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# It takes 3/5 feet of ribbon to make a pin. You have 9 feet of ribbon. How many pins can you make? Dec 8, 2016 $15$. #### Explanation: We have $9$ feet of ribbon to be cut into pieces measuring $\frac{3}{5}$ foot each. To find out how many pieces $\left(x\right)$ we can get, we divide the total length by the length of the piece. $x = 9 \div \frac{3}{5}$ This is the same as: $x = 9 \times \frac{5}{3}$ $x = 3 \cancel{9} \times \frac{5}{\cancel{3}}$ $x = 3 \times 5$ $x = 15$
# How Do You Prove Continuity? ## How do you prove continuity over an interval? A function is said to be continuous on an interval when the function is defined at every point on that interval and undergoes no interruptions, jumps, or breaks. If some function f(x) satisfies these criteria from x=a to x=b, for example, we say that f(x) is continuous on the interval [a, b].. ## What are the 3 conditions of continuity? Key ConceptsFor a function to be continuous at a point, it must be defined at that point, its limit must exist at the point, and the value of the function at that point must equal the value of the limit at that point.Discontinuities may be classified as removable, jump, or infinite.More items… ## How do you define continuity? Continuity, in mathematics, rigorous formulation of the intuitive concept of a function that varies with no abrupt breaks or jumps. A function is a relationship in which every value of an independent variable—say x—is associated with a value of a dependent variable—say y. ## How do you prove the continuity of a function? Definition: A function f is continuous at x0 in its domain if for every ϵ > 0 there is a δ > 0 such that whenever x is in the domain of f and \|x − x0\| < δ, we have \|f(x) − f(x0)\| < ϵ. Again, we say f is continuous if it is continuous at every point in its domain. ## How do you prove differentiability implies continuity? If a function f(x) is differentiable at a point x = c in its domain, then f(c) is continuous at x = c. f(x) – f(c)=0. This will be useful. ## What is meant by continuity of a function? In mathematics, a continuous function is a function that does not have any abrupt changes in value, known as discontinuities. More precisely, sufficiently small changes in the input of a continuous function result in arbitrarily small changes in its output. If not continuous, a function is said to be discontinuous. ## Why continuity test is needed? A continuity test is a quick check to see if a circuit is open or closed. Only a closed, complete circuit (one that is switched ON) has continuity. During a continuity test, a digital multimeter sends a small current through the circuit to measure resistance in the circuit. ## What is another word for continuity? In this page you can discover 45 synonyms, antonyms, idiomatic expressions, and related words for continuity, like: continuation, unity, continuousness, cut, intermittence, dissipation, desultoriness, duration, endurance, continue and connectedness. ## Does differentiability mean continuity? We see that if a function is differentiable at a point, then it must be continuous at that point. There are connections between continuity and differentiability. Differentiability Implies Continuity If is a differentiable function at , then is continuous at . … If is not continuous at , then is not differentiable at . ## What does lack of continuity mean? n uninterrupted connection or union. Antonyms: discontinuity. lack of connection or continuity. Type of: coherence, coherency, cohesion, cohesiveness. the state of cohering or sticking together. ## How do you show continuity? In calculus, a function is continuous at x = a if – and only if – all three of the following conditions are met:The function is defined at x = a; that is, f(a) equals a real number.The limit of the function as x approaches a exists.The limit of the function as x approaches a is equal to the function value at x = a. ## What is the continuity of a function? A function is said to be continuous on the interval [a,b] if it is continuous at each point in the interval. Note that this definition is also implicitly assuming that both f(a) and limx→af(x) lim x → a ⁡ exist. If either of these do not exist the function will not be continuous at x=a . ## What is the difference between differentiability and continuity? A continuous function is a function whose graph is a single unbroken curve. A discontinuous function then is a function that isn’t continuous. A function is differentiable if it has a derivative. You can think of a derivative of a function as its slope. ## Does continuity guarantee differentiability? No, continuity does not imply differentiability. For instance, the function ƒ: R → R defined by ƒ(x) = \|x\| is continuous at the point 0 , but it is not differentiable at the point 0 . ## What is the formal definition of continuity? The formal definition of continuity at a point has three conditions that must be met. A function f(x) is continuous at a point where x = c if. exists. f(c) exists (That is, c is in the domain of f.) ## What is Movie Continuity? About this video. Continuity in filmmaking is the practice of ensuring that details in a shot are consistent from shot to shot within a film scene. When there is continuity between shots, then audiences have a greater suspension of disbelief and will be more engaged in the film. ## Which is the continuity equation? The continuity equation reflects the fact that mass is conserved in any non-nuclear continuum mechanics analysis. The equation is developed by adding up the rate at which mass is flowing in and out of a control volume, and setting the net in-flow equal to the rate of change of mass within it.
# what is the least common multiple of 3 and 9 LCM of 3 and 9LCM of 3 and 9 is the smallest of all common multiples of 3 and 9. The first multiples of 3 and 9 are (3, 6, 9, 12, 15,…) and ( 9, 18, 27, 36, 45, 54, 63,…). There are 3 commonly used methods to find the LCM of 3 and 9 – by division, by prime factors, and by enumerating multiples. 1. LCM of 3 and 9 2. List of methods 3. Solved examples 4. FAQAnswer: The LCM of 3 and 9 is 9. Read: what is the most common multiple of 3 and 9 Explanation: Read more: What are healing foods? ) without any residuals. Let’s see the different methods to find the LCM of 3 and 9. • According to the division method • According to the prime factorization method • By multiples of the list ### LCM of 3 and 9 by Division Method To calculate the LCM of 3 and 9 using the division method, we divide the numbers (3, 9) by their prime factors (preferably common). The product of these divisors for the LCM is 3 and 9. • Step 1: Find the smallest prime number that is a factor of at least one of the numbers, 3 and 9. Write this prime (3) to the left of the given numbers (3 and 9), separated by a ladder arrangement. . • Step 2: If a given number (3, 9) is a multiple of 3 then divide it by 3 and write the quotient below. Bring down any number that is not divisible by primes. • Step 3: Continue the steps until there are only 1s left in the last row. The LCM of 3 and 9 is the product of all the primes to the left, i.e. LCM (3, 9) by division = 3 × 3 = 9. See Also Tan 30 Degrees \| Top Q&A ### LCM of 3 and 9 in prime factors Read more: Remove Amazon Assistant aa.hta Virus \| Instructions for completely uninstalling the virus The factorization of 3 and 9 is (3) = 31 and (3 × 3) = 32, respectively. The LCM of 3 and 9 can be obtained by multiplying integer factors. factors are raised to their highest powers, i.e. 32 = 9. Hence, the LCM of 3 and 9 by multiplying the prime factor is 9. ### LCM of 3 and 9 in List Multiples To calculate the LCM of 3 and 9 by enumerating common multiples, we can follow these steps: • Step 1: List some multiples of 3 (3, 6, 9, 12, 15,…) and 9 (9, 18, 27, 36, 45, 54, 63,…) • Step 2: The common multiples from the multiples of 3 and 9 are 9, 18 ,. . . • Step 3: The least common multiple of 3 and 9 is 9. ∴ Least common multiple of 3 and 9 = 9.☛ Also check:Read more: What is the factor of 27 • LCM of 6 and 7 – 42 • LCM of 4 and 10 – 20 • LCM of 25 and 100 – 100 • LCM of 3 and 7 – 21 • LCM of 56 and 72 – 504 • LCM of 13 and 15 – 195 • LCM of 8 and 42 – 168 Last, Wallx.net sent you details about the topic “what is the least common multiple of 3 and 9❤️️”.Hope with useful information that the article “what is the least common multiple of 3 and 9” It will help readers to be more interested in “what is the least common multiple of 3 and 9 [ ❤️️❤️️ ]”. Posts “what is the least common multiple of 3 and 9” posted by on 2021-08-22 16:54:03. Thank you for reading the article at wallx.net Rate this post
# Question Video: Adding and Converting Decimal Numbers to Mixed Numbers Mathematics • 5th Grade The given table shows the quantities of some ingredients used to bake a cake. What is the total weight of the ingredients as a mixed number? 01:58 ### Video Transcript The given table shows the quantities of some ingredients used to bake a cake. What is the total weight of the ingredients as a mixed number? And we’ve got nought point five five pounds of flour, nought point four five pounds of milk, nought point three five pounds of sugar, and naught point one five pounds of butter. Now we’ve got to add up all those weights and then convert our answer to a mixed number. Now I’m quite happy adding with decimal numbers, so that’s what I’m gonna do to calculate the total weight. And to calculate the total weight, we could do this as a column addition. So just writing in all of our numbers like this, we’ve got five plus five plus five plus five is twenty. So that’s zero in the unit’s column carry two. And then in the next column, we’ve got five plus four is nine plus three is twelve plus one is thirteen plus two is fifteen. So that’s a five in that column carry one. And then we’ve got zero plus zero plus zero plus zero plus one is one. So the answer is one point five zero pounds or one point five pounds. But another quick mental method for doing that is to notice that nought point five five and nought point four five added together make one whole one. And nought point three five and nought point one five added together make nought point five. So we could have spotted quite easily one plus nought point five is equal to one point five. Anyway, whichever method you do it is not a problem. We’ve got our answer, one point five, but we’ve got to convert that to a mixed number. Well one point five is one whole number plus nought point five, a half of a whole number. So that’s gonna be one plus a half, which is equal to one and a half. So our answer is gonna be one and a half pounds.
# How do you find the quotient of four-ninths divided by eight? Mar 7, 2018 See a solution process below; #### Explanation: We can write this problem as: $\frac{\frac{4}{9}}{8}$ which can be rewritten as: $\frac{\frac{4}{9}}{\frac{8}{1}}$ We can then use this rule for dividing fractions to evaluate the expression: $\frac{\frac{\textcolor{red}{a}}{\textcolor{b l u e}{b}}}{\frac{\textcolor{g r e e n}{c}}{\textcolor{p u r p \le}{d}}} = \frac{\textcolor{red}{a} \times \textcolor{p u r p \le}{d}}{\textcolor{b l u e}{b} \times \textcolor{g r e e n}{c}}$ $\frac{\frac{\textcolor{red}{4}}{\textcolor{b l u e}{9}}}{\frac{\textcolor{g r e e n}{8}}{\textcolor{p u r p \le}{1}}} \implies$ $\frac{\textcolor{red}{4} \times \textcolor{p u r p \le}{1}}{\textcolor{b l u e}{9} \times \textcolor{g r e e n}{8}} \implies$ $\frac{\cancel{\textcolor{red}{4}} \textcolor{red}{1} \times \textcolor{p u r p \le}{1}}{\textcolor{b l u e}{9} \times \cancel{\textcolor{g r e e n}{8}} \textcolor{g r e e n}{2}} \implies$ $\frac{\textcolor{red}{1} \times \textcolor{p u r p \le}{1}}{\textcolor{b l u e}{9} \times \textcolor{g r e e n}{2}} \implies$ $\frac{1}{18}$
# 12 Surface Area and Volume Size: px Start display at page: Transcription 1 12 Surface Area and Volume 12.1 Three-Dimensional Figures 12.2 Surface Areas of Prisms and Cylinders 12.3 Surface Areas of Pyramids and Cones 12.4 Volumes of Prisms and Cylinders 12.5 Volumes of Pyramids and Cones 12.6 Surface Areas and Volumes of Spheres 12.7 Spherical Geometry Earth (p. 692) Tennis alls (p. 685) Khafre's Pyramid (p. 674) SEE the ig Idea Great lue Hole (p. 669) Traffic fic Cone (p. 656) 6) Mathematical Thinking: Mathematically proficient students can apply the mathematics they know to solve problems arising in everyday life, society, and the workplace. 2 Maintaining Mathematical Proficiency Finding the Area of a Circle (7.9.) Example 1 Find the area of the circle. A = πr 2 Formula for area of a circle = π 8 2 Substitute 8 for r. = 64π Simplify Use a calculator. The area is about square inches. 8 in. Find the area of the circle ft 6 m 20 cm Finding the Area of a Composite Figure (7.9.C) Example 2 Find the area of the composite figure. The composite figure is made up of a rectangle, a triangle, and a semicircle. Find the area of each figure. 16 ft 17 ft 32 ft 30 ft Area of rectangle Area of triangle Area of semicircle A = w A = 1 2 bh A = πr2 2 = 32(16) = 1 2 (30)(16) = π(17)2 2 = 512 = So, the area is about = square feet. Find the area of the composite figure m 5. 6 in cm 20 m 7 m 7 m 10 in. 3 in. 5 in. 6 cm 3 3 cm 9 cm 6 cm 7. ASTRACT REASONING A circle has a radius of x inches. Write a formula for the area of the circle when the radius is multiplied by a real number a. 637 3 Mathematical Thinking Creating a Coherent Representation Core Concept Mathematically profi cient students create and use representations to organize, record, and communicate mathematical ideas. (G.1.E) Nets for Three-Dimensional Figures A net for a three-dimensional figure is a two-dimensional pattern that can be folded to form the three-dimensional figure. w lateral face base lateral face w w lateral face lateral face h w h base Drawing a Net for a Pyramid Draw a net of the pyramid. 20 in. The pyramid has a square base. Its four lateral faces are congruent isosceles triangles. 19 in. 19 in. 19 in. 19 in. 20 in. Monitoring Progress Draw a net of the three-dimensional figure. Label the dimensions ft 2. 5 m in. 2 ft 4 ft 8 m 12 m 10 in. 10 in. 638 Chapter 12 Surface Area and Volume 4 12.1 Three-Dimensional Figures TEXAS ESSENTIAL KNOWLEDGE AND SKILLS G.10.A Essential Question What is the relationship between the numbers of vertices V, edges E, and faces F of a polyhedron? A polyhedron is a solid that is bounded by polygons, called faces. edge Each vertex is a point. Each edge is a segment of a line. Each face is a portion of a plane. face vertex Analyzing a Property of Polyhedra Work with a partner. The five Platonic solids are shown below. Each of these solids has congruent regular polygons as faces. Complete the table by listing the numbers of vertices, edges, and faces of each Platonic solid. tetrahedron cube octahedron dodecahedron icosahedron Solid Vertices, V Edges, E Faces, F tetrahedron cube MAKING MATHEMATICAL ARGUMENTS To be proficient in math, you need to reason inductively about data. octahedron dodecahedron icosahedron Communicate Your Answer 2. What is the relationship between the numbers of vertices V, edges E, and faces F of a polyhedron? (Note: Swiss mathematician Leonhard Euler ( ) discovered a formula that relates these quantities.) 3. Draw three polyhedra that are different from the Platonic solids given in Exploration 1. Count the numbers of vertices, edges, and faces of each polyhedron. Then verify that the relationship you found in Question 2 is valid for each polyhedron. Section 12.1 Three-Dimensional Figures 639 5 12.1 Lesson What You Will Learn Core Vocabulary polyhedron, p. 640 face, p. 640 edge, p. 640 vertex, p. 640 cross section, p. 641 solid of revolution, p. 642 axis of revolution, p. 642 Previous solid prism pyramid cylinder cone sphere base Classify solids. Describe cross sections. Sketch and describe solids of revolution. Classifying Solids A three-dimensional figure, or solid, is bounded by flat or curved surfaces that enclose a single region of space. A polyhedron is a solid that is bounded by polygons, called faces. An edge of a polyhedron is a line segment formed by the intersection of two faces. A vertex of a polyhedron is a point where three or more edges meet. The plural of polyhedron is polyhedra or polyhedrons. Core Concept Types of Solids Polyhedra vertex Not Polyhedra edge face prism cylinder cone pyramid sphere Pentagonal prism ases are pentagons. To name a prism or a pyramid, use the shape of the base. The two bases of a prism are congruent polygons in parallel planes. For example, the bases of a pentagonal prism are pentagons. The base of a pyramid is a polygon. For example, the base of a triangular pyramid is a triangle. Triangular pyramid ase is a triangle. Classifying Solids Tell whether each solid is a polyhedron. If it is, name the polyhedron. a. b. c. a. The solid is formed by polygons, so it is a polyhedron. The two bases are congruent rectangles, so it is a rectangular prism. b. The solid is formed by polygons, so it is a polyhedron. The base is a hexagon, so it is a hexagonal pyramid. c. The cone has a curved surface, so it is not a polyhedron. 640 Chapter 12 Surface Area and Volume 6 Monitoring Progress Help in English and Spanish at igideasmath.com Tell whether the solid is a polyhedron. If it is, name the polyhedron Describing Cross Sections Imagine a plane slicing through a solid. The intersection of the plane and the solid is called a cross section. For example, three different cross sections of a cube are shown below. square rectangle triangle Describing Cross Sections Describe the shape formed by the intersection of the plane and the solid. a. b. c. d. e. f. a. The cross section is a hexagon. b. The cross section is a triangle. c. The cross section is a rectangle. d. The cross section is a circle. e. The cross section is a circle. f. The cross section is a trapezoid. Monitoring Progress Help in English and Spanish at igideasmath.com Describe the shape formed by the intersection of the plane and the solid Section 12.1 Three-Dimensional Figures 641 7 Sketching and Describing Solids of Revolution A solid of revolution is a three-dimensional figure that is formed by rotating a two-dimensional shape around an axis. The line around which the shape is rotated is called the axis of revolution. For example, when you rotate a rectangle around a line that contains one of its sides, the solid of revolution that is produced is a cylinder. Sketching and Describing Solids of Revolution Sketch the solid produced by rotating the figure around the given axis. Then identify and describe the solid. a. 9 b a. 9 b. 4 5 The solid is a cylinder with a height of 9 and a base radius of 4. 2 The solid is a cone with a height of 5 and a base radius of 2. Monitoring Progress Help in English and Spanish at igideasmath.com Sketch the solid produced by rotating the figure around the given axis. Then identify and describe the solid Chapter 12 Surface Area and Volume 8 12.1 Exercises Tutorial Help in English and Spanish at igideasmath.com Vocabulary and Core Concept Check 1. VOCAULARY A(n) is a solid that is bounded by polygons. 2. WHICH ONE DOESN T ELONG? Which solid does not belong with the other three? Explain your reasoning. Monitoring Progress and Modeling with Mathematics In Exercises 3 6, match the polyhedron with its name In Exercises 11 14, describe the cross section formed by the intersection of the plane and the solid. (See Example 2.) A. triangular prism. rectangular pyramid C. hexagonal pyramid D. pentagonal prism In Exercises 7 10, tell whether the solid is a polyhedron. If it is, name the polyhedron. (See Example 1.) In Exercises 15 18, sketch the solid produced by rotating the figure around the given axis. Then identify and describe the solid. (See Example 3.) Section 12.1 Three-Dimensional Figures 643 9 19. ERROR ANALYSIS Describe and correct the error in identifying the solid. The solid is a rectangular pyramid. 28. ATTENDING TO PRECISION The figure shows a plane intersecting a cube through four of its vertices. The edge length of the cube is 6 inches. 20. HOW DO YOU SEE IT? Is the swimming pool shown a polyhedron? If it is, name the polyhedron. If not, explain why not. a. Describe the shape formed by the cross section. b. What is the perimeter of the cross section? c. What is the area of the cross section? REASONING In Exercises 29 34, tell whether it is possible for a cross section of a cube to have the given shape. If it is, describe or sketch how the plane could intersect the cube. 29. circle 30. pentagon 31. rhombus 32. isosceles triangle 33. hexagon 34. scalene triangle In Exercises 21 26, sketch the polyhedron. 21. triangular prism 22. rectangular prism 23. pentagonal prism 24. hexagonal prism 25. square pyramid 26. pentagonal pyramid 27. MAKING AN ARGUMENT Your friend says that the polyhedron shown is a triangular prism. Your cousin says that it is a triangular pyramid. Who is correct? Explain your reasoning. 35. REASONING Sketch the composite solid produced by rotating the figure around the given axis. Then identify and describe the composite solid. a b THOUGHT PROVOKING Describe how Plato might have argued that there are precisely five Platonic Solids (see page 639). (Hint: Consider the angles that meet at a vertex.) Maintaining Mathematical Proficiency Decide whether enough information is given to prove that the triangles are congruent. If so, state the theorem you would use. (Sections 5.3, 5.5, and 5.6) 37. AD, CD 38. JLK, JLM 39. RQP, RTS A J Reviewing what you learned in previous grades and lessons Q R S D C K L M P T 644 Chapter 12 Surface Area and Volume 10 12.2 TEXAS ESSENTIAL KNOWLEDGE AND SKILLS G.10. G.11.C Surface Areas of Prisms and Cylinders Essential Question How can you find the surface area of a prism or a cylinder? Recall that the surface area of a polyhedron is the sum of the areas of its faces. The lateral area of a polyhedron is the sum of the areas of its lateral faces. Finding a Formula for Surface Area APPLYING MATHEMATICS To be proficient in math, you need to analyze relationships mathematically to draw conclusions. Work with a partner. Consider the polyhedron shown. a. Identify the polyhedron. Then sketch its net so that the lateral faces form a rectangle with the same height h as the polyhedron. What types of figures make up the net? b. Write an expression that represents the perimeter P of the base of the polyhedron. Show how you can use P to write an expression that represents the lateral area L of the polyhedron. height, h a b c c. Let represent the area of a base of the polyhedron. Write a formula for the surface area S. Finding a Formula for Surface Area Work with a partner. Consider the solid shown. a. Identify the solid. Then sketch its net. What types of figures make up the net? b. Write an expression that represents the perimeter P of the base of the solid. Show how you can use P to write an expression that represents the lateral area L of the solid. radius, r height, h c. Write an expression that represents the area of a base of the solid. d. Write a formula for the surface area S. Communicate Your Answer 3. How can you find the surface area of a prism or a cylinder? 4. Consider the rectangular prism shown. a. Find the surface area of the rectangular prism by drawing its net and finding the sum of the areas of its faces. b. Find the surface area of the rectangular prism by using the formula you wrote in Exploration 1. c. Compare your answers to parts (a) and (b). What do you notice? Section 12.2 Surface Areas of Prisms and Cylinders 645 11 12.2 Lesson What You Will Learn Core Vocabulary lateral faces, p. 646 lateral edges, p. 646 surface area, p. 646 lateral area, p. 646 net, p. 646 right prism, p. 646 oblique prism, p. 646 right cylinder, p. 647 oblique cylinder, p. 647 Previous prism bases of a prism cylinder composite solid Find lateral areas and surface areas of right prisms. Find lateral areas and surface areas of right cylinders. Use surface areas of right prisms and right cylinders. Finding Lateral Areas and Surface Areas of Right Prisms Recall that a prism is a polyhedron with two congruent faces, called bases, that lie in parallel planes. The other faces, called lateral faces, are parallelograms formed by connecting the corresponding vertices of the bases. The segments connecting these vertices are lateral edges. Prisms are classified by the shapes of their bases. base base lateral edges lateral faces The surface area of a polyhedron is the sum of the areas of its faces. The lateral area of a polyhedron is the sum of the areas of its lateral faces. Imagine that you cut some edges of a polyhedron and unfold it. The two-dimensional representation of the faces is called a net. The surface area of a prism is equal to the area of its net. The height of a prism is the perpendicular distance between its bases. In a right prism, each lateral edge is perpendicular to both bases. A prism with lateral edges that are not perpendicular to the bases is an oblique prism. height height Right rectangular prism Oblique triangular prism Core Concept Lateral Area and Surface Area of a Right Prism For a right prism with base perimeter P, base apothem a, height h, and base area, the lateral area L and surface area S are as follows. Lateral area L = Ph Surface area S = 2 + L = ap + Ph P h 646 Chapter 12 Surface Area and Volume 12 Finding Lateral Area and Surface Area Find the lateral area and the surface area of the right pentagonal prism ft 6 ft Find the apothem and perimeter of a base. 9 ft a = = P = 5(7.05) = ft a 6 ft ATTENDING TO PRECISION Throughout this chapter, round lateral areas, surface areas, and volumes to the nearest hundredth, if necessary. Find the lateral area and the surface area. L = Ph ft ft = (35.25)(9) Substitute. = Multiply. S = ap + Ph = ( ) (35.25) Substitute Formula for lateral area of a right prism Formula for surface area of a right prism Use a calculator. The lateral area is square feet and the surface area is about square feet. Monitoring Progress Help in English and Spanish at igideasmath.com 1. Find the lateral area and the surface area of a right rectangular prism with a height of 7 inches, a length of 3 inches, and a width of 4 inches. height right cylinder height oblique cylinder Finding Lateral Areas and Surface Areas of Right Cylinders Recall that a cylinder is a solid with congruent circular bases that lie in parallel planes. The height of a cylinder is the perpendicular distance between its bases. The radius of a base is the radius of the cylinder. In a right cylinder, the segment joining the centers of the bases is perpendicular to the bases. In an oblique cylinder, this segment is not perpendicular to the bases. The lateral area of a cylinder is the area of its curved surface. For a right cylinder, it is equal to the product of the circumference and the height, or 2πrh. The surface area of a cylinder is equal to the sum of the lateral area and the areas of the two bases. Core Concept Lateral Area and Surface Area of a Right Cylinder For a right cylinder with radius r, r 2 πr 2 πr height h, and base area, the lateral area L and surface area S are as follows. Lateral area Surface area L = 2πrh S = 2 + L = 2πr 2 + 2πrh h r lateral area A = 2 rh π base area A = πr 2 h base area A = r 2 π Section 12.2 Surface Areas of Prisms and Cylinders 647 13 Finding Lateral Area and Surface Area Find the lateral area and the surface area of the right cylinder. 4 m Find the lateral area and the surface area. L = 2πrh Formula for lateral area of a right cylinder = 2π(4)(8) Substitute. = 64π Simplify Use a calculator. S = 2πr 2 + 2πrh Formula for surface area of a right cylinder = 2π(4) π Substitute. = 96π Simplify Use a calculator. 8 m The lateral area is 64π, or about square meters. The surface area is 96π, or about square meters. Solving a Real-Life Problem You are designing a label for the cylindrical soup can shown. The label will cover the lateral area of the can. Find the minimum amount of material needed for the label. 9 cm Find the radius of a base. r = 1 (9) = Find the lateral area. L = 2πrh = 2π(4.5)(12) Substitute. = 108π Simplify Formula for lateral area of a right cylinder Use a calculator. 12 cm You need a minimum of about square centimeters of material. Monitoring Progress Help in English and Spanish at igideasmath.com 2. Find the lateral area and the surface area of the right cylinder. 10 in. 18 in. 3. WHAT IF? In Example 3, you change the design of the can so that the diameter is 12 centimeters. Find the minimum amount of material needed for the label. 648 Chapter 12 Surface Area and Volume 14 Using Surface Areas of Right Prisms and Right Cylinders Finding the Surface Area of a Composite Solid 3 m 4 m Find the lateral area and the surface area of the composite solid. 12 m Lateral area of solid = Lateral area of cylinder + Lateral area of prism = 2πrh + Ph = 2π(6)(12) + 14(12) 6 m = 144π Surface area of solid = Lateral area of solid + 2 ( Area of a base of the cylinder Area of a base of the prism ) = 144π (πr 2 w) = 144π [π(6) 2 4(3)] = 216π The lateral area is about square meters and the surface area is about square meters. Changing Dimensions in a Solid Describe how doubling all the linear dimensions affects the surface area of the right cylinder. 2 ft efore change After change Dimensions r = 2 ft, h = 8 ft r = 4 ft, h = 16 ft 8 ft Surface area S = 2πr 2 + 2πrh = 2π(2) 2 + 2π(2)(8) = 40π ft 2 S = 2πr 2 + 2πrh = 2π(4) 2 + 2π(4)(16) = 160π ft 2 2 mm Doubling all the linear dimensions results in a surface area that is 160π 40π = 4 = 22 times the original surface area. 8 mm 10 mm 6 mm Monitoring Progress Help in English and Spanish at igideasmath.com 4. Find the lateral area and the surface area of the composite solid at the left. 5. In Example 5, describe how multiplying all the linear dimensions by 1 affects the 2 surface area of the right cylinder. Section 12.2 Surface Areas of Prisms and Cylinders 649 15 12.2 Exercises Tutorial Help in English and Spanish at igideasmath.com Vocabulary and Core Concept Check 1. VOCAULARY Sketch a right triangular prism. Identify the bases, lateral faces, and lateral edges. 2. WRITING Explain how the formula S = 2 + L applies to finding the surface area of both a right prism and a right cylinder. Monitoring Progress and Modeling with Mathematics In Exercises 3 and 4, find the surface area of the solid formed by the net in. 8 cm 13. MODELING WITH MATHEMATICS The inside of the cylindrical swimming pool shown must be covered with a vinyl liner. The liner must cover the side and bottom of the swimming pool. What is the minimum amount of vinyl needed for the liner? (See Example 3.) 24 ft 10 in. 20 cm 4 ft In Exercises 5 8, find the lateral area and the surface area of the right prism. (See Example 1.) ft 8 ft 3 ft 3 m 8 m 9.1 m 7. A regular pentagonal prism has a height of 3.5 inches and a base edge length of 2 inches. 8. A regular hexagonal prism has a height of 80 feet and a base edge length of 40 feet. In Exercises 9 12, find the lateral area and the surface area of the right cylinder. (See Example 2.) in in. 16 cm 8 cm 14. MODELING WITH MATHEMATICS The tent shown has fabric covering all four sides and the floor. What is 4 ft the minimum amount of fabric needed to 6 ft construct the tent? In Exercises 15 18, find the lateral area and the surface area of the composite solid. (See Example 4.) cm 4 cm 1 cm cm 8 cm 4 ft 6 ft 5 ft 8 ft 7 ft 4 ft 11. A right cylinder has a diameter of 24 millimeters and a height of 40 millimeters in m 7 m 12. A right cylinder has a radius of 2.5 feet and a height of 7.5 feet. 11 in. 9 m 6 m 15 m 5 in. 650 Chapter 12 Surface Area and Volume 16 19. ERROR ANALYSIS Describe and correct the error in finding the surface area of the right cylinder. 6 cm 8 cm S = 2π (6) 2 + 2π(6)(8) = 168π cm ERROR ANALYSIS Describe and correct the error in finding the surface area of the composite solid. 16 ft 7 ft 27. MATHEMATICAL CONNECTIONS A cube has a surface area of 343 square inches. Write and solve an equation to find the length of each edge of the cube. 28. MATHEMATICAL CONNECTIONS A right cylinder has a surface area of 108π square meters. The radius of the cylinder is twice its height. Write and solve an equation to find the height of the cylinder. 29. MODELING WITH MATHEMATICS A company makes two types of recycling bins, as shown. oth types of bins have an open top. Which recycling bin requires more material to make? Explain. 6 in. 20 ft 18 ft S = 2(20)(7) + 2(18)(7) + 2π (8)(7) + 2[(18)(20) + π (8) 2 ] ft 2 36 in. 36 in. In Exercises 21 24, describe how the change affects the surface area of the right prism or right cylinder. (See Example 5.) 21. doubling all the linear dimensions 17 in. 5 in. 4 in. 22. multiplying all the linear dimensions by mm 10 in. 12 in. 30. MODELING WITH MATHEMATICS You are painting a rectangular room that is 13 feet long, 9 feet wide, and 8.5 feet high. There is a window that is 2.5 feet wide and 5 feet high on one wall. On another wall, there is a door that is 4 feet wide and 7 feet high. A gallon of paint covers 350 square feet. How many gallons of paint do you need to cover the four walls with one coat of paint, not including the window and door? 23. tripling the radius 2 yd 7 yd 24 mm 24. multiplying the base edge lengths by 1 4 and the height by 4 2 m 8 m 16 m 31. ANALYZING RELATIONSHIPS Which creates a greater surface area, doubling the radius of a cylinder or doubling the height of a cylinder? Explain your reasoning. 32. MAKING AN ARGUMENT You cut a cylindrical piece of lead, forming two congruent cylindrical pieces of lead. Your friend claims the surface area of each smaller piece is exactly half the surface area of the original piece. Is your friend correct? Explain your reasoning. In Exercises 25 and 26, find the height of the right prism or right cylinder. 25. S = 1097 m S = 480 in. 2 h 8.2 m 8 in. h 15 in. 33. USING STRUCTURE The right triangular prisms shown have the same surface area. Find the height h of prism. 20 cm Prism A 24 cm 20 cm 3 cm 6 cm Prism 8 cm h Section 12.2 Surface Areas of Prisms and Cylinders 651 17 34. USING STRUCTURE The lateral surface area of a regular pentagonal prism is 360 square feet. The height of the prism is twice the length of one of the edges of the base. Find the surface area of the prism. 35. ANALYZING RELATIONSHIPS Describe how multiplying all the linear dimensions of the right rectangular prism by each given value affects the surface area of the prism. 38. THOUGHT PROVOKING You have 24 cube-shaped building blocks with edge lengths of 1 unit. What arrangement of blocks gives you a rectangular prism with the least surface area? Justify your answer. 39. USING STRUCTURE Sketch the net of the oblique rectangular prism shown. Then find the surface area. 4 ft h 8 ft 7 ft a. 2 b. 3 c. 1 2 d. n 36. HOW DO YOU SEE IT? An open gift box is shown. a. Why is the area of the net of the box larger than the minimum amount of wrapping paper needed dd to cover the closed box? b. When wrapping the box, why would you want to use more than the minimum amount of paper needed? 37. REASONING Consider a cube that is built using 27 unit cubes, as shown. a. Find the surface area of the solid formed when the red unit cubes are removed from the solid shown. b. Find the surface area of the solid formed when the blue unit cubes are removed from the solid shown. c. Explain why your answers are different in parts (a) and (b). w 15 ft 40. WRITING Use the diagram to write a formula that can be used to find the surface area S of any cylindrical ring where 0 < r 2 < r 1. r USING STRUCTURE The diagonal of a cube is a segment whose endpoints are vertices that are not on the same face. Find the surface area of a cube with a diagonal length of 8 units. 42. USING STRUCTURE A cuboctahedron has 6 square faces and 8 equilateral triangular faces, as shown. A cuboctahedron can be made by slicing off the corners of a cube. a. Sketch a net for the cuboctahedron. b. Each edge of a cuboctahedron has a length of 5 millimeters. Find its surface area. r 2 h Maintaining Mathematical Proficiency Reviewing what you learned in previous grades and lessons Find the area of the regular polygon. (Section 11.3) in m 9 in. 8 cm 6 cm 652 Chapter 12 Surface Area and Volume 18 12.3 TEXAS ESSENTIAL KNOWLEDGE AND SKILLS G.10. G.11.C Surface Areas of Pyramids and Cones Essential Question How can you find the surface area of a pyramid or a cone? A regular pyramid has a regular polygon for a base and the segment joining the vertex and the center of the base is perpendicular to the base. In a right cone, the segment joining the vertex and the center of the base is perpendicular to the base. APPLYING MATHEMATICS To be proficient in math, you need to analyze relationships mathematically to draw conclusions. Finding a Formula for Surface Area Work with a partner. Consider the polyhedron shown. a. Identify the polyhedron. Then sketch its net. slant height, height, h What types of figures make up the net? b. Write an expression that represents the perimeter P of the base of the polyhedron. Show how you can use P to write an expression that represents the lateral area L of the polyhedron. base edge length, b c. Let represent the area of a base of the polyhedron. Write a formula for the surface area S. Finding a Formula for Surface Area Work with a partner. Consider the solid shown. a. Identify the solid. Then sketch its net. What types of figures make up the net? b. Write an expression that represents the area of the base of the solid. c. What is the arc measure of the lateral surface of the solid? What is the circumference and area of the entire circle that contains the lateral surface of the solid? Show how you can use these three measures to find the lateral area L of the solid. slant height, radius, r d. Write a formula for the surface area S. Communicate Your Answer 3. How can you find the surface area of a pyramid or cone? 4. Consider the rectangular pyramid shown. a. Find the surface area of the rectangular pyramid by drawing its net and finding the sum of the areas of its faces. 12 ft b. Find the surface area of the rectangular pyramid by using the formula you wrote in Exploration 1. c. Compare your answers to parts (a) and (b). What do you notice? 8 ft Section 12.3 Surface Areas of Pyramids and Cones 653 19 12.3 Lesson What You Will Learn Core Vocabulary vertex of a pyramid, p. 654 regular pyramid, p. 654 slant hieght of a regular pyramid, p. 654 vertex of a cone, p. 655 right cone, p. 655 oblique cone, p. 655 slant height of a right cone, p. 655 lateral surface of a cone, p. 655 Previous pyramid cone composite solid Find lateral areas and surface areas of regular pyramids. Find lateral areas and surface areas of right cones. Use surface areas of regular pyramids and right cones. Finding Lateral Areas and Surface Areas of Regular Pyramids vertex A pyramid is a polyhedron in which the base height is a polygon and the lateral faces are triangles with a common vertex, called the vertex of the pyramid. The intersection of two lateral faces is a lateral edge. The intersection of the base and a lateral face is a base edge. The height of the pyramid is the perpendicular distance base between the base and the vertex. height Regular pyramid slant height Core Concept lateral faces Pyramid A regular pyramid has a regular polygon for a base and the segment joining the vertex and the center of the base is perpendicular to the base. The lateral faces of a regular pyramid are congruent isosceles triangles. The slant height of a regular pyramid is the height of a lateral face of the regular pyramid. A nonregular pyramid does not have a slant height. Lateral Area and Surface Area of a Regular Pyramid For a regular pyramid with base perimeter P, slant height, and base area, the lateral area L and surface area S are as follows. lateral edge base edge Lateral area L = 1 2 P Surface area S = + L = P P Finding Lateral Area and Surface Area 14 ft 10 ft 5 3 ft Find the lateral area and the surface area of the regular hexagonal pyramid. The perimeter P of the base is 6 10 = 60, feet and the apothem a is 5 3 feet. The slant height of a face is 14 feet. Find the lateral area and the surface area. L = 1 P Formula for lateral area of a regular pyramid 2 = 1 (60)(14) Substitute. 2 = 420 Simplify. S = + 1 P Formula for surface area of a regular pyramid 2 = 1 2 ( 5 3 ) (60) Substitute. = Simplify Use a calculator. The lateral area is 420 square feet and the surface area is about square feet. 654 Chapter 12 Surface Area and Volume 20 4.8 m 5.5 m Monitoring Progress Help in English and Spanish at igideasmath.com 1. Find the lateral area and the surface area of the regular pentagonal pyramid. slant height base lateral surface 8 m vertex r Right cone height vertex lateral surface height Finding Lateral Areas and Surface Areas of Right Cones A cone has a circular base and a vertex that is not in the same plane as the base. The radius of the base is the radius of the cone. The height is the perpendicular distance between the vertex and the base. In a right cone, the segment joining the vertex and the center of the base is perpendicular to the base. In an oblique cone, this segment is not perpendicular to the base. The slant height of a right cone is the distance between the vertex and a point on the edge of the base. An oblique cone does not have a slant height. The lateral surface of a cone consists of all segments that connect the vertex with points on the edge of the base. Core Concept Lateral Area and Surface Area of a Right Cone For a right cone with radius r, slant height, and base area, the lateral area L and surface area S are as follows. Lateral area L = πr r base Oblique cone Surface area S = + L = πr 2 + πr r Finding Lateral Area and Surface Area Find the lateral area and the surface area of the right cone. 6 m Use the Pythagorean Theorem (Theorem 9.1) to find the slant height. 8 m = = 10 Find the lateral area and the surface area. L = πr Formula for lateral area of a right cone = π(6)(10) Substitute. = 60π Simplify S = πr 2 + πr Use a calculator. Formula for surface area of a right cone = π(6) π Substitute. = 96π Simplify Use a calculator. The lateral area is 60π, or about square meters. The surface area is 96π, about square meters. Section 12.3 Surface Areas of Pyramids and Cones 655 21 Solving a Real-Life Problem The traffic cone can be approximated by a right cone with a radius of 5.7 inches and a height of 18 inches. Find the lateral area of the traffic cone. Use the Pythagorean Theorem (Theorem 9.1) to find the slant height. = (5.7) 2 = Find the lateral area. L = πr = π(5.7) ( ) Substitute Formula for lateral area of a right cone Use a calculator. The lateral area of the traffic cone is about square inches Monitoring Progress Help in English and Spanish at igideasmath.com 2. Find the lateral area and the surface area of the right cone. 3. WHAT IF? The radius of the cone in Example 3 is 6.3 inches. Find the lateral area. 8 ft 15 ft Using Surface Areas of Regular Pyramids and Right Cones Finding the Surface Area of a Composite Solid Find the lateral area and the surface area of the composite solid. 5 cm Lateral area of solid = Lateral area of cone = πr + 2πrh + = π(3)(5) + 2π(3)(6) = 51π Lateral area of cylinder 3 cm 6 cm Surface area of solid = Lateral area of solid = 51π + πr 2 = 51π + π(3) 2 = 60π Area of a base of the cylinder 656 Chapter 12 Surface Area and Volume The lateral area is about square centimeters and the surface area is about square centimeters. 22 Changing Dimensions in a Solid Describe how the change affects the surface area of the right cone. a. multiplying the radius by m b. multiplying all the linear dimensions by 3 2 ANALYZING MATHEMATICAL RELATIONSHIPS Notice that while the surface area does not scale by a factor of 3, the lateral 2 surface area does scale by a factor or π(15)(26) π(10)(26) = 3 2. a. efore change After change Dimensions r = 10 m, = 26 m r = 15 m, = 26 m Surface area S = πr 2 + πr = π(10) 2 + π(10)(26) = 360π m 2 Multiplying the radius by 3 2 the original surface area. S = πr 2 + πr = π(15) 2 + π(15)(26) = 615π m 2 10 m 615π results in a surface area that is 360π = times b. efore change After change Dimensions r = 10 m, = 26 m r = 15 m, = 39 m Surface area S = 360π m 2 S = πr 2 + πr = π(15) 2 + π(15)(39) = 810π m 2 Multiplying all the linear dimensions by 3 results in a surface area that 2 is 810π 360π = 9 4 ( = 3 2) 2 times the original surface area. Monitoring Progress Help in English and Spanish at igideasmath.com 4. Find the lateral area and the surface area of the composite solid. 10 ft 6 ft 5 ft 5. Describe how (a) multiplying the base edge lengths by 1 and (b) multiplying all the linear 2 dimensions by 1 affects the surface area of the 2 square pyramid. 6 m 5 m Section 12.3 Surface Areas of Pyramids and Cones 657 23 12.3 Exercises Tutorial Help in English and Spanish at igideasmath.com Vocabulary and Core Concept Check 1. WRITING Describe the differences between pyramids and cones. Describe their similarities. 2. DIFFERENT WORDS, SAME QUESTION Which is different? Find both answers. Find the slant height of the regular pyramid. A 5 in. Find A. Find the height of the regular pyramid. 6 in. Find the height of a lateral face of the regular pyramid. Monitoring Progress and Modeling with Mathematics In Exercises 3 6, find the lateral area and the surface area of the regular pyramid. (See Example 1.) in mm 7.2 mm 11. ERROR ANALYSIS Describe and correct the error in finding the surface area of the regular pyramid. 4 ft 5 ft S = + 12Pℓ = (24)(4) = 84 ft2 6 ft 5 in. 5. A square pyramid has a height of 21 feet and a base edge length of 40 feet. finding the surface area of the right cone. 6. A regular hexagonal pyramid has a slant height of 15 centimeters and a base edge length of 8 centimeters. In Exercises 7 10, find the lateral area and the surface area of the right cone. (See Example 2.) ERROR ANALYSIS Describe and correct the error in cm 16 in. 11 cm 10 cm 8 cm S = 𝛑r 2 + 𝛑r 2ℓ = 𝛑 (6)2 + 𝛑 (6)2(10) 6 cm = 396𝛑 cm2 13. MODELING WITH MATHEMATICS You are making cardboard party hats like the one shown. About how much cardboard 5.5 in. do you need for each hat? (See Example 3.) 8 in. 3.5 in. 9. A right cone has a radius of 9 inches and a height of 12 inches. 10. A right cone has a diameter of 11.2 feet and a height of 9.2 feet. 658 Chapter 12 Surface Area and Volume 14. MODELING WITH MATHEMATICS A candle is in the shape of a regular square pyramid with a base edge length of 16 centimeters and a height of 15 centimeters. Find the surface area of the candle. 24 In Exercises 15 18, find the lateral area and the surface area of the composite solid. (See Example 4.) yd 4 yd 8 yd in. 5 in. 5 in cm 5 mm 7 mm 5 in 7.5 in. 24. USING STRUCTURE The sector shown can be rolled 150 to form the lateral surface area of a right cone. The lateral surface area of the cone is 20 square meters. a. Use the formula for the area of a sector to find the slant height of the cone. Explain your reasoning. b. Find the radius and the height of the cone. In Exercises 25 and 26, find the missing dimensions of the regular pyramid or right cone. 25. S = 864 in S = cm 2 12 cm h 15 in. h 4 mm In Exercises 19 22, describe how the change affects the surface area of the regular pyramid or right cone. (See Example 5.) 19. doubling the radius 20. multiplying the base edge lengths by 4 5 and the slant height by in. 4 mm x 8 in. 27. WRITING Explain why a nonregular pyramid does not have a slant height. 28. WRITING Explain why an oblique cone does not have a slant height. 29. ANALYZING RELATIONSHIPS In the figure, AC = 4, A = 3, and DC = 2. 3 in. 10 mm A 21. tripling all the linear dimensions 22. multiplying all the linear dimensions by 4 3 D E 4 m 3.6 ft 2 m 2.4 ft 23. PROLEM SOLVING Refer to the regular pyramid and right cone a. Which solid has the base with the greater area? b. Which solid has the greater slant height? c. Which solid has the greater lateral area? 4 3 a. Prove AC DEC. b. Find C, DE, and EC. c. Find the surface areas of the larger cone and the smaller cone in terms of π. Compare the surface areas using a percent. 30. REASONING To make a paper drinking cup, start with a circular piece of paper that has a 3-inch radius, then follow the given steps. How does the surface area of the cup compare to the original paper circle? Find m AC. 3 in. fold C fold A C open cup Section 12.3 Surface Areas of Pyramids and Cones 659 26 What Did You Learn? Core Vocabulary polyhedron, p. 640 face, p. 640 edge, p. 640 vertex, p. 640 cross section, p. 641 solid of revolution, p. 642 axis of revolution, p. 642 lateral faces, p. 646 lateral edges, p. 646 surface area, p. 646 lateral area, p. 646 net, p. 646 right prism, p. 646 oblique prism, p. 646 right cylinder, p. 647 oblique cylinder, p. 647 vertex of a pyramid, p. 654 regular pyramid, p. 654 slant height of a regular pyramid, p. 654 vertex of a cone, p. 655 right cone, p. 655 oblique cone, p. 655 slant height of a right cone, p. 655 lateral surface of a cone, p. 655 Core Concepts Section 12.1 Types of Solids, p. 640 Cross Section of a Solid, p. 641 Solids of Revolution, p. 642 Section 12.2 Lateral Area and Surface Area of a Right Prism, p. 646 Lateral Area and Surface Area of a Right Cylinder, p. 647 Section 12.3 Lateral Area and Surface Area of a Regular Pyramid, p. 654 Lateral Area and Surface Area of a Right Cone, p. 655 Mathematical Thinking 1. In Exercises on page 644, describe the steps you took to sketch each polyhedron. 2. Sketch and label a diagram to represent the situation described in Exercise 32 on page In Exercise 13 on page 658, you need to make a new party hat using 4 times as much cardboard as you previously used for one hat. How should you change the given dimensions to create the new party hat? Explain your reasoning. Study Skills Form a Final Exam Study Group Form a study group several weeks before the final exam. The intent of this group is to review what you have already learned while continuing to learn new material. 661 27 Quiz Tell whether the solid is a polyhedron. If it is, name the polyhedron. (Section 12.1) Sketch the composite solid produced by rotating the figure around the given axis. Then identify and describe the composite solid. (Section 12.1) Find the lateral area and the surface area of the right prism or right cylinder. (Section 12.2) ft 7. 5 in. 7 in. 9 in. 7 ft 10 cm 9 m 12 m 10 m 8. Find the lateral area and the surface area of the composite solid. (Section 12.2) 12 cm 32 cm 10 cm Find the lateral area and the surface area of the regular pyramid or right cone. (Section 12.3) cm ft m 4 3 m 8 cm 16 ft 8 m 12 ft 12. You are replacing the siding and the roofing on the house shown. You have 900 square feet of siding, 500 square feet of roofing material, and 2000 square feet of tarp, in case it rains. (Section 12.3) 12 ft 18 ft 18 ft a. Do you have enough siding to replace the siding on all four sides of the house? Explain. b. Do you have enough roofing material to replace the entire roof? Explain. c. Do you have enough tarp to cover the entire house? Explain. 