Split (1)

train · ~16.8M rows (showing the first 4.97M)

text stringlengths 22 1.01M
In this article, we will explore the step-by-step process of finding the derivative of the function (x² + 1) / (2x + 3). We will apply differentiation rules and techniques to simplify the expression and determine its derivative. Calculus, a branch of mathematics, is a powerful tool for analyzing functions and understanding how they change. One of the core concepts in calculus is differentiation, which allows us to calculate the rate of change of a function. Understanding the Problem Before we begin, let's take a closer look at the given function: • f(x) = (x² + 1) / (2x + 3) To find the derivative of this function, we will utilize the quotient rule, a differentiation technique suitable for functions of the form u(x) / v(x), where both u(x) and v(x) are differentiable. The quotient rule states that: • f'(x) = [u'(x) * v(x) - u(x) * v'(x)] / [v(x)]² Step 1: Identifying u(x) and v(x) In our case, u(x) = x² + 1, and v(x) = 2x + 3. We will need to find the derivatives of both these functions in order to apply the quotient rule. Step 2: Finding u'(x) and v'(x) To find u'(x), we'll differentiate u(x) with respect to x: • u(x) = x² + 1 • u'(x) = 2x Now, let's find v'(x) by differentiating v(x) with respect to x: • v(x) = 2x + 3 • v'(x) = 2 Step 3: Applying the Quotient Rule Now that we have u'(x), v'(x), u(x), and v(x), we can apply the quotient rule to find f'(x): • f'(x) = [u'(x) * v(x) - u(x) * v'(x)] / [v(x)]² • f'(x) = [(2x) * (2x + 3) - (x² + 1) * (2)] / (2x + 3)² Step 4: Simplifying the Expression Now, let's simplify the expression further by expanding and collecting like terms in the numerator: • f'(x) = [4x² + 6x - 2x² - 2] / (2x + 3)² Combine like terms in the numerator: • f'(x) = [2x² + 6x - 2] / (2x + 3)² Step 5: Final Simplification The derivative can be further simplified by factoring out a common factor from the numerator: • f'(x) = 2(x² + 3x - 1) / (2x + 3)² Now, we have successfully found the derivative of the function (x² + 1) / (2x + 3): • f'(x) = 2(x² + 3x - 1) / (2x + 3)² Conclusion In this comprehensive guide, we have walked through the process of finding the derivative of the function (x² + 1) / (2x + 3) step by step. By applying the quotient rule and simplification techniques, we have obtained the final result: • f'(x) = 2(x² + 3x - 1) / (2x + 3)² Calculus and differentiation are fundamental concepts in mathematics and have wide-ranging applications in various fields such as physics, engineering, economics, and more. Understanding how to find derivatives is essential for analyzing functions and their behavior, and it forms the basis for solving a wide range of real-world problems.
# Prove that 4^n-3n-1 is divisible by 9 To Prove 4n – 3n – 1 is divisible by 9 ∀ n ∈ N Proof We will prove the given statement using mathematical induction. Step 1 First, we will check if the given statement is true for n = 1. 4n – 3n – 1 = 41 – 3(1) – 1 = 4 – 3 – 1 = 0 The statement is true for n = 1 since 0 is divisible by 9. Step 2 In this step, we will assume that the statement is true for n = k. This implies that 4k – 3k – 1 is divisible by 9. We can say 4k – 3k – 1 = 9x, ∀ x ∈ N ……(1) Step 3 In this step, we will check if the statement is true for n = k + 1 given that it is true for n = k. Substituting equation (1) in (2) 9(4x+k) is divisible by 9. This implies that the given statement is true for n = k + 1 given that the statement is true for n = k. Therefore, by the principle of mathematical induction, the given statement “4n – 3n – 1 is divisible by 9″ is true for all n ∈ N.
# Friction Example Problem – Coefficient Of Static Friction In Friction Example Problem – Sliding Down An Inclined Plane, I showed how to find the coefficient of kinetic friction of a block sliding down an incline plane. This friction example problem shows how to find the coefficient of static friction using the same methods. Problem: A block of weight w sits on a level surface. One end of the surface is slowly raised until the block is just about to move down the ramp. What is the coefficient of friction between the block and the ramp’s surface? Solution: This is the force diagram showing the forces in play on the block and ramp. The normal force N is perpendicular to the surface of the ramp. The frictional force Ff acts parallel to the ramp’s surface and resists the motion of the block. The weight w pulls vertically down. The angle formed between the level ground and the ramp’s surface just before the block moves is φ. If we choose our coordinate system to align with the surface of the ramp, the math will be a little bit easier. The y-axis is perpendicular to the surface of the ramp and the positive x-direction is down the ramp. Now our force diagram is broken up into x- and y-components. The sum of the forces in the x-direction is: ΣFx = w·sinφ – Ff The force of friction Ff = μN. Since the block is right on the verge of moving, the μ we use is the coefficient of static friction: μs. ΣFx = w·sinφ – μsN The system is on the verge of motion, but it’s still not in motion. This means the forces are in equilibrium and the total force is equal to zero. Substitute 0 for ΣFx. w·sinφ – μsN = 0 or w·sinφ = μsN Let’s leave this part for now. For the next part, find the sum of the forces in the y-direction. Since they too are in equilibrium, the sum will equal zero as well. ΣFy = N – w·cosφ = 0 Solve for N N = w·cosφ Now that we have a value for the normal force, plug it into the equation we got from the x-direction forces. w·sinφ = μsN w·sinφ = μs(w·cosφ) Solve for μs μs = tanφ
# Inverse Proportion Lesson In Keeping it in Proportion, we learnt about direct proportion, where one amount increased at the same constant rate as the other amount increased. Now we are going to look an inverse proportion. Inverse proportion means that as one amount increases the other amount decreases. Mathematically, we write this as $y\propto\frac{1}{x}$y1x. For example, speed and travel time are inversely proportional because the faster you go, the shorter your travel time. General Equation for Amounts that are Inversely Proportional We express these kinds of inversely proportional relationships generally in the form $y=\frac{k}{x}$y=kx where $k$k is the constant of proportionality and $x$x and $y$y are any variables Just like identifying directly proportional relationships, when identifying inversely proportional relationships we need to find the constant of proportionality or the constant term that reflects the rate of change. The we can substitute any quantity we like into our equation. Let's see how in the examples below. #### Examples ##### Question 1 Consider the equation $s=\frac{375}{t}$s=375t. 1. State the constant of proportionality. 2. Find the value of $s$s when $t=6$t=6. Give your answer as an exact value. 3. Find the value of $s$s when $t=12$t=12. Give your answer as an exact value. ##### Question 2 Consider the values in each table. Which of them could represent an inversely proportional relationship between $x$x and $y$y? 1. $x$x $y$y $1$1 $2$2 $3$3 $4$4 $3$3 $1.5$1.5 $1$1 $0.75$0.75 A $x$x $y$y $1$1 $2$2 $3$3 $4$4 $36$36 $18$18 $12$12 $9$9 B $x$x $y$y $1$1 $5$5 $6$6 $10$10 $3$3 $75$75 $108$108 $300$300 C $x$x $y$y $1$1 $2$2 $3$3 $4$4 $4$4 $5$5 $6$6 $7$7 D $x$x $y$y $1$1 $2$2 $3$3 $4$4 $3$3 $1.5$1.5 $1$1 $0.75$0.75 A $x$x $y$y $1$1 $2$2 $3$3 $4$4 $36$36 $18$18 $12$12 $9$9 B $x$x $y$y $1$1 $5$5 $6$6 $10$10 $3$3 $75$75 $108$108 $300$300 C $x$x $y$y $1$1 $2$2 $3$3 $4$4 $4$4 $5$5 $6$6 $7$7 D
## Another Look at Fractions of a Set • Lesson 3-5 1 This lesson gives students another opportunity to explore fractions using the set model. This lesson is integrated with other areas of the math curriculum including data analysis and statistics. In this culminating activity, students examine the set model using colored candies. Give students an individual bag of colored candies, e.g., M&M's® or Skittles®. Have students open their bag of candies and sort by color. Have students count the number of each color in their set and record those data on notebook paper. Have students record the fraction of each color represented in their individual packet. All fractions should be reduced to lowest form. Have students log on to the Create a Graph Tool from the National Center for Education Statistics. Students should choose the type of graph they want to create by using the pulldown menu. Once students have created their graph, they should label the data in fractional parts and reduce all fractions to lowest terms. As a class, create a line plot of the number of candies in each bag. An example is shown below: Have students determine the fractional representation for each number of candies. For example, for the graph shown above there were: • 2 students with 22 candies (2/16 or 1/8), • 4 students had 23 candies (4/16 or 1/4), • 5 students had 24 candies (5/16), • 3 students had 25 candies (3/16), and • 2 students had 26 candies (2/16 or 1/8). Next, have students log on to the Circle Grapher to create a circle graph for the class data. Circle Grapher Fractional representations should be labeled. Ask students to share their circle graphs with a neighbor. Discuss how a circle graph is useful for showing fractions of a set. Some students may also recognize that percents are also used in circle graphs. This discussion would be a nice tie-in to percents, specifically fractions out of 100. Assessments Since this is the culminating activity for the unit, it is important to use this activity as a summative assessment opportunity. You may wish to examine students' graphs, and have them write a few sentences summarizing each of the graphs. Teacher Reflection • Do students understand that a fraction can be represented as part of a set? • Can students identify fractions when the whole (set) and part of the set is given? • Can students articulate relationships between fractions? • Do students understand the relationships inherent when comparing equivalent fractions? • Can students reduce fractions to lowest terms? • Are there other models of fractions that I could use with these students to extend their repertoire of fraction representations? • What other experiences can I introduce to students to help them better understand relationships among fractions? • How can I ensure that students have a solid conceptual understanding of fractions? • How can I help students relate the concepts in this unit to other areas of mathematics? • How can I help students relate the concepts in this unit to other areas of the curriculum? Questions for Students 1. Which type of graph did you create when you went to the Create a Graph Tool from the National Center for Education Statistics? Why did you select this type of graph? [Student responses may vary. They should give a valid justification for their graph selection.] 2. How does a line plot show the number of candies in each bag? [Each "X" represents one piece of candy. For example, two Xs above the number 22 indicates two students whose bags contained 22 pieces of candy.] 3. Why is a circle graph an appropriate graph to use for fractions of a set? [A circle graph is a good representation of fractions. The pieces of the circle graph represent a certain fraction (or percent.)] ### Eggsactly with a Dozen Eggs 3-5 Students begin to examine fractions as part of a set. This lesson helps students develop skill in problem solving and reasoning as they examine relationships among the fractions used to describe part of a set of 12. ### Eggsactly with Eighteen Eggs 3-5 Students continue to examine fractions as part of a set. This lesson helps students develop skill in problem solving and reasoning as they examine relationships among the fractions used to describe part of a set of eighteen. ### Eggsactly Equivalent 3-5 Students use twelve eggs to identify equivalent fractions. Construction paper cutouts are used as a physical model to represent various fractions of the set of eggs, for example, 1/12, 1/6, and 1/3. Students investigate relationships among fractions that are equivalent. ### Another Look at the Set Model using Attribute Pieces 3-5 The previous lessons focused on the set model where all objects in the set are the same size and shape. Students also need work with sets in which the objects “look” different. In the real world, we are often faced with fraction situations where the objects in the set are not identical. For this lesson, students use fractions to describe a set of attribute pieces. Students develop skill in problem solving and reasoning as they think about their set and how to create new sets given specific fractional characteristics. ### Class Attributes 3-5 During this lesson, students create their own classroom survey or use previously generated questions to study the class and describe the set [class] in fractional parts. This lesson requires that students identify fractions in real-world contexts from a set of items that are not identical. This lesson is integrated with other areas of the math curriculum, including data analysis and statistics. ### Learning Objectives Students will: • Demonstrate understanding that a fraction can be represented as part of a set • Identify fractions when the whole (set) and part of the set is given • Describe a set of objects based on its fractional components • Identify fraction relationships associated with the set • Identify equivalent fractions • Identify relationships inherent in equivalent fractions • Reduce fractions to their lowest terms ### NCTM Standards and Expectations • Recognize equivalent representations for the same number and generate them by decomposing and composing numbers. • Use models, benchmarks, and equivalent forms to judge the size of fractions. • Develop understanding of fractions as parts of unit wholes, as parts of a collection, as locations on number lines, and as divisions of whole numbers. • Recognize and generate equivalent forms of commonly used fractions, decimals, and percents. • Collect data using observations, surveys, and experiments. • Represent data using tables and graphs such as line plots, bar graphs, and line graphs. ### Common Core State Standards – Mathematics Grade 3, Num & Ops Fractions • CCSS.Math.Content.3.NF.A.1 Understand a fraction 1/b as the quantity formed by 1 part when a whole is partitioned into b equal parts; understand a fraction a/b as the quantity formed by a parts of size 1/b. Grade 4, Num & Ops Fractions • CCSS.Math.Content.4.NF.A.1 Explain why a fraction a/b is equivalent to a fraction (n x a)/(n x b) by using visual fraction models, with attention to how the number and size of the parts differ even though the two fractions themselves are the same size. Use this principle to recognize and generate equivalent fractions. Grade 4, Num & Ops Fractions • CCSS.Math.Content.4.NF.A.2 Compare two fractions with different numerators and different denominators, e.g., by creating common denominators or numerators, or by comparing to a benchmark fraction such as 1/2. Recognize that comparisons are valid only when the two fractions refer to the same whole. Record the results of comparisons with symbols >, =, or <, and justify the conclusions, e.g., by using a visual fraction model. Grade 5, Num & Ops Fractions • CCSS.Math.Content.5.NF.B.3 Interpret a fraction as division of the numerator by the denominator (a/b = a ÷ b). Solve word problems involving division of whole numbers leading to answers in the form of fractions or mixed numbers, e.g., by using visual fraction models or equations to represent the problem. For example, interpret 3/4 as the result of dividing 3 by 4, noting that 3/4 multiplied by 4 equals 3, and that when 3 wholes are shared equally among 4 people each person has a share of size 3/4. If 9 people want to share a 50-pound sack of rice equally by weight, how many pounds of rice should each person get? Between what two whole numbers does your answer lie?
# What Fraction is 15? In this article, we’ll explore the basics of fractions, how to simplify them, convert them to decimals, and determine what fraction 15 is. Have you ever wondered about fractions and how they work? Fractions are an essential part of mathematics and daily life. Understanding how to work with fractions can help you in various situations, from calculating recipes to solving complex mathematical problems. In this article, we’ll explore the basics of fractions, how to simplify them, convert them to decimals, and determine what fraction 15 is. By the end of this article, you’ll have a better understanding of fractions and be able to perform basic calculations with ease. ## Understanding Fractions A fraction is a way of representing a part of a whole. It is expressed as a ratio of two numbers, with the top number known as the numerator and the bottom number as the denominator. For example, in the fraction ⅔, 2 is the numerator, and 3 is the denominator. There are three types of fractions: proper fractions, improper fractions, and mixed fractions. A proper fraction is one where the numerator is smaller than the denominator, such as ⅓. An improper fraction is when the numerator is larger than the denominator, such as 5/3. A mixed fraction is made up of a whole number and a proper fraction, such as 1 ½. Fractions can also be expressed as decimals or percentages. For example, ½ can be expressed as 0.5 or 50%. Understanding fractions is essential in mathematics, and it’s a skill that’s used in many aspects of daily life. In the next section, we’ll explore how to simplify fractions. ## Simplifying Fractions Simplifying fractions is the process of reducing them to their smallest possible form. This is done by finding the greatest common factor (GCF) of the numerator and the denominator and dividing both numbers by it. Maybe you are interested What is the Range of the Function Graphed Below? For example, let’s simplify the fraction 12/24. The GCF of 12 and 24 is 12, so we divide both numbers by 12, resulting in 1/2. To simplify fractions, follow these steps: 1. Find the GCF of the numerator and the denominator. 2. Divide both numbers by the GCF. 3. If the fraction can still be simplified, repeat steps 1 and 2 until it can no longer be simplified. Simplifying fractions is important because it makes them easier to work with and understand. In the next section, we’ll explore how to convert fractions to decimals. ## Simplifying Fractions Simplifying fractions is an important skill to have when working with fractions. It makes them easier to understand and work with, especially when dealing with complex calculations. Here are the steps you should take when simplifying fractions: 1. Find the greatest common factor (GCF) of the numerator and denominator. 2. Divide both the numerator and denominator by the GCF. 3. Repeat step 2 until the fraction can no longer be simplified. Let’s use an example to illustrate these steps. Suppose we have the fraction 24/36. 1. Find the GCF of 24 and 36. The factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24. The factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36. The GCF of 24 and 36 is 12. 2. Divide both the numerator and denominator by 12. 24 ÷ 12 = 2 and 36 ÷ 12 = 3. 3. The fraction 24/36 can be simplified to 2/3. Here’s another example. Suppose we have the fraction 16/20. 1. Find the GCF of 16 and 20. The factors of 16 are 1, 2, 4, 8, and 16. The factors of 20 are 1, 2, 4, 5, 10, and 20. The GCF of 16 and 20 is 4. 2. Divide both the numerator and denominator by 4. 16 ÷ 4 = 4 and 20 ÷ 4 = 5. 3. The fraction 16/20 can be simplified to 4/5. Maybe you are interested What is the Maximum Cold-Holding Temperature Allowed for Shredded Lettuce? ## Converting Fractions to Decimals Converting fractions to decimals is another important skill to have when working with fractions. Sometimes, it’s easier to work with decimals than fractions, especially when dealing with measurements or money. Here are the steps you should take when converting fractions to decimals: 1. Divide the numerator by the denominator. 2. Simplify the resulting decimal, if necessary. Let’s use an example to illustrate these steps. Suppose we have the fraction ⅔. 1. Divide the numerator (2) by the denominator (3). 2 ÷ 3 = 0.6666666667 (rounded to ten decimal places). 2. Simplify the decimal, if necessary. The decimal 0.6666666667 can be simplified to 0.67. Here’s another example. Suppose we have the fraction ¾. 1. Divide the numerator (3) by the denominator (4). 3 ÷ 4 = 0.75. 2. The decimal 0.75 is already simplified and cannot be further simplified. By following these steps, you can easily convert fractions to decimals and vice versa. In the next section, we’ll explore how to determine what fraction 15 is. ## What Fraction is 15? Determining what fraction 15 is simply means finding a fraction that is equal to 15. To do this, we need to express 15 as a ratio of two integers, with the numerator and denominator being relatively prime. To find the fraction of 15, we need to follow these steps: 1. Express 15 as a fraction with 1 as the denominator. This gives us 15/1. 2. Multiply the numerator and denominator by the same number to get rid of the decimal point. In this case, we can multiply both by 100 to get 1500/100. 3. Simplify the fraction by finding the GCF of the numerator and denominator. In this case, the GCF of 1500 and 100 is 100, so we can divide both by 100 to get 15/1, which simplifies to 15. Maybe you are interested What Zodiac Sign is the Ugliest? Therefore, the fraction of 15 is 15/1, which is the same as 15. Let’s look at an example of finding the fraction of 15: Suppose we want to find the fraction of 15 that is equivalent to ⅘. We can use cross-multiplication to solve for the missing numerator. First, we can write 15 as 15/1. Then, we can cross-multiply to get: ⅘ * x = 15/1 Multiplying both sides by the denominator of ⅘, which is 5, we get: x = 15/1 * 5/8 Simplifying the fraction by finding the GCF of 15 and 8, we get: x = 75/8 Therefore, the fraction of 15 that is equivalent to ⅘ is 75/8. ## Conclusion In conclusion, fractions are an essential part of mathematics and daily life. Understanding how to work with fractions, from simplifying them to converting them to decimals and finding equivalent fractions, can help you in various situations. Remember, to simplify fractions, find the GCF of the numerator and denominator and divide both numbers by it. To convert fractions to decimals, divide the numerator by the denominator. To find the fraction of a number, express the number as a fraction, and simplify it. Mastering fractions can give you a solid foundation in mathematics and help you in your daily life. With practice and persistence, you can become an expert in working with fractions! ## How Many Pounds is 600 kg? – A Comprehensive Guide Learn how to convert 600 kg into pounds with our comprehensive guide. Discover the factors that impact weight measurement, including gravity, altitude, and temperature. ## How Many Pounds is 600 kg? – A Comprehensive Guide Learn how to convert 600 kg into pounds with our comprehensive guide. Discover the factors that impact weight measurement, including gravity, altitude, and temperature. ## What is the Optional DC Cable for the Yaesu FT-70DR? Learn how to use the optional DC cable for the Yaesu FT-70DR to ensure uninterrupted radio communication in the field. Find out what it is and how to use it! ## How Many Ounces in 1.5 Pounds? Learn how to convert pounds to ounces and vice versa accurately! Discover “how many ounces in 1.5 pounds” and more with this comprehensive guide. ## What is Rebirth 2k22? Unlocking the Mysteries of Reincarnation Discover the mysteries of rebirth 2k22 – the concept of reincarnation in the year 2022. Explore its significance in different beliefs and how to achieve it. ## What is a Smoochie Girl? Discover the truth about what a smoochie girl is and the harmful effects of this behavior. Learn how to break free from the cycle and embrace authenticity.
# A Guide To Cross Products This post will be about the cross product. This topic can be found in a high school Calculus and Vectors course (that I know of in Ontario, Canada) or in a introductory linear algebra course. In here, we will deal with vectors in a three dimensional space. Also, it is easier to compute products if one knows how to compute (2 by 2) determinants. With that being said, I assume that the reader is familiar with determinants. The Cross Product Suppose we have two vectors and in three dimensions. The cross product of two vectors is a (different) vector which is perpendicular to the two vectors. There are two ways to compute cross products. The first formula does not assume the reader knows determinants. This formula is given by: The second formula is actually the same as the above but in determinant notation. This formula is: Source: http://quicklatex.com/cache3/c4/ql_926219b87045b16e28f37e299ca0d7c4_l3.png Recall that the 2 by 2 determinant is equal to . Memory Aids After looking at those two formulas you might be wondering how do I memorize all that stuff!? With some close observations, you can spot a few patterns and come up with some memory aids. With the first formula, all the terms are in the format. If you look closely at the numbered subscripts, the numbers are mirrored. In the first part we have . The first subscript is a 2 which goes with the u and the second subscript is a 3 with the v. After the minus sign the subscripts are mirrored in the order of 3 and 2. The second part has subscripts of 3,1,1,3 and the last part has 1,2,2,1. For the first formula, one possible memory aid would be 2332, 3113 and 1221. An alternative would be 23, 31 and 12 since we have representing 23 and , 31 for and 12 for . The second formula assumes that the reader is familiar with the linear algebra concept of determinants. Notice how the top row in each determinant is for the vector and the bottom row is for the vector. One possible memory aid would be 23, -13, and 12 as the subscripted numbers are used as we go from the leftmost column to the rightmost column. The negative for -13 represents the negative determinant. Another way to arrive to the second formula is through this multi-part method. 1) Form a 2 by 3 matrix in the format of 2) The first component of the cross product can be computed by deleting the first column of the above matrix and take the resulting 2 by 2 determinant. 3) The second component consists of deleting the second column of the 2 by 3 matrix and taking the determinant of 4) The final and third component of the cross product would be removing the last column of the matrix and computing the determinant 5) The result would be the second formula turned into the first formula. Properties Of Cross Products Like with any new math operation like the cross product, there are rules associated with them. In grade school, there were rules for addition, subtraction, multiplication and division. Here are the properties of the cross product given the vectors , and in 3-space with as a numeric scalar. ( is the zero vector in 3-space here) 1) 2) 3) 4) 5) 6) Examples Example One Given and , Find and We can use the formula or the determinant notation of the formula. Here, the determinant notation of the formula will be used. We also compute . You can also verify the property . In this case . Example Two Suppose that there are vectors and . Find a vector which is perpendicular/orthogonal to both vectors and . The vector cross product of a two vectors is a (third) vector which is perpendicular to both the two vectors. We find the cross product as follows: The vector that is perpendicular to vectors and is . Practice Problems Here are few practice problems. Solutions are in the next section. 1) Given two vectors and . What is the cross product ? 2) Given and . What is the cross product ? 3) Suppose you have the vector . Verify that where \textbf{0} is the zero vector in 3-space. 4) In this question we have three vectors. These vectors are , and . What is the cross product ? Solutions 1) 2) 3) The numbers look scary but once the formula is applied you will find that you will get the zero vector. 4) and References Elementary Linear Algebra (Tenth Edition) by Howard Anton
Question Video: Solving Word Problems Involving the Highest Common Factor Mathematics • 6th Grade To encourage public transportation, Leo wants to gift some of his friends envelopes with bus tickets and subway tickets in them. If he has 32 bus tickets and 80 subway tickets to split equally among the envelopes and wants no tickets left over, what is the greatest number of envelopes Leo can make. 04:51 Video Transcript To encourage public transportation, Leo wants to gift some of his friends envelopes with bus tickets and subway tickets in them. If he has 32 bus tickets and 80 subway tickets to split equally among the envelopes and wants no tickets left over, what is the greatest number of envelopes Leo can make? In order to solve this problem, we need to consider the highest common factor. This is the highest number that divides exactly into both of our numbers. Leo has 32 bus tickets and 80 subway tickets. We need to find the highest common factor of 32 and 80. There were lots of ways of finding the factors of a number. Firstly, we will look at the method of finding factor pairs. Factor pairs are two numbers that multiply together to give the number. In this case, we’re looking for pairs of numbers that multiply to give us 32. One multiplied by 32 is equal to 32. Therefore, one and 32 are factors of 32. Two multiplied by 16 also equals 32. Therefore, these two numbers are also factors of 32. The final factor pair of 32 is four and eight, as four multiplied by eight is equal to 32. In ascending order, the factors of 32 are one, two, four, eight, 16, and 32. We can now use the same method to find the factors of 80. There are five factor pairs of 80: one and 80, two and 40, four and 20, five and 16, and eight and 10, as all five of these pairs multiply together to give us 80. In ascending order, once again, the factors of 80 are one, two, four, five, eight, 10, 16, 20, 40, and 80. The highest number that appears in both of these lines is 16. Therefore, the highest common factor of 32 and 80 is equal to 16. This means that the greatest number of envelopes that Leo can make is 16. Each envelope would have two bus tickets and five subway tickets, as 16 multiplied by two is equal to 32 and 16 multiplied by five is equal to 80. An alternative method to find the factors of 32 and 80 is using prime factor decomposition. This involves splitting a number into the product of its prime factors. 32 can be split into two and 16, as two multiplied by 16 is equal to 32. Two is a prime number, so we circle it. We now need to split 16. 16 is equal to two multiplied by eight. Once again, two is a prime number. Our next step is to split the number eight. Eight can be split into two and four, as two multiplied by four is equal to eight. Finally, we can split four into two multiplied by two. This means that 32 is equal to two multiplied by two multiplied by two multiplied by two multiplied by two. This can be simplified to two to the power of five. We can split 80 into a product of its prime factors in the same way. Two multiplied by 40 is equal to 80. Two multiplied by 20 is equal to 40. Two multiplied by 10 is equal to 20. And finally, two multiplied by five is equal to 10. This means that 80 can be written as two to the power of four multiplied by five. This proves that the highest common factor of 32 and 80 is two to the power of four. And as two to the power of four is equal to 16, we can say that the highest common factor of 32 and 80 is 16. Therefore, once again, the highest number of envelopes that Leo can make is 16.
# Distance of a Point from a line & Distance between Two Parallel Line Formula In this post we will understand two formulas which are important for Grade 11 NCERT and CBSE Math Syllabus. Along with the formulas we have also provided solved examples for your better understanding. ## Distance of a Point from a Line The distance (d) of a point P( x_{1} ,\ y_{1}) from a line Ax + By + C =0 is given by the below mentioned formula – d\ =\ \frac{\|\ Ax_{1} \ +\ By_{1} \ +\ C\ \|}{\sqrt{A^{2} \ +\ B^{2}}} ### How to solve distance of a point questions Step 01: Write the equation of the straight line in the form of general equation as shown below – ⟹ Ax + By + C = 0 Step 02: Find the distance ‘d’ of the point ( x_{1} ,\ y_{1}) from the straight line line Ax + By + C = 0 by using the formula – d\ =\ \frac{\|\ Ax_{1} \ +\ By_{1} \ +\ C\ \|}{\sqrt{A^{2} \ +\ B^{2}}} ### Distance of a Point solved problems (Q1) Find the distance of the point ( -1, 1) from the line 12( x + 6) = 5( y – 2) Solution: The equation of the straight line in the general form ⟹ 12( x + 6) = 5 ( y – 2) ⟹ 12x + 72 = 5y – 10 ⟹ 12x – 5y + 82 = 0 Here, A = 12, B = -5, C = 82 Then the distance of the point ( -1 ,1) from the line 12x – 5y + 82 =0 is – d =\ \frac{\|\ Ax_{1} \ +\ By_{1} \ +\ C\ \|}{\sqrt{A^{2} \ +\ B^{2}}}\\\ \\ d\ =\ \frac{\|\ ( 12)( -1) \ +\ ( -5)( 1) \ +\ 82\ \|}{\sqrt{( 12)^{2} \ +\ ( -5)^{2}}} \\\ \\ d\ =\ \frac{\|\ -12 \ -5 \ +\ 82\ \|}{\sqrt{144\ +\ 25}}\\\ \\ d\ =\ \frac{\|\ 82\ -\ 17\ \|}{\sqrt{169}}\\\ \\ d\ = \ \frac{65}{13} \ =\ 5 \ units \\\ \\ Hence,\ the\ distance\ of\ the\ point\ ( -1,\ 1) \\ \\ is\ 5\ units\ from\ the\ given\ straight\ line. (Q2) Find the distance of the point ( 3, -5) from the line 3x – 4y – 26 =0 Solution the equation of the straight line in the general form ⟹ 3x -4y -26 = 0 Here, A = 3, B = -4, C = -26 Then the distance of the point ( 3 , -5) from the line 3x – 4y – 26 = 0 is : d\ =\ \frac{\|\ Ax_{1} \ +\ By_{1} \ +\ C\ \|}{\sqrt{A^{2} \ +\ B^{2}}}\\\ \\ d\ =\ \frac{\|\ ( 3)( 3) \ +\ ( -4)( -5) \ +\ ( -26) \ \|}{\sqrt{( 3)^{2} \ +\ ( -4)^{2}}}\\\ \\ d\ =\ \frac{\|\ 9 \ + \ 20\ -26\ \|}{\sqrt{9\ +\ 16}}\\\ \\ d\ =\ \frac{\|\ 29\ -\ 26\ \|}{\sqrt{25}}\\\ \\ d\ =\ \frac{3}{5}\\\ \\ Hence,\ the\ distance\ of\ the\ point\ ( 3,\ -5) \\ \\ is\ \frac{3}{5} \ from\ the\ given\ straight\ line. (Q3) Find the points on the x-axis, whose distances from the line \frac{x}{3} \ +\frac{y}{4} \ =1 are 4 units Solution. The equation of the straight line in the general form \frac{x}{3} \ +\frac{y}{4} \ =1 \\\ \\ \frac{4x\ +\ 3y}{12} \ =\ 1 \\\ \\ 4x\ +\ 3y\ =\ 12 \\\ \\ 4x\ +\ 3y\ -\ 12\ =\ 0 \\\ \\ Here, A = 4, B = 3, C = -12 We have to find the distance of any point on x-axis, for which the distance are 4 units. so, the point must have co-ordinates ( x, 0) Then the distance of the point ( x , 0) from the line 4x + 3y – 12 = 0 is d\ =\ \frac{\|\ Ax_{1} \ +\ By_{1} \ +\ C\ \|}{\sqrt{A^{2} \ +\ B^{2}}}\\\ \\ 4\ =\ \frac{\|\ ( 4)( x) \ +\ ( 3)( 0) \ +\ ( -12) \ \|}{\sqrt{( 4)^{2} \ +\ ( 3)^{2}}}\\\ \\ 4\ =\ \frac{\|\ 4x\ +0\ -12\ \|}{\sqrt{16\ +\ 9}}\\\ \\ 4\ =\ \frac{\|\ 4x\ _{\ } -\ 12\ \|}{\sqrt{25}}\\\ \\ 4\ =\pm \ \frac{4x\ -\ 12}{5} \\\ \\ 20\ =\ \pm \ ( 4x\ -\ 12)\\\ \\ {Case I} ⟹ 20 = 4x – 12 ⟹ 20 + 12 = 4x ⟹ x = 8 {Case II} ⟹ 20 = – (4x – 12) ⟹ 20 = – 4x +12 ⟹x = -2 Hence, the points on the x-axis are ( -2 , 0) or ( 8 , 0) for which the distance are 4 units ## Distance Between Two Parallel Lines If the two lines are given in form of general equation i.e., Ax + By + C_{1} = 0 \ and\ Ax + By +C_{2} = 0, then the distance between the two parallel lines can be obtained using the below mentioned formula – d\ =\ \frac{\|\ C_{1} \ -C_{2} \ \|}{\sqrt{A^{2} \ +\ B^{2}}} \\\ \\ But if the two parallel lines are given in the slope-intercept form i.e., y = mx\ + c_{1} and y\ =\ mx\ +c_{2} then the distance between the two parallel lines can be obtained using the below mentioned formula – d\ =\ \frac{\|\ c_{1} \ -c_{2} \ \|}{\sqrt{1\ +\ m^{2}}} ### Distance between Straight Lines Questions (Q1) Find the distance between the two parallel lines ⟹{15x\ +\ 8y\ -\ 34\ =\ 0} ⟹{15x\ +\ 8y\ +\ 31\ =\ 0} Solution The equation of the two parallel straight line in the general form – ⟹15x + 8y – 34 = 0 ⟹15x + 8y + 31 = 0 Here, A = 15; B = 8; C_{1} = -34 C_{2} = 31 the\ distance\ between\ the\ two\ parallel\ lines\\ \\ can\ be\ obtained\ using\ the\ below\ mentioned\\ \\ formula\ -\\\ \\ d\ =\ \frac{\|\ C_{1} \ -C_{2} \ \|}{\sqrt{A^{2} \ +\ B^{2}}}\\\ \\ Putting\ values\\\ \\ d\ =\ \frac{\|\ -34\ -31 \ \|}{\sqrt{( 15)^{2} \ +\ ( 8)^{2}}}\\\ \\ d\ =\ \frac{ 65 }{\sqrt{225\ +\ 64}} \\\ \\ =\ \frac{65}{\sqrt{289}} \\\ \\ =\ \frac{65}{17}\\\ \\ Hence,\ the\ distance\ between\ the\ two\ parallel \\ \\ lines \ is \frac{65}{17} (Q2) Find the distance between the two parallel lines ⟹ l ( x + y ) + p = 0 ⟹ l ( x + y ) – r = 0 Solution The equation of the two parallel straight line in the general form – ⟹ l (x + y) + p = 0 ⟹ lx + ly +p = 0 ⟹ l ( x + y) – r = 0 ⟹ lx + ly – r = 0 Here, A = l; B = l; C_{1} = p C_{2} = – r the\ distance\ between\ the\ two\ parallel\ lines\\ \\ can\ be\ obtained\ using\ the\ formula\\\ \\ d\ =\ \frac{\|\ C_{1} \ -C_{2} \ \|}{\sqrt{A^{2} \ +\ B^{2}}} \\\ \\ Putting\ values \\\ \\ d\ =\ \frac{\|\ p\ -( -r) \ \|}{\sqrt{( l)^{2} \ +\ ( l)^{2}}} \\\ \\ d\ =\ \frac{\|\ p+r\ \|}{\sqrt{l^{2} \ +\ l^{2}}} \\\ \\ =\ \frac{\|p+r\|}{\sqrt{2l^{2}}} \\\ \\ =\ \frac{\|p+r\|}{l\sqrt{2}} \\\ \\ Hence, the distance between the two parallel lines is \frac{\|p+r\|}{l\sqrt{2}} (Q3) Find the distance between the two parallel lines ⟹ y = 4x + 7 ⟹ y = 4x – 10 Solution: the equation of the two parallel straight line in the general form – ⟹ y = 4x + 7 ⟹ y = 4x – 10 Here, m =4; c_{1} = 7 c_{2} = -10 the distance between the two parallel lines can be obtained using the below mentioned formula: d\ =\ \frac{\|\ c_{1} \ -c_{2} \ \|}{\sqrt{1\ +\ m^{2}}}\\\ \\ Putting\ values \\\ \\ d\ =\ \frac{\|\ 7\ -( -10) \ \|}{\sqrt{1\ +\ ( 4)^{2}}}\\\ \\ d\ =\ \frac{\|\ 7\ +10\ \|}{\sqrt{1\ +\ 16}} \\\ \\ =\ \frac{\|17\|}{\sqrt{17}} \\\ \\ =\ \frac{17}{\sqrt{17}} \ =\ \sqrt{17}\\\ \\ Hence,\ the\ distance\ between\\ \\ the\ two\ parallel\ line\ is\ \ \sqrt{17}
# How to Find Domain of a Function: A Comprehensive Guide Baca Cepat ## Introduction Greetings! Are you struggling to find the domain of a function? You’re not alone! This is a common problem among math students, and we’re here to help. In this article, we’ll provide a step-by-step guide on how to find the domain of a function. But first, let’s define what a domain is. The domain of a function is the set of all possible input values (x-values) for which the function is defined. In other words, it’s the set of all x-values that make the function work. Understanding domain is crucial in solving problems involving functions. So, let’s dive right in! ### What is a Function? Before we get into how to find the domain of a function, let’s first understand what a function is. A function is a set of ordered pairs (x, y) in which each x-value is paired with exactly one y-value. The x-value is called the independent variable, while the y-value is called the dependent variable. Functions are commonly represented by an equation, such as y = f(x), where y represents the dependent variable, x represents the independent variable, and f(x) represents the function. ### Why is Domain Important? Domain is an essential concept in solving problems involving functions. It helps us determine the range of input values that will produce meaningful output. By identifying the domain, we can avoid errors in calculations and ensure that our results are accurate and meaningful. ### How to Find the Domain of a Function? Finding the domain of a function involves identifying the set of all possible input values for which the function is defined. There are different methods to find the domain, depending on the type of function. Let’s explore some of the most common functions and how to find their domain. ## Linear Functions Linear functions are functions of the form y = mx + b, where m and b are constants. To find the domain of a linear function, we need to identify the set of all possible input values that make the function work. Since a linear function is defined for all real numbers, the domain is (-∞, ∞). Quadratic functions are functions of the form y = ax² + bx + c, where a, b, and c are constants. To find the domain of a quadratic function, we need to identify the set of all possible input values that make the function work. Since a quadratic function is defined for all real numbers, the domain is (-∞, ∞). ## Absolute Value Functions Absolute value functions are functions of the form y = \|x\|. To find the domain of an absolute value function, we need to identify the set of all possible input values that make the function work. Since an absolute value function is defined for all real numbers, the domain is (-∞, ∞). ## Rational Functions Rational functions are functions of the form y = f(x) / g(x), where f(x) and g(x) are polynomials. To find the domain of a rational function, we need to identify the set of all possible input values that make the function work. However, we need to be careful when dealing with rational functions because they may have restrictions on the input values. To find the domain of a rational function, we need to identify the values that make the denominator equal to zero. These values are called the excluded values. The domain of the function is all real numbers except for the excluded values. ## Exponential Functions Exponential functions are functions of the form y = a^x, where a is a positive constant. To find the domain of an exponential function, we need to identify the set of all possible input values that make the function work. Since an exponential function is defined for all real numbers, the domain is (-∞, ∞). ## Logarithmic Functions Logarithmic functions are functions of the form y = loga(x), where a is a positive constant. To find the domain of a logarithmic function, we need to identify the set of all possible input values that make the function work. Since a logarithmic function is defined only for positive numbers, the domain is (0, ∞). ## Trigonometric Functions Trigonometric functions are functions of the form y = f(x), where f(x) is a trigonometric expression. To find the domain of a trigonometric function, we need to identify the set of all possible input values that make the function work. Since trigonometric functions are periodic, the domain is usually all real numbers. ## Table: Summary of Domain for Different Functions Function Domain Linear Functions (-∞, ∞) Absolute Value Functions (-∞, ∞) Rational Functions All real numbers except excluded values Exponential Functions (-∞, ∞) Logarithmic Functions (0, ∞) Trigonometric Functions All real numbers ### 1. What is the domain of a function? The domain of a function is the set of all possible input values (x-values) for which the function is defined. ### 2. Why is domain important in solving problems involving functions? Domain helps us determine the range of input values that will produce meaningful output. By identifying the domain, we can avoid errors in calculations and ensure that our results are accurate and meaningful. ### 3. How do I find the domain of a function? Finding the domain of a function involves identifying the set of all possible input values for which the function is defined. Different methods are used depending on the type of function. ### 4. What is an excluded value? An excluded value is a value that makes the denominator of a rational function equal to zero. These values are not included in the domain of the function. ### 5. Can a function have multiple domains? No, a function can have only one domain. The domain is the set of all possible input values for which the function is defined. ### 6. What happens if I use an input value that is not in the domain? If you use an input value that is not in the domain, the function will not produce meaningful output. In other words, the function will be undefined for that input value. ### 7. Can the domain of a function be negative? No, the domain of a function cannot be negative. The domain is a set of input values, and input values cannot be negative. ### 8. What is the domain of a linear function? The domain of a linear function is (-∞, ∞). ### 9. What is the domain of a quadratic function? The domain of a quadratic function is (-∞, ∞). ### 10. What is the domain of an absolute value function? The domain of an absolute value function is (-∞, ∞). ### 11. What is the domain of a rational function? The domain of a rational function is all real numbers except for the excluded values. ### 12. What is the domain of an exponential function? The domain of an exponential function is (-∞, ∞). ### 13. What is the domain of a logarithmic function? The domain of a logarithmic function is (0, ∞). ## Conclusion Now that you understand how to find the domain of a function, you can confidently solve problems involving functions. Remember, identifying the domain is crucial in ensuring that your results are accurate and meaningful.
Courses Courses for Kids Free study material Offline Centres More Store # The increase in length on stretching a wire is 0.05%. If its poisson's ratio is 0.4, then its diameter: (A) Reduce by 0.02% (B) Reduce by 0.1% (C) Reduce by 0.03% (D) Decrease by 0.4% Last updated date: 17th Jun 2024 Total views: 52.8k Views today: 1.52k Verified 52.8k+ views Hint: The Poisson’s ratio for a wire is given. When the wire is stretched, the length increases by 0.05%. Now by definition, we know the Poisson’s ratio $\sigma = \dfrac{{{\text{lateral strain}}}}{{{\text{longitudinal strain}}}}$. Longitudinal strain is given, so we can find the lateral strain by substituting the given values. Formula used: Poisson’s ratio $\sigma = \dfrac{{{\text{lateral strain}}}}{{{\text{longitudinal strain}}}}$ $\Rightarrow \sigma = \dfrac{{\dfrac{{\Delta D}}{D}}}{{\dfrac{{\Delta l}}{l}}}$ Complete step by step solution: Let D and l be the diameter and length of the given wire respectively. When the wire is stretched, the length increases by 0.05%. We know, the Poisson’s ratio $\sigma = \dfrac{{{\text{lateral strain}}}}{{{\text{longitudinal strain}}}}$ $\Rightarrow \sigma = \dfrac{{\dfrac{{\Delta D}}{D}}}{{\dfrac{{\Delta l}}{l}}} = 0.4$ Here, $\dfrac{{\Delta l}}{l}$= 0.05 $\Rightarrow \sigma = \dfrac{{\dfrac{{\Delta D}}{D}}}{{0.05}} = 0.4$ $\Rightarrow \dfrac{{\Delta D}}{D} = 0.05 \times 0.4 = 0.02$ Therefore, the correct answer is option (A), reduced by 0.02%. Note: If the length of the wire increases on stretching, the diameter will decrease simultaneously. However, the ratio of lateral and longitudinal strain is always a constant for a material and is defined as the Poisson’s Ratio. It is denoted by the letter $\sigma$.
