{\pi}{6} + 2\pi k \end{array} \right.\\ \left\{ \begin{array}{l} x + y = \frac{\pi}{2} + 2\pi n\\ x - y = -\frac{5\pi}{6} + 2\pi k \end{array} \right. \end{array} \right.,$ where $$n,k \in \mathbb{Z}.$$ Solve now each system with respect to $$x$$ and $$y.$$ To do this, add and subtract the equations again: $\left[ \begin{array}{l} \left\{ \begin{array}{l} 2x = \frac{\pi}{2} -\frac{\pi}{6} + 2\pi\left({n + k}\right)\\ 2y = \frac{\pi}{2} +\frac{\pi}{6} +2\pi\left({n - k}\right) \end{array} \right.\\ \left\{ \begin{array}{l} 2x = \frac{\pi}{2} -\frac{5\pi}{6} + 2\pi\left({n + k}\right)\\ 2y = \frac{\pi}{2} +\frac{5\pi}{6} +2\pi\left({n - k}\right) \end{array} \right. \end{array} \right., \Rightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x = \frac{\pi}{6} + \pi\left({n + k}\right)\\ y = \frac{\pi}{3} + \pi\left({n - k}\right) \end{array} \right.\\ \left\{ \begin{array}{l} x = -\frac{\pi}{6} + \pi\left({n + k}\right)\\ y = \frac{2\pi}{3} + \pi\left({n - k}\right) \end{array} \right. \end{array} \right..$ $\left({\frac{\pi}{6} + \pi\left({n + k}\right),\,\frac{\pi}{3} + \pi\left({n - k}\right)}\right),\left({-\frac{\pi}{6} + \pi\left({n + k}\right),\,\frac{2\pi}{3} + \pi\left({n - k}\right)}\right), \;n,k\in\mathbb{Z}.$ ## Multiplication and Division of Equations Sometimes you can come to a solution by multiplying or dividing the equations by each other. ### Example Solve the system: $\left\{ \begin{array}{l} \sin x + \sin y = 1\\ \cos x - \cos y = \sqrt{3} \end{array} \right..$ Solution. Using sum-to-product identities, we write the system as $\left\{ \begin{array}{l} 2\sin\frac{x+y}{2}\cos\frac{x-y}{2} = 1\\ -2\sin\frac{x+y}{2}\sin\frac{x-y}{2} = \sqrt{3} \end{array} \right..$ Denote $$\alpha = \frac{x+y}{2},$$ $$\beta = \frac{x-y}{2}$$ and rewrite the system in the following form: $\left\{ \begin{array}{l} 2\sin\alpha\cos\beta = 1\\ -2\sin\alpha\sin\beta = \sqrt{3} \end{array} \right..$ It is clear that $$\sin\alpha \ne 0.$$ Divide the second equation by the first: $-\tan\beta = \sqrt{3}, \Rightarrow \tan\beta = -\sqrt{3}, \Rightarrow \beta = \arctan\left({-\sqrt{3}}\right) + \pi n = -\frac{\pi}{3} + \pi n, \;n\in\mathbb{Z}.$ The angle $$\beta$$ takes two values on the unit circle. Represent them in separate formulas: $\beta_1 = -\frac{\pi}{3} + 2\pi n,\; \beta_2 = \frac{2\pi}{3} + 2\pi n,\;n\in\mathbb{Z}.$ Calculate the corresponding values of $$\alpha:$$ $\beta_1 = -\frac{\pi}{3} + 2\pi n, \Rightarrow \cos\beta_1 = \cos\left({-\frac{\pi}{3}}\right) = \frac{1}{2}, \Rightarrow 2\sin\alpha_1\cos\beta_1 = 1, \Rightarrow \sin\alpha_1 = 1, \Rightarrow \alpha_1 = \frac{\pi}{2} + 2\pi k,\;k\in\mathbb{Z}.$ $\beta_2 = \frac{2\pi}{3} + 2\pi n, \Rightarrow \cos\beta_2 = \cos\left({\frac{2\pi}{3}}\right) = -\frac{1}{2}, \Rightarrow 2\sin\alpha_2\cos\beta_2 = 1, \Rightarrow \sin\alpha_2 = -1, \Rightarrow \alpha_2 = -\frac{\pi}{2} + 2\pi k,\;k\in\mathbb{Z}.$ So we get the following solutions for $$\alpha$$ and $$\beta:$$ $\left[ \begin{array}{l} \left\{ \begin{array}{l} \alpha_1 = \frac{\pi}{2} + 2\pi k\\ \beta_1 = -\frac{\pi}{3} + 2\pi n \end{array} \right.\\ \left\{ \begin{array}{l} \alpha_2 = -\frac{\pi}{2} + 2\pi k\\ \beta_2 = \frac{2\pi}{3} + 2\pi n \end{array} \right. \end{array} \right.,\;k,n\in\mathbb{Z}.$ Hence, $\left[ \begin{array}{l} \left\{ \begin{array}{l} \frac{x+y}{2} = \frac{\pi}{2} + 2\pi k\\ \frac{x-y}{2} = -\frac{\pi}{3} + 2\pi n \end{array} \right.\\ \left\{ \begin{array}{l} \frac{x+y}{2} = -\frac{\pi}{2} + 2\pi k\\ \frac{x-y}{2} = \frac{2\pi}{3} + 2\pi n \end{array} \right. \end{array} \right., \Rightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x = \frac{\pi}{2} - \frac{\pi}{3} + 2\pi\left({k + n}\right)\\ y = \frac{\pi}{2} + \frac{\pi}{3} + 2\pi\left({k - n}\right) \end{array} \right.\\ \left\{ \begin{array}{l} x = -\frac{\pi}{2} + \frac{2\pi}{3} + 2\pi\left({k + n}\right)\\ y = -\frac{\pi}{2} - \frac{2\pi}{3} + 2\pi\left({k - n}\right) \end{array} \right. \end{array} \right., \Rightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x = \frac{\pi}{6} + 2\pi\left({k + n}\right)\\ y = \frac{5\pi}{6} + 2\pi\left({k - n}\right) \end{array} \right.\\ \left\{ \begin{array}{l} x = \frac{\pi}{6} + 2\pi\left({k + n}\right)\\ y = -\frac{7\pi}{6} + 2\pi\left({k - n}\right) \end{array} \right. \end{array} \right.,$ where $$k,n \in \mathbb{Z}.$$ $\left({\frac{\pi}{6} + 2\pi\left({k + n}\right),\,\frac{5\pi}{6} + 2\pi\left({k - n}\right)}\right),\left({\frac{\pi}{6} + 2\pi\left({k + n}\right),\,-\frac{7\pi}{6} + 2\pi\left({k - n}\right)}\right), \;k,n\in\mathbb{Z}.$
Sales Toll Free No: 1-855-666-7446 # Domain and Range Top Sub Topics Functions is a very important topic in mathematics. These are relation which give list of output corresponding to a set of inputs. While working with functions, students come across the two very useful concepts domain and range often. Domain of the function means the group coming from all x values or even all inputs. These input values when applied to the function, must produce corresponding output values. Domain of the function is the set of all the so-called values or volumes that go inside function to present outputs. The set of all the output values is termed as a range. We can say how the range of a function will be the set of many possible output beliefs corresponding to permissible input values. The range is usually calculated as hundreds of values that go toward Y axis. The number is an interval among the uppermost number for the minor number. ## Domain and Range of a Function For any function y = f (x), the domain is the set of all the possible values of x, where x is the independent variable. Range of a function is the set of all the corresponding values of y, for every value of x, Given below are some examples: Example 1: Find the domain value of the function given below F(x) = $\frac{1}{x + 15}$ Solution: Equate the denominator to zero for the above function x + 15 = 0 x =  - 15 Set of all variable of x is the domain function f, the interval is the (- $\infty$, - 15) $\cup$  (- 15, $\infty$) Therefore the domain value of the given function is (-$\infty$, - 15) $\cup$ (-15, $\infty$) ## Domain and Range of a Circle Circle is the two dimensional closed geometrical figure where the distance from the centre point to the locus coming from all points is constant. Domain of circle: All the feasible x values are considered to be domain of circle. Range on the circle: All the feasible y values are considered to be range of circle. The standard way of the circle is actually (x - a)$^2$ + (y - b)$^2$ = r$^2$ Here r is the radius of circle (a, b) is the centre of the circle. Given below is an example: Example 1:  Find the domain and range of the circle : x$^{2}$ + y$^{2}$ = 49 Solution: The general form of a circle is (x - a)$^{2}$ + (y - b)$^{2}$ = r$^{2}$ Now, x$^{2}$ + y$^{2}$ = 49 $\rightarrow$ (x - 0)$^{2}$ + (y - 0)$^{2}$ = 7$^{2}$ so that x$^{2}$ + y$^{2}$ = 49  is a circle with its center at the origin and radius 7. If you construct that circle, it's easy to see that x runs from - 7 to  + 7 and y runs from - 7 to  + 7. Therefore, DOMAIN = {x│ - 7 $\leq$ x $\leq$ 7 } RANGE = {y│ -7 $\leq$ y $\leq$ 7 } ## Domain and Range of Exponential and Logarithmic Functions Exponential Functions: In general, the domain of the function y = a$^{x}$ is the possible x values and the range is the possible y values of that function. For a function y = a$^{x}$ , where a > 1 (or) 0 < a < 1, The domain consists of all the real numbers which can be written in the interval form as $( -\infty , \infty)$. The Range consists of all positive real numbers $(0 , \infty)$. We can also observe that $\lim_{x \to \infty} a^{x} = \infty$ And, for no value of x, a$^{x}$ is 0. The graph will never meet the x-axis. It is likely to meet the x - axis at a very small value of x when a > 1, and for a very large value of x, when a< 1. Hence, for the above function, the horizontal asymptote is x-axis. Logarithmic Functions: Log is the inverse function of exponentiation Consider a function y=log(x−3) We are trying to find the Domain and the Range of this function, recalling that: Domain: Includes all values of x for which the function is defined. Range: Includes all values y for which there is some x such that y=log(x−3). It makes no sense to write y = log(a) when a $\leq$ 0 because log(a) is defined only for positive a. So in this problem, y = log(x - 3), is defined if and only if x − 3 > 0 $\rightarrow$   x > 3, and that gives domain x  $\in$ (3, $\infty$). Range of y is all of R. Therefore the domain is $\in$ (3, $\infty$). Range : y $\in$ R
# What Is the St. Petersburg Paradox? You’re on the streets of St. Petersburg, Russia, and an old man proposes the following game. He flips a coin (and will borrow one of yours if you don’t trust that his is a fair one). If it lands tails up then you lose and the game is over. If the coin lands heads up then you win one ruble and the game continues. The coin is tossed again. If it is tails, then the game ends. If it is heads, then you win an additional two rubles. The game continues in this fashion. For each successive head we double our winnings from the previous round, but at the sign of the first tail, the game is done. How much would you pay to play this game? When we consider the expected value of this game, you should jump at the chance, no matter what the cost is to play. However, from the description above, you probably wouldn’t be willing to pay much. After all, there is a 50% probability of winning nothing. This is what is known as the St. Petersburg Paradox, named due to the 1738 publication of Daniel Bernoulli Commentaries of the Imperial Academy of Science of Saint Petersburg. ## Some Probabilities Let's begin by calculating probabilities associated with this game. The probability that a fair coin lands heads up is 1/2. Each coin toss is an independent event and so we multiply probabilities possibly with the use of a tree diagram. • The probability of two heads in a row is (1/2)) x (1/2) = 1/4. • The probability of three heads in a row is (1/2) x (1/2) x (1/2) = 1/8. • To express the probability of n heads in a row, where n is a positive whole number we use exponents to write 1/2n. ## Some Payouts Now let's move on and see if we can generalize what the winnings would be in each round. • If you have a head in the first round you win one ruble for that round. • If there is a head in the second round you win two rubles in that round. • If there is a head in the third round, then you win four rubles in that round. • If you have been lucky enough to make it all the way to the nth round, then you will win 2n-1 rubles in that round. ## Expected Value of the Game The expected value of a game tells us what the winnings would average out to be if you played the game many, many times. To calculate the expected value, we multiply the value of the winnings from each round with the probability of getting to this round, and then add all of these products together. • From the first round, you have probability 1/2 and winnings of 1 ruble: 1/2 x 1 = 1/2 • From the second round, you have probability 1/4 and winnings of 2 rubles: 1/4 x 2 = 1/2 • From the first round, you have probability 1/8 and winnings of 4 rubles: 1/8 x 4 = 1/2 • From the first round, you have probability 1/16 and winnings of 8 rubles: 1/16 x 8 = 1/2 • From the first round, you have probability 1/2n and winnings of 2n-1 rubles: 1/2n x 2n-1 = 1/2 The value from each round is 1/2, and adding the results from the first n rounds together gives us an expected value of n/2 rubles. Since n can be any positive whole number, the expected value is limitless.
# Difference Between Permutation and Combination Edited by Diffzy | Updated on: September 29, 2023 Our articles are well-researched We make unbiased comparisons Our content is free to access We are a one-stop platform for finding differences and comparisons We compare similar terms in both tabular forms as well as in points ## Introduction Mathematics is surrounded by many branches of vast areas like statistics, combinatorics, numerical analysis, probability, geometry, and many more. One of the known areas in mathematics is Combinatorics. Combinatorics deals with the arrangement and selection within a finite system or structure. It includes two most important parts- permutation and combination. Permutation is how a group of objects can be arranged or selected. The order of the arrangements matters with the permutation. It can be understood by different examples and activities. Say a teacher provides a set of letters and asks the students to arrange them in various orders. The arrangement of the letters by the students in all possible ways is permutation. While in combination, the order of selection does not matter. It is the way that determines the number of possible arrangements in a group of objects. Say, we have a set of 3 numbers P, Q, and R. The selection of the numbers from each set in as many as possible ways is a combination. Permutation and combination are two different ways of grouping items of each set into sub-sets. In some scenarios, they share the same properties and ways. But based on terms, sequence, and concepts, they are different from each other. Let's understand the difference between them in detail. ## Permutation vs Combination The concepts of permutation and combination varying from each other are full of conundrums. The formulas and the meaning of both of these terms are a little confusing and seem like a riddle to solve. Permutation simply means ordering and counting the arrangements of the objects. It focuses on the lists where order matters. The factorials of permutation involve all possible results of an event. The keywords of the concepts are- arrange, line up, and order. It uses the fundamental counting principle. On the other hand, the combination means selecting objects from the group of the sets. It focuses on a selection of items in such a way that its ordering does not matter. In combination, the order of an arrangement is not important. It is used when the same type of items or objects are to be sorted. The keywords included in the combination are- choose, select, and pick. Permutation and combination are the most fundamental concepts in Mathematics, which introduced the students to a new branch i.e., Combinatorics. So, it is necessary to differentiate them based on formula usage, value, key terms, and sample problems as well. ## What is Permutation? In simple words, Permutation is the arrangement of a set of objects in a particular way or order. It concerns the selection and arrangement of the objects. That is why permutation is also known as an ordered combination. Permutation is denoted in many ways:- • P (n,r) • Pnr • nPr • nPr • Pn , r The elements of the sets are arranged in a sequence or linear order. A permutation is the choice of specific items from a set of items without replacement. In the permutation, the order of the objects matters and is important. It is used in every branch of mathematics and other fields of science. A permutation is used for analyzing and sorting algorithms in computer science. It is used to describe the states of the particles in quantum physics, and it is used to describe RNA sequences in biology. Permutation also comes in great use in daily life. There are many real-life examples of permutation, like arranging people, picking first, second, and third place, passwords, arranging digits, alphabets, colors, etc. Permutation can be calculated using various formulas. The main formula is:- P (n,r) = n ! ÷ (n-r)! , where n is the total items in the set, r is the items taken for permutation, and ! denotes the factorial. The simple question that generalizes the expression of the formula is: How many ways can you arrange 'r' from a set of 'n' if the order matters? Permutation can also be calculated without formula, by writing all possible permutations. The easiest approach is visualizing the permutation in many ways. A sequence of a 3-digit keypad can be arranged. Using the digits 0 through 9 and using a specific digit only once on the keypad, the number of permutations are:- P (10,3) = 10 ! ÷ (10-3) ! = 10 ! ÷ 7 ! = 10×9×8 = 720 The formula of permutation can be derived in the following way:- By the fundamental counting principle, P (n,r) = n . (n-1) . (n-2) . (n-3) ---------- (n-(r-1)) ways = n . (n-1) . (n-2) . (n-3) ----------- (n-r+1) --------- (1) Multiplying and dividing (1) by (n-r) (n-r-1) (n-r-2) --------- 3. 2. 1 P (n,r) = [n . (n-1) . (n-2) . (n-3) ---------- (n-r+1)] = [(n-r) (n-r-1) (n-r-2) -------- 3. 2. 1] [(n-r) (n-r-1) (n-r-2) -------- 3 . 2. 1 ] P (n,r) = (n !) / (n-r) ! Hence, derived. There are five formulas for permutation in total, which are used in different situations:- • Permutation formula without repetition- When ‘r’ thing from ‘n’ things have to be arranged without repetition. nP r = (n !) / (n-r) ! • Permutation formula with repetition- When 'r' from 'n' has to be arranged in the repetition. n×n×n× --------- ×n (r times) = nr • Taken all at a time- When several ways of arranging 'n' among themselves are nothing. nP n = (n !) / (n-n) ! = n! o! = n! / 1 = n ! • Same sets of data- When all of 'n' things are not different and some of them are the same. n! / (s1 ! × s2 ! × ------- × sn ! ) • Circular permutation formula- When several ways of arranging 'n' different numbers of things in a circle. (n-1) ! ### Examples 1. How many 4 letter words, with or without meaning, can be formed out of the letters of the word SWING when repetition of letters is not allowed? So, total permutations = P(n,r) = 5 ! 5-4! = 5×4×3×2×1×1 = 120 Thus, 120 permutations are possible for the word SWING. 2. How many 2 letter words with or without meaning can be formed out of the letters of the word SMOKE when repetition of words is allowed? r = 2 P = nr = 52 = 25 Thus, 25 permutations are possible for the word SMOKE. The fundamental counting principle states that if one operation can be performed in 'm' ways and there are 'n' ways of performing a second operation, then the number of ways of performing the two operations together is m×n. This principle can be extended to the situation in which different operations are done in m, n, p -------- ways. In this situation, the number of ways of performing all the operations one after the other is m×n×p×---------- and so on. Some examples of the sample problems of permutations:- 1. Find the number of words, with or without meaning, that can be formed with the letters of the word PARK. Answer: Total words= P (n,r) = 4! = 4×3×2×1 = 24 words Thus, 24 words can be formed with the letters of the word PARK. 2. A zip code contains 5 digits. How many different zip codes can be made with the digits 0-9 if no digit is used more than once and the first digit is not 0? Answer: Since 0 is not allowed, For the First possible total possible choices = 9 For the next 4 positions, P = 9×P (9,4) = 9× 9 ! 9-4! = 9× 9 !5 ! = 9×9×8×7×6×55 ! = 27, 216 Thus, 27,216 zip codes can be made. 3. In how many ways can 5 different books be arranged on a shelf? = 5 ! / (5-5) ! = 5 ! / 0 ! = 5×4×3×2×1 = 120 Thus, in 120 ways, books can be arranged. ## What is Combination? Combination is a method of arrangement of the items where the order in which the items or the objects are chosen or selected does not matter. In other words, it means selecting things with no importance of their order. Apart from combinatorics, combinations are also studied in other fields like mathematics, finance, computer programming, probability theory, and genetics. In real life also, it is used in the selection of menus, making a cup of coffee, picking a table and selecting members, lottery games, selection of clothes, etc. Combinations are also referred to as selections. The number of unique combinations out of a group of n objects is denoted by nCr. The general combination formula of the number of C of n different things taken r at a time is given:- nCr = n ! r ! n-r! , where 0≤ r ≤ n Another formula of combination is by using permutation:- C (n,r) = P (n,r) / r ! where r is the size of each permutation, n is the size of the set, n,r are non-negative integers, and ! is the factorial expression. The formula shows the number of ways a sample of 'r' items can be yielded from a larger set of 'n' distinguishable items. The formula of combination can be derived in the following ways:- C (n,r) = Total number of permutations/ number of ways to arrange r different objects By the fundamental principle of counting, Several ways to arrange r different objects in r ways = r! C (n,r) = P (n,r) / r ! = n ! (n-r) ! r! Hence derived. There are two types of combinations where the order of the objects does not matter:- 1. Combinations with Repetitions- Where repetition is allowed and the number of the items chosen is unlimited. (n + r -1r) 2. Combinations without repetitions- Where order of selection does not matter, and each object can be selected only once. rCn = n ! r ! n-r! ### Examples 1. A father asks his son to choose 4 items from the table. If the table has 18 items to choose from, how many different answers could the son give? n = 18 C (n,r) = n ! r ! n-r! = 18 !4 !18-4! = 18 !14 ! ×4! Thus, the son could give 3060 answers. 2. On the plane, there are 6 different points ( no 3 of them are lying on the same line). How many segments do you get by joining all the points? = 6!4 ! 2 ! = 6.52 = 15 Thus, by joining all the points, we will get 15 points. 3. If 18Cr = 18Cr +2 , find rC5. 18C18-r = 18Cr +2 18-r = r + 2 2r = 18-2 r = 16/2 = 8 Then, 8C5 = 8 !5 !8-5! = 8 ×7×63×2×1 = 56 ## Main Difference Between Permutation and Combination (In Points ) • Permutation is a way in which the ordering and counting of a set of items is done in a sequence. Combination is a way in which the selection of objects is done from the large sets. • French mathematician and engineer Augustin–Louis Cauchy played a huge role in developing the theory of permutation. On the other hand, French mathematicians Blaise Pascal and Pierre de Femat contributed to the theory of combination. • The word permutation is derived from the Anglo-French word 'permatacion', which means exchange or transformation. Whereas, the word combination is derived from the Late Latin word 'combinacyoun' which means the act of uniting in a whole. • Permutations are used for things of different kinds. While combinations are used for things of similar kinds. • Permutations are used for creating passwords, alphabets, and different seating arrangements. On the other hand, combinations are used for the selection of people, formation of teams, and groups of objects. • The keywords like arrangement, order, and unique indicate permutation. Whereas, the keywords like selection, choose, and pick indicate combination. • Permutation allows the repetition of objects. While combination allows the repetition of objects only in rare cases. • In science and fictional novels, permutation can be symbolized as an example to depict the arranging of a gadget or others. On the other hand, the combination can be represented as the layers of various traits in an individual or others. ## Conclusion In short, permutation and combination are two different methods to take a set of items or options and create subsets. While permutation deals with arranging and ordering items, combination is the selection and choosing of items. Sometimes, these two terms become confusing. So, it is important to note that the problem of determining whether something is a permutation or combination depends on the given problems or conditions. ### Category Styles: #### MLA Style Citation "Difference Between Permutation and Combination." Diffzy.com, 2024. Fri. 09 Aug. 2024. <https://www.diffzy.com/article/difference-between-permutation-and-combination>. Edited by Diffzy
# Into Math Grade 8 Module 13 Lesson 4 Answer Key Apply Volume We included HMH Into Math Grade 8 Answer Key PDF Module 13 Lesson 4 Apply Volume to make students experts in learning maths. ## HMH Into Math Grade 8 Module 13 Lesson 4 Answer Key Apply Volume I Can use the formulas for the volumes of cones, cylinders, and spheres to solve real-world problems. Step It Out Question 1. A youth sports club is holding its yearly tri-county tournament. The club is selling fruit cups in two different sizes. Fill in the missing information in the table after completing Parts A and B. A. Find the volume of the large fruit cup to the nearest hundredth. Use the given diameter to find the radius. Use 3.14 for π. V = πr2h r = $$\frac{d}{2}$$ = = (_______) = π (___________)2 (__________) ≈ (_________) (_________) (________) ≈ __________ Given dimensions: height = 7 inches diameter = 3 inches radius = d/2 = 3/2 = 1.5 Its volume is, V = πr2h V = 3.14 x 1.5 x 1.5 x 7 V = 49.45 The nearest hundred value is 50. B. Find the volume of the regular fruit cup to the nearest hundredth. Substitute in the given values and simplify. Use 3.14 for π. V = πr2h r = $$\frac{d}{2}$$ = = (_______) = π (___________)2 (__________) ≈ (_________) (_________) (________) ≈ __________ diameter = 2.8 r = d/2 = 2.8/2 = 1.4 The radius of the cylinder is, r = 1.4 inches Its height is, h = 6 inches Its volume is, V = πr2h V = 3.14 x 1.4 x 1.4 x 6 V = 36.92 The nearest value is 37 cubic inches. Turn and Talk Which is the better buy? Why? According to all measurements, we would think that the large fruit cup is better to buy. Only small differences are there in inches. in height 1 inch difference and in radius .5 inch difference. So if we see the measurements large one is better. Question 2. A paper cone cup has a height of 3.6 inches and a volume of 2.028π cubic inches. A. What is the radius of the cone? Use the formula for the volume of a cone. Substitute values for the volume V and the height h. The radius of the cone is _________ inches. The formula to calculate the volume of a cone, given the height and its base radius is: V = (1/3)πr2h cubic units The given dimensions are; height = 3.6 in volume = 2.028π substitute the values in the formula 2.028π = 1/3 x π x r^2 x 3.6 2.028π = 1.2 x π x r^2 2.028π/1.2π = r^2 1.69 = r^2 r = 1.3 inches Therefore, the radius of the cone is 1.3 inches. B. The diameter of the cone is ___________ inches. To find the diameter of the circle when the radius is given, you need to multiply the radius by 2. we get r = 1.3 Diameter = 2 * 1.3 Diameter = 2.6 inches Therefore, the diameter of the cone is 2.6 inches C. Estimate the volume of the cup in cubic inches. Use 3.14 for π and round to four decimal places. ___________ cubic inches The formula to calculate the volume of a cone, given the height and its base radius is: V = (1/3)πr2h cubic units V = 1/3 x 3.14 x 1.3 x 1.3 x 3.6 V = 6.36792 Therefore, the volume of the cup is 6.36792 cubic inches. D. One fluid ounce of water has a volume of 1.8 cubic inches. About how many fluid ounces will the cup hold? Round your answer to the nearest tenth. __________ fluid ounces volume = 1.8 cubic inches The number of fluid ounces will the cup hold = F for the volume we got above is 6.3 1 fluid ounce = 1.8 cubic inch Therefore, the cup will hold 3.5 fluid ounces Turn and Talk Why did you need to find a square root in Part A? What tools did you use to find the square root? We need to find out the r-value. The formula: V = (1/3)πr2h cubic units – Suppose, x is the square root of y, then it is represented as x=√y or we can express the same equation as x2 = y. Here,’√’ is the radical symbol used to represent the root of numbers. The positive number, when multiplied by itself, represents the square of the number. The square root of the square of a positive number gives the original number. – A square root of a number is another number that when multiplied by itself gives back the original number. Methods to find square root: 1. The method of repeated subtraction 2. Prime factorization method 3. Long division method. There are certain square root rules that need to be followed while calculating the square root. The symbol used to denote the root of a number is called radical. Question 3. Holly has an empty cylindrical container. She places four tennis balls, each with a diameter of 2.6 inches, inside the container. What is the approximate volume of air remaining inside the container? A. Find the volume of the cylindrical container to the nearest hundredth. Use 3.14 for π. V = πr2h = π (___________)2 (__________) ≈ (__________) (__________) (__________) ≈ _________ in3 the volume of a cylinder of height ‘h’ and base radius ‘r’ is given as Volume of a Cylinder = πr2h Above-given dimensions are: height = 10.5 substitute the values in the formula. V = 3.14 x 1.5 x 1.5 x 10.5 V = 74.18 Therefore, the volume of a cylinder is 74.18 cubic inches. B. Find the volume taken up by the four tennis balls to the nearest hundredth. Use 3.14 for π. V = $$\frac{4}{3}$$ πr3 • 4 = $$\frac{4}{3}$$ π (___________)3 • 4 ≈ $$\frac{4}{3}$$ (_________) (_________) 4 ≈ (__________) 4 ≈ __________ in3 The formula to find the volume of a sphere is given: The volume of the sphere = 4/3 πr3 Cubic units We know that r = 1.5 V = 4/3 x 1.5 x 1.5 x 1.5 V = 4.5cubic inches Therefore, the volume of the sphere is 4.5 cubic inches. C. Subtract to find the difference. D = volume of cylinder – volume of sphere D = 74.18 – 4.5 D = 69.68 Check Understanding Question 1. Keisha is making cone-shaped snack holders for a party. The cones are 11 centimetres tall and have a radius of 6 centimetres. What is the volume of each cone? Use 3.14 for π. The given dimensions are: height = 11 cm We need to find out the volume of the cone. The formula to calculate the volume of a cone, given the height and its base radius is: V = (1/3)πr2h cubic units substitute the values in the formula V = 1/3 x 3.14 x 36 x 11 V = 3.14 x 12 x 11 V = 414.48 Therefore, the volume of the cone is 414.48 cubic centimetres. Question 2. The interior of a small barrel used to store rice has a radius of 4 inches and a height of 7 inches. A. What is the interior volume of the barrel in cubic inches? Give your answer in terms of π. The given dimensions are: height = 7 inches It is a cylindrical shape. the volume of a cylinder of height ‘h’ and base radius ‘r’ is given as V = πr2h V = π x 16 x 7 V = 112π Therefore, the volume of the barrel in terms of π is 112π cubic inches. B. How many scoops of rice can the barrel hold if each scoop is a hemisphere with a radius of 1 inch? A plane through the centre of the sphere cuts into two equal parts. Each part is called a hemisphere. The volume of a hemisphere is half the volume of a sphere, therefore, it is expressed as, The volume of the hemisphere = 2/3πr3, where r is the radius of the hemisphere. r = 1 V = 2/3 x 3.14 x 1 V = 2.09 Thefeore, 2 scoops of rice a barrel can hold. Question 3. The Westhafen tower in Germany is in the shape of a cylinder. It has a height of 109 meters and a diameter of 38 meters. What is the volume of the building? Express your answer in terms of π. The given dimensions are: height = 109 m diameter = 38 m radius = d/2 = 38/2 = 19 m the volume of a cylinder of height ‘h’ and base radius ‘r’ is given as V = πr2h V = π x 19 x 19 x 109 V = 39349π Therefore, the volume of the building in terms of π is 39349π. Question 4. Reason The conical cup, cylindrical cup, and hemispherical bowl shown have the same unknown radius, r. If the height of each container is the same as that radius, which of the containers holds the most liquid? Explain. The volume of the cone = V = (1/3)πr2h cubic units The volume of the cylinder = V = πr2h The volume of the hemisphere = V = 2/3πr3, h = r If we substitute r in the place of h then The volume of the cone = 1/3 πr^3 The volume of the cylinder = πr^3 The volume of the hemisphere = 2/3πr^3 I think a hollow sphere can hold most liquid. Question 5. Attend to Precision Brady has a beach ball with a diameter of 16 inches when it is inflated. What is the volume of the beach ball, to the nearest hundredth of a cubic inch? Use 3.14 for π. If the sphere’s radius formed is r and the volume of the sphere is V. Then, the volume of the sphere is given by: Volume of Sphere, V = (4/3)πr3 diameter = 16 inches r = d/2 = 16/2 = 8 inches V = 4/3 x 3.14 x 8 x 8 x 8 V = 2143.57333 The nearest hundred value is 2143 cubic inches. Therefore, the volume of the beach ball is 2143 cubic units. Question 6. Attend to Precision A conical container can hold up to 654 cubic centimetres of sand. If the radius of the cone is 5 centimetres, what is the height of the cone? Round your answer to the nearest whole centimetre. Use 3.14 for π. The given dimensions are: volume = 654 cubic cms. we need to find out the height of the cone. The formula to calculate the volume of a cone, given the height and its base radius is: V = (1/3)πr2h cubic units Substitute the values in the formula 654 = 1/3 x 3.14 x 25 x h 654 = 27 x h 654/h = 27/1 h x 27 = 654 h = 654/27 h = 24.22 Therefore, the height of cone is 24.22 cms Question 7. Attend to Precision A cylindrical water tank is 7 feet tall and has a diameter of 12 feet. If the tank is currently half full, how much more water can be poured into the tank? Use $$\frac{22}{7}$$ for π and round your answer to the nearest cubic foot. The given dimensions are: height = 7 feet diameter = 12 radius = d/2 = 12/2 =6 feet The volume of the cylinder = V = πr2h V = 3.14 x 36 x 7 V = 791.28 The tank is filled half already. The remaining place we need to find out. Let it be X. X = 791.28/2 X = 395.64 Approximately 396 cubic foot need to fill. Question 8. Attend to Precision Brittany makes dough and packages it in cylindrical containers that each have a height of 4 inches. What is the radius of each container if a pack of 6 containers contains 169.56 cubic inches of dough? Use 3.14 for π and round the radius to the nearest tenth of an inch. The given dimensions are: height = 4 inches V = 169.56 cubic inches The volume of the cylinder = V = πr2h 169.56 = 3.14 x r^2 x 4 169.56 = 12.56 x r^2 r^2 = 169.56/12.56 r^2 = 13.5 r = 3.67 inches for 1 container , the radius is 3.67 inches for 6 containers, the radius is R inches Question 9. Attend to Precision A large spherical helium balloon has a diameter of 20 feet. What is the volume of the balloon? Use 3.14 for π and express your answer in scientific notation, with the first factor rounded to the nearest hundredth. If the sphere’s radius formed is r and the volume of the sphere is V. Then, the volume of the sphere is given by: Volume of Sphere, V = (4/3)πr3 diameter = 20 feet r = d/2 = 20/2 = 10 feet V = 4/3 x 3.14 x 10 x 10 x 10 V = 4186.666 The approximate value is 4187 cubic foot The scientific notation is 4.18666667 × 103 Question 10. Mr. Jonas has a set of cylindrical containers. Fill in the missing information to complete the table. Round each volume to the nearest hundredth of a cubic inch. Use 3.14 for π. The volume of the cylinder = V = πr2h for small cylinder: height = 6.5 diameter = 2 * 2.5 = 5 in V = 3.14 x 6.25 x 6.5 V = 127.56 approximately 128 cubic inches. for medium cylinder: diameter = 6 r = d/2 = 6/2 = 3 height = 8 V = 3.14 x 9 x 8 V = 226.08 cubic inches for large cylinder: diameter = 7 r = d/2 = 7/2 = 3.5 in height = 9.5 V = 3.14 x 3.5 x 3.5 x 9.5 V = 365.41 Question 11. STEM One of the most common types of volcanoes is called a cinder cone volcano. These types of volcanoes are the smallest type of volcano, ranging between 300 feet and 1200 feet tall, and are in the shape of a cone. Find the volume of a cinder cone volcano with a height of 350 feet and a diameter of 1100 feet. Use 3.14 for π and round your answer to the nearest cubic foot. The formula to calculate the volume of a cone, given the height and its base radius is: V = (1/3)πr2h cubic units given units: height = 350 feet diameter = 1100 feet r = d/2 = 1100/2 = 550 V = 1/3 x 3.14 x 302500 x 350 V = 110815833 The nearest number is 110000000. Therefore, the volume of the cinder cone volcano is 110000000 cubic foot. Question 12. Construct Arguments Corrie wants to buy cylindrical containers to package leftover soup. The containers need to each have a volume of 66 cubic inches. To fit on a shelf, they can be no more than 6 inches tall. Would containers with a volume of 66 cubic inches and a radius of 2 inches fit? Explain. The given dimensions: V = 66 cubic inches h = 6 inches The volume of the cylinder = V = πr2h 66 = 3.14 x r^2 x 6 66 = 18.84 x r^2 r^2 = 18.84/66 r^2 = 0.2854 r = 0.534 The radius of 2 inches is more than enough to fit. Question 13. Critique Reasoning A cylindrical container is three times as tall as a cone with the same diameter. Jeff says the volume of the cylinder will be three times the volume of the cone. Is Jeff correct? Explain. the formula for the volume of a cone is obtained from the volume of a cylinder. Volume of cone = (1/3) × Volume of cylinder = (1/3) × πr2h = (1/3)πr2h. Question 14. What is the volume of the volleyball rounded to the nearest cubic inch? Use 3.14 for π. If the sphere’s radius formed is r and the volume of the sphere is V. Then, the volume of the sphere is given by: Volume of Sphere, V = (4/3)πr3 diameter = 8.15 r = d/2 = 8.15/2 = 4.075 in V = 4/3 x 3.14 x 68 V = 285 (approximately) Therefore, the volume of volleyball is 285 cubic inches. Question 15. What is the volume of the snow globe to the nearest hundredth of a cubic centimeter? Use 3.14 for π. If the sphere’s radius formed is r and the volume of the sphere is V. Then, the volume of the sphere is given by: Volume of Sphere, V = (4/3)πr3 r = 7 cm V = 4/3 x 3.14 x 343 V = 1436. Therefore, the volume of the snow globe is 1436 cubic cms. Question 16. A small cylinder-shaped jar has a volume of 3.768 × 105 mm3 and a height of 75 millimetres. What is the radius of the jar? Use 3.14 for π. The volume of the cylinder = V = πr2h volume = 3.768 x 10^5 height = 75 mm 3.768 x 10^5 = 3.14 x r^2 x 75 376800 = 235.5 x r^2 r^2 = 376800/235.5 r^2 = 1600 r = 40 m Therefore, the radius of the jar is 40 m Question 17. Bethany is installing some new water pipes. One pipe has a diameter of 1.5 inches and is 23 inches long. The other pipe has the same diameter but a length of 30 inches. How much more water can the larger pipe hold? Use 3.14 for π and round your answer to the nearest cubic inch. The volume of the cylinder = V = πr2h diameter = 1.5 r = d/2 = 1.5/2 = 0.75 h = 23 V = 3.14 x 0.5625 x 23 V = 40.62 cubic inches Therefore, the volume of the first pipe is 40.62 cubic inches. The volume of the second pipe: r = 0.75 h = 30 V = 3.14 x 0.5625 x 30 V = 53 (approximately) Now subtract both volumes to know how much water did larger pipe can hold. Let it be X. X = 53 – 40.62 X = 12.38 Therefore, the second pipe hold 12.38 more than first. Question 18. Lu Chen makes a large paper hat in the shape of a cone. Approximate the volume of the hat to the nearest cubic inch. Use $$\frac{22}{7}$$ for π. The formula to calculate the volume of a cone, given the height and its base radius is: V = (1/3)πr2h cubic units height = 21 in V = 1/3 x 3.14 x 16 x 21 V = 351.68 approximately 352 Therefore, the volume of the hat is 352 cubic inches. Question 19. Attend to Precision Sonia fills half of the spherical bowl shown with sand using the cylindrical scoop shown. How many scoops of sand will it take to fill half of the bowl? Question 20. A swimming pool has a radius of 8 feet and a height of 5 feet. The swimming pool is shaped like a cylinder. What is the volume of the swimming pool, to the nearest cubic foot? Use 3.14 for π. The given dimensions: height = 5 feet The volume of the cylinder = V = πr2h V = 3.14 x 64 x 5 V = 1004.8 approximately 1005 cubic feet. Therefore, the volume of a swimming pool is 1005 cubic feet. Lesson 13.4 More Practice/Homework Question 1. Critique Reasoning Eddie measures and finds the volume of the baseball shown. Does the approximate volume look correct? Explain. The diameter is given 3in r = d/2 = 3/2 = 1.5 If the sphere’s radius formed is r and the volume of the sphere is V. Then, the volume of the sphere is given by: Volume of Sphere, V = (4/3)πr3 V = 4/3 x 3.14 x 1.5^3 V = 14.13 cubic inches. Eddie calculates the volume is wrong. Question 2. Math on the Spot Some party hats are shaped like cones. Use a calculator to find the volume of a party hat to the nearest hundredth of a cubic centimeter. If the radius of the base Is 7 centimeters and the height is 20 centimeters. The formula to calculate the volume of a cone, given the height and its base radius is: V = (1/3)πr2h cubic units the given dimensions: h = 20cm V = 1/3 x 3.14 x 49 x 20 V = 1025.733 The nearest hundred’s place value is 1026 cubic cms. Therefore, the volume of a party hat is 1026 cubic cms. Question 3. A grain silo is in the shape of a cylinder. The area of the circular roof is 803.84 square feet. If 13,665.28 cubic feet of grain fits in the silo, what is the height of the silo? The volume of the cylinder = V = πr2h The given dimensions: area of circle (πr^2) = 803.84 Volume = 13,665.28 cubic feet we need to find out the height of grain. 13665.28 = 803.84 x h h = 13665.28/803.84 h = 17 ft Therefore, the height of the silo is 17 ft. Question 4. A funnel in the shape of a cone has a diameter of 4 centimetres and a height of 9 centimetres. What is the volume of the funnel, to the nearest cubic centimetre? Use 3.14 for π. The formula to calculate the volume of a cone, given the height and its base radius is: V = (1/3)πr2h cubic units diameter = 4 cm r = d/2 = 4/2 = 2 cm height = 9cm V = 1/3 x 3.14 x 4 x 9 V = 1/3 x 113.04 V = 37.68 approximately 38 Therefore, the volume of the funnel is 38 cubic centimetres. Question 5. Art Ha-joon is a potter who makes sculptures of cylinders, spheres, and cones from clay. The diameter of each clay sphere, the diameter of the base of each clay cylinder and cone, and the height of each clay cylinder and cone are all 6 centimeters. A. Find the volume of the clay sphere, clay cylinder, and clay cone. Leave the answers in terms of π. sphere: ___________ cylinder: ___________ cone: ___________ The given dimension: height = 6cm diameter = 6 cm r = d/2 = 6/2 = 3cm If the sphere’s radius formed is r and the volume of the sphere is V. Then, the volume of the sphere is given by: Volume of Sphere, V = (4/3)πr3 V = 4/3 x π x 27 V = 36π The volume of the cylinder = V = πr2h V = π x 9 x 6 V = 54π The formula to calculate the volume of a cone, given the height and its base radius is: V = (1/3)πr2h cubic units V = 1/3 x π x 9 x 6 V = 18π B. How many clay spheres can Ha-Joon make if he wants to use the same amount of clay as two clay cylinders? How many clay cones can Ha-Joon make if he wants to use the same amount of clay as two clay cylinders? The volume of the sphere is 36π The volume of the cylinder is 54π The volume of the cone is 18π For two spheres we use: 54 x 2 = 108 1 volume of sphere = 36π the number of spheres we can make is 108/32 = 3 Therefore, we can make 3 spheres with the same amount of clay as two cylinders. 1 volume of a cone is 18π the number of cones we can make is 108/18 = 6 Therefore, we can make 6 cones with the same amount of clay as two cylinders. Test Prep Question 6. Aspherical fishbowl has a diameter of 4 inches. What is the volume of the fishbowl, to the nearest cubic inch? Use 3.14 for π. The given dimensions: diameter = 4 inches r = d/2 = 4/2 = 2 If the sphere’s radius formed is r and the volume of the sphere is V. Then, the volume of the sphere is given by: Volume of Sphere, V = (4/3)πr3 V= 4/3 x 3.14 x 8 V = 33.49 approximately 34 therefore, the volume of the fishbowl is 34 cubic inches. Question 7. Alissa has an 8.2-inch-tall water bottle with a radius of 1.5 inches. Find the volume of the water bottle, rounded to the nearest hundredth of a cubic inch. Use 3.14 for π. The given dimensions: height = 8.2 in The volume of the cylinder = V = πr2h v = 3.14 x 2.25 x 8.2 V = 57.933 approximately 60. Therefore, the volume of water bottle is 60 cubic inches. Question 8. A waffle cone has a volume of 31.25π cubic centimetres and a radius of 2.5 centimetres. What is the approximate height of the cone? (A) 5 cm (B) 10 cm (C) 15 cm (D) 30 cm The given dimensions: volume = 31.25π The formula to calculate the volume of a cone, given the height and its base radius is: V = (1/3)πr2h cubic units 31.25π = 1/3 x π x 6.25 x h 31.25 = 2.08 h h = 31.25/2.08 h = 15cm Spiral Review Question 9. Simplify. $$\frac{7^{4} \cdot 7^{3}}{7^{5}}$$ = ___________ Question 10. Find the unknown side length of the right triangle. By applying Pythagoras theorem on the cone, we can find the relation between volume and slant height of the cone. We know, h2 + r2 = L2 The given dimensions: l = 5; r = 3 Now substitute the values in the formula: h^2 + 9 = 25 h^2 = 25 -9 h^2 = 16 h = 4 Therefore, the other side of the right triangle is 4. Question 11. Adrienne graphs the line y – x = 7. At what point will the line intersect the y-axis?
# Ch10 - Chapter 10 Statistical Inference about Means and... This preview shows pages 1–6. Sign up to view the full content. 10 - 1 Chapter 10 Statistical Inference about Means and Proportions with Two Populations Learning Objectives 1. Be able to develop interval estimates and conduct hypothesis tests about the difference between two population means when 1 σ and 2 are known. 2. Know the properties of the sampling distribution of 12 x x . 3. Be able to use the t distribution to conduct statistical inferences about the difference between two population means when 1 and 2 are unknown. 4. Learn how to analyze the difference between two population means when the samples are independent and when the samples are matched. 5. Be able to develop interval estimates and conduct hypothesis tests about the difference between two population proportions. 6. Know the properties of the sampling distribution of pp . This preview has intentionally blurred sections. Sign up to view the full version. View Full Document Chapter 10 10 - 2 Solutions: 1. a. xx 12 = 13.6 - 11.6 = 2 b. /2 .05 zz α == 1.645 22 1.645 nn σσ −± + (2.2) (3) 21 . 6 4 5 50 35 ±+ 2 ± .98 (1.02 to 2.98) c. .025 1.96 (2.2) (3) . 9 6 50 35 2 ± 1.17 (.83 to 3.17) 2. a. () 0 (25.2 22.8) 0 2.03 (5.2) 6 40 50 xx D z −− = + + b. p -value = .5000 - .4788 = .0212 c. p -value .05, reject H 0 . 3. a. 0 2 2 (104 106) 0 1.53 (8.4) (7.6) 80 70 z = + + b. p -value = 2(.5000 - .4370) = .1260 c. p -value > .05, do not reject H 0 . 4. a. x x = 2.04 - 1.72 = .32 b. .025 z + (.10) (.08) 1.96 1.96(.0208) .04 40 35 += = c. .32 ± .04 (.28 to .36) Statistical Inference about Means and Proportions with Two Populations 10 - 3 5. a. 12 x x = 14.9 - 10.3 = 4.6 years b. 22 /2 (5.2) (3.8) 1.96 1.3 100 85 z nn α σσ += + = c. 4.6 ± 1.3 (3.3 to 5.9) 6. 1 µ = Mean loan amount for 2002 2 = Mean loan amount for 2001 H 0 : 0 µµ −≤ H a : 0 −> () 0 2 2 (175 165) 0 2.17 55 50 270 250 xx D z −− == = + + p -value = .5000 - .4850 = .0150 p -value .05; reject H 0 . The mean loan amount has increased between 2001 and 2002. 7. a. H 0 : 0 −= H a : 0 −≠ b. 0 2 2 (40 35) 0 2.41 91 0 36 49 z = + + p -value = 2(.5000 - .4920) = .0160 p -value .05; reject H 0 . There is a difference between the population mean ages at the two stores. 8. a. 2 2 0 (69.95 69.56) 0 1.08 2.5 2.5 112 84 xx z = + + b. p -value = 2(.5000 - .3599) = .2802 c. p -value > .05; do not reject H 0 . Cannot conclude that there is a difference between the population mean scores for the two golfers. This preview has intentionally blurred sections. Sign up to view the full version. View Full Document Chapter 10 10 - 4 9. a. 12 x x = 22.5 - 20.1 = 2.4 b. 2 2 22 222 2 2 1122 2.5 4.8 20 30 45.8 1 2.5 1 4.8 11 19 20 29 30 ss nn df nnnn  + +   == = + + −− Use df = 45. c. t .025 = 2.014 .025 2.5 4.8 2.014 2.1 20 30 t += + = d. 2.4 ± 2.1 (.3 to 4.5) 10. a. () 2 2 0 (13.6 10.1) 0 2.18 5.2 8.5 35 40 xx t = + + b. 2 2 2 2 5.2 8.5 35 40 65.7 15 . 2 18 . 5 34 35 39 40 df + + =   + +   Use df = 65 c. Using t table, area in tail is between .01 and .025 two-tail p -value is between .02 and .05. Actual p -value = .0329 d. p -value .05, reject H 0 . 11. a. 1 54 9 6 x 2 42 7 6 x b. 2 1 1 1 2.28 1 i s n Σ− 2 2 2 2 1.79 1 i s n c. x x = 9 - 7 = 2 Statistical Inference about Means and Proportions with Two Populations 10 - 5 d. 2 2 22 12 222 2 2 1122 2.28 1.79 66 9.5 . 2 8 11 . 7 9 56 ss nn df nnnn   + +    == =    + +    −− Use df = 9, t .05 = 1.833 2.28 1.79 1.833 xx −± + 2 ± 2.17 (-.17 to 4.17) 12. a. x x = 22.5 - 18.6 = 3.9 b. 2 2 2 2 8.4 7.4 50 40 87.1 1 8.4 1 7.4 49 50 39 40 df + + =   + +   Use df = 87, t .025 = 1.988 8.4 7.4 3.9 1.988 50 40 ±+ 3.9 ± 3.3 (.6 to 7.2) 13. Computer used to obtain the following: Miami Los Angeles n 1 = 50 n 2 = 50 1 x = 6.34 2 x = 6.72 s 1 = 2.16 s 2 = 2.37 x x = 6.34 - 6.72 = -.38 Los Angeles is slightly higher rated. This preview has intentionally blurred sections. Sign up to view the full version. View Full Document This is the end of the preview. Sign up to access the rest of the document. ## This note was uploaded on 03/01/2012 for the course ECON 371 taught by Professor Staff during the Spring '08 term at UVA. ### Page1 / 20 Ch10 - Chapter 10 Statistical Inference about Means and... This preview shows document pages 1 - 6. Sign up to view the full document. View Full Document Ask a homework question - tutors are online
#### Transcript Slide 1 ```Chapter 2 Section 4 2.4 1 2 3 4 5 An Introduction to Applications of Linear Equations Learn the six steps for solving applied problems. Solve problems involving unknown numbers. Solve problems involving sums of quantities. Solve problems involving supplementary and complementary angles. Solve problems involving consecutive integers. Objective 1 Learn the six steps for solving applied problems. Slide 2.4 - 3 Learn the six steps for solving applied problems. We now look at how algebra is used to solve applied problems. While there is no one specific method that enables you to solve all kinds of applied problems, the following six-step method is often applicable. Step 1: Read the problem carefully until you understand what is given and what is to be found. Step 2: Assign a variable to represent the unknown value, using diagrams or tables as needed. Write down what the variable represents. If necessary, express any other unknown values in terms of the variable. Step 3: Write an equation using the variable expression(s). Step 4: Solve the equation. Step 5: State the answer. Does it seem reasonable? Step 6: Check the answer in the words of the original problem. Slide 2.4 - 4 Learn the six steps for solving applied problems. (cont’d) The third step in solving an applied problem is often the hardest. To translate the problem into an equation, write the given phrases as mathematical expressions. Replace any words that mean equals or same with an = sign. Other forms of the verb “to be,” such as is, are, was, and were, also translate this way. The = sign leads to an equation to be solved. Slide 2.4 - 5 Objective 2 Solve problems involving unknown numbers. Slide 2.4 - 6 EXAMPLE 1 Finding the Value of an Unknown Number If 5 is added to the product of 9 and a number, the result is 19 less than the number. Find the number. Solution: 9 x  5  x  x  19  x Let x = the number. 8 x  5  5  19  5 9 x  5  x  19 8x 8  24 8 x  3 The number is −3. When solving an equation, use solution set notation to write the answer. When solving an application, state the answer in a sentence. Slide 2.4 - 7 Objective 3 Solve problems involving sums of quantities. Slide 2.4 - 8 Solve problems involving sums of quantities. A common type of problem in elementary algebra involves finding two quantities when the sum of the quantities is known. In Example 9 of Section 2.3, we prepared for this type of problem, by writing mathematical expressions for two related unknown quantities. PROBLEM-SOLVING HINT To solve problems involving sums of quantities, choose a variable to represent one of the unknowns. Represent the other quantity in terms of the same variable, using information from the problem. Write an equation based on the words of the problem. Slide 2.4 - 9 EXAMPLE 2 Finding Numbers of Olympics Medals In the 2006 Winter Olympics in Torino, Italy, the United States won 6 more medals than Norway. The two countries won a total of 44 medals. How many medals did each country win? (Source: U.S. Olympic Committee.) Solution: Let x = the number of medals Norway won. Let x + 6 = the number of medals the U.S. won. 44  x   x  6  44  6  2 x  6  6 38 2 x  2 2 x  19 19  6  25 Norway won 19 medals and the U.S. won 25 medals. Slide 2.4 - 10 Solve problems involving sums of quantities. (cont’d) The problem in example 2 could also be solved by letting x represent the number of medals the United States won. Then x − 6 would represent the number of medals Norway won. The equation would be 44  x   x  6  The solution to this equation is 25, which is the number of medals the U.S. won. The number of Norwegian medals would be 25 − 6 = 19. The answers are the same, whichever approach is used. The nature of the applied problem restricts the set of possible solutions. 1 For example, an answer such as −33 medals or 25 2 medals should be recognized as inappropriate. Slide 2.4 - 11 EXAMPLE 3 Finding the Number of Orders for Croissants On that same day, the owner of Terry’s Coffeehouse 1 found that the number of orders for croissants was 6 the number of muffins. If the total number for the two breakfast rolls was 56, how many orders were placed for croissants? 336  6 x  x Solution: 336 7 x Let 1 x = the number of muffins.  7 7 Then 6 x = the number of croissants. x  48 56  x  1 x 6 1 56  6   x  6   x  6  6 1 6  48   8 8 croissants were ordered. Slide 2.4 - 12 Finding the Number of Orders for Croissants (cont’d) PROBLEM SOLVING HINT In Example 3, it was easier to let the variable represent the quantity that was not specified. This required extra work in Step 5 to find the number of orders for croissants. In some cases, this approach is easier than letting the variable represent the quantity that we are asked to find. Slide 2.4 - 13 EXAMPLE 4 Analyzing the Mixture of a Computer User Group At a meeting of the local computer user group, each member brought two nonmembers. If a total of 27 people attended, how many were nonmembers? Solution: Let x = number of members. Then 2x = number of nonmembers. 2 x  x  27 3x  2 9   18 27 3 3 x9 There were 9 members and 18 nonmembers at the meeting. Slide 2.4 - 14 Dividing a Board into Pieces PROBLEM SOLVING HINT Sometimes it is necessary to find three unknown quantities in an applied problem. Frequently, the three unknowns are compared in pairs. When this happens it is usually easiest to let the variable represent the unknown found in both pairs. Slide 2.4 - 15 EXAMPLE 5 Dividing a Pipe into Pieces A piece of pipe is 50 in. long. It is cut into three pieces. The longest piece is 10 in. more than the middle-sized piece, and the shortest piece measures 5 in. less than the middle-sized piece. Find the lengths of the three pieces. Solution: Let x = the length of the middle-sized piece, then x +10 = longest piece, 15  10  25 and x − 5 = shortest piece. x   x  10    x  5   50 3 x  5  5  50  5 3x  45 3 3 x  15 15  5  10 The shortest piece is 10 in., the middle-size piece is 15 in., and the longest is 25 in. Slide 2.4 - 16 Objective 4 Solve problems involving supplementary and complementary angles. Slide 2.4 - 17 Solve problems involving supplementary and complementary angles. An angle can be measured by a unit called the degree (°), which is 3 61 0 of a complete rotation. Two angles whose sum is 90° are said to be complementary, or complements of each other. An angle that measures 90° is a right angle. Two angles who sum is 180° are said to be supplementary, or supplements of each other. One angle supplements the other to form a straight angle of 180°. Slide 2.4 - 18 Solve problems involving supplementary and complementary angles. (cont’d) If x represents the degree measure of an angle, then 90 − x represents the degree measure of its complement, and 180 − x represents the degree measure of is supplement. Slide 2.4 - 19 EXAMPLE 6 Finding the Measure of an Angle Find the measure of an angle such that the sum of the measures of its complement and its supplement is 174°. Solution: Let x = the degree measure of the angle. Then 90 − x = the degree measure of its complement, and 180 − x = the degree measure of its supplement.  90  x   180  x   174 270  2 x  270  174  270 2 x 2 The measure of the angle is 48°.  96 2 x  48 Slide 2.4 - 20 Objective 5 Solve problems involving consecutive integers. Slide 2.4 - 21 Solve problems involving consecutive integers. Two integers that differ by 1 are called consecutive integers. For example, 3 and 4, 6 and 7, and −2 and −1 are pairs of consecutive integers. In general, if x represents an integer, x + 1 represents the next larger consecutive integer. Consecutive even integers, such as 8 and 10, differ by 2. Similarly, consecutive odd integers, such as 9 and 11, also differ by 2. In general if x represents an even integer, x + 2 represents the larger consecutive integer. The same holds true for odd integers; that is if x is an odd integer, x + 2 is the larger odd integer. Slide 2.4 - 22 Solve problems involving consecutive integers. (cont’d) PROBLEM SOLVING HINT In solving consecutive integer problems, if x = the first integer, then, for any two consecutive integers, use two consecutive even integers, use two consecutive odd integers, use x , x  1; x , x  2; x , x  2. Slide 2.4 - 23 EXAMPLE 7 Finding Consecutive Integers Two back-to-back page numbers in this book have a sum of 569. What are the page numbers? Solution: Let x = the lesser page number. Then x + 1= the greater page number. x   x  1  5 6 9 2 x  1  1  569  1 2x 2  568 2 x  284 284  1  285 The lesser page number is 284, and the greater page number is 285. It is a good idea to use parentheses around x + 1, (even though they are not necessary here). Slide 2.4 - 24 EXAMPLE 8 Finding Consecutive Even Integers Find two consecutive even integers such that six times the lesser added to the greater gives a sum of 86. Solution: Let x = the lesser integer. Then x + 2 = the greater integer. 6 x   x  2   86 12  2  14 7 x  2  2  86  2 7x 7  84 7 x  12 The lesser integer is 12 and the greater integer is 14.
6.5 Multiplication of decimals  (Page 2/2) Page 2 / 2 Practice set b Use a calculator to find each product. If the calculator will not provide the exact product, round the result to four decimal places. $5\text{.}\text{126}\cdot \text{4}\text{.}\text{08}$ 20.91408 $0\text{.}\text{00165}\cdot \text{0}\text{.}\text{04}$ 0.000066 $0\text{.}\text{5598}\cdot \text{0}\text{.}\text{4281}$ 0.2397 $0\text{.}\text{000002}\cdot \text{0}\text{.}\text{06}$ 0.0000 Multiplying decimals by powers of 10 There is an interesting feature of multiplying decimals by powers of 10. Consider the following multiplications. Multiplication Number of Zeros in the Power of 10 Number of Positions the Decimal Point Has Been Moved to the Right $\text{10}\cdot 8\text{.}\text{315274}=\text{83}\text{.}\text{15274}$ 1 1 $\text{100}\cdot 8\text{.}\text{315274}=\text{831}\text{.}\text{5274}$ 2 2 $1,\text{000}\cdot 8\text{.}\text{315274}=8,\text{315}\text{.}\text{274}$ 3 3 $\text{10},\text{000}\cdot 8\text{.}\text{315274}=\text{83},\text{152}\text{.}\text{74}$ 4 4 Multiplying a decimal by a power of 10 To multiply a decimal by a power of 10, move the decimal place to the right of its current position as many places as there are zeros in the power of 10. Add zeros if necessary. Sample set c Find the following products. $\text{100}\cdot \text{34}\text{.}\text{876}$ . Since there are 2 zeros in 100, Move the decimal point in 34.876 two places to the right. $1,\text{000}\cdot 4\text{.}\text{8058}$ . Since there are 3 zeros in 1,000, move the decimal point in 4.8058 three places to the right. $\text{10},\text{000}\cdot \text{56}\text{.}\text{82}$ . Since there are 4 zeros in 10,000, move the decimal point in 56.82 four places to the right. We will have to add two zeros in order to obtain the four places. Since there is no fractional part, we can drop the decimal point. Practice set c Find the following products. $\text{100}\cdot \text{4}\text{.}\text{27}$ 427 $\text{10,000}\cdot \text{16}\text{.}\text{52187}$ 165,218.7 $\left(\text{10}\right)\left(0\text{.}\text{0188}\right)$ 0.188 $\left(\text{10,000,000,000}\right)\left(\text{52}\text{.}7\right)$ 527,000,000,000 Multiplication in terms of “of” Recalling that the word "of" translates to the arithmetic operation of multiplica­tion, let's observe the following multiplications. Sample set d Find 4.1 of 3.8. Translating "of" to "×", we get Thus, 4.1 of 3.8 is 15.58. Find 0.95 of the sum of 2.6 and 0.8. We first find the sum of 2.6 and 0.8. $\begin{array}{c}\hfill 2.6\\ \hfill \underline{+0.8}\\ \hfill 3.4\end{array}$ Now find 0.95 of 3.4 Thus, 0.95 of $\left(2\text{.}\text{6}+\text{0}\text{.}8\right)$ is 3.230. Practice set d Find 2.8 of 6.4. 17.92 Find 0.1 of 1.3. 0.13 Find 1.01 of 3.6. 3.636 Find 0.004 of 0.0009. 0.0000036 Find 0.83 of 12. 9.96 Find 1.1 of the sum of 8.6 and 4.2. 14.08 Exercises For the following 30 problems, find each product and check each result with a calculator. $3\text{.}4\cdot 9\text{.}2$ 31.28 $4\text{.}5\cdot 6\text{.}1$ $8\text{.}0\cdot 5\text{.}9$ 47.20 $6\text{.}1\cdot 7$ $\left(0\text{.}1\right)\left(1\text{.}\text{52}\right)$ 0.152 $\left(1\text{.}\text{99}\right)\left(0\text{.}\text{05}\right)$ $\left(\text{12}\text{.}\text{52}\right)\left(0\text{.}\text{37}\right)$ 4.6324 $\left(5\text{.}\text{116}\right)\left(1\text{.}\text{21}\right)$ $\left(\text{31}\text{.}\text{82}\right)\left(0\text{.}1\right)$ 3.182 $\left(\text{16}\text{.}\text{527}\right)\left(9\text{.}\text{16}\right)$ $0\text{.}\text{0021}\cdot 0\text{.}\text{013}$ 0.0000273 $1\text{.}\text{0037}\cdot 1\text{.}\text{00037}$ $\left(1\text{.}6\right)\left(1\text{.}6\right)$ 2.56 $\left(4\text{.}2\right)\left(4\text{.}2\right)$ $0\text{.}9\cdot 0\text{.}9$ 0.81 $1\text{.}\text{11}\cdot 1\text{.}\text{11}$ $6\text{.}\text{815}\cdot 4\text{.}3$ 29.3045 $9\text{.}\text{0168}\cdot 1\text{.}2$ $\left(3\text{.}\text{5162}\right)\left(0\text{.}\text{0000003}\right)$ 0.00000105486 $\left(0\text{.}\text{000001}\right)\left(0\text{.}\text{01}\right)$ $\left(\text{10}\right)\left(4\text{.}\text{96}\right)$ 49.6 $\left(\text{10}\right)\left(\text{36}\text{.}\text{17}\right)$ $\text{10}\cdot \text{421}\text{.}\text{8842}$ 4,218.842 $\text{10}\cdot 8\text{.}\text{0107}$ $\text{100}\cdot 0\text{.}\text{19621}$ 19.621 $\text{100}\cdot 0\text{.}\text{779}$ $\text{1000}\cdot 3\text{.}\text{596168}$ 3,596.168 $\text{1000}\cdot \text{42}\text{.}\text{7125571}$ $\text{1000}\cdot \text{25}\text{.}\text{01}$ 25,010 $\text{100},\text{000}\cdot 9\text{.}\text{923}$ $\left(4\text{.}6\right)\left(6\text{.}\text{17}\right)$ Actual product Tenths Hundreds Thousandths Actual product Tenths Hundreds Thousandths 28.382 28.4 28.38 28.382 $\left(8\text{.}\text{09}\right)\left(7\text{.}1\right)$ Actual product Tenths Hundreds Thousandths $\left(\text{11}\text{.}\text{1106}\right)\left(\text{12}\text{.}\text{08}\right)$ Actual product Tenths Hundreds Thousandths Actual product Tenths Hundreds Thousandths 134.216048 134.2 134.22 134.216 $0\text{.}\text{0083}\cdot 1\text{.}\text{090901}$ Actual product Tenths Hundreds Thousandths $7\cdot \text{26}\text{.}\text{518}$ Actual product Tenths Hundreds Thousandths Actual product Tenths Hundreds Thousandths 185.626 185.6 185.63 185.626 For the following 15 problems, perform the indicated operations Find 5.2 of 3.7. Find 12.03 of 10.1 121.503 Find 16 of 1.04 Find 12 of 0.1 1.2 Find 0.09 of 0.003 Find 1.02 of 0.9801 0.999702 Find 0.01 of the sum of 3.6 and 12.18 Find 0.2 of the sum of 0.194 and 1.07 0.2528 Find the difference of 6.1 of 2.7 and 2.7 of 4.03 Find the difference of 0.071 of 42 and 0.003 of 9.2 2.9544 If a person earns $8.55 an hour, how much does he earn in twenty-five hundredths of an hour? A man buys 14 items at$1.16 each. What is the total cost? $16.24 In the problem above, how much is the total cost if 0.065 sales tax is added? A river rafting trip is supposed to last for 10 days and each day 6 miles is to be rafted. On the third day a person falls out of the raft after only $\frac{2}{5}$ of that day’s mileage. If this person gets discouraged and quits, what fraction of the entire trip did he complete? 0.24 A woman starts the day with$42.28. She buys one item for $8.95 and another for$6.68. She then buys another item for sixty two-hundredths of the remaining amount. How much money does she have left? Calculator problems For the following 10 problems, use a calculator to determine each product. If the calculator will not provide the exact product, round the results to five decimal places. $0.019\cdot 0.321$ 0.006099 $0.261\cdot 1.96$ $4.826\cdot 4.827$ 23.295102 ${\left(9.46\right)}^{2}$ ${\left(0.012\right)}^{2}$ 0.000144 $0.00037\cdot 0.0065$ $0.002\cdot 0.0009$ 0.0000018 $0.1286\cdot 0.7699$ $0.01\cdot 0.00000471$ 0.0000000471 $0.00198709\cdot 0.03$ Exercises for review ( [link] ) Find the value, if it exists, of $\text{0}÷\text{15}$ . 0 ( [link] ) Find the greatest common factor of 210, 231, and 357. ( [link] ) Reduce $\frac{\text{280}}{2,\text{156}}$ to lowest terms. $\frac{10}{77}$ ( [link] ) Write "fourteen and one hundred twenty-one ten-thousandths, using digits." ( [link] ) Subtract 6.882 from 8.661 and round the result to two decimal places. 1.78 can someone help me with some logarithmic and exponential equations. 20/(×-6^2) Salomon okay, so you have 6 raised to the power of 2. what is that part of your answer I don't understand what the A with approx sign and the boxed x mean it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared Salomon I'm not sure why it wrote it the other way Salomon I got X =-6 Salomon ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6 oops. ignore that. so you not have an equal sign anywhere in the original equation? Commplementary angles hello Sherica im all ears I need to learn Sherica right! what he said ⤴⤴⤴ Tamia what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks. a perfect square v²+2v+_ kkk nice algebra 2 Inequalities:If equation 2 = 0 it is an open set? or infinite solutions? Kim The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined. Al y=10× if |A| not equal to 0 and order of A is n prove that adj (adj A = |A| rolling four fair dice and getting an even number an all four dice Kristine 2*2*2=8 Differences Between Laspeyres and Paasche Indices No. 7x -4y is simplified from 4x + (3y + 3x) -7y is it 3×y ? J, combine like terms 7x-4y im not good at math so would this help me yes Asali I'm not good at math so would you help me Samantha what is the problem that i will help you to self with? Asali how do you translate this in Algebraic Expressions Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)= . After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight? 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 what is nanomaterials​ and their applications of sensors. what is nano technology what is system testing? preparation of nanomaterial Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it... what is system testing what is the application of nanotechnology? Stotaw In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google Azam anybody can imagine what will be happen after 100 years from now in nano tech world Prasenjit after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments Azam name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world Prasenjit how hard could it be to apply nanotechnology against viral infections such HIV or Ebola? Damian silver nanoparticles could handle the job? Damian not now but maybe in future only AgNP maybe any other nanomaterials Azam can nanotechnology change the direction of the face of the world At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light. the Beer law works very well for dilute solutions but fails for very high concentrations. why? how did you get the value of 2000N.What calculations are needed to arrive at it Privacy Information Security Software Version 1.1a Good 7hours 36 min - 4hours 50 min
Lesson 1-3: Mult/Div Real #s Presentation on theme: "Lesson 1-3: Mult/Div Real #s"— Presentation transcript: Lesson 1-3: Mult/Div Real #s The first slides here review adding, subtracting, multiplying, and dividing fractions. You do not have to take notes on these slides. We will do a few practice problems in class. Find the common denominator. Make equivalent fractions using the new common denominator. Add/sub the numerators. Denominator stays the same. Simplify/reduce. Leave improper fractions as is. Hooray! Examples 1 2 Multiply Fractions & Mixed #s Change any mixed #s into improper fractions. Multiply numerators. Multiply denominators. Simplify. (You can also simplify before you multiply.) Leave improper fractions as is. Examples 1 2 2 1 Divide Fractions & Mixed #s Change any mixed #s into improper fractions. Find the reciprocal of the 2nd fraction (the divisor), rewriting the problem as a multiplication problem. Multiply. Simplify. (You can also simplify before you multiply.) Leave improper fractions as is. Examples 1 2 8 5 Decimal Operations - Reminders Multiplication Line up digits as in whole # mult. After multiplying as usual, count up total places behind decimal point, and move decimal that number of places. Decimal Operations - Reminders Division: Shift decimal in the divisor (outside #) to the right so that you are dividing by a whole #. Shift the decimal in the dividend the same # of places. Now divide as usual. Keep dividing until it ends or repeats. Now in your Know It Notes: Reciprocal: numerator and denominator change place (fraction flipped over) Multiplicative inverse: a number and its reciprocal are called multiplicative inverses A number times its reciprocal = 1 Inverse Prop. Of Multiplication The product of a real # (but not zero) and its reciprocal is 1. Algebraically: For a≠0, Integer Rules Positive # • Positive # = Positive # Negative # • Negative # = Positive # (Same signs = positive answer) Positive # • Negative # = Negative # Negative # • Positive # = Negative # (Opposite signs = neg answer) Same rules for division.