662 Chapter 12 Surface Area and Volume 28 12.4 Volumes of Prisms and Cylinders TEXAS ESSENTIAL KNOWLEDGE AND SKILLS G.10. G.11.D Essential Question How can you find the volume of a prism or cylinder that is not a right prism or right cylinder? Recall that the volume V of a right prism or a right cylinder is equal to the product of the area of a base and the height h. right prisms right cylinder V = h Finding Volume USING PRECISE MATHEMATICAL LANGUAGE To be proficient in math, you need to communicate precisely to others. Work with a partner. Consider a stack of square papers that is in the form of a right prism. a. What is the volume of the prism? b. When you twist the stack of papers, as shown at the right, do you change the volume? Explain your reasoning. 8 in. c. Write a carefully worded conjecture that describes the conclusion you reached in part (b). d. Use your conjecture to find the volume of the twisted stack of papers. 2 in. 2 in. Finding Volume Work with a partner. Use the conjecture you wrote in Exploration 1 to find the volume of the cylinder. a. 2 in. b. 5 cm 3 in. 15 cm Communicate Your Answer 3. How can you find the volume of a prism or cylinder that is not a right prism or right cylinder? 4. In Exploration 1, would the conjecture you wrote change if the papers in each stack were not squares? Explain your reasoning. Section 12.4 Volumes of Prisms and Cylinders 663 29 12.4 Lesson What You Will Learn Core Vocabulary volume, p. 664 Cavalieri s Principle, p. 664 Previous prism cylinder composite solid Find volumes of prisms and cylinders. Use volumes of prisms and cylinders. Finding Volumes of Prisms and Cylinders The volume of a solid is the number of cubic units contained in its interior. Volume is measured in cubic units, such as cubic centimeters (cm 3 ). Cavalieri s Principle, named after onaventura Cavalieri ( ), states that if two solids have the same height and the same cross-sectional area at every level, then they have the same volume. The prisms below have equal heights h and equal cross-sectional areas at every level. y Cavalieri s Principle, the prisms have the same volume. h Core Concept Volume of a Prism The volume V of a prism is V = h where is the area of a base and h is the height. h h Find the volume of each prism. Finding Volumes of Prisms a. 4 cm 3 cm b. 3 cm 14 cm 2 cm 5 cm 6 cm a. The area of a base is = 1 2 (3)(4) = 6 cm2 and the height is h = 2 cm. V = h Formula for volume of a prism = 6(2) Substitute. = 12 Simplify. The volume is 12 cubic centimeters. b. The area of a base is = 1 2 (3)(6 + 14) = 30 cm2 and the height is h = 5 cm. V = h = 30(5) Substitute. = 150 Simplify. Formula for volume of a prism The volume is 150 cubic centimeters. 664 Chapter 12 Surface Area and Volume 30 Consider a cylinder with height h and base radius r and a rectangular prism with the same height that has a square base with sides of length r π. h r π r π r The cylinder and the prism have the same cross-sectional area, πr 2, at every level and the same height. y Cavalieri s Principle, the prism and the cylinder have the same volume. The volume of the prism is V = h = πr 2 h, so the volume of the cylinder is also V = h = πr 2 h. Core Concept Volume of a Cylinder The volume V of a cylinder is V = h = πr 2 h where is the area of a base, h is the height, and r is the radius of a base. r h r h Finding Volumes of Cylinders Find the volume of each cylinder. a. 9 ft 6 ft b. 4 cm 7 cm a. The dimensions of the cylinder are r = 9 ft and h = 6 ft. V = πr 2 h = π(9) 2 (6) = 486π The volume is 486π, or about cubic feet. b. The dimensions of the cylinder are r = 4 cm and h = 7 cm. V = πr 2 h = π(4) 2 (7) = 112π The volume is 112π, or about cubic centimeters. Monitoring Progress Find the volume of the solid m 5 m 8 m Help in English and Spanish at igideasmath.com 2. 8 ft 14 ft Section 12.4 Volumes of Prisms and Cylinders 665 31 Using Volumes of Prisms and Cylinders Modeling with Mathematics You are building a rectangular chest. You want the length to be 6 feet, the width to be 4 feet, and the volume to be 72 cubic feet. What should the height be? V = 72 ft 3 h 6 ft 4 ft 1. Understand the Problem You know the dimensions of the base of a rectangular prism and the volume. You are asked to find the height. 2. Make a Plan Write the formula for the volume of a rectangular prism, substitute known values, and solve for the height h. 3. Solve the Problem The area of a base is = 6(4) = 24 ft 2 and the volume is V = 72 ft 3. V = h Formula for volume of a prism 72 = 24h Substitute. 3 = h Divide each side by 24. The height of the chest should be 3 feet. 4. Look ack Check your answer. V = h = 24(3) = 72 Monitoring Progress Help in English and Spanish at igideasmath.com 3. WHAT IF? In Example 3, you want the length to be 5 meters, the width to be 3 meters, and the volume to be 60 cubic meters. What should the height be? Changing Dimensions in a Solid ANALYZING MATHEMATICAL RELATIONSHIPS Notice that when all the linear dimensions are multiplied by k, the volume is multiplied by k 3. V original = h = wh V new = (k )(kw)(kh) = (k 3 ) wh Describe how doubling all the linear dimensions affects the volume of the rectangular prism. 6 ft 4 ft efore change After change Dimensions = 4 ft, w = 3 ft, h = 6 ft = 8 ft, w = 6 ft, h = 12 ft V = h V = h Volume = (4)(3)(6) = (8)(6)(12) = 72 ft 3 = 576 ft 3 3 ft = (k 3 )V original Doubling all the linear dimensions results in a volume that is = 8 = 2 3 times the original volume. 666 Chapter 12 Surface Area and Volume 32 Changing a Dimension in a Solid Describe how tripling the radius affects the volume of the cylinder. 3 cm 6 cm efore change After change Dimensions r = 3 cm, h = 6 cm r = 9 cm, h = 6 cm Volume V = πr 2 h = π(3) 2 (6) = 54π cm 3 V = πr 2 h = π(9) 2 (6) = 486π cm 3 Tripling the radius results in a volume that is 486π 54π = 9 = 32 times the original volume. Monitoring Progress Help in English and Spanish at igideasmath.com 4. In Example 4, describe how multiplying all the linear dimensions by 1 affects the 2 volume of the rectangular prism. 5. In Example 4, describe how doubling the length and width of the bases affects the volume of the rectangular prism. 6. In Example 5, describe how multiplying the height by 2 affects the volume of the 3 cylinder. 7. In Example 5, describe how multiplying all the linear dimensions by 4 affects the volume of the cylinder. Finding the Volume of a Composite Solid 0.39 ft Find the volume of the concrete block ft 0.33 ft To find the area of the base, subtract two times the area of the small rectangle from the large rectangle ft 0.66 ft 0.66 ft 3 ft 10 ft 6 ft = Area of large rectangle 2 Area of small rectangle = 1.31(0.66) 2(0.33)(0.39) = Using the formula for the volume of a prism, the volume is V = h = (0.66) The volume is about 0.40 cubic foot. Monitoring Progress 8. Find the volume of the composite solid. Help in English and Spanish at igideasmath.com Section 12.4 Volumes of Prisms and Cylinders 667 33 12.4 Exercises Tutorial Help in English and Spanish at igideasmath.com Vocabulary and Core Concept Check 1. VOCAULARY In what type of units is the volume of a solid measured? 2. COMPLETE THE SENTENCE Cavalieri s Principle states that if two solids have the same and the same at every level, then they have the same. Monitoring Progress and Modeling with Mathematics In Exercises 3 6, find the volume of the prism. (See Example 1.) 12. A pentagonal prism has a height of 9 feet and each base edge is 3 feet cm 1.8 cm 2.3 cm 2 cm 4. 4 m 1.5 m 2 m In Exercises 13 18, find the missing dimension of the prism or cylinder. (See Example 3.) 13. Volume = 560 ft Volume = 2700 yd in. 10 in. u v 5 in. 14 m 7 ft 8 ft 12 yd 15 yd 15. Volume = 80 cm Volume = in. 3 6 m 11 m In Exercises 7 10, find the volume of the cylinder. (See Example 2.) 8 cm 5 cm w 2 in. x 7. 3 ft cm 17. Volume = 3000 ft Volume = m ft 9.8 cm 9.3 ft y z 15 m 9. 5 ft m 8 ft 18 m 19. ERROR ANALYSIS Describe and correct the error in finding the volume of the cylinder. 60 In Exercises 11 and 12, make a sketch of the solid and find its volume. 11. A prism has a height of 11.2 centimeters and an equilateral triangle for a base, where each base edge is 8 centimeters. 4 ft 3 ft V = 2πrh = 2π(4)(3) = 24π So, the volume of the cylinder is 24π cubic feet. 668 Chapter 12 Surface Area and Volume ### Area of Parallelograms, Triangles, and Trapezoids (pages 314 318) Area of Parallelograms, Triangles, and Trapezoids (pages 34 38) Any side of a parallelogram or triangle can be used as a base. The altitude of a parallelogram is a line segment perpendicular to the base ### SURFACE AREA AND VOLUME SURFACE AREA AND VOLUME In this unit, we will learn to find the surface area and volume of the following threedimensional solids:. Prisms. Pyramids 3. Cylinders 4. Cones It is assumed that the reader has ### Area of Parallelograms (pages 546 549) A Area of Parallelograms (pages 546 549) A parallelogram is a quadrilateral with two pairs of parallel sides. The base is any one of the sides and the height is the shortest distance (the length of a perpendicular ### SA B 1 p where is the slant height of the pyramid. V 1 3 Bh. 3D Solids Pyramids and Cones. Surface Area and Volume of a Pyramid Accelerated AAG 3D Solids Pyramids and Cones Name & Date Surface Area and Volume of a Pyramid The surface area of a regular pyramid is given by the formula SA B 1 p where is the slant height of the pyramid. ### Geometry Unit 6 Areas and Perimeters Geometry Unit 6 Areas and Perimeters Name Lesson 8.1: Areas of Rectangle (and Square) and Parallelograms How do we measure areas? Area is measured in square units. The type of the square unit you choose ### Shape Dictionary YR to Y6 Shape Dictionary YR to Y6 Guidance Notes The terms in this dictionary are taken from the booklet Mathematical Vocabulary produced by the National Numeracy Strategy. Children need to understand and use ### 12 Surface Area and Volume CHAPTER 12 Surface Area and Volume Chapter Outline 12.1 EXPLORING SOLIDS 12.2 SURFACE AREA OF PRISMS AND CYLINDERS 12.3 SURFACE AREA OF PYRAMIDS AND CONES 12.4 VOLUME OF PRISMS AND CYLINDERS 12.5 VOLUME ### PERIMETER AND AREA. In this unit, we will develop and apply the formulas for the perimeter and area of various two-dimensional figures. PERIMETER AND AREA In this unit, we will develop and apply the formulas for the perimeter and area of various two-dimensional figures. Perimeter Perimeter The perimeter of a polygon, denoted by P, is the ### Area of a triangle: The area of a triangle can be found with the following formula: You can see why this works with the following diagrams: Area Review Area of a triangle: The area of a triangle can be found with the following formula: 1 A 2 bh or A bh 2 You can see why this works with the following diagrams: h h b b Solve: Find the area of ### Geometry and Measurement The student will be able to: Geometry and Measurement 1. Demonstrate an understanding of the principles of geometry and measurement and operations using measurements Use the US system of measurement for ### ACTIVITY: Finding a Formula Experimentally. Work with a partner. Use a paper cup that is shaped like a cone. 8. Volumes of Cones How can you find the volume of a cone? You already know how the volume of a pyramid relates to the volume of a prism. In this activity, you will discover how the volume of a cone relates ### Platonic Solids. Some solids have curved surfaces or a mix of curved and flat surfaces (so they aren't polyhedra). Examples: Solid Geometry Solid Geometry is the geometry of three-dimensional space, the kind of space we live in. Three Dimensions It is called three-dimensional or 3D because there are three dimensions: width, ### Geometry Notes PERIMETER AND AREA Perimeter and Area Page 1 of 57 PERIMETER AND AREA Objectives: After completing this section, you should be able to do the following: Calculate the area of given geometric figures. Calculate the perimeter ### Solids. Objective A: Volume of a Solids Solids Math00 Objective A: Volume of a Solids Geometric solids are figures in space. Five common geometric solids are the rectangular solid, the sphere, the cylinder, the cone and the pyramid. A rectangular ### Geometry Notes VOLUME AND SURFACE AREA Volume and Surface Area Page 1 of 19 VOLUME AND SURFACE AREA Objectives: After completing this section, you should be able to do the following: Calculate the volume of given geometric figures. Calculate ### 56 questions (multiple choice, check all that apply, and fill in the blank) The exam is worth 224 points. 6.1.1 Review: Semester Review Study Sheet Geometry Core Sem 2 (S2495808) Semester Exam Preparation Look back at the unit quizzes and diagnostics. Use the unit quizzes and diagnostics to determine which ### Area of a triangle: The area of a triangle can be found with the following formula: 1. 2. 3. 12in Area Review Area of a triangle: The area of a triangle can be found with the following formula: 1 A 2 bh or A bh 2 Solve: Find the area of each triangle. 1. 2. 3. 5in4in 11in 12in 9in 21in 14in 19in 13in ### MENSURATION. Definition MENSURATION Definition 1. Mensuration : It is a branch of mathematics which deals with the lengths of lines, areas of surfaces and volumes of solids. 2. Plane Mensuration : It deals with the sides, perimeters ### Teacher Page Key. Geometry / Day # 13 Composite Figures 45 Min. Teacher Page Key Geometry / Day # 13 Composite Figures 45 Min. 9-1.G.1. Find the area and perimeter of a geometric figure composed of a combination of two or more rectangles, triangles, and/or semicircles ### 12-1 Representations of Three-Dimensional Figures Connect the dots on the isometric dot paper to represent the edges of the solid. Shade the tops of 12-1 Representations of Three-Dimensional Figures Use isometric dot paper to sketch each prism. 1. triangular ### Area is a measure of how much space is occupied by a figure. 1cm 1cm Area Area is a measure of how much space is occupied by a figure. Area is measured in square units. For example, one square centimeter (cm ) is 1cm wide and 1cm tall. 1cm 1cm A figure s area is the number ### Angle - a figure formed by two rays or two line segments with a common endpoint called the vertex of the angle; angles are measured in degrees Angle - a figure formed by two rays or two line segments with a common endpoint called the vertex of the angle; angles are measured in degrees Apex in a pyramid or cone, the vertex opposite the base; in ### GAP CLOSING. Volume and Surface Area. Intermediate / Senior Student Book GAP CLOSING Volume and Surface Area Intermediate / Senior Student Book Volume and Surface Area Diagnostic...3 Volumes of Prisms...6 Volumes of Cylinders...13 Surface Areas of Prisms and Cylinders...18 ### Activity Set 4. Trainer Guide Geometry and Measurement of Solid Figures Activity Set 4 Trainer Guide Mid_SGe_04_TG Copyright by the McGraw-Hill Companies McGraw-Hill Professional Development GEOMETRY AND MEASUREMENT OF SOLID FIGURES ### 1. Kyle stacks 30 sheets of paper as shown to the right. Each sheet weighs about 5 g. How can you find the weight of the whole stack? Prisms and Cylinders Answer Key Vocabulary: cylinder, height (of a cylinder or prism), prism, volume Prior Knowledge Questions (Do these BEFORE using the Gizmo.) [Note: The purpose of these questions is ### Algebra Geometry Glossary. 90 angle lgebra Geometry Glossary 1) acute angle an angle less than 90 acute angle 90 angle 2) acute triangle a triangle where all angles are less than 90 3) adjacent angles angles that share a common leg Example: ### B = 1 14 12 = 84 in2. Since h = 20 in then the total volume is. V = 84 20 = 1680 in 3 45 Volume Surface area measures the area of the two-dimensional boundary of a threedimensional figure; it is the area of the outside surface of a solid. Volume, on the other hand, is a measure of the space ### Chapter 4: Area, Perimeter, and Volume. Geometry Assessments Chapter 4: Area, Perimeter, and Volume Geometry Assessments Area, Perimeter, and Volume Introduction The performance tasks in this chapter focus on applying the properties of triangles and polygons to ### 9 Area, Perimeter and Volume 9 Area, Perimeter and Volume 9.1 2-D Shapes The following table gives the names of some 2-D shapes. In this section we will consider the properties of some of these shapes. Rectangle All angles are right ### FCAT FLORIDA COMPREHENSIVE ASSESSMENT TEST. Mathematics Reference Sheets. Copyright Statement for this Assessment and Evaluation Services Publication FCAT FLORIDA COMPREHENSIVE ASSESSMENT TEST Mathematics Reference Sheets Copyright Statement for this Assessment and Evaluation Services Publication Authorization for reproduction of this document is hereby ### SOLIDS, NETS, AND CROSS SECTIONS SOLIDS, NETS, AND CROSS SECTIONS Polyhedra In this section, we will examine various three-dimensional figures, known as solids. We begin with a discussion of polyhedra. Polyhedron A polyhedron is a three-dimensional ### Angles that are between parallel lines, but on opposite sides of a transversal. GLOSSARY Appendix A Appendix A: Glossary Acute Angle An angle that measures less than 90. Acute Triangle Alternate Angles A triangle that has three acute angles. Angles that are between parallel lines, ### Show that when a circle is inscribed inside a square the diameter of the circle is the same length as the side of the square. Week & Day Week 6 Day 1 Concept/Skill Perimeter of a square when given the radius of an inscribed circle Standard 7.MG:2.1 Use formulas routinely for finding the perimeter and area of basic twodimensional ### Unit 3 Practice Test. Name: Class: Date: Multiple Choice Identify the choice that best completes the statement or answers the question. Name: lass: ate: I: Unit 3 Practice Test Multiple hoice Identify the choice that best completes the statement or answers the question. The radius, diameter, or circumference of a circle is given. Find ### VOLUME of Rectangular Prisms Volume is the measure of occupied by a solid region. Math 6 NOTES 7.5 Name VOLUME of Rectangular Prisms Volume is the measure of occupied by a solid region. The formula for the volume of a rectangular prism is: l = length w = width h = height Study Tip: ### Characteristics of the Four Main Geometrical Figures Math 40 9.7 & 9.8: The Big Four Square, Rectangle, Triangle, Circle Pre Algebra We will be focusing our attention on the formulas for the area and perimeter of a square, rectangle, triangle, and a circle. ### CHAPTER 8, GEOMETRY. 4. A circular cylinder has a circumference of 33 in. Use 22 as the approximate value of π and find the radius of this cylinder. TEST A CHAPTER 8, GEOMETRY 1. A rectangular plot of ground is to be enclosed with 180 yd of fencing. If the plot is twice as long as it is wide, what are its dimensions? 2. A 4 cm by 6 cm rectangle has ### Perimeter, Area, and Volume Perimeter, Area, and Volume Perimeter of Common Geometric Figures The perimeter of a geometric figure is defined as the distance around the outside of the figure. Perimeter is calculated by adding all ### of surface, 569-571, 576-577, 578-581 of triangle, 548 Associative Property of addition, 12, 331 of multiplication, 18, 433 Absolute Value and arithmetic, 730-733 defined, 730 Acute angle, 477 Acute triangle, 497 Addend, 12 Addition associative property of, (see Commutative Property) carrying in, 11, 92 commutative property ### Finding Volume of Rectangular Prisms MA.FL.7.G.2.1 Justify and apply formulas for surface area and volume of pyramids, prisms, cylinders, and cones. MA.7.G.2.2 Use formulas to find surface areas and volume of three-dimensional composite shapes. ### Calculating Area, Perimeter and Volume Calculating Area, Perimeter and Volume You will be given a formula table to complete your math assessment; however, we strongly recommend that you memorize the following formulae which will be used regularly ### Conjectures. Chapter 2. Chapter 3 Conjectures Chapter 2 C-1 Linear Pair Conjecture If two angles form a linear pair, then the measures of the angles add up to 180. (Lesson 2.5) C-2 Vertical Angles Conjecture If two angles are vertical ### Surface Area and Volume Cylinders, Cones, and Spheres Surface Area and Volume Cylinders, Cones, and Spheres Michael Fauteux Rosamaria Zapata CK12 Editor Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) To access a customizable ### Student Outcomes. Lesson Notes. Classwork. 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During testing students will have access to the Algebra I/Geometry EOC Assessments Reference ### How does one make and support a reasonable conclusion regarding a problem? How does what I measure influence how I measure? Middletown Public Schools Mathematics Unit Planning Organizer Subject Mathematics Grade/Course Grade 7 Unit 3 Two and Three Dimensional Geometry Duration 23 instructional days (+4 days reteaching/enrichment) ### CCGPS UNIT 3 Semester 1 ANALYTIC GEOMETRY Page 1 of 32. Circles and Volumes Name: GPS UNIT 3 Semester 1 NLYTI GEOMETRY Page 1 of 3 ircles and Volumes Name: ate: Understand and apply theorems about circles M9-1.G..1 Prove that all circles are similar. M9-1.G.. 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Revision - Areas Chapter 8 Volumes The formulae for calculating the areas of quadrilaterals, circles and triangles should already be known :- SQUARE RECTANGE RHOMBUS KITE B dd d D D Area = 2 Area = x B ### MD5-26 Stacking Blocks Pages 115 116 MD5-26 Stacking Blocks Pages 115 116 STANDARDS 5.MD.C.4 Goals Students will find the number of cubes in a rectangular stack and develop the formula length width height for the number of cubes in a stack. ### Area. Area Overview. Define: Area: Define: Area: Area Overview Kite: Parallelogram: Rectangle: Rhombus: Square: Trapezoid: Postulates/Theorems: Every closed region has an area. If closed figures are congruent, then their areas are equal. ### DISCOVERING 3D SHAPES . DISCOVERING 3D SHAPES WORKSHEETS OCTOBER-DECEMBER 2009 1 . Worksheet 1. Cut out and stick the shapes. SHAPES WHICH ROLL SHAPES WHICH SLIDE 2 . Worksheet 2: COMPLETE THE CHARTS Sphere, triangle, prism, ### GAP CLOSING. 2D Measurement GAP CLOSING. 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Thursday, August 13, 2015 8:30 to 11:30 a.m., only. GEOMETRY The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY Thursday, August 13, 2015 8:30 to 11:30 a.m., only Student Name: School Name: The possession or use of any communications ### Georgia Online Formative Assessment Resource (GOFAR) AG geometry domain AG geometry domain Name: Date: Copyright 2014 by Georgia Department of Education. Items shall not be used in a third party system or displayed publicly. Page: (1 of 36 ) 1. Amy drew a circle graph to represent ### The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY. Tuesday, August 13, 2013 8:30 to 11:30 a.m., only. GEOMETRY The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY Tuesday, August 13, 2013 8:30 to 11:30 a.m., only Student Name: School Name: The possession or use of any communications ### 1. A plane passes through the apex (top point) of a cone and then through its base. 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Assessment Pizza! Pizza! Assessment 1. A local pizza restaurant sends pizzas to the high school twelve to a carton. If the pizzas are one inch thick, what is the volume of the cylindrical shipping carton for the ### Applications for Triangles Not drawn to scale Applications for Triangles 1. 36 in. 40 in. 33 in. 1188 in. 2 69 in. 2 138 in. 2 1440 in. 2 2. 188 in. 2 278 in. 2 322 in. 2 none of these Find the area of a parallelogram with the given ### The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY. Student Name: GEOMETRY The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY Thursday, June 17, 2010 1:15 to 4:15 p.m., only Student Name: School Name: Print your name and the name of your ### The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY. Student Name: GEOMETRY The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY Wednesday, August 18, 2010 8:30 to 11:30 a.m., only Student Name: School Name: Print your name and the name of ### Geometry Regents Review Name: Class: Date: Geometry Regents Review Multiple Choice Identify the choice that best completes the statement or answers the question. 1. If MNP VWX and PM is the shortest side of MNP, what is the shortest ### WEIGHTS AND MEASURES. Linear Measure. 1 Foot12 inches. 1 Yard 3 feet - 36 inches. 1 Rod 5 1/2 yards - 16 1/2 feet WEIGHTS AND MEASURES Linear Measure 1 Foot12 inches 1 Yard 3 feet - 36 inches 1 Rod 5 1/2 yards - 16 1/2 feet 1 Furlong 40 rods - 220 yards - 660 feet 1 Mile 8 furlongs - 320 rods - 1,760 yards 5,280 feet ### GEOMETRY COMMON CORE STANDARDS 1st Nine Weeks Experiment with transformations in the plane G-CO.1 Know precise definitions of angle, circle, perpendicular line, parallel line, and line segment, based on the undefined notions of point, ### Grade 7 & 8 Math Circles Circles, Circles, Circles March 19/20, 2013 Faculty of Mathematics Waterloo, Ontario N2L 3G Introduction Grade 7 & 8 Math Circles Circles, Circles, Circles March 9/20, 203 The circle is a very important shape. In fact of all shapes, the circle is ### YOU MUST BE ABLE TO DO THE FOLLOWING PROBLEMS WITHOUT A CALCULATOR! 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U.S. CUSTOMARY SYSTEM OF MEASUREMENT LENGTH The standard U.S. Customary System units of length are inch, foot, yard, and mile. MEASUREMENTS A measurement includes a number and a unit. 3 feet 7 minutes 12 gallons Standard units of measurement have been established to simplify trade and commerce. TIME Equivalences between units ### Geometry. Higher Mathematics Courses 69. Geometry The fundamental purpose of the course is to formalize and extend students geometric experiences from the middle grades. This course includes standards from the conceptual categories of and Statistics and ### 43 Perimeter and Area 43 Perimeter and Area Perimeters of figures are encountered in real life situations. For example, one might want to know what length of fence will enclose a rectangular field. 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Through daily activities, we will explore the relationship between flat, two-dimensional figures and solid, three-dimensional ### The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY GEOMETRY The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY Wednesday, June 20, 2012 9:15 a.m. to 12:15 p.m., only Student Name: School Name: Print your name and the name ### Grade 8 Mathematics Geometry: Lesson 2 Grade 8 Mathematics Geometry: Lesson 2 Read aloud to the students the material that is printed in boldface type inside the boxes. Information in regular type inside the boxes and all information outside ### TEKS TAKS 2010 STAAR RELEASED ITEM STAAR MODIFIED RELEASED ITEM 7 th Grade Math TAKS-STAAR-STAAR-M Comparison Spacing has been deleted and graphics minimized to fit table. (1) Number, operation, and quantitative reasoning. The student represents and uses numbers in
# Positive and Negative Numbers ## Presentation on theme: "Positive and Negative Numbers"— Presentation transcript: Positive and Negative Numbers by Monica Yuskaitis Definition Positive number – a number greater than zero. 1 2 3 4 5 6 Definition Negative number – a number less than zero. -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 Definition Opposite Numbers – numbers that are the same distance from zero in the opposite direction -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 Definition Integers – Integers are all the whole numbers and all of their opposites on the negative number line including zero. 7 opposite -7 Negative Numbers Are Used to Measure Temperature Negative Numbers Are Used to Measure Under Sea Level 30 20 10 -10 -20 -30 -40 -50 Negative Numbers Are Used to Show Debt Let’s say your parents bought a car but had to get a loan from the bank for \$5,000. When counting all their money they add in -\$5.000 to show they still owe the bank. Positive & Negative Numbers Working with positive and negative numbers is like driving in traffic. You have to watch the traffic signs and follow the driving rules. Hint If you don’t see a negative or positive sign in front of a number it is positive. 9 + One Way to Add Integers Is With a Number Line When the number is positive count to the right. When the number is negative count to the left. - + 1 2 3 4 5 6 -1 -2 -3 -4 -5 -6 One Way to Add Integers Is With a Number Line = -2 + 1 2 3 4 5 6 -1 -2 -3 -4 -5 -6 - One Way to Add Integers Is With a Number Line = +2 + 1 2 3 4 5 6 -1 -2 -3 -4 -5 -6 - One Way to Add Integers Is With a Number Line = -4 + 1 2 3 4 5 6 -1 -2 -3 -4 -5 -6 - One Way to Add Integers Is With a Number Line = +4 - 1 2 3 4 5 6 -1 -2 -3 -4 -5 -6 + One Way to Add Integers Is With a Number Line = +4 - 1 2 3 4 5 6 -1 -2 -3 -4 -5 -6 + One Way to Add Integers Is With a Number Line = +6 - 1 2 3 4 5 6 -1 -2 -3 -4 -5 -6 + One Way to Add Integers Is With a Number Line = -3 - 1 2 3 4 5 6 -1 -2 -3 -4 -5 -6 + One Way to Add Integers Is With a Number Line = -6 - 1 2 3 4 5 6 -1 -2 -3 -4 -5 -6 - One Way to Add Integers Is With a Number Line = -5 - 1 2 3 4 5 6 -1 -2 -3 -4 -5 -6 - One Way to Add Integers Is With a Number Line = -5 - 1 2 3 4 5 6 -1 -2 -3 -4 -5 -6 - One Way to Add Integers Is With a Number Line = -5 - 1 2 3 4 5 6 -1 -2 -3 -4 -5 -6 - One Way to Add Integers Is With a Number Line 4 + 1 = 5 + 1 2 3 4 5 6 -1 -2 -3 -4 -5 -6 + One Way to Add Integers Is With a Number Line 3 + 2 = 5 + 1 2 3 4 5 6 -1 -2 -3 -4 -5 -6 + One Way to Add Integers Is With a Number Line 1 + 5 = 6 + 1 2 3 4 5 6 -1 -2 -3 -4 -5 -6 + Solve the Problems below using a number line. = 4 + 7 = = -8 11 -1 Solve the Problems below using a number line. = 5 + 9 = = -13 14 -12 Rule #1 – If the signs are the same, add the numbers and keep the original sign. 9 + 5 = 14 = -14
# How do you integrate int (4x)/sqrt(x^2-14x+45)dx using trigonometric substitution? Feb 29, 2016 $4 \sqrt{{\left(\frac{x - 7}{2}\right)}^{2} + 1} + 7 {\cosh}^{-} 1 \left(\frac{x - 7}{2}\right) + C$ #### Explanation: We can begin by taking ${x}^{2} - 14 x + 45$ an re writing it in in completed square/ vertex form giving us: ${\left(x - 7\right)}^{2} - 4$ Putting this into the integral will give: $\int \frac{4 x}{\sqrt{{\left(x - 7\right)}^{2} - 4}} \mathrm{dx}$ We can factor the $4$ out to the front then play with the numerator a little and we get: $= 4 \int \frac{x - 7 + 7}{\sqrt{{\left(x - 7\right)}^{2} - 4}} \mathrm{dx}$ $= 4 \int \frac{x - 7}{\sqrt{{\left(x - 7\right)}^{2} - 4}} + \frac{7}{\sqrt{{\left(x - 7\right)}^{2} - 4}} \mathrm{dx}$ Now lets consider the substitution (we'll have to use a hyperbolic function, not a trig function): $2 \cosh \left(u\right) = x - 7$ $2 \sinh \left(u\right) \mathrm{du} = \mathrm{dx}$ Now putting this substitution in we get: $4 \int \left(\frac{2 \cosh \left(u\right)}{\sqrt{4 {\cosh}^{2} \left(u\right) - 4}} + \frac{7}{\sqrt{4 {\cosh}^{2} \left(u\right) - 4}}\right) \sinh \left(u\right) \mathrm{du}$ Simplify this a little by factoring the $4$ from the square root: $= 4 \int \frac{\sinh \left(u\right) \cosh \left(u\right)}{\sqrt{{\cosh}^{2} \left(u\right) - 1}} + \frac{7 \sinh \left(u\right)}{2 \sqrt{{\cosh}^{2} \left(u\right) - 1}} \mathrm{du}$ Now at this point we can use the identity: ${\cosh}^{2} \left(u\right) - 1 = {\sinh}^{2} \left(u\right)$ and then do a bit of cancelling. That will give us: $= 4 \int \frac{\sinh \left(u\right) \cosh \left(u\right)}{\sqrt{{\sinh}^{2} \left(u\right)}} + \frac{7 \sinh \left(u\right)}{2 \sqrt{{\sinh}^{2} \left(u\right)}} \mathrm{du}$ $= 4 \int \frac{\sinh \left(u\right) \cosh \left(u\right)}{\sinh} \left(u\right) + \frac{7 \sinh \left(u\right)}{2 \sinh \left(u\right)} \mathrm{du}$ $= 4 \int \cosh \left(u\right) + \frac{7}{2} \mathrm{du}$ Now evaluate the integral: $4 \left(\sinh \left(u\right) + \frac{7}{2} u\right) + C = 4 \sinh \left(u\right) + 14 u + C$ Now we can use the identity that we used earlier to convert $\sinh$ into $\cosh$: $= 4 \left(\sqrt{{\cosh}^{2} \left(u\right) + 1} + \frac{7}{2} u\right) + C$ Now reverse the substitution and we get: $4 \sqrt{{\left(\frac{x - 7}{2}\right)}^{2} + 1} + 7 {\cosh}^{-} 1 \left(\frac{x - 7}{2}\right) + C$ At this point you can re write the function in terms of its exponential and logarithmic definitions but I believe this maybe satisfactory>
or Find what you need to study Light Find what you need to study # 5.1 Using the Mean Value Theorem In the previous unit, we learned all about applying derivatives to different real-world contexts. What else are derivatives useful for? Turns out, we can also use derivatives to determine and analyze the behaviors of functions! 👀 ## 📈 Mean Value Theorem The Mean Value Theorem states that if a function f is continuous over the interval $[a, b]$ and differentiable over the interval $(a, b)$, then there exists a point $c$ within that open interval $(a,b)$ where the instantaneous rate of change of the function at $c$ equals the average rate of change of the function over the interval $(a, b)$. In other words, if a function f is continuous over the interval $[a, b]$ and differentiable over the interval $(a, b)$, there exists some $c$ on $(a,b)$ such that $f'(c)=\frac{f(b)-f(a)}{b-a}$. Image Courtesy of Sumi Vora and Ethan Bilderbeek Yet another way to phrase this theorem is that if the stated conditions of continuity and differentiability are satisfied, there is a point where the slope of the tangent line is equivalent to the slope of the secant line between $a$ and $b$. Image Courtesy of Math.net Remember from Unit 1, to be continuous over $[a, b]$ means that there are no holes, asymptotes, or jump discontinuities between points a and b. Because the interval contains closed brackets, the graph must also be continuous at points $a$ and $b$. Additionally, if we recall from previous guides, to be differentiable over $(a, b)$ means that the function is continuous over the interval and that for any point $c$ over the interval, $\lim_{x\to c} \frac{f(x) - f(c)}{x - c}$ exists. ### ✏️ Mean Value Theorem: Walkthrough We can use the Mean Value Theorem to justify conclusions about functions by applying it over an interval. For example: Let $f$ be a differentiable function. The table gives its selected values: $x$ $3$ $9$ $11$ $f(x)$ $20$ $44$ $67$ Can we use the Mean Value Theorem to say the equation $f'(x)=5$ has a solution where $3< x<9$? Since it is given that $f$ is differentiable, we can apply the Mean Value Theorem (MVT) on the interval $(3,9)$. This is what we should write out! Since $f$ is continuous and differentiable on $(3,9)$, MVT can be applied. The MVT states that there exists a $c$ on $(3,9)$ such that $f'(c)=\frac{f(9)-f(3)}{9-3}=\frac{44-20}{9-3}=\frac{24}{6}=4.$ $4 \neq 5$ so MVT cannot be used to say that $f'(x)=5$ has a solution. ## 📝 Mean Value Theorem: Practice Problems Now, it’s time for you to do some practice on your own! 🍀 ### ❓ Mean Value Theorem: Practice #### Question 1: Mean Value Theorem Let $h(x)=x^3+3x^2$ and let $c$ be the number that satisfies the Mean Value Theorem for $h$ on the interval $[-3,0]$. What is $c?$ #### Question 2: Mean Value Theorem Let $f$ be a differentiable function. The table gives its selected values: $x$ $2$ $7$ $9$ $f(x)$ $14$ $43$ $35$ Can we use the Mean Value Theorem to say the equation $f'(x)=2$ has a solution where $2? ### ✅ Mean Value Theorem: Answers and Solutions #### Question 1: Mean Value Theorem Since $h$ is a polynomial, $h$ is continuous on $[-3,0]$ and differentiable on $(-3,0)$. Therefore, the Mean Value Theorem can be applied. By the Mean Value Theorem, there exists a $c$ on $(-3,0)$ such that… $f'(c)=\frac{f(0)-f(-3)}{0-(-3)}=\frac{0-0}{3}=0.$ To find $c$, we need to differentiate $f(x)$ and find $c$ such that $f'(c)=0.$ $f'(x)=3x^2+6x$ $3x^2+6x=0$ By the quadratic formula, we have $x=-2,0$. Since only $x=-2$ is in the interval $(-3,0)$, $c=-2.$ Great work! Make sure you remember to check if the value(s) you get are in the given interval. #### Question 2: Mean Value Theorem Since it is given that $f$ is differentiable, we can apply the Mean Value Theorem (MVT) on the interval $(2,9)$. The MVT states that there exists a $c$ on $(2,9)$ such that $f'(c)=\frac{f(9)-f(2)}{9-2}=\frac{35-14}{9-2}=\frac{21}{7}=3.$ $3 \neq 2$ so MVT cannot be used to say that $f'(x)=2$ has a solution. ## 💫 Closing The Mean Value Theorem states that for any continuous function on a closed interval, there exists a value c in the interval such that the value of the derivative of the function at c is equal to the average rate of change of the function over the interval. By using this theorem, we can find the mean value of a function on a given interval, which can provide useful information about the behavior of the function. 🧐 # 5.1 Using the Mean Value Theorem In the previous unit, we learned all about applying derivatives to different real-world contexts. What else are derivatives useful for? Turns out, we can also use derivatives to determine and analyze the behaviors of functions! 👀 ## 📈 Mean Value Theorem The Mean Value Theorem states that if a function f is continuous over the interval $[a, b]$ and differentiable over the interval $(a, b)$, then there exists a point $c$ within that open interval $(a,b)$ where the instantaneous rate of change of the function at $c$ equals the average rate of change of the function over the interval $(a, b)$. In other words, if a function f is continuous over the interval $[a, b]$ and differentiable over the interval $(a, b)$, there exists some $c$ on $(a,b)$ such that $f'(c)=\frac{f(b)-f(a)}{b-a}$. Image Courtesy of Sumi Vora and Ethan Bilderbeek Yet another way to phrase this theorem is that if the stated conditions of continuity and differentiability are satisfied, there is a point where the slope of the tangent line is equivalent to the slope of the secant line between $a$ and $b$. Image Courtesy of Math.net Remember from Unit 1, to be continuous over $[a, b]$ means that there are no holes, asymptotes, or jump discontinuities between points a and b. Because the interval contains closed brackets, the graph must also be continuous at points $a$ and $b$. Additionally, if we recall from previous guides, to be differentiable over $(a, b)$ means that the function is continuous over the interval and that for any point $c$ over the interval, $\lim_{x\to c} \frac{f(x) - f(c)}{x - c}$ exists. ### ✏️ Mean Value Theorem: Walkthrough We can use the Mean Value Theorem to justify conclusions about functions by applying it over an interval. For example: Let $f$ be a differentiable function. The table gives its selected values: $x$ $3$ $9$ $11$ $f(x)$ $20$ $44$ $67$ Can we use the Mean Value Theorem to say the equation $f'(x)=5$ has a solution where $3< x<9$? Since it is given that $f$ is differentiable, we can apply the Mean Value Theorem (MVT) on the interval $(3,9)$. This is what we should write out! Since $f$ is continuous and differentiable on $(3,9)$, MVT can be applied. The MVT states that there exists a $c$ on $(3,9)$ such that $f'(c)=\frac{f(9)-f(3)}{9-3}=\frac{44-20}{9-3}=\frac{24}{6}=4.$ $4 \neq 5$ so MVT cannot be used to say that $f'(x)=5$ has a solution. ## 📝 Mean Value Theorem: Practice Problems Now, it’s time for you to do some practice on your own! 🍀 ### ❓ Mean Value Theorem: Practice #### Question 1: Mean Value Theorem Let $h(x)=x^3+3x^2$ and let $c$ be the number that satisfies the Mean Value Theorem for $h$ on the interval $[-3,0]$. What is $c?$ #### Question 2: Mean Value Theorem Let $f$ be a differentiable function. The table gives its selected values: $x$ $2$ $7$ $9$ $f(x)$ $14$ $43$ $35$ Can we use the Mean Value Theorem to say the equation $f'(x)=2$ has a solution where $2? ### ✅ Mean Value Theorem: Answers and Solutions #### Question 1: Mean Value Theorem Since $h$ is a polynomial, $h$ is continuous on $[-3,0]$ and differentiable on $(-3,0)$. Therefore, the Mean Value Theorem can be applied. By the Mean Value Theorem, there exists a $c$ on $(-3,0)$ such that… $f'(c)=\frac{f(0)-f(-3)}{0-(-3)}=\frac{0-0}{3}=0.$ To find $c$, we need to differentiate $f(x)$ and find $c$ such that $f'(c)=0.$ $f'(x)=3x^2+6x$ $3x^2+6x=0$ By the quadratic formula, we have $x=-2,0$. Since only $x=-2$ is in the interval $(-3,0)$, $c=-2.$ Great work! Make sure you remember to check if the value(s) you get are in the given interval. #### Question 2: Mean Value Theorem Since it is given that $f$ is differentiable, we can apply the Mean Value Theorem (MVT) on the interval $(2,9)$. The MVT states that there exists a $c$ on $(2,9)$ such that $f'(c)=\frac{f(9)-f(2)}{9-2}=\frac{35-14}{9-2}=\frac{21}{7}=3.$ $3 \neq 2$ so MVT cannot be used to say that $f'(x)=2$ has a solution. ## 💫 Closing The Mean Value Theorem states that for any continuous function on a closed interval, there exists a value c in the interval such that the value of the derivative of the function at c is equal to the average rate of change of the function over the interval. By using this theorem, we can find the mean value of a function on a given interval, which can provide useful information about the behavior of the function. 🧐
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # AA Similarity ## Two triangles are similar if two pairs of angles are congruent. 0% Progress Practice AA Similarity Progress 0% AA Similarity What if you were given a pair of triangles and the angle measures for two of their angles? How could you use this information to determine if the two triangles are similar? After completing this Concept, you'll be able to use the AA Similarity Postulate to decide if two triangles are similar. ### Watch This CK-12 Foundation: AA Similarity For additional help, first watch this video beginning at the 2:09 mark. James Sousa: Similar Triangles Then watch this video. James Sousa: Similar Triangles by AA ### Guidance By definition, two triangles are similar if all their corresponding angles are congruent and their corresponding sides are proportional. It is not necessary to check all angles and sides in order to tell if two triangles are similar. In fact, if you only know that two pairs of corresponding angles are congruent that is enough information to know that the triangles are similar. This is called the AA Similarity Postulate. AA Similarity Postulate: If two angles in one triangle are congruent to two angles in another triangle, then the two triangles are similar. If and , then . #### Example A Determine if the following two triangles are similar. If so, write the similarity statement. Compare the angles to see if we can use the AA Similarity Postulate. Using the Triangle Sum Theorem, and So, and and the triangles are similar. . #### Example B Determine if the following two triangles are similar. If so, write the similarity statement. Compare the angles to see if we can use the AA Similarity Postulate. Using the Triangle Sum Theorem, and . , So and are not similar. #### Example C by AA. Find and . Set up a proportion to find the missing sides. When two triangles are similar, the corresponding sides are proportional. But, what are the corresponding sides? Using the triangles from this example, we see how the sides line up in the diagram to the right. CK-12 Foundation: AA Similarity --> ### Guided Practice 1.Are the following triangles similar? If so, write the similarity statement. 2. Are the triangles similar? If so, write a similarity statement. 3. Are the triangles similar? If so, write a similarity statement. 1. Because and by the Alternate Interior Angles Theorem. By the AA Similarity Postulate, . 2. Yes, there are three similar triangles that each have a right angle. . 3. By the reflexive property, . Because the horizontal lines are parallel, (corresponding angles). So yes, there is a pair of similar triangles. . ### Explore More Use the diagram to complete each statement. 1. ______ 2. = ______ 3. = ______ 1. Name two similar triangles. How do you know they are similar? 2. Write a true proportion. 3. Name two other triangles that might not be similar. 4. If and , find . Be careful! Use the triangles to the left for questions 10-14. , and . 1. Are the two triangles similar? How do you know? 2. Write an expression for in terms of . 3. If , what is ? 4. Fill in the blanks: If an acute angle of a _______ triangle is congruent to an acute angle in another ________ triangle, then the two triangles are _______. 5. Writing How do congruent triangles and similar triangles differ? How are they the same? Are the following triangles similar? If so, write a similarity statement. ### Answers for Explore More Problems To view the Explore More answers, open this PDF file and look for section 7.4. ### Vocabulary Language: English Spanish similar triangles similar triangles Two triangles where all their corresponding angles are congruent (exactly the same) and their corresponding sides are proportional (in the same ratio). AA Similarity Postulate AA Similarity Postulate If two angles in one triangle are congruent to two angles in another triangle, then the two triangles are similar. Dilation Dilation To reduce or enlarge a figure according to a scale factor is a dilation. Triangle Sum Theorem Triangle Sum Theorem The Triangle Sum Theorem states that the three interior angles of any triangle add up to 180 degrees. Rigid Transformation Rigid Transformation A rigid transformation is a transformation that preserves distance and angles, it does not change the size or shape of the figure.
# Where is the mistake in the calculation of $y'$ if $y = \Bigl( \dfrac{x^2+1}{x^2-1} \Bigr)^{1/4}$? Plase take a look here. If $y = \Bigl( \dfrac{x^2+1}{x^2-1} \Bigr)^{1/4}$ \begin{eqnarray} y'&=& \dfrac{1}{4} \Bigl( \dfrac{x^2+1}{x^2-1} \Bigr)^{-3/4} \left \{ \dfrac{2x(x^2-1) - 2x(x^2+1) }{(x^2-1)^2} \right \}\\ &=& \Bigl( \dfrac{x^2+1}{x^2-1} \Bigr)^{-3/4} \dfrac{-x}{(x^2-1)^2}. \end{eqnarray} By the other hand, we have $$\log y = \dfrac{1}{4} \left \{ \log (x^2+1) - \log (x^2-1) \right \}$$ Then, \begin{eqnarray} \dfrac{dy}{dx} &=& y \dfrac{1}{4} \left \{ \dfrac{2x}{(x^2+1)} -\dfrac{ 2x}{(x^2-1)} \right \} \\ &=& \dfrac{1}{4} \dfrac{x^2+1}{x^2-1} \cdot 2x \dfrac{(x^2-1) - (x^2+1)}{(x^2+1)(x^2-1)} \\ &=& \dfrac{x^2+1}{x^2-1} \dfrac{-x}{(x^2+1)(x^2-1)} \\ &=& \dfrac{-x}{(x^2-1)^2}. \end{eqnarray} But this implies, $$\dfrac{-x}{(x^2-1)^2} = \Bigl( \dfrac{x^2+1}{x^2-1} \Bigr)^{-3/4} \dfrac{-x}{(x^2-1)^2}.$$ Where is the mistake? - It's recommendable that you use LaTeX in the exponents. Instead of $x²$, use $x^2$ x^2. Also, there's a typo in title "calculation". Better $y'$ than $y´$. – Américo Tavares Dec 4 '12 at 19:49 @AméricoTavares : I was about to post the same comment about squares. We had that same discussion several years ago on Wikipedia, about the style manual for typesetting in math articles. – Michael Hardy Dec 4 '12 at 20:20 @MichaelHardy I saw your post on meta meta.math.stackexchange.com/questions/6717/…. – Américo Tavares Dec 4 '12 at 20:22 I believe you forgot a power 1/4 when substituting for $y$ (in the calculation using logarithms). I think no, look $y = \Bigl( \dfrac{x^2+1}{x^2-1} \Bigr)^{1/4} \Rightarrow \log y = \dfrac{1}{4} \log \left \{ \dfrac{x^2+1}{x^2-1} \right \} = \dfrac{1}{4} \left \{ \log (x²+1) - \log (x²-1) \right \}$ – user29999 Dec 4 '12 at 19:53 When substituting for $y$. This is from the first to the second line of the calculation starting with $\frac{dy}{dx}$. – Daan Michiels Dec 4 '12 at 19:54
Courses Courses for Kids Free study material Offline Centres More Store # Square Prism Reviewed by: Last updated date: 06th Sep 2024 Total views: 397.5k Views today: 10.97k ## Square Based Prism As per the square prism definition, it is a three-dimensional geometric solid object that has a base of a square. In a square prism, the sides and angles opposite of each other are congruent. Thus, when a minimum of two of the sides and angles are equal in measurement, it can also be termed as a square prism. You can also consider cuboid whose base is square as a square prism. In the illustration below, you can see that the bases are squared and thus it is a square prism. ### The Volume of a Square Prism The volume of a square-based prism is a computation of the inhabited units of the solid. The volume of a square prism is the number of units that are used to fill a cube. It is represented in the form of cubic units. We have the formula to calculate the volume of a square prism, i.e. V= a²h cubic units Where, V = volume A = side of the base H = height of the prism ### How to Calculate Surface Area of a Square Prism The surface area of a square prism is a quantification of the total area of a surface of a 3-D solid object. For a square-based prism, the surface area is defined as the sum total of twice the base area and the lateral surface area (LSA) of the prism. Its measure is represented in square units. Formula to find the surface area of a square prism is as follows: The surface area of the square prism i.e. SA = 2a² + 4ah square units In which, a = side of the square prism h = height of the square prism Is Square Prism a Cube? A cube is a unique case of a square prism where the lengths in all the three dimensions are identical. Hence, all cubes make square prisms but not all square prisms are cubes. ### Types of Prisms There are different prisms that are categorized based on the cross-section. These are as given: • Square Prism • Rectangular Prism • Triangular Prism • Oblique Prism • Cubical Prism • Pentagonal Prism Note: prism with rectangular base is also sometimes called a square prism. ### Cross Section in a Prism A cross section is a shape obtained by cutting straight through an object. For example, in the image below, the cross-section of the object is a triangle and also contains Identical cross-section all along its length, therefore it is a triangular prism. ### Solved Examples on Prisms Example: Find the surface area of a prism where the base area is 20 m², the base perimeter is 22 m, and the length is 10 m. Solution: Surface Area of A Prism = 2 × (Base Area + Base Perimeter) × Length Thus, we get = 2 × 20 m² + 22 m × 10 m = 40 m² + 220 m² = 260 m² Example: Calculate the volume and the surface area of a square prism whose side is 7 cm and height is 11 cm. Solution: Given: Side, a = 7 cm Height, h = 11 cm The Volume a Square Prism, V = a² h cubic units. V = (7)² (11) V = 49 (11) V = 539 cm³ Hence, the volume of a square prism is 539 cm³ The surface area of the square prism, SA = 2a² + 4ah square units Substituting the given values, we get SA = 2(7)² + 4(7)(11) cm² SA = 2(49) + 4(77) cm² SA = 98 + 308 cm² SA = 406 cm² Thus, the surface area of a given square prism is 406 cm². ## FAQs on Square Prism Q1. Why do we Call a Prism a Square Prism? Answer: As there are a variety of shapes that can serve as the base of a prism, there are different forms of prisms that can be created. Prisms are basically named for the shape of their bases, so a square prism attained its name due to the squares in its bases. Q2. What is an Example of a Square Prism? Answer: Think about the rolling dice. You will get a clear idea of how a square prism is. Dice is an enclosed 3-dimensional shape, which is based on two squares. For a fact, regular dice are cubed; all the faces of this solid object are square. So cubes are actually a square prism regardless of how they are looked at! Another common real-world example is the famous Rubik's Cube. Also, imagine diced cubes of cottage cheese, those are square prisms too. Q3. How can we Spot a Square Prism Easily? Answer: When finding out a square prism, make sure that you do not have to look for all the faces that are squares. As long as the solid object has two square faces, meaning its bases are squares, it's a square prism. Q4. What are the Properties of a Prism? Answer: A prism is a solid 3-dimensional figure with: • Flat faces. • Uniform and similar ends. • Same cross-section all along its length!