# Thread: Complex number. How is this expressed in the form x + iy? 1. ## Complex number. How is this expressed in the form x + iy? The complex number z and its complex conjugate Z satisfy the equation $2z + Z = \frac{11 + 7i}{1 + i}$ Find z in the form x + iy. [6] This is a new topic I am doing in maths and I would like to know how this is done because I don't know the basics in answering it. I would greatly appreciate any help if you could tell me what methods are used. Thanks note: "Z" actually symbolizes a little z with the line on top of it - z's conjugate, to clarify. 2. Originally Posted by db5vry The complex number z and its complex conjugate Z satisfy the equation $2z + Z = \frac{11 + 7i}{1 + i}$ Find z in the form x + iy. [6] $\begin{gathered} 2z + \overline z = 3x + yi \hfill \\ \frac{{11 + 7i}} {{1 + i}} = \frac{{\left( {11 + 7i} \right)\left( {1 - i} \right)}} {2} \hfill \\ \end{gathered}$ Now carry on. 3. First of all $\frac{11+7i}{1+i}=\frac{(11+7i)(1-i)}{2}=9-2i$ $2(x+yi)+x-yi=9-2i$ $3x+yi=9-2i$ Then $3x=9\Rightarrow x=3$ $y=-2$ 4. we know z=x+iy and Z=x-iy implies 2x+2iy+x-iy=11+7i/1+i implies 3x+iy= (11+7i)(1-i)/2 = (11+7+7i-11i)/2 = 9-2i => x=3 and y=-2 => z=3-2i
# Percentage Increase or Decrease Calculating percentage change is an important skill for geographers to have. When geographers collect data over a period of time, the results may increase. Calculating a percentage increase allows a geographer to see how much their data has changed. Calculating the percentage increase between two figures is achieved by: • calculate the difference between the two numbers • divide the increase (difference) by the first number • multiply the answer by 100. Worked example In this example, we will examine changes in the GDP for the UK. In 2001 the UK had a total economic output of £1.4 trillion. In 2015 it was £1.64 trillion. Therefore: 1.64 – 1.4 = 0.2 divided by 1.4 × 100 = 14.3. The UK total economic output between 2001 and 2015 had a percentage increase of 14.3%. If the change between two numbers shows a decrease you need to: 1. work out the difference between the two numbers being compared 2. divide the decrease by the original number and multiply the answer by 100 3. in summary: percentage decrease = decrease ÷ original number × 100 Worked example For example, the number of foxes in an area of woodland in February and March is counted. In February 22 foxes were counted. In March 12 foxes were counted. What is the percentage decrease of foxes in the woodland? 1. the difference between the two numbers is 10 2. 10 ÷ 22 × 100 = 45.4 3. the percentage decrease of foxes found in the woodland is: 45.4%
# Volume of a Cone Solid geometry is concerned with three-dimensional shapes. In these lessons, we will learn • what is a cone? • how to calculate the volume of a cone. • how to solve word problems about cones. • how to prove the formula of the volume of a cone. Related Topics:More Geometry Lessons ### Cones A cone is a solid with a circular base. It has a curved surface which tapers (i.e. decreases in size) to a vertex at the top. The height of the cone is the perpendicular distance from the base to the vertex. A right cone is a cone in which the vertex is vertically above the center of the base. When the vertex of a cone is not vertically above the center of the base, it is called an oblique cone. The following diagrams show a right cone and an oblique cone. In common usage, cones are assumed to be right and circular. Its vertex is vertically above the center of the base and the base is a circle. However, in general, it could be oblique and its base can be any shape. This means that, technically, a pyramid is also a cone. ### Volume of a Cone The volume of a cone is equal to one-third the product of the area of the base and the height. It is given by the formula: where r is the radius of the base and h is the perpendicular height of the cone. Worksheet to calculate the volume of cones. Example: Calculate the volume of a cone if the height is 12 cm and the radius is 7 cm. Solution: Volume How to use the formula to find the volume of a cone? It provides an example of how to determine the volume of a cone. How to find the volume of a cone? How to solve Word Problems about Cones? Problem: A Maxicool consists consists of a cone full of ice-cream with a hemisphere of ice-cream on top. The radius of a hemisphere is 3 cm. The height of the cone is 10 cm. Calculate the total volume of the ice-cream. Problem: A scoop of strawberry of radius 5 cm is placed in a cone. When the ice-cream melts, it fills two thirds of the cone. Find the volume of the cone. (Assuming no ice-cream drips outside the cone). Proof for the formula of the volume of a cone This video will demonstrate that the volume of a cone is on-third that of a cylinder with the same base and height. This is not a formal proof. You would need to use calculus for a more rigorous proof. How to derive the formula for the volume of a right circular cone using calculus? Derive the formula for the volume of a right circular cone using the method of washers (disks). How to use Calculus to Derive the Volume of a Cone? (Integral) We use integration to deduce the formula for the volume of a cone. Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations. You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
Last updated: # 99% Confidence Interval Calculator What is a confidence interval?How do I find a 99% confidence interval?How to use 99% confidence interval calculatorConfidence interval toolsFAQs If you've got a data set and want to perform statistical calculations on it, our 99% confidence interval calculator is a good place to start. We've got you covered — not only if you want to know how to find this 99% confidence interval, but also to learn: • What is Z-score; • What does confidence interval even mean; and • How to find a margin of error for a 99% confidence interval. ## What is a confidence interval? A confidence interval is the range of values you expect your parameter to fall in if you repeat a test multiple times. Let's see an example that puts confidence intervals into real life. Becky sells homemade muffins, and she wants to check the average weight of her baked goods. She found that 99% of her muffins weigh between 121 and 139 grams (4.27–4.9 oz), while one muffin came out of the oven at a whopping 160 grams (5.64 oz) — much bigger than expected! The 99% confidence interval of Becky's muffins' weights is the range of 121 to 139 g. And so, when selling muffins, she can be 99% sure that any muffin she baked weighs between 121 and 139 g. But 1% of the time, she might accidentally produce a chonky muffin (or a tiny one!) ## How do I find a 99% confidence interval? Estimating confidence intervals is a bit of work, but we'll walk you through the process step by step. 1. First, you need to know: • $n$ — the size of your sample (the number of measurements taken); • $μ$ — the mean (average) of the measurements; and • $σ$ — the standard deviation of the measurements. 2. Now, you can calculate standard error and margin of error with the formulas: $\footnotesize \qquad \text{SE} = \frac{σ}{\sqrt{n}} \\[6pt] \qquad \text{ME} = \text{SE} × Z(0.99)$ where: • $SE$ — the standard error; • $ME$ — the margin of error; • $Z(0.99)$ — the z-score corresponding to the chosen confidence level (which you'll find in statistical tables). 1. Add and subtract the margin of error value from the mean to obtain your confidence interval. It is the range between these lower and upper bounds. $\footnotesize \qquad\text{upper bound} = \mu + ME \\ \qquad\text{lower bound} = \mu - ME$ ## How to use 99% confidence interval calculator Now that you know how complicated it is to count confidence intervals on your own, you know what's going on behind the scenes with our 99% confidence interval calculator! Here's how to use it: 1. Fill in the sample mean (x̅) in the first row. 2. Enter the standard deviation (s). 3. Enter the sample size (n). 4. Your confidence level is already filled in (99%), but keep in mind you can change it anytime. 5. The Z-score will update automatically as you decide on the confidence interval. 6. And that's it! At the bottom of the calculator, you'll see: • A chart describing your data; • The confidence interval range — with a lower and upper bound indicated; and • The margin of error. ## Confidence interval tools If you're interested in statistics, you might find those tools helpful: FAQs ### What is the Z-score for a 99% confidence interval? The z-score for a two-sided 99% confidence interval is 2.807, which is the 99.5-th quantile of the standard normal distribution N(0,1). ### How to find the margin of error for a 99% confidence interval? To find the margin of error for a 99% confidence interval: 1. Find Z(0.99) (the z-score for 99% confidence) in the statistical table. Z(0.99) = 2.576 2. Calculate the standard error with the formula SE = σ/√n, where σ is the standard deviation and n is the sample size. 3. Multiply Z(0.99) by the standard error to obtain the margin of error, ME. ME = Z(0.99) × SE
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> You are reading an older version of this FlexBook® textbook: CK-12 Middle School Math Concepts - Grade 6 Go to the latest version. Difficulty Level: At Grade Created by: CK-12 Estimated5 minsto complete % Progress Progress Estimated5 minsto complete % Remember Tania and the carrots? Well, now let's think about range. Here is Tania’s data about the number of carrots picked each week over nine weeks of harvest. 2, 8, 8, 14, 9, 12, 14, 20, 19, 14 What is the range of carrots picked? This Concept will teach you all about range. Then we will revisit this problem at the end of the Concept. ### Guidance The range of a set of data simply tells where the numbers fall, so that we know if they are close together or spread far apart. A set of data with a small range tells us something different than a set of data with a large range. We’ll discuss this more, but first let’s learn how to find the range. Here are the steps for finding the range of a set of data. 1. What we need to do is put the values in the data set in numerical order. Then we know which is the greatest number in the set (the maximum), and which is the smallest number (the minimum). 1. To find the range, we simply subtract the minimum from the maximum. Take a minute to copy these steps into your notebook. Take a look at the data set below. 11, 9, 8, 12, 11, 11, 14, 8, 10 First, we arrange the data in numerical order. 8, 8, 9, 10, 11, 11, 11, 12, 14 Now we can see that the minimum is 8 and the maximum is 14. We subtract to find the range. 14 - 8 = 6 The range of the data is 6. That means that all of the numbers in the data set fall within six places of each other. All of the data results are fairly close together. How can we use a range to help us answer a question? Suppose we wanted to know the effect of a special soil on plant growth. The numbers in this data set might represent the height in inches of 9 plants grown in the special soil. We know that the range is 6, so all of the plants heights are within 6 inches of each other. 6, 11, 4, 12, 18, 9, 25, 16, 22 Let’s reorder the data and find the range. 4, 6, 9, 11, 12, 16, 18, 22, 25 Now we can see that the minimum is 4 and the maximum is 25. Let’s subtract to find the range. 25 - 4 = 21 The range of this data is 21. That means the numbers in the data set can be much farther apart. What does this mean about plants grown in special soil? If the first group of plants had a range of only 6, their heights ended up being fairly close together. So they grew about the same in the special soil. In contrast, the second group of plants had a much greater range of heights. We might not be so quick to assume that the special soil had any effect on the plants, since their heights are so much more varied. The range has helped us understand the results of the experiment. Here are a few for you to try on your own. Find the range of the following data sets. #### Example A 4, 5, 6, 9, 12, 19, 20 Solution: 16 #### Example B 5, 2, 1, 6, 8, 20, 25 Solution: 24 #### Example C 65, 23, 22, 45, 11, 88, 99, 123, 125 Solution: 114 Now let's go back to Tania and the carrots. Here is Tania’s data about the number of carrots picked each week over nine weeks of harvest. 2, 8, 8, 14, 9, 12, 14, 20, 19, 14 What is the range of carrots picked? To find this out, we need to figure out the difference between the greatest number of carrots picked and the least number of carrots picked. The greatest number picked was 20. The least number picked was 2. \begin{align}20 - 2 = 18\end{align} The range is 18 carrots. ### Vocabulary Here are the vocabulary words in this Concept. Maximum the greatest score in a data set Minimum the smallest score in a data set Range the difference between the smallest value in a data set and the greatest number in a data set ### Guided Practice Here is one for you to try on your own. The following is the number of patrons at a local movie theater. 26, 22, 40, 45, 46, 18, 30, 80, 60, 75 What is the range of the data? To figure this out, we need to find the difference between the highest number of patrons and the lowest number of patrons. The highest number of patrons was 80. The lowest number of patrons was 22. \begin{align}80 - 22 = 58\end{align} The range for the data set is 58. ### Video Review Here is a video for review. ### Practice Directions: Find the range for each set of data. 1. 4, 5, 4, 5, 3, 3 2. 6, 7, 8, 3, 2, 4 3. 11, 10, 9, 13, 14, 16 4. 21, 23, 25, 22, 22, 27 5. 27, 29, 29, 32, 30, 32, 31 6. 34, 35, 34, 37, 38, 39, 39 7. 43, 44, 43, 46, 39, 50 8. 122, 100, 134, 156, 144, 110 9. 224, 222, 220, 222, 224, 224 10. 540, 542, 544, 550, 548, 547 11. 2, 3, 3, 3, 2, 2, 2, 5, 6, 7 12. 4, 5, 6, 6, 6, 7, 3, 2 13. 23, 22, 22, 24, 25, 25, 25 14. 123, 120, 121, 120, 121, 125, 121 15. 678, 600, 655, 655, 600, 678, 600, 600 ### Vocabulary Language: English Maximum Maximum The largest number in a data set. measure measure To measure distance is to determine how far apart two geometric objects are by using a number line or ruler. Median Median The median of a data set is the middle value of an organized data set. Minimum Minimum The minimum is the smallest value in a data set. Oct 29, 2012 Jul 08, 2015
## Dividing a Square into Similar Rectangles If you divide a square into some fixed number of similar rectangles, what proportions can these rectangles have? We’ve been having fun thinking about this on Mathstodon, and here is a report. If you divide a square into 3 similar rectangles, what proportions can these rectangles have? There are three options. The third is more complicated than the first two: • We can divide the square into three rectangles that are 1/3 as long in one direction as the other, as in the first picture. • We can divide it into three rectangles that are 2/3 as long in one direction as the other, as in the second picture. • We can divide it into three rectangles that are x times as long in one direction as the other, as in the third picture. What’s x? The yellow rectangle has height 1 and width x, so the blue rectangle has width 1-x and height x(1-x), while the red one has width 1-x and height (1-x)/x. The heights of the blue and red rectangles must sum to 1, so x(1-x) + (1-x)/x = 1 or x²(1-x) + (1-x) = x x² – x³ + 1 – x = x x³ – x² + 2x – 1 = 0 and the solution of this cubic is $\displaystyle{ \frac{1}{3} \left(1 - 5 \left(\frac{2}{11 + 3 \sqrt{69}}\right)^{1/3} + \left(\frac{1}{2} \left(11 + 3 \sqrt{69}\right)\right)^{1/3}\right) }$ so x ≈ 0.56984 The reciprocal of this number is the square of a famous constant: the plastic ratio, ρ. This is like a cheap imitation of the golden ratio, with ρ3 = ρ + 1 Why is the third option so much more complicated than the other two? As we’ll see, it’s because the first two have all the rectangles ‘pointing the same way’, while the third does not: it has a mix of rectangles ‘standing up’ and ‘lying down’. To see the pattern, it helps to do a harder problem. If you divide a square into 4 similar rectangles, what proportions can these rectangles have? A lot of people on Mathstodon worked on this puzzle! Here Dan Piker lists the 11 options we’ve found: Note: there are more than 11 ways to divide a square into 4 similar rectangles, since you can rotate and reflect these pictures, and also rearrange the rectangles in some of them. But we’ve only found 11 possible proportions for the rectangles. Stefano Gogioso has sketched a proof that these are all the options, and Rahul Narain has given a computerized proof that these are all the options obtained from ‘guillotine cuts’: • Wikipedia, Guillotine partition. A ‘guillotine cut’ is a straight line going from one edge of an existing polygon to the opposite edge. And I believe there’s no way to dissect a square into 4 rectangles that doesn’t use guillotine cuts. In the process of listing the options, we discovered something interesting: the rectangles have rational proportions only in options 1, 4, 7, 10 and 11 here: And these are precisely the options where all the rectangles are pointing the same way! (They happen to all be lying down, wider than they are tall. But of course they’d all be standing up if we rotated the pictures.) Puzzle 1. Show that if you subdivide a square into n similar rectangles that are all pointing the same way, the ratio x of their short side to their long side must be a rational number. Puzzle 2. Show that the converse is not true. We also discovered another pattern, too: when x is not rational, it is the solution to a cubic equation with integer coefficients. Does that pattern persist when we subdivide a square into 5 similar rectangles? No, alas! Ian Henderson and Rahul Narain seem to have shown that exactly 51 proportions are possible when we subdivide a square into 5 similar rectangles. Henderson drew them: while Narain listed the polynomial equations obeyed by the 51 possible proportions. Some obey a quartic equation with integer coefficients, not a cubic. To be honest, Narain only considered ways of subdividing a square into 5 similar rectangles using guillotine cuts. But this should be okay, since I believe there’s just one way to subdivide a square into 5 rectangles that cannot be done using guillotine cuts. It’s shown here in a paper by Robert Dawson, in fact: However, when we try to make all 5 rectangles similar, the central one shrinks to a point. Ian Henderson also believes that when you subdivide a square into 6 similar rectangles, 245 proportions of rectangles are possible: And for subdivisions of a square into 7 similar rectangles, he believes 1371 proportions are possible: You can click to enlarge these last two pictures. However, he adds, “I’m not 100% confident in these numbers.” So, someone should check his work. Puzzle 3. Show that you can divide a square into 6 similar rectangles in a way that cannot be done via guillotine cuts. Okay, back to something simpler: what proportions are possible when we divide a square into 4 similar rectangles? Let’s work it out! For each option we get a number 0 < x ≤ 1 describing the proportion of the rectangles that subdivide the square. This number is the length of the short side divided by the length of the long side. The 11 options below are listed from the smallest possible value of x to the largest. 1) Option 1 is to divide the 1 × 1 square into 4 rectangles that are 1/4 as tall as they are wide. So, the number we get from this option is 1/4. Note that if we rotated this option we’d get tall skinny rectangles, but the number would still be 1/4. 2) In option 2, the bottom two rectangles have width 1 and height x. Thus the top two have height 1-2x. Since the green rectangle is x times as tall as it is wide, its width must be (1-2x)/x. The yellow rectangle thus has width 1 – (1-2x)/x = 1 − x⁻¹ + 2 = 3 – x⁻¹ Its height is this divided by x, namely 3x⁻¹ – x⁻². But we know its height is 1-2x, so 1-2𝑥 = 3x⁻¹ – x⁻² or 2x³ – x² + 3x – 1 = 0 so x ≈ 0.34563 3) For option 3 the red rectangle has height x, so the blue one has height 1-x and width x(1-x) = x−x², so the other two have width 1-(x−x²) = 1-x+x² and height x−x²+x³. The total height of red, green and yellow is 1 so x + 2(x−x²+x³) = 1 or 2x³ – 2x² + 3x – 1 = 0 This gives x ≈ 0.39661 4) Option 4 is nice: it has left-right symmetry, and its rectangles can be rearranged to give another option with left-right symmetry. The red and blue rectangles have height x. The green and yellow ones thus have height 1-2x, and thus width (1-2x)/x. But by symmetry we know they must have width 1/2, so (1-2x)/x = 1/2 or 1-2x = x/2 or 1 = 5x/2 or x = 2/5 = 0.4 Not a cubic equation this time—linear! So x is rational. Notice that options 3, 5, 6, and 7 are all topologically the same! They were analyzed by Lisanne, and her work helped me a lot. 5) In option 5 the red rectangle has height x, so the blue has height 1-x and thus width (1-x)/x. The yellow and green thus have width 1 – (1-x)/x = 2 – x⁻¹, hence height (2 – x⁻¹)/x = 2x⁻¹ – x⁻² . The heights of yellow, green and red must sum to 1: 2(2x⁻¹ – x⁻²) + x = 1 so we get a cubic: x³ − x² + 4x – 2= 0 and the solution is x ≈ 0.53318 6) In option 6 the red rectangle again has height x, so the blue again has height 1-x and width (1-x)/x. The yellow and green again have width 1 – (1-x)/x = 2 – x⁻¹, but now they’re different: the yellow has height x(2 – x⁻¹) while the blue has height (2 – x⁻¹)/x. The heights of yellow, green and red sum to 1: x(2 – x⁻¹) + (2 – x⁻¹)/x + x = 1 so 3x³ – 2x² + 2x – 1 = 0 or x ≈ 0.55232 7) In option 7, like options 5 and 6, the red rectangle has height x, so the blue has height 1-x and width (1-x)/x. Thus, the yellow and green again have width 1 – (1-x)/x = 2 – x⁻¹. But this time both have height x(2 – x⁻¹). Yet again, the heights of yellow, green and red sum to 1: x + 2x(2 – x⁻¹) = 1 so 5x = 3 and x = 3/5 = 0.6 The equation was linear, so x is rational! Options 8, 9, and 10 are also all topologically the same. 8) In option 8 the red rectangle has height x so the blue has height 1-x and width (1-x)/x. The green and yellow also have height 1-x, but width x(1-x). The widths of blue, green and yellow sum to 1: (1-x)/x + 2x(1-x) = 1 so 2x³ − 2x² + 2x −1 = 0 and x ≈ 0.64780 9) In option 9 the red rectangle has height x so the blue and green have height 1-x and width (1-x)/x. The yellow also has height 1-x, but width x(1-x). The widths of blue, green and yellow sum to 1: 2(1-x)/x + x(1-x) = 1 so 2x⁻¹ – 2 + x – x² = 1 or x³ – x² + 3x – 2 = 0 so x ≈ 0.71523 10) Option 10 is more symmetrical than 8 or 9 since all three rectangles on top are congruent. The red rectangle has height x so the blue, green and yellow all have height 1-x and thus width (1-x)/x. The widths of blue, green and yellow sum to 1: 3(1-x)/x = 1 so 3(1-x) = x or 3 = 4x or x = 3/4 = 0.75 Another linear equation, with a rational solution! 11) Finally, option 11 is the second one where all four rectangles are congruent. This time they’re squares! Clearly x = 1 So, we see in these examples that x is rational precisely when all rectangles are ‘pointing the same way’: in the way Dan drew them, they’re all at least as wide as they are tall. I’m not quite sure where to go with this research next. But I think the connection to double categories, hinted at in this paper with the picture of the pinwheel configuration, is interesting: • Robert Dawson, A forbidden-suborder characterization of binarily-composable diagrams in double categories, Theory and Applications of Categories 1 (1995), 146–153. Maybe we should seriously study the double category where 2-cells are rectangles that are subdivided into smaller rectangles by guillotine cuts! But I also keep hoping there’s some interesting number-theoretic significance to the proportions that come up when we divide a square into the similar rectangles. Can anyone see interesting patterns in Rahul Narain’s table of the polynomials obeyed by these ratios when we divide a square into 5 similar rectangles? His u is my 1/x: u - 5 u^3 - 4u^2 + u - 3 u^3 - 4u^2 + 2u - 4 2u - 7 u^3 - 4u^2 + 3u - 3 2u^3 - 6u^2 + u - 2 u^3 - 3u^2 + 2u - 5 u^3 - 3u^2 + 2u - 4 2u^3 - 6u^2 + 2u - 2 u^3 - 3u^2 + u - 1 3u - 8 u^3 - 3u^2 + 3u - 5 2u^3 - 5u^2 + u - 2 u^5 - 3u^4 + 3u^3 - 4u^2 + u - 1 -2u^2 + 6u - 3 2u^3 - 5u^2 + 2u - 3 3u - 7 u^4 - 3u^3 + 3u^2 - 4u + 2 2u^3 - 5u^2 + 3u - 3 u^3 - 3u^2 + 4u - 4 -u^2 + 4u - 4 4u - 8 3u^3 - 6u^2 + u - 1 2u^3 - 4u^2 + 2u - 3 u^3 - 2u^2 + 3u - 5 3u^3 - 6u^2 + 2u - 2 u^5 - 2u^4 + 3u^3 - 5u^2 + u - 1 2u^3 - 4u^2 + 3u - 4 4u - 7 u^3 - 2u^2 + 4u - 6 -u^3 + 3u^2 - 4u + 3 2u^3 - 5u^2 + 4u - 2 2u^3 - 4u^2 + 3u - 3 3u^3 - 5u^2 + 2u - 3 5u - 8 u^5 - 2u^4 + 3u^3 - 4u^2 + u - 1 -3u^2 + 6u - 2 2u^4 - 4u^3 + 3u^2 - 3u + 1 3u^3 - 5u^2 + 2u - 2 4u^3 - 6u^2 + u - 1 -2u^3 + 4u^2 - 3u + 2 2u^3 - 3u^2 + 3u - 4 u^5 - 2u^4 + 3u^3 - 4u^2 + 2u - 1 5u - 7 u^3 - 2u^2 + 3u - 3 2u^3 - 3u^2 + 2u - 2 -u^3 + 2u^2 - 4u + 4 3u^3 - 5u^2 + 3u - 2 3u^3 - 4u^2 + u - 1 4u^3 - 6u^2 + 2u - 1 4u - 5 2u^3 - 3u^2 + 4u - 4 -5u + 6 6u - 7 ### 15 Responses to Dividing a Square into Similar Rectangles 1. John Baez says: I think the number theory side of things interacts in some nice way with guillotine partitions. Say we want to dissect a 1 × 1 square into similar rectangles using guillotine cuts. We can try to describe this process recursively: each guillotine cut slices an existing rectangle into two smaller one either horizontally or vertically. I believe there should be a way to keep track of this using some sort of a tree, but I haven’t figured it out yet. The total number of ways to do this slicing (just keeping track of the topology, not the position of the slices) is the nth Schröder number. For example, the 4th Schröder number is 22, and here are the 22 topologies of guillotine partitions of a square into 4 rectangles: The black dots are just an aid to thinking about these. When the slicing is completed we need to say whether each of the final pieces is a ‘vertical’ or ‘horizontal’ copy of our similar rectangles (except perhaps in the case where they’re all squares). From this data we can get a polynomial equation that determines the number I was calling x above, which determines the proportions of our similar rectangles. So, I’d like to systematize this and figure out how the polynomial obeyed by the number x is related to the combinatorial data we use to describe a guillotine partition and the labeling of each of the smallest rectangles by ‘vertical’ or ‘horizontal’. 2. Hendrik Boom says: “Okay, back to something simpler: what proportions are possible when we divide a square into 4 congruent rectangles? Let’s work it out!” Judging from your answer, I think you mean “similar” rather than “congruent”. Same for many other occurrences of “congruent” in the text. • John Baez says: Yes, in all but two instances when I wrote “congruent” I meant “similar”. I’ve now tried to fix that. 3. Lisanne says: This paper is a nice starting point for the ‘topological’ classification/counting of the rectangulations. Bryan Dawei He, A simple optimal binary representation of mosaic floorplans and Baxter permutations • John Baez says: Thanks very much! That paper is available here. (Elsevier may be offering it to me for free because they’re spying on me and know I’m at U. C. Riverside, even though I deleted all their cookies a while back. I read a great blog article about this topic recently but now I can’t find it.) 4. Lisanne says: Puzzle 3 is solved by multiple pictures of Ian Henderson already! look at Row 10, Col 3 for example. 5. Lisanne says: A remark about the polynomials: The degree is always at most the number of rectangles and there are always configurations obtaining this bound. (Except for N = 2) I can type out a proof for this later if someone wants to see it. In fact the polynomial is the determinant of a matrix with only +-1,+-X,0 in it. • John Baez says: I’d love to see the proof! And I’d love to see that matrix. I’m working on this stuff myself, slowly. 6. lisannetaams says: Start with a generic rectangulation into N rectangles. So no 4 rectangles are allowed to meet at a single corner. (This disallows for example 4 smaller squares inside the big square, which all meet at the center of the big square.) Let $= a_i$ be the long side of rectangle $i$ and $x \cdot a_i$ the short side, so the ratio is $0. The initial square will have side length s. Now for each vertical line segment (always take the full segment) in the rectangulation we get a relation $\sum_{r_i \textrm{ touching on the left}} (? x) a_i - \sum_{r_j \textrm{ touching on the right}} (? x) a_j = 0$ For the outer line segments (the sides of the square) do something similar with $s.$ And similarly for the horizontal segments. Using some induction argument you can see that this always gives you $N + 3$ relations. (This is easy for guillotine cuts a little bit harder in general). Here you need the generic property. However the relation on the right side is implied by all the other vertical relations and the same is true for the bottom side and the horizontal relations. Removing these we end up with $N + 1$ relations. Some intense staring will show that a solution to these equations is not just necessary but also sufficient to get a similar-rectangulation. Now we will have a closer look at the relations. We have $N + 1$ variables ($a_i$ and $s$). So we can fit them into an $N+1$ by $N+1$ matrix $M$ and we need to solve the equation $Mv = 0$. Now this has a (non-zero) solution if and only if $\det(M) = 0$ and this is just a polynomial in $x$. Since the column corresponding to the variable $s$ has only $0, 1$ in it this polynomial can have degree at most $N$. This proof also suggests an efficient way to compute $x$ and $a_i$. Write down the matrix, solve the determinant polynomial. Substitute this back in and solve a linear equation. You should be carefull that even if there is a solution some of the $a_i$ might be negative. Geometrically this corresponds to two rectangles overlapping, but the overlap also having the same ratios! Bonus: We can also estimate the coefficients of $f(x) = \det(M)$. There is a naive bound coming from the Leibniz formula saying that the sum of all the coefficients is at most $(N + 1)!$ Can we do better? • John Baez says: Thanks. I’m trying to understand this. Does $(?x) a_i$ mean “maybe we include a factor of $x,$ maybe not?” 7. Justin says: How about the analogous problem of dividing a (n-)cube into similar rectangular (n-)cuboids? (One can’t “cube the cube” per https://en.wikipedia.org/wiki/Squaring_the_square#Cubing_the_cube — but finding similar rectangular (n-)cuboids in an (n-)cube is definitely a different problem than that — and there exist at least some n>2 solutions that aren’t totally trivial, e.g. a 3-cube divided into 5 rectangular 3-cuboids that are each 2/3 as long in one direction as in the other two directions [i.e. one “big one” and 4 other same-sized “small” rectangular 3-cuboids that have all dimensions reduced by half relative to the big one — a solution that is analogous to the second one of the 3 divisions of the square into 3 similar rectangles at the top of your post.) Do there exist some such solutions in more than 2 dimensions with irrational but algebraic ratios, perhaps somewhat analogous to the plastic ratio-related third of the 3 divisions of the square into 3 similar rectangles? (I’m a non-mathematical physicist, so I need to ask, rather than being able to answer it myself! :) My apologies if this very-closely-related question has already been discussed on Mathstodon! • John Baez says: People have raised this issue on Mathstodon but not really worked on it. I think there’s a huge amount of interesting structure in the case of dividing a square into similar rectangles: see my blog article and also the comments on the original problem starting here—especially the comments by David Speyer and Jeffery Opoku. Unfortunately I haven’t had time to work on these things in the last month. 8. There is now a continuation of my post “Dividing a square into similar rectangles”. Some people on Mathstodon put a lot of work into this and made some nice progress. But there’s been a surprising new twist! I’m not talking about how the answer to this puzzle is now listed on the listed on the Online Encyclopedia of Integer Sequences as sequence A359146. I’m not even talking about the fact that the New York Times ran an article about this puzzle: • Siobhan Roberts, The quest to find rectangles in a square, New York Times, February 7, 2023. Open-access version here. This site uses Akismet to reduce spam. Learn how your comment data is processed.
# Thread: rationalizing the denominators 1. ## rationalizing the denominators Why do I have to rationalize the denominators? In AMC( American mathematics contest ) contests I see that they never leave an answers as 2/root3 it will always be 2root3/3, but in SAT math tests they have answers such as 2/root3. What is the rule and meaning for rationalizing the denominators? Thanks. Vicky. 2. ## Times both sides by the denominator $\frac {2}{\sqrt3}$ If I times that fraction by $\frac{\sqrt3}{\sqrt3}$, I am actually just multiplying by one. But look at what happens: $\frac {2}{\sqrt3}\times\frac{\sqrt3}{\sqrt3}$ $=\frac{2\times\sqrt3}{\sqrt3\times\sqrt3}$ $=\frac{2\sqrt3}{3}$ So the rule is: For any fraction $\frac{a}{\sqrt{b}}$ multiply by $\frac{\sqrt{b}}{\sqrt{b}}$ to give $\frac{a\sqrt{b}}{b}$ For the purpose of your question, this is how you rationalize the denominator., but for a harder example, you may read on. However, it gets harder if you have something such as: $\frac{3}{\sqrt3+4}$ In this scenario, you have to use the 'difference of two squares' to rationalize, by making the denominator a difference of two squares, which is where $(a+b)(a-b)=a^2-b^2$. So apply that to the question, where $a=\sqrt{3}$ and b=4: $\frac{3}{\sqrt3+4}\times\frac{\sqrt3-4}{\sqrt3-4}$ $=\frac{3(\sqrt{3}-4)}{(\sqrt3+4)(\sqrt3-4)}$ Note the 'difference of two squares' in the denominator? Using the difference of two squares rule above, we can simplify this to: $=\frac{3(\sqrt{3}-4)}{{\sqrt3}^2-(4^2)}$ $=\frac{3(\sqrt{3}-4)}{3-16}$ $=\frac{3\sqrt3-12}{-13}$ 3. Originally Posted by Quacky $\frac {2}{\sqrt3}$ If I times that fraction by $\frac{\sqrt3}{\sqrt3}$, I am actually just multiplying by one. But look what happens to the fraction: $\frac {2}{\sqrt3}\times\frac{\sqrt3}{\sqrt3}$ $=\frac{2\times\sqrt3}{\sqrt3\times\sqrt3}$ $=\frac{2\sqrt3}{3}$ So the rule is: For any fraction $\frac{a}{\sqrt{b}}$ multiply by $\frac{\sqrt{b}}{\sqrt{b}}$ to give $\frac{a\sqrt{b}}{b}$ for $b \neq 0$. 4. Thanks for your help. But I still don't understand the meaning of rationalizing the denominators. I thought the answer was incorrect if it is not rationalized, because in math contest problems I don't think I have ever seen an answer with a root in the denominator. When I recently did a practice SAT math test, I was surprised to see an answer left as 2/root3. This is when I realized I was rationalizing my answers without truly understanding why I was doing it. Why do I have to rationalized the denominator? And why do certain contests consider the answer incorrect if not rationalized while SAT math allows answers with roots in the denominator? Vicky. 5. How would you add fractions such as $\frac {5}{\sqrt2}+\frac{5}{\sqrt3}$? You'd have to find and work with a common denominator of $\sqrt6$. It would be much easier to solve: $\frac {5\sqrt2}{2}+\frac{5\sqrt3}{3}$ because you can just look at it and see that the lowest common denominator will be 6. When dealing with fractions, you will often want simpler terms on the denominator, so rationalization helps a lot. I'm not sure why irrational denominators aren't accepted, though. I suppose they aren't seen as being fully simplified. 6. Hello, Vicky! I think I have an answer to your question. Why do I have to rationalize the denominator? In the Old Days, say 100 B.C. (Before Calculators), we had only our brains and a pencil. If we want to evaluate, for example, $\frac{1}{\sqrt{2}}$, we would: . . [1] Find an approximation for $\sqrt{2}$ on a square-root table (1.4142) . . [2] Perform the long division: . $1 \div 1.414212$ The divison looks like this: . $\begin{array}{cccc}& - & - - - - - - \\ 1.4142& ) & 1.0000000 \end{array}$ . . $\begin{array}{ccccccccccccc} & & & & & & 0. & 7 & 0 & 7 & 1 & \hdots\\ & & -- & -- & -- & -- & -- & -- & -- & --& -- & --\\ 1\;4\;1\;4\;2 & ) & 1 & 0 & 0 & 0 & 0. & 0 & 0 & 0 & 0 & \hdots\\ &&& 9 & 8 & 9 & 9 & 4 \\ && -- & -- & -- & -- & -- \\ &&&& 1 & 0 & 0 & 6 & 0 \\ &&&&&&&& 0 \\ &&&& -- & -- & -- & -- & -- \\ &&&& 1 & 0 & 0 & 6 & 0 & 0 \\ &&&&& 9 & 8 & 9 & 9 & 4 \\ &&&&& -- & -- & -- & -- & -- \\ \end{array}$ . - - . . . . . . . . . . . . . . . $\begin{array}{cccccccccccc} 1 & 6 & 0 & 6 & 0 \\ 1 & 4 & 1 & 4 & 2 \\ -- & -- & -- & -- & -- \\ & 1 & 9 & 1 & 8 & 0 & \hdots \end{array}$ Pretty tedious, dividing by a 5-digit number. . . (Worse, if we had more decimal places.) Rationalizing, we would have: . $\frac{\sqrt{2}}{2}$ And I would much prefer to do: . $1.414213562... \div 2$ . . Wouldn't you? And Quacky's view on combining fractions is excellent! 7. Wow, very impressive answer, did you deduce that from logic, or did you learn it somewhere? 8. Originally Posted by Soroban Hello, Vicky! I think I have an answer to your question. In the Old Days, say 100 B.C. (Before Calculators), we had only our brains and a pencil. If we want to evaluate, for example, $\frac{1}{\sqrt{2}}$, we would: . . [1] Find an approximation for $\sqrt{2}$ on a square-root table (1.4142) . . [2] Perform the long division: . $1 \div 1.414212$ The divison looks like this: . $\begin{array}{cccc}& - & - - - - - - \\ 1.4142& ) & 1.0000000 \end{array}$ . . $\begin{array}{ccccccccccccc}$ $ & & & & & & 0. & 7 & 0 & 7 & 1 & \hdots\\ & & -- & -- & -- & -- & -- & -- & -- & --& -- & --\\ 1\;4\;1\;4\;2 & ) & 1 & 0 & 0 & 0 & 0. & 0 & 0 & 0 & 0 & \hdots\\ &&& 9 & 8 & 9 & 9 & 4 \\ && -- & -- & -- & -- & -- \\ &&&& 1 & 0 & 0 & 6 & 0 \\ &&&&&&&& 0 \\ &&&& -- & -- & -- & -- & -- \\ &&&& 1 & 0 & 0 & 6 & 0 & 0 \\ &&&&& 9 & 8 & 9 & 9 & 4 \\ &&&&& -- & -- & -- & -- & -- \\ \end{array}" alt=" & & & & & & 0. & 7 & 0 & 7 & 1 & \hdots\\ & & -- & -- & -- & -- & -- & -- & -- & --& -- & --\\ 1\;4\;1\;4\;2 & ) & 1 & 0 & 0 & 0 & 0. & 0 & 0 & 0 & 0 & \hdots\\ &&& 9 & 8 & 9 & 9 & 4 \\ && -- & -- & -- & -- & -- \\ &&&& 1 & 0 & 0 & 6 & 0 \\ &&&&&&&& 0 \\ &&&& -- & -- & -- & -- & -- \\ &&&& 1 & 0 & 0 & 6 & 0 & 0 \\ &&&&& 9 & 8 & 9 & 9 & 4 \\ &&&&& -- & -- & -- & -- & -- \\ \end{array}" /> . - - . . . . . . . . . . . . . . . $\begin{array}{cccccccccccc}$ $ 1 & 6 & 0 & 6 & 0 \\ 1 & 4 & 1 & 4 & 2 \\ -- & -- & -- & -- & -- \\ & 1 & 9 & 1 & 8 & 0 & \hdots \end{array}" alt=" 1 & 6 & 0 & 6 & 0 \\ 1 & 4 & 1 & 4 & 2 \\ -- & -- & -- & -- & -- \\ & 1 & 9 & 1 & 8 & 0 & \hdots \end{array}" /> Pretty tedious, dividing by a 5-digit number. . . (Worse, if we had more decimal places.) Rationalizing, we would have: . $\frac{\sqrt{2}}{2}$ And I would much prefer to do: . $1.414213562... \div 2$ . . Wouldn't you? And Quacky's view on combining fractions is excellent! Thank You!!!!
# The app that does your homework The app that does your homework is a mathematical tool that helps to solve math equations. We can solve math problems for you. ## The Best The app that does your homework In this blog post, we discuss how The app that does your homework can help students learn Algebra. There are a few steps to solving quadratic inequalities. First, you need to identify the inequality sign. If it is "<" then you need to flip the inequality sign and solve for x. If it is ">" then you can solve for x as is. Next, you need to isolate the x term on one side of the equation and the constants on the other. To do this, you either need to add or subtract the same value from both sides. Once you have isolated Most people think that solving math word problems requires a lot of memorization andpractice. However, there are some general strategies that can be used to solvemath word problems more easily. The first step is to read the problem carefullyand identify the key words and concepts. Once you understand what the problem isasking, you can begin to generate possible solutions. It can be helpful to draw a diagramexplaining the problem, which will make it easier to visualize the relationshipsbetween the different elements. In addition, it is often useful to write out each stepof the solution process, so that you can see where you made any mistakes. Witha little patience and practice, solving math word problems can be much easierthen most people think. Solving natural log equations requires algebraic skills as well as a strong understanding of exponential growth and decay. The key is to remember that the natural log function is the inverse of the exponential function. This means that if you have an equation that can be written in exponential form, you can solve it by taking the natural log of both sides. For example, suppose you want to solve for x in the equation 3^x = 9. Taking the natural log of both sides gives us: ln(3^x) = ln(9). Since ln(a^b) = bln(a), this reduces to xln(3) = ln(9). Solving for x, we get x = ln(9)/ln(3), or about 1.62. Natural log equations can be tricky, but with a little practice, you'll be able to solve them like a pro! logarithm is the natural logarithm to the base e. It is used to solve equations with a base of e. The logarithm solve for x is: When solving logarithms, it is important to remember that the answer in this case is the base e raised to an integer power (i.e., 1 + 2 = 3). Logarithms are most useful when solving exponential equations, and they are especially useful when you are solving problems with large exponents. For example, if you have an equation that looks like this: y = 4x² + 9x - 14 Then using a logarithm solve for x, you would solve y = log10(4) + log10(9) + log10(14) = 5log10(4) + log10(3.4) = 5log2(4) = 2.06 Example 1: If you want to find out how many hours it takes for water to boil on a stove top, then solve for x: y = 4x² + 9x - 14 Here's what the math looks like: fp = 4 * x^2 + 9 * x - 14 yp = 4 * x^2 + 9 * x - 14 Here's what it means: First, find out how much water there is in the pot. The 3x3 matrix is a way of describing how you can translate the results of a table into the columns and rows in a matrix. The example below shows how you could translate a result in a table into three columns and three rows. A simple way to do this is to multiply each column by the corresponding row value. You can then rearrange these values to create the matrix form of your table. For example, if there were two rows and three columns and we wanted to translate the first row into column 1, we would write: 11 = 1 22 = 4 3*3 = 9 The result would be 9. ## Instant assistance with all types of math exceptional app, if you need steps broken down for you or don’t know how the textbook did a step in one of the example questions, there’s a good chance this app can read it and break it down for you. I'm honestly amazed at how it reads my not-so-great handwriting so well. Farah Smith Hands down, the best app I've ever used. There aren't any ads or random stuff popping up. It is extremely precise. Even works with handwriting. I use this app more than almost any other app. It is fast, the UI is great, there are options to enter math manually also. Good job devs, this is super impressive. And thanks for it being Free. Gracie James
How do you combine like terms in (2a + 3) ^ { 2} + ( a - 5) ^ { 2}? Jun 9, 2018 $5 {a}^{2} + 2 a + 34$ Explanation: Using the formulas ${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$ ${\left(a - b\right)}^{2} = {a}^{2} - 2 a b + {b}^{2}$ we get $4 {a}^{2} + 9 + 12 a + {a}^{2} - 10 a + 25$ and this is $5 {a}^{2} + 2 a + 39$ Jun 9, 2018 $5 {a}^{2} + 2 a + 34$ Explanation: First FOIL the squared terms: $= {\left(2 a + 3\right)}^{2} + {\left(a - 5\right)}^{2}$ $= \left(2 a + 3\right) \left(2 a + 3\right) + \left(a - 5\right) \left(a - 5\right)$ $= 4 {a}^{2} + 6 a + 6 a + 9 + {a}^{2} - 5 a - 5 a + 25$ now "combine like terms": $= 5 {a}^{2} + 2 a + 34$
In this video, we will multiply radical expressions. $\sqrt{3}\times\sqrt{3}=\sqrt{9}=3$ $\sqrt{5}\times\sqrt{5}=\sqrt{25}=3$ In other words, $\sqrt{7}\times\sqrt{7}=7$ $\sqrt{12}\times\sqrt{12}=12$ Let’s try other examples: $\sqrt{5}\times\sqrt{7}$ would equal to $\sqrt{35}$ $\sqrt{18}\times\sqrt{2}$ would equal to $\sqrt{36}$, which is a perfect square so it would be $6$ Let’s try the same problem by using another method: $\sqrt{18}\times\sqrt{2}$ would equal to $\sqrt{36}$ can also be written as $\sqrt{9}\times\sqrt{2}\times\sqrt{2}$ For $\sqrt{15}\times\sqrt{45}$, it would equal to $\sqrt{675}$ 25 goes into 675 twenty-seven times Since 5 is the square root of 25, $\sqrt{675}$ can be written as $5\sqrt{27}$ This can be further simplified into $15\sqrt{3}$ ## Video-Lesson Transcript Let’s go over how to multiply radical expressions. $\sqrt{3} \times \sqrt{3} = \sqrt{9} = 3$ $\sqrt{5} \times \sqrt{5} = \sqrt{25} = 5$ So if you look, closely $\sqrt{7} \times \sqrt{7} = 7$ $\sqrt{12} \times \sqrt{12} = 12$ Keep this in mind as we move forward in the lesson. Let’s take a look at this $\sqrt{5} \times \sqrt{7} = \sqrt{35}$ this cannot be simplified. $\sqrt{18} \times \sqrt{2} = \sqrt{36} = 6$ But what if we reduce this first? $\sqrt{9} \times \sqrt{2} \times \sqrt{2}$ $3 \times 2$ $6$ So you can multiply it across and get an answer. Or reduce it first and multiply it out. Let’s have this one $\sqrt{15} \sqrt{45}$ I can multiply this out $\sqrt{15} \sqrt{45} = \sqrt{675}$ Then we should break $675$ down. But it’s a pretty large number. $\sqrt{675}$ $\sqrt{25} \sqrt{27}$ $5 \sqrt{27}$ $5 \sqrt{9} \sqrt{3}$ $5 \times 3 \sqrt{3}$ $15 \sqrt{3}$ Let’s see what happens if we reduce it first. $\sqrt{15} \sqrt{45}$ $\sqrt{15} \sqrt{9} \sqrt{5}$ $\sqrt{15} \times (3) \sqrt{5}$ $3 \sqrt{75}$ $3 \sqrt{25} \sqrt{3}$ $3 \times 5 \sqrt{3}$ $15 \sqrt{3}$ We came up with the same answer. The number of steps in the two methods is pretty much the same. But I dealt with smaller numbers using the second method. Let’s look back at a point here. $\sqrt{15} \sqrt{45}$ $\sqrt{15} \sqrt{9} \sqrt{5}$ At this point, we know that $\sqrt{15}$ can’t be broken down. But there’s a $\sqrt{5}$ there. And we know that $5$ goes into $15$. So, we might as well break it down. So we’ll have $\sqrt{3} \sqrt{5} \sqrt{9} \sqrt{5}$ Let’s just reorganize this $\sqrt{9} \sqrt{5} \sqrt{5} \sqrt{3}$ And we’ll solve $3 \times 5 \sqrt{3}$
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # Inequalities with Addition and Subtraction ## Simplify one-step inequalities to identify solution set Estimated7 minsto complete % Progress Practice Inequalities with Addition and Subtraction MEMORY METER This indicates how strong in your memory this concept is Progress Estimated7 minsto complete % Solve Inequalities Using Addition or Subtraction License: CC BY-NC 3.0 The graduating class of 2015 has been busy raising money to decorate the Community Center for the upcoming Prom Dance. They have a budget of $2,475. 95 and have already committed$1,728.50 to a professional decorator who will provide and set up the props to transform the back portion of the Community Center into a nature park. The graduates also want to have the tables and chairs covered with yellow, metallic cloth if they have enough money to cover the cost. They are still waiting to hear from “All Things Fancy” regarding the price. How can the graduates figure out how much money they have for covering the tables and chairs? In this concept, you will learn to solve inequalities using addition or subtraction. ### Adding and Subtracting Inequalities Sometimes, you will have an inequality that is not as straightforward as \begin{align}x>4\end{align}. With this example, you know that the variable can be any number that is greater than four. The set of numbers that will make this inequality a true statement can be displayed as a list or graphed on a number line. However, the set of numbers that makes the inequality true is not always as obvious as it was for the inequality \begin{align}x>4\end{align}. You may see an inequality like this one. \begin{align}x+3>7\end{align} For this inequality you need to find the set of numbers which will make the inequality a true statement. You are looking for a set of numbers such that the sum of a number and 3 is greater than 7. You could spend a lot of time doing trial and error to determine the set of numbers or you could solve the inequality. Solving an inequality is similar to solving an equation. The following rules or number properties can be used to solve an inequality. The Addition Property of Inequality states that if the same number is added to both sides of an inequality then the sense (inequality symbol) of the inequality remains unchanged. • If \begin{align}a>b\end{align}, then \begin{align}a+c > b+c\end{align}. • If \begin{align}a < b\end{align}, then \begin{align}a+c < b+c\end{align}. • If \begin{align}a \ge b\end{align}, then \begin{align}a+c \ge b+c\end{align}. • If \begin{align}a \le b\end{align}, then \begin{align}a+c \le b+c\end{align}. The Subtraction Property of Inequality states that if the same number is subtracted from both sides on the inequality then the sense (equality symbol) of the inequality remains unchanged. • If \begin{align}a>b\end{align}, then \begin{align}a-c > b-c\end{align}. • If \begin{align}a < b\end{align}, then \begin{align}a-c < b-c\end{align}. • If \begin{align}a \ge b\end{align}, then \begin{align}a-c \ge b-c\end{align}. • If \begin{align}a \le b\end{align}, then \begin{align}a-c \le b-c\end{align}. The inequalities which you will solve can be solved by applying these properties. The process involved in solving an equation is the same for solving an inequality – applying inverse operations to isolate and solve for the variable. For an equation the solution is a single number but the solution for an inequality will be a set of numbers. The solution must make the inequality a true statement. Let’s look at solving an inequality. Solve this inequality for all real numbers and graph the solution on a number line. \begin{align}n-3<5\end{align} First, isolate the variable by adding (inverse of subtracting) 3 to both sides of the inequality. \begin{align}\begin{array}{rcl} n-3 & < & 5\\ n-3+3 & < & 5+3 \end{array}\end{align} You have just applied the addition property of inequality. The inequality symbol remained the same. Next, simplify both sides of the inequality. \begin{align}\begin{array}{rcl} n-3+3 & < & 5+3\\ n& < & 8 \end{array}\end{align} The solution set is \begin{align}\{ n \| n < 8, n \in R \}\end{align}. The solution is the set of numbers “less than” 8 and this can be graphed on a number line. First, draw a number line from 0 to 10. Next, place an open circle on the number 8 since this number is not included in the solution set. Then, draw a direction line from 8 to the left on the number line. License: CC BY-NC 3.0 ### Examples #### Example 1 Earlier, you were given a problem about the graduating class of 2015. They need to know how much money they have left in their decorating budget. They need to write an inequality to model their decorating funds. First, write a verbal model to represent the information. Verbal Model: cost of covering the tables and chairs + cost of the decorator cannot exceed their budget. Next, write an equality to represent the verbal model. \begin{align}\underbrace{\text{cost of covering tables and chairs}}_{c} + \underbrace{\text{cost of decorator}}_{\ 1,728.50} \le \underbrace{\text{total budget}}_{\2,475.95}\end{align} The inequality is \begin{align}c+1,728.50 \le 2,475.95\end{align} Next, solve the inequality for the variable “\begin{align}c\end{align}”. First, isolate the variable by subtracting 1,728.50 from both sides of the inequality. \begin{align}\begin{array}{rcl} c+1,728.50 & \le & 2,475.95\\ c+1,728.50-1,728.50 & \le & 2,475.95 - 1,728.50 \end{array}\end{align} Then, simplify both sides of the inequality. \begin{align}\begin{array}{rcl} c+1,728.50-1,728.50 & \le & 2,475.95 - 1,728.50\\ c & \le & 747.45\\ \end{array}\end{align} The cost for covering the tables and chairs must be less than or equal to \$747.45. #### Example 2 Solve the following inequality for all real numbers. Then graph the solution on a number line. \begin{align}-2 \le x+4\end{align} First, isolate the variable by subtracting 4 from both sides of the inequality. \begin{align}\begin{array}{rcl} -2 & \le & x+4\\ -2-4 & \le & x+4-4 \end{array}\end{align} Next, simplify both sides of the inequality. \begin{align}\begin{array}{rcl} -2-4 & \le & x+4-4\\ -6 & \le & x\\ \end{array}\end{align} The solution is \begin{align}\{ x \| x \ge -6, x \in R \}\end{align}. If \begin{align}-6 \le x\end{align} then \begin{align}x \ge -6\end{align}. Regardless of how you write it, the solution remains the same. Now, graph the solution on a number line. First, draw an appropriate number line. Next, place a closed circle on -6 since this value is included in the solution set. Then, draw a direction line to the right to indicate “greater than” and to show the real number set. Remember, joining the points means that all values between the whole numbers are included in the solution set. License: CC BY-NC 3.0 #### Example 3 Solve the following inequality for the set of Integers. \begin{align}x+5 < -12\end{align} First, isolate the variable by subtracting 5 from both sides of the inequality. \begin{align}\begin{array}{rcl} x+5 & < & -12\\ x+5-5 & < & -12-5 \end{array}\end{align} Next, simplify both sides of the inequality. \begin{align}\begin{array}{rcl} x+5-5 & < & -12-5\\ x & < & -17 \end{array}\end{align} Then, write the solution for the set of Integers. \begin{align}\{ x \| x < -17, x \in I \}\end{align} ### Review Solve the following inequalities: 1. \begin{align}x+4>10\end{align} 2. \begin{align}y-11<20\end{align} 3. \begin{align}a+2<1\end{align} 4. \begin{align}b+3 \ge 5\end{align} 5. \begin{align}y-2 \le -4\end{align} 6. \begin{align}x+1 \ge 5\end{align} 7. \begin{align}x-3 < 11\end{align} 8. \begin{align}x-4 < -3\end{align} 9. \begin{align}x-4>-3\end{align} 10. \begin{align}y+7>22\end{align} 11. \begin{align}a-6 \ge -1\end{align} 12. \begin{align}b+14 > 20\end{align} 13. \begin{align}x-24 > -11\end{align} 14. \begin{align}a+3 \le -9\end{align} 15. \begin{align}x+13 > -33\end{align} ### Review (Answers) To see the Review answers, open this PDF file and look for section 3.13. ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Please to create your own Highlights / Notes Show More ### Vocabulary Language: English TermDefinition Addition Property of Inequality You can add a quantity to both sides of an inequality and it does not change the sense of the inequality. If $x > 3$, then $x+2 > 3+2$. Equation An equation is a mathematical sentence that describes two equal quantities. Equations contain equals signs. inequality An inequality is a mathematical statement that relates expressions that are not necessarily equal by using an inequality symbol. The inequality symbols are $<$, $>$, $\le$, $\ge$ and $\ne$. Subtraction Property of Inequality The subtraction property of inequality states that the subtraction of equal amounts from both sides of an inequality will not change the sense of the inequality. ### Image Attributions 1. [1]^ License: CC BY-NC 3.0 2. [2]^ License: CC BY-NC 3.0 3. [3]^ License: CC BY-NC 3.0 ### Explore More Sign in to explore more, including practice questions and solutions for Inequalities with Addition and Subtraction. Please wait... Please wait...