Last Updated: May 27, 2023 Medium # Kruskal's Algorithm Author Shreya Deep ## Introduction A Graph is a data structure consisting of edges connecting vertices. When all the vertices of a graph are connected such that there is no cycle and the number of edges is (v-1)  ( where v is the number of vertices of the graph), the weighted tree formed is called a spanning tree A single graph can have more than one spanning tree, and the spanning tree with the minimum sum of all edge weights is called the minimum spanning tree There are various algorithms to find the minimum spanning tree of a graph. One such algorithm is Prims Algorithm. In this article, we’ll learn about another algorithm to find the minimum spanning tree. This algorithm is known as Kruskal’s minimum spanning tree. So let’s get started! ## Problem Statement The problem states that “Given a connected and undirected graph, find its minimum spanning tree using Kruskal’s minimum spanning tree algorithm.” Note: A graph having bidirectional edges is known as an undirected graph. Such graphs can be traversed in either direction. Let’s understand the problem statement with the help of an example. Given a graph having Number of vertices(V) = 5, and Number of edges(E) = 6. Cost of an edge and the vertices having the edge are(cost, vertex1, vertex2): ( 2,0,1) , (1,4,3), (3,1,2), (2,4,2), (3,2,3), (1,1,4). The graph contains five vertices and six edges. So the minimum spanning tree will have four edges. The diagram below shows the graph and its minimum spanning tree. Above is the graph Above is the corresponding minimum spanning tree Edges of the minimum spanning trees are: (1,4), (4,3), (0,1), (4,2). And, the total weight of the tree will be 6. Now, let's move on to the solution approach of this problem. ## Solution Approach Kruskal’s minimum spanning tree algorithm is a greedy approach(see Greedy Algorithms)because we pick the smallest edge. We're making that choice because we want our spanning tree to have a minimum total sum of edges. A graph may or may not contain a cycle. But in MSP, we need to ensure that there is no cycle. So, in this algorithm, we're using a disjoint data structure to find if two nodes are in the same subset or not. This will help detect a cycle Want to know what a disjoint set data structure is? • A disjoint set data structure is used to track the division of elements into different disjoint subsets. • The union-find algorithm performs two operations on this data structure, namely, find and union. For more details, you can refer to this blog. The steps of Kruskal’s algorithm are as follows : ``````1. Sort all the edges in ascending order based on their weights. 2. Draw the edge with the least weight. Check if it generates a cycle with the spanning-tree formed till now using a union-find algorithm. If the cycle is not formed, include this edge. Else, discard it and move to the next. 3. Repeat step 2 until there are (v-1) edges in the spanning tree. `````` Let’s understand this algorithm using the above example : The graph contains five vertices and six edges. So, the minimum spanning tree will have (5-1) = four edges. We'll be performing one step at a time and understand how that'll work. Step 1:  After sorting the edges in increasing order of their weights, we get : Step 2: Now, we start picking edges one by one until we get four edges of our spanning tree. Currently, our spanning tree has no edges. • Pick edge 1 - 4: It doesn't form any cycle, so add it. • Pick edge 4-3: It doesn't form any cycle, so add it. • Pick edge 0-1: It doesn't form any cycle, so add it. • Pick edge 4-2: It doesn't form any cycle, so add it. At this point, we've already taken four edges in our spanning tree, and therefore, we stop, and our spanning tree is complete and has the minimum weight. We have taken the edges 1 - 4, 4 - 3, 0 - 1, and 4 - 2. Thus, the total weight of the minimum spanning tree is 1+1+2+2 = 6 You’ve now understood the working and the reasoning behind using Kruskal’s minimum spanning tree algorithm to find the minimum spanning tree. Let’s work on the implementation now! Before directly jumping to the solution, we suggest you try and solve this problem on CodeStudio. ### Implementation Let’s see the implementation of the above approach. ``````#include<bits/stdc++.h> using namespace std; const int MAX = 1e6-1; int parent[MAX]; int find(int a){  //function to find the parent of the subset this a belongs to while(parent[a]!=a){ parent[a] = parent[parent[a]]; a = parent[a]; } return a; } void union_(int a,int b){  //function to merge two subsets int d = find(a); int e = find(b); parent[d] = parent[e]; } int main(){ int vertices; cin>>vertices; int edges; cin>>edges; vector<pair<int,pair<int,int>>>adj;  // vector to store the edges in the form - > {weight, {source, destination}} for(int i=0;i<edges;i++){ int weight; int src,destination; cin>>weight>>src>>destination; adj.push_back({weight,{src,destination}}); // pushing back the edges one by one } for(int i = 0;i<MAX;i++){ parent[i] = i;  // initialising the parent of each node as itself } vector<pair<int,int>>tree_edges;  // vector for storing the edges of the minimum spanning tree int totalweight = 0; //initialising the total weight to 0 int a = x.second.first; int b = x.second.second; int cost = x.first; if(find(a)!=find(b)){  // if the two vertices are in different subsets, merge them into one totalweight+=cost; union_(a,b); tree_edges.push_back({a,b}); } } cout<<"Edges are : "<<endl; for(auto x:tree_edges){  // printing the edges of the minimum spanning tree cout<<x.first<<" "<<x.second<<endl; } cout<<"Total weight of MST = "; cout<<totalweight<<endl;  //printing the total weight of the minimum spanning tree return 0; }`````` Input ``````//entering the number of vertices 5 //entering the number of edges 6 //entering edges one by one in the format: cost, v1, v2 2 0 1 1 4 3 3 1 2 2 4 2 3 2 3 1 1 4`````` Output ``````//Output Edges are : 1 4 4 3 0 1 4 2 Total weight of MST = 6`````` ### Complexity Analysis Time Complexity The time complexity of Kruskal's minimum spanning tree algorithm is O(E*logV), where E is the number of edges in the graph and V is the number of vertices. Reason: Since we're sorting the edges, which takes O(E*logE) time. Then we check for each edge whether to add it or not by using the union-find algorithm, which takes at most O(logV) time for every edge E, Hence total O(ElogV). Thus the overall complexity will be O(ElogE + ElogV), which is approximately O(ElogV). Space Complexity The space complexity of Kruskal's minimum spanning tree algorithm is O(|E| + |V|), where E is the number of edges in the graph and V is the number of vertices. Reason: Since the disjoint set structure takes O(|V|) space to store the parents of vertices and O(|E|), we’re additionally storing the edges of the graph. If you've made it this far, congratulations, Champ! You've understood the Kruskal’s minimum spanning tree algorithm”. If you haven't already submitted it to CodeStudio. Without further ado, have it accepted as early as possible. You must watch this video for the conceptual understanding and proper code implementation of the “Kruskal’s Algorithm”. ### What is Kruskal’s minimum spanning tree? The tree formed using Kruskal's minimum spanning tree algorithm on a graph is called the minimum spanning tree. ### What is a spanning tree? A spanning tree of a Graph G is its subset, having all the vertices covered with the minimum possible number of edges. Thus, a spanning tree has v-1 edges. ### What is a minimum spanning tree? A spanning tree with a minimum sum of all edges among all the spanning trees of a graph is called the minimum spanning tree. ### What is Kruskal’s algorithm used for? Kruskal’s algorithm is used for finding the minimum spanning tree of a graph. ### What are minimum spanning trees used for? For network architectures, minimum spanning trees are utilized (i.e. telephone or cable networks). They're also utilized to develop approximate solutions to complex mathematical issues like the Traveling Salesman Problem. ## Conclusion In this article, we've discussed Kruskal’s minimum spanning tree algorithm here, along with its approach and implementation in C++. Other similar algorithms are Prim’s algorithm using a minimum spanning tree and The traveling salesman problem. Don't forget to try it out, as it'll help you empower your understanding. To practice more such problems, Codestudio is a one-stop destination. This platform will help you acquire effective coding techniques and overview student interview experience in various product-based companies such as Amazon, Google, Adobe, etc. Also check out some of the Guided Paths on topics such as Data Structure and Algorithms, Competitive Programming, etc. as well as some Contests and Interview Experiences curated by top Industry Experts only on  CodeStudio. Happy Coding! Codekaze-June23 India's Biggest Tech Hiring Challenge is LIVE! Go on top
# Samson straight line Simson straight The Samson straight line is an object of triangular geometry . If the base points from a point of precipitated solder on the (possibly extended) sides of a triangle on a common straight line, then this straight line as a Simson line or wallacesche straight line and the point as its pole , respectively. This is exactly the case if lies in the perimeter of . ${\ displaystyle P}$ ${\ displaystyle \ triangle ABC}$${\ displaystyle P}$${\ displaystyle P}$${\ displaystyle \ triangle ABC}$ The Samson straight line is erroneously named after the mathematician Robert Simson (1687–1768), in whose work, however, no work on the Samson straight line can be found. In reality, it was discovered in 1797 by William Wallace (1768–1843). ## Other properties ### Parallels to the Simson straight line Simson straight line is parallel to GC Every Simson straight line of a triangle has three particular parallels, which each run through one of the three corner points of the triangle. More precisely, the following theorem applies: A triangle , a point P on its circumference and the associated Simson straight line are given. If G AB is the intersection of the perpendicular from P to AB with the circumference, then the straight line CG is parallel to the Simson straight line.${\ displaystyle \ triangle ABC}$ ### Intersection angle between Simson lines ${\ displaystyle 2 \ alpha = \ beta}$ If you look at two different points on the circumference of a triangle, you get two different Simson lines. The angle of intersection of these two Simson straight lines is exactly half the size of the angle that the two points form with the center of the circumference. Let and be two points on the perimeter of with a center . Furthermore, let the intersection angle of the two associated Simson lines and . Then applies .${\ displaystyle P_ {1}}$${\ displaystyle P_ {2}}$${\ displaystyle \ triangle ABC}$${\ displaystyle O}$${\ displaystyle \ alpha}$${\ displaystyle \ beta = \ angle P_ {1} OP_ {2}}$${\ displaystyle 2 \ alpha = \ beta}$ ### Simson straight line as bisector ${\ displaystyle | HD | = | DP |}$ If you connect the vertical intersection of a triangle with a point on the circumference of the triangle, this connecting line is halved by the associated Simson straight line. A triangle , a point on its circumference and the associated Simson straight line are given. H is the orthocenter of , the Simson straight the route cuts in and it is . Also lies on the Feuerbachkreis .${\ displaystyle \ triangle ABC}$${\ displaystyle P}$${\ displaystyle \ triangle ABC}$${\ displaystyle {\ overline {HP}}}$${\ displaystyle D}$${\ displaystyle | HG | = | GP |}$${\ displaystyle G}$ ### Set of straight lines Simson lines as tangents of a deltoid If the Samson pole is allowed to move on the circle, the family of Simson straight lines thus created has a deltoid , also referred to as Steiner's hypocycloid , as an envelope curve . ${\ displaystyle P}$ ### Others If two triangles have the same circumcircle and their associated Simson lines have the same pole, the intersection angle of the two Simson lines is independent of the choice of pole. In other words: For all points on the common circumference of the two triangles, the intersection angle of the two associated Simson lines is the same. ${\ displaystyle P}$ ## proof Sketch to prove the collinearity of the base points It is proven: If lies on the perimeter of , the base points lie on a common straight line. One shows that it is true. ${\ displaystyle P}$${\ displaystyle \ triangle ABC}$${\ displaystyle \ angle EFP + \ angle PFD = 180 ^ {\ circ}}$ The base points and are above the Thaleskreis . Since circumferential angles (peripheral angles) are equal over the same arc, it follows ${\ displaystyle E}$${\ displaystyle F}$${\ displaystyle [PA]}$ {\ displaystyle {\ begin {aligned} \ angle EFP & = 90 ^ {\ circ} + \ angle EFA = 90 ^ {\ circ} + \ angle EPA \\ & = 90 ^ {\ circ} + (90 ^ {\ circ} - \ angle PAC) = 180 ^ {\ circ} - \ angle PAC \ end {aligned}}}. On the other hand , a chordal quadrilateral is assumed . The opposite angles and this square therefore complement each other . So overall it results ${\ displaystyle PBCA}$${\ displaystyle \ angle PAC}$${\ displaystyle \ angle CBP}$${\ displaystyle 180 ^ {\ circ}}$ ${\ displaystyle \ angle EFP = \ angle CBP}$. The points and are above the Thales circle , so that there is also a quadrilateral tendon. Similar to before you close . Because of it you get ${\ displaystyle D}$${\ displaystyle F}$${\ displaystyle [PB]}$${\ displaystyle PBDF}$${\ displaystyle \ angle PFD = \ angle DBP}$${\ displaystyle \ angle DBP + \ angle CBP = 180 ^ {\ circ}}$ {\ displaystyle {\ begin {aligned} \ angle PFD & = 90 ^ {\ circ} - \ angle DFB = 90 ^ {\ circ} - (90 ^ {\ circ} - \ angle PBD) \\ & = \ angle PBD = 180 ^ {\ circ} - \ angle CBP \ end {aligned}}}. So is with ${\ displaystyle \ angle EFP + \ angle PFD = \ angle CBP + (180 ^ {\ circ} - \ angle CBP) = 180 ^ {\ circ}}$ the claim proved. Note: The proof given relates to the position of the height base points shown in the sketch. If these are different, the reason must be varied accordingly. ## Individual evidence 1. H. SM Coxeter, SL Greitzer: Simson Lines . §2.5 in Geometry Revisited. In: Math. Assoc. Amer., Washington DC 1967, p. 41. 2. a b Eric W. Weisstein : Simson straight line . In: MathWorld (English). ## literature • Max Koecher , Aloys Krieg : level geometry . 3. Edition. Springer-Verlag, Berlin 2007, ISBN 978-3-540-49327-3 , pp. 170-172 • Coxeter, HSM , SL Greitzer: Timeless geometry . Klett, Stuttgart 1983 • Roger A. Johnson: Advanced Euclidean Geometry . Dover 2007, ISBN 978-0-486-46237-0 , pp. 137 ff., 206 ff., 243, 251 (first published in 1929 by the Houghton Mifflin Company (Boston) under the title Modern Geometry ) • Ross Honsberger : Episodes in Nineteenth and Twentieth Century Euclidean Geometry . MAA, 1995, pp. 43-48, 82-83, 121, 128-136
# The Absolute-Value Inequality: Definition & Example An error occurred trying to load this video. Try refreshing the page, or contact customer support. Coming up next: The Quadratic Formula: Definition & Example ### You're on a roll. Keep up the good work! Replay Your next lesson will play in 10 seconds • 0:02 The Absolute Value • 1:31 The Absolute Value Inequality • 2:28 Setting Up Our Problem • 3:27 Solving Our Problem • 4:51 Lesson Summary Save Save Want to watch this again later? Timeline Autoplay Autoplay Speed Speed #### Recommended Lessons and Courses for You Lesson Transcript Instructor: Yuanxin (Amy) Yang Alcocer Amy has a master's degree in secondary education and has taught math at a public charter high school. After watching this video lesson, you will be able to solve absolute value inequalities. Learn how you can easily set up your problem to successfully solve your problem each and every time. ## The Absolute Value Before we talk about what an absolute value inequality is, let's talk about what an absolute value is. When we take the absolute value of something, we are trying to find the distance that value is from 0. For example, the absolute value of 5 is 5 since it is 5 away from 0. How about the absolute value of -5? What do you think that is? It is 5 as well because -5 is also 5 away from 0. Since we're talking math here, we also have a symbol for the absolute value and that is a vertical line on either side of the value as in |5| and |-5|. We can solve these absolute values easily since we know that both are 5 away from 0. So, we know that |5| = 5 and |-5| = 5. If our problem was |x| = 5, then we already know that our answer is 5 and -5. Here's a trick question: Can we have |x| = -5? What do you think? Is it possible that the absolute value of something could equal a negative number? No, it's not, because we don't have a negative distance. So, if you ever see an absolute value equaling a negative number, you can right away say that the problem is false and can't be solved since the absolute value of anything can never be negative. Now let's talk about absolute value inequalities. ## The Absolute Value Inequality An absolute value inequality is a problem with absolute values as well as inequality signs. We can have problems like |x + 3| > 1. We have four different inequality signs to choose from. We have less than, greater than, less than or equal to and greater than or equal to. So, our absolute value inequalities can have any one of these four signs. The process to solve them is the same regardless of the sign. But you do have to pay attention to what type of sign it is, as you will be taking that into account when you solve. Just like our regular absolute values, we can't have our absolute value being a negative number. So, a problem such as |x + 3| < 0 or less than a negative number will not be possible and cannot be solved. ## Setting Up Our Problem Now let's see about solving an absolute value inequality. We will work with the inequality |x + 3| > 1. To set up our problem for solving, we need to write two problems from our one absolute value inequality. We do this by writing one problem exactly as we see the absolute value inequality but without the absolute value signs. The next one we will write will be almost the same except we are flipping the inequality sign around and changing the right side to the negative version. So, for our absolute value inequality, our two problems are x + 3 > 1 and x + 3 < -1. Notice how I've flipped the inequality around in the second problem, and I've changed my right side to the negative version. If the right side is already negative, then we write the positive since the negative of a negative is a positive. Once we've done this, we are ready to proceed and find our answers. To unlock this lesson you must be a Study.com Member. ### Register to view this lesson Are you a student or a teacher? ### Unlock Your Education #### See for yourself why 30 million people use Study.com ##### Become a Study.com member and start learning now. Back What teachers are saying about Study.com ### Earning College Credit Did you know… We have over 200 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level.
Upcoming SlideShare × # Standard Deviation Calculation Process 2 1,264 Published on 1 Like Statistics Notes • Full Name Comment goes here. Are you sure you want to Yes No • Be the first to comment Views Total Views 1,264 On Slideshare 0 From Embeds 0 Number of Embeds 0 Actions Shares 0 0 0 Likes 1 Embeds 0 No embeds No notes for slide ### Transcript of "Standard Deviation Calculation Process 2" 1. 1. MARKETING EMPIRE Standard Deviation and Variance Deviation just means how far from the normalStandard DeviationThe Standard Deviation is a measure of how spread out numbers are.Its symbol is σ (the greek letter sigma)The formula is easy: it is the square root of the Variance. So now you ask, "What is theVariance?"VarianceThe Variance is defined as: The average of the squared differences from the Mean.To calculate the variance follow these steps: Work out the Mean (the simple average of the numbers) Then for each number: subtract the Mean and then square the result (the squared difference). Then work out the average of those squared differences.ExampleYou and your friends have just measured the heights of your dogs (in millimeters): www.marketingempire.org 2. 2. MARKETING EMPIREThe heights (at the shoulders) are: 600mm, 470mm, 170mm, 430mm and 300mm.Find out the Mean, the Variance, and the Standard Deviation.Your first step is to find the Mean:Answer: 600 + 470 + 170 + 430 + 300 1970 Mean = = = 394 5 5so the mean (average) height is 394 mm. Lets plot this on the chart:Now, we calculate each dogs difference from the Mean:To calculate the Variance, take each difference, square it, and then average the result: 2062 + 762 + (-224)2 + 362 + (-94)2 108,520 2 Variance: σ = = = 21,704 5 5So, the Variance is 21,704.And the Standard Deviation is just the square root of Variance, so: www.marketingempire.org 3. 3. MARKETING EMPIREStandard Deviation: σ = √21,704 = 147.32... = 147 (to the nearest mm)And the good thing about the Standard Deviation is that it is useful. Now we can showwhich heights are within one Standard Deviation (147mm) of the Mean:So, using the Standard Deviation we have a "standard" way of knowing what is normal, andwhat is extra large or extra small.Rottweillers are tall dogs. And Dachsunds are a bit short ... but dont tell them!*Note: Why square ?Squaring each difference makes them all positive numbers (to avoid negatives reducing theVariance)And it also makes the bigger differences stand out. For example 1002=10,000 is a lot biggerthan 502=2,500.But squaring them makes the final answer really big, and so un-squaring the Variance (bytaking the square root) makes the Standard Deviation a much more useful number. SEE BELOW www.marketingempire.org 4. 4. MARKETING EMPIRE MAKE IT MORE CLEAR QUESTION ANSWERwww.marketingempire.org
# Is f(x)=e^xsqrt(x^2-x) increasing or decreasing at x=3? Jun 25, 2016 $f \left(x\right) = {e}^{x} \sqrt{{x}^{2} - x}$ is increasing at $x = 3$ #### Explanation: A function $f \left(x\right)$ is increasing at $x = a$ if $f ' \left(a\right) > 0$ and is decreasing at $x = a$ if $f ' \left(a\right) < 0$. As $f \left(x\right) = {e}^{x} \sqrt{{x}^{2} - x}$, using product rule $f ' \left(x\right) = {e}^{x} \sqrt{{x}^{2} - x} + {e}^{x} \times \frac{1}{2 \sqrt{{x}^{2} - x}} \times \left(2 x - 1\right)$ = ${e}^{x} \sqrt{{x}^{2} - x} + \frac{{e}^{x} \left(2 x - 1\right)}{2 \sqrt{{x}^{2} - x}}$ and at $x = 3$ $f ' \left(x\right) = {e}^{3} \sqrt{{3}^{2} - 3} + \frac{{e}^{3} \left(2 \times 3 - 1\right)}{2 \sqrt{{3}^{2} - 3}}$ = ${e}^{3} \sqrt{6} + \frac{5 {e}^{3}}{2 \sqrt{6}} = {e}^{3} \sqrt{6} \left(1 + \frac{5}{12}\right) = {e}^{3} \sqrt{6} \times \frac{17}{12}$ which is clearly positive. Hence, $f \left(x\right) = {e}^{x} \sqrt{{x}^{2} - x}$ is increasing at $x = 3$.
# Lesson A - Natural Exponential Function and Natural Logarithm Functions Save this PDF as: Size: px Start display at page: ## Transcription 1 A- Lesson A - Natural Exponential Function and Natural Logarithm Functions Natural Exponential Function In Lesson 2, we explored the world of logarithms in base 0. The natural logarithm has a base of e. e is approximately The number e was discovered by a great 8th century mathematician named Euler. This famous irrational number is useful for determining rates of growth and decay. The derivation of e is beyond the scope of this course, but the expresson + n closer and closer to e, as the value substituted for n gets larger and larger. n produces results The natural exponential function is f(x) ex. Your scientific calculator should have a key for finding ex. You can also use the approximate value of e for computations. Below is the graph for f(x) ex f(x) ex x ex Rules of exponents apply to the exponential function. Study the examples of factoring. Example e2x - ex ex ex (ex - ) ex ex - Sometimes it is easier to factor by making a substitution. Let y ex. Example 2 4e2x - 3ex 2ex Substituting yields 4y2-3y 2y y(4y - 3) 2y 4y y - 3 Don t forget to substitute back again. 2 4ex Practice Problems Factor the following. ) e2x - 2) e2x - ex - 6 3) e2x - ex - 2 4) 2e2x - 5ex - 3 Solutions ) e2x - 2) e2x - ex - 6 3) e2x - ex - 2 4) 2e2x - 5ex - 3 y2 - y2 - y - 6 y2 - y - 2 2y2-5y - 3 (y - )(y + ) (ex -)(ex + ) (y - 3)(y + 2) (ex - 3)(ex + 2) (y - 2)(y + ) (ex - 2)(ex + ) (2y + )(y - 3) (2ex + )(ex - 3) 2 A-2 The natural exponential function is used in every area of science. Example 3 Suppose that the number of bacteria present in a culture is given by N(t) 000 e (.)t. How many bacteria would be in the culture when t 4 hours? 000 e(.)(4) 000 e(.4) 49.8 rounds to,492 bacteria Practice Problems ) How many bacteria would be in the culture above when t 0 hours? 2) How long will it be before the number of bacteria reaches 0,000? Try to solve this by plugging in values for t. We will learn how to solve this a different way soon. Solutions ) 000 e(.)(0) 000 e() 2,78.3 rounds to 2,78 bacteria 2) try t 5 hours 000 e(.)(5) 000 e(.5) rounds to 4482 bacteria try t 20 hours 000 e(.)(20) 000 e(2) 7389 bacteria try t 24 hours 000 e(.)(24) 000 e(2.4),023 bacteria Natural Logarithm Function It is the inverse of the exponential function, which is f(x) ex. You may have tried different values, but you should have determined that the time to reach 0,000 bacteria is somewhat less than 24 hours. The natural logarithm function is f(x) ln(x). In stands for natural log. There are several properties and laws of the natural log function which you need to memorize. Use the directions that came with your scientific calculator to find and use the natural log key. (It may be marked LN.) Check the following relationships with your calculator, choosing any value you like for the variables. ) ln 0 2) ln e 3) ln ex X 4) elnx x when x 0 5) ln xy ln x + ln y 6) ln x/y In x - ln y 7) ln xa a In x Example 4 Determine the value of ln 2e in simplest terms. ln 2e ln 2 + ln e Law #5 above.69 + Use your calculator to find the natural log of 2. You should know the natural log of e from #2 above. ln 2e.69 Add to find the solution. 3 A-3 Example 6 Determine the value of 2 ln 3-3 ln 2 in simplest terms. 2 ln 3-3 ln 2 ln 32 - ln 23 Law #7 above ln 9 - ln 8 Simplify radicals ln 9/8 ln.25.8 Law #6 above Divide and find the ln. The answer is rounded. Example 7 Determine the value of ln 5 + ln 3 - ln 5 in simplest terms. ln 5 + ln 3 - ln 5 ln (5 3) - ln Law #5 above 5 ln 5 - ln 5 Multiply 0 Subtract Practice Problems Determine the value of each expression in simplest terms. Round your answers to the nearest hundredth. ) In 4e2 2) In 8 - In 2 + In 5 3) 4 In x + 2 In y 4) In 5 Solutions ) In 4 e2 ln 4 + ln e2 2) In 8 - In 2 + In 5 ln 8/2 + ln ln e ln e () ) 4 In x + 2 In y ln x4 + ln y2 4) In 5 ln 52 ln x4 y2 ln x4y2 /2 ln 5 ln 4 + ln 5 ln 4 5 ln /2 (.609).80 Example 8 Recall that on page 22-2, you were asked to find out how long it would be before the number of bacteria reached 0,000. Let s work that problem a different way using the natural logarithm function. 000 e.t 0,000 e.t 0 In (e.t) In (0).t In 0 Taking the natural logarithm of both sides. t In 0 23 hours. 4 Exponential and logarithmic functions can be manipulated in algebraic equations. Remember that as long as we do the same thing to both sides of an equation, we do not change the value of the equation. In the examples below, find the natural log of each side in order to simplify exponents and put the equation in a form that is easier to manipulate. A-4 Example 9 4 X2 9 ln 9 ln 4 X2 X 2 ln 4 ln 9 X 2 ln 9 ln X.26 Example 0 4x+2 72x- In (4x+2 ) In (72x- ) (X + 2) In 4 (2X - ) In 7 X In In 4 2X In 7 - In 7 X In 4-2X In 7-2 In 4 - In 7 X (In 4-2 In 7) -2 In 4 - In 7 X -2 In 4 - In 7 In 4-2 In X 5 B- Lesson B - Limits Limits in life are boundary points. Usually they denote the highest or top number allowable. Speed its are the highest speed allowable which cars may tavel without breaking the law. A supermarket it of 2 quarts of strawberries on their store special means that you can buy up to 2 quarts per person and no more. In science an instrument might be accurate within /00th of an inch. In each case, the it is a real number. Mathematical its are similar, but are not identical, to the its we see in everyday life. In math, we refer to the it of a function. A function s it may or may not exist. If the it does exist, then the it will be a real number and it will be unique. Example Consider the following sequence:, /2, /3, /4, /5, /6,... /000000,... As we look at these numbers, what is happening to them? They are getting smaller and smaller, right? How small will they get? Will they reach 0? No, but they will get so close to zero as to be indistinguishable. Therefore, the mathematical it of these numbers is 0. In this case, the it exists and is uniquely 0. It is not approaching any other number except 0. Example 2 Now let s look at another sequence:, 2, 3, 4, 5, 6,... 0,000,... 00,000, What is happening to these numbers? They are getting larger and larger and are approaching infinity. Is there a it? NO! If you say that the it is 00,000,000,000,003, then I can always find a bigger number in the sequence. Therefore, the it does not exist. Example 3 In geometry, consider a circle which has a polygon inscribed inside. If you begin with a 3-sided polygon (triangle), and continue to inscribe polygons with larger and larger numbers of sides, what happens? Notice that the area of the inscribed polygon gets closer and closer to the area of the circle. Does the inscribed polygon become a circle? No, but it gets close enough as to be indistinguishable. Therefore, the it exists and is uniquely the area of the circle. Historically, Archimedes used this exact reasoning to determine the area of circles before the discovery of pi. Now let s look at some functions on a graph, and see if we can understand its more fully. Consider the following function. f(x) Example 4 f(x) X + 6 B-2 A it in a graphical sense represents the anticipated height of a function. What would be the it of f(x) as we allow our X values to get closer and closer to 0? We could approach 0 by using negative numbers or positive numbers. A chart below shows numbers which are approaching X 0 from both perspectives. X f(x) X f(x) looking at positive values for X looking at negative values for X Either way we look, we see that as the X value of this function approaches 0, the f(x) value or the height of the function will be approaching. Indeed, f(0)! Thereore, the it exists and it is uniquely. Example 5 f(x) X + for all X except x 0 Graphically, we now have: f(x) Now what is the it as X approaches 0? It is still. Even though the value at X 0 is undefined, we can get SO close to X 0, and it still appears that f(x) is SO close to. How close is close enough? That is a discussion for Caluculus. f(x) Let s try a few more: Example 6 f(x) X2 What is the it of f(x) as X approaches -2? This one is simple. The graph will show the exact value of 4 when x -2. You could look at values on both sides of -2, but your answer will be the same. The it exists and it is uniquely equal to 4. 7 B-3 Example 7 X 2 X 2 X 2 The graph looks like: f(x) What is the it of f(x) as X approaches 2? We will need a table here, because the value of 2 cannot be substituted into the function. Therefore, the it of f(x) 3. Notice that the numerator of this function can be factored into (X + )(X - 2). Then, you can cancel the common factors, and you get (X ). The graph looks like f(x) X +, except that there is a hole at X 2. In fact, if you substitute x 2 into the new equation, f(x) X +, you get f(2) Now, let s work with mathematical notation to accurately describe what we have done so far. The appropriate it notation looks like the following: f(x) B X A where X A means that X is approaching a and the value of the it is B. X 2 X 2 In Example 7, we would express the it as follows: 3 X A X 2 8 B-4 Example 8 below: What about more serious holes and gaps in a function? Refer to the graph f(x) This type of function is commonly referred to as a step funcition. What is the it as X approaches 2 of this step function? X f(x) Approaching from the left of X 2 X f(x) Approaching from the right of X 2 As you can see, it depends on whether you approach X 2 from the right or from the left, as to what you would expect the height of the function to be. From the right, it appears that the it would be 2, and from the left, it appears that the it would be. Remember that we said that in order for the it to exist, that it had to be unique? Because of this, the it as X approaches 2, does not exist. Written in appropriate it notation, we get: f(x) Does not exist (DNE) x 2 For completeness, let s write Examples, 2, 4, and 6 in proper it notation. Example : Example 2: Example 4: Example 6: x X 0 x Does not exist x 0 X + X 2 4 x 2 Let s review:. Some its exist and some do not. 2. If the it exists, then it is a real number and it is unique. 3. The proper nototation for a it is: f(x) B x A 4. To evaluate the it, first try substituting A into the function, (i.e., try f(a).) 5. If you cannot evaluate the it by substituting, then try factoring. 6. If neither method works, then graph the function. 7. Finally, the it is simply the anticipated height of f(x) when X A. 9 A-A Find the value of y for several values of x and graph each equation. ) y e 2x x y ) y 2 ln x x y Factor. 3) ln 2 x - ln x - 2 4) 2 ln 3 x + 3 ln 2 x 5) 3e 2x + 5e x - 2 6) e 2x - e x - Solve for x. 7) 2 5x 3 2x 8) 2 x 4 x-2 9) e 2x ln 5 Read the information given and answer the questions. The number of bacteria present in a culture is given by N(t) 2000 e.3t. 0) How many bacteria will be in the culture when t 2 hours? ) How many hours would it take for the bacteria count to reach 00,000? 10 A-B Find the value of y for several values of x and graph each equation. ) y e -2x x y ) y ln 2x x y Factor. 3) 2ln3 x - 8ln x (ln x - 2) 4) 2 ln 5 x + 5 ln 4 x 5) 6e 2x + 4e x - 2 6) e 2x - ln 2 x Solve for x. 7) e x + 2 -x 3 Hint: First multiply both sides 8) 2 5+x 3 2-x 9) e -3x ln 4 through by e x, then substitute y for e x. Read the information given and answer the questions. The amount of money which has accumulated after t years at interest rate r under continuous compounding is M Pe rt, where P principle(original money invested), r interest rate, and t time in years. 0) If P \$600, r.09, and t 0, compute M. ) If \$000 was invested at 5% over 20 years, find M. 2) How long would it take \$000 to become \$500 if it was invested at 6% interest? 11 A-C Find the value of y for several values of x and graph each equation. ) y 2e x x y x y 2) y ln x Factor. 3 3) 3e 2x - 7e x + 2 4) 5e 2x - 4e x ) ln 2 x - 4 ln x +2 6) 2 ln 3 x + 6 ln x Solve for x. 7) 3 2x 4 x-2 8) e x - e -x 4e -x 9) log 4 (ln 5) X Read the information given and answer the questions. The half-life of iodine-3 is 8 days. That is the time it takes for one half of the substance to decay or change into another kind of matter. The first question is done as an example. 0) How much of 6 grams of iodine-3 will remain after 0 days? The formula needed is the decay formula: Q(t) I e -kt where I equals initial amount of the substance, t time, and k decay constant. The decay constant is different for different substances. For Iodine-3 it is approximately.087. Q(t) I e -kt Q(0) 6(e (-.087)(0) ) Q(0) 6(e -.87 ) Q(0) 6(.49) 2.5 g remaining after 0 days ) How much iodine would be left after 2 days? 2) How many days would it take for only gram of iodine to remain? 12 A-D Find the value of y for several values of x and graph each equation. ) y e x x y 2) y ln x x y Factor. 3) 5e 2x + 3e x - 6 e x + 3 4) 6 e 2x - 2 e x 5) ln 2 x - ln 2 x + 2 ln x + 6) 3 ln 3 x + 6 ln x Solve for x. 7) e 2x ) ln 2 x - 5 ln x ) e 3x ln 2 Read the information given and answer the questions. The half-life of sodium-24 is 5 hours. When you are given the half-life af a substance, you can find the decay constant as follows. Start with the decay formula, and plug in the values you know. Q(t) Se -kt /2 e -k(5) Set the size of the original sample to, and the size of the sample after 5 hours to /2. ln /2 ln e -5k ln /2-5k ln /2-5 k.046 0) A 0 gram quantity of sodium-24 was studied. How much sodium-24 was left after 20 hours? ) How many days would it take for only 2 grams of sodium-24 to remain? 13 B-A Evaluate the following its from the given graph. ) g(x) 2) r(x) g(x) x 0 r(x) x 2 r(x) x Evaluate the following its by drawing a graph. 3) x r(x) X - 2 4) x 2x Evaluate the following its by factoring. 5) X 2 + 3X 0 x 2 X 2 6) x 3 X 2 + X 6 X 2 + 2X 3 Evaluate the following its using any appropriate method. 7) x 2X2-6 8) x X2 + 2X 9) x π/2 cos X sin 2 X 0) x 0 0 cot 2 θ csc 2 θ s ec 2 θ Hint: Simplify and then evaluate. 14 B-B Evaluate the following its from the given graph. ) w(x) 2) q(x) w(x) x 2 q(x) x q(x) x 2 Evaluate the following its by drawing a graph. 3) x x 4) θ π/2 sin θ Evaluate the following its by factoring. 5) 2X 2 6 x 3 X 3 6) X 2 + 9X 20 x 4 X 4 Evaluate the following its using any appropriate method. 7) y 0 log y 8) x 0 2e x (e x ) e 2x e x Hint: Factor, then evaluate. 9) x 2 X 2 0) x 3 X 3 3 X 15 B-C Evaluate the following its from the given graph. ) s(x) 2) p(x) s(x) x 2 p(x) x Evaluate the following its by drawing a graph. 3) X 2 X x x 4) θ π/2 cos θ Evaluate the following its by factoring. 5) x X 2 x 6) 3X 7X 2 x 0 X Evaluate the following its using any appropriate method. 7) θ 90 sec 2 θ 3sec 2 θ 8) x π x 9) x 2 X 2 7X + 0 X 2 4 (X + 3) 0) 3 27 x 0 X Hint: Use Pascal s triangle to expand (X+ 3)3. 16 B-D Evaluate the following its from the given graph. ) v(x) 2) s(x) 3π/2 π/2 π 2π v(x) x 2 s(x) x π v(x) x 4 Evaluate the following its by drawing a graph. 3) x e x 4) θ 80 2 tan θ Evaluate the following its by factoring. 5) 3X 3 27X x 3 X 3 6) x 4 4 X X 2 6 Evaluate the following its using any appropriate method. 7) x 2 4X 7 + X 8) x π/2 cos x cotx 9) θ 90 sinθ cos θ csc θ + cotθ Hint: Multiply through by the conjugate for - cosθ. 0) x 0 e 2x + e x 2 e x Hint: Factor the numerator. 17 Test A ) The natural logarithm function, ln x, is the inverse of A) sin X B) cos X C) e x D) log x 2) The graph of the e function lies in which quadrants? A) I only B) II only C) I and II D) III only 3) The value of e is approximately A) 27 B) 2.7 C) 270 D).27 9) If the number of bacteria present is given by N (t) 5000 e -3t, how many bacteria will be present when t 5 hours? A) 6,749 B) 22,408 C) 6,600 D).6 x 0 0 0) Use the formula from #9 to find how long it will be until there are 50,000 or more bacteria. A) nearly 0 hours B) nearly 6 hours C) nearly 7 hours D) nearly 8 hours 4) The factorization of ln 2 x - is A) (ln x -)(ln x - ) B) (ln x - )(ln x + ) C) 2 ln x (-x) D) log x 2A log x B 5) The factorization of 2e 2x + 5e x - 2 is A) (2e 2x - 3) (e x + 4) B) (2e x + 3) (e x - 4) C) (2e x - 6) (e x + 2) D) (2e x - 3) (e x + 4) ) A triangle has side c 28, m A 34 and m C 72. Use the law of sines to find the length of side a. A) 3.2 B) 6.5 C) 52.8 D) 28 2) What are the reference angle and tan for 405? A) 45, 2 B) 45, C) 60, 3 D) 35, 6) If e 2x 8, then x A).03 B) 2.08 C).69 D).35 3) cos(90 - q) csc q A) cos 2 q B) tan q C) sin q cos q D) sin 2 q 7) If e 2+x e ln 3+, then x A). B). C) 4. D) 3. 8) If ln (4e x ) -6, then x A).4 B) 2.4 C) 7.4 D) 4.6 4) A triangle has sides a 8, b 4 and c. Use the law of cosines to solve for angle A. A) 90.7 B) 85.4 C) 33.9 D) ) Change r to rectangular form. 3 cos q - sin q A) 3X - Y 6 B) X - Y 2 C) X - 3Y 6 D) 3X 2 - Y 2 2 18 Test B ) Which of the following methods can be used to compute a it? I) graphing II) factoring III) the Pythagorean theorem A) III nd IV B) I only 2) Which of the following statements is not true? A) Some its exist. B) A it is a complex number. C) A it is the anticipated height of a function. 3) x 2π cos θ A) 0 B) IV) quadratic formula C) I and II only D) I, II and III D) Limits are unique. C) - D) does not exist 4) -X 2 + 3X + 0 x 5 X - 5 A) 7 B) does not exist C) -3 D) -7 5) The graph is Y cot θ. 8) 2 ln x + 3 ln x x ln x A) 3 B) -3 C) 0 D) does not exist 9) x 4 X 4 - X A) /254 B) 254 C) 0 D) does not exist 0) - x 2X + 7 A) 7 B) -2 C) 0 D) does not exist ) If X + < -2, then the solutions for X are A) X > B) X < -3 C) X > -3 D) no solutions 2) If X + < 7, then the solutions for X are A) X < 48 B) X > 48 C) X < 49 D) X < 6 π 2 cot θ x 2π π 3π 2π 2 A) -00 B) 00 C) 0 D) does not exist 6) e -2x + 2e -x x e -x A) 0 B) C) 2 D) does not exist cos 7) 2 θ - θ 0 sin θ(sin θ - ) A) 0 B) C) - D) does not exist 3) - B 3 3 A) B2 B) B 3 3 C) B4 D) B8 4) 3( - cos 2 θ) is equal to which of the cos(90 - θ) following when θ 90? A) 0 B) 3 C) D) -3 5) B 2 B -3 C 4 C 2 B A) C 6 /B 2 B) B 2 /C 6 C) C 4 B -2 D) C -4 B 2 19 Solutions A - A ) y e2x x y ) y 2 ln x x y ) ln2 x - ln x - 2 Y2 - Y - 2 (Y - 2)(Y + ) (ln x - 2)(ln x + ) 4) 2 ln3 x + 3 ln2 x 2Y3 + 3Y2 Y2(2Y + 3) ln2 X(2 ln X + 3) 5) 3e2x + 5ex - 2 3Y2 + 5Y - 2 (3Y - )(Y + 2) (3ex - )(ex + 2) Y 2 6) Y ( Y + ) ( Y ) ( Y ) Y + e x + 0) N t 2000e ) 2 5X 3 2X ln ( 3 2X ) ln 2 5X 5X ln 2 2X ln 3 5X ln 2 2X ln 3 0 X ( 5 ln 2 2 ln 3) 0 X 0 8) 2 X 4 X 2 ln 2 X ln 4 X 2 ( X ) ( ln 2 ) ( X 2 ) ( ln 4 ) X X 2 ln 4 ln 2 X X 2 2 X 2 ( X 2 ) X 2X 4 X 4 X 4 9) e 2X ln 5 ln ln 5 ln e 2X 2X ln ( ln 5 ) ln ( ln 5 ) X.24 2 ) 00,000 2,000e.3t t ln 50.3t ln 50.3 t t 3 20 Solutions A - B ) y e-2x x y ) y ln 2x x y 3) 2Y 3 8Y Y 2 2Y ( Y2 4 ) Y 2 4) 2Y 5 + 5Y 4 Y 4 ( 2Y + 5 ) ln 4 X ( 2 lnx + 5 ) Y ( Y + 2 ) ( Y 2 ) 2 lnx ( lnx + 2 ) Y ) 6Y 2 + 4Y 2 2 ( 3Y 2 + 2Y ) 2 ( 3Y ) ( Y + ) 2 ( 3e X ) ( e X + ) 6) e 2X ln 2 X ( e X ln X ) ( e X + ln X ) (difference of 2 squares) 7) e X + 2 X 3 e 2X + 2e 0 3X e 2X 3e X Y 2 3Y ( Y ) ( Y 2 ) 0 ( e X ) e X 2 0 e X ln X ln X 0 e X 2 ln e X ln 2 X ln ) 2 5+X 3 2 X 5 ln 2 + X ln 2 2 ln 3 X ln 3 X ln 2 + X ln 3 2 ln 3 5 ln 2 X ( ln 2 + ln 3) 2 ln 3 5 ln 2 X 2 ln 3 5 ln 2 ln 2 + ln ) e 3X ln 4 ln ln 4 ln e 3X 3X ln ( ln 4 ) ln ( ln 4 ) X. 3 ( 0) M 600e.09 )( 0 ) M 600e.9 M, ( ) M 000e.05 )( 20 ) M 000e M 2, ) M Pe rt e.06t.5 e.06t ln.5 ln e.06t ln.5.06t t ln or about years 21 Solutions A - C ) y 2ex 2) y ln x x y x y ) 3Y 2 7Y + 2 ( 3Y ) ( Y 2 ) ( 3e X ) e X 2 4) 5Y 2 4Y 3 ( 5Y + ) ( Y 3) ( 5e X + ) e X 3 5) Y 2 4 Y + 2 ( Y + 2 ) ( Y 2 ) Y + 2 Y 2 ln X 2 6) 2Y 3 + 6Y 2Y Y ln X ln 2 X + 3 7) 3 2X 4 X 2 ln 3 2X ln 4 X 2 2X ln 3 ( X 2 ) ln 4 2X ln 3 X ln 4 2 ln 4 2X ln 3 X ln 4 2 ln 4 X ( 2 ln 3 ln 4 ) 2 ln 4 X 2 ln 4 2 ln 3 ln ) e X e X 4e X e 2X e 0 4e 0 multiply by e X e 2X 4 e 2X 5 ln e 2X ln 5 2X ln 5 X ( ln 5 / 2 ).8 9) log 4 ( ln 5 ) X 4 X ln 5 ln 4 X ln ( ln 5 ) X ln 4 ln ( ln 5 ) ln ( ln 5 ) X ln ) done ) Q ( t ) Ie kt Q ( 2 ) 2 (.087 ) 6e 6e g 2) Q ( t ) Ie kt 6e.087 ( t ) 6 e.087 ( t ) ln ( / 6 ) ln e.087 ( t ) ln ( / 6 ).087 ( t ) ln ( / 6 ).087 t 20.6 days 22 Solutions A - D ) y ex x y ) y ln x x y ) 5Y 2 + 3Y 6 Y + 3 ( Y + 3) ( 5Y 2 ) Y + 3 5Y 2 5e x 2 4) 6Y 2 2Y 6Y ( Y 2 ) 6e x e x 2 5) Y 2 Y 2 + 2X + ( Y + ) ( Y ) ( Y + ) ( Y + ) Y Y + lnx lnx + 6) 3Y 3 + 6Y 3Y Y ln X ln 2 X + 2 7) Y ( Y + 2 ) ( Y 2 ) 0 Y 2 Y 2 e X 2 e X 2 X ln 2 X ln ( 2 ) X.69 X is undefined 8) Y 2 5Y ( Y 2 ) ( Y 3) 0 Y 2 Y 3 lnx 2 ln X 3 X e 2 X e 3 X 7.39 X ) e 3X ln 2 ln e 3X ln ( ln 2 ) 3X ln ( ln 2 ) ln ( ln 2 ) X.2 3 0) Q ( t ) Se kt.046 ( 20 ) 0e 0e.92 0 (.4 ) 4 g ) Q ( t ) Se kt 2 0e.046 ( t ) 5 e.046t ln ( / 5 ).046t ln ( / 5 ).046 t 35 hours days 23 Sol B-A and B-B ) g ( X ) 0 X 0 From the left and from the right, the values for g ( X ) get closer to 0 as X approaches 0. 2) r ( X ) DNE (does not exist) X 2 From the left the it -2. From the right the it 0 r ( X ) 0 X 3) ) W ( X ) 0 X 2 2) q ( X ) DNE X 3) q X X 2 3 X X 0 X 2 2 X 4) 4) X 2X 0 X 2 + 3X 0 ( X 2 ) ( X + 5 ) 5) X 2 X 2 X 2 X 2 ( X + 5 ) X 2 6) X 3 X 3 X 2 + X 6 X 2 + 2X 3 ( X 2 ) ( X + 3 ) X 3 ( X + 3) ( X ) ( X 2 ) ( X ) ) X 2X ) X 2 + 2X DNE These numbers will grow X without bound. 9) X π/ 2 cos X sin 2 X cos π / 2 sin 2 π / cos 2 θ cot 2 θ csc 2 θ 0) X 0º sec 2 sin 2 θ sin 2 θ θ X 0º cos 2 θ X 0º cos 2 θ sin 2 θ cos 2 θ X 0º sin 2 θ cos2 θ cos2 θ sin 2 θ X 0º sin 2 θ cos2 θ X 0º cos2 θ cos 2 0º 2X 2 6 5) X 3 X 3 2 ( X 3 ) X 3 X 3 X 2 + 9X 20 6) X 4 X 4 X 4 X 5 X 4 X 4 7) log y log0 Y 0 sin θ sin π / 2 θ π/ 2 2 X 2 9X + 20 X 4 X 4 2e X e X 2e X e X 8) X 0 e 2X e X X 0 e X ( e X ) 9) X 2 X X 3 0) X 3 3 X ( 3 X ) 3 X ( X 5 ) ( 4 5 ) X 4 2 24 Sol B-C and B- D ) s ( X ) DNE X 2 2) p ( X ) 0 X 3) X 2 X X X ( X ) X X X X X 4) cos θ 0 θ π/ 2 ) V ( X ) DNE X 2 V ( X ) 2 X 4 2) s ( X ) 0 X π 3) 4) DNE X 5) X X 2 X X + 2 X 3X 7X 2 6) X 0 X 3 7 ( 0 ) 3 X ( X ) ( X + ) X X X 0 ( 3 7X ) sec 2 θ 7) θ 90º 3 sec 2 θ tan 2 θ θ 90º 3 sec 2 θ θ 90º sin 2 90º 3 sin 2 θ cos 2 θ 3 cos 2 θ 3 8) X π X DNE X sin 2 θ θ 90º 3 ( X 5 ) X 2 7X + 0 X 2 9) X 2 X 2 4 X 2 ( X + 2 ) X 2 X 5 X 2 X ( X + 3 ) ) X 0 X X 0 ( X 3 + 9X X + 27 ) 27 X X 3 + 9X X X 0 X X 2 + 9X X 0 3X 2 27X 5) X 3 X 3 X 3 3X X + 3 X 3 3 ( 3) ( 3) + 3 6) X 4 X 4 7) X 2 8) X π/ 2 9) θ 90º θ 90º θ 90º θ 90º θ 90º X X X 2 9 X 3 X 3 4 X X 2 6 X 4 ( X + 4 ) X 4 3X ( X + 3) X 3 X X 7 + X 4 ( 2 ) 7 + ( 2 ) 8 3 cos X cot X X π/ 2 cos X cos X sin X X π/2 sin X sin θ cos θ csc θ + cot θ sin θ + cos θ cos θ + cos θ csc θ + cot θ sin θ + sin θ cos θ cos 2 θ csc θ + cot θ sin θ + sin θ cos θ sin 2 θ csc θ + cot θ csc θ + cot θ csc θ + cot θ e X + 2 e 2X + e X 2 e X 0) X 0 e X X 0 e X e X X 0 2 tan θ 2 ( 0 ) 0 θ 80º 25 Test Solutions Test A ) C 2) C 3) B 4) B: y ln x 5) D: y e x 6) A: e 2x 8 y 2 - (y + )(y - ) (ln x + )(ln x - ) 2y2 + 5y - 2 (2y - 3)(y + 4) 2x ln 8 x ln ) A: e 2 +x ln (e ln 3 + ) 2 + x ln 3 + x ln 3 - x. 8) C: ln (4ex) ln 4 - ln e x ln x ln ) B: N(t) 5000 e 3(5) 5000 e.5 22,408 0) D: 50,000 5,000e.3t 0 e.3 t t ln hours or nearly 8 hours ) where are - 5? Test B ) C 2) B 3) B X 2 3X 0 4) D: x 5 X 5 5) D ( X + 2 ) ( ) 7 x 5 ( X + 2 ) ( x 5 ) x 5 ( X 5 ) e 2X + 2e X e X e X + 2 6) C: x e X x e X 2 x e X + 2 sin 2 θ 7) A: θ 0º sin θ ( sin θ ) sin θ θ 0º sin θ 0 θ 0º 0 0 lnx ( 2 lnx + 3 ) 8) A: x lnx x 2 lnx ) A: x 4 X 4 X ) A: (the first term gets closer and x 2X closer to 0) ) D: Absolute value cannot be negative. 2) A: Solving the equality gives us X 48. Substituting 0 for X in the original inequality gives a true result, so the solution is X < ) C: B 3 B 2 B 4 4) B: 3 cos 2 θ cos ( 90º θ) 3 sin2 θ 3 sin θ; 3 sin90º 3 sin θ 5) A B 2 B 3 C 4 C 2 B B C 4 C 2 B C2 4 C B 2 C 6 B 2 ### ALGEBRA 2/TRIGONOMETRY ALGEBRA /TRIGONOMETRY The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION ALGEBRA /TRIGONOMETRY Thursday, January 9, 015 9:15 a.m to 1:15 p.m., only Student Name: School Name: The possession ### Math 120 Final Exam Practice Problems, Form: A Math 120 Final Exam Practice Problems, Form: A Name: While every attempt was made to be complete in the types of problems given below, we make no guarantees about the completeness of the problems. Specifically, ### Pre-Calculus Review Problems Solutions MATH 1110 (Lecture 00) August 0, 01 1 Algebra and Geometry Pre-Calculus Review Problems Solutions Problem 1. Give equations for the following lines in both point-slope and slope-intercept form. (a) The ### Algebra 2/Trigonometry Practice Test Algebra 2/Trigonometry Practice Test Part I Answer all 27 questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. For each question, write on the separate ### Friday, January 29, 2016 9:15 a.m. to 12:15 p.m., only ALGEBRA /TRIGONOMETRY The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION ALGEBRA /TRIGONOMETRY Friday, January 9, 016 9:15 a.m. to 1:15 p.m., only Student Name: School Name: The possession ### 2. Simplify. College Algebra Student Self-Assessment of Mathematics (SSAM) Answer Key. Use the distributive property to remove the parentheses College Algebra Student Self-Assessment of Mathematics (SSAM) Answer Key 1. Multiply 2 3 5 1 Use the distributive property to remove the parentheses 2 3 5 1 2 25 21 3 35 31 2 10 2 3 15 3 2 13 2 15 3 2 ### Higher Education Math Placement Higher Education Math Placement Placement Assessment Problem Types 1. Whole Numbers, Fractions, and Decimals 1.1 Operations with Whole Numbers Addition with carry Subtraction with borrowing Multiplication ### Algebra. Exponents. Absolute Value. Simplify each of the following as much as possible. 2x y x + y y. xxx 3. x x x xx x. 1. Evaluate 5 and 123 Algebra Eponents Simplify each of the following as much as possible. 1 4 9 4 y + y y. 1 5. 1 5 4. y + y 4 5 6 5. + 1 4 9 10 1 7 9 0 Absolute Value Evaluate 5 and 1. Eliminate the absolute value bars from ### ALGEBRA 2/ TRIGONOMETRY The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION ALGEBRA 2/ TRIGONOMETRY Friday, June 14, 2013 1:15 4:15 p.m. SAMPLE RESPONSE SET Table of Contents Practice Papers Question 28....................... ### Review for Calculus Rational Functions, Logarithms & Exponentials Definition and Domain of Rational Functions A rational function is defined as the quotient of two polynomial functions. F(x) = P(x) / Q(x) The domain of F is the set of all real numbers except those for ### Chapter 7 Outline Math 236 Spring 2001 Chapter 7 Outline Math 236 Spring 2001 Note 1: Be sure to read the Disclaimer on Chapter Outlines! I cannot be responsible for misfortunes that may happen to you if you do not. Note 2: Section 7.9 will ### 1. At what interest rate, compounded monthly, would you need to invest your money so that you have at least \$5,000 accumulated in 4 years? Suppose your grandparents offer you \$3,500 as a graduation gift. However, you will receive the gift only if you agree to invest the money for at least 4 years. At that time, you hope to purchase a new ### Dear Accelerated Pre-Calculus Student: Dear Accelerated Pre-Calculus Student: I am very excited that you have decided to take this course in the upcoming school year! This is a fastpaced, college-preparatory mathematics course that will also ### Complex Numbers Basic Concepts of Complex Numbers Complex Solutions of Equations Operations on Complex Numbers Complex Numbers Basic Concepts of Complex Numbers Complex Solutions of Equations Operations on Complex Numbers Identify the number as real, complex, or pure imaginary. 2i The complex numbers are an extension ### Trigonometric Functions: The Unit Circle Trigonometric Functions: The Unit Circle This chapter deals with the subject of trigonometry, which likely had its origins in the study of distances and angles by the ancient Greeks. The word trigonometry ### Chapter 8. Exponential and Logarithmic Functions Chapter 8 Exponential and Logarithmic Functions This unit defines and investigates exponential and logarithmic functions. We motivate exponential functions by their similarity to monomials as well as their ### ALGEBRA 2/TRIGONOMETRY ALGEBRA /TRIGONOMETRY The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION ALGEBRA /TRIGONOMETRY Friday, June 14, 013 1:15 to 4:15 p.m., only Student Name: School Name: The possession ### Section 6-3 Double-Angle and Half-Angle Identities 6-3 Double-Angle and Half-Angle Identities 47 Section 6-3 Double-Angle and Half-Angle Identities Double-Angle Identities Half-Angle Identities This section develops another important set of identities ### Definition of Vertical Asymptote The line x = a is called a vertical asymptote of f (x) if at least one of the following is true: f (x) = Vertical Asymptotes Definition of Vertical Asymptote The line x = a is called a vertical asymptote of f (x) if at least one of the following is true: lim f (x) = x a lim f (x) = lim x a lim f (x) = x a ### ALGEBRA 2/TRIGONOMETRY ALGEBRA /TRIGONOMETRY The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION ALGEBRA /TRIGONOMETRY Tuesday, January 8, 014 1:15 to 4:15 p.m., only Student Name: School Name: The possession ### SAT Subject Math Level 2 Facts & Formulas Numbers, Sequences, Factors Integers:..., -3, -2, -1, 0, 1, 2, 3,... Reals: integers plus fractions, decimals, and irrationals ( 2, 3, π, etc.) Order Of Operations: Arithmetic Sequences: PEMDAS (Parentheses ### Week 13 Trigonometric Form of Complex Numbers Week Trigonometric Form of Complex Numbers Overview In this week of the course, which is the last week if you are not going to take calculus, we will look at how Trigonometry can sometimes help in working ### Inverse Circular Function and Trigonometric Equation Inverse Circular Function and Trigonometric Equation 1 2 Caution The 1 in f 1 is not an exponent. 3 Inverse Sine Function 4 Inverse Cosine Function 5 Inverse Tangent Function 6 Domain and Range of Inverse ### y = rsin! (opp) x = z cos! (adj) sin! = y z = The Other Trig Functions MATH 7 Right Triangle Trig Dr. Neal, WKU Previously, we have seen the right triangle formulas x = r cos and y = rsin where the hypotenuse r comes from the radius of a circle, and x is adjacent to and y ### CHAPTER FIVE. Solutions for Section 5.1. Skill Refresher. Exercises CHAPTER FIVE 5.1 SOLUTIONS 265 Solutions for Section 5.1 Skill Refresher S1. Since 1,000,000 = 10 6, we have x = 6. S2. Since 0.01 = 10 2, we have t = 2. S3. Since e 3 = ( e 3) 1/2 = e 3/2, we have z = ### Math Placement Test Practice Problems Math Placement Test Practice Problems The following problems cover material that is used on the math placement test to place students into Math 1111 College Algebra, Math 1113 Precalculus, and Math 2211 ### Mathematical Procedures CHAPTER 6 Mathematical Procedures 168 CHAPTER 6 Mathematical Procedures The multidisciplinary approach to medicine has incorporated a wide variety of mathematical procedures from the fields of physics, ### 1 Review of complex numbers 1 Review of complex numbers 1.1 Complex numbers: algebra The set C of complex numbers is formed by adding a square root i of 1 to the set of real numbers: i = 1. Every complex number can be written uniquely ### ALGEBRA REVIEW LEARNING SKILLS CENTER. Exponents & Radicals ALGEBRA REVIEW LEARNING SKILLS CENTER The "Review Series in Algebra" is taught at the beginning of each quarter by the staff of the Learning Skills Center at UC Davis. This workshop is intended to be an ### Find the length of the arc on a circle of radius r intercepted by a central angle θ. Round to two decimal places. SECTION.1 Simplify. 1. 7π π. 5π 6 + π Find the measure of the angle in degrees between the hour hand and the minute hand of a clock at the time shown. Measure the angle in the clockwise direction.. 1:0. ### Solving Logarithmic Equations Solving Logarithmic Equations Deciding How to Solve Logarithmic Equation When asked to solve a logarithmic equation such as log (x + 7) = or log (7x + ) = log (x + 9), the first thing we need to decide ### Differentiation and Integration This material is a supplement to Appendix G of Stewart. You should read the appendix, except the last section on complex exponentials, before this material. Differentiation and Integration Suppose we have ### 6 Rational Inequalities, (In)equalities with Absolute value; Exponents and Logarithms AAU - Business Mathematics I Lecture #6, March 16, 2009 6 Rational Inequalities, (In)equalities with Absolute value; Exponents and Logarithms 6.1 Rational Inequalities: x + 1 x 3 > 1, x + 1 x 2 3x + 5 ### Math Review. for the Quantitative Reasoning Measure of the GRE revised General Test Math Review for the Quantitative Reasoning Measure of the GRE revised General Test www.ets.org Overview This Math Review will familiarize you with the mathematical skills and concepts that are important ### Algebra Unpacked Content For the new Common Core standards that will be effective in all North Carolina schools in the 2012-13 school year. This document is designed to help North Carolina educators teach the Common Core (Standard Course of Study). NCDPI staff are continually updating and improving these tools to better serve teachers. Algebra ### 2.3. Finding polynomial functions. An Introduction: 2.3. Finding polynomial functions. An Introduction: As is usually the case when learning a new concept in mathematics, the new concept is the reverse of the previous one. Remember how you first learned ### EXPONENTS. To the applicant: KEY WORDS AND CONVERTING WORDS TO EQUATIONS To the applicant: The following information will help you review math that is included in the Paraprofessional written examination for the Conejo Valley Unified School District. The Education Code requires ### Utah Core Curriculum for Mathematics Core Curriculum for Mathematics correlated to correlated to 2005 Chapter 1 (pp. 2 57) Variables, Expressions, and Integers Lesson 1.1 (pp. 5 9) Expressions and Variables 2.2.1 Evaluate algebraic expressions PRE-CALCULUS GRADE 12 [C] Communication Trigonometry General Outcome: Develop trigonometric reasoning. A1. Demonstrate an understanding of angles in standard position, expressed in degrees and radians. ### MATH 65 NOTEBOOK CERTIFICATIONS MATH 65 NOTEBOOK CERTIFICATIONS Review Material from Math 60 2.5 4.3 4.4a Chapter #8: Systems of Linear Equations 8.1 8.2 8.3 Chapter #5: Exponents and Polynomials 5.1 5.2a 5.2b 5.3 5.4 5.5 5.6a 5.7a 1 ### Logarithmic and Exponential Equations 11.5 Logarithmic and Exponential Equations 11.5 OBJECTIVES 1. Solve a logarithmic equation 2. Solve an exponential equation 3. Solve an application involving an exponential equation Much of the importance ### Thinkwell s Homeschool Algebra 2 Course Lesson Plan: 34 weeks Thinkwell s Homeschool Algebra 2 Course Lesson Plan: 34 weeks Welcome to Thinkwell s Homeschool Algebra 2! We re thrilled that you ve decided to make us part of your homeschool curriculum. This lesson ### Variable. 1.1 Order of Operations. August 17, evaluating expressions ink.notebook. Standards. letter or symbol used to represent a number 1.1 evaluating expressions ink.notebook page 8 Unit 1 Basic Equations and Inequalities 1.1 Order of Operations page 9 Square Cube Variable Variable Expression Exponent page 10 page 11 1 Lesson Objectives ### What are the place values to the left of the decimal point and their associated powers of ten? The verbal answers to all of the following questions should be memorized before completion of algebra. Answers that are not memorized will hinder your ability to succeed in geometry and algebra. (Everything ### Mathematics Pre-Test Sample Questions A. { 11, 7} B. { 7,0,7} C. { 7, 7} D. { 11, 11} Mathematics Pre-Test Sample Questions 1. Which of the following sets is closed under division? I. {½, 1,, 4} II. {-1, 1} III. {-1, 0, 1} A. I only B. II only C. III only D. I and II. Which of the following ### REVIEW EXERCISES DAVID J LOWRY REVIEW EXERCISES DAVID J LOWRY Contents 1. Introduction 1 2. Elementary Functions 1 2.1. Factoring and Solving Quadratics 1 2.2. Polynomial Inequalities 3 2.3. Rational Functions 4 2.4. Exponentials and ### The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION MATHEMATICS B. Thursday, January 29, 2004 9:15 a.m. to 12:15 p.m. The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION MATHEMATICS B Thursday, January 9, 004 9:15 a.m. to 1:15 p.m., only Print Your Name: Print Your School s Name: Print your name and ### ALGEBRA 2/TRIGONOMETRY ALGEBRA /TRIGONOMETRY The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION ALGEBRA /TRIGONOMETRY Tuesday, June 1, 011 1:15 to 4:15 p.m., only Student Name: School Name: Print your name ### Prentice Hall: Middle School Math, Course 1 2002 Correlated to: New York Mathematics Learning Standards (Intermediate) New York Mathematics Learning Standards (Intermediate) Mathematical Reasoning Key Idea: Students use MATHEMATICAL REASONING to analyze mathematical situations, make conjectures, gather evidence, and construct ### FINAL EXAM SECTIONS AND OBJECTIVES FOR COLLEGE ALGEBRA FINAL EXAM SECTIONS AND OBJECTIVES FOR COLLEGE ALGEBRA 1.1 Solve linear equations and equations that lead to linear equations. a) Solve the equation: 1 (x + 5) 4 = 1 (2x 1) 2 3 b) Solve the equation: 3x ### CHAPTER 8: ACUTE TRIANGLE TRIGONOMETRY CHAPTER 8: ACUTE TRIANGLE TRIGONOMETRY Specific Expectations Addressed in the Chapter Explore the development of the sine law within acute triangles (e.g., use dynamic geometry software to determine that ### Prentice Hall Mathematics: Algebra 2 2007 Correlated to: Utah Core Curriculum for Math, Intermediate Algebra (Secondary) Core Standards of the Course Standard 1 Students will acquire number sense and perform operations with real and complex numbers. Objective 1.1 Compute fluently and make reasonable estimates. 1. Simplify ### 2 Integrating Both Sides 2 Integrating Both Sides So far, the only general method we have for solving differential equations involves equations of the form y = f(x), where f(x) is any function of x. The solution to such an equation ### Chapter 2 Formulas and Decimals Chapter Formulas and Decimals Section A Rounding, Comparing, Adding and Subtracting Decimals Look at the following formulas. The first formula (P = A + B + C) is one we use to calculate perimeter of a ### Florida Math Correlation of the ALEKS course Florida Math 0022 to the Florida Mathematics Competencies - Lower and Upper Florida Math 0022 Correlation of the ALEKS course Florida Math 0022 to the Florida Mathematics Competencies - Lower and Upper Whole Numbers MDECL1: Perform operations on whole numbers (with applications, ### Unit 6 Trigonometric Identities, Equations, and Applications Accelerated Mathematics III Frameworks Student Edition Unit 6 Trigonometric Identities, Equations, and Applications nd Edition Unit 6: Page of 3 Table of Contents Introduction:... 3 Discovering the Pythagorean ### MATH 21. College Algebra 1 Lecture Notes MATH 21 College Algebra 1 Lecture Notes MATH 21 3.6 Factoring Review College Algebra 1 Factoring and Foiling 1. (a + b) 2 = a 2 + 2ab + b 2. 2. (a b) 2 = a 2 2ab + b 2. 3. (a + b)(a b) = a 2 b 2. 4. (a ### Math 0980 Chapter Objectives. Chapter 1: Introduction to Algebra: The Integers. Math 0980 Chapter Objectives Chapter 1: Introduction to Algebra: The Integers. 1. Identify the place value of a digit. 2. Write a number in words or digits. 3. Write positive and negative numbers used ### Chapter 4 -- Decimals Chapter 4 -- Decimals \$34.99 decimal notation ex. The cost of an object. ex. The balance of your bank account ex The amount owed ex. The tax on a purchase. Just like Whole Numbers Place Value - 1.23456789 ### South Carolina College- and Career-Ready (SCCCR) Pre-Calculus South Carolina College- and Career-Ready (SCCCR) Pre-Calculus Key Concepts Arithmetic with Polynomials and Rational Expressions PC.AAPR.2 PC.AAPR.3 PC.AAPR.4 PC.AAPR.5 PC.AAPR.6 PC.AAPR.7 Standards Know ### Review of Fundamental Mathematics Review of Fundamental Mathematics As explained in the Preface and in Chapter 1 of your textbook, managerial economics applies microeconomic theory to business decision making. The decision-making tools ### X On record with the USOE. Textbook Alignment to the Utah Core Algebra 2 Name of Company and Individual Conducting Alignment: Chris McHugh, McHugh Inc. A Credential Sheet has been completed on the above company/evaluator and is ### Algebra I Credit Recovery Algebra I Credit Recovery COURSE DESCRIPTION: The purpose of this course is to allow the student to gain mastery in working with and evaluating mathematical expressions, equations, graphs, and other topics, ### In mathematics, there are four attainment targets: using and applying mathematics; number and algebra; shape, space and measures, and handling data. MATHEMATICS: THE LEVEL DESCRIPTIONS In mathematics, there are four attainment targets: using and applying mathematics; number and algebra; shape, space and measures, and handling data. Attainment target ### Click on the links below to jump directly to the relevant section Click on the links below to jump directly to the relevant section What is algebra? Operations with algebraic terms Mathematical properties of real numbers Order of operations What is Algebra? Algebra is ### Pre-Session Review. Part 2: Mathematics of Finance Pre-Session Review Part 2: Mathematics of Finance For this section you will need a calculator with logarithmic and exponential function keys (such as log, ln, and x y ) D. Exponential and Logarithmic Functions ### 5.2 Unit Circle: Sine and Cosine Functions Chapter 5 Trigonometric Functions 75 5. Unit Circle: Sine and Cosine Functions In this section, you will: Learning Objectives 5..1 Find function values for the sine and cosine of 0 or π 6, 45 or π 4 and ### Roots and Coefficients of a Quadratic Equation Summary Roots and Coefficients of a Quadratic Equation Summary For a quadratic equation with roots α and β: Sum of roots = α + β = and Product of roots = αβ = Symmetrical functions of α and β include: x = and ### Trigonometric Functions and Triangles Trigonometric Functions and Triangles Dr. Philippe B. Laval Kennesaw STate University August 27, 2010 Abstract This handout defines the trigonometric function of angles and discusses the relationship between ### Solutions to Homework 10 Solutions to Homework 1 Section 7., exercise # 1 (b,d): (b) Compute the value of R f dv, where f(x, y) = y/x and R = [1, 3] [, 4]. Solution: Since f is continuous over R, f is integrable over R. Let x MTN Learn Mathematics Grade 10 radio support notes Contents INTRODUCTION... GETTING THE MOST FROM MINDSET LEARN XTRA RADIO REVISION... 3 BROADAST SCHEDULE... 4 ALGEBRAIC EXPRESSIONS... 5 EXPONENTS... 9 ### Mathematics Placement Examination (MPE) Practice Problems for Mathematics Placement Eamination (MPE) Revised August, 04 When you come to New Meico State University, you may be asked to take the Mathematics Placement Eamination (MPE) Your inital ### Some Notes on Taylor Polynomials and Taylor Series Some Notes on Taylor Polynomials and Taylor Series Mark MacLean October 3, 27 UBC s courses MATH /8 and MATH introduce students to the ideas of Taylor polynomials and Taylor series in a fairly limited ### Math Placement Test Study Guide. 2. The test consists entirely of multiple choice questions, each with five choices. Math Placement Test Study Guide General Characteristics of the Test 1. All items are to be completed by all students. The items are roughly ordered from elementary to advanced. The expectation is that ### Practice Math ACT Test 1. Notes on the Practice Exam Notes on the Practice Exam This practice exam has three major sections to it: test, answer key with hints, and scoring guide. You ll get the most out of this practice if you make the experience as authentic ### Solve each right triangle. Round side measures to the nearest tenth and angle measures to the nearest degree. Solve each right triangle. Round side measures to the nearest tenth and angle measures to the nearest degree. 42. The sum of the measures of the angles of a triangle is 180. Therefore, The sine of an angle ### Lies My Calculator and Computer Told Me Lies My Calculator and Computer Told Me 2 LIES MY CALCULATOR AND COMPUTER TOLD ME Lies My Calculator and Computer Told Me See Section.4 for a discussion of graphing calculators and computers with graphing ### Dimensional Analysis and Exponential Models MAT 42 College Mathematics Module XP Dimensional Analysis and Exponential Models Terri Miller revised December 3, 200. Dimensional Analysis The purpose of this section is to convert between various types ### 2. Right Triangle Trigonometry 2. Right Triangle Trigonometry 2.1 Definition II: Right Triangle Trigonometry 2.2 Calculators and Trigonometric Functions of an Acute Angle 2.3 Solving Right Triangles 2.4 Applications 2.5 Vectors: A Geometric ### Section 4-7 Exponential and Logarithmic Equations. Solving an Exponential Equation. log 2. 3 2 log 5. log 2 1.4406 314 4 INVERSE FUNCTIONS; EXPONENTIAL AND LOGARITHMIC FUNCTIONS Section 4-7 Exponential and Logarithmic Equations Exponential Equations Logarithmic Equations Change of Base Equations involving exponential ### Equations and Inequalities Rational Equations Overview of Objectives, students should be able to: 1. Solve rational equations with variables in the denominators.. Recognize identities, conditional equations, and inconsistent equations. Chapter Additional Topics in Math In addition to the questions in Heart of Algebra, Problem Solving and Data Analysis, and Passport to Advanced Math, the SAT Math Test includes several questions that are ### Parallel and Perpendicular. We show a small box in one of the angles to show that the lines are perpendicular. CONDENSED L E S S O N. Parallel and Perpendicular In this lesson you will learn the meaning of parallel and perpendicular discover how the slopes of parallel and perpendicular lines are related use slopes ### 3. Exponential and Logarithmic functions 3. ial and s ial and ic... 3.1. Here are a few examples to remind the reader of the definitions and laws for expressions involving exponents: 2 3 = 2 2 2 = 8, 2 0 = 1, 2 1 = 1 2, 2 3 = 1 2 3 = 1 8, 9 1/2 ### 2312 test 2 Fall 2010 Form B 2312 test 2 Fall 2010 Form B 1. Write the slope-intercept form of the equation of the line through the given point perpendicular to the given lin point: ( 7, 8) line: 9x 45y = 9 2. Evaluate the function ### 1. The algebra of exponents 1.1. Natural Number Powers. It is easy to say what is meant by a n a (raised to) to the (power) n if n N. CHAPTER 3: EXPONENTS AND POWER FUNCTIONS 1. The algebra of exponents 1.1. Natural Number Powers. It is easy to say what is meant by a n a (raised to) to the (power) n if n N. For example: In general, if ### MATHS LEVEL DESCRIPTORS MATHS LEVEL DESCRIPTORS Number Level 3 Understand the place value of numbers up to thousands. Order numbers up to 9999. Round numbers to the nearest 10 or 100. Understand the number line below zero, and ### Midterm 2 Review Problems (the first 7 pages) Math 123-5116 Intermediate Algebra Online Spring 2013 Midterm Review Problems (the first 7 pages) Math 1-5116 Intermediate Algebra Online Spring 01 Please note that these review problems are due on the day of the midterm, Friday, April 1, 01 at 6 p.m. in Exponents and Radicals (a + b) 10 Exponents are a very important part of algebra. An exponent is just a convenient way of writing repeated multiplications of the same number. Radicals involve the use of ### How do you compare numbers? On a number line, larger numbers are to the right and smaller numbers are to the left. The verbal answers to all of the following questions should be memorized before completion of pre-algebra. Answers that are not memorized will hinder your ability to succeed in algebra 1. Number Basics ### MATH 304 Linear Algebra Lecture 24: Scalar product. MATH 304 Linear Algebra Lecture 24: Scalar product. Vectors: geometric approach B A B A A vector is represented by a directed segment. Directed segment is drawn as an arrow. Different arrows represent ### 12.6 Logarithmic and Exponential Equations PREPARING FOR THIS SECTION Before getting started, review the following: Section 1.6 Logarithmic and Exponential Equations 811 1.6 Logarithmic and Exponential Equations PREPARING FOR THIS SECTION Before getting started, review the following: Solve Quadratic Equations (Section ### Sometimes it is easier to leave a number written as an exponent. For example, it is much easier to write 4.0 Exponent Property Review First let s start with a review of what exponents are. Recall that 3 means taking four 3 s and multiplying them together. So we know that 3 3 3 3 381. You might also recall ### Biggar High School Mathematics Department. National 5 Learning Intentions & Success Criteria: Assessing My Progress Biggar High School Mathematics Department National 5 Learning Intentions & Success Criteria: Assessing My Progress Expressions & Formulae Topic Learning Intention Success Criteria I understand this Approximation ### Study Guide 2 Solutions MATH 111 Study Guide 2 Solutions MATH 111 Having read through the sample test, I wanted to warn everyone, that I might consider asking questions involving inequalities, the absolute value function (as in the suggested ### Infinite Algebra 1 supports the teaching of the Common Core State Standards listed below. Infinite Algebra 1 Kuta Software LLC Common Core Alignment Software version 2.05 Last revised July 2015 Infinite Algebra 1 supports the teaching of the Common Core State Standards listed below. High School ### MyMathLab ecourse for Developmental Mathematics MyMathLab ecourse for Developmental Mathematics, North Shore Community College, University of New Orleans, Orange Coast College, Normandale Community College Table of Contents Module 1: Whole Numbers and ### 4.1 Radian and Degree Measure Date: 4.1 Radian and Degree Measure Syllabus Objective: 3.1 The student will solve problems using the unit circle. Trigonometry means the measure of triangles. Terminal side Initial side Standard Position
# Domain of cos x | Range of cos x | Period of cos x Note that the cosine function cos(x) is a trigonometric function. Here, we will learn about the domain, range, and period of cos(x). ## Domain of cos x One knows that for any value of x, the cosine function cos(x) is defined. Thus, by the definition of the domain of a function, we can say that the domain of cos(x) is the set of all real numbers. In other words, The domain of cos(x) = $(-\infty, \infty)$ For any real number $a$, the function cos(ax) is defined for all real numbers. So the domain of cos(ax) is the set of all real numbers. For example, • Domain of cos(2x) = $(-\infty, \infty)$ • Domain of cos(3x) = $(-\infty, \infty)$ • Domain of cos(4x) = $(-\infty, \infty)$ ## Range of cos x From the graph y=cos(x), one can see that the value of cos(x) oscillates between -1 and 1. That is, we have $-1 \leq \cos x \leq 1$ for all real numbers x. Thus, the range of cos(x) is the closed interval [-1, 1]. Range of cos(x) = [-1, 1] The range of cos(ax) is the closed interval [-1, 1]. For example, • Range of cos(2x) = [-1, 1] • Range of cos(3x) = [-1, 1] • Range of cos(4x) = [-1, 1] ## Period of cos x For any natural number n, it is known that $\cos(2n\pi+x)=\cos x$. From this, we see that $2\pi$ is the smallest number (corresponding to n=1) such that $\cos(2\pi+x)=\cos x$. Thus, $2\pi$ is the period of cos(x). Question 1: Find the period of cos(ax). Solution: The period of cos(ax) is $\dfrac{2\pi}{|a|}$. For example, the period of cos(3x) is equal to $\dfrac{2\pi}{3}$. Question 2: Find the period of cos(8x)-cos(2x). Solution: The period of cos(8x) is $\dfrac{2\pi}{8}$ $=\frac{\pi}{4}$ and the period of cos(2x) is $\dfrac{2\pi}{2}=\pi$. As the period of the difference of two periodic functions is the least common multiple (LCM) of their periods, we conclude that the period of cos(8x)-cos(2x) is = LCM $\{\dfrac{\pi}{4}, \pi \}$ = $\pi$. Therefore, the period of cos(8x)-cos(2x) is $\pi$.