Uploaded by David Pinto # Fraction review & summary advertisement ```Middle School Math Huddle Fractions Review & Summary Factors and Multiples 1∙12 2∙6 3∙4 The key to fractions is understanding factors and multiples. 12______ 12 24 36 ______18______ 1∙18 18 2∙9 36 3∙6 54 The GCF of 12 and 18 is 6 The LCM of 12 and 18 is 36 The Greatest Common Factor (GCF) of two number is the largest number that is a factor of both. The Least Common Multiple (LCM) of two numbers is the smallest number that is a multiple of both Simplifying Fractions To simplify a fraction, find the GCF of the numerator and denominator and factor it out. Then cancel the common factor. 𝟏𝟐 𝟏𝟖 = 𝟔∙𝟐 𝟔∙𝟑 = 𝟐 𝟏𝟓 𝟑 𝟐𝟏 = 𝟑∙𝟓 𝟓 = 𝟑∙𝟕 𝟕 Adding and Subtracting Fraction To add or subtract fractions, find the LCM of the two denominators. Then multiply each fraction so that the LCM is the denominator of both. 𝟑 𝟏𝟐 𝟑 ∙ 𝟑 𝟑 𝟏𝟐 𝟗 𝟑𝟔 + + + 𝟒 𝟏 𝟏𝟖 𝟒 𝟒 𝟏 − 𝟔 𝟐 𝟑 𝟏 𝟏 𝟐 𝟏𝟖 𝟐 𝟑 𝟒 𝟔 𝟐 ∙ ∙ − ∙ 𝟖 𝟑 𝟑𝟔 𝟏𝟐 − 𝟏𝟕 𝟏 𝟑𝟔 𝟏𝟐 𝟐 𝟏𝟐 Multiplying and Dividing Fractions To multiply fractions, multiply the numerators and denominators. Simplify if possible. 𝟐 𝟑 𝟔 𝟑 𝟒 𝟏𝟐 ∙ = = 𝟏 𝟒 𝟐 𝟖 𝟐 𝟕 𝟑 𝟐𝟏 ∙ = To divide fractions, invert (flip) the second fraction and multiply. 𝟑 𝟖 𝟐 𝟑 𝟓 𝟏𝟓 𝟓 𝟖 𝟐 𝟏𝟔 ÷ = ∙ = Middle School Math Huddle Fraction Review Practice Pages Instructions: 1) Prepare a solution page by writing “Fraction Review” and your name at the top of the page. 2) For each problem: a) Number a space on your solution page and neatly copy the problem. b) Solve the problem on your White Board, showing all of the steps. c) Neatly copy your solution onto the solution page. d) Leave plenty of space for each problem and use extra pages as needed. Find the Least Common Multiple (LCM) 1) 10 12 2) 12 24 4) 7 8 5) 10 25 3) 4 11 6) 8 9 Simplify (be sure to show GCF and cancellation) 7) 10) 𝟔 8) 𝟖 𝟏𝟒 𝟒𝟐 𝟏𝟓 𝟗 9) 𝟏𝟖 11) 𝟏𝟎 14) 𝟏 𝟐𝟕 𝟏𝟐 12) 𝟓𝟎 𝟖 Find the sum or difference: 13) 𝟏 16) 𝟒 𝟑 𝟗 + − 𝟐 𝟓 𝟏 𝟔 17) 𝟐 − 𝟓 𝟏𝟐 𝟏 𝟑 + 𝟓 𝟏𝟎 15) 𝟑 18) 𝟒 21) 𝟑 𝟏 𝟒 𝟗 + − Find the product or quotient: 19) 𝟒 𝟑 ∙ 𝟗 𝟓 20) 𝟓 𝟖 ÷ 𝟑 𝟒 ∙ 𝟕 𝟐 𝟏 𝟖 𝟏 𝟔 ```
# Holt CA Course 1 11-4 Solving Equations with Variables on Both Sides Preview of Algebra 1 5.0 Students solve multistep problems, including word problems, ## Presentation on theme: "Holt CA Course 1 11-4 Solving Equations with Variables on Both Sides Preview of Algebra 1 5.0 Students solve multistep problems, including word problems,"— Presentation transcript: Holt CA Course 1 11-4 Solving Equations with Variables on Both Sides Preview of Algebra 1 5.0 Students solve multistep problems, including word problems, involving linear equations and linear inequalities in one variable and provide justification for each step. California Standards Holt CA Course 1 11-4 Solving Equations with Variables on Both Sides Group the terms with variables on one side of the equal sign, and simplify. Additional Example 1: Using Inverse Operations to Group Terms with Variables A. 60 – 4y = 8y 60 – 4y + 4y = 8y + 4y 60 = 12y B. –5b + 72 = –2b –5b + 72 = –2b –5b + 5b + 72 = –2b + 5b 72 = 3b Add 4y to both sides. Simplify. 60 – 4y = 8y Add 5b to both sides. Simplify. Holt CA Course 1 11-4 Solving Equations with Variables on Both Sides Group the terms with variables on one side of the equal sign, and simplify. A. 40 – 2y = 6y 40 – 2y + 2y = 6y + 2y 40 = 8y B. –8b + 24 = –5b –8b + 24 = –5b –8b + 8b + 24 = –5b + 8b 24 = 3b Add 2y to both sides. Simplify. 40 – 2y = 6y Add 8b to both sides. Simplify. Check It Out! Example 1 Holt CA Course 1 11-4 Solving Equations with Variables on Both Sides Solve. Additional Example 2A: Solving Equations with Variables on Both Sides 7c = 2c + 55 5c = 55 5 5 c = 11 Subtract 2c from both sides. Simplify. Divide both sides by 5. Check 7c = 2c + 55 7(11) = 2(11) + 55 ? 77 = 77 – 2c Substitute 11 for c. 11 is the solution. Holt CA Course 1 11-4 Solving Equations with Variables on Both Sides Additional Example 2B: Solving Equations with Variables on Both Sides Solve. 49 – 3m = 4m + 14 49 = 7m + 14 35 = 7m 7 7 5 = m Add 3m to both sides. Simplify. Subtract 14 from both sides. Divide both sides by 7. + 3m – 14 Holt CA Course 1 11-4 Solving Equations with Variables on Both Sides Additional Example 2C: Solving Equations with Variables on Both Sides 2525 x = 1515 x– 12 2525 x =x = 1515 x 1515 x = 1515 x(5)(–12) = (5) x = –60 Subtract 1515 x from both sides. Simplify. Multiply both sides by 5. 1515 –x 1515 –x Holt CA Course 1 11-4 Solving Equations with Variables on Both Sides Solve. 8f = 3f + 65 5f = 65 5 5 f = 13 Subtract 3f from both sides. Simplify. Divide both sides by 5. Check It Out! Example 2A –3f Holt CA Course 1 11-4 Solving Equations with Variables on Both Sides Solve. 54 – 3q = 6q + 9 54 = 9q + 9 45 = 9q 9 9 5 = q Add 3q to both sides. Simplify. Subtract 9 from both sides. Divide both sides by 9. Check It Out! Example 2B + 3q – 9 Holt CA Course 1 11-4 Solving Equations with Variables on Both Sides 2323 w = 1313 w– 9– 9 2323 w =w = 1313 w– 9 1313 w = 1313 w(3)(–9) = (3) w = –27 Subtract 1313 w from both sides. Simplify. Multiply both sides by 3. Check It Out! Example 2C Solve. 13 13 –w –w 13 13 Holt CA Course 1 11-4 Solving Equations with Variables on Both Sides Christine can buy a new snowboard for \$136.50. She will still need to rent boots for \$8.50 a day. She can rent a snowboard and boots for \$18.25 a day. How many days would Christine need to rent both the snowboard and the boots to pay as much as she would if she buys the snowboard and rents only the boots for the season? Additional Example 3: Consumer Math Application Holt CA Course 1 11-4 Solving Equations with Variables on Both Sides Additional Example 3 Continued 18.25d = 136.5 + 8.5d 9.75d = 136.5 9.75 d = 14 Let d represent the number of days. Subtract 8.5d from both sides. Simplify. Divide both sides by 9.75. Christine would need to rent both the snowboard and the boots for 14 days to pay as much as she would have if she had bought the snowboard and rented only the boots. – 8.5d Holt CA Course 1 11-4 Solving Equations with Variables on Both Sides Check It Out! Example 3 A local telephone company charges \$40 per month for services plus a fee of \$0.10 a minute for long distance calls. Another company charges \$75.00 a month for unlimited service. How many minutes does it take for a person who subscribes to the first plan to pay as much as a person who subscribes to the unlimited plan? Download ppt "Holt CA Course 1 11-4 Solving Equations with Variables on Both Sides Preview of Algebra 1 5.0 Students solve multistep problems, including word problems," Similar presentations
# Properties of 0° Right triangle There are three fundamental properties of a right triangle when its angle is zero degrees. 1. The length of opposite side is zero. 2. The lengths of adjacent side and hypotenuse are equal. 3. The angle of right angled triangle is zero and the other two angles are right angles. ## Theorem Geometrically, it is not possible to construct a zero degrees right triangle understandably but it can be imagined theoretically. $\Delta POQ$ is a right triangle basically and the $\angle POQ$ represents the angle of this triangle. If the angle $POQ$ is equal to zero, then the triangle $POQ$ is called a zero degrees right angled triangle. In this triangle, $\angle POQ$, $\angle OPQ$ and $\angle OQP$ are three angles but the $\angle POQ \,=\, 0^°$ and $\angle OQP \,=\, 90^°$ According to the angle sum property of a triangle, the sum of three interior angles in a triangle is $180^°$. $\angle POQ + \angle OPQ + \angle OQP = 180^°$ $\implies 0^° + \angle OPQ + 90^° = 180^°$ $\implies \angle OPQ = 180^° – 90^°$ $\,\,\, \therefore \,\,\,\,\,\, \angle OPQ = 90^°$ Theoretically, it is proved that the two angles in a right angles are right angles when the angle of a right angled triangle is zero degrees. Assume that the length of adjacent side of this triangle is denoted by $d$. The zero angle is possible geometrically in a right triangle when the length of opposite side is equal to zero. $Length \, of \, Opposite \, side$ $\,=\,$ $0$ $\implies$ $PQ$ $\,=\,$ $0$ Due to zero length of opposite side, the length of hypotenuse is absolutely equal to the length of hypotenuse. So, remember this property when angle of right triangle is zero. $\,\,\, \therefore \,\,\,\,\,\,$ $Length \, of \, Adjacent \, side$ $\,=\,$ $Length \, of \, Hypotenuse$ $\,=\,$ $d$ ### Proof A right triangle is required to construct with zero angle for proving all its properties geometrically. 1. Draw a horizontal line from point $\small P$ in a plane. 2. Identify zero angle from point $\small P$ and draw zero angle line from point $\small P$ but it appears on horizontal line. 3. Set compass to a length (for example $7 \, cm$) by a ruler. Now, draw an arc on zero angle line and they are intersected at point $\small Q$. 4. It’s not possible to draw a perpendicular line from point $\small Q$ to horizontal line because they both lie on same line. So, assume a perpendicular line is drawn from point $\small Q$ to horizontal line and it intersects the horizontal line at point $\small R$. A right triangle ($\Delta RPQ$) is constructed with zero degrees angle. Now, it is time to study the properties of right angled triangle when its angle is zero. 1. $\small \overline{QR}$ is opposite side (or perpendicular). There is no distance between points $\small Q$ and $\small R$. So, the length of opposite side is zero. 2. $\small \overline{PQ}$ is hypotenuse. It is $7 \, cm$ in this case. 3. $\small \overline{PR}$ is adjacent side (or base). Due to zero length of opposite side, the length of adjacent side is equal to length of hypotenuse exactly. So, the length of adjacent side $\small \overline{PR}$ is also $7 \, cm$. Therefore, it is proved that the length of opposite side is zero and the length of adjacent side is equal to length of hypotenuse when angle of right triangle is zero. 1. There is no angle between sides $\small \overline{PR}$ and $\small \overline{PQ}$. So, the angle of right triangle is zero. It means $\small \angle RPQ = 0^°$. 2. $\small \overline{QR}$ is a perpendicular line to side $\small \overline{PR}$ at point $\small R$. So, $\small \angle PRQ = 90^°$, which means second angle of right triangle is a right angle. 3. The length of perpendicular line ($\small \overline{QR}$) (known as opposite side) is zero. The hypotenuse and adjacent side are appeared in horizontal line position. Therefore, $\small \angle PQR = 90^°$. In other words, the third angle of right triangle is also a right angle. ###### Math Questions The math problems with solutions to learn how to solve a problem. Learn solutions Practice now ###### Math Videos The math videos tutorials with visual graphics to learn every concept. Watch now ###### Subscribe us Get the latest math updates from the Math Doubts by subscribing us.
## Thinking Mathematically (6th Edition) Base five Numeral is ${{12344}_{\text{five}}}$ First write the given Roman numeral as Hindu Arabian Numeral CMLXXIV$\,=\left( 1000-100 \right)+\left( 50+10 \right)+10+\left( 5-1 \right)=974$ Now let’s find out the base five numeral using base ten Hindu Arabic numeral, The Place values in base five numeral are ${{5}^{1}},\,{{5}^{2}},{{5}^{3}},{{5}^{4}},\ldots$ or $5,25,125,625,\ldots$ Using highest divisor from these place values which is less than dividend for the first time and then continue in same manner, we get \begin{align} 974=625\overset{1}{\overline{\left){974}\right.}} & \\ \underline{625} & \\ 349 & \\ \end{align} Then, \begin{align} 125\overset{2}{\overline{\left){349}\right.}} & \\ \underline{250} & \\ 99 & \\ \end{align} Then, \begin{align} 25\overset{3}{\overline{\left){99}\right.}} & \\ \underline{75} & \\ 24 & \\ \end{align} Then, \begin{align} 245\overset{4}{\overline{\left){24}\right.}} & \\ \underline{20} & \\ 4 & \\ \end{align} So, ${{974}_{\text{ten}}}=\,{{12344}_{\text{five}}}$ Using the four quotients and the last remainder, we can directly write the Base Five Numeral as above.
# Linear Functions The Linear Function provides a convenient method to determine behavior of a sloping straight line on a Cartesian Coordinate System before drawing a graph using rectangular coordinates. It is function math for a Linear Equation. y = 0 = 2x + 3 then x = −3/2 The Linear Function defines sloped straight lines as: f(x) = ax + b and a ≠ 0, for all real numbers. a is slope. b is the y intercept, the value of y when x = 0. When a is: 1. > 0; the graph line is rising, a positive slope. 2. < 0; the graph line is falling, a negative slope. When b is: 1. > 0; the graph line is above a Cartesian coordinate origin, (0, 0), when intersecting the y axis. When x = zero then y > zero. 2. < 0; the graph line is below a Cartesian coordinate origin when intersecting the y axis. When x = zero then y < zero. 3. = 0; the graph line passes through a Cartesian coordinate origin. When b ≠ 0 there exists an x intercept when y = 0. The Linear Function does not define: 1. A horizontal line; When a = 0, then f(x) = b, a constant. f(x) = c is the Constant Function. 2. A vertical line; Slope a would be infinite. ## Linear Function Examples f(x) = ax1 + b g(x) = ax2 + b h(x) = ax3 + b Where x1x2x3 f(x), g(x) and h(x) are different points on the same line as unique (x, y) coordinates. Their a and b values of all linear functions are the same. Let a = 2 and b = 3; y intercept is value of b and graph coordinate (0, 3). Replace f(x), g(x) and h(x) math functions by y, then for all values of x; y = 0 = 2x + 3 2x = −3 x = −3/2 x intercept when y = 0 is: −1 ½ f(x) = ax + b1 g(x) = ax + b2 Where b1b2, Lines f(x) and g(x) are parallel separated by the absolute value of any perpendicular between b1 and b2. f(x) = a1x + b g(x) = a2x + b Where a1a2, Lines f(x) and g(x) intersect at exactly one point (0, b). It is the only value for x that provides the same y for both linear functions: f(x) = g(x) when x = 0 f(x) = a1x + b g(x) = a2x + b Where a1a2 and a2 < 0 Line f(x) has positive slope and line g(x) has negative slope. f(x) and g(x) intersect at exactly one point; (0, b). It is the only value for x that provides the same y for both functions: f(x) = g(x) when x = 0 ## Linear Function Identity When a = 1 and b = 0 then y = f(x) = ax Its graph is a straight line passing through a Cartesian coordinate origin, bisecting at 45 degree right angles the first and third quadrants. Top of Page
# A triangle has sides A,B, and C. If the angle between sides A and B is (7pi)/8, the angle between sides B and C is pi/12, and the length of B is 12, what is the area of the triangle? Apr 7, 2018 color(green)("Area of " Delta " " A_t = 54.62 " sq units" #### Explanation: $\hat{A} = \frac{\pi}{12} , \hat{C} = \frac{7 \pi}{8} , b = 12 , \text{ To find the area of } \Delta$ $\hat{B} = \pi - \hat{A} - \hat{C} = \pi - \frac{\pi}{12} - \frac{7 \pi}{8} = \frac{\pi}{24}$ Applying the Law of Sines, $\frac{a}{\sin} \left(\frac{\pi}{12}\right) = \frac{12}{\sin} \left(\frac{\pi}{24}\right) = \frac{c}{\sin} \left(\frac{7 \pi}{8}\right)$ $a = \frac{12 \cdot \sin \left(\frac{\pi}{12}\right)}{\sin} \left(\frac{\pi}{24}\right) = 23.79$ Knowing two sides a,b and the included angle C, to find the area we can use the formula color(crimson)(A_t = (1/2) a b sin C A_t = (1/2) * 23.79 * 12 * sin((7pi)/8) = color(purple)(54.62 " sq units"
# Quick Answer: What Is The Rule For Adding Two Negative Numbers? ## What is a negative times a positive? Rule 2: A negative number times a positive number equals a negative number. When you multiply a negative number to a positive number, your answer is a negative number. It doesn’t matter which order the positive and negative numbers are in that you are multiplying, the answer is always a negative number.. ## Why do you add when you subtract negative numbers? The addition/subtraction tells us which way to face, and the positive/negative tells us if our steps will be forward or backward (regardless of the way we’re facing). ## What is the rule for adding positive and negative numbers? The Rules:RuleExample+(+)Two like signs become a positive sign3+(+2) = 3 + 2 = 5−(−)6−(−3) = 6 + 3 = 9+(−)Two unlike signs become a negative sign7+(−2) = 7 − 2 = 5−(+)8−(+2) = 8 − 2 = 62 more rows ## Do you add or subtract two negative numbers? Rule 3: Subtracting a negative number from a negative number – a minus sign followed by a negative sign, turns the two signs into a plus sign. So, instead of subtracting a negative, you are adding a positive. Basically, – (-4) becomes +4, and then you add the numbers. ## When the signs are the same you add? Same Sign – Add the absolute values and give the answer the same sign. Subtracting Integers – Same as adding the opposite of that integer number. ## What is the rule for subtracting two negative numbers? Subtracting a number is the same as adding the opposite of the number. Note: In both examples, we change subtraction to addition and changed the sign of the second number to the opposite sign. ## When two positive integers are added what we get? When two positive integers are added the result will be a positive integer. When two negative integers are added the result will be a negative integer. ## What are the four rules of integers? Integers are whole numbers, both positive and negative. You can perform four basic math operations on them: addition, subtraction, multiplication, and division. When you add integers, remember that positive integers move you to the right on the number line and negative integers move you to the left on the number line. ## Why is a double negative a positive in math? When you multiply a negative by a negative you get a positive, because the two negative signs are cancelled out. ## What is the rule for adding two numbers with different signs? To add integers with different signs, keep the sign of the number with the largest absolute value and subtract the smallest absolute value from the largest. Subtract an integer by adding its opposite. ## When 2 negative integers are added we get? Summary: Adding two positive integers always yields a positive sum; adding two negative integers always yields a negative sum. To find the sum of a positive and a negative integer, take the absolute value of each integer and then subtract these values. ## What is a positive minus a negative? So, subtracting a positive number is like adding a negative; you move to the left on the number line. Subtracting a negative number is like adding a positive; you move to the right on the number line. ## What is the rule for adding and subtracting integers? When a positive and a negative integer are added, the numbers are subtracted without signs and the answer is assigned the sign of the larger integer. For example, to add 10 + (-15) = -5, the larger number in this case is 15 without the sign. ## Why do two minuses make a plus? Subtracting a negative is the same as… Again, before too long, the consensus is that subtracting a negative is the same as adding a positive. … Either way it reinforces the big idea that subtracting a negative is the same as adding a positive. This is so much better than saying ‘two minuses make a plus’.
# SAT Practice Test 3, Section 8: Questions 11 - 16 This is for SAT in Jan 2016 or before. The following are worked solutions for the questions in the math sections of the SAT Practice Tests found in the The Official SAT Study Guide Second Edition. It would be best that you go through the SAT practice test questions in the Study Guide first and then look at the worked solutions for the questions that you might need assistance in. Due to copyright issues, we are not able to reproduce the questions, but we hope that the worked solutions will be helpful. Given: To find: The possible values of n Solution: Topic(s): Symbol problems We then test the given statements for the values of n: Given: 20% of x = 80% of y To find: y in terms of x Solution: Topic(s): Percent Translate ‘of” as ‘×’ Convert between percent and decimal. Given: x + y is even (x + y) 2 + x + z is odd x, y and z are positive integers To find: The statement that is true Solution: Topic(s): Odd and even numbers You were given that x + y is even. Applying the rule: Even × Even = Even (x + y)2 = (x + y) × (x + y) = Even You were also given that (x + y)2 + x + z is odd. Applying the rule: Even + Odd = Odd Since (x + y)2 is even, so, x + z is odd Applying the same rule again: Even + Odd = Odd If z is even, then x must be odd. Answer: (C) If z is even, then x must be odd. Given: 0 < x < 1 To find: The statements that must be true Solution: Topic(s): Fractions Given 0 < x < 1, we can say the x is a positive fraction with a value less than 1. If a positive fraction with a value less than 1 is raised to a positive integer exponent then the larger the exponent the smaller the result. This means that conditions I. x 2 > x 3 and III. x > x 3 are true. Half of a positive integer is less than the original integer. This means that condition II. is true. Answer: (E) I, II and III Given: A scatter plot To find: The function that models the relationship between t, the number of seconds to complete the maze and p, the number of practices. Solution: If we were to draw a line to reflect the approximate values of the data on the graph, we would get a horizontal line that intersects the y-axis between 40 and 50. A horizontal line would have an equation of the form t(p) = c, where c is a constant. Out of all the answers given only (A) t(p) = 40, satisfies that condition. Given: The pattern consisting of 5 L by W rectangles To find: The number of L by W rectangles needed to cover a region 12L by 10L Solution: Since the region to be covered is given is in terms of L, we would want to get the width and length of the above pattern to be in terms of L as well. The above pattern forms a rectangle. From the two vertical sides, we know that: The width of the rectangle is then The above pattern is a rectangle with dimensions: In order to cover the region without any leftovers, we need to place the width of the rectangle along the side that is 10L. The number of rectangles that we can place along the side 10L is: The length of the rectangle is 2L. The number of rectangles that can be placed along the side 12L is 12L ÷ 2L = 6. So, we would need 6 × 6 = 36 of rectangles with the above pattern. Be careful! The question wants the number of rectangles of dimensions L by W. The above pattern has 5 rectangles of dimensions L by W. So, the answer is 36 × 5 = 180 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. [?] Subscribe To This Site
# Video: Finding a Volume of Revolution About the π₯-Axis Using the Washer Method The region π in the figure shown is bounded by π¦ = βsec (π₯) and π¦ = β3. What is the volume of the solid formed when π is rotated about the π₯-axis? 06:18 ### Video Transcript The region π in the figure shown is bounded by π¦ is equal to negative sec π₯ and π¦ is negative three. What is the volume of the solid formed when π is rotated about the π₯-axis? Weβre asked to find the volume of the solid formed when we rotate our region π about the π₯-axis. In order to do this, we need to define the boundaries of our region, sketch the solid so we have a graphical representation of what weβre looking for, sketch and determine the cross-sectional area. And finally, for the volume, we integrate the cross-sectional area between the boundaries. To find the boundaries, letβs look at our region π. Our region is bounded by the function π¦ is equal to negative sec π₯ and the line π¦ is negative three. Remember that sec π₯ is the secant of π₯, which is one of the cos of π₯. Where our function negative sec π₯ crosses the π¦-axis, π₯ is equal to zero. And when π₯ is zero, negative sec of zero is negative one over cos of zero which is negative one. So our function negative sec π₯ meets the π¦-axis at negative one. We need also to find the π₯-values where our function negative sec π₯ intersects the line π¦ is at negative three. So we need to solve the equation negative sec π₯ is negative three. The negatives cancel each other out so that sec π₯ is equal to three. This means that the inverse sec of three is equal to π₯. We can either solve this on our calculator or note that one of our cos π₯ is equal to three, which means that cos π₯ is equal to one over three so that the inverse cos of one over three is equal to π₯. And taking the inverse cos of one over three on our calculator, we get π₯ is equal to 1.23096 to five decimal places. Since our function is symmetric about the line π₯ is equal to zero, which is the π¦-axis, then our boundary to the left of π₯ is equal to zero is negative 1.23096. When we come to find our volume via integration, these will be our limit of integration. Our next step is to try and sketch the solid formed when this region is rotated about the π₯-axis. Our initial region is in the π₯π¦-plane, bounded by the function π¦ is equal to negative sec π₯ and the line π¦ is negative three. And we want to rotate this area about the π₯-axis. If we continue in this way, we end up with a solid that resembles a tyre. Our next step is to look at our cross sections. If we slice vertically through the solid, the cross section is in this shape of a washer. We use the washer method to calculate the volume of the solid when the axis of rotation is not part of the boundary of our initial region. In our case, the axis of rotation is the π₯-axis which is not part of our region π. Our cross section has an inner radius π in which is the radius of the hollow centre circle. We also have an outer radius, which is the radius from the axis of rotation to the outer circumference of the solid. To calculate the volume, we move through the solid summing the area of each of the infinitesimally thin cross sections via integration. The area of each cross section is π times the outer radius squared minus the inner radius squared. If we look again at our region π, the outer radius is the distance from the π₯-axis to π¦ is negative three, which is the outside edge of the solid. And thatβs equal to three. Our inner radius, on the other hand, depends on π₯ because this is the distance from the π₯-axis to the function π¦ is negative sec π₯. So our inner radius is sec π₯. For example, if we take π₯ equal to zero, the inner radius is one because itβs the distance from zero to negative one. Thatβs the absolute value of negative sec π₯ at π₯ is equal to zero. Remember that radius is a distance and distance is always positive. So we can now work out the area of our cross section. Our area, which is a function of π₯, is π times the outer radius squared, which is three squared, minus the inner radius squared, which is sec squared π₯, which is π times nine minus sec squared π₯. And finally, to find the volume of our solid, we integrate the area between the boundaries of our region. Since π is a constant, we can take this outside. So the volume of our solid is π times the integral between negative 1.23096 and positive 1.23096 of nine minus sec squared π₯ with respect to π₯. Before we integrate, itβs worth noting that if the axis of rotation was vertical, then our area would be a function of π¦. The limit of integration would then be π¦-values. In our case, however, the axis of rotation is the π₯-axis. So our limits are π₯-values. Since our region is symmetric about π₯ is equal to zero, we can actually take our lower limit as π₯ is equal to zero and multiply the integral by two. This gives us a volume of two π times the integral between zero and 1.23096 of nine minus sec squared π₯ with respect to π₯. We can do this integral using a calculator or by hand. If we perform the integral by hand, since integration is additive, our volume is then two π times the integral between zero and 1.23096 of nine with respect to π₯ minus two π times the integral between zero and 1.23096 of sec squared π₯ with respect to π₯. Our first integral gives us two π multiplied by nine π₯ evaluated between zero and 1.23096. And for our second integral, we can use the standard integral. The integral of sec squared π₯ with respect to π₯ is tan π₯ plus constant so that our second integral is two π times tan π₯ evaluated between zero and 1.23096. This gives us two π times nine times 1.23096 minus two π times the tan of 1.23096 minus the tan of zero. Nine multiplied by 1.23096 is equal to 11.078635. And the tan of 1.23096 radiance is 2.828427. The tan of zero is equal to zero. Evaluating this gives us a volume of 51.837 to three decimal places. And remember that volume is measured in units cubed.
# Question 3e542 Aug 30, 2016 The Curves intersect at two pts., $O \left(0 , 0\right) , \mathmr{and} , A \left(1 , 5\right)$ At $O$, the $X$-axis is their common tgt, &, hence, the $\angle$ btwn them is ${0}^{\circ}$. At $A$, the $\angle$ btwn them is ${1.8965}^{\circ}$. #### Explanation: Let the curves be ${C}_{1} : y = 5 {x}^{3} , \mathmr{and} , {C}_{2} : y = 5 {x}^{2}$. To find their pt. of intersection, we have to solve their eqns. $5 {x}^{3} = y = 5 {x}^{2} \Rightarrow 5 {x}^{2} \left(x - 1\right) = 0 \Rightarrow x = 0 , x = 1$. $y = 5 {x}^{2} , x = 0 , x = 1 \Rightarrow y = 0 , y = 5$. Thus, we have, ${C}_{1} \cap {C}_{2} = \left\{O \left(0 , 0\right) , A \left(1 , 5\right)\right\}$. To find the slopes of tgts. to the curves ${C}_{1} , \mathmr{and} , {C}_{2}$, at the pts. $O , \mathmr{and} , A$, we will find $\frac{\mathrm{dy}}{\mathrm{dx}}$ as it denotes the reqd. slopes. For C_1, dy/dx=15x^2, &, "for" C_2, dy/dx=10x $\therefore f \mathmr{and} {C}_{1} , \mathmr{and} , {C}_{2} , {\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]}_{O} = 0.$ This means that, both tgts. to C_1,&,C_2# have the same slope, namely $0$, and, both pass thro. the same pt. $O \left(0 , 0\right)$. Clearly, (1) the $X$-axis is the common tgt. to ${C}_{1} \mathmr{and} {C}_{2}$ touching at the Origin. (2) The angle btwn. them, at $O$ is ${0}^{\circ}$. $\text{At the pt." A(1,5), dy/dx=15, "for" C_1, and, "for} {C}_{2} , \frac{\mathrm{dy}}{\mathrm{dx}} = 10$ :. ${m}_{1} = 15 , \mathmr{and} {m}_{2} = 10 ,$ are the respective slopes of tgts. $\therefore \alpha = t h e \angle \left({C}_{1} , {C}_{2}\right) \Rightarrow \tan \alpha = \| \frac{{m}_{1} - {m}_{2}}{1 + {m}_{1} {m}_{2}} \|$ $= \| \frac{15 - 10}{1 + 150} \| = \frac{5}{151} \approx 0.0331$ $\therefore \alpha = a r c \tan 0.0331 = {1.8965}^{\circ}$. Enjoy Maths.!
# Boundary\|Definition & Meaning ## Definition A boundary is a line that outlines an object’s shape or a polygon. The total length of the boundary is called the perimeter. The boundary is different for all shapes. A boundary allows us to identify the shape of an object or image and is necessary for separating one shape from another. ## Visualizing a Boundary Figure 1 below shows a rectangle ABCD. The lines between the points ABCD is known as the boundary. With the help of the boundary, we can easily distinguish the shape. Figure 1 – Boundary of a rectangle. Figure 2 below shows the points EFG; as in the previous figure, these points are connected with a boundary that gives it its shape. The figure is called a triangle. Figure 2 – Boundary of a triangle. ## What Is a Perimeter? The perimeter of any two-dimensional geometric shape is the length of its outline or boundary. Depending on the dimensions, the perimeters of different figures can be equal in size. Consider a triangle made of a wire that is 12 cm long. If all the sides are the same length, the same wire can be used to create a square. Figure 3 – Boundary of shapes. Now that we know the perimeter of a geometric shape refers to its outer boundary, how would one determine its value? Let us look at an example. To prevent his cattle from straying off the farm, Jake wants to erect a fence around it. He is curious about the amount of wire required to fence his farm. His farm has a rectangular shape, which means: • The farm has four sides. • The length of the opposing sides is equal. • Each angle is 90 degrees. Now to find out the perimeter of his farm, Jack only needs to measure all four sides and add them together to get the perimeter of his farm. ### Units of a Perimeter The length that the shape’s perimeter covers are known as the perimeter. Therefore, the perimeter’s units will be the same as its length units. Perimeter is one-dimensional, as we can say. As a result, it can be measured in meters, kilometers, centimeters, etc. Inches, feet, yards, and miles are some additional perimeter measurement units that are recognized on a global scale. ### The Formula of a Perimeter The formula for calculating a perimeter changes according to our provided shape. The general formula for calculating a perimeter is as follows: Perimeter = Sum of all borders However, different formulas exist for various shapes. Below we can look at some of the commonly used formulas to calculate a perimeter: • Square: The perimeter of the square = 4 x L, where L is the length of one boundary • Rectangle: The perimeter of the rectangle = 2 x (L + B), where L and B are the Length and Breath of the rectangle. • Triangle: The perimeter of a triangle = l + m + n, where l, m, and n are the lengths of a triangle. • Circle: The perimeter of a Circle = $2 \pi r$, where $\pi$ is a constant and r is the circle’s radius. ## Area of a Shape The area of a 2-dimensional figure is the amount of space it takes up. In other words, it refers to the number of unit squares covering the surface of a shape enclosed by a boundary. The length and width of a shape are used to calculate its area. The units used to measure length are unidimensional and include meters (m), centimeters (cm), inches (in), etc. However, a shape’s area can only be measured in two dimensions. It is measured in square units, for example, meters or (m2), square centimeters or (cm2), square inches or (in2), and so on. The majority of the objects or shapes have edges and corners. These edges’ length and width are considered when calculating the area of a specific shape. ### Calculating the Area of a Rectangle The area is the space inside the boundary of a shape. The following figure shows a rectangle with a length of 5cm and a width of 3 cm. This formula can calculate the area of the rectangle: Area of Rectangle = l x w, where l is the length of the rectangle and w is the width. Figure 4 – Perimeter of a rectangle. Using the formula given, we can find the area of the rectangle. The area calculated is 3 cm x 5 cm = 15 cm. ### Calculating the Area of a Circle The shape of a circle is curved. The amount of space within a circle’s boundary is known as its area. The formula: $\pi$r2, where r is the circle’s radius, and $\pi$ is a mathematical constant with a value roughly 3.14 or 22/7, is used to determine the area of a circle. Looking at the figure below, we can see that the circle’s radius is 10cm. Figure 6 – Radius of a circle. We can use the formula above to find the area of the circle. The area is calculated as shown below: Area of Circle = $\pi$r2 Area of Circle = $\pi$ x (10 cm)2 Area of Circle = 100 x $\pi$ cm2 $\approx$ 314.16 cm2 ## Example of Boundary The following example on boundaries will help you understand the concept more easily. In the following figures, distinguish the rectangular shape and find its perimeter. Figure 6 – Representation of boundaries. ### Solution We must find the rectangle from the given shapes in the first step. After closer inspection of the shapes, we can conclude that the shape ABCD is a rectangle. Now to calculate the perimeter of the rectangle, we use the following formula: Perimeter of rectangle = 2(l x w) Now we know the length of the rectangle is 6cm and the width is 2cm, we put these values in the formula. Perimeter of rectangle = 2(6 cm + 2 cm) After solving this formula, we get 18 cm2. All images/tables are created using GeoGebra.
# Thread: Arithmatic Progression Word Problem 1. ## Arithmatic Progression Word Problem 1) Find three numbers in A.P such that their sum is 21 and sum of their product is 280. 2) Also, what the general idea for assuming "any set of numbers" required by the condition? For eg. if the problem demands four numbers, the assumed ones are (a-3d),(a-d),(a+d),(a+3d). Where "d" is the common difference. How do you frame them? 2. Hello saberteeth Originally Posted by saberteeth 1) Find three numbers in A.P such that their sum is 21 and sum of their product is 280. Suppose that the first number is $\displaystyle a$, and that the common difference is $\displaystyle d$. Then the next two numbers are $\displaystyle (a+d)$ and $\displaystyle (a+2d)$. So we have: Sum: $\displaystyle a+(a+d)+(a+2d) =21$ $\displaystyle \Rightarrow 3a+3d=21$ $\displaystyle \Rightarrow a+d=7$ $\displaystyle \Rightarrow d = 7-a$ (1) Product: $\displaystyle a(a+d)(a+2d)=280$ So, substituting for $\displaystyle d$: $\displaystyle 7a(a+14-2a)=280$ $\displaystyle \Rightarrow a(14-a)=40$ $\displaystyle \Rightarrow 14a-a^2=40$ $\displaystyle \Rightarrow a^2-14a+40=0$ $\displaystyle \Rightarrow (a-4)(a-10)=0$ $\displaystyle \Rightarrow a=4,10$ Plugging these into (1) gives: $\displaystyle d =3,-3$ Each of these results gives the same set of numbers: $\displaystyle 4, 7, 10$. 2) Also, what the general idea for assuming "any set of numbers" required by the condition? For eg. if the problem demands four numbers, the assumed ones are (a-3d),(a-d),(a+d),(a+3d). Where "d" is the common difference. How do you frame them? You could use the ones you suggest here, since you have written 4 numbers with a common difference (in this case $\displaystyle 2d$). Or, more staightforwardly, you could do as I have in question (1), and assume that the first is $\displaystyle a$, common difference $\displaystyle d$, and write them as $\displaystyle a, (a+d), (a+2d), (a+3d)$. Either way, you'll have two unknowns, and you'll therefore need to be able set up two equations in order to solve a particular problem.
# Class 12 Maths MCQ – Evaluation of Definite Integrals by Substitution This set of Class 12 Maths Chapter 7 Multiple Choice Questions & Answers (MCQs) focuses on “Evaluation of Definite Integrals by Substitution”. 1. Evaluate the integral $$\int_0^{\frac{π^2}{4}} \frac{9 sin⁡\sqrt{x}}{2\sqrt{x}} dx$$. a) 9 b) -9 c) $$\frac{9}{2}$$ d) –$$\frac{9}{2}$$ Explanation: I=$$\int_0^{\frac{π^2}{4}} \frac{9 sin⁡\sqrt{x}}{2\sqrt{x}} dx$$ Let $$\sqrt{x}$$=t Differentiating both sides w.r.t x, we get $$\frac{1}{2\sqrt{x}} dx=dt$$ The new limits are When x=0 , t=0 When x=$$\frac{π^2}{4}, t=\frac{π}{2}$$ ∴$$\int_0^{\frac{π^2}{4}} \frac{9 sin⁡\sqrt{x}}{2\sqrt{x}} dx=9\int_0^{π/2} sin⁡t \,dt$$ =$$9[-cos⁡t]_0^{π/2}$$=-9(cos⁡ π/2-cos⁡0)=-9(0-1)=9 2. Find $$\int_0^1 20x^3 e^{x^4}$$ dx. a) (e-1) b) 5(e+1) c) 5e d) 5(e-1) Explanation: I=$$\int_0^1 20x^3 e^{x^4}$$ dx Let x4=t Differentiating w.r.t x, we get 4x3 dx=dt ∴The new limits When x=0, t=0 When x=1,t=1 ∴$$\int_0^1 \,20x^3 \,e^{x^4} \,dx=\int_0^1 5e^t dt$$ $$=5[e^t]_0^1=5(e^1-e^0)$$=5(e-1). 3. Find $$\int_{-1}^1 \frac{5x^4}{\sqrt{x^5+3}} dx$$. a) 4-$$\sqrt{2}$$ b) 4+2$$\sqrt{2}$$ c) 4-2$$\sqrt{2}$$ d) 1-2$$\sqrt{2}$$ Explanation: I=$$\int_{-1}^1 \frac{5x^4}{\sqrt{x^5+3}} dx$$ Let x5+3=t Differentiating w.r.t x, we get 5x4 dx=dt The new limits when x=-1,t=2 when x=1,t=4 ∴$$\int_{-1}^1 \frac{5x^4}{\sqrt{x^5+3}} dx=\int_2^4 \frac{dt}{\sqrt{t}}$$ =$$[2\sqrt{t}]_2^4=2(\sqrt{4}-\sqrt{2})=4-2\sqrt{2}$$ 4. Find $$\int_0^{\frac{\sqrt{π}}{2}} 2x \,cos⁡ x^2 \,dx$$. a) 1 b) $$\frac{1}{\sqrt{2}}$$ c) –$$\frac{1}{\sqrt{2}}$$ d) $$\sqrt{2}$$ Explanation: I=$$\int_0^{\frac{\sqrt{π}}{2}} \,2x \,cos⁡ x^2 \,dx$$ Let x2=t Differentiating w.r.t x, we get 2x dx=dt The new limits When x=0,t=0 When $$x={\frac{\sqrt{π}}{2}}, t=\frac{π}{4}$$ ∴$$\int_0^{\frac{\sqrt{π}}{2}} \,2x \,cos⁡ x^2 \,dx=\int_0^{\frac{π}{4}} \,cos⁡t \,dt$$ $$I =[sin⁡t]_0^{\frac{π}{4}}=sin⁡ \frac{π}{4}-sin⁡0=1/\sqrt{2}$$. 5. Evaluate the integral $$\int_1^6 \frac{\sqrt{x}+3}{\sqrt{x}} \,dx$$. a) 9 b) $$\frac{9}{2}$$ c) –$$\frac{9}{2}$$ d) $$\frac{4}{5}$$ Explanation: I=$$\int_1^4 \frac{\sqrt{x}+3}{\sqrt{x}} \,dx$$ Let $$\sqrt{x}+3=t$$ Differentiating w.r.t x, we get $$\frac{1}{2\sqrt{x}} \,dx=dt$$ $$\frac{1}{\sqrt{x}} \,dx=2 \,dt$$ The new limits When x=1,t=4 When x=4,t=5 ∴$$\int_1^4 \frac{\sqrt{x}+3}{\sqrt{x}} dx=\int_4^5 \,t \,dt$$ =$$[\frac{t^2}{2}]_4^5=\frac{5^2-4^2}{2}=\frac{9}{2}$$ Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now! 6. Find $$\int_1^2 \frac{12 \,log⁡x}{x} \,dx$$. a) -12 log⁡2 b) 24 log⁡2 c) 12 log⁡2 d) 24 log⁡4 Explanation: I=$$\int_1^2 \frac{12 log⁡x}{x} \,dx$$ Let log⁡x=t Differentiating w.r.t x, we get $$\frac{1}{x} \,dx=dt$$ The new limits When x=1,t=0 When x=2,t=log⁡2 $$\int_1^2 \frac{12 log⁡x}{x} dx=12\int_0^{log⁡2} \,t \,dt$$ =$$12[t^2]_0^{log⁡2}=12((log⁡2)^2-0)$$ =12 log⁡4=24 log⁡2(∵(log⁡2)2=log⁡2.log⁡2=log⁡4=2 log⁡2) 7. Find $$\int_0^{π/4} \frac{5 \,sin⁡(tan^{-1}⁡x)}{1+x^2} \,dx$$. a) 5-$$\frac{1}{\sqrt{2}}$$ b) 5+$$\frac{5}{\sqrt{2}}$$ c) -5+$$\frac{5}{\sqrt{2}}$$ d) 5-$$\frac{5}{\sqrt{2}}$$ Explanation: I=$$\int_0^1 \frac{5 \,sin⁡(tan^{-1)}x}{1+x^2} \,dx$$ Let tan-1⁡x=t Differentiating w.r.t x, we get $$\frac{1}{1+x^2} \,dx=dt$$ The new limits When x=0, t=tan-1⁡0=0 When x=1, t=tan-1)1=π/4 ∴$$\int_0^1 \frac{5 \,sin⁡(tan^{-1}⁡x)}{1+x^2} \,dx=\int_0^{π/4} \,5 \,sin⁡t \,dt$$ =$$5[-cos⁡t]_0^{π/4}=-5[cos⁡t]_0^{π/4}$$ $$=-5(cos⁡ \frac{π}{4}-cos⁡0)=-5(\frac{1}{\sqrt{2}-1})=5-\frac{5}{\sqrt{2}}$$ 8. Find $$\int_{-1}^1 \,7x^6 \,(x^7+8)dx$$ a) -386 b) –$$\frac{386}{3}$$ c) $$\frac{386}{3}$$ d) 386 Explanation: I=$$\int_{-1}^1 \,7x^6 \,(x^7+8)dx$$ Let x7+8=t Differentiating w.r.t x, we get 7x6 dx=dt The new limits When x=-1,t=7 When x=1,t=9 ∴$$\int_{-1}^1 \,7x^6 \,(x^7+8)dx=\int_7^9 \,t^2 \,dt$$ =$$[\frac{t^3}{3}]_7^9=\frac{1}{3} (9^3-7^3)=\frac{386}{3}$$. 9. Evaluate $$\int_{\sqrt{2}}^2 \,14x \,log⁡ x^2 \,dx$$ a) 14(3 log⁡2-1) b) 14(3 log⁡2+1) c) log⁡2-1 d) 3 log⁡2-1 Explanation: I=$$\int_{\sqrt{2}}^2 \,14x \,log⁡ x^2 \,dx$$ Let x2=t Differentiating w.r.t x, we get 2x dx=dt The new limits When x=$$\sqrt{2}$$, t=2 When x=2, t=4 ∴$$\int_{\sqrt{2}}^2 \,14x \,log⁡ x^2 \,dx =\int_2^4 \,7 \,log⁡ t \,dt$$ Using integration by parts, we get $$\int_2^4 \,7 \,log⁡ t \,dt=7(log⁡ t\int dt-\int (log⁡t)’ \int \,dt)$$ =7 (t log⁡t-t)24 =7(4 log⁡4-4-2 log⁡2+2) =7(6 log⁡2-2)=14(3 log⁡2-1) 10. Find $$\int_2^3 \,2x^2 \,e^{x^3} \,dx$$. a) $$e^{27}-e^8$$ b) $$\frac{2}{3} (e^{27}-e^8)$$ c) $$\frac{2}{3} (e^8-e^{27})$$ d) $$\frac{2}{3} (e^{27}+e^8)$$ Explanation: I=$$\int_2^3 \,2x^2 \,e^{x^3} \,dx$$ Let x3=t Differentiating w.r.t x, we get 3x2 dx=dt x2 dx=$$\frac{dt}{3}$$ The new limits When x=2, t=8 When x=3, t=27 ∴$$\int_2^3 \,2x^2 \,e^{x^3} \,dx=\frac{2}{3} \int_8^{27} \,e^t \,dt$$ =$$\frac{2}{3} [e^t]_8^{27}=\frac{2}{3} (e^{27}-e^8).$$ Sanfoundry Global Education & Learning Series – Mathematics – Class 12. To practice all chapters and topics of class 12 Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers. If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]
# multiplying 54 results multiplying - to increase in number especially greatly or in multiples 1 multiplied by b squared multiplied by c squared 1 multiplied by b squared multiplied by c squared b squared means we raise b to the power of 2: b^2 c squared means we raise c to the power of 2: c^2 b squared multiplied by c squared b^2c^2 1 multiplied by b squared multiplied by c squared means we multiply 1 by b^2c^2 1b^2c^2 Multiplying by 1 can be written by [U][I]removing[/I][/U] the 1 since it's an identity multiplication: [B]b^2c^2[/B] 11 to the power of 6 multiply 11 to the power of 3 11 to the power of 6 multiply 11 to the power of 3 Take this in parts. [U]Step 1: 11 to the power of 6 means we raise 11 to the 6th power using exponents:[/U] 11^6 [U]Step 2: 11 to the power of 3 means we raise 11 to the 3rd power using exponents:[/U] 11^3 [U]Step 3: Multiply each term together:[/U] 11^6 * 11^3 [U]Step 4: Simplify[/U] Because we have 2 numbers that are the same, in this case, 11, we can add the exponents together when multiplying: 11^(6 + 3) [B]11^9 [MEDIA=youtube]gCxVq7LqyHk[/MEDIA][/B] 3 salads, 4 main dishes, and 2 desserts 3 salads, 4 main dishes, and 2 desserts Total meal combinations are found by multiplying each salad, main dish, and dessert using the fundamental rule of counting. The fundamental rule of counting states, if there are a ways of doing one thing, b ways of doing another thing, and c ways of doing another thing, than the total combinations of all the ways are found by a * b * c. With 3 salads, 4 main dishes, and 2 desserts, our total meal combinations are: 3 * 4 * 2 = [B]24 different meal combinations.[/B] 4 divided by sin60 degrees 4 divided by sin60 degrees. We can write as 4/sin(60). [URL='https://www.mathcelebrity.com/anglebasic.php?entry=60&coff=&pl=sin']Using our trigonometry calculator[/URL], we see sin(60) = sqrt(3)/2. So we have 4/sqrt(3)/2. Multiplying by the reciprocal we have: 42/sqrt(3) [B]8/sqrt(3)[/B] a = v^2/r for r a = v^2/r for r Start by cross multiplying to get r out of the denominator: ar = v^2 Divide each side of the equation by a to isolate r: ar/a = v^2/a Cancel the a's on the left side, and we get: r = [B]v^2/a[/B] A boy is 10 years older than his brother. In 4 years he will be twice as old as his brother. Find th A boy is 10 years older than his brother. In 4 years he will be twice as old as his brother. Find the present age of each? Let the boy's age be b and his brother's age be c. We're given two equations: [LIST=1] []b = c + 10 []b + 4 = 2(c + 4) [/LIST] Substitute equation (1) into equation (2): (c + 10) + 4 = 2(c + 4) Simplify by multiplying the right side through and grouping like terms: c + 14 = 2c + 8 [URL='https://www.mathcelebrity.com/1unk.php?num=c%2B14%3D2c%2B8&pl=Solve']Type this equation into our search engine[/URL] and we get: c = [B]6[/B] Now plug c = 6 into equation (1): b = 6 + 10 b = [B]16[/B] A committee of 6 students are being selected from a class of 10 girls and 8 boys. How many committee A committee of 6 students are being selected from a class of 10 girls and 8 boys. How many committees are possible if three must be girls and 3 must be boys? We want combinations. How many ways can we choose 3 boys from 8 boys: [URL='https://www.mathcelebrity.com/permutation.php?num=8&den=3&pl=Combinations']8 choose 3[/URL] = 56 We want combinations. How many ways can we choose 3 girls from 10 girls: [URL='https://www.mathcelebrity.com/permutation.php?num=10&den=3&pl=Combinations']10 choose 3[/URL] = 120 Our total choices are found by multiplying each event: Total committees = (8 boys choose 3) (10 girls choose 3) Total committees = 56 * 120 Total committees = [B]6,720[/B] A culture of bacteria doubles every hour. If there are 500 bacteria at the beginning, how many bacte A culture of bacteria doubles every hour. If there are 500 bacteria at the beginning, how many bacteria will there be after 9 hours? Assumptions and givens; [LIST] []h is the number of hours. []B(h) is the number of bacteria at time h []B(0) is the starting bacteria amount []Doubling means multiplying by 2, so we have: [/LIST] B(h) = B(0) * 2^h We want h = 9, so we have: B(9) = 500 * 2^9 B(9) = 500 * 512 B(9) = [B]256,000[/B] A person has \$13,000 invested in stock A and stock B. Stock A currently sells for \$20 a share and A person has \$13,000 invested in stock A and stock B. Stock A currently sells for \$20 a share and stock B sells for \$90 a share. If stock B triples in value and stock A goes up 50%, his stock will be worth \$33,000. How many shares of each stock does he own? Set up the given equations, where A is the number of shares for Stock A, and B is the number of shares for Stock B [LIST=1] []90A + 20B = 13000 []3(90A) + 1.5(20B) = 33000 <-- [I]Triple means multiply by 3, and 50% gain means multiply by 1.5[/I] [/LIST] Rewrite (2) by multiplying through: 270A + 30B = 33000 Using our simultaneous equations calculator, we get [B]A = 100 and B = 200[/B]. Click the links below to solve using each method: [LIST] [][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=90A+%2B+20B+%3D+13000&term2=270A+%2B+30B+%3D+33000&pl=Substitution']Substitution Method[/URL] [][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=90A+%2B+20B+%3D+13000&term2=270A+%2B+30B+%3D+33000&pl=Elimination']Elimination Method[/URL] [][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=90A+%2B+20B+%3D+13000&term2=270A+%2B+30B+%3D+33000&pl=Cramers+Method']Cramers Method[/URL] [/LIST] Check our work using equation (1) 90(100) + 20(200) ? 13,000 9000 + 4000 ? 