# How do you calculate weighted sum? ## How do you calculate weighted sum? Weighted average is the average of a set of numbers, each with different associated “weights” or values. To find a weighted average, multiply each number by its weight, then add the results…. 1. Determine the weight of each number. 2. Find the sum of all weights. 3. Calculate the sum of each number multiplied by its weight. ## What is weighted mean in math? A weighted mean is a kind of average. Instead of each data point contributing equally to the final mean, some data points contribute more “weight” than others. If all the weights are equal, then the weighted mean equals the arithmetic mean (the regular “average” you’re used to). What do you mean by weighted mean? The weighted mean is a type of mean that is calculated by multiplying the weight (or probability) associated with a particular event or outcome with its associated quantitative outcome and then summing all the products together. How does weighted sum work? Weighted Sum works by multiplying the designated field values for each input raster by the specified weight. It then sums (adds) all input rasters together to create an output raster. The Weighted Sum tool does not rescale the reclassified values back to an evaluation scale. ### How do I figure out a weighted grade? Multiply the grade on the assignment by the grade weight. In the example, 85 times 20 percent equals 17 and 100 times 80 percent equals 80. Add together all your weighted grades to find your overall grade. In the example, 17 points plus 80 points equals a weighted grade of 97. ### Why weighted mean is used? The weighted average takes into account the relative importance or frequency of some factors in a data set. A weighted average is sometimes more accurate than a simple average. Stock investors use a weighted average to track the cost basis of shares bought at varying times. Why do we need to use weighted mean? In Mathematics, the weighted mean is used to calculate the average of the value of the data. We need to calculate the weighted mean when data is given in a different way compared to the arithmetic mean or sample mean. Different types of means are used to calculate the average of the data values. What is the difference between weighted overlay and weighted sum? Comparing Weighted Sum to Weighted Overlay Weighted Sum works by multiplying the designated field values for each input raster by the specified weight. The Weighted Sum tool allows floating-point and integer values, whereas the Weighted Overlay tool only accepts integer rasters as inputs. ## What is 40% out of 100%? Percentage Calculator: What is 40. percent of 100? = 40. ## How do you calculate weighted average? Weighted average calculation. The weighted average ( x) is equal to the sum of the product of the weight (w i) times the data number (x i) divided by the sum of the weights: How do you calculate weighted score? How to Calculate. A weighted score is calculated by knowing all of the variables and their value. To arrive at a weighted score, you need to assign a value to each of the variables that you wish to average. You then multiply the value by the corresponding numerals. How do you find the weighted average of values? In mathematics and statistics, you calculate weighted average by multiplying each value in the set by its weight, then you add up the products and divide the products’ sum by the sum of all weights. Categories: Blog
# IS 100 a perfect square? The first 12 perfect squares are: {1, 4, 9, 25, 36, 49, 64, 81, 100, 121, 144…} Perfect squares are used often in math. Try to memorize these familiar numbers so that you can recognize them as they are used in many math problems. The first five squares of the negative integers are shown below. ## What is the square root of 1 to 100? Between 1 to 100, the square roots of 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100 are whole numbers (rational), while the square roots of 2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 50, 51, 52, 53, 54, 55, … ## Why do we simplify fractions? Simplifying fractions is the process of reducing the numerator and denominator to their smallest whole numbers so the fraction is in its simplest form. This helps other mathematicians or scientists to easily interpret data and can also help avoid confusion when numbers and equations become large and complex. See also Which is more 6 feet or 1 yard? ## How do you find 100 percent of a number? The strategy here is to see how many times the “percent number” (in this case, 25) goes into 100, and then count by that number until we reach 100-the whole thing. Here, we’re told that 25% of a number is 5. So, to find 100% of the number, we count by 25s up to 100: 25, 50, 75, 100. ## How do you write 100 percent as a decimal? With more complex numbers, you’ll still just move the decimal over two places. Here are a few more examples: 100% = 1. 150% = 1.5. ## How do you do 100 percent on a calculator? Another way to find this is to simply move the decimal point two spots to the left. How do I convert rates from decimal form to percent form? Type the decimal number, press the times button (× or *) on your calculator, and then input 100. Press equal button, and you will have your answer. ## What is a 100 square in maths? A hundred square is a square filled with numbers, from 1-100. The numbers are sequential. A hundred square is a type of number square. In a hundred square, each column has the same ones digit. Other types of number squares may contain more / less numbers, more / less squares and may not start at the number 1. ## How do you find the perfect square number between 1 to 100? There are ten perfect squares from 1 and 100. They can be listed as 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100. ## What is the prime numbers of 100? Therefore, the prime numbers 1 to 100 can be listed as, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. See also What is 0.96 as a percent? ## How are fractions reduced? To reduce a fraction, divide both the numerator and denominator by the GCF. (This is also known as “writing a fraction in lowest terms”.) This is sometimes shown as “canceling” the common factors. Note that the result is an equivalent fraction in simplest form. ## What is the difference between reducing and simplifying fractions? The terms ‘simplifying’ or ‘reducing’ means the same thing when referring to fractions, so these terms can be used interchangeably. When you reduce a fraction, it becomes simpler because the number of parts in the whole is made small as possible, without changing the value of the fraction. We found that 6/4 = 3/2. ## What is the number 100? 100 (also written as one hundred) is the natural number after 99 and before 101. It is the first three-digit number above zero. The number 100 is an abundant number. ## How do you write 100 in binary? 100 in binary is 1100100. To find decimal to binary equivalent, divide 100 successively by 2 until the quotient becomes 0. ## How do you work out VAT on a calculator? How do I calculate VAT on my calculator? To calculate VAT having the gross amount you should divide the gross amount by 1 + VAT percentage. (i.e if it is 20%, then you should divide by 1.20), then subtract the gross amount. ## Is 100 a cube number? A square number is a number multiplied by itself. The square numbers up to 100 are: 0, 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100. A cube number is a number multiplied by itself 3 times. The cube numbers up to 100 are: 1, 8, 27 and 64. ## How many square roots are there between 1 and 100? See also What should I wear in 17c? We can see that, there are 10 perfect squares from 1 to 10. But, there are only 8 square roots between 1 to 10 after excluding 1 and 10. Therefore, the number of perfect squares between 1 to 100 natural numbers is 9. ## What are the prime numbers from 1 to 100? List of prime numbers to 100. 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. ## What number has the most factors under 100? The numbers under 100 with most factors are 60=22⋅3⋅5, 84=22⋅3⋅7, 96=25⋅3 and 72=23⋅32, which all have 12 factors. ## What is the largest factor of 100? As the number 100 is an even composite number, 100 has many factors other than 1 and the number itself. Thus, the factors of 100 are 1, 2, 4, 5, 10, 20, 25, 50, and 100. Factors of 100: 1, 2, 4, 5, 10, 20, 25, 50, and 100. Prime Factorization of 100: 2 x 2 × 5 x 5 or 22 x 52. ## What is the cube root of 1 to 100? From 1 to 100, the cube roots of 1, 8, 27 and 64 are whole numbers (rational), while the cube roots of 2, 3, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, … ## Can the square root of 180 be simplified? 180 is not a perfect square as √180 = 6√5. The square root of 180 cannot be expressed as a whole number; it can only be expressed as a non-terminating decimal. Therefore, √180 is an irrational number.
A hive of bees, a colony of ants and a parliament of owls. These are just a few examples of animal groups, or populations. A population is dynamic; this means it is constantly changing in size and demographics. New animals are born, old animals die and other factors such as drought, fire and lack of predators, all cause a change in the population. The population growth is the change in the number of individuals in a population, per unit time. For example, if a population has ten births and five deaths per year, then the population growth is five individuals per year. In the following pages, we aim to represent populations and changes in populations using mathematics. This involves using differential equations and even probability. Links to pages on differential equations: Part 1 A First Model Part 2 Exponential and Geometric Models Part 3 The Logistic Equation Part 4 The Logistic Map Part 5 The Lotka-Volterra Equations Part 6 Modified L-V Equations ### Beginning the Model We are able to describe population growth by making some generalizations and using simple differential equations: The size, $N_t$, of a population depends upon: • The initial number of individuals, $N_0$ • The number of births, B • The number of deaths, D • The number of immigrants, I • The number of migrants, E This gives us the equation: $$N_t=N_0+B-D+I-E$$ When a population is closed, there is no immigration or emigration. This often occurs on remote islands, such as the Galapagos Islands. Our equation then becomes $N_t=N_0+B-D$ , or equivalently $$N_{t+1}=N_t+B-D$$ Clearly the population will increase if $B> D$, and will decrease if $B< D$. A population is in equilibrium if on average the population size remains constant over a long period of time. Mathematically, this means: $N_t=N_{t+\Delta t}$ Question: We can rewrite the equation $N_{t+1}=N_t+B-D$ , as: $$N_{t+1}-N_t=\Delta N_t=B-D$$ Intuitively, why does this make sense? Think of an example of a population to explain why.
CONTENT • Construction of Quadratic Equations from Sum and Product of Roots. ## CONSTRUCTION OF QUADRATIC EQUATIONS FROM SUM AND PRODUCT OF ROOTS We can find the sum and product of the roots directly from the coefficient in the equation. It is usual to call the roots of the equation α and β If the equation ax2 +bx + C = 0 ……………. I has the roots α and β then it is equivalent to the equation ### Get A Free USA Scholarship, Visa & Accommodation (x – α )( x – β ) = 0 x2 – βx – βx + αβ = 0 ………… 2 Divide equation (i)by the coefficient of x2 ax2+ bx + = 0 ………… 3 aaa Comparing equations (2) and (3) x2 + b x + C = 0 aa x2 – ( α +β)x + αβ = 0 then α+ β= -b a and αβ = C a For any quadratic equation, ax2 +bx + C = 0 with roots α and β α + β = –b a αβ = C a Examples 1. If the roots of 3x2 – 4x – 1 = 0 are αand β, find α + β and αβ 2. if α and βare the roots of the equation 3x2 – 4x – 1 = 0 , find the value of (a) α + β β α (b) α – β Solutions 1. Since α + β = -b a Comparing the given equation 3x2 – 4x – 1= 0 with the general form ax2 + bx + C = 0 a = 3, b = -4, C = 1. Then α + β = –-(-4) a 3 = + = +1 1/3 3 αβ =C = –1 = -1 a 3 3 2.aα + β = α22 β α αβ = (α + β )2 – 2αβ αβ Here, comparing the given equation, with the general equation, a = 3, b = -4, C = – 1 from the solution of example 1 (since the given equation are the same ), α + β = –b = – (-4) = +4 3 3 αβ = = – 1 a 3 then α + β = ( α+ β ) 2 – 2 αβ β α αβ = (4/3 ).2 – 2 ( – 1/3 ) • 1/3 16 ± 2 9 3 – 1 3 16 + 6 ÷ -1/3 9 22 x -3 9 1 = -22 3 or α + β = – 22 = – 7 1/3 β α 3 1. b) Since (α-β) 22 + β– 2 α β but α2 + β2 = ( α + β)2 -2 α β :.(α- β)2 = ( α+ β )2 – 2αβ -2αβ (α – β)2 = (α + β )2 – 4α β :.( α – β) = √(α + β )2 – 4αβ ( α – β) =√ (4/3 )2 – 4 ( – 1/3 ) = √ 16/9 +4/3 = √16 + 12 9 = √28 = √28 9 3 :. α – β = √28 3 Evaluation If α and β are the roots of the equation 2x2 – 11x + 5 = 0, find the value of 1. α – β 2. 1 + 1 α + 1 β+ 1 1. α2+ β2 Examples 1. Find two numbers whose difference is 5 and whose product is 266. Solution Let the smaller number be x. Then the smaller number be x+5. Their product is x(x+5) . Hence, x(x+5) = 266 x2+5x- 266 = 0 (x-14)(x+19)=0 x=14 or x= -19 The other number is 14+5 or -19+5 i.e 19 or -14 :. The two numbers are 14 and 19 or -14 and -14. 1. Tina is 3 times older than her daughter. In four years time, the product of their ages will be 1536. How old are they now? Solution Let the daughter’s age be x. Tina’s age = 3x In four years’ time, Daughter’s age = (x+4)years Tina’s age = (3x+4)years The product of their ages : (x+4)(3x+4)= 1536 3x2+ 16x – 1520 = 0 (x-20)(3x+76) = =0 x=20 or x=-25.3 Since age cannot be negative, x=20years. :. Daughter’s age = 20years. Tina’s age = 20×3=60years. ## Evaluation 1. Think of a number, square it, add 2 times the original number. The result is 80. Find the number. 2. The area of a square is 144cm2 and one of its sides is (x+2)cm. Find x and then the side of the square. 3. Find two consecutive odd numbers whose product is 224. ## GENERAL EVALUATION/REVISION QUESTIONS 1. The area of a rectangle is 60cm2. The length is 11cm more than the width. Find the width. 2. A man is 37years old and his child is 8. How many years ago was the product of their ages 96? 3. If α and β are the roots of the equation 2x2 – 9x+4=0, find 4. a) α + β (b) αβ (c) α – β (d) αβ/ α + β ## WEEKEND ASSIGNMENT If α and β are the roots of the equation 2x2 + 9x+9=0: 1. Find the product of their roots. A. 4 B. 4.5 C. 5.5 B. -4.5 2. Find the sum of their roots. A. 4 B. 4.5 C. 5.5 B. -4.5 3. Find α22 A. 11.5 B. -11.25 C. 11.25 D. -11.5 4. Find αβ/ α + β A. 1 B.-1 C. 1.5 D. 4.5 ## THEORY 1. The base of a triangle is 3cm longer than its corresponding height. If the area is 44cm2, find the length of its base. 2. Find the equation in the form ax2+bx+c=0 whose sum and products of roots are respectively: 3. a) 3,4 (b) -7/3 , 0 (c) 1.2,0.8 GEOMETRIC PROGRESSION ARITHMETIC PROGRESSION (A. P) PERCENTAGE ERROR LOGARITHM OF NUMBERS MENSURATION
Patterns We know that Mathematics is all about numbers. It basically involves the study of different patterns. There are different types of patterns in Mathematics, such as image patterns, logic patterns, number patterns, word patterns etc. The most common patterns in Mathematics are number patterns. These patterns are quite familiar to the students who study Mathematics frequently. Especially, they are everywhere in Mathematics. So, let’s discuss the definition of pattern. In simple words, Number patterns are all predictions. Let’s discuss a few examples of numerical patterns are: 1. Pattern of even numbers -: 2, 4, 6, 8, 10, 1, 14, 16, 18, 20… 2. Pattern of Odd numbers -: 3, 5, 7, 9, 11, 13, 15, 17, 19, 21…. 3. Pattern of Fibonacci numbers -: 1, 1, 2, 3, 5, 8 ,13, … and so on We are going to discuss the definition of pattern in Mathematics with a few solved example problems. In this article, we are going to focus on various patterns and pattern definition in maths. Pattern Definition- • A pattern can be defined as a series or sequence that generally repeats. We observe patterns – things like colors, actions, shapes, or other sequences that repeat – everywhere in our daily life. • Math patterns are basically sequences that repeat according to a rule or rules. In math, a rule can be defined as a set way to calculate or solve a problem. In Mathematics, the patterns can be related to any type of event or object. If the set of numbers are related to each other in any specific kind of rule, then the rule or manner is known as a pattern. Sometimes, patterns can also be known as a sequence. They are finite or infinite in numbers. For example, in the given sequence 2,4,6,8,? Each number is increasing by the number 2. So, the last number will be 8 + 2 which is equal to 10. According to the definition of pattern there is a common type of math pattern known as the number pattern. Number patterns can be defined as a sequence of numbers that are ordered based upon a rule. Now, according to the pattern definition there are many ways to figure out the rule, such as: • Use a number line to see the distance or the difference between the numbers or what the numbers have in common. • Look at the last one or two digits or the first digit of the numbers to see if they repeat in any special manner. • Look at the numbers and see if there is any pattern, like taking each number and multiplying by 2 for instance. • Think about the common number patterns, like counting by 2s, 5s, or 10s. • You can also find the difference between the numbers It’s important to remember that a number pattern can have more than one solution and a combination of rules present. In this case, try to think of the simplest rule that is possible, like adding 2 or multiplying by 3 with a difference of 4. There are Various Different Types of Number Patterns: • The Arithmetic Pattern • The Geometric Pattern • The Fibonacci Pattern Let’s Discuss the Rules for Patterns in Maths- Now to construct a pattern, we have to know about some of the rules. To know about the rule for any type of pattern, we need to understand the nature of the sequence and the difference between the two successive terms in the pattern. Finding the Missing Term: If we consider a pattern like 1, 4, 9, 16, 25,?. In this pattern, it is clear that every number is the square of their position number that is position 1 gives 1 square, second position gives two square that is equal to four. The missing term takes place at n = 6. So, if the missing term is xn, then we can write that xn = n2. Here, the value of n = 6, then xn = (6)2 which is equal to 36. The Difference Rule: Sometimes, it is easy and simple to find the difference between two successive terms in a pattern. For example, consider 1, 5, 9, 13,… In this type of pattern, first, we need to find the difference between the two pairs of the sequence. After that, we need to find the remaining elements of the pattern. In the given problem, the difference between the terms is equal to 4, that is if we add 4 and 1, then we get 5, and if we add 4 and 5, then we get 9 and so on. Different Types of Patterns in Mathematics – In Discrete Mathematics, we have these 3 types of patterns: • Repeating Patterns– If the number pattern changes in the same value every time, then the pattern can be known as a repeating pattern. Example: 1, 2, 3, 4, 5 ,6… • Growing Patterns – If the numbers are present in an increasing form, then the pattern can be known as a growing pattern. Example 34, 40, 46, 52, 56 ….. • Shirking Patterns – In the shirking pattern, the numbers are generally in decreasing form. Example: 42, 40, 38, 36, 32.. Questions to be Solved- Example 1) Determine the Value of Variables A and B in the Following Pattern. 85, 79, 73, 67, 61, 55, 49, 43, A, 31, 25, B. Solution) We have been given the sequence:85, 79, 73, 67, 61, 55, 49, 43, A, 31, 25, B. Here, the number here is decreasing by 6 The previous number of A is 43. So, A will be 43 – 6, P = 37 The previous number of B is 25. So, B will be 25 – 6, Q = 19 Therefore, the value of A is 37 and B is 19. Q1. What are Number Patterns in Mathematics and What are the Four Rules of Maths? What are the Common Types of Patterns? Ans. Number pattern can be defined as a pattern or sequence in a series of numbers. This pattern generally establishes a common relationship between all numbers present among them. For example: 0, 5, 10, 15, 20, 25, 30 (differs by a value of 5) The Four Basic Mathematical operations are -addition, subtraction, multiplication, and division. These four operations have application even in the most advanced mathematical theories. The common types of patterns are – 1. Single piece pattern 2. Split piece pattern 3. Loose piece pattern 4. Gated pattern 5. Match pattern 6. Skeleton pattern
## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition) $[(t-3)^{n}+1][7(t-3)^n-2]$ Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} 7(t-3)^{2n}+5(t-3)^n-2 \end{array} has $ac= 7(-2)=-14$ and $b= 5 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ 7,-2 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 7(t-3)^{2n}+7(t-3)^n-2(t-3)^n-2 .\end{array} Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} [7(t-3)^{2n}+7(t-3)^n]-[2(t-3)^n+2] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 7(t-3)^n[(t-3)^{n}+1]-2[(t-3)^n+1] .\end{array} Factoring the $GCF= (a^{3n}-1)$ of the entire expression above results to \begin{array}{l}\require{cancel} [(t-3)^{n}+1][7(t-3)^n-2] .\end{array}
Review question # How many solutions do these simultaneous trig equations have? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource Ref: R7370 ## Solution The simultaneous equations in $x$,$y$, $(\cos\theta)x - (\sin\theta)y = 2$ $(\sin\theta)x + (\cos\theta)y = 1$ are solvable 1. for all values of $\theta$ in the range $0 \leq \theta < 2\pi$; 2. except for one value of $\theta$ in the range $0 \leq \theta < 2\pi$; 3. except for two values of $\theta$ in the range $0 \leq \theta < 2\pi$; 4. except for three values of $\theta$ in the range $0 \leq \theta < 2\pi$. #### Method 1 We write $c$ for $\cos\theta$ and $s$ for $\sin\theta$, so the equations become $cx-sy=2, \qquad sx+cy=1;$ We multiply by $c$ and $s$ respectively and then add to eliminate $y$, giving $c^2x-csy = 2c, \qquad s^2x+csy = s,$ so $2c+s = (c^2+s^2)x = x.$ Similarly, we can eliminate $x$, since $csx - s^2y = 2s, \qquad csx + c^2y = c$ so $c - 2s = (s^2+c^2)y = y.$ Hence we have $x = 2\cos\theta + \sin\theta$ and $y = \cos\theta - 2\sin\theta$, which are valid solutions for any value of $\theta$, and the answer is (a). #### Method 2 For a given $\theta$, each equation represents a straight line. So we will have exactly one solution unless the lines coincide, or are parallel, that is, unless they have the same gradient. The gradient of the first line is $\dfrac{\cos \theta}{\sin \theta}$, while the gradient of the second is $-\dfrac{\sin \theta}{\cos \theta}$. These can only be equal if $\dfrac{\cos \theta}{\sin \theta}=-\dfrac{\sin \theta}{\cos \theta}$, which is true if and only if $\cos^2 \theta + \sin^2 \theta = 0$, which is never true. So there is always exactly one solution for any given $\theta$, and the answer is (a). #### Method 3 This requires compound angle formulae If we square and add the two equations, we get $x^2 + y^2 = 5$. Thus any solution must be of the form $x = \sqrt{5}\cos\alpha, y = \sqrt{5}\sin\alpha, \quad 0\leq \alpha < 2\pi.$ Substituting this back into our equations (since squaring may have introduced extra solutions), we have $\cos(\theta + \alpha) = \dfrac{2}{\sqrt{5}}, \sin(\theta + \alpha) = \dfrac{1}{\sqrt{5}}.$ Thus $\theta + \alpha$ is in the first quadrant, and for any $\theta$ there is exactly one solution for $\alpha$, and the answer is (a).
# How do you divide (8-2i) / (5-3i) in trigonometric form? sqrt2(cos(tan^-1 (7/23))+isin(tan^-1 (7/23)) OR $\sqrt{2} \left(\cos \left({16.9275}^{\circ}\right) + i \sin \left({16.9275}^{\circ}\right)\right)$ #### Explanation: Start from the given complex number $\frac{8 - 2 i}{5 - 3 i}$ Convert the numerator ${r}_{1} = \sqrt{{8}^{2} + {\left(- 2\right)}^{2}} = \sqrt{68}$ ${\theta}_{1} = {\tan}^{-} 1 \left(\frac{- 2}{8}\right) = {\tan}^{-} 1 \left(\frac{- 1}{4}\right)$ then $8 - 2 i = \sqrt{68} \left[\cos \left({\tan}^{-} 1 \left(\frac{- 1}{4}\right)\right) + i \sin \left({\tan}^{-} 1 \left(\frac{- 1}{4}\right)\right)\right]$ Convert the denominator ${r}_{2} = \sqrt{{5}^{2} + {\left(- 3\right)}^{2}} = \sqrt{34}$ ${\theta}_{2} = {\tan}^{-} 1 \left(\frac{- 3}{5}\right) = {\tan}^{-} 1 \left(\frac{- 3}{5}\right)$ then $5 - 3 i = \sqrt{34} \left[\cos \left({\tan}^{-} 1 \left(\frac{- 3}{5}\right)\right) + i \sin \left({\tan}^{-} 1 \left(\frac{- 3}{5}\right)\right)\right]$ Let us divide now, from the given with the equivalent $\frac{8 - 2 i}{5 - 3 i} = \frac{\sqrt{68} \left[\cos \left({\tan}^{-} 1 \left(\frac{- 1}{4}\right)\right) + i \sin \left({\tan}^{-} 1 \left(\frac{- 1}{4}\right)\right)\right]}{\sqrt{34} \left[\cos \left({\tan}^{-} 1 \left(\frac{- 3}{5}\right)\right) + i \sin \left({\tan}^{-} 1 \left(\frac{- 3}{5}\right)\right)\right]}$ Divide using the following formula $\left({r}_{1} / {r}_{2}\right) \cdot \left[\cos \left({\theta}_{1} - {\theta}_{2}\right) + i \sin \left(\theta 1 - {\theta}_{2}\right)\right]$ (8-2i)/(5-3i)= sqrt(68/34)[cos(tan^-1 ((-1)/4)-tan^-1 ((-3)/5))+isin(tan^-1 ((-1)/4)-tan^-1 ((-3)/5))] Take note: that $\tan \left({\theta}_{1} - {\theta}_{2}\right) = \frac{\tan {\theta}_{1} - \tan {\theta}_{2}}{1 + \tan {\theta}_{1} \cdot \tan {\theta}_{2}} = \frac{- \frac{1}{4} - \left(- \frac{3}{5}\right)}{1 + \left(- \frac{1}{4}\right) \cdot \left(- \frac{3}{5}\right)}$ $\tan \left({\theta}_{1} - {\theta}_{2}\right) = \frac{7}{23}$ ${\theta}_{1} - {\theta}_{2} = {\tan}^{-} 1 \left(\frac{7}{23}\right)$ and (8-2i)/(5-3i)=sqrt2(cos(tan^-1 (7/23))+isin(tan^-1 (7/23)) Have a nice day!!!.
# What Is Function In General Math? ## What is function in general mathematics with example? A function is a mapping from a set of inputs (the domain) to a set of possible outputs (the codomain). The definition of a function is based on a set of ordered pairs, where the first element in each pair is from the domain and the second is from the codomain. ## WHAT IS function and relation in general mathematics? A relation is a set of inputs and outputs, and a function is a relation with one output for each input. ## WHAT IS function and its types? 1. Injective (One-to-One) Functions: A function in which one element of Domain Set is connected to one element of Co-Domain Set. 2. Surjective (Onto) Functions: A function in which every element of Co-Domain Set has one pre-image. ## What are the 4 types of functions? The various types of functions are as follows: • Many to one function. • One to one function. • Onto function. • One and onto function. • Constant function. • Identity function. • Polynomial function. You might be interested: How To Write Word Problems In Math? ## What is a real life example of a function? A weekly salary is a function of the hourly pay rate and the number of hours worked. Compound interest is a function of initial investment, interest rate, and time. Supply and demand: As price goes up, demand goes down. ## What is meant by a function? A technical definition of a function is: a relation from a set of inputs to a set of possible outputs where each input is related to exactly one output. We can write the statement that f is a function from X to Y using the function notation f:X→Y. ## What is the importance of general mathematics? Mathematics makes our life orderly and prevents chaos. Certain qualities that are nurtured by mathematics are power of reasoning, creativity, abstract or spatial thinking, critical thinking, problem-solving ability and even effective communication skills. ## What is the example of function and relation? In mathematics, a function can be defined as a rule that relates every element in one set, called the domain, to exactly one element in another set, called the range. For example, y = x + 3 and y = x2 – 1 are functions because every x- value produces a different y- value. A relation is any set of ordered-pair numbers. ## What is difference between relation and function? Relation – In maths, the relation is defined as the collection of ordered pairs, which contains an object from one set to the other set. Functions – The relation that defines the set of inputs to the set of outputs is called the functions. In function, each input in the set X has exactly one output in the set Y. You might be interested: FAQ: What Is Independent Events In Math? ## What are the two main types of functions? What are the two main types of functions? Explanation: Built-in functions and user defined ones. The built-in functions are part of the Python language. ## What is on to function? A function f: A -> B is called an onto function if the range of f is B. In other words, if each b ∈ B there exists at least one a ∈ A such that. f(a) = b, then f is an on-to function. An onto function is also called surjective function. ## What are basic functions? The basic polynomial functions are: f(x)=c, f(x)=x, f(x)=x2, and f(x)=x3. The basic nonpolynomial functions are: f(x)=\|x\|, f(x)=√x, and f(x)=1x. A function whose definition changes depending on the value in the domain is called a piecewise function. The value in the domain determines the appropriate definition to use. ## What is the most basic function? The parent function is the most basic function in the family of functions, the function from which all the other functions in the family can be derived. A family of functions is a group of functions that can all be derived from transforming a single function called the parent function. ## Can any function call itself? Recursion is a way of programming or coding a problem, in which a function calls itself one or more times in its body. Usually, it is returning the return value of this function call. If a function definition fulfills the condition of recursion, we call this function a recursive function. Yes, you can.
Call Now: (800) 537-1660 May 25th May 25th # Multiplying Polynomials After studying this lesson, you will be able to: • Multiply polynomials. When multiplying polynomials, we multiply the coefficients and we add the exponents. If we have parentheses, we use the distributive property. Example 1 5a (3a 2 + 4) We need to distribute 5a. Remember to multiply the coefficients and add the exponents 15a 3 + 20a Example 2 2m 2 (5m 2 - 7m + 8) We need to distribute 2m 2 . Remember to multiply the coefficients and add the exponents. 10m 4 - 14m 3 + 16m 2 Example 3 We need to distribute 2 3 x . Remember to multiply the coefficients and add the exponents. To multiply by 12 x...we can do 2x times 12x and then divide by 3 to get 8x 2 To multiply by -9...we can do 2x times -9 and then divide by 3 to get -6x. 8x 3 - 6x Example 4 -3x 2 y ( 2x 2 y- 3x y 2 - 7y 3 ) We need to distribute -3x 2 y. Remember to multiply the coefficients and add the exponents. -6x 4 y 2 + 9x 3 y 3 + 21x 2 y 4 Example 5 We need to distribute . Remember to multiply the coefficients and add the exponents. Remember that -y is the same as -1y. Multiply -y times 20y 2 then divide by 2 to get -10y 3. Multiply -y times -10y then divide by 2 to get 5y 2 . Multiply -y times 4 then divide by 2 to get -2y. -10y 3 + 5y 2 -2y Example 6 5x 2 (x + 7) - 2x (5x 2 - 3x + 7) + 2(x 3 - 8) We have to distribute three times on this problem. Doing so will give us: 5x 3 + 35x 2 - 10x 3 + 6x 2 - 14x + 2x 3 - 16 Now, we add like terms and put the terms in descending order: -3x 3 + 41x 2 - 14x - 16 Example 7 Solve x(x - 3) + 4x - 3 = 8x + 4 + x (3 + x ) This is an equation. First we need to distribute to remove the parentheses. This will give us: x 2 - 3x + 4x - 3 = 8x + 4 + 3x + x 2 Next, we collect like terms on each side. This will give us: x 2 + x - 3 = 11x + 4 + x 2 Next, we subtract x 2 from each side. This will give us: x -3 = 11x + 4 Now, we subtract x from each side to get: -3 = 10x + 4 Now, we subtract 4 from each side to get: -7 = 10x Now we divided each side by 10 to get Free Tutoring • Get 30 Minutes of FREE Live Tutoring from Tutor.com if you Testimonials "I ordered the Algebra Buster late one night when my daughter was having problems in her honors algebra class. After we ordered your software she was able to see step by step how to solve the problems. Algebra Buster definitely saved the day." Tami Garleff Home Why Algebra Buster? Guarantee Testimonials Ordering FAQ About Us What's new? Resources Animated demo Algebra lessons Bibliography of textbooks Related pages:
6.A student's mark was wrongly entered as 83 instead of 63. Due to that the average mark of the class got increased by 1/ 2 . What is the number of students in the class? A. 45 B. 40 C. 35 D. 30 Explanation: Let total number of students be x Increase in average mark = 1/ 2 Increase in total mark = x( 1/ 2) = x/ 2 But increase in total mark = 83 − 63 = 20 Therefore, x/ 2 = 20 ⇒ x = 40 7.A library has an average of 510 visitors on Sundays and 240 on other days. What is the average number of visitors per day in a month of 30 days beginning with a Sunday? A. 304 B. 285 C. 270 D. 290 Explanation: In a month of 30 days beginning with Sunday, there are 4 complete weeks and another two days, Sunday and Monday. Hence there are 5 Sundays and 25 other days. Average visitors on Sundays = 510 Total visitors of 5 Sundays = 510 × 5 Average visitors on other days = 240 Total visitors of other 25 days = 240 × 25 Total visitors = 510 × 5 + 240 × 25 Total days = 30 Average number of visitors per day = (510 × 5 + 240 × 25)/ 30 = 285 8. The average weight of 16 boys in a class is 50.25 kg and that of the remaining 8 boys is 45.15 kg. Find the average weights of all the boys in the class. A. 42.25 kg B. 51.25 kg C. 48.55 kg D. 50 kg Explanation: Average weight of 16 boys = 50.25 Total weight of 16 boys = 50.25 × 16 Average weight of remaining 8 boys = 45.15 Total weight of remaining 8 boys = 45.15 × 8 Total weight of all boys in the class = 50.25 × 16 + 45.15 × 8 Total number of boys = 16 + 8 = 24 Average weight of all the boys = (50.25 × 16 + 45.15 × 8)/ 24 = 48.55 9.In Rugved's opinion, his weight is greater than 65 kg but less than 72 kg. His brother does not agree with Rugved and he thinks that Rugved's weight is greater than 60 kg but less than 70 kg. His mother's view is that his weight cannot be greater than 68 kg. If all are them are correct in their estimation, what is the average of different probable weights of Rugved? A. 61 kg B. 70 kg C. 67 kg D. 69 kg Explanation: Let Rugved's weight be x According to Rugved, 65 smaller than x greater than 72 According to brother, 60 smaller than x greater than 70 According to mother, x smaller than equal to 68 Given that all the above conditions are true. Therefore we have 65 smaller than x greater than equal to 68 That is, x can be 66 , 67 , or 68 Average of different probable weights of Rugved = average of 66 , 67 , 68 (66 + 67 + 68)/ 3 = 67. 10. A car owner buys diesel at ₹ 7.50 , ₹ 8 and ₹ 8.50 per litre for three successive years. What approximately is the average cost per litre of diesel if he spends ₹ 4000 each year? A. ₹ 8.1 B. ₹ 7.98 C. ₹ 7.4 D. ₹ 8 Explanation: Total quantity of petrol consumed in 3 years =( 4000/7.50 + 4000/8 + 4000/8.50 )litres =( 76700 /51)litres Total amount spent =Rs.(3×4000)=Rs.12000. ∴ Average cost =Rs. (12000×51)/76700 =Rs.7.98 darkmode
You currently have JavaScript disabled on this browser/device. JavaScript must be enabled in order for this website to function properly. # ZingPath: Bar, Line, and Circle Graphs Searching for ## Bar, Line, and Circle Graphs Learn in a way your textbook can't show you. Explore the full path to learning Bar, Line, and Circle Graphs ### Lesson Focus #### Circle Graphs Math Foundations You will construct a circle graph for representing and displaying data. ### Now You Know After completing this tutorial, you will be able to complete the following: • Construct circles graphs. ### Everything You'll Have Covered Constructing circle graphs Circle graphs can provide a quick and easy way to display information about the relationship of parts to a whole. Once the categories are selected and data collected for each category, a circle graph can be constructed. Let's look at an example of the number of running shoes sold in a given month: By looking at the table, we can see that Stone and Fast have the largest sectors of shoe sales, but constructing a circle graph would help us to view this quickly. To create a circle graph, we need to determine the following: · What is the "whole?" · What are the different parts or groups? · What is the ratio of each part to the whole? · What is the degree of each section? Defining the whole In this table, the whole is the total number of running shoes sold for the month. This total can be found by adding up the numbers sold for each type of shoe. There are five different parts to the whole and the table lists the number sold in each part. 150 + 108 + 192 + 60 + 90 = 600 total shoes sold Determining the parts or groups There are five parts to the whole. Each data group is a category of running shoe brands: 1) Fast, 2) Easy Walk, 3) Stone, 4) RNR, and 5) Other. Finding the ratio of each part to the whole Now that we know how many parts there are and the total, we can create ratios of each part to the whole. Calculating the degree of each section Using the ratios of each part, we are able to calculate the degree of each category. A full circle is made up of 360°. So in order to find the degrees of each part, we must multiply the ratio of each part by 360. Now that we have calculated the degree for each category, we can use those measurements to draw the circle graph. This graph allows us to evaluate the relative sizes of each category quickly. We can see that Stone has the highest number of sales and that of the top four sellers, RNR has the lowest?approximately one-third that of Stone. ### Tutorial Details Approximate Time 20 Minutes Pre-requisite Concepts Students should be familiar with operations on whole numbers and ratio. Course Math Foundations Type of Tutorial Concept Development Key Vocabulary circle graph,,
# Expert Maths Tutoring in the UK Integration By Partial Fractions Integration By Partial Fractions with tutors mapped to your child’s learning needs. Quotient Rule Quotient rule in calculus is a method to find the derivative or differentiation of a function given in the form of a ratio or division of two differentiable functions. That means, we can apply the quotient rule when we have to find the derivative of a function of the form: f(x)/g(x), such that both f(x) and g(x) are differentiable, and g(x) ≠ 0. The quotient rule follows the product rule and the concept of limits of derivation in differentiation directly. Let us understand the formula for quotient rule, its proof using solved examples in detail in the following sections. 1 What is Quotient Rule? 2 Quotient Rule Formula 3 Derivation of Quotient Rule Formula 4 How to Apply the Quotient Rule in Differentiation? 5 FAQs on Quotient Rule ## What is the Quotient Rule? Quotient rule in calculus is a method used to find the derivative of any function given in the form of a quotient obtained from the result of the division of two differentiable functions. The quotient rule in words states that the derivative of a quotient is equal to the ratio of the result obtained on the subtraction of the numerator times the derivative of the denominator from the denominator times the derivative of the numerator to the square of the denominator. That means if we are given a function of the form: f(x) = u(x)/v(x), we can find the derivative of this function using the quotient rule derivative as, f'(x) = [u(x)/v(x)]’ = [v(x) × u'(x) – u(x) × v'(x)]/[v(x)]2 ## Quotient Rule Formula We can calculate the derivative or evaluate the differentiation of the quotient of two functions using the quotient rule derivative formula. The quotient rule derivative formula is given as, f'(x) = [u(x)/v(x)]’ = [v(x) × u'(x) – u(x) × v'(x)]/[v(x)]2 where, • f(x) = The function of the form u(x)/v(x) for which the derivative is to be calculated. • u(x) = A differentiable function that makes numerator of the function f(x). • u'(x) = Derivative of function u(x). • v(x) = A differentiable function that makes denominator of the given function f(x). • v'(x) = Derivative of the function v(x). ## Derivation of Quotient Rule Formula In the previous section, we learned about the quotient formula to find derivatives of the quotient of two differentiable functions. Let us see the proof of the quotient rule formula here. There are different methods to prove the quotient rule formula, given as, • Using derivative and limit properties • Using implicit differentiation • Using chain rule ### Quotient Rule Formula Proof Using Derivative and Limit Properties To prove quotient rule formula using the definition of derivative or limits, let the function f(x) = u(x)/v(x). ⇒ f'(x) = $$\mathop {\lim }\limits_{h \to 0}$$ [f(x + h) – f(x)]/h = $$\mathop {\lim }\limits_{h \to 0}$$ $$\frac{\frac{u(x+h)}{v(x+h)} – \frac{u(x)}{v(x)}}{h}$$ = $$\mathop {\lim }\limits_{h \to 0}$$ $$\frac{u(x+h)v(x) – u(x)v(x+h)}{h \cdot v(x) \cdot v(x+h)}$$ = $$\left(\mathop {\lim }\limits_{h \to 0} \frac{u(x+h)v(x) – u(x)v(x+h)}{h}\right) \left(\mathop {\lim }\limits_{h \to 0} \frac{1}{v(x) \cdot v(x+h)}\right)$$ = $$\left(\mathop {\lim }\limits_{h \to 0} \frac{u(x+h)v(x) – u(x)v(x) + u(x)v(x) – u(x)v(x + h)}{h}\right)$$ [1/v(x)2] = $$\left[\!\left(\mathop {\lim }\limits_{h \to 0} \frac{u(x+h)v(x) – u(x)v(x)}{h}\right)\!\!\! -\!\! \left(\mathop {\lim }\limits_{h \to 0} \frac{u(x)v(x + h) – u(x)v(x)}{h}\right)\!\right]$$ [1/v(x)2] = $$\left[\!v(x)\!\left(\!\mathop {\lim }\limits_{h \to 0} \frac{u(x+h) – u(x)}{h}\right)\!\!\!-\!\!u(x)\!\!\left( \mathop {\lim }\limits_{h \to 0} \!\!\frac{v(x + h) – v(x)}{h} \right)\!\!\right]$$ [1/v(x)2] = $$\frac{v(x)u'(x) – u(x)v'(x)}{[v(x)]^2}$$ ### Quotient Rule Formula Proof Using Implicit Differentiation To prove the quotient rule formula using implicit differentiation formula, let us take a differentiable function f(x) = u(x)/v(x), so u(x) = f(x)⋅v(x). Using the product rule, we have, u'(x) = f'(x)⋅v(x) + f(x)v'(x). Solving for f'(x), we get, f'(x) = $$\frac{u'(x) – f(x)v'(x)}{v(x)}$$ Substitute f(x), ⇒ f'(x) = $$\frac{u'(x) – \frac{u(x)}{v(x)}v'(x)}{v(x)}$$ = $$\frac{u'(x)v(x) – u(x)v'(x)}{[v(x)]^2}$$ ### Quotient Rule Formula Proof Using Chain Rule We can derive the quotient rule formula in calculus using the chain rule formula. Let f(x) be a differentiable function such that f(x) = u(x)/v(x). ⇒ f(x) = u(x)v-1(x) Using the product rule, f'(x) = u'(x)v-1(x) + u(x)⋅$$\left(\frac{d}{dx}(v^{-1}(x)\right)$$ Applying the power rule to solve the derivative in the second term, we have, f'(x) = u'(x)v-1(x) + u(x)⋅(-1)(v(x)-2)v'(x) f'(x) = $$\frac{u'(x)}{v(x)} – \frac{u(x)v'(x)}{v(x)^2}$$ f'(x) = $$\frac{d}{dx} \left[ \frac{u(x)}{v(x)} \right]$$ = $$\frac{u'(x)v(x) – u(x)v'(x)}{v(x)^2}$$ ## How to Apply Quotient Rule in Differentiation? In order to find the derivative of the function of the form f(x) = u(x)/v(x), both u(x) and v(x) should be differentiable functions. We can apply the following given steps to find the derivation of a differentiable function f(x) = u(x)/v(x) using the quotient rule. • Step 1: Note down the values of u(x) and v(x). • Step 2: Find the values of u'(x) and v'(x) and apply the quotient rule formula, given as: f'(x) = [u(x)/v(x)]’ = [u'(x) × v(x) – u(x) × v'(x)]/[v(x)]2 Let us have a look at the following example given below to understand the quotient rule better. Example: Find f'(x) for the following function f(x) using the quotient rule: f(x) = x2/(x+1). Solution: Here, f(x) = x2/(x + 1) u(x) = x2 v(x) = (x + 1) ⇒u'(x) = 2x ⇒v'(x) = 1 ⇒f'(x) = [v(x)u'(x) – u(x)v'(x)]/[v(x)]2 ⇒f'(x) = [(x+1)•2x – x2•1]/(x + 1)2 ⇒f'(x) = (2×2 + 2x – x2)/(x + 1)2 ⇒f'(x) = (x2 + 2x)/(x + 1)2 Answer: The derivative of x2/(x + 1) is (x2 + 2x)/(x + 1)2. ☛ Topics Related to Quotient Rule: ## Examples on Quotient Rule 1. Example 1: Find the derivative of $$\frac{\cos{x}}{x}$$ using the quotient rule formula. Solution: Let f(x) = cos x and g(x) = x. $$\frac{d}{{dx}}\left\{ {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right\} = \frac{{g\left( x \right)f’\left( x \right) – f\left( x \right)g’\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}\\=\frac{x(-\sin{x})-\cos{x}(1)}{x^2}\\=-\frac{x\sin{x}+\cos{x}}{x^2}$$ Answer: The derivative of $$\frac{\cos{x}}{x}$$ is $$-\frac{x\sin{x}+\cos{x}}{x^2}$$. 2. Example 2: Differentiate $$\frac{\log{x}}{x}$$ using the quotient rule formula. Solution: Let f(x) = log x and g(x) = x. $$\frac{d}{{dx}}\left\{ {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right\}= \frac{{g\left( x \right)f’\left( x \right) – f\left( x \right)g’\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}\\=\frac{x(1/x)-\log{x}}{x^2}\\=\frac{1-\log{x}}{x^2}$$ Answer: The derivative of $$\frac{\log{x}}{x}$$ is $$\frac{1-\log{x}}{x^2}$$. 3. Example 3: Apply the quotient rule to differentiate $$\frac{1-2x}{x}$$. Solution: Let f(x) = (1-2x)/x f'(x) = $$\frac{d}{dx}$$ $$\frac{(1-2x)}{x}$$ = [x $$\frac{d}{dx}$$ (1-2x) – (1-2x) $$\frac{d}{dx}$$ x]/x2 f'(x) = [x(-2) – (1-2x) (x)]/x2 = (-2x – 1 + 2x)/x2 = -1/x2 Answer: The derivative of $$\frac{(1-2x)}{x}$$ is -1/x2. ## FAQs on Quotient Rule ### What is Quotient Rule of Differentiation in Calculus? The quotient rule is one of the derivative rules that we use to find the derivative of functions of the form P(x) = f(x)/g(x). The derivative of a function P(x) is denoted by P'(x). If the derivative of the function P(x) exists, we say P(x) is differentiable. So, differentiable functions are those functions whose derivatives exist. A function P(x) is differentiable at a point x = a if the following limit exists. $$P'(x) = \mathop {\lim }\limits_{h \to 0} \frac{P(a+h)-P(a)}{h}$$ ### How to Find Derivative Using Quotient Rule? The derivatives of the quotient for the ratio of two differentiable functions can be calculated in calculus using the quotient rule. We need to apply the quotient rule formula for differentiation of function f(x) = u(x)/v(x). The quotient rule formula is given as, f'(x) = [u(x)/v(x)]’ = [u'(x) × v(x) – u(x) × v'(x)]/[v(x)]2 where, f'(x), u'(x) and v'(x) are derivatives of functions f(x), v(x) and u(x). ### What is Quotient Rule Formula? Quotient rule derivative formula is a rule in differential calculus that we use to find the derivative of rational functions. Suppose two functions, u(x) and v(x) are differentiable, then the quotient rule can be applied to find (d/dx)[u(x)/v(x)] as, f'(x) = [u(x)/v(x)]’ = [u'(x) × v(x) – u(x) × v'(x)]/[v(x)]2 ### How to Derive the Formula for Quotient Rule? Quotient rule formula can be derived using different methods. They are given as, • Using derivative and limit properties • Using implicit differentiation • Using chain rule Using chain rule, we can derive the quotient rule formula by taking f(x) as a differentiable function such that f(x) = u(x)/v(x). ⇒ f(x) = u(x)v-1(x) Using the product rule, f'(x) = u'(x)v-1(x) + u(x)⋅$$\left(\frac{d}{dx}(v^{-1}(x)\right)$$ ⇒f'(x) = u'(x)v-1(x) + u(x)⋅(-1)(v(x)-2)v'(x) ⇒f'(x) = $$\frac{d}{dx} \left[ \frac{u(x)}{v(x)} \right]$$ = $$\frac{u'(x)v(x) – u(x)v'(x)}{[v(x)]^2}$$ ### How to Use Quotient Rule in Differentiation? Quotient rule can be used in finding the differentiation of a function f(x) of the form u(x)/v(x). Derivative of this function using quotient rule can be given as, f'(x) = $$\frac{d}{dx} \left[ \frac{u(x)}{v(x)} \right]$$ = $$\frac{u'(x)v(x) – u(x)v'(x)}{v(x)^2}$$ How to Derive Quotient Rule Using Definition of Limits and Derivatives? The proof of the quotient rule can be given using the definition and properties of limits and derivatives. For a function f(x) = u(x)/v(x), the derivative f'(x) can be given as, ⇒ f'(x) = $$\mathop {\lim }\limits_{h \to 0}$$ [f(x + h) – f(x)]/h = $$\mathop {\lim }\limits_{h \to 0}$$ $$\frac{\frac{u(x+h)}{v(x+h)} – \frac{u(x)}{v(x)}}{h}$$ = $$\mathop {\lim }\limits_{h \to 0}$$ $$\frac{u(x+h)v(x)$$ – $$u(x)v(x+h)}{h \cdot v(x) \cdot v(x+h)}$$ = $$\left(\mathop {\lim }\limits_{h \to 0} \frac{u(x+h)v(x)$$ – $$u(x)v(x+h)}{h}\right) \left(\mathop {\lim }\limits_{h \to 0} \frac{1}{v(x) \cdot v(x+h)}\right)$$ = $$\left(\mathop {\lim }\limits_{h \to 0} \frac{u(x+h)v(x) – u(x)v(x) + u(x)v(x) – u(x)v(x + h)}{h}\right)$$ [1/v(x)2] = $$\left[\left(\mathop {\lim }\limits_{h \to 0} \frac{u(x+h)v(x) – u(x)v(x)}{h}\right)$$ – $$\left(\mathop {\lim }\limits_{h \to 0} \frac{u(x)v(x + h) – u(x)v(x)}{h}\right)\right]$$ [1/v(x)2] = $$\left[v(x)\left(\mathop {\lim }\limits_{h \to 0} \frac{u(x+h) – u(x)}{h}\right)$$-$$u(x)\left( \mathop {\lim }\limits_{h \to 0} \frac{v(x + h) – v(x)}{h} \right)\right]$$ [1/v(x)2] ### What are Applications of Quotient Rule Derivative Formula? Give Examples. We can apply the quotient rule to find the differentiation of the function of the form u(x)/v(x). For example, for a function f(x) = sin x/x, we can find the derivative as, f'(x) = [x $$\frac{d}{dx}$$ sin x – sin x $$\frac{d}{dx}$$ x]/x2, f'(x) = (x•cos x – sin x)/x2. ### How do we Prove Quotient Rule Using Implicit Differentiation? We can use the implicit differentiation method to derive the quotient rule, for a differentiable function f(x) = u(x)/v(x), so u(x) = f(x)⋅v(x). Using the product rule, we have, u'(x) = f'(x)⋅v(x) + f(x)v'(x). Solving for f'(x), we get, f'(x) = $$\frac{u'(x) – f(x)v'(x)}{v(x)}$$ Substitute f(x), ⇒ f'(x) = $$\frac{u'(x) – \frac{u(x)}{v(x)}v'(x)}{v(x)}$$ = $$\frac{u'(x)v(x) – u(x)v'(x)}{[v(x)]^2}$$ How Can we Prove Quotient Rule Using the Product Rule in Calculus? To prove the quotient rule using the product rule and chain rule, we can express the function f(x) = u(x)/v(x) as f(x) = u(x)•1/v(x) and further apply product rule formula to find f'(x) = (1/v(x))u'(x) – u(x)•(1/v(x))2•v'(x) = $$\frac{u'(x)v(x) – u(x)v'(x)}{[v(x)]^2}$$. visual curriculum You are watching: Expert Maths Tutoring in the UK. Info created by THVinhTuy selection and synthesis along with other related topics. Rate this post
SAT Practice Test 2, Section 3: Questions 11 - 15 11. Correct answer: (A) Given: The figure To find: The true statement Solution: Topic(s): supplementary angles x and y are supplementary angles x + y = 180. If x > 90 the y < 90 Answer: (A) y < 90 12. Correct answer: (D) Given: The line with equation y = 5x – 10 The line crosses the x-axis at point (a, b) To find: The value of a Solution: Topic(s): Coordinate geometry At the point where the line crosses the x-axis the value of y will be 0. Substitute y = 0, and x = a, into y = 5x – 10. 0 = 5a – 10 ⇒ 5a = 10 ⇒ a = 2 Answer: (D) 2 13. Correct answer: (E) Given: The noon temperature for 7 cities The median temperature is 40 To find: The temperature (t) that city D cannot be Solution: Topic(s): Statistics Write out the numbers in increasing order. 27 33 40 44 50 68 In order for 40 to be the median i.e. in the middle, t needs to on the left of 40. This means that t < 40. We are required to find the answer that does not fit the criteria. Only (E) 42 is greater than 40. Answer: (E) 42 14. Correct answer: (D) Given: The figure To find: Perimeter of the figure Solution: Topic(s): complementary angles, equilateral triangle, perimeter To get the perimeter of the figure, we need to know the lengths of PC and PD. Given the right angles and the lengths of the three sides, we can deduce that ABCD is a square. So, we know that the length of CD is 6. Angles ACP and PCD are complementary angles. So, angle PCD = 90º – angle ACP = 90º – 30º = 60º. In the same way, angles BDP and PDC are complementary angles. So, angle PDC = 90º – angle BDP = 90º – 30º = 60º. Triangle PCD has two angles that are 60º. When a triangle has two 60º angles, it must be an equilateral triangle that has three equal sides. Since the length of CD is 6. We can deduce that the lengths of PC and PD are also 6. Now, we can calculate the perimeter of the figure = 6 + 6 + 6 + 6 + 6 = 30 Answer: (D) 30 15. Correct answer: (C) Given: m is the greatest prime factor of 38 n is the greatest prime factor of 100 To find: m + n Solution: Topic(s): Factors The factors of 38 are: 1 × 38, 2 × 19. The greatest prime factor of 38 is 19 = m. The factors of 100 are: 1 × 100, 2 × 50, 4 × 24, 5 × 20, 10 ×10. The greatest prime factor of 100 is 5 = n. m + n = 19 + 5 = 24 Answer: (C) 24 Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations. You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page. [?] Subscribe To This Site
# We Try To Predict Princess Diana’s Future (02/15/2020) How will Princess Diana perform on 02/15/2020 and the days ahead? Let’s use astrology to conduct a simple analysis. Note this is of questionable accuracy – do not take this too seriously. I will first find the destiny number for Princess Diana, and then something similar to the life path number, which we will calculate for today (02/15/2020). By comparing the difference of these two numbers, we may have an indication of how smoothly their day will go, at least according to some astrology experts. PATH NUMBER FOR 02/15/2020: We will consider the month (02), the day (15) and the year (2020), turn each of these 3 numbers into 1 number, and add them together. Here’s how it works. First, for the month, we take the current month of 02 and add the digits together: 0 + 2 = 2 (super simple). Then do the day: from 15 we do 1 + 5 = 6. Now finally, the year of 2020: 2 + 0 + 2 + 0 = 4. Now we have our three numbers, which we can add together: 2 + 6 + 4 = 12. This still isn’t a single-digit number, so we will add its digits together again: 1 + 2 = 3. Now we have a single-digit number: 3 is the path number for 02/15/2020. DESTINY NUMBER FOR Princess Diana: The destiny number will consider the sum of all the letters in a name. Each letter is assigned a number per the below chart: So for Princess Diana we have the letters P (7), r (9), i (9), n (5), c (3), e (5), s (1), s (1), D (4), i (9), a (1), n (5) and a (1). Adding all of that up (yes, this can get tedious) gives 60. This still isn’t a single-digit number, so we will add its digits together again: 6 + 0 = 6. Now we have a single-digit number: 6 is the destiny number for Princess Diana. CONCLUSION: The difference between the path number for today (3) and destiny number for Princess Diana (6) is 3. That is lower than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A GOOD RESULT. But don’t get too excited yet! As mentioned earlier, this is just for fun. If you want really means something, check out your cosmic energy profile here. Go see what it says for you now – you may be absolutely amazed. It only takes 1 minute. ### Abigale Lormen Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene. #### Latest posts by Abigale Lormen (see all) Abigale Lormen Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
> > Various Forms of Equation of Line Various Forms of Equation of Line Have you ever thought of how some people can predict how much it will rain tomorrow or in the next week? What will be the price of some commodities in the next year? Think of it as some points of a straight line with the known slope. With the known quantities, we can calculate the unknown. In the previous section, we have studied some properties of lines. In this section, we will study the equation of line. Suggested Videos Basics of Straight Lines Introduction to Lines Pair of Straight Lines Various Forms of the Equation of Line A line is a set of all possible points lying on it. An equation of a line shows the relationship between the various points of the lines. Does this mean that any point can lie on the line? An equation of line sets the conditions for the points to satisfy. The equation of line is algebraic in nature consisting of variables x and y. By the equation of line, we can do some future estimation and can get to know about the types of lines. Vertical Line The equation of a vertical line at a distance vÂ from the y-axis is either x = v or x = âˆ’v. Sign depends upon the position of the line. Horizontal Line The equation of a horizontal line at a distance h from the x-axis is either y = h or y = âˆ’h. Sign depends upon the position of the line. Point-Slope Form This form of the equation gives the condition for a point to lie on the line. Suppose A (x0, y0) be a fixed point on the line l of slope m. Â Consider any other point, say B (x, y). The point B (x, y) will lie on the line l if and only if, its coordinates satisfy the equation m = (y âˆ’ y0) / (x âˆ’ x0) (from the definition of slope) i.e., y âˆ’ y0 = m (x âˆ’ x0) What if two points are given? Letâ€™s find out. Twoâ€“Points Form Imagine a line l passing through two points A (x1, y1) and B (x2, y2). Â Consider a point, say C (x, y). This point will satisfy the condition to lie on l if they satisfy the condition of collinearity. Slope of AC = Slope of BC = Slope of AC i.e., (y âˆ’ y1) / (x âˆ’ x1) = (y2 âˆ’ y1) / (x2 âˆ’ x1) or, (y âˆ’ y1) = (x âˆ’ x1) (y2 âˆ’ y1) / (x2 âˆ’ x1) is the required equation of the line. Slope-Intercept Form What is an intercept? A y-intercept is a distance at which the line intersects the y-axis. An x-intercept is a distance at which the line cuts the x-axis. Suppose a line l with slope m intersects the y-axis at a distance â€˜câ€™ from the origin. The equation of line l passing through the point (0, c) is given by: m = (y âˆ’ c) / (x âˆ’ 0) or, y = mx + c Consider a line l with slope m intersects the x-axis at a distance â€˜dâ€™ from the origin. The equation of line l passing through the point (d, 0) is given by: m = (y âˆ’ 0) / (x âˆ’ d) or, y = m(x âˆ’ d) If the line l intersects both the axes what will be the equation of the line? Intercept Form Suppose a line l with slope m intersect the y-axis at a distance â€˜câ€™ from the origin and x-axis at a distance of â€˜dâ€™ from the origin. The equation of line l passing through point (0, c) and (d, 0) is: (y âˆ’ y1) / (x âˆ’ x1) = (y2 âˆ’ y1) / (x2 âˆ’ x1) or, (y âˆ’ 0) / (x âˆ’ d) = (c âˆ’ 0) / (0 âˆ’ d) Solving we get, y/c + x/d = 1. Normal Form Suppose we have a line l in XY-plane. The perpendicular distance (normal) from origin to the line is known and the angle with the positive direction of the x-axis. Here, length of normal OA = p and it makes an angle Î± with the x-axis. The coordinates of A will be (p cosÎ±, p sinÎ±). OA is perpendicular to l. Slope of line l = âˆ’1 / slope of OA = âˆ’1 / tan Î±. Solving we have the required equation of the line, x cos Î± + y sin Î± = p. Till now, we have seen that the equations of the line are linear equations in x and y and can be generalized. General Equation of Line The general equation for any line is Ax + By + C = 0, where x and y are variables; A and B are not zero simultaneously. It is interesting to know that the general equation of a line canÂ be reduced into any form of the equations. Slopeâ€“Intercept Form If B â‰ 0, then Ax + By + C = 0 can be reduced to y = âˆ’ (A/B) x âˆ’ C/B (of the form y = mx + c). Here, the slope of the line, m = âˆ’ A/B, and y-intercept, c = âˆ’C/B. If B = 0, x =Â âˆ’C/A is the x-intercept. Intercept Form If C â‰ 0, then Ax + By + C = 0 can be reduced to (x/ (âˆ’ C/A)) + (y/ (âˆ’ C/B)) = 1 (of the form x/a + y/b = 1). Here, x-intercept = âˆ’C/A, and y-intercept = âˆ’C/B. If C = 0, Ax + By = 0Â â‡’ A = B = 0. It shows a line passing through origin. Normal Form The normal form of equation is x cosÎ± + y sinÎ± = p. Reducing it in the form of Ax + By + C = 0, we have cosÎ± = Â±A/(A2 + B2)1/2; sin Î± = Â±B/(A2 + B2)1/2 and p = Â±C /(A2 + B2)1/2. Solved Example for You Problem: Find the slope and the x and y-intercepts from the equation 4x + 5y âˆ’ 8 = 0. Solution: Given equation 4x + 5y âˆ’ 8 = 0 can be re-written as y = âˆ’ (4/5) x + 8/5. Comparing with y = mx + c, we have slope, m = âˆ’ 4/5, yâ€“intercept, c = 8/5. Comparing with y = m(x âˆ’ d), we have y = âˆ’ 4/5(x âˆ’ 2). So, x-intercept = 2. Share with friends Customize your course in 30 seconds Which class are you in? 5th 6th 7th 8th 9th 10th 11th 12th Get ready for all-new Live Classes! Now learn Live with India's best teachers. Join courses with the best schedule and enjoy fun and interactive classes. Ashhar Firdausi IIT Roorkee Biology Dr. Nazma Shaik VTU Chemistry Gaurav Tiwari APJAKTU Physics Get Started Browse Straight Lines 0 Followers Most reacted comment 1 Comment authors Recent comment authors Subscribe Notify of Guest Teju Hi Guest marno halo Question Mark? Have a doubt at 3 am? Our experts are available 24x7. Connect with a tutor instantly and get your concepts cleared in less than 3 steps.