# Distance Between any Two Parallel Lines ## Introduction We can easily calculate the distance between any two parallel lines by the distance of a point from a line. Besides, it is equivalent to the length of the vertical distance from any point on one of the lines to another line. In this article, we discuss about the distance between the two lines’ derivation and formulas. Moreover, there are many benefits of this process to calculating the distance between two lines. ## Key Takeaways: Distance between two lines, Formulas, derivations ## Distance Between Point and Line Derivation We can express a line with a equation by Ax + By + C = 0. Consider a point P in the Cartesian plane having the coordinates (x1,y1). Besides, there are many things to calculate the distance from the point to the line. Moreover, in the Cartesian system, we can calculate the length of the perpendicular between the point and line. The distance between the point P and the line L can be calculated by figuring out the length of the perpendicular as the figure given below. The coordinate points for different points are as follows: Point P (x1, y1), Point N (x2, y2), Point R (x3,y3) The line L makes intercepts on both the x-axis and y – axis at the points N and M respectively. The coordinates of these points are M(0,−C/B)M(0,−C/B) and N (−C/A,0)N (−C/A,0). The area of Δ MPN can be given as: Area of Δ MPN = 1/2 × Base × Height ⇒Area of Δ MPN⇒Area of Δ MPN = 1/2 × PQ × MN ⇒PQ⇒PQ = 2 × Area of Δ MPN/MN ………………………(i) In terms of Co-ordinate Geometry, the area of the triangle is given as: Area of Δ MPN = 1/2[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]1/2[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)] ## Distance Between Two Parallel Lines Furthermore, the distance between two parallel lines is equal to the perpendicular distance between the two lines. Thus, we know that the slopes of two parallel lines are the same. Therefore, the equation of two parallel lines can be given as: y = mx + c1 and y = mx + c2 Besides, Point A is the intersection point of the second line on the x-axis. The perpendicular distance would be the required distance between two lines The distance between the point A and the line y = mx + c2 can be given by using the formula: ## Shortest Distance Between Two Parallel Lines Basically, the shortest distance between the two parallel lines can be determined using the length of the perpendicular segment between the lines. Besides, it does not matter which perpendicular line you are choosing, as long as two points are on the line. Thus, we can now easily calculate the distance between two parallel lines and the distance between a point and a line. ## Things to Remember: • We can determine the distance between two lines through perpendicular between the point and line • The perpendicular line can be used for the identification of the length of the distance between two lines Question: How can we calculate the distance between two parallel lines? Ans: We can calculate the distance between two lines using a perpendicular between the point and line. It is the best method in the Cartesian system. Question: What is the Role of a perpendicular line in calculating the distance between two parallel lines? Answer: The distance between point A and the line y = mx + c2 can be given by using the formula: Best Web Hosting in the USA ### How to Apply For this Job Post 1. Open the official website mentioned in the important link section 2. Register yourself on the official website of the department 3. Fill in all the details including personal photo and signature 4. Pay as per the job post you have to apply for. 5. Take a print of the confirmation details [table id=5 /] [table id=6 /] [table id=30 /]
1. ## Affine transformation Having a little trouble with this question: f is an affine transformation. The transformation f is a reflection in the line y=-x-2. By first translating an appropriate point to the origin find f in the form f(x)=Ax+a 2. Originally Posted by offahengaway and chips Having a little trouble with this question: f is an affine transformation. The transformation f is a reflection in the line y=-x-2. By first translating an appropriate point to the origin find f in the form f(x)=Ax+a The line y= -x-2 is a line with slope -1 passing through the point (0, -2). The translation (x,y)-> (x,y+2) maps that point into (0, 0) and maps everypoint on the line into a point on y= -x. A reflection in that line is (x,y)->(-y, -x), which would be represented by the matrix equation $\displaystyle A(x,y)= \begin{bmatrix}0 & -1 \\ -1 & 0\end{bmatrix}\begin{bmatrix}x \\y\end{bmatrix}= \begin{bmatrix}-y \\ -x\end{bmatrix}$. Finally, the translation (x,y)->(x, y-2) maps back to the original position. So we are doing A((x,y)+ (0,2))+ (0,-2)= A(x,y)+ A(0,2)+ (0,-2). $\displaystyle A(0,2)= \begin{bmatrix}0 & -1 \\ -1 & 0\end{bmatrix}\begin{bmatrix}0 \\ 2 \end{bmatrix}= \begin{bmatrix}-2 \\ 0\end{bmatrix}$ So A(x,y)+ a is $\displaystyle \begin{bmatrix}0 & -1 \\ -1 & 0\end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix}+ \begin{bmatrix}-2 \\ -2\end{bmatrix}$.
# If a rectangle measures 8 feet on the short side and 9.8 feet on the long side.What is the area of the rectangle in square feet? Apr 20, 2018 $A = 78.4$ ${\text{ft}}^{2}$ #### Explanation: Multiply the length and width of the rectangle to find its area: $A = 9.8 \cdot 8$ $A = 78.4$ ${\text{ft}}^{2}$ Apr 20, 2018 If the decimal is giving you a problem consider this approach. $78.4 \textcolor{w h i t e}{.} f {t}^{2}$ #### Explanation: We have: $8 \text{ feet"xx9.8" feet}$ But 9.8 is the same as $98 \times \frac{1}{10}$ So re-write as: $8 \times 98 \times \frac{1}{10} {\text{ ft}}^{2}$ $\textcolor{w h i t e}{\text{d}} \textcolor{w h i t e}{.} 9 \textcolor{w h i t e}{. .} 8$ ul(color(white)("d")color(white)(9999)8larr" Multiply") $\underline{7 \textcolor{w h i t e}{.} 8 \textcolor{w h i t e}{. .} 4}$ $\textcolor{w h i t e}{\text{dd}} 6$ Now we multiply by $\frac{1}{10}$ to give: $784 \textcolor{w h i t e}{.} f {t}^{2} \times \frac{1}{10} = 78.4 \textcolor{w h i t e}{.} f {t}^{2}$
# 1232. Algorithm - Sqrt DecompositionSqrt Decomposition Implement Sqrt Decomposition for range search problems. ## 1. Range Query ### 1.1 Problem Description Given an integer array with n elements, find the sum of some elements in range from i to j. This array is mutable, meaning, any element in the array can be modified. So, we have two following operations query and update. The query method should always work even if the array is updated. • query(i, j) - Find the sum of the elements between indices i and j (i ≤ j), inclusive. • update(i, val) - Modify the array by updating the element at index i to val. ### 1.2 Solutions There are two solutions for such range query problem. The first solution for this problem is segment tree. The second solution is Sqrt Decomposition. ## 2. Sqrt Decomposition Square root(Sqrt) decomposition allows us to answer queries in $O(\sqrt{k})$ time and the implementation is usually simpler than a segment tree. ### 2.1 Building Blocks Decompose the array into small blocks. Suppose the array has 9 elements, it can be split into $\sqrt{9}$ = 3 blocks, and we can easily calculate the sum of each block. • If the length of the array n is not a perfect square, then we will have one more block. ### 2.2 Query There are two cases for the query. If range is on block boundaries, we can get the sum from blocks directly. For example, if the range is from 3 to 8, then block[1] and block[2] perfectly cover this range. If range crosses blocks partially, we cannot get the sum from blocks directly. For example, if the range is from 2 to 7, we can get the sum for array[3,4,5] from block[1], however, we have to get other sums from the array. ### 2.3 Update If we update the array, we need to update the block accordingly. For example, update(2,7) will modify the array[2] from 1 to 7, and update block[0] from 9 to 15. ## 3. Implementation Implement Sqrt Decomposition with three methods. • build • query • update public class RangeSearchSum { private int[] nums; private long[] sumBlocks; private int sqrt; public RangeSearchSum(int[] arr) { if (arr != null && arr.length > 0) { build(arr); } } private void build(int[] arr) { this.sqrt = (int) Math.ceil(Math.sqrt(arr.length)); this.nums = new int[sqrt * sqrt]; System.arraycopy(arr, 0, nums, 0, arr.length); // the tail items in nums may be zero this.sumBlocks = new long[sqrt]; for (int i = 0; i < sumBlocks.length; i++) { int startIndex = i * sqrt; for (int j = 0; j < sqrt; j++) { sumBlocks[i] += nums[startIndex + j]; } } } // update value by index public void update(int index, int value) { int blockIndex = index / sqrt; sumBlocks[blockIndex] = sumBlocks[blockIndex] - nums[index] + value; nums[index] = value; } // query with left and right indexes public long query(int left, int right) { int startBlockIndex = left / sqrt; int endBlockIndex = right / sqrt; long sum = 0; if (startBlockIndex == endBlockIndex) { // in the same block for (int i = left; i <= right; i++) { sum += nums[i]; } } else { // in the different blocks // overlap for (int i = startBlockIndex + 1; i < endBlockIndex; i++) { sum += sumBlocks[i]; } // left non-overlap int startIndex = left % sqrt; for (int i = startIndex; i < sqrt; i++) { sum += nums[startBlockIndex * sqrt + i]; } // right non-overlap int endIndex = right % sqrt; for (int i = 0; i <= endIndex; i++) { sum += nums[endBlockIndex * sqrt + i]; } } return sum; } } Usage. RangeSearchSum sd = new RangeSearchSum(new int[] {3,5,1,5,6,10,4,2,8}); sd.query(0, 1); // 8 sd.query(0, 2); // 9 sd.query(3, 5); // 21 sd.query(6, 8); // 14 sd.query(3, 7); // 27 sd.query(2, 7); // 28 sd.query(0, 8); // 44 sd.update(2, 7); // {3,5,7,5,6,10,4,2,8} sd.query(0, 2); // 15 sd.query(2, 7); // 34 sd.query(6, 7); // 6 sd.query(7, 8); // 10
Education.com Try Brainzy Try Plus # Combination of Like Terms Help (not rated) By McGraw-Hill Professional Updated on Sep 26, 2011 ## Combining Like Terms Two or more terms are alike if they have the same variables and the exponents (or roots) on those variables are the same: 3 x 2 y and 5 x 2 y are like terms but 6 xy and 4 xy 2 are not. Constants are terms with no variables. The number in front of the variable(s) is the coefficient —in 4 x 2 y 3 , 4 is the coefficient. If no number appears in front of the variable, then the coefficient is 1. Add or subtract like terms by adding or subtracting their coefficients. #### Examples 3 x 2 + 4 xy – 8 xy 2 – (2 x 2 – 3 xy – 4 xy 2 + 6) = 3 x 2 + 4 xy – 8 xy 2 – 2 x 2 + 3 xy + 4 xy 2 – 6 = 3 x 2 – 2 x 2 – 8 xy 2 + 4 xy 2 + 4 xy + 3 xy – 6 = (3 – 2) x 2 + (–8 + 4) xy 2 + (4 + 3) xy – 6 = x 2 – 4 xy 2 + 7 xy – 6 ## Combining Like Terms Practice Problems #### Practice 1. 3 xy + 7 xy = 2. 4 x 2 – 6 x 2 = 5. 2 xy 2 – 4 x 2 y – 7 xy 2 + 17 x 2 y = 6. 14 x + 8 – (2 x – 4) = 7. 16 x –4 + 3 x –2 – 4 x + 9 x –4x –2 + 5 x – 6 = 9. x 2 y + xy 2 + 6 x + 4 – (4 x 2 y + 3 xy 2 – 2 x + 5) = #### Solutions 1. 3 xy + 7 xy = (3 + 7) xy = 10 xy 2. 4 x 2 – 6 x 2 = (4 – 6) x 2 = –2 x 2 5. 2 xy 2 – 4 x 2 y – 7 xy 2 + 17 x 2 y = 2 xy 2 – 7 xy 2 – 4 x 2 y + 17 x 2 y = (2 – 7) xy 2 +(–4 + 17) x 2 y = –5 xy 2 + 13 x 2 y 6. 14 x + 8 – (2 x – 4) = 14 x + 8 – 2 x + 4 = 14 x – 2 x + 8 + 4 = (14 – 2) x + 12 = 12 x + 12 7. 16 x –4 + 3 x –2 – 4 x + 9 x –4x –2 + 5 x – 6 = 16 x –4 + 9 x –4 + 3 x –2x –2 – 4 x + 5 x – 6 = (16 + 9) x –4 +(3 – 1) x –2 +(–4 + 5) x – 6 = 25 x –4 + 2 x –2 + x – 6 9. x 2 y + xy 2 + 6 x + 4 – (4 x 2 y + 3 xy 2 – 2 x + 5) = x 2 y + xy 2 + 6 x + 4 – 4 x 2 y – 3 xy 2 + 2 x – 5 = x 2 y – 4 x 2 y + xy 2 – 3 xy 2 + 6 x + 2 x + 4 – 5 = (1 – 4) x 2 y +(1 – 3) xy 2 + (6 + 2) x – 1 = –3 x 2 y – 2 xy 2 + 8 x – 1 Practice problems for this concept can be found at: Algebra Factoring Practice Test. 150 Characters allowed ### Related Questions #### Q: See More Questions ### Today on Education.com Top Worksheet Slideshows
 GreeneMath.com - Proportions Test #4 # In this Section: In this section, we learn how to determine if two ratios or two rates represent a proportion. We start with determining if two fractions are equal in value. To do this, we utilize a procedure known as the “equality test for fractions”. This tells us that if two fractions are equal in value, their cross products must be equal. For example, one - fourth is equal to two - eighths. The cross products would be the same in each case: eight. Once we have mastered this concept, we apply the same logic to working with ratios and rates. To determine if two ratios or two rates represent a proportion, we simply check to see if their cross products are equal. We only think about the number parts, we do not look at the units. In the case of a rate, we must ensure that the same units are in the numerator and denominator of each. Sections: # In this Section: In this section, we learn how to determine if two ratios or two rates represent a proportion. We start with determining if two fractions are equal in value. To do this, we utilize a procedure known as the “equality test for fractions”. This tells us that if two fractions are equal in value, their cross products must be equal. For example, one - fourth is equal to two - eighths. The cross products would be the same in each case: eight. Once we have mastered this concept, we apply the same logic to working with ratios and rates. To determine if two ratios or two rates represent a proportion, we simply check to see if their cross products are equal. We only think about the number parts, we do not look at the units. In the case of a rate, we must ensure that the same units are in the numerator and denominator of each.
Edit Article # How to Do Math Proofs Mathematical proofs can be difficult, but can be conquered with the proper background knowledge of both mathematics and the format of a proof. Unfortunately, there is no quick and easy way to learn how to construct a proof. You must have a basic foundation in the subject to come up with the proper theorems and definitions to logically devise your proof. By reading example proofs and practicing on your own, you will be able to cultivate the skill of writing a mathematical proof. ### Method 1 Understanding the Problem 1. 1 Identify the question. You must first determine exactly what it is you are trying to prove. This question will also serve as the final statement in the proof. In this step, you also want to define the assumptions that you will be working under. Identifying the question and the necessary assumptions gives you a starting point to understanding the problem and working the proof. 2. 2 Draw diagrams. When trying to understand the inner working of a math problem, sometimes the easiest way is to draw a diagram of what is happening. Diagrams are particularly important in geometry proofs, as they help you visualize what you are actually trying to prove. • Use the information given in the problem to sketch a drawing of the proof. Label the knowns and unknowns. • As you work through the proof, draw in necessary information that provides evidence for the proof. 3. 3 Study proofs of related theorems. Proofs are difficult to learn to write, but one excellent way to learn proofs is to study related theorems and how those were proved. • Realize that a proof is just a good argument with every step justified. You can find many proofs to study online or in a textbook.[1] 4. 4 Ask questions. It’s perfectly okay to get stuck on a proof. Ask your teacher or fellow classmates if you have questions. They might have similar questions and you can work through the problems together. It’s better to ask and get clarification than to stumble blindly through the proof. • Meet with your teacher out of class for extra instruction. ### Method 2 Formatting a Proof 1. 1 Define mathematical proofs. A mathematical proof is a series of logical statements supported by theorems and definitions that prove the truth of another mathematical statement.[2] Proofs are the only way to know that a statement is mathematically valid. • Being able to write a mathematical proof indicates a fundamental understanding of the problem itself and all of the concepts used in the problem. • Proofs also force you to look at mathematics in a new and exciting way. Just by trying to prove something you gain knowledge and understanding even if your proof ultimately doesn’t work. 2. 2 Know your audience. Before writing a proof, you need to think about the audience that you’re writing for and what information they already know. If you are writing a proof for publication, you will write it differently than writing a proof for your high school math class.[3] • Knowing your audience allows you to write the proof in a way that they will understand given the amount of background knowledge that they have. 3. 3 Identify the type of proof you are writing. There are a few different types of proofs and the one you choose depends on your audience and the assignment. If you’re unsure which version to use, ask your teacher for guidance. In high school, you may not be expected to write a formal proof.[4] • A two-column proof is a setup that puts givens and statements in one column and the supporting evidence next to it in a second column. They are very commonly used in geometry. • A formal proof uses grammatically correct statements and no symbols. At higher levels, you should always use a formal proof. 4. 4 Write the two-column proof as an outline. You can use a two-column proof to as an outline of your formal proof before you write it. The two-column proof is an easy way to organize your thoughts and think through the problem. Draw a line down the middle of the page and write all givens and statements on the left side. Write the corresponding definitions/theorems on the right side, next to the givens they support. • For example:[5] • Angle A and angle B form a linear pair. Given. • Angle ABC is straight. Definition of a straight angle. • Angle ABC measures 180°. Definition of a line. • Angle A + Angle B = Angle ABC. Angle addition postulate. • Angle A + Angle B = 180°. Substitution. • Angle A supplementary to Angle B. Definition of supplementary angles. • Q.E.D. 5. 5 Convert the two-column proof to a formal written proof. Using the two-column proof as a foundation, write the formal paragraph form of your proof without too many symbols and abbreviations. • For example: Let angle A and angle B be linear pairs. By hypothesis, angle A and angle B are supplementary. Angle A and angle B form a straight line because they are linear pairs. A straight line is defined as having an angle measure of 180°. Given the angle addition postulate, angles A and B sum together to form line ABC. Through substitution, angles A and B sum together to 180°, therefore they are supplementary angles. Q.E.D. ### Method 3 Writing the Proof 1. 1 Learn the vocabulary of a proof. There are certain statements and phrases that you will see over and over in a mathematical proof. These are phrases that you need to be familiar with and know how to use properly when writing your own proof.[6] • “If A, then B” statements mean that you must prove whenever A is true, B must also be true.[7] • “A if and only if B” means that you must prove that A and B are true and false at the same time. Prove both “if A, then B” and “if not A, then not B”. • “A only if B” is equivalent to “if A, then B”, so it is not frequently used. It is good to be aware of it though, in case you see it. • When composing the proof, avoid using “I”, but use “we” instead.[8] 2. 2 Write down all givens. When composing a proof, the first step is to identify and write down all of the givens. This is the best place to start because it helps you think through what is known and what information you will need to complete the proof. Read through the problem and write down each given. • For example: Prove that two angles (angle A and angle B) forming a linear pair are supplementary.[9] • Givens: angle A and angle B are a linear pair • Prove: angle A is supplementary to angle B 3. 3 Define all variables. In addition to writing the givens, it is helpful to define all of the variables. Write the definitions at the beginning of the proof to avoid confusion for the reader. If variables are not defined, a reader can easily get lost when trying to understand your proof. • Don’t use any variables in your proof that haven’t been defined. • For example: Variables are the angle measure of angle A and measure of angle B. 4. 4 Work through the proof backwards. It’s often easiest to think through the problem backwards. Start with the conclusion, what you’re trying to prove, and think about the steps that can get you to the beginning.[10] • Manipulate the steps from the beginning and the end to see if you can make them look like each other. Use the givens, definitions you have learned, and proofs that are similar to the one you’re working on. • Ask yourself questions as you move along. "Why is this so?" and "Is there any way this can be false?" are good questions for every statement or claim. • Remember to rewrite the steps in the proper order for the final proof. • For example: If angle A and B are supplementary, they must sum to 180°. The two angles combine together to form line ABC. You know they make a line because of the definition of a linear pairs. Because a line is 180°, you can use substitution to prove that angle A and angle B add up to 180°. 5. 5 Order your steps logically. Start the proof at the beginning and work towards the conclusion. Although it is helpful to think about the proof by starting with the conclusion and working backwards, when you actually write the proof, state the conclusion at the end.[11] It needs to flow from one statement to the other, with support for each statement, so that there is no reason to doubt the validity of your proof. • Start by stating the assumptions you are working with. • Include simple and obvious steps so a reader doesn’t have to wonder how you got from one step to another. • Writing multiple drafts for your proofs is not uncommon. Keep rearranging until all of the steps are in the most logical order. • Angle A and angle B form a linear pair. • Angle ABC is straight. • Angle ABC measures 180°. • Angle A + Angle B = Angle ABC. • Angle A + Angle B = Angle 180°. • Angle A is supplementary to Angle B. 6. 6 Avoid using arrows and abbreviations in the written proof. When you are sketching out the plan for your proof, you can use shorthand and symbols, but when writing the final proof, symbols such as arrows can confuse the reader. Instead, use words like “then” or “therefore”.[12] • Exceptions to using abbreviations include, e.g. (for example) and i.e. (that is), but be sure that you are using them properly.[13] 7. 7 Support all statements with a theorem, law, or definition. A proof is only as good as the evidence used. You cannot make a statement without supporting it with a definition. Reference other proofs that are similar to the one you are working on for example evidence. • Try to apply your proof to a case where it should fail, and see whether it actually does. If it doesn’t fail, rework the proof so that it does. • Many geometric proofs are written as a two-column proof, with the statement and the evidence. A formal mathematical proof for publication is written as a paragraph with proper grammar. 8. 8 End with a conclusion or Q.E.D. The last statement of the proof should be the concept you were trying to prove. Once you have made this statement, ending the proof with a final concluding symbol such as Q.E.D. or a filled-in square indicates that the proof is completely finished.[14] • Q.E.D. (quod erat demonstrandum, which is Latin for "which was to be shown"). • If you're not sure if your proof is correct, just write a few sentences saying what your conclusion was and why it is significant.
Math # The 8 Laws Of Indices In Maths Explained Index (indices) in Maths is the exponent which is raised to a number. For example, in number 42, 2 is the index or power of 4. The plural form of index is indices. Also, a number of the form xn where x is a real number, x is multiplied by itself n times i.e. xn = x*x*x*x*x*——(n times). The number x is called the base and the super script n is called the index or power or exponent. In this article, you will learn the laws of the indices along with formulas and solved examples. ## 8 Laws of Indices 1st law Any base variable raise to zero (0) is one (1) i.e. A0 = 1. For example, 20 = 1 2nd Law If a base variable is raised to a negative number, then it will be equal to inverse of the base variable raised to a positive number i.e. A-m = 1/Am 3rd Law If a base number (which is a fraction) is raised to a number (m), then it will be equal to the numerator raised to the number (m) and the denominator raised to the same number (m) i.e. (A/B)m = Am/Bm Example, (2/3)2 = 22/32 = 4/9 4th Law If a base number is raised to a number (m) and multiply a base number of the same value raised to a number (n), then it will be equal to the base number raised to the sum of the exponents (m + n) i.e. Am X An = Am+n Example, y5 x y4 = y5+4 = y9 5th Law If a base number is raised to a number (m) and is divided by a base number of the same value raised to a number (n), then it will be equal to the base number raised to the difference between the exponents (m – n) i.e. Am -:- An = Am-n Example, 54 -:- 52 = 54-2 = 52 = 5 x 5 = 25 6th Law If a base variable is raised to a number (which is a fraction (x/y)), then it will be equal to the yth root of the base number raised to x i.e. Ax/y = y√Ax Example, 272/3 = 3√272 = 32 = 9 7th Law If a base varaible is raised to a number (n) and the entire number raised to power (m), then it is equal to the base number raised to the multiplication of the two exponents (m*n) i.e. (Am)n = Amn Example, (33)2 = 33*2 = 36 = 3*3*3*3*3*3 = 729 8th Law When two base variables with different bases, but same indices are multiplied together, we have to multiply the two bases and raise the same index to multiplied variables i.e. An x Bn = (A.B)n Example, 32 x 22 = (3*2)2 = 62 = 36 If you have any question concerning laws of indices, you can drop a comment in the box below. ### Bolarinwa Olajire A tutor with a demonstrated history of working in the education industry. Skilled in analytical skills. Strong education professional with a M. SC focused in condensed matter. You can follow me on Twitter by clicking on the icon below to ask questions.
# 2020 AMC 12B Problems/Problem 5 ## Problem Teams $A$ and $B$ are playing in a basketball league where each game results in a win for one team and a loss for the other team. Team $A$ has won $\tfrac{2}{3}$ of its games and team $B$ has won $\tfrac{5}{8}$ of its games. Also, team $B$ has won $7$ more games and lost $7$ more games than team $A.$ How many games has team $A$ played? $\textbf{(A) } 21 \qquad \textbf{(B) } 27 \qquad \textbf{(C) } 42 \qquad \textbf{(D) } 48 \qquad \textbf{(E) } 63$ ## Solution 1 (One Variable) Suppose team $A$ has played $g$ games in total so that it has won $\frac23g$ games. It follows that team $B$ has played $g+14$ games in total so that it has won $\frac23g+7$ games. We set up and solve an equation for team $B$'s win ratio: \begin{align*} \frac{\frac23g+7}{g+14}&=\frac58 \\ \frac{16}{3}g+56&=5g+70 \\ \frac13g&=14 \\ g&=\boxed{\textbf{(C) } 42}. \end{align*} ~MRENTHUSIASM ## Solution 2 (Two Variables) If we consider the number of games team $B$ has played as $x$ and the number of games that team $A$ has played as $y$, then we can set up the following system of equations: \begin{align*} \frac{5}{8}x &= \frac{2}{3}y+7, \\ \frac{3}{8}x &= \frac{1}{3}y+7. \end{align*} The first system equated the number of wins of each team, while the second system equates the number of losses by each team. By multiplying the second equation by $2$ and solving the system, we get $y = 42$ or answer choice $\boxed{\textbf{(C) } 42}.$ ~AnkitAMC ## Solution 3 (Two Variables) First, let us assign some variables. Let $$A_w=2x, \ A_l=x, \ A_g=3x,$$ $$B_w=5y, \ B_l=3y, \ B_g=8y,$$ where $X_w$ denotes number of games won, $X_l$ denotes number of games lost, and $X_g$ denotes total games played for $X\in \{A, B\}$. Using the given information, we can set up the following two equations: \begin{align*} B_w=A_w+7&\implies 5y=2x+7, \\ B_l=A_l+7&\implies 3y=x+7. \end{align*} We can solve through substitution, as the second equation can be written as $x=3y-7$, and plugging this into the first equation gives $5y=6y-7\implies y=7$, which means $x=3(7)-7=14$. Finally, we want the total number of games team $A$ has played, which is $A_g=3(14)=\boxed{\textbf{(C) } 42}$. ~Argonauts16 ## Solution 4 (Answer Choices: Substitutions) Using the information from the problem, we can note that team $A$ has lost $\frac{1}{3}$ of their matches. Using the answer choices, we can construct the following list of possible win-lose scenarios for $A,$ represented in the form $(w, l)$ for convenience: \begin{align*} \textbf{(A)} &\implies (14, 7) \\ \textbf{(B)} &\implies (18, 9) \\ \textbf{(C)} &\implies (28, 14) \\ \textbf{(D)} &\implies (32, 16) \\ \textbf{(E)} &\implies (42, 21) \end{align*} Thus, we have $5$ matching $B$ scenarios, simply adding $7$ to $w$ and $l.$ We can then test each of the five $B$ scenarios for $\frac{w}{w+l} = \frac{5}{8}$ and find that $(35, 21)$ fits this description. Then working backwards and subtracting $7$ from $w$ and $l$ gives us the point $(28, 14),$ making the answer $\boxed{\textbf{(C) } 42}.$ ## Solution 5 (Answer Choices: Observations) Let's say that team $A$ plays $n$ games in total. Therefore, team $B$ must play $n + 14$ games in total (7 wins, 7 losses) Since the ratio of $A$ is $$\frac{2}{3} \implies n \equiv 0 \pmod{3}$$ Similarly, since the ratio of $B$ is $$\frac{5}{8} \implies n + 14 \equiv 0 \pmod{8}$$ Now, we can go through the answer choices and see which ones work: \begin{align*} \textbf{(A) } 21 &\implies 21 + 14 = 35 \not \equiv 0\pmod{8} \\ \textbf{(B) } 27 &\implies 27 + 14 = 41 \not \equiv 0\pmod{8} \\ \textbf{(C) } 42 &\implies 42 + 14 = 56 \equiv 0\pmod{8} \\ \textbf{(D) } 48 &\implies 48 + 14 = 62 \not \equiv 0\pmod{8} \\ \textbf{(E) } 63 &\implies 63 + 14 = 77 \not \equiv 0\pmod{8} \\ \end{align*} So we can see $\boxed{\textbf{(C) } 42}$ is the only valid answer. ~herobrine-india ## Video Solution (HOW TO THINK CREATIVELY!!!) ~Education, the Study of Everything
# How many feet are in 1/2 of a mile? Mar 6, 2018 See a solution process below: #### Explanation: We know the conversion factor of miles to feet is: $1 \text{ mile" = 5280" feet}$ To find how many feet in a $\frac{1}{2}$ mile we can multiply each side of the equation by $\textcolor{red}{\frac{1}{2}}$ giving: $\textcolor{red}{\frac{1}{2}} \times 1 \text{ mile" = color(red)(1/2) xx 5280" feet}$ $\frac{1}{2} \text{ mile" = 5280/2" feet}$ $\frac{1}{2} \text{ mile" = 2640" feet}$ There are 2,640 feet in $\frac{1}{2}$ mile Mar 6, 2018 2640’ make $\left(\frac{1}{2}\right)$ a mile. #### Explanation: 12” make 1’ 3’ make 1 yard $220 y a r \mathrm{ds}$ make $1 f u r l o n g$ $8 f u r l o n g s$ make 1 mile Or 5280’ make $1$ mile Hence 2640’# make $\left(\frac{1}{2}\right)$ a mile.
Courses Courses for Kids Free study material Offline Centres More Store An open box of length 1.5 m, breadth 1 m and height 1 m is to be made for use on a trolley for carrying garden waste. How much sheet metal will be required to make this box? The inside and outside surface of the box is to be painted with rust-proof paint. At a rate of 150 rupees per sqm, how much will it cost to paint the box? Last updated date: 13th Jul 2024 Total views: 449.4k Views today: 5.49k Verified 449.4k+ views Hint: First of all, get the metal sheet required by finding the total surface area of the open box that is S = 2 (bh + lh) + lb and then multiply twice of this area with cost to paint per ${{\text{m}}^{2}}$ to get the total cost of the paint required. We are given an open box whose length, breadth and height are 1.5 m, 1 m and 1 m respectively. We have to find the metal sheet required to make this box. Also, the inside and outside surface area of the box is painted at a rate of 150 rupees/${{\text{m}}^{2}}$. We have to find the cost to paint the box. First all, we have to find the sheet required to make this box. To get the total sheet required, we have to find the total surface area of the given box. Let us consider the surface area of the box to be S. We know that this box is in the shape of a cuboid. So, the total surface area of cuboid = 2 (lb + bh + hl) where l, b and h are the length, breadth and height of the cuboid. But as we are given that this box is an open box that is it is open at the top. So, we must subtract the area of the upper face from the total surface area of the cuboid to get the required surface area of the given box. So, we get Surface area of box (S) = 2 (lb + bh + hl) – (Area of the upper face) We know that area of the upper face of box = l x b So, we get, S = 2(lb + bh + hl) – lb Therefore, we get surface area of the box (S) = 2 (bh + hl) + lb …..(1) Now, we know that the metal sheet required = Surface area of the box So now, we substitute the length of the box = 1.5 m, breadth of box = 1 m and height of the box = 1 m in equation (1) to get the metal sheet required. So, we get, S = Metal sheet required $=2\left[ \left( 1\times 1+1\times 1.5 \right)+\left( 1\times 1.5 \right) \right]\text{ }{{\text{m}}^{2}}$ $S=\left[ 2\left( 1+1.5 \right)+1.5 \right]\text{ }{{\text{m}}^{2}}$ $S=6.5\text{ }{{\text{m}}^{2}}$ So, we get the metal sheet required to make the given box equal to 6.5 sq.m. Now, we are given that the inside and outside surface area of the box is painted. So, we get, The total area of the box to be painted = Inside the surface area of the box + outside surface area of the box As we know that the inside surface area = outside surface area = S for this box, so we get, Total area of the box to be painted = S + S = 2S. By substituting the value of $S=6.5{{\text{m}}^{2}}$, we get, The total area of the box to be painted = $2\times 6.5{{\text{m}}^{2}}=13{{\text{m}}^{2}}$. We are given that cost to paint the box = $\text{Rs}\text{.150/}{{\text{m}}^{2}}$ So, we get the cost to paint the outside and inside surface area of the box $=\text{Rs}\left( 150\times 13 \right)=\text{Rs}.1950$ Therefore, we get the total cost to paint the box equal to Rs.1950. Note: Here, many students forget to subtract the area of the upper face of the cuboid box. So, this must be kept in mind whenever the box is open. Also, students must note that we have to paint both inside and outside of the box. So, we have to add the surface area twice to get the area to be painted. So, students should keep this point in mind.
# AP Statistics Curriculum 2007 Hypothesis Proportion (Difference between revisions) Revision as of 18:46, 19 November 2008 (view source)IvoDinov (Talk | contribs)m ← Older edit Current revision as of 00:16, 24 April 2013 (view source)IvoDinov (Talk | contribs) (7 intermediate revisions not shown) Line 3: Line 3: === Background=== === Background=== [[AP_Statistics_Curriculum_2007_Estim_Proportion |Recall that for large samples]], the sampling distribution of the sample proportion $\hat{p}$ is approximately Normal, by [[AP_Statistics_Curriculum_2007_Limits_CLT |CLT]], as the sample proportion may be presented as a [[AP_Statistics_Curriculum_2007_Limits_Norm2Bin |sample average or Bernoulli random variables]]. When the sample size is small, the normal approximation may be inadequate. To accommodate this, we will modify the '''sample-proportion''' $\hat{p}$ slightly and obtain the '''corrected-sample-proportion''' $\tilde{p}$: [[AP_Statistics_Curriculum_2007_Estim_Proportion |Recall that for large samples]], the sampling distribution of the sample proportion $\hat{p}$ is approximately Normal, by [[AP_Statistics_Curriculum_2007_Limits_CLT |CLT]], as the sample proportion may be presented as a [[AP_Statistics_Curriculum_2007_Limits_Norm2Bin |sample average or Bernoulli random variables]]. When the sample size is small, the normal approximation may be inadequate. To accommodate this, we will modify the '''sample-proportion''' $\hat{p}$ slightly and obtain the '''corrected-sample-proportion''' $\tilde{p}$: - : $\hat{p}={y\over n} \longrightarrow \tilde{y}={y+0.5z_{\alpha \over 2}^2 \over n+z_{\alpha \over 2}^2},$ + : $\hat{p}={y\over n} \longrightarrow \tilde{p}={y+0.5z_{\alpha \over 2}^2 \over n+z_{\alpha \over 2}^2},$ where [[AP_Statistics_Curriculum_2007_Normal_Critical | $z_{\alpha \over 2}$ is the normal critical value we saw earlier]]. where [[AP_Statistics_Curriculum_2007_Normal_Critical | $z_{\alpha \over 2}$ is the normal critical value we saw earlier]]. Line 10: Line 10: === Hypothesis Testing about a Single Sample Proportion=== === Hypothesis Testing about a Single Sample Proportion=== - * Null Hypothesis: $H_o: p=p_o$ (e.g., 0), where ''p'' is the population proportion of interest. + * Null Hypothesis: $H_o: p=p_o$ (e.g., $p_o={1\over 2}$), where ''p'' is the population proportion of interest. * Alternative Research Hypotheses: * Alternative Research Hypotheses: - ** One sided (uni-directional): $H_1: p >p_o$, or $H_o: pp_o$, or $H_1: pZ_o)< 1.9\times 10^{-8}$. This small p-values provides extremely strong evidence to reject the null hypothesis that there are no differences between the proportions of mothers that had a girl as a second child but had either boy or girl as their first child. Hence there is strong statistical evidence implying that genders of siblings are not independent. + * $P_{value} = P(Z>Z_o)< 1.9\times 10^{-8}$. This small p-values provides extremely strong evidence to reject the null hypothesis that there are no differences between the proportions of mothers that had a girl as a second child but had either boy or girl as their first child. Hence there is strong statistical evidence implying that genders of siblings are not independent. * '''Practical significance''': The practical significance of the effect (of the gender of the first child on the gender of the second child, in this case) can only be assessed using [[AP_Statistics_Curriculum_2007#Estimating_a_Population_Proportion |confidence intervals]]. A 95% $CI (p_1- p_2) =[0.033; 0.070]$ is computed by $p_1-p_2 \pm 1.96 SE(p_1 - p_2)$. Clearly, this is a practically negligible effect and no reasonable person would make important prospective family decisions based on the gender of their (first) child. * '''Practical significance''': The practical significance of the effect (of the gender of the first child on the gender of the second child, in this case) can only be assessed using [[AP_Statistics_Curriculum_2007#Estimating_a_Population_Proportion |confidence intervals]]. A 95% $CI (p_1- p_2) =[0.033; 0.070]$ is computed by $p_1-p_2 \pm 1.96 SE(p_1 - p_2)$. Clearly, this is a practically negligible effect and no reasonable person would make important prospective family decisions based on the gender of their (first) child. Line 75: Line 75: - ===References=== + ===[[EBook_Problems_Hypothesis_Proportion|Problems]]=== + + ===Notes=== + * [http://sciencenow.sciencemag.org/cgi/content/full/2009/1030/1 Read this Science discussion "Mission Improbable: A Concise and Precise Definition of P-Value"]. + * Use these data to investigate if there are significant gender effects on mortality rates in [[SOCR_Data_AMI_NY_1993_HeartAttacks|this study of heart attacks/ Acute Myocardial Infarction]]. ## General Advance-Placement (AP) Statistics Curriculum - Testing a Claim about Proportion ### Background Recall that for large samples, the sampling distribution of the sample proportion $\hat{p}$ is approximately Normal, by CLT, as the sample proportion may be presented as a sample average or Bernoulli random variables. When the sample size is small, the normal approximation may be inadequate. To accommodate this, we will modify the sample-proportion $\hat{p}$ slightly and obtain the corrected-sample-proportion $\tilde{p}$: $\hat{p}={y\over n} \longrightarrow \tilde{p}={y+0.5z_{\alpha \over 2}^2 \over n+z_{\alpha \over 2}^2},$ The standard error of $\hat{p}$ also needs a slight modification $SE_{\hat{p}} = \sqrt{\hat{p}(1-\hat{p})\over n} \longrightarrow SE_{\tilde{p}} = \sqrt{\tilde{p}(1-\tilde{p})\over n+z_{\alpha \over 2}^2}.$ ### Hypothesis Testing about a Single Sample Proportion • Null Hypothesis: Ho:p = po (e.g., $p_o={1\over 2}$), where p is the population proportion of interest. • Alternative Research Hypotheses: • One sided (uni-directional): H1:p > po, or H1:p < po • Double sided: $H_1: p \not= p_o$ • Test Statistics: $Z_o={\tilde{p} -p_o \over SE_{\tilde{p}}} \sim N(0,1).$ ### Example Suppose a researcher is interested in studying the effect of aspirin in reducing heart attacks. He randomly recruits 500 subjects with evidence of early heart disease and has them take one aspirin daily for two years. At the end of the two years, he finds that during the study only 17 subjects had a heart attack. Use α = 0.05 to formulate a test a research hypothesis that the proportion of subject on aspirin treatment that have heart attacks within 2 years of treatment is po = 0.04. $\tilde{p} = {17+0.5z_{0.025}^2\over 500+z_{0.025}^2}== {17+1.92\over 500+3.84}=0.038$ $SE_{\tilde{p}}= \sqrt{0.038(1-0.038)\over 500+3.84}=0.0085$ And the corresponding test statistics is $Z_o={\tilde{p} - 0.04 \over SE_{\tilde{p}}}={0.002 \over 0.0085}=0.2353$ The p-value corresponding to this test-statistics is clearly insignificant. ### Genders of Siblings Example Is the gender of a second child influenced by the gender of the first child, in families with >1 kid? Research hypothesis needs to be formulated first before collecting/looking/interpreting the data that will be used to address it. Mothers whose 1st child is a girl are more likely to have a girl, as a second child, compared to mothers with boys as 1st children. Data: 20 yrs of birth records of 1 Hospital in Auckland, New Zealand. Second Child Male Female Total First Child Male 3,202 2,776 5,978 Female 2,620 2,792 5,412 Total 5,822 5,568 11,390 Let p1=true proportion of girls in mothers with girl as first child, p2=true proportion of girls in mothers with boy as first child. The parameter of interest is p1p2. • Hypotheses: Ho:p1p2 = 0 (skeptical reaction). H1:p1p2 > 0 (research hypothesis). Second Child Number of births Number of girls Proportion Group 1 (Previous child was girl) n1 = 5412 2792 $\hat{p}_1=0.516$ 2 (Previous child was boy) n2 = 5978 2776 $\hat{p}_2=0.464$ • Test Statistics: $Z_o = {Estimate-HypothesizedValue\over SE(Estimate)} = {\hat{p}_1 - \hat{p}_2 - 0 \over SE(\hat{p}_1 - \hat{p}_2)} = {\hat{p}_1 - \hat{p}_2 - 0 \over \sqrt{{\hat{p}_1(1-\hat{p}_1)\over n_1} + {\hat{p}_2(1-\hat{p}_2)\over n_2}}} \sim N(0,1)$ and Zo = 5.4996. • $P_{value} = P(Z>Z_o)< 1.9\times 10^{-8}$. This small p-values provides extremely strong evidence to reject the null hypothesis that there are no differences between the proportions of mothers that had a girl as a second child but had either boy or girl as their first child. Hence there is strong statistical evidence implying that genders of siblings are not independent. • Practical significance: The practical significance of the effect (of the gender of the first child on the gender of the second child, in this case) can only be assessed using confidence intervals. A 95% CI(p1p2) = [0.033;0.070] is computed by $p_1-p_2 \pm 1.96 SE(p_1 - p_2)$. Clearly, this is a practically negligible effect and no reasonable person would make important prospective family decisions based on the gender of their (first) child.
# 8.6 Parametric equations  (Page 5/6) Page 5 / 6 ## Verbal What is a system of parametric equations? A pair of functions that is dependent on an external factor. The two functions are written in terms of the same parameter. For example, $\text{\hspace{0.17em}}x=f\left(t\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y=f\left(t\right).$ Some examples of a third parameter are time, length, speed, and scale. Explain when time is used as a parameter. Explain how to eliminate a parameter given a set of parametric equations. Choose one equation to solve for $\text{\hspace{0.17em}}t,\text{\hspace{0.17em}}$ substitute into the other equation and simplify. What is a benefit of writing a system of parametric equations as a Cartesian equation? What is a benefit of using parametric equations? Some equations cannot be written as functions, like a circle. However, when written as two parametric equations, separately the equations are functions. Why are there many sets of parametric equations to represent on Cartesian function? ## Algebraic For the following exercises, eliminate the parameter $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ to rewrite the parametric equation as a Cartesian equation. $\left\{\begin{array}{l}x\left(t\right)=5-t\hfill \\ y\left(t\right)=8-2t\hfill \end{array}$ $y=-2+2x$ $\left\{\begin{array}{l}x\left(t\right)=6-3t\hfill \\ y\left(t\right)=10-t\hfill \end{array}$ $\left\{\begin{array}{l}x\left(t\right)=2t+1\hfill \\ y\left(t\right)=3\sqrt{t}\hfill \end{array}$ $y=3\sqrt{\frac{x-1}{2}}$ $\left\{\begin{array}{l}x\left(t\right)=3t-1\hfill \\ y\left(t\right)=2{t}^{2}\hfill \end{array}$ $\left\{\begin{array}{l}x\left(t\right)=2{e}^{t}\hfill \\ y\left(t\right)=1-5t\hfill \end{array}$ $x=2{e}^{\frac{1-y}{5}}\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}y=1-5ln\left(\frac{x}{2}\right)$ $\left\{\begin{array}{l}x\left(t\right)={e}^{-2t}\hfill \\ y\left(t\right)=2{e}^{-t}\hfill \end{array}$ $\left\{\begin{array}{l}x\left(t\right)=4\text{log}\left(t\right)\hfill \\ y\left(t\right)=3+2t\hfill \end{array}$ $x=4\mathrm{log}\left(\frac{y-3}{2}\right)$ $\left\{\begin{array}{l}x\left(t\right)=\text{log}\left(2t\right)\hfill \\ y\left(t\right)=\sqrt{t-1}\hfill \end{array}$ $\left\{\begin{array}{l}x\left(t\right)={t}^{3}-t\hfill \\ y\left(t\right)=2t\hfill \end{array}$ $x={\left(\frac{y}{2}\right)}^{3}-\frac{y}{2}$ $\left\{\begin{array}{l}x\left(t\right)=t-{t}^{4}\hfill \\ y\left(t\right)=t+2\hfill \end{array}$ $\left\{\begin{array}{l}x\left(t\right)={e}^{2t}\hfill \\ y\left(t\right)={e}^{6t}\hfill \end{array}$ $y={x}^{3}$ $\left\{\begin{array}{l}x\left(t\right)={t}^{5}\hfill \\ y\left(t\right)={t}^{10}\hfill \end{array}$ ${\left(\frac{x}{4}\right)}^{2}+{\left(\frac{y}{5}\right)}^{2}=1$ $\left\{\begin{array}{l}x\left(t\right)=3\mathrm{sin}\text{\hspace{0.17em}}t\hfill \\ y\left(t\right)=6\mathrm{cos}\text{\hspace{0.17em}}t\hfill \end{array}$ ${y}^{2}=1-\frac{1}{2}x$ $\left\{\begin{array}{l}x\left(t\right)=\mathrm{cos}\text{\hspace{0.17em}}t+4\\ y\left(t\right)=2{\mathrm{sin}}^{2}t\end{array}$ $\left\{\begin{array}{l}x\left(t\right)=t-1\\ y\left(t\right)={t}^{2}\end{array}$ $y={x}^{2}+2x+1$ $\left\{\begin{array}{l}x\left(t\right)=-t\\ y\left(t\right)={t}^{3}+1\end{array}$ $\left\{\begin{array}{l}x\left(t\right)=2t-1\\ y\left(t\right)={t}^{3}-2\end{array}$ $y={\left(\frac{x+1}{2}\right)}^{3}-2$ For the following exercises, rewrite the parametric equation as a Cartesian equation by building an $x\text{-}y$ table. $\left\{\begin{array}{l}x\left(t\right)=2t-1\\ y\left(t\right)=t+4\end{array}$ $\left\{\begin{array}{l}x\left(t\right)=4-t\\ y\left(t\right)=3t+2\end{array}$ $y=-3x+14$ $\left\{\begin{array}{l}x\left(t\right)=2t-1\\ y\left(t\right)=5t\end{array}$ $\left\{\begin{array}{l}x\left(t\right)=4t-1\\ y\left(t\right)=4t+2\end{array}$ $y=x+3$ For the following exercises, parameterize (write parametric equations for) each Cartesian equation by setting $x\left(t\right)=t$ or by setting $\text{\hspace{0.17em}}y\left(t\right)=t.$ $y\left(x\right)=3{x}^{2}+3$ $y\left(x\right)=2\mathrm{sin}\text{\hspace{0.17em}}x+1$ $\left\{\begin{array}{l}x\left(t\right)=t\hfill \\ y\left(t\right)=2\mathrm{sin}t+1\hfill \end{array}$ $x\left(y\right)=3\mathrm{log}\left(y\right)+y$ $x\left(y\right)=\sqrt{y}+2y$ $\left\{\begin{array}{l}x\left(t\right)=\sqrt{t}+2t\hfill \\ y\left(t\right)=t\hfill \end{array}$ For the following exercises, parameterize (write parametric equations for) each Cartesian equation by using $x\left(t\right)=a\mathrm{cos}\text{\hspace{0.17em}}t$ and $\text{\hspace{0.17em}}y\left(t\right)=b\mathrm{sin}\text{\hspace{0.17em}}t.\text{\hspace{0.17em}}$ Identify the curve. $\frac{{x}^{2}}{4}+\frac{{y}^{2}}{9}=1$ $\frac{{x}^{2}}{16}+\frac{{y}^{2}}{36}=1$ $\left\{\begin{array}{l}x\left(t\right)=4\mathrm{cos}\text{\hspace{0.17em}}t\hfill \\ y\left(t\right)=6\mathrm{sin}\text{\hspace{0.17em}}t\hfill \end{array};\text{\hspace{0.17em}}$ Ellipse ${x}^{2}+{y}^{2}=16$ ${x}^{2}+{y}^{2}=10$ $\left\{\begin{array}{l}x\left(t\right)=\sqrt{10}\mathrm{cos}t\hfill \\ y\left(t\right)=\sqrt{10}\mathrm{sin}t\hfill \end{array};\text{\hspace{0.17em}}$ Circle Parameterize the line from $\text{\hspace{0.17em}}\left(3,0\right)\text{\hspace{0.17em}}$ to $\text{\hspace{0.17em}}\left(-2,-5\right)\text{\hspace{0.17em}}$ so that the line is at $\text{\hspace{0.17em}}\left(3,0\right)\text{\hspace{0.17em}}$ at $\text{\hspace{0.17em}}t=0,\text{\hspace{0.17em}}$ and at $\text{\hspace{0.17em}}\left(-2,-5\right)\text{\hspace{0.17em}}$ at $\text{\hspace{0.17em}}t=1.$ Parameterize the line from $\text{\hspace{0.17em}}\left(-1,0\right)\text{\hspace{0.17em}}$ to $\text{\hspace{0.17em}}\left(3,-2\right)\text{\hspace{0.17em}}$ so that the line is at $\text{\hspace{0.17em}}\left(-1,0\right)\text{\hspace{0.17em}}$ at $\text{\hspace{0.17em}}t=0,\text{\hspace{0.17em}}$ and at $\text{\hspace{0.17em}}\left(3,-2\right)\text{\hspace{0.17em}}$ at $\text{\hspace{0.17em}}t=1.$ $\left\{\begin{array}{l}x\left(t\right)=-1+4t\hfill \\ y\left(t\right)=-2t\hfill \end{array}$ Parameterize the line from $\text{\hspace{0.17em}}\left(-1,5\right)\text{\hspace{0.17em}}$ to $\text{\hspace{0.17em}}\left(2,3\right)$ so that the line is at $\text{\hspace{0.17em}}\left(-1,5\right)\text{\hspace{0.17em}}$ at $\text{\hspace{0.17em}}t=0,\text{\hspace{0.17em}}$ and at $\text{\hspace{0.17em}}\left(2,3\right)\text{\hspace{0.17em}}$ at $\text{\hspace{0.17em}}t=1.$ Parameterize the line from $\text{\hspace{0.17em}}\left(4,1\right)\text{\hspace{0.17em}}$ to $\text{\hspace{0.17em}}\left(6,-2\right)\text{\hspace{0.17em}}$ so that the line is at $\text{\hspace{0.17em}}\left(4,1\right)\text{\hspace{0.17em}}$ at $\text{\hspace{0.17em}}t=0,\text{\hspace{0.17em}}$ and at $\text{\hspace{0.17em}}\left(6,-2\right)\text{\hspace{0.17em}}$ at $\text{\hspace{0.17em}}t=1.$ $\left\{\begin{array}{l}x\left(t\right)=4+2t\hfill \\ y\left(t\right)=1-3t\hfill \end{array}$ ## Technology For the following exercises, use the table feature in the graphing calculator to determine whether the graphs intersect. yes, at $t=2$ For the following exercises, use a graphing calculator to complete the table of values for each set of parametric equations. $\left\{\begin{array}{l}{x}_{1}\left(t\right)=3{t}^{2}-3t+7\hfill \\ {y}_{1}\left(t\right)=2t+3\hfill \end{array}$ $t$ $x$ $y$ –1 0 1 $\left\{\begin{array}{l}{x}_{1}\left(t\right)={t}^{2}-4\hfill \\ {y}_{1}\left(t\right)=2{t}^{2}-1\hfill \end{array}$ $t$ $x$ $y$ 1 2 3 $t$ $x$ $y$ 1 -3 1 2 0 7 3 5 17 $\left\{\begin{array}{l}{x}_{1}\left(t\right)={t}^{4}\hfill \\ {y}_{1}\left(t\right)={t}^{3}+4\hfill \end{array}$ $t$ $x$ $y$ -1 0 1 2 ## Extensions Find two different sets of parametric equations for $\text{\hspace{0.17em}}y={\left(x+1\right)}^{2}.$ Find two different sets of parametric equations for $\text{\hspace{0.17em}}y=3x-2.$ Find two different sets of parametric equations for $\text{\hspace{0.17em}}y={x}^{2}-4x+4.$ how fast can i understand functions without much difficulty what is set? a colony of bacteria is growing exponentially doubling in size every 100 minutes. how much minutes will it take for the colony of bacteria to triple in size I got 300 minutes. is it right? Patience no. should be about 150 minutes. Jason It should be 158.5 minutes. Mr ok, thanks Patience 100•3=300 300=50•2^x 6=2^x x=log_2(6) =2.5849625 so, 300=50•2^2.5849625 and, so, the # of bacteria will double every (100•2.5849625) = 258.49625 minutes Thomas what is the importance knowing the graph of circular functions? can get some help basic precalculus What do you need help with? Andrew how to convert general to standard form with not perfect trinomial can get some help inverse function ismail Rectangle coordinate how to find for x it depends on the equation Robert yeah, it does. why do we attempt to gain all of them one side or the other? Melissa whats a domain The domain of a function is the set of all input on which the function is defined. For example all real numbers are the Domain of any Polynomial function. Spiro Spiro; thanks for putting it out there like that, 😁 Melissa foci (–7,–17) and (–7,17), the absolute value of the differenceof the distances of any point from the foci is 24. difference between calculus and pre calculus? give me an example of a problem so that I can practice answering x³+y³+z³=42 Robert dont forget the cube in each variable ;) Robert of she solves that, well ... then she has a lot of computational force under her command .... Walter what is a function? I want to learn about the law of exponent explain this
## Algebraic expressions With example Problems and also Interactive Exercises Use the following examples and also interactive exercises to learn about Writing Algebraic Expressions. You are watching: The sum of x and 9 Problem: Ms. Jensen likes to divide her class into groups of 2. Usage mathematical symbols to stand for all the students in she class. Solution: Let g represent the variety of groups in Ms. Jensen"s class. Then 2 · g, or 2g deserve to represent "g teams of 2 students". In the trouble above, the change g represents the variety of groups in Ms. Jensen"s class. A variable is a symbol provided to stand for a number in an expression or an equation. The worth of this number can vary (change). Let"s watch at an instance in i m sorry we use a variable. Example 1: Write each expression as a mathematics expression. Phrase Expression the sum of nine and eight 9 + 8 the sum of nine and also a number x 9 + x The expression 9 + 8 represents a solitary number (17). This expression is a numerical expression, (also called an arithmetic expression). The expression 9 + x represents a value that deserve to change. If x is 2, climate the expression 9 + x has a value of 11. If x is 6, then the expression has actually a worth of 15. So 9 + x is an algebraic expression. In the next few examples, we will be working solely with algebraic expressions. Example 2: Write each phrase as an algebraic expression. Phrase Expression nine boosted by a number x 9 + x fourteen diminished by a number p 14 - p seven much less than a number t t - 7 the product of 9 and also a number n 9 · n or 9n thirty-two divided by a number y 32 ÷ y or In instance 2, every algebraic expression had one number, one operation and also one variable. Let"s look at at an instance in i m sorry the expression consists of an ext than one number and/or operation. Example 3: Write each phrase as an algebraic expression utilizing the change n. Phrase Expression five an ext than double a number 2n + 5 the product that a number and 6 6n seven split by twice a number 7 ÷ 2n or three times a number lessened by 11 3n - 11 Example 4: A little company has \$1000 to distribution to the employees as a bonus. Exactly how much money will certainly each employee get? Solution: Let e represent the number of employees in the company. The quantity of money every employee will gain is stood for by the adhering to algebraic expression: Example 5: An electrician dues \$45 per hour and spends \$20 a work on gasoline. Write an algebraic expression to stand for his income for one day. Solution: Let x stand for the variety of hours the electrician functions in one day. The electrician"s earnings have the right to be stood for by the complying with algebraic expression: Solution: 45x - 20 Summary: A variable is a symbol provided to stand for a number in one expression or one equation. The value of this number deserve to change. An algebraic expression is a mathematics expression that is composed of variables, numbers and also operations. The value of this expression can change. ### Exercises Directions: pick the algebraic expression that effectively represents the expression provided. Select your price by clicking its button. Feedback to her answer is provided in the results BOX. If you do a mistake, pick a different button. 1. Fifteen less than twice a number 15 - 2x2x + 152x - 15None the the above.RESULTS BOX: 2. Three times a number, raised by seventeen 3(a + 17)3a + 17(17 + 3)aNone that the above.RESULTS BOX: 3. The product the nine and also a number, diminished by six 9(m - 6)6(9 - m)9m - 6None of the above.RESULTS BOX: 4. Thirty split by seven times a number 30 + 7n 30(7n) None that the above.See more: Does Ham Go Bad? How To Tell If Ham Is Spoiled How To Know When Ham Has Gone Bad RESULTS BOX: 5. Jenny earns \$30 a work working part time in ~ a supermarket. Create an algebraic expression to stand for the amount of money she will certainly earn in d days. 30d 30 + d None that the above RESULTS BOX:
Unit 2 - Algebra & Factoring Q. Why do we test you? • To check for understanding • Students (feedback) • Parents (feedback) • To help prepare for the exam... college, university... life? • Pinpoint errors in my teaching or your learning - so we can improve Q. What would we assess on the Unit 2 test? • Knowledge of the Unit • Expanding & Simplifying Example: (x - 7)(x + 12) • Common Factors What is "common" between 18ax^2 and 6bx? • GCF - looking for the GREATEST common factor. • Factor 27xyz + 15x = 3x( 9yz + 5 ) • What about 4x - 8y + 3x - 6y • Grouping! Choice - what should you group by? (4x - 8y)+(3x - 6y) = 4(x - 2y) + 3(x - 2y) = (x - 2y)(4 + 3) = (x - 2y)(7) • Factoring "Quadratics": ax^2 + bx + c • Also called a Trinomial • Monic is when a = 1 • Non-Monic when a is not 1 • MONIC: • Is there a GCF? If yes, factor it out. • What two numbers multiply to 'c' and add to 'b'? • NON-MONIC: • Is there a GCF? If yes, factor it out. • Factor by "decomposition" where we break-up the 'bx' into two pieces and then factor by grouping. • Multiply a(c) to get a new factor "m" • Check for factors of m that add to b • Break up b into two pieces and factor by grouping • Difference of Squares • Actually decomposition when the b is zero. 64x^2 - 81 = 64^2 + 0x - 81 = (8x + 9)(8x - 9) • Area of a Rectangle = (length)(width)
Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12 # Find the Time Period of the Oscillation of Mass M in Figures 12−E4 A, B, C. What is the Equivalent Spring Constant of the Pair of Springs in Each Case? - Physics Sum Find the time period of the oscillation of mass m in figures  a, b, c. What is the equivalent spring constant of the pair of springs in each case?] #### Solution (a) Spring constant of a parallel combination of springs is given as, K = k1 + k2  (parallel) Using the relation of time period for S.H.M. for the given spring-mass system, we have : $T = 2\pi\sqrt{\frac{m}{K}} = 2\pi\sqrt{\frac{m}{k_1 + k_2}}$ (b) Let be the displacement of the block of mass m, towards left. Resultant force is calculated as, F = F1 + F2 = (k1 + k2)x Acceleration $\left( a \right)$ is given by, $a = \left( \frac{F}{m} \right) = \frac{\left( k_1 + k_2 \right)}{m}x$ Time period $\left( T \right)$ is given by , $T = 2\pi\sqrt{\frac{\text { displacement }}{\text { acceleration }}}$ $\text { On substituting the values of displacement and acceleration, we get: }$ $T = 2\pi\sqrt{\frac{x}{x\frac{\left( k_1 + k_2 \right)}{m}}}$ $= 2\pi\sqrt{\frac{m}{k_1 + k_2}}$ Required spring constant, K = k1 + k2 (c) Let K be the equivalent spring constant of the series combination. $\frac{1}{K} = \frac{1}{k_1} + \frac{1}{k_2} = \frac{k_2 + k_1}{k_1 k_2}$ $\Rightarrow K = \frac{k_1 k_2}{k_1 + k_2}$ $\text { Time period is given by, }$ $T = 2\pi\sqrt{\frac{m}{K}}$ $\text { On substituting the respective values, we get: }$ $T = 2\pi\sqrt{\frac{m\left( k_1 + k_2 \right)}{k_1 k_2}}$ Is there an error in this question or solution? #### APPEARS IN HC Verma Class 11, 12 Concepts of Physics 1 Chapter 12 Simple Harmonics Motion Q 17 | Page 253
# Do You Know What is the Value of x if mc020-1.jpg mc020-2.jpg mc020-3.jpg mc020-4.jpg mc020-5.jpg ## What is the Value of x if mc020-1.jpg mc020-2.jpg mc020-3.jpg mc020-4.jpg mc020-5.jpg When it comes to understanding the value of X, we need to delve into the world of equations and algebra. X is typically used to represent an unknown value that we are trying to solve for. It is used in mathematical equations to represent an element that is not yet known. Let’s break it down further. ### X as an Unknown Value In mathematics, equations are used to represent relationships between different variables. In these equations, X is often used to represent the unknown value that we are trying to find. By solving for X, we can determine the value that makes the equation true. ### Solving for X To find the value of X, we use various methods and techniques depending on the type of equation we are dealing with. These methods can involve simplifying equations, factoring, expanding, or even using more advanced formulas. ### Importance of Finding X The value of X is crucial in many real-life applications. It helps us solve problems, make predictions, and understand patterns and relationships. Whether it’s finding the value of an angle, determining the roots of an equation, or calculating probabilities, X plays a fundamental role in mathematical reasoning. ### Practical Examples Let’s take a look at a couple of practical examples to understand the relevance of X in real-life scenarios. 1. Suppose you have a budget and want to calculate how much money you can spend on each item. If X represents the total amount of money you have, you can divide it by the number of items to find out the value of X for each item. 2. In physics, X can represent a physical quantity that you are trying to measure, such as the velocity of an object or the amount of force applied. By understanding the relationship between variables and solving for X, we can calculate and predict various physical phenomena. Remember, finding the value of X is not just about solving equations. It helps us make sense of the world around us and enables us to make informed decisions based on data and mathematical reasoning. So, the next time you come across an equation with X, embrace the challenge and unlock the hidden value behind it. ## Solving the Equation When it comes to solving equations, finding the value of **x** is an essential step. In math, equations are used to represent relationships between different variables or quantities. Solving the equation means finding the specific value or values that make the equation true. There are various methods and techniques to solve equations, depending on the complexity and type of the equation. Here are a few common techniques that can be used to find the value of **x**: 1. Simplifying and isolating x: Start by simplifying the equation by combining like terms and grouping constants on one side of the equation. Then, isolate the variable **x** by performing inverse operations. For example, if there is an addition or subtraction involving **x**, use the inverse operation of addition or subtraction to remove it from the equation. 2. Using the multiplication or division property: If there is a multiplication or division involving **x**, use the multiplication or division property to isolate **x**. Remember to perform the same operation on both sides of the equation to maintain equality. 3. Factoring: If the equation is a quadratic equation, it can be solved by factoring. Factoring involves finding two binomials that, when multiplied, give the original quadratic equation. Once the equation is factored, you can set each factor equal to zero and solve for **x**. 4. Quadratic formula: If factoring is not possible, the quadratic formula can be used to solve quadratic equations. The quadratic formula provides the solutions for **x** in terms of the coefficients of the quadratic equation. 5. Graphing: Another way to find the value of **x** is by graphing the equation. Graphing allows you to visually see the points where the equation intersects the x-axis, which represents the value of **x**. It’s important to remember that there may be multiple solutions or no solutions at all for a given equation. Solving the equation provides valuable information and answers to mathematical problems, scientific inquiries, and real-life applications. By finding the value of **x**, we can determine unknown quantities, make predictions, and solve real-world problems that rely on mathematical reasoning. So, whether you’re tackling a simple equation or a complex one, understanding these techniques will help you confidently solve for **x** and unlock the answers hidden within the equations.