13,000 13000 = 13000 A rectangular room is 3 times as long as it is wide, and its perimeter is 56 meters. A rectangular room is 3 times as long as it is wide, and its perimeter is 56 meters. We're given the following: [LIST] []l = 3w [/LIST] We know the Perimeter (P) of a rectangle is: P = 2l + 2w Substituting l = 3w and P = 56 into this equation, we get: 2(3w) + 2w = 56 Multiplying through, we get: 6w + 2w = 56 (6 +2)w = 56 8w = 56 [URL='https://www.mathcelebrity.com/1unk.php?num=8w%3D56&pl=Solve']Typing this equation into our search engine[/URL], we get: [B]w = 7[/B] Substitute w = 7 into l = 3w, we get: l = 3(7) [B]l = 21[/B] A retired couple invested \$8000 in bonds. At the end of one year, they received an interest payment A retired couple invested \$8000 in bonds. At the end of one year, they received an interest payment of \$584. What was the simple interest rate of the bonds? For simple interest, we have: Balance * interest rate = Interest payment 8000i = 584 Divide each side of the equation by 8000 to isolate i: 8000i/8000 = 584/8000 Cancelling the 8000's on the left side, we get: i = 0.073 Most times, interest rates are expressed as a percentage. Percentage interest = Decimal interest * 100% Percentage interest = 0.073 * 100% Multiplying by 100 is the same as moving the decimal point two places right: Percentage interest = [B]7.3%[/B] Aaron is staying at a hotel that charges \$99.95 per night plus tax for a room. A tax of 8% is applie Aaron is staying at a hotel that charges \$99.95 per night plus tax for a room. A tax of 8% is applied to the room rate, and an additional onetime untaxed fee of \$5.00 is charged by the hotel. Which of the following represents Aaron’s total charge, in dollars, for staying [I]x[/I] nights? [LIST] []The Room cost equals 99.95 times x where x is the number of rooms []We express an 8% tax by multiplying the room cost by 1.08 []Finally, we add on \$5, which is [I]untaxed[/I] [/LIST] [I][/I] Take this in pieces: Room Cost: 99.95x Tax on Room 1.08(99.95x) Add on \$5 which is untaxed: [B]1.08(99.95x) + 5[/B] An angle is 30 degrees less than 5 times it's complement. Find the angle. An angle is 30 degrees less than 5 times it's complement. Find the angle. Let the angle be a. The complement of a is 90 - a. We're given the following equation: a = 5(90 - a) - 30 <-- Less means we subtract Multiplying though, we get: a = 450 - 5a - 30 a = 420 - 5a [URL='https://www.mathcelebrity.com/1unk.php?num=a%3D420-5a&pl=Solve']Typing this equation into our search engine[/URL], we get: a =[B] 70[/B] Ande has 8 pints of milk. If he drinks 1/4 of a pint of milk each day, how long will the 8 pints of Ande has 8 pints of milk. If he drinks 1/4 of a pint of milk each day, how long will the 8 pints of milk last him? Milk Days = Total Pints of Milk / pints drank per day Milk Days = 8 / 1/4 Dividing by a fraction is the same as multiplying by it's reciprocal. The [URL='https://www.mathcelebrity.com/fraction.php?frac1=1%2F4&frac2=3%2F8&pl=Reciprocal']reciprocal of[/URL] 1/4 is 4/1, so we have: 8 4/1 = [B]32 days[/B] Basic Math Operations Free Basic Math Operations Calculator - Given 2 numbers, this performs the following arithmetic operations: * Subtraction (Subtracting) (-) * Multiplication (Multiplying) (x) * Long division (Dividing) with a remainder (÷) * Long division to decimal places (÷) * Partial Sums (Shortcut Sums) * Short Division * Duplication and Mediation Bob finished reading his book in x days. Each day, he read 4 pages. His book has 28 pages Bob finished reading his book in x days. Each day, he read 4 pages. His book has 28 pages Our equation for this is found by multiplying pages per day times number of days; 4x = 28 To solve for x, [URL='https://www.mathcelebrity.com/1unk.php?num=4x%3D28&pl=Solve']we type the equation into our search engine[/URL] and we get: x = [B]7[/B] Danny's mom ate 1/6 of an ice cream cake. Danny and his sister want to split the remainder of it. Ho Danny's mom ate 1/6 of an ice cream cake. Danny and his sister want to split the remainder of it. How much of the cake would each get? If Danny's mom ate 1/6 of the cake, then we have: 1 - 1/6 of the cake left. We [URL='https://www.mathcelebrity.com/fraction.php?frac1=1&frac2=1%2F6&pl=Subtract']use our fraction subtraction calculator[/URL] for 1 - 1/6 to get: 5/6 If Danny and his sister split the remainder, then we divide 5/6 by 2. It's also the same as multiplying 5/6 by 1/2: We [URL='https://www.mathcelebrity.com/fraction.php?frac1=5%2F6&frac2=1%2F2&pl=Multiply']use our fraction multiplication calculator[/URL] to get: [B]5/12 for Danny and his sister[/B] Derek must choose a 4 digit PIN. Each Digit can be chosen from 0 to 9. Derek does not want to reuse Derek must choose a 4 digit PIN. Each Digit can be chosen from 0 to 9. Derek does not want to reuse any digits. He also only wants an even number that begins with 5. How many possible PINS could he choose from? [LIST=1] []First digit must begin with 5. So we have 1 choice []We subtract 1 possible digit from digit 3 to have 8 - 1 = 7 possible digits []This digit can be anything other than 5 and the even number in the next step. So we have 0-9 is 10 digits - 2 = 8 possible digits []Last digit must end in 0, 2, 4, 6, 8 to be even. So we have 5 choices [/LIST] Our total choices from digits 1-4 are found by multiplying each possible digit choice: 1 * 7 * 8 * 5 = [B]280 possible PINS[/B] Divide 73 into two parts whose product is 402 Divide 73 into two parts whose product is 40 Our first part is x Our second part is 73 - x The product of the two parts is: x(73 - x) = 40 Multiplying through, we get: -x^2 + 73x = 402 Subtract 40 from each side, we get: -x^2 + 73x - 402 = 0 This is a quadratic equation. To solve this, we type it in our search engine, choose "solve Quadratic", and we get: [LIST=1] []x = [B]6[/B] []x = [B]67[/B] [/LIST] f - g = 1/4b for b f - g = 1/4b for b Multiply each side of the equation by 4 to remove the 1/4 and isolate b: 4(f - g) = 4/4b 4/4 = 1, so we have: b = [B]4(f - g)[/B] [I]the key to this problem was multiplying by the reciprocal of the constant[/I] Faith is 1/5 her mother's age. Their combined ages are 30. How old is faith? Faith is 1/5 her mother's age. Their combined ages are 30. How old is faith? Let Faith's age be f. Let her mother's age be m. We're given: [LIST=1] []f = m/5 []f + m = 30 [/LIST] Rearrange (1) by cross-multiplying: m = 5f Substitute this into equation (2): f + 5f = 30 [URL='https://www.mathcelebrity.com/1unk.php?num=f%2B5f%3D30&pl=Solve']Type this equation into our search engine[/URL] and we get: f = [B]5[/B] Farah rolls a fair dice and flips a fair coin. What is the probability of obtaining a 5 and a head? Farah rolls a fair dice and flips a fair coin. What is the probability of obtaining a 5 and a head? Give your answer in its simplest form. Probability of a 5 is 1/6 Probability of a head is 1/2 Since each event is independent, we get the total probability by multiplying both together: P(5,H) = 1/6 * 1/2 P(5,H) = [B]1/12[/B] Fractions and Mixed Numbers Free Fractions and Mixed Numbers Calculator - Given (improper fractions, proper fraction, mixed numbers, or whole numbers), this performs the following operations: * Subtraction (Subtracting) * Positive Difference (Absolute Value of the Difference) * Multiplication (Multiplying) * Division (Dividing: complex fraction division is included) * Compare Fractions * Simplifying of proper and improper fractions as well as mixed numbers. Fractions will be reduced down as far as possible (Reducing Fractions). * Reciprocal of a Fraction * Find all fractions between two fractions * reduce a fraction Given y= 4/3x what is the constant of proportionality Given y= 4/3x what is the constant of proportionality Direct variation means the constant of proportionality is y/x. Cross multiplying, we get: y/x = [B]4/3[/B] How many 1/4 sheets are there in 5 sheets How many 1/4 sheets are there in 5 sheets We divide 5 sheets by 1/4 sheets: 5/1/4 However, when we divide by a fraction, it's the same as multiplying by the reciprocal of the fraction: The reciprocal of 1/4 is 4/1, so we have: 5 * 4/1 = 20/1 = [B]20[/B] How many one-fifths are there in 200? How many one-fifths are there in 200? Using the rule of dividing by a fraction is the same as multiplying by the reciprocal, we have: 200 / 1/5 = 200 * 5 = [B]1000[/B] If 25% of 30% of x is 9, what is x? If 25% of 30% of x is 9, what is x? Convert percentages to decimals when multiplying: 25% = 0.25 30% = 0.3 0.25 * 0.3 * x = 9 0.075x = 9 Using our math engine, we [URL='https://www.mathcelebrity.com/1unk.php?num=0.075x%3D9&pl=Solve']type this equation in[/URL] and we get: x = [B]120 [MEDIA=youtube]5EwNxiBdLu0[/MEDIA][/B] If half the number is added to twice the number, the answer is 50 If half the number is added to twice the number, the answer is 50. Let the number be n. Half is also written as 0.5, and twice is written by multiplying by 2. We have: 0.5n + 2n = 50 [URL='https://www.mathcelebrity.com/1unk.php?num=0.5n%2B2n%3D50&pl=Solve']Plugging this equation into our search engine[/URL], we get: [B]n = 20[/B] In 2000 a company increased its workforce by 50%. In 2001 it decreased its workforce by 50%. How doe In 2000 a company increased its workforce by 50%. In 2001 it decreased its workforce by 50%. How does the size of its workforce at the end of 2001 compare with the size of the workforce at the beginning of 2000? Let w be the size of the workforce before any changes. We have: [LIST] []w(2000) = w(1999) 1.5 [I](50% increase is the same as multiplying by 1.5)[/I] []w(2001) = w(2000)/1.5 [I](50% decrease is the same as dividing by 1.5)[/I] [/LIST] Substitute the first equation back into the second equation w(2001) = w(1999) 1.5/1.5 Cancel the 1.5 on top and bottom w(2001) = w(1999) This means the workforce had [B]zero net change[/B] from the beginning of 2000 to the end of 2001. In this class of 4/5 students are right handed. if there are 20 right handed students, what is the t In this class of 4/5 students are right handed. if there are 20 right handed students, what is the total number of students in this class? Let x be the total number of students in the class. We have: 4/5x = 20 Cross multiplying or using our [URL='http://www.mathcelebrity.com/1unk.php?num=4x%3D100&pl=Solve']equation calculator[/URL], we get: 4x = 100 Divide each side by 4 [B]x = 25[/B] Jerry, an electrician, worked 7 months out the year. What percent of the year did he work? Jerry, an electrician, worked 7 months out the year. What percent of the year did he work? We know that there are 12 months in a year. Percentage worked = Months worked in a year / months in a year * 100% Percentage worked = 7/12 * 100% Percentage worked = 0.5833333 * 100% Multiplying by 100 means we shift the decimal place 2 spaces to the right: Percentage worked = [B]58.33%[/B] Kendra is half as old as Morgan and 3 years younger than Lizzie. The total of their ages is 39. How Kendra is half as old as Morgan and 3 years younger than Lizzie. The total of their ages is 39. How old are they? Let k be Kendra's age, m be Morgan's age, and l be Lizzie's age. We're given: [LIST=1] []k = 0.5m []k = l - 3 []k + l + m = 39 [/LIST] Rearranging (1) by multiplying each side by 2, we have: m = 2k Rearranging (2) by adding 3 to each side, we have: l = k + 3 Substituting these new values into (3), we have: k + (k + 3) + (2k) = 39 Group like terms: (k + k + 2k) + 3 = 39 4k + 3 = 39 [URL='https://www.mathcelebrity.com/1unk.php?num=4k%2B3%3D39&pl=Solve']Type this equation into the search engine[/URL], and we get: [B]k = 9 [/B] Substitute this back into (1), we have: m = 2(9) [B]m = 18 [/B] Substitute this back into (2), we have: l = (9) + 3 [B][B]l = 12[/B][/B] Layla buys 2 1/2pounds of chocolate for 3.50 how much is she paying for a pound of chocolate Layla buys 2 1/2pounds of chocolate for 3.50 how much is she paying for a pound of chocolate? [URL='https://www.mathcelebrity.com/fraction.php?frac1=2%261%2F2&frac2=3%2F8&pl=Simplify']Using our mixed fraction converter[/URL], 2&1/2 = 5/2 Cost per pound = 3.50 / 5/2 pounds Dividing by 5/2 is the same as multiplying by the reciprocal 2/5: 3.50 2/5 7/5 [B]\$1.40 per pound[/B] M/n = p-6 for m M/n = p-6 for m Solve this literal equation by multiplying each side by n to isolate M: Mn/n = n(p - 6) Cancelling the n terms on the left side, we get: [B]M = n(p - 6)[/B] multiply 9 by the quotient of 4 and z multiply 9 by the quotient of 4 and z Quotient of 4 and z is written as: 4/z Multiply this quotient by 9: 9(4)/z Multiplying the top, we get: [B]36/z[/B] Multiply a number by 6 and subtracting 6 gives the same result as multiplying the number by 3 and su Multiply a number by 6 and subtracting 6 gives the same result as multiplying the number by 3 and subtracting 4. Find the number The phrase [I]a number [/I]means an arbitrary variable, let's call it x. multiply a number by 6 and subtract 6: 6x - 6 Multiply a number by 3 and subtract 4: 3x - 4 The phrase [I]gives the same result[/I] means an equation. So we set 6x - 6 equal to 3x - 4 6x - 6 = 3x - 4 To solve this equation for x, we type it in our search engine and we get: x = [B]2/3[/B] Multiplying a number by 6 is equal to the number increased by 9 Multiplying a number by 6 is equal to the number increased by 9. The phrase [I]a number[/I] means an arbitrary variable, let's call it x. Multiply it by 6 --> 6x We set this equal to the same number increased by 9. Increased by means we add: [B]6x = x + 9 <-- This is our algebraic expression [/B] To solve this equation, we [URL='https://www.mathcelebrity.com/1unk.php?num=6x%3Dx%2B9&pl=Solve']type it into the search engine [/URL]and get x = 1.8. Penny bought a new car for \$25,000. The value of the car has decreased in value at rate of 3% each Penny bought a new car for \$25,000. The value of the car has decreased in value at rate of 3% each year since. Let x = the number of years since 2010 and y = the value of the car. What will the value of the car be in 2020? Write the equation, using the variables above, that represents this situation and solve the problem, showing the calculation you did to get your solution. Round your answer to the nearest whole number. We have the equation y(x): y(x) = 25,000(0.97)^x <-- Since a 3 % decrease is the same as multiplying the starting value by 0.97 The problem asks for y(2020). So x = 2020 - 2010 = 10. y(10) = 25,000(0.97)^10 y(10) = 25,000(0.73742412689) y(10) = [B]18,435.60[/B] Sally and Adam works a different job. Sally makes \$5 per hour and Adam makes \$4 per hour. They each Sarah makes \$9 per hour working at a daycare center and \$12 per hour working at a restaurant. Next Sarah makes \$9 per hour working at a daycare center and \$12 per hour working at a restaurant. Next week, Sarah is scheduled to work 8 hours at the daycare center. Which of the following inequalities represents the number of hours (h) that Sandra needs to work at the restaurant next week to earn at least \$156 from these two jobs? Set up Sarah's earnings function E(h) where h is the hours Sarah must work at the restaurant: 12h + 9(8) >= 156 <-- The phrase [I]at least[/I] means greater than or equal to, so we set this up as an inequality. Also, the daycare earnings are \$9 per hour * 8 hours Multiplying through and simplifying, we get: 12h + 72 >= 156 We [URL='https://www.mathcelebrity.com/1unk.php?num=12h%2B72%3E%3D156&pl=Solve']type this inequality into the search engine[/URL], and we get: [B]h>=7[/B] Solve 100 / 1/2 Solve 100 / 1/2 Dividing by a fraction is the same as multiplying by the reciprocal of the fraction: [URL='https://www.mathcelebrity.com/fraction.php?frac1=1%2F2&pl=Reciprocal']Reciprocal of 1/2[/URL] = 2 100 * 2 = [B]200[/B] The average height of a family of 6 is 6 feet. After the demise of the mother, the average height re The average height of a family of 6 is 6 feet. After the demise of the mother, the average height remained the same. What is the height of the mother? [LIST] []Let the height of the family without the mom be f. Let the height of the mother be m. []Averages mean we add the heights and divide by the number of people who were measured. [/LIST] We're given two equations: [LIST=1] [](f + m)/6 = 6 []f/5 = 6 [/LIST] Cross multiplying equation (2), we get: f = 5 * 6 f = 30 Plug f = 30 into equation (1), we get: (30 + m)/6 = 6 Cross multiplying, we get: m + 30 = 6 * 6 m + 30 = 36 To solve this equation for m, we [URL='https://www.mathcelebrity.com/1unk.php?num=m%2B30%3D36&pl=Solve']type it in our search engine[/URL] and we get: m = [B]6[/B] [SIZE=3][FONT=Arial][COLOR=rgb(34, 34, 34)][/COLOR][/FONT][/SIZE] The dimensions of a rectangle are 30 cm and 18 cm. When its length decreased by x cm and its width i The dimensions of a rectangle are 30 cm and 18 cm. When its length decreased by x cm and its width is increased by x cm, its area is increased by 35 sq. cm. a. Express the new length and the new width in terms of x. b. Express the new area of the rectangle in terms of x. c. Find the value of x. Calculate the current area. Using our [URL='https://www.mathcelebrity.com/rectangle.php?l=30&w=18&a=&p=&pl=Calculate+Rectangle']rectangle calculator with length = 30 and width = 18[/URL], we get: A = 540 a) Decrease length by x and increase width by x, and we get: [LIST] []length = [B]30 - x[/B] []width = [B]18 + x[/B] [/LIST] b) Our new area using the lw = A formula is: (30 - x)(18 + x) = 540 + 35 Multiplying through and simplifying, we get: 540 - 18x + 30x - x^2 = 575 [B]-x^2 + 12x + 540 = 575[/B] c) We have a quadratic equation. To solve this, [URL='https://www.mathcelebrity.com/quadratic.php?num=-x%5E2%2B12x%2B540%3D575&pl=Solve+Quadratic+Equation&hintnum=+0']we type it in our search engine, choose solve[/URL], and we get: [B]x = 5 or x = 7[/B] Trying x = 5, we get: A = (30 - 5)(18 + 5) A = 25 * 23 A = 575 Now let's try x = 7: A = (30 - 7)(18 + 7) A = 23 * 25 A = 575 They both check out. So we can have The mean height of a class of 20 children is 1.27 the mean height of 12 boys in the class is 1.29 wh The mean height of a class of 20 children is 1.27 the mean height of 12 boys in the class is 1.29 what is the mean height of the girls in the class? The mean of sums is the sum of means. So we have: Total Height / 20 = 1.27 Cross multiplying, we get: Total Height = 20 * 1.27 Total Height = 25.4 Boys Height / 12 = 1.29 Cross multiplying, we get: Boys Height = 12 * 1.29 Boys Height = 15.48 The Problem asks for mean height for girls. The formula is: Girls Height / # of Girls = Mean of Girls Height # of Girls = Total children - # of boys # of Girls = 20 - 12 # of Girls = 8 Girls Height = Total Height - Boys Height Girls Height = 25.4 - 15.48 Girls Height = 9.92 Plugging this into the Mean of girls height, we get: 9.92 /8 = [B]1.24[/B] The perimeter of a poster is 20 feet. The poster is 6 feet tall. How wide is it? The perimeter of a poster is 20 feet. The poster is 6 feet tall. How wide is it? [U]Assumptions and givens:[/U] [LIST] []The poster has a rectangle shape []l = 6 []P = 20 []The perimeter of a rectangle (P) is: 2l + 2w = P [/LIST] Plugging in our l and P values, we get: 2(6) + 2w = 20 Multiplying through and simplifying, we get: 12 + 2w = 20 To solve for w, we [URL='https://www.mathcelebrity.com/1unk.php?num=12%2B2w%3D20&pl=Solve']type this equation into our search engine [/URL]and we get: w = [B]4[/B] the result of quadrupling a number is 80 the result of quadrupling a number is 80 Let our number be x. Quadrupling any number means multiplying it by 4. We have: 4x = 80 [URL='https://www.mathcelebrity.com/1unk.php?num=4x%3D80&pl=Solve']Typing this problem into our search engine[/URL], we get: [B]x = 20[/B] The sum of the ages of levi and renee is 89 years. 7 years ago levi's age was 4 times renees age. Ho The sum of the ages of levi and renee is 89 years. 7 years ago levi's age was 4 times renees age. How old is Levi now? Let Levi's current age be l. Let Renee's current age be r. Were given two equations: [LIST=1] []l + r = 89 []l - 7 = 4(r - 7) [/LIST] Simplify equation 2 by multiplying through: [LIST=1] []l + r = 89 []l - 7 = 4r - 28 [/LIST] Rearrange equation 1 to solve for r and isolate l by subtracting l from each side: [LIST=1] []r = 89 - l []l - 7 = 4r - 28 [/LIST] Now substitute equation (1) into equation (2): l - 7 = 4(89 - l) - 28 l - 7 = 356 - 4l - 28 l - 7 = 328 - 4l To solve for l, we [URL='https://www.mathcelebrity.com/1unk.php?num=l-7%3D328-4l&pl=Solve']type the equation into our search engine[/URL] and we get: l = [B]67[/B] triple the sum of y and six The sum of y and six is denoted as: y + 6 We triple that sum by multiplying it by 3 3(y + 6) two numbers have an average of 2100 and one number is \$425 more than the other number. What are the two numbers have an average of 2100 and one number is \$425 more than the other number. What are the numbers Let the first number be x and the second number be y. We're given two equations: [LIST=1] [](x + y)/2 = 2100 (Average) []y = x + 425 [/LIST] Rearrange equation (1) by cross multiplying x + y = 2 * 2100 x + y = 4200 So we have our revised set of equations: [LIST=1] []x + y = 4200 []y = x + 425 [/LIST] Substituting equation (2) into equation (1) for y, we get: x + (x + 425) = 4200 Combining like terms, we get: 2x + 425 = 4200 Using our [URL='https://www.mathcelebrity.com/1unk.php?num=2x%2B425%3D4200&pl=Solve']equation solver[/URL], we get: x = [B]1887.5[/B] Which means using equation (2), we get y = 1887.5 + 425 y = [B]2312.5[/B] what integer is tripled when 9 is added to 3 fourths of it? what integer is tripled when 9 is added to 3 fourths of it? Let the integer be n. Tripling an integer means multiplying it by 3. We're given: 3n = 3n/4 + 9 Since 3 = 12/4, we have: 12n/4 = 3n/4 + 9 Subtract 3n/4 from each side: 9n/4 = 9 [URL='https://www.mathcelebrity.com/prop.php?num1=9n&num2=9&den1=4&den2=1&propsign=%3D&pl=Calculate+missing+proportion+value']Typing this equation into the search engine[/URL], we get: [B]n = 4[/B] What is the inverse of dividing by 3 What is the inverse of dividing by 3 [B]Multiplying by 3[/B] Suppose we have 2 divided by 3: 2/3 To undo this operation to get to 2 again, we'd multiply by 3: 2/3 * 3 = 2 What is the value of 998^2 – 2^2? A) 988,036 B) 990,000 C) 995,988 D) 996,000 E) 1,000,000 This is a difference of squares. The formula for 2 numbers a and b is: a^2 - b^2 = (a + b)(a - b) In our problem, we have a = 998 and b = 2: 998^2 – 2^2 = (998 + 2)(998 - 2) 998^2 – 2^2 = 1000(996) Multiplying by 1000 means we move the decimal place of the other number 3 places to the right: 998^2 – 2^2 = [B]996,000 or Answer D [MEDIA=youtube]IeKLs8Ds-No[/MEDIA][/B] When five people are playing a game called hearts, each person is dealt ten cards and the two remain When five people are playing a game called hearts, each person is dealt ten cards and the two remaining cards are put face down on a table. Because of the rules of the game, it is very important to know the probability of either of the two cards being a heart. What is the probability that at least one card is a heart? Probability that first card is not a heart is 3/4 since 4 suits in the deck, hearts are 1/4 of the deck. Since we don't replace cards, the probability of the next card drawn without a heart is (133 - 1)/51 = 38/51 Probability of both cards not being hearts is found by multiplying both individual probabilities: 3/4 38/51 = 114/204 Having at least one heart is found by subtracting this from 1 which is 204/204: 204/204 - 114/204 = 90/204 [URL='https://www.mathcelebrity.com/search.php?q=90%2F204&x=0&y=0']This reduces to[/URL] [B]15/34[/B] Zalika thinks of a number. She subtracts 6 then multiplies the result by 5. The answer is the same a Zalika thinks of a number. She subtracts 6 then multiplies the result by 5. The answer is the same as subtracting 5 from the number then multiplying by 4. The phrase [I]a number[/I] means an arbitrary variable, let's call it x. We're given two expressions in relation to this number (x): [U]She subtracts 6 then multiplies the result by 5[/U] [LIST] []Subtract 6: x - 6 []Multiply the result by 5: 5(x - 6) [/LIST] [U]She subtracts 5 from the number then multiplying by 4[/U] [LIST] []Subtract 6: x - 5 []Multiply the result by 5: 4(x - 5) [/LIST] Finally, the expression [I]is the same as[/I] means an equation, so we set the first expression equal to the second expression to make the following equation: 5(x - 6) = 4(x - 5) Now, let's solve the equation for x. To do this, we [URL='https://www.mathcelebrity.com/1unk.php?num=5%28x-6%29%3D4%28x-5%29&pl=Solve']type this equation into our search engine [/URL]and we get: x = [B]10[/B]
8.1: Trigonometric Ratios: This branch of mathematics helps us in finding height of tall buildings, height of temples, width of river, height of mountains, towers etc without actually measuring them. In the lesson 8.3 we will be solving few problems related to these.Different branches of Engineering use trigonometry and its functions extensively. Using trigonometry we can find sides and angles of triangles when sufficient data is given about triangles. Trigonometry deals with three (tri) angles (gonia) and measures (metric) namely triangles. Ancient Indians were aware of the sine function and it is believed that modern trigonometry migrated from Hindus to Europe through Arabs. The Indian mathematicians who contributed to the development of Trigonometry are Aryabhata(6th Century AD), Brahmagupta(7th century AD) and Neelakantha Somayaji(15th Century AD). We measure an angle in degrees from 0 to 3600. The angles are also measured using a unit called radians. The relationship between degree and radii is given by 2 radians = 3600. Hence, we have the table which gives relationship for various values of degree. Degree >> 1800 900 600 450 360 300 150 Radian >> /2 /3 /4 /5 /6 /12 Since any triangle can be split into 2 right angled triangle, in trigonometry we study right angled triangles only. The three sides of a right triangle are called Opposite Side with respect to an angle( The side opposite to angle): In the figure: SA,TB,UC and XP in respect of Y) Adjacent Side with respect to an angle(The side on which angle rests) : In the figure: YA,YB,YC and YP in respect of Y) Hypotenuse (the side opposite to the right angle): In the figure YS, YT, YU and YX) Since sum of angles in a triangle is 1800 and one angle is 900, the other two angles in a right angled triangle have to be necessarily acute(<900)angles. The acute angles (two in number) are normally denoted by Greek letters alpha (), beta (), gamma (), theta (), phi (). In the adjoining figure XPY is a right angles triangle with XPY = 900 We also notice that SAY \|\|\| TBY \|\|\| UCY \|\|\|XPY. Thus by similarity property of SAY and TBY, YA/YB =YS/YT=AS/BT YA/YS=YB/YT= Adjacent Side /Hypotenuse YA/AS=YB/BT= Adjacent Side /Opposite Side AS/YS=BT/YT= Opposite Side / Hypotenuse Since these ratios are constant irrespective of length of the sides obviously, why not we represent these ratios by some standard names? Thus, we have definitions of sine, cosine and other terms: Since the right triangle has three sides we can have six different ratios of their sides as given in the following table: No Name Short form Ratio of sides In the Figure Remarks 1 sine Y sin Y Opposite Side /Hypotenuse =PX/YX (OH) 2 cosine Y cos Y Adjacent Side /Hypotenuse =YP/YX (AH) 3 tangent Y tan Y Opposite Side /Adjacent Side =PX/YP =sin Y /cos Y,(OA) 4 cosecant Y cosec Y Hypotenuse/Opposite Side =YX/PX =1/sin Y 5 secant Y sec Y Hypotenuse/Adjacent Side =YX/YP =1/cos Y 6 cotangent Y cot Y Adjacent Side /Opposite Side =YP/PX =1/tanY=cosY/sinY Notes: 1. Last three ratios (4, 5 and 6) are reciprocals (inverse) of the first three ratios, hence for any angle 1. sin Cosec =1 2. cos Sec =1 3. tanCot =1 2. Naming (Identification) of Adjacent Side and Opposite Side sides are interchangeable depending upon the angle opposite to the sides (With respect to X, the Adjacent Side is XP and Opposite Side is PY. With respect to Y, the Adjacent Side is YP and Opposite Side is PX). PX is also called ‘Perpendicular’ of Y and YP is also called ‘Base’ of Y) 3. Trigonometric ratios are numbers without units. Exercise: Name the ratios with respect to the angle X. 8.1 Problem 1: From the adjacent figure find the value of sin B, tan C, sec2B - tan2B and sin2C + cos2C Solution: By Pythagoras theorem BA2 = BD2+AD2 AD2 = BA2-BD2 = 132-52 = 169 -25 = 144 = 122 AD = 12 By Pythagoras theorem AC2 = AD2+DC2 = 122+162 = 144 +256 = 400 = 202 AC = 20 By definition 1. sin B = Opposite Side /Hyp. = AD/AB= 12/13 2. tan C =Opposite Side /Adjacent Side = AD/DC = 12/16 = 3/4 3. sec2B - tan2B = (AB/BD)2 – (AD/BD)2 = (AB2 - AD2)/ BD2 = (132 - 122)/ 52 =(169-144)/25 =1 4. sin2C + cos2C = (AD/AC)2+ (DC/AC)2 = (AD2 +DC2)/ AC2 = (122 +162)/ 202 = (144+256)/400 =1 8.1 Problem 2: If 5 tan = 4 find the value of (5 sin -3 cos)/(5 sin +2 cos) Solution: tan = 4/5 (It is given that 5 tan = 4) In the adjacent figure, tan = Opposite Side /Adjacent Side =BC/AB. Let the sides be multiples of x units. (For example, Let x be 3 cm so that BC = 12(43) cm and AB =15(53) and hence BC/AB = 12/15 =4/5) We can say BC = 4x and AB= 5x 5 sin -3 cos = 5BC/AC – 3AB/AC = (5BC-3AB)/AC 5 sin +2 cos = 5BC/AC + 2AB/AC = (5BC+2AB)/AC (5 sin -3 cos)/(5 sin +2 cos) = {(5BC-3AB)/AC}/{(5BC+2AB)/AC} = (5BC-3AB)/(5BC+2AB) = (54x- 35x)/(54x+25x) (By substituting values for BC and AB) = (20x-15x)/(20x+10x) = 5x/30x = 1/6 8.1 Problem 3: Given sin = p/q, find sin + cos in terms of p and q. Solution: By definition sin = Opposite Side /Hyp.= BC/AC Since it is given that sin = p/q, we can say BC =px and AC=qx By Pythagoras theorem AC2 = AB2+BC2 AB2 = AC2-BC2 = (qx)2-(px)2 = x2(q2-p2) AB = x By definition cos = AB/AC = (x )/qx = ()/q sin + cos = p/q +()/q = (p+)/q 8.1 Problem 4: Using the measurements given in the adjacent figure 1. find the value of sin and tan 2. Write an expression for AD in terms of Solution: Construction: Draw a line parallel to BC from D to meet BA at E. By Pythagoras theorem BD2 = BC2+CD2 CD2 = BD2-BC2 = 132-122 = 169 -144 = 25 = 52 CD = 5 Since BA \|\| CD and BC\|\|DE, BE=CD(=5) EA = BA-BE = 14-5 =9 By Pythagoras theorem AD2 = AE2+ED2 = 92+122 = 81+144= 225 = 152 AD = 15 By definition 1. sin = 5/13 2. tan = 12/9 = 4/3 3. cos = 9/AD AD = 9/cos = 9 sec 8.1 Problem 5: Given 4 sin = 3 cos Find the value ofsin , cos , cot2- cosec2. Solution: Since it is given that 4 sin = 3 cos , by simplifying we get sin /cos =3/4 By definition tan = Opposite Side / Adjacent Side = BC/AB =3/4 Thus we can say BC = 3x and AB = 4x By Pythagoras theorem AC2 = BC2+AB2= (3x)2+(4x)2 = 9x2+16x2 = 25x2 = (5x)2 AC = 5x sin = BC/AC = 3x/5x = 3/5 cos = AB/AC= 4x/5x = 4/5 cot2- cosec2 = (AB/BC)2-(AC/BC)2 = (4x/3x)2-(5x/3x)2 = (4/3)2-(5/3)2 = 16/9 -25/9 = (16-9)/9 = -9/9 = -1 8.1 Problem 6: In the given figure AD is perpendicular to BC, tan B = 3/4, tan C = 5/12 and BC= 56cm, calculate the length of AD Solution: By definition tan B = Opposite Side /Adjacent Side =AD/BD, and it is given that tan B = 3/4 AD/BD = 3/4 i.e. 4AD = 3BD i.e. 12AD = 9BD ----à(1) tan C = Opposite Side /Adjacent Side = AD/DC and it is given that tan C = 5/12 AD/ DC = 5/12 i.e. 12AD = 5DC ----à(2) Equating (1) and (2), we get 9BD = 5DC ----à(3) It is given that BD+DC = 56 and hence DC = 56-BD Substituting this value in (3) we get 9BD = 5(56-BD) = 280-5BD 9BD+5BD = 280 (By transposition) BD = 280/14 = 20 DC = 56-BD = 56-20 = 36 AD = (3/4)BD = (3/4)20 = 15cm 8.1 Summary of learning No Points studied 1 sine= Opposite Side /hypotenuse(OH) 2 cosine= Adjacent Side /hypotenuse(AH) 3 tangent= Opposite Side /Adjacent Side (OA) 4 cosecant is reciprocal of sin 5 secant is reciprocal of cos 6 cotangent is reciprocal of tan
FutureStarr A 2 Out of 16 As a Percentage 2 Out of 16 As a Percentage via GIPHY I was trying to figure out what to wear and had about 3 outfits. When I got the auto-suggestions, my first response was “2 out of 16? That must be a coincidence”. I was wrong. Percentage I've seen a lot of students get confused whenever a question comes up about converting a fraction to a percentage, but if you follow the steps laid out here it should be simple. That said, you may still need a calculator for more complicated fractions (and you can always use our calculator in the form below). Let's give ourselves a little bit of practice with percentages. So let's ask ourselves, what percent of-- I don't know, let's say what percent of 16 is 4? And I encourage you to pause this video and to try it out yourself. So when you're saying what percent of 16 is 4, percent is another way of saying, what fraction of 16 is 4? And we just need to write it as a percent, as per 100. So if you said what fraction of 16 is 4, you would say, well, look, this is the same thing as 4/16, which is the same thing as 1/4. But this is saying what fraction 4 is of 16. You'd say, well, 4 is 1/4 of 16. But that still doesn't answer our question. What percent? So in order to write this as a percent, we literally have to write it as something over 100. Percent literally means "per cent." The word "cent" you know from cents and century. It relates to the number 100. So it's per 100. So you could say, well, this is going to be equal to question mark over 100, the part of 100. And there's a bunch of ways that you could think about this. You could say, well, look, if in the denominator to go from 4 to 100, I have to multiply by 25. In the numerator to go from-- I need to also multiply by 25 in order to have an equivalent fraction. So I'm also going to multiply by 25. So 1/4 is the same thing as 25/100. And another way of saying 25/100 is this is 25 per 100, or 25%. So this is equal to 25%. Now, there's a couple of other ways you could have thought about it. You could have said well, 4/16, this is literally 4 divided by 16. Well, let me just do the division and convert to a decimal, which is very easy to convert to a percentage. So let's try to actually do this division right over here. So we're going to literally divide 4 by 16. Now, 16 goes into 4 zero times. 0 times 16 is 0. You subtract, and you get a 4. And we're not satisfied just having this remainder. We want to keep adding zeroes to get a decimal answer right over here. So let's put a decimal right over here. We're going into the tenths place. And let's throw some zeroes right over here. The decimal makes sure we keep track of the fact that we are now in the tenths, and in the hundredths, and in the thousandths place if we have to go that far. But let's bring another 0 down. 16 goes into 40 two times. 2 times 16 is 32. If you subtract, you get 8. And you could bring down another 0. And we have 16 goes into 80. Let's see, 16 goes into 80 five times. 5 times 16 is 80. You subtract, you have no remainder, and you're done. 4/16 is the same thing as 0.25. Now, 0.25 is the same thing as twenty-five hundredths. Or, this is the same thing as 25/100, which is the same thing as 25%. (Source: www.khanacademy.org) Related Articles • 20 Percent of 35: August 11, 2022 \| M aqib • Calculator With Tape Roll August 11, 2022 \| sheraz naseer • What Percentage Is 2 Out of 15 OR August 11, 2022 \| Muhammad Waseem • Is Love Calculator Real OR August 11, 2022 \| Jamshaid Aslam • 2 Out of 9 Percentage in 2022 August 11, 2022 \| Jamshaid Aslam • A Google Calculator Showing Fractions August 11, 2022 \| Shaveez Haider • What Percent Is 33 Out of 40 August 11, 2022 \| Muhammad Waseem • The Significance of Euler's Number August 11, 2022 \| Future Starr • How to find an area of a triangleor August 11, 2022 \| Muhammad basit • How many min in a year August 11, 2022 \| m basit • 20 Percent of 17 August 11, 2022 \| Faisal Arman • Online Fraction Calculator That Shows Work August 11, 2022 \| Muhammad Umair • Astrology Love Calculator August 11, 2022 \| sheraz naseer • 5 14 As a Percent August 11, 2022 \| sheraz naseer • A 6 17 As a Percentage August 11, 2022 \| Shaveez Haider
# Video: Using the Operators of Vectors and Dot Product Between them to Find the Value of an Expression 03:31 ### Video Transcript Find the value of the magnitude of π¨ cross π© squared plus the magnitude of π¨ dot π© squared divided by two times the magnitude of π¨ squared times the magnitude of π© squared. Here, we know that π¨ and π© are vectors. But beyond that, we donβt know anything about them. In that sense, theyβre completely general. We donβt know anything about their components. As a start point to evaluate this expression though, we see that weβre crossing as well as dotting these two vectors. In general, if we cross two vectors, called π¨ and π©, then that cross product equals the magnitude of the first vector times the magnitude of the second vector times the sine of the angle between the two vectors, called π here. And this whole result points in a direction that is normal or perpendicular both to vector π¨ and to vector π©. If we then take the absolute value of this cross product, weβre no longer calculating a vector. So we donβt have a direction associated with our result. And whatever our angle π is, the sine of that angle must be a nonnegative number. This is so that overall the magnitude of π¨ cross π© is itself nonnegative. In our given expression, we see that weβre working not with the magnitude of π¨ cross π© but the magnitude of π¨ cross π© squared. That would give us this expression. And if we consider the square of the magnitude of the sin of π, we find that either one of these operations, taking the absolute value or squaring sin π, would make the result nonnegative. To make this value positive or zero then, we only need to apply one of these two operations. Weβll choose to remove the absolute value bars but continue to square this value. This way, weβll end up with a nonnegative result that reflects the fact that weβre squaring the sin of π. So instead of the magnitude of π¨ cross π© squared, we can write this. When it comes to the magnitude of π¨ dot π© squared, letβs recall that, in general, the dot product of two vectors is equal to the product of their magnitudes multiplied by the cosine of the angle between the two vectors. So then, the magnitude of π¨ dot π© equals the magnitude of π¨ times the magnitude of π© times the magnitude of the cos of π. And therefore, when we square this quantity, we get the square of the magnitude of π¨ times the square of the magnitude of π© times the square of the magnitude of cos π. Once again, itβs unnecessary both to take the absolute value and to square our trigonometric term. And so we can remove the absolute value bars and not change this result. Looking back at our given expression, we see that we are adding together the magnitude of π¨ cross π© squared and the magnitude of π¨ dot π© squared. Based on what we found so far, that gives us this expression on the right-hand side here. And notice that the value of the magnitude of π¨ squared times the magnitude of π© squared is common to both of these terms. If we factor it out, then we see something very interesting. We have the sine squared of an angle plus the cosine squared of that same angle. Whenever this happens, whenever we have the sine squared of one angle and the cosine squared of that same angle being added together, the result equals one. And so the right-hand side of our expression simplifies to the magnitude of π¨ squared times the magnitude of π© squared. And letβs recall that weβre adding together these two terms, which form the entire numerator of our fraction. Writing out our entire fraction then, we get this result. We see that here the magnitude of vector π¨ squared cancels in numerator and denominator, as does the magnitude of vector π© squared. This whole expression then simplifies to one over two. For two general vectors π¨ and π© then, the magnitude of π¨ cross π© squared plus the magnitude of π¨ dot π© squared divided by two times the magnitude of π¨ squared times the magnitude of π© squared equals one-half.
# How do you simplify (9sqrt25)/sqrt50? Apr 27, 2018 $\frac{9}{\sqrt{2}}$ #### Explanation: $\frac{9 \sqrt{25}}{\sqrt{50}}$ $= \frac{9 \times 5}{\sqrt{50}}$ $\sqrt{50}$ can be simplified to $\sqrt{25 \times 2} = 5 \sqrt{2}$ $\frac{45}{5 \sqrt{2}}$ = $\frac{9}{\sqrt{2}}$ Apr 27, 2018 $\frac{9 \sqrt{2}}{2}$ #### Explanation: $9 \sqrt{25}$ = $9 \times 5$ = 45 $\sqrt{50} = \sqrt{25} \times 2 = 5 \sqrt{2}$ so $\frac{9 \sqrt{25}}{\sqrt{50}} = \frac{45}{5 \sqrt{2}}$ dividing by top and bottom by 5 leaves $\frac{9}{\sqrt{2}}$ If we multiply top and bottom by $\sqrt{2}$ this will rationalise it (remove the surd from the denominator) $\frac{9}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$ =$\frac{9 \sqrt{2}}{2}$ Apr 27, 2018 $\frac{9 \sqrt{25}}{\sqrt{50}} = \frac{9 \sqrt{2}}{2}$ #### Explanation: $\sqrt{a b} = \sqrt{a} \cdot \sqrt{b}$ So, $\frac{9 \sqrt{25}}{\sqrt{50}} = \frac{9 \sqrt{25}}{\sqrt{25} \cdot \sqrt{2}} = \frac{9}{\sqrt{2}}$ $\frac{9}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{9 \sqrt{2}}{2}$
# Conjugate Surds: Definition, Examples, Properties In this section, we will discuss about conjugate surds. For the basics of surds, please visit the page an introduction to surds. ## Definition of Conjugate Surds Mathematically, if x=a+√b where a and b are rational numbers but √b is an irrational number, then a-√b is called the conjugate of x. Conjugate surds are also known as complementary surds. For example, 1+√2 and 1-√2 are conjugate surds of each other. Thus we can define conjugate surds as follows: A surd is said to be a conjugate surd to another surd if they are the sum and difference of two simple quadratic surds. In other words, the sum and the difference of two simple quadratic surds are conjugate to each other. Examples of Conjugate Surds: we consider two simple quadratic surds $\sqrt{2}$ and $7\sqrt{3}.$ According to the above definition, the two binomial surds $\sqrt{2}+7\sqrt{3}$ and $\sqrt{2}-7\sqrt{3}$ are conjugate (or complementary) to each other. In a similar way, we have the following examples of conjugate surds: (i) 5√2 +2√7 and 5√2 – 2√7 (ii) 1+√3 and 1-√3 Note: In general, surds of the forms (a – y√b) and (a + y√b) are complementary/conjugate to each other. ## Properties of Conjugate Surds • The general form of two conjugate surds are $a\sqrt{x}+b\sqrt{y}$ and $a\sqrt{x}-b\sqrt{y}.$ • The product of two conjugate surds is always a rational number. Proof. The product of two general conjugate surds is given by $(a\sqrt{x}+b\sqrt{y})$$(a\sqrt{x}-b\sqrt{y})$ $=(a\sqrt{x})^2-(b\sqrt{y})^2$ $[\because (m+n)(m-n)=m^2-n^2]$ $=a^2x-b^2y,$ which is a rational number. The sum of two conjugate surds is always a rational number. Proof. As conjugate surds have the form a+√b and a-√b where both a and b are rational numbers, the sum will be equal to (a+√b)+(a-√b) = a+√b+a-√b = 2a, which is a rational number. The difference of two conjugate surds is not a rational number. Proof. The difference of two conjugate surds a+√b and a-√b where both a and b are rational numbers is (a+√b)-(a-√b) = a+√b-a+√b = 2√b, which is an irrational number. ## Importance of Conjugate Surds To rationalize the denominator of a fraction containing surds, we need to take the help of conjugate surds. In this case, we need to multiply the denominator with its conjugate surd. For example, Question: Rationalize the denominator of $\dfrac{1}{1+\sqrt{2}}$ See that the denominator 1+√2 is not a rational number. To rationalize the denominator we have to multiply with the conjugate of 1+√2 which is 1-√2. By doing so we get that $\dfrac{1}{1+\sqrt{2}}=\dfrac{1-\sqrt{2}}{(1+\sqrt{2})(1-\sqrt{2})}$ $=\dfrac{1-\sqrt{2}}{1^2-\sqrt{2}^2}$ $[\because (a-b)(a+b)=a^2-b^2]$ $=\dfrac{1-\sqrt{2}}{1-2}$ $=-(1-\sqrt{2})=-1+\sqrt{2}$ Note: Conjugate of surds are also very useful in many branches of Mathematics; like Calculus, Geometry, etc. ## Question Answer on Conjugate Surds Question 1: Find the conjugate surd of √5-√2. Solution: See that the surd √5-√2 is the difference of the two surds √5 and √2. So the desired conjugate surd will be the sum of the two surds. Hence, the conjugate of √5-√2 is √5+√2. Question 2: Find the conjugate surd of √3 -2. Solution: Note that the surd √3 -2 can be written as -2+√3. As we know that the conjugate surd of a+√b (a rational and √b irrational) is a-√b, so the conjugate of -2+√3 will be -2-√3. Question 3: Find the conjugate surd of √5-2. Solution: We have √5-2 = -2+√5. The conjugate of a-√b is a+√b. Thus, the conjugate of -2+√5 is -2-√5. Related Topics ## FAQs on Conjugate Surds Q1: What are conjugate surds? Answer: A pair of surds is called conjugate of each other if their forms are x√a-y√b and x√a+y√b where a, b, x, and y are rational numbers and both a, b are square-free. Q2: Give an example of conjugate surds. Answer: The conjugate surd of √2 is -√2; and vice-versa. Share via:
# What is a formula that demonstrates the law of multiple proportions? Apr 21, 2014 You need two formulas to illustrate the Law of Multiple Proportions, for example, $\text{CO}$ and ${\text{CO}}_{2}$. #### Explanation: The Law of Multiple Proportions deals with elements that form more than one compound. It states that the masses of one element that combine with a fixed mass of the second element are in a small whole number ratio. For example, carbon and oxygen react to form two compounds. In the first compound (A), 42.9 g of $\text{C}$ react with 57.1 g of $\text{O}$. In the second compound (B), 27.3 g of $\text{C}$ react with 72.7 g of $\text{O}$. Let's calculate the mass of $\text{O}$ in each compound that reacts with 1 g ("a fixed mass") of $\text{C}$. In Compound A, $\text{mass of O" = 1 color(red)(cancel(color(black)("g C"))) × ("57.1 g O")/(42.9 color(red)(cancel(color(black)( "g C")))) = "1.33 g O}$ In Compound B, $\text{mass of O" = 1 color(red)(cancel(color(black)("g C"))) × ("72.7 g O")/(27.3 color(red)(cancel(color(black)("g C")))) = "2.66 g O}$ "Ratio" = ("Mass of O in B")/("Mass of O in A") = (2.66 color(red)(cancel(color(black)("g"))))/(1.33 color(red)(cancel(color(black)("g")))) = 2.00 ≈ 2 The masses of oxygen that combine with a fixed mass of carbon are in a 2:1 ratio. The whole-number ratio is consistent with the Law of Multiple Proportions. The simplest formulas that fit are $\text{CO}$ and ${\text{CO}}_{2}$.
Paul's Online Notes Home / Calculus I / Derivatives / Chain Rule Show Mobile Notice Show All Notes Hide All Notes Mobile Notice You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width. ### Section 3.9 : Chain Rule 30. The position of an object is given by $$s\left( t \right) = \sin \left( {3t} \right) - 2t + 4$$. Determine where in the interval $$\left[ {0,3} \right]$$ the object is moving to the right and moving to the left. Show All Steps Hide All Steps Start Solution We’ll first need the derivative because we know that the derivative will give us the velocity of the object. Here is the derivative. $\require{bbox} \bbox[2pt,border:1px solid black]{{s'\left( t \right) = 3\cos \left( {3t} \right) - 2}}$ Show Step 2 Next, we need to know where the object stops moving and so all we need to do is set the derivative equal to zero and solve. $3\cos \left( {3t} \right) - 2 = 0\hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\hspace{0.25in}\cos \left( {3t} \right) = \frac{2}{3}$ A quick calculator computation tells us that, $3t = {\cos ^{ - 1}}\left( {\frac{2}{3}} \right) = 0.8411$ Recalling our work in the Review chapter and a quick check on a unit circle we can see that either $$3t = 0.8411$$ or $$3t = 2\pi - 0.8411 = 5.4421$$ can be used for the second angle. Note that either will work, but we’ll use the second simply because it is the positive angle. Putting all of this together and dividing by 3 we can see that the derivative will be zero at, \begin{align}3t & = 0.8411 + 2\pi n\hspace{0.25in}{\mbox{and}}\hspace{0.25in}3t = 5.4421 + 2\pi n\hspace{0.25in}n = 0, \pm 1, \pm 2, \pm 3, \ldots \\ t & = 0.2804 + \frac{{2\pi n}}{3}\hspace{0.25in}{\mbox{and}}\hspace{0.25in}t = 1.8140 + \frac{{2\pi n}}{3}\hspace{0.25in}n = 0, \pm 1, \pm 2, \pm 3, \ldots \end{align} Finally, all we need to do is plug in some $$n$$’s to determine which solutions fall in the interval we are working on, $$\left[ {0,3} \right]$$. So, in the interval $$\left[ {0,3} \right]$$, the object stops moving at the following three points. $t = 0.2804,\,\,\,\,\,1.8140,\,\,\,\,2.3748$ Show Step 3 To get the answer to this problem all we need to know is where the derivative is positive (and hence the object is moving to the right) or negative (and hence the object is moving to the left). Because the derivative is continuous we know that the only place it can change sign is where the derivative is zero. So, as we did in this section a quick number line will give us the sign of the derivative for the various intervals. Here is the number line for this problem. From this we get the following moving right/moving left information. $\require{bbox} \bbox[2pt,border:1px solid black]{\begin{array}{ll}{\mbox{Moving Right :}} & \,\,0 \le t < 0.2804,\,\,\,\,1.8140 < t < 2.3748\\ {\mbox{Moving Left :}} & \,\,0.2804 < t < 1.8140,\,\,\,\,2.3748 < t \le 3\end{array}}$ Note that because we’ve only looked at what is happening in the interval $$\left[ {0,3} \right]$$ we can’t say anything about the moving right/moving left behavior of the object outside of this interval.
mathleaks.com mathleaks.com Start chapters home Start History history History expand_more Community Community expand_more {{ filterOption.label }} {{ item.displayTitle }} {{ item.subject.displayTitle }} arrow_forward {{ searchError }} search {{ courseTrack.displayTitle }} {{ printedBook.courseTrack.name }} {{ printedBook.name }} The general form of a quadratic equation is ax2+bx+c=0, where a, b, and c are real numbers, and a0. There are many ways to solve such equations. One way is by using the quadratic formula, which is derived by completing the square. ### Rule To solve a quadratic equation written in standard form ax2+bx+c=0, the Quadratic Formula can be used. In this formula, the discriminant b24ac determines the number of real solutions of the quadratic equation. ### Proof The Quadratic Formula can be derived by completing the square given the standard form of the quadratic equation ax2+bx+c=0. This method will be used to isolate the x-variable. To complete the square, there are five steps to follow. ### 1 Factor Out the Coefficient of x2 It is easier to complete the square when the quadratic expression is written in the form x2+bx+c. Therefore, the coefficient a should be factored out. Since the equation is quadratic, the coefficient a is not equal to 0. Therefore, both sides of the equation can be divided by a. ### 2 Identify the Constant Needed to Complete the Square The next step is to rewrite the equation by moving the existing constant to the right-hand side. To do so, will be subtracted from both sides of the equation. The constant needed to complete the square can now be identified by focusing on the x-term, while ignoring the rest. One way to find this constant is by squaring half the coefficient of the x-term, which in this case is Note that leaving the constant as a power makes the next steps easier to perform. ### 3 Complete the Square The square can now be completed by adding the found in Step 2 to both sides of the equation. The first three terms form a perfect square trinomial, which can be factored as the square of a binomial. The other two terms do not contain the variable x. Therefore, their value is constant. ### 4 Factor the Perfect Square Trinomial The perfect square trinomial can now be factored and rewritten as the square of a binomial. The process of completing the square is now finished. ### 5 Simplify the Equation Finally, the right-hand side of the equation can be simplified. Simplify right-hand side Now, there is only one x-term. To isolate x, it is necessary to take square roots on both sides of the equation. This results in both a positive and a negative term on the right-hand side. Now, the equation can be further simplified to isolate x. Solve for x Finally, the Quadratic Formula has been obtained. fullscreen Exercise Use the quadratic formula to solve the equation. 2x24x16=0 Show Solution Solution Notice that the given equation is written in standard form. Thus, it can be solved using the quadratic formula. To begin, it's necessary to note the values of a,b, and c. It can be seen that To solve the equation, we can substitute these values into the formula and simplify. Remember, since a quadratic function can have 0,1, or 2 roots, this equation can have 0,1, or 2 real solutions. 2x24x16=0 The solutions to the equation are x=-2 and x=4. ## Discriminant In the Quadratic Formula, the expression b24ac, which is under the radical symbol, is called the discriminant. A quadratic equation can have two, one, or no real solutions. Since the discriminant is under the radical symbol, its value determines the number of real solutions of a quadratic equation. Value of the Discriminant Number of Real Solutions b24ac>0 2 b24ac=0 1 b24ac<0 0 Moreover, the discriminant determines the number of x-intercepts of the graph of the related quadratic function. fullscreen Exercise Determine the number of real solutions to the equations without solving them. Show Solution Solution It's possible to use the discriminant of the quadratic formula to determine the number of solutions a quadratic equation has. We'll first focus on x22x+9=0. Since the equation is written in standard form, we can see that a=1,b=-2, and c=9. We can substitute these values into the discriminant and simplify. b24ac (-2)2419 436 -32 Since -32<0, the first quadratic equation has 0 real solutions. We can perform the same process on the second equation where a=1,b=-4, and c=4. b24ac (-4)2414 1616 0 Since 0=0, the second quadratic equation has 1 real solution.
Question Solve the equations: x â€“ y = âˆ’1 and 2y+3x = 12 using a graphical method. Hint- We should be remembering that the graph of a line represents every point that is a possible solution for the equation of that line. So when the graphs of two equations cross, the point of intersection lies on both lines, meaning that it is a possible solution for both equations. x âˆ’ y = âˆ’1 --- (1) 2y+3x = 12 --- (2) From equation (1) assume the value of x and y to satisfy the equation to zero. xâˆ’y+1=0 ---- (3) Put x=2,y=3 in equation (3) 2âˆ’3+1=0 âˆ’1+1=0 0=0 , Hence Point (2,3) satisfy the line equation so we can say that Point (2.3) lies on straight line x âˆ’ y = âˆ’1 Again put x=4,y=5 in equation (3) 4âˆ’5+1=0 âˆ’1+1=0 0=0 Hence Point (4,5) satisfy the line equation so we can say that Point (4,5) lies on straight line x âˆ’ y = âˆ’1 Now plotting (2,3), (4,5) on graph and joining them, we get a straight line. Shown in below figure From equation (2) assume the value of x and y to satisfy the equation to zero. 2y + 3xâˆ’12=0 --- (4) Put x=2,y=3 in equation (4) 2(3)+3(2)âˆ’12=0 6+6âˆ’12=0 12âˆ’12=0 0=0 Hence Point (2,3) satisfy the line equation so we can say that Point (2.3) lies on straight line 2y + 3xâˆ’12=0 Again put x=4,y=0 in equation (4) 2(0)+3(4)âˆ’12=0 0+12âˆ’12=0 12âˆ’12=0 0=0 Hence Point (4,0) satisfy the line equation so we can say that Point (2.3) lies on straight line 2y + 3xâˆ’12=0 Plotting (2,3),(4,0) on graph and joining them, we get another straight line shown in below figure These lines intersect at the point (2, 3) and therefore the solution of the equation is x=2, y=3 (see fig.) because the graph of a line represents every point that is a possible solution for the equation of that line. So (2,3) is the possible solution of equations: x â€“ y = âˆ’1 and 2y+3x = 12.