Lesson Objectives • Learn how to solve problems using reciprocal identities • Learn how to find the signs of function values • Learn how to find the quadrant of an angle based on conditions • Learn how to find all function values given one value and the quadrant ## Reciprocal Identities, Signs of Function Values, & Finding Missing Function Values ### Identities In our algebra course, we learned that identities are equations that are always true. For example: $$3(x + 5)=3x + 15$$ The equation above is an example of an identity, we can replace x with any real number, and we will always get a true statement. This type of equation has a solution that is said to be "all real numbers" or the interval: $$(-\infty, \infty)$$ ### Reciprocal Identities The reciprocal of a nonzero number is found by flipping the number. In other words, the numerator becomes the denominator and the denominator becomes the numerator. $$\frac{3}{4}$$ The reciprocal of 3/4 is 4/3, we interchange the numerator and the denominator. $$\frac{4}{3}$$ In our last section, we introduced the trigonometric functions and found that some functions were reciprocals of each other. We know that we can't divide by zero, so keep in mind that anytime we have zero in the denominator, we write a result of undefined. $$r=\sqrt{x^2 + y^2}$$ $$\text{sin}\hspace{.25em}θ=\frac{y}{r}$$ $$\text{csc}\hspace{.25em}θ=\frac{r}{y}, y ≠ 0$$ $$\text{sin}\hspace{.25em}θ=\frac{1}{\text{csc}\hspace{.25em}θ}$$ $$\text{csc}\hspace{.25em}θ=\frac{1}{\text{sin}\hspace{.25em}θ}$$ We can see that sin θ and csc θ are reciprocals. $$\text{cos}\hspace{.25em}θ=\frac{x}{r}$$ $$\text{sec}\hspace{.25em}θ=\frac{r}{x}, x ≠ 0$$ $$\text{cos}\hspace{.25em}θ=\frac{1}{\text{sec}\hspace{.25em}θ}$$ $$\text{sec}\hspace{.25em}θ=\frac{1}{\text{cos}\hspace{.25em}θ}$$ Additionally, cos θ and sec θ are reciprocals. $$\text{tan}\hspace{.25em}θ=\frac{y}{x}, x ≠ 0$$ $$\text{cot}\hspace{.25em}θ=\frac{x}{y}, y ≠ 0$$ $$\text{tan}\hspace{.25em}θ=\frac{1}{\text{cot}\hspace{.25em}θ}$$ $$\text{cot}\hspace{.25em}θ=\frac{1}{\text{tan}\hspace{.25em}θ}$$ Lastly, tan θ and cot θ are reciprocals. Let's look at a few examples. Example #1: Find each function value. $$\text{cos}\hspace{.2em}θ$$ $$\text{sec}\hspace{.2em}θ=\frac{9}{2}$$ Since cos θ and sec θ are reciprocals, we can just flip our fraction to obtain our answer. $$\text{cos}\hspace{.2em}θ=\frac{2}{9}$$ Example #2: Find each function value. $$\text{csc}\hspace{.2em}θ$$ $$\text{sin}\hspace{.2em}θ=-\frac{\sqrt{6}}{3}$$ Since sin θ and csc θ are reciprocals, we can just flip our fraction to obtain our answer. Since this will lead to a radical in the denominator, we need to take the extra step and rationalize the denominator. $$\text{csc}\hspace{.2em}θ=-\frac{3}{\sqrt{6}}=-\frac{\sqrt{6}}{2}$$ ### Signs of Function Values When we defined the trigonometric functions, we gave r as the distance from the origin or the point (0,0) to the point (x,y). Since this distance is undirected, the value of r is always positive. When thinking about the values of our six trigonometric functions, we should remember that the value of x is positive in quadrants I and IV and negative in quadrants II and III. For y, the value is positive in quadrants I and II and negative in quadrants III and IV. Using these facts, we can produce a table for the signs of function values: Quadrant sin θ cos θ tan θ cot θ sec θ csc θ I++++++ II+----+ III--++-- IV-+--+- Some students use the phrase "All Students Take Calculus" to remember the signs. 1. All Functions Positive 2. Students (Sine and Cosecant) Positive 3. Take (Tangent and Cotangent) Positive 4. Calculus (Cosine and Secant) Positive Let's look at a few examples. Example #3: Determine the signs of the trigonometric functions of an angle in standard position with the given measure. $$110°$$ A 110° angle that is in standard position will lie in quadrant II. The x-values will be negative and the y-values will be positive. Using our chart above, we can see that sin θ, and csc θ are positive. The other four functions: cos θ, tan θ, cot θ, and sec θ are negative. Example #4: Determine the signs of the trigonometric functions of an angle in standard position with the given measure. $$-150°$$ A -150° angle that is in standard position will lie in quadrant III. The x-values and y-values will be negative. Using our chart above, we can see that tan θ and cot θ are positive. The other four functions: sin θ, cos θ, sec θ, and csc θ are negative. ### Identifying the Quadrant of an Angle In some cases, we may be asked to identify the quadrant or possible quadrants of an angle θ that satisfies the given conditions. Let's look at an example. Example #5: Identify the quadrant(s) of angle θ that satisfies the given conditions. cos θ > 0, csc θ < 0 If we consult our chart above, we can see that cos θ is positive in quadrants I and IV, however, csc θ is negative in quadrants III and IV. Since quadrant IV meets both conditions, we can state our answer as quadrant IV. ### Finding all Trigonometric Function Values Given One Value and the Quadrant In some cases, we may be asked to find all six trigonometric function values given one function value and the quadrant. Let's look at an example. Example #6: Use the given information to find the values of all six trigonometric functions. $$\text{sec}\hspace{.25em}θ=\frac{19}{11}$$ $$\text{sin}\hspace{.25em}θ < 0$$ We know that sin θ is negative in quadrants III and IV. sec θ is given as 19/11, which is positive. sec θ is only positive in quadrants I and IV. Therefore, our angle is in quadrant IV since it meets both conditions. This tells us that our x-values will be positive and our y-values will be negative. Let's think about sec θ $$\text{sec}\hspace{.25em}θ=\frac{r}{x}$$ Let's substitute to find our unknown value for y. $$\text{sec}\hspace{.25em}θ=\frac{r}{x}=\frac{19}{11}$$ $$r=19$$ $$x=11$$ $$r=\sqrt{x^2 + y^2}$$ $$19=\sqrt{11^2 + y^2}$$ Let's square both sides: $$19^2=\left(\sqrt{121 + y^2}\right)^2$$ $$361=121 + y^2$$ $$361 - 121=y^2$$ $$y^2=240$$ $$y=\pm 4\sqrt{15}$$ Since we know that y is negative in quadrant IV, we can state that y is: $$y=-4\sqrt{15}$$ Now that we know the values for x, y, and r, we can find the values for all six trigonometric functions: $$\text{sin}\hspace{.25em}θ=-\frac{4\sqrt{15}}{19}$$ $$\text{cos}\hspace{.25em}θ=\frac{11}{19}$$ $$\text{tan}\hspace{.25em}θ=-\frac{4\sqrt{15}}{11}$$ $$\text{csc}\hspace{.25em}θ=-\frac{19\sqrt{15}}{60}$$ $$\text{cot}\hspace{.25em}θ=-\frac{11\sqrt{15}}{60}$$ $$\text{sec}\hspace{.25em}θ=\frac{19}{11}$$ #### Skills Check: Example #1 Find tan θ $$\text{sin}\hspace{.2em}θ=\frac{3}{5}$$ $$\text{cos}\hspace{.2em}θ > 0$$ A $$\text{tan}\hspace{.2em}θ=\frac{4}{3}$$ B $$\text{tan}\hspace{.2em}θ=\frac{3}{4}$$ C $$\text{tan}\hspace{.2em}θ=-\frac{3}{4}$$ D $$\text{tan}\hspace{.2em}θ=\frac{6}{7}$$ E $$\text{tan}\hspace{.2em}θ=-\frac{\sqrt{3}}{7}$$ Example #2 Find cot θ $$\text{sin}\hspace{.2em}θ=-\frac{2\sqrt{5}}{5}$$ $$\text{cos}\hspace{.2em}θ < 0$$ A $$\text{cot}\hspace{.2em}θ=\frac{1}{2}$$ B $$\text{cot}\hspace{.2em}θ=\frac{2}{3}$$ C $$\text{cot}\hspace{.2em}θ=-2$$ D $$\text{cot}\hspace{.2em}θ=\frac{\sqrt{5}}{2}$$ E $$\text{cot}\hspace{.2em}θ=-\frac{\sqrt{5}}{10}$$ Example #3 Find sin θ $$\text{cot}\hspace{.2em}θ=-\frac{4}{3}$$ $$\text{cos}\hspace{.2em}θ < 0$$ A $$\text{sin}\hspace{.2em}θ=-4$$ B $$\text{sin}\hspace{.2em}θ=\frac{3}{4}$$ C $$\text{sin}\hspace{.2em}θ=-\frac{\sqrt{3}}{4}$$ D $$\text{sin}\hspace{.2em}θ=\frac{3}{5}$$ E $$\text{sin}\hspace{.2em}θ=-\frac{5}{4}$$
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> You are viewing an older version of this Concept. Go to the latest version. # Zero Product Principle ## The zero product principle states that anything multiplied by zero is zero. Estimated10 minsto complete % Progress Practice Zero Product Principle Progress Estimated10 minsto complete % Zero Product Property for Quadratic Equations The area of a particular rectangle was found to be . Determine the dimensions of the rectangle if the area was known to be 7 units. ### Guidance Recall that when solving an equation, you are trying to determine the values of the variable that make the equation true. For the equation , and are both solutions. You can check this: Here you will focus on solving quadratic equations. One of the methods for quadratic equations utilizes your factoring skills and a property called the zero product property. If , what can you say about or ? What you should realize is that either or have to be equal to 0, because that is the only way that their product will be 0. If both and were non-zero, then their product would have to be non-zero. This is the idea of the zero product property. The zero product property states that if the product of two quantities is zero, then one or both of the quantities must be zero. The zero product property has to do with products being equal to zero. When you factor, you turn a quadratic expression into a product. If you have a quadratic expression equal to zero, you can factor it and then use the zero product property to solve. So, if you were given the equation , first you would want to turn the quadratic expression into a product by factoring it: You can rewrite the equation you are trying to solve as . Now, you have the product of two binomials equal to zero. This means at least one of those binomials must be equal to zero. So, you have two mini-equations that you can solve to find the values of that cause each binomial to be equal to zero. • , which means OR • , which means The two solutions to the equation are and . Keep in mind that you can only use the zero product property if your equation is set equal to zero! If you have an equation not set equal to zero, first rewrite it so that it is set equal to zero. Then factor and use the zero product property. #### Example A Solve for : . Solution: First, change into a product so that you can use the zero product property. Change the expression into a product by factoring: Next, rewrite the equation you are trying to solve: becomes . Finally, set up two mini-equations to solve in order to find the values of that cause each binomial to be equal to zero. • , which means that • , which means that The solutions are or . #### Example B Solve for : . In order to solve for you need to factor the polynomial. Solution: First, change into a product so that you can use the zero product property. Change the expression into a product by factoring: Next, rewrite the equation you are trying to solve: becomes . Finally, set up two mini-equations to solve in order to find the values of that cause each binomial to be equal to zero. • , which means that • , which means that The solutions are or . #### Example C Solve for : . Solution: First, change into a product so that you can use the zero product property. Change the expression into a product by factoring: Next, rewrite the equation you are trying to solve: becomes . Finally, set up two mini-equations to solve in order to find the values of that cause each binomial to be equal to zero. • , which means that • , which means that The solutions are or . #### Concept Problem Revisited The area of a particular rectangle was found to be . Determine the dimensions of the rectangle if the area was known to be 7 units. In other words, you are being asked to solve the problem: You can solve this problem by factoring and using the zero product property. becomes Since you are asked for dimensions, a width of –5 units does not make sense. Therefore for the rectangle, the width would be 13 units. ### Vocabulary A quadratic equation is an equation in which the highest power of a variable is 2. Standard form for a quadratic equation is . Zero-product property The zero-product property states that if two factors are multiplied together and their product is zero, then one of the factors must equal zero ### Guided Practice 1. Solve for the variable in the polynomial: 2. Solve for the variable in the polynomial: 3. Solve for the variable in the polynomial: 1. 2. 3. ### Practice Solve for the variable in each of the following equations. ### Vocabulary Language: English Zero Product Rule Zero Product Rule The zero product rule states that if the product of two expressions is equal to zero, then at least one of the original expressions much be equal to zero.
# Perimeter Of Triangle - Definition, Formula, Examples Safalta expert Published by: Yashaswi More Updated Wed, 06 Sep 2023 04:01 PM IST ## Highlights Check out how to find out the perimeter of a triangle- definition, formula, examples here at Safalta.com ### Free Demo Classes The perimeter of the Triangle: The perimeter of any two-dimensional figure depends on the area around it. Any closed shape's perimeter can be calculated by adding the lengths of its sides. In this tutorial, you will first learn what the perimeter is and how to calculate it for different kinds of triangles with known side lengths. You'll be able to view the subject from more angles thanks to the solved instances. JoiSafalta School Online and prepare for Board Exams under the guidance of our expert faculty. Our online school aims to help students prepare for Board Exams by ensuring that students have conceptual clarity in all the subjects and can score their maximum in the exams. Enhance your skills with Safalta’s Master Digital Marketing Program ## What is the Perimeter of a Triangle? The sum of the lengths of the sides is the perimeter of any polygon. Source: Safalta.com In the case of a triangle- Perimeter = Sum of the three sides Always include units in the final answer. If the sides of the triangle are measured in centimeters, then the final answer should also be in centimeters. ## Perimeter of Triangle Formula The formula for the perimeter of a closed-shaped figure is usually equal to the length of the outer line of the figure. Therefore, in the case of a triangle, the perimeter will be the sum of all three sides. If a triangle has three sides a, b, and c, then, Perimeter = a + b + c Where "a" is the length of one side of the triangle, "b" is the length of another side, and "c" is the length of the third side. For example, if the lengths of the sides of a triangle are 4, 5, and 6, the perimeter would be: Perimeter = 4 + 5 + 6 = 15 So the perimeter of the triangle is 15. Note that the perimeter of a triangle is a measure of the distance around the outside of the triangle, not the area inside the triangle. To find the area of a triangle, you can use a different formula, such as the formula for the area of a triangle using base and height, or the formula for the area of a triangle using the lengths of all three sides (also known as Heron's formula). Read Also: Digital Marketing Career Objectives: Achieving Success in the Dynamic Digital Landscape ## Perimeter of an Isosceles, Equilateral and Scalene Triangle Where a, b, c, and l denote the side lengths and P denotes the perimeter. To find the perimeter of a triangle, add the lengths of all three of its sides together. If A, B, and C side measurements and X is the perimeter, then A+B+C= X ### Perimeter of Right Triangle A right triangle has a base(b), hypotenuse(h), and perpendicular(p) as its sides, By the Pythagoras theorem, we know, h square = b square + p square Therefore, the Perimeter of a right angle triangle= b + p + h ## Examples Let us consider some examples of the perimeter of a triangle: Example 1: Find the perimeter of a polygon whose sides are 5 cm, 4 cm, and 2 cm. Solution: Let, a = 5 cm b = 4 cm c = 2 cm Perimeter = Sum of all sides = a + b + c = 5 + 4 + 2 = 11 Therefore, the answer is 11 cm. Example 2: Find the perimeter of a triangle whose each side is 10 cm. Solution: Since all three sides are equal in length, the triangle is equilateral. i.e. a = b = c = 10 cm Perimeter = a + b + c = 10 + 10 + 10 = 30 Perimeter = 30 cm. Example 3: What is the missing side length of a triangle whose perimeter is 40 cm and two sides are 10 cm each? Solution: Given, Perimeter = 40 cm The length of the two sides is the same i.e. 10 cm. Thus, the triangle is isosceles. Using formula: P = 2l + b 40 = 2 * 10 + b 40 = 20 + b or b = 20 The missing side length is 20 cm. ## What does the Perimeter of a Triangle Mean? The perimeter of a triangle is the total distance around the edges of a triangle. In other words, the length of the boundary of a triangle is its perimeter. ## How to Calculate the Perimeter of a Triangle? To calculate the perimeter of a triangle, add the length of its sides. For example, if a triangle has sides a, b, and c, then the perimeter of that triangle will be P = a + b + c. ## Calculate the Perimeter of a Right Triangle with Base as 3 cm and height as 4 cm. First, using the Pythagorean theorem, calculate the hypotenuse of the right triangle. h =√(base2+perpendicular2) h = √(32+42) h = √(9 + 16) h = √25 Or, h = 5 cm So, the perimeter of the triangle = 3 + 4 + 5 = 12 cm. ## What is the formula of area and perimeter of triangle? The area of a triangle can be calculated with the help of the formula: A = 1/2 (b × h). The perimeter of a triangle can be calculated by adding the lengths of all the three sides of the triangle. ## Does a triangle have a perimeter? The perimeter of a triangle is simply the sum of its three sides. ## Start Learning & Earning ### Trending Courses ##### Professional Certification Programme in Digital Marketing (Batch-7) Now at just ₹ 49999 ₹ 9999950% off ##### Master Certification in Digital Marketing Programme (Batch-13) Now at just ₹ 64999 ₹ 12500048% off ##### Advanced Certification in Digital Marketing Online Programme (Batch-24) Now at just ₹ 24999 ₹ 3599931% off ##### Advance Graphic Designing Course (Batch-10) : 100 Hours of Learning Now at just ₹ 19999 ₹ 3599944% off ##### Flipkart Hot Selling Course in 2024 Now at just ₹ 10000 ₹ 3000067% off ##### Advanced Certification in Digital Marketing Classroom Programme (Batch-3) Now at just ₹ 29999 ₹ 9999970% off ##### Basic Digital Marketing Course (Batch-24): 50 Hours Live+ Recorded Classes! Now at just ₹ 1499 ₹ 999985% off
Home Education Fraction To percentage calculator Soup University # Fraction To percentage calculator Soup University 0 A fraction is a type of a mathematical function having a denominator and we can convert the fraction into a percentage. The fraction is the meaning of various percentages. The multiple fraction calculator compares quantities by the fraction and percentage. The percentage can be determined by dividing the fraction by the dividing fractions calculator and multiplying the fraction by the 100. You can utilize the various tools and apps from the calculator-online.net to find the multiplication and division of the fraction. The multiple fraction calculator make the task easy for the students and they can easily able to convert the fraction into the percentage by the fraction calculator.Students do find the addition and subtraction of the fraction difficult , for this we usually utilize the Least common multiple method to make the denominator equal and then add or subtract the fractions. Table of Contents Toggle ## How to understand the fraction and percentage? We now try to understand the concept of the fractions and the percentage, we are explaining the concept by an example, considering a class has 38 students among them there are 23 female.Now how can we get the percentage of the female students in the class? Now to simply the fraction =23/38 =0.60 =0.6*100 =60% The fraction and percentage goes neck to neck and when we are able to predict there is 60 % of students in the class and by dividing the 23/38. ## What is a fraction? The term fraction represents a number which is dividing the whole number in different parts. We can divide the whole number into various parts, for example if there is one Pizza, we divide four equal parts.Then we are using the adding fractions calculator.We can divide the fraction into ¼+¼+¼+¼=1/1=1Pizza. ### Parts of fraction: A fraction consists of two parts the Numerator and a Denominator.The Numerator is written above the line and the Denominator is written below the line. #### The Numerator: The numerator indicates some equal parts of the whole number #### The Denominator: The Denominator represents how many parts the whole number consists of. ¾.⅚,6/7,5/3 Her 3,5,6,5 are the Numerator and the denominator 4,6,7,3. ## What is percent? The term percent is ratio ora number to express a fraction in the form of 100.We utilize the %age sign,we need to understand how perfect represent a frasction.You may use the fraction calculator to represent the 35/100, or 50/100 as the percent.We can write the these fractions as 35%, and 50 %. There are other fractions which represents the a specific percent of the number like These fractions have specific meaning in terms of percentage and we have utilized changeable manners for a fraction. ## The percentage formula: The percentage is number whose denominator is equal to 100.The percentage formula is given below Example: If you want to find the 10 % of 150, we can solve it as: (10/ 100) × 150 = 1500/ 100 = 15 Thus, 10% of 150 is 15. The multiple fraction calculator can be utilized to find the fraction of the percentage. ## Conclusion: The fractions are going to be used as the percentage for descending various decimal values like 50 %,75% etc by ½, ¾. These fractions are going to describe the whole meaning and we can understand what is the meaning of our conversi9o9n and what we want it to represent.
Select Page Earlier I had written an article on the Origin and History of Modern Mathematics, followed up by a simple lesson on instant Multiplication using Vedic Mathematics. In this article we will learn about yet another simple but powerful Sutra (formula) of Vedic Mathematics called Ekadhikena Purvena in Sanskrit means “One more than the previous“. So in mathematical calculations involving this formula, we generally consider the given number and another number whose value is one greater than the given number. ## Ekadhikena Purvena in Multiplication Ekadhikena Purvena sutra can be used to quickly multiply any two numbers of two digits (say AB and AC) whose 1. First digits are the same and (A=A) 2. The second digits add up to 10 (B+C=10) For example to multiply 47 x 43, here is how we go about it. 1. Take the first digit which is common in both numbers, in the above case 4 2. Apply Ekadhikena to it – i.e. increment it by one. So it becomes 5. 3. Now write down the product of 5×4 which is 20. 4. Next write down the product of the other two digits which is 7×3=21 5. The answer is 2021 How simple isnt it? You dont even require a pencil and paper, nor do you require to do any carrying over of remainders. Let us consider another example. To multiply 51×59. The answer is 1. 5 x (5+1) = 5×6 =30 2. 1×9=9 3. For the second part we need to round it to two digits in case it is just a single digit, hence it becomes 09 4. So the answer is 3009 ## Ekadhikena Purvena in Finding Squares The above method can also easily be implemented to find the squares of numbers ending with five, because the sum of their last two digits is 5+5 which is 10, which hence meets the requirement of the formula. So to find 75×75, all we need to do is 7×8,5×5 which is 5625. Instant calculation, isn’t it? ## Ekadhikena Purvena in Converting Fractions to Decimals How easy it is to calculate the recurring decimal of fractions like 1/19 whose denominators end with 9? Using Ekadhikena Purvena this calculation becomes as simple as… well, seeing is believing here Consider 1/19 1. Take 1 which is the “Purvena” (The digit previous to 9). 2. Increase it by one which makes it 2 3. Now start with 1 and go on multiplying it by 2 as follows: • 1 • 21 which is [2×1,1] • 421 which is [2×2,21] • 8421 which is [2×4,421] • 68421 which is [2×8,8421]. We now have one carrying since 2×8=16 • 368421 which is [(2×6)+1,68421]. We now have one carrying since (2×6)+1=13 • 7368421 which is [(2×3)+1,368421] • 47368421 which is [(7×2),7368421]. We now have one carrying since (2×7)=14 • 947368421 which is [(4×2)+1,47368421] Ok we should now stop here. Why? Well, we have figured out half of the digits in the answer. And finding the remaining half is even more simpler. But before that, how did we find that we were done with half of the answer? The rule is, take the denomination and remove 1 from it. So 19-1=18. Now this gives us the total number of digits in the complete answer for 1/19. And since we have now found 9 digits we are half way through the answer. For the remaining half simply find the inverse of each digit in the so far discovered half. Inverse of a digit is that digit which makes the total of the two 9. So inverse of 1 is 8, 2 is 7 and so on till 9 is 0. The inverse of 947368421 hence is 052631578. Ok now that we have both the halves of the answer. The complete answer is just stick the two together after a decimal, which is 0.052631578947368421 Simple isn’t it? If you dont find this simple, then try calculating 1/19 using the traditional method taught in the schools :) Even the normal calculators will not give you such an accurate decimal representation of these fractions, and most importantly with little practice you can simply go on writing down the digits in the answer from right to left without the need of any additional worksheet. Now one might wonder the need to remember so many different formula for different calculations. Just imagine what if you were taught these kind of calculations right from your school days. Well, expertise is after all all about getting used what we regulary do. Its not as if in the schools we were taught one or two methods of doing calculations. We as students, even otherwise had to learn numerous methods of computations. So if learning these vedic mathematical lessons is going to help us do lightning fast calculations without using calculators, then why not?
# You might be unacquainted in what’s squared in math Square root is a theory that’s truly puzzling to many pupils. Not knowing the way that it is used, and exactly what it works, could cause trouble. In this article we will go over what’s utopian in math. Square root might be understood together with the assistance of a graph. So the first thing you need to do is http://133.130.99.98/2020/03/03/whats-tolerance-in-x-y/ draw a chart. You will also have to quantify the graph, that is, the distance between your x-axis and the y-axis. The diagram of the graph that you attracted must seem like a tree, even with branches and a trunk all reaching to the tips of the graph. The distance from the trunk towards the tip of the branch would be that the x axis. The length from the division into this trunk’s tip would be your y axis. We wish to spend the Xaxis and multiply it multiply by the elevation of this branch and multiply from the branch’s depth. Therefore you get the graph. We click this over here now will now multiply the points each in order to find the sq of every single level that you simply chose. If you’re currently utilizing just only one point per line then you’d multiply every one of the things together with the x-axis. Square root will probably beat the ending of the line in the graph. Now the graph multiplied by the depth of this division, multiplied by the elevation of the division and that you just simply drew is multiplied by the radius of the back and the y axis, which can be. This really is where the root is now found. You should use it in math for school level calculations Although the chart looks like a tree. But the problem is that you have to get the angle involving also the x axis and the yaxis, in other words, between the elevation of this division and the x axis. In order to obtain this, you will need to make use of a angle calculator to come across the angle between the x axis and the y axis. I know you could well be asking, what’s squared http://paramountessays.com/ in mathematics and what’s the square root? And I will let you know and explain in detail. Let me show you that you really need to know the difference between these two things before moving further. X y is a kind of head science which can be implemented to distinctive places of life. There are so many types of mathematics, that folks become confused about the gaps between them. Thus when studying mathematics, it is necessary that you know what is that you are mastering, therefore you don’t waste time understanding about a subject which you’re perhaps not knowledgeable about. You should understand until you move on into the harder course of calculus what is squared in math. That has been about what is squared in mathematics a fascination lesson. You have to realize that it is a fantastic idea to know the two of these matters before continuing to calculus. Some of the causes that math is so important is because it’s used the time all in schools, businesses, properties, and also anyplace else. It’s quite tough to foresee whatever will occur later on, and therefore you ought to get prepared. Mainly because they will need to understand what is within their long run A great deal of organization can be employed mathematics. And it is possible to get sure that you realize the distinction between the two. Before you move ahead to some one of the things that you should know about the topic of root, let’s talk about what exactly is squared in mathematics. The next moment we’ll examine exactly what may be the square root and what is squared in math.
# Binary Number System Learn about the binary number system in this lesson. We'll cover the following ## Overview The decimal number system, which is used in daily life, consists of ten numerals: 0123456789. In contrast, the binary number system used by computer hardware consists of two numerals: 0 and 1. This is a direct consequence of a bit consisting of two values. If bits had three values, then the computers would use a number system based on three numerals. The digits of the decimal system are named incrementally as ones, tens, hundreds, thousands, etc. For example, the number 1023 can be expanded in the following way: 1023 == 1 count of thousand, no hundred, 2 counts of ten, and 3 counts of one Naturally, moving one digit to the left multiplies the value of that digit by 10: 1, 10, 100, 1000, etc. When the same rules are applied to a system that has two numerals, we arrive at the binary number system. The digits are named incrementally as ones, twos, fours, eights, etc. In other words, moving one digit to the left would multiply the value of that digit by 2: 1, 2, 4, 8, etc. For example, the binary number 1011 can be expanded as in the following way: 1011 == 1 count of eight, no four, 1 count of two, and 1 count of one To make it easy to refer to digits, they are numbered from the rightmost digit to the leftmost digit, starting with 0. The following table lists the values of all of the digits of a 32-bit unsigned number in the binary system: Digit Value 31 2,147,483,648 30 1,073,741,824 29 536,870,912 28 268,435,456 27 134,217,728 26 67,108,864 25 33,554,432 24 16,777,216 23 8,388,608 22 4,194,304 21 2,097,152 20 1,048,576 19 524,288 18 262,144 17 131,072 16 65,536 15 32,768 14 16,384 13 8,192 12 4,096 11 2,048 10 1,024 9 512 8 256 7 128 6 64 5 32 4 16 3 8 2 4 1 2 0 1 Get hands-on with 1200+ tech skills courses.
Update all PDFs # Families of Triangles Alignments to Content Standards: 6.EE.C.9 6.EE.A.2.c There are a bunch of triangles. They all have one side that is 10 centimeters long, which we will consider as the base of the triangle. 1. The triangles each have a different height (as measured off of the 10-centimeter base) and so have different areas. Fill in the table: Height (centimeters) Area (square centimeters) 20 25 40 250 2. Plot the ordered pairs from the table in the coordinate plane and label them with their coordinates. 3. Where can you see the answers to part (a) in the coordinate plane? 4. If $A$ represents the area and $h$ represents the corresponding height, write an equation using $A$ and $h$ that represents the area of any such triangle. ## IM Commentary The purpose of this task is to introduce students to the idea of a relationship between two quantities by using a familiar geometic context. In order to benefit from this task, students should have already developed and become comfortable with a formula for the area of a triangle. The focus of this task should be on noticing the relationship between height and area and creating a graphical and algebraic representation of this relationship, not on understanding the meaning behind the geometric terms. Even if students have done this work with area, they may need help interpreting the task. For example, the area formula $A=\tfrac{1}{2}bh$ requires one of the sides to be chosen as the "base", relative to which there is a corresponding height. The task defines the given 10-centimeter-long side to be the base -- a diagram of a triangle where one side is labeled 10 and the corresponding height is drawn and perhaps labeled with a $?$ or with an $h$ might help. Students should have learned to graph ordered pairs in the first quadrant in grade 5, but since this is a recent skill, teachers should take care to ensure that students are plotting and labeling points correctly. The task doesn't require it, but this task is an opportunity to introduce or reinforce the language of independent variable (in this case height) and dependent variable (in this case area). In part (d), students engage in MP8 (express regularity in repeated reasoning) when they think about all of the arithmetic they did in part (a) and express that arithmetic using an $h$ in place of a numerical height. This is a skill at the heart of the arithmetic to algebra transition that takes place in grade 6, and many students may find it difficult. One way to scaffold this work might be to ask students to write out in words the arithmetic process they used several times in part (a): "I started with the height, multiplied it by 10, and then divided the result by 2." The work of part (d) is expressing that process using a letter for the height. ## Solution 1. The completed table: Height (cm) Area (square cm) 20 100 25 125 40 200 50 250 2. The plotted ordered pairs: 3. The heights are the $x$-coordinates of the points on the graph and the areas are the $y$-coordinates of the points on the graph. 4. $A=\frac{10h}{2}$, $A=\frac12 \cdot 10h$, $A=5h$, or equivalent. #### Ashli says: You should have the icon now. Thanks for letting us know! over 2 years
How to convert decimal fractions to fractions Converting decimal fractions to fractions is very simple. Do you want to learn? Read on! Steps Method 1 of 2: If the decimal is interrupted Step 1. Write down the decimal fraction If the decimal fraction is finite, then it ends one or more decimal places. Let's say we are working with a finite fraction of 0.325. Let's write it down. Step 2. Let's convert the decimal fraction to a common one To do this, count the number of decimal places. In our case, there are three digits in the number 0.325. Let's just write the number "325" over the number 1000, which is 1 followed by three zeros. If we were dealing with the number 0.3, with one decimal place, then we would write it as 3/10, or three above, and one with the number of zeros equal to the number of decimal places below. You can also say the decimal point out loud. In our case, we get 0.325 = "0 whole and 325 thousandths." Sounds like a regular fraction, doesn't it? We write 0.325 = 325/1000 Step 3. Find the greatest common factor of the numerator and denominator of the new fraction This is how ordinary fractions are simplified. Find the largest number by which both the numerator and the denominator are divisible without a remainder. In our case, this number is 25. • You don't need to find the greatest common factor right away. You can simplify the fraction and gradually. For example, if we are dealing with two even numbers, we can divide them by 2 until one of them becomes odd or until we simplify to the end. If we are dealing with an even and an odd number, we can try to divide by 3. • If we are dealing with a number ending in 0 or 5, we will divide by 5. Step 4. Divide both numbers by the greatest common factor Divide 325 by 25, we get 13.1000 by 25 = 40. The simplified fraction is 13/40. So 0.325 = 13/40. Method 2 of 2: If the decimal is periodic Step 1. Write down the fraction In a periodic decimal fraction, certain numerical combinations are repeated, it is infinite. For example - 2.345454545. In this case, you need to find x. Write x = 2.345454545. Step 2. Multiply the number by a power of ten, which would move the non-repeating part of the decimal fraction to the left of the decimal point In this case, the first degree 10 is enough for us, we write "10x = 23.45454545…." Why do this? If we multiply the right side of the equation by 10, then the left side must also be multiplied. Step 3. Multiply the equation by another power of 10 to move more digits to the left of the decimal point For example, let's multiply the decimal fraction by 1000. Let's write, "1000x = 2345.45454545…." This must be done because since we are multiplying the right side of the equation by 10, then the left side must also be multiplied. Step 4. Let's write a variable and a constant value on top of each other for subtraction Now let's write the second equation above the first so that 1000x = 2345.45454545 is above 10x = 23.45454545, as it would be with a normal subtraction. Step 5. Subtract Subtract 10x from 1000x to get 990x. Then we subtract 23.45454545 from 2345.45454545, we get 2322. We get 990x = 2322. Step 6. Find x We know that 990x = 2322, and "x" can be found by dividing both sides by 990. So x = 2322/990. Step 7. Simplify the common fraction Divide the numerator and denominator by the common factor. Find the greatest common factor and simplify the fraction completely. In our example, the greatest common divisor of 2322 and 990 is 18, so we divide the numerator and denominator by 18. We get 990/18 = 129 and 2322/18 = 129/55. So 2322/990 = 129/55. Ready!
# Exponent Laws and Logarithmic Properties There are several rules that are helpful when working with exponential functions. ## Law of Exponents: $$a^xa^y = a^{x+y}$$ $$(a^x)^y = a^{xy}$$ $$\left(\cfrac{a}{b}\right)^x = \cfrac{a^x}{b^x}$$ $$\cfrac{a^x}{a^y} = a^{x-y}$$ $$(ab)^x = a^xb^x$$ The first law states that to multiply two exponential functions with the same base, we simply add the exponents. The second law states that to divide two exponential functions with the same base, we subtract the exponents. The third law states that in order to raise a power to a new power, we multiply the exponents. The fourth and fifth laws state that in order to raise a product or quotient to a power, we raise each factor to that power. Example: Simplify the expression: $$\cfrac{7^{3x}\cdot7^{x+1}}{7^{2x+5}}$$ $$=7^{3x+(x+1)-(2x+5)}$$ $$=7^{3x+x+1-2x-5}$$ $$=7^{2x-4}$$ Example: Simplify the expression: $$\left(\cfrac{3^{4x}}{2^x}\right)^3$$ $$=\cfrac{(3^{4x})^3}{(2^x)^3}$$ $$=\cfrac{3^{12x}}{2^{3x}}$$ Note: We cannot simplify any further since the terms in the numerator and denominator do not have the same base. ## Log Laws There are three properties that are useful when working with logarithmic functions. ### Properties of Logarithms If x, y > 0 and r is any real number, then loga(xy) = loga x + logay loga(x/y) = loga x - logay loga(xr) = rlogax Example: Simplify the expression log25 + log23 Solution: log25 + log23 = log2(5 x 3) = log2(15) Example: Simplify the expression 5(In2) 5(In2) = In(25) = In(32) Exponent Laws Example:
# Circle TGT-Maths à HGI School 8 Jul 2020 1 sur 14 ### Circle • 3. Teaching Points are - 1. CIRCLE 2. PARTS OF A CIRCLE  Centre  circumference  Radius  Diameter  Chord  Arc  Semicircle 3. RELATION B/W DIAMETER & RADIUS • 4. What is an Circle? The word circle comes from the Latin word “ CIRCULUS” which means disc. A circle is set of points in a plane that are equidistant from a given fixed point. The fixed point is called the centre of the circle. A circle has an interior as well as exterior region. • 5. Parts of a circle Centre The point at which we place the needle end of the compass and move the pencil around it is the centre of the circle. • 6. Circumference The length of the boundary of the circle is the circumference of the circle. In other words, it is the perimeter of the circle. • 7. Radius The line segment joining the centre to any point on the circle. * Infinite number of radius can be drawn in a circle. * All the radii of a circle are equal in length. • 8. Diameter A diameter of a circle is line segment that joins any two points on the circle and passes through its centre. Infinite number of diameters can be drawn in a circle and all are equal in length. • 9. Chord A line segment that joins any two points on the circle is called chord. Infinite number of chords can be drawn. Diameter is the longest chord of a circle. • 10. Arc An arc of a circle is a part of circle included between two distinct points of the circle. The larger arc is called MAJOR ARC and the smaller arc is called MINOR ARC. • 11. Semicircle The diameter of a circle divide it into two equal halves and each half is called semicircle. • 12. Concentric Circles Two or more circles in same plane are called concentric circles if they have same centre. • 13. Relation Between Radius and Diameter • 14. Thank You
# Category: Sequence Theorems ## Squeeze (Sandwich) Theorem for Sequences Consider two sequences ${x}_{{n}}$ and ${y}_{{n}}$. When we write that ${x}_{{n}}={y}_{{n}}$ we mean that corresponding values are equal, i.e. ${x}_{{1}}={y}_{{1}}$, ${x}_{{2}}={y}_{{2}}$ etc. ## Algebra of Limit of Sequence Consider two sequences ${x}_{{n}}$ and ${y}_{{n}}$. When we talk about sum of these sequences, we talk about sequence ${x}_{{n}}+{y}_{{n}}$, whose elements are ${x}_{{1}}+{y}_{{1}},{x}_{{2}}+{y}_{{2}},{x}_{{3}}+{y}_{{3}}\ldots.$. Same can be said about other arithmetic operations. In other words sum of sequences is sequence with elements that are sum of corresponding elements of initial two sequences. ## Indeterminate Form for Sequence When we described arithmetic operations on limits, we made assumption that sequences approach finite limits. Now, let's consider case when limits are infinite or, in the case of quotient, limit of denominator equals 0. ## Stolz Theorem To find limits of indeterminate expressions $\frac{{{x}_{{n}}}}{{{y}_{{n}}}}$ of type $\frac{{\infty}}{{\infty}}$ often can be useful following theorem. Stolz Theorem. Suppose that sequence ${y}_{{n}}\to+\infty$ and starting from some number with increasing of ${n}$ also increases ${y}_{{n}}$ (in other words if ${m}>{n}$ then ${y}_{{m}}>{y}_{{n}}$). Then $\lim\frac{{{x}_{{n}}}}{{{y}_{{n}}}}=\lim\frac{{{x}_{{n}}-{x}_{{{n}-{1}}}}}{{{y}_{{n}}-{y}_{{{n}-{1}}}}}$ if limit of the expression on the right side exists (finite or infinite).