$$\require{cancel}$$ # Linear Momentum The scientific definition of linear momentum is consistent with most people’s intuitive understanding of momentum: a large, fast-moving object has greater momentum than a smaller, slower object. Linear momentum is defined as the product of a system’s mass multiplied by its velocity. In symbols, linear momentum is expressed as $p = mv$ Momentum is directly proportional to the object’s mass and also its velocity. Thus the greater an object’s mass or the greater its velocity, the greater its momentum. Momentum $$p$$ is a vector having the same direction as the velocity $$v$$. The SI unit for momentum is $$kg \cdot m/s.$$ Linear Momentum Linear momentum is defined as the product of a system’s mass multiplied by its velocity: $p = mv$ Example 8.2.1: Calculating Momentum: A Football Player and a Football (a) Calculate the momentum of a 110-kg football player running at 8.00 m/s. (b) Compare the player’s momentum with the momentum of a hard-thrown 0.410-kg football that has a speed of 25.0 m/s. Strategy No information is given regarding direction, and so we can calculate only the magnitude of the momentum, $$p$$ (As usual, a symbol that is in italics is a magnitude, whereas one that is italicized, boldfaced, and has an arrow is a vector.) In both parts of this example, the magnitude of momentum can be calculated directly from the definition of momentum given in the equation, which becomes $p = mv$ when only magnitudes are considered. Solution for (a) To determine the momentum of the player, substitute the known values for the player’s mass and speed into the equation. $p_{player} = (110 \space kg)(8.00 \space m/s) = 880 \space kg \cdot m/s$ Solution for (b) To determine the momentum of the ball, substitute the known values for the ball’s mass and speed into the equation. $p_{ball} = (0.410 \space kg)(25.0 \space m/s) = 10.3 \space kg \cdot m/s$ The ratio of the player’s momentum to that of the ball is $\dfrac{p_{player}}{p_{ball}} = \dfrac{880}{10.3} = 85.0$ Discussion Although the ball has greater velocity, the player has a much greater mass. Thus the momentum of the player is much greater than the momentum of the football, as you might guess. As a result, the player’s motion is only slightly affected if he catches the ball. We shall quantify what happens in such collisions in terms of momentum in later sections. # Momentum and Newton’s Second Law The importance of momentum, unlike the importance of energy, was recognized early in the development of classical physics. Momentum was deemed so important that it was called the “quantity of motion.” Newton actually stated his second law of motion in terms of momentum: The net external force equals the change in momentum of a system divided by the time over which it changes. Using symbols, this law is $F_{net} = \dfrac{ \Delta p}{\Delta t}$ where $$F_{net}$$ is the net external force, $$\Delta p$$ is the change in momentum, and $$\Delta t$$ is the change in time. Newton’s Second Law of Motion in Terms of Momentum The net external force equals the change in momentum of a system divided by the time over which it changes. $F_{net} = \dfrac{\Delta p}{\Delta t}$ ### Making Connections: Force and Momentum Force and momentum are intimately related. Force acting over time can change momentum, and Newton’s second law of motion, can be stated in its most broadly applicable form in terms of momentum. Momentum continues to be a key concept in the study of atomic and subatomic particles in quantum mechanics. This statement of Newton’s second law of motion includes the more familiar $$F_{net} = ma$$ as a special case. We can derive this form as follows. First, note that the change in momentum $$\Delta p$$ is given by $\Delta p = \Delta (mv)$ If the mass of the system is constant, then $\Delta (mv) = m\Delta v.$ So that for constant mass, Newton’s second law of motion becomes $F_{net} = \dfrac{\Delta p}{\Delta t} = \dfrac{m \Delta v}{\Delta t}.$ Because $$\frac{\Delta v}{\Delta t} = a,$$ we get the familiar equation $F_{net} = ma$ when the mass of the system is constant. Newton’s second law of motion stated in terms of momentum is more generally applicable because it can be applied to systems where the mass is changing, such as rockets, as well as to systems of constant mass. We will consider systems with varying mass in some detail; however, the relationship between momentum and force remains useful when mass is constant, such as in the following example. Example 8.2.2: Calculating Force: Venus Williams’ Racquet During the 2007 French Open, Venus Williams hit the fastest recorded serve in a premier women’s match, reaching a speed of 58 m/s (209 km/h). What is the average force exerted on the 0.057-kg tennis ball by Venus Williams’ racquet, assuming that the ball’s speed just after impact is 58 m/s, that the initial horizontal component of the velocity before impact is negligible, and that the ball remained in contact with the racquet for 5.0 ms (milliseconds)? Strategy This problem involves only one dimension because the ball starts from having no horizontal velocity component before impact. Newton’s second law stated in terms of momentum is then written as $F_{net} = \dfrac{\Delta p}{\Delta t}$ As noted above, when mass is constant, the change in momentum is given by $\Delta p = m\Delta v = m(v_f - v_i).$ In this example, the velocity just after impact and the change in time are given; thus, once $$\Delta p$$ s calculated, $$F_{net} = \frac{\Delta p}{\Delta t}$$ can be used to find the force. Solution To determine the change in momentum, substitute the values for the initial and final velocities into the equation above. $\Delta p = m(v_f - v_i)$ $= (0.057 \space kg)(58 \space m/s - 0 \space m/s)$ $= 3.306 \space kg \cdot m/s = 3.3 \space kg \cdot m/s$ Now the magnitude of the net external force can determined by using $$F_{net} = \frac{\Delta p}{\Delta t}$$ $F_{net} = \dfrac{\Delta p}{\Delta t} = \dfrac{3.306 \space kg}{5.0 \times 10^{-3}}$ $= 661 \space N,$ where we have retained only two significant figures in the final step. Discussion This quantity was the average force exerted by Venus Williams’ racquet on the tennis ball during its brief impact (note that the ball also experienced the 0.56-N force of gravity, but that force was not due to the racquet). This problem could also be solved by first finding the acceleration and then using $$F = ma$$ but one additional step would be required compared with the strategy used in this example. # Section Summary • Linear momentum (momentum for brevity) is defined as the product of a system’s mass multiplied by its velocity. • In symbols, linear momentum $$p$$ is defined to be $p = mv$ where $$m$$ is the mass of the system and $$v$$ is its velocity. • The SI unit for momentum is $$kg \cdot m/s.$$ • Newton’s second law of motion in terms of momentum states that the net external force equals the change in momentum of a system divided by the time over which it changes. • In symbols, Newton’s second law of motion is defined to be $$F_{net} = \frac{\Delta p}{\Delta t}$$, $$F_{net}$$ is the net external force, $$\Delta p$$ s the change in momentum, and $$\Delta t$$ is change in time. # Conceptual Questions Exercise 8.2.1: An object that has a small mass and an object that has a large mass have the same momentum. Which object has the largest kinetic energy? Exercise 8.2.2: An object that has a small mass and an object that has a large mass have the same kinetic energy. Which mass has the largest momentum? Exercise 8.2.3: Professional Application Football coaches advise players to block, hit, and tackle with their feet on the ground rather than by leaping through the air. Using the concepts of momentum, work, and energy, explain how a football player can be more effective with his feet on the ground. Exercise 8.2.4: How can a small force impart the same momentum to an object as a large force? # Problems & Exercises Exercise 8.2.5: (a) Calculate the momentum of a 2000-kg elephant charging a hunter at a speed of $$7.50 \space m/s$$ (b) Compare the elephant’s momentum with the momentum of a 0.0400-kg tranquilizer dart fired at a speed of $$600 \space m/s$$ (c) What is the momentum of the 90.0-kg hunter running at $$7.40 \space m/s$$ after missing the elephant? [Hide Solution] (a) $$1.50 \times 10^4 \space kg \cdot m/s$$ (b) 625 to 1 (c) 6.66 \times 10^2 \space kg \cdot m/s\) Exercise 8.2.6: (a) What is the mass of a large ship that has a momentum of $$1.60 \times 10^9 \space kg \space m/s$$, when the ship is moving at a speed of $$48 \space km/h$$? (b) Compare the ship’s momentum to the momentum of a 1100-kg artillery shell fired at a speed of $$1200 \space m/s$$. Exercise 8.2.7: (a) At what speed would a $$2.00 \times 10^4 \space kg$$ airplane have to fly to have a momentum of $$1.60 \times 10^9 \space kg \cdot m/s$$ (the same as the ship’s momentum in the problem above)? (b) What is the plane’s momentum when it is taking off at a speed of $$60 m/s$$? (c) If the ship is an aircraft carrier that launches these airplanes with a catapult, discuss the implications of your answer to (b) as it relates to recoil effects of the catapult on the ship. (a) $$8.00 \times 10^4 \space m/s$$ (b) $$1.20 \times 10^6 \space kg \cdot m/s$$ (c) Because the momentum of the airplane is 3 orders of magnitude smaller than of the ship, the ship will not recoil very much. The recoil would be - 0.0100 \space m/s\), which is probably not noticeable. Exercise 8.2.8: (a) What is the momentum of a garbage truck that is $$1.2 \times 10^4 \space kg$$ and is moving at $$10 \space m/s$$. (b) At what speed would an 8.00-kg trash can have the same momentum as the truck? Exercise 8.2.9: A runaway train car that has a mass of 15,000 kg travels at a speed of $$5.4 \space m/s$$ down a track. Compute the time required for a force of 1500 N to bring the car to rest. [Hide Solution] 54 s Exercise 8.2.10: The mass of Earth is $$5.972 \times 10^{24} \space kg$$ and its orbital radius is an average of$$1.496 \times 10^{11} \space m$$. Calculate its linear momentum. ## Glossary linear momentum the product of mass and velocity second law of motion physical law that states that the net external force equals the change in momentum of a system divided by the time over which it changes
# Show that Question: $e^{2 x} \sin x$ Solution: Let $I=\int e^{2 x} \sin x d x$   ...(1) Integrating by parts, we obtain $I=\sin x \int e^{2 x} d x-\int\left\{\left(\frac{d}{d x} \sin x\right) \int e^{2 x} d x\right\} d x$ $\Rightarrow I=\sin x \cdot \frac{e^{2 x}}{2}-\int \cos x \cdot \frac{e^{2 x}}{2} d x$ $\Rightarrow I=\frac{e^{2 x} \sin x}{2}-\frac{1}{2} \int e^{2 x} \cos x d x$ Again integrating by parts, we obtain $I=\frac{e^{2 x} \cdot \sin x}{2}-\frac{1}{2}\left[\cos x \int e^{2 x} d x-\int\left\{\left(\frac{d}{d x} \cos x\right) \int e^{2 x} d x\right\} d x\right.$ $\Rightarrow I=\frac{e^{2 x} \sin x}{2}-\frac{1}{2}\left[\cos x \cdot \frac{e^{2 x}}{2}-\int(-\sin x) \frac{e^{2 x}}{2} d x\right]$ $\Rightarrow I=\frac{e^{2 x} \cdot \sin x}{2}-\frac{1}{2}\left[\frac{e^{2 x} \cos x}{2}+\frac{1}{2} \int e^{2 x} \sin x d x\right]$ $\Rightarrow I=\frac{e^{2 x} \sin x}{2}-\frac{e^{2 x} \cos x}{4}-\frac{1}{4} I$    [From (1)] $\Rightarrow I+\frac{1}{4} I=\frac{e^{2 x} \cdot \sin x}{2}-\frac{e^{2 x} \cos x}{4}$ $\Rightarrow \frac{5}{4} I=\frac{e^{2 x} \sin x}{2}-\frac{e^{2 x} \cos x}{4}$ $\Rightarrow I=\frac{4}{5}\left[\frac{e^{2 x} \sin x}{2}-\frac{e^{2 x} \cos x}{4}\right]+\mathrm{C}$ $\Rightarrow I=\frac{e^{2 x}}{5}[2 \sin x-\cos x]+\mathrm{C}$
# 6.6 Inverse Functions An Alaskan business executive is heading to Calgary, Alberta in Canada for a conference. They ask their assistant, Niketa, what 75 degrees Fahrenheit is in Celsius and after a quick search on Google, she finds the formula . Using this formula, she calculates Celsius. The next day, the executive sends Niketa the week’s weather forecast for Calgary: At first, Niketa might consider the formula she has already found to do the conversions. After all, she knows her algebra and can easily solve the equation for after substituting a value for . For example, to convert 25 C, she could write: After considering this option for a moment, she realizes that solving the equation for each of the temperatures would get awfully tedious, and realizes that since evaluation is easier than solving, it would be much more convenient to have a different formula, one which takes the Celsius temperature and outputs the Fahrenheit temperature.  This is the idea of an inverse function, where the input becomes the output and the output becomes the input. Inverse Function If , then a function is an inverse of if for all values, a, in the domain and all values, b, in the range of . The inverse of is typically notated , which is read “f inverse of x”, so equivalently, if then Important: The raised -1 used in the notation for inverse function is simply a notation, and does NOT designate an exponent or power of -1. Example of Inverse Functions If, for a particular function, , what do we know about the inverse? The inverse function reverses which quantity is input and which quantity is output, so if , then . Alternatively, if you want to re-name the inverse function g(x), then g(4) = 2. Try it Now 1 Given that , what do we know about the original function h(x)? Notice that original function and the inverse function undo each other.  If , then , returning us to the original input.  More simply put, if you compose these functions together you get the original input as your answer. Since the outputs of the function are the inputs to , the range of is also the domain of .  Likewise, since the inputs to  are the outputs of , the domain of  is the range of . Basically, like how the input and output values switch, the domain & ranges switch as well.  But be careful, because sometimes a function doesn’t even have an inverse function, or only has an inverse on a limited domain.  For example, the inverse of is , since a square “undoes” a square root, but it is only the inverse of on the domain , since that is the range of . Example of the Domain and Range of Inverses The function  has domain and range , what would we expect the domain and range of  to be? We would expect   to swap the domain and range of , so   would have domain  and range . Example of Inverse Functions with a Function Defined by a Table A function f(t) is given as a table below, showing distance in miles that a car has traveled in t minutes.  Find and interpret . t (minutes) 30 50 70 90 f(t) (miles) 20 40 60 70 The inverse function takes an output of f and returns an input for f.  So in the expression , the 70 is an output value of the original function, representing 70 miles.  The inverse will return the corresponding input of the original function f, 90 minutes, so .  Interpreting this, it means that to drive 70 miles, it took 90 minutes. Alternatively, recall the definition of the inverse was that if  then .  By this definition, if you are given  then you are looking for a value a so that .  In this case, we are looking for a t so that , which is when t = 90. Try it Now 2 Using the table below t (minutes) 30 50 70 90 f(t) (miles) 20 40 60 70 Find and interpret the following: a. b. Example of an Inverse with a Function Defined as a Graph A function is given as a graph below.  Find and To evaluate , we find 3 on the horizontal axis and find the corresponding output value on the vertical axis. The point (3, 1) tells us that . To evaluate , recall that by definition means .  By looking for the output value 3 on the vertical axis we find the point (5, 3) on the graph, which means , so by definition . Try it Now 3 Using the graph in the example above a. find b. estimate Example – Back to the Celsuis and Fahrenheit Temperature Problem Returning to our executive’s assistant, find a formula for the inverse function that gives Fahrenheit temperature given a Celsius temperature. A quick Google search would find the inverse function, but alternatively, Niketa might look back at how she solved for the Fahrenheit temperature for a specific Celsius value, and repeat the process in general: By solving in general, we have uncovered the inverse function.  If Then In this case, we introduced a function to represent the conversion since the input and output variables are descriptive, and writing could get confusing. It is important to note that not all functions will have an inverse function.  Since the inverse  takes an output of f and returns an input of f, in order for to itself be a function, then each output of f (input to ) must correspond to exactly one input of f (output of ) in order for to be a function.  This is the definition of a one-to-one function. One-to-One Function If each output of a function, corresponds exactly one input, then the function is said to be a one-to-one function.  A function is one-to-one if and only if its inverse, is also a function. In some cases, it is desirable to have an inverse for a function even though the function is not one-to-one.  In those cases, we can often limit the domain of the original function to an interval on which the function is one-to-one, then find an inverse only on that interval. Example Finding Inverse of a Function that is NOT one-to-one on Limited Domain The quadratic function is not one-to-one.  Find a domain on which this function is one-to-one, and find the inverse on that domain. We can limit the domain to to restrict the graph to a portion that is one-to-one, and find an inverse on this limited domain. You may have already guessed that since we undo a square with a square root, the inverse of on this domain is . You can also solve for the inverse function algebraically.  If , we can introduce the variable to represent the output values, allowing us to write .  To find the inverse we solve for the input variable. To solve for x we take the square root of each side.   and get , so  .  We have restricted x to being non-negative, so we’ll use the positive square root, or . In cases like the example above where the variables are not descriptive, it is common to see the inverse function rewritten with the variable x.  Rewriting the inverse using the variable x is often required for graphing inverse functions using calculators or computers. Note that the domain and range of the square root function do correspond with the range and domain of the quadratic function on the limited domain.  In fact, if we graph on the restricted domain and on the same axes, we can notice symmetry: the graph of is the graph of reflected over the line . This is the case with all functions and their inverses. Example of a Function and its Inverse Graphed Given the graph of shown, sketch a graph of . This is a one-to-one function, so we will be able to sketch an inverse.  Note that the graph shown has an apparent domain of and range of , so the inverse will have a domain of and range of Reflecting this graph of the line , the point (1, 0) reflects to (0, 1), and the point (4, 2) reflects to (2, 4).  Sketching the inverse on the same axes as the original graph: #### Finding Inverse Functions We’ve already discussed that inverse functions undo the original function.  However, consider when you have a function and you want to know its inverse. You will have to undo everything. When you are working with a computer program and use the undo button, the computer undoes the last thing you did, if you click it again, it undoes the thing before that, and so on. The same can be done with an inverse function. We simply look at all the steps a function takes and undo each step in reverse order. Just remember all of your order of operations rules. Examples Finding Inverses Using Undo a. Given , find What does the function actually do in this case?  It takes a value, and: 1. Multiplies 3 2. Subtracts 5 So the undo button should start by taking a value, we can call it also, at the last thing done and: 2. Divide 3 So our inverse b. Given , find So we will start by consider what does, it takes a value : 1. Cubes it 2. Multiplies 8 So the inverse, will undo these in the opposite order, take the value and: 1. Subtract 11 2. Divide 8 3. Take the cube root So we have Another way to find inverses is algebraically solving for the other variable. This is best seen in applications. Example of Finding Inverses in Applications A beverage company has decided a new fun marketing tool would be to put their product in spherical containers.  They know the volumes of the products they want, but they need to know the radii of the spheres to give the specifications to the container manufacturers.  They need a way to use the formula for the volume of a sphere, to enter as a formula into their software to give the radius for any volume they decide they want. We can approach this using an algebraic approach and solve for r instead of V: so we have the inverse function: Since the variables here are meaningful, we do not switch them, but we would introduce another function name, like and . Try it Now 4 a. Given that find b. Given that (With the restricted domain ). Find 1. a. . In 60 minutes, 50 miles are traveled. b. . To travel 60 miles, it will take 70 minutes. 2. a. b. (this is an approximation – answers may vary slightly) 3. a. b.
# Lesson: Unit 6 Lesson 4: Dilations 11641 Views 29 Favorites ### Lesson Objective Students will be able to… 1. Describe a dilation in the coordinate plane 2. Draw the image of a figure under a dilation ### Lesson Plan Lesson Plan Print Unit 6 Lesson 4 Dilations guided notes packet. Students follow along by writing notes. (Text from the notes is in italics). Students will need rulers during this lesson to connect the dots. Opening 1)     Show short clip of “Honey, I Shrunk the Kids!” 2)     Take a quick moment to discuss what students think the word “dilate” means and where they have heard or seen it (Example might be dilating pupils at the doctor’s office) 3)     Remember, in Math the word dilate means to expand or shrink a figure. When you expand or shrink something, the scale factor tells you by how much you need to expand or shrink it. When scale factors are greater than 1, your figure gets bigger. Examples: 1, 5, 10,000, 2.5, 50/2, etc. When scale factors are less than 1, your figure gets smaller. Examples: 0.5, 1/10, ¾, etc. Vocabulary: 1. Coordinate Grid/Coordinate Plane 2. Point 3. Image 4. Transformation 5. Dilation 6. Scale Factor 8. Origin Example 1: Dilate a figure with a scale factor > 1 (Expand) 1. 1.     Identify and record the points (J, K, L) of the original figure. 2. 2.     Have students identify the scale factor in the problem and predict whether it will expand or shrink given what they know about scale factors > or < 1. 3. 3.     Show the multiplication by the scale factor in the space between J and J’, K and K’, and L and L’ and record the new points 4. 4.     Have students plot the points and connect the dots 5. 5.     Ask students if their dilation matches their prediction and if it makes sense. 6. 6.     Have students complete You Try 1 1. The second problem in you try 1 has a scale factor of 1.5. Students tend to have trouble multiplying by decimals, and this is a great opportunity for students to practice taking half of a number and adding it to the original 2. Be aware that Z’ is in the negatives and that you have to take the 1.5 of the absolute value otherwise the negative will throw students off! Common Student Mistakes: 1. 1.     Students plot their points incorrectly, mixing up their x coordinates with their y coordinates 2. 2.     Students multiply incorrectly 3. 3.     Students commit transcription errors (get the answer but write it in the wrong place) 4. 4.     Students plot a point incorrectly, resulting in a figure that is NOT similar. The first step of making a prediction and the last step of asking if their answer makes sense helps build the metacognitive piece. 5. 5.     Students forget to label their points with appropriate letters and primes! Example 2: Dilate a figure with a scale factor < 1 (Shrink) Teacher’s Note:  Shrinking tends to be more difficult as students often have trouble with fraction operations. It’s a good idea to incorporate multiplying by fractions into a Do Now or previous homework. 1. 1.      Identify and record the points (J, K, L) of the original figure. 2. 2.     Have students identify the scale factor in the problem and predict whether it will expand or shrink given what they know about scale factors > or < 1. 3. 3.     Show the multiplication by the scale factor in the space between J and J’, K and K’, and L and L’ and record the new points 4. 4.     Have students plot the points and connect the dots 5. 5.     Ask students if their dilation matches their prediction and if it makes sense. 6. 6.     Have students complete You Try 2 1. The first problem has a scale factor of 1/3. I recommend letting students struggle with the scale factor, and providing guiding questions that tie back to the 0.5 scale factor of Example 2. 2. A question I found myself asking was, “When you get ½ of a pizza, how much do you have? How did you figure that out? So when you get 1/3 of a pizza, how much do you have? How did you figure that out? Interesting, so what conclusions can you draw about multiplying by 1/3 and dividing by 3?” Example 3: Identify and Describe a Dilation 1. 1.     Ask students how they can tell the difference between an original figure and a new image 2. 2.     Have students describe the first dilation. Encourage them to find the coordinates of the points and figure it out the change like they did in examples 1 and 2. This skill leads perfectly into the table dilation problem 3. 3.     Ask students how they can tell the difference between an original figure and a new image in a table 4. 4.     Go through the other answer choices and discuss why each one would not work 1. What would the answer choices have to look like for A, C and D to work? 5. 5.     Have students complete You Try 3 Sample Test Questions 2. 2.     What kind of blunders were you making during the lesson? 3. 3.     How can you avoid a similar fate? Common Blunders: Students should articulate mistakes they were making during their guided practice. Answers may vary but may include: 1. Forgetting primes 2. Multiplying the scale factor incorrectly 3. Mixing up their x and y coordinates 4. Students label the points wrong (A instead of B) 5. Mixing dilations up with rotations, reflections, and/or translations. 6. Students draw shapes that are not similar Independent Practice Have students work through questions 1 – 9 and continue onto the homework. The IP is more focused on describing dilations wherease the homework is more focused on actually drawing them. Closing Have students share out and summarize what they learned today. Assessment Have students complete the Exit Ticket Reflection: What works: The coordinate method, multiplying by the scale factor really worked for my students. As students evolved in their understanding for dilations, I also challenged to them to dilate visually (by counting the number of grid boxes horizontally and vertically). There’s a decent mix of problems, and ample opportunity for students to make mistakes and learn from them. What didn't work: I did not anticipate the difficult students would have multiplying by a scale factor smaller than one. Whether it was multiplying by ½ or 1/3 or 0.5, students had a lot of trouble applying what they know about the connection between multiplying fractions and dividing into pieces. I strongly recommend priming them on this idea through another structure like Do Now’s, home works, mental math, or cranium crunches leading up the lesson. I thought that teaching them the “half plus original” trick for 1.5 also worked will to address this. Lastly, students were not very reflective about their shape getting bigger or smaller and maintaining similarity. If a shape looked totally off, they did not stop and say, hey! This looks totally off! So, as you go through the lesson I strongly recommend reiterating that dilations result in similar figures and have students reflect if their answer makes sense given their predictions. ### Lesson Resources Unit 6 Lesson 4 Dilations HW.docx 1,587 Unit 6 Lesson 4 Dilations.docx 2,409 Memorable-Scene_-Honey,-I-Shrunk-the-Kids-The-Shrinking-Scene[www.savevid.com].mp4 1,867
# How many 1/4 cups does it make 2/3 cup? Posted by Hey there, kitchen adventurers and math enthusiasts! Ever found yourself in the midst of a baking spree, only to realize your measuring cup set is incomplete? Or maybe you’re just a curious soul pondering the math behind cooking measurements. Today, we’re cracking the code on how many 1/4 cups it takes to make 2/3 of a cup. This might seem like simple math, but understanding these conversions can be a real game-changer in the kitchen. So, grab your aprons and calculators, and let’s dive in! Related Posts ## The Math Behind the Measurements At the heart of this culinary conundrum is a bit of fraction arithmetic. Converting measurements in recipes is essential, especially when your measuring cup collection is missing a few members, or you’re scaling recipes up or down. So, how do we solve our specific puzzle? To find out how many 1/4 cups make up 2/3 of a cup, we’re looking at a simple division problem. But instead of diving straight into the numbers, let’s understand the principle behind it. ### Breaking Down the Fractions • 1/4 Cup: This measurement is a quarter of a cup, a common measurement in cooking for both liquid and dry ingredients. • 2/3 Cup: A bit trickier, this one represents two-thirds of a cup, often seen in recipes for its ability to balance proportions just right. The question at hand is how many times does 1/4 fit into 2/3? Or, put mathematically, how many quarters are there in two-thirds? ### Crunching the Numbers To find the answer, we divide 2/3 by 1/4. In the language of fractions, dividing is akin to multiplying by the reciprocal. So, we turn 1/4 into its reciprocal (4/1) and multiply: 2/3 * 4/1 = 8/3 But wait, 8/3 doesn’t quite answer our question in a straightforward way, does it? It tells us something interesting though—1/4 cup fits into 2/3 cup more than twice but not quite three times. To make sense of this, let’s convert 8/3 into a mixed number or another format easier to understand in the context of cooking. ### Practical Kitchen Translation After simplifying, we understand that 2/3 of a cup is equivalent to approximately 2 and 2/3 of 1/4 cups. However, since we can’t easily measure 2/3 of a 1/4 cup in most kitchens, the practical takeaway is knowing that a little more than 2 and a half times of 1/4 cup will give you roughly 2/3 of a cup. This approximation can come in handy when you’re measuring ingredients and you don’t have a 2/3 cup measure. ## Why This Matters in Cooking and Baking Cooking and baking are as much art as they are science. While there’s room for creativity in flavors and textures, measurements require precision. The balance of ingredients affects everything from taste to texture and structural integrity, especially in baking. Knowing how to convert measurements allows you to: • Adapt recipes based on the tools you have. • Scale recipes up or down with accuracy. • Make substitutions with confidence. Understanding how many 1/4 cups make 2/3 of a cup is more than just a math exercise—it’s about empowering yourself in the kitchen. With this knowledge, you’re better equipped to tackle recipes with confidence, even when your kitchen tools are less than complete. So, the next time you come across a measurement conundrum, remember: a little math can lead to delicious results. Happy cooking (and calculating)!
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # Clearing Denominators in Rational Equations ## Multiply both sides by a common multiple Estimated16 minsto complete % Progress Practice Clearing Denominators in Rational Equations MEMORY METER This indicates how strong in your memory this concept is Progress Estimated16 minsto complete % Workers' Comp Teacher Contributed ## Real World Applications – Algebra I ### Topic How can we lower workers’ compensation in companies using Algebra? ### Student Exploration Most companies have a Human Resources department that handles all of the logistics about the company’s workers. One of their biggest responsibilities is working with employees when they get hurt on the job. When this happens, the company offers workers’ compensation. For one particular company, the human resources advisor created algebraic relationships to represent the number of people that were hurt at one of the branches of the company. He concluded that the function that represents this relationship is I0(x)=500x\begin{align*}I_0(x)= \frac{500}{x}\end{align*}. The advisor also created a plan to implement after the initial training in attempts to reduce the number of injuries at work. With additional extensive training and check-ups with employees, he concluded that this model would produce I1(x)=500(x+1)\begin{align*}I_1(x)= \frac{500}{(x + 1)}\end{align*} less injuries. We can represent the difference between the two models to see if the number of injuries would be effective or not. We can represent this revised model by taking the difference between the two relationships as one rational expression. I0I1=500x500x+1\begin{align*}I_0-I_1= \frac{500}{x} - \frac{500}{x+1}\end{align*} Since we’re subtracting two rational expressions, we have to make sure we have a common denominator, and then subtract. In this case, our common denominator is x(x+1)\begin{align*}x(x + 1)\end{align*}, so we do the following: In the picture above, we found the common denominator by making the denominators the same, and then we simplified the numerator. What does this resulting expression mean? We can input some values into a table for this function and make graph the relationship. It would look like this: For this graph, the x\begin{align*}x\end{align*} values represent the number of hours of training, and the y\begin{align*}y\end{align*} values represent the number of injuries per year. This graph clearly shows that that number of injuries has been drastically reduced over time. What if we were to compare the two graphs together separately? What would that graph tell us? See below. The graphs are really close together, so it’s a little difficult to compare. So, let’s take a look at the tables that represent the two relationships. # of hours of training (x)\begin{align*}(x)\end{align*} Without plan With plan 5 100 83.33333333 10 50 45.45454545 15 33.33333333 31.25 20 25 23.80952381 25 20 19.23076923 30 16.66666667 16.12903226 50 10 9.803921569 100 5 4.95049505 200 2.5 2.487562189 Looking at this table, we can see that the plan will definitely make a difference. For every number of hours of training, we can see that the rightmost column has significantly lower numbers. The plan will work! We can also use the formula I0I1=500x500x+1\begin{align*}I_0-I_1= \frac{500}{x}- \frac{500}{x+1}\end{align*} or 500x2+x\begin{align*}\frac{500}{x^2+x}\end{align*} to make predictions of the model, assuming that it’s been proven effective for the past and present. Let’s use 500x2+x\begin{align*}\frac{500}{x^2+x}\end{align*} and find out when this would equal 5 incidents over the course of a year. We can find out how many trainings would be needed so that we shouldn’t have no more than 5 injuries. We set our expression 500x2+x\begin{align*}\frac{500}{x^2+x}\end{align*} equal to 5 and then solve for x\begin{align*}x\end{align*}. When we set this equal to 5, we can look at solving this two different ways. We can look at this in the sense like we’re solving a proportion and 5 is the same as 51\begin{align*}\frac{5}{1}\end{align*}. 500x2+x=51\begin{align*}\frac{500}{x^2+x}=\frac{5}{1}\end{align*} Or, we can solve it like we’re rationalizing the equation and trying to clear the denominator in this rational equation. We would take 500x2+x=5\begin{align*}\frac{500}{x^2+x}=5\end{align*} and multiply both sides of the equation by x2+x\begin{align*}x^2+x\end{align*}. Essentially, either of these two methods will yield the same next step. Let’s work with the proportion 500x2+x=51\begin{align*}\frac{500}{x^2+x}=\frac{5}{1}\end{align*} and multiply both sides of the equation by 1 and x2+x\begin{align*}x^2+x\end{align*}. We now have 500=5x2+5x\begin{align*}500=5x^2+5x\end{align*}. It’s a quadratic! Since we’re trying to find our x\begin{align*}x-\end{align*}values, we can solve this using the quadratic formula after setting the equation equal to zero. Check all of the steps below: After using the quadratic formula, we’ve found two answers. Would both answers make sense in this situation? And what do these answers mean? Since we’re trying to find out how many trainings it would take to ensure that the number of incidents are no more than 5, 9.5 would make sense. Offering -10.5 trainings wouldn’t make sense! ### Extension Investigation Research different companies that would represent a similar type of model. Why would this company have this type of model? Why did you choose this company? Would this model be limited to injuries? What about production of certain goods? ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes
Dropped Text First recall that in Chapter 14 we demonstrated (with some effort it must be said) that the generic form of a 2 × 2 Unitary matrix, M, is: Where a and b ∈ ℂ, θ is an angle with 0 ≤ θ < 2π and |a|2 + |b|2 = 1. We can use this to form our modified matrix, M – λI as follows: We then recall the requirement that the determinant of this matrix is zero: Again the determinant of a 2 × 2 matrix with entries {a, b, c, d} is simply ad – bc, so we have: (a – λ)(eiθ a – λ) + b eiθ b = 0 Multiplying out, we get: a eiθ a – aλ – λ eiθ a + λ2 + b eiθ b = 0 Regrouping: λ2 – λ(a + eiθ a) + (a eiθ a + b eiθ b) = 0 Or: λ2 – λ(a + a eiθ) + (a a eiθ + b b eiθ) = 0          (1) From the definition of Unitary matrices, we know that |a|2 + |b|2 = 1 and from Chapter 14 we can recall that for any complex number, z, we have: z z = |z|2 This means that: a a + b b = 1 Or: b b = 1 – a a Substituting this in (1) above, we get: λ2 – λ(a + a eiθ) + (a a eiθ + (1 – a a) eiθ) = 0 Multiplying out the bracket again we get: λ2 – λ(a + a eiθ) + (a a eiθ + eiθ – a a eiθ) = 0 Cancelling the ± a a eiθ, we get: λ2 – λ(a + a eiθ) + eiθ = 0          (2) Once more in Chapter 14 we also established that for a complex number, z, which forms an angle φ with the x-axis, we have: z = |z| eiφ and z = |z| e– iφ If we substitute both a and a with this formulation in (2), we get: λ2 – λ(|a| eiφ + |a| e– iφ eiθ) + eiθ = 0 Which we can simplify to: λ2 – λ|a|(eiφ + ei(θ – φ)) + eiθ = 0 FROM CHAPTER 21 Back in Chapter 17, we showed that the Cayley-Hamilton Theorem for a 3 × 3 matrix, M3, implies that: M33 – tr(M3)M32 + ½[tr(M3)2 – tr(M32)]M3 – det(M3)I3 = 0 Where I3 is the 3 × 3 identity matrix. If we consider a matrix S ∈ su(3), then by definition its trace is zero, so the terms involving tr(S) in the above drop out and we get: S3 – ½tr(S2)S – det(S)I3 = 0 or S3 = ½tr(S2)S + det(S)I3          (1) Back in Chapter 18, we noted that: tr(M + N) = tr(M) + tr(N)          (2) and tr(αM) = α tr(M)          (3) Where α is a scalar. If we take the trace of both sides of (1), we get: tr(S3) = tr[½tr(S2)S + det(S)I3] Using (2) we note that: tr(S3) = tr[½tr(S2)S] + tr[det(S)I3] Using (3) and noting that the trace of a matrix is a scalar we have that: tr(S3) = ½tr(S2)tr(S) + det(S)tr(I3) Again tr(S) is zero and obviously tr(I3) = 3, so: tr(S3) = 3det(S) or det(S) = tr(S3)/3
A divides b relationship Example 5 - R = {(a, b) : 2 divides a-b} is equivalence relation Given: Two odd numbers a and b. Prove: a+b is even. From definition of odd number, a=2n+1 and b=2k+1 for integers n, k. a+b = 2n+1+2k+1 = 2n + 2k+ 2. Figure given below shows a relationship between the steps P and Q. Write this relation (i) in Let R be the relation on A defined by {(a, b): a ∈ A, a divides b}. 5 Set Theory Exercise 1; 6 Relationships between Sets .. On the left, the sets A and B are disjoint, because the loops don't overlap. . like 'add', 'subtract', ' multiply' and 'divide', so we can combine two sets to form a third set in various ways. Points inside the rectangle represent elements that are in the universal set; points outside represent things not in the universal set. You can think of this rectangle, then, as a 'fence' keeping unwanted things out - and concentrating our attention on the things we're talking about. Other sets are represented by loops, usually oval or circular in shape, drawn inside the rectangle. Again, points inside a given loop represent elements in the set it represents; points outside represent things not in the set. On the right A is a subset of B, because the loop representing set A is entirely enclosed by loop B. Worked Examples[ edit ] Venn diagrams: Note that the rectangle representing the universal set is divided into four regions, labelled i, ii, iii and iv. What can be said about the sets A and B if it turns out that: So A is a subset of B, and the diagram should be re-drawn like Fig 2 above. definition - What does it mean to say "a divides b" - Mathematics Stack Exchange The diagram should then be re-drawn like Fig 1 above. Example 2 a Draw a Venn diagram to represent three sets A, B and C, in the general case where nothing is known about possible relationships between the sets. In each case, the Venn diagram can be re-drawn so that empty regions are no longer included. If region ii is empty, the loop representing A should be made smaller, and moved inside B and C to eliminate region ii. If regions ii, iii and iv are empty, make A and B smaller, and move them so that they are both inside C thus eliminating all three of these regionsbut do so in such a way that they still overlap each other thus retaining region vi. If regions iii and vi are empty, 'pull apart' loops A and B to eliminate these regions, but keep each loop overlapping loop C. Drawing Venn diagrams for each of the above examples is left as an exercise for the reader. Example 3 The following sets are defined: The technique is as follows: Draw a 'general' 3-set Venn diagram, like the one in Example 2. Go through the elements of the universal set one at a time, once only, entering each one into the appropriate region of the diagram. Discrete Mathematics/Set theory Re-draw the diagram, if necessary, moving loops inside one another or apart to eliminate any empty regions. Don't begin by entering the elements of set A, then set B, then C — you'll risk missing elements out or including them twice! The key was that we had played together for a long time. There were a lot of injuries in the buildup — David Beckham was coming back from a metatarsal, Kieron Dyer and Steven Gerrard were both affected by injury and replacements were flying in and out. You want to get rid of all that razzmatazz — a player needs to have played at least a couple of games at the end of the season, and the manager needs to be strong enough to make that call. • ‘On the pitch there was a divide’: tales behind England’s World Cup failures • Example 5 - Chapter 1 Class 12 Relation and Functions • Equivalence Relation on Set I have good memories of that World Cup. I think it was the most well-organised tournament I went to. There were some big distances to travel but the facilities were fantastic, the pitches were first class and I think every detail was taken care of perfectly. My family came out, and nobody had a murmur of complaint. Plus we beat Argentina. Jermaine Jenas Knocked out by Portugal on penalties in quarter-finals We were staying in Baden-Baden and a lot of us had friends and relatives nearby. It felt good at the time. But the way the British media reported the fact that our families were out there and enjoying themselves, the way they blew everything out of proportion, I thought was out of order. My mum was having conversations with journalists, having meals with them — they were staying at the same hotel — but then these reporters were stabbing people in the back and sneaking secret cameras around. Since then the FA have gone down a different route.
# What is 25 percent of 80 + Solution With Free Steps 25 percent of 80 results in 20. The solution may be found by multiplying 80 by the factor 0.25. The calculation of 25 percent of 80 is crucial in many applications. Let us suppose, for example, that you fancied a very stylish hat on an online selling platform. The initial listing of the hat is 80 dollars. Your friend tells you that they are going to offer a 25% discount on the hat in a couple of months time. Now since you know that 25 percent of 80 equals 20, you would have a fair idea that you would save 20 dollars on the purchase. This will help you make an informed decision. In addition to the above hat example, there are many more examples where knowing 25 percent of 80 may be helpful. This article focuses on computation of the 25 percent of 80. ## What Is 25 percent of 80? 25% of 80 is equal to the number 20. This result will be obtained by multiplying the factor 0.25 by 80. By dividing the fractional value of 25/100 by the number 80, the answer to the problem 25 percent of 80 may be calculated. The result, 20, can then be obtained by further reducing 80 x 25/100. Calculating 25 percent of 80 may be done using these straightforward arithmatic procedures. ## How To Calculate 25 percent of 80? The simple technique provided below can be employed to find 25 percent of 80. ### Step 1 Writing 25 percent of 80 in mathematical form: 25 percent of 80 = 25% x 80 ### Step 2 Substitute the % symbol in 25% x 80 with the fraction 1/100: 25 percent of 80 = ( 25 x 1/100 ) x 80 ### Step 3 Rearranging the expression ( 25 x 1/100 ) x 80 gives: 25 percent of 80 = ( 25 x 80) / 100 ### Step 4 Multiplying 25 with 80: 25 percent of 80 = ( 2000 ) / 100 ### Step 5 Dividing 2000 by 100: 25 percent of 80 = 20 Therefore, the 25 percent of 80 is equivalent to 20. We can see that 25 percent of 80 is equal to 20 by using the following graph. Figure 1: Pie Chart of 25 percent of 80 The pie chart above displays two sections. The 25 percent of 80 is represented by the pink slice in the pie chart, which is equal to 20 on an absolute scale. The remaining 75 percent of 80, which is equal to 60 on an absolute scale, is represented by the yellow slice of the pie chart. The 100 percent of 80 is represented by the chart’s overall surface area. Percentage is a way of expressing any value on any scale as a number between zero and hundred. Using different numbers to compare is a wonderful tactic. All the Mathematical drawings/images are created using GeoGebra.
# To calculate: The points needed on the final to avetage the points to 75. Given information: Grades in three tests in College Algebra are 87, 59 and 73. The average after final is 75. To calculate: The points needed on the final to avetage the points to 75. Given information: Grades in three tests in College Algebra are 87, 59 and 73. The average after final is 75. • Questions are typically answered in as fast as 30 minutes ### Solve your problem for the price of one coffee • Math expert for every subject • Pay only if we can solve it Alix Ortiz Calculation: Let the grade in the final test be x Average of n numbers $$\displaystyle{a}_{{{1}}}+{a}_{{{2}}}+{a}_{{{3}}}+\ldots+{a}_{{{n}}}$$ is given by $$\displaystyle{A}={\frac{{{a}_{{{1}}}+{a}_{{{2}}}+{a}_{{{3}}}+\ldots+{a}_{{{n}}}}}{{{n}}}}$$ Plugging the values, $$\displaystyle{A}={\frac{{{87}+{59}+{73}+{x}}}{{{4}}}}$$ Given average is 75 Therefore, $$\displaystyle{75}={\frac{{{87}+{59}+{73}+{x}}}{{{4}}}}$$ Multiplying both sides by 4 $$\displaystyle{75}\cdot{4}={\frac{{{87}+{59}+{73}+{x}}}{{{4}}}}\cdot{4}$$ $$\displaystyle{300}={219}+{x}$$ Subtracting 219 from both the sides, $$\displaystyle{300}—{219}={219}+{x}-{219}$$ $$\displaystyle{81}={x}$$ $$\displaystyle{x}={81}$$ Therefore the score required in final test to get an average of 75 is 81.
Documentation/Calc Functions/ACOS/pt-br This page is a translated version of the page Documentation/Calc Functions/ACOS and the translation is 0% complete. Outdated translations are marked like this. ACOS Mathematical Resumo: Retorna o arco cosseno (inverso do cosseno trigonométrico) de um número. A função calcula o valor principal. ACOS(Number) Retorna: Essa função retorna o cosseno trigonométrico inverso (arco cosseno) de Número, isto é, o ângulo (em radianos) cujo cosseno é Número. O ângulo retornado está entre 0 e PI [0,π]. Argumentos: Number is a real number in the range [-1, 1], or a reference to a cell containing that number, whose inverse trigonometric cosine is to be calculated. • If Number is non-numeric, then ACOS reports a #VALOR! error. • If Number lies outside the range [-1, 1], then ACOS reports a #NÚM! error. • To return the angle in degrees, use the GRAUS function. • Alternatively multiply by 180/PI() to convert the result from radians to degrees. The formula for the inverse trigonometric cosine of x is: $\displaystyle{ \arccos(x) ~=~ \frac {\pi} {2} ~-~ \sum_{n=0}^{\infty } \left ( \frac{(2n)!}{2^{2n}(n!)^2(2n+1)} \right )x^{2n+1} }$ The figure below illustrates the function ACOS. ACOS function Examples: Formula Description Returns =ACOS(−1) Inverse trigonometric cosine of -1. 3,14159265358979 (π) =ACOS(D1) where cell D1 contains the number 0.5. Inverse trigonometric cosine of 0.5. 1,0471975511966 (π/3) =GRAUS(ACOS(0,5)) Inverse trigonometric cosine of 0.5, converted to degrees. 60 =ACOS(−0,5)*180/PI() Inverse trigonometric cosine of -0.5, converted to degrees. 120 =GRAUS(ACOS(COS(RADIANOS(60))) An angle of 60° is converted to radians, its cosine is taken, then the inverse cosine of that value is calculated, and finally the result is converted back to degrees. 60 ACOS
# The Line of Intersection of Two Planes ## Homework Statement Find a set of parametric equations for the line of intersection of the planes. 6x-3y+z=5 and -x+y+5z=5[/B] ## Homework Equations The cross product formula The formula for the parametric equations of a line in three dimensional space: x=x1+at, y=y1+bt, z=z1+ct Knowing the fact that the coefficients in front of the variables in the equation of a plane are the components for the normal vector of the plane ## The Attempt at a Solution [/B] I found the cross product between the normal of the two plane. The resulting vector is going to function as the direction vector for the intersection line. a= the normal vector of the first plane b=the normal vector of the second plane axb=-16i-31j+3k finding a point on the intersection line: We could do this by forming a system of equations. We get rid of on of the variables by saying that at some point the value of z in both planes will be the same and at some point it will be zero. This will give us the system of equations: 6x-3y=5 -x+y=5 Would solve for one of the variable by using the method of elimination. We could multiply the second equation by 6. This will give us: 6x-3y=5 -6x+6y=30 We could solve for y: y=35/3 we could substitute the value of y into the second equation and that will give us a value for x: -x+(35/3)=5 x=20/3 Now, we have the point ((20/3), (35/3), 0) From this point it is really easy to plug the numbers into the parametric equations of a line. I referred to the solution of this problem. We both had the same values for the direction vector, but the point on the line was different. This makes the parametric equations of the intersection line have different values for x1, y1, z1. I was wondering if there is a unique answer to this problem? My thought is that there isn't, because we could choose any point on the line and use it to write the parametric equations of the intersection line. Am at right on this? RUber Homework Helper You are correct that any point plus a direction vector defines a unique line. The point itself is in no way unique, and may only be an indicator of which technique was used to find it. If you are looking for additional or alternative methods, it may be worth trying to understand how the textbook solution was derived. HallsofIvy
## Recursive expectation (part 2) This post follows the previous recursive expectation post. ### Problem: There is an equilateral triangle ABC and the bug is in the vertex A. The bug can move only along edges of the triangle. Once it starts going along an edge, it doesn’t change its moving direction until it reaches the next vertex. It starts walking from the vertex A. It takes bug 1 minute to walk between any two vertices. Every time it comes to a new vertex, it makes a random choice about which of other two vertices to go to next. What is the expected time for the bug to return to the starting vertex A? ### Solution 1: First, completely intuitive solution, without (much) mathematics. It takes one minute for the bug to come to the edge BC. From that edge, each minute there is 1/2 chance that the bug will return to A. Basically we start the experiment that has 1/2 chance of success each time. So, according to the previous post, expected time to get back to A from BC will be 2 minutes, so all together with the first minute, we get the final result – 3 minutes. ### Solution 2: And now, let’s try to do it pure mathematical way, without much thinking. Possible solutions for number of minutes (length of trip) to return back to A are 2, 3, 4, … until infinity. If we take one of the possible solutions, let’s say n minutes, how many different paths do we have for this to happen? Basically, we can go from A to B and then go n-2 times between B and C and then the last time, we go back to A, either from B or from C. Another path would be from A to C and then n-2 times between B and C and then the last time, we go back to A. So, for each n, number of possible paths is always 2. Probability for each path of length n is: $\left ( \frac{1}{2} \right )^n$ So, probability for n-minutes solution to happen is: $P\left ( n \right ) = \left ( \frac{1}{2} \right )^n \cdot 2 = \left ( \frac{1}{2} \right )^{n-1}$ Now, the solution for the expectation becomes: $\begin{array}{rcl} E\left ( X \right ) & = & \sum_{n=2}^{\infty }n\cdot\left ( \frac{1}{2}\right )^{n-1} \\ & = & \sum_{n=0}^{\infty }\left [n\cdot\left ( \frac{1}{2}\right )^{n-1} \right ] - 1 \\ & = & \frac{1}{\left ( 1 - \frac{1}{2}\right )^2} - 1 \\ & = & 4 - 1 \\ & = & 3 \end{array}$ which requires some mathematical knowledge, but doesn’t require much of the intuition. ### Solution 3: This one will actually use recursive expectation (please see the previous post). Expected time from A back to A is equal to expected time from A to edge BC + expected time from edge BC back to A. Expected time from A to BC is 1. Expected time from BC to A (following the logic from the previous post) is: $\begin{array}{rcl} E\left ( BC\rightarrow A \right ) & = & \frac{1}{2}\cdot1+\frac{1}{2}\left ( 1 + E\left ( BC\rightarrow A \right )\right ) \\ \frac{1}{2} \cdot E\left ( BC\rightarrow A \right ) & = & 1 \\ E\left ( BC\rightarrow A \right ) & = & 2 \end{array}$ So, the overall expectation is: $\begin{array}{rcl} E\left ( A\rightarrow A \right ) & = & E\left ( A \rightarrow BC \right ) + E\left ( BC\rightarrow A \right ) \\ & = & 1 + 2 \\ & = & 3 \end{array}$ Recursive expectation once again, provides much easier solution than the mathematical one. Additionally, it’s much more intuitive. ## Monster fly Now I understand my daughter when she asks me: “Daddy, why is your face looking like a monster while swimming?” ## Rolling dice expectation The day after my previous expectation related post, I kind of realised what lies behind intuitive answer to the question: “What is the expected number of times we need to roll a dice to get ‘6’ for the first time?” The question above is equivalent to the following. Let’s roll the dice large number of times. We will get long sequence, like: 3, 4, 6, 5, 6, 1, 2, 4, 3, 5, 1, 4, 5, 6, 6, … Next, let’s cut the tail off after the last ‘6’. Let’s mark the length of the remaining sequence with N. Now we need to count “distances” between every two consecutive ‘6’s. In our example above, we will get 3, 2, 9, 1, … Distance is defined as the number of rolls between two consecutive ‘6’s. Since ‘6’s are at positions: 3, 5, 14, 15, … in the sequence above, distances are 3, 5-3=2, 14-5=9, 15-14=1, … The answer to the question, about the expected number of times, is the same as the average distance between every two consecutive ‘6’s. Let’s mark distances with the sequence: $d_1, d_2, d_3, \dots d_m$. $m$ is the total number of distances. Basically, $m$ will correspond to the number of experiments performed. The cut off tail is unfinished last experiment, which we decided to ignore. If X is the number of tries to get ‘6’ for the first time, we can write: $E\left ( X \right ) = \frac{1}{m}\sum_{i=1}^{m}d_i$ Furthermore, sum of all distances must be equal to the total number of tries – N. $\sum_{i=1}^{m}d_i=N$ Since $m$ is the number of distances, it is also the number of ‘6’s in the whole sequence. Since there is no reason for any number on the dice to appear (by expectation) different number of times than any other number (on the dice), we expect for $m$ to be: $m = \frac{N}{6}$ If we replace last two equations in the formula for E(X), we get: $\begin{array}{rcl} E\left ( X \right ) & = & \frac{1}{\frac{N}{6}} \cdot N \\ & = & \frac{1}{\frac{1}{6}} \\ & = & 6 \end{array}$ and no mathematics knowledge was needed for this proof. So, maybe this is what lies behind the instinctive answer. If we write down the results of each experiment (experiment is said to be finished once we get ‘6’ for the first time): $\begin{array}{rl} E_1: & 3, 4, 6 \\ E_2: & 5, 6 \\ E_3: & 1, 2, 4, 3, 5, 1, 4, 5, 6\\ E_4: & 6 \\ \cdots \\ E_m: & 1, 5, 3, 4, 6 \end{array}$ Since there are $m$ experiments, we have exactly $m$ 6’s, but we also expect to have each other dice number also $m$ times, so the sum of all lengths of all experiments is expected to be $6 \cdot m$, which makes average length to be $6$. # High hips Some months ago, I was so desperate since I couldn’t push my hips above the water (after first kick). In fact, it (hips up) was happening occasionally, but most of the time they would come up very close to the surface, but wouldn’t break it. Then, completely by chance, I did a video of myself where I swim fly with much higher pace than usually. Usually, my stroke cycle is longer than 2 seconds, sometimes even 3 seconds when I swim really slowly. I knew it was much slower than race pace which is slightly above 1 second per stroke cycle. So, I made a video and I was very surprised. My hips were so unexpectedly moving nicely (for my standards) above the water :). So, I had to come up with some conclusions. My first guess was that there was a relationship between the time distance between the second and the first kick and hips height. During the recovery, swimmer’s center of mass reaches its highest point. In this phase of the stroke, especially if it’s a no breathing one, arms, head and shoulders are above the water surface, so buoyancy is weaker than the gravity force, which means that at some moment, center of mass will start falling down. Usually it happens during recovery, that’s why recovery cannot be slowed down significantly. This falling down will not happen in a moment, it is a gradual process. And it will even continue, by inertia, below the equilibrium point. So, if we wait for too long, center of mass could dive deep and we wouldn’t be able to raise our hips above the water surface. But if we kick while we didn’t sink completely yet, we could still have a chance to put our hips above. This also can explain why it’s much easier to keep the hips high when swimming only with legs. Or one arm drill. When swimming one arm drill, only one arm is out of water, half of head, one shoulder, plus extended arm can create some upforce as well by moving it downwards together with the upper body, or even relative to the upper body if we need more help. It’s harder with the arm next to the body, since it can’t create upforce in that position, unless moved in the opposite direction than during the real stroke. But this theory doesn’t stand since I learnt how to swim butterfly and lift my hips above the water even when swimming very slow pace. In this “drill” I try to emphasize hips breaking the surface by stopping the moment when they are above the surface. I can even feel it on my lower back when I break the surface. Talking about slow pace butterfly, I gradually came up with a version of butterfly with slightly modified timing that could literally be swum effortlessly for very long time. This version of butterfly allows even for the recovery to be prolonged. It allows for the swimmer to highly concentrate on every single part of the stroke. And, what is very important, it’s very easy. This is the same exercise underwater: The point is to take the arms slowly out of the water after the head is already back into the water. So, there is no moment when both head and arms are above the water which keep our buoyancy in “good” range. Recovery could be very slow too. Second kick is very close to the first one. Basically, both of them happen while the arms are fully extended in gliding position. There is a variation when gliding is quite short and second kick happens almost at the right time, which is slightly harder. Head goes out of water while arms are doing underwater stroke, thus giving necessary lift for the head. I experimented also with another version of this exercise, with two first kicks. First first kick happens when head enters the water, the second first kick happens when arms enter the water. There is only one second kick. It’s harder since first first kick brings the upper body down (since there are no extended arms to prevent it), so recovery somehow starts with shoulders little bit deeper in the water than usually. Although it could be good exercise for shoulders flexibility In conclusion, it could be that all of above have something to do with how high hips go. Firstly, if we don’t go too high above the water when breathing, we will sink less afterwards. So, soft head and arms entry is very helpful in that aspect. Secondly, if the pace is high, body doesn’t have time to sink as much as during the low pace butterfly. Thirdly, most probably, lower back flexibility helps it a bit. And what I think helped me a lot was my shallow water drills since they taught me how to swim softly. Shallow water exercises taught me how not to kick too deep (with legs), how to “jump back” into the water softly and how to swim butterfly (almost) effortlessly. ## Recursive expectation I’ll start this with an example: What is the expected number of tries to get ‘6’ when rolling dice? Or even simpler, experiment is performed where the dice is thrown as many times as necessary to get ‘6’ for the first time. Save the number of throws for this experiment. Repeat the same experiment many times. What is the average value of throws in all experiments? Some people would immediately say: “6”. Why 6? Because the probability to get ‘6’ on each throw is 1/6, so the expected number of throws will be: $\frac{1}{\frac{1}{6}}=6$ Although intuitively everything looks fine, formal proof is far from trivial: $\begin{array}{rcl} E\left ( X \right ) & = & \sum_{n=1}^{\infty}n\cdot P\left ( X=n \right ) \\ & = & \sum_{n=1}^{\infty}n\cdot\frac{1}{6}\cdot\left ( \frac{5}{6}\right )^{n-1} \\ & = & \frac{1}{5}\sum_{n=1}^{\infty}n\cdot\left ( \frac{5}{6}\right )^n \\ & = & \frac{1}{5}\frac{\frac{5}{6}}{\left ( 1-\frac{5}{6}\right )^2} \\ & = & 6 \end{array}$ but one must be more than intermediate mathematician to be able to go through this proof. I will not go through the details since it’s not the purpose of this post. There is also a very intuitive solution as well, for which you don’t need to be some mathematician: $\begin{array}{rcl} E\left ( X \right ) & = & \frac{1}{6}\cdot 1+ \frac{5}{6}\cdot \left ( 1+E\left ( X \right ) \right )\end{array}$ and it’s fairly easy to come to solution from this simple equation. What does this equation say? It says that there is 1/6 chance that our experiment will be finished after only one throw (we get 6 from the first try), or there is 5/6 chance that it will be longer. In this case it will be 1 throw longer than what we expected before the first throw. This is probably the hardest part to imagine. If we don’t get 6 from the first throw, we are basically at the same place where we were before we even started the experiment. The only difference is that counter is already on 1 and not on zero as it was before we started the experiment. To put it even simpler. Let’s say that we do this experiment 600,000 times. 1/6 of them will be finished after first throw, because 1 in 6 times, we’ll get 6 in the first attempt. So, there will be 100,000 experiments out of 600,000 times for which the result will be 1 throw. The rest of 500,000 experiments will have, by average, one throw more than the average in all 600,000 experiments. What is then the average for all 600,000 experiments. $E=\frac{100,000 \cdot 1+500,000 \cdot \left ( E+1 \right )}{600,000}$ which is basically the same as formula above. It’s very easy to get the result $E=\6$. ## Generalization If we keep on trying succeeding the event that has probability p, how many times is expected to try it until the first success? $\begin{array}{rcl} E\left ( X \right ) & = & p \cdot 1 + \left ( 1 - p \right ) \cdot \left ( E\left ( X \right ) + 1 \right ) \\ E\left ( X \right ) \cdot \left ( 1 - \left ( 1 - p \right ) \right ) & = & p + 1 - p \\ E\left ( X \right ) & = & \frac{1}{p} \end{array}$ ## Programming solution Following c# code will prove the dice problem using Monte Carlo simulation. using System; namespace RecursiveExpectation { class Program { static void Main() { var r = new Random((int)DateTime.Now.Ticks); var total = 0; // repeat experiment 10000 times for (var game = 1; game <= 10000; game++) { // number of throws in the current game var throws = 0; // current number on the dice var dice = 0; // roll the dice until we get '6' while (dice != 6) { // choose the random number between 1 and 6 dice = r.Next(1, 7); // increase the number of throws throws++; } // total sum of throws in all games total += throws; // calculate average number of throws in // all experiments (games) var avg = (double)total / game; Console.WriteLine(avg); } Console.WriteLine("Finished");
Imagine you lived in a city which was laid out in a grid system, a bit like some American cities are. If you wanted to get to your friend’s house you might measure the distance, not as the crow flies, but as the distance you have to walk along the streets. If you were to do this you would be using a special kind of geometry called ‘Taxicab Geometry’. The normal kind of geometry we use at school is called Euclidean Geometry. In Euclidean Geometry you measure the distance between two points as being the direct distance as the crow flies, whereas in Taxicab Geometry you are confined to moving along the lines of a grid. Look at the diagram below. Using traditional Euclidean Geometry and Pythagoras’ theorem we measure the distance between A and B (the green line) as being 8.49 (2dp). If however we can only walk along the streets between our points, then the Taxicab distance between A and B is 12 units. Some possible paths of 12 units in length are illustrated below in red, blue and yellow. Taxicab geometry gets its name from the fact that taxis can only drive along streets, rather than moving as the crow flies. Euclidian Distance between A and B as the crow flies: 8.49units (Green). Taxicab Distance between A and B: 12 units (Red,Blue and Yellow). ### Formal definition of the Taxicab Distance. Suppose you have two points and then: Taxicab Distance between and The function which is shown with two straight lines is called the modulus and means that we take the positive value of whatever is inside it. So for example the value of is just 3 and the value of is also just 3. ### Example The Taxicab distance between and would be ### The Taxicab World Now we can imagine that we live in a world where everything is measured using Taxicab distances. What would familiar shapes look like? It is quite surprising when we start to investigate. ### The Circle in the Taxicab world We can define a circle to be the set of points which are a constant distance from a centre. For example if we take all the points which are a distance of 4 units from a point A, then we have a circle of radius 4 with a centre at point A. In the Taxicab world this turns out not to look like a circle but a square! If you look at the red points on the diagram on the right then they are all 4 Taxicab units from the blue centre point using the Taxicab distance. ### Value of π in the Taxicab world If we take π to be circumference diameter then in the Taxicab circle above we would calculate the circumference to be 32 and the diameter to be 8, giving a value for π as 4. This turns out to hold for all Taxicab circles! ### The perpendicular line bisector When using the Euclidean measure of distance the perpendicular line bisector is simple to draw. It is defined as being the set of points which are an equal distance from two points. In the diagram on the left, all the points on the dotted line are an equal distance from points A and B: Using the Taxicab measure, the perpendicular line bisector is a really surprising shape. The points on the line on the right are an equal distance from the two points using the Taxicab measure of distance. However the configuration above is just one possibility. Have a play around with the points A and B in the geogebratube interactivity below and you should find a surprising result. There are some places for A and B which produce a perpendicular line bisector which is not just a line, it also includes a whole region of points! Can you find them? Created with GeoGebra – Shared by Steve PhelpsShare or copy One of the wonderful things about Taxicab geometry is that you can keep on investigating all manner of shapes and geometrical properties. You might want to investigate what an ellipse might look like or even a parabola within the Taxicab world. If you would like to have a go at completing a worksheet on Taxicab Geometry, one can be found here. There are even some useful applications to Taxicab Geometry with some researchers using it when modelling the spread of fire within a grid based system. Article by Hazel Lewis
# Math Snap ## Question 18 1 Point A wall $3.12 \mathrm{~m}$ high is $3.75 \mathrm{~m}$ away from a building. What is the length in meters of the shortest ladder that can reach the building with one end resting on the ground outside the wall. Express the answer in $m$ rounded to two decimal places. Do not include the unit. Add your answer #### STEP 1 Assumptions 1. The height of the wall is $3.12 \, \text{m}$. 2. The distance from the wall to the building is $3.75 \, \text{m}$. 3. The ladder rests on the ground outside the wall and reaches the top of the building. 4. The ladder forms a right triangle with the wall and the ground. #### STEP 2 To find the length of the shortest ladder, we need to calculate the hypotenuse of a right triangle formed by the height of the wall, the distance from the wall to the building, and the ladder itself. $\text{Hypotenuse}^2 = \text{Height}^2 + \text{Distance}^2$ #### STEP 3 Now, plug in the given values for the height of the wall and the distance from the wall to the building to calculate the hypotenuse. $\text{Hypotenuse}^2 = (3.12 \, \text{m})^2 + (3.75 \, \text{m})^2$ #### STEP 4 Calculate the squares of the height and the distance. $\text{Hypotenuse}^2 = (3.12)^2 + (3.75)^2$ $\text{Hypotenuse}^2 = 9.7344 + 14.0625$ #### STEP 5 Add the squared values. $\text{Hypotenuse}^2 = 9.7344 + 14.0625 = 23.7969$ #### STEP 6 Take the square root of the sum to find the length of the ladder. $\text{Hypotenuse} = \sqrt{23.7969}$ #### STEP 7 Calculate the square root. $\text{Hypotenuse} = \sqrt{23.7969} \approx 4.8787$ ##### SOLUTION Round the result to two decimal places. $\text{Hypotenuse} \approx 4.88$ The shortest ladder that can reach the building with one end resting on the ground outside the wall is approximately $4.88 \, \text{m}$.