Mean and Variance Go back to 'Probability' It is in most cases very useful to talk about the mean of a random variable X. For example, in the experiment above of four tosses of a coin, someone might want to know the average number of heads obtained. Now the reader may wonder about what meaning to attach to the phrase “average number of heads”. After all, we are doing the experiment only once and we’ll obtain a particular value for the number of Heads, say 0 or 1 or 2 or 3 or 4, but what is then this “average number of heads”? By the average number of heads we mean this: repeat the experiment an indefinitely large number of times. Each time you’ll get a certain number of Heads. Take the average of the number of Heads obtained in each repetition of the experiment. For example, if in 5 repetitions of this experiment, you obtain 2, 2, 3, 1, 1 Heads respectively, the average number of Heads would be (2 + 2 + 3 + 1 + 1) / 5 = 1.9. This is not a natural number, which shouldn’t worry you since it is an average over the 5 repetitions. To calculate the true average, you have to repeat the experiment an indefinitely large number of times. An alert reader might have realized that the average value of a RV is easily calculable through its PD. For example, let us calculate the true average number of heads in the experiment of Example - 18. The PD is reproduced below: X 0 1 2 3 4 P(X) \begin{align}\frac{1}{16}\end{align} \begin{align}\frac{1}{4}\end{align} \begin{align}\frac{3}{8}\end{align} \begin{align}\frac{1}{4}\end{align} \begin{align}\frac{1}{16}\end{align} Thus, for example,\begin{align}P(X=1)=\frac{1}{4}\end{align} , which means that if the experiment is repeated an indefinitely large number of times, we’ll obtain Heads exactly once, (about) \begin{align}{{\frac{1}{4}}^{th}}\end{align} of the time. Similarly, (about) \begin{align}{{\frac{3}{8}}^{th}}\end{align}of the time, Heads will be obtained exactly twice, and so on. Let us denote the number of repetition of the experiment by N, where $$N\to \infty$$, Thus, the average number of Heads per repetition would be (< > denotes average) \begin{align}& <\text{Heads}>\ =\frac{\text{Total no}\text{. of Heads in N repetitions}}{\text{N}} \\ &\qquad\qquad\quad \;\;=\frac{0\times \frac{N}{16}+1\times \frac{N}{4}+2\times \frac{3N}{8}+3\times \frac{N}{4}+4\times \frac{N}{16}}{N} \\ & \qquad\qquad\quad \;\;=0\times \frac{1}{16}+1\times \frac{1}{4}+2\times \frac{3}{8}+3\times \frac{1}{4}+4\times \frac{1}{16} \\ & \qquad\qquad\quad\;=\sum \left( \text{Value of the RV} \right)\text{ }\times \text{ }\left( \text{Corresponding Probability of this value} \right) \\ \end{align} Thus, we see that if a RV X has possible values x1, x2, ....xn with respective probabilities p1, p2, ...., pn, the mean of X, denote by $$\left\langle X \right\rangle$$ , is simply given by $\left\langle X \right\rangle =\sum\limits_{i=1}^{n}{{{x}_{i}}\,{{p}_{i}}}\qquad\qquad\qquad.......\left( 1 \right)$ As another example, recall the experiment of rolling two dice where the RV X was the sum of the numbers on the two dice. The PD of X is given in the table on Page - 42, and the average value of X is \begin{align}& \left\langle X \right\rangle \ =2\times \frac{1}{36}+3\times \frac{1}{18}+4\times \frac{1}{12}+5\times \frac{1}{9}+6\times \frac{5}{36}+7\times \frac{1}{6} \\ & \quad\qquad+8\times \frac{5}{36}+9\times \frac{1}{9}+10\times \frac{1}{12}+11\times \frac{1}{18}+12\times \frac{1}{36} \\ &\quad\quad =7 \\ \end{align} The average value is also called the expected value, which signifies that it is what we can expect to obtain by averaging the RV’s value over a large number of repetitions of the experiment. Note that the value itself may not be expected in the general sense - the “expected value” itself may be unlikely or even impossible. For example, in the rolling of a fair die, the expected value of the number that shows up is 3.5 (verify), which in itself can never be a possible outcome. Thus, you must take care while interpreting the expected value - see it as an average of the RV’s values when the experiment is repeated indefinitely. Another quantity of great significance associated with any RV X is its variance, denoted by Var(X). To understand this properly, consider two RVs X1 and X2 and their PDs shown in graphical form below. Both the RVs have an expected value of 3 (verify), but it is obvious that there is a significant difference between the two distributions. What is this difference? Can you put it into words? And more importantly, can you quantify it? It turns out that we can, in a way very simple to understand. The ‘data’ or the PD of X1 is more widely spread than that of X2. This is what is obvious visually, but we must now assign a numerical value to this spread. So what we’ll do is measure the spread of the PD about the mean of the RV. For both X1 and X2, the mean is 3, but the PD of X1 is spread more about 3 than that of X2. We now quantity the spread in X1. Observe that the various value of $$X-\left\langle X \right\rangle$$ tell us how far the corresponding values of X are from the mean (which is fixed). One way that may come to your mind to measure the spread is sum all these distances, i.e. $'Spread'\text{ }={{\sum\limits_{\begin{smallmatrix} \text{For all} \\ \text{values } \\ \text{of }X \end{smallmatrix}}{\left(\text{X}-\left\langle X \right\rangle \right)}}}$ However, a little thinking should immediately make it obvious to you that the right hand side is always 0, because the data is spread in such a way around the mean that positive contributions to the sum from those X values greater than $$\left\langle X \right\rangle$$ and negative contributions from those X values smaller than $$\left\langle X \right\rangle$$ exactly cancel out. Work it out yourself. So what we do is use the sum of the squares of these distances: $'Spread'\text{ }={{\sum\limits_{\begin{smallmatrix} \text{For all} \\ \text{values } \\ \text{of }X \end{smallmatrix}}{\left(\text{X}-\left\langle X \right\rangle \right)}}^{2}}$ However, there is still something missing. To understand what consider the following PD: Although the PD seems visually widespread here, the probabilities of those X values far from the mean are extremely low, which means that their contribution to the spread must take into account how probable they are and so on. This is simply accomplished by multiplying the value of $${{\left( X-\left\langle X \right\rangle \right)}^{2}}$$ with the probability of the corresponding value of X. Thus, if X can take the values x1, x2, ......., xn with probabilities p1, p2, ..., pn, the spread in the PD of X can be appropriately represented by $Spread\text{ }={{\sum\limits_{i\ =\ 1}^{n}{\left( {{x}_{i}}-\left\langle X \right\rangle \right)}}^{2}}{{p}_{i}}$ This definition of spread is termed the variance of X, and is denoted by Var(X). Statisticians defined another quantify for spread, called the standard deviation, denote by $$\sigma _{x}^{2}$$ , and related to the variance by $Var(X)=\sigma _{X}^{2}$ Note that the expected value of X was $\left\langle X \right\rangle =\ \sum\limits_{i=1}^{n}{{{x}_{i}}{{p}_{i}}}$ Similarly, variance is nothing but the expected value of $${{({{x}_{i}}-\left\langle X \right\rangle )}^{2}}$$ $Var\left( X \right)=\left\langle {{\left( {{x}_{i}}-\left\langle X \right\rangle \right)}^{2}} \right\rangle \ =\ \sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}-\left\langle X \right\rangle \right)}^{2}}{{p}_{i}}}$ Coming back to Fig-16, the variance in X1 is \begin{align}& Var({{X}_{1}})=\ {{(1-3)}^{2}}\cdot \frac{1}{10}+{{(2-3)}^{2}}\cdot \frac{1}{5}+{{(3-3)}^{2}}\cdot \frac{2}{5} \\ &\qquad\qquad +{{(4-3)}^{2}}\cdot \frac{1}{5}+{{(5-3)}^{2}}\cdot \frac{1}{10} \\ & \qquad\quad\;\;=\frac{4}{10}+\frac{1}{5}+0+\frac{1}{5}+\frac{4}{10} \\ & \qquad\quad\;\; =1.2 \\ \end{align} Similarly, the variance in X2 is \begin{align} & Var({{X}_{2}})={{(2-3)}^{2}}\cdot \frac{1}{4}+{{(3-3)}^{2}}\cdot \frac{1}{2}+{{(4-3)}^{2}}\cdot \frac{1}{4} \\ & \qquad\quad\;\;=\frac{1}{4}+0+\frac{1}{4} \\ &\qquad\quad\;\; =0.5 \\ \end{align} which confirms our visual observation that the PD of X1 is more widely spread than of X2, because Var(X1) > var(X2). Example – 19 Show that $$\text{Var}(X)=\left\langle {{X}^{2}} \right\rangle -{{\left\langle X \right\rangle }^{2}}$$ Solution: \begin{align}& \text{Var}(X)=\left\langle {{\left( X-\left\langle X \right\rangle \right)}^{2}} \right\rangle \ \\ & \qquad\quad=\sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}-\left\langle X \right\rangle \right)}^{2}}{{p}_{i}}}\left\{ \text{where the symbols have their usual meanings} \right\} \\ &\qquad\quad =\sum\limits_{i=1}^{n}{\left( x_{i}^{2}+{{\left\langle X \right\rangle }^{2}}-2{{x}_{i}}\left\langle X \right\rangle \right){{p}_{i}}} \\ & \qquad\quad=\sum\limits_{i=1}^{n}{{{p}_{i}}x_{i}^{2}}+\sum\limits_{i=1}^{n}{{{p}_{i}}{{\left\langle X \right\rangle }^{2}}}-2\ \sum\limits_{i=1}^{n}{{{x}_{i}}{{p}_{i}}\left\langle X \right\rangle } \\ \end{align} Since $$\left\langle X \right\rangle \,\text{and}\,{{\left\langle X \right\rangle }^{2}}$$ are constants, they can be taken out of the summation in the second and third terms. Also, note that $\sum\limits_{i=1}^{n}{{{p}_{i}}x_{i}^{2}}=\left\langle {{X}^{2}} \right\rangle ,\ \ \sum\limits_{i=1}^{n}{{{p}_{i}}=1},\ \ \sum\limits_{i=1}^{n}{{{p}_{i}}{{x}_{i}}}=\left\langle X \right\rangle \$ so that, \begin{align} & \text{Var}(X)=\ \left\langle {{X}^{2}} \right\rangle +{{\left\langle X \right\rangle }^{2}}-2{{\left\langle X \right\rangle }^{2}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\,\left\langle {{X}^{2}} \right\rangle -{{\left\langle X \right\rangle }^{2}} \\ \end{align} which proves the assertion. The relation is important and useful since it gives us the variance directly in terms of $$\left\langle X \right\rangle \,\,and\,\,\left\langle {{X}^{2}} \right\rangle$$. You are urged to try using this relation to calculate variance in the examples of variance we’ve discussed earlier. Example – 20 Two cards are drawn simultaneously from a well-shuffled deck of 52 cards. Find the mean and variance of the number of kings. Solution: The number of kings is the random variable here. Call it X. The values that X can take are 0, 1, 2. The probabilities of the various values are easily calculated: \begin{align}& P(X=0)=\frac{^{48}{{C}_{2}}}{^{52}{{C}_{2}}}=\frac{188}{221} \\ & P(X=1)=\frac{^{4}{{C}_{1}}\times {{\,}^{48}}{{C}_{1}}}{^{52}{{C}_{2}}}=\frac{32}{221} \\ & P(X=2)=\frac{^{4}{{C}_{2}}}{^{52}{{C}_{2}}}=\frac{1}{221} \\ \end{align} The PD of X is therefore X $$0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2$$ P(X) $$\frac{188}{221}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{32}{221}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{1}{221}$$ The mean is simply \begin{align} &\; \left\langle X \right\rangle \ =0\times \frac{188}{221}+1\times \frac{32}{221}+2\times \frac{1}{221} \\ & \qquad\; =\frac{34}{221} \\ \end{align} To calculate the variance, we first calculate $$\left\langle {{X}^{2}} \right\rangle$$ : \begin{align}& \left\langle {{X}^{2}} \right\rangle \ =\sum\limits_{i=1}^{n}{x_{i}^{2}}{{p}_{i}} \\ & \qquad\quad={{0}^{2}}\times \frac{188}{221}+{{1}^{2}}\times \frac{32}{221}+{{2}^{2}}\times \frac{1}{221}=\frac{36}{221} \\ \end{align} Thus, the variance is \begin{align} & \text{Var}(X)=\ \left\langle {{X}^{2}} \right\rangle -{{\left\langle X \right\rangle }^{2}} \\ & \quad\qquad\;=\frac{36}{221}-{{\left( \frac{34}{221} \right)}^{2}} \\ & \quad\qquad\; =\frac{6800}{48841} \\ \end{align}
# Thread: log deriv. 1. ## log deriv. Use logarithmic differentiation to find the derivative of the function. y = x^sinx How do you do this? Help please. 2. Originally Posted by kwivo Use logarithmic differentiation to find the derivative of the function. y = x^sinx How do you do this? Help please. begin by taking the log of both sides. $y = x^{\sin x}$ $\Rightarrow \ln y = \ln x^{\sin x}$ $\Rightarrow \ln y = \sin x \ln x$ differentiating implicitly with repsect to x, we get: $\frac {y'}y = \cos x \ln x + \frac {\sin x}x$ $\Rightarrow y' = y \left( \cos x \ln x + \frac {\sin x}x\right) = x^{\sin x}\left( \cos x \ln x + \frac {\sin x}x\right)$ 3. Hello, kwivo! Use logarithmic differentiation to differentiate: . $y \:=\:x^{\sin x}$ We have: . $y \;=\;x^{\sin x}$ Take logs: . $\ln y \:=\:\ln\left(x^{\sin x}\right)\quad\Rightarrow\quad\ln y \:=\:\sin x\cdot\ln x$ Differentiate implicitly: . $\frac{1}{y}\cdot\frac{dy}{dx} \;=\;\sin x\cdot\frac{1}{x} + \cos x\cdot\ln x$ Multiply by $y\!:\;\;\frac{dy}{dx}\;=\;y\left[\frac{\sin x}{x} + \cos x\!\cdot\!\ln x\right]$ Replace $y\!:\;\;\frac{dy}{dx}\;=\;x^{\sin x}\left[\frac{\sin x}{x} + \cos x\!\cdot\!\ln x\right]$ 4. Originally Posted by kwivo Use logarithmic differentiation to find the derivative of the function. y = x^sinx I'll use the never well weighed trick $a=e^{\ln a},\,\forall a>0,$ then $y=e^{\sin x\cdot\ln x}$ Now it remains to compute $(\sin x\cdot\ln x)',$ and yields the desired.
# Circular arc A circular arc is the arc of a circle between a pair of distinct points. If the two points are not directly opposite each other, one of these arcs, the minor arc, subtends an angle at the center of the circle that is less than π radians (180 degrees); and the other arc, the major arc, subtends an angle greater than π radians. The arc of a circle is defined as the part or segment of the circumference of a circle. A straight line that connects the two ends of the arc is known as a chord of a circle. If the length of an arc is exactly half of the circle, it is known as a semicircular arc. ## Length The length (more precisely, arc length) of an arc of a circle with radius r and subtending an angle θ (measured in radians) with the circle center — i.e., the central angle — is ${\displaystyle L=\theta r.}$ This is because ${\displaystyle {\frac {L}{\mathrm {circumference} }}={\frac {\theta }{2\pi }}.}$ Substituting in the circumference ${\displaystyle {\frac {L}{2\pi r}}={\frac {\theta }{2\pi }},}$ and, with α being the same angle measured in degrees, since θ = α/180π, the arc length equals ${\displaystyle L={\frac {\alpha \pi r}{180}}.}$ A practical way to determine the length of an arc in a circle is to plot two lines from the arc's endpoints to the center of the circle, measure the angle where the two lines meet the center, then solve for L by cross-multiplying the statement: measure of angle in degrees/360° = L/circumference. For example, if the measure of the angle is 60 degrees and the circumference is 24 inches, then {\displaystyle {\begin{aligned}{\frac {60}{360}}&={\frac {L}{24}}\\[6pt]360L&=1440\\[6pt]L&=4.\end{aligned}}} This is so because the circumference of a circle and the degrees of a circle, of which there are always 360, are directly proportional. The upper half of a circle can be parameterized as ${\displaystyle y={\sqrt {r^{2}-x^{2}}}.}$ Then the arc length from ${\displaystyle x=a}$ to ${\displaystyle x=b}$ is ${\displaystyle L=r{\Big [}\arcsin \left({\frac {x}{r}}\right){\Big ]}_{a}^{b}.}$ ## Sector area The area of the sector formed by an arc and the center of a circle (bounded by the arc and the two radii drawn to its endpoints) is ${\displaystyle A={\frac {r^{2}\theta }{2}}.}$ The area A has the same proportion to the circle area as the angle θ to a full circle: ${\displaystyle {\frac {A}{\pi r^{2}}}={\frac {\theta }{2\pi }}.}$ We can cancel π on both sides: ${\displaystyle {\frac {A}{r^{2}}}={\frac {\theta }{2}}.}$ By multiplying both sides by r2, we get the final result: ${\displaystyle A={\frac {1}{2}}r^{2}\theta .}$ Using the conversion described above, we find that the area of the sector for a central angle measured in degrees is ${\displaystyle A={\frac {\alpha }{360}}\pi r^{2}.}$ ## Segment area The area of the shape bounded by the arc and the straight line between its two end points is ${\displaystyle {\frac {1}{2}}r^{2}(\theta -\sin \theta ).}$ To get the area of the arc segment, we need to subtract the area of the triangle, determined by the circle's center and the two end points of the arc, from the area ${\displaystyle A}$. See Circular segment for details. Using the intersecting chords theorem (also known as power of a point or secant tangent theorem) it is possible to calculate the radius r of a circle given the height H and the width W of an arc: Consider the chord with the same endpoints as the arc. Its perpendicular bisector is another chord, which is a diameter of the circle. The length of the first chord is W, and it is divided by the bisector into two equal halves, each with length W/2. The total length of the diameter is 2r, and it is divided into two parts by the first chord. The length of one part is the sagitta of the arc, H, and the other part is the remainder of the diameter, with length 2r − H. Applying the intersecting chords theorem to these two chords produces ${\displaystyle H(2r-H)=\left({\frac {W}{2}}\right)^{2},}$ whence ${\displaystyle 2r-H={\frac {W^{2}}{4H}},}$ so ${\displaystyle r={\frac {W^{2}}{8H}}+{\frac {H}{2}}.}$ The arc, chord, and sagitta derive their names respectively from the Latin words for bow, bowstring, and arrow.
This is also sometimes called “subtraction using a blank numberline” and I’ve even heard it called “that nonsensical modern method”. This latter is a real misnomer since it is neither nonsensical nor modern. In fact it’s a method that dates back to before I was born, in days before we had calculators and electronic tills. It’s also a really useful method involving counting forwards, which is always easier than counting backwards – even for maths geniuses! Let’s return to those old-fashioned little shops. I buy some sweets for 24p and hand over a £5. To work out my change, the shopkeeper needs to calculate £5 – 24p. Now she could count £4.99, £4.98, £4.97 until she had subtracted 24p, but what you actually would have heard is this: 24 and 6 makes 30, and 20 makes 50 and another 50 makes £1. 2, 3, 4, £5. And while doing this they counted the change (£4.76) into your hand. This is the method that schools have returned to. To begin with, Children use a “blank numberline” – that is, a line that they can write the numbers on themselves. They then write the lower number at one end, and see what they need to add to make the higher number. Here’s an example: 96-38 The children first of all use their knowledge of number bonds to add to the next 10 (38 + 2 = 40). They then use their ability to add multiples of 10. In the example above the child has done 40+10 = 50 and then 50 + 40 = 90. They may have been able to see straight away 40 + 50 = 90 and done this as one step, or they may have needed to do 40 + 10 = 50, 50 + 10 = 60, 60 + 10 = 70 and so on up to 90. The method isn’t about having to complete it in a certain number of steps, it’s about each child breaking it down into the smallest number of steps that they can manage. When they reach the multiple of 10 before the higher number, they add on whatever units are needed to make the higher number, in this case it was +6 to make 96. Finally, they add up all the numbers they added on to find the answer: 2+10+40+6 = 58 so 96 – 38 = 58. When they are confident with this, they move on to jotting down only the numbers they are adding on, and they keep the tally in their head, until eventually they develop their working memory enough to hold all of the numbers in their head and write down just the answer. ## Learning Chinese As readers of this blog will know, I love learning new things. Last summer I spotted an advert for a course in Chinese for primary school teachers, and as MFL (modern foreign languages) is my specialist subject, I decided to sign up. Throwing myself in at the deep end, I promised my new school that I would set up a lunchtime Chinese club, so I had to make sure I really did learn some! I must confess, I was a bit worried. I mean – Chinese is really difficult, right? It’s doesn’t even have an alphabet, just thousands of characters. But it actually turned out to be a lot easier than I imagined. Obviously, it takes years to learn to speak a language fluently, so I have only learnt the basics, but this is what I discovered: – It’s a subject-verb-object language, so the word order is the same as English. This already makes it easier than some languages. – The verbs don’t conjugate (i.e. there are no different endings depending on who is doing it – like he lives, they live in English, or il habite, ils habitent in French. – There are no articles (English has ‘a’ and ‘the’; French has un, une, des, le, la and les; Spanish has un, una, unos, unas, el, la, los and las; Chinese has nothing) – There are no tenses. In Chinese, the verb remains exactly the same and you know whether it’s past, present or future from the context. This simplicity actually makes it ideal for primary school children to learn. Like any language, it does have its peculiarities and difficulties, such as the tones (the way your voice goes up or down for certain words) but this is no more challenging than getting children to understand the concept of nouns having genders (Chinese doesn’t have those) or that ‘you are’ might be ‘tu es’ but might be ‘vous êtes’ depending on who and/or how many people you are talking to. Of course the characters are tricky but the children in my club really enjoy drawing and practising them, and they have the advantage that children are not influenced by how the word is written, so in general their pronunciation is better right from the start. The fact that the language isn’t written with an English alphabet doesn’t faze them at all. (In fact, I also run an Ancient Greek club and the children there are also fascinated by the fact that language can be written using different symbols.) We all enjoy making up little stories to help remember the characters. On the course I did, we learned a little about how the characters are made up, with radicals giving an indication of meaning and a phonetic element indicating pronunciation. And there is far more vocabulary in some topic areas. For example, English has mum, dad, brother, sister, grandma, granddad, while Chinese has different words depending on whether it’s an older or younger brother, a maternal or paternal grandmother etc. But for the moment the primary aged children I am teaching only need to learn the ones they require for their own family. The children and I are really enjoying learning together, and although I will never be fluent in Mandarin, you never know – one of the children I am teaching may be inspired to study it further and become fluent in the future.
# Class 9 Maths Chapter 11 Constructions NCERT Solutions for Class 9 Maths Chapter 11 Constructions. Exercise 11.1 Question 1. Construct an angle of 90° at the initial point of a given ray and justify the construction. Solution: Steprf of Construction: Step I : Draw . Step II : Taking O as centre and having a suitable radius, draw a semicircle, which cuts at B. Step III : Keeping the radius same, divide the semicircle into three equal parts such that . Step IV : Draw and . Step V : Draw , the bisector of ∠COD. Thus, ∠AOF = 90° Justification: ∵ O is the centre of the semicircle and it is divided into 3 equal parts. ⇒ ∠BOC = ∠COD = ∠DOE [Equal chords subtend equal angles at the centre] And, ∠BOC + ∠COD + ∠DOE = 180° ⇒ ∠BOC + ∠BOC + ∠BOC = 180° ⇒ 3∠BOC = 180° ⇒ ∠BOC = 60° Similarly, ∠COD = 60° and ∠DOE = 60° is the bisector of ∠COD ∴ ∠COF = ∠COD = (60°) = 30° Now, ∠BOC + ∠COF = 60° + 30° ⇒ ∠BOF = 90° or ∠AOF = 90° Question 2. Construct an angle of 45° at the initial point of a given ray and justify the construction. Solution: Steps of Construction: Stept I : Draw . Step II : Taking O as centre and with a suitable radius, draw a semicircle such that it intersects . at B. Step III : Taking B as centre and keeping the same radius, cut the semicircle at C. Now, taking C as centre and keeping the same radius, cut the semicircle at D and similarly, cut at E, such that Step IV : Draw and . Step V : Draw , the angle bisector of ∠BOC. Step VI : Draw , the ajngle bisector of ∠FOC. Thus, ∠BOG = 45° or ∠AOG = 45° Justification: ∴ ∠BOC = ∠COD = ∠DOE [Equal chords subtend equal angles at the centre] Since, ∠BOC + ∠COD + ∠DOE = 180° ⇒ ∠BOC = 60° is the bisector of ∠BOC. ∴ ∠COF = ∠BOC = (60°) = 30° …(1) Also, is the bisector of ∠COF. ∠FOG = ∠COF = (30°) = 15° …(2) Adding (1) and (2), we get ∠COF + ∠FOG = 30° + 15° = 45° ⇒ ∠BOF + ∠FOG = 45° [∵ ∠COF = ∠BOF] ⇒ ∠BOG = 45° Question 3. Construct the angles of the following measurements (i) 30° (ii) 22 (iii) 15° Solution: (i) Angle of 30° Steps of Construction: Step I : Draw . Step II : With O as centre and having a suitable radius, draw an arc cutting at B. Step III : With centre at B and keeping the same radius as above, draw an arc to cut the previous arc at C. Step IV : Join which gives ∠BOC = 60°. Step V : Draw , bisector of ∠BOC, such that ∠BOD = ∠BOC = (60°) = 30° Thus, ∠BOD = 30° or ∠AOD = 30° (ii) Angle of 22 Steps of Construction: Step I : Draw . Step II : Construct ∠AOB = 90° Step III : Draw , the bisector of ∠AOB, such that ∠AOC = ∠AOB = (90°) = 45° Step IV : Now, draw OD, the bisector of ∠AOC, such that ∠AOD = ∠AOC = (45°) = 22 Thus, ∠AOD = 22 (iii) Angle of 15° Steps of Construction: Step I : Draw . Step II : Construct ∠AOB = 60°. Step III : Draw OC, the bisector of ∠AOB, such that ∠AOC = ∠AOB = (60°) = 30° i.e., ∠AOC = 30° Step IV : Draw OD, the bisector of ∠AOC such that ∠AOD = ∠AOC = (30°) = 15° Thus, ∠AOD = 15° Question 4. Construct the following angles and verify by measuring them by a protractor (i) 75° (ii) 105° (iii) 135° Solution: Step I : Draw . Step II : With O as centre and having a suitable radius, draw an arc which cuts at B. Step III : With centre B and keeping the same radius, mark a point C on the previous arc. Step IV : With centre C and having the same radius, mark another point D on the arc of step II. Step V : Join and , which gives ∠COD = 60° = ∠BOC. Step VI : Draw , the bisector of ∠COD, such that ∠COP = ∠COD = (60°) = 30°. Step VII: Draw , the bisector of ∠COP, such that ∠COQ = ∠COP = (30°) = 15°. Thus, ∠BOQ = 60° + 15° = 75°∠AOQ = 75° (ii) Steps of Construction: Step I : Draw . Step II : With centre O and having a suitable radius, draw an arc which cuts at B. Step III : With centre B and keeping the same radius, mark a point C on the previous arc. Step IV : With centre C and having the same radius, mark another point D on the arc drawn in step II. Step V : Draw OP, the bisector of CD which cuts CD at E such that ∠BOP = 90°. Step VI : Draw , the bisector of such that ∠POQ = 15° Thus, ∠AOQ = 90° + 15° = 105° (iii) Steps of Construction: Step I : Draw . Step II : With centre O and having a suitable radius, draw an arc which cuts at A Step III : Keeping the same radius and starting from A, mark points Q, R and S on the arc of step II such that . StepIV :Draw , thebisector of which cuts the arc at T. Step V : Draw , the bisector of . Thus, ∠POQ = 135° Question 5. Construct an equilateral triangle, given its side and justify the construction. Solution: pt us construct an equilateral triangle, each of whose side = 3 cm(say). Steps of Construction: Step I : Draw . Step II : Taking O as centre and radius equal to 3 cm, draw an arc to cut at B such that OB = 3 cm Step III : Taking B as centre and radius equal to OB, draw an arc to intersect the previous arc at C. Step IV : Join OC and BC. Thus, ∆OBC is the required equilateral triangle. Justification: ∵ The arcs and are drawn with the same radius. = ⇒ OC = BC [Chords corresponding to equal arcs are equal] ∵ OC = OB = BC ∴ OBC is an equilateral triangle. Exercise 11.2 Question 1. Construct a ∆ ABC in which BC = 7 cm, ∠B = 75° and AB + AC = 13 cm. Solution: Steps of Construction: Step I : Draw . Step II : Along , cut off a line segment BC = 7 cm. Step III : At B, construct ∠CBY = 75° Step IV : From , cut off BD = 13 cm (= AB + AC) Step V : Join DC. Step VI : Draw a perpendicular bisector of CD which meets BD at A. Step VII: Join AC. Thus, ∆ABC is the required triangle. Question 2. Construct a ABC in which BC = 8 cm, ∠B = 45° and AB – AC = 35 cm. Solution: Steps of Construction: Step I : Draw . Step II : Along , cut off a line segment BC = 8 cm. Step III : At B, construct ∠CBY = 45° Step IV : From , cut off BD = 3.5 cm (= AB – AC) Step V : Join DC. Step VI : Draw PQ, perpendicular bisector of DC, which intersects at A. Step VII: Join AC. Thus, ∆ABC is the required triangle. Question 3. Construct a ∆ ABC in which QR = 6 cm, ∠Q = 60° and PR – PQ = 2 cm. Solution: Steps of Construction: Step I : Draw . Step II : Along , cut off a line segment QR = 6 cm. Step III : Construct a line YQY’ such that ∠RQY = 60°. Step IV : Cut off QS = 2 cm (= PR – PQ) on QY’. Step V : Join SR. Step VI : Draw MN, perpendicular bisector of SR, which intersects QY at P. Step VII: Join PR. Thus, ∆PQR is the required triangle. Question 4. Construct a ∆ XYZ in which ∠Y = 30°, ∠Y = 90° and XY + YZ + ZX = 11 cm. Solution: Steps of Construction: Step I : Draw a line segment AB = 11 cm = (XY+YZ + ZX) Step II : Construct ∠BAP = 30° Step III : Construct ∠ABQ = 90° Step IV : Draw AR, the bisector of ∠BAP. Step V : Draw BS, the bisector of ∠ABQ. Let AR and BS intersect at X. Step VI : Draw perpendicular bisector of , which intersects AB at Y. Step VII: Draw perpendicular bisector of , which intersects AB at Z. Step VIII: Join XY and XZ. Thus, ∆XYZ is the required triangle. Question 5. Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm. Solution: Steps of Construction: Step I : Draw BC = 12 cm. Step II : At B, construct ∠CBY = 90°. Step III : Along , cut off a line segment BX = 18 cm. Step IV : Join CX. Step V : Draw PQ, perpendicular bisector of CX, which meets BX at A. Step VI : Join AC. Thus, ∆ABC is the required triangle.
# FREE 5th Grade MEAP Math Practice Test Welcome to our FREE 5th Grade MEAP Math practice test, with answer key and answer explanations. This practice test’s realistic format and high-quality practice questions can help your student succeed on the 5th Grade MEAP Math test. Not only does the test closely match what students will see on the real MEAP, but it also comes with detailed answer explanations. For this practice test, we’ve selected 20 real questions from past exams for your student’s MEAP Practice test. Your student will have the chance to try out the most common 5th Grade MEAP Math questions. For every question, there is an in-depth explanation of how to solve the question and how to avoid mistakes next time. Use our free 5th Grade MEAP Math practice tests and study resources (updated for 2021) to help your students ace the 5th Grade MEAP Math test! Make sure to follow some of the related links at the bottom of this post to get a better idea of what kind of mathematics questions students need to practice. ## 10 Sample 5th Grade MEAP Math Practice Questions 1- How long is the line segment shown on the number line below? A. 6 B. 7 C. 8 D. 9 2- If a rectangle is 30 feet by 45 feet, what is its area? A. 1350 B. 1250 C. 1000 D. 870 3- If a vehicle is driven 32 miles on Monday, 35 miles on Tuesday, and 29 miles on Wednesday, what is the average number of miles driven each day? A. 32 B. 33 C. 34 D. 35 4- Peter traveled 120 miles in 4 hours and Jason traveled 160 miles in 8 hours. What is the ratio of the average speed of Peter to average speed of Jason? A. 3 : 2 B. 2 : 3 C. 5 : 9 D. 5 : 6 5- If $$x=- 8$$, which equation is true? A. $$x(2x-4)=120$$ B. $$8 (4-x)=96$$ C. $$2 (4x+6)=79$$ D. $$6x-2=-46$$ 6- A circle has a diameter of 8 inches. What is its approximate circumference? ($$π = 3.14$$) A. 6.28 inches B. 25.12 inches C. 34.85 inches D. 35.12 inches 7- A woman owns a dog walking business. If 3 workers can walk 9 dogs, how many dogs can 5 workers walk? A. 13 B. 15 C. 17 D. 19 8- What are the coordinates of the intersection of $$x$$–axis and the $$y$$–axis on a coordinate plane? A. $$(5, 5)$$ B. $$(1, 1)$$ C. $$(0, 0)$$ D. $$(0, 11)$$ 9- Jack added 19 to the product of 16 and 26. What is this sum? A. 61 B. 330 C. 435 D. 135 10- Joe makes $4.75 per hour at his work. If he works 8 hours, how much money will he earn? A.$32.00 B. $34.75 C.$36.50 D. $38.00 11- Which of the following is an obtuse angle? A. 89$$^\circ$$ B. 55$$^\circ$$ C. 143$$^\circ$$ D. 235$$^\circ$$ 12- What is the value of $$6 – 3 \frac{4}{9}$$? A. $$\frac{23}{9}$$ B. $$3\frac{4}{9}$$ C. $$-\frac{1}{9}$$ D. $$\frac{42}{9}$$ 13- The bride and groom invited 220 guests for their wedding. 190 guests arrived. What percent of the guest list was not present? A. $$90\%$$ B. $$20\%$$ C. $$23.32\%$$ D. $$13.64\%$$ 14- Frank wants to compare these two measurements. $$18.023 kg \space …….. \space 18,023 g$$ Which symbol should he use? A. $$<$$ B. $$>$$ C. $$≠$$ D. $$=$$ 15- Aria was hired to teach three identical 5th grade math courses, which entailed being present in the classroom 36 hours altogether. At$25 per class hour, how much did Aria earn for teaching one course? A. $50 B.$300 C. $600 D.$1400 16- In a classroom of 60 students, 22 are male. What percentage of the class is female? A. $$51\%$$ B. $$59\%$$ C. $$63\%$$ D. $$73\%$$ 17- In a party, 6 soft drinks are required for every 9 guests. If there are 171 guests, how many soft drinks are required? A. 9 B. 27 C. 114 D. 171 18- While at work, Emma checks her email once every 90 minutes. In 9 hours, how many times does she check her email? A. 4 Times B. 5 Times C. 6 Times D. 7 Times 19- In a classroom of 44 students, 18 are male. About what percentage of the class is female? A. $$63\%$$ B. $$51\%$$ C. $$59\%$$ D. $$53\%$$ 20- A florist has 516 flowers. How many full bouquets of 12 flowers can he make? A. 40 B. 41 C. 43 D. 45 ## Best 5th Grade MEAP Math Exercise Resource for 2021 1- D The line segment is from 1 to$$-8$$. Therefore, the line is 9 units. $$1 –(-8)= 1+8=9$$ 2- A Use area of rectangle formula. Area $$=$$ length $$×$$ width $$⇒ A = 30 × 45 ⇒ A = 1,350$$ 3- A $$average (mean) = \frac{sum \space of \space terms}{number \space of \space terms}⇒ average= \frac{32+35+29}{3}⇒ average = 32$$ 4- A Peter’s speed $$= \frac{120}{4}= 30$$ Jason’s speed $$= \frac{160}{8}=20$$ $$\frac{The \space average \space speed \space of \space peter}{The \space average \space speed \space of \space Jason}=\frac{30}{20}$$ equals to: $$\frac{3}{2}$$or 3 : 2 5- B Plug in $$x=- 8$$ in each equation. $$x(2x-4)=120→(-8)(2(-8)-4)=(-8)×(-16-4)=160$$ $$8 (4-x)=96→8(4-(-8)=8(12)=96$$ $$2 (4x+6)=79→2(4(-8)+6)=2(-32+6)=-52$$ $$6x-2=-46→6(-8)-2=-48-2=-50$$ Only option B is correct. 6- B The diameter of the circle is 8 inches. Therefore, the radius of the circle is 4 inches. Use circumference of circle formula. $$C = 2πr ⇒ C = 2 × 3.14 × 4 ⇒ C = 25.12$$ 7- B 3 workers can walk 9 dogs ⇒ 1 workers can walk 3 dogs. 5 workers can walk $$(5 × 3) 15$$ dogs. 8- C The horizontal axis in the coordinate plane is called the x-axis. The vertical axis is called the y-axis. The point at which the two axes intersect is called the origin. The origin is at 0 on the x-axis and 0 on the y-axis. 9- C $$19 + (16 × 26) = 19 + 416 = 435$$ 10- D 1 hour: $$4.75$$ 8 hours: $$8 × 4.75 = 38$$ 11- C An obtuse angle is an angle of greater than 90$$^\circ$$ and less than 180$$^\circ$$. From the options provided, only option C (143 degrees) is an obtuse angle. 12- A $$6 – 3\frac{4}{9}=\frac{54}{9}-\frac{31}{9}=\frac{23}{9}$$ 13- D The number of guests that are not present are $$(220 – 190) 30$$ out of $$220 =\frac{30}{220}$$ Change the fraction to percent: $$\frac{30}{220}×100\%=13.64\%$$ 14- D Each kilogram is 1,000 grams. 18,023 grams $$= (\frac{18,023}{1,000}) =18.023$$ kilograms. Therefore, two amounts provided are equal. 15- B Aria teaches 36 hours for three identical courses. Therefore, she teaches 12 hours for each course. Aria earns $25 per hour. Therefore, she earned$300 ($$12 × 25$$) for each course. 16- C The number of female students in the class is $$(60 – 22) 38$$ out of $$60 = \frac{38}{60}$$ Change the fraction to percent: $$\frac{38}{60} 3 ×100\%=63\%$$ 17- C Write a proportion and solve. $$\frac{6 \space soft \space drinks}{9 \space guests}=\frac{x}{171 \space guests}$$ $$x =\frac{171×6}{9}⇒x=114$$ 18- C Every 90 minutes Emma checks her email. In 9 hours (540 minutes), Emma checks her email $$(540 ÷ 90) 6$$ times. 19- C There are 44 students in the class. 18 of the are male and 26 of them are female. 26 out of 44 are female. Then: $$\frac{26}{44}=\frac{x}{100}→2,600=44x→x=2,600÷44≈59\%$$ 20- C Divide the number flowers by $$12: 516 ÷ 12 = 43$$ ## How Does It Work? ### 1. 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# How to Find The Distance Between Two Points On a Curve Print Many students have difficulty finding the distance between two points on a straight line, it is more challenging for them when they have to find the distance between two points along a curve. This article, by the way of an example problem will show how to find this distance. To find the distance between two points A(x1,y1) and B(x2,y2) on a straight line on the xy-plane, we use the Distance Formula, which is... d(AB) = √[(x1-y1)^2+(x2-y2)^2]. We will now demonstrate how this formula works by an example problem. Please click on the image to see how this is done. Now we will find the distance between two points A and B on a curve defined by a function f(x) on a closed interval [a,b]. To find this distance we should use the formula s =The integral, between the lower limit, a, and the upper limit, b, of the integrand √(1 +[f'(x)]^2) in respect to variable of integration, dx. Please click on the image for a better view. The function that we will be using as an example problem, over the closed Interval, [1,3], is... f(x)= (1/2)[(x+4)√[(x+4)^2-1]-ln[(x+4)+√[(x+4)^2-1]]]. the derivative of this function, is... f'(x)=√[(x+4)^2-1], we will now square both sides of the function of the derivative. That is [f'(x)]^2 = [√[(x+4)^2-1]]^2, which gives us [f'(x)]^2 = (x + 4)^2 - 1. We now substitute this expression into the arc length formula/Integral of, s. then Integrate. Please click on the image for a better understanding. Then by substitution, we have the following: s =The integral, between the lower limit, 1, and the upper limit, 3, of the integrand √(1 +[f'(x)]^2) = the integrand √(1 + (x + 4)^2 - 1). which is equal to √((x + 4)^2). By performing the antiderivative on this Integrand, and By the Fundamental Theorem of Calculus, we get... {[(x^2)/2] + 4x} in which we first substitute the upper limit, 3, and from this result, we Subtract the result of the substitution of the lower limit, 1. That is {[(3^2)/2] + 4(3)} - {[(1^2)/2] + 4(1)} which is equal to {[(9/2) + 12]} - {[(1/2) + 4]} = {(33/2) - (9/2)} which is equal to (24/2) = 12. So the Arclength/distance of the function/curve over the Interval [1,3], is, 12 units. • Paper and • Pencil
# 2006 AMC 10B Problems/Problem 11 (diff) ← Older revision \| Latest revision (diff) \| Newer revision → (diff) ## Problem What is the tens digit in the sum $7!+8!+9!+...+2006!$ $\textbf{(A) } 1\qquad \textbf{(B) } 3\qquad \textbf{(C) } 4\qquad \textbf{(D) } 6\qquad \textbf{(E) } 9$ ## Solution Since $10!$ is divisible by $100$, any factorial greater than $10!$ is also divisible by $100$. The last two digits of all factorials greater than $10!$ are $00$, so the last two digits of $10!+11!+...+2006!$ are $00$. () So all that is needed is the tens digit of the sum $7!+8!+9!$ $7!+8!+9!=5040+40320+362880=408240$ So the tens digit is $\boxed{\textbf{(C) }4}$. () A slightly faster method would be to take the $\pmod {100}$ residue of $7! + 8! + 9!.$ Since $7! = 5040,$ we can rewrite the sum as $$5040 + 8\cdot 5040 + 72\cdot 5040 \equiv 40 + 8\cdot 40 + 72\cdot 40 = 40 + 320 + 2880 \equiv 40 \pmod{100}.$$ Since the last two digits of the sum is $40$, the tens digit is $\boxed{\textbf{(C) }4}$.
Courses Courses for Kids Free study material Offline Centres More Store # From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet. Take $\left( {\pi = 3.14} \right)$ Last updated date: 22nd May 2024 Total views: 433.2k Views today: 12.33k Verified 433.2k+ views Hint – In this question from the main circular sheet of radius 4 cm, a smaller circle of radius 3 cm is removed. So using the basic formula for area of circle, we can easily remove the smaller circular area from the larger circular area. This concept will give us the required remaining area. Area of the circle of radius r is given as ${\text{A = }}\pi {r^2}$……………………. (1) Now the radius of the larger circle centered at O2 is having radius r1=4cm……………. (2) The radius of the smaller circle that is being removed from the larger circle is centered at O1 and has the radius r2=3cm……………………. (3) Now putting the values in equation (1) Area of larger circle centered at 02, ${{\text{A}}_2} = \pi {\left( 4 \right)^2} = 16\pi {\text{c}}{{\text{m}}^2}$…………………………… (4) Now putting the values in equation (1) again The area of smaller circle centered at O1, ${{\text{A}}_1} = \pi {\left( 3 \right)^2} = 9\pi {\text{c}}{{\text{m}}^2}$……………………… (5) Now the required remaining area which is shaded in diagram is equal to the difference of larger circle centered at 02 and the area of smaller circle centered at O1. Thus using equation (4) and (5) we can say that ${\text{Area req = 16}}\pi {\text{ - 9}}\pi \\ \Rightarrow 7\pi \\$ Using $\pi = 3.14$ we get Area required = $7 \times 3.14 = 21.98{\text{c}}{{\text{m}}^2}$ Note – Whenever we face such types of problems the key concept is to think of the diagrammatic representation using the data provided in the problem. This will give you the actual understanding about which area has to be removed from which area in order to reach the required area.
## 6.6 Example: How Many Numbers Needed? Let us pause for a review. So far our examples have involved estimating a probability: the chance for a particular event to occur. We have accomplished this with Monte Carlo simulation: • We repeated the chance process a very large number of times • On each repetition, we recorded whether or not the event occurred. • After we finished all of the repetitions, we computed the proportion of those repetitions in which the event did occur. • This proportion served as our estimate of the probability for the event to occur. • We know that the more repetitions we arrange for, the closer our estimate is liable to be to the actual probability for the event to occur. Let us now switch gears and take up the task of estimating the expected value of a random variable. Recall that a random variable is a number whose value is the result of some chance process. The expected value of the random variable is the average—the mean—of its values over many, many repetitions of that chance process. Let us consider the following chance process: • Pick a real number $$x_1$$ between 0 and 1 at random. • Pick another real number $$x_2$$ at random (also between 0 and 1). Add it to $$x_1$$. • If the sum $$x_1 + x_2$$ exceeds 1, then stop. • Otherwise, pick a third real number $$x_3$$, and add it to the sum of the previous two. • If the sum $$x_1+x_2+x_3$$ exceeds 1, then stop. • Otherwise, pick a fourth real number … The idea is to keep on going in this way until the sum of the numbers you have picked exceeds 1. The random variable we are interested in is the number of numbers we have to pick to make the sum of those numbers exceed 1. Let us call this random variable $$X$$. As you can see, this number could be 2, or 3, or 4, or even more depending on how small the random numbers you pick happen to be. In order to fully grasp what is going on, it will be helpful to go through the process of computing $$X$$. The following function will do this for us: numberNeeded <- function(verbose = FALSE) { mySum <- 0 count <- 0 while( mySum < 1 ) { number <- runif(1) mySum <- mySum + number count <- count + 1 if ( verbose ) { cat("Just now picked ", number, ".\n", sep = "") cat(count, " number(s) picked so far.\n", sep = "") cat("Their sum is ", mySum, ".\n\n", sep = "") } } count } Call the function a few times, like this: numberNeeded(verbose = TRUE) ## Just now picked 0.2869393. ## 1 number(s) picked so far. ## Their sum is 0.2869393. ## ## Just now picked 0.7092682. ## 2 number(s) picked so far. ## Their sum is 0.9962075. ## ## Just now picked 0.7524099. ## 3 number(s) picked so far. ## Their sum is 1.748617. ## [1] 3 As you can see, it’s quite common to need 2 or 3 numbers to make the sum exceed 1, but sometimes, you need 4, 5 or even more numbers. Since we are interested in the expected value of $$X$$, we should repeat the process a large number of times and compute the mean of the $$X$$-values that we find. The following code repeats the process 1000 times: needed <- numeric(1000) for (i in 1:1000 ) { needed[i] <- numberNeeded() } cat("The expected number needed is about ", mean(needed), ".\n", sep = "") ## The expected number needed is about 2.697. Of course the expected value indicates what $$X$$ will be on average. By itself it doesn’t tell us about the variability in $$X$$—how much $$X$$ tends to “hop around” from one repetition to another. In order to get a sense of the variability of $$X$$, we can make a table of the results of our 1000 repetitions: table(needed) ## needed ## 2 3 4 5 6 7 ## 528 303 123 37 8 1 Sure enough, we usually got 2 or 3, but higher values are certainly possible. For the most part we are less interested in the number of times we got each possible value than we are in the proportion of times we got each possible value. For this all we need is prop.table(): prop.table(table(needed)) ## needed ## 2 3 4 5 6 7 ## 0.528 0.303 0.123 0.037 0.008 0.001 The proportions serve as estimates of the probability of getting each of the possible values of $$X$$. Such a table of proportions gives us a sense of the distribution of $$X$$—the chances for $$X$$ to assume each of its possible values.20 Now that we have the basic idea of how to accomplish the simulation, it remains to package up our code into a nice function that will enable a user to set the number of repetitions, set a seed, and choose reporting options. Along the way we will generalize a bit, allowing the user to set a target sum other than 1. We’ll also return our estimate of the expected number needed invisibly, so that the function can be used in the context of larger-scale investigations (see the exercises from this Chapter). numberNeededSim <- function(target = 1, reps = 1000, seed = NULL, report = FALSE) { if ( !is.null(seed) ) { set.seed(seed) } # define helper function numberNeeded <- function(target) { mySum <- 0 count <- 0 while( mySum < target ) { number <- runif(1) mySum <- mySum + number count <- count + 1 } count } #perform simulation needed <- numeric(reps) for (i in 1:reps ) { needed[i] <- numberNeeded(target) } # report results (if desired) if ( report ) { print(prop.table(table(needed))) cat("\n") cat("The expected number needed is about ", mean(needed), ".\n", sep = "") } # return estimate of the expected number needed: invisible(mean(needed)) } Let’s test the function with ten thousand repetitions: numberNeededSim(target = 1, reps = 10000, seed = 4848, report = TRUE) ## needed ## 2 3 4 5 6 7 8 ## 0.4937 0.3350 0.1313 0.0317 0.0071 0.0010 0.0002 ## ## The expected number needed is about 2.7273. Does the estimate of the expected number of numbers look a bit familiar? Mathematicians have proven that the exact expected value—what you would get for an average if you could do more and more repetitions, without limit—is the famous number $$e$$: $\mathrm{e} \approx 2.71828 \ldots$ In the exercises you will be asked to investigate the expected value when the target is a number other than 1. 1. Of course not all of the possible values for $$X$$ appear on the table: the only values appearing are those that happened to occur in our simulation. The values that did not occur probably have a very small (but nonzero) chance of occurring.↩︎
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: Arithmetic>Unit 15 Lesson 5: Dividing fractions by fractions # Dividing fractions: 2/5 ÷ 7/3 Dividing fractions is a breeze when you follow these simple steps! First, multiply the first fraction by the reciprocal of the second fraction. Then, multiply the numerators and denominators separately. Voila! You've successfully divided two fractions and found your answer. Keep practicing to master this essential math skill. Created by Sal Khan. ## Want to join the conversation? • is there another method to do this • (Prepare for a long answer) Srivish, To answer your question, there IS another way to solve it. Cross multiplication. Here are some steps to help solve it 1: First take your Equation, and if you've already written an answer. Erase it, then write a new set of brackets to have your fractions in. 2: Secondly, take your equation. Let's ignore that division sign for now. So you have your two pairs of fractions, Correct? 3: TO do this, you will have to multiply a bit. First, take the numerator of your first fraction, and multiply it with the denominator of the second Fraction. This will turn into your new Numerator. Now cross off the fractions you just used. However, if your Equation with variables, such as X, the whole question will change. In order to adapt to this, you will have to change the way you solve it. Here is a link detailing how to. https://www.wikihow.com/Cross-Multiply Hope this helps! • at 0.29 i didn't understand why we flipped it • Because dividing is the same thing as multiplying by the reciprocal of a number. A reciprocal is basically just a number flipped upside down. Example: 4 ÷ 2 is the same thing as 4 * 1/2. • How the heck does a reciprocal even work?! I'm mind blown. Can somebody tell me why? • A reciprocal is when you flip the numerator and the denominator to get it. So 2/4 would become 4/2, 5/6 would become 6/5, and so forth. • Do you still have to do the extra math to the side to get the actual answer. • not if you feel like torchuring yourself lol (no seriously, if you dont, it'll take forever to do it). • When dividing two fractions how many methods are there and what are they? • I think there is only one method, but it is shown or named in several ways. KCF (keep the first fraction, change the sign from / to *, and flip the second fraction) is a common mnemonic device to help remember what to do. The words are the same method of reciprocating the denominator and multiplying. (1 vote) • What if the fraction is improper • oh, my gosh, i was hoping Sal would show us on the number line as is the previous video. i have no problem with solving the problem, but i have no intuitive understanding! in the last video he worked on 8/3 divided by 1/3. it meant we were breaking up 8/3 into segments of 1/3 of a unit each. so we ended up with 8 segments of 1/3 unit, total. i have no idea how to follow that logic for 2/5 divided by 7/3... help, someone? thank you! • Same idea except this time the segment is larger than the initial amount. If we have segments of 7/3 (2 1/3) how many of them make 2/5. In this case you need less than one of them. 6/35 of one in fact. This is because 6/35 of 7/3 equals 2/5.