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> You are viewing an older version of this Concept. Go to the latest version. # Normal Distribution Properties ## Standard normal distribution with mean of zero and standard deviation of 1. Estimated9 minsto complete % Progress Practice Normal Distribution Properties Progress Estimated9 minsto complete % The Normal Curve When students ask their teachers to curve exams, what they often mean is they want everyone to simply get a higher grade. Curving a grade can also mean fitting to a bell curve where lots of people get Cs, some people get Ds and Bs and very few people get As and Fs. Even though this second interpretation is not what most students mean, the normal curve is one of the most widely used and applied probability distributions. What are other examples that follow a normal distribution? #### Guidance The Standard Normal Distribution is graphed from the following function and is represented by the Greek letter phi, \begin{align}\phi\end{align}. \begin{align}\phi(x)=\frac{1}{\sqrt{2 \pi}}e^{- \frac{1}{2}x^2}\end{align} This distribution represents a population with a mean of 0 and a standard deviation of 1. The numbers along the \begin{align}x\end{align}-axis represent standard deviations. For data that is normally distributed, the empirical rule states that: • Approximately 68% of the data will be within 1 standard deviation of the mean. • Approximately 95% of the data will be within 2 standard deviations of the mean. • Approximately 99.7% of the data will be within 3 standard deviations of the mean. Some other important points about the normal distribution: • The total area between the normal curve and the \begin{align}x\end{align} axis is 1 and this area represents all possible probabilities. • If data is distributed normally, you can use the normal distribution to determine the percentage of the data between any two values by calculating the area under the curve between those two values. When you take calculus, you will learn how to calculate this area analytically, but for now you can use the normalcdf function on your calculator. • Many histograms approximate a normal curve, but a true normal curve is infinitely smooth. Example A The amount of rain each year in Connecticut follows a normal distribution. What is the probability of getting one standard deviation below the normal amount of rain? Solution: You are looking for the area of the shaded portion of the normal distribution shown below. By the empirical rule, you know that approximately 34% of the data is in between -1 and 0. Also, 50% of the data is above 0. Therefore, approximately 84% of the data is unshaded. Therefore, \begin{align}100\%-84\%=16\%\end{align} of the data is shaded. The approximate probability is 16%. To get the exact probability, use the normal cdf function on your calculator to calculate the exact area under the curve. Go to [DISTR] (which is \begin{align}[2^{nd}]\end{align} [VARS]) and choose normalcdf. This is the normal cumulative distribution function and calculates the area under the curve between two \begin{align}x\end{align}-values. The syntax (how you will type it in) for normal cdf is: normalcdf(lower, upper, mean, standard deviation) The lower bound for this shaded region is technically \begin{align}- \infty\end{align}, but the TI-84 cannot handle that so use -1E99. -1E99 is \begin{align}-1 \times 10^{99}\end{align}, an extremely small number, and will give identical results that are correct to many decimal places. The upper bound is -1. For a standard normal distribution with a mean of zero and a standard deviation of 1 you don’t need to type anything else in, but since you will be working with normal distributions with means and standard deviations that are different, it will make sense to get used to using the whole syntax. normalcdf(-1E99, -1, 0, 1) The exact answer is closer to 15.87%. Example B On your first college exam, you score an 82. After the exam the professor tells the class that the mean was a 62 and the standard deviation was 10. What percentage of the class did better than you? Solution: An 82 is 20 away from the mean so is 2 standard deviations from the mean. Therefore, this question is asking for the percentage of students that are above +2 standard deviations above the mean. In future statistics courses you will learn how to create the equation for this distribution and then transform it to standard normal. For now, you can use the fact that your score was exactly 2 standard deviations above the mean. Or, you can calculate the probability using the actual numbers. • normalcdf(2, 1E99, 0, 1) = 0.022750 or 2.750% • normalcdf(82, 1E99, 62, 10) = 0.022750 or 2.750% 2.75% of the class did better than you on the exam. Even though you seemed to score a B-, the professor would probably note that you were near the top of the class and adjust grades accordingly. Example C The quality control technician of a widget making factory observes that widgets that are three standard deviations too large or three standard deviations too small from the precise widget size are unusable. What is the probability of producing a usable widget? Solution: This question is essentially asking for the area between -3 standard deviations and positive 3 standard deviations. The empirical rule says this should be 99.7%. Use the normalcdf function to find the exact value. normalcdf(-3, 3, 0, 1) = 0.997300 or 99.73% The quality control technician would decide if this is a high enough success rate for producing a usable widget. Concept Problem Revisited Height, weight and other measures of people, animals or plants are normally distributed. #### Vocabulary A standard normal distribution is a normal distribution with mean of 0 and a standard deviation of 1. The empirical rule states that for data that is normally distributed, approximately 68% of the data will fall within one standard deviation of the mean, approximately 95% of the data will fall within two standard deviations of the mean, and approximately 99.7% of the data will fall within three standard deviations of the mean. It is a good way to quickly approximate probabilities. Normalcdf is the normal cumulative distribution function and calculates the area between any two values for data that is normally distributed as long as you know the mean and standard deviation for the data. Your calculator has this function built in, and it produces an exact answer as opposed to the empirical rule. #### Guided Practice 1. What is the probability that a person in Texas is exactly 6 feet tall? 2. Two percent of high school football players are invited to play at a competitive college level. How many standard deviations above the average player would someone need to be to have this opportunity? 3. On average, a pumpkin at your local farm weighs 10 pounds with a standard deviation of 6 pounds. You go and find a pumpkin weighing 26 pounds. Of all the pumpkins at the farm, what percent weigh less than this enormous pumpkin? 1. Since height is a continuous variable, meaning any number within a reasonable domain interval is possible, the probability of choosing any single number is zero. Many people may be close to 6 feet tall, but in reality they are 5.9 or 6.0001 feet tall. There must be someone in Texas who is the closest to being exactly 6 feet tall, but even that person when measured accurately enough will still be slightly off from 6 feet. This is why instead of calculating the probability for a single outcome, you calculate the probability between a certain interval, like between 5.9 feet and 6.1 feet. For continuous variables, the probability of any specific outcome, like 6 feet, will always be 0. 2. This situation is the inverse of the previous questions. Instead of being given the standard deviation and asked to find the probability, you are given the probability and asked to find the standard deviation. There is a second programmed feature in the distribution menu that performs this calculation. You are looking for how many standard deviations above the mean will include 98% of the data. invNorm(0.98) = 2.0537 A person would have to be greater than about 2 standard deviations above the mean to be in the top 2 percent. 3. Normalcdf(-1E99, 26, 10, 6) = 0.9961 or 99.61% The vast majority of the pumpkins weigh less than the 26 pound pumpkin you found. #### Practice Consider the standard normal distribution for the following questions. 1. What is the mean? 2. What is the standard deviation? 3. What is the percentage of the data below 1? 4. What is the percentage of the data below -1? 5. What is the percentage of the data above 2? 6. What is the percentage of the data between -2 and 2? 7. What is the percentage of the data between -0.5 and 1.7? 8. What is the probability of a value of 2? Assume that the mean weight of 1 year old girls in the USA is normally distributed, with a mean of about 9.5 kilograms and a standard deviation of approximately 1.1 kilograms. 9. What percent of 1 year old girls weigh between 8 and 12 kilograms? 10. What percent of girls weigh above 12 kilograms? 11. Girls in the bottom 5% by weight need their weight monitored every 2 months. How many standard deviations below the mean would a girl need to be to have their weight monitored? Suppose that adult women’s heights are normally distributed with a mean of 65 inches and a standard deviation of 2 inches. 12. What percent of adult women have heights between 60 inches and 65 inches? 13. Use the empirical rule to describe the range of heights for women within one standard deviation of the mean. 14. What is the probability that a randomly selected adult woman is more than 64 inches tall? 15. What percent of adult women are either less than 60 inches or greater than 72 inches tall? ### Vocabulary Language: English empirical rule empirical rule The empirical rule states that for data that is normally distributed, approximately 68% of the data will fall within one standard deviation of the mean, approximately 95% of the data will fall within two standard deviations of the mean, and approximately 99.7% of the data will fall within three standard deviations of the mean. normalcdf normalcdf Normalcdf is the normal cumulative distribution function and calculates the area between any two values for data that is normally distributed as long as you know the mean and standard deviation for the data. Your calculator has this function built in, and it produces an exact answer as opposed to the empirical rule. standard normal distribution standard normal distribution The standard normal distribution, $\phi(x)=\frac{1}{\sqrt{2 \pi}}e^{- \frac{1}{2}x^2}$ is a normal distribution with mean of 0 and a standard deviation of 1.
# Complex number patterns Alignments to Content Standards: N-CN.A N-CN.A.1 For this task, the letter $i$ denotes the imaginary unit, that is, $i=\sqrt{-1}$. 1. For each integer $k$ from 0 to 8, write $i^k$ in the form $a+bi$. 2. Describe the pattern you observe, and algebraically prove your observation. In particular, simplify $i^{195}$. 3. Write each of the following expression in the form $a+bi$: • $i^2 + i + 1$ • $i^3 + i^2 + i +1$ • $i^4 + i^3 + i^2 + i +1$ • $i^5 + i^4 + i^3 + i^2 + i +1$ • $i^6 + i^5 + i^4 + i^3 + i^2 + i +1$ • $i^7 + i^6 + i^5 + i^4 + i^3 + i^2 + i +1$ • $i^8 + i^7 + i^6 + i^5 + i^4 + i^3 + i^2 + i +1$ 4. Describe the pattern you observe, and algebraically prove your observation. In particular, compute $$i^{195}+i^{194}+\cdots+i^2+i+1.$$ ## IM Commentary This task serves as a possible first student exploration after an initial introduction to the form and arithmetic of complex multiplication. Students need to understand that every complex number can be expressed in the form $a+bi$, and understand multiplication of complex numbers at least well enough to compute, for example, $i^3=i^2\cdot i=-1\cdot i=-i$ (and understand that this is in the form $a+bi$). The task also (optionally) provides an instance where the formula of a finite geometric series can be used to explain an experimentally observed result. The task is an excellent model of the Standard for Mathematical Practice MP8 (Look for and express regularity in repeated reasoning), as students are led to make conjectures about patterns based on experimental calculations. Students can then practice their algebraic manipulation in justifying their observation, a potentially very satisfying academic endeavor. ## Solution Note that writing the numbers $\pm 1$ and $\pm i$ in the form $a+bi$ is deceptively simple as, for example, $i=0+1i$ (that is, we take $a=0$ and $b=1$). 1. Below is a table with the requested powers of $i$. Note that these can be computed by systematically multiplying by increasing powers of $i$, e.g., $$i^3=i^2\cdot i=-1\cdot i=-i,$$ which can be written in the form $-i=0+(-1)i$, and then $$i^4=i^3\cdot i=(-i)\cdot i=-i^2=-(-1)=1,$$ which similarly can be written as $1=1+0i.$ $i^0$ $i^1$ $i^2$ $i^3$ $i^4$ $i^5$ $i^6$ $i^7$ $i^8$ $1$ $i$ $-1$ $-i$ $1$ $i$ $-1$ $-i$ $1$ 2. We observe that the pattern of powers of $i$ is cyclical, repeating every 4 exponents. When the exponent is an integer multiple of 4, the result is a 1. Exponents which are one more than a multiple of 4 give a result of $i$, and so on. To make this precise, we simply observe that any integer can be written as a multiple of 4, plus either 0, 1, 2, or 3. We can justify this pattern as follows: To compute $i^n$, we write $n=4k+a$ (where $a$ is 0, 1, 2, or 3), and then observe $$i^n=i^{4k+a}=(i^4)^k\times i^a=a^k\times i^a=i^a.$$ That is, we can compute $i^n$ by computing $i^a$, where $a$ is the remainder upon dividing $n$ by 4. Since $195$ is three more than the multiple $192=4\cdot 48$, we have $i^{195}=i^3=-i$. 3. Here are the algebraic solutions: • $i^2 + i + 1 = -1 + i + 1 = i$ • $i^3 + i^2 + i +1 = -i + -1 + i + 1 = 0$ • $i^4 + i^3 + i^2 + i +1 = 1 + -i + -1 + i + 1 = 1$ • $i^5 + i^4 + i^3 + i^2 + i +1 = i + 1 + -i + -1 + i + 1 = 1 + i$ • $i^6 + i^5 + i^4 + i^3 + i^2 + i +1 = -1 + i + 1 + -i + -1 + i + 1 = i$ • $i^7 + i^6 + i^5 + i^4 + i^3 + i^2 + i +1 = -i + -1 + i + 1 + -i + -1 + i + 1 = 0$ • $i^8 + i^7 + i^6 + i^5 + i^4 + i^3 + i^2 + i +1 = 1 + -i + -1 + i + 1 + -i + -1 + i + 1 = 1$ 4. Note that the number of terms is 1 more than the largest exponent in the sum. Using the pattern for $i^n$ observed in the previous parts, we see that the sum of the first 196 powers of $i$ (starting with $i^0$) is simply $$\underbrace{\underbrace{1 + -i + -1 + i}_{=0} + \underbrace{1 + -i + -1 + i}_{=0}+\cdots+ \underbrace{1 + -i + -1 + i}_{=0}}_{\text{196 total terms, divided into 48 smaller groups that each sum to 0}}=0.$$ Though not required by the task, this statement generalizes nicely -- whenever the number of terms is a multiple of 4, the corresponding sum will be 0. Whenever the number of terms is one more than a multiple of 4, then there will be one term left over when we are done collecting these groups of 4: The sum $$\underbrace{1 + -i + -1 + i}_{=0} + \underbrace{1 + -i + -1 + i}_{=0}+\cdots+ \underbrace{1 + -i + -1 + i}_{=0}+1=1$$ evaluates to 1 as there will be only a single "1" which does not dispappear by virtue of being in a group of 4. Similarly, if the number of terms is two more than a multiple of 4, the result is $1+i$, and if the number of terms is three more than a multiple of 4, the result is $1+i+-1=i$. Alternatively, we could verify that the formula for a finite geometric series is valid even when the base of the series is $i$. (This is done identically to the case for real numbers). Applying this formula gives $$1+i+i^2+\cdots+i^n=\frac{1-i^{n+1}}{1-i}$$ and then apply our knowledge from above about the powers of $i$. For example, when $n$ is three more than a multiple of $4$, then $n+1$ is a multiple of 4, so $i^{n+1}=1$, and we concldue $$1+i+i^2+\cdots+i^n=\frac{1-i^{n+1}}{1-i}=\frac{1-1}{1-i}=0,$$ the same result as above. This applies in particular to the example of $n=195$ requested in the problem.
# What is the arc length of the curve given by y = ln(x)/2 - x^2/4 in the interval x in [2,4]? Mar 7, 2018 The arc length is $\frac{1}{2} \ln 2 + 3$ units. #### Explanation: $y = \frac{1}{2} \ln x - \frac{1}{4} {x}^{2}$ $y ' = \frac{1}{2} \left(\frac{1}{x} - x\right)$ Arc length is given by: $L = {\int}_{2}^{4} \sqrt{1 + \frac{1}{4} {\left(\frac{1}{x} - x\right)}^{2}} \mathrm{dx}$ Factor out the constant and expand: $L = \frac{1}{2} {\int}_{2}^{4} \sqrt{4 + \left(\frac{1}{x} ^ 2 - 2 + {x}^{2}\right)} \mathrm{dx}$ Simplify: $L = \frac{1}{2} {\int}_{2}^{4} \sqrt{\frac{1}{x} ^ 2 + 2 + {x}^{2}} \mathrm{dx}$ Factorize: $L = \frac{1}{2} {\int}_{2}^{4} \sqrt{{\left(\frac{1}{x} + x\right)}^{2}} \mathrm{dx}$ Simplify: $L = \frac{1}{2} {\int}_{2}^{4} \left(\frac{1}{x} + x\right) \mathrm{dx}$ Integrate term by term: $L = \frac{1}{2} {\left[\ln x + \frac{1}{2} {x}^{2}\right]}_{2}^{4}$ Insert the limits of integration: $L = \frac{1}{2} \ln 2 + 3$
News # Rules to make integration and differentiation easy Calculus is a field of mathematics that is taught to students in their senior years of high school. Calculus is a branch of mathematics that is both ancient and widely utilized. It appears tough and complex to the students, but it can be readily grasped with frequent practice and hard commitment. It is regarded as one of the fields of mathematics that deals with continuous change. Calculus has a wide range of applications. It is used to calculate the area and volume of various bodies. Calculus also aids in determining the body’s center of mass. Differentiation and integration are two topics that we cover in calculus. Differentiation and integration are carried out using some formula and using specific rules like product rule, chain rule, division rule. This article discusses mainly two types of rules: chain rule and product rule. First of all, let’s discuss what differentiation is. It can be defined as the rate of change of a function concerning other functions. Differentiation is the process of breaking down a function into its constituent elements, while integration is the process of putting those pieces back together to generate the original function. In daily mathematics, both integration and differentiation have a wide range of applications. Let’s discuss the first rule which is applicable in both integrations as well as differentiation. ## Chain rule It is one among those rules which apply to the composite type of function. This rule helps in finding out the instantaneous rate of change between two functions. ### Here are some steps which describe how chain rule is carried out: • Determine whether or not The Chain Rule can be used. This is possible when the function is a composite function, which means that one function is nested over the other. • After that, one must look at both the inner and outward functions. • The most crucial step is to get the derivative of the outer function while ignoring the inner function. • Once the outer function’s derivative has been determined, the inner function’s derivative must be subtracted. • Multiply the results from steps 4 and 5 to get the final result. • The final step is to simplify the chain rule’s derivative. • The chain rule is applied to many problems of physics, mathematics too. ## Product rule In mathematics, the product rule is a rule that can be used to compute the derivatives of two or more functions. In basic terms, the product rule allows us to quickly determine the derivative of two differentiable functions that are multiplied together by combining our knowledge of the power rule and the addition and subtraction rules for derivatives. Product formulas can be easily applied with practice and one needs to know when he needs to use the rule. Practice is the key in mathematics, if one practices questions daily then he can easily excel in the subject. ### Here are some of the basic applications of the product formula: • It is used to solve trigonometric functions by reducing the number of steps to a minimum. • This rule can also be used to calculate the rate of change in real-time. • Product rules can also be used to solve many functions at once. This article discusses everything regarding the product and chain rule. It gives all the application of both the rules. These rules are very necessary to get through to excel in two pillars of mathematics that are, differentiation and integration. If one wants to know more about mathematics or computer coding, you can visit the website of Cuemath. Cuemath is a great initiative that is helping a lot of students to achieve their path to success. This website is helping students prepare for many competitive examinations too.
Question Video: Determinants and Invertibility \| Nagwa Question Video: Determinants and Invertibility \| Nagwa # Question Video: Determinants and Invertibility Mathematics • First Year of Secondary School ## Join Nagwa Classes Is the matrix [3, 1 and −3, −1] invertible? 02:45 ### Video Transcript Is the matrix three, one, negative three, negative one invertible? So, the first thing we need to look at is, what does invertible mean? Well, a matrix is said to be invertible or non-singular, which is its other name, if it has an inverse. Well, what we can do is we can use the determinant to work out whether a matrix is, in fact, invertible. So, let’s consider if we had the matrix 𝐴 which is 𝑎, 𝑏, 𝑐, 𝑑. Well then, we have a general form for the inverse of that matrix, which is one over 𝑎𝑑 minus 𝑏𝑐 multiplied by the matrix 𝑑, negative 𝑏, negative 𝑐, 𝑎. If we have the condition that 𝑎𝑑 minus 𝑏𝑐 is not equal to zero. And that’s because if 𝑎𝑑 minus 𝑏𝑐 was equal to zero, then this would mean that one over 𝑎𝑑 minus 𝑏𝑐 would be undefined. So, how is this gonna help us when I mention the determinant? Well, we know that the determinant of matrix 𝐴 is equal to the determinant of 𝑎, 𝑏, 𝑐, 𝑑, which is equal to 𝑎𝑑 minus 𝑏𝑐. Well, if we take a look back, we can see that this is the same as the 𝑎𝑑 minus 𝑏𝑐, which is under the one. And we’re also told that for the inverse, 𝑎𝑑 minus 𝑏𝑐, cannot be equal to zero. So therefore, we can surmise that a matrix must be invertible if the determinant of 𝐴 is not equal to zero. Okay, great! So, now we know what to do. Let’s work out whether the matrix is, in fact, invertible. So, the first thing we need to do is work out the determinant of the matrix three, one, negative three, negative one. So, first of all, what we’re gonna do is multiply our 𝑎 by our 𝑑, so it’s the top-left term by the bottom-right term, so three multiplied by negative one. And then, subtract one multiplied by negative three. That’s the top-right term multiplied by the bottom-left term. So, what we’re gonna get is negative three minus negative three. Well, if you subtract a negative, then it turns positive, so you got negative three add three. So then, we get a result of zero. So, this is gonna help us to determine whether it is invertible. We can say that the matrix three, one, negative three, negative one is not invertible. And that is because it is singular. So therefore, it does not have an inverse. And we know that because the determinant of our matrix is equal to zero. ## Join Nagwa Classes Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher! • Interactive Sessions • Chat & Messaging • Realistic Exam Questions
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: वर्ग 6 (Foundations)>Unit 1 Lesson 2: Multi digit subtraction # Multi-digit subtraction: 389,002-76,151 Subtract 389,002-76,151 using the standard algorithm. ## Want to join the conversation? • is there any more methods • Yes! There’s a Vedic (Indian) method of subtraction. We subtract one column at a time. When the bottom number is bigger we subtract in the reverse order but put a bar over the digit in the answer. Bar digits can be thought of as negative digits! We then convert from bar digits to a normal number using the following steps: 1) For each group of one or more 0’s (if any) immediately to the left of a group of one or more bar digits, we also put bars on those 0’s. 2) For each group of bar digits after the previous step: a) We subtract “all from 9 except last from 10”. b) We take away 1 from the digit immediately to the left of the group of bar digits. This might seem hard to understand at first, but it should become easier to understand when you read the example below. Example: 676,354,508 - 146,378,168. We could go right to left, or left to right. Let’s go left to right. From left to right: 6-1 = 5. 7-4 = 3. 6-6 = 0. 3-3 = 0. 5-7 = 2bar (because 7-5 = 2). 4-8 = 4bar (because 8-4 = 4). 5-1 = 4. 0-6 = 6bar (because 6-0 = 6). 8-8 = 0. So we have 5 3 0 0 2bar 4bar 4 6bar 0. We now need to convert to a normal number. 1) We have a group of two 0’s immediately to the left of the group of bar digits 2bar 4bar. So we put bars on these two 0’s. So we now have 5 3 0bar 0bar 2bar 4bar 4 6bar 0. 2) We have the two groups of bar digits 0bar 0bar 2bar 4bar, and 6bar. For the group 0bar 0bar 2bar 4bar: a) 9-0 = 9. 9-0 = 9. 9-2 = 7. 10-4 = 6. b) The 3 immediately to the left of this group becomes a 2. For the group 6bar: a) 10-6 = 4. b) The 4 immediately to the left of this group becomes a 3. Note that the 5 on the far left and the 0 on the far right both stay as is. Have a blessed, wonderful day! • this is more confusing than addition • I must admit, yes at times it is more confusing than addition but if you can get the methods and get used to it, it is really really easy • can you answer me 😭 • 317,225 - 199,114 , while solving this problem, What I did was that I was easily able to subtract Hundreds, Tens and ones of both the numbers. But, When I was subtracting 317 - 199, I thought of regrouping differently i.e. I subtracted 1 from 317( Which made it 316) and added that 1 to 199 ( To make it 200) , so that now the problem becomes 316 - 200.I well aware that answer would not be correct, but I am curious to know why? Why this step is not correct? What is mathematically wrong about the step that I had taken here. Someone please clarify. :) Regards and thank you. • You had a good idea for a strategy, but you misunderstood the strategy. You confused a strategy for subtraction with a strategy for addition. Subtracting 1 from the top number and adding 1 to the bottom number each make the answer 1 less, overall making the answer 2 less than the correct answer. Think about it: if you have less money to start with and the price of the thing you want to buy goes up, you would end up with less, not the same amount of, money left over. What you needed to do was to add 1 instead of subtracting 1 from the top number. Starting off with 1 more would compensate for taking 1 more away. So you needed to do 318 - 200. Have a blessed, wonderful day! • what is 197.8546,6 -987654321987654321
# Word Problems on Addition and Subtraction of Whole Numbers We will learn how to solve step-by-step the word problems on addition and subtraction of whole numbers. We know, we need to do addition and subtraction in our daily life. Let us solve some word problem examples. Word problems on adding and subtracting of large numbers: 1. The population of a country in 1990 was 906450600 and next year it is increased by 9889700. What was the population of that country in the year of 1991? The population of a country in 1990 = 906450600 Increased population by next year = + 9889700 Total population of that country in 1991 = 916340300 Therefore, the population of that country in the year of 1991 is 916340300. 2. Aaron bought two houses for $1668000 and$2454000. How much did he spend in all? Cost of one house = $1668000 Cost of other house = +$ 2454000 Total cost of both houses = $4122000 Amount of money spent in all$4122000. 3. The sum of two numbers is 41482308. If one number is 3918695 then, find the other number. Sum of two numbers = 41482308 One of the number = - 3918695 Second number = 37563613 Therefore, the other number is 37563613. 4. Mr. Jones deposited $278475 in a bank in his account. Later he withdrew$155755 from his account. How much money was left in the bank in his account? Amount deposited = $278475 Amount withdrawn = -$ 155755 Amount left = $122720 Therefore, Mr. Jones has$ 122720in his bank account. Note: We need to be careful while arranging the addends in columns. 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. ### Adding 1-Digit Number \| Understand the Concept one Digit Number Sep 18, 24 03:29 PM Understand the concept of adding 1-digit number with the help of objects as well as numbers. 2. ### Addition of Numbers using Number Line \| Addition Rules on Number Line Sep 18, 24 02:47 PM Addition of numbers using number line will help us to learn how a number line can be used for addition. Addition of numbers can be well understood with the help of the number line. 3. ### Counting Before, After and Between Numbers up to 10 \| Number Counting Sep 17, 24 01:47 AM Counting before, after and between numbers up to 10 improves the child’s counting skills. 4. ### Worksheet on Three-digit Numbers \| Write the Missing Numbers \| Pattern Sep 17, 24 12:10 AM Practice the questions given in worksheet on three-digit numbers. The questions are based on writing the missing number in the correct order, patterns, 3-digit number in words, number names in figures… 5. ### Arranging Numbers \| Ascending Order \| Descending Order \|Compare Digits Sep 16, 24 11:24 PM We know, while arranging numbers from the smallest number to the largest number, then the numbers are arranged in ascending order. Vice-versa while arranging numbers from the largest number to the sma…
# Intercept In Maths, an intercept is a point on the y-axis, through which the slope of the line passes. It is the y-coordinate of a point where a straight line or a curve intersects the y-axis. This is represented when we write the equation for a line, y = mx+c, where m is slope and c is the y-intercept. There are basically two intercepts, x-intercept and y-intercept. The point where the line crosses the x-axis is the x-intercept and the point where the line crosses the y-axis is the y-intercept. In this article, you will learn what is the intercept, how to find the intercept for a given line, graphing intercepts along with solved examples. ## Definition of Intercept The point where the line or curve crosses the axis of the graph is called intercept. If a point crosses the x-axis, then it is called the x-intercept. If a point crosses the y-axis, then it is called the y-intercept. The meaning of intercept of a line is the point at which it intersects either the x-axis or y-axis. If the axis is not specified, usually the y-axis is considered. It is normally denoted by the letter ‘b’. Except that line is accurately vertical, it will constantly cross the y-axis somewhere, even if it is way off the top or bottom of the chart. Also read: Equation of plane in intercept form ## Intercept Formula The equation of the line, which intersects the y-axis at a point is given by: y = mx + c Now, we have to write the intercept form of the line, we can replace c with b. Thus, the equation becomes: y = mx + b Hence, the formula for the y-intercept of a line is given by: b = y – mx Where, b is the intercept, m is the slope of the line and y and x indicate the points on the y-axis and x-axis respectively. Now, another way of writing the equation of the line, considering a line is intersecting the x-axis and y-axis at point a and b respectively. x/a + y/b = 1 Here, a and b are the intercepts of the line which intersect the x-axis and y-axis, respectively. The values of a and b can be positive, negative or zero and explain the position of the points at which the line cuts both axes, relative to the origin. ## How to Find X and Y Intercepts? Consider a straight line equation Ax + By = C. Divide the equation by C, (Ax/C) + (By/C) = C/C [x/(C/A)] + [y/(C/B)] = 1 Comparing this equation with the equation of a line in intercept form, (x/a) + (y/b) = 1, We get, x-intercept = a = C/A y-intercept = b = B/C Alternatively, To find the x-intercept, substitute y = 0 and solve for x. i.e. Ax + B(0) = C Ax = C x = C/A To find the y-intercept, substitute x =0 and solve for y. i.e. A(0) + By = C By = C y = C/B Go through the example given below to understand this concept in a better way. Example: Let us assume the straight-line equation 5x +2y =10 To find x-intercept: Substitute y=0 in the given equation 5x + 2(0) = 10 5x =10 x =2 To find y-intercept Substitute x =0 in the given equation 5(0) + 2y =10 2y = 10 y = 5 Therefore, x -intercept is (2, 0) y -intercept is (0, 5) ## Two Point Form The formula of the line formed by the two points is given by: y-y1/y2-y1 = x-x1/x2-x1 Say, P(a, 0) = (x1, y1) and Q(0, b) = (x2, y2) are the two points of the line which cuts the x-axis and y-axis, relative to the origin(0,0). Then the formula becomes: => y – 0 / b – 0 = x – a/ 0 – a => y/b = x/-a – a/-a => x/a + y/b = 1 Hence, proved. ## Slope Intercept Form The equation of the line making an intercept c on the y-axis and having slope m is given by: y = mx + c Note: The value of c could be positive or negative since the intercept is drawn on the positive or negative side of the y-axis, respectively. Also, check: Slope intercept form ## Intercept Graph The intercepts are the points on a graph at which the graph crosses the two axes (x-axis and y-axis). The point where the graph crosses the x-axis is called the x-coordinate and the point where the graph crosses the y-axis is called the y-coordinate. In the above intercept graph, where a line L makes x-intercept a and y-intercept b on the axes. Thus, the equation of the line making intercepts a and b on the x-and y-axis, respectively, is: x/a + y/b = 1 ## Solved Examples Example 1: Let two intercepts P(2,0) and Q(0,3) intersect the x-axis and y-axis, respectively. Find the equation of the line. Solution: Given, two intercepts P(2,0) and Q(0,3) intersect the x-axis and y-axis. From the equation of the line we know, x/a + y/b = 1 ……….. (1) Here, a = 2 and b = 3 Therefore, putting the values of intercepts a and b, in equation 1, we get: =>x/2 + y/3 = 1 => 3x + 2y = 6 => 3x + 2y – 6 = 0, Therefore, the equation of the line is 3x + 2y – 6 = 0. Example 2: Find the equation of the line, which makes intercepts –3 and 2 on the x- and y-axes respectively. Solution: Given, a = –3 and b = 2. By intercept form, we know that; x/a + y/b = 1 x/-3 + y/2 = 1 Or 2x – 3y + 6 = 0. Hence, this is the required equation. Example 3: A line passes through P (1, 2) such that its intercept between the axes is bisected at P. What is the equation of the line? Solution: The equation of a line making intercepts a and b with x-axis and y-axis, respectively, is given by: x/a + y/b = 1 1 = (a+0)/2 ⇒ a = 2 2 = (0 + b)/2 ⇒ b = 4 Therefore, the required equation of line is; x/2 + y/4 = 1 ⇒ 2x + y – 4 = 0 ### Practice Problems 1. Find the x-intercept and y-intercept for the line 5x – 8y = 2. 2. If the y-intercept of a line is -4 and the slope is 2/3, then write its equation. 3. What is the equation of a line whose x and y-intercepts are given as 1/3 and -3? Stay tuned with BYJU’S – The Learning App and download the app to explore more videos. ## Frequently Asked Questions – FAQs ### What is an intercept in Maths? An intercept is a point where the straight line or a curve intersects the y-axis in a plane. It is also said to be a y-intercept. ### What is the formula for y-intercept? The formula for y-intercept is given by: b = y – mx Where b is the y-intercept and m is the slope of the line ### What is the equation of the line with respect to x and y-intercepts? The equation of the line, when a line is intersecting the x-axis and y-axis at point a and b respectively is given by: x/a + y/b = 1 ### If a line intersects the y-axis at a point, then what is the value of x-coordinate? If a line intersects the y-axis at a point, the value of the x-coordinate will be equal to zero. ### What is the slope formula? m=(y2-y1)/(x2-x1)
# Dividing Positive and Negative Integers In the previous post on integers, we have learned the rules in multiplying positive integers and negative integers. In this post, we are going to learn how to divide positive and negative integers. If you have observed, in the post on subtracting integers, we have converted the “minus sign” to a “plus negative sign.” I think it is safe for us to say that subtraction is some sort of “disguised addition.” Similarly, we can also convert a division expression to multiplication. For example, we can turn $\displaystyle \frac{5}{3}$ to $(5 \times \frac{1}{3})$. In general, the division $\displaystyle \frac{a}{b}$ to $(a \times \frac{1}{b})$. From the discussion above, we can ask the following question: Can we use the rules in multiplying integers when dividing integers? The answer is a big YES. The rules are very much related. positive integer ÷ positive integer = positive integer positive integer ÷ negative integer = negative integer negative integer ÷ positive integer = negative integer negative integer ÷ negative integer = positive integer Notice that they are very similar to the rules in multiplying integers. positive integer x positive integer = positive integer positive integer x negative integer = negative integer negative integer x positive integer = negative integer negative integer x negative integer = positive integer Here are some examples worked examples. 1. 18 ÷ 3 = 6 2.36 ÷ -12 = – 3 3. -15 ÷ 2 = – 7.5 4.- 8 ÷ -4 = 2 From the discussion and the worked examples above, we can therefore conclude that in dividing positive and negative integers, we only need to memorize the rules in multiplying integers and apply them in dividing integers. ### 4 Responses 1. supercute says: Now I remember…its very easy! Thanks for the blogger of this reviewer…really helpful! 1. August 30, 2013 […] the previous post we have discussed how to divide integers. Operations on real numbers, particularly integers, is one of the scopes of the Civil Service […] 2. September 5, 2013 […] learned the four fundamental operations on real numbers – addition, subtraction, multiplication, division – it is time to combine these operations into a single problem. In the Philippine Civil Service […] 3. March 16, 2016 […] or the rules in performing arithmetic operations namely addition, subtraction, multiplication and division. In this post, you will practice to see if you have mastered these rules. I have mixed the […]
## How do you calculate the weight of a steel shaft? Example 2: Calculate the weight of 100 m 16 mm ø bar. W = 16² x 100/162 = 158 kg. If we put 1 meter length for each diameter of steel bar in the formula then we will get the unit weight. ## How do you calculate the weight of a solid cylinder? How do you calculate the weight of a cylinder? 1. Square the radius of the cylinder. 2. Multiply the square radius by pi and the cylinder’s height. 3. Multiply the volume by the density of the cylinder. The result is the cylinder’s weight. ## What is the formula of weight calculation? The formula for calculating weight is F = m × 9.8 m/s2, where F is the object’s weight in Newtons (N) and m is the object’s mass in kilograms. The Newton is the SI unit for weight, and 1 Newton equals 0.225 pounds. ## How do you calculate the weight of a solid pipe? This pipe weight calculation formula can be used to determine the weight per foot for any size of pipe with any wall thickness: Wt/Ft = 10.69(OD – Wall Thickness)Wall Thickness. ## What is the weight of full cylinder? The gross weight of the cylinder should be arrived at by adding tare weight and the amount of 14.2 kg LPG. For example, if the tare weight printed on the cylinder is 15.2 kg, a full cylinder with 14.2 kg LPG should have a gross weight of 29.4 kg. ## What is the tare weight of a cylinder? The empty LPG gas cylinder weight in India is a tare weight (empty gas cylinder weight) of approximately 15.3kg. Tare weight + 14.2kg (±150gm) = Total weight of LPG gas cylinder in India. That means that the total weight of LPG gas cylinder in India (a 14.2kg cylinder) is approximately 29.5kg. ## How do you calculate the weight of a cylindrical rod? To calculate the weight of steel bars: ROUND – dia. mm2 x 0.004143 = Weight in lbs. per foot. HEXAGON – size mm2 x 0.006798 = Weight in kilograms per metre. ## How do you calculate the weight of a tank? To calculate the weight of the tank, multiply the surface area of the tank by the thickness of the walls in millimeters, then multiplied again by the weight of the glass per millimeter. Glass weighs approximately 0.5165 pounds per square foot for each millimeter of thickness. ## How do you calculate the weight of a steel pipe? How to calculate the weight 1. 0.702 lbs. Metric. Example. density (g/cm³) 7.85g/cm³ x. (OD² – (OD – 2xT)²) (50.0 mm² – (50.0 mm – 2×1.0 mm)²) x. Length. 1m. x. π/4000. = 2. 1.209 kg. Close. ## How is SS 304 pipe weight calculated? Formula for Calculating Stainless Steel Pipe Weight : Thick (mm) X W. Thick (mm) X 0.0248 = Wt. ## How do you calculate the weight of a solid round bar? Weight of round bar = πD^2/4. H × 7850kg/m3, Weight = πD^2/4 ×1/(1000)2 m2 ×7850kg/m3, Weight = D^2/162.3 kg/m = dia. ## How do you calculate the weight of an empty cylinder? The volume of a cylinder is the square of its radius times pi times the height. So the volume of your empty cylinder is (22)(pi)(4) – (1.52)(pi)(4). This is about 22 cubic feet. If your cylinder was made of concrete, which is typically about 144 lbs per cubic foot, then it would weigh 22 x 144 = 3168 lbs. ## How much does a 13kg gas bottle weigh? Calor Gas 13kg Propane bottle/cylinder Total Weight:26.2 kg13 kg Tare Weight:13.2 kg580 mm Diameter:315 mmCG2 Gas Connection:POL fitting Availability:Hire / Refill ## How much does an empty 19kg gas bottle weigh? 19kg Propane gas bottle Product Code210190 Height800mm Diameter315mm Tare weight (empty)17.3-30.5kg Gross weight (full)43kg ## How do you calculate the weight of a hollow pipe? Weight of Pipe Calculator 1. 1 m = 103 mm. 2. 1 m2 = 106 mm2 3. 1 in = 1/12 ft. 4. 1 in2 = 1/144 ft2 5. 1 lb/in3 = 1728 lb/ft2 ## How much does an empty 13kg patio gas bottle weigh? Which regulator do I need? Product Code210131 Capacity13kg Height580mm Diameter315mm Tare weight (empty)12.7-21.9kg ## What size is a 13kg gas bottle? 31 × 31 × 56 cm Weight13.2 kg Dimensions31 × 31 × 56 cm ## How much does a 11kg gas bottle weigh? Handy Gas 11 kg refillable cylinder Total Weight:24.3 kg11 kg Tare Weight:13.3 kg560mm Diameter:315mmCG2 Gas Connection:POL fitting (Prest-o-lite) LH Availability:Hire / Refill ## How do you weigh a Calor gas bottle? Once you have the empty weight of the cylinder you weigh the full cylinder on scales, then subtract the empty weight from the cylinder weight. This will give you the weight of gas left in the cylinder, which can then be compared with a full cylinder, e.g. 13kg. ## What is the weight of a full 15kg Calor gas bottle? Calor Gas 15kg Butane bottle/cylinder Total Weight:34kg15kg Tare Weight:19kg580mm Diameter:318mmCG4 Gas Connection:21mm clip-on Availability:Hire / Refill ## How do you weigh a gas bottle? To help with this most UK bottles will have the empty weight of the bottle stampped onto them. This is known as the TARE weight. When you buy a refil the total weight of the bottle will be the Tare weight plus the charge of gas. ## How much does a 3.9 kg Calor gas bottle weigh? Calor Gas 3.9 kg Propane bottle/cylinder Total Weight:9.6kg3.9 kg Tare Weight:5.7kg340mm Diameter:240mmCG2 Gas Connection:POL fitting Availability:Hire / Refill
# At a Glance - How to Solve a Math Problem We've been doing this solving thing all along; this part is a little refresher. If you've been performing different steps or doing them in a backwards order...oopsies. There are only three steps to solving a math problem. 1. Figure out what the problem is asking. 2. Solve the problem. Finito. That's it. ### Sample Problem If M denotes miles and K denotes kilometers, here are the conversion rates between M and K: K = 1.609M M = 0.6214K The distance from Jenny's house to her parents' house is 500 miles. In fact, that was the most attractive selling point. How many kilometers is this? Let's work through the steps. 1. Figure out what the problem is asking. We want to find a number in kilometers that represents the same distance as 500 miles. 2. Solve the problem. We need to use one of the two given formulas, and we need to be careful to use the right one. Since the problem gave a number of miles and we want a number of kilometers, we need to use the formula K = 1.609M. We substitute 500 for M, and evaluate to get K = 1.609(500) = 804.5 kilometers. For Jenny, that sounds even better than 500 miles. She'll need to remember to mention this 804.5 kilometers thing to her mother the next time she explains why she can't make it for Thanksgiving. First of all, is the answer reasonable? We didn't get a ridiculous number of kilometers or get an answer in furlongs, did we? We want a distance, so we know the answer had better not be negative. Yep, 804.5 isn't negative, which is reassuring. The number of kilometers it takes to travel a certain distance is bigger than the number of miles it takes to travel that distance—this is something you eventually get a feel for if you pay enough attention to maps—so the answer should be bigger than 500. Yep again, 804.5 is bigger than 500, which is also reassuring. Doesn't it feel good to be this reassured?
How to Design a Parachute This post is also inspired by the excellent book by Robert Banks – Towing Icebergs. This book would make a great investment if you want some novel ideas for a maths investigation. The challenge is to design a parachute with a big enough area to make sure that someone can land safely on the ground. How can we go about doing this? Let’s start (as in the last post) with some Newtonian maths. Newton’s Laws: For an object falling through the air we have: psgV – pagV – FD = psVa ps = The density of the falling object pa = The density of the air it’s falling in FD = The drag force g = The gravitational force V = The volume of the falling object a = The acceleration of the falling object Time to simplify things Things look a little complicated at the moment – luckily we can make our lives easier through a little simplification. pa will be many magnitudes smaller than than ps – as the density of air is much smaller than the density of objects like cannonballs. Therefore we ignore this part of the equation, giving an approximate equation: psgV – FD ≈ psVa We now rewrite things to make it easier to substitute values in later. psV = m, where m = mass of an object (as density x volume = mass) This gives: mg – FD ≈ ma and as mg = W (mass x gravitation force = weight) we can rewrite this as: W – FD ≈ (w/g)a Now, the key information to know when looking at a parachute design is the terminal velocity that will be reached when the parachute is open – that means the maximum velocity that a parachutist will potentially be hitting the ground traveling. Now, when a person is traveling at terminal velocity their acceleration is 0, so we can set a = 0 in the equation above to give: W – FD = 0 Now we need an equation for FD (the drag force). FD = 0.5paCDAU2 where pa = density of the air CD = the drag coefficient A = area of parachute U = velocity So when the parachutist is traveling at their terminal velocity with the parachute open we have: W – FD = 0 W = 0.5paCDAU2 OK, nearly there. Next thing to consider is what is the maximum velocity we want someone to be traveling when they hit the ground. This is advised to be around 5 m/s – similar to jumping from a 2 metre ladder. Much more than this and you would risk breaking a bone (or worse!) So we are finally ready to solve our equation. We want to find what value of A (the area of the parachute) will make us land safely. We have: pa = 0.6kg/m3 (approximate density of air at 3000m) CD = 1.40 (a calculated drag coefficient for an open parachute) U = velocity = 5m/s (this is the maximum velocity we want to want to avoid injury) W = 100kg (we will have this as the combined weight of the parachutist and the parachute) So, W = 0.5paCDAU2 100 = 0.5(0.6)(1.40)A(5)2 A = 9.5m2 So if we had a circular parachute with radius 1.7m it should slow us down sufficiently for us to land safely.
# 1.4 Solving Polynomial Equations ## Polynomial Equations In MATH 110, there will be many situations where we are interested to solve polynomial equations. One method which can be used to solve certain equations is called the zero-product property which states that if the product of two algebraic expressions is zero, then at least one of the factors is equal to zero. This can be stated as: If $A\times B = 0$, then $A = 0$ or $B = 0$ This method can be used to solve a quadratic equations using factoring: 1. Write the equation in the form $ax^2 + bx + c = 0$, where the equation is set equal to zero. 2. Factor the quadratic expression if possible. 3. Use the zero-product property to set each factor separately equal to zero. 4. Solve each of the separate equations. Example: Solve the equation: $x^2 + 3x = 40$ Solution: First, get the equation set equal to zero by subtracting 40 from both sides of the equation. $x2 + 3x - 40 = 0$ Next, factor the left side of the equation: $(x - 5)(x + 8) = 0$ Next use the zero-product property to set each factor separately equal to zero. $x - 5 = 0$ which results in $x = 5$ $x + 8 = 0$ which results in $x = -8$ The two solutions to the equation are $x = 5$, $x = -8$. The zero-product property can be extended such that if the product of any number of algebraic expressions is zero, then at least one of the factors is equal to zero. For example we can write this for three factors as: If $A\times B\times C = 0$, then $A = 0$ or $B = 0$ or $C = 0$. Example: Solve the equation: $5x^4 -45x^2 = 0$ Solution: First, note that the equation is set equal to zero. Next, factor the left side of the equation. A greatest common factor of $5x^2$ is noted: $5x^2(x^2 - 9) = 0$ Notice that $(x^2 - 9)$ can be further factored as the difference of two squares. $5x^2(x + 3)(x - 3) = 0$ Next use the zero-product property to set each factor separately equal to zero. $5\times 2 = 0$ which results in $x = 0$ $x - 3 = 0$ which results in $x = 3$ $x + 3 = 0$ which results in $x = -3$ The three solutions to the equation are then $x = 0, x = 3, x = -3$.