 Cramer's Rule 3x3 Step by Step # Example of Solving a System of Linear Equations by Cramer's Rule. ### This solution was made using the calculator presented on the site. It is necessary to solve the system of linear equations using Cramer's rule. - 2 x1 + x2 + x3 = -13 - x1 + x2 + 2 x3 = -9 3 x1 + x2 + x3 = 12 Let's write the Cramer's rule: x1 = det A1 / det A x2 = det A2 / det A x3 = det A3 / det A It is impossible to divide by zero. Therefore, if the determinant of A is zero, then it is impossible to use Cramer's rule. The determinant A consists of the coefficients of the left side of the system. - 2 x1 + x2 + x3 = -13 - x1 + x2 + 2 x3 = -9 3 x1 + x2 + x3 = 12 det A = -2 1 1 = -1 1 2 3 1 1 The elements of row 1 multiplied by -1 are added to the corresponding elements of row 3.   more info -2 1 1 -1 1 2 3 + ( -2) * ( -1) 1 + 1 * ( -1) 1 + 1 * ( -1) This elementary transformation does not change the value of the determinant. = -2 1 1 = -1 1 2 5 0 0 -2 1 1 -1 1 2 5 0 0 Row number 3 Column number 1 Element Row 3 and column 1 have been deleted ( -1) 3 + 1 * 5 * 1 1 1 2 -2 1 1 -1 1 2 5 0 0 Row number 3 Column number 2 Element Row 3 and column 2 have been deleted ( -1) 3 + 2 * 0 * -2 1 -1 2 -2 1 1 -1 1 2 5 0 0 Row number 3 Column number 3 Element Row 3 and column 3 have been deleted ( -1) 3 + 3 * 0 * -2 1 -1 1 Products are summed. If the element is zero than product is zero too. = ( -1) 3 + 1 * 5 * 1 1 = 1 2 = 5 * 1 1 = 1 2 = 5 * ( 1 * 2 - 1 * 1 ) = = 5 * ( 2 - 1 ) = = 5 The determinant A is not zero. It is possible to use the Cramer's rule. It is necessary to change column 1 in determinant A to the column of the right side of the system. System det A det A1 - 2 x1 + x2 + x3 = -13 - x1 + x2 + 2 x3 = -9 3 x1 + x2 + x3 = 12 -2 1 1 -1 1 2 3 1 1 -13 1 1 -9 1 2 12 1 1 det A1 = -13 1 1 = -9 1 2 12 1 1 The elements of row 1 multiplied by -1 are added to the corresponding elements of row 3.   more info -13 1 1 -9 1 2 12 + ( -13) * ( -1) 1 + 1 * ( -1) 1 + 1 * ( -1) This elementary transformation does not change the value of the determinant. = -13 1 1 = -9 1 2 25 0 0 -13 1 1 -9 1 2 25 0 0 Row number 3 Column number 1 Element Row 3 and column 1 have been deleted ( -1) 3 + 1 * 25 * 1 1 1 2 -13 1 1 -9 1 2 25 0 0 Row number 3 Column number 2 Element Row 3 and column 2 have been deleted ( -1) 3 + 2 * 0 * -13 1 -9 2 -13 1 1 -9 1 2 25 0 0 Row number 3 Column number 3 Element Row 3 and column 3 have been deleted ( -1) 3 + 3 * 0 * -13 1 -9 1 Products are summed. If the element is zero than product is zero too. = ( -1) 3 + 1 * 25 * 1 1 = 1 2 = 25 * 1 1 = 1 2 = 25 * ( 1 * 2 - 1 * 1 ) = = 25 * ( 2 - 1 ) = = 25 It is necessary to change column 2 in determinant A to the column of the right side of the system. System det A det A2 - 2 x1 + x2 + x3 = -13 - x1 + x2 + 2 x3 = -9 3 x1 + x2 + x3 = 12 -2 1 1 -1 1 2 3 1 1 -2 -13 1 -1 -9 2 3 12 1 det A2 = -2 -13 1 = -1 -9 2 3 12 1 -2 -13 1 -1 -9 2 3 + ( -2) 12 + ( -13) 1 + 1 This elementary transformation does not change the value of the determinant. = -2 -13 1 = -1 -9 2 1 -1 2 The elements of row 3 multiplied by 2 are added to the corresponding elements of row 1.   more info -2 + 1 * 2 -13 + ( -1) * 2 1 + 2 * 2 -1 -9 2 1 -1 2 This elementary transformation does not change the value of the determinant. = 0 -15 5 = -1 -9 2 1 -1 2 0 -15 5 -1 + 1 -9 + ( -1) 2 + 2 1 -1 2 This elementary transformation does not change the value of the determinant. = 0 -15 5 = 0 -10 4 1 -1 2 0 -15 5 0 -10 4 1 -1 2 Row number 1 Column number 1 Element Row 1 and column 1 have been deleted ( -1) 1 + 1 * 0 * -10 4 -1 2 0 -15 5 0 -10 4 1 -1 2 Row number 2 Column number 1 Element Row 2 and column 1 have been deleted ( -1) 2 + 1 * 0 * -15 5 -1 2 0 -15 5 0 -10 4 1 -1 2 Row number 3 Column number 1 Element Row 3 and column 1 have been deleted ( -1) 3 + 1 * 1 * -15 5 -10 4 Products are summed. If the element is zero than product is zero too. = ( -1) 3 + 1 * 1 * -15 5 = -10 4 = -15 5 = -10 4 = -15 * 4 - 5 * ( -10) = = -60 + 50 = = -10 It is necessary to change column 3 in determinant A to the column of the right side of the system. System det A det A3 - 2 x1 + x2 + x3 = -13 - x1 + x2 + 2 x3 = -9 3 x1 + x2 + x3 = 12 -2 1 1 -1 1 2 3 1 1 -2 1 -13 -1 1 -9 3 1 12 det A3 = -2 1 -13 = -1 1 -9 3 1 12 The elements of row 2 multiplied by -1 are added to the corresponding elements of row 1.   more info -2 + ( -1) * ( -1) 1 + 1 * ( -1) -13 + ( -9) * ( -1) -1 1 -9 3 1 12 This elementary transformation does not change the value of the determinant. = -1 0 -4 = -1 1 -9 3 1 12 The elements of row 2 multiplied by -1 are added to the corresponding elements of row 3.   more info -1 0 -4 -1 1 -9 3 + ( -1) * ( -1) 1 + 1 * ( -1) 12 + ( -9) * ( -1) This elementary transformation does not change the value of the determinant. = -1 0 -4 = -1 1 -9 4 0 21 -1 0 -4 -1 1 -9 4 0 21 Row number 1 Column number 2 Element Row 1 and column 2 have been deleted ( -1) 1 + 2 * 0 * -1 -9 4 21 -1 0 -4 -1 1 -9 4 0 21 Row number 2 Column number 2 Element Row 2 and column 2 have been deleted ( -1) 2 + 2 * 1 * -1 -4 4 21 -1 0 -4 -1 1 -9 4 0 21 Row number 3 Column number 2 Element Row 3 and column 2 have been deleted ( -1) 3 + 2 * 0 * -1 -4 -1 -9 Products are summed. If the element is zero than product is zero too. = ( -1) 2 + 2 * 1 * -1 -4 = 4 21 = -1 -4 = 4 21 = -1 * 21 - ( -4) * 4 = = -21 + 16 = = -5 Result: x1 = det A1 / det A = 25/5 = 5 x2 = det A2 / det A = -10/5 = -2 x3 = det A3 / det A = -5/5 = -1
# Differentiate the following functions with respect to x : Question: Differentiate the following functions with respect to $x$ : $\log \left\{\cot \left(\frac{\pi}{4}+\frac{x}{2}\right)\right\}$ Solution: Let $y=\log \left\{\cot \left(\frac{\pi}{4}+\frac{x}{2}\right)\right\}$ On differentiating $y$ with respect to $x$, we get $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left[\log \left\{\cot \left(\frac{\pi}{4}+\frac{\mathrm{x}}{2}\right)\right\}\right]$ We know $\frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{x})=\frac{1}{\mathrm{x}}$ $\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\cot \left(\frac{\pi}{4}+\frac{x}{2}\right)} \frac{\mathrm{d}}{\mathrm{dx}}\left[\cot \left(\frac{\pi}{4}+\frac{\mathrm{x}}{2}\right)\right]$ [using chain rule] $\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\tan \left(\frac{\pi}{4}+\frac{\mathrm{x}}{2}\right) \frac{\mathrm{d}}{\mathrm{dx}}\left[\cot \left(\frac{\pi}{4}+\frac{\mathrm{x}}{2}\right)\right]$ We have $\frac{\mathrm{d}}{\mathrm{dx}}(\cot \mathrm{x})=-\operatorname{cosec}^{2} \mathrm{x}$ $\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\tan \left(\frac{\pi}{4}+\frac{\mathrm{x}}{2}\right)\left[-\operatorname{cosec}^{2}\left(\frac{\pi}{4}+\frac{\mathrm{x}}{2}\right) \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\pi}{4}+\frac{\mathrm{x}}{2}\right)\right]$ $\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\tan \left(\frac{\pi}{4}+\frac{\mathrm{x}}{2}\right) \operatorname{cosec}^{2}\left(\frac{\pi}{4}+\frac{\mathrm{x}}{2}\right)\left[\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\pi}{4}\right)+\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{x}}{2}\right)\right]$ $\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\tan \left(\frac{\pi}{4}+\frac{\mathrm{x}}{2}\right) \operatorname{cosec}^{2}\left(\frac{\pi}{4}+\frac{\mathrm{x}}{2}\right)\left[\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\pi}{4}\right)+\frac{1}{2} \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})\right]$ However, $\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})=1$ and derivative of a constant is 0 . $\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\tan \left(\frac{\pi}{4}+\frac{\mathrm{x}}{2}\right) \operatorname{cosec}^{2}\left(\frac{\pi}{4}+\frac{\mathrm{x}}{2}\right)\left[0+\frac{1}{2} \times 1\right]$ $\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{1}{2} \tan \left(\frac{\pi}{4}+\frac{\mathrm{x}}{2}\right) \operatorname{cosec}^{2}\left(\frac{\pi}{4}+\frac{\mathrm{x}}{2}\right)$ $\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{1}{2} \times \frac{\sin \left(\frac{\pi}{4}+\frac{x}{2}\right)}{\cos \left(\frac{\pi}{4}+\frac{x}{2}\right)} \times \frac{1}{\sin ^{2}\left(\frac{\pi}{4}+\frac{x}{2}\right)}$ $\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{1}{2 \sin \left(\frac{\pi}{4}+\frac{\mathrm{x}}{2}\right) \cos \left(\frac{\pi}{4}+\frac{\mathrm{x}}{2}\right)}$ $\Rightarrow \frac{d y}{d x}=-\frac{1}{\sin \left[2\left(\frac{\pi}{4}+\frac{x}{2}\right)\right]}[\because \sin 2 \theta=2 \sin \theta \cos \theta]$ $\Rightarrow \frac{d y}{d x}=-\frac{1}{\sin \left(\frac{\pi}{2}+x\right)}$ $\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{1}{\cos \mathrm{x}}\left[\because \sin \left(90^{\circ}+\theta\right)=\cos \theta\right]$ $\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=-\sec \mathrm{x}$ Thus, $\frac{d}{d x}\left[\log \left\{\cot \left(\frac{\pi}{4}+\frac{x}{2}\right)\right\}\right]=-\sec x$
# 9. POLYNOMIALS. Example 1: The expression a(x) = x 3 4x 2 + 7x 11 is a polynomial in x. The coefficients of a(x) are the numbers 1, 4, 7, 11. To view this video please enable JavaScript, and consider upgrading to a web browser that supports HTML5 video Save this PDF as: Size: px Start display at page: Download "9. POLYNOMIALS. Example 1: The expression a(x) = x 3 4x 2 + 7x 11 is a polynomial in x. The coefficients of a(x) are the numbers 1, 4, 7, 11." ## Transcription 1 9. POLYNOMIALS 9.1. Definition of a Polynomial A polynomial is an expression of the form: a(x) = a n x n + a n-1 x n a 1 x + a 0. The symbol x is called an indeterminate and simply plays the role of a place marker. The role of the x is to provide positions in the expression which can be replaced (substituted) by a value. The numbers a 0, a 1,... are called the coefficients of the polynomial. Example 1: The expression a(x) = x 3 4x 2 + 7x 11 is a polynomial in x. The coefficients of a(x) are the numbers 1, 4, 7, 11. The powers of x in a polynomial must be non-negative integers and there must be a finite number of terms. Other expressions have the appearance of being polynomials but, because some powers are negative or fractional, or because there are infinitely many terms, they are not considered to be polynomials. Example 2: The expressions x + 1 x 1 x + x 2 and 1 + x + x2 2! + x3 3! +... are not polynomials even though they are built up from powers of x. The coefficients can be rational numbers (forming the set Q), real numbers (forming the set R), or complex numbers (forming the set C). These systems, called fields, have the property that x 1 exists for every non-zero x. (This somewhat loose description will do for our present purposes. An exact definition comprises 11 separate properties, or axioms.) Sometimes we shall restrict our attention to polynomials with integer coefficients even though the integers don't form a field. Notice that a rational polynomial can always be converted to an integer polynomial by multiplying by a common denominator. Example 3: The rational polynomial x3 2 x is equal to 88x3 70x It is also possible to consider polynomials whose coefficients are integers modulo p for some prime number. These systems are fields and the theory of polynomials over these finite fields works nicely even if some of the results look a bit strange. For example over Z 2, the field of integers mod 2 (where there are just two elements, 0 and 1 with 1 + 1=0) the factorisation of x = (x + 1) 2, since the 2x term is equal to 2 9.2. Degree of a Polynomial The degree of a polynomial is the largest power of x that occurs with a non-zero coefficient. That is, if a(x)=a n x n + a n 1 x n a 1 x + a 0, the degree of a(x) is n (provided that a n 0). We can write deg a(x) = n. Examples 4: Polynomials of degree 2 are quadratics, of the form ax 3 + bx+c (where a 0). Polynomials of degree 1 are the so-called linear polynomials such as 2x + 3 and ½x ¼. Polynomials of degree 0 are the non-zero constant polynomials. It might seem strange for numbers such as 3 or ½ to be considered as polynomials, but they can. If it makes you feel better you can write 3 as 0x 2 + 0x + 3. (Remember that perhaps you once found it strange to call 3 a fraction until you learnt that it could be written as 3 1.) There is one polynomial for which the degree remains undefined. It is the zero constant polynomial, 0. (Some books define this degree to be for certain technical reasons but we shall leave it undefined.) The coefficient of x n, for a polynomial of degree n, is called its leading coefficient. For example the leading coefficient of 3x 2 x + 5 is 3. But beware. The leading coefficient does not always come first the leading coefficient of 1 x 2 is 1, not 1. A monic polynomial is one where the leading coefficient is 1. Clearly every non-zero polynomial can be made monic by dividing it by its leading coefficient. Example 5: The polynomial 4x 3 8x +1 has degree 3. Its leading coefficient is 4 and so it is not monic. However it can be expressed as 4 times the monic polynomial x 3 2x + ¼. If F is a field (for example F might be Q, the system of rational numbers) we denote the set of all polynomials with coefficients coming from F by the symbol F[x]. Examples 6: For example R[x] contains the polynomial x 2 2x + 3 but not x 2 + i. However x 2 + i is in C[x] Addition and Multiplication of Polynomials Polynomials are added (and subtracted) in the usual way just add (or subtract) the corresponding coefficients. Multiplication is somewhat more complicated to describe abstractly, yet it just amounts to expanding the product of the expressions in the way you have always done. For example (ax 2 + bx + c)(dx + e) = adx 3 + (ae + bd)x 2 + (be + cd)x + ce. With these operations the system F[x] behaves very much like a field itself, but with one very important difference. In a field nearly every number has an inverse under multiplication (in fact 0 is the only exception). Most polynomials, on the other hand, do not have inverses. For example, since 1 x and 1 1 x are not polynomials, the polynomials x and 1 x do not have (polynomial) inverses. In fact the only polynomials with inverses are the non-zero constant polynomials such as 2 (whose inverse is the constant polynomial ½). 102 3 1 Now it is possible to write 1 x as 1 + x + x but although this looks like a polynomial it has infinitely many terms while a polynomial, by definition, has only finitely many. An expression like 1 + x + x is called a power series. The system F[x] of polynomials over a field, in fact, behaves much more like the system of integers (where only ±1 have integer inverses). Theorem 1: For polynomials a(x), b(x) F [x], where F is a field: deg [a(x)b(x)] = deg a(x) + deg b(x) Proof: This follows from the fact that (a m x m +... )(b n x n +... ) = am b n x m+n +..., and the fact that if a m and b n are non zero then a m b n is also non-zero. Thus the degree of a product is the sum of the degrees. Does this remind you of something? The degree function behaves like the logarithm function. Theorem 2: For polynomials a(x), b(x) F[x] deg [a(x) + b(x)] max[deg a(x), deg b(x)). There is nothing really surprising in this result except for the question as to why lessthan-or-equals rather than just equals. The answer is that when we add two polynomials of the same degree the leading coefficients can cancel thereby producing a polynomial of lower degree. Example 7: If a(x) = 2x 2 x + 7 and b(x) = 2x 2 + 4x + 1 then a(x) + b(x) = 3x + 8, which has smaller degree than either a(x) or b(x) Division and Remainders As mentioned earlier, one polynomial does not usually divide exactly into another. Like the system of integers we are usually left with a remainder. We get exact divisibility precisely when the remainder is zero. Furthermore this remainder, when it is not zero, is in some sense smaller than whatever we are dividing by. For polynomials, smaller means of smaller degree. The process of obtaining the remainder on dividing one polynomial by another is very similar to the familiar long-division algorithm. Example 8: 2x + 4 x 2 2x + 7 ) 2x 3 + 5x 3 2x 3 4x 2 +14x 4x 2 9x 3 4x 2 8x + 28 x 4 From this calculation we compute the remainder on dividing 2x 3 + 5x 3 by x 2 2x + 7 to be x 31. Note that the remainder has lower degree than x 2 2x + 7, the polynomial we are dividing by. Note also how we write the terms neatly underneath others of the same degree. The result of the calculation can also be expressed as: 2x 3 + 5x 3 = (x 2 2x +7)(2x + 4) + ( x 31). Theorem 3 (Division Algorithm): If a(x), b(x) are polynomials and b(x) is non-zero then a(x) = b(x)q(x) + r(x) for some polynomials q(x) and r(x) where either r(x) = 0 or deg r(x) < deg b(x). The polynomial q(x) is called the quotient and r(x) is called the remainder. If the remainder on dividing a(x) by b(x) is zero we say that b(x) divides a(x), or that a(x) is a multiple of b(x). If we can't be bothered saying it in words we just write b(x) a(x) and read it as b(x) divides a(x). Example 9: Since x 2 5x + 6 = (x 2)(x 3) it is true that x 2 x 2 5x + 6. But x 2 2x + 7 does not divide 2x 3 + 5x Substitution and the Remainder Theorem If f(x) F[x], in other words f(x) is a polynomial in x with coefficients coming from the field F, and F we define f( ) to be the number, in F, that results from replacing, or substituting, x in the polynomial by the value. For example if f(x) = x 2 + x 2 then f(2) = = 4, f(0) = 2 and f(1)=0. The following theorem connects the ideas of substitution and remainder. Theorem 4 (Remainder Theorem): The remainder on dividing f(x) by x is f( ). Proof: By the Division Algorithm, f(x) = (x )q(x) + r(x) for some polynomials q(x), r(x) and the remainder r(x) is either zero or has degree less than 1. In other words r(x) must be a constant polynomial, so we can drop the (x) and just call it r. Now substituting x = into the equation f(x) = (x )q(x) + r, we get f( ) = r. Corollary: The polynomial f(x) is divisible by x if and only if f( ) = 0. Example 11: We saw that if f(x) = x 2 + x 2 then f(2) = 4. The Remainder Theorem concludes that 4 must be the remainder on dividing f(x) by x 2. Numbers which produce zero when substituted into a polynomial f(x) are just the solutions of the polynomial equation f(x) = 0. They are called the roots of the polynomial and they are quite important features of the polynomial. 104 5 9.6. Roots of Polynomials A root of a polynomial f(x) is a number,, so that f( )=0. Solving a polynomial equation f(x) = 0 therefore means finding all its roots. But where do we look for potential roots? From the coefficient field. But here we have to be a little careful. Does the polynomial f(x) = x have any roots? That depends. The coefficients are real numbers so we could consider f(x) as belonging to the set of real polynomials, R[x]. If so, there are no roots. But we can just as validly consider f(x) as belonging to the set of complex polynomials C[x] (remember the set C includes all of the real numbers). Then we would have two roots for f(x), namely ± i. Frequently we switch from one field to another. So we can say that x has no real roots, but two complex roots. The corollary to the Remainder Theorem can be expressed by saying that the number is a root of f( ) if and only if x is one of its factors. Now the polynomial x has degree 1 and it is called a linear polynomial. So there is a connection between linear factors and roots of a polynomial. Theorem 5: A polynomial has a zero if and only if it has a linear factor. Proof: Whenever we have a root,, we have a linear factor x. Conversely having a linear factor bx + c for a polynomial means that we have a root x = c/b If we know one root of a polynomial we can use the remainder theorem and divide by the corresponding linear factor. The other roots will then be roots of the quotient. Example 12: Given that x = 2 is a root of x 3 x 2 x 2, find the other two roots. Solution: By the remainder theorem x 2 is a factor. Dividing the cubic by x 2 we get the other factor which we proceed to solve. So x 3 x 2 x 2 = (x 2)(x 2 + x + 1). The other roots are the roots of x 2 + x + 1, viz ½( 1 ± 3i) Sum and Product of Roots If and are the roots of the quadratic ax 2 + bx + c then: + = b a (sum of the roots) and = c a (product of the roots). Example 13: If and are the roots of x 2 + 3x + 7, find: (i) + ; (ii) ; (iii) ; (iv) 6 Solution: (i) + = 3; (ii) = 7; (iii) = ( + ) 2 2 = 9 14 = 5 (if you are worried by getting a negative sum of squares just remember that all this means is that the roots are non-real) ; (iv) = + = 3 7. Example 14: If and are the roots of 2x 2 5x + 1, find (i) ; (ii) +. Solution: + = 5/2 and = ½. We must now express each expression in terms of these. (i) = ( + ) = ( + ) 3 3 ( + ) = = (ii) ( + ) 2 = = = Hence + = 2. (The quadratic does have positive real roots and we are taking positive square roots throughout.) For the roots,, of the cubic ax 3 + bx 2 + cx + d we get similar results, but there is an additional combination of the roots we need to consider. + + = b a (sum of the roots); + + = c a (sum of the roots taken two at a time); = d a (product of the roots). Example 15: If,, are the roots of the cubic x 3 + 5x 2 2x 3, find the value of Solution: = + + = 2 3. The three expressions + +, + + and are symmetric in, and. This means that, and can be permuted in any order and the value of these expressions is unchanged. It is obvious that any expression that we can build up from these three expressions must also be symmetric. What is not obvious, but is nevertheless true, is the converse. Every symmetric expression in, and can be expressed in terms of these three special ones. They are known as the elementary symmetric functions. Example 16: Express the following symmetric functions in terms of the elementary symmetric ones: (i) + + ; (ii) ; (iii) Solution: (i) + + = = ( + + ) = ( + + )2 2 ( + + ) ; (ii) = ( + + )( + + ) 3 ; 106 7 (iii) = ( + + ) = ( + + ) 3 3( ) = ( + + ) 3 3( + + )( + + ) = ( + + ) 3 3( + + )( + + ) + 3. The above results extend to polynomials of higher degree. If f(x) = a n x n + a n 1 x n a 1 x + a 0 is a polynomial of degree n with roots 1, 2,, n then 1 2 r, the sum of all products of roots taken r at a time, is ( 1) r a n r a n. Moreover every symmetric function in the i can be expressed in terms of these elementary symmetric functions. Example 17: If,,, are the roots of x 4 + 5x 3 3x 2 3x 7 find Solution: = ( ) 2 2( ) = 2 = 5 2( 3) = Real and Complex Polynomials A real polynomial can be graphed in the usual way and real roots correspond to places where the curve cuts the x-axis. A real polynomial has at least one real root if its graph crosses the x-axis. Theorem 6: Real polynomials of odd degree have at least one real root. Proof: If a(x) has odd degree, then a(x) as x and a(x) as x. It therefore takes both positive and negative values and since it moves continuously between positive and negative values it must cross the x-axis in at least one place. Real polynomials of even degree can have real roots too, but there are plenty which don't (such as x 2 + 1). We can't graph a polynomial with complex coefficients without using 4 dimensions. However a very important theorem, easy to state but not so easy to prove, states that every complex polynomial (apart from constants) has complex roots. Theorem 7 (Fundamental Theorem of Algebra): If f(x) C[x] and deg f(x) 1, then f( ) = 0 for some C. Example 17: You will know that every real quadratic ax 2 + bx + c has roots in C from the quadratic equation formula: x = b b2 4ac 2a with real roots if b² 4ac and non-real roots if b 2 < 4ac. 107 8 But from this formula we can compute the roots of a quadratic over any field (except those, such as Z 2 where 1+1=0). And since it is possible, and not very difficult, to find the square roots of any complex number we can find roots for any complex quadratic. Example 18: Find the zeros of the quadratic x 2 + 2ix + i. Solution: x = 2i 4 4i 2 and since 4 4i = 32(cos 5 i/4 + i sin 5 i/4) we can express its square roots as 32 ¼ (cos 5 i/8 + i sin 5 i/8). The roots of the quadratic are thus 32 ¼ cos 5 i/8 + (32 ¼ sin 5 i/8 1)i. The Fundamental Theorem states that not only complex quadratics but complex cubics, complex quartics,... have complex roots. Because of this we say that C is algebraically closed. No new numbers need to be invented. The history of number consisted of mathematicians continuing to extend the number system so as to provide roots for polynomials which would otherwise have none. God, it has been said, gave us the natural numbers 1,2,3,... and man created the rest. When we just had the natural numbers 1,2,3,... we had to say that the equation 2x = 1 has no solutions. So fractions were invented. Then 2x = 1 could be solved but x 2 = 2 could not. So along come the irrational numbers. Since 2 now exists, x 2 = 2 now has a solution (just one because so far there is no such thing as a negative number). This all happened back in classical times. But despite all these extra numbers that had been invented, an equation such as x + 2 = 1 still had no solution. Negative numbers took another thousand years or so arrive. Once they did then x + 2 = 1 was OK. But x = 0 could not be solved. There were no roots for x The last stage was the extension from the real numbers to the complex numbers which took place in the 17th century. Now a polynomial equation such as x = 0 could be solved, but what about x 2 + i = 0, or x 5 +ix 3 = 1 + 2i? Might we not have to go on inventing more and more numbers? Might not our number system spring new leaks whenever we try to patch up existing ones? After all, whenever we extend our number system we introduce new equations. But the marvellous thing is that having arrived at the system of complex numbers we had, in a sense, reached perfection. No new numbers are necessary. We can solve all polynomial equations, with complex coefficients, entirely from within the complex number system itself. There is a complex number x such that x 2 + i = 0 (in fact two of them). There is a complex number x such that x 5 +ix 3 = 1 + 2i (in fact five such solutions). The Fundamental Theorem shows that we need not extend our number system any further. As a result of the Fundamental Theorem of Algebra every polynomial with complex coefficients can be factorised completely into linear factors. If the coefficients are real, and we insist that the factors have real coefficients, we cannot expect to be able to factorise the polynomial quite so completely. However we can factorise any real polynomial into real factors which, at worst, are quadratic. This comes about because the non-real roots of a real polynomial split into conjugate pairs. 108 9 Theorem 8 (Conjugate Pairs): The non-real roots of a real polynomial come in conjugate pairs. Proof: Suppose is a root of the real polynomial a(x) = a n x n + an 1 x n a1 x + a 0. Then a( ) = 0. Hence a n n + an 1 n a1 + a 0 = 0. Taking conjugates of both sides we see that the conjugate of is also a root. Example 19: The five roots of f(x) = x 5 2x 4 + 6x 3 2x 2 + 5x are 0, 1 2i and i. The factorisation of f(x) over C is f(x) = x(x i)(x+i)(x 1 2i)(x+1+2i). Grouping the linear factors which correspond to conjugate roots this becomes f(x) = x(x 2 + 1)(x 2 2x + 5), a factorisation over R. root. Theorem 8 can be useful in factorising certain real polynomials if we are given a non-real Example 20: Assuming that 2 + i is a root of x 4 6x 3 22x x 175 factorise it completely over the rationals. Solution: Since the coefficients are real, 2 i is also a root and hence (x 2 i)(x 2 + i) = x 2 4x + 5 is a factor. Dividing by this quadratic we get the other factor: z 2 2x Formulae for Higher Degree Polynomials We all know how to solve a quadratic equation ax 2 + bx + c = 0. Unless the factors are obvious it is probably quicker to use the quadratic equation formula: b b 2 4ac 2a. There is a similar, but more complicated, formula for cubics. Every cubic equation can be expressed in the form x 3 + bx 2 + cx + d = 0 by dividing through by the leading coefficient. Then by substituting y = x + b 3 we can eliminate the squared term and write y as a solution to a cubic of the form y 3 + cy + d = 0. For such a cubic the three solutions for y are 3 d 2 + d c d 2 d c3 27. There are three cube roots for every non-zero complex number, so this formula appears to give 9 solutions altogether. However the cube roots have to be paired up in a special way, giving only three solutions in all. Even if the roots turn out eventually to be real the process involves working with non-real numbers. This is why, despite the misgivings mathematicians had about whether square roots of 1 really exist, they could not ignore the fact that even if they didn t exist they were really useful. You will notice that the cubic formula is of the same type as the quadratic one. You start with the coefficients and then carry out certain arithmetic operations to find the solutions. These 109 10 operations are addition, subtraction, multiplication and division and the extraction of roots. In the case of the cubic we need to take square roots as well as cube roots. There exists an exact formulae for the quartic, along the same lines. The cubic and the quartic fomulae were discovered in 1515 and 1545 respectively and, as you would expect, the search was on for a quintic equation formula. You might ask who would really want such formulae. Well, don t forget that this was before calculus, so that Newton s method, or other iterative process based on derivatives were not possible. But what is more to the point was that solving polynomial equations became a fairground stunt. A moderately educated member of the public would choose some numbers, usually integers, and construct the polynomial that has these as its zeros. The operator of the stall would wager money that he could solve the polynomial in a certain time! The quintic formula was never found in the two centuries following the discovery of the quartics formula. There had to be a solution, it was just proving to be too complicated. Or, so everyone thought. In fact the question was finally settled in the early part of the nineteenth century when the Norwegian mathematician, Abel, showed that no such formula exists for degree 5 or more. But it was Evariste Galois, a young teenager, who was killed in a duel at the age of 20, who devised a wonderfully powerful theory that determines which polynomials of degree five or more can be solved in this way. A solution involving just the four basic arithmetic operations plus the extraction of roots is called a solution by radicals. Galois invented a whole new branch of mathematics, which we call group theory and applied this to permutations of the zeros of a polynomial in a way that we now call Galois theory, to determine whether a given polynomial is soluble by radicals. All this by an unknown teenage French boy, who scribbled most of his mathematics in prison (he was a student rebel), by the age of 19. If you ever get a chance to read his story, do so. It is the most dramatic story of any mathematician in history! Solving Systems of Polynomial Equations in One Variable Example 21: Solve the system x 3 1= 0 x 4 + x 3 x 2 + x 2 = 0. In theory we can always solve the first equation and test them in the second. That would indeed work in this case, because we know that the cube roots of 1 are 1, and 2, where = e 2 i/3. But in general this would not be at all feasible. The secret is to find the greatest common divisor of the two polynomials. This will be the monic (leading coefficient 1) of largest degree that divides both polynomials. The common solutions to the two polynomial equations will simply be all the solutions of the GCD. The process of finding the GCD, though it can sometimes be a little messy, is really not that hard. We still have to solve the equation for the GCD, but it is often the case that the degree of the GCD is quite small. A greatest common divisor (or GCD) of f(x) and g(x) is a polynomial of largest degree which divides them both. 110 11 Example22: The polynomial x 1 divides both x 2 1 and x 2 2x + 1. No polynomial of higher degree divides both so x 1 is a greatest common divisor. But so are 2x 2 and ¼x ¼. What distinguishes x 1 from these others is that it is monic its leading coefficient is 1. We would like to be able to define the GCD to be the monic one. But isn t it conceivable that there is more than one? Imagine that we had two polynomials f(x) and g(x). Suppose further that both are divisible by x 2, both are divisible by x + 5, and that no quadratic, or polynomial of higher degree divides them both. If such a situation was possible, what would be the GCD. of f(x) and g(x)? Would the GCD be x 2 or would it be x + 5? Both would be common divisors of larger degree that divide both and both are monic. The answer is that such a situation cannot arise. If x 2 and x + 5 were both common divisors, their product (x 2)(x + 5) would also have to be a common divisor. But the product of two divisors is not always a divisor. For example both x 2 1 and x 2 2x + 1 are divisors of the cubic (x 1) 3 (x + 1) but their product, (x 2 1)(x 2 2x + 1), has degree 4 and so can t possibly be a divisor of the cubic. To help clarify this situation we show that there is another characterisation of GCDs Theorem 9: Let S be the set of all polynomials which are expressible in the form: a(x)h(x) + b(x)k(x). Then any polynomial in S of lowest degree is a greatest common divisor of a(x) and b(x). Proof: S includes, of course, the zero polynomial, where h(x) = k(x) = 0, and it includes both a(x) and b(x) themselves. As well it includes polynomials such as a(x)(x 2 + 1) b(x)x 3 which generally have higher degree than a(x) and b(x) themselves but which, because of cancellation, could have lower degree. Choose a polynomial m(x) from S with lowest degree. Coming from S, m(x) = a(x)h(x) + b(x)k(x) for some polynomials h(x) and k(x). We now show that m(x) is a common divisor of a(x) and b(x). Let a(x) = m(x)q(x) + r(x) where r(x) = 0 or has lower degree than m(x). Making r(x) the subject of the formula we get: r(x) = a(x) m(x)q(x) = a(x)(1 h(x)q(x)) + b(x)( k(x)q(x)). This means that r(x) belongs to S, but this would contradict the fact that m(x) had lowest degree of any element of S (remember that deg r(x) < deg m(x)). The only way out is for r(x) to be zero, that is, for m(x) to divide a(x). Similarly m(x) must divide b(x) and so it must be a common divisor. But is m(x) a greatest common divisor? Can there be a common divisor having larger degree than that of m(x)? Suppose that d(x) is any common divisor. Since d(x) divides both a(x) and b(x), d(x) must divide m(x) = a(x)h(x) + b(x)k(x). This means that deg d(x) deg m(x). So no common divisor can have larger degree than m(x) and so m(x) is a greatest common divisor. Corollary: All GCDs of a given pair of polynomials are just constant multiples of one another. There is just one monic one among them. 111 12 Proof: Let m(x) = a(x)h(x) + b(x)k(x) be the G.C.D. of a(x) and b(x) obtained as above. Let d(x) be any G.C.D. of a(x) and b(x). Since d(x) divides both a(x) and b(x) it must divide m(x). So deg d(x) deg m(x). But since d(x) is a greatest common divisor, we must have deg d(x) = deg m(x). Now for one polynomial to divide another of the same degree, it must be that they are just constant multiples of one another. Hence all GCDs are constant mulktiples of m(x) and hence of each other. Example 23: The GCD of x 2 1 and x 2 2x + 1 is x 1. The above theorem shows that we can express x 1 in the form (x 2 1)h(x) + (x 2 2x + 1)k(x). Clearly the x 2 terms have to cancel out. A suitable expression is x 1 = (x 2 1)(½) + (x 2 2x + 1)( ½). Here the h(x) and k(x) are constant polynomials, but in more complicated cases they would have higher degree. Theorem 10: For all a, b, q (either integers or polynomials) GCD(a, b) = GCD(a bq, b). Proof: Let d 1 = GCD(a, b) and d 2 = GCD(a bq, b). Then d 1 a bq and d 1 b so d 1 d 2. And since a = (a bq) + bq, d 2 a and d 2 b so d 2 d 1 and d 2 d 1. Since d 1, d 2 are both positive (in the case of integers) or both monic (in the case of polynomials), d 1 = d The Euclidean Algorithm You may recall the Euclidean Algorithm as it applies to finding the GCD of two integers. The same process can be easily adapted to polynomials. It s just that the algebra, when working with polynomials, can get a little messier than the arithmetic, when working with integers. THE EUCLIDEAN ALGORITHM To find GCD(f(x), g(x)): Let c(x) be the remainder on dividing f(x) by g(x). Replace f(x) by g(x). Replace g(x) by c(x). Continue from the top until g(x) 0. When g(x) = 0, make the last non-zero remainder monic. This is the required GCD. Example 24: Find the G.C.D. of the polynomials x 3 1 and x 4 + x 3 x 2 + x 2. Solution: We divide the cubic into the quartic. (Doing it the other way round wouldn t be wrong but all the first division would achieve would be to swap them.) x + 1 x 3 1 ) x 4 + x 3 x 2 + x 2 x 4 x x 3 x 2 + 2x 2 x 3 1 x 2 + 2x 1 112 13 Since we are going to make the answer monic at the end we are entitled to multiply or divide any remainder by any non-zero constant. So we will replace our remainder by x 2 2x + 1. x + 2 x 2 2x + 1 ) x 3 1 x 3 2x 2 + x 2x 2 x 1 2x 2 4x + 2 3x 3 We replace 3x 3 by x 1. x 1 x 1 ) x 2 2x + 1 x 2 x x + 1 x The last non-zero remainder is x 1. This is the required GCD. Example 21 (revisited): The solution to the system of polynomial equations in Example 21 is simply x = 1. Example 25: Find the G.C.D. of the polynomials x and x 4 + x 3 x 2 + x. Solution: x + 1 x ) x 4 + x 3 x 2 + x x 4 + x x 3 x 2 x x 2 1 We change this remainder to x x x ) x x 3 + x x + 1 We change this to x 1. x + 1 x 1 ) x x 2 x x + 1 x 14 The last non-zero remainder (made monic) is 1 so the GCD of these two polynomials is 1. We say that the polynomials have no common factor (which is not strictly true we actually mean that there is no common factor other than constants), or that they are coprime. EXERCISES FOR CHAPTER 9 Exercise 1: Find the degree, the leading coefficient and the roots of the polynomial f(x) = 5x 6 6x 5 x 7. Exercise 2: Which of the following are polynomials in x? (a) x 5 + x; (b) x + x 1 ; (c) 1 + x 2 + x ; (d) 42; (e) x 2 + x + 1. Exercise 3: Find the degree of the polynomial f(x) = (2 n 8)x 2n+1 + 4x 5 + x in terms of n. Exercise 4: Find the quotient and remainder on dividing f(x) = x 4 + 7x + 2 by x 3. Exercise 5: Find the remainder on dividing x 7 + x 4 + x by x 2 3. Exercise 6: Show that x 3 is a factor of the cubic f(x) = x 3 7x x 24. Hence find the roots of f(x). Exercise 7: Given that 1 + i is a root of the polynomial x 4 4x x 2 38x + 34, find its other three roots. Exercise 8: If and are the roots of 4x 2 12x + 7 find the values of: (i) + ; (ii) ; (iii) ( ) 2 ; (iv). Exercise 9: If, and are the roots of the cubic f(x) = x 3 7x x 4, find the value of E = ( + )( + )( + ). Exercise 10: Find the remainder on dividing f(x) = x x 6 5x 2 + x 2 by x 1. Exercise 11: Find the remainder on dividing x 5 2x 3 + 5x 2 7 by x 2 + x + 2. Exercise 12: Given that x = 2 is a root of x 3 6x 2 + 9x 2, find the other two roots. 114 15 Exercise 13: Given that x = i is a root of 6x 4 x 3 + 4x 2 x 2, find the other three roots. Exercise 14: If and are the roots of the quadratic 2x 2 3x + 8, find the values of: (i) ; (ii) ; (iii) Exercise 15: If,, are the roots of x 3 + 5x 2 + 2x 3 find the values of: (i) ; (ii) ; (iii) + +. Exercise 16: If f(x) = x 9 + (k + 1)x 8 + (k 2 + k + 1)x find the sum of the square of the roots of f(x) and hence show that f(x) has at least one none-real root. Exercise 17: Find the GCD of 2x 4 + 7x 3 + 2x 2 + 9x + 7 and 2x 3 + 9x 2 + 9x + 7. Exercise 18: The cubic f(x) = x 3 13x x 80 has a repeated zero. Find it. Exercise 19: Find the GCD of a(x) = 10x 2 + x 21 and b(x) = 8x x Exercise 20: Find the GCD of x and x Exercise 21: Find any multiple roots of the polynomial f(x) = x 4 + 3x 3 + 4x 2 + 3x + 1. Exercise 22: Show that the polynomial f(x) = x 4 2x 3 + 5x 2 4x + 4 is a perfect square (i.e. the square of a polynomial). [HINT: Look for multiple roots.] 115 16 Exercise 1: f(x) = x 7 + 5x 6 6x 5 are 0, 2, 3. SOLUTIONS FOR CHAPTER 9 = x 5 (x 2)(x 3), so deg f(x) = 7; the leading coefficient is 1 and the roots Exercise 2: (a) and (d) are polynomials; the others are not. Exercise 3: For n = 1, f(x) = 6x 3 + 4x 5 + x and so deg f(x) = 5; For n = 2, f(x) = 4x 5 + 4x 5 + x = x so deg f(x) = 2; For n = 3, f(x) = 0x 7 + 4x 5 + x and so deg f(x) = 5; For n 4, deg f(x) = 2n+1. Exercise 4: x 3 + 3x 2 + 9x + 34 x 3 ) x 4 + 7x + 2 x 4 3x 3 3x 3 + 7x + 2 3x 3 9x 2 9x 2 + 7x + 2 9x 2 27x 34x x Hence the quotient is x 3 + 3x 2 + 9x + 34 and the remainder is 104. If we just wanted the reminder it would be much easier to use the Remainder Theorem. The remainder is f(3) = = 104. Exercise 5: x 5 + 3x 3 + x 2 + 9x + 4 x 2 3 ) x 7 + x 4 + x x 7 3x 5 3x 5 + x 4 + x x 5 9x 3 x 4 + 9x 3 + x x 4 3x 2 The remainder is 27x x 3 + 4x x 3 27x 4x x + 1 4x x 17 Exercise 6: We could verify the fact that x 3 is a factor by checking that f(3) = 0. However, since we will need to find the quotient we must carry out the long division process. x 2 4x + 4 x 3 ) x 3 7x x 24 x 3 3x 2 4x x 24 4x x 8x 24 8x 24 0 The quotient is x 2 4x + 8. Solving this quadratic we get x = 2 2i. So the root of f(x) are 3, 2 2i. Exercise 7: Since the polynomial has real coefficients, 1 i the conjugate of 1 + i, must also be a root. Hence x (1 + i) and x (1 i) are factors. Thus (x 1 i)(x 1 + i) = x 2 2x + 2 is a factor. We now divide this into the quartic to obtain the other quadratic factor. x 2 2x + 17 x 2 2x + 2 ) x 4 4x x 2 38x + 34 x 4 2x 3 + 2x 2 2x x 2 38x x 3 + 4x 2 4x 17x 2 34x x 2 34x Hence the polynomial factorises as (x 2 2x + 2)(x 2 2x + 17). Solving x 2 2x + 17 = 0 gives 1 4i. So the other three roots are 1 i, 1 4i. Exercise 8: + = 3, = 7/4. (i) + = = ( + )2 2 = 22 7 ; (ii) = ( + ) = 21/4; (iii) ( ) 2 = ( + ) 2 4 = 9 7 = 2; (iv) = 2. Exercise 9: + + = 7 etc. Now rather than expand E note that + = ( + + ) = 7, and similarly for the others. So E = (7 )(7 )(7 ). Now before you get carried away expanding this out, just note that this f(x) = (x )(x )(x ) and so E is simply f(7) = 73. Exercise 10: The remainder is f(1) = 18 Exercise 11: x 3 x 2 3x + 10 x 2 + x + 2 ) x 5 2x 3 + 5x 2 7 x 5 + x 4 + 2x 3 x 4 4x 3 + 5x 2 7 x 4 x 3 2x 2 Exercise 12: x 2 4x + 1 x 2 ) x 3 6x 2 + 9x 2 x 3 2x 2 4x 2 + 9x 2 4x 2 + 8x x 2 x 2 0 3x 3 + 7x 2 7 3x 3 3x 2 6x 10x 2 6x 7 10x x x 27 Exercise 13: Since i is a root, so is i. Hence (x i)(x + i) = x is a factor. 6x 2 x 2 x ) 6x 4 x 3 + 4x 2 x 2 6x 4 + 6x 2 x 3 2x 2 x 2 x 3 x 2x 2 2 2x Exercise 14: + = 3/2, = 4 (i) = ( + ) = ( + ) = = 19 Exercise 15: + + = 5, + + = 2, = 3. (i) = ( + + ) = 15; (ii) = ( + + ) 2 2( + + ) = 25 4 = 21; (iii) + + = = Exercise 16: = (k + 1) and = k 2 + k + 1. Hence 2 = ( ) 2 2 = (k + 1) 2 2(k 2 + k + 1) = (k 2 + 1). If all the roots were real then 2 0, whereas (k 2 + 1) < 0. Exercise 17: x 1 2x 3 + 9x 2 + 9x + 7 ) 2x 4 + 7x 3 + 2x 2 + 9x + 7 2x 4 + 9x 3 + 9x 2 + 7x 2x 3 7x 2 + 2x + 7 2x 3 9x 2 9x 7 2x x + 14 x 1 2x x + 14 ) 2x 3 + 9x 2 + 9x + 7 2x x x 2x 2 5x + 7 2x 2 11x 14 6x + 21 At the end we will be making the last non-zero remainder monic, so at any stage we can multiply or divide by a non-zero constant to simplify the calculations. So we shall use 2x + 7 instead of 6x x + 2 2x + 7 ) 2x x x 2 + 7x 4x x The last non-zero remainder is 6x + 21 (or equivalently 2x + 7). Making this monic we deduce that the GCD is x + 7/2. 119 20 Exercise 18: A repeated root will be a zero of f (x) = 3x 2 26x + 56 and so will be a zero of GCD(f(x), f (x)). For simplicity let s replace f(x) by 3f(x). x 13/3 3x 2 26x + 56 ) 3x 3 39x x 240 3x 3 26x x 13x x x /3x 728/3 2/3x + 8/3 Replace this remainder by the multiple x 4 = ( 3/2)( 2/3x + 8/3). 3x 14 x 4 ) 3x 2 26x x 2 12x 14x x Exercise 19: Rather than divide we can first subtract. This will simplify the arithmetic. GCD(a(x), b(x)) = GCD(a(x) b(x), b(x)) = GCD(2x 2 29x 48, 8x x + 27). = GCD(2x 2 29x 48, (8x x + 27) 4(2x 2 29x 48)) = GCD(2x 2 29x 48, 146x + 219) = GCD(2x 2 29x 48, 2x + 3) since 146x = 73(2x + 3). x 16 2x + 3 ) 2x 2 29x 48 2x 2 + 3x 32x 48 32x 48 0 Exercise 20: x x ) x x 5 + x x + 1 Replace this remainder by x 21 x 3 + x 2 + x x 1 ) x x 4 x 3 x x 3 x 2 x x Making this monic we see that the GCD is 1. Exercise 21: f (x) = 4x 3 + 9x 2 + 8x + 3. For convenience replace f(x) by by 4f(x). x + 3/4 4x 3 + 9x 2 + 8x + 3 ) 4x x x x + 4 4x 4 + 9x 3 + 8x 2 + 3x 3x 3 + 8x 2 + 9x + 4 3x 3 + (27/4)x 2 + 6x + 9/4 (5/4)x 2 + 3x + 7/4 (4/5)x 3/25 5x x + 7 ) 4x 3 + 9x 2 + 8x + 3 4x 3 + (48/5)x 2 + (28/5)x ( 3/5)x 2 + (12/5)x + 3 ( 3/5)x 2 (36/25)x 21/25 (96/25)x + 96/25 5x + 7 x + 1 ) 5x x + 7 5x 2 + 5x 7x + 7 7x Exercise 22: f (x) = 4x 3 6x x x x 3 3x 2 + 5x 2) x 4 2x 3 + 5x 2 4x + 4 x 4 1.5x x 2 x 0.5x x 2 3x x x x x x Replace this by x 2 x 22 2x 1 x 2 x + 2 ) 2x 3 3x 2 + 5x 2 2x 3 2x 2 + 4x x 2 + x 2 x 2 + x ### SOLVING POLYNOMIAL EQUATIONS C SOLVING POLYNOMIAL EQUATIONS We will assume in this appendix that you know how to divide polynomials using long division and synthetic division. If you need to review those techniques, refer to an algebra ### Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm. Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm. We begin by defining the ring of polynomials with coefficients in a ring R. After some preliminary results, we specialize ### PROBLEM SET 6: POLYNOMIALS PROBLEM SET 6: POLYNOMIALS 1. introduction In this problem set we will consider polynomials with coefficients in K, where K is the real numbers R, the complex numbers C, the rational numbers Q or any other ### 6 EXTENDING ALGEBRA. 6.0 Introduction. 6.1 The cubic equation. Objectives 6 EXTENDING ALGEBRA Chapter 6 Extending Algebra Objectives After studying this chapter you should understand techniques whereby equations of cubic degree and higher can be solved; be able to factorise ### a 1 x + a 0 =0. (3) ax 2 + bx + c =0. (4) ROOTS OF POLYNOMIAL EQUATIONS In this unit we discuss polynomial equations. A polynomial in x of degree n, where n 0 is an integer, is an expression of the form P n (x) =a n x n + a n 1 x n 1 + + a 1 x ### JUST THE MATHS UNIT NUMBER 1.8. ALGEBRA 8 (Polynomials) A.J.Hobson JUST THE MATHS UNIT NUMBER 1.8 ALGEBRA 8 (Polynomials) by A.J.Hobson 1.8.1 The factor theorem 1.8.2 Application to quadratic and cubic expressions 1.8.3 Cubic equations 1.8.4 Long division of polynomials ### Copy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any. Algebra 2 - Chapter Prerequisites Vocabulary Copy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any. P1 p. 1 1. counting(natural) numbers - {1,2,3,4,...} ### 3.6 The Real Zeros of a Polynomial Function SECTION 3.6 The Real Zeros of a Polynomial Function 219 3.6 The Real Zeros of a Polynomial Function PREPARING FOR THIS SECTION Before getting started, review the following: Classification of Numbers (Appendix, ### Lagrange Interpolation is a method of fitting an equation to a set of points that functions well when there are few points given. Polynomials (Ch.1) Study Guide by BS, JL, AZ, CC, SH, HL Lagrange Interpolation is a method of fitting an equation to a set of points that functions well when there are few points given. Sasha s method ### it is easy to see that α = a 21. Polynomial rings Let us now turn out attention to determining the prime elements of a polynomial ring, where the coefficient ring is a field. We already know that such a polynomial ring is a UF. Therefore ### Elementary Number Theory We begin with a bit of elementary number theory, which is concerned CONSTRUCTION OF THE FINITE FIELDS Z p S. R. DOTY Elementary Number Theory We begin with a bit of elementary number theory, which is concerned solely with questions about the set of integers Z = {0, ±1, ### The Division Algorithm for Polynomials Handout Monday March 5, 2012 The Division Algorithm for Polynomials Handout Monday March 5, 0 Let F be a field (such as R, Q, C, or F p for some prime p. This will allow us to divide by any nonzero scalar. (For some of the following, ### CHAPTER SIX IRREDUCIBILITY AND FACTORIZATION 1. BASIC DIVISIBILITY THEORY January 10, 2010 CHAPTER SIX IRREDUCIBILITY AND FACTORIZATION 1. BASIC DIVISIBILITY THEORY The set of polynomials over a field F is a ring, whose structure shares with the ring of integers many characteristics. ### PUTNAM TRAINING POLYNOMIALS. Exercises 1. Find a polynomial with integral coefficients whose zeros include 2 + 5. PUTNAM TRAINING POLYNOMIALS (Last updated: November 17, 2015) Remark. This is a list of exercises on polynomials. Miguel A. Lerma Exercises 1. Find a polynomial with integral coefficients whose zeros include ### Factoring Polynomials Factoring Polynomials Hoste, Miller, Murieka September 12, 2011 1 Factoring In the previous section, we discussed how to determine the product of two or more terms. Consider, for instance, the equations ### 3.4 Complex Zeros and the Fundamental Theorem of Algebra 86 Polynomial Functions.4 Complex Zeros and the Fundamental Theorem of Algebra In Section., we were focused on finding the real zeros of a polynomial function. In this section, we expand our horizons and ### SECTION 0.6: POLYNOMIAL, RATIONAL, AND ALGEBRAIC EXPRESSIONS (Section 0.6: Polynomial, Rational, and Algebraic Expressions) 0.6.1 SECTION 0.6: POLYNOMIAL, RATIONAL, AND ALGEBRAIC EXPRESSIONS LEARNING OBJECTIVES Be able to identify polynomial, rational, and algebraic ### Factoring Polynomials Factoring Polynomials Sue Geller June 19, 2006 Factoring polynomials over the rational numbers, real numbers, and complex numbers has long been a standard topic of high school algebra. With the advent ### Application. Outline. 3-1 Polynomial Functions 3-2 Finding Rational Zeros of. Polynomial. 3-3 Approximating Real Zeros of. Polynomial and Rational Functions Outline 3-1 Polynomial Functions 3-2 Finding Rational Zeros of Polynomials 3-3 Approximating Real Zeros of Polynomials 3-4 Rational Functions Chapter 3 Group Activity: ### 3.3. Solving Polynomial Equations. Introduction. Prerequisites. Learning Outcomes Solving Polynomial Equations 3.3 Introduction Linear and quadratic equations, dealt within Sections 3.1 and 3.2, are members of a class of equations, called polynomial equations. These have the general ### 7. Some irreducible polynomials 7. Some irreducible polynomials 7.1 Irreducibles over a finite field 7.2 Worked examples Linear factors x α of a polynomial P (x) with coefficients in a field k correspond precisely to roots α k [1] of ### Quotient Rings and Field Extensions Chapter 5 Quotient Rings and Field Extensions In this chapter we describe a method for producing field extension of a given field. If F is a field, then a field extension is a field K that contains F. ### Functions and Equations Centre for Education in Mathematics and Computing Euclid eworkshop # Functions and Equations c 014 UNIVERSITY OF WATERLOO Euclid eworkshop # TOOLKIT Parabolas The quadratic f(x) = ax + bx + c (with a,b,c ### Math Review. for the Quantitative Reasoning Measure of the GRE revised General Test Math Review for the Quantitative Reasoning Measure of the GRE revised General Test www.ets.org Overview This Math Review will familiarize you with the mathematical skills and concepts that are important ### H/wk 13, Solutions to selected problems H/wk 13, Solutions to selected problems Ch. 4.1, Problem 5 (a) Find the number of roots of x x in Z 4, Z Z, any integral domain, Z 6. (b) Find a commutative ring in which x x has infinitely many roots. ### March 29, 2011. 171S4.4 Theorems about Zeros of Polynomial Functions MAT 171 Precalculus Algebra Dr. Claude Moore Cape Fear Community College CHAPTER 4: Polynomial and Rational Functions 4.1 Polynomial Functions and Models 4.2 Graphing Polynomial Functions 4.3 Polynomial ### Introduction to Finite Fields (cont.) Chapter 6 Introduction to Finite Fields (cont.) 6.1 Recall Theorem. Z m is a field m is a prime number. Theorem (Subfield Isomorphic to Z p ). Every finite field has the order of a power of a prime number ### UNCORRECTED PAGE PROOFS number and and algebra TopIC 17 Polynomials 17.1 Overview Why learn this? Just as number is learned in stages, so too are graphs. You have been building your knowledge of graphs and functions over time. ### FACTORISATION YEARS. A guide for teachers - Years 9 10 June 2011. The Improving Mathematics Education in Schools (TIMES) Project 9 10 YEARS The Improving Mathematics Education in Schools (TIMES) Project FACTORISATION NUMBER AND ALGEBRA Module 33 A guide for teachers - Years 9 10 June 2011 Factorisation (Number and Algebra : Module ### Cubic Functions: Global Analysis Chapter 14 Cubic Functions: Global Analysis The Essential Question, 231 Concavity-sign, 232 Slope-sign, 234 Extremum, 235 Height-sign, 236 0-Concavity Location, 237 0-Slope Location, 239 Extremum Location, ### 2.3. Finding polynomial functions. An Introduction: 2.3. Finding polynomial functions. An Introduction: As is usually the case when learning a new concept in mathematics, the new concept is the reverse of the previous one. Remember how you first learned ### Basics of Polynomial Theory 3 Basics of Polynomial Theory 3.1 Polynomial Equations In geodesy and geoinformatics, most observations are related to unknowns parameters through equations of algebraic (polynomial) type. In cases where ### Some Polynomial Theorems. John Kennedy Mathematics Department Santa Monica College 1900 Pico Blvd. Santa Monica, CA 90405 rkennedy@ix.netcom. Some Polynomial Theorems by John Kennedy Mathematics Department Santa Monica College 1900 Pico Blvd. Santa Monica, CA 90405 rkennedy@ix.netcom.com This paper contains a collection of 31 theorems, lemmas, ### Unique Factorization Unique Factorization Waffle Mathcamp 2010 Throughout these notes, all rings will be assumed to be commutative. 1 Factorization in domains: definitions and examples In this class, we will study the phenomenon ### Algebra I Vocabulary Cards Algebra I Vocabulary Cards Table of Contents Expressions and Operations Natural Numbers Whole Numbers Integers Rational Numbers Irrational Numbers Real Numbers Absolute Value Order of Operations Expression ### RESULTANT AND DISCRIMINANT OF POLYNOMIALS RESULTANT AND DISCRIMINANT OF POLYNOMIALS SVANTE JANSON Abstract. This is a collection of classical results about resultants and discriminants for polynomials, compiled mainly for my own use. All results ### Continued Fractions and the Euclidean Algorithm Continued Fractions and the Euclidean Algorithm Lecture notes prepared for MATH 326, Spring 997 Department of Mathematics and Statistics University at Albany William F Hammond Table of Contents Introduction ### Copyrighted Material. Chapter 1 DEGREE OF A CURVE Chapter 1 DEGREE OF A CURVE Road Map The idea of degree is a fundamental concept, which will take us several chapters to explore in depth. We begin by explaining what an algebraic curve is, and offer two ### 3.6. The factor theorem 3.6. The factor theorem Example 1. At the right we have drawn the graph of the polynomial y = x 4 9x 3 + 8x 36x + 16. Your problem is to write the polynomial in factored form. Does the geometry of the ### 3 1. Note that all cubes solve it; therefore, there are no more Math 13 Problem set 5 Artin 11.4.7 Factor the following polynomials into irreducible factors in Q[x]: (a) x 3 3x (b) x 3 3x + (c) x 9 6x 6 + 9x 3 3 Solution: The first two polynomials are cubics, so if ### U.C. Berkeley CS276: Cryptography Handout 0.1 Luca Trevisan January, 2009. Notes on Algebra U.C. Berkeley CS276: Cryptography Handout 0.1 Luca Trevisan January, 2009 Notes on Algebra These notes contain as little theory as possible, and most results are stated without proof. Any introductory ### Zeros of Polynomial Functions Review: Synthetic Division Find (x 2-5x - 5x 3 + x 4 ) (5 + x). Factor Theorem Solve 2x 3-5x 2 + x + 2 =0 given that 2 is a zero of f(x) = 2x 3-5x 2 + x + 2. Zeros of Polynomial Functions Introduction ### COLLEGE ALGEBRA. Paul Dawkins COLLEGE ALGEBRA Paul Dawkins Table of Contents Preface... iii Outline... iv Preliminaries... Introduction... Integer Exponents... Rational Exponents... 9 Real Exponents...5 Radicals...6 Polynomials...5 ### 3.2 The Factor Theorem and The Remainder Theorem 3. The Factor Theorem and The Remainder Theorem 57 3. The Factor Theorem and The Remainder Theorem Suppose we wish to find the zeros of f(x) = x 3 + 4x 5x 4. Setting f(x) = 0 results in the polynomial ### Problem Set 7 - Fall 2008 Due Tuesday, Oct. 28 at 1:00 18.781 Problem Set 7 - Fall 2008 Due Tuesday, Oct. 28 at 1:00 Throughout this assignment, f(x) always denotes a polynomial with integer coefficients. 1. (a) Show that e 32 (3) = 8, and write down a list ### MATH 65 NOTEBOOK CERTIFICATIONS MATH 65 NOTEBOOK CERTIFICATIONS Review Material from Math 60 2.5 4.3 4.4a Chapter #8: Systems of Linear Equations 8.1 8.2 8.3 Chapter #5: Exponents and Polynomials 5.1 5.2a 5.2b 5.3 5.4 5.5 5.6a 5.7a 1 ### Modern Algebra Lecture Notes: Rings and fields set 4 (Revision 2) Modern Algebra Lecture Notes: Rings and fields set 4 (Revision 2) Kevin Broughan University of Waikato, Hamilton, New Zealand May 13, 2010 Remainder and Factor Theorem 15 Definition of factor If f (x) ### minimal polyonomial Example Minimal Polynomials Definition Let α be an element in GF(p e ). We call the monic polynomial of smallest degree which has coefficients in GF(p) and α as a root, the minimal polyonomial of α. Example: We ### Breaking The Code. Ryan Lowe. Ryan Lowe is currently a Ball State senior with a double major in Computer Science and Mathematics and Breaking The Code Ryan Lowe Ryan Lowe is currently a Ball State senior with a double major in Computer Science and Mathematics and a minor in Applied Physics. As a sophomore, he took an independent study ### Integer roots of quadratic and cubic polynomials with integer coefficients Integer roots of quadratic and cubic polynomials with integer coefficients Konstantine Zelator Mathematics, Computer Science and Statistics 212 Ben Franklin Hall Bloomsburg University 400 East Second Street ### An Introduction to Galois Fields and Reed-Solomon Coding An Introduction to Galois Fields and Reed-Solomon Coding James Westall James Martin School of Computing Clemson University Clemson, SC 29634-1906 October 4, 2010 1 Fields A field is a set of elements on ### Polynomial Expressions and Equations Polynomial Expressions and Equations This is a really close-up picture of rain. Really. The picture represents falling water broken down into molecules, each with two hydrogen atoms connected to one oxygen ### Discrete Mathematics Lecture 3 Elementary Number Theory and Methods of Proof. Harper Langston New York University Discrete Mathematics Lecture 3 Elementary Number Theory and Methods of Proof Harper Langston New York University Proof and Counterexample Discovery and proof Even and odd numbers number n from Z is called ### Winter Camp 2011 Polynomials Alexander Remorov. Polynomials. Alexander Remorov alexanderrem@gmail.com Polynomials Alexander Remorov alexanderrem@gmail.com Warm-up Problem 1: Let f(x) be a quadratic polynomial. Prove that there exist quadratic polynomials g(x) and h(x) such that f(x)f(x + 1) = g(h(x)). ### Homework 5 Solutions Homework 5 Solutions 4.2: 2: a. 321 = 256 + 64 + 1 = (01000001) 2 b. 1023 = 512 + 256 + 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = (1111111111) 2. Note that this is 1 less than the next power of 2, 1024, which ### Zeros of a Polynomial Function Zeros of a Polynomial Function An important consequence of the Factor Theorem is that finding the zeros of a polynomial is really the same thing as factoring it into linear factors. In this section we ### 5.4 The Quadratic Formula Section 5.4 The Quadratic Formula 481 5.4 The Quadratic Formula Consider the general quadratic function f(x) = ax + bx + c. In the previous section, we learned that we can find the zeros of this function ### The Method of Partial Fractions Math 121 Calculus II Spring 2015 Rational functions. as The Method of Partial Fractions Math 11 Calculus II Spring 015 Recall that a rational function is a quotient of two polynomials such f(x) g(x) = 3x5 + x 3 + 16x x 60. The method ### Revised Version of Chapter 23. We learned long ago how to solve linear congruences. ax c (mod m) Chapter 23 Squares Modulo p Revised Version of Chapter 23 We learned long ago how to solve linear congruences ax c (mod m) (see Chapter 8). It s now time to take the plunge and move on to quadratic equations. ### ON GALOIS REALIZATIONS OF THE 2-COVERABLE SYMMETRIC AND ALTERNATING GROUPS ON GALOIS REALIZATIONS OF THE 2-COVERABLE SYMMETRIC AND ALTERNATING GROUPS DANIEL RABAYEV AND JACK SONN Abstract. Let f(x) be a monic polynomial in Z[x] with no rational roots but with roots in Q p for ### Lecture Notes on Polynomials Lecture Notes on Polynomials Arne Jensen Department of Mathematical Sciences Aalborg University c 008 Introduction These lecture notes give a very short introduction to polynomials with real and complex ### The Notebook Series. The solution of cubic and quartic equations. R.S. Johnson. Professor of Applied Mathematics The Notebook Series The solution of cubic and quartic equations by R.S. Johnson Professor of Applied Mathematics School of Mathematics & Statistics University of Newcastle upon Tyne R.S.Johnson 006 CONTENTS ### 1 Lecture: Integration of rational functions by decomposition Lecture: Integration of rational functions by decomposition into partial fractions Recognize and integrate basic rational functions, except when the denominator is a power of an irreducible quadratic. ### Solving Cubic Polynomials Solving Cubic Polynomials 1.1 The general solution to the quadratic equation There are four steps to finding the zeroes of a quadratic polynomial. 1. First divide by the leading term, making the polynomial ### calculating the result modulo 3, as follows: p(0) = 0 3 + 0 + 1 = 1 0, Homework #02, due 1/27/10 = 9.4.1, 9.4.2, 9.4.5, 9.4.6, 9.4.7. Additional problems recommended for study: (9.4.3), 9.4.4, 9.4.9, 9.4.11, 9.4.13, (9.4.14), 9.4.17 9.4.1 Determine whether the following polynomials ### Rational Polynomial Functions Rational Polynomial Functions Rational Polynomial Functions and Their Domains Today we discuss rational polynomial functions. A function f(x) is a rational polynomial function if it is the quotient of ### CM2202: Scientific Computing and Multimedia Applications General Maths: 2. Algebra - Factorisation CM2202: Scientific Computing and Multimedia Applications General Maths: 2. Algebra - Factorisation Prof. David Marshall School of Computer Science & Informatics Factorisation Factorisation is a way of ### 1.5. Factorisation. Introduction. Prerequisites. Learning Outcomes. Learning Style Factorisation 1.5 Introduction In Block 4 we showed the way in which brackets were removed from algebraic expressions. Factorisation, which can be considered as the reverse of this process, is dealt with ### Understanding Basic Calculus Understanding Basic Calculus S.K. Chung Dedicated to all the people who have helped me in my life. i Preface This book is a revised and expanded version of the lecture notes for Basic Calculus and other ### expression is written horizontally. The Last terms ((2)( 4)) because they are the last terms of the two polynomials. This is called the FOIL method. A polynomial of degree n (in one variable, with real coefficients) is an expression of the form: a n x n + a n 1 x n 1 + a n 2 x n 2 + + a 2 x 2 + a 1 x + a 0 where a n, a n 1, a n 2, a 2, a 1, a 0 are ### 8 Divisibility and prime numbers 8 Divisibility and prime numbers 8.1 Divisibility In this short section we extend the concept of a multiple from the natural numbers to the integers. We also summarize several other terms that express ### Mathematics Course 111: Algebra I Part IV: Vector Spaces Mathematics Course 111: Algebra I Part IV: Vector Spaces D. R. Wilkins Academic Year 1996-7 9 Vector Spaces A vector space over some field K is an algebraic structure consisting of a set V on which are ### z = i ± 9 2 2 so z = 2i or z = i are the solutions. (c) z 4 + 2z 2 + 4 = 0. By the quadratic formula, 91 Homework 8 solutions Exercises.: 18. Show that Z[i] is an integral domain, describe its field of fractions and find the units. There are two ways to show it is an integral domain. The first is to observe: ### SECTION 2.5: FINDING ZEROS OF POLYNOMIAL FUNCTIONS SECTION 2.5: FINDING ZEROS OF POLYNOMIAL FUNCTIONS Assume f ( x) is a nonconstant polynomial with real coefficients written in standard form. PART A: TECHNIQUES WE HAVE ALREADY SEEN Refer to: Notes 1.31 ### Congruences. Robert Friedman Congruences Robert Friedman Definition of congruence mod n Congruences are a very handy way to work with the information of divisibility and remainders, and their use permeates number theory. Definition ### 5. Factoring by the QF method 5. Factoring by the QF method 5.0 Preliminaries 5.1 The QF view of factorability 5.2 Illustration of the QF view of factorability 5.3 The QF approach to factorization 5.4 Alternative factorization by the ### Algebra 1 Course Title Algebra 1 Course Title Course- wide 1. What patterns and methods are being used? Course- wide 1. Students will be adept at solving and graphing linear and quadratic equations 2. Students will be adept ### Introduction to Diophantine Equations Introduction to Diophantine Equations Tom Davis tomrdavis@earthlink.net http://www.geometer.org/mathcircles September, 2006 Abstract In this article we will only touch on a few tiny parts of the field ### 1.3 Algebraic Expressions 1.3 Algebraic Expressions A polynomial is an expression of the form: a n x n + a n 1 x n 1 +... + a 2 x 2 + a 1 x + a 0 The numbers a 1, a 2,..., a n are called coefficients. Each of the separate parts, ### CONTINUED FRACTIONS AND FACTORING. Niels Lauritzen CONTINUED FRACTIONS AND FACTORING Niels Lauritzen ii NIELS LAURITZEN DEPARTMENT OF MATHEMATICAL SCIENCES UNIVERSITY OF AARHUS, DENMARK EMAIL: niels@imf.au.dk URL: http://home.imf.au.dk/niels/ Contents ### Vocabulary Words and Definitions for Algebra Name: Period: Vocabulary Words and s for Algebra Absolute Value Additive Inverse Algebraic Expression Ascending Order Associative Property Axis of Symmetry Base Binomial Coefficient Combine Like Terms ### CORRELATED TO THE SOUTH CAROLINA COLLEGE AND CAREER-READY FOUNDATIONS IN ALGEBRA We Can Early Learning Curriculum PreK Grades 8 12 INSIDE ALGEBRA, GRADES 8 12 CORRELATED TO THE SOUTH CAROLINA COLLEGE AND CAREER-READY FOUNDATIONS IN ALGEBRA April 2016 www.voyagersopris.com Mathematical ### Some facts about polynomials modulo m (Full proof of the Fingerprinting Theorem) Some facts about polynomials modulo m (Full proof of the Fingerprinting Theorem) In order to understand the details of the Fingerprinting Theorem on fingerprints of different texts from Chapter 19 of the ### Some Notes on Taylor Polynomials and Taylor Series Some Notes on Taylor Polynomials and Taylor Series Mark MacLean October 3, 27 UBC s courses MATH /8 and MATH introduce students to the ideas of Taylor polynomials and Taylor series in a fairly limited ### Module MA3411: Abstract Algebra Galois Theory Appendix Michaelmas Term 2013 Module MA3411: Abstract Algebra Galois Theory Appendix Michaelmas Term 2013 D. R. Wilkins Copyright c David R. Wilkins 1997 2013 Contents A Cyclotomic Polynomials 79 A.1 Minimum Polynomials of Roots of ### Polynomials and Factoring Lesson 2 Polynomials and Factoring A polynomial function is a power function or the sum of two or more power functions, each of which has a nonnegative integer power. Because polynomial functions are built ### 1.7. Partial Fractions. 1.7.1. Rational Functions and Partial Fractions. A rational function is a quotient of two polynomials: R(x) = P (x) Q(x). .7. PRTIL FRCTIONS 3.7. Partial Fractions.7.. Rational Functions and Partial Fractions. rational function is a quotient of two polynomials: R(x) = P (x) Q(x). Here we discuss how to integrate rational ### 3. INNER PRODUCT SPACES . INNER PRODUCT SPACES.. Definition So far we have studied abstract vector spaces. These are a generalisation of the geometric spaces R and R. But these have more structure than just that of a vector space. ### a = bq + r where 0 r < b. Lecture 5: Euclid s algorithm Introduction The fundamental arithmetic operations are addition, subtraction, multiplication and division. But there is a fifth operation which I would argue is just as fundamental ### It is time to prove some theorems. There are various strategies for doing CHAPTER 4 Direct Proof It is time to prove some theorems. There are various strategies for doing this; we now examine the most straightforward approach, a technique called direct proof. As we begin, it ### POLYNOMIAL FUNCTIONS POLYNOMIAL FUNCTIONS Polynomial Division.. 314 The Rational Zero Test.....317 Descarte s Rule of Signs... 319 The Remainder Theorem.....31 Finding all Zeros of a Polynomial Function.......33 Writing a ### Decomposing Rational Functions into Partial Fractions: Prof. Keely's Math Online Lessons University of Phoenix Online & Clark College, Vancouver WA Copyright 2003 Sally J. Keely. All Rights Reserved. COLLEGE ALGEBRA Hi! Today's topic is highly structured and ### 5 Indefinite integral 5 Indefinite integral The most of the mathematical operations have inverse operations: the inverse operation of addition is subtraction, the inverse operation of multiplication is division, the inverse ### 3 Factorisation into irreducibles 3 Factorisation into irreducibles Consider the factorisation of a non-zero, non-invertible integer n as a product of primes: n = p 1 p t. If you insist that primes should be positive then, since n could ### 5.1 The Remainder and Factor Theorems; Synthetic Division 5.1 The Remainder and Factor Theorems; Synthetic Division In this section you will learn to: understand the definition of a zero of a polynomial function use long and synthetic division to divide polynomials ### Zeros of Polynomial Functions Zeros of Polynomial Functions Objectives: 1.Use the Fundamental Theorem of Algebra to determine the number of zeros of polynomial functions 2.Find rational zeros of polynomial functions 3.Find conjugate ### Mth 95 Module 2 Spring 2014 Mth 95 Module Spring 014 Section 5.3 Polynomials and Polynomial Functions Vocabulary of Polynomials A term is a number, a variable, or a product of numbers and variables raised to powers. Terms in an expression ### FIRST YEAR CALCULUS. Chapter 7 CONTINUITY. It is a parabola, and we can draw this parabola without lifting our pencil from the paper. FIRST YEAR CALCULUS WWLCHENW L c WWWL W L Chen, 1982, 2008. 2006. This chapter originates from material used by the author at Imperial College, University of London, between 1981 and 1990. It It is is
If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. ### Course: Geometry (FL B.E.S.T.)>Unit 4 Lesson 1: Transformations & congruence Finding missing triangle angle measures and identifying parallel lines from angle measures on transversals help prepare us to learn about congruence. Let’s refresh some concepts that will come in handy as you start the congruence unit of the high school geometry course. You’ll see a summary of each concept, along with a sample item, links for more practice, and some info about why you will need the concept for the unit ahead. This article only includes concepts from earlier courses. There are also concepts within this high school geometry course that are important to understanding congruence. If you have not yet mastered the Intro to Euclidean geometry lesson, the Rigid transformations overview lesson, or the Properties & definitions of transformations lesson, it may be helpful for you to review those before going farther into the unit ahead. ## Using angle relationships ### What is this, and why do we need it? When lines intersect, especially when a transversal intersects a pair of parallel lines, the intersections form angles with special relationships. We will use these angle relationships to explain how to construct parallel or perpendicular lines, which in turn help us to bisect angles and line segments. We will also use parallel sides to reveal more properties of parallelograms. ### Practice Problem 1 Exactly two of the lines in the following figure are parallel. The figure is not to scale. Which two lines are parallel? For more practice, go to Angle relationships with parallel lines. ### Where will we use this? Here are a few of the exercises where reviewing angle relationships might be helpful: ## Finding missing angle measures in triangles ### What is this, and why do we need it? The three angle measures in any triangle add up to $180\mathrm{°}$. We'll use congruence along with other concepts, like the fact that the interior angle measures of a triangle sum to $180\mathrm{°}$, to find missing measurements. ### Practice Problem 2 Find the value of $x$ in the triangle shown below. $x=$ $\mathrm{°}$ For more practice, go to Find angles in triangles. ### Where will we use this? Here are the first few exercises where reviewing angle measures in triangles might be helpful. ## Want to join the conversation? • For the first practice question,how do we know that line a and line d have alternate interior angles and corresponding angles,and why does it not work for line b and line c? • Any pair of lines have these angles, so the question is are they congruent (if you use alternate interior or alternate exterior or corresponding)? or are they supplementary (if you use same side interior or same side exterior)? For any one line and the transversal, you can use the fact that adjacent angle are supplementary and vertical angles are congruent. Thus, for b and c, alternate interior angles would be 88 degrees (180-92) and 91 degrees which are not congruent, thus not parallel. For corresponding angles, you would get the same results of 88 and 91 degrees which are not congruent. • what exactly does congruence mean? • congruence is to figures as equal is to expressions. If two things are congruent, they are the same size and/or the same shape. • As a freshman who started high school yesterday, I am really going to need this as I am in an honors course • same bro • How do you know which one is parallel and which one is not? (1 vote) • If you fine two congruent angles, you know it is parallel. Also, you can use a compass. If you're still confused, the next few videos shall help. -Duskpin the Avatar • I am still confused as to how a and d were parallel. line d was 89 degrees and line a was 91. • the angles add up to 180 degrees and so it makes them parallel. hope that helps. • i am a little confused here. what is congruence exactly?? • Congruence compares two line segments, shapes, or 2-d figures. It two things are exactly the same shape and same size, then they are congruent. If two line segments are congruent, that would mean their lengths would be the same. If two figures are congruent (like congruent triangles), then their angles are the same and their side lengths are the same. • I don't get this. could i get help?