## How do you use a number line to put integers from greatest to least? How would you use a number line to put integers in order from greatest to least? A Graph the integers, then read themfrom left to right. B Graph the integers, then read them from right to left. C Graph the absolute values of the integers, then read them from left to right. ## How does the number line help you in arranging real numbers? A real number line, or simply number line, allows us to visually display real numbers by associating them with unique points on a line. The real number associated with a point is called a coordinate. A point on the real number line that is associated with a coordinate is called its graph. Number lines provide a mental strategy for addition and subtraction. Research has shown they’re important because they promote good mental arithmetic strategies. Learners soon graduate from using simple number lines to visualising one in their mind – this is when the humble number line has done its job! ## How do you do order the numbers from least to greatest? Look at the ones place first, ignoring the decimal. The order from greatest to least is 2, 4, 5, 6, and 8. Because none of the numbers in the ones place are the same, it doesn’t matter what the decimal will be on these numbers- the order will remain the same. The order is 2.1, 4.8, 5.2, 6.9, 8.5. ## What is the greatest difference in math? You only have the digits 1,2,3,4,5 so the largest 3 digit number that you can make is 543 and the smallest 2 digit number you can make is 12. Thus the largest possible difference you can make is between 543 and 12 that is 543 – 12 = 531. ## How do you use a digit card? Say a number (age appropriate) and the child creates that number using the digit cards. They, he or she reads the number back to you. You can also ask place value questions like “What is the value of the 4 in your number?” or “What place is the 7 in?” Math Facts: Have the children lay out the digit cards. 99
# How To Find Domain And Range Of A Circle ## Introduction How To Find Domain And Range Of A Circle: Finding the domain and range of a circle involves determining the set of possible input and output values that the circle’s equation represents. The domain represents the set of x-values that correspond to valid points on the circle, while the range represents the set of y-values that the circle covers. By understanding the equation and properties of a circle, we can easily identify its domain and range. To find the domain of a circle, we consider the x-coordinates of all the points on the circle. Since a circle is defined by its center and radius, the domain consists of all x-values that fall within the range of the center’s x-coordinate minus and plus the radius. This accounts for all possible horizontal positions on the circle. Similarly, to determine the range of a circle, we examine the y-coordinates of the points on the circle. The range consists of all y-values that fall within the range of the center’s y-coordinate minus and plus the radius. This encompasses all possible vertical positions on the circle. By applying these principles and analyzing the equation of a circle, we can effectively find its domain and range, which provide valuable insights into the set of valid inputs and outputs for the circle’s coordinates. ## What is domain and range for circle?. Lesson. Practice. Lesson. Share. The diagram below shows that the domain of a circle consists of all \$\$ x -values within the interval \$\$ a − r ≤ x ≤ a + r and the range of of a circle consists of all \$\$ y -values within the interval \$\$ b − r ≤ y ≤ b + r . The domain and range of a circle depend on its equation and the coordinate system being used. For a circle centered at the point (h, k) with a radius r, the domain represents all possible x-values of the circle’s coordinates. In this case, the domain is given by the interval [h – r, h + r]. This means that the x-coordinate of any point on the circle falls between h – r and h + r. The range, on the other hand, represents all possible y-values of the circle’s coordinates. The range is given by the interval [k – r, k + r], where k is the y-coordinate of the center of the circle. This means that the y-coordinate of any point on the circle falls between k – r and k + r. In summary, for a circle with center (h, k) and radius r, the domain is [h – r, h + r] and the range is [k – r, k + r]. These intervals encompass all possible x and y-values of the circle’s coordinates. ## What is the formula for range of a circle? According to the distance formula, this is √(x−0)2+(y−0)2=√x2+y2. A point (x,y) is at a distance r from the origin if and only if √x2+y2=r, or, if we square both sides: x2+y2=r2. This is the equation of the circle of radius r. The formula for the range of a circle depends on the coordinate system being used and the properties of the circle. In general, for a circle centered at the point (h, k) with a radius r, the formula for the range of the circle is: Range = [k – r, k + r] Here, k represents the y-coordinate of the center of the circle, and r represents the radius of the circle. The range is given by the interval [k – r, k + r], which indicates that the y-coordinate of any point on the circle falls between k – r and k + r. This formula holds for circles in both Cartesian (x, y) coordinate systems and polar coordinate systems. It provides a concise way to express the range of a circle and represents the set of all possible y-values that the circle’s coordinates can take. It’s important to note that this formula assumes a standard circle where the center is at (h, k) and the radius is constant. If the circle is transformed or has other modifications, the formula for the range may change accordingly. ## How do I calculate domain and range? To find the domain and range of an equation y = f(x), determine the values of the independent variable x for which the function is defined. To calculate the range of the function, we simply express the equation as x = g(y) and then find the domain of g(y). To calculate the domain and range of a function, follow these steps: 1. Identify the given function: Determine the specific function you are working with. It could be an algebraic function, a trigonometric function, or any other type of mathematical function. 2. Understand the nature of the function: Analyze the properties and restrictions of the function to identify any limitations on the input values (domain) and output values (range). Consider factors such as division by zero, square roots of negative numbers, or logarithms of non-positive values. 3. Determine the domain: Find the set of all valid input values for the function. Look for any restrictions or limitations mentioned in the function’s definition or implied by its properties. Exclude any values that would make the function undefined or result in non-real outputs. The remaining set of valid input values represents the domain. 4. Find the range: Examine the output values produced by the function for the valid input values in the domain. Identify the set of all possible output values that the function can generate. This set represents the range. ## What is a domain for a circle? The domain is the values for x so you subtract the radius from the centre coordinate and you add the radius to it. The range is the values for y so you do the same to the y coordinate. In the context of a circle, the term “domain” is not typically used. The concept of domain is more commonly associated with functions, where it refers to the set of input values for which the function is defined. However, if you are referring to the set of possible x-values or the range of x-coordinates for points on a circle, it can be understood as the domain for the circle. In this case, the domain would consist of all valid x-values that fall within the range of the circle’s x-coordinates. For example, consider a circle centered at the point (h, k) with a radius r. The x-coordinates of the points on the circle would range from h – r to h + r, including all values in between. Therefore, the domain for the circle would be expressed as [h – r, h + r], indicating the range of valid x-values for the points on the circle. It’s important to note that the use of the term “domain” in the context of a circle may vary depending on the specific context or application. ## Which one is domain and range? The domain of a function is the set of values that we are allowed to plug into our function. This set is the x values in a function such as f(x). The range of a function is the set of values that the function assumes. This set is the values that the function shoots out after we plug an x value in. In the context of functions, the domain refers to the set of input values for which the function is defined, while the range represents the set of output values that the function can produce. To provide a clearer understanding: 1. Domain: The domain is the set of all valid input values of a function. It consists of the x-values or inputs for which the function is defined and meaningful. The domain can include a specific range of numbers, such as all real numbers (-∞, +∞), or it can have restrictions based on the nature of the function. For example, a square root function may have a domain that includes only non-negative numbers. 2. Range: The range is the set of all possible output values that the function can produce. It represents the y-values or outputs that the function can take on. The range can be a specific set of numbers or have restrictions based on the behavior of the function. For instance, a quadratic function with a minimum point will have a range that includes all y-values greater than or equal to the minimum value. In summary, the domain refers to the valid input values, while the range represents the set of possible output values for a function. ## How do you find the domain and range of a function in Class 11? The range of a set of numbers is the difference between its highest and lowest values. To find it, subtract the lowest number from the highest in the distribution. Ans : Domain of the function f(x)=\|x-1\| is R, Whereas range=[0,∞]. In Class 11, when learning about finding the domain and range of a function, the following steps can be followed: 1. Identify the function: Start by determining the specific function you are working with. It could be an algebraic function, trigonometric function, exponential function, or any other type of function. 2. Understand the restrictions: Analyze the properties and limitations of the function. Look for any specific rules or conditions that restrict the values the function can take. For example, consider the presence of square roots, fractions, or logarithms, which may impose certain constraints. 3. Determine the domain: Find the set of all valid input values for the function. Exclude any values that result in undefined expressions, such as division by zero, square roots of negative numbers, or logarithms of non-positive values. The remaining set of values represents the domain of the function. 4. Find the range: Examine the output values generated by the function for the valid input values in the domain. Identify the set of all possible output values that the function can produce. This set represents the range of the function. ## What are the four domains of circle? The circular figure is divided into four domains: ecology, economics, politics and culture. Each of these domains is divided in seven subdomains, with the names of each of these subdomains read from top to bottom in the lists under each domain name. A circle does not have four domains. Instead, a circle can be described by its equation or properties, which determine its characteristics. The domain of a circle is typically understood as the range of x-values that correspond to valid points on the circle. For a circle with a center at the point (h, k) and a radius r, the domain consists of all the x-values within the range of (h – r, h + r). In other words, it includes all the possible horizontal positions of the points on the circle. It’s important to note that a circle has a continuous curve and does not have distinct domains. Rather, the domain refers to the range of x-values that fall within the circle. The circle itself can be visualized as a set of infinitely many points with specific x and y-coordinates satisfying its equation. ## Is domain a closed circle? At the endpoints of the domain, we draw open circles to indicate where the endpoint is not included because of a less-than or greater-than inequality; we draw a closed circle where the endpoint is included because of a less-than-or-equal-to or greater-than-or-equal-to inequality. No, the domain of a function is not represented by a closed circle. The domain of a function refers to the set of all valid input values or x-values for which the function is defined. It is typically expressed using interval notation, set notation, or a combination of both. In interval notation, a closed interval is represented by using square brackets [ ] to indicate that the endpoints are included. For example, [a, b] represents a closed interval where both the values a and b are included in the domain. However, the notation [ ] used for interval notation should not be confused with a closed circle, which is used to represent the boundary of a circle. The boundary of a circle is a closed curve, while the domain of a function is a set of input values. Therefore, the domain of a function is not represented by a closed circle, but rather by a set of valid input values or intervals, depending on the function and its constraints. ## Conclusion Finding the domain and range of a circle involves understanding the equation, properties, and coordinates of the circle. By considering the x and y-coordinates of the points on the circle, we can determine its domain and range. The domain of a circle consists of all valid x-values that fall within the range of the center’s x-coordinate plus and minus the radius. This defines the set of possible horizontal positions on the circle. On the other hand, the range of a circle encompasses all valid y-values that fall within the range of the center’s y-coordinate plus and minus the radius. This represents the set of possible vertical positions on the circle. By analyzing the equation of a circle and utilizing the properties of its center and radius, we can easily find the domain and range. These insights provide valuable information about the valid input and output values for the coordinates of the circle. Understanding the domain and range of a circle is crucial for various applications in geometry, mathematics, and real-world scenarios where circles are encountered.
# How to Find the Diameter and Exact Center of a Circle For many kinds of do-it-yourself repair and building projects you need to find the exact center and diameter of a circular component. The geometric center of a circle is its center of gravity, while the diameter of the circle gives the widest width and allows you to compute the circumference and area. You can use simple geometry to find two intersecting diameters of a circle. And where the two diameter lines cross is the exact center of the circle. All you need is a straight edge, a right angle, and either a pencil or string and tape to mark the lines. Step 1: Draw a chord across the circle, somewhere away from the center. Step 2: Draw a second chord perpendicular to the first chord. Step 3: Draw a third chord perpendicular to the second and parallel to the first. Step 4: Draw a line connecting the ends of the first and second chords. This line is the diameter of the circle. Step 5: Draw a line connecting the ends of the second and third chords. This forms another diameter. The intersection of the two diameters is the center of the circle. ### Why does this method work? The underlying geometric principle is that whenever you circumscribe a right triangle, the hypotenuse of the triangle forms the diameter of the circle. Put another way, the circumcenter of a right triangle is the midpoint of the hypotenuse, so half the hypotenuse is the radius of the circle. The method above works by making two perpendicular chords to form the legs of a right triangle. When you draw the third side of the triangle (the hypotenuse), it is precisely the diameter. Also, by definition, a diameter is a line that passes through the center of a circle, so two diameters must cross at the center.
# Area of a Triangle by formula (Coordinate Geometry) Given the coordinates of the three vertices of a triangle ABC, the area can be foiund by the formula below. Try this Drag any point A,B,C. The area of the triangle ABC is continuously recalculated using the above formula. You can also drag the origin point at (0,0). Given the coordinates of the three vertices of any triangle, the area of the triangle is given by: where Ax and Ay are the x and y coordinates of the point A etc.. This formula allows you to calculate the area of a triangle when you know the coordinates of all three vertices. It does not matter which points are labelled A,B or C, and it will work with any triangle, including those where some or all coordinates are negative. Looking at the formula above, you will see it is enclosed by two vertical bars like this: The two vertical bars mean "absolute value". This means that it is always positive even if the formula produced a negative result. Polygons can never have a negative area. ## The 'handedness' of point B If you perform this calculation but omit the last step where you take the absolute value, the result can be negative. If it is negative, it means that the 2nd point (B) is to the left of the line segment AC. Here, we mean 'left' in the sense that if you were to stand on point A looking at C, then B is on your left. ## If the area is zero If the area comes out to be zero, it means the three points are collinear. They lie in a straight line and do not form a triangle. You can drag the points above to create this condition. ## You can also use Heron's Formula Heron's Formula allows you to calculate the area of a triangle if you know the length of all three sides. (See Heron's Formula). In coordinate geometry we can find the distance between any two points if we know their coordinates, and so we can find the lengths of the three sides of the triangle, then plug them into Heron's Formula to find the area. ## If one side is vertical or horizontal In the triangle above, the side AC is vertical (parallel to the y axis). In this case it is easy to use the traditional "half base times height" method. See Area of a triangle - conventional method. Here, AC is chosen as the base and has a length of 8, found by subtracting the y coordinates of A and C. Similarly the altitude is 11, found by subtracting the x-coordinates of B and A. So the area is half of 8 times 11, or 44. ## The box method You can also use the box method, which actually works for any polygon. For more on this see Area of a triangle - box method (Coordinate Geometry) ## Things to try 1. In the diagram at the top of the page, Drag the points A, B or C around and notice how the area calculation uses the coordinates. Try points that are negative in x and y. You can drag the origin point to move the axes. 2. Click "hide details". Drag the triangle to some random new shape. Calculate its area and then click "show details" to see if you got it right. 3. After the above, estimate the area by counting the grid squares inside the triangle. (Each square is 5 by 5 so has an area of 25). Once you have done the above, you can click on "print" and it will print the diagram exactly as you set it. ## Limitations In the interest of clarity in the applet above, the coordinates are rounded off to integers and the lengths rounded to one decimal place. This can cause calculatioons to be slightly off. For more see Teaching Notes
# Proving recursive function via mathematical induction (EASY, but need a boost) • Nov 11th 2012, 06:30 PM Proving recursive function via mathematical induction (EASY, but need a boost) This was a sample problem discussed in class that I already have the solution to, but I am having trouble understanding the highlighted section. Please be very detailed with your explanation. Thank you in advance! Let f be defined recursively by: Initial condition: f(0)=1 Recursive step: f(n)=2f(n-1)+1 For the function f above, prove that for all n, f(n)=2n+1-1. ---------------------------------------------------------------- SOLUTION: Proof by induction: 1) Basis step: Prove that f(0)=2n+1-1 f(0)=20+1-1 is true because 1=1 2) Inductive step: Inductive hypothesis: Assume that for some n>=0, f(n)=2n+1-1. We need to use this hypothesis to prove that f(n+1)=2n+2-1. Proof: Assume that the inductive hypothesis is true, (I DO NOT UNDERSTAND HOW THE RIGHT SIDE OF THE BELOW EQUATION (2f(n)+1) WAS FOUND) then f(n+1)=2f(n)+1 (via Recursive definition of f) =2(2n+1-1)+1 (via Inductive hypothesis) =2n+2-2+1 =2n+1-1 # END OF PROOF • Nov 11th 2012, 08:49 PM MarkFL Re: Proving recursive function via mathematical induction (EASY, but need a boost) After having shown the base case is true, I would state the induction hypothesis $P_n$: $f(n)=2^{n+1}-1$ Using the recursive definition, we may state: $f(n+1)=2f(n)+1$ $f(n+1)=2(2^{n+1}-1)+1$ $f(n+1)=2(2^{n+1}-1)+1$ $f(n+1)=2^{(n+1)+1}-1$ We have derived $P_{n+1}$ from $P_n$, thereby completing the proof by induction. • Nov 12th 2012, 03:42 PM ModusPonens Re: Proving recursive function via mathematical induction (EASY, but need a boost) You compose the functions on each side of the equation of the recursive step with g(k)=k+1 and obtain the second equation in red with k instead of n. • Jan 10th 2013, 05:42 PM
# Real Numbers Ex-1.4 Ch-1 [Solutions] NCERT Maths Class 10th Exercise 1.4 Q1. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion: (i) 13/3125 (ii) 17/8 (iii) 64/455 (iv) 15/1600 (v) 29/343 (vi) 23/23 × 52 (vii) 129/22 × 57 × 75 (viii) 6/15 (ix) 35/50 (x) 77/210 (i) 13/3125 Factorize the denominator we get 3125 =5 × 5 × 5 × 5 × 5 = 55 So denominator is in form of 5m so it is terminating . (ii) 17/8 Factorize the denominator we get 8 =2 × 2 × 2 = 23 So denominator is in form of 2m so it is terminating . (iii) 64/455 Factorize the denominator we get 455 =5 × 7 × 13 There are 7 and 13 also in denominator so denominator is not in form of 2m × 5n . so it is not terminating. (iv) 15/1600 Factorize the denominator we get 1600 =2 × 2 × 2 ×2 × 2 × 2 × 5 × 5 = 26 × 52 so denominator is in form of 2m × 5n Hence it is terminating. (v) 29/343 Factorize the denominator we get 343 = 7 × 7 × 7 = 73 There are 7 also in denominator so denominator is not in form of 2m × 5n Hence it is non-terminating. (vi) 23/(23 × 52) Denominator is in form of 2m × 5n Hence it is terminating. (vii) 129/(22 × 57 × 75 ) Denominator has 7 in denominator so denominator is not in form of 2m × 5n Hence it is none terminating. (viii) 6/15 divide nominator and denominator both by 3 we get 2/5 Denominator is in form of 5m so it is terminating. (ix) 35/50 divide denominator and nominator both by 5 we get 7/10 Factorize the denominator we get 10=2 × 5 So denominator is in form of 2m × 5n so it is terminating. (x) 77/210 simplify it by dividing nominator and denominator both by 7 we get 11/30 Factorize the denominator we get 30=2 × 3 × 5 Denominator has 3 also in denominator so denominator is not in form of 2m × 5n Hence it is none terminating. Q2. Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions. (i) 13/3125 = 13/55 = 13×25/55×25 = 416/105 = 0.00416 (ii) 17/8 = 17/23 = 17×53/23×53 = 17×53/103 = 2125/103 = 2.125 (iv) 15/1600 = 15/24×102 = 15×54/24×54×102 = 9375/106 = 0.009375 (vi) 23/2352 = 23×53×22/23 52×53×22 = 11500/105 = 0.115 (viii) 6/15 = 2/5 = 2×2/5×2 = 4/10 = 0.4 (ix) 35/50 = 7/10 = 0.7. Q3. The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form p , q you say about the prime factors of q? (i) 43.123456789 (ii) 0.120120012000120000… (iii) 43.123456789
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # 1.8: Right Triangle Trigonometry Difficulty Level: At Grade Created by: CK-12 Pacing Day 1 Day 2 Day 3 Day 4 Day 5 The Pythagorean Theorem Investigation 8-1 The Pythagorean Theorem Converse Quiz 1 Start Similar Right Triangles Finish Similar Right Triangles Special Right Triangles Investigation 8-2 Day 6 Day 7 Day 8 Day 9 Day 10 Finish Special Right Triangles Investigation 8-3 Quiz 2 Start Tangent, Sine, and Cosine Ratios Finish Tangent, Sine, and Cosine Ratios Solving Right Triangles Finish Solving Right Triangles Day 11 Day 12 Day 13 Quiz 3 Start Review of Chapter 8 Review Chapter 8 Chapter 8 Test Start Chapter 9 Be sure to take your time with this chapter. Remember that the Pacing Guide is merely a suggestion. Students can get caught up with vocabulary and theorems in this lesson. ## The Pythagorean Theorem Goal This lesson introduces the Pythagorean Theorem. It has several applications which we will explore throughout this chapter. Relevant Review Thoroughly review the Simplifying and Reducing Radicals subsection in this lesson. In this and subsequent lessons, answers will be given in simplest radical form. Make sure students know how to add, subtract, multiply, divide, and reduce radicals before moving on. Present anything under the radical like a variable. Students know they cannot add 2x+5y\begin{align}2x + 5y\end{align}. Therefore, they cannot combine 23+52\begin{align}2 \sqrt{3} + 5 \sqrt{2}\end{align}. This should make it easier for students to understand. Teaching Strategies Investigation 8-1 provides one proof of the Pythagorean Theorem. First, lead students through steps 1-3, then have them finish step 4 individually while you circle around to answer questions. Once students are done with this investigation, either take students to the computer lab or if you have an LCD screen, go to the site in the FlexBook to see two additional proofs of the Pythagorean Theorem. Both of these proofs are animated and provide another viewpoint. There are several applications of the Pythagorean Theorem. Students need to be familiar with as many as possible. Go over each example to make sure students understand the range of questions that can be asked. Each of the examples, even though worded differently, are completed the same way. The directions for a given problem can be one place where students can have confusion. Going over the different types of directions for the same type of problem can be very helpful. Finally the Distance Formula is proven in this lesson. Students might already have an idea of how it works. Explain that it is a variation on the Pythagorean Theorem, especially the before we solve for d\begin{align}d\end{align}, when the equation looks like d2=(x1x2)2+(y1y2)2\begin{align}d^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2\end{align}. Notice in this section, the Distance Formula looks different than it did when it was introduced in Chapter 3. Recall that order does not matter, as long as the corresponding x\begin{align}x\end{align} or y\begin{align}y\end{align} value is first (x1\begin{align}x_1\end{align} and y1\begin{align}y_1\end{align} are first or x2\begin{align}x_2\end{align} and y2\begin{align}y_2\end{align} are first). ## Converse of the Pythagorean Theorem Goal This lesson applies the converse of Pythagorean’s Theorem to determine whether triangles are right, acute, or obtuse. Relevant Review Make sure students are comfortable squaring and simplifying radical numbers. Teaching Strategies The Converse of the Pythagorean Theorem basically says if the sides of a triangle do not satisfy the Pythagorean Theorem, then the triangle is not a right triangle. Theorems 8-3 and 8-4 extend this concept to determine if the non-right triangle is acute or obtuse. To help students remember which is which, tell them to think opposite. When a2+b2>c2\begin{align}a^2 + b^2 > c^2\end{align} (a2+b2\begin{align}a^2 + b^2\end{align} is greater than c2\begin{align}c^2\end{align}) the triangle is acute (all angles are less than 90\begin{align}90^\circ\end{align}). When a2+b2<c2\begin{align}a^2 + b^2 < c^2\end{align} (a2+b2\begin{align}a^2 + b^2\end{align} is less than c2\begin{align}c^2\end{align}) the triangle is obtuse (one angle is greater than 90\begin{align}90^\circ\end{align}). Also, when in doubt, have them draw the triangle, as best they can, to scale. Then, they can see what the triangle should be and they can do the Pythagorean Theorem to confirm or deny. Remind students that the Triangle Inequality Theorem still holds. So, if they are given lengths like 5, 7, and 15, they need to be able to recognize that these lengths do not make a triangle at all. 5+7>15\begin{align}5 + 7 > 15\end{align}, so no triangle can be formed. If students do not see this, they will be doing unnecessary work, not to mention think that the triangle is obtuse. Review the Triangle Inequality Theorem as you are going through the examples in the text. No one likes to do extra work, if they do not have to. Additional Example: Do the following lengths form a triangle? If so, is it acute, right, or obtuse? a) 10, 15, 20 b) 7, 14, 21 c) 82,46,414\begin{align}8\sqrt{2}, 4 \sqrt{6}, 4 \sqrt{14}\end{align} Solution: First, check all the lengths to see if they make a triangle. b) does not, 7+14=21\begin{align}7 + 14 = 21\end{align}, so those lengths cannot make a triangle. Let’s see what type of triangles a) and c) are. a) 102+152=202100+225<400obtuse triangle\begin{align}&10^2 + 15^2 = 20^2\\ &100 + 225 < 400\\ &\text{obtuse triangle}\end{align} c) (82)2+(46)2=(414)2642+166=1614128+96=224right triangle\begin{align}&(8\sqrt{2})^2 + (4\sqrt{6})^2 = (4 \sqrt{14})^2\\ &64 \cdot 2 + 16 \cdot 6 = 16 \cdot 14\\ &128 + 96 = 224\\ &\text{right triangle}\end{align} Additional Example: Find an integer such that 9, 12, ____ represent an acute or obtuse triangle. Solution: 9, 12, 15 would be a right triangle (this is a multiple of the Pythagorean triple, 3-4-5). So 8, 9, 10, 11, 12, 13, or 14 would work. If the integer is less than 8, then the triangle would be obtuse, with 12 as the longest side. For an obtuse triangle, the third side could be less than 8, but greater than 3(9+3=12)\begin{align}3 (9 + 3 = 12)\end{align}. And, it could also be greater than 15, but less than 21(9+12=21)\begin{align}21 (9 + 12 = 21)\end{align}. So, the possibilities are 4, 5, 6, 7, 16, 17, 18, 19, or 20. ## Using Similar Right Triangles Goal Students will review similar triangles, primarily right triangles. Then, the concept of the geometric mean is introduced and applied to right triangles. Relevant Review Review similar triangles and their properties briefly before introducing Theorem 8-5. The two triangles formed by the altitude from the right angle in a right triangle are similar to the larger right triangle by AA (see Example 1). Show students how the three triangles fit together and which angles are congruent to each other and which sides are proportional (Examples 1 and 2). Also, remind students that answers should be in simplest radical form. You may need to add a few simplifying and reducing radical questions in the Review Queue. Teaching Strategies Example 3 introduces the geometric mean as it applies to right triangles. You can choose to use this example before discussing the geometric mean or after it, after Example 7. It might be easier for students to see the geometric in its literal, algebraic form (without being applied to triangles) and practice it that way for a few examples, and then apply it to a right triangle. In Examples 5 and 6, it might be helpful to show students the corresponding proportions for the geometric mean, 24x=x36\begin{align}\frac{24}{x} = \frac{x}{36}\end{align} and 18x=x54\begin{align}\frac{18}{x} = \frac{x}{54}\end{align}. Students will like the short cut, x=ab\begin{align}x = \sqrt{ab}\end{align}, but they should be shown the proportion first. The proportion directly relates to its application to right triangles. Examples 3 and 7 apply the geometric mean to the right triangle. The set-up of these proportions, using similar triangles, is the same as the geometric mean. So, students can always use similar triangles, rather than memorize the geometric mean. So often in math, there are two or even three ways to solve one problem. In Example 8, there are two ways presented to solve for y\begin{align}y\end{align}. Encourage students to try the geometric mean, but they can also use the Pythagorean Theorem. Show students both methods and then brainstorm why one method could be preferred over the other. Students should be allowed to use which ever method they feel more comfortable with. ## Special Right Triangles Goal The purpose of this lesson is to encourage the use of ratios to find values of the sides of special right triangles. These triangles are extremely useful in trigonometry. Relevant Review Special right triangle ratios are extended ratios. Here is an additional Review Queue question: Additional Review Queue: The sides of a triangle are in the extended ratio 3:7:9. If the shortest side is 21, find the length of the other two sides. Solution: We will rewrite the ratio as 3x:7x:9x\begin{align}3x: 7x: 9x\end{align}. So, 3x=21\begin{align}3x = 21\end{align}, which means x=7\begin{align}x = 7\end{align}. Therefore, the other two sides are 77=49\begin{align}7 \cdot 7 = 49\end{align} and 97=63\begin{align}9 \cdot 7 = 63\end{align}. Teaching Strategies Students should complete Investigation 8-2 independently. Lead students through step 1, but then allow them to complete the Pythagorean Theorem in steps 2 and 3 on their own. Ask students if they see a pattern among the hypotenuses. Then, go over how to solve an isosceles triangle given various sides and with various square roots. They should know how to solve an isosceles right triangle when the legs are not whole numbers. To find the legs, always divide the hypotenuse by 2\begin{align}\sqrt{2}\end{align} and then simplify the radical. (Students might also notice there is a pattern here too. The leg will be the hypotenuse divided by 2 and multiplied by 2\begin{align}\sqrt{2}\end{align}. For example, if the hypotenuse is 20, the legs will be 20÷22=102\begin{align}20 \div 2 \cdot \sqrt{2} = 10 \sqrt{2}\end{align}. This is a little short cut.) To find the hypotenuse, multiply the leg by 2\begin{align}\sqrt{2}\end{align}. Students may need to simplify the square root. Also, the diagonal of a square will always split the square into two 45-45-90 triangles. Make sure students notice this connection. With 30-60-90 triangles, students must make the connection that it is half an equilateral triangle. In Investigation 8-3, lead students through steps 1-3, then allow the students complete steps 4 and 5 on their own. Students will have a hard time remembering this extended ratio. Some students will think that x3\begin{align}x\sqrt{3}\end{align} should be the hypotenuse because 3 is bigger than 2. However, explain that 31.73\begin{align}\sqrt{3} \approx 1.73\end{align}, which is smaller than 2, so 2x\begin{align}2x\end{align} will always be the longest side, which is the hypotenuse. To solve 30-60-90 triangles, students will need to find the shortest leg, if they are not given it. If they are given the hypotenuse, divide by 2, then they can multiply by 3\begin{align}\sqrt{3}\end{align} to get the longer leg. If they are given the longer leg, they will need to divide by 3\begin{align}\sqrt{3}\end{align} (or use the shortcut as applied to the 45-45-90 triangle above, use 3 and 3\begin{align}\sqrt{3}\end{align} where appropriate) to get the shorter leg, then multiply that by 2. Students will also get these two ratios confused. One way to help them remember is for a 45-45-90 triangle there are 2 45\begin{align}45^\circ\end{align} angles, so the hypotenuse is x2\begin{align}x\sqrt{2}\end{align}. For a 30-60-90 triangle all the angles are divisible by 3, so the 3\begin{align}\sqrt{3}\end{align} is in the radical for this ratio. The best way for students to become comfortable with these ratios is to have them do lots of problems. Students can also make flashcards for these ratios. If students have trouble remembering these special shortcuts, encourage them to use Pythagorean’s Theorem and simplify the answer. The resulting answer will equal the shortcut. ## Tangent, Sine and Cosine Goal This lesson introduces the trigonometric functions; sine, cosine and tangent. Teaching Strategies Before introducing the trig ratios, make sure students understand what adjacent and opposite mean and which angles they are in reference to. c\begin{align}c\end{align} will always be the hypotenuse, but a\begin{align}a\end{align} and b\begin{align}b\end{align} can be either opposite or adjacent, depending on which acute angle we are using. At this point, do not overwhelm students with the fact that the trig functions can be applied to any angle; focus on acute angles in triangles. Encourage students to make flash cards for the sine, cosine and tangent ratios and to use the pneumonic SOH-CAH-TOA (in FlexBook). Both of these things will help students internalize the ratios. In the types of problems in this lesson, it will be very common that two of the three sides are given and students will need to use the Pythagorean Theorem to find the third side. This should always be done first, and then they can apply the ratios. Students will also need to reduce ratios and simplify any radicals. Show students several different orientations of the triangles (rotated, flipped, etc) so they are familiar with where an angle is and which sides are adjacent and opposite. Have each student check to ensure their calculator is set to degrees (DEG), not radians (RAD). Having a calculator in radians will provide incorrect answers. When checking homework at the beginning of the class period, check the mode of students’ calculators as well. When introducing how to find the sides of a right triangle, using the trig ratios, draw the triangle from Example 5 on the board with only one variable, \begin{align}a\end{align}. This will isolate \begin{align}a\end{align} and students should be able to see that the cosine ratio will solve for \begin{align}a\end{align}. Use the arrow to help illustrate that \begin{align}a\end{align} is adjacent to \begin{align}22^\circ\end{align} and 30 is the hypotenuse. After \begin{align}a\end{align} is found, redraw the triangle so \begin{align}b\end{align} is the only variable. Now, \begin{align}b\end{align} is isolated and students will be able to recognize that the sine ratio will solve for it. Again, use arrows, if needed. Stress to students that they should only use information that they are given in the problem. Using “solved for” information will not give them the most accurate answer or it could be completely wrong (if the “solved for” answer used is incorrect). To help illustrate the angles of elevation and depression, see the picture below. Show this to students and ask what the angle of elevation from the cow to the goat is. Then, ask what the angle of depression from the sheep down to the cow is. Students should notice it is the same measurement and alternate interior angles. Fill in the angle of depression/elevations with any measurement. ## Inverse Trigonometric Ratios Goal In the previous lesson, students used the special trigonometric values to determine approximate angle measurements. This lesson enables students to “cancel” a trigonometric function by applying its inverse to accurately find an angle measurement. Relevant Review Begin by listing several mathematical operations on the board in one column. In a second column, title it “Inverse.” Be sure students understand what an inverse means (an inverse cancels an operation, leaving the original value undisturbed). Operation Inverse Example Multiplication Squaring Sine Cosine Tangent The first four are typically easy for students (Subtraction, square root, multiplication, and addition). You may have to lead students a little more on the last two (inverse tangent and inverse sine). Students may say, “Un-tangent it.” Use the correct terminology here, but also use their wording, if at all possible. Students will be able to cancel the trigonometric function using the inverse of that function, even though they may use incorrect terminology. Go over Example 1 thoroughly. Make sure every student understands how to input inverse trig functions into their calculator. Remind students that the set-up for inverse problems is the same as those from the previous lesson. However, instead of being given an angle measure, we leave it as a variable. Students need to solve for the angle and then put everything into the calculator at the same time. Get them in this habit so they will produce the most accurate answers. For example: \begin{align}\underline{\text{Yes}}: \sin^{-1} \left (\frac{2}{3}\right ) && \underline{\text{No}}: \sin^{-1} (0.666)\end{align} Example 5 is a special right triangle. Students will probably go through the motions and not notice that they can use the ratios learned in the Special Right Triangles lesson. Either way, students will still arrive at the correct answer, but point out to them to not go blindly into each question. Read a problem, re-read, and then decide how to answer. Like the last lesson, real-life situations are a major application. At the end of this lesson, create a word problem as a class. Use the names of students or find the height of a local building. Make the problem personal. Then, add it to the test as an extra question or a bonus. ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Show Hide Details Description Tags: Subjects: Date Created: Feb 22, 2012 Aug 21, 2014 Image Detail Sizes: Medium \| Original CK.MAT.ENG.TE.1.Geometry-Basic.1.8
# Finding Determinants ## Matrix -> Determinants -> Finding Determinants Determinants are an important concept in linear algebra, particularly when working with matrices. In this tutorial, we will explore how to find determinants of matrices. We will cover the theory behind determinants, the steps involved in finding determinants, and provide code snippets and examples to illustrate the process. ### What is a Determinant? A determinant is a scalar value that can be calculated from a square matrix. It provides important information about the matrix, such as whether it is invertible or singular. The determinant of a matrix is denoted by the symbol \|A\| or det(A), where A represents the matrix. ### Steps to Find Determinants To find the determinant of a matrix, we follow a specific set of steps. Let's go through each step in detail: #### Step 1: Check Matrix Size Before finding the determinant, we need to ensure that the matrix is square, i.e., it has an equal number of rows and columns. Determinants can only be calculated for square matrices. #### Step 2: Identify the Order of the Matrix The order of a matrix refers to the number of rows (or columns) it has. For example, a matrix with 3 rows and 3 columns is a 3x3 matrix. The order of the matrix is important as it determines the number of steps required to find the determinant. #### Step 3: Define the Base Case For a 2x2 matrix, finding the determinant is straightforward. We can use the formula: ``````\|A\| = (a * d) - (b * c) `````` Where a, b, c, and d represent the elements of the matrix. #### Step 4: Expand the Matrix For matrices larger than 2x2, we need to expand the matrix to simplify the calculation. We can expand the matrix along any row or column. Let's consider expanding along the first row for simplicity. #### Step 5: Calculate the Determinant To calculate the determinant, we need to find the sum of the products of the elements and their corresponding cofactors. The cofactor of an element is the determinant of the submatrix obtained by removing the row and column containing that element. We can express the determinant calculation as follows: ``````\|A\| = a * C1 + b * C2 + c * C3 + ... `````` Where a, b, c, etc., represent the elements of the first row, and C1, C2, C3, etc., represent their corresponding cofactors. #### Step 6: Recursion To find the cofactors, we need to recursively find the determinants of the submatrices. We repeat the process until we reach a 2x2 matrix, for which we can use the base case formula. ### Example Let's work through an example to solidify our understanding. Consider the following 3x3 matrix: ``````\| 2 4 6 \| \| 1 3 5 \| \| 7 8 9 \| `````` #### Step 1: Check Matrix Size The matrix is 3x3, which is a square matrix. #### Step 2: Identify the Order of the Matrix The matrix is a 3x3 matrix. #### Step 3: Define the Base Case Since the matrix is larger than 2x2, we need to expand it. #### Step 4: Expand the Matrix Expanding along the first row, we have: ``````\| 2 4 6 \| \| 1 3 5 \| \| 7 8 9 \| `````` Expanding the first element, we obtain the submatrix: ``````\| 3 5 \| \| 8 9 \| `````` Expanding the second element, we obtain the submatrix: ``````\| 1 5 \| \| 7 9 \| `````` Expanding the third element, we obtain the submatrix: ``````\| 1 3 \| \| 7 8 \| `````` #### Step 5: Calculate the Determinant Using the formula mentioned earlier, we can calculate the determinant as follows: ``````\|A\| = 2 * \|3 5\| - 4 * \|1 5\| + 6 * \|1 3\| \|8 9\| \|7 9\| \|7 8\| `````` Calculating the determinants of the submatrices: ``````\|3 5\| = (3 * 9) - (5 * 8) = 27 - 40 = -13 \|1 5\| = (1 * 9) - (5 * 7) = 9 - 35 = -26 \|1 3\| = (1 * 8) - (3 * 7) = 8 - 21 = -13 `````` Substituting the determinants back into the main equation: ``````\|A\| = 2 * (-13) - 4 * (-26) + 6 * (-13) = -26 + 104 - 78 = 0 `````` Hence, the determinant of the given matrix is 0. ### Conclusion In this tutorial, we explored the concept of determinants and learned how to find determinants of matrices. We covered the necessary steps, including checking matrix size, identifying the order of the matrix, expanding the matrix, calculating the determinant, and using recursion for submatrices. We also worked through an example to illustrate the process. Determinants are a fundamental concept in linear algebra and have various applications in mathematics and computer science. Now that you have a solid understanding of finding determinants, you can apply this knowledge to solve more complex problems involving matrices. Happy coding!
# Make the Smallest and Largest Three-Digit Numbers In this worksheet, students will be given five digits from which they must identify the smallest and largest three-digit numbers. Key stage: KS 2 Curriculum topic: Number: Number and Place Value Curriculum subtopic: Order/Compare Numbers to 1000 Difficulty level: #### Worksheet Overview In this activity, we are going to make the largest and smallest three-digit numbers from five given digits. Let's pretend we can only use the digits 2, 3, 4, 5 and 6. To make the largest number: This is 6. Then comes the next largest digit down in the tens column. This is 5. Lastly, comes the next largest digit in the ones column. This is 4. The largest number is 654 Did you notice that we didn't need to use the two smallest numbers, 2 and 3, to make this number? To make the smallest number: This is 2. Then comes the next largest digit in the tens column. This is 3. Lastly, comes the next largest digit in the ones column. This is 4. The smallest number is 234. Did you notice that we didn't need to use the two largest numbers, 5 and 6 this time, to make this number? Let's try an example question together using the same method. Example Using any three of the digits from: 68671 make the smallest and largest three-digit numbers that you can make. First, we should put the digits in order: 1, 6, 6, 7, 8 To find the smallest three-digit number we do the following: 1 is the smallest digit. 6 is the next largest digit. 6 is the next largest digit, even though it is the same as the last one. So, the smallest number will be 166. Now to find the largest possible three-digit number: 8 is the largest digit. 7 is the next digit down. 6 is the next digit down. So, the largest number will be 876. Now that you've seen a good method, it's time to try some questions on your own. ### What is EdPlace? We're your National Curriculum aligned online education content provider helping each child succeed in English, maths and science from year 1 to GCSE. With an EdPlace account you’ll be able to track and measure progress, helping each child achieve their best. We build confidence and attainment by personalising each child’s learning at a level that suits them. Get started
Courses Courses for Kids Free study material Offline Centres More Store # Find the solution of the equation given below $\operatorname{Sin} {10^0} + \operatorname{Sin} {20^0} + \operatorname{Sin} {30^0} + ............ + \operatorname{Sin} {360^0}$ Last updated date: 17th Jul 2024 Total views: 451.2k Views today: 7.51k Verified 451.2k+ views Hint: By using the given trigonometric formula which is given below. The terms can be brought in the form of some series after manipulation. $= \operatorname{Sin} a + \operatorname{Sin} (a + d) + \operatorname{Sin} (a + 2d) + ......... + \operatorname{Sin} (a + (n - 1)d) \\ = \dfrac{{(\operatorname{Sin} (a + (n - 1)\dfrac{d}{2})\operatorname{Sin} (\dfrac{{nd}}{2})}}{{\operatorname{Sin} \left( {\dfrac{d}{2}} \right)}} \\$ Given that: $= \operatorname{Sin} {10^0} + \operatorname{Sin} {20^0} + \operatorname{Sin} {30^0} + ............ + \operatorname{Sin} {360^0} \\ = \operatorname{Sin} {10^0} + \operatorname{Sin} ({10^0} + {10^0}) + \operatorname{Sin} ({10^0} + 2 \times {10^0}) + ........... + \operatorname{Sin} ({10^0} + 35 \times {10^0}) \\$ …………………………..(1) As we know that$\left[ {\operatorname{Sin} a + \operatorname{Sin} (a + d) + \operatorname{Sin} (a + 2d) + ......... + \operatorname{Sin} (a + (n - 1)d) = \dfrac{{(\operatorname{Sin} (a + (n - 1)\dfrac{d}{2})\operatorname{Sin} (\dfrac{{nd}}{2})}}{{\operatorname{Sin} \left( {\dfrac{d}{2}} \right)}}} \right]$ By comparing equation (1) with the above formula, we get $a = {10^0},d = {10^0} \\ n - 1 = 35 \\ n = 36 \\$ Using the values of a, d, n and substituting these values in the above formula, we get $\Rightarrow \dfrac{{\operatorname{Sin} (10 + (36 - 1)\dfrac{{10}}{2})\operatorname{Sin} (\dfrac{{36 \times 10}}{2})}}{{\operatorname{Sin} (\dfrac{{10}}{2})}} \\ \Rightarrow \dfrac{{\operatorname{Sin} (10 + 35 \times 5)\operatorname{Sin} (36 \times 5)}}{{\operatorname{Sin} 5}} \\$ After simplifying further $\Rightarrow \dfrac{{\operatorname{Sin} ({{10}^0} + {{175}^0})\operatorname{Sin} ({{180}^0})}}{{\operatorname{Sin} {5^0}}}$ Since, the value of $\operatorname{Sin} {180^0} = 0$ $\Rightarrow \dfrac{{\operatorname{Sin} ({{185}^0}) \times 0}}{{\operatorname{Sin} {5^0}}} \\ \Rightarrow 0 \\$ Hence, after simplifying the given trigonometric equation the final result is 0. Note: For these types of problems, remember all important trigonometric identities and the values of trigonometric functions. Also be aware of the concept of quadrants, range and domain of these functions. Solving these types of problems will become simple if you remember all trigonometric expressions.
## Engage NY Eureka Math 8th Grade Module 1 Lesson 8 Answer Key ### Eureka Math Grade 8 Module 1 Lesson 8 Example Answer Key Example 1. In 1723, the population of New York City was approximately 7,248. By 1870, almost 150 years later, the population had grown to 942,292. We want to determine approximately how many times greater the population was in 1870 compared to 1723. The word approximately in the question lets us know that we do not need to find a precise answer, so we approximate both populations as powers of 10. → Population in 1723: 7248<9999<10000=104 → Population in 1870: 942 292<999 999<1 000 000=106 We want to compare the population in 1870 to the population in 1723: $$\frac{10^{6}}{10^{4}}$$ Now we can use what we know about the laws of exponents to simplify the expression and answer the question: $$\frac{10^{6}}{10^{4}}$$ =102. Therefore, there were approximately 100 times more people in New York City in 1870 compared to 1723. Example 2. Let’s compare the population of New York City to the population of New York State. Specifically, let’s find out how many times greater the population of New York State is compared to that of New York City. The population of New York City is 8,336,697. Let’s round this number to the nearest million; this gives us 8,000,000. Written as single-digit integer times a power of 10: 8 000 000=8×106. The population of New York State is 19,570,261. Rounding to the nearest million gives us 20,000,000. Written as a single-digit integer times a power of 10: 20 000 000=2×107. To estimate the difference in size we compare state population to city population: $$\frac{2 \times 10^{7}}{8 \times 10^{6}}$$ Now we simplify the expression to find the answer: $$\frac{2 \times 10^{7}}{8 \times 10^{6}}$$ = $$\frac{2}{8}$$ × $$\frac{10^{7}}{10^{6}}$$By the product formula = $$\frac{1}{4}$$×10 By equivalent fractions and the first law of exponents =0.25×10 =2.5 Therefore, the population of the state is 2.5 times that of the city. Example 3. There are about 9 billion devices connected to the Internet. If a wireless router can support 300 devices, about how many wireless routers are necessary to connect all 9 billion devices wirelessly? Because 9 billion is a very large number, we should express it as a single-digit integer times a power of 10. 9 000 000 000=9×109 The laws of exponents tells us that our calculations will be easier if we also express 300 as a single-digit integer times a power of 10, even though 300 is much smaller. 300=3×102 We want to know how many wireless routers are necessary to support 9 billion devices, so we must divide $$\frac{9 \times 10^{9}}{3 \times 10^{2}}$$ Now, we can simplify the expression to find the answer: $$\frac{9 \times 10^{9}}{3 \times 10^{2}}$$ = $$\frac{9}{3}$$ × $$\frac{10^{9}}{10^{2}}$$ By the product formula =3×107 By equivalent fractions and the first law of exponents =30 000 000 About 30 million routers are necessary to connect all devices wirelessly. Example 4. The average American household spends about $40,000 each year. If there are about 1×〖10〗8 households, what is the total amount of money spent by American households in one year? Let’s express$40,000 as a single-digit integer times a power of 10. 40000=4×〖10〗4 The question asks us how much money all American households spend in one year, which means that we need to multiply the amount spent by one household by the total number of households: (4×104 )(1×108 )=(4×1)(104×108 ) By repeated use of associative and commutative properties =4×1012 By the first law of exponents Therefore, American households spend about $4,000,000,000,000 each year altogether! ### Eureka Math Grade 8 Module 1 Lesson 8 Exercise Answer Key Exercise 1. The Federal Reserve states that the average household in January of 2013 had$7,122 in credit card debt. About how many times greater is the U.S. national debt, which is \$16,755,133,009,522? Rewrite each number to the nearest power of 10 that exceeds it, and then compare. Household debt=7122<9999<10000=104. U.S.debt =16 755 133 009 522<99 999 999 999 999<100 000 000 000 000=1014. $$\frac{10^{14}}{10^{4}}$$ = 1014-4=1010. The U.S. national debt is 1010 times greater than the average household’s credit card debt. Exercise 2. There are about 3,000,000 students attending school, kindergarten through Grade 12, in New York. Express the number of students as a single-digit integer times a power of 10. 3 000 000=3×106 The average number of students attending a middle school in New York is 8×〖10〗2. How many times greater is the overall number of K–12 students compared to the average number of middle school students? $$\frac{3 \times 10^{6}}{8 \times 10^{2}}$$= $$\frac{3}{8}$$×$$\frac{10^{6}}{10^{2}}$$ = $$\frac{3}{8}$$×$$\frac{10^{6}}{10^{2}}$$ =0.375×〖10〗4 =3750 There are about 3,750 times more students in K–12 compared to the number of students in middle school. Exercise 3. A conservative estimate of the number of stars in the universe is 6×1022. The average human can see about 3,000 stars at night with his naked eye. About how many times more stars are there in the universe compared to the stars a human can actually see? $$\frac{6 \times 10^{22}}{3 \times 10^{3}}$$ = $$\frac{6}{3}$$ ×$$\frac{10^{22}}{10^{3}}$$= 2×1022-3 = 2×1019 There are about 2×1019 times more stars in the universe compared to the number we can actually see. Exercise 4. The estimated world population in 2011 was 7×109. Of the total population, 682 million of those people were left-handed. Approximately what percentage of the world population is left-handed according to the 2011 estimation? 682 000 000≈700 000 000=7×108 $$\frac{7 \times 10^{8}}{7 \times 10^{9}}$$=$$\frac{7}{7}$$×$$\frac{10^{8}}{10^{9}}$$ = 1×$$\frac{1}{10}$$ =$$\frac{1}{10}$$ About one-tenth of the population is left-handed, which is equal to 10%. Exercise 5 The average person takes about 30,000 breaths per day. Express this number as a single-digit integer times a power of 10. 30000=3×104 If the average American lives about 80 years (or about 30,000 days), how many total breaths will a person take in her lifetime? (3×104 )×(3×104 )=9×108 ### Eureka Math Grade 8 Module 1 Lesson 8 Problem Set Answer Key Students practice estimating size of quantities and performing operations on numbers written in the form of a single-digit integer times a power of 10. Question 1. The Atlantic Ocean region contains approximately 2×1016 gallons of water. Lake Ontario has approximately 8,000,000,000,000 gallons of water. How many Lake Ontarios would it take to fill the Atlantic Ocean region in terms of gallons of water? 8 000 000 000 000=8×1012 $$\frac{2 \times 10^{16}}{8 \times 10^{12}}$$=$$\frac{2}{8}$$×$$\frac{10^{16}}{10^{12}}$$ =$$\frac{1}{4}$$×104 =0.25×104 =2500 2,500 Lake Ontario’s would be needed to fill the Atlantic Ocean region. Question 2. U.S. national forests cover approximately 300,000 square miles. Conservationists want the total square footage of forests to be 300,000〗2 square miles. When Ivanna used her phone to do the calculation, her screen showed the following: a. What does the answer on her screen mean? Explain how you know. The answer means 9×1010. This is because: (300 000)2=(3×105)2 =32×(105 )2 =9×1010 b. Given that the U.S. has approximately 4 million square miles of land, is this a reasonable goal for conservationists? Explain. 4 000 000=4×106. It is unreasonable for conservationists to think the current square mileage of forests could increase that much because that number is greater than the number that represents the total number of square miles in the U.S, 9×1010>4×106. Question 3. The average American is responsible for about 20,000 kilograms of carbon emission pollution each year. Express this number as a single-digit integer times a power of 10. 20 000=2×104 Question 4. The United Kingdom is responsible for about 1× 104 kilograms of carbon emission pollution each year. Which country is responsible for greater carbon emission pollution each year? By how much? 2× 104>1×104 America is responsible for greater carbon emission pollution each year. America produces twice the amount of the U.K. pollution. ### Eureka Math Grade 8 Module 1 Lesson 8 Exit Ticket Answer Key Most English-speaking countries use the short-scale naming system, in which a trillion is expressed as 1,000,000,000,000. Some other countries use the long-scale naming system, in which a trillion is expressed as 1,000,000,000,000,000,000,000. Express each number as a single-digit integer times a power of ten. How many times greater is the long-scale naming system than the short-scale? 1 000 000 000 000=1012 1 000 000 000 000 000 000 000= 1021 $$\frac{10^{21}}{10^{12}}$$ = 109. The long-scale is about 109 times greater than the short-scale. ### Eureka Math Grade 8 Module 1 Lesson 8 Sprint Answer Key Applying Properties of Exponents to Generate Equivalent Expressions—Round 1 Directions: Simplify each expression using the laws of exponents. Use the least number of bases possible and only positive exponents. When appropriate, express answers without parentheses or as equal to 1. All letters denote numbers.