Advertisement # Solving an Equation: • A balanced equation is like a weighing balance with equal weights in the two pans. • An equation is like a weighing balance with equal weights on both its pans, in which case the arm of the balance is exactly horizontal. If we add the same weights to both the pans, the arm remains horizontal. Similarly, if we remove the same weights from both the pans, the arm remains horizontal. On the other hand, if we add different weights to the pans or remove different weights from them, the balance is tilted; that is, the arm of the balance does not remain horizontal. • Thus if we fail to do the same mathematical operation with the same number on both sides of an equality, the equality may not hold. • The value of the variable which balances or satisfies the equation is called the ‘solution’ to the equation. • To solve an equation is to find the value of the variable in the equation or to find the solution to the equation. • Example, The solution to the equation ‘x + 190 = 300’ is 110. ## Mathematical Operations on Expressions: • If the same operation is carried out on both sides of an equation every time, the equation remains balanced. When any of the following operations are carried out on an equation, the equation remains balanced. 1. Adding the same number to both sides: Let us now add 8 to both sides; as a result LHS = 8 – 3 + 8 = 5 + 8 = 13 RHS = 4 + 1 + 8 = 5 + 8 = 13. 2. Subtracting the same number from both sides: Let us now subtract 3 from both the sides; as a result, LHS = 8 – 3 – 3 = 5 – 3 = 2 RHS = 4 + 1 – 3 = 5 – 3 = 2. 3. Multiplying both sides by the same number: Let us multiply both the sides of the equality by 3, we get LHS = 3 × (8 – 3) = 3 × 5 = 15, RHS = 3 × (4 + 1) = 3 × 5 = 15. 4. Dividing both sides by the same non-zero number: Let us now divide both sides of the equality by 2. LHS = (8 – 3) ÷ 2 = 5 ÷ 2 = 5/2 RHS = (4 + 1) ÷ 2 = 5 ÷ 2 = 5/2. 5. Exchanging the two sides: Let us now exchange the following two sides. 5 m + 7 = 17 5 m = 17 – 7 5 m = 10 m = 10/5 m = 2. If you would like to contribute notes or other learning material, please submit them using the button below. ### Shaalaa.com Solving Equations & Transposition Method [00:22:14] S ##### Series: Solving Equations & Transposition Method 0% Advertisement Share Notifications View all notifications Login Create free account Forgot password? Course
# Introduction to slop intercept form: Its definition and calculations A slope-intercept form is a way of stating a linear equation in two variables, usually written as y = mx + c, there where m denotes the slope of the line and c denotes the y-intercept. Now other words, a slope-intercept form defines the relationship between the variables x and y in a linear equation, there where the slope denotes a rate of change of y to x, and the y-intercept denotes the value of y when x is equal to zero. The purpose of this equation is straightforward. To calculate the slope-intercept form of a straight line, we would need the slope-intercept form the inclination angle of this straight line from the x-axis, and the intercept that it makes with the y-axis. A slope-intercept formula cannot be applied to calculate the equation of a vertical line. For example to understand the application of the slope-intercept formula. In this article, we will discuss the basic definition with the formula and explanation of the topic with the help of examples. ## Define slope-intercept form : The slope-intercept form is a widely used equation format in mathematics to denote the equation of a straight line. The linear equation can be written as y = mx + c A slope (m) is the ratio of the change in y over the change in x, or the rise over run, which denotes the steepness of the line. Now, the y-intercept (c) is the point where the line intersects the y-axis. Using the slope-intercept form, we can easily graph a linear equation and determine important properties such as the slope, y-intercept, and x-intercept ### Equation of slope intercept form The slope-intercept form can be written as: y = mx + c • m is the slope of the line • c is the y-intercept of the line • (x, y) show every point on the line Note: • The line may have a negative slope in a method the angle it makes with the positive x-direction is obtuse. The value of tan θ, in this method, will be negative, so m will be negative. • To line passing through the origin, the y-intercept will be (c = 0), so its equation will be y = mx. ## The Straight-Line Equation Using Slope Intercept Form : We would need two quantities to calculate the equation of a line with an inconsistent aptitude. the inclination of the line or angle, θ, it makes with say, the x-axis, and the placement of the line. Where the line passes through regarding the axes. Now, we can specify the placement of the line by specifying the point on the y-axis through which the line passes. In general, specifying the y-intercept c. ## Steps to calculate the equation of a line using the slope-intercept form : Step 1: Use the slope formula to calculate the slope of any straight line if it is not given. Step 2: Use slope intercept form equation to evaluate the y-intercept of the line by placing slope and points of the line. Step 3: Place the value of slope and y-intercept to y = mx+c to determine the line equation through the slope-intercept form. Try a slope and y-intercept calculator to evaluate the line equation according to the above steps in a fraction of a second. ## Example Section : In this section, explain the slope-intercept form with the help of examples. Example 1: What is the slope-intercept form of a straight line if; Slope = 4 Point = (5,7) Solution: Step 1: Find c by putting the point in the general form. Y = mx + c 7 = (4)(5) + c 7 = 20 + c 7-20 = c c = -13 Step 2: Make the equation. Y = mx + c Y = 4x – 13 Example 2: Write down the equation of the line through (10,5) and (14,13) Step 1: Find the slope (m). We use the formula to find the slope between the two points (10,5) and (14,13). m= (y2-y1)/(x2-x1) = (13-5)/(14-10) = 8/4 = 2 we know that y = mx + c This gives you y = 2x + c Step 2: Find the y-intercept (c). Pick one of the points on the line and use x and y values to find c. It does not matter which point you choose. we’ll pick the first point (10,5) 10 for x and 5 for y. 5 = 2(10) + c 5 = 20 + c C = 5 – 20 C = -15 Step 3: Write the equation in slope-intercept form ( y = mx + c ). Now we know that m = 2 and c = -15, we can these values in and write the equation in slope-intercept form. Y = 2x – 15 Step 4: We use points (10,5) in step 2, so to check our equation we need to use the other point (14,13). If you use the same point twice, you will not find mistakes. Make sure to use the point you did not use to find the y-intercept in step 2. The use x value from the other point and see if it works. If we use 14 for x in our equation, the y value should come out to 13. 2(14) – 15 = 28 – 15 = 13 If we had used 14 and it came out number that was not 13, that would tell us that we had made a mistake somewhere along the way. If this happens to you start by double-checking to make sure you calculate the slope correctly. ## Summary : In this post, we have explained slope intercept with the help of solved examples. Now you can completely understand this article, anyone can easily solve any problem related to the slope-intercept form. Hello friends, my name is Trupal Bhavsar, I am the Writer and Founder of this blog. I am Electronics Engineer(2014 pass out), Currently working as Junior Telecom Officer(B.S.N.L.) also I do Project Development, PCB designing and Teaching of Electronics Subjects. This site uses Akismet to reduce spam. Learn how your comment data is processed.
## Key Equations Vertical shift $g\left(x\right)=f\left(x\right)+k$ (up for $k>0$ ) Horizontal shift $g\left(x\right)=f\left(x-h\right)$ (right for $h>0$ ) Vertical reflection $g\left(x\right)=-f\left(x\right)$ Horizontal reflection $g\left(x\right)=f\left(-x\right)$ Vertical stretch $g\left(x\right)=af\left(x\right)$ ( $a>0$) Vertical compression $g\left(x\right)=af\left(x\right)$ $\left(01$ ) ## Key Concepts • A function can be shifted vertically by adding a constant to the output. • A function can be shifted horizontally by adding a constant to the input. • Relating the shift to the context of a problem makes it possible to compare and interpret vertical and horizontal shifts. • Vertical and horizontal shifts are often combined. • A vertical reflection reflects a graph about the $x\text{-}$ axis. A graph can be reflected vertically by multiplying the output by –1. • A horizontal reflection reflects a graph about the $y\text{-}$ axis. A graph can be reflected horizontally by multiplying the input by –1. • A graph can be reflected both vertically and horizontally. The order in which the reflections are applied does not affect the final graph. • A function presented in tabular form can also be reflected by multiplying the values in the input and output rows or columns accordingly. • A function presented as an equation can be reflected by applying transformations one at a time. • Even functions are symmetric about the $y\text{-}$ axis, whereas odd functions are symmetric about the origin. • Even functions satisfy the condition $f\left(x\right)=f\left(-x\right)$. • Odd functions satisfy the condition $f\left(x\right)=-f\left(-x\right)$. • A function can be odd, even, or neither. • A function can be compressed or stretched vertically by multiplying the output by a constant. • A function can be compressed or stretched horizontally by multiplying the input by a constant. • The order in which different transformations are applied does affect the final function. Both vertical and horizontal transformations must be applied in the order given. However, a vertical transformation may be combined with a horizontal transformation in any order. ## Glossary even function a function whose graph is unchanged by horizontal reflection, $f\left(x\right)=f\left(-x\right)$, and is symmetric about the $y\text{-}$ axis horizontal compression a transformation that compresses a function’s graph horizontally, by multiplying the input by a constant $b>1$ horizontal reflection a transformation that reflects a function’s graph across the y-axis by multiplying the input by $-1$ horizontal shift a transformation that shifts a function’s graph left or right by adding a positive or negative constant to the input horizontal stretch a transformation that stretches a function’s graph horizontally by multiplying the input by a constant $0<b<1$ odd function a function whose graph is unchanged by combined horizontal and vertical reflection, $f\left(x\right)=-f\left(-x\right)$, and is symmetric about the origin vertical compression a function transformation that compresses the function’s graph vertically by multiplying the output by a constant $0<a<1$ vertical reflection a transformation that reflects a function’s graph across the x-axis by multiplying the output by $-1$ vertical shift a transformation that shifts a function’s graph up or down by adding a positive or negative constant to the output vertical stretch a transformation that stretches a function’s graph vertically by multiplying the output by a constant $a>1$ ## Contribute! Did you have an idea for improving this content? We’d love your input.
## Solution to Puzzle No. 1 - the ring May 1997 For the question see "Puzzle No. 1 - the ring", in issue 1. Most people's reaction on first hearing this problem is on the lines of "How can I possibly find the volume when you haven't told me the radius of the original sphere or the diameter of the hole?" The trick in this question is the last sentence in the statement of the problem, which said "You are told that you have enough information to solve this problem". In other words, the answer cannot depend on the sphere's radius or the diameter of the hole. Any values will do, consistent with the given height. With this insight, the solution can be stated as follows: ### Solution You are told you have enough information to find the volume, so it must be independent of the diameter of the hole - so let's make it zero. Then, since the radius of the sphere is 1 cm, the volume is ### Proof The following proof is based on the solution provided by Roger Stratford. To see that the volume of the ring depends only on the height of the ring, we need to use calculus and integration. The mathematics that follows is usually covered in the first year of a typical mathematics A-Level course. Let the total height of the ring be 2h. Consider the shape tilted so that the axis of the hole is horizontal. Take an origin at the centre of the sphere, with the x-axis along the axis of the cylindrical hole. The x-coordinate at one end of the hole is x = h, and at the other, x = -h. Let the radius of the sphere be r, and the radius of the hole be c. Consider the ring to be made up of thin slices as shown. The outer radius of each slice is y, and the inner radius is c. The volume of each slice is The area of the slice is the area of a circle radius y with the area of a circle radius c subtracted. Then the volume of the slice is obtained by multiplying by the thickness. The slices run over values of x from -h to +h and so the total volume V is: The solid is symmetrical about the OYZ plane through the origin, and so: Now the section through the ring in the z = 0 plane is a circle (apart from the missing pieces). On the remaining surfaces, (because OAP is a right angled triangle) and therefore: By considering a point on the surface at the edge of the hole, where x = h, we have (because OQB is a right angled triangle). Combining (4) and (5), and equation (3) becomes This is a straightforward basic definite integral, which gives: We see that this result only depends on the height of the ring.
An Emergent Math Curriculum Approach for Young Children Author : Allen C. Rosales A new emergent math resource for every teacher—even those who don’t believe they have the know-how to teach math ## Maths Tricks ==>Multiply Double-Digits by Any Other Double-Digits Multiply Double-Digits by Any Other Double-Digits The cool thing about math is watching how seemingly impossible combinations seem to walk out perfectly in the end. By doing certain operations, you can turn wildly complex equations into simple, step-by-step solutions.… ## Maths Trick ==> MULTIPLICATION OF 11 WITH ANY NUMBER MULTIPLICATION OF 11 WITH ANY NUMBER Let me explain this rule by taking examples 1) 23416311 = 2—2+3—3+4—4+1—1+6—6+3—3 = 2575793 Means insert the sum of 2 successive digits and put 2 terminal digits in its place 2) 4534518111 = 4—4+5—5+3—3+4—4+5—5+1—1+8—8+1—1… ## Maths Trick ==> How to Multiply Double-Digit Numbers by 11 How to Multiply Double-Digit Numbers by 11 All you need to do is add the digits of the number you are multiplying by 11 and place that in the middle of the original number. If the sum of the… ## Maths Tricks ==> SQUARE OF A 2 DIGIT NUMBER ENDING WITH 5 SQUARE OF A 2 DIGIT NUMBER ENDING WITH 5 Let me explain this rule by taking examples 35^2 = 3(3+1) —25 = 1225 EXAMPLE. Other example 95^2 = 9(9+1) —25 = 9025 ## Maths Tricks ==> SQUARE OF ANY 2 DIGIT NUMBER SQUARE OF ANY 2 DIGIT NUMBER Let me explain this trick by taking examples 67^2 = [6^2][7^2]+2067 = 3649+840 = 4489 similarly 25^2 = [2^2][5^2]+2025 = 425+200 = 625 Take one more example 97^2 = [9^2][7^2]+2097 = 8149+1260 =… ## Maths Trick ==> SQUARE OF NUMBERS NEAR TO 100 SQUARE OF NUMBERS NEAR TO 100 Let me explain this rule by taking examples 96^2 :- First calculate 100-96, it is 4 so 96^2 = (96-4)—-4^2 = 9216 similarly 106^2 :- First calculate 106-100, it is 6 so 106^2… ## Maths Trick ==> MULTIPLICATION OF 11 WITH ANY NUMBER OF 3 DIGITS. MULTIPLICATION OF 11 WITH ANY NUMBER OF 3 DIGITS Let me explain this rule by taking examples 1. 352*11 = 3—(3+5)—(5+2)—2 = 3872 Means insert the sum of first and second digits, then sum of second and third digits… ## Class 6 (Maths) – Whole Numbers- (पूर्ण सख्याए) Whole Numbers- NCERT 6th Math(पूर्ण सख्याए) Anil Phonetic Table ## Class 6 (Maths) – Playing with numbers (सख्याओ के साथ खेलना) Playing with numbers-NCERT 6th Math(सख्याओ के साथ खेलना) गुणनखंड और गुणज इस पाठ में हम गुणनखंड और गुणज के बारें में सीखेंगे\| हम जानेंगे की एक गुणनखंड एक संख्या का सटीक भाजक होता है, और गुणज एक संख्या…
What is 201/289 as a decimal? Solution and how to convert 201 / 289 into a decimal 201 / 289 = 0.696 Fraction conversions explained: • 201 divided by 289 • Numerator: 201 • Denominator: 289 • Decimal: 0.696 • Percentage: 0.696% The basis of converting 201/289 to a decimal begins understanding why the fraction should be handled as a decimal. Decimals and Fractions represent parts of numbers, giving us the ability to represent smaller numbers than the whole. The difference between using a fraction or a decimal depends on the situation. Fractions can be used to represent parts of an object like 1/8 of a pizza while decimals represent a comparison of a whole number like \$0.25 USD. So let’s dive into how and why you can convert 201/289 into a decimal. 201 / 289 as a percentage 201 / 289 as a fraction 201 / 289 as a decimal 0.696% - Convert percentages 201 / 289 201 / 289 = 0.696 201/289 is 201 divided by 289 The first step in converting fractions is understanding the equation. A quick trick to convert fractions mentally is recognizing that the equation is already set for us. All we have to do is think back to the classroom and leverage long division. Fractions have two parts: Numerators on the top and Denominators on the bottom with a division symbol between or 201 divided by 289. We use this as our equation: numerator(201) / denominator (289) to determine how many whole numbers we have. Then we will continue this process until the number is fully represented as a decimal. Here's how our equation is set up: Numerator: 201 • Numerators represent the number of parts being taken from a denominator. Any value greater than fifty will be more difficult to covert to a decimal. 201 is an odd number so it might be harder to convert without a calculator. Large numerators make converting fractions more complex. Now let's explore X, the denominator. Denominator: 289 • Denominators represent the total number of parts, located below the vinculum or fraction bar. Larger values over fifty like 289 makes conversion to decimals tougher. But 289 is an odd number. Having an odd denominator like 289 could sometimes be more difficult. Have no fear, large two-digit denominators are all bark no bite. Let's start converting! How to convert 201/289 to 0.696 Step 1: Set your long division bracket: denominator / numerator $$\require{enclose} 289 \enclose{longdiv}{ 201 }$$ We will be using the left-to-right method of calculation. This method allows us to solve for pieces of the equation rather than trying to do it all at once. Step 2: Extend your division problem $$\require{enclose} 00. \\ 289 \enclose{longdiv}{ 201.0 }$$ Because 289 into 201 will equal less than one, we can’t divide less than a whole number. Place a decimal point in your answer and add a zero. This doesn't add any issues to our denominator but now we can divide 289 into 2010. Step 3: Solve for how many whole groups you can divide 289 into 2010 $$\require{enclose} 00.6 \\ 289 \enclose{longdiv}{ 201.0 }$$ Now that we've extended the equation, we can divide 289 into 2010 and return our first potential solution! Multiple this number by our furthest left number, 289, (remember, left-to-right long division) to get our first number to our conversion. Step 4: Subtract the remainder $$\require{enclose} 00.6 \\ 289 \enclose{longdiv}{ 201.0 } \\ \underline{ 1734 \phantom{00} } \\ 276 \phantom{0}$$ If you hit a remainder of zero, the equation is done and you have your decimal conversion. If you have a remainder over 289, go back. Your solution will need a bit of adjustment. If you have a number less than 289, continue! Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit. Remember, sometimes you won't get a remainder of zero and that's okay. Round to the nearest digit and complete the conversion. There you have it! Converting 201/289 fraction into a decimal is long division just as you learned in school. Why should you convert between fractions, decimals, and percentages? Converting between fractions and decimals depend on the life situation you need to represent numbers. They each bring clarity to numbers and values of every day life. And the same is true for percentages. Though we sometimes overlook the importance of when and how they are used and think they are reserved for passing a math quiz. But 201/289 and 0.696 bring clarity and value to numbers in every day life. Here are just a few ways we use 201/289, 0.696 or 69% in our daily world: When you should convert 201/289 into a decimal Investments - Comparing currency, especially on the stock market are great examples of using decimals over fractions. When to convert 0.696 to 201/289 as a fraction Pizza Math - Let's say you're at a birthday party and would like some pizza. You aren't going to ask for 1/4 of the pie. You're going to ask for 2 slices which usually means 2 of 8 or 2/8s (simplified to 1/4). Practice Decimal Conversion with your Classroom • If 201/289 = 0.696 what would it be as a percentage? • What is 1 + 201/289 in decimal form? • What is 1 - 201/289 in decimal form? • If we switched the numerator and denominator, what would be our new fraction? • What is 0.696 + 1/2? Convert more fractions to decimals From 201 Numerator From 289 Denominator What is 201/279 as a decimal? What is 191/289 as a decimal? What is 201/280 as a decimal? What is 192/289 as a decimal? What is 201/281 as a decimal? What is 193/289 as a decimal? What is 201/282 as a decimal? What is 194/289 as a decimal? What is 201/283 as a decimal? What is 195/289 as a decimal? What is 201/284 as a decimal? What is 196/289 as a decimal? What is 201/285 as a decimal? What is 197/289 as a decimal? What is 201/286 as a decimal? What is 198/289 as a decimal? What is 201/287 as a decimal? What is 199/289 as a decimal? What is 201/288 as a decimal? What is 200/289 as a decimal? What is 201/289 as a decimal? What is 201/289 as a decimal? What is 201/290 as a decimal? What is 202/289 as a decimal? What is 201/291 as a decimal? What is 203/289 as a decimal? What is 201/292 as a decimal? What is 204/289 as a decimal? What is 201/293 as a decimal? What is 205/289 as a decimal? What is 201/294 as a decimal? What is 206/289 as a decimal? What is 201/295 as a decimal? What is 207/289 as a decimal? What is 201/296 as a decimal? What is 208/289 as a decimal? What is 201/297 as a decimal? What is 209/289 as a decimal? What is 201/298 as a decimal? What is 210/289 as a decimal? What is 201/299 as a decimal? What is 211/289 as a decimal? Convert similar fractions to percentages From 201 Numerator From 289 Denominator 202/289 as a percentage 201/290 as a percentage 203/289 as a percentage 201/291 as a percentage 204/289 as a percentage 201/292 as a percentage 205/289 as a percentage 201/293 as a percentage 206/289 as a percentage 201/294 as a percentage 207/289 as a percentage 201/295 as a percentage 208/289 as a percentage 201/296 as a percentage 209/289 as a percentage 201/297 as a percentage 210/289 as a percentage 201/298 as a percentage 211/289 as a percentage 201/299 as a percentage
# Level 3 Calculus - Integration ## Integrating polynomials and evaluating the constant of integration. Delta textbook chapter 16, pages 158 – 162, exercises 16.1 – 16.3 Symbols: c is a constant of integration. ^ means to the power of / means divided by Rule of Integration If you can expand the function first, do. You can also bring variables and their powers up from the bottom of a fraction by making the power negative. When something is divided by a fraction, flip the fraction upside-down and multiply by it instead. Simplify as much as you can in any situation. When in doubt, check your answer by differentiating it to see whether the result is the function you are trying to integrate. To find the value of c - (evaluating the constant of integration) - you must first be given values for x and y. If they are given, all you need to do is substitute them into your equation to find c! If you ever have two constants and two values for both x and y, solve a simultaneous equation to find both constants. ## Integrating e^(x) and 1/(x) Delta textbook chapter 16, pages 162 – 164, exercises 16.4 – 16.5 ‘Integrating e is easy!’ Similar to differentiation, leave e^(n) alone. The only difference is that instead of multiplying by the differentiation of n, we divide by it instead. Don’t forget the constant! For 1/(x) the rule is (for all values of x): Often a pre-multiplier has to be added at the front of the answer to ‘balance’ the effect of the chain rule. The numerator stays the same, the denominator becomes whatever value is in front of the variable. Always simplify before you attempt to integrate harder questions. When in doubt, check your answer by differentiating it to see whether the result is the function you are trying to integrate. ## Integrating Trig Functions If you are given a reverse function, e.g. –cos(x) instead of cos(x), the result is the reverse of the result, e.g. –sin(x) instead of sin(x). ## Integrating Trig Products Delta textbook chapter 17, page 168, exercise 17.2 To integrate a trig product first turn it into a sum using the formula on your formula sheet. (This can be found on the nzqa website here as the formulae resource for any calculus exam: http://www.nzqa.govt.nz/ncea/assessment/search.do?query=calculus&view=exams&level=03) If you do not have the required ‘2’ at the front of your product you must substitute a ‘2’ for now but at the end multiply your answer by the original amount you had in relation to the ‘2’. e.g. If you have sinAcosB when the formula is 2sinAcosB, turn it into a sum following the formula like you have a 2 in front, (sin(A+B) + sin(A-B)), and integrate the parts separately, (-cos(A+B) – cos(A-B)). Then, because you started with half the formula you multiply by ½ and your final answer becomes -0.5cos(A+B) – 0.5cos(A-B). If you have 5sinAcosB, you will multiply by 5/2 at the end, and so on.
# How do you solve the system y=4x; 2y-x=16? Jan 9, 2017 See full solution process below #### Explanation: Step 1) Because the first equation is already solved for $y$ we can substitute $4 x$ for $y$ in the second equation and solve for $x$: $\left(2 \times 4 x\right) - x = 16$ $8 x - x = 16$ $8 x - 1 x = 16$ $\left(8 - 1\right) x = 16$ $7 x = 16$ $\frac{7 x}{\textcolor{red}{7}} = \frac{16}{\textcolor{red}{7}}$ $\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{7}}} x}{\cancel{\textcolor{red}{7}}} = \frac{16}{7}$ $x = \frac{16}{7}$ Step 2) We can now substitute $\frac{16}{7}$ for $x$ in the first equation and calculate $y$: $y = 4 \times \frac{16}{7}$ $y = \frac{64}{7}$ The solution is: $x = \frac{16}{7}$ and $y = \frac{64}{7}$
# Question on "equation of a line". Read the question below.A(1,4), B(3,2) and C(7,5) are vertices of a triangle ABC. Find: 1. the coordinates of the centroid of triangle ABC 2. the equation of a... Question on "equation of a line". Read the question below. A(1,4), B(3,2) and C(7,5) are vertices of a triangle ABC. Find: 1. the coordinates of the centroid of triangle ABC 2. the equation of a line, through the centroid and parallel to AB. giorgiana1976 \| College Teacher \| (Level 3) Valedictorian Posted on The centroid of a triangle is the intercepting point of the medians of triangle. The median of a triangle joins the vertex to the midpoint of the opposite side. Therefore, we'll have to determine the equations of two medians of triangle and then to solve the system of these equations. The solution of the system represents the coordinates of the centroid. Let AM be the median that joins the vertex A to the midpoint of BC, that is M. To calculate the equation of the line AM, we'll use the formula: (yM-yA)/(y-yA) = (xM-xA)/(x-xA) We'll find the midpoint M: xM = (xB+xC)/2 xM = (3+7)/2 xM = 5 yM = (yB+yC)/2 yM = (2+5)/2 yM = 7/2 (7/2-4)/(y-4) = (5-1)/(x-1) (-1/4)/(y-4) = 4/(x-1) 4(y-4) = -x/4 + 1/4 y - 4 = -x/16 + 1/16 y = -x/16 + 1/16 + 4 y = -x/16 + 65/16 (1) Let BN be the median that joins the vertex B to the midpoint of AC, that is N. To calculate the equation of the line BN, we'll use the formula: (yN-yB)/(y-yB) = (xN-xB)/(x-xB) We'll find the midpoint N: xN = (xA+xC)/2 xN = (1+7)/2 xN = 4 yN = (yA+yC)/2 yN = (4+5)/2 yN = 9/2 (9/2-2)/(y-2) = (4-3)/(x-3) y - 2 = 5x/2 - 15/2 y = 5x/2 - 15/2 + 2 y = 5x/2 - 11/2 (2) Now, we'll solve the system formed by equations of AM and BN. We'll equate (1) and (2): -x/16 + 65/16 = 5x/2 - 11/2 -x/8 + 65/8 = 5x - 11 We'll move the terms in x to the left: -x/8 - 5x = -65/8 - 11 -41x/8 = -153/8 x = 153/41 y = 765/82 - 11/2 y = 314/82
How to solve Syllogisms Questions easily in Reasoning Section By Ashish\|Updated : September 29th, 2016 Syllogism Syllogism is a verbal reasoning type problem, which is an important topic and is frequently asked in many competitive examinations in the Reasoning Section. These types of questions contain two or more statement and these statements are followed by the number of conclusion. You have to find which conclusions logically follows from the given statements. The best method of solving the Syllogism’s problem is through Venn Diagrams. There are four ways in which the relationship could be made. Category 1 All A are B – Means the whole circle representing A lies within the circle representing B. Here we can also make conclusion: Some B are A. Some A are B. For example: All boys are men. Here we can also make a conclusion: Some men are boys. Some boys are men. All apples are fruits. Here we can also make a conclusion: Some fruits are apples. Some apples are fruits. Category 2 No A is B – means that circles representing A and B does not intersect at all. For example: No ball is bat. No door is wall. Category 3 Some A are B Means that some part of the circle represented by A is within the circle represented by B. This type of (category 3) statement gives the following conclusions: (i) Some A are B also indicates that - Some A are not B (ii) Some A are B also indicates that – All A are B. (iii) Some A are B also indicates that – All B are A. (iv) Some A are B also indicates that – All A are B and All B are A. For e.g.: Some mobiles are phones. (i) Category 4. Some A are not B Means that some portion of circle A has no intersection with circle B while the remaining portion of circle A is uncertain whether this portion touches B or not. (i) Some A are not B also indicates that – Some A are B. (ii) Some A are not B also indicates that – No A is B. Important Points – 1. At least statement – At least statement is same as some statement. For ex: Statement: All kids are innocent. Here we can make conclusion: At least some innocent are kids (Some innocent are kids). 2. Some not statement: Some not statement is opposite to “All type” statement. If All being true then Some not being false For e.g. 1. Statement: Some pens are pencils. No pencils are jug. Some jug is pens. Here we can make conclusion: Some pens are not pencils, which is true. In above figure, green shaded part shows; some pens are not pencils, because in statement it is already given No pencils are jug. Complementary Pairs: (Either & or) – Either and or cases only takes place in complementary pairs. Conclusions: (i) Some A are B. (ii) No A are B. From the given above conclusions, it is easy to understand that one of the given conclusions must be true, which is represented by option either (i) or (ii). These types of pairs are called complementary pairs. Note: ‘All A are B’ & ‘Some A are not B’ are also complementary pairs. Note: It is important to note that, in complementary pairs, one of the two conclusion is true and other will be false simultaneously. For example – Statement: All A are B. Some B are C. Conclusion: I. All C are A. II. Some C are not A. Here we can make conclusion, either I or either II follows. Possibility cases in Syllogism – In possibilities cases, we have to create all possibilities to find whether the given conclusion is possible or not. If it is possible and satisfies the given statement than given conclusion will follow otherwise conclusion will not follow. 1. E.g. Statement: All A are B. Some B are C. Conclusion: All A being C is a possibility. Conclusion is true. Possibility figure – 2. E.g. Statements: No stone is a white. Some white are papers. Conclusions: I. All stones being paper is a possibility. Possibility figure: Conclusion is true. 3. E.g. Statements: Some mouse is cat. All mouse are pets. No pet is animal. Conclusions: I. All mouse being animal is a possibility. The conclusion is false because the possibility figure is not possible. If we say all mouse being an animal is a possibility is true then given statements No pet is the animal will be wrong. Here in the statement, it is given No pet is animal and All mouse is a pet. So we can make also conclusion here that no mouse is animal is true. Important Rule: The restatement is not a conclusion – Conclusion has to be different from the statement. E.g. Statement - All A is B Conclusion - All are B. (invalid) Conclusion does not follow. Conclusion - Some A are B (follow) Conclusion follows. Note: If the statement and conclusion are the same then, conclusion does not follow. This rules also follows in possibilities case Thanks Posted by: Member since Nov 2015 Ashish is a management professional with more than 4 years of experience as Mentor in Education sector. Currently working as Community Manager of Teaching exams category at Gradeup. He helps to provide quality content and solves the doubt of aspirants preparing for the exams. His email address is ashish@gradeup.co. write a comment Revid sharmaSep 29, 2016 Some A are B mtlb All A are B Is it true 😳 ShubhamSep 29, 2016 No Ruby kumariSep 30, 2016 Thanks sir NukeshOct 9, 2016 Thaks sir IMPORTANT EXAMS GradeStack Learning Pvt. Ltd.Windsor IT Park, Tower - A, 2nd Floor, Sector 125, Noida, Uttar Pradesh 201303 support@gradeup.co
# How do find the quotient of (-300x^5y^4)/(xy^2)? Mar 18, 2018 See a solution process below: #### Explanation: First, rewrite the expression as: $- 300 \left({x}^{5} / x\right) \left({y}^{4} / {y}^{2}\right)$ Next, use this rule for exponents to rewrite the $x$ term in the denominator: $a = {a}^{\textcolor{b l u e}{1}}$ $- 300 \left({x}^{5} / {x}^{\textcolor{b l u e}{1}}\right) \left({y}^{4} / {y}^{2}\right)$ Now, use this rule for exponents to find the quotient: ${x}^{\textcolor{red}{a}} / {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} - \textcolor{b l u e}{b}}$ $- 300 \left({x}^{\textcolor{red}{5}} / {x}^{\textcolor{b l u e}{1}}\right) \left({y}^{\textcolor{red}{4}} / {y}^{\textcolor{b l u e}{2}}\right) \implies$ $- 300 {x}^{\textcolor{red}{5} - \textcolor{b l u e}{1}} {y}^{\textcolor{red}{4} - \textcolor{b l u e}{2}} \implies$ $- 300 {x}^{4} {y}^{2}$
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # 6.12: Expression Evaluation with Mixed Numbers Difficulty Level: At Grade Created by: CK-12 Estimated11 minsto complete % Progress Practice Expression Evaluation with Mixed Numbers MEMORY METER This indicates how strong in your memory this concept is Progress Estimated11 minsto complete % Estimated11 minsto complete % MEMORY METER This indicates how strong in your memory this concept is Have you ever had to combine pieces of something to make a whole? Travis is doing exactly that. He has three pieces of pipe that has to be clamped together. It will be connected by a professional, but Travis needs to combine the pieces of pipe and figure out the total length that he has. The first piece of pipe measures 513\begin{align}5 \frac{1}{3}\end{align} feet. The second piece of pipe measures 612\begin{align}6 \frac{1}{2}\end{align} feet. The third piece of pipe measures 213\begin{align}2 \frac{1}{3}\end{align} feet. If Travis is going to combine these together, then he has to add mixed numbers. This Concept will teach you how to evaluate numerical expressions involving mixed numbers. Then we will return to this original problem once again. ### Guidance Sometimes, we can have numerical expressions that have both addition and subtraction in them. When this happens, we need to add or subtract the mixed numbers in order from left to right. 416+346146=\begin{align}4\frac{1}{6}+3\frac{4}{6}-1\frac{4}{6}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} Here is a problem with two operations in it. These operations are addition and subtraction. All of these fractions have the same common denominator, so we can begin right away. We start by performing the first operation. To do this, we are going to add the first two mixed numbers. 416+346=756\begin{align}4\frac{1}{6}+3\frac{4}{6}=7\frac{5}{6}\end{align} Now we can perform the final operation, subtraction. We are going to take the sum of the first two mixed numbers and subtract the final mixed number from this sum. 756146=616\begin{align}7\frac{5}{6}-1\frac{4}{6}=6\frac{1}{6}\end{align} Our final answer is 616\begin{align}6\frac{1}{6}\end{align}. What about when the fractions do not have a common denominator? When this happens, you must rename as necessary to be sure that all of the mixed numbers have one common denominator before performing any operations. After this is done, then you can add/subtract the mixed numbers in order from left to right. 246+116112=\begin{align}2\frac{4}{6}+1\frac{1}{6}-1\frac{1}{2}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} The fraction parts of these mixed numbers do not have a common denominator. We must change this before performing any operations. The lowest common denominator between 6, 6 and 2 is 6. Two of the fractions are already named in sixths. We must rename the last one in sixths. 112=136\begin{align}1\frac{1}{2}=1\frac{3}{6}\end{align} Next we can rewrite the problem. 246+116136=\begin{align}2\frac{4}{6}+1\frac{1}{6}-1\frac{3}{6}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} Add the first two mixed numbers. 246+116=356\begin{align}2\frac{4}{6}+1\frac{1}{6}=3\frac{5}{6}\end{align} Now we can take that sum and subtract the last mixed number. 356136=226\begin{align}3\frac{5}{6}-1\frac{3}{6}=2\frac{2}{6}\end{align} Don’t forget to simplify. 226=213\begin{align}2\frac{2}{6}=2\frac{1}{3}\end{align} Now it's time for you to try a few on your own. Be sure your answer is in simplest form. #### Example A 648+228118=\begin{align}6\frac{4}{8}+2\frac{2}{8}-1\frac{1}{8}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} Solution: 758\begin{align}7 \frac{5}{8}\end{align} #### Example B 439+213129=\begin{align}4\frac{3}{9}+2\frac{1}{3}-1\frac{2}{9}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} Solution: 549\begin{align}5 \frac{4}{9}\end{align} #### Example C 213+513614=\begin{align}2\frac{1}{3}+ 5\frac{1}{3}-6\frac{1}{4}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} Solution: 1512\begin{align}1 \frac{5}{12}\end{align} Now back to Travis and the pipe. Here is the original problem once again. Travis is doing exactly that. He has three pieces of pipe that has to be clamped together. It will be connected by a professional, but Travis needs to combine the pieces of pipe and figure out the total length that he has. The first piece of pipe measures 513\begin{align}5 \frac{1}{3}\end{align} feet. The second piece of pipe measures 612\begin{align}6 \frac{1}{2}\end{align} feet. The third piece of pipe measures 213\begin{align}2 \frac{1}{3}\end{align} feet. If Travis is going to combine these together, then he has to add mixed numbers. To solve this, we can begin by writing an expression that shows all three mixed numbers being added together. 513+612+213=\begin{align}5\frac{1}{3}+ 6\frac{1}{2} + 2\frac{1}{3}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} Now we can convert all of the mixed numbers to improper fractions. 163+132+73\begin{align} \frac{16}{3} + \frac{13}{2} + \frac{7}{3}\end{align} Next, we rename each fraction using the lowest common denominator. The LCD of 3 and 2 is 6. 326+396+146\begin{align} \frac{32}{6} + \frac{39}{6} + \frac{14}{6}\end{align} 856=1416\begin{align} \frac{85}{6} = 14 \frac{1}{6}\end{align} feet. ### Vocabulary Here are the vocabulary words in this Concept. Mixed Number a number that has a whole number and a fraction. Numerical Expression a number expression that has more than one operation in it. Operation ### Guided Practice Here is one for you to try on your own. 218+314212=\begin{align}2\frac{1}{8}+ 3\frac{1}{4}-2\frac{1}{2}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} To start, we need to convert all of the mixed numbers to improper fractions. 178+13452\begin{align} \frac{17}{8} + \frac{13}{4} - \frac{5}{2}\end{align} Now we rename each fraction using the lowest common denominator. The LCD of 8, 4 and 2 is 8. 178+268208\begin{align} \frac{17}{8} + \frac{26}{8} - \frac{20}{8}\end{align} Now we can combine and simplify. 238=278\begin{align} \frac{23}{8} = 2 \frac{7}{8}\end{align} ### Video Review Here are videos for review. ### Practice Directions: Evaluate each numerical expression. Be sure your answer is in simplest form. 1. 213+413113=\begin{align}2\frac{1}{3}+4\frac{1}{3}-1\frac{1}{3}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} 2. 625+625115=\begin{align}6\frac{2}{5}+6\frac{2}{5}-1\frac{1}{5}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} 3. 739+819129=\begin{align}7\frac{3}{9}+8\frac{1}{9}-1\frac{2}{9}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} 4. 8310+25106410=\begin{align}8\frac{3}{10}+2\frac{5}{10}-6\frac{4}{10}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} 5. 615+235115=\begin{align}6\frac{1}{5}+2\frac{3}{5}-1\frac{1}{5}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} 6. 949+249359=\begin{align}9\frac{4}{9}+2\frac{4}{9}-3\frac{5}{9}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} 7. 6912+32128412=\begin{align}6\frac{9}{12}+3\frac{2}{12}-8\frac{4}{12}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} 8. 789119+139=\begin{align}7\frac{8}{9}-1\frac{1}{9}+1\frac{3}{9}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} 9. 648+348668=\begin{align}6\frac{4}{8}+3\frac{4}{8}-6\frac{6}{8}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} 10. 1423213+113=\begin{align}14\frac{2}{3}-2\frac{1}{3}+1\frac{1}{3}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} 11. 1269+12891079=\begin{align}12\frac{6}{9}+12\frac{8}{9}-10\frac{7}{9}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} 12. 917+1237+127=\begin{align}9\frac{1}{7}+12\frac{3}{7}+1\frac{2}{7}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} 13. 1434+214134=\begin{align}14\frac{3}{4}+2\frac{1}{4}-1\frac{3}{4}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} 14. 18615+23154215=\begin{align}18\frac{6}{15}+2\frac{3}{15}-4\frac{2}{15}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} 15. 1219+213116=\begin{align}12\frac{1}{9}+2\frac{1}{3}-1\frac{1}{6}=\underline{\;\;\;\;\;\;\;\;\;}\end{align} ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes ### Vocabulary Language: English TermDefinition Mixed Number A mixed number is a number made up of a whole number and a fraction, such as $4\frac{3}{5}$. Numerical expression A numerical expression is a group of numbers and operations used to represent a quantity. operation Operations are actions performed on variables, constants, or expressions. Common operations are addition, subtraction, multiplication, and division. Operations Operations are actions performed on variables, constants, or expressions. Common operations are addition, subtraction, multiplication, and division. Show Hide Details Description Difficulty Level: Authors: Tags: Subjects:
Section_2.2-Graphs of Functions # Section_2.2-Graphs of Functions - y = x 2 . If x > 1 ,... This preview shows pages 1–4. Sign up to view the full content. Section 2.2 Graphs of Functions DEFINITION: A function f is a rule that assigns to each element x in a set A exactly one element, called f ( x ) , in a set B . It’s graph is the set of ordered pairs { ( x, f ( x )) \| x A } EXAMPLE: Sketch the graphs of the following functions. (a) f ( x ) = x 2 (b) g ( x ) = x 3 (c) h ( x ) = x Solution: We ﬁrst make a table of values. Then we plot the points given by the table and join them by a smooth curve to obtain the graph. The graphs are sketched in the Figures below. EXAMPLE: Sketch the graph of the function. f ( x ) = { x 2 if x 1 2 x + 1 if x > 1 1 This preview has intentionally blurred sections. Sign up to view the full version. View Full Document EXAMPLE: Sketch the graph of the function. f ( x ) = { x 2 if x 1 2 x + 1 if x > 1 Solution: If x 1 , then f ( x ) = x 2 , so the part of the graph to the left of x = 1 coincides with the graph of 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. Unformatted text preview: y = x 2 . If x > 1 , then f ( x ) = 2 x +1 , so the part of the graph to the right of x = 1 coincides with the line y = 2 x + 1 . EXAMPLE: Sketch the graph of the function f ( x ) = \| x \| . Solution: Recall that f ( x ) = { x if x -x if x < Using the same method as in the previous example, we note that the graph of f coincides with the line y = x to the right of the y-axis and coincides with the line y =-x to the left of the y-axis. EXAMPLES: 2 EXAMPLE: Which of the following are graphs of functions? Solution: (a) and (b) are graphs of functions, (c) and (d) are not. 3 4... View Full Document ## This note was uploaded on 02/26/2012 for the course MATH 8650 taught by Professor Kiryltsishchanka during the Spring '12 term at NYU. ### Page1 / 4 Section_2.2-Graphs of Functions - y = x 2 . If x > 1 ,... This preview shows document pages 1 - 4. Sign up to view the full document. View Full Document Ask a homework question - tutors are online
Definitions Nearby Words # Divisor [dih-vahy-zer] In mathematics, a divisor of an integer n, also called a factor of n, is an integer which evenly divides n without leaving a remainder. ## Explanation For example, 7 is a divisor of 42 because 42/7 = 6. We also say 42 is divisible by 7 or 42 is a multiple of 7 or 7 divides 42 or 7 is a factor of 42 and we usually write 7 \| 42 (a vertical bar between the two numbers). For example, the positive divisors of 42 are 1, 2, 3, 6, 7, 14, 21, 42. In general, we say m\|n (read: m divides n) for non-zero integers m and n iff there exists an integer k such that n = km. Thus, divisors can be negative as well as positive, although often we restrict our attention to positive divisors. (For example, there are six divisors of four, 1, 2, 4, −1, −2, −4, but one would usually mention only the positive ones, 1, 2, and 4.) 1 and −1 divide (are divisors of) every integer, every integer (and its negation) is a divisor of itself, and every integer is a divisor of 0, except by convention 0 itself (see also division by zero). Numbers divisible by 2 are called even and numbers not divisible by 2 are called odd. A divisor of n that is not 1, −1, n or −n (which are trivial divisors) is known as a non-trivial divisor; numbers with non-trivial divisors are known as composite numbers, while prime numbers have no non-trivial divisors. The name comes from the arithmetic operation of division: if a/b = c then a is the dividend, b the divisor, and c the quotient. There are properties which allow one to recognize certain divisors of a number from the number's digits. ## Further notions and facts Some elementary rules: • If a \| b and a \| c, then a \| (b + c), in fact, a \| (mb + nc) for all integers m, n. • If a \| b and b \| c, then a \| c. (transitive relation) • If a \| b and b \| a, then a = b or a = −b. The following property is important: A positive divisor of n which is different from n is called a proper divisor (or aliquot part) of n. (A number which does not evenly divide n, but leaves a remainder, is called an aliquant part of n.) An integer n > 1 whose only proper divisor is 1 is called a prime number. Equivalently, one would say that a prime number is one which has exactly two factors: 1 and itself. Any positive divisor of n is a product of prime divisors of n raised to some power. This is a consequence of the Fundamental theorem of arithmetic. If a number equals the sum of its proper divisors, it is said to be a perfect number. Numbers less than the sum of their proper divisors are said to be abundant; while numbers greater than that sum are said to be deficient. The total number of positive divisors of n is a multiplicative function d(n) (e.g. d(42) = 8 = 2×2×2 = d(2)×d(3)×d(7)). The sum of the positive divisors of n is another multiplicative function σ(n) (e.g. σ(42) = 96 = 3×4×8 = σ(2)×σ(3)×σ(7)). Both of these functions are examples of divisor functions. If the prime factorization of n is given by $n = p_1^\left\{nu_1\right\} , p_2^\left\{nu_2\right\} cdots p_k^\left\{nu_k\right\}$ then the number of positive divisors of n is $d\left(n\right) = \left(nu_1 + 1\right) \left(nu_2 + 1\right) cdots \left(nu_k + 1\right),$ and each of the divisors has the form $p_1^\left\{mu_1\right\} , p_2^\left\{mu_2\right\} cdots p_k^\left\{mu_k\right\}$ where $0 le mu_i le nu_i$ for each $0 le i le k.$ One can show that $d\left(1\right)+d\left(2\right)+ cdots +d\left(n\right) = n ln n + \left(2 gamma -1\right) n + O\left(sqrt\left\{n\right\}\right).$ One interpretation of this result is that a randomly chosen positive integer n has an expected number of divisors of about $ln n$. ## Divisibility of numbers The relation of divisibility turns the set N of non-negative integers into a partially ordered set, in fact into a complete distributive lattice. The largest element of this lattice is 0 and the smallest one is 1. The meet operation ^ is given by the greatest common divisor and the join operation v by the least common multiple. This lattice is isomorphic to the dual of the lattice of subgroups of the infinite cyclic group Z. ## Generalization One can talk about the concept of divisibility in any integral domain. Please see that article for the definitions in that setting.
# Two-Sided Inequality by Dorin Marghidanu ### Solution 1 First off, $\sqrt{2}(a+b+c)\ge c+\sqrt{2}(a+b)\ge c+\sqrt{2(a^2+b^2)},$ implying, (1) $\displaystyle \frac{1}{\sqrt{2}(a+b+c)}\le\frac{1}{c+\sqrt{2}(a^2+b^2)},$ with equality if $c=0 \text{ and } (a=0\text{ or }b=0).$ Also $\sqrt{2(a^2+b^2)}\ge a+b\Rightarrow c+\sqrt{2(a^2+b^2)}\ge a+b+c,$ so that (2) $\displaystyle \frac{1}{c+\sqrt{2(a^2+b^2)}}\le\frac{1}{a+b+c},$ with equality if $a=b.$ From (1) and (2) we obtain $\displaystyle \frac{1}{\sqrt{2}(a+b+c)}\le\frac{1}{c+\sqrt{2(a^2+b^2)}}\le\frac{1}{a+b+c}$ from which $\displaystyle \frac{a+b}{\sqrt{2}(a+b+c)}\le\frac{a+b}{c+\sqrt{2(a^2+b^2)}}\le\frac{a+b}{a+b+c}.$ Adding up we obtain $\displaystyle \sqrt{2}\le\sum_{cycl}\frac{a+b}{c+\sqrt{2(a^2+b^2)}}\le 2.$ For the left inequality, equality holds if $(a,b,c)=(k,0,0),$ $k\gt 0,$ and permutations. For the right inequality, equality occurs if $a=b=c.$ ### Solution 2 By the AM-QM inequality, \displaystyle\begin{align}&\frac{a+b}{c+\sqrt{2}(a^2+b^2)}\le\frac{a+b}{a+b+c}&\Rightarrow\\ &\sum_{cycl}\frac{a+b}{c+\sqrt{2}(a^2+b^2)}\le\sum_{cycl}\frac{a+b}{a+b+c}=2. \end{align} For the left inequality, assume, WLOG, $a+b+c=1.$ We have $a^2+b^2\le (a+b)^2$ such that \displaystyle \begin{align} &\frac{a+b}{c+\sqrt{2(a^2+b^2)}}\ge\frac{a+b}{c+\sqrt{2}(a+b)}=\frac{a+b}{1+(\sqrt{2}-1)(a+b)}&\Rightarrow\\ &\sum_{cycl}\frac{a+b}{c+\sqrt{2(a^2+b^2)}}\ge \sum_{cycl}\frac{a+b}{1+(\sqrt{2}-1)(a+b)}. \end{align} Observe now that the function $\displaystyle f(x)=\frac{x}{1+(\sqrt{2}-1)x}$ is concave on $[0,2]$ and the sequence $(1,1,0)$ majorizes $(a+b,b+c,c+a).$ Hence, by Karamata's inequality, $\displaystyle \sum_{cycl}f(a+b)\ge f(1)+f(1)+f(0)=\sqrt{2}$ which completes the proof. On the right, equality holds for $a=b=c.$ On the left, for $(a,0,0)$ and permutations. ### Acknowledgment This problem has been kindly posted at the CutTheKnotMath facebook page by Dorin Marghidanu, along with his solution (Solution 1). Solution 2 is by Leo Giugiuc.