Facts that Matter: Introduction to Trigonometry # Facts that Matter: Introduction to Trigonometry | Mathematics (Maths) Class 10 PDF Download ## Trigonometric Ratios The certain ratios involving the sides of a right-angled triangle are called trigonometric ratios. Look at the adjoining right triangle ABC, right-angled at B. The Trigonometric Ratios of angle A are: Reciprocals of the above T-ratios are: NOTE: I. ‘sin θ ’ is a single symbol and ‘sin’ cannot be detached from ‘θ ’. sin θ ≠ sin × θ. This remark is true for other t-ratios also. II. The values of the trigonometric ratios of an angle depend only on the magnitude of the angle and not on the lengths of the sides of the triangle. III. In a right angle, the hypotenuse is the longest side, therefore, the value of sin A or cos A is always less than 1 or at the most equal to 1. ## Trigonometric Ratios of Some Specific Angles The specific angles are 0°, 30°, 45°, 60° and 90°. Trigonometric ratios of these angles are given in the following table: NOTE: I. The value of sin A increases from 0 to 1, as A increases from 0° to 90°. II. The value of cos A decreases from 1 to 0, as A increases from 0° to 90°. III. The value of tan A increases from 0 to ∞ , as A increases from 0° to 90°. IV. √2 = 1.414 and √3 = 1.732. ### Trigonometric Identities Since an equation is called an identity when it is true for all the values of the variables involved. So, an equation involving trigonometric ratios of an angle is called a trigonometric identity, if it is true for all values of the angles involved. Some of the useful trigonometric identities: (i) cos2A + sin2A = 1 (ii) 1 + tan2A = sec2A (iii) cot2A + 1 = cosec2A The document Facts that Matter: Introduction to Trigonometry | Mathematics (Maths) Class 10 is a part of the Class 10 Course Mathematics (Maths) Class 10. All you need of Class 10 at this link: Class 10 ## Mathematics (Maths) Class 10 120 videos|463 docs|105 tests ## FAQs on Facts that Matter: Introduction to Trigonometry - Mathematics (Maths) Class 10 1. What are trigonometric ratios? Ans. Trigonometric ratios are mathematical functions that relate the angles of a right triangle to the ratio of the lengths of its sides. The three primary trigonometric ratios are sine (sin), cosine (cos), and tangent (tan). 2. How can trigonometric ratios be used to find missing side lengths in a right triangle? Ans. Trigonometric ratios can be used to find missing side lengths in a right triangle through the use of inverse trigonometric functions. For example, if we know the angle and one side length, we can use the sine function to find the length of another side. 3. What are the trigonometric ratios for the angles 0°, 30°, 45°, 60°, and 90°? Ans. The trigonometric ratios for the angles 0°, 30°, 45°, 60°, and 90° are as follows: - For 0°: sin(0°) = 0, cos(0°) = 1, tan(0°) = 0. - For 30°: sin(30°) = 1/2, cos(30°) = √3/2, tan(30°) = √3/3. - For 45°: sin(45°) = √2/2, cos(45°) = √2/2, tan(45°) = 1. - For 60°: sin(60°) = √3/2, cos(60°) = 1/2, tan(60°) = √3. - For 90°: sin(90°) = 1, cos(90°) = 0, tan(90°) = undefined. 4. Can trigonometric ratios be negative? Ans. Yes, trigonometric ratios can be negative. The sign of a trigonometric ratio depends on the quadrant in which the angle is located. In the first quadrant, all trigonometric ratios are positive. In the second quadrant, only the sine ratio is positive. In the third quadrant, only the tangent ratio is positive. In the fourth quadrant, only the cosine ratio is positive. 5. How are trigonometric ratios used in real-life applications? Ans. Trigonometric ratios have various applications in real life, including: - Navigation and GPS systems use trigonometry to calculate distances and angles. - Architects and engineers use trigonometry to design and construct buildings, bridges, and other structures. - Astronomers use trigonometry to study celestial objects and calculate distances between them. - Trigonometry is used in physics and engineering to analyze and model periodic phenomena, such as waves and vibrations. - Trigonometry is also applied in fields like computer graphics, music, and biology. ## Mathematics (Maths) Class 10 120 videos|463 docs|105 tests ### Up next Explore Courses for Class 10 exam ### Top Courses for Class 10 Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests. 10M+ students study on EduRev Track your progress, build streaks, highlight & save important lessons and more! Related Searches , , , , , , , , , , , , , , , , , , , , , ;
# Math Comic More Options: Make a Folding Card no #### Storyboard Text • Systems of Linear Equations • Hello class, today we will be learning different linear equations and how to solve them. • Elimination Method:  -24x + 8y = 4 -6x + 2y = 17 To solve this equation Elimination Method would be the easiest because they are lined up and can be added. After you add them you turn the result into y intercept form, and you end up with y = 3x + 7 • How to solve Linear Equations Elimination Method and Substitution Method • How to solve Linear Equations Elimination Method and Substitution Method • Substitution Method:  3x - 5y = 44 x = 2y + 13 For the substitution method requires you to plug the second equation into the first so you can solve. After plugging it in we get 3(2y + 13) - 5y = 44 6y + 36 - 5y = 44 y + 36 = 44 44 -  36 = 8 y = 8 • 12x - 3y = 9 4x - y = 3 These equations can be solved with the elimination method, they can be divided by the same number and you get  y = -4x + 3 as your final equaiton • One Solution • y = 3x + 8 y = 3x + 6 You can tell these are No solution because they have the same slope, the only difference is the y intercept. • No Solution • 2x - y = 4 6x - 3y = 12 If you try to solve this equation you will end up with the same number on both sides meaning they can be any number. • Infinite Solution More Storyboards By c344a09e Explore Our Articles and Examples ### Teacher Resources Lesson Plans Worksheet TemplatesPoster Templates ### Film Resources Film and Video Resources Video Marketing
# Warm UP Step 1: Write the whole numbers from 2 through 48. Step 2: Circle 2 and cross out all multiples of 2 other than 2 (the first row below is shown. ## Presentation on theme: "Warm UP Step 1: Write the whole numbers from 2 through 48. Step 2: Circle 2 and cross out all multiples of 2 other than 2 (the first row below is shown."— Presentation transcript: Warm UP Step 1: Write the whole numbers from 2 through 48. Step 2: Circle 2 and cross out all multiples of 2 other than 2 (the first row below is shown for you) Step 3: Then go to the next remaining number after 2, circle it, and cross out all its multiples other than itself. Step4: Repeat until every number is either circled or crossed out. Prime and Composite Numbers A prime number is a number greater than 1 whose only whole number factors are and itself. A composite number is a whole number greater than 1 that has whole number other than 1 and itself. The number is neither prime nor composite. 1 factors 1 Writing Factors of a Number Field Trip: A class of 36 students is on a field trip at the aquarium. The teacher wants to break the class into groups of the same size. Find all possible group sizes by writing all of the of 36: factors Identifying Prime and Composite Numbers Ex: Tell whether the number is prime or composite: 1.56 1.11 List the Factors: 56 = 1 x 56 = 2 x 28 = 4 x 14 = 7 x 8 List the Factors: 11 = 1 x 11 Prime Factorization & Factor Trees Prime Factorization: to factor a whole number as a of prime numbers. You can use a tree. Use an when a prime number factor appears more than once. Ex: Use a factor tree to write the prime factorization of 54: product factor exponent Both factors trees = 2 x 3 x 3 x 3 = 2 x 3 3 Greatest Common Factor (GCF) Common Factor : a whole number that is a factor of two or more whole numbers Greatest Common Factor (GCF): the of the common factors, which can be found two ways: 1. Making a list of all factors: Ex: Find the GCF of 48, 24, and 36: nonzero greatest Greatest Common Factor (GCF) 2. Use prime factorization: Ex: Find the GCF of 180 & 126 Guided Practice 1. Write all the factors of the number and tell whether it is prime or composite: a) 35 b) 65 c) 47 2. Use a factor tree to write the prime factorization of the number: 48 3. Find the GCF of 16 and 28 by listing the factors: 4. Find the GCF of 90 and 150 by using prime factorization 1, 5, 7, 35; Composite 1, 5, 13, 65; Composite1, 47; Prime 2 4 x 3 16: 1, 2, 4, 8, 16 28: 1, 2, 4, 7, 14, 28 The GCF is 4 30 Download ppt "Warm UP Step 1: Write the whole numbers from 2 through 48. Step 2: Circle 2 and cross out all multiples of 2 other than 2 (the first row below is shown." Similar presentations
tgt ## Basic Concepts of Probability ### Terminology Usually, probabilities are descriptions of the likelihood of some event occurring (ranging from 0 to 1) The probability of two events occurring will be termed independent if knowledge of the occurrence of non-occurrence of the first event provide does not effect estimates of the probability that the second event will occur Two events are termed mutually exclusive if the occurrence of one event precludes occurrence of the other event A set of events are termed exhaustive if they embody all of the possible outcomes in some situation ### Basic Laws of Probability The additive law of probabilities: given a set of mutually exclusive events, the probability of occurrence of one event or another event is equal to the sum of their separate probabilities Example: Place 100 marbles in a bag; 35 blue, 45 red, and 20 yellow. P(blue)=.35 P(red)=.45 P(yellow)=.20 What is the probability of choosing either a red or a yellow marble from the bag? P(red or yellow) = P(red)+ P(yellow) = .45+.20 = .65 The multiplicative law of probabilities: The probability of the joint occurrence of two or more independent events is the product of their individual probabilities Example: Say that the probability that I am in my office at any given moment of the typical school day is .65 Also, say that the probability that someone is looking for me in my office at any given moment of the school day is .15 What is the probability that that during some particular moment, I am in my office and someone looks for me there? P(in office and someone looks) P(in office) * P(someone looks) = .65 x .15 = .0975 Question: Say than Fred takes the car into work with a probability of .50, walks with a probability of .20, and takes public transit with a probability of .30 Barney, on the other hand, drives into work with a probability of .20, walks with a probability of .65, and takes public transit with a probability of .15 What is the probability that Fred walked or drove to work and Barney walked or took public transit to work, assuming Fred and Barney's behaviour to be independent? ### Joint and Conditional Probabilities The joint probability of two events A & B is the likelihood that both events will occur and is denoted as P(A,B) When the two events are independent, the joint probability simply follows the multiplicative rule thus, P(A,B) = P(A) x P(B) When they are not independent, it gets a little trickier ... but we won't worry about that for now conditional probability is the probability that some event (A) will occur, given that some other even (B) has occurred denoted as P(A|B) An Example: Drinking & Driving Accident No Accident Total Drinking 7 23 30 Not Drinking 6 64 70 Total 13 87 100 P(Drinking) = 30/100 = 0.3000 P(Accident) = 13/100 = 0.1300 P(Drinking, Accident) P(Drinking) x P(Accident) = 0.30 x 0.13 = 0.0390 P(Drinking | Accident) = 7/13 = 0.5385 P(Accident | Drinking) = 7/30 = 0.2333 ### Factorials! We will soon discuss the concepts of permutations and combinations Prior to that, it is necessary to understand another mathematical symbol, the symbol ! (read `factorial') N! = (N) x (N-1) x (N-2) x ... x (1) 5! = 5 x 4 x 3 x 2 x 1 = 120 3! = 3 x 2 x 1 = 6 Note: 0! = 1 ### Permutations If 3 people (p1, p2, & p3) entered a race, how many different finishing orders are possible? p1, p2, p3 p1, p3, p2 p2, p1, p3 p2, p3, p1 p3, p1, p2 p3, p2, p1 each of these is called a permutation of the three people, taken three at a time In permutation notation, this problem would be represented as  and the answer could be solved using the following formula: Another Permutation Example: Say that 5 people entered the previous race (p1 thru p5), but only the first two get prizes. How many different orderings of those first two positions are possible? p1, p2 p1, p3 p1, p4 p1, p5 p2, p1 p2, p3 p2, p4 p2, p5 p3, p1 p3, p2 p3, p4 p3, p5 p4, p1 p4, p2 p4, p3 p4, p5 p5, p1 p5, p2 p5, p3 p5, p4 Note: When doing permutations, order is important! Think of the word "permutations" as "orderings" ### Combinations Sometimes, we are not concerned about ordering but only in how many ways certain things can be combined into groups For example, let's say we again have five people and we want to form a team of two people. How many different teams of two people can we from our original five? 1&2 1&3 1&4 1&5 2&3 2&4 2&5 3&4 3&5 4&5 Other Examples: Say you had seven people and you wanted to form a committee of four people who would work together (on equal footing) to solve some problem. How many different committees are possible? What if the above committees were set up such that one person was to be president, the next vice president, the third treasurer, and the fourth secretary. Now how many committees are possible? ### The Binomial Distribution The binomial distribution occurs in situations in which each of a number of independent trials (termed Bernouli trials) results in one of two mutually exclusive outcomes e.g., coin tosses The mathematical description of the binomial distribution is the following: where: p(X) = The probability of X successes N = The number of trials p = The probability of success on any given trial q = The probability of failure on any given trial (i.e., 1-p) = The number of combinations of N things taken X at a time Examples: Suppose a batter (in baseball) gets a hit with a probability of 0.3, and gets out the rest of the time. What is the probability of that batter getting 0 hits in 10 at bats? What is the probability of flipping a fair coin eight times and getting only two heads? #### Plotting Binomial Distributions Given that we can obtain probabilities for any value of X associated with some level of p, we can also use this probabilities to create a probability distribution For example, if we toss a fair coin ten times, the following table represents the probabilities associated with the indicated outcomes: Number Heads Probability 0 .001 1 .010 2 .044 3 .117 4 .205 5 .246 6 .205 7 .117 8 .044 9 .010 10 .001 If these values were plotted as a distribution, they would look like: Note 3 things; 1) these probabilities are discreet, 2) plotting probabilities, not frequency, 3) mathematically derived data, not empirically acquired Mean of a binomial = Variance of a binomial = Standard Deviation = Testing Hypotheses Given all this, we can now ask questions like the following ... Let's say a person is performing a true/false exam. How many questions out of 10 does a person have to get correct for us to reject the notion that they are just guessing? Number Correct Probability 0 .001 1 .010 2 .044 3 .117 4 .205 5 .246 6 .205 7 .117 8 .044 9 .010 10 .001 ## Sampling Distributions and Hypothesis Testing ### Statistics is arguing Typically, we are arguing either 1) that some value (or mean) is different from some other mean, or 2) that there is a relation between the values of one variable, and the values of another. Thus, following Steve's in-class example, we typically first produce some null hypothesis (i.e., no difference or relation) and then attempt to show how improbably something is given the null hypothesis. ### Sampling Distributions Just as we can plot distributions of observations, we can also plot distributions of statistics (e.g., means) These distributions of sample statistics are called sampling distributions For example, if we consider the 48 students in my class who estimated my age as a population, their guesses have a  of 30.77 and an  of 4.43 (  = 19.58) If we repeatedly sampled groups of 6 people, found the  of their estimates, and then plotted the  s, the distribution might look like ### Hypothesis Testing What I have previously called "arguing" is more appropriately called hypothesis testing Hypothesis testing normally consists of the following steps: • some research hypothesis is proposed (or alternate hypothesis) - H1 • the null hypothesis is also proposed - H0 • the relevant sampling distribution is obtained under the assumption that H0 is correct • I obtain a sample representative of H1 and calculate the relevant statistic (or observation) • Given the sampling distribution, I calculate the probability of observing the statistic (or observation) noted in step 4, by chance • On the basis of this probability, I make a decision ### The beginnings of an example One of the students in our class guessed my age to be 55. I think that said student was fooling around. That is, I think that guess represents something different that do the rest of the guesses H0 - the guess is not really different H1 - the guess is different • obtain a sampling distribution of 0 • calculate the probability of guessing 55, given this distribution • Use that probability to decide whether this difference is just chance, or something more ### A Touch of Philosophy Some students new to this idea of hypothesis testing find this whole business of creating a null hypothesis and then shooting it down as a tad on the weird side, why do it that way? This dates back to a philosopher guy named Karl Popper who claimed that it is very difficult to prove something to be true, but no so difficult to prove it to be untrue So, it is easier to prove H0 to be wrong, than to prove HA to be right In fact, we never really prove H1 to be right. That is just something we imply (similarly H0) ### Using the Normal Distribution to Test Hypotheses The "Steve's Age" example begun earlier is an example of a situation where we want to compare one observation to a distribution of observations This represents the simplest hypothesis-testing situation because the sampling distribution is simply the distribution of the individual observations Thus, in this case we can use the stuff we learned about z-scores to test hypotheses that some individual observation is either abnormally high (or abnormally low) That is, we use our mean and standard deviation to calculate the a z-score for the critical value, then go to the tables to find the probability of observing a value as high or higher than (or as low or lower than) the one we wish to test Finishing the example = 30.77 Critical = 55 = 4.43 ( = 19.58) From the z-table, the area of the portion of the curve above a z of 3.21 (i.e., the smaller portion) is approximately .0006 Thus, the probability of observing a score as high or higher than 55 is .0006. ### Making decisions given probabilities It is important to realize that all our test really tells us is the probability of some event given some null hypothesis It does not tell us whether that probability is sufficiently small to reject H0, that decision is left to the experimenter In our example, the probability is so low, that the decision is relatively easy. There is only a .06% chance that the observation of 55 fits with the other observations in the sample. Thus, we can reject H0 without much worry But what if the probability was 10% or 5%? What probability is small enough to reject H0? It turns out there are two answers to that: #### or Type I and Type II errors First some terminology... The probability level we pick as our cut-off for rejecting H0 is referred to as our rejection level or our significance level Any level below our rejection or significance level is called our rejection region OK, so the problem is choosing an appropriate rejection level In doing so, we should consider the four possible situations that could occur when we're hypothesis testing H0 true H0 false Reject H0 Type I error Correct Fail to Correct Type II error Reject H0Type I Error Type I Error Type I error is the probability of rejecting the null hypothesis when it is really true e.g., saying that the person who guessed I was 55 was just screwing around when, in fact, it was an honest guess just like the others We can specify exactly what the probability of making that error was, in our example it was .06% Usually we specify some "acceptable" level of error before running the study • then call something significant if it is below this level This acceptable level of error is typically denoted as Before setting some level of  it is important to realize that levels of  are also linked to type II errors Type II Error Type II error is the probability of failing to reject a null hypothesis that is really false e.g., judging OJ as not guilty when he is actually guilty The probability of making a type II error is denoted as Unfortunately, it is impossible to precisely calculate  because we do not know the shape of the sampling distribution under H1 It is possible to "approximately" measure  , and we will talk a bit about that in Chapter 8 For now, it is critical to know that there is a trade-off between  and  , as one goes down, the other goes up Thus, it is important to consider the situation prior to setting a significance level While issues of type I versus type II error are critical in certain situations, psychology experiments are not typically among them (although they sometimes are) As a result, psychology has adopted the standard of accepting  =.05 as a conventional level of significance It is important to note, however, that there is nothing magical about this value (although you wouldn't know it by looking at published articles) ### One versus Two Tailed Tests Often, we are interested in determining if some critical difference (or relation) exists and we are not so concerned about the direction of the effect That situation is termed two-tailed, meaning we are interested in extreme scores at either tail of the distribution Note, that when performing a two-tailed test we must only consider something significant if it falls in the bottom 2.5% or the top 2.5% of the distribution (to keep  at 5%) If we were interested in only a high or low extreme, then we are doing a one-tailed or directional test and look only to see if the difference is in the specific critical region encompassing all 5% in the appropriate tail Two-tailed tests are more common usually because either outcome would be interesting, even if only one was expected ### Other Sampling Distributions The basics of hypothesis testing described in this chapter do not change All that changes across chapters is the specific sampling distribution (and its associated table of values) The critical issue will be to realize which sampling distribution is the one to use in which situation ## The Normal Distribution In Chapter 2, we spent a lot of time plotting distributions and calculating numbers to represent the distributions This raises the obvious question: ### Why Bother? Answer: because once we know (or assume) the shape of the distribution and have calculated the relevant statistics, we are then able to make certain inferences about values of the variable In the current chapter, this will be show how this works using the Normal Distribution ### Why the Normal Distribution? As shown by Galton (19th century guy), just about anything you measure turns out to be normally distributed, at least approximately so That is, usually most of the observations cluster around the mean, with progressively fewer observations out towards the extremes Thus, if we don't know how some variable is distributed, our best guess is normality #### A Cautionary Note Although most variables are normally distributed, it is not the case that all variables are normally distributed As examples, consider the following: Values of a dice roll Flipping a coin We will encounter some of these critters (i.e. distributions) later in the course The Relation Between Histograms and Line Graphs Any Histogram: Can be represented as a line graph: Example: Pop Quiz #1 ### Why line graphs? Line graphs make it easier to talk of the "area under the curve" between two points where: area = proportion (or percent) = probability That is, we could ask what proportion of our class scored between 7 & 9 on the quiz If we assume that the total area under the curve equals one, then the area between 7 & 9 equals the proportion of our class that scored between 7 & 9 and also indicates our best guess concerning the probability that some new data point would fall between 7 & 9 The problem is that in order to calculate the area under a curve, you must either: • use calculus (find the integral), or • use a table that specifies the areas associated with given values of your variable The good news is that a table does exist, thereby allowing you to avoid calculus. The bad news is that in order to use it you must: • assume that your variable is normally distributed • use your mean and standard deviation to convert your data into z-scores such that the new distribution has a mean of 0 and a standard deviation of 1 - standard normal distribution or N(0,1) ### The Standard Normal Distribution Mean to Larger Smaller z z Portion Portion .98 .3365 .8365 .1635 .99 .3389 .8389 .1611 1.00 .3413 .8413 .1587 1.01 .3438 .8438 .1562 ### Converting data into Z-scores It would be too much work to provide a table of area values for every possible mean and standard deviation Instead, a table was created for the standard normal distribution, and the dataset of interest is converted to a standard normal before using the table How do we get our mean equal to zero? Simple, subtract the mean from each data point What about the standard deviation? Well, if we divide all values by a constant, we divide the standard deviation by a constant. Thus, to make the standard deviation 1, we just divide each new value by the standard deviation In computational form then, where z is the z-score for the value of X we enter into the above equation Once we have calculated a z-score, we can then look at the z table in Appendix Z to find the area we are interested in relevant to that value • as we'll see, the z table actually provides a number of areas relevant to any specific z-score • What percent of students scored better than 9.2 out of 10 on the quiz, given that the mean was 7.6 and the standard deviation was 1.6 Because we are interested in the area greater than z=1, we look at the "smaller portion" part of the z table and find the value .1587 Thus, 15.87% of the students scored better than 9.2 on the quiz How percent of students scored between 7 & 9 on the quiz?
Mathematical Series Crib Sheets Hit = 351 DigitalDan Contact Site Fibonacci Series The Fibonacci Series is a set of numbers that regularly occur in nature. They can be calculated by starting with the numbers 0 and 1. Keep adding the last two numbers to find the next number in the series. example 0,1 (0 + 1 = 1) so next number is 1 0,1,1 (1 + 1 = 2) so next number is 2 0,1,1,2 (1 + 2 = 3) so next number is 3 0,1,1,2,3 (2 + 3 = 5) so next number is 5 0,1,1,2,3,5 ....... Here are the first few numbers in the Fibonacci Series 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, .... Lucas Series There is another similar series of numbers called Lucas Numbers. These are calculated in exactly the same way, but you start with 2,1 (instead of 0,1) Here are the first few numbers in the Lucas Series 2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843, 1364, 2207, 3571, 5778, 9349, 15127, 24476, 39603, 64079, 103682, 167761, 271443, 439204, 710647, 1149851, 1860498, 3010349, 4870847, .... Pascal's Triangle Each number in the Pascal Triangle is calculated by adding together the two numbers directly above it. (Every line starts and ends with 1.) The numbers in the Pascal Triangle appear in calculations like (x + 1)n e.g. (x + 1)6 uses the numbers from the last line of the triangle (x + 1)6 = 1x6 + 6x5 + 15x4 + 20x3 + 15x2 + 6x + 1 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 Triangular Number Series Start with the number 0. Add 1 for the next number, add 2 for the next number, add 3 ... You now get 0, 0+1, 0+1+2, 0+1+2+3, .... The Triangular series starts 0, 1, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, 231, 253, 276, 300, 325, 351, 378, 406, 435, 465, ... www.DigitalDan.co.uk
# Vertical Angles – Explanation & Examples In this article, we are going to learn what vertical angles are and how to calculate them. Before we begin, let’s first familiarize ourselves with the following concepts about lines. ## What are intersecting and parallel lines? Intersecting lines are straight lines that meet or crosses each other at a certain point. The figure below shows the illustration of intersecting lines. Line PQ and line ST meet at point Q. Therefore, the two lines are intersecting lines. Parallel lines are lines that do not meet at any point in a plane. Line AB and line CD are parallel lines because, they not intersect at any point. ## What are Vertical Angles? Vertical angles are pair angles formed when two lines intersect. Vertical angles are sometimes referred to as vertically opposite angles because the angles are opposite to each other. Real-life settings where vertical angles are used include; railroad crossing sign, letter “X’’, open scissors pliers etc. The Egyptians used to draw two intersecting lines and always measure the vertical angles to confirm that both of them are equal. Vertical angles are always equal to one another. In general, we can say that, 2 pairs of vertical angles are formed when two lines intersect. See the diagram below. In the diagram above: • ∠a and ∠b are vertical opposite angles. The two angles are also equal i.e. ∠a = ∠ • ∠c and ∠d make another pair of vertical angles and they are equal too. • We can also say that the two vertical angles share a common vertex (the common endpoint of two or more lines or rays). Proof of the Vertical Angle Theorem We can prove in the diagram above that. We know that angle b and angle d are supplementary angles i.e. We also know that angle a and angle d are supplementary angles i.e. We can re-arrange the above equations: Comparing the two equations, we have: Hence, proved. Vertical angles are supplementary angles when the lines intersect perpendicularly. For example, ∠W and ∠ Y are vertical angles which are also supplementary angles. Similarly, ∠X and ∠Z are vertical angles which are supplementary. ## How to Find Vertical Angles? There is no specific formula for calculating vertical angles, but you can identify unknown angles by relating different angles as shown the examples below. Example 1 Calculate the unknown angles in the following figure. Solution ∠ 470 and ∠ b are vertical angles. Therefore, ∠ b is also 470 (vertical angles are congruent or equal). ∠470 and ∠ a are supplementary angles. Therefore, ∠a = 1800 – 470 ⇒∠a = 1330 a and ∠c are vertical angles. Hence, ∠ c = 1330 Example 2 Determine the value of θ in the diagram shown below. Solution From the diagram above, ∠ (θ + 20)0 and ∠ x are vertical angles. Therefore, ∠ (θ + 20)0 = ∠ x But 1100 + x = 1800 (supplementary angles) x = (180 – 110)0 = 700 Substitute x = 700 in the equation; ⇒ ∠ (θ + 20)0 = ∠ 700 ⇒ θ = 700 – 200 = 500 Therefore, the value of θ is 50 degrees. Example 3 Calculate the value of angle y in the figure shown below. Solution 1400 + z = 1800 z = 1800 – 1400 z = 400 But (x + y) + z = 1800 (x + y) + 400 = 1800 x + y = 1400 900 + y = 1400 y = 500 Example 4 If 1000 and (3x + 7) ° are vertical angles, find the value of x. Solution Vertical angles are equal, therefore; (3x + 7)0 = 100 0 3x = 100 – 7 3x = 93 x = 310 Hence, the value of x is 31 degrees. Applications of Vertical Angles (h3) Vertical angles have many applications that we see or experience in our daily lives. • The roller coasters are being set on a certain angle for proper operation. These angles are so important that if they displaced a degree above or below, there would be a chance of an accident. The maximum vertical angle set for a roller coaster (Mumbo Jumbo, Flamingo Land’s) is 112 degrees. • At an airshow, we experience two vapor trails that crosses each other and make vertical angles. • Railroad crossing signs (X) placed on the roads for safety of vehicles. • A kite, where two wooden sticks crosses and hold the kite. • The dartboard has 10 pairs of vertical angles, where the bull’s eye is a virtual vertex. ### Practice Questions 1. Using the image shown above, which of the following shows the angle measure of $\angle b$? 2. Using the image shown above, which of the following shows the angle measure of $\angle c$? 3. Using the image shown above, what is the value of $x$ if $\angle b = (3x – 12)^{\circ}$ and $\angle d= (2x – 48)^{\circ}$ 4. Using the image shown above, what is the value of $x$ if $\angle a = (5x – 25)^{\circ}$ and $\angle c = 75^{\circ}$ 5. Using the image shown above, what is the value of $x$ if $\angle b = (5x – 20)^{\circ}$ and $\angle d = 50^{\circ}$?
Search # Variation About the Mean Part D: Deviations from the Mean (30 minutes) In This Part: Tallying Excesses and Deficits In Part B, we looked at excesses and deficits when we moved coins in the stacks to obtain the fair allocation. In Part C, we used a line plot to represent these excesses and deficits. We are now going to explore a new way to consider excesses and deficits. Let’s look at another line plot: For this line plot, here is the corresponding allocation of our 45 coins: Remember that the total of the excesses above the mean must equal the total of the deficits below the mean. In this case, each adds up to 7. If you denote the values of excesses as positive numbers and deficits as negative numbers, then the total of the excesses is: (+2) + (+2) + (+3) = +7 The total of the deficits is: (-4) + (-2) + (-1) = -7 Statisticians refer to these excesses and deficits as deviations from the mean. For this allocation, the deviations from the mean are recorded in the table below. Note that the deviations always sum to 0 because the total excesses (positive deviations) must be the same as the total deficits (negative deviations). Problem D1 Here is another allocation of our 45 coins divided into 9 stacks: a. Draw the corresponding line plot. Indicate the deviation for each dot as was done in the line plot that opened Part D b. Complete the following table of deviations In This Part: Line Plot Representations Use the Interactive Activity or your paper/poster board to display the line plots below (again, for 9 stacks of coins, with a mean of 5). Problem D2 Create a line plot with these deviations from the mean = 5: (-4), (-3), (-2), (-1), (0), (+1), (+2), (+3), (+4) Problem D3 Create a line plot with these deviations from the mean = 5: (-4), (-2), (-2), (-1), (0), (+1), (+2), (+2), (+4) Problem D4 Create a line plot with these deviations from the mean = 5, and specify a set of four remaining values: (-4), (-3), (-3), (-1), (-1) Problem D5 How would the line plots you created in Problems D2-D4 change if you were told that the mean was 6 instead of 5? Would this change the degree of fairness of these allocations (as described in Problem B2)? When the positive and negative deviations are added together, the total is always 0. This property illustrates another way to interpret the mean: The mean is the balance point of the distribution when represented in a line plot, since the total deviation above the mean must equal the total deviation below the mean. ### Solutions Problem D1 a. Here is the corresponding line plot, with the deviation for each dot included under the line: b. Here is the completed table: Problem D2 Here is the line plot: Problem D3 Here is the line plot: Problem D4 The total of the deficits is (-4) + (-3) + (-3) + (-1) + (-1) = -12, so the remaining 4 stacks will need a total excess of +12 to maintain the balance required by the mean. One possible answer is for the remaining four values to be (+1), (+3), (+4), (+4) although many other answers are also possible, which corresponds to the following line plot: Problem D5 The line plots would all move one unit to the right. This would not change the degree of fairness, since the degree of fairness relies on the deviations from the mean. Since the mean will move right along with the values in the line plot, the deviations will not change, and likewise the degree of fairness will not change.
# Difference between revisions of "2000 AMC 12 Problems/Problem 25" ## Problem Eight congruent equilateral triangles, each of a different color, are used to construct a regular octahedron. How many distinguishable ways are there to construct the octahedron? (Two colored octahedrons are distinguishable if neither can be rotated to look just like the other.) $[asy] import three; import math; unitsize(1.5cm); currentprojection=orthographic(2,0.2,1); triple A=(0,0,1); triple B=(sqrt(2)/2,sqrt(2)/2,0); triple C=(sqrt(2)/2,-sqrt(2)/2,0); triple D=(-sqrt(2)/2,-sqrt(2)/2,0); triple E=(-sqrt(2)/2,sqrt(2)/2,0); triple F=(0,0,-1); draw(A--B--E--cycle); draw(A--C--D--cycle); draw(F--C--B--cycle); draw(F--D--E--cycle,dotted+linewidth(0.7)); [/asy]$ $\text {(A)}\ 210 \qquad \text {(B)}\ 560 \qquad \text {(C)}\ 840 \qquad \text {(D)}\ 1260 \qquad \text {(E)}\ 1680$ ## Solutions ### Solution 1 Since the octahedron is indistinguishable by rotations, without loss of generality fix a face to be red. $[asy] size(8cm); defaultpen(0.5); import three; import math; currentprojection=orthographic(2,0.2,1); triple A=(0,0,1); triple B=(sqrt(2)/2,sqrt(2)/2,0); triple C=(sqrt(2)/2,-sqrt(2)/2,0); triple D=(-sqrt(2)/2,-sqrt(2)/2,0); triple E=(-sqrt(2)/2,sqrt(2)/2,0); triple F=(0,0,-1); draw(A--B--E--cycle); draw(A--C--D--cycle); draw(F--C--B--cycle); draw(F--D--E--cycle,dotted+linewidth(0.7)); draw(surface(A--B--C--cycle),rgb(1,.6,.6),nolight);[/asy]$ There are $7!$ ways to arrange the remaining seven colors, but there still are three possible rotations about the fixed face, so the answer is $7!/3 = 1680$. $[asy] size(8cm); defaultpen(0.5); import three; import math; currentprojection=orthographic(2,0,1); triple A=(0,0,1); triple B=(sqrt(2)/2,sqrt(2)/2,0); triple C=(sqrt(2)/2,-sqrt(2)/2,0); triple D=(-sqrt(2)/2,-sqrt(2)/2,0); triple E=(-sqrt(2)/2,sqrt(2)/2,0); triple F=(0,0,-1); triple right=(0,1,0); picture p = new picture, r = new picture, s = new picture; draw(p,A--B--E--cycle); draw(p,A--C--D--cycle); draw(p,F--C--B--cycle); draw(p,F--D--E--cycle,dotted+linewidth(0.7)); draw(p,surface(A--B--C--cycle),rgb(1,.6,.6),nolight); draw(p,surface(A--B--E--cycle),rgb(1,1,.6),nolight); add(scale3(2.2)*p); draw(r,A--B--E--cycle); draw(r,A--C--D--cycle); draw(r,F--C--B--cycle); draw(r,F--D--E--cycle,dotted+linewidth(0.7)); draw(r,surface(A--B--C--cycle),rgb(1,.6,.6),nolight); draw(r,surface(A--C--D--cycle),rgb(1,1,.6),nolight); add(scale3(2.2)*shift(2*right)*r); draw(s,A--B--E--cycle); draw(s,A--C--D--cycle); draw(s,F--C--B--cycle); draw(s,F--D--E--cycle,dotted+linewidth(0.7)); draw(s,surface(A--B--C--cycle),rgb(1,.6,.6),nolight); draw(s,surface(B--C--F--cycle),rgb(1,1,.6),nolight); add(scale3(2.2)*shift(4*right)*s); [/asy]$ ### Solution 2 We consider the dual of the octahedron, the cube; a cube can be inscribed in an octahedron with each of its vertices at a face of the octahedron. So the problem is equivalent to finding the number of ways to color the vertices of a cube. Select any vertex and call it $A$; there are $8$ color choices for this vertex, but this vertex can be rotated to any of $8$ locations. After fixing $A$, we pick another vertex $B$ adjacent to $A$. There are seven color choices for $B$, but there are only three locations to which $B$ can be rotated to (since there are three edges from $A$). The remaining six vertices can be colored in any way and their locations are now fixed. Thus the total number of ways is $\frac{8}{8} \cdot \frac{7}{3} \cdot 6! = 1680 \Rightarrow \mathrm{(E)}$. ### Solution 3 There are 8! ways to place eight colors on a fixed octahedron. An octahedron has six vertices, of which one can face the top, and for any vertex that faces the top, there are four different triangles around that vertex that can be facing you. Thus there are 6*4 = 24 ways to orient an octahedron, and $8!/24 = 1680 \Rightarrow \mathrm{(E)}$
# Can someone teach how to solve probability ## Stochastic independence Here you will find answers to your questions on the topic stochastic independence. This article covers the Independence from events based on an illustrative Example. We also calculate the probabilities with the associated one formula. Our Video on the topic explains briefly everything you need to know about Independence from events should know without having to read this article! ### Independence from events The stochastic independence of events implies that the occurrence of one event has no effect on the likelihood of occurrence of the other event. The event A is called stochastically independent of the event B if the probability P (A) is not influenced by it. It does not matter whether the second event occurs or not. For example, the likelihood that someone will have blue eyes is not related to the likelihood that that person will pass the statistics exam. ### Stochastic independence formula As can be seen, mathematically expressed for independent events, the following applies: The conditional probability so corresponds exactly to that unconditional probability. ### Stochastically independent That is also logical, since the occurrence of B by definition has no influence on the occurrence of A and vice versa. With this assumption, the probability can be calculated with this formula: ### Stochastically dependent But be careful! This formula can only be used for independent events. If the events are dependent, you have to use the following formula: In order to solve tasks related to stochastic independence, you can also use various aids. With the help of this one can map the given information in a structured way. This makes the calculation easier afterwards. A simple four-field board or a Venn diagram enable a better overview of the task with little effort. ### Independence in the tree diagram A tree diagram is also ideal for thisIndependence from events to illustrate. This would look like this, for example: ### Stochastic independence example Now let's look at a suitable example on the subject. Imagine a die is thrown once. We define “odd number” as event A and “number less than 5” as event B. Now you should determine whether the events A and B are different from each other dependent or independent are. #### Calculate stochastic independence First we need to determine the probability of the two events. Since the event A comprises three elements and the result B four, there is a probability of in each case or. . Next, we have to think about how many elements there are at the intersection of A and B, i.e. how many elements appear in both A and B. These are the numbers 1 and 3. Accordingly, the intersection of A and B results in a probability of . #### Check stochastic independence Now we can simply use the formula from before to check whether the events are interdependent or not. The following must apply to independent events: In our case: The events A and B are thus statistically independent of one another. ### Stochastic and causal dependency In conclusion, it is important to point out that stochastic dependency is not the same as causal dependency that you may know from your everyday life. Two events can be stochastically dependent, even if they are not related to each other in cause and effect. Here are the formulas that are important in connection with independent events: The following applies to independent events:
Folding a Square into Thirds Alignments to Content Standards: 8.EE.B 8.EE.C.8 Suppose we take a square piece of paper and fold it in half vertically and diagonally, leaving the creases shown below: Next we make a fold that joins the top of the vertical crease to the bottom right corner, leaving the crease shown below. The point $P$ is the intersection of this new crease with the first diagonal fold. 1. Place the lower left corner of the square at (0,0) on a coordinate grid with the upper right corner at (1,1) as pictured below: The lines $\ell$ and $m$ labelled in the picture contain the two diagonal folds. Find equations defining $\ell$ and $m$ and use these to calculate the coordinates of the point $P$. 2. Explain how to use part (a) in order to fold the square into thirds. IM Commentary The purpose of this task is to find and solve a pair of linear equations which can be used to understand a common method of folding a square piece of origami paper into thirds. It is important for students to do the actual paper folding: if square shaped paper is not available then they will have an extra step to do, constructing a square from a rectangular piece of paper. The teacher may wish to emphasize the connection between reflections and folding a piece of paper: each fold makes a crease along a line segment and the ''fold'' is closely linked with the refection (of the plane containing the paper) along that line. An alternative geometric solution based on similar triangles is presented in www.illustrativemathematics.org/illustrations/1572. Much of this other task is aligned with 8th grade geometry standards (such as 8.G.5), but at this level the arguments involving similarity are supposed to be informal. The teacher may wish to discuss the geometric argument informally. The two reflection lines associated to the folds of the square have relatively simple equations: students may be able to identify an equation defining $m$ right away, although finding an equation defining $\ell$ requires more careful manipulation. Students must both find, using the given graphs, the equations defining the two lines (8.EE.C) and then solve a simultaneous pair of linear equations (8.EE.C.8). The concrete setting and relatively simple coefficients should help facilitate work on this task, but it does have multiple parts and students need to be given ample time to complete it. One practice standard closely related to this task is MP.2, Reason Abstractly and Quantitatively. Students have to translate back and forth between the physical paper folding and the mathematical representation of the paper in the coordinate plane. Another relevant standard of practice is MP.5, Use Appropriate Tools Strategically. This applies both to the fact that the task involves paper folding and to the fact that a computer could be helpful for drawing pictures and verifying the accuracy of the solution. Solution 1. For the line $m$ we can see that this is defined by the equation $y = x$: it has slope 1 since it passes through (0,0) and (1,1) and the $y$-intercept is 0 since it passes through (0,0). For the line $\ell$, we can calculate the slope using the information given. Since $\ell$ contains the points $\left(\frac{1}{2},1\right)$ and $(1,0)$ its slope is $$\frac{1-0}{\frac{1}{2} - 1} = \frac{1}{-\frac{1}{2}} = -2.$$ To find the $y$-intercept of this line, note that going from (1,0) to $\left(\frac{1}{2},1\right)$ we have moved left $\frac{1}{2}$ and up 1. If we move left $\frac{1}{2}$ and up 1 a second time we will be at the point $(0,2)$. This means that the $y$-intercept of $\ell$ is 2. Putting together what we know about the slope and $y$-intercept, an equation defining $\ell$ is $$y = -2x + 2.$$ To solve $y = x$ and $y = -2x + 2$ simultaneously we can first plug in $y = x$ to $y = -2x + 2$ giving $$x = -2x + 2.$$ Solving for $x$ gives $x = \frac{2}{3}$. Plugging this into either of our equations gives $y = \frac{2}{3}$. So the point $P$ in common to $\ell$ and $m$ is $\left(\frac{2}{3},\frac{2}{3}\right)$. 2. If we fold the left side of the square over to $P$ so that the left edge remains parallel to the right edge we will have the additional vertical crease shown in the figure below. This crease will be $\frac{1}{3}$ of the width of the square since it is halfway from the left edge to $P$ and, from part (a), the $x$-coordinate of $P$ is $\frac{2}{3}$: We could then fold the right hand side of the square over to the purple dotted line to get a vertical crease going through $P$, dividing the square into three equal vertical sections, each of width $\frac{1}{3}$.
# NCERT Solutions for Class 7 Maths Exercise 5.2 Chapter 5 Lines and Angles NCERT Solutions for Class 7 Maths Exercise 5.2 Chapter 5 Lines and Angles in simple PDF are given here. The exercise of NCERT Solutions for Class 7 Maths Chapter 5 has topics related to pairs of lines such as, intersecting lines, transversal and angles made by a transversal, transversal of parallel lines, checking of parallel lines. A line that intersects two or more lines at distinct points is a transversal line. Two lines intersect if they have a point in common. Students are suggested to try solving the questions from NCERT Solutions for Class 7 Maths Chapter 5 Lines and Angles and then refer to these solutions to identify the most competent way of approaching different problems. ## Download the PDF of NCERT Solutions For Class 7 Maths Chapter 5 Lines and Angles – Exercise 5.2   ### Access answers to Maths NCERT Solutions for Class 7 Chapter 5 – Lines and Angles Exercise 5.2 1. State the property that is used in each of the following statements? (i) If a ∥ b, then ∠1 = ∠5. Solution:- Corresponding angles property is used in the above statement. (ii) If ∠4 = ∠6, then a ∥ b. Solution:- Alternate interior angles property is used in the above statement. (iii) If ∠4 + ∠5 = 180o, then a ∥ b. Solution:- Interior angles on the same side of transversal are supplementary. 2. In the adjoining figure, identify (i) The pairs of corresponding angles. Solution:- By observing the figure, the pairs of corresponding angle are, ∠1 and ∠5, ∠4 and ∠8, ∠2 and ∠6, ∠3 and ∠7 (ii) The pairs of alternate interior angles. Solution:- By observing the figure, the pairs of alternate interior angle are, ∠2 and ∠8, ∠3 and ∠5 (iii) The pairs of interior angles on the same side of the transversal. Solution:- By observing the figure, the pairs of interior angles on the same side of the transversal are ∠2 and ∠5, ∠3 and ∠8 (iv) The vertically opposite angles. Solution:- By observing the figure, the vertically opposite angles are, ∠1 and ∠3, ∠5 and ∠7, ∠2 and ∠4, ∠6 and ∠8 3. In the adjoining figure, p ∥ q. Find the unknown angles. Solution:- By observing the figure, ∠d = ∠125o … [∵ corresponding angles] We know that, Linear pair is the sum of adjacent angles is 180o Then, = ∠e + 125o = 180o … [Linear pair] = ∠e = 180o – 125o = ∠e = 55o From the rule of vertically opposite angles, ∠f = ∠e = 55o ∠b = ∠d = 125o By the property of corresponding angles, ∠c = ∠f = 55o ∠a = ∠e = 55o 4. Find the value of x in each of the following figures if l ∥ m. (i) Solution:- Let us assume other angle on the line m be ∠y, Then, By the property of corresponding angles, ∠y = 110o We know that Linear pair is the sum of adjacent angles is 180o Then, = ∠x + ∠y = 180o = ∠x + 110o = 180o = ∠x = 180o – 110o = ∠x = 70o (ii) Solution:- By the property of corresponding angles, ∠x = 100o 5. In the given figure, the arms of two angles are parallel. If ∠ABC = 70o, then find (i) ∠DGC (ii) ∠DEF Solution:- (i) Let us consider that AB ∥ DG BC is the transversal line intersecting AB and DG By the property of corresponding angles, ∠DGC = ∠ABC Then, ∠DGC = 70o (ii) Let us consider that BC ∥ EF DE is the transversal line intersecting BC and EF By the property of corresponding angles, ∠DEF = ∠DGC Then, ∠DEF = 70o 6. In the given figures below, decide whether l is parallel to m. (i) Solution:- Let us consider the two lines l and m, n is the transversal line intersecting l and m. We know that the sum of interior angles on the same side of transversal is 180o. Then, = 126o + 44o = 170o But, the sum of interior angles on the same side of transversal is not equal to 180o. So, line l is not parallel to line m. (ii) Solution:- Let us assume ∠x be the vertically opposite angle formed due to the intersection of the straight line l and transversal n, Then, ∠x = 75o Let us consider the two lines l and m, n is the transversal line intersecting l and m. We know that the sum of interior angles on the same side of transversal is 180o. Then, = 75o + 75o = 150o But, the sum of interior angles on the same side of transversal is not equal to 180o. So, line l is not parallel to line m. (iii) Solution:- Let us assume ∠x be the vertically opposite angle formed due to the intersection of the Straight line l and transversal line n, Let us consider the two lines l and m, n is the transversal line intersecting l and m. We know that the sum of interior angles on the same side of transversal is 180o. Then, = 123o + ∠x = 123o + 57o = 180o ∴The sum of interior angles on the same side of transversal is equal to 180o. So, line l is parallel to line m. (iv) Solution:- Let us assume ∠x be the angle formed due to the intersection of the Straight line l and transversal line n, We know that Linear pair is the sum of adjacent angles is equal to 180o. = ∠x + 98o = 180o = ∠x = 180o – 98o = ∠x = 82o Now, we consider ∠x and 72o are the corresponding angles. For l and m to be parallel to each other, corresponding angles should be equal. But, in the given figure corresponding angles measures 82o and 72o respectively. ∴Line l is not parallel to line m. ### Access other exercises of NCERT Solutions For Class 7 Chapter 5 – Lines and Angles Exercise 5.1 Solutions
# Shortcut Method for Time and Work Problems for Bank PO FREE PDF Download Time And Work Problems For Sbi PO, Shortcut Tricks To Solve Time And Work Problems, Time And Work Problems Tricks, Time And Work Problems Pdf, Time And Work Questions For Bank PO, Time And Work Problems With Solutions For Bank Exams Pdf, How To Solve Time And Work Problems In Easy Way, Time and Work Important Formulas, Important Formula for Time and Work Problems, Time and Work – Aptitude Questions and Answers IBPS PO Time and Work Questions with Answers FREE PDF Download- IBPS PO is one the primary bank exams of all times. Competitive exam preparation has always been a crucial task involving dedication and commitment. In this article, we also provide you Important Short Tricks on Time & Work Questions which are asked in bank exams like IBPS. IBPS PO time and work questions with answers FREE PDF download will help you solve maximum number of questions in minimum amount of time. Candidates must take lot of online IBPS PO mock test for practice. Time and work problems based on this concept are not only a little complicated but also asked very frequently asked in competitive exams. So let’s establish a relation between terms and formulas that help to solve such time and work formulas. ## Shortcut Method for Time and Work Problems for Bank PO FREE PDF Download 1. A can do a work in 15 days and B in 20 days. If they work on it together for 4 days, then the fraction of the work that is left is : A. 1 4 B. 1 10 C. 7 15 D. 8 15 2. A is thrice as good as workman as B and therefore is able to finish a job in 60 days less than B. Working together, they can do it in: A.20 days B. 22 1 days 2 C.25 days D.30 days 3. A can finish a work in 18 days and B can do the same work in 15 days. B worked for 10 days and left the job. In how many days, A alone can finish the remaining work? A.5 B. 5 1 2 C.6 D.8 4. A and B can together finish a work 30 days. They worked together for 20 days and then B left. After another 20 days, A finished the remaining work. In how many days A alone can finish the work? A. 40 B. 50 C. 54 D. 60 5. 10 women can complete a work in 7 days and 10 children take 14 days to complete the work. How many days will 5 women and 10 children take to complete the work? A. 3 B. 5 C. 7 D. Cannot be determined E. None of these Time and work is an important topic in IBPS exams, including clerks, PO and specialist officers. In order to make the chapter easy for you all, we provide IBPS PO time and work questions with answers PDF with important short tricks to solve time & work questions will surely make the chapter easy for you all. 6. X and Y can do a piece of work in 20 days and 12 days respectively. X started the work alone and then after 4 days Y joined him till the completion of the work. How long did the work last? A. 6 days B. 10 days C. 15 days D. 20 days 7. A, B and C can complete a piece of work in 24, 6 and 12 days respectively. Working together, they will complete the same work in: A. 1 day 24 B. 7 day 24 C. 3 3 days 7 D.4 days 8. A and B can complete a work in 15 days and 10 days respectively. They started doing the work together but after 2 days B had to leave and A alone completed the remaining work. The whole work was completed in : A. 8 days B. 10 days C. 12 days D. 15 days Each of the questions given below consists of a question followed by three statements. You have to study the question and the statements and decide which of the statement(s) is/are necessary to answer the question. 9. In how many days can 10 women finish a work? I. 10 men can complete the work in 6 days. II. 10 men and 10 women together can complete the work in 3 3 days 7 III. If 10 men work for 3 days and thereafter 10 women replace them, the remaining work in completed in 4 days. A. Any two of the three B. I and II only C. II and III only D. I and III only E. None of these 10. How many workers are required for completing the construction work in 10 days? I. 20% of the work can be completed by 8 workers in 8 days. II. 20 workers can complete the work in 16 days. III. One-eighth of the work can be completed by 8 workers in 5 days. A. I only B. II and III only C. III only D. I and III only E. Any one of the three 1. A’s 1 day’s work = 1 ; 15 B’s 1 day’s work = 1 ; 20 (A + B)’s 1 day’s work = 1 + 1 = 7 . 15 20 60 (A + B)’s 4 day’s work = 7 x 4 = 7 . 60 15 Therefore, Remaining work = 1 – 7 = 8 . 15 15 2. Ratio of times taken by A and B = 1 : 3. The time difference is (3 – 1) 2 days while B take 3 days and A takes 1 day. If difference of time is 2 days, B takes 3 days. If difference of time is 60 days, B takes 3 x 60 = 90 days. 2 So, A takes 30 days to do the work. A’s 1 day’s work = 1 30 B’s 1 day’s work = 1 90 (A + B)’s 1 day’s work = 1 + 1 = 4 = 2 30 90 90 45 A and B together can do the work in 45 = 22 1 days. 2 2 3. B’s 10 day’s work = 1 x 10 = 2 . 15 3 Remaining work = 1 – 2 = 1 . 3 3 Now, 1 work is done by A in 1 day. 18 1 work is done by A in 18 x 1 = 6 days. 3 3 4. (A + B)’s 20 day’s work = 1 x 20 = 2 . 30 3 Remaining work = 1 – 2 = 1 . 3 3 Now, 1 work is done by A in 20 days. 3 Therefore, the whole work will be done by A in (20 x 3) = 60 days 5. 1 woman’s 1 day’s work = 1 70 1 child’s 1 day’s work = 1 140 (5 women + 10 children)’s day’s work = 5 + 10 = 1 + 1 = 1 70 140 14 14 7 5 women and 10 children will complete the work in 7 days. 6. Work done by X in 4 days = 1 x 4 = 1 . 20 5 Remaining work = 1 – 1 = 4 . 5 5 (X + Y)’s 1 day’s work = 1 + 1 = 8 = 2 . 20 12 60 15 Now, 2 work is done by X and Y in 1 day. 15 So, 4 work will be done by X and Y in 15 x 4 = 6 days. 5 2 5 Hence, total time taken = (6 + 4) days = 10 days 7. Formula: If A can do a piece of work in n days, then A’s 1 day’s work = 1 . n (A + B + C)’s 1 day’s work = 1 + 1 + 1 = 7 . 24 6 12 24 Formula: If A’s 1 day’s work = 1 , then A can finish the work in n days. n So, all the three together will complete the job in 24 days = 3 3 days. 7 7 8. (A + B)’s 1 day’s work = 1 + 1 = 1 . 15 10 6 Work done by A and B in 2 days = 1 x 2 = 1 . 6 3 Remaining work = 1 – 1 = 2 . 3 3 Now, 1 work is done by A in 1 day. 15 2 work will be done by a in 15 x 2 = 10 days. 3 3 Hence, the total time taken = (10 + 2) = 12 days. 9. I. (10 x 6) men can complete the work in 1 day. 1 man’s 1 day’s work = 1 60 II. 10 x 24 men + 10 x 24 women can complete the work in 1 day. 7 7 240 men’s 1 day work + 240 women’s 1 day work = 1. 7 7 240 x 1 + 240 women’s 1 day’s work = 1. 7 60 7 240 women’s 1 day’s work = 1 – 4 = 3 7 7 7 10 women’s 1 day’s work = 3 x 7 x 10 = 1 7 240 8 So, 10 women can finish the work in 8 days. III. (10 men’s work for 3 days) + (10 women’s work for 4 days) = 1 (10 x 3) men’s 1 day’s work + (10 x 4) women’s 1 day’s work = 1 30 men’s 1 day’s work + 40 women’s 1 day’s work = 1 Thus, I and III will give us the answer. And, II and III will give us the answer. Correct answer is (A). 10. I. 20 work can be completed by (8 x 8) workers in 1 day. 100 Whole work can be completed by (8 x 8 x 5) workers in 1 day. = 8 x 8 x 5 workers in 10 days = 32 workers in 10 days. 10 II. (20 x 16) workers can finish it in 1 day. (20 x 16) workers can finish it in 10 days. 10 32 workers can finish it in 10 days. III. 1 work can be completed by (8 x 5) workers in 1 day. 8 Whole work can be completed by (8 x 5 x 8) workers in 1 day. = 8 x 5 x 8 workers in 10 days = 32 workers in 10 days. 10 Any one of the three gives the answer. Correct answer is (E).
# Numerical integration# Numerical Methods Numerical integration involves finding the integral of a function. While SymPy can be used to do analytical integration, there are many functions for which finding an analytical solution to integration is very difficult, and numerical integration is used instead. To understand how to perform numerical integration, we first need to understand what exactly is the purpose of integration. For a 1D function, integration means finding the area underneath the curve. However, integration can also be extended for a 2D function and even a 3D function. Quadrature is the term used for numerical evaluation of a definite (i.e. over a range $$[a,b]$$) integral, or in 1D finding the area under a curve. Source: Wikipedia Wondered why the symbol of integration looks like this $$\int$$? The symbol of integration actually comes from the cursive of capital letter $$S$$, meaning summation. But why would it come from the idea of summation? Well, integration is very much related to summation. We know that for a 1D function, which will be our primary topic of discussion today, if we evaluate an integral, we find the area under the function, as illustrated in the figure above. If your curve was a straight line (remember, straight lines are but a subset of curves which has no curves), then it would be rather straightforward to calculate the area under the curve, since you would either have a rectangle if your line was a horizontal straight line, or a trapezoid if your line was not a horizontal straight line. Yet, what happens if your line got a bunch of bend and curves? For example, we could have a very simple equation like $f(x) = \sin(x) +5.$ This equation is a sinusoidal function, of much more complicated shape that is not as easy to get the area through a rectangle or a trapezoid. Therefore, integration comes in to help you find the area. But how does the integration actually get you the area underneath the curve? Well, what integration essentially does is basically breaking the area into smaller and smaller parts, evaluating the area of each part, and then summing each small part together. The small part of the area can be approximated to be a rectangle, a trapezoid, or some other weird shape if you find it suitable. A very simple example, using rectangles are shown below: Source: Wikipedia As each rectangle slice becomes thinner and thinner, the summed area from the rectangles become more and more closely fitting to the area under the curve. It should be understood that if the rectangle slice becomes infinitely thin, then the summed area from the rectangle would become so close to the area underneath the curve that the two would be essentially the same. You may find the mathematics for the Riemann integral on Wikipedia. Of course, integration has come a long way since the Riemann integral, and other integrals were developed to deal with the deficiencies with the Riemann integral. The choice of approximation method, as well as the size of the intervals, will control the error. Better methods as well as smaller (i.e. more to cover our total interval of interest: $$[a,b]$$) sub-intervals will lead to lower errors, but will generally cost more to compute. Here the following quadrature methods will be covered in the context of a simple function: • Midpoint rule (also known as the rectangle method) • Trapezoid rule • Simpson’s rule • Composite Simpson’s rule • Weddle’s rule. ## Example# Let’s begin with a simple function to demonstrate some of the most basic methods for performing numerical integration: $f\left ( x \right ) := \sin \left ( x \right ),$ and assume that we want to know what the area under the, $$\sin$$ function between 0 and $$\pi$$, i.e. $$[a,b]=[0,\pi]$$. The indefinite integral (or anti-derivative) of $$\sin \left ( x \right )$$ is of course $$-\cos \left ( x \right )$$ (plus a constant of integration, $$C$$, which we can simply ignore as we saw above as it drops out as soon as we perform a definite integral). Since we know the indefinite integral exactly in this case, we can perform the definite integration (i.e. find the area under the curve) ourselves exactly by hand: $I := \int_{0}^{\pi} \sin \left ( x \right ) = \left [ -\cos\left ( x \right )+ C \right ]_{0}^{\pi} =-\cos\left ( \pi \right ) - (-\cos\left ( 0 \right )) =-\cos\left ( \pi \right ) + \cos\left ( 0 \right ) = -(-1) + 1 = 2.$ We included the constant $$C$$ here to just to emphasise again the fact that it’s present doesn’t matter - we can just not write it down in this type of expression. Let’s start by plotting the function between these points. import numpy as np import matplotlib.pyplot as plt # Set up the figure fig = plt.figure(figsize=(10, 4)) ax1 = plt.subplot(111) # Get the value of pi from numpy and generate 100 equally spaced values from 0 to pi. x = np.linspace(0, np.pi, 100) # Calculate sin at these points. y = np.sin(x) # plot ax1.plot(x, y, 'b') # Set x axis limits between 0 and pi. ax1.set_xlim([0, np.pi]) ax1.set_ylim([0, 1.1]) # Label axis. ax1.set_xlabel('$x$', fontsize=14) ax1.set_ylabel('$f(x)=\sin(x)$', fontsize=14) ax1.set_title('An example function we wish to integrate', fontsize=14) # Overlay a grid. ax1.grid(True) plt.show() ## Midpoint rule (rectangle method)# The midpoint rule is perhaps the simplest quadrature rule. For reasons you will see below it is sometimes also called the rectangle method. Consider one of the subintervals $$[x_i, x_{i+1}]$$. The midpoint rule approximates the integral over this (the $$i$$-th) subinterval by the area of a rectangle, with a base of length $$(x_{i+1}-x_i)$$ and a height given by the value of $$f(x)$$ at the midpoint of that interval (i.e. at $$x=(x_{i+1}+x_i)/2$$): $I_M^{(i)} := (x_{i+1}-x_i) \times f \left ( \frac {x_{i+1}+x_i} {2} \right ), \quad\text{for} \quad 0\le i \le n-1.$ The midpoint estimate of $$I$$ then simply involves summing up over all the subintervals: $I_M := \sum_{i=0}^{n-1} \, f \left ( \frac {x_{i+1}+x_i} {2} \right )\, (x_{i+1}-x_i).$ Midpoint rule: 1. Divide the interval you want to calculate the area under the curve for into smaller pieces, each will be called a subinterval. 2. Assume that the interval begins at $$x_0$$ and ends at $$x_n$$. We can pick a random subinterval $$[x_i,x_{i+1}]$$ where $$0\le i \le n-1$$. For example, your 1st interval will be $$[x_0,x_1]$$ and your $$i=0$$, and your last interval will be $$[x_{n-1},x_n]$$ and your $$i=n-1$$. 3. For every subinterval, we approxiamte the slice to be a rectangle. To find the area of the rectangle we need to find the width and the height. The width of the rectangle is simply the width of the subinterval. The height of the rectangle can be estimated as the value of the function at the midpoint of the subinterval, so $$f \left ( \frac {x_{i+1}+x_i} {2} \right )$$. To find the area we simply multiply the width of the rectangle by the height of the rectangle. Width of the rectangle: $x_{i+1} - x_i.$ Height of the rectangle: $f \left ( \frac {x_{i+1}+x_i} {2} \right ).$ Area of the rectangle: $(x_{i+1} - x_i) f \left ( \frac {x_{i+1}+x_i} {2} \right ).$ Generalizing the above for all slices, where $$I_M^{(i)}$$ is simply the area of the subinterval $$[x_i,x_{i+1}]$$. $$M$$ subscript here denotes the use of the midpoint method. $I_M^{(i)} := (x_{i+1}-x_i) \times f \left ( \frac {x_{i+1}+x_i} {2} \right ), \quad\text{for} \quad 0\le i \le n-1.$ 4. To find the area under the curve, we need to sum up all of the areas from the subinterval, so we are going to use the summation symbol. We know that the subinterval index goes from the first subinterval where $$i=0$$ to the last subinterval where $$i=n-1$$, thus we arrive at $I_M := \sum_{i=0}^{n-1} f \left ( \frac {x_{i+1}+x_i} {2} \right ) (x_{i+1}-x_i).$ Note that we dropped $$i$$ prefix from $$I$$ because now it is the whole area under the curve, not just the area from one subinterval. Let’s write some code to plot the idea as well as compute an estimate of the integral using the midpoint rule. # this is a matplotlib function that allows us to easily plot rectangles # which will be useful for visualising what the midpoint rule does from matplotlib.patches import Rectangle def f(x): """The function we wish to integrate""" return np.sin(x) # Get the value of pi from numpy and generate equally spaced values from 0 to pi. x = np.linspace(0, np.pi, 100) y = f(x) # Plot fig = plt.figure(figsize=(10, 4)) ax1 = plt.subplot(111) ax1.plot(x, y, 'b', lw=2) ax1.margins(0.1) # Label axis. ax1.set_xlabel('x', fontsize=14) ax1.set_ylabel('$f(x)=\sin(x)$', fontsize=14) # Overlay a grid. ax1.grid(True) number_intervals = 5 xi = np.linspace(0, np.pi, number_intervals+1) I_M = 0.0 for i in range(number_intervals): f((xi[i+1]+xi[i])/2), fill=False, ls='--', color='k', lw=2)) I_M += f((xi[i+1]+xi[i])/2)*(xi[i+1] - xi[i]) ax1.set_title('The sum of the areas of the rectangles is $I_M =$ {:.12f}.'.format(I_M), fontsize=14) plt.show() A more complex example is shown below, where the red line shows the original function we wish to compute the integral of, and the blue rectangles approximate the area under that function for a number of sub-intervals: ### Implementation# Clearly the sum of the areas of all the rectangles provides an estimate of the true integral. In the case above we observe an error of around 1.5%. As we are going to compare different rules below, let’s implement a midpoint rule function. def midpoint_rule(a, b, function, number_intervals=10): """ Our implementation of the midpoint quadrature rule. a and b are the end points for our interval of interest. 'function' is the function of x \in [a,b] which we can evaluate as needed. number_intervals is the number of subintervals/bins we split [a,b] into. Returns the integral of function(x) over [a,b]. """ interval_size = (b - a)/number_intervals # Some examples of some asserts which might be useful here - # you should get into the habit of using these sorts of checks as much as is possible/sensible. assert interval_size > 0 assert type(number_intervals) == int # Initialise to zero the variable that will contain the cumulative sum of all the areas I_M = 0.0 # Find the first midpoint -- i.e. the centre point of the base of the first rectangle mid = a + (interval_size/2.0) # and loop until we get past b, creating and summing the area of each rectangle while (mid < b): # Find the area of the current rectangle and add it to the running total # this involves an evaluation of the function at the subinterval midpoint I_M += interval_size * function(mid) # Move the midpoint up to the next centre of the interval mid += interval_size # Return our running total result return I_M # Check the function runs if it agrees with our first # version used to generate the schematic plot of the method above: print('midpoint_rule(0, np.pi, np.sin, number_intervals=5) = ', midpoint_rule(0, np.pi, np.sin, number_intervals=5)) midpoint_rule(0, np.pi, np.sin, number_intervals=5) = 2.033281476926104 Now let’s test the midpoint function: print("The exact area found by direct integration = 2") for i in (1, 2, 10, 100, 1000): area = midpoint_rule(0, np.pi, np.sin, i) print("Area %g rectangle(s) = %g (error=%g)"%(i, area, abs(area-2))) The exact area found by direct integration = 2 Area 1 rectangle(s) = 3.14159 (error=1.14159) Area 2 rectangle(s) = 2.22144 (error=0.221441) Area 10 rectangle(s) = 2.00825 (error=0.00824841) Area 100 rectangle(s) = 2.00008 (error=8.22491e-05) Area 1000 rectangle(s) = 2 (error=8.22467e-07) Observations: • With one rectangle, we are simply finding the area of a box of shape $$\pi \times 1$$, where $$\pi$$ is the width of the rectangle and $$1$$ is the value of the function evaluated at the midpoint, $$\pi/2$$. So of course the result is $$\pi$$. • As we increase the number of subintervals, or rectangles, we increase the accuracy of our area. We can observe from the slope of the log-log plot of error against number of subintervals that the error is a quadratic function of the inverse of the number of subintervals (or equivalently is quadratically dependent on the spacing between the points - the interval size). This demonstrates that (for this particular example at least), the method demonstrates second-order accuracy - if we halve the interval size the error goes down by a factor of 4! • The simplicity of this method is its weakness, as rectangles (i.e. a flat top) are rarely a good approximation for the shape of a smooth function. • We want to use as few shapes as possible to approximate our function, because each additional rectangle is one extra time round the loop, which includes its own operations as well as an extra evaluation of the function, and hence increases the overall computational cost. ## Trapezoid rule# As previously stated, the slices we use do not have to be rectangles, they can also be trapezoids. Rectangle rule is very similar to the trapezoid rule except for one small difference. For the trapezoid rule, the width of the subinterval will be multiplied by $\frac{f(x_i) + f(x_{i+1})}{2}.$ For the trapezoid rule, the subscipt we will use the subscript $$T$$. If we change the shape of the rectangle to a trapezoid (i.e. the top of the shape now being a linear line fit defined by the values of the function at the two end points of the subinterval, rather than the constant value used in the midpoint rule), we arrive at the trapezoid, or trapezoidal rule. The trapezoid rule approximates the integral by the area of a trapezoid with base $$(x_{i+1}-x_i)$$ and the left- and right-hand-sides equal to the values of the function at the two end points. In this case the area of the shape approximating the integral over one subinterval, is given by: $I_T^{(i)} := (x_{i+1}-x_i) \times \left( \frac {f\left ( x_{i+1}\right ) + f \left (x_{i} \right )} {2} \right) \quad\text{for} \quad 0\le i \le n-1.$ The trapezoidal estimate of $$I$$ then simply involves summing up over all the subintervals: $I_T := \sum_{i=0}^{n-1}\left(\frac{f(x_{i+1}) + f(x_{i})}{2}\right )(x_{i+1}-x_i).$ Let’s write some code to plot the idea and compute an estimate of the integral. # This is a matplotlib function that allows us to plot polygons from matplotlib.patches import Polygon # Get the value of pi from numpy and # generate equally spaced values from 0 to pi. x = np.linspace(0, np.pi, 100) y = f(x) # plot fig = plt.figure(figsize=(10, 4)) ax1 = plt.subplot(111) ax1.plot(x, y, 'b', lw=2) ax1.margins(0.1) # Label axis. ax1.set_xlabel('$x$', fontsize=14) ax1.set_ylabel('$\sin(x)$', fontsize=14) ax1.set_title('Approximating function with trapezoids', fontsize=14) # Overlay a grid. ax1.grid(True) number_intervals = 5 xi = np.linspace(0, np.pi, number_intervals+1) I_T = 0.0 for i in range(number_intervals): xi[i+1], f(xi[i+1])], [xi[i+1], 0]]), closed=True, fill=False, ls='--', color='k', lw=2)) I_T += ((f(xi[i+1]) + f(xi[i]))/2)*(xi[i+1] - xi[i]) ax1.set_title('The sum of the areas of the trapezoids is $I_T =$ {:.12f}.'.format(I_T), fontsize=14) plt.show() For our pictorial example used above, the approximation looks like it should be more accurate than the midpoint rule: The tops of the shapes (now trapezoids) are approximating the variation of the function with a linear function, rather than a flat (constant) function. This looks like it should give more accurate results, but see below. Note that numpy has a function for the trapezoid rule, numpy.trapz, but we’ll make our own that works in a similar way to our midpoint rule function. def trapezoidal_rule(a, b, function, number_intervals=10): """Our implementation of the trapezoidal quadrature rule. Note that as discussed in the lecture this version of the implementation performs redundant function evaluations - see the composite implementation in the homework for a more efficient version. """ interval_size = (b - a)/number_intervals assert interval_size > 0 assert type(number_intervals) == int I_T = 0.0 # Loop to create each trapezoid # note this function takes a slightly different approach to Midpoint # (a for loop rather than a while loop) to achieve the same thing for i in range(number_intervals): # Set the start of this interval this_bin_start = a + (interval_size * i) # Find the area of the current trapezoid and add it to the running total I_T += interval_size * \ (function(this_bin_start)+function(this_bin_start+interval_size))/2.0 # Return our running total result return I_T We can test the function in a similar way: print("The exact area found by direct integration = 2") for i in (1, 2, 10, 100, 1000): area = trapezoidal_rule(0, np.pi, np.sin, i) print("Area %g trapezoid(s) = %g (error=%g)"%(i, area, abs(area-2))) The exact area found by direct integration = 2 Area 1 trapezoid(s) = 1.92367e-16 (error=2) Area 2 trapezoid(s) = 1.5708 (error=0.429204) Area 10 trapezoid(s) = 1.98352 (error=0.0164765) Area 100 trapezoid(s) = 1.99984 (error=0.000164496) Area 1000 trapezoid(s) = 2 (error=1.64493e-06) ## Error analysis# It is important to understand the errors of the numerical integration method we are using. If there are a limited number of slices, the area covered by the slices from trapezoid rule or the midpoint rule will not be slightly different from the actual area under the curve. Of course, if we increase the number of slices, the error becomes smaller; however, we also increase the computational power required. A good numerical integration method should be able to have few slices, meaning using little computational power, but still be able to have a small error only. Thus, to know if the numerical integration method we used is a good or a bad method, we need to analyse the errors, or more exactly the change in the errors with numbers of slices, of our numerical integration method. A good method will have a rapid decrease in the error when increasing the number of slices, while a bad method will have a slow decrease in the error when increasing the number of slices. The accuracy of a quadrature, i.e. the mid point rule, the trapezoidal rule etc. is predicted by examining its behaviour in relationship with polynominals. We say that the degree of accuracy or the degree of precision of a quadrature rule is equal to $$M$$ if it is exact for all polynomials of degree up to and including $$M$$, but not exact for some polynomial of degree $$M+1$$. Clearly both the midpoint and trapezoid rules will give the exact result for both constant and linear functions, but they are not exact for quadratics. Therefore, they have a degree of precision of 1 (remember that through the error analysis, we found that it is 2nd order accurate, which is a different term to degree of precision!!). ### Concave-down functions# The first half of a sine wave is concave-down and we notice from the plot that trapezoidal rule consistently underestimate the area under the curve as line segments are always under the curve. In contrast, the mid-point rule will have parts of each rectangle above and below the curve, hence to a certain extent the errors will cancel each other out. This is why, for this particular example, the errors in the mid-point rule turn out to be approximately half those in the trapezoidal rule. While this result turns out to be generally true for smooth functions, we can always come up with (counter) examples where the trapezoid rule will win. Taylor series analysis can be used to formally construct upper bounds on the quadrature error for both methods. We know that the error when integrating constant and linear functions is zero for our two rules, so let’s first consider an example of integrating a quadratic polynomial. We know analytically that $\int_{0}^{1} x^{2}\,dx = \left.\frac{1}{3}x^3\right|_0^1=\frac {1}{3}.$ Numerically, the midpoint rule on a single interval gives $I_M = 1 \left(\frac {1}{2}\right)^{2} = \frac {1}{4},$ while the trapezoidal rule gives $I_T = 1 \frac {0+1^{2}}{2} = \frac {1}{2}.$ The error for $$I_M$$ is therefore $$1/3 - 1/4 = 1/12$$, while the error for $$I_T$$ is $$1/3 - 1/2 = -1/6$$. Therefore, the midpoint rule is twice as accurate as the trapezoid rule: $|E_M| = \frac{1}{2} |E_T|,$ where $$|E|$$ indicates the error (the absolute value of the difference from the exact solution). This is the case for this simple example, and we can see from the actual error values printed above that it also appears to be approximately true for the sine case (which is not a simple polynomial) as well. ## Simpson’s rule# For our half sine wave, the rectangle method overestimates it by about 0.4%, while the trapezoid method underestimates it for 0.9%. We notice that in this situation, the rectangle method overestimates while the trapezoid method underestimates about twice the overestimate of the rectangle method. Could we combine this to obtain something more accurate? Knowing the error estimates from the two rules explored so far opens up the potential for us to combine them in an appropriate manner to create a new quadrature rule, generally more accurate than either one separately. Suppose $$I_S$$ indicates an unknown, but more accurate, estimate of the integral over an interval. Then, as seen above, as $$I_T$$ has an error that is approximately $$-2$$ times the error in $$I_M$$, the following relation must hold approximately: $I_S - I_T \approx -2 \left ( I_S - I_M\right ).$ This follows from the fact that $$I - I_T \approx -2 \left ( I - I_M\right )$$, provided that $$I_S$$ is closer to $$I$$ than either of the other two estimates. Replacing this approximately equals sign with actual equality defines $$I_S$$ for us in terms of things we know. We can rearrange this to give an expression for $$I_S$$ that yields a more accurate estimate of the integral than either $$I_M$$ or $$I_T$$: $I_S := \frac{2}{3}I_M + \frac{1}{3}I_T.$ We combined twice the overestimate from the rectangle method and once the underestimate from the trapezoid method, and then divided everything by 3 to obtain something more accurate! What we’re doing here is using the fact that we know something about (the leading order behaviour of the) two errors, and we can therefore combine them to cancel this error to a certain extent. This estimate will generally be more accurate than either $$M$$ or $$T$$ alone. The error won’t be zero in general as we’re only cancelling out the leading order term in the error, but a consequence is that we will be left with higher-degree terms in the error expansion of the new quadrature rule which should be smaller (at least in the asymptotic limit), and converge faster. The resulting quadrature method in this case is known as Simpson’s rule. Let’s expand the Simpsons rule by substituting in what we know about the rectangle rule and the trapezoid rule: $I_S = \frac{2}{3}I_M + \frac{1}{3}I_T$ $= \frac{2}{3} (b-a)f\left ( \frac{a+b}{2}\right ) + \frac{1}{3}(b-a)\frac{(f(a) + f(b))}{2}$ $= \frac{(b-a)}{6}\left( f \left ( a\right ) + 4f \left ( c\right ) + f\left ( b\right )\right)$ where $$a$$ and $$b$$ are the end points of an interval and $$c = \left ( a+b\right )/2$$ is the midpoint. Note that an alternate derivation of the same rule involves fitting a quadratic function (i.e. $$P_2(x)$$ rather than the constant and linear approximations already considered) that interpolates the integral at the two end points of the interval, $$a$$ and $$b$$, as well as at the midpoint, $$c = \left ( a+b\right )/2$$, and calculating the integral under that polynomial approximation. Let’s plot what this method is doing and compute the integral for our sine case. # Get the value of pi from numpy and generate equally spaced values from 0 to pi. x = np.linspace(0, np.pi, 100) y = f(x) # plot fig = plt.figure(figsize=(10, 4)) ax1 = plt.subplot(111) ax1.plot(x, y, 'b', lw=2) ax1.margins(0.1) # Label axis. ax1.set_xlabel('x', fontsize=16) ax1.set_ylabel('sin(x)', fontsize=16) # Overlay a grid. ax1.grid(True) number_intervals = 5 xi = np.linspace(0, np.pi, number_intervals+1) I_S = 0.0 for i in range(number_intervals): # Use a non-closed Polygon to visualise the straight sides of each interval ax1.add_patch(Polygon(np.array([[xi[i], f(xi[i])], [xi[i], 0], [xi[i+1], 0], [xi[i+1], f(xi[i+1])]]), closed=False, fill=False, ls='--', color='k', lw=2)) poly_coeff = np.polyfit((xi[i], (xi[i] + xi[i+1])/2.0, xi[i + 1]), (f(xi[i]), f((xi[i] + xi[i+1])/2.0), f(xi[i+1])), 2) # Plot the quadratic using 20 plotting points within the interval ax1.plot(np.linspace(xi[i], xi[i+1], 20), f(np.linspace(xi[i], xi[i+1], 20)), ls='--', color='k', lw=2) # qdd in the area of the interval shape to our running total using Simpson's formula I_S += ((xi[i+1] - xi[i])/6.) * (f(xi[i]) + 4 * f((xi[i] + xi[i+1])/2.0) + f(xi[i+1])) ax1.set_title("The Simpson's rule approximation is $I_s =$ {:.12f}.".format(I_S), fontsize=14) plt.show() It looks much closer to the actual function: Let’s make a function to test it out. def simpsons_rule(a, b, function, number_intervals=10): """ Function to evaluate Simpson's rule. Note that this implementation takes the function as an argument, and evaluates this at the midpoint of subintervals in addition to the end point. Hence additional information is generated and used through This is different to the function/implementation available with SciPy where discrete data only is passed to the function. Bear this in mind when comparing results - there will be a factor of two in the definition of "n" we need to be careful about! Also note that this version of the function performs redundant function evaluations - see the **composite** implementation below. """ interval_size = (b - a)/number_intervals assert interval_size > 0 assert type(number_intervals) == int I_S = 0.0 # Loop to valuate Simpson's formula over each interval for i in range(number_intervals): # Find a, c, and b this_bin_start = a + interval_size * (i) this_bin_mid = this_bin_start + interval_size/2 this_bin_end = this_bin_start + interval_size # Calculate the rule and add to running total. I_S += (interval_size/6) * (function(this_bin_start) + 4 * function(this_bin_mid) + function(this_bin_end)) # Return our running total result return I_S Let’s test the function: print("The area found by direct integration = 2") for i in (1, 2, 10, 100, 1000): area = simpsons_rule(0, np.pi, np.sin, i) print("Area %g Simpson's interval(s) = %g (error=%g)"%(i, area, abs(area-2))) The area found by direct integration = 2 Area 1 Simpson's interval(s) = 2.0944 (error=0.0943951) Area 2 Simpson's interval(s) = 2.00456 (error=0.00455975) Area 10 Simpson's interval(s) = 2.00001 (error=6.78444e-06) Area 100 Simpson's interval(s) = 2 (error=6.76471e-10) Area 1000 Simpson's interval(s) = 2 (error=6.79456e-14) For this simple function you should find far smaller errors, and which drop much more rapidly with smaller $$h$$ (or more sub-intervals). Observations: • The errors are lower than for the midpoint and trapezoidal rules, and the method converges more rapidly - i.e. the relative improvement only gets better for more subintervals. • This expression now integrates up to cubics exactly (by construction), so it is of order 4 (if we halve the interval size, the error goes by a factor of $$2^4=16$$). The convergence can be confirmed in the plot below: • The degree of accuracy or precision of this method is 3. • Simpson’s rule integrates to cubics exactly, so since it’s integrating exactly, cubics cannot contribute to the error. Only the quartic (4th) order terms contribute to the error, so it’s 4th order accurate. • We’re getting down to errors close to the machine precision now when we use 1000 subintervals. Remember: your average consumer grade hardware can only handle that many decimal points, and you will need some rather expensive hardware to have even higher levels of precisions. Continuing with 1000 subintervals is actually not helpful as the error that you will get stops decreasing as it is so small that your computer stops being able to discriminate it from 0. Remember we may well either have a relatively small number of data points, or want to minimise the number of function evaluations well below this relatively high number. This will mean that for problems with lots of variation, and/or in higher dimensions, that we still work to do in improving our quadrature methods. • As was the case with our first trapezoidal implementation, we are performing unnecessary function evaluations here; we can fix this issue through the implementation of a so-called composite version of the rule, which still gives the same result as your Simpson’s rule, but makes it but easier for the computer. The composite Simpson’s rule still does many evaluations, but fewer evaluations than your standard Simpson’s rule. ## Composite Simpson’s rule# If we assume that our interval $$[a,b]$$ has been split up into $$n$$ intervals (or $$n+1$$ data points) we can save some function evaluations by writing Simpson’s rule in the following form: $I_{S} = \frac{\Delta x}{3}\left[ f \left ( x_0\right ) + 4f \left ( x_1\right ) + 2f\left ( x_2\right ) + 4f \left ( x_3\right ) + \cdots + 2 f \left ( x_{n-2}\right ) + 4 f \left ( x_{n-1}\right ) + f \left ( x_{n}\right ) \right]$ $= \frac{\Delta x}{3}\left[ f \left ( x_0\right ) + 2\sum_{i=1}^{n/2 - 1} f\left(x_{2i}\right) + 4\sum_{i=1}^{n/2} f\left(x_{2i-1}\right) + f \left ( x_{n}\right ) \right].$ This is known as the composite Simpson’s rule, or more precisely the composite Simpson’s 1/3 rule. You can find a version of Simpson’s rule implemented by SciPy - scipy.interpolate.simps. Note that this way of formulating Simpson’s rule (where we do not allow additional function evaluations at the midpoints of intervals - we assume we are only in a position to use the given data points) requires that $$n$$ be even. This way of writing the composite form in the case of $$n=2$$ is equivalent to the formula over $$[a,b]$$ that introduced the additional midpoint location $$c$$. Let’s implement this rule: def simpsons_composite_rule(a, b, function, number_intervals=10): """Function to evaluate the composite Simpson's rule only using function evaluations at (number_intervals + 1) points. This implementation requires that the number of subintervals (number_intervals) be even """ assert number_intervals % 2 == 0, "number_intervals is not even" interval_size = (b - a) / number_intervals I_cS2 = function(a) + function(b) # Add in those terms with a coefficient of 4 for i in range(1, number_intervals, 2): I_cS2 += 4 * function(a + i * interval_size) # And those terms with a coefficient of 2 for i in range(2, number_intervals-1, 2): I_cS2 += 2 * function(a + i * interval_size) return I_cS2 * (interval_size / 3.0) Let’s test the rule: print("The area found by direct integration = 2") for i in (2, 10, 100, 1000): area = simpsons_composite_rule(0, np.pi, np.sin, i) print("Area %g rectangle(s) = %g (error=%g)"%(i, area, abs(area-2))) The area found by direct integration = 2 Area 2 rectangle(s) = 2.0944 (error=0.0943951) Area 10 rectangle(s) = 2.00011 (error=0.000109517) Area 100 rectangle(s) = 2 (error=1.08245e-08) Area 1000 rectangle(s) = 2 (error=1.07869e-12) This is a slight improvement for a simple function like $$\sin$$, but will be much more of an improvement for functions which oscillate more, in a relative sense compared to the size of our bins. ## Weddle’s rule# We noted above that Simpson’s rule is fourth-order accurate. Suppose we take an approximation to $$I$$ using $$n$$ subintervals with Simpson’s rule and call the result $$I_S$$, and then apply Simpson’s rule with double the number of intervals ($$2n$$) and call the result $$I_{S_2}$$. Then we have two estimates for the integral where we expect $$I_{S_2}$$ to be approximately $$2^4=16$$ times more accurate than $$S$$. In particular, we expect the lowest (i.e. the leading) order error term in $$I_{S_2}$$ to be precisely one sixteenth that of $$I_S$$. Similar to how we derived Simpson’s rule by combining what we knew of the error for the midpoint and trapezoidal rules, with this knowledge we can combine the two estimates from Simpson’s rule to derive an even more accurate estimate of $$I$$. Let’s call this more accurate rule $$I_W$$, which we can find by solving: $I_W - I_S = 16 \left ( I_W - I_{S_2} \right ),$ for $$I_W$$. With a bit of manipulation, $I_W - I_S = 16 \left ( I_W - I_{S_2} \right )$ $\implies \;\;\; I_W - I_S = 16 I_W - 16 I_{S_2}$ $\implies \;\;\; 15 I_W = 16 I_{S_2} - I_S$ $\implies \;\;\; 15 I_W = 15 I_{S_2} + (I_{S_2} - I_S)$ we get this expression $I_W = I_{S_2} + \frac {\left (I_{S_2} - I_S \right )}{15}.$ This is known as Weddle’s rule, or the extrapolated Simpson’s rule because it uses two different values for the interval size and extrapolates from these two to obtain an even more accurate result. Making a function for this rule is easy as we can just call our Simpson’s rule functions with two values for the number of intervals. ### Implementation# We can implement this by calling already created functions for composite Simpson’s rule: def weddles_rule(a, b, function, number_intervals=10): """ Function to evaluate Weddle's quadrature rule using appropriate calls to the composite_simpson function """ S = simpsons_composite_rule(a, b, function, number_intervals) S2 = simpsons_composite_rule(a, b, function, number_intervals*2) return S2 + (S2 - S)/15. We can test it in a similar way: for i in (2, 10, 100, 1000): area = weddles_rule(0, np.pi, np.sin, i) print("Area with %g Weddle's interval(s) = %g (error=%g)"%(i, area, abs(area-2))) Area with 2 Weddle's interval(s) = 1.99857 (error=0.00142927) Area with 10 Weddle's interval(s) = 2 (error=6.44164e-08) Area with 100 Weddle's interval(s) = 2 (error=6.23945e-14) Area with 1000 Weddle's interval(s) = 2 (error=8.88178e-16) Our final result is much more accurate for fewer required bins: ## Other rules# Note that the above technique of using the same rule, but with different values for the interval size, $$h$$, to derive a more accurate estimate of the integral is an example of what is more generally called Richardson extrapolation. Performing this approach using the trapezoid rule as the starting point leads to what is termed Romberg integration. Taking the idea behind Simpson’s rule which fits a quadratic Lagrange interpolating polynomial to equally spaced points in the interval, end extending to any order Lagrange polynomial leads to the Newton-Cotes family of quadrature rules. Note finally, that even wider families exist where the function being integrated is evaluated at non-equally-spaced points. And of course for practical application these ideas need to be extended to more than one dimension.