# Powers of 10 When we talked about exponents, we said that raising $a$ to $b$-th power is $a^b=\underbrace{a\cdot a\cdot a\cdot a\cdot...\cdot a}_{b}$. $a$ is called base, $b$ is exponent (power). We assumed that $b$ is positive integer. Here we will talk about special case $10^b$, i.e. when the base equals $10$. So, for example, ${{10}}^{{3}}={10}\cdot{10}\cdot{10}={1000}$ and ${{10}}^{{4}}={10}\cdot{10}\cdot{10}\cdot{10}={10000}$. We can see a pattern here. Exponent shows how many zeros will be there in result. ${{10}}^{{{\color{red}{{{7}}}}}}={1}{\color{red}{{{0000000}}}}$ : exponent equals 7, so in result there will be 1 with seven zeros. This makes it easier to calculate powers of ten. Now, let's see how powers of 10 are related to decimals. For this we need to recall negative exponents. So, if ${b}$ is positive integer then ${{a}}^{{-{b}}}=\frac{{1}}{{{{a}}^{{b}}}}$. In case of 10 we have that ${{10}}^{{-{b}}}=\frac{{1}}{{{{10}}^{{b}}}}$. For example, ${{10}}^{{-{2}}}=\frac{{1}}{{{{10}}^{{2}}}}=\frac{{1}}{{100}}$. There is special notation for such fraction: $\frac{{1}}{{100}}={0.01}$. In general, ${{10}}^{{-{b}}}$ is a decimal that has ${b}$ digits in decimal part and all digits except last equals 0. Last digit is 1. For example, ${{10}}^{{-{\color{red}{{{4}}}}}}={0}.{\color{red}{{{0001}}}}$ (4 digits in decimal part; all zeros except last). These things relate decimals and mixed numbers, meaning that mixed number is a decimal written in another form and vice versa. Rule. If we multiply by 10, we move decimal point one postion to the right, if we divide by 10, we move decimal point to the left. Suppose, you have number 500. It can be written as ${50}\cdot{10}={50}\times{{10}}^{{1}}$ or it can be written as ${5}\cdot{100}={5}\times{{10}}^{{2}}$. We can even further move decimal point to the left and write, that ${500}={0.5}\cdot{1000}={0.5}\times{{10}}^{{3}}$. Same works with negative exponent ${0.7}={7}\cdot\frac{{1}}{{10}}={7}\times{{10}}^{{-{1}}}={70}\cdot\frac{{1}}{{100}}={7}\times{{10}}^{{-{2}}}$. Moving in another direction is also applicable: ${0.7}={0.07}\cdot{10}={0.07}\times{{10}}^{{1}}$. Now, let's practice a bit. Exercise 1. Convert ${{10}}^{{-{{3}}}}$ to decimal. Answer: ${{10}}^{{-{{3}}}}={0.001}$. Next exercise. Exercise 2. Convert ${{10}}^{{-{5}}}$ to decimal. Answer: ${{10}}^{{-{5}}}={0.00001}$. Last exercise. Exercise 3. Convert ${0.0001}$ into fraction. Answer: ${0.0001}=\frac{{1}}{{{10000}}}={{10}}^{{-{4}}}$.
How To Divide The Circle Into 5 Parts \| The Science The Science # How to divide the circle into 5 parts Circumferential length can not be accurately measured with a ruler, and therefore its division into equal parts is not an easy task, especially when these parts is an odd number. Divide the circle into five parts is carried out using a conventional compass or protractor. Divide the circle into five parts, inscribed in it a regular pentagon. ### You will need: Ruler without division, a compass, a protractor ## Instruction how to divide the circle into 5 parts Step 1: Construct a circle centered at a point about an arbitrary radius. Through the center of the circle holding its diameter, its name, such as AB. Draw another diameter of the circle, the diameter perpendicular to AB. To do this, swipe from the points A and B are two circles with radii larger than the radius of a circle. A point at the intersection of them, and through the point O draw the diameter perpendicular to the diameter AB. Let's call it a CD. With the help of such a construct, a circle is drawn from the points A and D, construct a point E, which is the midpoint of AB. Radius CE, from the center to the point E, draw a circle and find its point of intersection with the segment AB. At the intersection point put F. Step 2: The resulting segment CF and a side of the pentagon, which is inscribed in a circle undertaken. Compasses take segment CF. Let the first dividing point is C. Conduct of her CF radius circle to the intersection with the circle divisible. From the resulting point again draw a circle with the same radius, to a new intersection with the circle. Repeat this step two more times. As a result, the circle will have five points, which are vertices inscribed in it a regular pentagon. Arc between points obtained will be equal, and hence, the circle is divided into five equal parts. You can then divide the circle. To do this, from the point O to draw segments, dividing the circle points. The result is a five sectors of the same area, which divide the circle into equal parts. Step 3: To divide the circle into five equal parts, use a protractor. Spend the radius of the circle and the center of the radius and angle of 36º aside. Angle describes the sector, an area which is equal to 1/5 of the area of the circle. Repeat this operation three more times, getting five equal sectors, which will divide the circle into five equal parts.
# How do you solve 8(x-3)=96 using the distributive property? Nov 19, 2017 x = 15 #### Explanation: Given $8 \left(x - 3\right) = 96$ Solve for x 1) Clear the parentheses by distributing the 8 $8 x - 24 = 96$ 2) Add 24 to both sides to isolate the $8 x$ term $8 x = 120$ 3) Divide both sides by 8 to isolate $x$ $x = 15$ ← answer $x = 15$ ........................ Check Using the original equation, sub in 15 in the place of $x$ to see if the equation is still true $8 \left(x - 3\right) = 96$ Sub in 15 in the place of $x$ $8 \left(15 - 3\right) = 96$ This equation should still really equal 96 Distribute the 8 $8 \left(10 + 5\right) - 8 \left(3\right)$ should still equal 96 $80 + 40 - 24$ should still equal 96 $120 - 24$ should still equal 96 96 does equal 96 Check!
# 6-3 Warm Up Problem of the Day Lesson Presentation ## Presentation on theme: "6-3 Warm Up Problem of the Day Lesson Presentation"— Presentation transcript: 6-3 Warm Up Problem of the Day Lesson Presentation The Pythagorean Theorem Warm Up Problem of the Day Lesson Presentation Pre-Algebra Pre-Algebra 6-3 The Pythagorean Theorem Warm Up Graph the figures with the given verticals and find the area. 1. (–1, –1), (–1, 3), (6, –1) 2. (2, 1), (8, 1), (6, –3) 3. (3, –2), (15, –2), (14, 6), (4, 6) 14 units2 12 units2 88 units2 Problem of the Day A side of a square A is 5 times the length of square B. How many times as great is the area of square A than the area of square B? 25 6-3 The Pythagorean Theorem Learn to use the Pythagorean Theorem and its converse to solve problems. Vocabulary Pythagorean Theorem leg hypotenuse a2 + b2 = c2 Additional Example 1A: Find the the Length of a Hypotenuse Find the length of the hypotenuse. c A. 4 5 a2 + b2 = c2 Pythagorean Theorem = c2 Substitute for a and b. = c2 Simplify powers. 41 = c2 41 = c Solve for c; c = c2. 6.40  c Additional Example 1B: Find the the Length of a Hypotenuse Find the length of the hypotenuse. triangle with coordinates B. (1, –2), (1, 7), and (13, –2) a2 + b2 = c2 Pythagorean Theorem = c2 Substitute for a and b. = c2 Simplify powers. 225 = c Solve for c; c = c2. 15 = c Find the length of the hypotenuse. Try This: Example 1A Find the length of the hypotenuse. c A. 5 7 a2 + b2 = c2 Pythagorean Theorem = c2 Substitute for a and b. = c2 Simplify powers. 74 = c Solve for c; c = c2. 8.60  c Find the length of the hypotenuse. Try This: Example 1B Find the length of the hypotenuse. B. triangle with coordinates (–2, –2), (–2, 4), and (3, –2) x y (–2, 4) The points form a right triangle. a2 + b2 = c2 Pythagorean Theorem = c2 Substitute for a and b. = c2 Simplify powers. (–2, –2) (3, –2) 61 = c Solve for c; c = c2. 7.81  c Additional Example: 2 Finding the Length of a Leg in a Right Triangle Solve for the unknown side in the right triangle. a2 + b2 = c2 Pythagorean Theorem 25 b 72 + b2 = 252 Substitute for a and c. 49 + b2 = 625 Simplify powers. – –49 b2 = 576 7 b = 24 576 = 24 Try This: Example 2 Solve for the unknown side in the right triangle. a2 + b2 = c2 Pythagorean Theorem 12 b 42 + b2 = 122 Substitute for a and c. 16 + b2 = 144 Simplify powers. – –16 4 b2 = 128 b  11.31 128  11.31 Additional Example 3: Using the Pythagorean Theorem to Find Area Use the Pythagorean Theorem to find the height of the triangle. Then use the height to find the area of the triangle. a2 + b2 = c2 Pythagorean Theorem a = 62 Substitute for b and c. a = 36 6 6 a a2 = 20 a = units ≈ 4.47 units 4 4 Find the square root of both sides. A = hb = (8)( 20) = units2  units2 1 2 Try This: Example 3 Use the Pythagorean Theorem to find the height of the triangle. Then use the height to find the area of the triangle. a2 + b2 = c2 Pythagorean Theorem a = 52 Substitute for b and c. 5 5 a2 + 4 = 25 a a2 = 21 a = units ≈ 4.58 units 2 2 Find the square root of both sides. A = hb = (4)( 21) = units2  4.58 units2 1 2 Lesson Quiz Use the figure for Problems 1-3. 1. Find the height of the triangle. 8m 2. Find the length of side c to the nearest meter. 10 m c h 12m 3. Find the area of the largest triangle. 60m2 6 m 9 m 4. One leg of a right triangle is 48 units long, and the hypotenuse is 50 units long. How long is the other leg? 14 units
# What is the formula for finding terms in an arithmetic sequence? ## What is the formula for finding terms in an arithmetic sequence? Finding the number of terms in an arithmetic sequence might sound like a complex task, but it’s actually pretty straightforward. All you need to do is plug the given values into the formula tn = a + (n – 1) d and solve for n, which is the number of terms. What’s the missing number in this sequence? Solution: The missing number found in the following sequence is 13. It is because all the given numbers in the sequence 1, 3, 5, 7, 11, 17, 19 are prime numbers. The numbers given in the sequence are prime numbers as they can be divided only by 1 and itself….Solved Examples on Missing Numbers. 8 3 21 12 2 ? ### How do you find the nth term? How to find the nth term 1. To find the nth term, first calculate the common difference, d . 2. Next multiply each term number of the sequence (n = 1, 2, 3, …) by the common difference. 3. This will give you the n th term term in the form an + b where a and b are unknown values that we will have calculated. How do you find terms in a sequence? Solution: To find a specific term of an arithmetic sequence, we use the formula for finding the nth term. Step 1: The nth term of an arithmetic sequence is given by an = a + (n – 1)d. So, to find the nth term, substitute the given values a = 2 and d = 3 into the formula. ## What are the missing numbers in the sequence 5/8 11 14? The next number in the list of numbers 2, 5, 8, 11, 14, . . . is 17. Notice that the difference between each consecutive term in this sequence is 3. What is the nth term examples? The nth term is a formula used to generate any term of a sequence. To find a given term, substitute the corresponding value of n into the nth term formula. For example, if the nth term is 3n + 2, the 10th term of the sequence can be found by substituting n = 10 into the nth term. ### What is an arithmetic sequence in math? An arithmetic sequence is a sequence where each term increases by adding/subtracting some constant k. This is in contrast to a geometric sequence where each term increases by dividing/multiplying some constant k. Example: a1 = 25. a(n) = a(n-1) + 5. ## How do I manually install EGit? How do I manually install EGit? Installing EGit in Eclipse you can look in the “All Available Sites” drop down panel if EGit is existing there or add… ## Does Walmart still offer site to store? Does Walmart still offer site to store? Shop Online: Customers can access Site to Store at www.walmart.com/sitetostore or search for Site to Store on the Walmart.com homepage. After… ## What is a heat stable allergen? What is a heat stable allergen? Some allergens or, more properly, some allergenic foods, are described as heat stable (e.g. milk, egg, fish, peanuts, and products thereof), while… ## How can I contact Nick Jenkins? How can I contact Nick Jenkins? How to hire Nick Jenkins. Contact the Champions Speakers agency to provisionally enquire about Nick Jenkins for your event today. Simply call… ## What is a Cas9 Nickase? What is a Cas9 Nickase? A Cas9 nickase variant can be generated by alanine substitution at key catalytic residues within these domains: the RuvC mutant D10A produces a… ## How accurate is kinetic inRide? How accurate is kinetic inRide? Using the inRide pod and a magnet in the resistance unit roller, we take speed at the wheel and translate that into power…
GeeksforGeeks App Open App Browser Continue # Class 10 NCERT Solutions- Chapter 6 Triangles – Exercise 6.3 \| Set 2 ### Question 11. In the following figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ~ ΔECF. Solution: Given, ABC is an isosceles triangle. Since, the sides of an isosceles triangle are equal, we have, ∴ AB = AC ⇒ ∠ABD = ∠ECF In ΔABD and ΔECF, Since, each of the following angles are 90°. By AA similarity criterion, we have, ∴ ΔABD ~ ΔECF ### Question 12. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ΔPQR (see Fig 6.41). Show that ΔABC ~ ΔPQR. Solution: Given that in ΔABC and ΔPQR, AB is proportional to PQ BC is proportional to QR That is, AB/PQ = BC/QR = AD/PM We know, Since, D is the midpoint of BC and M is the midpoint of QR By SSS similarity criterion, we have, ⇒ ΔABD ~ ΔPQM Now, since the corresponding angles of two similar triangles are equal, we obtain, ∴ ∠ABD = ∠PQM ⇒ ∠ABC = ∠PQR Now, In ΔABC and ΔPQR AB/PQ = BC/QR ………….(i) ∠ABC = ∠PQR ……………(ii) From equation (i) and (ii), we get, By SAS similarity criterion, we have, ΔABC ~ ΔPQR ### Question 13. D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA2 = CB.CD Solution: Given, D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. ∠ACD = ∠BCA (Common angles) By AA similarity criterion, we have, Since, the corresponding sides of similar triangles are in proportion, we obtain, ∴ CA/CB = CD/CA That is, CA2 = CB.CD. Hence, proved. ### Question 14. Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ΔABC ~ ΔPQR. Solution: Given, Two triangles ΔABC and ΔPQR in which AD and PM are medians such that; Now, Construction, Produce AD further to E so that AD = DE. Now, join CE. Similarly, produce PM further until N such that PM = MN. Join RN. In ΔABD and ΔCDE, we have From the construction done, Now, since AP is the median, BD = DC and, ∠ADB = ∠CDE [Vertically opposite angles are equal] By SAS criterion, ∴ ΔABD ≅ ΔCDE By CPCT, we have, ⇒ AB = CE……………..(i) In ΔPQM and ΔMNR, From the construction done, PM = MN Now, since PM is the median, QM = MR and, ∠PMQ = ∠NMR [Vertically opposite angles are equal] By SAS criterion, ∴ ΔPQM = ΔMNR By CPCT, ⇒ PQ = RN …………………(ii) Now, AB/PQ = AC/PR = AD/PM From equation (i) and (ii), we conclude, ⇒ CE/RN = AC/PR = 2AD/2PM We know, 2AD = AE and 2PM = PN. ⇒ CE/RN = AC/PR = AE/PN By SSS similarity criterion, ∴ ΔACE ~ ΔPRN Thus, ∠2 = ∠4 And, ∠1 = ∠3 ∴ ∠1 + ∠2 = ∠3 + ∠4 ⇒ ∠A = ∠P ………………….(iii) Now, in ΔABC and ΔPQR, we have AB/PQ = AC/PR [Given] From equation (iii), we have, ∠A = ∠P By SAS similarity criterion, ∴ ΔABC ~ ΔPQR ### Question 15. A vertical pole of a length 6 m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower. Solution: Given, Length of the vertical pole = 6m Length of shadow of the tower = 28 m Shadow of the pole = 4 m Let us assume the height of tower to be h m. In ΔABC and ΔDEF, ∠C = ∠E (By angular elevation of sum) Since, the following angles are equivalent to 90° ∠B = ∠F By AA similarity criterion, we have, ∴ ΔABC ~ ΔDEF Since, if two triangles are similar corresponding sides are proportional ∴ AB/DF = BC/EF Substituting values, ∴ 6/h = 4/28 ⇒h = (6×28)/4 ⇒ h = 6 × 7 ⇒ h = 42 m Hence, the height of the tower specified is 42 m. ### Question 16. If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQR prove that AB/PQ = AD/PM. Solution: Given, ΔABC ~ ΔPQR Since, the corresponding sides of similar triangles are in proportion. ∴ AB/PQ = AC/PR = BC/QR……………(i) And, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R ………….…..(ii) Since, AD and PM are the medians, they will divide their opposite sides correspondingly. ∴ BD = BC/2 and QM = QR/2 ………….(iii) From equations (i) and (iii), we obtain, AB/PQ = BD/QM …………….(iv) In ΔABD and ΔPQM, From equation (ii), we have ∠B = ∠Q From equation (iv), we have, AB/PQ = BD/QM By SAS similarity criterion, we have, ∴ ΔABD ~ ΔPQM That is, AB/PQ = BD/QM = AD/PM My Personal Notes arrow_drop_up Related Tutorials
# Lesson Video: Quadratic Functions in Different Forms Mathematics • 9th Grade In this video, we will learn how to evaluate and write a quadratic function in different forms. 17:39 ### Video Transcript In this video, we will learn how to evaluate and write a quadratic function in different forms. We recall, firstly, that a quadratic expression is one in which the highest power of the variable that appears is two. So, the expression includes a squared term and no terms with higher powers, such as cubed or to the power of four. For example, the expression π₯ squared plus two π₯ minus three is a quadratic expression, as is two π¦ squared minus three. The expression π minus one all squared is also a quadratic expression. Because once weβve distributed the parentheses, we see that this is equivalent to π squared minus two π plus one, and so the expression includes an π squared term. There are a number of different key forms in which a quadratic function can be written. The first of these is the general or expanded form. π of π₯ equals ππ₯ squared plus ππ₯ plus π, where π, π, and π are constants and π must not be equal to zero. Itβs usual to write the terms in order of descending powers of the variable. So, we have the π₯ squared term first, then the π₯ term if there is one, and finally the constant term if there is one. Although this isnβt essential. The second form is completed square or vertex form. π of π₯ equals π multiplied by π₯ plus π all squared plus π, where π, π, and π are all constants and, again, π must be nonzero. The form we use depends on which features of the quadratic function or its graph weβre particularly interested in. In this video, weβll mostly be focusing on the completed square form because itβs useful for determining the coordinates of the vertex and the equation of the axis of symmetry of a quadratic graph. The final form in which we can express a quadratic function is its factored or factorized form. π of π₯ equals ππ₯ plus π multiplied by ππ₯ plus π, where π, π, π, and π are constants. Although, it isnβt possible to express every quadratic in this form. Letβs now consider some examples. In our first example, weβll see how we can use the completed square or vertex form of a quadratic function to determine the coordinates of the vertex of its graph. Find the vertex of the graph of π¦ equals five multiplied by π₯ plus one all squared plus six. Here, we have a quadratic function. And itβs been expressed in its completed square or vertex form. In general, this form is π multiplied by π₯ plus π all squared plus π. So, comparing the general form with our quadratic, we see that the value of π is five, the value of π is one, and the value of π is six. We were asked to use this form to find the vertex or turning point of the graph of this quadratic function. Now, we should recall that all quadratic functions have the same general shape, which is called a parabola. When the value of π is positive, the parabola will curve upwards. And when the value of π is negative, the parabola will curve downwards. The vertex of the graph is its minimum point in the first instance and its maximum point in the second. Our value of π in this question is five, which is positive. So, weβre in this first instance here. Weβre looking for the coordinates of the minimum point of this graph. Letβs consider then how we can minimize this quadratic function and for what value of π₯ this minimum will occur. And then, weβll see how we can generalize. Letβs consider each part of the function in turn, starting with π₯ plus one squared. Well, this is a square. And we know that squares are always nonnegative. Whatever value of π₯ we choose, once we have added one and then squared it, the result will be zero or above. So, the minimum value of π₯ plus one all squared is zero. We then multiply this by five. But, of course, anything multiplied by zero is still zero. So, the minimum value of five multiplied by π₯ plus one all squared is still zero. We then add six, so the entire function increases by six units, which means that its minimum value overall is six. This gives the minimum value of the function or the minimum value of π¦. So, it is the π¦-coordinate of the vertex. We then need to determine what value of π₯ causes this to be the case. Well, itβs the value of π₯ such that π₯ plus one all squared takes its minimum value of zero. Itβs, therefore, the value of π₯ such that the expression inside the parentheses, thatβs π₯ plus one, is equal to zero. The solution to this simple linear equation is π₯ equals negative one. So, we find that the π₯-coordinate of the vertex is negative one. The coordinates of the vertex of this graph then are negative one, six. Now, letβs consider how we can generalize. The π₯-coordinate of the vertex was the value that made the expression inside the parentheses zero. In the case of the general expression π₯ plus π, this will be the value negative π because negative π plus π give zero. So, this is the π₯-coordinate of vertex. The π¦-coordinate of six was the value that was added on to this squared expression. So, in the general form, that will be the value π. So, we can generalize. If a quadratic function is in its completed square or vertex form π¦ equals π multiplied by π₯ plus π all squared plus π, then the coordinates of its vertex are negative π, π. Now, the quadratic function in this question was given to us in its completed square form. In our next example, weβll remind ourselves how to take a quadratic in its general form and write it in its completed square form. In completing the square for the quadratic function π of π₯ equals π₯ squared plus 14π₯ plus 46, you arrive at the expression π₯ minus π all squared plus π. What is the value of π? Here, weβre given a quadratic in two forms, its general or expanded form and its completed square form. These two expressions describe the same quadratic, and therefore theyβre equal. We can, therefore, form an equation π₯ squared plus 14π₯ plus 46 is equal to π₯ minus π all squared plus π. Now, there are two approaches that we can take to answering this question. The two approaches involve working in opposite directions. In our first approach, weβll manipulate the expression on the right-hand side of our equation to bring it into its expanded form. Using whatever method weβre most comfortable for squaring a binomial, we see that π₯ minus π all squared is equivalent to π₯ squared minus two ππ₯ plus π squared. So, the expression on the right-hand side becomes π₯ squared minus two ππ₯ plus π squared plus π. As these expressions are equivalent for all values of π₯, we can now compare coefficients on the two sides of the equation. The coefficients of π₯ squared on each side are one. And then, if we compare the coefficients of π₯, we have 14π₯ on the left-hand side and negative two ππ₯ on the right-hand side, giving the equation 14 equals negative two π. We can solve this equation for π by dividing each side by negative two, giving π equals negative seven. Now, we have actually completed the problem because all we weβre asked for was the value of π. But suppose weβd also been asked to determine the value of π. We could do this by comparing the constant terms on the two sides of the equation. On the left, we have positive 46. And on the right, we have π squared plus π. So, that gives us a second equation; π squared plus π is equal to 46. We know that π is equal to negative seven and negative seven squared is 49. So, substituting this value into our equation, we have 49 plus π equals 46. And subtracting 49 from each side, we find that π is equal to negative three. This means that the completed square or vertex form of our quadratic, using π equals negative seven and π equals negative three, is π₯ minus negative seven all squared minus three. Of course, π₯ minus negative seven is better written as π₯ plus seven. So, we can express this as π₯ plus seven all squared minus three. So, thatβs our first method in which we expanded to the expression on the right-hand side. In our second method, weβll see how we can work the other way. So, weβll start on the left-hand side and bring it into its completed square form. We already know what weβre working towards. Itβs π₯ plus seven all squared minus three. Now, notice that the value inside the parentheses of positive seven is exactly half the coefficient of π₯ in our original equation. And this is no coincidence. This will always be the case. So, we begin by writing π₯ squared plus 14π₯ plus 46 as π₯ plus seven all squared. Weβve halved the coefficient of π₯ to give the value inside the parentheses. The trouble is, though, π₯ plus seven all squared isnβt just equivalent to π₯ squared plus 14π₯; itβs equivalent to π₯ squared plus 14π₯ plus 49. So, weβve introduced an extra 49. We, therefore, need to subtract this so that the expression we have is still equivalent to π₯ squared plus 14π₯. This new expression of π₯ plus seven all squared minus 49 is, therefore, exactly equivalent to π₯ squared plus 14π₯. We also need to include the positive 46 so that the two sides of the equation are the same. Now, that value 49 is, of course, the square of seven. So, what weβre doing is subtracting the square of the value inside our parentheses. The final step is just to simplify. We have negative 49 plus 46, which is equivalent to negative three. And so, weβve found that this quadratic, in its completed square form, is π₯ plus seven all squared minus three, which is the same as we found using our first method. Comparing this then to the given form of π₯ minus π all squared plus π, we would see that positive seven is equal to negative π. Dividing or, indeed, multiplying both sides of this equation by negative one, we find that π is equal to negative seven. Using two methods then, thatβs working in both directions, weβve found that the value of π is negative seven. Completing the square is a little bit more complicated when the value of π, thatβs the coefficient of π₯ squared in the general form, is not equal to one. So, letβs consider an example of this. Which of the following is the vertex form of the function π of π₯ equals two π₯ squared plus 12π₯ plus 11. And weβre given five answer options. So, we have a quadratic function in its general or expanded form, and weβre asked to find its vertex form. Thatβs π of π₯ equals π multiplied by π₯ plus π all squared plus π, where π, π, and π are constants we need to find. To answer to this question then, we need to write our quadratic function in its completed square form. Now, this is slightly trickier than usual because the coefficient of π₯ squared in our function isnβt one. It is two. Now, we can deal with this by factoring by this coefficient. You can either factor from all three terms giving two multiplied by π₯ squared plus six π₯ plus 11 over two. Or we can simply factor this coefficient from the first two terms, giving two multiplied by π₯ squared plus six π₯ plus 11. And personally, I think the second method is easier. What weβre now going to do is complete the square on the expression within the parentheses. Thatβs π₯ squared plus six π₯. We begin by halving the coefficient of π₯ to give the number inside the parentheses. So, we have π₯ plus three all squared. And then, we need to subtract the square of this value. So, we have π₯ plus three all squared minus nine. A quick check of redistributing these parentheses confirms that π₯ plus three all squared minus nine is indeed equal to π₯ squared plus six π₯. Now, we need to be very careful here. All of this expression of π₯ squared plus six π₯ was being multiplied by two. So, we need to put a large set of brackets or parentheses around π₯ plus three all squared minus nine, multiply it by two. And then, we still have the 11 that we were adding on. So, we found that our function π of π₯ is equivalent to two multiplied by π₯ plus three all squared minus nine plus 11. Next, we need to distribute the two. So, we have two multiplied by π₯ plus three all squared and then two multiplied by negative nine, which is negative 18, and then plus 11. Remember that 11 is not being multiplied by two. Finally, we just simplify negative 18 plus 11 is negative seven. So, we have our quadratic in its vertex form two multiplied by π₯ plus three all squared minus seven. Looking carefully at the five answer options we were given because theyβre all quite similar, we see that this is answer option (a). Now, in this question, we just worked through the completing-the-square process ourselves and then determined the correct answer option. It would actually have been possible to eliminate a couple of the options straightaway though. If we look at the vertex form of the function, we see that within the parentheses we have just π₯ plus π all squared. We could, therefore, have eliminated options (b) and (c) as they are not the vertex form of any function because inside the parentheses they each have two π₯. We could also have eliminated option (e) because we can see that there is no two involved in this option. And as the coefficient of π₯ squared in our original function was two, weβd need a factor of two outside the parentheses as we have in answer options (a) and (d). The only difference between options (a) and (d) is the sign inside the parentheses. We have positive three for option (a) and negative three for option (d). We should remember, though, that the sign here is always the same as the sign of the coefficient of π₯ in the original function. So, in this case, itβs positive. In any case, we have our answer though. The vertex form of this function is π of π₯ equals two multiplied by π₯ plus three all squared minus seven. Quadratic functions can be used to model practical situations such as the path followed by an object or a worded problem. In our final example, weβll see how we can form a quadratic function from a description. Two siblings are three years apart in age. Write an equation for π, the product of their ages, in terms of π, the age of the youngest sibling. So, weβre told in this question to use the letter π to represent the age of the younger sibling. We now want to find an expression in terms of π for the age of the older sibling. Well, weβre told that the siblings are three years apart in age. So, the older sibling is three years older than the younger. An expression for the age of the older sibling is, therefore, π plus three. The product of their ages means we need to multiply them together. So, we take our two expressions of π and π plus three and multiply them. This is a quadratic equation. Because if we were to distribute the parentheses, we have π multiplied by π giving π squared and π multiplied by three giving three π. Either of these two forms would be acceptable for the answer, but weβll give our answer as π equals π multiplied by π plus three. Letβs now review some of the key points from this video. Firstly, one of the key forms in which a quadratic can be written is its general or expanded form. π of π₯ equals ππ₯ squared plus ππ₯ plus π, where π, π, and π are constants and π must be nonzero. We can also express quadratics in their completed square or vertex form. π of π₯ equals π multiplied by π₯ plus π all squared plus π, where π, π, and π are constants and, again, π must be nonzero. Weβve also seen that for a quadratic function expressed in its completed square or vertex form, then its vertex, which will be a minimum when the value of π is positive and a maximum when the value of π is negative, is at the point with coordinates negative π, π. Through our examples, we saw how we can convert between these two key forms either by following the completing-the-square process or by distributing the parentheses and simplifying the resulting expression. Although we didnβt need to use it in this video, we also know that some quadratic functions can be written in a factored or factorized form.
Math Access for Teachers and Home Child Care Providers In a partnered game, children will flip a penny and collect points depending on whether the penny lands heads up or tails up. Content Area Standard Target • Data Analysis and Probability • Formulate questions that can be addressed with data and collect, organize, and display relevant data to answer them • Develop and evaluate inferences and predictions that are based on data • Sort and classify objects according to their attributes and organize data about objects • Discuss events related to students’ experiences as likely or unlikely Obtain the Materials Note: Small parts create a choking hazard for children. Make sure that all materials you choose to use for an activity or lesson with children meet safety requirements. Small parts are not appropriate for children who are 5 years of age or younger. Introduce the Activity 1. Demonstrate flipping a coin and calling “Heads or Tails” before the coin lands on the surface. 2. Play a quick game of Heads or Tails before explaining probability and how partners can play the game. Have all the children stand in a circle around you as you flip the coin. Ask how many children think the coin will land on heads? Those children will sit down after the coin is flipped and the children who thought that the coin would land on tails will stay standing. Play this game several times. 3. Introduce probability - the chance that something is going to happen. Explain to the children that they are going to explore the concept of probability by flipping a coin and collecting the data. 4. Ask the children whether or not they can control which side of the coin lands face up. Why or why not? “Heads or Tails” is a game of chance. When you play the game you are going to guess which side will probably land face up. That’s called probability.” 5. Give the children examples of probability in real life scenarios – “The weather man said it will probably rain today but, it may not.” Engage the Children 1. Explain that the game will be played in pairs and that one player will be heads and one player will be tails. Say: “You will play 2 rounds with 10 penny tosses. You will each get a chance to be heads and will also get a chance to be tails.” 2. Explain the recording sheet. Say: “Player 1 will go first. Before the first roll, Player 1 will decide if they want heads or tails. Let’s say, they call heads. If the penny lands on heads, they get a point and record the point in the Player 1 column. If the penny lands on tails, Player 2 gets a point and records it in the Player 2 column.” Show the exact columns you are referring to as you are explaining the directions. You may want to model a couple of rounds so that the children get the idea of one player playing for 10 coin tosses and then Player 2 taking their turn. 3. When they are all done, ask who won each game. "Did you think it was a fair game?" "Why or why not?" "Do you think you get practice to get better at this game?" "Why or why not?" • Play with 2 coins. Again, the player calls either heads or tails. Both coins need to be the side the player calls in order for the player to receive a point. If both coins land of the opposite sides then the other player gets a point and if the two coins land on head and tails, no one gets a point. Encourage Vocabulary • Probability – The chance that something is going to happen (e.g., "What is the probability that the coin will land on heads?") Glossary of MATH vocabulary Supporting Children at Different Levels Toddlers Pre-K Toddlers may: • Not yet grasp the concept of probability. • Need help taking turns. • Need the game simplified. Pre-K Children may: • Be able to increase their knowledge of probability by adding another variable into the game. Home child care providers may: • Adapt the game to a one player game. Have just one child count the number of times the coin landed on heads. As they begin to understand the concept of probability, add tails as an counting option as well. Home child care providers may: • Play with 2 coins. Again, the player calls either heads or tails. Both coins need to be the side the player callsin order for the player to receive a point. If both coins land of the opposite sides then the other player gets a point and if the two coins land on head and tails, no one gets a point. Books • That's a Possibility!: A Book About What Might Happen by Bruce Goldstone (New York: Henry Holt and Co, 2013) • Do You Wanna Bet?: Your Chance to Find Out About Probability by Jean Cushman (New York: HMH Books for Young Readers, 2007) • A Very Improbable Story: A Math Adventure by Edward Finhorn (Boston, MA: Charlesbridge Publishing, 2008) Music and Movement • Many students learn through music. Witness the number of adults who easily sing the ABC song with their own children. Composers are beginning to develop math-specific songs to help students learn math concepts and skills. Whether teachers use a song to introduce or reinforce a concept, or as a regular part of calendar time, students are bound to benefit from the multi-sensory experience. http://www.mathwire.com/music/music.html#mdata Outdoor Connections Web Resources # Comment on this lesson To report a problem with the site, please email us. © 2011. M.A.T.H.
Conic Sections Circle and Line Line circle intersection A line and a circle in a plane can have one of the three positions in relation to each other, depending on the distance d of the center S(p, q) of the circle (x - p)2 + (y - q)2 = rfrom the line Ax + By + C = 0, where the formula for the distance: If the distance of the center of a circle from a line is such that: d < r, then the line intersects the circle in two points, d = r, the line touches the circle at only one point, d > r, the line does not intersect the circle, and they have no common points. Example: At which points the line x + 5y + 16 = 0 intersects the circle x2 + y2 - 4x + 2y - 8 = 0. Solution: To find coordinates of points at which the line intersects the circle solve the system of equations: So, the line intersects the circle at points, A(4, -4) and B(-1, -3). Example: Find equation of a circle with the center at S(1, 20) which touches the line 8x + 15y - 19 = 0. Solution: If a line touches a circle then the distance between the tangency point and the center of the circle d = DS = r i.e., thus, equation of the circle (x - 1)2 + (y - 20)2 = 289. We can use another method to solve this problem. Since, the normal n through the center is perpendicular to the tangent t then the direction vector sn is perpendicular to the direction vector st . Therefore, as mt = sy/sx = -8/15 then so, equation of the normal is As the tangency point D is the common point of the tangent and the normal then, putting coordinates of the radius vector of the normal into equation of the tangent determines a value of the parameter l to satisfy that condition, as then, these variable coordinates of the radius vector put into equation of the tangent follows 8 · (1 + 8l) + 15 · (20 + 15l) - 19 = 0 => 289l = 289 => l = - 1 so, the radius vector of the tangency point therefore the tangency point D(-7, 5). The radius of the circle, since This result we can check by plugging the coordinates of the tangency point into equation of the circle, that is D(-7, 5) => (x - 1)2 + (y - 20)2 = 289, (-7 - 1)2 + (5 - 20)2 = 289 => (-8)2 + (- 15)2 = 289 therefore, the tangency point is the point of the circle. Equation of a tangent at a point of a circle with the center at the origin The direction vector of the tangent at the point P1 of a circle and the radius vector of P1 are perpendicular to each other so their scalar product is zero. Points, O, P1 and P in the right figure, determine vectors, the scalar product written in the components gives, x1x + y1y = r2 This is equation of a tangent at the point P1(x1, y1) of a circle with the center at the origin College algebra contents E
Home » Math Theory » Decimals » Dividing with decimals Note: this page contains legacy resources that are no longer supported. You are free to continue using these materials but we can only support our current worksheets, available as part of our membership offering. Dividing with decimals When students are first introduced to the division of decimals numbers they can think of partitioning into equal groups in a similar way to how they have done when dividing whole numbers. In doing this they will apply their understanding of decimal place value and the important role of the decimal point as an extension of whole number place value. Before moving on to algorithmic methods of division, students should model division into equal groups in a “hands-on” manner. Base-10 blocks can be used to do this – use a cube or a flat to represent one whole unit and the rods and units to represent the smaller units. Cuisenaire rods can be similarly re-purposed with an orange (ten) representing one whole unit and a white (one) representing one tenth. The division examples below use a place value chart with counters to illustrate the process. Dividend and Divisor Whole Numbers – Decimal Quotient 14 ÷ 4 Students will already have experienced division with whole number dividends and divisors, and this will most likely have included examples, like the one above, where the quotient would have a remainder. Using decimals instead of a whole number remainder is an important progression. Highlight this difference to your children. Whole Numbers Divisor – Decimal Dividend (no regrouping) 4.82 ÷ 2 Regrouping different place value units is often a key step in performing multi-digit arithmetic operations although its importance can sometimes lead students to forget that it is not always required as the example above shows. Rounding, Estimating, and Checking Encourage your children to estimate the answer when dividing with decimals as a check for the reasonableness of their calculation. Rounding the decimal and doing a quick mental check is one way of doing this. Dividing by powers of 10 Discuss with your children any patterns in the location of the decimal point and the number of zeros. Do not encourage the thought that the decimal point is moving. Instead stress that the values of the digits are decreasing by the power of 10 and, it is the digits, if anything that are moving. There is more on multiplying and dividing decimals by powers off 10 here. Dividing Decimals – Algorithm As with multiplying decimals, division of decimals requires care in the placement of the decimal point. If there is a remainder, keep adding zeros to the right of the dividend and continue to divide. Worksheets There is a dividing decimals worksheet generator. It provides an unlimited number of questions with various options to change the type of question.
# What is the HCF of 630? ## What is the HCF of 630? HCF of 630,1188 by prime factorization it’s answer is a 6. What are the factors of 315? Factors of 315 • All Factors of 315: 1, 3, 5, 7, 9, 15, 21, 35, 45, 63, 105 and 315. • Prime Factors of 315: 3, 5, 7. • Prime Factorization of 315: 32 × 51 × 71 • Sum of Factors of 315: 624. What are the factors of 810? Factors of 810 • All Factors of 810: 1, 2, 3, 5, 6, 9, 10, 15, 18, 27, 30, 45, 54, 81, 90, 135, 162, 270, 405 and 810. • Prime Factors of 810: 2, 3, 5. • Prime Factorization of 810: 21 × 34 × 51 • Sum of Factors of 810: 2178. ### What is the HCF of 630 and 1560? The GCF of 630 and 1560 is 30. What is the HCF of 144252 and 630? The HCF is 18. Step-by-step explanation: Given : Number 144, 252, 630. What are the factors of 264? Factors of 264 • All Factors of 264: 1, 2, 3, 4, 6, 8, 11, 12, 22, 24, 33, 44, 66, 88, 132 and 264. • Prime Factors of 264: 2, 3, 11. • Prime Factorization of 264: 23 × 31 × 111 • Sum of Factors of 264: 720. ## What are the prime factors of 385? The Prime Factors of 385 are 5 × 7 × 11 and its Factors in Pairs are (1, 385), (5, 77), (7, 55), (11, 35). What are factors of 176? Factors of 176 • Factors of 176: 1, 2, 4, 8, 11, 16, 22, 44, 88, 176. • Prime Factorization of 176: 176 = 24 × 11. What are the prime factors of 72? For example, we can write the number 72 as a product of prime factors: 72 = 2 3 ⋅ 3 2 . The expression 2 3 ⋅ 3 2 is said to be the prime factorization of 72. ### What are factors of 405? Factors of 405 • All Factors of 405: 1, 3, 5, 9, 15, 27, 45, 81, 135 and 405. • Negative Factors of 405: -1, -3, -5, -9, -15, -27, -45, -81, -135 and -405. • Prime Factors of 405: 3, 5. • Prime Factorization of 405: 34 × 51 • Sum of Factors of 405: 726.
# Unit -2: Relation and Function Let’s start by saying that a relation is simply a set or collection of ordered pairs. Nothing really special about it. An ordered pair, commonly known as a point, has two components which are the x and y coordinates. This is an example of an ordered pair. ## Main Ideas and Ways How to Write or Represent Relations As long as the numbers come in pairs, then that becomes a relation. If you can write a bunch of points (ordered pairs) then you already know how a relation looks like. For instance, here we have a relation that has five ordered pairs. Writing this in set notation using curly braces. Relation in set notation However, aside from set notation, there are other ways to write this same relation. We can show it in a table, plot it on the xy-axis, and express it using a mapping diagram. • Relation in table • Relation in graph • Relation in mapping diagram • The domain is the set of all x or input values. We may describe it as the collection of the first values in the ordered pairs. • The range is the set of all y or output values. We may describe it as the collection of the second values in the ordered pairs. So then in the relation below our domain and range are as follows: When listing the elements of both domain and range, get rid of duplicates and write them in increasing order. ### Relations and Functions Let’s start by saying that a relation is simply a set or collection of ordered pairs. Nothing really special about it. An ordered pair, commonly known as a point, has two components which are the x and y coordinates. This is an example of an ordered pair. Relation in set notation However, aside from set notation, there are other ways to write this same relation. We can show it in a table, plot it on the xy-axis, and express it using a mapping diagram. • Relation in table • Relation in graph • Relation in mapping diagram • The domain is the set of all x or input values. We may describe it as the collection of the first values in the ordered pairs. • The range is the set of all y or output values. We may describe it as the collection of the second values in the ordered pairs. So then in the relation below our domain and range are as follows: When listing the elements of both domain and range, get rid of duplicates and write them in increasing order. ## What Makes a Relation a Function? On the other hand, a function is actually a “special” kind of relation because it follows an extra rule. Just like a relation, a function is also a set of ordered pairs; however, every x-value must be associated to only one y-value. Suppose we have two relations written in tables, • A relation that is not a function Since we have repetitions or duplicates of x-values with different y-values, then this relation ceases to be a function. • A relation that is a function This relation is definitely a function because every x-value is unique and is associated with only one value of y. # Relations and Functions Let’s start by saying that a relation is simply a set or collection of ordered pairs. Nothing really special about it. An ordered pair, commonly known as a point, has two components which are the x and y coordinates. This is an example of an ordered pair. Relation in set notation However, aside from set notation, there are other ways to write this same relation. We can show it in a table, plot it on the xy-axis, and express it using a mapping diagram. • Relation in table • Relation in graph • Relation in mapping diagram • The domain is the set of all x or input values. We may describe it as the collection of the first values in the ordered pairs. • The range is the set of all y or output values. We may describe it as the collection of the second values in the ordered pairs. So then in the relation below our domain and range are as follows: When listing the elements of both domain and range, get rid of duplicates and write them in increasing order. ## What Makes a Relation a Function? On the other hand, a function is actually a “special” kind of relation because it follows an extra rule. Just like a relation, a function is also a set of ordered pairs; however, every x-value must be associated to only one y-value.https://tpc.googlesyndication.com/safeframe/1-0-37/html/container.htm Suppose we have two relations written in tables, • A relation that is not a function Since we have repetitions or duplicates of x-values with different y-values, then this relation ceases to be a function. • A relation that is a function This relation is definitely a function because every x-value is unique and is associated with only one value of y. So for a quick summary, if you see any duplicates or repetitions in the x-values, the relation is not a function. How about this example though? Is this not a function because we have repeating entries in x? Be very careful here. Yes, we have repeating values of x but they are being associated with the same value of y. The point (1,5) shows up twice, and while the point (3,-8) is written three times. This table can be cleaned up by writing a single copy of the repeating ordered pairs. The relation is now clearly a function! ### Examples of How to Determine if a Relation is also a Function Let’s go over a few more examples by identifying if a given relation is a function or not. Example 1: Is the relation expressed in the mapping diagram a function? Each element of the domain is being traced to one and only element in the range. However, it is okay for two or more values in the domain to share a common value in the range. That is, even though the elements 5 and 10 in the domain share the same value of 2 in the range, this relation is still a function.null Example 2: Is the relation expressed in the mapping diagram a function? What do you think? Does each value in the domain point to a single value in the range? Absolutely! There’s nothing wrong when four elements coming from the domain are sharing a common value in the range. This is a great example of a function as well. Example 3: Is the relation expressed in the mapping diagram a function? Messy? Yes! Confusing? Not really. The only thing I am after is to observe if an element in the domain is being “greedy” by wanting to be paired with more than one element in the range. The element 15 has two arrows pointing to both 7 and 9. This is a clear violation of the requirement to be a function. A function is well behaved, that is, each element in the domain must point to one element in the range. Therefore, this relation is not a function. Example 4: Is the relation expressed in the mapping diagram a function? If you think example 3 was “bad”, this is “worse”. A single element in the domain is being paired with four elements in the range. Remember, if an element in the domain is being associated with more than one element in the range, the relation is automatically disqualified to be a function. Thus, this relation is absolutely not a function. Example 5: Is the mapping diagram a relation, or function? Let me show you this example to highlight a very important idea about a function that is usually ignored. Your teacher may give you something like this just to check if you pay attention to the details of the definition of a function.null So far it looks normal. But there’s a little problem. The element “2” in the domain is not being paired with any element in the range. Here’s the deal! Every element in the domain must have some kind of correspondence to the elements in the range for it to be considered a relation, at least. Since this is not a relation, it follows that it can’t be a function. So, the final answer is neither a relation nor a function. ## Definition of One-to-One Functions A function has many types and one of the most common functions used is the one-to-one function or injective function. Also, we will be learning here the inverse of this function. One-to-One functions define that each element of one set say Set (A) is mapped with a unique element of another set, say, Set (B). Or Itcould be defined as each element of Set A has a unique element on Set B. In brief, let us consider ‘f’ is a function whose domain is set A. The function is said to be injective if for all x and y in A, Whenever f(x)=f(y), then x=y And equivalently, if x ≠ y, then f(x) ≠ f(y) ## Onto Function Definition (Surjective Function) Onto function could be explained by considering two sets, Set A and Set B which consist of elements. If for every element of B there is at least one or more than one element matching with A, then the function is said to be onto function or surjective function. The term for the surjective function was introduced by Nicolas Bourbaki. Onto Function In the first figure, you can see that for each element of B there is a pre-image or a matching element in Set A, therefore, its an onto function. But if you see in the second figure, one element in Set B is not mapped with any element of set A, so it’s not an onto or surjective function. Into Function : Let f : A —-> B be a function. There exists even a single element in B having no pre-image in A, then f is said to be an into function. The figure given below represents a one-one function. ## Into Function – Practice Problems Problem 1 : Let f : A —-> B. A, B and f are defined as A = {1, 2, 3} B = {5, 6, 7, 8} f = {(1, 5), (2, 8), (3, 6)} Is f into function? Explain. Solution : Write the elements of f (ordered pairs) using arrow diagram as shown below In the above arrow diagram, all the elements of A have images in B and every element of A has a unique image. That is, no element of A has more than one image. So, f is a function. There exists an element “7” in B having no pre-image in A. Therefore, f is into function. # Introduction to trigonometry Trigonometry is primarily a branch of mathematics that deals with triangles, mostly right triangles. In particular the ratios and relationships between the triangle’s sides and angles. # Trigonometry functions – introduction There are six functions that are the core of trigonometry. There are three primary ones that you need to understand completely: • Sine (sin) • Cosine (cos) • Tangent (tan) The other three are not used as often and can be derived from the three primary functions. Because they can easily be derived, calculators and spreadsheets do not usually have them. • Secant (sec) • Cosecant (csc) • Cotangent (cot) All six functions have three-letter abbreviations (shown in parentheses above). ## Definitions of the six functions Consider the right triangle above. For each angle P or Q, there are six functions, each function is the ratio of two sides of the triangle. The only difference between the six functions is which pair of sides we use. In the following table • a is the length of the side adjacent to the angle (x) in question. • o is the length of the side opposite the angle. • h is the length of the Hypotenuse. ## Trigonometric Functions in terms of Sine and Cosine Functions Let’s use sine and cosine functions to determine the other trigonometric functions. • cosec = 1/sin x, where ≠ nπ • sec x = 1/cos where ≠ (2n +1)π/ 2 • tan = sin x /cos x , where x ≠ (2n +1)π/ 2 • cot = cos x/ sin x, where ≠ nπ In all the above functions, n is an integer. For all the real values of x, we already know that, • sin2 x + cos2 x = 1, This lets us know that, • 1 + tan2 x = sec2 x 1 + cot2 x = cosec2 x We already know the values of trigonometric ratios for the angles of 0°, 30°, 45°,60° and 90°. We use the same values for trigonometric functions as well. The values of trigonometric functions thus are as shown in the table below: For knowing the values of cosec x, sec x and cot x we reciprocate the values of sin x, cos x and tan x, respectively. From the above discussion, we can now calculate the values of the various trigonometric functions by using the respective trigonometric ratios, as stated in the table above.