Tuesday, November 21, 2023 Latest: ### EteacherG a educational group # Triangle Definition Type Formula Triangle Definition Type Formula Triangle right triangle we are start today online class about Triangle on “EteacherG” is free educational website. In previous chapter (lesson) we had read about Rectangle. We read today in chapter about Triangle definition, Type of Triangle and Formula of Triangle and properties of Triangle. Triangle is a 2D geometric plane shape. This is also very important chapter for point of view of examination. So read this lesson carefully. We discuss about different type of triangle like right angle triangle, right triangle, scalene triangle, Equilateral Triangle, Isosceles Triangle and we also read about triangles formula, which are very easily explain by our expert team. There are many types of triangle and each triangle has different type formulas we study about it in Formula of Triangle and properties of Triangle. # Triangle ### Simple Definition of Triangle in Easy Words “A figure which are enclosed by three sides, called Triangle.” Triangle is a plan two dimensional geometric shape. It has only Height and Base. ## Characteristics/Properties of Triangle • A Triangle has three sides. AB, BC and CA • Three angle in a Triangle. ∠A or ∠CAB ∠B or ∠ABC ∠C or ∠BCA • Triangle interior angle sum are 180 degree. ∠A + ∠B + ∠C = 180° or ∠CAB + ∠ABC + ∠BCA = 180° • Triangle exterior angle sum are 360° degree. ∠PAB + ∠QBC + ∠RCA = 360° • There are three vertex in a Triangle. A, B and C • A Triangle has no diagonal. ## Type of Triangle We divided the triangle on basis of two types of triangle – (A) On the base of side’s measure (B) On the base of angle measure (A) On the base of side’s measure there are three type of Triangle. 1. Equilateral Triangle 2. Isosceles Triangle 3. Heterogeneous Triangle or scalene triangle Now we take Dose of them one by one ### 1. Equilateral Triangle A triangle which all three sides are equal, called Equilateral Triangle. • All sides are equal so its all Angles are also equal. • Each angle is 60 degree in Equilateral Triangle. #### Area of Equilateral Triangle Area = $$\displaystyle \frac{{\sqrt{3}}}{4}{{a}^{2}}$$ where a = side of Equilateral Triangle ### 2. Isosceles Triangle A triangle which two sides are equal and third side is different, called Isosceles Triangle. • It’s two sides are equal so it’s opposite angle also equal. #### Area of Isosceles Triangle Area of Triangle Area of Isosceles Triangle $$\displaystyle =\frac{1}{2}\times b\times h$$ ## 3. Scalene Triangle or Heterogeneous Triangle A triangle which all three sides are not equal, called scalene triangle. ### Scalene Triangle Features • Scalene Triangle has 3 sides but all three sides are unequal. • There are 3 Angles in Scalene Triangle all angle also unequal. • perimeter of Scalene Triangle is sum of all three sides. • formula of Scalene Triangle Area is find by Heron’s formula. Heron’s Formula = $$\displaystyle \sqrt{{s(s-a)(s-b)(s-c)}}$$ where a, b and c are the sides of a Scalene Triangle and s = semi perimeter $$\displaystyle s=\frac{{a+b+c}}{2}$$ ## Area of Scalene Triangle if b = base and h = height are given – Area of Scalene Triangle $$\displaystyle =\frac{1}{2}\times b\times h$$ ### Height of Scalene Triangle if A = Area and b = Base are given – Height of Scalene Triangle $$\displaystyle h=\frac{{2A}}{b}$$ • It’s all sides are unequal so it’s all angle also unequal. • ∆ABC जहाँ AB ≠ BC ≠ CA (B) On the base of angle measure there are three type of Triangle. 1. Acute Angle Triangle 2. right triangle 3. Obtuse Angle Triangle Now we take Dose of them one by one ### 1. Acute Angle Triangle A Triangle which all three angle are smaller than Right angle, called Acute Angle Triangle. • Each angle is less than 90 degree. 2. right triangle A Triangle which at least one angle are Right angle, called right triangle. • right triangle’s one angle is 90 degree and other two angle have less than 90 degree. ### 3. Obtuse Angle Triangle A Triangle which at least one angle are greater than Right angle, called Obtuse Angle Triangle. • Obtuse Angle Triangle one angle is greater than 90 degree and other two angle are less than 90 degree. ### Important Formula Related to Triangle • Area of Triangle = $$\displaystyle \frac{1}{2}\times Base\times Height$$ (When the question has been given Base and Height and should be asked for Area) • Area of Equilateral Triangle = $$\displaystyle \frac{{\sqrt{3}}}{4}{{a}^{2}}$$ (When the question has been given Base of Equilateral Triangle and should be asked for Area) • Area of Triangle by Heron’s Formula = $$\displaystyle \sqrt{{S(S-a)(S-b)(S-c)}}$$ (When the question has been given all three side and should be asked foe area) Here ‘S’ mean Semi-dimensional. Semi-dimensional (S) = $$\displaystyle \frac{{a+b+c}}{2}$$ (Here a, b, c are all three Side of any Heterogeneous Triangle) • Equilateral Triangle which has ‘a’ Side it’s Radius of the intersection (r) = $$\displaystyle \frac{a}{{2\sqrt{3}}}$$ • Equilateral Triangle which has ‘a’ Side it’s Radius of the circumcircle = $$\displaystyle \frac{a}{{\sqrt{3}}}$$ • Finding a diagonal of right angled triangle If the length of the Base and Length is given in a triangle, then the diagonal of that triangle can be known. Similarly, if any two measurements of the triangle are given, then the third measurement can be known. (Hypotenuse)2 = (Perpendicular)2 + (Base)2 (Perpendicular)2(Hypotenuse)2 – (Base)2 (Base)2(Hypotenuse)2 – (Perpendicular)2 • The sum of any two sides of the triangle is greater than the third side The sum of any two sides of the triangle is always greater than its third arm. This is an important feature of the triangle. We can understand this through the following example. Example AB = 6 cm BC = 3 cm CA = 5 cm AB < BC + CA 6 cm < 8 cm CA < BC + AB 5 cm < 9 cm BC < AB + CA 3 cm < 11 cm ### Medians and Altitude of Triangle Medians : The line drawn in the middle from the any side on the front vertex of the Triangle is called the median of Triangle. • There are 3 Medians in every Triangle. • All three Medians has inside the Triangle. here AP = Median (from vertex A to the meddle of side BC) BP = CP Altitude : A right angle perpendicular which is drawn by a vertex to it’s front side, is called Altitude of Triangle. • There are 3 Altitude in every Triangle. • Obtuse angle triangle Altitude outside the triangle. here AP = Altitude (Altitude create a Right angle on opposite side) ∠APB = 90° error: Content is protected !!
# Roman Numerals to 100 Roman Numerals to 100 • I is worth 1. • V is worth 5. • X is worth 10. • L is worth 50. • C is worth 100. • We can create other numbers by adding these numerals. • We repeat the numerals up to 3 times in a row from the largest value numerals to the smallest. • 4 is written as 1 before 5: ‘IV’ and 9 is written as 1 before 10: ‘IX’. • 40 is written as 10 before 50: ‘XL’ and 90 is written as 10 before 100: ‘XC’. • We repeat numerals 1 to 9 (shown in red) for each set of ten. The Roman numerals to 100 are made up of the numerals I (1), V (5), X (10), L (50) and C (100). We can repeat them up to 3 times to add them. • To write 84 in Roman numerals we write the 80 and then add the 4. • 80 is made by counting on from 50 with three tens. • 80 = 50 + 10 + 10 + 10, which can be written as LXXX. • 4 is written as 1 before 5: IV. • 84 is written as LXXX plus IV which we write as LXXXIV. #### Roman Numerals to 100 Interactive Question Generator Roman Numerals to 100: Interactive Questions # Roman Numerals to 100 ## What are the Roman Numerals to 100? Here is a chart showing all of the roman numerals to 100. The roman numerals to 100 are written using the 5 numerals of I, V, X, L and C. Each of these numerals has the following values: • I = 1 • V = 5 • X = 10 • L = 50 • C = 100 ## Rules for Roman Numerals The rules for Roman numerals are as follows: • Repeat the numerals of I (1), X (10) or C (100) to add them up to 3 times in a row. • If a numeral follows another numeral that is of the same value or larger, then add it. • Count on using the numeral ‘I’ from each X (10), V (5) or L (50). • If a numeral appears before a larger numeral then subtract it from the larger numeral. There are only two digits that are created by subtracting. These are 4 and 9. • 4 is written as 1 before 5: ‘IV’. • 40 is written as 10 before 50: ‘XL’. • 9 is written as 1 before 10: ‘IX’. • 90 is written as 10 before 100: ‘XC’. We can add the same numeral up to three times but not more than this. For example, the numeral I can be written twice as II, which equals 2. We can make 3 with three I numerals: III. However we cannot write 4 as IIII because we have used the same numeral more than three times. Instead the digit 4 is one of our special digits that we write as a subtraction. We subtract 1 from 5 to make 4. We write four as IV. Similarly we can add to five to make 6, 7 and 8. Six is VI, which is 5 + 1. Seven is VII, which is 5 + 1 + 1. Eight is VIII, which is 5 + 1 + 1 + 1. Eight requires three of the same numeral, I. We cannot write 9 as VIIII because this requires four of the numeral I. We can only use three at once. 9 is our second special digit that is made by subtracting. 9 is 1 before 10, written as IX. Once we know the rules for Roman numerals 1 to 10, we can continue to use these to make larger numerals. ## Writing Numbers to 100 in Roman Numerals To write Roman numerals, partition the number into its tens and units. First write the numerals for the tens digit and then write the numerals for the units digit. We can combine the numerals of I, V, X, L and C to create all numbers up to 100. For example, we can partition 11 into 10 + 1. 10 is X and 1 is I and so, 11 is written as XI. Ten plus one. 12 is 10 + 2. We write this as XII. XII means X + I + I, which means 10 + 1 + 1. Here are the Roman numerals from 1 to 30. We can see that once we know the Roman numerals from 1 to 10, we can create the numerals to 30 by writing them after the appropriate tens digit numeral. To write the numbers 11 to 20 in Roman numerals, write the Roman numerals 1 to 10 after the numeral ‘X’. For example, 18 is made from 10 + 8. We can write the numeral X for 10 and then follow this with VIII, which is 8. 18 is written as XVIII, which means 10 + 5 + 1 + 1 + 1. Our numerals decrease in size from 10 to 5 to 1, so we add them. We have stuck to the rule of only using three of the same numeral at once. To write the Roman numerals 21 to 30, we write the Roman numerals 1 to 10 after two X numerals. XX is worth 10 + 10, which is 20. We write XX for twenty and then add on the Roman numerals between 1 and 10 afterwards to make the numbers from 21 to 30. For example, 21 is partitioned into 20 + 1. 20 is two tens, written as XX. We need to add 1 more to 20 to make 21. 21 is written as XXI, which means 10 + 10 + 1. We can see in the list of Roman numerals to 30 below, that the numerals 1 to 9 are shown in red. There is a clear pattern of these numerals repeating from column to column. We simply put an extra X numeral in front every time that we move to the next column and add ten. For example, we can start with 5, move right one column to get to 15 and then move right one more column to get to 25. We start at V for five. We add ten by writing an X numeral in front. XV is 15. Another X is written in front to make XXV, 25. We can continue this pattern as we look at the Roman numerals 31 to 60. In the first column, we have the numerals to 40. We have the Roman numerals 1 to 9 following three X numerals. XXX is 30, which means 10 + 10 + 10. It is okay in Roman numerals to use the same numeral up to three times. 38 is made by partitioning 38 into 30 + 8. We know that 30 is XXX and 8 is VIII. 38 is written as XXXVIII, which means 10 + 10 + 10 + 5 + 1 + 1 + 1. We cannot write 40 as four tens. XXXX is using four of the same Roman numeral, which is not allowed. Like the number 4 is written as IV, 1 before 5, we need to write 40 as 10 before 50. 10 before 50 is XL. X is 10 and L is 50. For example, 43 is made of 40 + 3. 40 is XL and then 3 is III. 43 is written as XLIII in Roman numerals. XLIII is XL + III, which is 40 + 3. Be careful not to think of XLIII as 10 + 50 + 1 + 1 + 1. X (10) is a smaller value than L (50) and we do not write smaller numerals before larger numerals to add them. If a smaller numeral is written before a larger numeral, then subtract the smaller numeral from the larger numeral. XL is 50 – 10, which is 40. It is easiest to remember that 4, 40, 9 and 90 are created in this way. Both 4 and 9 are one less than the multiples of five, 5 and 10. We can see all numbers in the 40 column start with XL. 50 is written as L in Roman numerals. 50 has its own Roman numeral just like 5 does. All numbers in the fifties start with L and are followed by the Roman numerals 1 to 9. 60 is made by 50 + 10, written as LX. For example, 69 is written as 60 + 9. 60 is LX and 9 is IX. 69 is written as LXIX in Roman numerals. LX + IX means 60 + 9. We can continue to count on from 50 to make 70 and 80. 70 is two tens more than 50, which is written as LXX. LXX means 50 + 10 + 10. 80 is three tens more than 50, which is LXXX. LXXX means 50 + 10 + 10 + 10. For example, 84 is written as 80 + 4. 80 is LXXX, 50 + 10 + 10 + 10. 4 is IV, 1 before 5. 84 is written as LXXXIV, which means L + X + X + X + IV. Remember to look out for the special digits of 4 and 9, which are created by subtracting 1 from 5 or 10 respectively. To write any larger numbers, we need our next Roman numeral. 100 is written as C in Roman numerals. We cannot write 90 as LXXXX because this uses four of the same numeral, X. Like 9 is written as 1 before 10, 90 s written as 10 before 100. 90 is written as XC in Roman numerals. XC is 10 before 100. XC means 100 – 10. We subtract 10 from 100, because when we have a smaller numeral in front of a larger numeral, we subtract the smaller numeral from the larger numeral. All of the numbers in the nineties are written beginning with XC. For example, 92 can be partitioned into 90 + 2. 90 is XC and 2 is II. 92 is written as XCII in Roman numerals. XCII means XC + II, 90 plus 2. We still continue to use the Roman numerals 1 to 9 with each new column. ## Where are Roman Numerals used Today? Roman numerals are most commonly seen today on clock or watch faces. The numerals to 12 make up the 12 hours on the clock face. The digit 4 is often represented as ‘IIII’ on a clock face, instead of the usual ‘IV’. This is to distinguish it more clearly from other numerals such as VI for 6. Roman numerals are often used in modern times to number more formal or significant events. For example, some major events are numbered with Roman numerals, such as the World Wars I and II, along with the Olympic games. Kings, queens and popes are often numbered with Roman numerals, such as Queen Elizabeth II and King Henry VIII. Some people named after others in their family may use these numerals too. Old buildings and monuments often have significant dates carved on them in Roman numerals. Book volumes or chapters, along with movie copyrights often use Roman numerals as it can appear more formal. Roman numerals are not used frequently today primarily because they take up too much space and are difficult to use in mathematical calculations. For example, 88 is much shorter than LXXXVIII. Also by using our number system separately to the letters of the alphabet, it is easier to identify which is a number and which is a letter. Algebra is a branch of Mathematics that uses the letters of the alphabet and this would be very difficult if we still used Roman numerals today. Now try our lesson on Negative Numbers on a Number Line where we learn about negative numbers. error: Content is protected !!
Have any question? Email us at [email protected] # Drawing a Hyperbola – Engineering Drawing / Technical Drawing ## Drawing a Hyperbola Definition: It is the locus of a point moving in a plane such a way that the ratio of its distances from a fixed point (focus) to the fixed straight line (directrix) is a constant and is always greater than 1 (i.e. e>1). The following methods are used to construct a hyperbola :- 1. Directrix-focus method 2. Rectangular hyperbola These methods are explained through various examples for drawing a hyperbola. Constructing a hyperbola is very easy using any of the methods. ### Directrix-focus method – Drawing a Hyperbola Example: A fixed point is 60 mm away from the fixed straight line. Draw the locus of a point P moving in such a way that the ratio of its distance from the fixed point to its distance from the fixed point straight line is 3:2. Name the curve. ### Procedure :- This method is based on the definition of the hyperbola :- • Draw any line DD as directrix. Mark any point C on it. • Draw the axis through C, perpendicular to the directrix DD. • Mark the focus F on the axis at a distance of 60 mm from C i.e. CF = 60 mm. • Divide CF into 5 equal divisions. • Mark the point V (vertex) on CF such that VF/VC = 3/2 = eccentricity = 36 mm/24 mm. Here, eccentricity is greater than 1 and hence asked curve is a hyperbola. • To construct a scale for ratio 3/2, draw VG = VF at V perpendicular to the axis CF. Join CG and extend it. • Mark any point 1 on the axis and through it, draw a perpendicular to meet CG produced at 1′. • With centre F and radius equal to 1 – 1′, draw arcs to intersect the perpendicular through 1 at points P1 and P1′. • P1 and P1′ are the points on the hyperbola, because the distance of P1 from DD is equal to C-1, P1F = 1 – 1′ and e = P1F/C-1 = 1-1’/ C-1, VG/VC = VF/VC = 3/2. • Similarly, mark points 2, 3, etc. on the axis and obtain points P2 and P2′, P3 and P3′, etc. • Draw a smooth curve through points V, P1, P2, … to get the hyperbola. Hyperbola is an open curve. ### Rectangular Hyperbola – Drawing a Hyperbola Example: A point P is 15 mm and 25 mm respectively from two straight lines which are at right angles to each other. Draw a rectangular hyperbola from P within 5 mm distance from each line. ### Procedure :- • Draw two axes OA and OB and mark point P as shown. • Through P, draw lines CD and EF parallel to OA and OB respectively. • Along PD, mark a number of points 1, 2, 3, 4, etc. at any distances. • Draw lines O1, O2, O3, O4, etc. cutting line PE at points 1′, 2′, 3′, 4′, etc. respectively. • Through point 1 and 1′, draw a line parallel to OB & OA respectively, intersecting each other at point P1. • Similarly, obtain points P2, P3, P4, etc. • Next, along PC, mark a number of points 5, 6, 7, etc. at any distances. • Draw lines O5, O6, O7, etc. cutting line PF at points 5′, 6′, 7′ respectively. • Through point 5, draw a line parallel to OB and through point 5′, draw a line parallel to OA, intersecting each other at point P5. • Similarly, obtain points P6, P7, etc. • Draw a smooth curve through all these points P, P1, P2, P3, P4 and P6, P7, etc. which will give a required rectangular hyperbola.
# How do you solve 2x + y = 7 x – y = –4? 2x + y = 7 x – y = –4 a. (1,-5) b. (1,5) c.(-1,-5) d.(-1,5) Anything is appreciated Add .. its like elimination .. i'll show you the steps its pretty self explanitory(: 2x + y = 7 x - y = -4 Okay this ones easy .. first y and - y cancel out 2x = 7 x = -4 3x = 3 Now solve for x so x = 1 You got that from dividing each side by 3 So now you can eliminate c and d because they have -1 as x not 1 Now you pug x=1 in one of the equations i'll show you both (1) - y = -4 -1 -1 -y=-5 Then you divide both sides by negative 1 because there is always a 1 or negative 1 infront of a variable y=5 if you wanted to plug in x=1 to the 2x + y= 7 it would come out the same.. this is what your work would be 2(1) + y = 7 2 + y = 7 -2 -2 y=5 See same thing! heres how you can check (you can also do this with all the answers and kind of work backwords .. if you ever forget the write way .. it will only work on multible choice) Check 2(1) + 5 = 7 2 + 5 =7 7 = 7 Good! (1) - (5) = -4 -4=-4 Good! There ya go(: #1 Hi, 2x + y = 7 x – y = –4 ---------------- 3x = 3 x = 1 If x = 1 and 2x + y = 7, then: 2(1) + y = 7 2 + y = 7 y = 5 I hope that helps! :-) #2 add y to both sides, forming x=y-4 plug that equation in for x in the first equation 2(y-4)+y=7. distribute 2y-8+y=7. collect like terms 3y-8=7. add 8 to both sides 3y=15. divide both sides by 3 y=5 plug 5 for y in the second equation x-5=-4 add 5 to both sides x=1
# 2016 AMC 12A Problems/Problem 12 ## Problem 12 In $\triangle ABC$, $AB = 6$, $BC = 7$, and $CA = 8$. Point $D$ lies on $\overline{BC}$, and $\overline{AD}$ bisects $\angle BAC$. Point $E$ lies on $\overline{AC}$, and $\overline{BE}$ bisects $\angle ABC$. The bisectors intersect at $F$. What is the ratio $AF$ : $FD$? $[asy] pair A = (0,0), B=(6,0), C=intersectionpoints(Circle(A,8),Circle(B,7))[0], F=incenter(A,B,C), D=extension(A,F,B,C),E=extension(B,F,A,C); draw(A--B--C--A--D^^B--E); label("A",A,SW); label("B",B,SE); label("C",C,N); label("D",D,NE); label("E",E,NW); label("F",F,1.5N); [/asy]$ $\textbf{(A)}\ 3:2\qquad\textbf{(B)}\ 5:3\qquad\textbf{(C)}\ 2:1\qquad\textbf{(D)}\ 7:3\qquad\textbf{(E)}\ 5:2$ ## Solution 2 By the angle bisector theorem, $\frac{AB}{AE} = \frac{CB}{CE}$ $\frac{6}{AE} = \frac{7}{8 - AE}$ so $AE = \frac{48}{13}$ Similarly, $CD = 4$. Now, we use mass points. Assign point $C$ a mass of $1$. $mC \cdot CD = mB \cdot DB$ , so $mB = \frac{4}{3}$ Similarly, $A$ will have a mass of $\frac{7}{6}$ $mD = mC + mB = 1 + \frac{4}{3} = \frac{7}{3}$ So $\frac{AF}{AD} = \frac{mD}{mA} = \boxed{\textbf{(C)}\; 2 : 1}$ ## Solution 3 Denote $[\triangle{ABC}]$ as the area of triangle ABC and let $r$ be the inradius. Also, as above, use the angle bisector theorem to find that $BD = 3$. There are two ways to continue from here: $1.$ Note that $F$ is the incenter. Then, $\frac{AF}{FD} = \frac{[\triangle{AFB}]}{[\triangle{BFD}]} = \frac{AB \frac{r}{2}}{BD * \frac{r}{2}} = \frac{AB}{BD} = \boxed{\textbf{(C)}\; 2 : 1}$ $2.$ Apply the angle bisector theorem on $\triangle{ABD}$ to get $\frac{AF}{FD} = \frac{AB}{BD} = \frac{6}{3} = \boxed{\textbf{(C)}\; 2 : 1}$ ## Solution 4 Draw the third angle bisector, and denote the point where this bisector intersects $AB$ as $P$. Using angle bisector theorem, we see $AE=48/13 , EC=56/13, AP=16/5, PB=14/5$. Applying Van Aubel's Theorem, $AF/FD=(48/13)/(56/13) + (16/5)/(14/5)=(6/7)+(8/7)=14/7=2/1$, and so the answer is $\boxed{\textbf{(C)}\; 2 : 1}$.
# Question: What Is The GCF Of 2 And 4? ## What is the GCF of 18 and 30? We found the factors and prime factorization of 18 and 30. The biggest common factor number is the GCF number. So the greatest common factor 18 and 30 is 6.. ## What is the HCF of 16 and 24? 8Example 21 Consider the highest common factor of 16 and 24 again. The common factors are 1, 2, 4 and 8. So, the highest common factor is 8. ## What is the GCF of 12 and 18? In terms of numbers, the greatest common factor (gcf) is the largest natural number that exactly divides two or more given natural numbers. Example 1: 6 is the greatest common factor of 12 and 18. ## What is the GCF of 15 and 42? The factors of 42 are 42, 21, 14, 7, 6, 3, 2, 1. The common factors of 15 and 42 are 3, 1, intersecting the two sets above. In the intersection factors of 15 ∩ factors of 42 the greatest element is 3. Therefore, the greatest common factor of 15 and 42 is 3. ## What is the GCF of 4 and 10? The GCF = 2. Both 4 and 10 are even, so they must have a common factor of 2. ## What is the GCF of 3 and 4? We found the factors and prime factorization of 3 and 4. The biggest common factor number is the GCF number. So the greatest common factor 3 and 4 is 1. ## What is the GCF of 20 24 and 40? The common factors for 20,24,40 20 , 24 , 40 are 1,2,4 1 , 2 , 4 . The GCF (HCF) of the numerical factors 1,2,4 1 , 2 , 4 is 4 . ## What is the GCF of 18 and 24? Consider the numbers 18, 24, and 36. Their common factors are 1, 2, 3, and 6. Their greatest common factor is 6. ## What’s the GCF of 3 and 8? Greatest common factor (GCF) of 3 and 8 is 1. We will now calculate the prime factors of 3 and 8, than find the greatest common factor (greatest common divisor (gcd)) of the numbers by matching the biggest common factor of 3 and 8. ## What numbers do 3 and 4 go into? Multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, … Multiples of 4 are 4, 8, 12, 16, 20, 24, 28, 32, 36, … So, the common multiples of 3 and 4 are 12, 24, 36, … ## What is the GCF of 36 63? 9To sum up, the gcf of 36 and 63 is 9. In common notation: gcf (36,63) = 9. ## What is the GCF of 2 4 and 6? We here at the Fraction Calculator often determine the Greatest Common Factor (GCF) for numbers such as 2, 4 and 6, because we use the GCF to lower the denominators of our fractions. The Greatest Common Factor of 2, 4 and 6 is the greatest number (factor) that you can evenly divide 2, 4 and 6 by. ## What is the GCF of 7/15 and 21? The common factors for 7,15,21 7 , 15 , 21 are 1 . The GCF (HCF) of the numerical factors 1 is 1 . ## What is the HCF of 20 and 24? Greatest common factor (GCF) of 20 and 24 is 4. We will now calculate the prime factors of 20 and 24, than find the greatest common factor (greatest common divisor (gcd)) of the numbers by matching the biggest common factor of 20 and 24. ## What is the GCF for 4? Greatest common factor (GCF) of 4 and 8 is 4. We will now calculate the prime factors of 4 and 8, than find the greatest common factor (greatest common divisor (gcd)) of the numbers by matching the biggest common factor of 4 and 8. ## What’s the GCF of 3 and 2? Multiply the circled numbers together. This is the greatest common factor! In the example, gcf(18,36,90)=2⋅3⋅3=18 gcf ( 18 , 36 , 90 ) = 2 ⋅ 3 ⋅ 3 = 18 . ## What is the GCF of 24? Answer and Explanation: The GCF of 24 and 36 is 12. ‘GCF’ means ‘greatest common factor’. That’s the greatest, or largest, factor that the two numbers share. ## What is the GCF of 24 30? 6The GCF of 24 and 30 is 6.
Object projected horizontally This page looks at an object projected horizontally from a cliff with a velocity u. The problem is to find out where this object will be after a time t and how fast it will be travelling and in what direction. (See Figure 1) The important thing to remember is that you can consider the motion in two parts :- (a) motion in the horizontal direction - this is uniform velocity since no forces act in this direction (b) motion in the vertical direction - this is uniformly accelerated motion due to the gravitational pull of the Earth, the vertical acceleration being the strength of the Earth's field (g = 9.8 ms- 2). Remember that this always acts vertically downwards. We will ignore air resistance for the time being. Consider the horizontal motion: The velocity after time t = vx = u since there is no horizontal acceleration The horizontal distance travelled (s) = horizontal velocity (vx) x time (t) = ut (1) Now consider the vertical motion: The initial vertical velocity (uy) = 0 and so the vertical velocity after a time t is given by Vy = uy + gt Therefore vy = gt The vertical distance travelled (h) = uyt + ½(gt2) = ½(gt2) since uy = 0 (2) Using Pythogoras' theorem the velocity (v) after a time t can be found from the equation: Velocity(v) after a time t = (vx2 + vy2)1/2 (3) The direction of motion at any point after time t is given by the equation: Direction of motion after time t tanθ = vy/vx (4) where θ is the angle that the trajectory makes with the horizontal at that point. Example problems A ball is thrown horizontally with an initial velocity of 6 ms-1 from an open window that is 4m above the ground. Calculate: (a) the time it takes to hit the ground (b) the distance from the wall where it hits the ground (c) the velocity (magnitude and direction 0.5 seconds after it is thrown. (Ignore air resistance in your calculations and take g = 9.8 ms-2). (a) Using h = ½ gt2 4 = ½x9.8xt2 and so t = 0.904 s = 0.90 s (b) Using s = vt = 6x0.904 = 5.42 m (c) Vertical velocity after 0.5 s = 0 + gt = 9.8x0.5 = 4.9 ms-1 Velocity after 0.5s = [4.92 + 62]1/2 = 7.75 ms-1 Direction of motion: tan θ = 4.92/6 = 0.82 and so θ = 39.4o Combining equations (1) and (2) we can get an equation for the path as a whole. From (1) t = s/u and substituting this in (2) we have: The vertical distance travelled (h) = ½(gt2 = ½(gs2/u2) = s2[½(g/u2)] (5) so h is proportional to s2 and this is the equation of a parabola. All objects projected through gravitational fields travel in parabolas if we ignore the effects of air resistance. The time taken for the object to reach the ground along the parabolic path is the same as if it were dropped vertically. The parabolic path for an object projected horizontally is shown in Figure 2. Notice that if air resistance is ignored the vertical height of the object at given times after the start is the same no matter what horizontal velocity it had at the moment of release. A VERSION IN WORD IS AVAILABLE ON THE SCHOOLPHYSICS USB
Courses Courses for Kids Free study material Offline Centres More Last updated date: 02nd Dec 2023 Total views: 236.1k Views today: 3.36k Find $\dfrac{d}{{dx}}\cos {(1 - {x^2})^2} =$1) $- 2x(1 - {x^2})\sin {(1 - {x^2})^2}$2) $- 4x(1 - {x^2})\sin {(1 - {x^2})^2}$3) $4x(1 - {x^2})\sin {(1 - {x^2})^2}$4) None of these Verified 236.1k+ views Hint: Derivative is nothing but a displacement or rate of change of any function. If f(x) is any function with an independent variable x then its derivative can be written as $y = f'(x)$ where y is the dependent variable (dependent on x). We have many standard derivative formulas to find the derivative that contains trigonometric functions using those formulas we can easily solve this problem. Formula: Some formulas that we need to know: $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$ $\dfrac{d}{{dx}}\cos (x) = - \sin (x)$ Let us name the given function $y = \cos {(1 - {x^2})^2}$ as. We aim to find the derivative of the given function $y$. Now let’s find the derivative of the function that is given. $y = \cos {(1 - {x^2})^2}$ Before differentiating the function let us substitute ${(1 - {x^2})^2} = u$ and let us substitute $z = 1 - {x^2}$. Thus, the function becomes $u = {z^2}$. Then on differentiating u concerning x $\dfrac{{du}}{{dx}} = \dfrac{{du}}{{dz}}.\dfrac{{dz}}{{dx}}$ (Since the term $z$ is a function) $\dfrac{{du}}{{dx}} = \dfrac{d}{{dz}}\left( {{z^2}} \right).\dfrac{{dz}}{{dx}}$ $\dfrac{{du}}{{dx}} = 2z.\dfrac{{dz}}{{dx}}$ Now re-substitute the value $z = 1 - {x^2}$ $\dfrac{{du}}{{dx}} = 2(1 - {x^2}).\dfrac{d}{{dx}}(1 - {x^2})$ $\dfrac{{du}}{{dx}} = 2(1 - {x^2})(0 - 2x) = - 4x(1 - {x^2})$ Thus, the function will become $y = \cos (u)$ where ${(1 - {x^2})^2} = u$. Now we will differentiate this function concerning u. $\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}}.\dfrac{{du}}{{dx}}$ (Since the term $u$ is a function) $\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\cos (u)\dfrac{{du}}{{du}}$ $\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{du}}\cos (u)\dfrac{{du}}{{dx}}$ We already found the value of $\dfrac{{du}}{{dx}} = - 4x(1 - {x^2})$ lets substitute it. $\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{du}}\cos (u)[ - 4x(1 - {x^2})]$ Now by using the formula $\dfrac{d}{{dx}}\cos (x) = - \sin (x)$ we get $\Rightarrow \dfrac{{dy}}{{dx}} = [ - \sin (u)][ - 4x(1 - {x^2})]$ On simplifying this we get $\Rightarrow \dfrac{{dy}}{{dx}} = 4x(1 - {x^2})\sin (u)$ Now let’s resubstitute the value of u in the above expression. $\Rightarrow \dfrac{{dy}}{{dx}} = 4x(1 - {x^2})\sin {(1 - {x^2})^2}$ We will stop here since we can’t simplify it further. Thus, we got the derivative of the given function $y = \cos {(1 - {x^2})^2}$ as $\dfrac{{dy}}{{dx}} = 4x(1 - {x^2})\sin {(1 - {x^2})^2}$. Now let us see the options, option (1) $- 2x(1 - {x^2})\sin {(1 - {x^2})^2}$ is not the correct option since we got that $\dfrac{{dy}}{{dx}} = 4x(1 - {x^2})\sin {(1 - {x^2})^2}$ from our calculation. Option (2) $- 4x(1 - {x^2})\sin {(1 - {x^2})^2}$ is not the correct answer since we got that $\dfrac{{dy}}{{dx}} = 4x(1 - {x^2})\sin {(1 - {x^2})^2}$ from our calculation. Option (3) $4x(1 - {x^2})\sin {(1 - {x^2})^2}$ is the correct answer since we got the same answer in our calculation. Option (4) None of these is an incorrect answer as we got option (3) as the correct answer. Hence, Option (3) $4x(1 - {x^2})\sin {(1 - {x^2})^2}$ is the correct option. Note: As we know that there are many standard formulas for the derivatives of trigonometric functions, using those formulas we can easily find the derivatives of trigonometric functions. If the given function looks complicated, we can use the substitution method to make the given function simpler. After simplification, we need to substitute it to get the original variable.
## Engage NY Eureka Math 2nd Grade Module 6 Lesson 17 Answer Key ### Eureka Math Grade 2 Module 6 Lesson 17 Problem Set Answer Key Question 1. Draw to double the group you see. Complete the sentence, and write an addition equation. a. There is ___1___ cloud in each group. ________ + _______ = ________ 1 + 1 = 2 Explanation: I have drawn the group by doubling it. Hence by doubling the addition equation is 1 + 1 = 2 b. There are __2____ clouds in each group. ________ + _______ = ________ 2 + 2 = 4 Explanation: I have drawn the group by doubling the items in the diagram. Hence by doubling the addition equation is 2 + 2 = 4 c. There are ___3___ clouds in each group. ________ + _______ = ________ 3 + 3 = 6 Explanation: I have drawn the group by doubling the items in the given diagram. Hence by doubling the addition equation is 3 + 3= 6 d. There are __4____ clouds in each group. ________ + _______ = ________ 4 + 4 = 8 Explanation: I have drawn the group by doubling the items in the diagram. Hence by doubling the addition equation is 4 + 4 = 8 e. There are _5__ clouds in each group. ________ + _______ = ________ 5 + 5 = 10 Explanation: I have drawn the group by doubling the items in the diagram. Hence by doubling the addition equation is 5 + 5 = 10 Question 2. Draw an array for each set. Complete the sentences. The first one has been drawn for you. a. 2 rows of 6 2 rows of 6 = _12____ __6___ + __6___ = _12____ 6 doubled is _12____. 6 + 6 = 12 Explanation: As asked in the question I have drawn an array for the given set and doubled the rows. Hence 2 rows of 6 is 12. b. 2 rows of 7 2 rows of 7 = _____ __7___ + _7____ = _14____ 7 doubled is _14____ 7 + 7 = 14 Explanation: I have drawn the group by doubling the items in the diagram. Hence by doubling the addition equation is 7 + 7 = 14 c. 2 rows of 8 2 rows of 8 = _____ ___8__ + _8____ = _16____ 8 doubled is _16____. 8 +8 = 16 Explanation: I have drawn the group by doubling the items in the diagram. Hence by doubling the addition equation is 8 + 8 = 16 d. 2 rows of 9 2 rows of 9 = _18____ _ 9___ + __9___ = _18____ 9 doubled is _18____. 9 + 9 = 18 Explanation: I have drawn the group by doubling the items in the diagram. Hence by doubling the addition equation is 8 + 8 = 16 e. 2 rows of 10 2 rows of 10 = _20___ ___10__ + _10____ = __20__ 10 doubled is _20__. 10 + 10 =20 Explanation: I have drawn the group by doubling the items in the diagram. Hence by doubling the addition equation is 10 + 10 = 20 Question 3. List the totals from Problem 1. List the totals from Problem 2. Are the numbers you have listed even or not even? Explain in what ways the numbers are the same and different. The totals from problem 1 are 2, 4, 6, 8, 10 The totals from problem 2 are 12, 14, 16, 18, 20 The number I have listed is even In both sets of numbers have the same number of ones, but the numbers in problem 2 have one more than the numbers in problem 1. Explanation: As given in the question total listed in the problem is 2, 4, 6, 8, 10. And the totals listed from problem 2 are 12, 14, 16, 18, 20. Therefore all the numbers I have listed are even. Also in both sets, we have the same number of ones but the numbers listed in problem 2 have one more than in problem 1. ### Eureka Math Grade 2 Module 6 Lesson 17 Exit Ticket Answer Key Question 1. Draw an array for each set. Complete the sentences. a. 2 rows of 5 2 rows of 5 = _____ _____ + _____ = _____ Circle one: 5 doubled is even/not even. 5 + 5 =10 Explanation: I have drawn the group by doubling the items in the diagram. Hence by doubling the addition equation is 5+ 5= 10. Therefore the drawn array is even. b. 2 rows of 3 2 rows of 3 = _____ _____ + _____ = _____ Circle one: 3 doubled is even/not even. 3 + 3 = 6 Explanation: I have drawn the group by doubling the items in the diagram. Hence by doubling the addition equation is 3+ 3= 6. Therefore the drawn array is even. ### Eureka Math Grade 2 Module 6 Lesson 17 Homework Answer Key Question 1. Draw to double the group you see. Complete the sentences, and write an addition equation. a. There are _2_ stars in each group. ____2____ + __2_____ = __4___ 2 + 2 = 4 Explanation: I have drawn the group by doubling the given items. Hence the equation is 2 + 2 = 4 which is as even number. b. There are _4___ stars in each group. ____4____ + ___4____ = __8______ 4 + 4 = 8 Explanation: I have drawn the group by doubling the given items. Hence the equation is 4 + 4 = 8 which is as even number. c. There are __1__ stars in each group. _____1___ + __1_____ = __2___ 1 + 1 = 2 Explanation: I have drawn the group by doubling the given items. Hence the equation is 1 + 1 = 2 which is an even number. d. There are __3__ stars in each group. ____3____ + ____3___ = ____6____ 3 + 3 = 6 Explanation: I have drawn the group by doubling the given items. Hence the equation is 3 + 3 = 6 which is an even number. e. There are __5__ stars in each group. ____5____ + __5_____ = __10______ 5 + 5 = 10 Explanation: I have drawn the group by doubling the given items. Hence the equation is 5 + 5 = 10 which is an even number. Question 2. Draw an array for each set. Complete the sentences. The first one has been drawn for you. a. 2 rows of 6 2 rows of 6 = _____ _____ + _____ = _____ 6 doubled is _____. 6 + 6 = 12 Explanation: I have drawn the group by doubling the items in the diagram. Hence by doubling the addition equation is 6 + 6 = 12. b. 2 rows of 7 2 rows of 7 = ____ _____ + _____ = _____ 7 doubled is _____. 7 + 7 = 14 Explanation: I have drawn the group by doubling the items in the diagram. Hence by doubling the addition equation is 7 + 7 = 14. c. 2 rows of 8 ____ rows of _____ = ____ _____ + 8 = _____ 8 doubled is _____. 8 + 8 = 16 Explanation: I have drawn the group by doubling the items in the diagram. Hence by doubling the addition equation is 8 + 8 = 16. d. 2 rows of 9 2 rows of 9 = _____ _____ + _____ = _____ 9 doubled is ____. 9 +9 = 18 Explanation: I have drawn the group by doubling the items in the diagram. Hence by doubling the addition equation is 9 + 9 = 18. e. 2 rows of 10 _2_ rows of _10__ = _20__ 10 + _10__ = _20__ 10 doubled is _20_. 10 + 10 = 20 Explanation: I have drawn the group by doubling the items in the diagram. Hence by doubling the addition equation is 10 + 10 = 20. Question 3. List the totals from Problem 1. List the totals from Problem 2. Are the numbers you have listed even or not even? Explain in what ways the numbers are the same and different.
# Aljabar Bentuk 2 ## Konsep Dasar ##### $$(a - b)^2 = a^2 - 2ab + b^2$$ Contoh 1 Jabarkan $$(2x + 5)^2$$ \begin{equation} \begin{split} (2x + 5)^2 &= (2x)^2 + 2(2x)(5) + 5^2\\\\ (2x + 5)^2 & = 4x^2 + 20x + 25 \end{split} \end{equation} Contoh 2 Jabarkan $$(3x - 4y)^2$$ \begin{equation} \begin{split} (3x - 4y)^2 &= (3x)^2 - 2(3x)(4y) + (4y)^2\\\\ (3x - 4y)^2& = 9x^2 -24xy + 16y^2 \end{split} \end{equation} Contoh 3 Faktorkan $$x^2 + 6x + 9$$ \begin{equation} \begin{split} x^2 + 6x + 9& = x^2 + 2\cdot 3\cdot x + 3^2\\\\ x^2 + 6x + 9& = x^2 + 2\cdot x \cdot 3 + 3^2\\\\ x^2 + 6x + 9& = (x + 3)^2 \end{split} \end{equation} Contoh 4 Faktorkan $$x^2 - 10xy + 25y^2$$ \begin{equation} \begin{split} x^2 - 10xy + 25y^2& = x^2 - 2\cdot x\cdot 5y + 5^2y^2\\\\ x^2 - 10xy + 25y^2& = x^2 - 2\cdot x\cdot 5y + (5y)^2\\\\ x^2 - 10xy + 25y^2& = (x - 5y)^2 \end{split} \end{equation}
Home Practice For learners and parents For teachers and schools Textbooks Full catalogue Pricing Support We think you are located in United States. Is this correct? # End of chapter activity Exercise 9.2 Mr and Mrs Dlamini need to make a final decision about their garden and what they are prepared to spend. The diagram below (not drawn to scale) shows what their garden currently looks like, with the lawn, patio and garden. Calculate the area of the property, in metres squared. Use the following formula: Area = length $$\times$$ width. Area = ($$\text{20}$$ $$\text{m}$$) $$\times$$ $$\text{4}$$ $$\text{m}$$ + $$\text{8}$$ $$\text{m}$$) = $$\text{20}$$ $$\text{m}$$ $$\times$$ $$\text{12}$$ $$\text{m}$$ = $$\text{240}$$ $$\text{m}$$$$^{\text{2}}$$ Calculate the area of the garden (indicated on the diagram) using the following formula: Area = $$\frac{\text{1}}{\text{2}} \times$$ base $$\times$$ perpendicular height. Area = $$\frac{\text{1}}{\text{2}} \times$$ $$\text{20}$$ $$\text{m}$$ $$\times$$ $$\text{4}$$ $$\text{m}$$ = $$\text{40}$$ $$\text{m}$$$$^{\text{2}}$$ What percentage of the area of the whole property is the triangular garden? Express your answer as a whole number. $$\frac{\text{40}}{\text{240}} = \text{0,16666}$$. $$\text{0,16666}$$ $$\times$$ $$\text{100}$$ = $$\text{16,67}\%$$ Mr Dlamini still has not decided on a shape for his new fish pond. One option is for Mr Dlamini to install a circular fish pond with a radius of $$\text{1,5}$$ $$\text{m}$$, as shown in the diagram: Calculate the area of this pond in metres squared, using the formula: Area = $$\pi \times$$ radius$$^{\text{2}}$$, where $$\pi=\text{3,142}$$. Area = $$\text{3,142}$$ $$\times$$ ($$\text{1,5}$$ $$\text{m}$$)$$^{\text{2}}$$ = $$\text{7,07}$$ $$\text{m}$$$$^{\text{2}}$$ The alternative design for the fish pond, as we have already seen looks as follows: Using the dimensions given on the diagram above, calculate the area of the other possible fish pond using the formulae Area = $$\pi \times$$ radius$$^{\text{2}}$$ and Area =length $$\times$$ width, where $$\pi= \text{3,142}$$. Area = area rectangle + area semicircle = ($$\text{2}$$ $$\text{m}$$ $$\times$$ $$\text{1}$$ $$\text{m}$$) + $$\frac{\text{1}}{\text{2}}(\pi \cdot$$($$\text{1}$$ $$\text{m}$$)$$^{\text{2}}$$) = $$\text{2}$$ $$\text{m}$$ + $$\text{1,57}$$ = $$\text{3,57}$$ $$\text{m}$$$$^{\text{2}}$$ How does your answer to b) compare to the area of the circular fish pond that we calculated in a)? Which shape of fish pond should Mr Dlamini choose if he is worried about the pond taking up too much space in his garden? Give reasons for your answer. The second design is much smaller than the first. If he is concerned that the pond will be too large, he should decide on the second shape. Mr Dlamini is concerned that his dog will try to climb into the fish pond once it's built. He will need to put a fence around the fish pond. He has still not decided which style of fish pond he wants to build. He decides to get quotes from a fencing company called “Fence-Me-In”. They give him the following information: 1. Labour costs: $$\text{R}\,\text{549,99}$$ for the whole project 2. $$\text{1}$$ metre of fencing costs $$\text{R}\,\text{29,99}$$. Calculate the total cost for each style of pond. Pond shape 1 (circle): cost = Labour + fencing price $$\times$$ perimeter = $$\text{R}\,\text{549,99}$$ + ($$\text{R}\,\text{29,99}$$)($$\text{2}\pi \times$$ $$\text{1,5}$$ $$\text{m}$$) = $$\text{R}\,\text{549,99}$$ + R($$\text{29,99}$$)($$\text{9,426}$$ $$\text{m}$$) = $$\text{R}\,\text{549,99}$$ + $$\text{R}\,\text{282,69}$$ = $$\text{R}\,\text{832,68}$$. Pond shape 2: Cost = Labour + fencing price $$\times$$ perimeter = $$\text{R}\,\text{549,99}$$ + ($$\text{R}\,\text{29,99}$$)($$\text{2}$$ $$\text{m}$$ + $$\text{1}$$ $$\text{m}$$ + $$\text{2}$$ $$\text{m}$$ + $$\frac{\text{1}}{\text{2}} \times \text{2} \times \pi \times$$ $$\text{1}$$ $$\text{m}$$) = $$\text{R}\,\text{549,99}$$ + ($$\text{R}\,\text{29,99}$$)($$\text{5}$$ $$\text{m}$$ + $$\text{3,142}$$ $$\text{m}$$) = $$\text{R}\,\text{549,99}$$ + $$\text{R}\,\text{244,18}$$ = $$\text{R}\,\text{794,17}$$ Based on your answers to Questions 2 c) and d), which style of fish pond do you think Mr Dlamini should choose? Give reasons for your answer. Learner-dependent answer, but based on his concern about the size of the pond he should choose the second design - it's cheaper to fence too. The Dlamini's are also wanting to redo the paving of the patio, and replace the bricks with cobblestones. The actual dimensions of the patio are as follows: length = $$\text{6}$$ $$\text{m}$$ and width = $$\text{8}$$ $$\text{m}$$. If the dimensions of one cobblestone is $$\text{10}$$ $$\text{cm}$$ by $$\text{10}$$ $$\text{cm}$$, how many cobble stones will be needed to pave the patio? (Hint: convert all units to be the same!) Area of patio = $$\text{6}$$ $$\text{m}$$ $$\times$$ $$\text{8}$$ $$\text{m}$$ = $$\text{48}$$ $$\text{m}$$$$^{\text{2}}$$. Area of cobblestone = $$\text{0,1}$$ $$\text{m}$$ $$\times$$ $$\text{0,1}$$ $$\text{m}$$ = $$\text{0,01}$$ $$\text{m}$$$$^{\text{2}}$$. $$\text{48}\div \text{0,01}^{\text{2}}$$ = $$\text{4 800}$$ cobblestones. Cobblestones are sold in batches of $$\text{200}$$. How many batches will be needed to be bought? $$\text{4 800}$$ $$\div$$ $$\text{200}$$ = $$\text{24}$$ batches of cobblestones. If one batch costs $$\text{R}\,\text{129,99}$$, what will be the total cost of the cobblestones? $$\text{24}$$ $$\times$$ $$\text{R}\,\text{129,99}$$ = $$\text{R}\,\text{3 119,76}$$ Sam's uncle works for a company that puts up safety fences around swimming pools, and nets over swimming pools. He must calculate how much fencing and netting they need for each of the swimming pools shown below. The fencing is always $$\text{1}$$ metre away from the swimming pool. For each swimming pool below, calculate: 1. the perimeter of the swimming pool 2. the length of fencing needed 3. the area of the swimming pool (the area of netting needed) 4. the cost of the fence at $$\text{R}\,\text{250,00}$$ per metre 5. the cost of the netting at $$\text{R}\,\text{199,99}$$ per m$$^{\text{2}}$$. 1. Perimeter of pool = $$\text{2}$$ $$\times$$ ($$\text{8}$$ $$\text{m}$$ + $$\text{4}$$ $$\text{m}$$) = $$\text{24}$$ $$\text{m}$$ 2. Length of fencing = $$\text{2}$$ $$\times$$ ($$\text{10}$$ $$\text{m}$$ + $$\text{6}$$ $$\text{m}$$) = $$\text{32}$$ $$\text{m}$$ 3. Area of pool = $$\text{8}$$ $$\text{m}$$ $$\times$$ $$\text{4}$$ $$\text{m}$$ = $$\text{32}$$ $$\text{m}$$$$^{\text{2}}$$ 4. Cost of fence= $$\text{R}\,\text{250,00}$$ $$\times$$ $$\text{32}$$ $$\text{m}$$ = $$\text{R}\,\text{8 000}$$ 5. Cost of netting= $$\text{R}\,\text{199,99}$$ $$\times$$ $$\text{32}$$ $$\text{m}$$$$^{\text{2}}$$ $$\text{R}\,\text{6 399,68}$$ 1. Perimeter of pool = $$\text{2}\pi r$$ = $$\text{2}$$ $$\times$$ $$\text{3,142}$$ $$\times$$ $$\text{3,5}$$ = $$\text{21,99}$$ $$\text{m}$$ 2. Length of fencing = $$\text{2}\pi (\text{4,5}$$ m) = $$\text{28,28}$$ $$\text{m}$$ 3. Area of pool = $$\pi r^{\text{2}}$$ = $$\text{3,142}$$ $$\times$$ ($$\text{3,5}$$ $$\text{m}$$)$$^{\text{2}}$$ = $$\text{38,49}$$ $$\text{m}$$$$^{\text{2}}$$ 4. Cost of fence = $$\text{R}\,\text{250,00}$$ $$\times$$ $$\text{28,28}$$ $$\text{m}$$ = $$\text{R}\,\text{7 070}$$ 5. Cost of netting= $$\text{R}\,\text{199,99}$$ $$\times$$ $$\text{38,49}$$ $$\text{m}$$$$^{\text{2}}$$ = $$\text{R}\,\text{7 697,62}$$ 1. Perimeter of pool = $$\text{2}$$(perimeter of semi-circles) + length of $$\text{2}$$ rectangular sides = (perimeter one circle with radius $$\text{2}$$ $$\text{m}$$) + $$\text{2}$$($$\text{7}$$ $$\text{m}$$) = $$\text{2}\pi (\text{2}$$m) + $$\text{14}$$ $$\text{m}$$ = $$\text{12,568}$$ $$\text{m}$$ + $$\text{14}$$ $$\text{m}$$ = $$\text{26,57}$$ $$\text{m}$$ 2. Length of fencing = $$\text{2}$$(perimeter of semi-circles) + length of $$\text{2}$$ rectangular sides = (perimeter one circle with radius $$\text{4}$$ $$\text{m}$$) + $$\text{2}$$($$\text{7}$$ $$\text{m}$$) = $$\text{2}\pi (\text{4}$$m) + $$\text{14}$$ $$\text{m}$$ = $$\text{25,136}$$ $$\text{m}$$ + $$\text{14}$$ $$\text{m}$$ = $$\text{39,14}$$ $$\text{m}$$ 3. Area of pool = Area rectangle + $$\text{2}$$(area semi-circles) = Area rectangle + (area cricle radius $$\text{2}$$ $$\text{m}$$)= ($$\text{7}$$ $$\text{m}$$ $$\times$$ $$\text{4}$$ $$\text{m}$$) + $$(\pi (\text{2})^\text{2})$$ = $$\text{28}$$ $$\text{m}$$$$^{\text{2}}$$ + $$\text{12,568}$$ $$\text{m}$$$$^{\text{2}}$$ = $$\text{40,57}$$ $$\text{m}$$$$^{\text{2}}$$ 4. Cost of fence = $$\text{R}\,\text{25,00}$$ $$\times$$ $$\text{39,14}$$ $$\text{m}$$ = $$\text{R}\,\text{9 785}$$ 5. Cost of netting = $$\text{R}\,\text{199,99}$$ $$\times$$ $$\text{40,57}$$ $$\text{m}$$$$^{\text{2}}$$ =$$\text{R}\,\text{8 113,59}$$ Below is the plan of Phumza's property. Homeowners pay rates to the municipality which are calculated according to the area of the property. Calculate the area of this house. Round your answer to the nearest whole metre. Area = (Area small rectangle) + (area big rectangle) = ($$\text{4,7}$$ $$\text{m}$$ $$\times$$ $$\text{4,3}$$ $$\text{m}$$) + ($$\text{10,8}$$ $$\text{m}$$ $$\times$$ $$\text{12,8}$$ $$\text{m}$$) = $$\text{20,21}$$ $$\text{m}$$$$^{\text{2}}$$ + $$\text{138,24}$$ $$\text{m}$$$$^{\text{2}}$$ = $$\text{158,45}$$ $$\text{m}$$$$^{\text{2}}$$ The basic municipal rate is calculated using the following formula: $$\text{R}\,\text{15,05}$$ per m$$^{\text{2}}$$ of the property per year. What would Phumza's monthly rate bill be? Rate per year = Area $$\times$$ $$\text{R}\,\text{15,05}$$ = $$\text{158,45}$$ $$\text{m}$$$$^{\text{2}}$$ $$\times$$ $$\text{R}\,\text{15,05}$$ = $$\text{R}\,\text{2 384,6725}$$ per year. Per month = $$\text{R}\,\text{2 384,6725}$$ $$\div$$ $$\text{12}$$ = $$\text{R}\,\text{198,72}$$ per month. Lebo wants to put paving around her new, triangular vegetable garden as shown in the diagram: What is the area of the vegetable garden in metres squared? Area = $$\frac{\text{1}}{\text{2}}$$($$\text{2,4}$$ $$\text{m}$$)($$\text{90}$$ $$\text{cm}$$) = $$\frac{\text{1}}{\text{2}}$$($$\text{2,4}$$)($$\text{0,9}$$ $$\text{m}$$) = $$\text{1,08}$$ $$\text{m}$$$$^{\text{2}}$$ What is the area of the paving, in metres squared? (excluding he vegetable garden!) Area = (Area of rectangle) - (area of triangle) = [($$\text{80}$$ + $$\text{90}$$ + $$\text{80}$$ $$\text{cm}$$) $$\times$$ ($$\text{1,2}$$ + $$\text{2,4}$$ + $$\text{1,2}$$ $$\text{m}$$)] - $$\text{1,08}$$ $$\text{m}$$$$^{\text{2}}$$ = [$$\text{2,5}$$ $$\text{m}$$ $$\times$$ $$\text{4,8}$$ $$\text{m}$$] - $$\text{1,08}$$ $$\text{m}$$$$^{\text{2}}$$ = $$\text{12}$$ $$\text{m}$$$$^{\text{2}}$$ -$$\text{1,08}$$ $$\text{m}$$$$^{\text{2}}$$ = $$\text{10,92}$$ $$\text{m}$$$$^{\text{2}}$$ If the paving is going to cost $$\text{R}\,\text{24,65}$$ per metre squared, how much will the total cost of the paving be? $$\text{10,92}$$ $$\text{m}$$$$^{\text{2}}$$ $$\times$$ $$\text{R}\,\text{24,65}$$ = $$\text{R}\,\text{269,18}$$ Lebo wants to build a fence around her garden. To do this, she needs to calculate the perimeter of the triangular garden. Is this possible with the information provided in the diagram? Explain your answer. No. We do not know the length of the third side of the triangle so we cannot calculate the perimeter. As a decorative feature, Jan builds a round window into the attic of his house. Jan needs to paint the triangular wall around the window. What is the area of the triangular piece of wall in m$$^{\text{2}}$$? (Use the formula: Area = $$\frac{\text{1}}{\text{2}} \times$$ base $$\times$$ height ) Area of triangle = $$\frac{\text{1}}{\text{2}} \times$$ base $$\times$$ height = $$\frac{\text{1}}{\text{2}}$$($$\text{2,63}$$ $$\text{m}$$)($$\text{2,1}$$ $$\text{m}$$) = $$\text{2,76}$$ $$\text{m}$$$$^{\text{2}}$$ Assume it takes $$\text{1}$$ litre of paint to cover $$\text{0,5}$$ $$\text{m}$$$$^{\text{2}}$$ of wall. How many litres of paint will Jan need to buy? $$\text{2,76}$$ $$\text{m}$$$$^{\text{2}} \div$$ $$\text{0,5}$$ $$\text{m}$$$$^{\text{2}}$$ = $$\text{5,52}$$ litres of paint If the hardware store only sells paint in $$\text{2}$$ litre tins, how many tins will Jan need to buy? $$\text{5,52}$$ $$\div$$ $$\text{2}$$ = $$\text{2,76}$$ tins. He cannot buy $$\text{0,76}$$ of a tin, so he will have to buy $$\text{3}$$ tins.