# Powers and Roots of Complex Numbers ## De Moivre's Theorem using polar form. Estimated17 minsto complete % Progress Practice Powers and Roots of Complex Numbers MEMORY METER This indicates how strong in your memory this concept is Progress Estimated17 minsto complete % Powers and Roots of Complex Numbers Manually calculating (simplifying) a statement such as: \begin{align*}(14 - 17i)^5\end{align*} or \begin{align*}\sqrt[4]{(3 - 2i)}\end{align*} in present (rectangular) form would be a very intensive process at best. Fortunately you will learn in this lesson that there is an alternative: De Moivre's theorem. De Moivre's theorem is really the only practical method for finding the powers or roots of a complex number, but there is a catch... What must be done to a complex number before De Moivre's theorem can be utilized? ### Powers and Roots of Complex Numbers #### Powers of Complex Numbers How do we raise a complex number to a power? Let’s start with an example: \begin{align*}(-4 - 4i)^3=(-4 - 4i)\cdot (-4 - 4i)\cdot (-4 - 4i)\end{align*} In rectangular form, this can get very complex. What about in r cis θ form? \begin{align*}(-4 - 4i) = 4\sqrt{2}\ cis \left ( \frac{5\pi}{4} \right )\end{align*} So the problem becomes \begin{align*}4\sqrt{2} \ \mbox{cis} \left ( \frac{5\pi}{4} \right ) \cdot \ 4\sqrt{2} \ \mbox{cis}\left ( \frac{5\pi}{4} \right ) \cdot \ 4\sqrt{2} \ \mbox{cis}\left ( \frac{5\pi}{4} \right )\end{align*} and using our multiplication rule from the previous section, \begin{align*}(-4 - 4i)^3 = (4\sqrt{2})^3 \ \mbox{cis}\left ( \frac{15\pi}{4} \right )\end{align*} Notice, (a + bi)3 = r3 cis 3 θ In words: Raise the r-value to the same degree as the complex number is raised and then multiply that by cis of the angle multiplied by the number of the degree. Reflecting on the example above, we can identify De Moivre's Theorem: Let z = r(cos θ + i sin θ) be a complex number in rcisθ form. If n is a positive integer, znis zn = rn (cos() + i sin()) It should be clear that the polar form provides a much faster result for raising a complex number to a power than doing the problem in rectangular form. #### Roots of Complex Numbers You probably noticed long ago that when an new operation is presented in mathematics, the inverse operation often follows. That is generally because the inverse operation is often procedurally similar, and it makes good sense to learn both at the same time. This is no exception: The inverse operation of finding a power for a number is to find a root of the same number. 1. Recall from algebra that any root can be written as x1/n 2. Given that the formula for De Moivre’s theorem also works for fractional powers, the same formula can be used for finding roots: \begin{align*}z^{1/n} = (a + bi)^{1/n} = r^{1/n} cis \left ( \frac{\theta}{n} \right )\end{align*} ### Examples #### Example 1 Earlier, you were asked what should be done to a complex number before you can use De Moivre's theorem on it. A complex number operation written in rectangular form, such as: \begin{align*}(13 - 4i)^3\end{align*} must be converted to polar form before utilizing De Moivre's theorem. #### Example 2 Find the value of \begin{align*}(1 + \sqrt{3}i)^4\end{align*}. \begin{align*}r = \sqrt{(1)^2 + (\sqrt{3})^2} = 2\end{align*} \begin{align*}\mbox{tan} \ \theta_{ref} = \frac{\sqrt{3}}{1},\end{align*} and θ is in the 1st quadrant, so \begin{align*}\theta = \frac{\pi}{3}\end{align*} Using our equation from above: \begin{align*}z^4 = r^4\ cis\ 4\theta\end{align*} \begin{align*}z^4 = (2)^4\ cis\ 4\frac{\pi}{3}\end{align*} Expanding cis form: \begin{align*}z^4 = 16\left ( \mbox{cos} \left ( \frac{4\pi}{3} \right ) + i \ \mbox{sin}\left ( \frac{4\pi}{3} \right ) \right )\end{align*} \begin{align*}= 16\left ( (-0.5) - 0.866i \right )\end{align*} Finally we have z4 = -8 - 13.856i #### Example 3 Find \begin{align*}\sqrt{1 + i}\end{align*}. First, rewriting in exponential form: (1 + i)½ And now in polar form: \begin{align*}\sqrt{1 + i} = \left (\sqrt{2}\ cis \left (\frac{\pi}{4} \right ) \right )^{1/2}\end{align*} Expanding cis form, \begin{align*}= \left (\sqrt{2} \left ( \mbox{cos} \left ( \frac{\pi}{4} \right ) + i \ \mbox{sin} \left ( \frac{\pi}{4} \right ) \right ) \right )^{1/2}\end{align*} Using the formula: \begin{align*}=(2^{1/2})^{1/2} \left ( \mbox{cos} \left ( \frac{1}{2} \cdot \frac{\pi}{4} \right ) + i \ \mbox{sin} \left ( \frac{1}{2} \cdot \frac{\pi}{4} \right ) \right )\end{align*} \begin{align*}= 2^{1/4} \left ( \mbox{cos} \left ( \frac{\pi}{8} \right ) + i \ \mbox{sin} \left ( \frac{\pi}{8} \right ) \right )\end{align*} In decimal form, we get =1.189( 0.924 + 0.383i) =1.099 + 0.455i To check, we will multiply the result by itself in rectangular form: \begin{align*}(1.099 + 0.455i)\ \cdot \ (1.099 + 0.455\mbox{i}) = 1.099^2 + 1.099(0.455i) + 1.099(0.455i) \ +\end{align*} \begin{align*}(0.455i)^2\end{align*} \begin{align*}= 1.208 + 0.500i + 0.500i + 0.208i^2\end{align*} \begin{align*}= 1.208 + i - 0.208 \ \mbox{or}\end{align*} \begin{align*}= 1 + i\end{align*} #### Example 4 Find the value of \begin{align*}x : x^3 = \left (1 - \sqrt{3}i \right )\end{align*}. First we put \begin{align*}1 - \sqrt{3}i\end{align*} in polar form. Use \begin{align*}x = 1, \ y = -\sqrt{3}\end{align*} to obtain \begin{align*}r = 2, \ \theta = \frac{5\pi}{3}\end{align*} let \begin{align*}z = \left ( 1 - \sqrt{3}i \right )\end{align*} in rectangular form \begin{align*}z = 2\ \mbox{cis}\ \left ( \frac{5\pi}{3} \right )\end{align*} in polar form \begin{align*}x = \left (1 - \sqrt{3}i \right )^{1/3}\end{align*} \begin{align*}x = \left [2cis \left ( \frac{5\pi}{3} \right ) \right ]^{1/3}\end{align*} Use De Moivre’s theorem to find the first solution: \begin{align*}x_1 = 2^{1/3} cis \left ( \frac{5\pi /3}{3} \right )\end{align*} or \begin{align*}2^{1/3} cis \left ( \frac{5\pi}{9} \right )\end{align*} Leave answer in cis form to find the remaining solutions: n = 3 which means that the 3 solutions are \begin{align*}\frac{2\pi}{3}\end{align*} radians apart or \begin{align*}x_2 = 2^{1/3} cis \left ( \frac{5\pi}{9} + \frac{2\pi}{3} \right )\end{align*} and \begin{align*}x_3 = 2^{1/3} cis \left ( \frac{5\pi}{9} + \frac{2\pi}{3} + \frac{2\pi}{3} \right )\end{align*} NOTE: It is not necessary to add \begin{align*}\frac{2\pi}{3}\end{align*} again. Adding \begin{align*}\frac{2\pi}{3}\end{align*} three times equals 2π. That would result in rotating around a full circle and to start where it all began- that is the first solution. The three solutions are: \begin{align*}x_1 = 2^{1/3} cis \left (\frac{5\pi}{9} \right )\end{align*} \begin{align*}x_2 = 2^{1/3} cis \left (\frac{11\pi}{9} \right )\end{align*} \begin{align*}x_3 = 2^{1/3} cis \left (\frac{17\pi}{9} \right )\end{align*} Each of these solutions, when graphed will be \begin{align*}\frac{2\pi}{3}\end{align*} apart. Check any one of these solutions to see if the results are confirmed. Checking the second solution: \begin{align*}x_2 = 2^{1/3} cis \left (\frac{11\pi}{9} \right )\end{align*} \begin{align*}= 1.260 \left [ \mbox{cos} \left (\frac{11\pi}{9} \right ) + i \ \mbox{sin} \left (\frac{11\pi}{9} \right ) \right ]\end{align*} \begin{align*}= 1.260[-0.766 - 0.643i]\end{align*} \begin{align*}= -0.965 - 0.810i\end{align*} Does (-0.965 – 0.810i)3 or (-0.965 – 0.810i) (-0.965 – 0.810i) (-0.965 – 0.810i) \begin{align*}= \left (1-\sqrt{3}i\right )?\end{align*} #### Example 5 What are the two square roots of i? Let \begin{align*}z = \sqrt{0 + i}\end{align*}. \begin{align*}r = 1, \ \theta = \pi/2\end{align*} or \begin{align*}z = \left [1 \times \ cis \frac{\pi}{2} \right ]^{1/2}\end{align*} Utilizing De Moivre’s theorem: \begin{align*}z_1 = \left [1 \times \ cis \frac{\pi}{4} \right ]\end{align*} or \begin{align*}z_2 = \left [1 \times \ cis \frac{5\pi}{4} \right ]\end{align*} \begin{align*}z_1 = 1\left ( \mbox{cos}\frac{\pi}{4} + i \ \mbox{sin}\frac{\pi}{4} \right )\end{align*} or \begin{align*}z_2 = 1 \left (\mbox{cos} \frac{5\pi}{4} + i \ \mbox{sin} \frac{5\pi}{4} \right )\end{align*} \begin{align*}z_1 = 0.707 + 0.707i\end{align*} or \begin{align*}z_2 = -0.707 - 0.707i\end{align*} Check for z1 solution: (0.707 + 0.707i)2 = i? 0.500 + 0.500i + 0.500i + 0.500i2 = 0.500 + i + 0.500(-1) or i #### Example 6 Calculate \begin{align*}\sqrt[4]{(1 + 0i)}\end{align*}. What are the four fourth roots of 1? Let z = 1 or z = 1 + 0i. Then the problem becomes find z1/4 = (1 + 0i)1/4. Since \begin{align*}r = 1 \ \theta = 0, \ z^{1/4} = [1 \times cis \ 0]^{1/4}\end{align*} with \begin{align*}z_1 = 1^{1/4} \left (\mbox{cos}\ \frac{0}{4} + i \ \mbox{sin}\ \frac{0}{4} \right )\end{align*} or \begin{align*}1(1 + 0)\end{align*} or \begin{align*}1\end{align*} That root is not a surprise. Now use De Moivre’s to find the other roots: \begin{align*}z_2 = 1^{1/4} \left [\mbox{cos}\left (0 + \frac{\pi}{2}\right ) + i \ \mbox{sin}\left (0 + \frac{\pi}{2} \right ) \right ]\end{align*} Since there are 4 roots, dividing 2π by 4 yields 0.5π or 0 + i or just i \begin{align*}z_3 = 1^{1/4} \left [\mbox{cos}\left (0 + \frac{2\pi}{2}\right ) + i \ \mbox{sin}\left (0 + \frac{2\pi}{2} \right ) \right ]\end{align*} which yields z3 = -1 Finally, \begin{align*}z_4 = 1^{1/4} \left [\mbox{cos}\left (0 + \frac{3\pi}{2}\right ) + i \ \mbox{sin}\left (0 + \frac{3\pi}{2} \right ) \right ]\end{align*} or \begin{align*}z_4 = -i\end{align*} The four fourth roots of 1 are 1, i, -1 and -i. #### Example 7 Calculate \begin{align*}(\sqrt{3} + i)^7\end{align*}. To calculate \begin{align*}(\sqrt{3} + i)^7\end{align*} start by converting to \begin{align*}rcis\end{align*} form. First, find \begin{align*}r\end{align*}. Recall \begin{align*}r = \sqrt{\sqrt{3}^2 + 1^2}\end{align*}. \begin{align*}r = \sqrt{3 + 1}\end{align*} \begin{align*}r = 2\end{align*} If \begin{align*}cos \theta = \frac{\sqrt{3}}{2}\end{align*} and \begin{align*}sin \theta = \frac{1}{2}\end{align*} then \begin{align*}\theta = 30^o\end{align*} and is in quadrant I. Now that we have trigonometric form, the rest is easy: \begin{align*}(\sqrt{3} + i)^7 = [2 (cos 30^o + i sin 30^o)]^7\end{align*} ..... Write the original problem in \begin{align*}r cis\end{align*} form \begin{align*}2^7[ (cos (7 \cdot 30^o) + i sin(7 \cdot 30^o)]\end{align*} ..... De Moivre's theorem \begin{align*}128 [-\frac{\sqrt{3}}{2} + \frac{-1}{2}i]\end{align*} ..... Simplify \begin{align*}(\sqrt{3} + i)^7 = -64\sqrt{3} - 64i\end{align*} ..... Simplify again \begin{align*}\therefore (\sqrt{3} + i)^7 = -64\sqrt{3} - 64i\end{align*} ### Review Perform the indicated operation on these complex numbers: 1. Divide: \begin{align*}\frac{2 + 3i}{1 - i}\end{align*} 2. Multiply: \begin{align*}(-6 - i)(-6 + i)\end{align*} 3. Multiply: \begin{align*}\left (\frac{\sqrt{3}}{2} - \frac{1}{2} i \right )^2\end{align*} 4. Find the product using polar form: \begin{align*}(2 + 2i)(\sqrt{3} - i)\end{align*} 5. Multiply: \begin{align*}2(\mbox{cos} \ 40^\circ + i \ \mbox{sin} \ 40^\circ) \bullet 4(\mbox{cos} \ 20^\circ + i \ \mbox{sin} \ 20^\circ)\end{align*} 6. Multiply: \begin{align*}2 \left (\mbox{cos} \ \frac{\pi}{8} + i \ \mbox{sin} \ \frac{\pi}{8} \right ) \bullet 2 \left (\mbox{cos} \ \frac{\pi}{10} + i \ \mbox{sin} \ \frac{\pi}{10} \right )\end{align*} 7. Divide: \begin{align*}2(\mbox{cos} \ 80^\circ + i \ \mbox{sin} \ 80^\circ) \div 6(\mbox{cos} \ 200^\circ + i \ \mbox{sin} \ 200^\circ)\end{align*} 8. Divide: \begin{align*}3\ \mbox{cis}(130^\circ) \div 4\ \mbox{cis}(270^\circ)\end{align*} Use De Moivre’s theorem. 1. \begin{align*}[3(\mbox{cos} \ 80^\circ + i \ \mbox{sin} \ 80^\circ)]^3\end{align*} 2. \begin{align*}\left [\sqrt{2} \left (\mbox{cos}\ \frac{5\pi}{16} + i \ \mbox{sin} \ \frac{5\pi}{16} \right ) \right ]^4\end{align*} 3. \begin{align*}\left (\sqrt{3} - i \right )^6\end{align*} 4. Identify the 3 complex cube roots of \begin{align*}1 + i\end{align*} 5. Identify the 4 complex fourth roots of \begin{align*}-16i\end{align*} 6. Identify the five complex fifth roots of \begin{align*}i\end{align*} To see the Review answers, open this PDF file and look for section 4.10. ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Please to create your own Highlights / Notes ### Vocabulary Language: English TermDefinition complex number A complex number is the sum of a real number and an imaginary number, written in the form $a + bi$. De Moivre's Theorem De Moivre's theorem is the only practical manual method for identifying the powers or roots of complex numbers. The theorem states that if $z= r(\cos \theta + i \sin \theta)$ is a complex number in $r cis \theta$ form and $n$ is a positive integer, then $z^n=r^n (\cos (n\theta ) + i\sin (n\theta ))$.
# How Many Different Ways Can 6 Books Be Arranged On A Shelf? (TOP 5 Tips) As a result, there are 720 different ways to arrange six books on a shelf. ## How many ways can 6 items be arranged? As a result, the six letters can be made up of any combination of 654 321 letters, or any number of configurations. ## How many ways can 7 different books can be arranged on a shelf? = 5040 different ways. As a result, we may arrange seven different books on a shelf in a total of 5040 distinct ways. ## How many different ways can 5 books be arranged on a shelf? There are 120 different ways to arrange five books on a bookshelf, as a result. ## How many different ways can 6 books be arranged on a shelf if a dictionary and a thesaurus must not be together? There are a total of 720 options. ## How many combinations of 6 items are there? For any combination of elements, each item is either included or excluded from the combo depending on its position in the combo. This indicates that for any combination, there are two possible outcomes for each component. For a total of 6 things, the number of possible combinations would be = 26 = 64. ## How many ways can 7 books be arranged on a shelf 5 at a time? 7 books may be put on a shelf in 2520 different ways, 5 at a time. ## How many ways can you arrange 7? 7! =7! =7! =7! =7! =7! =7! =7! =7! =7! =7! This particular problem is a permutation of the previous problem. Remember that the distinction between permutations and combinations is that with permutations, the order of the elements is important. ## How many ways can you select 3 books from a collection of 7 books and then arrange them in shelf? (a) Any configuration is feasible, including the following: Seven books can be placed among themselves in a total of seven! = 5040 different ways. We recommend reading:  What Are The Primary Books Of Accounts? (Correct answer) ## How many ways can 5 different books be arrange on a shelf if one of the books is a math book and it must be on the first row? ways. If you stack five books one by one vertically on a shelf, there are 120 different ways to arrange them. ## How many ways can 5 different books be arranged on a shelf if two specified books must not be side by side? How many different ways may five distinct books be stacked on a shelf if two specific books are not allowed to be next to one other? There are 120 different ways to arrange five books on a bookshelf, as a result. ## How many different ways can 8 books be arranged on a shelf? 8 books may be arranged on a shelf in any of the following ways: 8*7*6*5*4*3*2*1 That is the number 40320. ## How many ways can 10 books be arranged 6 books at a time in a shelf? ANSWER: As a result, the total number of ways in which 10 books may be squeezed into 6 spots is 10*9*8*7*6*5 = 1,51,200. ## How many ways can 12 math textbooks be arranged on a shelf? Solution: There are three options for each of the twelve books. The correct answer is 312. ## How many ways can you arrange 4 of 6 trophies on a shelf? There are 360 different ways to arrange four of the six kids.
```Lesson 1: Generating Equivalent Expressions Vocabulary  Variable:  A symbol (like a letter) that represents a number. It’s like a placeholder for a number.  Examples: x, y, w are common. 3x=15  Numerical Expression:  A numerical expression is a number, or it is any combination of sums, differences, products, or divisions of numbers that evaluates to a number.  Examples: 3(5) + 2; 8; 15; 3(7–2); 25÷5 Vocabulary  Value of a Numerical Expression:  The value of a numerical expression is the number found by evaluating the expression.  Example: 1/3 • (2 + 4) – 7 is a numerical expression. The value of the numerical expression is calculated as –5.  Other words matching this definition:  Number computed  Computation  Calculation   Simplified value Vocabulary  Expression  An expression is a numerical expression, or it is the result of replacing some (or all) of the numbers in a numerical expression with variables.  Two ways to “build” an expression:  Start out with a numerical expression and replace one/some of the numbers with letters/variables. 1/3•(2 + 4) – 7; 1/3•(x + 4) – 7  Create an expression like x + x(y – z) and realize that if numbers are placed in it for the variables the result would be a numerical expression. Vocabulary  Equivalent Expressions: Two expressions are equivalent if both expressions evaluate to the same number.  Examples: 5(x+3)+4 and 5x + 15 + 4  Expression in Expanded Form: An expression that is written as sums (and/or differences) of products whose factors are numbers, variables, or variables raised to whole number powers.  Examples: 5x + 15 + 4; 8, 3x, x, 7x + 2x + 4 + 3 Vocabulary  Expression in Standard Form:  An expression that is in expanded form where all like terms have been collected. (Simplified expression).  Example: 5x + 19; 9x + 7 Opening Exercise – a  t = number of triangles. q = number of quadrilaterals.  Write an expression using t and q that represents the total number of sides in your envelope. Explain what the terms in your expression represent.  3t + 4q  Triangles have 3 sides, so there will be 3 sides for each triangle in the envelope. This is represented by 3t.  Quadrilaterals have 4 sides, so there will be 4 sides for each quadrilateral in the envelope This is represented by 4q.  Added together we have the total number of sides. Opening Exercise – b  t = number of triangles. q = number of  You and your partner have the same number of triangles and quadrilaterals in your envelopes. Write an expression that represents the total number of sides that you and your partner have. Write more than one expression to represent this total.  What did you get? Opening Exercise – c  Each envelope in the class has the same number of triangles and quadrilaterals. Write an expression that represents the total number of of sides in the room.  We have 15 students in this class.  15(3t + 4q) Opening Exercise – d  Use the given values of t and q and your expression from part a to determine the number of sides that should be found in your envelope.  3t + 4q  t=4 q=2 Opening Exercise – e  Use the same values for t and q and your expression from part b to determine the number of sides that should be contained in combined. Opening Exercise – f  Use the same values for t and q and your expression from part c to determine the number of sides that should be contained in all of the envelopes combined envelope combined. Opening Exercise – g  What do you notice about the various expressions in parts e and f? Example 1: a. Any Order, Any  Rewrite 5x + 3x and 5x – 3x by combining like terms. Write the original expressions and expand each term using addition. What are the new expressions equivalent to? 5x + 3x = x + x + x + x + x + x + x + x = 8x 5x – 3x = x + x + x + x + x = 2x Example 1: b. Any Order, Any  Find the sum of 2x + 1 and 5x  (2x + 1) + 5x original expression  2x + (1 + 5x) Associative property of addition (any grouping)  2x + (5x + 1) (any order)  (2x + 5x) + 1  (2 + 5)x + 1 Combined like terms -Distributive property  7x + 1 Equivalent expression to the given problem Example 1: c. Any Order, Any  Find the sum of –3a + 2 and 5a – 3 (–3a + 2) + (5a – 3) Original expression –3a + 2 + 5a + (–3) –3a + 5a + 2 + (–3) any order, any grouping 2a + (–1) Combine like terms – not simplified 2a – 1 Example 3: Any Order, Any Grouping Multiplication 3(2x)  3(2x) + 4y(5) (3•2)x (3•2)x + (4•5)y 6x 6x + 20y 4y(5)  3(2x) + 4y(5) + 4•2•z (4•5)y (3•2)x + (4•5)y + (4•2)z 20y 6x + 20y + 8z 4•2•z (4•2)z 8z Example 3: Any Order, Any Grouping Multiplication  Alexander says that 3x + 4y is equivalent to (3)(4) + xy because of any order, any grouping. Is he correct? 3x + 4y 3•4 + xy Summary  We can use any order, any grouping of  They are both associative and commutative.  When you move or group them the value of  With subtracting we need to change to Problem Set Examples Write an equivalent expression by combining like terms. Verify the equivalence of your expression and the given expression by evaluating each for the given values: a=2, b=5, c= –3. 1. 3a + 5a  Combine like terms: 8a  Plug in given values for variable and calculate:  8(2) = 16  3(2) + 5(2) = 6 + 10 = 16 Problem Set Examples Use any order, any grouping to write equivalent expressions by combining like terms. Then, erify the equivalence of your expression to the given expression by evaluating for the values given in each problem. 10. 3(6a); for a = 3  Combine like terms:   (3•6)a = 18a Evaluate for given value:  Substitute into 18a: 18(3) = 54  Substitute into 3(6a): 3(6•3) Homework
Related Articles • CBSE Class 12 Syllabus • CBSE Class 12 Maths Notes • CBSE Class 12 Physics Notes • CBSE Class 12 Chemistry Notes • CBSE Class 12 Accountancy Notes • CBSE Class 12 Computer Science (Self-Paced Course) Homogeneous Differential Equations • Difficulty Level : Medium • Last Updated : 08 Jun, 2021 In mathematics, a differential equation is an equation that relates one or more functions and their derivatives. In this article, we are going to discuss homogeneous equations but before jumping to the topic let’s understand homogeneous function first. Homogeneous Function A function f(x, y) in x and y is said to be a homogeneous function of the degree of each term is p. For example: f(x, y) = (x2 + y2 – xy) is a homogeneous function of degree 2 where p = 2. Similarly, g(x, y) = (x3 – 3xy2 + 3x2y + y3) is a homogeneous function of degree 3 where p = 3. In general, a homogeneous function ƒ(x, y) of degree n is expressible as: ƒ(x, y) = xn ƒ(y/x) Homogeneous Differential Equation An equation of the form dy/dx = f(x, y)/g(x, y), where both f(x, y) and g(x, y) are homogeneous functions of the degree n in simple word both functions are of the same degree, is called a homogeneous differential equation. For Example: dy/dx = (x2 – y2)/xy is a homogeneous differential equation. Solving a Homogeneous Differential Equation Let  dy/dx = f(x, y)/g(x, y) be a homogeneous differential equation. Now putting y = vx and dy/dx = (v + x dv/dx) in the given equation, we get v + x dy/dx = F(v) => ∫dv/{F(v) – v} = ∫dx/x => ∫dv/{F(v) – v} = log|x| + C Now, replace v by (y/x) to obtain the required solution. Lets look some examples. Example 1: Solve dy/dx = y2 – x2/2xy? Solution: Clearly, since each of the functions (y2 – x2) and 2xy is a homogeneous function of degree 2, the given equation is homogeneous. Putting y = vx and dy/dx = v + x dy/dx, the given equation becomes v + x dv/dx = (v2x2 – x2)/2vx2 => v + x dv/dx = v2 – 1/2v   [after dividing (v2x2/2vx2 – x2/2vx2)] => x dv/dx = ((v2 – 1/2v) – v) => x dv/dx = -(1 + v2)/2v => 2v/(1 + v2)dv = -1/x dx => ∫2v/(1 + v2)dv = -∫1/x dx [Integrating both the sides] => log | 1 + v2 | = -log | x | + log C => log | 1 + v2 | + log | x | = log C => log | x(1 + v2) | = log C => x(1 + v2) = ±C => x(1 + v2) = C1 => x(1 + y2/x2) = C1  [Putting the original value of v = y/x] => (x2 + y2) = xC1, which is the required solution Example 2: Solve (x√(x2 + y2) – y2)dx + xy dy = 0? Solution: The given equation may be written as dy/dx = y2 – x√(x2 + y2)/xy, which is clearly homogeneous Putting y = vx and dy/dx = v + x dv/dx in it, we get v + x dv/dx = {v2x2 – x√(x2 + v2y2)}/vx2 => x dv/dx = [{v2 – √(1 + v2)}/v – v] => x dv/dx = -√(1 + v2)/v => ∫v/√(1 + v2)dv = -∫dx/xc [Integrating both the sides] => √(1 + v2) = -log | x | + C => √(x2 + y2) + x log | x | = Cx, which is the required solution after putting the value of v = y/x. Example 3: Solve x dy/dx – y = √(x2 + y2)? Solution: The given equation may be written as dy/dx = {y + √(x2 + y2)}/x ,which is clearly homogeneous. Putting y = vx and dy/dx = v + x dv/dx in it, we get v + x dv/dx = {vx + √(x2 + v2x2)}/x => v + x dv/dx = v + √(1+v2) [After dividing the {vx + √(x2 + v2x2)}/x] => x dv/dx = √(1 + v2) [v on the both sides gets cancelled] => dv/√(1+v2) = 1/x dx [after rearranging] => ∫dv/√(1+v2) = ∫1/x dx [integrating both sides] => log | v | + √(1 + v2) | = log | x | + log C => log | {v + √(1 + v2)}/x | = log | C | => {v + √(1 + v2)}/x = ±C => v + √(1 + v2) = C1x, where C1 = ±C => y + √(x2 + y2) = C1x2, which is the required solution after putting the value of v = y/x Example 4: Solve (x cos(y/x))(y dx + x dy) = y sin(y/x)(x dy – y dx)? Solution: The given equation may be written as (x cos(y/x) + y sin(y/x))y – (y sin(y/x) – x cos (y/x)) x . dy/dx = 0 => dy/dx = {x cos (y/x) + y sin(y/x)}y / {y sin(y/x) – x cos(y/x)}x => dy/dx = {cos (y/x) + (y/x)sin(y/x)}(y/x) / {(y/x)sin(y/x) – cos(y/x)} [Dividing numerator and denominator by x2], which is clearly homogeneous ,being a function of (y/x). Putting y = vx and dy/dx = (v + x dv/dx) in it, we get v + x dv/dx = v(cos v + sin v)/(v sin v – cos v) => x dv/dx = [v(cos v + sin v)/(v sin v – cos v) -v] => x dv/dx = 2vcos v/(v sin v – cos v) =>∫{(v sin v – cos v)/2vcos v}dv = ∫x dx  [Integrating both sides] => ∫tan v dv – ∫ dv/v = ∫ 2/x dx => -log | cos v | – log | v | + log C = 2 log | x | => log | cos v | + log | v | + 2log | x | = log | C | => log | x2v cos v | = log | C | => | x2v cos v | = C [After cancelling log on the both sides] => x2v cos v = ± C => x2v cos v = C1 [here we taking ±C = C1] => xy cos(y/x) = C1, which is the required solution after putting the actual value of v = y/x My Personal Notes arrow_drop_up
# HCF Calculator ## What is the meaning of HCF? Before we begin with the calculations part of HCF, let us be clear of the fact that this HCF or the highest common factor is the other word for GCD or the greatest common divisor or GCF / greatest common factor or highest common divisor or GCM / greatest common measure. And as far this GCD is concerned then in mathematics, it turns out to be the greatest common divisor of either 2 or more than that integers only when at the least one of them does not happen to be a 0. It is also stated to be the largest positive integer which divides numbers without leaving a reminder. ## HCF Calculator Enter the X value: Enter the Y value: HCF/GCD: 1,601 total views, 4 views today $$E.g.\, GCD \, of \, 8\, and \ 12\,=\,4$$ ## How to find GCD / HCF of a number? Let us take the example of 54 to understand as to how GCD is calculated. Moreover, this number 54 can also be expressed in the form of 2 integers that too in quite some number of ways in the following manner – $$54 \, \times\, 1\,=\,27\, \times \,2\,=\,18\, \times \, 3\,=\,9\times 6$$ Hence the divisors of 54 are as follows -$$1,2,3,6,9,18,27,54$$ On the other hand, divisors of 24 have been mentioned here -$$1,2,3,4,6,8,12,24.$$ $$common \, divisors\, of\, 54\, and\, 24\,=\,1,2,3,6.$$ $$greatest \, common\, divisor\, of\, 54\, and\, 24\, is \,6$$ $$gcd(54,24)=6$$ Hence, in the above manner, one will be able to write the GCD of a given number. ## How to calculate HCF? One can make use of the prime factorization method in order to calculate HCF / GCD. To get to know as to how to calculate this HCF, we have taken 2 numbers which are 48 & 180 respectively & given below are the prime factorizations of these 2 numbers. $$48\, =\,2 \times 2 \times 2 \times 2 \times 3$$ $$180 \,= \,2 \times 2 \times 3 \times 3 \times 5$$ In the equation given above, the common number shared by 180 & 48 is nothing but 2 & 3. $$therefore,Greatest \,common\, divisor\, = \,2 \times 2 \times 3\, =\, 12$$ An individual will have to keep the simple fact in mind that though there are a number of methods to calculate the greatest common divisor or the highest common factor, still the most – commonly used method to know this GCD is prime factorization method. ### Formula for HCF There is another formula to know the greatest common divisor or the HCF of a number & the formula which is put in to use has been given below – $$H C F\,=\, \frac {H C F \,of\, numerators}{L C M \,of \,denominators}$$ ## How to use an online HCF calculator? To make use of the HCF calculator which is available on this site, the user will be required to follow the steps mentioned below. Firstly, the user will have to enter the number in the field provided & then click on the calculate tab. However, if you again wish to find the HCF or are not so sure about the numbers entered then click on reset tab to re – enter the numbers.
# 2. INEQUALITIES AND ABSOLUTE VALUES Save this PDF as: Size: px Start display at page: ## Transcription 1 2. INEQUALITIES AND ABSOLUTE VALUES 2.1. The Ordering of the Real Numbers In addition to the arithmetic structure of the real numbers there is the order structure. The real numbers can be represented by points on a line and if one point is to the left of another we say that the corresponding number is smaller, or less, than the other Alternatively we can call the numbers on the right of zero the positive numbers and define x < y if x = y + z for some positive z. Numbers to the left of zero are called negative. We define x y to mean that x < y or x = y. Of course if x and y were specific numbers we would know which is the case and would write x < y or x = y instead of the more uncertain x y. But often we do not know which is the case. Sometimes it is convenient to specify the larger number first and write x y instead of y x. Finally, x > y means the same as y < x. Example 1: 2 4. In fact we can be more specific by writing 2 < 4. These can be written alternatively as 4 2 and 4 > 2 respectively. Inequalities look rather like equations and in many ways we can operate with them the way we would equations. But beware. We will later see that there are some important differences. The relation has three important properties that are obvious when you consider the number line. Reflexive Property: x x for all x. Anti-Symmetric Property: If x y and y x then x = y. Transitive Property: If x y and y z then x z. The first of these is so obvious it is barely worth mentioning. The second is occasionally useful as a way of proving that two numbers are equal. The third is something we use all the time. There are other orderings in mathematics where these three properties hold. For example, if S and T are sets the statement S is a subset of T means that every element of S is an element of T. We write S T. For example the set of boys in a family is a subset of the set of children in the family. The analogy is made clear by the fact that the symbol we use for subset is very similar to the one we use for less-than-or-equals. Moreover the reflexive, anti-symmetric and transitive properties hold for subsets. 23 2 Reflexive Property: S S for all S. Anti-Symmetric Property: If S T and T S then S = T. Transitive Property: If S T and T W then S W. Another important property of the ordering of real numbers is the following. Total Order Property: For any two distinct real numbers either x < y or y < x. That is the real numbers are arranged linearly and any two such numbers can be compared. Notice that the subset ordering does not have this property. It is possible to have two distinct sets S, T where neither is a subset of the other. For example, the set of positive numbers is not a subset of the set of negative numbers, nor vice versa Inequalities An algebraic inequality is a statement in which there are two algebraic expressions separated by one of the following:, <,, >. Example 2: Examples of inequalities are the following: (1) 2x + 5 < x 2 ; (2) x 2 y x + 2y. Inequalities look very much like equations and we can work with them like equations in many ways. Theorem 1: If a b and c d then a + c b + d. Proof: Suppose a b and c d. Then b a 0 and d c 0. Then (b + d) (a + c) = (b a) + (d c) 0. Note that this does not work for subtraction. That is, just because a b and c d it does not follow that a c b d. [For example take a = 5, b = 3, c = 0, d = 1.] Similar results hold for the three other types of inequality. When it comes to multiplying or dividing inequalities we must be careful. We can multiply or divide both sides of an inequality by a positive number. Theorem 2: If a b and x > 0 then ax bx. Proof: Suppose a b and x > 0. Then b a 0 and so bx ax = (b a)x 0. Thus ax bx. [This is because if b > a then both b a and x are positive. If b = a then b a = 0.] Similar results hold for the other three types of inequality. Just remember that we must multiply or divide both sides by a positive number. Example 3: If 3 < 5 is 3x < 5x? Answer: If x is positive this is so, but suppose x = 2. Is 6 < 15? No! It is the other way around: 15 < 6. 24 3 When you multiply by a negative number the inequality reverses direction. And if you don t know whether x is positive or negative you are stuck. You can t do anything without considering cases. One of the standard things to do is to solve an inequality, that is to find the range of values of the variables for which the inequality holds. Here we consider only inequalities that inolve just one variable. Example 4: Solve the inequality 3x + 5 < 7x 19. Solution: Subtract 3x + 5 from both sides to get 0 < 4x 24, that is 4x > 24. We now divide both sides by 4 to get x > 6. So the solution is {x x > 6}. Example 5: Solve the inequality x x. Solution: Subtract 5x from both sides to get x 2 5x Now factorise x 2 5x + 6 to get (x 2)(x 3) 0. This is exactly what we might have done if we were solving the equation x = 5x. If (x 2)(x 3) = 0 then indeed we get x = 2 or 3, but what if (x 2)(x 3) < 0? Clearly this can only happen if the factors have opposite signs. So either x 2 > 0 and x 3 < 0, in which case 2 < x < 3, or x 2 < 0 and x 3 > 0 in which case x < 2 and x > 3. The latter case is impossible, A number cannot be less than 2 and at the same time be greater than 3. So we are left with just 2 < x < 3. Including the endpoints where we get equality and the complete solution is {x 2 x 3}. If we sketch the parabola x 2 5x + 6 = (x 2)(x 3) we can see clearly what is going on Intervals In each of the above cases the answer has been an interval. An interval on the real line is a subset S such that if x, y S with x < y then S contains every real number between them. The whole real line is clearly an interval, as is a set {a} containing just one number. (The definition says that if there are two different numbers in the set the set has to contain every number between them. It doesn t say that there have to be two distinct numbers in the set.) These are the extreme types of interval. All other intervals have to be one of the following eight types. [a, b] = {x a x b}; [a, b) = {x a x < b}; (a, b] = {x a < x b}; (a, b) = {x a < x < b}; (, a] = {x x a}; (, a) = {x x < a}; [a, ) = {x x a}; (a, ) = {x x > a}; 25 4 We can represent these by means of diagrams: a b [a, b] [a, b) (a, b] (a, b) (, a] (, a) [a, ) (a, ) The answer to Example 4 can be written as (6, ) and the answer to Example 5 as [2, 3] Harder Inequalities Example 6: Solve the inequality x + 1 2x + 6 > x + 2 3x + 7. Solution: If this was the equation x + 1 2x + 6 = x + 2 3x + 7 we would simple cross multiply, that is multiply both sides by the product of the denominators, to get (x + 1)(3x + 7) = (x + 2)(2x + 6) 3x x + 7 = 2x x + 12 x 2 = 5, giving x = 5. Can we do something similar here? The answer is no, at least not without a lot of tedious case-splitting. We are permitted to cross multiply but if the product of the denominators is negative the inequality changes from < to >. Let s see how this case-splitting would work. Case I: (2x + 6)(3x + 7) > 0: That will come about if both factors are positive or both are negative. Case IA: 2x + 6 > 0 and 3x + 7 > 0: Here x > 3 and x > 7 3. Since 3 < 7 3 both inequalities are satisfied when x > 7 3. In this case we can cross-multiply to get (x + 1)(3x + 7) > (x + 2)(2x + 6) 3x x + 7 > 2x x + 12 x 2 > 5, giving x < 5 or x > 5. We have to combine these with x > 7 3. Now 7 3 is about 2.33 and 5 is about 2.24 so plotting 5, 5 and 7 3 get on the number line we so if x > 7 3 and x < 5 we get the interval 7 3, 5 and if x > 7 3 and x > 5 we get the interval ( 5, ). 26 5 The solution in this case is the union of these intervals which we write as 7 3, 5 ( 5, ). Case IB: 2x + 6 < 0 and 3x + 7 < 0: Here x < 3 and x < 7 3. Since 3 < 7 3 both inequalities are satisfied when x < 3. In this case we can cross-multiply, again getting x 2 > 5, giving x < 5 or x > 5. We have to combine these with x < 3. Now x < 3 is incompatible with x > 5 and if x < 5 and x < 3 we simply have x < 3. So the solution in this case is x < 3. Case II: (2x + 6)(3x + 7) < 0: That will come about if one factors is positive and the other is negative. Case IIA: 2x + 6 > 0 and 3x + 7 < 0: Here x > 3 and x < 7 3. So in this case 3 < x < 7 3. In this case we can cross-multiply but we get (x + 1)(3x + 7) < (x + 2)(2x + 6) 3x x + 7 < 2x x + 12 x 2 < 5, giving x ( 5, 5). We have to combine this with x ( 3, 7 3 ). But this is disjoint with ( 5, 5). That is there are no numbers in common. So case IIA cannot arise. Case IIB: 2x + 6 < 0 and 3x + 7 > 0: Here x < 3 and x > 7 3. Again this case cannot arise. Putting all these cases together we get the solution 7 3, 5 ( 5, ) (, 3). Are you still with me? This method involves endless case-splitting and then having to combine the resulting inequalities, keeping track of which ones are ands and which are ors. There has to be a better way! And there is. Example 6 (again): Solve the inequality x + 1 2x + 6 > x + 2 3x + 7. Solution: We subtract to get x + 1 2x + 6 x + 2 3x + 7 > 0. Putting over a common denominator we get (3x x + 7) (2x x + 12) (2x + 6)(3x + 7) > 0. x 2 5 (2x + 6)(3x + 7) > 0. (x 5)(x + 5) (2x + 6)(3x + 7) > 0. Plotting the four number 5, 3 and 7 3 on the number line we get 6 If x > 5 all four factors are positive and the inequality holds. As x moves left, every time it passes one of these four marked points one of the factors becomes negative. The sign of (x 5)(x + 5) (2x + 6)(3x + 7) alternates between positive and negative as indicated below So the solution set is as before , 5 ( 5, ) (, 3) Theorem 3: If 0 a < b then a 2 < b 2. Proof: Suppose 0 a < b. Then b 2 a 2 = (b a)(b + a) > 0 since both factors are positive. Hence a 2 < b 2. Example 7: Solve the inequality 5x 6 < x. Solution: To begin with we must have x 6 5 for the square root to exist. Also, since denotes the non-negative square root we must have x > 0. If x 6 5 then the right hand side is positive so we can square both sides. 5x 6 < x 2 and so x 2 5x + 6 > 0, that is (x 2)(x 3) > 0. This is satisfied when x < 2 and also when x > 3. But since x 6 5 this reduces to [6/5, 2) (3, ) Example 8 (Hard): Find two real numbers x such that x 2 + x + 5 < 3. Solution: This seems to be a much easier question than finding all the solutions. Clearly x = 2 is one solution so all we have to do is use trial and error to find another. But it s not so easy as the solution set is very narrow. The smart thing would be to use continuity and try values of x very slightly bigger than 2, and yes, this will succeed. The value x = 2.1 works just. Alright, let s change the question to solving the inequality. For the square roots to exist we must have x 2. Since both sides will be positive we can square both sides, getting (x 2) + (x + 5) + 2 (x 2)(x + 5) < 9. (x 2)(x + 5) < 6 2x 2 = 3 x. We can conclude from this that x 3 but we can make the interval narrower. We are entitled here to square both sides again, getting (x 2)(x + 5) < x 2 6x + 9. x 2 + 3x 10 < x 2 6x x < 7 x < 19 9 = The solution set is [2, ). So two real numbers satisfying the inequality will be, for example x = 2 and x = Absolute Values The absolute value of a real number x is the magnitude of x, ignoring the sign. It is denoted by x. If x is positive then x = x and if x is negative then x = x. And, of course 0 = 0. We can define the absolute value of x very compactly as x 2, remembering that denotes the positive square root. But, of course, it would be silly to find the absolute value this way! Clearly xy = x. y for all real numbers x, y but things don t work out quite so neatly for sums. It is NOT true in general that x + y = x + y. If x, y have opposite signs then x + y will be less than x + y. For example. 3 + ( 1) = 2 while = 4. All we can say for sums is that x + y x + y. Theorem 4 (Triangle Inequality): x + y x + y for all real numbers x, y. Proof: We could prove this by examining various cases but the simplest proof is the following. First note that xy = x. y x. y. Hence (x + y) 2 = x 2 + y 2 + 2xy = x 2 + y 2 + 2xy x 2 + y x. y ( x + y ) 2. Taking positive square roots we get x + y x + y. We get equality unless x, y have opposite signs. The reason for calling this result the Triangle Inequality is that the corresponding result for complex numbers (where numbers live in a plane and z denotes the length of the line joining 0 to z) can be interpreted as saying that the length of any side of a triangle is less than or equal to the sum of the lengths of the other two sides. Example 9: Solve the equation 3 2x = x + 1. Solution: Here case splitting is not such a bad option. Case I: x 1: The equation becomes 3 2x = x 1, so x = 4. This lies outside the range for this case so we reject it. Case II: 1 < x < 3 2 : The equation is now 3 2x = x + 1, so 3x = 2 and so x = 2 3. Case III: x 3 : We solve 2x 3 = x + 1, getting x = 4, which this time we accept. 2 So the solutions are x = 2 3 and x = 4. 29 8 However a better option is to square both sides, removing the absolute value signs. (3 2x) 2 = (x + 1) 2, giving 4x 2 12x + 9 = x 2 + 2x + 1 or 3x 2 14x + 8 = 0. (3x 2)(x 4) = 0 and so x = 2 3 or 4. Example 10: Solve the inequality 3 2x > x + 1. Solution: Since both sides are positive we may square: (3 2x) 2 > (x + 1) 2. 4x 2 12x + 9 > x 2 + 2x x 2 14x + 8 > 0. (3x 2)(x 4) > 0. x > 2 3 and x > 4, in which case x > 4, or x < 2 3 and x < 4, in which case, x < 2 3. So the solution is (, 2/3) (4, ). Example 11: Solve the inequality 3 x > 2 x + 1. Solution: We can t square both sides because the right hand side could be negative. But we can rewrite the equation as x 3 + x + 1 > 2 and now it is permissible to square both sides getting (x 3) 2 + (x + 1) (x 3)(x + 1) > 4. 2x 2 4x (x 3)(x + 1) > 4. x 2 2x (x 3)(x + 1) > 0. We could square both sides again but that would give us an inequality involving x 4. Not pretty! Instead we fall back on good old case splitting. Case I: (x 3)(x + 1) 0: Clearly we have x 1 or x 3 in this case. The equation reduces to x 2 2x x 2 2x 3 > 0. 2x 2 4x > 0. x(x 2) > 0, which gives x < 0 or x > 2. We have to intersect this region with the region for the case. That is, x 1 or x 3 and x < 0 or x > 2. The former region lies within the latter so the solution in this case is x 1 or x 3. In other words it is (, 1] [3, ). Case II: (x 3)(x + 1) < 0: This is the case 1 < x < 3. The equation reduces to x 2 2x + 3 x 2 + 2x + 3 > 0. 6 > 0. But this is always true so the entire region for case II is in the solution set. This is the interval ( 1, 3). Combining the solutions for the two cases we get (, 1] ( 1, 3) [3, ). But this is the entire real line! Do you get the feeling that there has to be a simpler method? Let s try casesplitting from the very beginning. Write the inequality as x 3 + x + 1 > 2. Case I: x 1: The equation becomes 3 x x 1 > 2, that is, x < 0. So the entire case I is in the solution set. Case II: 1 < x < 3: The equation becomes 3 x + x + 1 > 2, that is, 2 > 0. So the entire case II is in the solution set. 30 9 Case III: x 3: The equation becomes x 3 + x + 1 > 2, that is, x > 0. So the entire case III is in the solution set. Clearly the solution set is the whole real line. Hmm! The might be an even simpler method. Consider (x 3) (x + 1). On the one hand this is 4 = 4. But the Triangle Inequality gives (x 3) (x + 1) x 3 + x + 1. So in fact x 3 + x for all x. Yes this is the solution. 31 10 32 ### Welcome to Analysis 1 Welcome to Analysis 1 Each week you will get one workbook with assignments to complete. Typically you will be able to get most of it done in class and will finish it off by yourselves. In week one there ### Algebra Revision Sheet Questions 2 and 3 of Paper 1 Algebra Revision Sheet Questions and of Paper Simple Equations Step Get rid of brackets or fractions Step Take the x s to one side of the equals sign and the numbers to the other (remember to change the ### 1.1 Solving a Linear Equation ax + b = 0 1.1 Solving a Linear Equation ax + b = 0 To solve an equation ax + b = 0 : (i) move b to the other side (subtract b from both sides) (ii) divide both sides by a Example: Solve x = 0 (i) x- = 0 x = (ii) ### 1.4. Arithmetic of Algebraic Fractions. Introduction. Prerequisites. Learning Outcomes Arithmetic of Algebraic Fractions 1.4 Introduction Just as one whole number divided by another is called a numerical fraction, so one algebraic expression divided by another is known as an algebraic fraction. ### Analysis MA131. University of Warwick. Term Analysis MA131 University of Warwick Term 1 01 13 September 8, 01 Contents 1 Inequalities 5 1.1 What are Inequalities?........................ 5 1. Using Graphs............................. 6 1.3 Case ### 1 Algebra - Partial Fractions 1 Algebra - Partial Fractions Definition 1.1 A polynomial P (x) in x is a function that may be written in the form P (x) a n x n + a n 1 x n 1 +... + a 1 x + a 0. a n 0 and n is the degree of the polynomial, ### 5 =5. Since 5 > 0 Since 4 7 < 0 Since 0 0 a p p e n d i x e ABSOLUTE VALUE ABSOLUTE VALUE E.1 definition. The absolute value or magnitude of a real number a is denoted by a and is defined by { a if a 0 a = a if a ### CHAPTER 3 Numbers and Numeral Systems CHAPTER 3 Numbers and Numeral Systems Numbers play an important role in almost all areas of mathematics, not least in calculus. Virtually all calculus books contain a thorough description of the natural, ### CHAPTER 2. Inequalities CHAPTER 2 Inequalities In this section we add the axioms describe the behavior of inequalities (the order axioms) to the list of axioms begun in Chapter 1. A thorough mastery of this section is essential ### Guide to SRW Section 1.7: Solving inequalities Guide to SRW Section 1.7: Solving inequalities When you solve the equation x 2 = 9, the answer is written as two very simple equations: x = 3 (or) x = 3 The diagram of the solution is -6-5 -4-3 -2-1 0 ### The Not-Formula Book for C1 Not The Not-Formula Book for C1 Everything you need to know for Core 1 that won t be in the formula book Examination Board: AQA Brief This document is intended as an aid for revision. Although it includes ### INTRODUCTION TO PROOFS: HOMEWORK SOLUTIONS INTRODUCTION TO PROOFS: HOMEWORK SOLUTIONS STEVEN HEILMAN Contents 1. Homework 1 1 2. Homework 2 6 3. Homework 3 10 4. Homework 4 16 5. Homework 5 19 6. Homework 6 21 7. Homework 7 25 8. Homework 8 28 ### It is time to prove some theorems. There are various strategies for doing CHAPTER 4 Direct Proof It is time to prove some theorems. There are various strategies for doing this; we now examine the most straightforward approach, a technique called direct proof. As we begin, it ### The Product Property of Square Roots states: For any real numbers a and b, where a 0 and b 0, ab = a b. Chapter 9. Simplify Radical Expressions Any term under a radical sign is called a radical or a square root expression. The number or expression under the the radical sign is called the radicand. The radicand ### 6 EXTENDING ALGEBRA. 6.0 Introduction. 6.1 The cubic equation. Objectives 6 EXTENDING ALGEBRA Chapter 6 Extending Algebra Objectives After studying this chapter you should understand techniques whereby equations of cubic degree and higher can be solved; be able to factorise ### Lecture 1 (Review of High School Math: Functions and Models) Introduction: Numbers and their properties Lecture 1 (Review of High School Math: Functions and Models) Introduction: Numbers and their properties Addition: (1) (Associative law) If a, b, and c are any numbers, then ( ) ( ) (2) (Existence of an ### 2. Simplify. College Algebra Student Self-Assessment of Mathematics (SSAM) Answer Key. Use the distributive property to remove the parentheses College Algebra Student Self-Assessment of Mathematics (SSAM) Answer Key 1. Multiply 2 3 5 1 Use the distributive property to remove the parentheses 2 3 5 1 2 25 21 3 35 31 2 10 2 3 15 3 2 13 2 15 3 2 ### STRAND B: Number Theory. UNIT B2 Number Classification and Bases: Text * * * * * Contents. Section. B2.1 Number Classification. B2. STRAND B: Number Theory B2 Number Classification and Bases Text Contents * * * * * Section B2. Number Classification B2.2 Binary Numbers B2.3 Adding and Subtracting Binary Numbers B2.4 Multiplying Binary ### Click on the links below to jump directly to the relevant section Click on the links below to jump directly to the relevant section What is algebra? Operations with algebraic terms Mathematical properties of real numbers Order of operations What is Algebra? Algebra is ### CHAPTER 6: RATIONAL NUMBERS AND ORDERED FIELDS CHAPTER 6: RATIONAL NUMBERS AND ORDERED FIELDS LECTURE NOTES FOR MATH 378 (CSUSM, SPRING 2009). WAYNE AITKEN 1. Introduction In this chapter we construct the set of rational numbers Q using equivalence ### MAT2400 Analysis I. A brief introduction to proofs, sets, and functions MAT2400 Analysis I A brief introduction to proofs, sets, and functions In Analysis I there is a lot of manipulations with sets and functions. It is probably also the first course where you have to take ### 2015 Junior Certificate Higher Level Official Sample Paper 1 2015 Junior Certificate Higher Level Official Sample Paper 1 Question 1 (Suggested maximum time: 5 minutes) The sets U, P, Q, and R are shown in the Venn diagram below. (a) Use the Venn diagram to list ### Polynomials can be added or subtracted simply by adding or subtracting the corresponding terms, e.g., if 1. Polynomials 1.1. Definitions A polynomial in x is an expression obtained by taking powers of x, multiplying them by constants, and adding them. It can be written in the form c 0 x n + c 1 x n 1 + c ### An inequality is a mathematical statement containing one of the symbols <, >, or. Further Concepts for Advanced Mathematics - FP1 Unit 3 Graphs & Inequalities Section3c Inequalities Types of Inequality An inequality is a mathematical statement containing one of the symbols , or. ### Chapter 1 Section 5: Equations and Inequalities involving Absolute Value Introduction The concept of absolute value is very strongly connected to the concept of distance. The absolute value of a number is that number s distance from 0 on the number line. Since distance is always ### 3(vi) B. Answer: False. 3(vii) B. Answer: True Mathematics 0N1 Solutions 1 1. Write the following sets in list form. 1(i) The set of letters in the word banana. {a, b, n}. 1(ii) {x : x 2 + 3x 10 = 0}. 3(iv) C A. True 3(v) B = {e, e, f, c}. True 3(vi) ### Math 113 HW #9 Solutions Math 3 HW #9 Solutions 4. 50. Find the absolute maximum and absolute minimum values of on the interval [, 4]. f(x) = x 3 6x 2 + 9x + 2 Answer: First, we find the critical points of f. To do so, take the ### Solving inequalities. Jackie Nicholas Jacquie Hargreaves Janet Hunter Mathematics Learning Centre Solving inequalities Jackie Nicholas Jacquie Hargreaves Janet Hunter c 6 Universit of Sdne Mathematics Learning Centre, Universit of Sdne Solving inequalities In these nots ### 3. Polynomials. 3.1 Parabola primer 3. Polynomials 3.1 Parabola primer A parabola is the graph of a quadratic (degree 2) polynomial, that is, a parabola is polynomial of the form: standard form: y = ax 2 + bx + c. When we are given a polynomial ### Teaching & Learning Plans. Quadratic Equations. Junior Certificate Syllabus Teaching & Learning Plans Quadratic Equations Junior Certificate Syllabus The Teaching & Learning Plans are structured as follows: Aims outline what the lesson, or series of lessons, hopes to achieve. ### 2.1.1 Examples of Sets and their Elements Chapter 2 Set Theory 2.1 Sets The most basic object in Mathematics is called a set. As rudimentary as it is, the exact, formal definition of a set is highly complex. For our purposes, we will simply define ### Solving Equations with One Variable Type - The Algebraic Approach. Solving Equations with a Variable in the Denominator - The Algebraic Approach 3 Solving Equations Concepts: Number Lines The Definitions of Absolute Value Equivalent Equations Solving Equations with One Variable Type - The Algebraic Approach Solving Equations with a Variable in ### ELEMENTARY NUMBER THEORY AND METHODS OF PROOF CHAPTER 4 ELEMENTARY NUMBER THEORY AND METHODS OF PROOF Copyright Cengage Learning. All rights reserved. SECTION 4.4 Direct Proof and Counterexample IV: Division into Cases and the Quotient-Remainder Theorem ### Basic Math Refresher A tutorial and assessment of basic math skills for students in PUBP704. Basic Math Refresher A tutorial and assessment of basic math skills for students in PUBP704. The purpose of this Basic Math Refresher is to review basic math concepts so that students enrolled in PUBP704: ### with "a", "b" and "c" representing real numbers, and "a" is not equal to zero. 3.1 SOLVING QUADRATIC EQUATIONS: * A QUADRATIC is a polynomial whose highest exponent is. * The "standard form" of a quadratic equation is: ax + bx + c = 0 with "a", "b" and "c" representing real numbers, ### 4. Number Theory (Part 2) 4. Number Theory (Part 2) Terence Sim Mathematics is the queen of the sciences and number theory is the queen of mathematics. Reading Sections 4.8, 5.2 5.4 of Epp. Carl Friedrich Gauss, 1777 1855 4.3. ### Domain of a Composition Domain of a Composition Definition Given the function f and g, the composition of f with g is a function defined as (f g)() f(g()). The domain of f g is the set of all real numbers in the domain of g such ### Chapter 8. Quadratic Equations and Functions Chapter 8. Quadratic Equations and Functions 8.1. Solve Quadratic Equations KYOTE Standards: CR 0; CA 11 In this section, we discuss solving quadratic equations by factoring, by using the square root property ### Simultaneous Equations Paper Reference(s) 666/01 Edexcel GCE Core Mathematics C1 Advanced Subsidiary Simultaneous Equations Calculators may NOT be used for these questions. Information for Candidates A booklet Mathematical Formulae ### 4.2 Algebraic Properties: Combining Expressions 4.2 Algebraic Properties: Combining Expressions We begin this section with a summary of the algebraic properties of numbers. Property Name Property Example Commutative property (of addition) Commutative ### Equations and Inequalities Rational Equations Overview of Objectives, students should be able to: 1. Solve rational equations with variables in the denominators.. Recognize identities, conditional equations, and inconsistent equations. ### Chapter 11 Number Theory Chapter 11 Number Theory Number theory is one of the oldest branches of mathematics. For many years people who studied number theory delighted in its pure nature because there were few practical applications ### Computing with Signed Numbers and Combining Like Terms Section. Pre-Activity Preparation Computing with Signed Numbers and Combining Like Terms One of the distinctions between arithmetic and algebra is that arithmetic problems use concrete knowledge; you know ### Continued fractions and good approximations. Continued fractions and good approximations We will study how to find good approximations for important real life constants A good approximation must be both accurate and easy to use For instance, our ### ELEMENTARY NUMBER THEORY AND METHODS OF PROOF CHAPTER 4 ELEMENTARY NUMBER THEORY AND METHODS OF PROOF SECTION 4.4 Direct Proof and Counterexample IV: Division into Cases and the Quotient-Remainder Theorem Copyright Cengage Learning. All rights reserved. ### a) x 2 8x = 25 x 2 8x + 16 = (x 4) 2 = 41 x = 4 ± 41 x + 1 = ± 6 e) x 2 = 5 c) 2x 2 + 2x 7 = 0 2x 2 + 2x = 7 x 2 + x = 7 2 Solving Quadratic Equations By Square Root Method Solving Quadratic Equations By Completing The Square Consider the equation x = a, which we now solve: x = a x a = 0 (x a)(x + a) = 0 x a = 0 x + a = 0 ### PRINCIPLES OF PROBLEM SOLVING PRINCIPLES OF PROBLEM SOLVING There are no hard and fast rules that will ensure success in solving problems. However, it is possible to outline some general steps in the problem-solving process and to ### Lecture 7 : Inequalities 2.5 3 Lecture 7 : Inequalities.5 Sometimes a problem may require us to find all numbers which satisfy an inequality. An inequality is written like an equation, except the equals sign is replaced by one of ### Calculate Highest Common Factors(HCFs) & Least Common Multiples(LCMs) NA1 Calculate Highest Common Factors(HCFs) & Least Common Multiples(LCMs) NA1 What are the multiples of 5? The multiples are in the five times table What are the factors of 90? Each of these is a pair of factors. ### Chapter 15 Introduction to Linear Programming Chapter 15 Introduction to Linear Programming An Introduction to Optimization Spring, 2014 Wei-Ta Chu 1 Brief History of Linear Programming The goal of linear programming is to determine the values of ### The x-intercepts of the graph are the x-values for the points where the graph intersects the x-axis. A parabola may have one, two, or no x-intercepts. Chapter 10-1 Identify Quadratics and their graphs A parabola is the graph of a quadratic function. A quadratic function is a function that can be written in the form, f(x) = ax 2 + bx + c, a 0 or y = ax ### 5. GRAPHS ON SURFACES . GRPHS ON SURCS.. Graphs graph, by itself, is a combinatorial object rather than a topological one. But when we relate a graph to a surface through the process of embedding we move into the realm of topology. ### Tangent and normal lines to conics 4.B. Tangent and normal lines to conics Apollonius work on conics includes a study of tangent and normal lines to these curves. The purpose of this document is to relate his approaches to the modern viewpoints ### Chapter 3. Algebra. 3.1 Rational expressions BAa1: Reduce to lowest terms Contents 3 Algebra 3 3.1 Rational expressions................................ 3 3.1.1 BAa1: Reduce to lowest terms...................... 3 3.1. BAa: Add, subtract, multiply, and divide............... 5 ### MINI LESSON. Lesson 5b Solving Quadratic Equations MINI LESSON Lesson 5b Solving Quadratic Equations Lesson Objectives By the end of this lesson, you should be able to: 1. Determine the number and type of solutions to a QUADRATIC EQUATION by graphing 2. ### Logic, Sets, and Proofs Logic, Sets, and Proofs David A. Cox and Catherine C. McGeoch Amherst College 1 Logic Logical Statements. A logical statement is a mathematical statement that is either true or false. Here we denote logical ### North Carolina Math 1 Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively 3. Construct viable arguments and critique the reasoning of others 4. ### Functions and Equations Centre for Education in Mathematics and Computing Euclid eworkshop # Functions and Equations c 014 UNIVERSITY OF WATERLOO Euclid eworkshop # TOOLKIT Parabolas The quadratic f(x) = ax + bx + c (with a,b,c ### 5.4 The Quadratic Formula Section 5.4 The Quadratic Formula 481 5.4 The Quadratic Formula Consider the general quadratic function f(x) = ax + bx + c. In the previous section, we learned that we can find the zeros of this function ### MATH 21. College Algebra 1 Lecture Notes MATH 21 College Algebra 1 Lecture Notes MATH 21 3.6 Factoring Review College Algebra 1 Factoring and Foiling 1. (a + b) 2 = a 2 + 2ab + b 2. 2. (a b) 2 = a 2 2ab + b 2. 3. (a + b)(a b) = a 2 b 2. 4. (a ### Math 113 HW #9 Solutions Math 3 HW #9 Solutions. Exercise 4..50. Find the absolute maximum and absolute minimum values of on the interval [, 4]. f(x) = x 3 6x + 9x + Answer: First, we find the critical points of f. To do so, take ### A fairly quick tempo of solutions discussions can be kept during the arithmetic problems. Distributivity and related number tricks Notes: No calculators are to be used Each group of exercises is preceded by a short discussion of the concepts involved and one or two examples to be worked out ### Math 4310 Handout - Quotient Vector Spaces Math 4310 Handout - Quotient Vector Spaces Dan Collins The textbook defines a subspace of a vector space in Chapter 4, but it avoids ever discussing the notion of a quotient space. This is understandable ### Adding and Subtracting Positive and Negative Numbers Adding and Subtracting Positive and Negative Numbers Absolute Value For any real number, the distance from zero on the number line is the absolute value of the number. The absolute value of any real number ### MULTIPLICATION AND DIVISION OF REAL NUMBERS In this section we will complete the study of the four basic operations with real numbers. 1.4 Multiplication and (1-25) 25 In this section Multiplication of Real Numbers Division by Zero helpful hint The product of two numbers with like signs is positive, but the product of three numbers with ### 3.1. RATIONAL EXPRESSIONS 3.1. RATIONAL EXPRESSIONS RATIONAL NUMBERS In previous courses you have learned how to operate (do addition, subtraction, multiplication, and division) on rational numbers (fractions). Rational numbers ### MATH 10034 Fundamental Mathematics IV MATH 0034 Fundamental Mathematics IV http://www.math.kent.edu/ebooks/0034/funmath4.pdf Department of Mathematical Sciences Kent State University January 2, 2009 ii Contents To the Instructor v Polynomials. ### Core Maths C1. Revision Notes Core Maths C Revision Notes November 0 Core Maths C Algebra... Indices... Rules of indices... Surds... 4 Simplifying surds... 4 Rationalising the denominator... 4 Quadratic functions... 4 Completing the ### Solutions to Homework 10 Solutions to Homework 1 Section 7., exercise # 1 (b,d): (b) Compute the value of R f dv, where f(x, y) = y/x and R = [1, 3] [, 4]. Solution: Since f is continuous over R, f is integrable over R. Let x ### Partial Fractions. (x 1)(x 2 + 1) Partial Fractions Adding rational functions involves finding a common denominator, rewriting each fraction so that it has that denominator, then adding. For example, 3x x 1 3x(x 1) (x + 1)(x 1) + 1(x + ### Solve addition and subtraction word problems, and add and subtract within 10, e.g., by using objects or drawings to represent the problem. Solve addition and subtraction word problems, and add and subtract within 10, e.g., by using objects or drawings to represent the problem. Solve word problems that call for addition of three whole numbers ### Theorem 2. If x Q and y R \ Q, then. (a) x + y R \ Q, and. (b) xy Q. Math 305 Fall 011 The Density of Q in R The following two theorems tell us what happens when we add and multiply by rational numbers. For the first one, we see that if we add or multiply two rational numbers ### Elementary Number Theory We begin with a bit of elementary number theory, which is concerned CONSTRUCTION OF THE FINITE FIELDS Z p S. R. DOTY Elementary Number Theory We begin with a bit of elementary number theory, which is concerned solely with questions about the set of integers Z = {0, ±1, ### Determining When an Expression Is Undefined Determining When an Expression Is Undefined Connections Have you ever... Tried to use a calculator to divide by zero and gotten an error Tried to figure out the square root of a negative number Expressions ### Markov Chains, part I Markov Chains, part I December 8, 2010 1 Introduction A Markov Chain is a sequence of random variables X 0, X 1,, where each X i S, such that P(X i+1 = s i+1 X i = s i, X i 1 = s i 1,, X 0 = s 0 ) = P(X ### ALGEBRA 2: 4.1 Graph Quadratic Functions in Standard Form ALGEBRA 2: 4.1 Graph Quadratic Functions in Standard Form Goal Graph quadratic functions. VOCABULARY Quadratic function A function that can be written in the standard form y = ax 2 + bx+ c where a 0 Parabola ### Quadratic Equations and Inequalities MA 134 Lecture Notes August 20, 2012 Introduction The purpose of this lecture is to... Introduction The purpose of this lecture is to... Learn about different types of equations Introduction The purpose ### POLYNOMIAL FUNCTIONS POLYNOMIAL FUNCTIONS Polynomial Division.. 314 The Rational Zero Test.....317 Descarte s Rule of Signs... 319 The Remainder Theorem.....31 Finding all Zeros of a Polynomial Function.......33 Writing a ### The Quadratic Formula Definition of the Quadratic Formula The Quadratic Formula uses the a, b and c from numbers; they are the "numerical coefficients"., where a, b and c are just The Quadratic Formula is: For ax 2 + bx + c This assignment will help you to prepare for Algebra 1 by reviewing some of the things you learned in Middle School. If you cannot remember how to complete a specific problem, there is an example at the ### Year 9 set 1 Mathematics notes, to accompany the 9H book. Part 1: Year 9 set 1 Mathematics notes, to accompany the 9H book. equations 1. (p.1), 1.6 (p. 44), 4.6 (p.196) sequences 3. (p.115) Pupils use the Elmwood Press Essential Maths book by David Raymer (9H ### Properties of Real Numbers 16 Chapter P Prerequisites P.2 Properties of Real Numbers What you should learn: Identify and use the basic properties of real numbers Develop and use additional properties of real numbers Why you should ### 1.5. Factorisation. Introduction. Prerequisites. Learning Outcomes. Learning Style Factorisation 1.5 Introduction In Block 4 we showed the way in which brackets were removed from algebraic expressions. Factorisation, which can be considered as the reverse of this process, is dealt with ### Determine If An Equation Represents a Function Question : What is a linear function? The term linear function consists of two parts: linear and function. To understand what these terms mean together, we must first understand what a function is. The ### Geometry Unit 1. Basics of Geometry Geometry Unit 1 Basics of Geometry Using inductive reasoning - Looking for patterns and making conjectures is part of a process called inductive reasoning Conjecture- an unproven statement that is based ### MATH 65 NOTEBOOK CERTIFICATIONS MATH 65 NOTEBOOK CERTIFICATIONS Review Material from Math 60 2.5 4.3 4.4a Chapter #8: Systems of Linear Equations 8.1 8.2 8.3 Chapter #5: Exponents and Polynomials 5.1 5.2a 5.2b 5.3 5.4 5.5 5.6a 5.7a 1 ### BEST METHODS FOR SOLVING QUADRATIC INEQUALITIES. BEST METHODS FOR SOLVING QUADRATIC INEQUALITIES. I. GENERALITIES There are 3 common methods to solve quadratic inequalities. Therefore, students sometimes are confused to select the fastest and the best ### Applications of Fermat s Little Theorem and Congruences Applications of Fermat s Little Theorem and Congruences Definition: Let m be a positive integer. Then integers a and b are congruent modulo m, denoted by a b mod m, if m (a b). Example: 3 1 mod 2, 6 4 ### Equivalence relations Equivalence relations A motivating example for equivalence relations is the problem of constructing the rational numbers. A rational number is the same thing as a fraction a/b, a, b Z and b 0, and hence ### Montana Common Core Standard Algebra 2 Grade Level: 10(with Recommendation), 11, 12 Length: 1 Year Period(s) Per Day: 1 Credit: 1 Credit Requirement Fulfilled: Mathematics Course Description This course covers the main theories in ### GRE MATH REVIEW #4. To find 30% of 200, convert 30% to.30. Then multiply 200 by.30, which results in 60. Hence, 60 is 30% of 200. GRE MATH REVIEW #4 Percentages A percent is just a shorthand way of expressing a fraction whose denominator is 100. Percent means per 100, out of 100, or divided by 100. For example, 25% = 25/100 = 0.25 ### LINE INTEGRALS OF VECTOR FUNCTIONS: GREEN S THEOREM. Contents. 2. Green s Theorem 3 LINE INTEGRALS OF VETOR FUNTIONS: GREEN S THEOREM ontents 1. A differential criterion for conservative vector fields 1 2. Green s Theorem 3 1. A differential criterion for conservative vector fields We ### 4.3 Limit of a Sequence: Theorems 4.3. LIMIT OF A SEQUENCE: THEOREMS 5 4.3 Limit of a Sequence: Theorems These theorems fall in two categories. The first category deals with ways to combine sequences. Like numbers, sequences can be added, ### A Second Course in Mathematics Concepts for Elementary Teachers: Theory, Problems, and Solutions A Second Course in Mathematics Concepts for Elementary Teachers: Theory, Problems, and Solutions Marcel B. Finan Arkansas Tech University c All Rights Reserved First Draft February 8, 2006 1 Contents 25 ### RELATIONS AND FUNCTIONS Chapter 1 RELATIONS AND FUNCTIONS There is no permanent place in the world for ugly mathematics.... It may be very hard to define mathematical beauty but that is just as true of beauty of any kind, we ### Algebra I Notes Review Real Numbers and Closure Unit 00a Big Idea(s): Operations on sets of numbers are performed according to properties or rules. An operation works to change numbers. There are six operations in arithmetic that "work on" numbers: addition, ### A Foundation for Geometry MODULE 4 A Foundation for Geometry There is no royal road to geometry Euclid 1. Points and Lines We are ready (finally!) to talk about geometry. Our first task is to say something about points and lines. ### LAKE ELSINORE UNIFIED SCHOOL DISTRICT LAKE ELSINORE UNIFIED SCHOOL DISTRICT Title: PLATO Algebra 1-Semester 2 Grade Level: 10-12 Department: Mathematics Credit: 5 Prerequisite: Letter grade of F and/or N/C in Algebra 1, Semester 2 Course Description:
Chapter 9: Separation of Variables ### Separation of Variables Separation of variables is a procedure which can turn a partial differential equation into a set of ordinary differential equations. The procedure only works in very special cases involving a high degree of symmetry. Remarkably, the procedure works for many important physics examples. Here, we will use the procedure on the wave equation. Step 1: Write the partial differential equation in appropriate coordinate system. For the wave equation we have: $$\frac{\partial^2 f}{\partial x^2}-\frac{1}{v^2}\frac{\partial^2 f}{\partial t^2} =0 \label{Wave}$$ Step 2: Assume that the solution $f(x,t)$ can be written as the product of functions, each of which depends on only one variable, in this case $x$ or $t$, i.e.\ assume $$f(x,t)=X(x)T(t) \label{separated}$$ This is a very strong assumption. Not all solutions will be of this form. However, it turns out that all of the solutions can be written as linear combinations of solutions of this form. The study of when and why this works is called Sturm-Liouville theory. Plug this assumed solution (\ref{separated}) into the partial differential equation (\ref{Wave}). Because of the special form for $f$, the partial derivatives each act on only one of the factors in $f$. $$T\frac{d^2 X}{d x^2}-\frac{1}{v^2}X\frac{d^2 T}{d t^2} =0 \label{wavesep}$$ Any partial derivatives that act only on a function of a single variable may be rewritten as total derivatives. Step 3: Divide by $f$ in the form of (\ref{separated}). Many, many students forget this step. Don't be one of them! The rest of the procedure doesn't work if you do. \begin{eqnarray} \frac{1}{X}\frac{d^2 X}{d x^2}-\frac{1}{v^2}\frac{1}{T}\frac{d^2 T}{d t^2} =0 \label{wavedivide} \end{eqnarray} Step 4: Isolate all of the dependence on one coordinate on one side of the equation. Do as much algebra as you need to do to achieve this. In our example, notice that in (\ref{wavedivide}), all of the $t$ dependence is already in one term while all of the dependence on the $x$ variable is in the other term. In this case, the $t$ dependence is trivially isolated by putting the $t$ term on the other side of the equation, without any other algebra on our part. \begin{eqnarray} \frac{1}{X}\frac{d^2 X}{d x^2}=\frac{1}{v^2}\frac{1}{T}\frac{d^2 T}{d t^2} \label{wavedivide2} \end{eqnarray} Step 5: Now imagine changing the isolated variable $t$ by a small amount. In principle, the right-hand side of (\ref{wavedivide2}) could change, but nothing on the left-hand side would. (This argument is the magic of the separation of variables procedure–compare it to arguments about constants of the motion from classical mechanics.) Therefore, if the equation is to be true for all values of $t$, the particular combination of $t$ dependence on the right-hand side must be constant. We will call this constant $A$. Note that we don't know (yet) whether the constant is positive or negative. $$\frac{1}{X}\frac{d^2 X}{d x^2}=\frac{1}{v^2}\frac{1}{T}\frac{d^2 T}{d t^2}\equiv A \label{Adefinition}$$ In this way we have broken our original partial differential equation up into a pair of equations, one of which is an ordinary differential equation involving only $t$, the other is an ordinary differential equation involving only $x$. \begin{eqnarray} \frac{1}{X}\frac{d^2 X}{d x^2}=A \label{wavexdivide}\\ \frac{1}{v^2}\frac{1}{T}\frac{d^2 T}{d t^2}= A \label{wavetdivide} \end{eqnarray} The separation constant $A$ appears in both equations. Step 6: Write each equation in standard form by multiplying each equation by its unknown function to clear it from the denominator. \begin{eqnarray} \frac{d^2 X}{d x^2}-AX=0 \label{wavex}\\ \frac{1}{T}\frac{d^2 T}{d t^2}-Av^2T=0 \label{wavet} \end{eqnarray} Notice that (\ref{wavex}) is an eigenvalue equation for the operator $\frac{d^2}{dx^2}$. At the moment, the eigenvalue $A$ could be anything. We will use the spatial boundary conditions for a problem that we want to solve to find the possible values of $A$. Once we have found the possible eigenvalues $A$, (\ref{wavet}) becomes a second order ordinary differential equation with constant coefficients which you should know how to solve. It is NOT an eigenvalue equation.