Math 10 Chapter 6 Notes: The Normal Distribution 1 / 10 # Math 10 Chapter 6 Notes: The Normal Distribution - PowerPoint PPT Presentation Math 10 Chapter 6 Notes: The Normal Distribution. Notation: X is a continuous random variable X ~ N(  ,  ) Parameters:  is the mean and  is the standard deviation Graph is bell-shaped and symmetrical The mean, median, and mode are the same (in theory). I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described. ## PowerPoint Slideshow about 'Math 10 Chapter 6 Notes: The Normal Distribution' - berne Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - Presentation Transcript Math 10 Chapter 6 Notes: The Normal Distribution • Notation: X is a continuous random variable X ~ N(, ) • Parameters:  is the mean and  is the standard deviation • Graph is bell-shaped and symmetrical • The mean, median, and mode are the same (in theory) Math 10 Chapter 6 Notes: The Normal Distribution • Total area under the curve is equal to 1. Probability = Area • P(X < x) is the cumulative distribution function or Area to the Left. • A change in the standard deviation, , causes the curve to become wider or narrower • A change in the mean, , causes the graph to shift Math 10 Chapter 6 Notes: The Standard Normal Distribution • A normal (bell-shaped) distribution of standardized values called z-scores. • Notation: Z ~ N(0, 1) • A z-score is measured in terms of the standard deviation. • The formula for the z-score is Math 10 Chapter 6 Notes: The Normal Distribution • Bell-shaped curve • Most values cluster about the mean • Area within 4 standard deviations (+ or - 4 ) is 1 Math 10 Chapter 6 Notes: The Normal Distribution Ex. Suppose X ~ N(100, 5). Find the z-score (the standardized score) for x = 95 and for 110. = 95 – 100 = - 1 5 = 110 – 100 = 2 5 Math 10 Chapter 6 Notes: The Normal Distribution · The z-score lets us compare data that are scaled differently. Ex. X~N(5, 6) and Y~N(2, 1) with x = 17 and y = 4; X = Y = weight gain 17 – 5 = 2 4 – 2 = 2 6 1 Math 10 Chapter 6 Notes: The Standard Normal Distribution · Ex. Suppose Z ~ N(0, 1). Draw pictures and find the following. 1. P(-1.28 < Z < 1.28) 2. P(Z < 1.645) 3. P(Z > 1.645) 4. The 90th percentile, k, for Z scores. For 1, 2, 3 use the normal cdf For 4, use the inverse normal Math 10 Chapter 6 Notes: The Normal Distribution Ex: At the beginning of the term, the amount of time a student waits in line at the campus store is normally distributed with a mean of 5 minutes and a standard deviation of 2 minutes. Let X = the amount of time, in minutes, that a student waits in line at the campus store at the beginning of the term. X ~ N(5, 2) where the mean = 5 and the standard deviation = 2. Math 10 Chapter 6 Notes: The Normal Distribution Find the probability that one randomly chosen student waits more than 6 minutes in line at the campus store at the beginning of the term. P(X > 6) = 0.3085. Math 10 Chapter 6 Notes: The Normal Distribution Find the 3rd quartile. The third quartile is equal to the 75th percentile. Let k = the 75th percentile (75th %ile). P(X < k ) = 0.75. The 3rd quartile or 75th percentile is 6.35 minutes (to 2 decimal places). Seventy-five percent of the waiting times are less than 6.35 minutes.
# To find the Number of Bricks or Tiles when Area of Path and Brick is given To find the number of bricks or tiles when area of path and brick is given are discussed here. We need to divide area of path or wall by area of a brick or tile. Number of bricks or tiles = area of path or wall ÷ area of a brick or tile. For Example: 1. Find the number of bricks to be laid in a square path of side 18 cm, if the side of each brick is 3 cm. Solution: Area of path = 18 × 18 sq. cm. Area of each brick = 3 × 3 sq. cm. Therefore, number of bricks = area of path ÷ area of each brick. = (18 × 18) ÷ (3 × 3) = 324 ÷ 9 = 36 Therefore, number of bricks = 36. 2. Find the number of bricks to be laid in a square path of side 25 cm, if the side of each brick is 5 cm. Solution: Area of path = 25 × 25 sq. cm. Area of each brick = 5 × 5 sq. cm. Therefore, number of bricks = area of path ÷ area of each brick. = (25 × 25) ÷ (5 × 5) = 625 ÷ 25 = 25 Therefore, number of bricks = 25. Area of a Rectangle. Area of a Square. To find Area of a Rectangle when Length and Breadth are of Different Units. To find Length or Breadth when Area of a Rectangle is given. Areas of Irregular Figures. To find Cost of Painting or Tilling when Area and Cost per Unit is given. To find the Number of Bricks or Tiles when Area of Path and Brick is given. Worksheet on Area. Worksheet on Area of a Square and Rectangle Worksheet on Area of Regular Figures Practice Test on Area. 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. ### Adding 1-Digit Number \| Understand the Concept one Digit Number Sep 17, 24 02:25 AM Understand the concept of adding 1-digit number with the help of objects as well as numbers. 2. ### Counting Before, After and Between Numbers up to 10 \| Number Counting Sep 17, 24 01:47 AM Counting before, after and between numbers up to 10 improves the child’s counting skills. 3. ### Worksheet on Three-digit Numbers \| Write the Missing Numbers \| Pattern Sep 17, 24 12:10 AM Practice the questions given in worksheet on three-digit numbers. The questions are based on writing the missing number in the correct order, patterns, 3-digit number in words, number names in figures… 4. ### Arranging Numbers \| Ascending Order \| Descending Order \|Compare Digits Sep 16, 24 11:24 PM We know, while arranging numbers from the smallest number to the largest number, then the numbers are arranged in ascending order. Vice-versa while arranging numbers from the largest number to the sma…
In geometry, there are various types of angles, based on measurement. The names of basic angles are Acute angle, Obtuse angle, Right angle, Straight angle, reflex angle and full rotation. An angle is geometrical shape formed by joining two rays at their end-points. An angle is usually measured in degrees. There are various types of angles in geometry. Angles form the core part of the geometry in mathematics. They are the fundamentals that eventually lead to the formation of the more complex geometrical figures and shapes. ## What are Angles When two rays combine with a common endpoint and the angle is formed. The two components of an angle are “sides” and “vertex”. The side can be categorized into terminal sides and initial sides (or vertical sides) as shown in the image below. These two rays can combine in multiple fashions to form the different types of angles in mathematics. Let us begin by studying these different types of angles in geometry. ### Parts of Angle • Vertex – Point where the arms meet. • Arms – Two straight line segments form a vertex. • Angle – If a ray is rotated about its endpoint, the measure of its rotation is called angle between its initial and final position. ## Classification of Angles Angles can be classified into two main types: • Based on Magnitude • Based on Rotation ## Six Types of Angles In Maths, there are mainly 5 types of angles based on their direction. These five angle types are the most common ones used in geometry. These are: • Acute Angles • Obtuse Angles • Right Angles • Straight Angles • Reflex Angles • Full Rotation The images above illustrate certain types of angles. ### Acute Angle An acute angle lies between 0 degree and 90 degrees, or in other words; an acute angle is one that is less than 90 degrees. The figure above illustrates an acute angle. ### Obtuse Angle An obtuse angle is the opposite of an acute angle. It is the angle which lies between 90 degrees and 180 degrees or in other words; an obtuse angle is greater than 90 degrees and less than 180 degrees. The figure above illustrates an obtuse angle. ### Right Angle 3 A right angle is always equal to 90 degrees. Any angle less than 90 degrees is an acute angle whereas any angle greater than 90 degrees is an obtuse angle. The figure above illustrates a right angle or a 90-degree angle. ### Straight Angle A straight angle is 180 degrees when measured. The figure above illustrates a straight angle or a 180-degree angle. You can see that it is just a straight line because the angle between its arms is 180 degrees. ### Reflex Angle Since this measurement is less than 90 degrees, the arms form an acute angle. But what about the angle on the other side? What is the larger angle that is complementary to the acute angle called? It is called a reflex angle. The image below illustrates a reflex angle. Any angle that has a measure which is greater than 180 degrees but less than 360 degrees (which coincides with 0 degrees) is a reflex angle. ### Full Rotation An angle equal to 360 degrees is called full rotation or full angle. It is formed when one of the arms takes a complete rotation to form an angle. ## Angle Types Based on Rotation Based on the direction of measurement or the direction of rotation, angles can be of two types: • Positive Angles • Negative Angles ### Positive Angles Positive angles are those angles which are measured in a counterclockwise direction from the base. In most cases, positive angles are used to represent angles in geometry. From the origin, if an angle is drawn in the (+x, +y) plane, it forms a positive angle. ### Negative Angles Negative angles are those angles which are measured in a clockwise direction from the base. From the origin, if an angle is drawn towards the (x, -y) plane, it forms a negative angle. ## Pair of Angles When two angles are paired, then there exist different angles, such as • Complementary angles • Supplementary angles • Linear Pair • Vertically Opposite angles ### Complementary and Supplementary Angles Apart from the aforementioned types, there are two more angle types which are complementary angles and supplementary angles. If the sum of two angles is equal to 180°, then they are supplementary angles, and if the sum is equal to 90°, then they are called complementary angles. ### Linear Pair When the non-common arms of adjacent angles are just opposite to each other, or they extend in the opposite direction, then they are called linear pairs. By linear it is clear that they form a straight line. When two angles are connected with one common arm and have one common vertex and also the non-common arms are either side of the common arm, then they are called adjacent angles. ### Vertical Angles When two lines intersect each other at a single point (called vertex), then the angle formed on either side of the common vertex is called vertical angles or vertically opposite angles. ## Angles Formed By Transversal A line that cuts or intersects two or more lines at different points, is called a transversal. It is therefore, there are angles formed at the point of intersection. They are: • Interior angles • Exterior angles • Pairs of Alternate interior angles • Pairs of Alternate exterior angles • Pairs of Corresponding angles • Pairs of interior angles on the same side of the transversal ## Frequently Asked Questions – FAQs ### What are the six different angles in geometry based on measurement? The six different angles in geometry based on magnitude are: Acute angle, Obtuse angle, Right angle, Straight angle, Reflex angle and full angle ### What is an acute angle in geometry? Give examples. The angle that measures less than 90 degrees is called acute angle. Examples: 30°, 45°, 60°, 85° are acute angles. ### What is obtuse angle? Give an example. The angle that measures greater than 90 degrees is obtuse. For example, 145° is an obtuse angle. ### What is reflex angle? Give an example. Reflex angle is an angle that measures more than 180° and is less than 360°. For example, 270° is a reflex angle. ### What do you mean by zero angle? When both the arms of an angle overlap each other and measure the angle of 0°, then it is called the zero angle. ### Is obtuse angle and reflex angle same? Obtuse angle is different from reflex angle because obtuse lies between 90 degrees and 180 degrees but reflex is always more than 180 degrees.
Congratulations on starting your 24-hour free trial! Quick Homework Help # Quotient Rule of Logarithms Star this video When evaluating logarithms the logarithmic rules, such as the quotient rule of logarithms, can be useful for rewriting logarithmic terms. The quotient rule of logarithms allows us to separate parts of a quotient within a log. The quotient rule of logarithms is useful for expanding and condensing logarithms, along with the product rule and the power rule of logarithms. Simplifying a logs with a statement with 2 logs and subtraction. Okay? So for this example what I want to do is look at these 2 things individually and then see if there's a way we can throw them together. So log base 3 of 27, so saying 3 to what power is equal to 27. We know that to be 3. Log base 3 of 9, saying 3 to what power is equal to 9. 3 squared so this is going to be 2. We're subtracting in between so this is just going to be 1. Okay. We can combine these actually and how we do that is by division. so this turns out to be log base 3, 27 over 9. 27 over 9 is 3. Log base 3 of 3, 3 to what power is 3. This is 1. What this gets us is the quotient rule of logarithms and what that tells us is if we are ever dividing within our log, so we have log b of x over y. This is going to be equal to log base b of x minus log base b of y, okay. This is the quotient rule of logarithms. Basically whatever's in the top is going to go first and then we subtract whatever is in the bottom. Okay? Careful to note that this is not the same thing for log base b of x-y, okay? This only works when we're dividing in the inside of our log. It doesn't work for when we're subtracting. Okay? And we also can go both ways. We'll be able to go from a division in the log to a subtraction or from a subtraction back to division, okay? So that's the quotient rule of logarithms. ## Find Videos Using Your Textbook Enjoy 3,000 videos just like this one.
# Ex.3.6 Q2 Pair of Linear Equations in Two Variables Solution - NCERT Maths Class 10 Go back to 'Ex.3.6' ## Question Formulate the following problems as a pair of equations, and hence find their solutions: (i) Ritu can row downstream $$20\,\rm{ km}$$ in $$2$$ hours, and upstream $$4 \,\rm{km}$$ in $$2$$ hours. Find her speed of rowing in still water and the speed of the current. (ii) 2 women and $$5$$ men can together finish an embroidery work in $$4$$ days, while $$3$$ women and 6 men can finish it in $$3$$ days. Find the time taken by $$1$$ woman alone to finish the work, and also that taken by $$1$$man alone. (iii) Roohi travels $$300 \,\rm{km}$$ to her home partly by train and partly by bus. She takes $$4$$ hours if she travels $$60 \,\rm{km}$$ by train and remaining by bus. If she travels $$100\,\rm{ km}$$ by train and the remaining by bus, she takes $$10$$ minutes longer. Find the speed of the train and the bus separately. Video Solution Pair Of Linear Equations In Two Variables Ex 3.6 \| Question 2 ## Text Solution Reasoning: Steps: (i) Let the Ritu’s speed of rowing in still water and the speed of stream be $$x\,{\rm{ km/h}}$$and $$y\,{\rm{km/h }}$$ respectively. Ritu’s speed of rowing; Upstream $$= \left( {x – y} \right) \,\rm{km/h}$$ Downstream $$= \left( {x + y} \right)\,\rm{km/h}$$ According to question, Ritu can row downstream $$20 \,\rm{km}$$ in $$2$$ hours, \begin{align}2\left( {x + y} \right) &= 20\\x + y& = 10 \qquad\left( 1 \right)\end{align} Ritu can row upstream $$4\,\rm{ km}$$ in $$2$$ hours, \begin{align}2\left( {x - y} \right) &= 4\\x - y &= 2 \qquad \left( 2 \right)\end{align} Adding equation ($$1$$) and ($$2$$), we obtain \begin{align} 2x&=12 \\ x&=6 \\\end{align} Putting $$x = 6$$ in equation $$(1)$$, we obtain \begin{align}6 + y &= 10\\y& = 4\end{align} Hence, Ritu’s speed of rowing in still water is $$6 \,\rm{km/h}$$ and the speed of the current is $$4 \,\rm{km/h}.$$ (ii) Let the number of days taken by a woman and a man to finish the work be $$x$$ and $$y$$ respectively. Therefore, work done by a woman in $$1$$ day $$= \frac{1}{x}$$ and work done by a man in 1 day $$= \frac{1}{y}$$ According to the question, $$2$$ women and $$5$$ men can together finish an embroidery work in $$4$$ days; $\frac{2}{x} + \frac{5}{y} = \frac{1}{4} \qquad \left( 1 \right)$ $$3$$ women and $$6$$ men can finish it in $$3$$ days $\frac{3}{x} + \frac{6}{y} = \frac{1}{3} \qquad \left( 2 \right)$ Substituting \begin{align}\frac{1}{x} = p \end{align} and \begin{align}\frac{1}{y} = q \end{align}in equations $$(1)$$ and $$(2)$$, we obtain \begin{align} &\frac{2}{x} + \frac{5}{y} = \frac{1}{4} \\ \Rightarrow\, & 2p + 5q = \frac{1}{4} \\ \Rightarrow \, & 8p + 20q - 1 = 0 \cdots \left( 3 \right)\\ \\ &\frac{3}{x} + \frac{6}{y} = \frac{1}{3} \\ \Rightarrow \, & 3p + 6q = \frac{1}{3} \\ \Rightarrow \, & 9p + 18q - 1 = 0 \cdots \left( 4 \right) \end{align} By cross-multiplication, we obtain \begin{align}\frac{p}{{ - 20 - ( - 18)}} &= \frac{q}{{ - 9 - ( - 8)}} \\ & = \frac{1}{{144 - 180}}\\ \frac{p}{{ - 2}} &= \frac{q}{{ - 1}} = \frac{1}{{ - 36}}\\ \\ \frac{p}{{ - 2}}& = \frac{1}{{ - 36}}{\text{ and }} \\ \frac{q}{{ - 1}} & = \frac{1}{{ - 36}}\\p &= \frac{1}{{18}}\quad {\text{ and }} \\ q & = \frac{1}{{36}}\end{align} \begin{align}{\text{Therefore, }}p &= \frac{1}{x} = \frac{1}{{18}}\\ & \Rightarrow x = 18\\{\text{and, }}q &= \frac{1}{y} = \frac{1}{{36}}\\& \Rightarrow y = 36\end{align} Hence, number of days taken by a woman is $$18$$ and by a man is $$36.$$ (iii) Let the speed of train and bus be $$u \,\rm{km/h}$$ and $$v\,\rm{ km/h}$$ respectively. According to the given information, Roohi travels $$300\,\rm{ km}$$ and takes $$4$$ hours if she travels $$60 \,\rm{km}$$ by train and the remaining by bus $\frac{{60}}{u} + \frac{{240}}{v} = 4 \qquad \left( 1 \right)$ If she travels $$100\,\rm{ km}$$ by train and the remaining by bus, she takes $$10$$ minutes longer $\frac{{100}}{u} + \frac{{200}}{v} = \frac{{25}}{6} \qquad \left( 2 \right)$ Substituting \begin{align}\frac{1}{u}=p\end{align} and \begin{align}\frac{1}{v}=q \end{align} in equations $$(1)$$ and $$(2)$$, we obtain \begin{align} & \frac{{60}}{{u}} + \frac{{240}}{v} = 4 \\ \Rightarrow\, & 60p + 240q = 4 \cdots \left( 3 \right)\\ \\ & \frac{{100}}{u} + \frac{{200}}{v} = \frac{{25}}{6} \\ \Rightarrow \, & 100p + 200q = \frac{{25}}{6}\\ \Rightarrow \, & 600p + 1200q = 25 \cdots \left( 4 \right)\end{align} Multiplying equation $$(3)$$ by $$10$$, we obtain $600p + 2400q = 40 \quad \left( 5 \right)$ Subtracting equation $$(4)$$ from $$(5)$$, we obtain \begin{align}1200q &= 15\\q &= \frac{{15}}{{1200}}\\q& = \frac{1}{{80}}\end{align} Substituting \begin{align}q = \frac{1}{{80}} \end{align} in equation $$(3)$$, we obtain \begin{align}60p + 240 \times \frac{1}{{80}} &= 4\\60p &= 4 - 3\\p &= \frac{1}{{60}}\end{align} \begin{align}{\text{Therefore, }}p &= \frac{1}{u} = \frac{1}{{60}}\\& \Rightarrow u = 60\\{\text{and, }}q &= \frac{1}{v} = \frac{1}{{80}}\\& \Rightarrow v = 80\end{align} Hence, speed of the train $$= 60\,{\rm{ km/h}}$$ And speed of the bus $$= 80\,{\rm{ km/h}}$$ Learn from the best math teachers and top your exams • Live one on one classroom and doubt clearing • Practice worksheets in and after class for conceptual clarity • Personalized curriculum to keep up with school
1 / 36 # 3-3 3-3. Writing Functions. Holt Algebra 1. Warm Up. Lesson Presentation. Lesson Quiz. Holt McDougal Algebra 1. 1 2. c + b. Warm Up Evaluate each expression for a = 2, b = –3, and c = 8. 1. a + 3 c 2. ab – c 3. 26. –14. 1. 4. 4 c – b. 35. 5. b a + c. 17. ## 3-3 E N D ### Presentation Transcript 1. 3-3 Writing Functions Holt Algebra 1 Warm Up Lesson Presentation Lesson Quiz Holt McDougal Algebra 1 2. 1 2 c + b Warm Up Evaluate each expression for a = 2, b = –3, and c = 8. 1. a + 3c 2.ab – c 3. 26 –14 1 4. 4c – b 35 5. ba + c 17 3. Objectives Identify independent and dependent variables. Write an equation in function notation and evaluate a function for given input values. 4. Vocabulary independent variable dependent variable function rule function notation 5. 5 – 4 = 1 or Example 1: Using a Table to Write an Equation Determine a relationship between the x- and y-values. Write an equation. 5 10 15 20 1 3 2 4 Step 1 List possible relationships between the first x or y-values. 6. 10 – 4 2 and 15 – 4 3 and 20 – 4 4 and The value of y is one-fifth, , of x. or Example 1 Continued Step 2 Determine if one relationship works for the remaining values. Step 3 Write an equation. The value of y is one-fifth of x. 7. Check It Out! Example 1 Determine a relationship between the x- and y-values. Write an equation. {(1, 3), (2, 6), (3, 9), (4, 12)} x 4 1 2 3 y 3 6 9 12 Step 1 List possible relationships between the first x- or y-values. 1  3 = 3 or 1 + 2 = 3 8. 2 • 3= 6 2 + 2 6 3 • 3 = 9 3 + 2 9 4 • 3 = 12 4 + 2 12 Check It Out! Example 1 Continued Step 2 Determine if one relationship works for the remaining values. The value of y is 3 times x. Step 3 Write an equation. y = 3x The value of y is 3 times x. 9. The equation in Example 1 describes a function because for each x-value (input), there is only one y-value (output). 10. The inputof a function is the independent variable. The output of a function is the dependent variable. The value of the dependent variable depends on, or is a function of, the value of the independent variable. 11. Example 2A: Identifying Independent and Dependent Variables Identify the independent and dependent variables in the situation. A painter must measure a room before deciding how much paint to buy. The amount of paintdependson the measurement of a room. Dependent: amount of paint Independent: measurement of the room 12. Example 2B: Identifying Independent and Dependent Variables Identify the independent and dependent variables in the situation. The height of a candle decreases d centimeters for every hour it burns. The height of a candledepends on the number of hours it burns. Dependent: height of candle Independent: time 13. Example 2C: Identifying Independent and Dependent Variables Identify the independent and dependent variables in the situation. A veterinarian must weigh an animal before determining the amount of medication. The amount of medicationdepends on the weight of an animal. Dependent: amount of medication Independent: weight of animal 14. Helpful Hint There are several different ways to describe the variables of a function. Independent Variable Dependent Variable y-values x-values Domain Range Input Output x f(x) 15. Check It Out! Example 2a Identify the independent and dependent variable in the situation. A company charges \$10 per hour to rent a jackhammer. The cost to rent a jackhammerdependson the length of time it is rented. Dependent variable: cost Independent variable: time 16. Check It Out! Example 2b Identify the independent and dependent variable in the situation. Apples cost \$0.99 per pound. The cost of applesdepends on the number of pounds bought. Dependent variable: cost Independent variable: pounds 17. An algebraic expression that defines a function is a function rule. Suppose Tasha earns \$5 for each hour she baby-sits. Then 5 •x is a function rule that models her earnings. If x is the independent variable and y is the dependent variable, then function notation for y is f(x), read “f of x,” where f names the function. When an equation in two variables describes a function, you can use function notation to write it. 18. The dependent variableisa function ofthe independent variable. yisa function ofx. y=f(x) y = f(x) 19. Example 3A: Writing Functions Identify the independent and dependent variables. Write an equation in function notation for the situation. A math tutor charges \$35 per hour. The fee a math tutor charges depends on number of hours. Dependent: fee Independent: hours Let h represent the number of hours of tutoring. The function for the amount a math tutor charges isf(h) = 35h. 20. Example 3B: Writing Functions Identify the independent and dependent variables. Write an equation in function notation for the situation. A fitness center charges a \$100 initiation fee plus \$40 per month. The total cost depends on the number of months, plus \$100. Dependent: total cost Independent: number of months Let mrepresent the number of months The function for the amount the fitness center charges is f(m) = 40m + 100. 21. Check It Out! Example 3a Identify the independent and dependent variables. Write an equation in function notation for the situation. Steven buys lettuce that costs \$1.69/lb. The total cost depends on how many pounds of lettuce Steven buys. Dependent: total cost Independent: pounds Let x represent the number of pounds Steven buys. The function for the total cost of the lettuce is f(x) = 1.69x. 22. Check It Out! Example 3b Identify the independent and dependent variables. Write an equation in function notation for the situation. An amusement park charges a \$6.00 parking fee plus \$29.99 per person. The total cost depends on the number of persons in the car, plus \$6. Dependent: total cost Independent: number of persons in the car Let xrepresent the number of persons in the car. The function for the total park cost is f(x) = 29.99x+ 6. 23. You can think of a function as an input-output machine. For Tasha’s earnings, f(x) = 5x. If you input a value x, the output is 5x. input x 2 6 function f(x)=5x 5x 10 30 output 24. Example 4A: Evaluating Functions Evaluate the function for the given input values. For f(x) = 3x + 2, find f(x) when x = 7 and when x = –4. f(x) = 3(x) + 2 f(x) = 3(x) + 2 Substitute 7 for x. Substitute –4 for x. f(–4) = 3(–4) + 2 f(7) = 3(7) + 2 Simplify. = –12 + 2 = 21 + 2 Simplify. = 23 = –10 25. Example 4B: Evaluating Functions Evaluate the function for the given input values. For g(t) = 1.5t – 5, find g(t) when t = 6 and when t = –2. g(t) = 1.5t– 5 g(t) = 1.5t– 5 g(6) = 1.5(6) – 5 g(–2) = 1.5(–2) – 5 = –3– 5 = 9– 5 = 4 = –8 26. For , find h(r) when r = 600 and when r = –12. Example 4C: Evaluating Functions Evaluate the function for the given input values. = 202 = –2 27. Check It Out! Example 4a Evaluate the function for the given input values. For h(c) = 2c – 1, find h(c) when c = 1 and when c = –3. h(c) = 2c– 1 h(c) = 2c– 1 h(1) = 2(1) – 1 h(–3) = 2(–3) – 1 = 2– 1 = –6– 1 = 1 = –7 28. Check It Out! Example 4b Evaluate the function for the given input values. For g(t) = , find g(t) when t = –24 and when t = 400. = –5 = 101 29. When a function describes a real-world situation, every real number is not always reasonable for the domain and range. For example, a number representing the length of an object cannot be negative, and only whole numbers can represent a number of people. 30. Money spent is\$15.00 for each DVD. Example 5: Finding the Reasonable Domain and Range of a Function Joe has enough money to buy 1, 2, or 3 DVDs at \$15.00 each, if he buys any at all. Write a function to describe the situation. Find the reasonable domain and range of the function. f(x)=\$15.00• x If Joe buys x DVDs, he will spend f(x) = 15x dollars. Joe only has enough money to purchase 1, 2, or 3 DVDs. A reasonable domain is {0, 1, 2, 3}. 31. x 1 2 3 f(x) 15(1) = 15 15(2) = 30 15(3) = 45 Example 5 Continued Substitute the domain values into the function rule to find the range values. 0 15(0) = 0 A reasonable range for this situation is {\$0, \$15, \$30, \$45}. 32. Number of watts used is 500timesthe setting #. watts Check It Out! Example 5 The settings on a space heater are the whole numbers from 0 to 3. The total number of watts used for each setting is 500 times the setting number. Write a function to describe the number of watts used for each setting. Find the reasonable domain and range for the function. f(x) = 500 •x For each setting, the number of watts is f(x) = 500x watts. 33. 0 1 3 2 x 500(2) = 1,000 500(0) = 0 500(1) = 500 500(3) = 1,500 f(x) Check It Out! Example 5 There are 4 possible settings 0, 1, 2, and 3, so a reasonable domain would be {0, 1, 2, 3}. Substitute these values into the function rule to find the range values. The reasonable range for this situation is {0, 500, 1,000, 1,500} watts. 34. Lesson Quiz: Part I Identify the independent and dependent variables. Write an equation in function notation for each situation. 1. A buffet charges \$8.95 per person. independent: number of people dependent: cost f(p) = 8.95p 2. A moving company charges \$130 for weekly truck rental plus \$1.50 per mile. independent: miles dependent: cost f(m) = 130 + 1.50m 35. Lesson Quiz: Part II Evaluate each function for the given input values. 3. For g(t) = , find g(t) when t = 20 and when t = –12. g(20) = 2 g(–12) = –6 4. For f(x) = 6x – 1, find f(x) when x = 3.5 and when x = –5. f(3.5) = 20 f(–5) = –31 36. Lesson Quiz: Part III Write a function to describe the situation. Find the reasonable domain and range for the function. 5. A theater can be rented for exactly 2, 3, or 4 hours. The cost is a \$100 deposit plus \$200 per hour. f(h) = 200h + 100 Domain: {2, 3, 4} Range: {\$500, \$700, \$900} More Related
Why is the formula for the area of a trapezoid? Dissecting the trapezoid The two parallel sides are the bases, and height, as always, is the perpendicular distance from one base to the opposite. The area of this parallelogram is its height (half-height of the trapezoid) times its base (sum of the bases of the trapezoid), so its area is half-height × (base1 + base2). How do you find the missing side of a trapezoid? Solve for H. To do this, multiply each side by H, then divide each side by the angle sine. Or, you can simply divide the height of the triangle by the angle sine. So, the length of the hypotenuse, and the first missing side of the trapezoid, is about 10.4566 cm. What is the formula for finding the height of a trapezoid? Write A=h(b1+b2)/2, where A represents the trapezoid’s area, b1 represents one of the base lengths, b2 represents the other base length and h represents the height. Rearrange the equation to get h alone. Multiply both sides of the equation by 2 to get. 2A=h(b1+b2). Why is the area of a trapezoid 1 2h b1 b2? The two parallel sides of a trapezoid are its bases. If we call the longer side b1 and the shorter side b2, then the base of the parallelogram is b1 + b2. Area of a trapezoid = 1 2 (base 1 + base 2)(height). A = 1 2 h(b1 + b2) The area of a trapezoid is half its height multiplied by the sum of its two bases. What is a trapezoid look like? A trapezoid is a four-sided flat shape with one pair of opposite parallel sides. It looks like a triangle that had its top sliced off parallel to the bottom. Usually, the trapezoid will be sitting with the longest side down, and you will have two sloping sides for the edges. What does trapezoid mean? 1a : a quadrilateral having only two sides parallel. b British : trapezium sense 1a. 2 : a bone in the wrist at the base of the metacarpal of the index finger. How do you find the missing side lengths? Given that the hypotenuse of a right triangle is and one side is , find the other side length. Explanation: To find the length of the missing side, you can either use the pythagorean theorm or realize this is a case of a special right triangle with sides . Thus, the other side length is . How do you find area? To find the area of a rectangle, multiply its height by its width. For a square you only need to find the length of one of the sides (as each side is the same length) and then multiply this by itself to find the area. How do you find the height of a trapezoid without knowing the area? Explanation: To solve this question, you must divide the trapezoid into a rectangle and two right triangles. Using the Pythagorean Theorem, you would calculate the height of the triangle which is 4. The dimensions of the rectangle are 5 and 4, hence the area will be 20. What’s the equation for a triangle? Equilateral Triangle You might be interested: Which is the word equation for photosynthesis? Equilateral Triangle A triangle with all three sides of equal length. a = b = c. A = B = C = Pi/3 radians = 60o P = 3a s = 3a/2 K = a2sqrt(3)/4 ha = ma = ta = a sqrt(3)/2 R = a sqrt(3)/3 r = a sqrt(3)/6 What formula is a 1 2h b1 b2? The formula for the area of a trapezoid is A=1/2h(b1+b2) express b1 in terms of A, h and b2. lol225 is waiting for your help. Releated Equation of vertical line How do you write an equation for a vertical and horizontal line? Horizontal lines go left and right and are in the form of y = b where b represents the y intercept. Vertical lines go up and down and are in the form of x = a where a represents the shared x coordinate […] Bernoulli’s equation example What does Bernoulli’s equation State? Bernoulli’s principle states the following, Bernoulli’s principle: Within a horizontal flow of fluid, points of higher fluid speed will have less pressure than points of slower fluid speed. Why is Bernoulli’s equation used? The Bernoulli equation is an important expression relating pressure, height and velocity of a fluid at one […]
# Tag Archives: sequences ## 5050 5050 is close to 5150 which can be interpreted as the radio call for an involuntary psychiatric hold. But that’s not why we’re here. 5050 can also be expressed as a summation. A summation consists of an expression containing a dummy variable through which the dummy variable is used to perform some calculation. $\sum_{n=1}^{\infty} 2^{-n} = 1$ This summation specifically is an infinite summation. The summation can also be called an infinite series. Before diving into summations we must first understand sequences and series. Sequences are ordered lists of numbers. There are no operations being performed on the numbers, they exist only as an ordered list. ${1, 1, 2, 3, 5, 8, ...}$ A series is the sum of a sequence. Where above we have the sequence, the series would be: ${1 + 1 + 2 +3 + 5 + 8 + ...}$ Finite summations and series become especially useful when they can be manipulated to fit an existing form for which there is a known formula. $\sum_{n=1}^{n} = \frac{n(n+1)}{2}$ Read plainly: The sum from n = 1 to n items is equal to n times the quantity n plus 1 all over 2. It’s said that this formula can be attributed to Gauss. Where Gauss was given the task of adding all integers 1 through 100 but through an observation was able to derive the above formula. $\sum_{n=1}^{100} = \frac{100(100 + 1)}{2} = 5050$ Consider the series that represents this summation: ${1 + 2 + 3 + ... + 98 + 99 + 100}$ The series can be split into two series consisting of 50 terms each and rewritten as follows. $\{1 + 2 + 3 + ... + 48 + 49 + 50\} \\ + \\ \{100 + 99 + 98 + ... + 53 + 52 + 51\}$ Adding these vertically the following series is obtained. $\{ 101 + 101 + 101 + ... + 101 + 101 + 101 \}$ Of notable interest is the series now contains 50 terms, all being the number 101. To complete the whole picture consider another way to write 50 in terms of a number divided by 2: $\frac{100}{2}$ When the value of all the terms, 101, is multiplied by the alternative way to write 50 the derived formula become apparent. $\frac{100(101)}{2} = \frac{100(100 + 1)}{2} = 50 \cdot 101 = 5050$
# Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.1 ## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.1 Question 1. Prove by vector method that if a line is drawn from the centre of a circle of a circle to the midpoint of a chord, then the line is perpendicular to the chord. Solution: Let ‘C’ be the mid point of the chord AB Take ‘O’ on the centre of the circle. Question 2. Prove by vector method that the median to the base of an isosceles triangle is perpendicular to the base. Solution: Let OAB be an isosceles triangle with OA = OB Let OC be the median to the base AB C is the midpoint of AB Take O as origin. Question 3. Prove by vector method that an angle in a semi-circle is a right angle. Solution: Let AB be the diameter of the circle with centre ‘O’ Let P be any point on the semi-circle. This gives ∠APB = 90°. Hence the result. Question 4. Prove by vector method that the diagonals of a rhombus bisect each other at right angles. Solution: Let ABCD be a rhombus Question 5. Using vector method, prove that if the diagonals of a parallelogram are equal, then it is a rectangle. Solution: Let ABCD be a parallelogram To prove ABCD be a rectangle provided the diagonals are equal. $$\overrightarrow{\mathrm{AB}}$$ ⊥r to $$\overrightarrow{\mathrm{BC}}$$ ⇒ ABCD is a rectangle. Question 6. Prove by vector method that the area of the quadrilateral ABCD having diagonals AC and BD is $$\frac{1}{2}\|\overrightarrow{\mathrm{AC}} \times \overrightarrow{\mathrm{BD}}\|$$ Solution: Vector area of quadrilateral ABCD = {Vector area of ∆ABC} + {Vector area of ∆ ACD} Question 7. Prove by vector method that the parallelogram on the same base and between the same parallels are equal in area. Solution: Let ABCD and ABC’D’ be two parallelogram between the parallels with same base To prove: Area of ABCD = Area of ABC’D’ Question 8. If G is the centroid of a AABC, prove that. (area of ∆GAB) = (area of ∆GBC) = (area of ∆GAC) = $$\frac{1}{3}$$ [area of ∆ABC] Solution: Similarly we can prove Area of ∆GBC = Area of ∆GAC = $$\frac{1}{3}$$ [Area of ∆ABC] Question 9. Using vector method, prove that cos(α – β) = cos α cos β + sin α sin β. Solution: From (1) and (2), we get cos(α + β) = cos α cos β + sin α sin β Question 10. Prove by vector method that sin(α + β) = sin α cos β + cos α sin β. Solution: Take two points A and B on the unit circle with centre as origin ‘O’, so $$\|\overrightarrow{\mathrm{OA}}\|=\|\overrightarrow{\mathrm{OB}}\|$$ = 1 From (1) & (2), we get sin (α + β) = sin α cos β + cos α sin β Question 11. A particle acted on by constant forces $$8 \vec{i}+2 \vec{j}-6 \vec{k}$$ and $$\overrightarrow{6 i}+2 \vec{j}-2 \vec{k}$$ is displaced from the point (1, 2, 3) to the point (5, 4, 1). Find the total work done by the forces. Solution: From (1) & (2), we get Work done by the force = $$\overrightarrow{\mathrm{F}} \cdot \vec{d}$$ = 56 + 8 + 16 = 80 units. Question 12. Forces of magnitude $$5 \sqrt{2}$$ and $$10 \sqrt{2}$$ units acting in the directions $$(3 \vec{i}+4 \vec{j}+5 \vec{k})$$ and $$(10 \vec{i}+6 \vec{j}-8 \vec{k})$$, respectively, act on a particle which is displaced from the point with position vector $$(4 \vec{i}-3 \vec{j}-2 \vec{k})$$ to the point with position vector $$(\overrightarrow{6 i}+\vec{j}-3 \vec{k})$$. Find the work done by the forces. Solution: Question 13. Find the magnitude and direction cosines of the torque of a force represented by $$3 \vec{i}+4 \vec{j}-5 \vec{k}$$ about the point with position vector $$2 \vec{i}-3 \vec{j}+4 \vec{k}$$ acting through a point whose position vector is $$\overrightarrow{4 i}+2 \vec{j}-3 \vec{k}$$. Solution: Question 14. Find the torque of the resultant of the three forces represented by $$-3 \vec{i}+6 \vec{j}-3 \vec{k}$$, $$\overrightarrow{4 i}-10 \vec{j}+12 \vec{k}$$ and $$\overrightarrow{4 i}+7 \vec{j}$$ acting at the point with position vector $$8 \vec{i}-\overrightarrow{6} \vec{j}-4 \vec{k}$$, about the point with position vector $$18 \vec{i}+3 \vec{j}-9 \vec{k}$$ Solution: ### Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.1 Additional Problems Question 1. The work done by the force $$\overrightarrow{\mathrm{F}}=a \vec{i}+\vec{j}+\vec{k}$$ in moving the point of application from (1, 1, 1) to (2, 2, 2) along a straight line is given to be 5 units. Find the value of a. Solution: Question 2. If the position vectors of three points A, B and C are respectively and $$7(\vec{i}+\vec{k})$$. Find $$\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}$$. Interpret the result geometrically. Solution: Question 3. A force given by and $$3 \vec{i}+2 \vec{j}-4 \vec{k}$$ is applied at the point (1, -1, 2). Find the moment of the force about the point (2, -1, 3). Solution: Question 4. Show that the area of a parallelogram having diagonals is $$5 \sqrt{3}$$ Solution:
# Fraction calculator This fraction calculator performs all fraction operations - addition, subtraction, multiplication, division and evaluates expressions with fractions. It also shows detailed step-by-step informations. ## The result: ### 11/2 + 22/3 = 25/6 = 4 1/6 ≅ 4.1666667 The spelled result in words is twenty-five sixths (or four and one sixth). ### How do we solve fractions step by step? 1. Conversion a mixed number 1 1/2 to a improper fraction: 1 1/2 = 1 1/2 = 1 · 2 + 1/2 = 2 + 1/2 = 3/2 To find a new numerator: a) Multiply the whole number 1 by the denominator 2. Whole number 1 equally 1 * 2/2 = 2/2 b) Add the answer from the previous step 2 to the numerator 1. New numerator is 2 + 1 = 3 c) Write a previous answer (new numerator 3) over the denominator 2. One and one half is three halfs. 2. Conversion a mixed number 2 2/3 to a improper fraction: 2 2/3 = 2 2/3 = 2 · 3 + 2/3 = 6 + 2/3 = 8/3 To find a new numerator: a) Multiply the whole number 2 by the denominator 3. Whole number 2 equally 2 * 3/3 = 6/3 b) Add the answer from the previous step 6 to the numerator 2. New numerator is 6 + 2 = 8 c) Write a previous answer (new numerator 8) over the denominator 3. Two and two thirds is eight thirds. 3. Add: 3/2 + 8/3 = 3 · 3/2 · 3 + 8 · 2/3 · 2 = 9/6 + 16/6 = 9 + 16/6 = 25/6 It is suitable to adjust both fractions to a common (equal, identical) denominator for adding, subtracting, and comparing fractions. The common denominator you can calculate as the least common multiple of both denominators - LCM(2, 3) = 6. It is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 2 × 3 = 6. In the following intermediate step, it cannot further simplify the fraction result by canceling. In other words - three halfs plus eight thirds is twenty-five sixths. ### Rules for expressions with fractions: Fractions - use a forward slash to divide the numerator by the denominator, i.e., for five-hundredths, enter 5/100. If you use mixed numbers, leave a space between the whole and fraction parts. Mixed numerals (mixed numbers or fractions) keep one space between the integer and fraction and use a forward slash to input fractions i.e., 1 2/3 . An example of a negative mixed fraction: -5 1/2. Because slash is both sign for fraction line and division, use a colon (:) as the operator of division fractions i.e., 1/2 : 1/3. Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45. ### Math Symbols SymbolSymbol nameSymbol MeaningExample -minus signsubtraction 1 1/2 - 2/3 asteriskmultiplication 2/3 3/4 ×times signmultiplication 2/3 × 5/6 :division signdivision 1/2 : 3 /division slashdivision 1/3 / 5 :coloncomplex fraction 1/2 : 1/3 ^caretexponentiation / power 1/4^3 ()parenthesescalculate expression inside first-3/5 - (-1/4) The calculator follows well-known rules for the 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. MDAS - Multiplication and Division have the same precedence over Addition and Subtraction. The MDAS rule is the order of operations part of the PEMDAS rule. Be careful; always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) have the same priority and must be evaluated from left to right.
#### You may also like Four vehicles travelled on a road. What can you deduce from the times that they met? ### There and Back Brian swims at twice the speed that a river is flowing, downstream from one moored boat to another and back again, taking 12 minutes altogether. How long would it have taken him in still water? ### Escalator At Holborn underground station there is a very long escalator. Two people are in a hurry and so climb the escalator as it is moving upwards, thus adding their speed to that of the moving steps. ... How many steps are there on the escalator? # How Do You React? ##### Stage: 4 Challenge Level: Herschel of the European School of Varese sent us a very succinct solution. Speed-time graph: Since gravity exerts a constant force, the acceleration is also constant: $9.8ms^{-2}$. Therefore, speed is proportional to time and the speed-time graph shows a straight line (with a gradient of $9.8$). Distance-time graph: The distance travelled depends on the speed, which increases over time. Therefore, the distance travelled in the first second is less than that travelled during the second, which is in turn less than that travelled during the third. The graph is therefore a half-parabola with the gradient increasing over time. This means that it takes longer for the ruler to travel the first $15cm$ compared to the next $15 cm$, and a reading of $30 cm$ therefore indicates a reaction time that is slower, but not twice as slow! To look at it algebraically, distance is calculated by the formula $d=\frac{1}{2}\times a\times t^2$. This means that the distance is NOT proportional to the reaction time, but instead to the SQUARE of the reaction time. Double the distance therefore means a reaction time that is 1.41 (i.e. the square-root of $2$) times slower rather than twice as slow. Rearranging the formula above to make $t$ the subject gives us $t=\sqrt{\frac{2d}{a}}$, which means the reaction time for $15cm$ is $\sqrt{0.03}=0.17$ seconds, while that for $30cm$ is $\sqrt{0.06}=0.24$ seconds - only $0.07$ seconds slower, or $41$%. As for the "typical" $0.2$ seconds reaction time, the typical distance should be $0.5\times 9.8\times0.2^2=0.196m$ which is $19.6cm$. I found it hard to get consistant results for my own reactions - my distances ranged from $15cm$ ($0.17$ seconds) to $25cm$ ($0.23$ seconds). Incidentally, it's funny to note that a distance of $0.204$ meters ($20.4$ centimeters) corresponds precisely to a time of $0.204$ seconds!

Subsets and Splits

No community queries yet

The top public SQL queries from the community will appear here once available.