# a swimming pool 50m by 25 m is to be built with a walkway w meters wide around it write the total area of the pool and walk way in the form ax exponet 2 +bx+c 9 months ago ## To find the total area of the pool and the walkway, let's break down the problem into smaller components. 1. Calculate the area of the pool: The pool measures 50m by 25m, so the area of the pool without the walkway is 50m * 25m = 1250 square meters (m²). 2. Calculate the dimensions of the pool including the walkway: Since the walkway surrounds the pool on all sides, the length and width of the pool will be increased by twice the width of the walkway. Therefore, the overall dimensions of the pool including the walkway will be (50m + 2w) by (25m + 2w). 3. Calculate the area of the expanded pool: To find the area of the expanded pool, we multiply its length by its width. So, the area of the expanded pool is (50m + 2w) * (25m + 2w). 4. Subtract the area of the original pool: To find the area of just the walkway, we subtract the area of the original pool from the area of the expanded pool. Therefore, the area of just the walkway is (50m + 2w) * (25m + 2w) - 1250m². Now, we have the area of the walkway and the original pool. To express the total area in the form ax² + bx + c, we need to simplify the equation. Let's expand the equation: = (50m + 2w) * (25m + 2w) - 1250m² = (50m * 25m) + (50m * 2w) + (25m * 2w) + (2w * 2w) - 1250m² = (1250m²) + (100mw) + (50mw) + (4w²) - 1250m² = 150mw + 4w² Therefore, the total area of the pool and walkway is given by the expression 150mw + 4w², where "a" equals 150m, "b" equals 0 (since there is no term without the variable), and "c" equals 4w².
# Divisibility rules This lesson presents divisibility rules for the numbers 2, 3, 4, 5, 6, 7, 8, 9, and 10. Divisibility rules of whole numbers are very useful because they help us to quickly determine if a number can be divided by 2, 3, 4, 5, 9, and 10 without doing long division. Divisibility means that you are able to divide a number evenly For instance, 8 can be divided evenly by 4 because 8/4 = 2. However, 8 cannot be divided evenly by 3. To illustrate the concept, let's say you have a cake and your cake has 8 slices, you can share that cake between you and 3 more people evenly. Each person will get 2 slices. However,if you are trying to share those 8 slices between you and 2 more people, there is no way you can do this evenly. One person will end up with less cake In general, a whole number x divides another whole number y if and only if you can find a whole number n such that x times n = y For instance, 12 can be divided by 3 because 3 times 4 = 12 When the numbers are large, use the following divisibility rules: Rule #1: divisibility by 2 A number is divisible by 2 if it's last digit is 0,2,4,6,or 8. For instance, 8596742 is divisible by 2 because the las t digit is 2. Rule # 2: divisibility by 3: A number is divisible by 3 if the sum of its digits is divisible by 3 For instance, 3141 is divisible by 3 because 3+1+4+1 = 9 and 9 is divisible by 3. Rule # 3: divisibility by 4 A number is divisible by 4 if the number represented by its last two digits is divisible by 4. For instance, 8920 is divisible by 4 because 20 is divisible by 4. Rule #4: divisibility by 5 A number is divisible by 5 if its last digit is 0 ot 5. For instance, 9564655 is divisible by 5 because the last digit is 5. Rule # 5: divisibility by 6 A number is divisible by 6 if it is divisible by 2 and 3. Be careful! it is not one or the other. The number must be divisible by both 2 and 3 before you can conclude that it is divisible by 6. Rule # 6: divisibility by 7 To ckeck divisibility rules for 7, study carefully the following two examples: Is 348 divisible by 7? Remove the last digit, which is 8. The number becomes 34. Then, Double 8 to get 16 and subtract 16 from 34. 34 − 16 = 18 and 18 is not divisible by 7. Therefore, 348 is not divisible by 7 Is 37961 divisible by 7? Remove the last digit, which is 1. The number becomes 3796. Then, Double 1 to get 2 and subtract 2 from 3796. 3796 − 2 = 3794, so still too big? Thus repeat the process. Remove the last digit, which is 4. The number becomes 379. Then, Double 4 to get 8 and subtract 8 from 379. 379 − 8 = 371, so still too big? Thus repeat the process. Remove the last digit, which is 1. The number becomes 37. Then, Double 1 to get 2 and subtract 2 from 37. 37 − 2 = 35 and 35 is divisible by 7. Therefore, 37961 is divisible by 7. Rule #7:divisibility by 8 A number is divisible by 8 if the number represented by its last three digits is divisible by 8. For instance, 587320 is divisible by 8 because 320 is divisible by 8. Rule #8: divisibility by 9 A number is divisible by 9 if the sum of its digits is divisible by 9. For instance, 3141 is divisible by 9 because the sum of its digits is divisible by 9. Rule # 9: divisibility by 10 A number is divisible by 10 if its last digits is 0 For instance, 522480 is divisible by 10 because the last digit is 0. Test your knowledge with the quiz below: Ask a Math Teacher Math teachers are online. Ask a Question, Get an Answer ASAP. [?] Subscribe To This Site |Are you a fan of this site? Support us |Our awards! |Our partners |About me |Disclaimer |Build your website! |Advertise on my site |Try our free toolbar |Like us on Facebook |Take our survey| Copyright © 2008. Basic-mathematics.com. All right reserved
## Selina Publishers Concise Mathematics Class 7 ICSE Solutions Chapter 10 Simple Interest Selina Publishers Concise Mathematics Class 7 ICSE Solutions Chapter 10 Simple Interest ### Simple Interest Exercise 10 – Selina Concise Mathematics Class 7 ICSE Solutions Question 1. Find the S.I. and amount on : (i) Rs. 150 for 4 years at 5% per year. (ii) Rs. 350 for 3$$\frac {1 }{ 2 }$$ years at 8% p.a. (iii) Rs. 620 for 4 months at 8 p. per rupee per month. (iv) Rs. 3,380 for 30 months at 4 $$\frac { 1 }{ 2 }$$ % p.a. (v) 600 from July 12 to Dec. 5 at 10% p.a. (vi) Rs. 850 from 10th March to 3rd August at 2 $$\frac { 1 }{ 2 }$$ % p.a. (vii) Rs. 225 for 3 years 9 months at 16% p.a. Solution: Question 2. On what sum of money does the S.I. for 10 years at 5% become Rs. 1,600 ? Solution: Question 3. Find the time in which Rs. 2,000 will amount to Rs. 2,330 at 11% p.a. ? Solution: Question 4. In what time will a sum of money double it self at 8% p.a ? Solution: Question 5. In how many years will be ₹870 amount to ₹1,044, the rate of interest being 2$$\frac { 1 }{ 2 }$$ % p.a ? Solution: Question 6. Find the rate percent if the S.I. on ₹275 is 2 years is ₹22. Solution: Question 7. Find the sum which will amount to ₹700 in 5 years at 8% rate p.a. Solution: Question 8. What is the rate of interest, if ₹3,750 amounts to ₹4,650 in 4 years ? Solution: Question 9. In 4 years, ₹6,000 amount to ₹8,000. In what time will ₹525 amount to ₹700 at the same rate ? Solution: Question 10. The interest on a sum of money at the end of 2$$\frac { 1 }{ 2 }$$ years is $$\frac { 4 }{ 5 }$$ of the sum. What is the rate percent ? Solution: Question 11. What sum of money lent out at 5% for 3 years will produce the same interest as Rs. 900 lent out at 4% for 5 years ? Solution: Question 12. A sum of Rs. 1,780 become Rs. 2,136 in 4 years, Find : (i) the rate of interest. (ii) the sum that will become Rs. 810 in 7 years at the same rate of interest ? Solution: Question 13. A sum amounts to Rs. 2,652 in 6 years at 5% p.a. simple interest. Find : (i) the sum (ii) the time in which the same sum will double itself at the same rate of interest. Solution: Question 14. P and Q invest Rs. 36,000 and Rs. 25,000 respectively at the same rate of interest per year. If at the end of 4 years, P gets Rs. 3,080 more interest than Q; find the rate of interest. Solution: Question 15. A sum of money is lent for 5 years at R% simple interest per annum. If the interest earned be one-fourth of the money lent, find the value of R. Solution: Question 16. The simple interest earned on a certain sum in 5 years is 30% of the sum. Find the rate of interest. Solution:
# Class 6 Maths Chapter 6 Exercise 6.1 Pdf Notes NCERT Solutions Class 6 Maths Chapter 6 Integers Exercise 6.1 pdf notes:- Exercise 6.1 Class 6 maths Chapter 6 Pdf Notes:- ## Ncert Solution for Class 6 Maths Chapter 6 Integers Exercise 6.1 Tips:- Introduction:- Sunitaā€™s mother has 8 bananas. Sunita has to go for a picnic with her friends. She wants to carry 10 bananas with her. Can her mother give 10 bananas to her? She does not have enough, so she borrows 2 bananas from her neighbour to be returned later. After giving 10 bananas to Sunita, how many bananas are left with her mother? Can we say that she has zero bananas? She has no bananas with her, but has to return two to her neighbour. So when she gets some more bananas, say 6, she will return 2 and be left with 4 only. Ronald goes to the market to purchase a pen. He has only `12 with him but the pen costs` 15. The shopkeeper writes `3 as due amount from him. He writes` 3 in his diary to remember Ronaldā€™s debit. But how would he remember whether ` 3 has to be given or has to be taken from Ronald? Can he express this debit by some colour or sign? Ruchika and Salma are playing a game using a number strip which is marked from 0 to 25 at equal intervals. To begin with, both of them placed a coloured token at the zero mark. Two coloured dice are placed in a bag and are taken out by them one by one. If the die is red in colour, the token is moved forward as per the number shown on throwing this die. If it is blue, the token is moved backward as per the number shown when this die is thrown. The dice are put back into the bag after each move so that both of them have equal chance of getting either die. The one who reaches the 25th mark first is the winner. They play the game. Ruchika gets the red die and gets four on the die after throwing it. She, thus, moves the token to mark four on the strip. Salma also happens to take out the red die and wins 3 points and, thus, moves her token to number 3. In the second attempt, Ruchika secures three points with the red die and Salma gets 4 points but with the blue die. Where do you think both of them should place their token after the second attempt? Ruchika moves forward and reaches 4 + 3 i.e. the 7th mark. Whereas Salma placed her token at zero position. But Ruchika objected saying she should be behind zero. Salma agreed. But there is nothing behind zero. What can they do? Salma and Ruchika then extended the strip on the other side. They used a blue strip on the other side.
```Question 276490 <pre><font size = 4 color = "indigo"><b> {{{drawing(360,400,-3,15,-3,17, locate(6,0,14), locate(9.5,7,15), locate(1,6,13), triangle(0,0,14,0,5,12) )}}} Let's draw a rectangle in which that triangle is circumscribed: {{{drawing(360,400,-3,15,-3,17, locate(6,0,14), locate(9.5,7,15), locate(1,6,13), rectangle(0,0,14,12), triangle(0,0,14,0,5,12) )}}} Let's draw an altitude h of that triangle {{{drawing(360,400,-3,15,-3,17, locate(8,0,14-x), locate(2,0,x), locate(9.5,7,15), locate(1,6,13), rectangle(0,0,14,12), triangle(0,0,5,0,5,12), locate(5.3,6,h), triangle(0,0,14,0,5,12) )}}} Now you can see that the original triangle's area is half of the large rectangle's area because each of the two right triangles w which the altitude divides the original triangle into are congruent to the right triangles above it with which it shares a hypotenuse. This would be true regardless of which side of the original triangle we chose for a side of the rectangle in which it is inscribed. So we know that the sheet of paper would have to have twice the area of the triangle. Since the base of that triangle is 14, we will call the left part of that base x, then the right part of it is 14-x. Applying the Pythagorean theorem to both right triangles into which the original triangle is split we have this system: {{{system(x^2+h^2=13^2,(14-x)^2+h^2=15^2)}}} {{{system(x^2+h^2=169,(14-x)^2+h^2=225)}}} Simplifying the second one: {{{(14-x)^2+h^2=225)}}} {{{196-28x+x^2+h^2=225}}} {{{x^2+h^2=29+28x}}} So our system is now: {{{system(x^2+h^2=169,x^2+h^2=29+28x)}}} and therefore {{{169=29+28x}}} since things equal to the same thing are equal to each other. {{{140=28x}}} {{{5=x}}} Substituting in {{{x^2+h^2=169}}} {{{5^2+h^2=169}}} {{{25+h^2=169}}} {{{h^2=144}}} {{{h=12}}} {{{drawing(360,400,-3,15,-3,17, locate(8,0,9), locate(2,0,5), locate(9.5,7,15), locate(1,6,13), rectangle(0,0,14,12), triangle(0,0,5,0,5,12), locate(5.3,6,12), triangle(0,0,14,0,5,12) )}}} And since the height of the rectangle equals the altitude, h, of the triangle, the area of the rectangle in which the original triangle is inscribed is found by {{{A=(base)(height)=(14)(12)=168}}} So the correct choice is (a) 168 Edwin</pre> ```
Monday, August 8, 2016 0 Whichever IB Mathematics level you are following, you need to understand the concept of number systems. The simplest system is the set of Natural Numbers, sometimes called the Counting Numbers. Using set notation, these numbers are: {1, 2, 3, 4 ……} and are called the natural numbers because they are the only ones which occur naturally. Negative numbers are an invention (you can’t have less than nothing), although we can easily use negative numbers to represent real things. For example, a temperature of -10°C is real enough, but only relates to an artificially created zero (the freezing point of water); it’s impossible to have temperatures below absolute zero. Similarly, it’s possible to represent an overdraft at the bank using negative numbers, but there are no such things as negative banknotes! Zero is itself a man-made number, but it has become common to include it in the set of natural numbers, so our set above should actually read {0, 1, 2, 3, 4…}. Before continuing, let’s have a look at the symbols for the different number systems: After Natural Numbers comes the set of Integers = {…-3, -2, -1, 0, 1, 2, 3 …}, in other words, all the whole numbers. You’ll notice that all Natural Numbers are also Integers, a fact we can express in set notation as: Rational Numbers are fractions (rational here meaning “to do with ratios” rather than “logical”). More specifically, the numerator and denominator of a rational number must both be integers, and the fraction must be in its simplest form. So, is not a rational number, nor is  since it is not in simplest form – in fact, it’s just another way of writing , rather than a number in its own right. When written as decimals, rational numbers will either terminate (eg 1.875) or recur (eg ): in fact, all recurring decimals can be written as fractions. Worth noting is that all integers are in fact rational, since any integer can be written as a fraction with denominator 1, and hence fits the rational definition. So any numbers which can’t be written as fractions in their simplest form – as decimals, they are non-terminating and non-recurring – aren’t rational, and these form the set of Irrational Numbers. These include the square roots of all integers which aren’t perfect squares (eg or ), and well-known numbers such as and e. However, don’t get the impression that there aren’t many irrational numbers: between any two rationals there will be an infinite number of irrationals; and between any two irrationals there will be an infinite number of rationals – mind-boggling! Put together all the rationals and all the irrationals and you will get the set of Real Numbers: basically, all the numbers we deal with on a daily basis in Mathematics. Unless you are an HL Student, in which case you will also have to cope with Complex Numbers, those based on . Incidentally, real numbers are a subset of complex numbers since any real number a can be written as a +0i. To tie this all together, I have drawn a Venn Diagram showing how all the number systems are related.
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # 1.2: Understanding and Interpreting Frequency Tables and Histograms Difficulty Level: At Grade Created by: CK-12 Estimated10 minsto complete % Progress Practice Histograms Progress Estimated10 minsto complete % Do you understand histograms? Did you know that histograms can be a useful way of explaining data? Jessie works at an ice cream stand. For one hour she recorded the ages of the children who came with their parents to buy ice cream. The histogram above shows this information. Looking at the histogram, can you determine how many persons of each age came to the ice cream stand? Pay attention and you will know how to answer this question by the end of the Concept. ### Guidance To understand what a frequency table is, let’s first look at the words themselves. The word frequency refers to how often something occurs. A table is a way of organizing information using columns. Therefore, a frequency table is way of summarizing data by depicting the number of times a data value occurs. To show this, a frequency table organizes the information into a table with three separate columns. How do we create a frequency table? First, you need to make a table with three separate columns. One column is designated for intervals. The number of intervals is determined by the range of data values. Intervals are equal in size and do not overlap. If the range in data values is close together, then the intervals will be small. If the range in data values is spread apart, then the intervals will be larger. It is important that the range of values in each interval is equal and that the values do not overlap from one interval to the next. Another column is created for tallied results. This is where you tally the number of times you see a data value from each interval. This is where you will see tally marks or lines that record the number of times a data value occurs. In the last column, add the tally marks to determine the frequency results. Let’s look at creating a frequency table. Create a frequency table to display the data below. 43,42,45,42,39,38,50,52,36,49,38,50,40,37,35\begin{align*}43, 42, 45, 42, 39, 38, 50, 52, 36, 49, 38, 50, 40, 37, 35\end{align*} Step 1: Make a table with three separate columns. • Intervals • Tallied results • Frequency results Since the range in data values is not that great, intervals will be in groups of five. Step 2: Looking at the data, tally the number of times a data value occurs. Step 3: Add the tally marks to record the frequency. Take a few minutes to write down the steps for creating a frequency table. Thinking about the frequency that an event occurs can help you to understand and predict certain trends. Think about how useful the trend of grades could be if you were a teacher thinking about a student’s progress. We can also create a histogram to display data. Histograms and bar graphs are often confused, but they are different. Let’s look at how. A histogram shows the frequency of data values on a graph. Like a frequency table, data is grouped in intervals of equal size that do not overlap. Like a bar graph, the height of each bar depicts the frequency of the data values. However, on a histogram the vertical columns have no space in between each other. Create a histogram to display the information on the frequency table. Here are the steps for creating a histogram from data organized in a frequency table. Step 1: Draw the horizontal (x)\begin{align*}(x)\end{align*} and vertical (y)\begin{align*}(y)\end{align*} axis. Step 2: Give the graph the title “Frequency Table Data.” Step 3: Label the horizontal axis “Hours.” List the intervals across the horizontal axis. Step 4: Label the vertical axis “Frequency.” Since the range in frequencies is not that great, label the axis by ones. Step 5: For each interval on the horizontal access, draw a vertical column to the appropriate frequency value. On a histogram, there is no space in between vertical columns. Looking at the histogram, you can see that data values between thirty-six and forty were most frequent. Data values between forty-one and forty-five and forty-six and fifty occurred an equal number of times. Use this histogram of scores earned on math exam to answer the following questions. #### Example A Where did the majority of the scores fall? Solution: Between eighty-six and ninety- five percent #### Example B What fraction of the students earned between seventy-six and eighty-five percent? Solution: One-fourth #### Example C Which scores were in the minority? Solution: Between ninety-six and one hundred and five percent Now back to the dilemma from the beginning of the Concept. Here is the histogram once again. Let's use it to answer these questions. What was the most popular age group at the ice cream stand? There were seven, seven year old children. How many one year old children came? None How many eight year old children came? Three How many ten year old children came? None Notice that you could write many questions and answers using this histogram. The histogram provides a wonderful visual display of the data. ### Vocabulary Data information that has been collected regarding an occurrence or an event. Bar Graph a graph that uses columns to compare quantities or amounts. Frequency Table a display that summarizes data by depicting the number of times that a data value occurs. Histogram a display that shows the frequency of data values on a graph. ### Guided Practice Here is one for you to try on your own. The data values below depict student scores (out of 100%) on a recent math exam. Organize the data into a frequency table. 92,88,75,82,95,99,84,89,90,79,68,71,88,93,87,92,77,68,71,85\begin{align*}92, 88, 75, 82, 95, 99, 84, 89, 90, 79, 68, 71, 88, 93, 87, 92, 77, 68, 71, 85\end{align*} Solution Step 1: Make a table with three separate columns. • Intervals • Tallied results • Frequency results Since the range in data is big (thirty-one), intervals will be in groups of ten. Step 2: Looking at the data, tally the number of times a data value occurs. Step 3: Add the tally marks to record the frequency. Now our work is complete. ### Practice Directions: Use the frequency table to answer the following questions. This frequency table shows scores from an history exam. Score (%) Tally Frequency 50-60 ||||\begin{align*}||||\end{align*} 4 60-70 |||| |\begin{align*}\cancel{||||} \ |\end{align*} 6 70-80 |||| |||| |\begin{align*}\cancel{||||} \ \cancel{||||} \ |\end{align*} 11 80-90 |||| |||\begin{align*}\cancel{||||} \ |||\end{align*} 8 90-100 ||||\begin{align*}||||\end{align*} 4 1. How many students total took the test? 2. How many students scored between 70% and 80%? 3. What fraction of the students scored between 70% and 80%? 4. What percent of the students would that be? 5. How many students scored between 90% and 100%? 6. What fraction of the students scored between 90% and 100%? 7. If failing is below 60%, how many students did not pass the test? 8. True or false. The same number of students received the highest scores as did not pass the test. Directions: Use this histogram on siblings to answer the following questions. 9. How many people surveyed have two siblings? 10. How many people surveyed have three siblings? 11. How many people are only children? 12. How many people have ten siblings? 13. How many people combined have four or five siblings? 14. How many people have only one sibling? 15. How many people have nine siblings? ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes ### Vocabulary Language: English bar graph A bar graph is a plot made of bars whose heights (vertical bars) or lengths (horizontal bars) represent the frequencies of each category, with space between each bar. Data Data is information that has been collected to represent real life situations, usually in number form. frequency density The vertical axis of a histogram is labelled frequency density. Frequency table A frequency table is a table that summarizes a data set by stating the number of times each value occurs within the data set. Histogram A histogram is a display that indicates the frequency of specified ranges of continuous data values on a graph in the form of immediately adjacent bars. Interval An interval is a range of data in a data set. Range The range of a data set is the difference between the smallest value and the greatest value in the data set. right-skewed distribution A right-skewed distribution has a peak to the left of the distribution and data values that taper off to the right. unimodal If a data set has only 1 value that occurs most often, the set is called  unimodal. Show Hide Details Description Difficulty Level: Authors: Tags: Subjects:
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # Graphs of Linear Equations ## Graph lines presented in ax+by = c form Estimated9 minsto complete % Progress Practice Graphs of Linear Equations MEMORY METER This indicates how strong in your memory this concept is Progress Estimated9 minsto complete % Graphs of Linear Equations Credit: CK-12, Larame Spence Source: Desmos Graphing Calculator Dana is collecting information about caterpillars for science class. She’s comparing the lengths and widths of several caterpillars. Dana puts the data she has so far into a table. Dana is convinced there is a pattern. Can organize this information as a set of ordered pairs, graph it on a coordinate plane and write an equation that could model this? x (width in cm)23456y (length in cm)246810\begin{align*}\begin{array}{|c|c|} \hline x \text{ (width in cm)} & y \text{ (length in cm)} \\\hline 2 & 2 \\\hline 3 & 4 \\\hline 4 & 6 \\\hline 5 & 8 \\\hline 6 & 10 \\\hline \end{array}\end{align*} In this concept, you will learn to graph linear functions on the coordinate plane. ### Graphing Linear Functions A linear function is a specific type of function. You may notice that the word “line” is part of the word “linear”. That fact can help you remember that when a linear function is graphed on a coordinate plane, its graph will be a straight line. You can represent a function as a set of ordered pairs, through a table, and as an equation. You can also take the information in ordered pairs or in a table and represent a function as a graph. Let’s look at an example. The table of values below represents a function on a coordinate plane. On a coordinate plane, graph the linear function that is represented by the ordered pairs in the table below. x-4-2024y531-1-3\begin{align*}\begin{array}{|c|c|} \hline x & y \\\hline \text{-}4 & 5 \\\hline \text{-}2 & 3 \\\hline 0 & 1 \\\hline 2 & \text{-}1 \\\hline 4 & \text{-}3 \\\hline \end{array}\end{align*} You can represent the information in this table as a set of ordered pairs {(4,5),(2,3),(0,1),(2,1),(4,3)}\begin{align*}\{(-4,5),(-2,3),(0,1),(2,-1),(4,-3)\}\end{align*}. Plot those five points on the coordinate plane. Then, connect them as shown below. Notice that the graph of this linear function is a straight line. You can also graph a linear function if you are given an equation for that function. This will involve a few more steps. When you have an equation, you can use the equation to create a table. Then, plot several of the ordered pairs in the table and connect them with a line. Here is another example. The equation y=2x1\begin{align*}y = 2x-1\end{align*} is a linear function. Graph that function on a coordinate plane. First, use the equation to create a table and find several ordered pairs for the function. It is a good idea to use some negative x\begin{align*}x\end{align*}-values, some positive x\begin{align*}x\end{align*}-values and 0. For example, you can create a table to find the values of y\begin{align*}y\end{align*} when x\begin{align*}x\end{align*} is equal to -2, -1, 0, 1, and 2. x\begin{align*}x\end{align*} y\begin{align*}y\end{align*} −2\begin{align*}-2\end{align*} −5\begin{align*}-5\end{align*} 2(−2)−1=−5\begin{align*}2(-2)-1=-5\end{align*} −1\begin{align*}-1\end{align*} −3\begin{align*}-3\end{align*} 2(−1)−1=−3\begin{align*}2(-1)-1=-3\end{align*} −1\begin{align*}-1\end{align*} 2(0)−1=−1\begin{align*}2(0)-1=-1\end{align*} 1\begin{align*}1\end{align*} 1\begin{align*}1\end{align*} 2(1)−1=1\begin{align*}2(1)-1=1\end{align*} 2\begin{align*}2\end{align*} 3\begin{align*}3\end{align*} 2(2)−1=3\begin{align*}2(2)-1=3\end{align*} The ordered pairs shown in the table are (2,5),(1,3),(0,1),(1,1)\begin{align*}(-2, -5), (-1, -3), (0, -1), (1, 1)\end{align*} and (2,3)\begin{align*}(2, 3)\end{align*}. Plot those five points on the coordinate plane. Then connect them as shown below. ### Examples #### Example 1 Earlier, you were given a problem about Dana’s project, which was comparing the lengths and widths of caterpillars. She’s put the data collected so far in a table (shown below). Can you plot these points and write the equation that models this information? x (width in cm)23456y (length in cm)246810\begin{align*}\begin{array}{|c|c|} \hline x \text{ (width in cm)}& y \text{ (length in cm)} \\ \hline 2 & 2 \\ \hline 3 & 4 \\ \hline 4 & 6 \\ \hline 5 & 8 \\ \hline 6 & 10 \\ \hline \end{array}\end{align*} First, represent this information as a set of ordered pairs so that you can plot the points {(2,2),(3,4),(4,6),(5,8),(6,10)}\begin{align*}\{(2,2),(3,4),(4,6),(5,8),(6,10)\}\end{align*}. Now, can you see a pattern in the table and then write the rule that describes it? Notice that as x\begin{align*}x\end{align*} increases by 1, y\begin{align*}y\end{align*} increases by 2. So, you know that 2x\begin{align*}2x\end{align*} is involved in the equation. But y\begin{align*}y\end{align*} is not quite 2x\begin{align*}2x\end{align*}. It is 2x2\begin{align*}2x -2\end{align*}. So the equation that models this information is 2x2\begin{align*}2x -2\end{align*}. Next, plot the points on the coordinate plane and draw a line through them. The graph is shown below. #### Example 2 The table below represents inputs and outputs of a linear function. Can you represent this information as ordered pairs, figure out the equation for this function, and then graph the function? x1234y5101520\begin{align*}\begin{array}{|c|c|} \hline x & y \\\hline 1 & 5 \\\hline 2 & 10 \\\hline 3 & 15 \\\hline 4 & 20 \\\hline \end{array}\end{align*} You can extract information from the table and represent the same information as a set of ordered pairs. The x\begin{align*}x\end{align*}-coordinate is the first value and the y\begin{align*}y\end{align*}-coordinate is the second value. {(1,5),(2,10),(3,15),(4,20)}\begin{align*}\{(1,5),(2,10),(3,15),(4,20)\}\end{align*} Next, looking at the information in the table, you can see that when you multiply the x\begin{align*}x\end{align*}-value by 5 you get the y\begin{align*}y\end{align*}-value. The rule is multiply x\begin{align*}x\end{align*} by 5 to get y\begin{align*}y\end{align*}. You can write this as an equation. y=5x\begin{align*}y=5x\end{align*} You can graph plot the coordinates {(1,5),(2,10),(3,15),(4,20)}\begin{align*}\{(1,5),(2,10),(3,15),(4,20)\}\end{align*} and draw a line through them to see the graph. #### Example 3 Is the function above increasing or decreasing? Notice that as x\begin{align*}x\end{align*} increases y\begin{align*}y\end{align*} increases. Notice that every time you increase x\begin{align*}x\end{align*} by 1, y\begin{align*}y\end{align*} will always increase. In this case, \begin{align*}y\end{align*} increases by two every time \begin{align*}x\end{align*} increases by 1. The answer is the function is increasing. #### Example 4 In the point \begin{align*}(-3, 4)\end{align*} is the \begin{align*}x\end{align*}-value positive or negative? The \begin{align*}x\end{align*}-value is the first value in the coordinate. It is a negative number. The answer is the \begin{align*}x\end{align*}-value is negative. #### Example 5 In \begin{align*}(-6, -7)\end{align*}, which value is \begin{align*}y\end{align*}-value? The \begin{align*}y\end{align*}-value is the second value in a coordinate, and it is equal to -7. The answer is the \begin{align*}y\end{align*}-value is -7. ### Review The information in the table represents points from a linear function. Plot the points in the table on a coordinate plane, and then draw a straight line through them to graph each function. Then identify the rule (equation) for the function. 1. Input Output 1 4 2 5 3 6 4 7 1. Input Output 2 4 3 6 4 8 5 10 1. Input Output 1 3 2 6 4 12 5 15 1. Input Output 9 7 7 5 5 3 3 1 1. Input Output 8 12 9 13 11 15 20 24 1. Input Output 3 21 4 28 6 42 8 56 1. Input Output 2 5 3 7 4 9 5 11 1. Input Output 4 7 5 9 6 11 8 15 1. Input Output 5 14 6 17 7 20 8 23 1. Input Output 4 16 5 20 6 24 8 32 ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes ### Vocabulary Language: English TermDefinition Cartesian Plane The Cartesian plane is a grid formed by a horizontal number line and a vertical number line that cross at the (0, 0) point, called the origin. Function A function is a relation where there is only one output for every input. In other words, for every value of $x$, there is only one value for $y$. Function Rule A function rule describes how to convert an input value ($x$) into an output value ($y$) for a given function. An example of a function rule is $f(x) = x^2 + 3$. Linear Function A linear function is a relation between two variables that produces a straight line when graphed. Slope Slope is a measure of the steepness of a line. A line can have positive, negative, zero (horizontal), or undefined (vertical) slope. The slope of a line can be found by calculating “rise over run” or “the change in the $y$ over the change in the $x$.” The symbol for slope is $m$
# Arithmetic Progression: #### Sequence: A sequence is a set of numbers that are in a particular order is called sequence. Example (1): 1, 3, 5, 7, 9, ....... Example (2): 2, 4, 6, 8, 10, ....... #### Series: When the terms of a sequence are written with the addition or subtraction operation then it is called a series. Example (1): 1 + 3 + 5 + 7 + 9 + ....... Example (2): 1 - 3 - 5 - 7 - 9 - ....... Example (3): 2 + 4 + 6 + 8 + 10 + ....... #### Arithmetic Progression: A sequence of numbers in which each term is driven from the preceding term by adding or subtracting a fixed number is called arithmetic progression and the fixed number which is added or subtracted by each term is called "common difference" and it is denoted by "d". The common difference can be positive, negative or zero. Example (1): 1, 3, 5, 7, 9, ...... Here 2 is added with each term to get the next term so 2 is a common difference (d). Example (2): 2, 3, 4, 5, 6, ...... Here 1 is added with each term to get the next term so 1 is a common difference (d). #### The general term of Arithmetic Progression: Let's consider the first term of a series is "a" and the common difference is "d" then the general term of arithmetic progression will be a, a+d, a+2d, a+3d, a+4d, ..... Here the common difference (d) in the second term is multiplied by 1, d in the third term is multiplied by 2, d in the fourth term is multiplied by 3 and so on. Hence it can be written as $$n^{th} \ term \ (l) = a + (n - 1) \ d$$ Example: If the nth term of a series is (2n - 3) then find the 7th term of that series? Solution: $$n^{th} \ term = (2n - 3)$$ $$7^{th} \ term = (2 \times 7 - 3)$$ $$= 14 - 3$$ $$= 11$$ Hence 7th term of that series is 11.
# 1. Introduction A general linear function has the form y = mx + c , where m and c are constants. Example 1: If x is the temperature in °C and y the temperature in °F then there is a simple rule relating the values of x and y. The table below illustrates this rule for various values of x and y. xC) yF) 0 32 freezing point of water 10 50 temperature on a cold day 25 77 temperature on a warm day 37 98.6 blood temperature 100 212 boiling point of water The general rule is: y = 9/5 x + 32 - so that m = 9 ⁄ 5 and c = 32 in this case. Example 2: A straight line passes through the two points P(x,y) and Q(x,y) with coordinates P(0, 2) Q(1, 5). Find the equation of this straight line. Solution: The general equation of a straight line is y = mx + c. Since the line passes through the points P, with coordinates x = 0, y = 2, and Q, with coordinates x = 1, y = 5, these coordinates must satisfy this equation, i.e. : 2 = m × 0 + c 1 = m × 5 + c (See the module on Simultaneous Equations.) Solving these equations gives c = 2 and m = 3, i.e. the equation of the line is: y = 3x + 2 Exercise 1: In each of the following, find the equation of the straight line through the given pairs of points: (a)The points P (0, −3) and Q(2, 1) Solution: The general equation of a straight line is y = mx + c. Since the line passes through the points P and Q, the coordinates of both points must satisfy this equation. The point P has coordinates x = 0, y = −3 and the point Q has coordinates x = 2, y = 1. These satisfy the pair of simultaneous equations: −3 = m × 0 + c 1 = m × 2 + c Solving these equations gives c = −3 and m = 2, i.e. the equation of the line is: y = 2x − 3 (b)The points P(0, 4) and Q(1, 3) Solution: The general equation of a straight line is y = mx + c. Since the line passes through the points P and Q, the coordinates of both points must satisfy this equation. The point P has coordinates x = 0, y = 4 and the point Q has coordinates x = 1, y = 3. These satisfy the pair of simultaneous equations: 4 = m × 0 + c 3 = m × 1 + c Solving these equations gives c = 4 and m = −3, i.e. the equation of the line is: y = −3x + 4 Click on questions to reveal solutions # 2. Gradient of a Straight Line The gradient of a straight line is defined as follows: Suppose that two points P, Q, on the line have coordinates P(x1, y1) and Q(x2, y2) (see figure on the right). The gradient of the line is given by: gradient = RQ/PR = y2 − y1/x2 − x1 Example 3: From the table given in Example 1, find the gradient of the line giving the relationship between x (◦C) and y (◦F). Solution: The boiling point, Q, of water is 100°C or 212°F, i.e. Q(100, 212). The freezing point of water is 0°C or 32°F , i.e. P(0, 32). The gradient is therefore: y2y1 / x2x1 = 212 − 32 / 100 − 0 = 180 / 100 = 9 / 5 The equation of the line is y = (9 ⁄ 5)x + 32. Comparing this with the general equation y = mx + c shows that m is the value of the gradient. Exercise 2: Find the gradient of the line through the points P, Q with the following coordinates: (a)P(3,9), Q(2,3) Solution: gradient = y2 − y1 /x2 − x1 = 3 − 9/2 − 3 = −6/−1 = 6 So that m = 6 in this case. (b)P(−1,2), Q(2,−1) Solution: gradient = y2 − y1 / x2 − x1 = −1 − 2/2 − (−1) = −1 So that m = −1 in this case. We shall interpret the negative gradient later in this module. (c) P(1,2), Q(4,3) Solution: gradient = y2 − y1 /x2 − x1 = 3 − 2/4 − 1 = 1/3 So that m = 1 ⁄ 3 in this case. Click on questions to reveal solutions # 3. Intercepts of a Straight Line By putting x = 0 into the equation y = mx + c, the point where the straight line crosses the y axis is found to be y = c. This is known as the intercept on the y axis. The intercept on the x axis, i.e. when y = 0, is at: 0 = mx + c −c = mx −c ⁄ m = x The intercepts are shown in the figure to the right. Example 3: By rearranging the equation 2y − 3x − 5 = 0, show that it is a straight line and find its gradient and intercept. Sketch the line. Solution: Rearranging the equation: 3y − 2x − 5 = 0 3y = 2x + 5 y = (2 ⁄ 3)x + (5 ⁄ 3) Exercise 3: Each of the following equations represent straight lines. By rearranging each of them find their gradient and the intercepts on the y and x axis. Sketch the straight lines they represent. (a)2y − 2x + 3 = 0 Solution: 2y − 2x + 3 = 0 2y = 2x − 3 y = x − 3⁄2 So that m = 1 and c = −3 ⁄ 2 The intercept on the x-axis is: cm = −(−3 ⁄ 2) ⁄ 1 = 3 ⁄ 2 (b)3y − 5x + 6 = 0 Solution: 3y − 5x + 6 = 0 3y = 5x − 6 y = (5⁄3)x − 2 So that m = 15 ⁄ 3 and c = −2 The intercept on the x-axis is: cm = −(−2 ⁄ 2) ⁄ (5 ⁄ 3) = 6 ⁄ 5 (c) 2y + 4x + 3 = 0 Solution: 2y + 4x + 3 = 0 2y = −4x − 3 y = −2x − 3⁄2 So that m = −2 and c = −3 ⁄ 2 The intercept on the x-axis is: cm = −(−3 ⁄ 2) ⁄ (−2) = −3 ⁄ 4 Click on questions to reveal solutions Quiz 1: A straight line has the equation 3x + y + 3 = 0. If P is the point where the line crosses the x axis and Q is the point where it crosses the y axis, which of the following pair is P, Q? (a)P(3, 0), Q(0, −1) Wrong - please try again! (b)P(−1, 0), Q(0, 3) Wrong - please try again! (c)P(−1, 0), Q(0, −3) Correct - well done! (d)P(−3, 0), Q(0, −1) Wrong - please try again! The line crosses the x-axis when y = 0. Putting this into the equation of the line, 3x + y + 3 = 0, gives: 3x + 0 + 3 = 0 3x = −3 x = −1 Thus P(−1,0) is the first point. The line crosses the y-axis when x = 0. Putting this into the equation of the line: 3(0) + y + 3 = 0 y + 3 = 0 y = −3 Thus, Q(0,−3) is the second point. Quiz 2: If the straight line 6x + 2y = 3 is written in the form y = mx + c, which of the following is the correct set of values for m and c? (a)m = 3, c = 3 ⁄ 2 Wrong - please try again! (b)m = 3, c = −3 ⁄ 2 Wrong - please try again! (c)m = −3, c = −3 ⁄ 2Wrong - please try again! (d)m = −3, c = 3 ⁄ 2 Correct - well done! The equation is rearranged as follows: 6x + 2y = 3 2y = −6x + 3 y = -3x + 3 ⁄ 2 So m = −3 and c = 3 ⁄ 2 # 4. Positive and Negative Gradients If a line has gradient m = 1 then, providing that the scales are the same for both axes, it makes an angle of 45° with the positive x-axis. If m > 1 then the gradient is steeper. If 0 < m < 1 then the line makes an angle between 0° and 45° with the positive x-axis. This is illustrated in the figure to the right. The figure on the left illustrates lines similar to those of the previous figure, except with negative gradients. They are the mirror images of the straight lines shown above, with the y axis acting as the mirror. Exercise 4: In each of the following either the coordinates of two points, P, Q are given, or the coordinates of a single point R and a gradient m. In each case, find the equation of the line: (a)P(1, 1), Q(2, −1) Solution: Let the line be y = mx + c. Since both P (1, 1) and Q(2,−1) lie on the line, both sets of coordinates must satisfy the equation. Thus we have: 1 = m × 1 + c ⇐ using the coordinates of P −1 = m × 2 + c ⇐ using the coordinates of Q or: m + c = 1 2m + c = −1 This is a set of simultaneous equations which can be solved to give m = −2 and c = 3. (See the module on simultaneous equations for the technique for solving them.) The required equation is thus: y = 2x − 3 Substituting the coordinates for P and then Q into this equation will confirm that this line passes through both of these points. (b)R(1, 2), m = 2 Solution: Since m = 2, the equation must have the form y = 2x + c and only the value of c remains to be found. The line passes through R(1, 2) so the coordinates of this point must satisfy the equation. Thus: y = 2x + c 2 = 2 × 1 + c ⇐ using the coordinates of R - giving c = 0. The equation of the line is now: y = 2x (c)P(−1, 2), Q(1, −3) Solution: Let the line be y = mx + c . Since both P(−1, 2) and Q(1, −3) lie on the line, both sets of coordinates must satisfy the equation. Thus we have: 2 = m × (−1) + c ⇐ using the coordinates of P −3 = m × 1 + c ⇐ using the coordinates of Q or: −m + c = 2 m + c = −3 This is a pair of simultaneous equations which can be solved to give m = −5 ⁄ 2 and c = −1 ⁄ 2. (See the module on simultaneous equations for the technique for solving them.) The required equation is thus: y = −5/2x1/2 Substituting the coordinates for P and then Q into this equation will confirm that this line passes through both of these points. (d)R(−2, 1), m = 4 Solution: Since m = 4, the equation must have the form y = 4x + c and only c remains to be found. The line passes through R(−2, 1) so the coordinates of this point must satisfy the equation. Thus: y = 4x + c 1 = 4 × (−2) + c ⇐ using the coordinates of R or: −8 + c = 1 c = 9 and the equation of the line is : y = 4x + 9 (e)P(−1, 2), Q(−3, 3) Solution: Let the line be y = mx + c. The coordinates P(−1, 2) and Q(−3, 3) both lie on the line so both sets of coordinates must satisfy the equation. We have: 2 = m × (−1) + c ⇐ using the coordinates of P 3 = m × (−3) + c ⇐ using the coordinates of Q or: −m + c = 2 −3m + c = 3 This set of simultaneous equations can be solved to give m = −1/2 and c = 3/2. (See the module on simultaneous equations for the technique for solving them.) The required equation is thus: y = −1/2x + 3/2 Substituting the coordinates for P and then Q into this equation will confirm that this line passes through both of these points. (f)P(1, 2), Q(−4, 7) Solution: Let the line be y = mx + c . Both P(1, 2) and Q(−4, 7) both lie on the line so both sets of coordinates must satisfy the equation. We have: 2 = m × (1) + c ⇐ using the coordinates of P 7 = m × (−4) + c ⇐ using the coordinates of Q or: m + c = 2 −4m + c = 7 The solution to this set of simultaneous equations is m = −1 and c = 3. (See the module on simultaneous equations for the technique for solving them.) The equation of the line is thus: y = −x + 3 Substituting the coordinates for P and then Q into this equation will confirm that this line passes through both of these points. Example 5: Two lines are described as follows: the first has gradient −1 and passes through the point R(2, 1); the second passes through the two points with coordinates P(2, 0) and Q(0, 4). Find the equation of both lines and find the coordinates of their point of intersection. Solution: The first line has gradient m = −1 so it must be y = (−1)x + c, i.e. y = −x + c, for some c. Since the line passes through the point R(2, 1) these values of x, y must satisfy the equation. Thus 2 = −(1) + c so c = 3. The first line therefore has equation y = −x + 3. For the second case both points lie on the line and so satisfy the equation. If the equation is y = mx + c then putting these values into the equation gives: y = mx + c 0 = 2m + 3 ⇐ using the coordinates of P 4 = c ⇐ using the coordinates of Q These equations yield m = −2 and c = 4. The second line thus has the equation y = −2x + 4. The equations of the two lines can now be rewritten as: y + x = 3 (1) y + 2x = 4 (2) - which is a pair of simultaneous equations. Subtracting equation (1) from equation (2) gives x = 1 and substituting this into the first equation then yields y = 2. The point of intersection thus has coordinates x = 1, y = 2. (By substituting these coordinates into equation (2) and verifying that they satisfy the equation, it can be checked that this is also a point on the second line. # 5. Some Useful Facts • Parallel lines have the same gradient. Thus, for example, the lines with equations y = 3x + 7 and y = 3x − 2 are parallel. • Lines parallel to the x-axis have equations of the form y = k, for some constant, k. • Lines parallel to the y-axis (when m = 0) have equations of the form x = k, for some constant, k. • The larger the absolute value of m, the ‘steeper’ the slope of the line. • If two lines intersect at right angles then the product of their gradients is −1. The lines y = −7x + 4 and y = (1 ⁄ 7)x + 5, for example, intersect each other at right angles. # 6. Quiz on Straight Line Graphs In each of the following, choose the solution from the options given: 1. What is the equation of the straight line through P (−5, 4) and Q(2, −3) ? (a)2xy = −14 (b)x + 2y = 17 (c)x + y − 1 = 0 (d)x + y + 1 = 0 2. What are the gradient m and intercept c of −2x + 3y + 6 = 0 ? (a)m = 2 ⁄ 3, c = −2 (b)m = −2 ⁄ 3, c = 2 (c)m = 3 ⁄2, c = −3 (d)m = −3 ⁄2, c = 3 3. What is the equation of the straight line with gradient m = −3 passing through R(−1, 3) ? (a)−3x + y = 0 (b)2y − 6x = 4 (c)y − 3x = 1 (d)y + 3x = 0 4. What is the point of intersection of the lines 2x + y = 1 and 3x − 2y = 5 ? (a)(−1, 1) (b)(1, −1) (c)(3, −5) (d)(2, −3)
# NCERT Exemplar Solution for Class 10 Mathematics: Circles (Part-IIIB) In this article you will get CBSE class 10 Mathematics NCERT Exemplar Problems and Solutions for chapter 9- Circles (Part-IIIB). In this part you will get solutions to question number 6 to 10 form exercise 9.3 of NCERT Exemplar for Mathematics chapter 9. It consists of Short Answer Type Questions only. Updated: Aug 9, 2017 11:03 IST Here you get the CBSE Class 10 Mathematics chapter 9, Circles: NCERT Exemplar Problems and Solutions (Part-IIIB). This part of the chapter includes solutions to Question Number 6 to 10 from Exercise 9.3 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Circles. This exercise comprises only the Short Answer Type Questions framed from various important topics in the chapter. Each question is provided with a detailed solution. NCERT Exemplar Solution for Class 10 Mathematics: Circles (Part-IIIA) NCERT exemplar problems are a very good resource for preparing the critical questions like Higher Order Thinking Skill (HOTS) questions. All these questions are very important to prepare for CBSE Class 10 Mathematics Board Examination 2017-2018 as well as other competitive exams. Find below the NCERT Exemplar problems and their solutions for Class 10 Mathematics Chapter, Circles: Exercise 9.3 Short Answer Type Questions (Q. No. 6-10) Question. 6 In figure, AB and CD are common tangents to two circles of equal radii. Prove that AB = CD. Solution. Given: Two circles of equal radii, two common tangents, AB and CD on circles C1 and C2 with centres O1 and O2. To prove: AB = CD Construction: Join AC, BD, O2A, O2C, O2B, O2D, O1A and O1C. Proof: Question. 7 In figure, common tangents AB and CD to two circles intersect at E. Prove that AB = CD. Question. 8 A chord PQ of a circle is parallel to the tangent drawn at a point R of the circle. Prove that R bisects the arc PRQ. Solution. Given: Chord PQ of a circle is parallel to tangent drawn at point R of that circle. Question. 9 Prove that the tangents drawn at the ends of a chord of a circle make equal angles with the chord. Solution. Given: Two tangents are drawn at the ends of the chord of a circle. To prove: ∠1 = ∠2 Proof: Let two tangents drawn at the ends of a chord AB intersect at point C. As, we know that tangents drawn from an external point to a circle are equal, ∴   AC = BC ⟹ ∠2 = ∠1           [Angles opposite to equal sides of a triangle are equal] Hence, tangents AC and BC make equal angles with chord AB. Hence, proved. Question. 10 Prove that a diameter AB of a circle bisects all those chords which are parallel to the tangent at the point A. Solution. Given: AB is a diameter of the circle with centre O and a chord CD is parallel to tangent MAN . Now, we know that a perpendicular drawn from centre of circle to chord bisects the chord. Thus, OE bisects CD. Similarly, the diameter AB bisects all the chords which are parallel to the tangent at the point A. CBSE Class 10 Mathematics Syllabus 2017-2018 CBSE Class 10 NCERT Textbooks & NCERT Solutions NCERT Solutions for CBSE Class 10 Maths NCERT Exemplar Problems and Solutions Class 10 Science: All Chapters रोमांचक गेम्स खेलें और जीतें एक लाख रुपए तक कैश ## Related Stories Comment (0) ### Post Comment 3 + 7 = Post Disclaimer: Comments will be moderated by Jagranjosh editorial team. Comments that are abusive, personal, incendiary or irrelevant will not be published. Please use a genuine email ID and provide your name, to avoid rejection.
# Fraction calculator The calculator performs basic and advanced operations with fractions, expressions with fractions combined with integers, decimals, and mixed numbers. It also shows detailed step-by-step information about the fraction calculation procedure. Solve problems with two, three, or more fractions and numbers in one expression. ## Result: ### 3/8 * 5 = 15/8 = 1 7/8 = 1.875 Spelled result in words is fifteen eighths (or one and seven eighths). ### How do you solve fractions step by step? 1. Multiple: 3/8 * 5 = 3 · 5/8 · 1 = 15/8 Multiply both numerators and denominators. Result fraction keep to lowest possible denominator GCD(15, 8) = 1. In the next intermediate step, the fraction result cannot be further simplified by canceling. In words - three eighths multiplied by five = fifteen eighths. #### Rules for expressions with fractions: Fractions - use the slash “/” between the numerator and denominator, i.e., for five-hundredths, enter 5/100. If you are using mixed numbers, be sure to leave a single space between the whole and fraction part. The slash separates the numerator (number above a fraction line) and denominator (number below). Mixed numerals (mixed fractions or mixed numbers) write as non-zero integer separated by one space and fraction i.e., 1 2/3 (having the same sign). An example of a negative mixed fraction: -5 1/2. Because slash is both signs for fraction line and division, we recommended use colon (:) as the operator of division fractions i.e., 1/2 : 3. Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45. The colon : and slash / is the symbol of division. Can be used to divide mixed numbers 1 2/3 : 4 3/8 or can be used for write complex fractions i.e. 1/2 : 1/3. An asterisk * or × is the symbol for multiplication. Plus + is addition, minus sign - is subtraction and ()[] is mathematical parentheses. The exponentiation/power symbol is ^ - for example: (7/8-4/5)^2 = (7/8-4/5)2 #### Examples: subtracting fractions: 2/3 - 1/2 multiplying fractions: 7/8 * 3/9 dividing Fractions: 1/2 : 3/4 exponentiation of fraction: 3/5^3 fractional exponents: 16 ^ 1/2 adding fractions and mixed numbers: 8/5 + 6 2/7 dividing integer and fraction: 5 ÷ 1/2 complex fractions: 5/8 : 2 2/3 decimal to fraction: 0.625 Fraction to Decimal: 1/4 Fraction to Percent: 1/8 % comparing fractions: 1/4 2/3 multiplying a fraction by a whole number: 6 * 3/4 square root of a fraction: sqrt(1/16) reducing or simplifying the fraction (simplification) - dividing the numerator and denominator of a fraction by the same non-zero number - equivalent fraction: 4/22 expression with brackets: 1/3 * (1/2 - 3 3/8) compound fraction: 3/4 of 5/7 fractions multiple: 2/3 of 3/5 divide to find the quotient: 3/5 ÷ 2/3 The calculator follows well-known rules for order of operations. The most common mnemonics for remembering this order of operations are: PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction. BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction. GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction. Be careful, always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) has the same priority and then must evaluate from left to right. ## Fractions in word problems: • Ratio 11 Simplify this ratio 10 : 1/4 • Andre Andre, Thomas, and Ivan split 88 postage stamps in a 2:5:4 ratio. How much did Thomas get? • Equation with x Solve the following equation: 2x- (8x + 1) - (x + 2) / 5 = 9 • Red diplomas The numbers of students with honors in 2013 and 2014 are in ratio 40:49. How big is the year-on-year percentage increase? • Percent calculation Calculate 8% if 44% is 32. • Average speed When the bus stops at a bus stop driving average speed is 45 km/h. If it did not stop, it would drive at a speed of 54 km/h. How many minutes of every hour does it spend at stops? • Day What part of the day are 23 hours 22 minutes? Express as a decimal number. • Family party Anna has 5 €, Anezka has 4,60 € and for all the money they want to buy desserts for a family party. They decide between cakes and pinwheel: The pinwheel is € 0.40 more expensive than the cake, and the cakes can be bought for all the money by a third more • Stones in aquarium In an aquarium with a length 2 m; width 1.5 m and a depth of 2.5 m is a water level up to three-quarters of the depth. Can we place stones with a volume of 2 m3 into the aquarium without water being poured out? • Pocket Mirka spent on a trip for gifts half pocket. For the third remain money bought a bagel. Six euros left her. How many euros had Mirka in pocket? • Scale of the map Determine the map's scale, which is the actual distance of 120 km l represented by a segment long 6 cm. • Three shapes 1/5 of a circle is shaded. The ratio of area if square to the sum of area of rectangle and that of the circle is 1:2. 60% of the square is shaded and 1/3 of the rectangle is shaded. What is the ratio of the area of circle to that of the rectangle? • In the cafeteria There are 18 students in Jacob’s homeroom. Six students bring their lunch to school. The rest eat lunch in the cafeteria. In simplest form, what fraction of students eat lunch in the cafeteria?
# What is 20 percent of 72000? WRITTEN BY: supportmymoto.com STAFF #### Resolution for What’s 20 % of 72000: 20 % *72000 = (20:100)*72000 = (20*72000):100 = 1440000:100 = 14400 Now now we have: 20 % of 72000 = 14400 Query: What’s 20 % of 72000? Step 1: Our output worth is 72000. Step 2: We characterize the unknown worth with {x}. Step 3: From step 1 above,{72000}={100%}. Step 4: Equally, {x}={20%}. Step 5: This leads to a pair of straightforward equations: {72000}={100%}(1). {x}={20%}(2). Step 6: By dividing equation 1 by equation 2 and noting that each the RHS (proper hand aspect) of each equations have the identical unit (%); now we have frac{72000}{x}=frac{100%}{20%} Step 7: Once more, the reciprocal of either side offers frac{x}{72000}=frac{20}{100} Rightarrow{x} = {14400} Due to this fact, {20%} of {72000} is {14400} #### Resolution for What’s 72000 % of 20: 72000 % *20 = (72000:100)*20 = (72000*20):100 = 1440000:100 = 14400 Now now we have: 72000 % of 20 = 14400 Query: What’s 72000 % of 20? Step 1: Our output worth is 20. Step 2: We characterize the unknown worth with {x}. Step 3: From step 1 above,{20}={100%}. Step 4: Equally, {x}={72000%}. Step 5: This leads to a pair of straightforward equations: {20}={100%}(1). {x}={72000%}(2). Step 6: By dividing equation 1 by equation 2 and noting that each the RHS (proper hand aspect) of each equations have the identical unit (%); now we have frac{20}{x}=frac{100%}{72000%} Step 7: Once more, the reciprocal of either side offers frac{x}{20}=frac{72000}{100} Rightarrow{x} = {14400} Due to this fact, {72000%} of {20} is {14400} NOTE : Please do not copy - https://supportmymoto.com