Applied Discrete Structures Section1.2Basic Set Operations Subsection1.2.1Definitions Definition1.2.1.Intersection. Let $$A$$ and $$B$$ be sets. The intersection of $$A$$ and $$B$$ (denoted by $$A \cap B$$) is the set of all elements that are in both $$A$$ and $$B\text{.}$$ That is, $$A \cap B = \{x:x \in A \textrm{ and } x \in B\}\text{.}$$ • Let $$A = \{1, 3, 8\}$$ and $$B = \{-9, 22, 3\}\text{.}$$ Then $$A \cap B = \{3\}\text{.}$$ • Solving a system of simultaneous equations such as $$x + y = 7$$ and $$x - y = 3$$ can be viewed as an intersection. Let $$A = \{(x,y): x + y = 7, x,y \in \mathbb{R}\}$$ and $$B = \{(x,y): x - y = 3, x,y\in \mathbb{R}\}\text{.}$$ These two sets are lines in the plane and their intersection, $$A \cap B = \{(5, 2)\}\text{,}$$ is the solution to the system. • $$\mathbb{Z}\cap \mathbb{Q}=\mathbb{Z}\text{.}$$ • If $$A = \{3, 5, 9\}$$ and $$B = \{-5, 8\}\text{,}$$ then $$A\cap B =\emptyset\text{.}$$ Definition1.2.3.Disjoint Sets. Two sets are disjoint if they have no elements in common. That is, $$A$$ and $$B$$ are disjoint if $$A \cap B = \emptyset\text{.}$$ Definition1.2.4.Union. Let $$A$$ and $$B$$ be sets. The union of $$A$$ and $$B$$ (denoted by $$A \cup B$$) is the set of all elements that are in $$A$$ or in $$B$$ or in both A and B. That is, $$A\cup B= \{x:x \in A\textrm{ or } x\in B\}\text{.}$$ It is important to note in the set-builder notation for $$A\cup B\text{,}$$ the word “or” is used in the inclusive sense; it includes the case where $$x$$ is in both $$A$$ and $$B\text{.}$$ • If $$A = \{2, 5, 8\}$$ and $$B = \{7, 5, 22\}\text{,}$$ then $$A \cup B = \{2, 5, 8, 7, 22\}\text{.}$$ • $$\displaystyle \mathbb{Z}\cup \mathbb{Q}=\mathbb{Q}.$$ • $$A \cup \emptyset = A$$ for any set $$A\text{.}$$ Frequently, when doing mathematics, we need to establish a universe or set of elements under discussion. For example, the set $$A = \{x : 81x^4 -16 = 0 \}$$ contains different elements depending on what kinds of numbers we allow ourselves to use in solving the equation $$81 x^4 -16 = 0\text{.}$$ This set of numbers would be our universe. For example, if the universe is the integers, then $$A$$ is empty. If our universe is the rational numbers, then $$A$$ is $$\{2/3, -2/3\}$$ and if the universe is the complex numbers, then $$A$$ is $$\{2/3, -2/3, 2i/3, - 2i/3\}\text{.}$$ Definition1.2.6.Universe. The universe, or universal set, is the set of all elements under discussion for possible membership in a set. We normally reserve the letter $$U$$ for a universe in general discussions. Subsection1.2.2Set Operations and their Venn Diagrams When working with sets, as in other branches of mathematics, it is often quite useful to be able to draw a picture or diagram of the situation under consideration. A diagram of a set is called a Venn diagram. The universal set $$U$$ is represented by the interior of a rectangle and the sets by disks inside the rectangle. $$A \cap B$$ is illustrated in Figure 1.2.8 by shading the appropriate region. The union $$A \cup B$$ is illustrated in Figure 1.2.9. In a Venn diagram, the region representing $$A \cap B$$ does not appear empty; however, in some instances it will represent the empty set. The same is true for any other region in a Venn diagram. Definition1.2.10.Complement of a set. Let $$A$$ and $$B$$ be sets. The complement of $$A$$ relative to $$B$$ (notation $$B - A$$) is the set of elements that are in $$B$$ and not in $$A\text{.}$$ That is, $$B-A=\{x: x\in B \textrm{ and } x\notin A\}\text{.}$$ If $$U$$ is the universal set, then $$U-A$$ is denoted by $$A^c$$ and is called simply the complement of $$A\text{.}$$ $$A^c=\{x\in U : x\notin A\}\text{.}$$ 1. Let $$U = \{1,2, 3, \text{...} , 10\}$$ and $$A = \{2,4,6,8, 10\}\text{.}$$ Then $$U-A = \{1, 3, 5, 7, 9\}$$ and $$A - U= \emptyset\text{.}$$ 2. If $$U = \mathbb{R}\text{,}$$ then the complement of the set of rational numbers is the set of irrational numbers. 3. $$U^c= \emptyset$$ and $$\emptyset ^c= U\text{.}$$ 4. The Venn diagram of $$B - A$$ is represented in Figure 1.2.11. 5. The Venn diagram of $$A^c$$ is represented in Figure 1.2.13. 6. If $$B\subseteq A\text{,}$$ then the Venn diagram of $$A- B$$ is as shown in Figure 1.2.14. 7. In the universe of integers, the set of even integers, $$\{\ldots , - 4,-2, 0, 2, 4,\ldots \}\text{,}$$ has the set of odd integers as its complement. Definition1.2.15.Symmetric Difference. Let $$A$$ and $$B$$ be sets. The symmetric difference of $$A$$ and $$B$$ (denoted by $$A\oplus B$$) is the set of all elements that are in $$A$$ and $$B$$ but not in both. That is, $$A \oplus B = (A \cup B) - (A \cap B)\text{.}$$ 1. Let $$A = \{1, 3, 8\}$$ and $$B = \{2, 4, 8\}\text{.}$$ Then $$A \oplus B = \{1, 2, 3, 4\}\text{.}$$ 2. $$A \oplus \emptyset = A$$ and $$A \oplus A = \emptyset$$ for any set $$A\text{.}$$ 3. $$\mathbb{R} \oplus \mathbb{Q}$$ is the set of irrational numbers. 4. The Venn diagram of $$A \oplus B$$ is represented in Figure 1.2.17. Subsection1.2.3SageMath Note: Sets To work with sets in Sage, a set is an expression of the form Set(list). By wrapping a list with Set( ), the order of elements appearing in the list and their duplication are ignored. For example, L1 and L2 are two different lists, but notice how as sets they are considered equal: L1=[3,6,9,0,3] L2=[9,6,3,0,9] [L1==L2, Set(L1)==Set(L2) ] The standard set operations are all methods and/or functions that can act on Sage sets. You need to evaluate the following cell to use the subsequent cell. A=Set(srange(5,50,5)) B=Set(srange(6,50,6)) [A,B] We can test membership, asking whether 10 is in each of the sets: [10 in A, 10 in B] The ampersand is used for the intersection of sets. Change it to the vertical bar, \|, for union. A & B Symmetric difference and set complement are defined as “methods” in Sage. Here is how to compute the symmetric difference of $$A$$ with $$B\text{,}$$ followed by their differences. [A.symmetric_difference(B),A.difference(B),B.difference(A)] Exercises1.2.4Exercises 1. Let $$A = \{0, 2, 3\}\text{,}$$ $$B = \{2, 3\}\text{,}$$ $$C = \{1, 5, 9\}\text{,}$$ and let the universal set be $$U = \{0, 1, 2, . . . , 9\}\text{.}$$ Determine: 1. $$\displaystyle A \cap B$$ 2. $$\displaystyle A \cup B$$ 3. $$\displaystyle B \cup A$$ 4. $$\displaystyle A \cup C$$ 5. $$\displaystyle A - B$$ 6. $$\displaystyle B - A$$ 7. $$\displaystyle A^c$$ 8. $$\displaystyle C^c$$ 9. $$\displaystyle A\cap C$$ 10. $$\displaystyle A\oplus B$$ 1. $$\displaystyle \{2,3\}$$ 2. $$\displaystyle \{0,2,3\}$$ 3. $$\displaystyle \{0,2,3\}$$ 4. $$\displaystyle \{0,1,2,3,5,9\}$$ 5. $$\displaystyle \{0\}$$ 6. $$\displaystyle \emptyset$$ 7. $$\displaystyle \{ 1,4,5,6,7,8,9\}$$ 8. $$\displaystyle \{0,2,3,4,6,7,8\}$$ 9. $$\displaystyle \emptyset$$ 10. $$\displaystyle \{0\}$$ 2. Let $$A\text{,}$$ $$B\text{,}$$ and $$C$$ be as in Exercise 1, let $$D = \{3, 2\}\text{,}$$ and let $$E = \{2, 3, 2\}\text{.}$$ Determine which of the following are true. Give reasons for your decisions. 1. $$\displaystyle A = B$$ 2. $$\displaystyle B = C$$ 3. $$\displaystyle B = D$$ 4. $$\displaystyle E=D$$ 5. $$\displaystyle A\cap B = B\cap A$$ 6. $$\displaystyle A \cup B = B \cup A$$ 7. $$\displaystyle A-B = B-A$$ 8. $$\displaystyle A \oplus B = B \oplus A$$ 3. Let $$U= \{1, 2, 3, . . . , 9\}\text{.}$$ Give examples of sets $$A\text{,}$$ $$B\text{,}$$ and $$C$$ for which: 1. $$\displaystyle A\cap (B\cap C)=(A\cap B)\cap C$$ 2. $$\displaystyle A\cap (B\cup C)=(A\cap B)\cup (A\cap C)$$ 3. $$\displaystyle (A \cup B)^c= A^c\cap B^c$$ 4. $$\displaystyle A \cup A^c = U$$ 5. $$\displaystyle A \subseteq A\cup B$$ 6. $$\displaystyle A\cap B \subseteq A$$ These are all true for any sets $$A\text{,}$$ $$B\text{,}$$ and $$C\text{.}$$ 4. Let $$U= \{1, 2, 3, . . . , 9\}\text{.}$$ Give examples to illustrate the following facts: 1. If $$A \subseteq B$$ and $$B \subseteq C\text{,}$$ then $$A\subseteq C\text{.}$$ 2. There are sets $$A$$ and $$B$$ such that $$A - B \neq B - A$$ 3. If $$U = A\cup B$$ and $$A \cap B = \emptyset\text{,}$$ it always follows that $$A = U - B\text{.}$$ 5. What can you say about $$A$$ if $$U = \{1, 2, 3, 4, 5\}\text{,}$$ $$B = \{2, 3\}\text{,}$$ and (separately) 1. $$\displaystyle A \cup B = \{1, 2, 3,4\}$$ 2. $$\displaystyle A \cap B = \{2\}$$ 3. $$\displaystyle A \oplus B = \{3, 4, 5\}$$ 1. $$\displaystyle \{1, 4\} \subseteq A \subseteq \{1, 2, 3, 4\}$$ 2. $$\displaystyle \{2\} \subseteq A \subseteq \{1, 2, 4, 5\}$$ 3. $$\displaystyle A = \{2, 4, 5\}$$ 6. Suppose that $$U$$ is an infinite universal set, and $$A$$ and $$B$$ are infinite subsets of $$U\text{.}$$ Answer the following questions with a brief explanation. 1. Must $$A^c$$ be finite? 2. Must $$A\cup B$$ be infinite? 3. Must $$A\cap B$$ be infinite? 7. Given that $$U$$ = all students at a university, $$D$$ = day students, $$M$$ = mathematics majors, and $$G$$ = graduate students. Draw Venn diagrams illustrating this situation and shade in the following sets: 1. evening students Let the sets $$D\text{,}$$ $$M\text{,}$$ $$G\text{,}$$ and $$U$$ be as in exercise 7. Let $$\lvert U \rvert = 16,000\text{,}$$ $$\lvert D \rvert = 9,000\text{,}$$ $$\|M\|= 300\text{,}$$ and $$\lvert G \rvert = 1,000\text{.}$$ Also assume that the number of day students who are mathematics majors is 250, 50 of whom are graduate students, that there are 95 graduate mathematics majors, and that the total number of day graduate students is 700. Determine the number of students who are:
# Can be divided into? Can be Divided Into: A Comprehensive Guide Division is a fundamental mathematical operation that involves breaking down a whole into equal parts or groups. It is a crucial concept that can be applied in various areas of life, from sharing food among friends to analyzing data in business. In this comprehensive guide, we will explore the different aspects of division, including its types, tools, techniques, and applications in real life. We will also discuss common mistakes to avoid when dividing things and provide tips and tricks for mastering the art of division. ## Understanding the Concept of Division Division is one of the four basic arithmetic operations, alongside addition, subtraction, and multiplication. It involves splitting a number or quantity into equal parts or groups. For example, if you have 12 apples and want to divide them equally among four people, you can use division to calculate that each person will get three apples. The symbol used to represent division is ÷ or /, and the result is called the quotient. ## The Different Types of Division There are two main types of division: exact division and division with remainder. Exact division occurs when the dividend (the number being divided) is evenly divisible by the divisor (the number dividing the dividend). For example, 12 ÷ 3 = 4, because 12 is evenly divisible by 3. Division with remainder, on the other hand, occurs when the divisor does not evenly divide the dividend. For example, 13 ÷ 3 = 4 with a remainder of 1, because 3 goes into 13 four times with one left over. ## When and Why Do We Need to Divide Things? Division is a useful tool in many situations, such as sharing resources, measuring quantities, solving equations, and analyzing data. For example, if you want to share a pizza equally among four people, you would use division to determine how many slices each person should get. If you want to measure the length of a rope, you can divide it into smaller units, such as inches or centimeters. In business, division can be used to calculate profit margins, market shares, and growth rates. ## The Tools and Techniques of Division There are several tools and techniques that can be used for division, depending on the context and level of precision required. One of the simplest methods is long division, which involves dividing the dividend by the divisor one digit at a time and carrying over the remainder to the next digit. Another method is short division, which involves dividing the dividend by the divisor without writing down all the steps. There are also various types of calculators and software programs that can perform division automatically. ## Common Mistakes to Avoid in Division Like any other mathematical operation, division can be prone to errors if not done carefully. Common mistakes include dividing by zero, forgetting to carry over remainders, confusing the dividend and divisor, and rounding off decimals incorrectly. To avoid these mistakes, it is important to double-check your work, use a calculator or other tools when necessary, and practice dividing different types of numbers. ## Applications of Division in Real Life Division is a ubiquitous concept in everyday life, from sharing food and expenses with friends to calculating percentages and discounts in shopping. It is also used in a wide range of professions, such as engineering, finance, science, and education. For example, civil engineers use division to calculate the quantities of materials needed for construction projects, while financial analysts use division to evaluate the performance of stocks and bonds. ## Mastering the Art of Division: Tips and Tricks To become proficient in division, it is important to practice regularly and master the basic skills, such as dividing by one-digit and two-digit numbers, decimals, and fractions. Some tips and tricks to make division easier include estimating the quotient before doing the actual division, using mental math and shortcuts, and breaking down the problem into smaller parts. It is also helpful to understand the properties of division, such as the commutative, associative, and distributive properties. In conclusion, division is a versatile and essential concept that plays a key role in many areas of life. By understanding the different types of division, its tools and techniques, and its applications in real life, you can become more confident and proficient in dividing things. Whether you are a student, a professional, or an everyday person, mastering the art of division can help you solve problems and make better decisions.
# Line Charts Line Graph is the innovative version of Bar Graph representation. If we connect the upper point of the first Bar to the upper point of the second Bar and then tie these dots, we will get a line. Repeating the procedure gives us the Line Graph representation. Line graph and bar graph are easy to comprehend. A Line Graph looks like this: ## Important Points: While solving the line chart questions, some points are very crucial to note these are as follows: 1. Understanding the various headings of DI table/graph/chart is very important. 2. Data Interpretation depends upon the type of questions asked. 3. Some questions are solved via reasoning process. 4. And solving some questions helps solving the other questions. This section comprises of questions in which the data collected in a particular discipline are represented by specific points joined together by straight lines. The points are plotted on a two-dimensional plane taking one parameter on the horizontal axis and the other on the vertical axis. The candidate is required to analyse the given information and thereafter answer the given questions on the basis of the analysis of data. ## TIPS on answering Aptitude Questions on Data Interpretation Tip #1: Use a physical calculator and a notepad Online proctored tests disallow candidates from using Calculator Software Applications in Desktop/ Laptop/ Mobile Device where the test is being given. However, a separate handheld calculator can be used. Use a notebook with a pencil to jot down relevant numbers to help you in making quick calculations. Tip #2: While calculating averages, it is easier to assume an approximate mean and calculation average of differences from that mean Question: What is the average marks scored by the student across all six periodical exams? Solution: You can assume the mean to be 380. The differences from the assumed mean are: -20, -15, -10, +5, +20, +25 Average of the differences = + 0.83333 Average Marks = Assumed Mean + Average Difference = 380.8333 Maximum Total Marks in each Periodical Exam = 500 Tip #3: Percentage of Change (Increase/ Decrease) = Difference/ Original Value * 100 Question: If the expenditure in 2000 is 25% more than expenditure in 1997, then the income in 1997 is what percent less than the income in 2000? Solution: Assume that Expenditure in 1997 is ₹ 100. Expenditure in 2000 = ₹ 125. Profits in 1997 = ₹ 45 (45% of expenditure) Income in 1997 = ₹ 145 Profits in 2000 = ₹ 75 (60% of expenditure) Income in 2000 = ₹200 Change in income from 2000 to 1997 = Change/ Original Value x 100 = 55/200 * 100 = 27.5% Percent Profit Earned by a Company Over the Years. %Profit = Income - Expenditure x 100 Expenditure LearnFrenzy provides you lots of fully solved "Line Charts" Questions and Answers with explanation.
Algebra and Trigonometry # 10.6Parametric Equations Algebra and Trigonometry10.6 Parametric Equations ## Learning Objectives In this section, you will: • Parameterize a curve. • Eliminate the parameter. • Find a rectangular equation for a curve defined parametrically. • Find parametric equations for curves defined by rectangular equations. Consider the path a moon follows as it orbits a planet, which simultaneously rotates around the sun, as seen in Figure 1. At any moment, the moon is located at a particular spot relative to the planet. But how do we write and solve the equation for the position of the moon when the distance from the planet, the speed of the moon’s orbit around the planet, and the speed of rotation around the sun are all unknowns? We can solve only for one variable at a time. Figure 1 In this section, we will consider sets of equations given by $x( t ) x( t )$ and $y( t ) y( t )$ where $t t$ is the independent variable of time. We can use these parametric equations in a number of applications when we are looking for not only a particular position but also the direction of the movement. As we trace out successive values of $t, t,$ the orientation of the curve becomes clear. This is one of the primary advantages of using parametric equations: we are able to trace the movement of an object along a path according to time. We begin this section with a look at the basic components of parametric equations and what it means to parameterize a curve. Then we will learn how to eliminate the parameter, translate the equations of a curve defined parametrically into rectangular equations, and find the parametric equations for curves defined by rectangular equations. ## Parameterizing a Curve When an object moves along a curve—or curvilinear path—in a given direction and in a given amount of time, the position of the object in the plane is given by the x-coordinate and the y-coordinate. However, both $x x$ and $y y$ vary over time and so are functions of time. For this reason, we add another variable, the parameter, upon which both $x x$ and $y y$ are dependent functions. In the example in the section opener, the parameter is time, $t. t.$ The $x x$ position of the moon at time, $t, t,$ is represented as the function $x(t), x(t),$ and the $y y$ position of the moon at time, $t, t,$ is represented as the function $y(t). y(t).$ Together, $x(t) x(t)$ and $y(t) y(t)$ are called parametric equations, and generate an ordered pair $( x(t),y(t) ). ( x(t),y(t) ).$ Parametric equations primarily describe motion and direction. When we parameterize a curve, we are translating a single equation in two variables, such as $x x$ and $y , y ,$ into an equivalent pair of equations in three variables, $x,y, x,y,$ and $t. t.$ One of the reasons we parameterize a curve is because the parametric equations yield more information: specifically, the direction of the object’s motion over time. When we graph parametric equations, we can observe the individual behaviors of $x x$ and of $y. y.$ There are a number of shapes that cannot be represented in the form $y=f(x), y=f(x),$ meaning that they are not functions. For example, consider the graph of a circle, given as $r 2 = x 2 + y 2 . r 2 = x 2 + y 2 .$ Solving for $y y$ gives $y=± r 2 − x 2 , y=± r 2 − x 2 ,$ or two equations: $y 1 = r 2 − x 2 y 1 = r 2 − x 2$ and $y 2 =− r 2 − x 2 . y 2 =− r 2 − x 2 .$ If we graph $y 1 y 1$ and $y 2 y 2$ together, the graph will not pass the vertical line test, as shown in Figure 2. Thus, the equation for the graph of a circle is not a function. Figure 2 However, if we were to graph each equation on its own, each one would pass the vertical line test and therefore would represent a function. In some instances, the concept of breaking up the equation for a circle into two functions is similar to the concept of creating parametric equations, as we use two functions to produce a non-function. This will become clearer as we move forward. ## Parametric Equations Suppose $t t$ is a number on an interval, $I. I.$ The set of ordered pairs, $( x(t), ( x(t),$ $y(t) ), y(t) ),$ where $x=f(t) x=f(t)$ and $y=g(t), y=g(t),$ forms a plane curve based on the parameter $t. t.$ The equations $x=f(t) x=f(t)$ and $y=g(t) y=g(t)$ are the parametric equations. ## Example 1 ### Parameterizing a Curve Parameterize the curve $y= x 2 −1 y= x 2 −1$ letting $x(t)=t. x(t)=t.$ Graph both equations. ### Analysis The arrows indicate the direction in which the curve is generated. Notice the curve is identical to the curve of $y= x 2 −1. y= x 2 −1.$ ## Try It #1 Construct a table of values and plot the parametric equations: $x( t )=t−3, x( t )=t−3,$ $y( t )=2t+4;−1≤t≤2. y( t )=2t+4;−1≤t≤2.$ ## Example 2 ### Finding a Pair of Parametric Equations Find a pair of parametric equations that models the graph of $y=1− x 2 , y=1− x 2 ,$ using the parameter $x( t )=t. x( t )=t.$ Plot some points and sketch the graph. ## Try It #2 Parameterize the curve given by $x= y 3 −2y. x= y 3 −2y.$ ## Example 3 ### Finding Parametric Equations That Model Given Criteria An object travels at a steady rate along a straight path $(−5,3) (−5,3)$ to $(3,−1) (3,−1)$ in the same plane in four seconds. The coordinates are measured in meters. Find parametric equations for the position of the object. ### Analysis Again, we see that, in Figure 5(c), when the parameter represents time, we can indicate the movement of the object along the path with arrows. ## Eliminating the Parameter In many cases, we may have a pair of parametric equations but find that it is simpler to draw a curve if the equation involves only two variables, such as $x x$ and $y. y.$ Eliminating the parameter is a method that may make graphing some curves easier. However, if we are concerned with the mapping of the equation according to time, then it will be necessary to indicate the orientation of the curve as well. There are various methods for eliminating the parameter $t t$ from a set of parametric equations; not every method works for every type of equation. Here we will review the methods for the most common types of equations. ### Eliminating the Parameter from Polynomial, Exponential, and Logarithmic Equations For polynomial, exponential, or logarithmic equations expressed as two parametric equations, we choose the equation that is most easily manipulated and solve for $t. t.$ We substitute the resulting expression for $t t$ into the second equation. This gives one equation in $x x$ and $y. y.$ ## Example 4 ### Eliminating the Parameter in Polynomials Given $x(t)= t 2 +1 x(t)= t 2 +1$ and $y(t)=2+t, y(t)=2+t,$ eliminate the parameter, and write the parametric equations as a Cartesian equation. ### Analysis This is an equation for a parabola in which, in rectangular terms, $x x$ is dependent on $y. y.$ From the curve’s vertex at $( 1,2 ), ( 1,2 ),$ the graph sweeps out to the right. See Figure 6. In this section, we consider sets of equations given by the functions $x( t ) x( t )$ and $y( t ), y( t ),$ where $t t$ is the independent variable of time. Notice, both $x x$ and $y y$ are functions of time; so in general $y y$ is not a function of $x. x.$ Figure 6 ## Try It #3 Given the equations below, eliminate the parameter and write as a rectangular equation for $y y$ as a function of $x. x.$ $x(t)=2 t 2 +6 y(t)=5−t x(t)=2 t 2 +6 y(t)=5−t$ ## Example 5 ### Eliminating the Parameter in Exponential Equations Eliminate the parameter and write as a Cartesian equation: $x(t)= e −t x(t)= e −t$ and $y(t)=3 e t , y(t)=3 e t ,$ $t>0. t>0.$ ### Analysis The graph of the parametric equation is shown in Figure 7(a). The domain is restricted to $t>0. t>0.$ The Cartesian equation, $y= 3 x y= 3 x$ is shown in Figure 7(b) and has only one restriction on the domain, $x≠0. x≠0.$ Figure 7 ## Example 6 ### Eliminating the Parameter in Logarithmic Equations Eliminate the parameter and write as a Cartesian equation: $x(t)= t +2 x(t)= t +2$ and $y(t)=log(t). y(t)=log(t).$ ### Analysis To be sure that the parametric equations are equivalent to the Cartesian equation, check the domains. The parametric equations restrict the domain on $x= t +2 x= t +2$ to $t>0; t>0;$ we restrict the domain on $x x$ to $x>2. x>2.$ The domain for the parametric equation $y=log(t) y=log(t)$ is restricted to $t>0; t>0;$ we limit the domain on $y=log ( x−2 ) 2 y=log ( x−2 ) 2$ to $x>2. x>2.$ ## Try It #4 Eliminate the parameter and write as a rectangular equation. $x(t)= t 2 y(t)=lntt>0 x(t)= t 2 y(t)=lntt>0$ ### Eliminating the Parameter from Trigonometric Equations Eliminating the parameter from trigonometric equations is a straightforward substitution. We can use a few of the familiar trigonometric identities and the Pythagorean Theorem. First, we use the identities: $x( t )=acost y( t )=bsint x( t )=acost y( t )=bsint$ Solving for $cost cost$ and $sint, sint,$ we have $x a =cost y b =sint x a =cost y b =sint$ Then, use the Pythagorean Theorem: $cos 2 t+ sin 2 t=1 cos 2 t+ sin 2 t=1$ Substituting gives $cos 2 t+ sin 2 t= ( x a ) 2 + ( y b ) 2 =1 cos 2 t+ sin 2 t= ( x a ) 2 + ( y b ) 2 =1$ ## Example 7 ### Eliminating the Parameter from a Pair of Trigonometric Parametric Equations Eliminate the parameter from the given pair of trigonometric equations where $0≤t≤2π 0≤t≤2π$ and sketch the graph. $x(t)=4cost y(t)=3sint x(t)=4cost y(t)=3sint$ ### Analysis Applying the general equations for conic sections (introduced in Analytic Geometry, we can identify $x 2 16 + y 2 9 =1 x 2 16 + y 2 9 =1$ as an ellipse centered at $( 0,0 ). ( 0,0 ).$ Notice that when $t=0 t=0$ the coordinates are $( 4,0 ), ( 4,0 ),$ and when $t= π 2 t= π 2$ the coordinates are $( 0,3 ). ( 0,3 ).$ This shows the orientation of the curve with increasing values of $t. t.$ ## Try It #5 Eliminate the parameter from the given pair of parametric equations and write as a Cartesian equation: $x(t)=2cost x(t)=2cost$ and $y(t)=3sint. y(t)=3sint.$ ## Finding Cartesian Equations from Curves Defined Parametrically When we are given a set of parametric equations and need to find an equivalent Cartesian equation, we are essentially “eliminating the parameter.” However, there are various methods we can use to rewrite a set of parametric equations as a Cartesian equation. The simplest method is to set one equation equal to the parameter, such as $x( t )=t. x( t )=t.$ In this case, $y( t ) y( t )$ can be any expression. For example, consider the following pair of equations. $x( t )=t y( t )= t 2 −3 x( t )=t y( t )= t 2 −3$ Rewriting this set of parametric equations is a matter of substituting $x x$ for $t. t.$ Thus, the Cartesian equation is $y= x 2 −3. y= x 2 −3.$ ## Example 8 ### Finding a Cartesian Equation Using Alternate Methods Use two different methods to find the Cartesian equation equivalent to the given set of parametric equations. $x(t)=3t−2 y(t)=t+1 x(t)=3t−2 y(t)=t+1$ ## Try It #6 Write the given parametric equations as a Cartesian equation: $x(t)= t 3 x(t)= t 3$ and $y(t)= t 6 . y(t)= t 6 .$ ## Finding Parametric Equations for Curves Defined by Rectangular Equations Although we have just shown that there is only one way to interpret a set of parametric equations as a rectangular equation, there are multiple ways to interpret a rectangular equation as a set of parametric equations. Any strategy we may use to find the parametric equations is valid if it produces equivalency. In other words, if we choose an expression to represent $x, x,$ and then substitute it into the $y y$ equation, and it produces the same graph over the same domain as the rectangular equation, then the set of parametric equations is valid. If the domain becomes restricted in the set of parametric equations, and the function does not allow the same values for $x x$ as the domain of the rectangular equation, then the graphs will be different. ## Example 9 ### Finding a Set of Parametric Equations for Curves Defined by Rectangular Equations Find a set of equivalent parametric equations for $y= ( x+3 ) 2 +1. y= ( x+3 ) 2 +1.$ ## Media Access these online resources for additional instruction and practice with parametric equations. ## 10.6 Section Exercises ### Verbal 1. What is a system of parametric equations? 2. Some examples of a third parameter are time, length, speed, and scale. Explain when time is used as a parameter. 3. Explain how to eliminate a parameter given a set of parametric equations. 4. What is a benefit of writing a system of parametric equations as a Cartesian equation? 5. What is a benefit of using parametric equations? 6. Why are there many sets of parametric equations to represent on Cartesian function? ### Algebraic For the following exercises, eliminate the parameter $t t$ to rewrite the parametric equation as a Cartesian equation. 7. ${ x( t )=5−t y( t )=8−2t { x( t )=5−t y( t )=8−2t$ 8. ${ x( t )=6−3t y( t )=10−t { x( t )=6−3t y( t )=10−t$ 9. ${ x( t )=2t+1 y( t )=3 t { x( t )=2t+1 y( t )=3 t$ 10. ${ x( t )=3t−1 y( t )=2 t 2 { x( t )=3t−1 y( t )=2 t 2$ 11. ${ x( t )=2 e t y( t )=1−5t { x( t )=2 e t y( t )=1−5t$ 12. ${ x( t )= e −2t y( t )=2 e −t { x( t )= e −2t y( t )=2 e −t$ 13. ${ x(t)=4log(t) y(t)=3+2t { x(t)=4log(t) y(t)=3+2t$ 14. ${ x(t)=log(2t) y(t)= t−1 { x(t)=log(2t) y(t)= t−1$ 15. ${ x( t )= t 3 −t y( t )=2t { x( t )= t 3 −t y( t )=2t$ 16. ${ x( t )=t− t 4 y( t )=t+2 { x( t )=t− t 4 y( t )=t+2$ 17. ${ x( t )= e 2t y( t )= e 6t { x( t )= e 2t y( t )= e 6t$ 18. ${ x( t )= t 5 y( t )= t 10 { x( t )= t 5 y( t )= t 10$ 19. ${ x(t)=4cost y(t)=5sint { x(t)=4cost y(t)=5sint$ 20. ${ x( t )=3sint y( t )=6cost { x( t )=3sint y( t )=6cost$ 21. ${ x(t)=2 cos 2 t y(t)=−sint { x(t)=2 cos 2 t y(t)=−sint$ 22. ${ x(t)=cost+4 y(t)=2 sin 2 t { x(t)=cost+4 y(t)=2 sin 2 t$ 23. ${ x(t)=t−1 y(t)= t 2 { x(t)=t−1 y(t)= t 2$ 24. ${ x(t)=−t y(t)= t 3 +1 { x(t)=−t y(t)= t 3 +1$ 25. ${ x(t)=2t−1 y(t)= t 3 −2 { x(t)=2t−1 y(t)= t 3 −2$ For the following exercises, rewrite the parametric equation as a Cartesian equation by building an $x-y x-y$ table. 26. ${ x(t)=2t−1 y(t)=t+4 { x(t)=2t−1 y(t)=t+4$ 27. ${ x(t)=4−t y(t)=3t+2 { x(t)=4−t y(t)=3t+2$ 28. ${ x(t)=2t−1 y(t)=5t { x(t)=2t−1 y(t)=5t$ 29. ${ x(t)=4t−1 y(t)=4t+2 { x(t)=4t−1 y(t)=4t+2$ For the following exercises, parameterize (write parametric equations for) each Cartesian equation by setting $x( t )=t x( t )=t$ or by setting $y(t)=t. y(t)=t.$ 30. $y( x )=3 x 2 +3 y( x )=3 x 2 +3$ 31. $y( x )=2sinx+1 y( x )=2sinx+1$ 32. $x( y )=3log( y )+y x( y )=3log( y )+y$ 33. $x( y )= y +2y x( y )= y +2y$ For the following exercises, parameterize (write parametric equations for) each Cartesian equation by using $x( t )=acost x( t )=acost$ and $y(t)=bsint. y(t)=bsint.$ Identify the curve. 34. $x 2 4 + y 2 9 =1 x 2 4 + y 2 9 =1$ 35. $x 2 16 + y 2 36 =1 x 2 16 + y 2 36 =1$ 36. $x 2 + y 2 =16 x 2 + y 2 =16$ 37. $x 2 + y 2 =10 x 2 + y 2 =10$ 38. Parameterize the line from $(3,0) (3,0)$ to $(−2,−5) (−2,−5)$ so that the line is at $(3,0) (3,0)$ at $t=0, t=0,$ and at $(−2,−5) (−2,−5)$ at $t=1. t=1.$ 39. Parameterize the line from $(−1,0) (−1,0)$ to $(3,−2) (3,−2)$ so that the line is at $(−1,0) (−1,0)$ at $t=0, t=0,$ and at $(3,−2) (3,−2)$ at $t=1. t=1.$ 40. Parameterize the line from $(−1,5) (−1,5)$ to $(2,3) (2,3)$ so that the line is at $(−1,5) (−1,5)$ at $t=0, t=0,$ and at $(2,3) (2,3)$ at $t=1. t=1.$ 41. Parameterize the line from $(4,1) (4,1)$ to $(6,−2) (6,−2)$ so that the line is at $(4,1) (4,1)$ at $t=0, t=0,$ and at $(6,−2) (6,−2)$ at $t=1. t=1.$ ### Technology For the following exercises, use the table feature in the graphing calculator to determine whether the graphs intersect. 42. 43. For the following exercises, use a graphing calculator to complete the table of values for each set of parametric equations. 44. ${ x 1 ( t )=3 t 2 −3t+7 y 1 ( t )=2t+3 { x 1 ( t )=3 t 2 −3t+7 y 1 ( t )=2t+3$ $t t$ $x x$ $y y$ –1 0 1 45. ${ x 1 ( t )= t 2 −4 y 1 ( t )=2 t 2 −1 { x 1 ( t )= t 2 −4 y 1 ( t )=2 t 2 −1$ $t t$ $x x$ $y y$ 1 2 3 46. ${ x 1 ( t )= t 4 y 1 ( t )= t 3 +4 { x 1 ( t )= t 4 y 1 ( t )= t 3 +4$ $t t$ $x x$ $y y$ -1 0 1 2 ### Extensions 47. Find two different sets of parametric equations for $y= ( x+1 ) 2 . y= ( x+1 ) 2 .$ 48. Find two different sets of parametric equations for $y=3x−2. y=3x−2.$ 49. Find two different sets of parametric equations for $y= x 2 −4x+4. y= x 2 −4x+4.$ Order a print copy As an Amazon Associate we earn from qualifying purchases.
how to find the missing side of a right triangle When we know 2 sides of the right triangle, use the Pythagorean theorem. Finding the length of a side of a non right angled triangle. Law of Sines $$\frac{sin(A)}{a} = \frac{sin(B)}{b} = \frac{sin(C)}{c}$$ Examples: Law of sines. Such a triangle can be solved by using Angles of a Triangle to find the other angle, and The Law of Sines to find each of the other two sides. Add your answer and earn points. Median response time is 34 minutes and may be longer for new subjects. In a right-angled triangle, the sides of the triangle are named as the opposite, adjacent, and hypotenuse sides. In the left triangle, the measure of the hypotenuse is missing. The Cosine Rule. To solve the missing side of a right triangle use Pythagorean Therom. Case I. He has helped many students raise their standardized test scores--and attend the colleges of their dreams. Step-by-step explanations are provided for each calculation. All values should be in positive values but decimals are allowed and valid. Finding the missing side of a right triangle is a pretty simple matter if two sides are known. Below is a brief of Pythagoras theorem. cos θ = Adjacent side / Hypotenuse side. - Choose either sin, cos, or tan by determining which side you know and which side you are looking for. The factors are the lengths of the sides and one of the two angles, other than the right angle. Mckenzie and Cara both tried to find the missing side of the right triangle. There are many ways to find the side length of a right triangle. Write the polynomials as a product of linear factors. 8 months ago. Finding the measurement of the third side of a triangle when you know the measurement of the other two sides only works if you have a right triangle or the measurement of at least one other angle. Preview. /. Pythagoras’ Theorem (The Pythagorean Theorem) The hypotenuse is the longest side of a right triangle, and is located opposite the right angle. 0. To find the length of the missing side of a right triangle we can use the following trigonometric ratios. We can do that math, 19 plus one more is 20. Solve the triangle in Figure $$\PageIndex{10}$$ for the missing side and find the missing angle measures to the nearest tenth. These are the four steps to follow: Step 1 Find the names of the two sides we are using, one we are trying to find and one we already know, out of Opposite, Adjacent and Hypotenuse. Is either of them correct? Sides "a" and "b" are the perpendicular sides and side "c" is the hypothenuse. The cosine rule can be used to find a missing side when all sides and an angle are involved in the question. Sal is given a right triangle with an acute angle of 65° and a leg of 5 units, and he uses trigonometry to find the two missing sides. Recall the three main trigonometric functions. A right triangle has two sides perpendicular to each other. A right ...” in Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions. Edit. A right triangle has two sides perpendicular to each other. Video. In this tutorial I show you how to find a length of a side of a non right-angled triangle by using the Cosine Rule. Free . sec θ = Hypotenuse side / Adjacent side You can do this if you are given the opposite angle and the two other sides. Our right triangle side and angle calculator displays missing sides and angles! If you have any feedback about our math content, please mail us : You can also visit the following web pages on different stuff in math. Mckenzie and Cara both tried to find the missing side of the right triangle. And then plus four more to 24, so one more to 20, and four more, that's a total of five more. Type in the given values. How do you calculate unknown or missing side of right triangle? How do you find the missing side of a triangle? cosec Î¸ = Hypotenuse side / Opposite side. How To Solve Similar Right Triangles. Your email address will not be published. See and . They help us to create proportions for finding missing side lengths! This will solve for the missing length and, if you have an HTML5 compatible web browser, redraw the triangle. The General Formula of . Step by step guide to finding missing sides and angles of a Right Triangle By using Sine, Cosine or Tangent, we can find an unknown side in a right triangle when we have one length, and one angle (apart from the right angle). Effortless Math: We Help Students Learn to LOVE Mathematics - Â© 2021. Find the length of the side marked x in the following triangle: Show Solution. Round answers to the nearest tenth. s i n ( 53) = x 12 x = 12 ⋅ s i n ( 53) x = 11.98. If you can’t immediately see which Pythagorean triple family a triangle belongs to, don’t worry: you can always use the following step-by-step method to pick the family and find the missing side. This was created as a revision aid for year 11s. There are three steps: 1. cos Î¸ = Adjacent side / Hypotenuse side, cosec Î¸ = Hypotenuse side / Opposite side, sec Î¸ = Hypotenuse side / Adjacent side. The formula is as follows: c 2 = a 2 + b 2. c = â(a 2 + b 2) What Is the Name of a Triangle With Two Equal Sides? It follows that any triangle in which the sides satisfy this condition is a right triangle. The cosine rule can be used to find a missing side when all sides and an angle are involved in the question. Use the law of sines to find the missing measurements of the triangles in these examples. And a test-prep expert who has been tutoring students since 2008 special right triangle where c the... 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# A Guide To Completing The Square This article will be about the algebra technique called completing the square. This technique is useful for finding the vertex form of a quadratic equation. Prerequisites One should be comfortable with arithmetic and algebra techniques such as factoring and expanding. What Is Completing The Square? Sometimes in order to succeed in mathematics, you need to know its secrets and tricks. Before I talk about the actual complete the square technique, I will first provide the motivation behind it. Say I have . The value 2 is stored in the y variable. If I did y = 2 + 0, I still have two. Likewise, y = 2 + 4 – 4 for example still gives me two. All I have is changed the form (or how it looks) without changing the actual value of y. Recall that is equal to . How do we go from a quadratic equation such as into a factored form such as ? We use the complete the square technique. The General Case and Method If you are not interested in the theory, you can skip to the examples section two sections below on how to use the complete the square technique. Let’s say we have a general quadratic equation such as: and we want it in factored form such as (k and j are used instead of a and b). We start with . Factoring the first two terms by gives us: We look for the coefficient that is with the which is . This is divided by 2 and its square is taken to obtain which is . Add and subtract by this amount as follows: The last line is the general formula for completing the square. You can either choose to memorize the last line or memorize the procedure to get to the last line. This method will make sense with examples. Examples In these examples, the complete the square technique is used to find the minimum or maximum point of a parabola. Example One Suppose you are given the function The coefficient with is . Half of two is 1 and one squared is 1. We add and subtract by 1 as follows: The roots of when are and . The vertex of this quadratic function is (-1, -1) which is a minimum. Example Two Here is another example with . Factoring the 2 from the first and second terms gives us: The coefficient with x is 3. Three divided by 2 is and is . Adding and subtracting by gives us: Expanding and simplifying gives: The vertex is ( , ) which is a minimum. Notes 1) You can add by “zero”. Remember that it is add and then subtract. 2) Completing the square may be needed to find an inverse of a function. 3) You don’t have to memorize the general formula. It is probably easier to memorize the procedures of completing the square. 4) I was taught to factor the first two terms and not the constant term. You can factor all three terms if you wish if you would like to do that variation. The featured square image is taken from http://www.freelearningchannel.com/l/Content/Materials/Mathematics/Algebra_I/textbooks/CK12_Algebra_I/html/10/4.html.
Question 1. # If (X – 2K) Is A Factor Of F(X), Which Of The Following Must Be True? When it comes to studying mathematics, there are certain topics that can cause confusion and can be difficult to understand. One such concept is factorization – the process of breaking down a polynomial into its constituent factors. The ability to identify which of a set of given numbers is a factor of another number is an important one for any student of mathematics, and this article will explore this concept in more detail. We’ll look at what it means when a number (X – 2K) is said to be a factor of another function F(x), and discuss which of the following must then be true. After reading this article, you will have gained a better understanding of this concept and feel more confident when it comes to solving problems with factorization. ## What is (X – 2K)? The quantity (X – 2K) is the difference between two given numbers X and K. This difference is a factor of the function F(X). Therefore, if (X – K) is a factor of F(X), then (X – 2K) must also be a factor of F(X). ## What is a factor of F(X)? A factor of F(X) is a polynomial that divides evenly into F(X). In other words, it is a number that can be multiplied by another number to produce F(X). ## Which of the following must be true? If (X – K) is a factor of F(X), then F(K) must be equal to 0. In other words, if you were to plug in the number K for X in the function F(X), the resulting output would have to be 0. ## Conclusion As we have seen, if (x – 2k) is a factor of f(x), then the constant k must be a zero. This means that the linear expression (x-2k) must simplify to x, and that any other factors which might exist in f(x) are not dependent on k. Additionally, we can also conclude that if there is no remainder when dividing f(x) by (x-2k), then this indicates that it is indeed a factor of F(X). In conclusion, understanding how to identify factors of polynomials is an essential skill for anyone studying mathematics at any level. 2. If (X – 2K) Is A Factor Of F(X), Which Of The Following Must Be True? Have you ever encountered a math problem that seemed to have no straightforward answer? If so, you may have asked yourself the question: “If (X – 2K) is a factor of F(X), which of the following must be true?” It’s a tricky question, but with a little bit of knowledge, it’s possible to figure out the answer. Let’s start by breaking down the question. In this type of problem, “X – 2K” is known as a polynomial. A polynomial is a mathematical expression that consists of multiple terms and is usually written in the form of a function. In this case, the function is F(X). The question then asks us to determine which of the following must be true if (X – 2K) is a factor of F(X). To answer this, we must first understand what a factor is. A factor is a number that divides into another number without leaving a remainder. In this case, if (X – 2K) is a factor of F(X), then it means that (X – 2K) divides evenly into F(X). So, the following must be true if (X – 2K) is a factor of F(X): 1. F(X) must be divisible by (X – 2K). 2. F(X) must have a factor equal to (X – 2K). 3. The degree of F(X) must be greater than or equal to the degree of (X – 2K). These are the three conditions that must be satisfied in order for (X – 2K) to be a factor of F(X). So there you have it: if (X – 2K) is a factor of F(X), then the following must be true.
Plotting points on a coordinate plane worksheet pdf in coordinate geometry is available here. In the worksheet on plotting points in the coordinate plane, the students can learn how to plot the points on graph paper. Practice plotting points on a coordinate plane worksheet answers help you to identify axes, quadrants, ordered pairs on the x-y plane. This plotting points on a graph worksheet pdf will help you to score better grades in the exams. ## Plotting Points in the Coordinate Plane Worksheet PDF with Answers Example 1. Plot the points (−2,3) in the coordinate plane. Solution: Given that the point is (-2,3) Here the x coordinate is -2 and the y coordinate is 3. Here the second coordinate is positive and the first coordinate is negative The x-coordinate is -2 moves two units to the left from the origin. The y-coordinate 3 moves three units up in the positive y-direction. Example 2. Plot the points (4,4) in the coordinate plane. Solution: Given that the point is (4,4) Here the x coordinate is 4 and the y coordinate is 4 both the coordinates are positive. The x-coordinate is 4 moves four units to the right. The y-coordinate is also 4 moves four units up in the positive y-direction. Example 3. Plot the points (0,-6) in the coordinate plane. Solution: Given that the points are (0,-6) Here the x-coordinate is 0 and the y-coordinate is -6. Here the x-coordinate is positive and the y-coordinate is negative at the point (0,-6) beginning at the origin. The x-coordinate is 0 not to move in either direction along the x-axis. The y-coordinate is -6 moves six units down in the negative y-direction. Example 4. Plot the point (-6,5) in the coordinate plane. Solution: In the given points -6 is the x coordinate and 5 is the y coordinate. Here the x coordinate is negative and the y coordinate is positive so the point lies in the second quadrant. We move 6 units to the left from the origin and then 5 units vertically up to plot the point (−6,5). Example 5. Plot the points (−4,6) in the coordinate plane. Solution: Given that the point is (-4,6) Here the x coordinate is -4 and the y coordinate is 6. Here the second coordinate is positive and the first coordinate is negative The x-coordinate is -4 moves four units to the left from the origin. The y-coordinate 6 moves six units up in the positive y-direction. Example 6. Plot the point (-9,-5) in the coordinate plane. Solution: In the given points -9 is the x coordinate and -5 is the y coordinate. Here the x coordinate is negative and the y coordinate is also negative so the point lies in the third quadrant. We move 9 units to the left from the origin and then 5 units vertically up to plot the point (−9,-5) Example 7. Plot the point (1,7) in the coordinate plane. Solution: In the given points 1 is the x coordinate and 7 is the y coordinate. Here the x coordinate is positive and the y coordinate is also positive so the point lies in the first quadrant. We move 1 unit to the left from the origin and then 7 units vertically up to plot the point (1,7) Example 8. Plot the point (9,-11) in the coordinate plane. Solution: In the given points 9 is the x coordinate and -11 is the y coordinate. Here the x coordinate is positive and the y coordinate is negative so the point lies in the fourth quadrant. We move 9 units to the left from the origin and then 11 units vertically up to plot the point (9,-11) Example 9. Plot the point (-1,-2) in the coordinate plane. Solution: In the given points -1 is the x coordinate and -2 is the y coordinate. Here the x coordinate is negative and the y coordinate is also negative so the point lies in the third quadrant. We move 1 unit to the left from the origin and then 2 units vertically up to plot the point (−1,-2) Example 10. Plot the point (-8,7) in the coordinate plane. Solution: In the given points -8 is the x coordinate and 7 is the y coordinate. Here the x coordinate is negative and the y coordinate is also negative so the point lies in the third quadrant. We move 8 units to the left from the origin and then 7 units vertically up to plot the point (−8,7)
# Show that Question: $x \frac{d y}{d x}-y+x \sin \left(\frac{y}{x}\right)=0$ Solution: $x \frac{d y}{d x}-y+x \sin \left(\frac{y}{x}\right)=0$ $\Rightarrow x \frac{d y}{d x}=y-x \sin \left(\frac{y}{x}\right)$ $\Rightarrow \frac{d y}{d x}=\frac{y-x \sin \left(\frac{y}{x}\right)}{x}$ ...(1) Let $F(x, y)=\frac{y-x \sin \left(\frac{y}{x}\right)}{x}$. $\therefore F(\lambda x, \lambda y)=\frac{\lambda y-\lambda x \sin \left(\frac{\lambda y}{\lambda x}\right)}{\lambda x}=\frac{y-x \sin \left(\frac{y}{x}\right)}{x}=\lambda^{0} \cdot F(x, y)$ Therefore, the given differential equation is a homogeneous equation. To solve it, we make the substitution as: $y=v x$ $\Rightarrow \frac{d}{d x}(y)=\frac{d}{d x}(v x)$ $\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$ Substituting the values of $y$ and $\frac{d y}{d x}$ in equation (1), we get: $v+x \frac{d v}{d x}=\frac{v x-x \sin v}{x}$ $\Rightarrow v+x \frac{d v}{d x}=v-\sin v$ $\Rightarrow-\frac{d v}{\sin v}=\frac{d x}{x}$ $\Rightarrow \operatorname{cosec} v d v=-\frac{d x}{x}$ Integrating both sides, we get: $\log \|\operatorname{cosec} v-\cot v\|=-\log x+\log \mathrm{C}=\log \frac{\mathrm{C}}{x}$ $\Rightarrow \operatorname{cosec}\left(\frac{y}{x}\right)-\cot \left(\frac{y}{x}\right)=\frac{\mathrm{C}}{x}$ $\Rightarrow \frac{1}{\sin \left(\frac{y}{x}\right)}-\frac{\cos \left(\frac{y}{x}\right)}{\sin \left(\frac{y}{x}\right)}=\frac{\mathrm{C}}{x}$ $\Rightarrow x\left[1-\cos \left(\frac{y}{x}\right)\right]=\mathrm{C} \sin \left(\frac{y}{x}\right)$ This is the required solution of the given differential equation.
# What is 1/375 as a decimal? ## Solution and how to convert 1 / 375 into a decimal 1 / 375 = 0.003 1/375 converted into 0.003 begins with understanding long division and which variation brings more clarity to a situation. Both are used to handle numbers less than one or between whole numbers, known as integers. Depending on the situation, decimals can be more clear. We don't say 1 and 1/2 dollar. We use the decimal version of \$1.50. Same goes for fractions. We will say 'the student got 2 of 3 questions correct'. After deciding on which representation is best, let's dive into how we can convert fractions to decimals. ## 1/375 is 1 divided by 375 Teaching students how to convert fractions uses long division. The great thing about fractions is that the equation is already set for us! Fractions have two parts: Numerators and Denominators. This creates an equation. To solve the equation, we must divide the numerator (1) by the denominator (375). Here's 1/375 as our equation: ### Numerator: 1 • Numerators are the parts to the equation, represented above the fraction bar or vinculum. Small values like 1 means there are less parts to divide into the denominator. The bad news is that it's an odd number which makes it harder to covert in your head. Values like 1 doesn't make it easier because they're small. Now let's explore the denominator of the fraction. ### Denominator: 375 • Denominators differ from numerators because they represent the total number of parts which can be found below the vinculum. Larger values over fifty like 375 makes conversion to decimals tougher. But the bad news is that odd numbers are tougher to simplify. Unfortunately and odd denominator is difficult to simplify unless it's divisible by 3, 5 or 7. Overall, two-digit denominators are no problem with long division. Let's start converting! ## Converting 1/375 to 0.003 ### Step 1: Set your long division bracket: denominator / numerator $$\require{enclose} 375 \enclose{longdiv}{ 1 }$$ We will be using the left-to-right method of calculation. This method allows us to solve for pieces of the equation rather than trying to do it all at once. ### Step 2: Extend your division problem $$\require{enclose} 00. \\ 375 \enclose{longdiv}{ 1.0 }$$ Because 375 into 1 will equal less than one, we canâ€™t divide less than a whole number. Place a decimal point in your answer and add a zero. Now 375 will be able to divide into 10. ### Step 3: Solve for how many whole groups you can divide 375 into 10 $$\require{enclose} 00.0 \\ 375 \enclose{longdiv}{ 1.0 }$$ We can now pull 0 whole groups from the equation. Multiple this number by our furthest left number, 375, (remember, left-to-right long division) to get our first number to our conversion. ### Step 4: Subtract the remainder $$\require{enclose} 00.0 \\ 375 \enclose{longdiv}{ 1.0 } \\ \underline{ 0 \phantom{00} } \\ 10 \phantom{0}$$ If your remainder is zero, that's it! If there is a remainder, extend 375 again and pull down the zero ### 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 1/375 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. This is also true for percentages. So we sometimes overlook fractions and decimals because they seem tedious or something we only use in math class. But 1/375 and 0.003 bring clarity and value to numbers in every day life. Here are just a few ways we use 1/375, 0.003 or 0% in our daily world: ### When you should convert 1/375 into a decimal Investments - Comparing currency, especially on the stock market are great examples of using decimals over fractions. ### When to convert 0.003 to 1/375 as a fraction Time - spoken time is used in many forms. But we don't say It's '2.5 o'clock'. We'd say it's 'half passed two'. ### Practice Decimal Conversion with your Classroom • If 1/375 = 0.003 what would it be as a percentage? • What is 1 + 1/375 in decimal form? • What is 1 - 1/375 in decimal form? • If we switched the numerator and denominator, what would be our new fraction? • What is 0.003 + 1/2?
AP State Syllabus AP Board 6th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.1 Textbook Questions and Answers. ## AP State Syllabus 6th Class Maths Solutions 11th Lesson Perimeter and Area Ex 11.1 Question 1. Find the perimeters of the following figures. i) Check whether the perimeter of Δ XYZ = 3 x Length of the side? ii) Check whether the perimeter of □ ABCD = 4 x Length of the side? iii) Check whether the perimeter of □ PQRS = 4 x Length of the side? i) In ΔXYZ, XY = 2cm, YZ = 2cm and XZ = 2cm Perimeter of ΔXYZ = XY + YZ + XZ = 2 + 2 + 2 = 6cm Perimeter of ΔXYZ = 6 cm By observing that perimeter of ΔXYZ = 6cm = 3 x 2cm = 3 x length of the side in ΔABC So, perimeter of ΔXYZ = 3 x length of the side ii) Given in quadrilateral ABCD, AB = 3cm, BC = 3cm, CD = 3cm and AD Perimeter of □ ABCD = AB + BC + CD + AD = 3 + 3 + 3 + 3 Perimeter of □ ABCD = 12 cm By observing that perimeter of □ ABCD = 12cm = 4 x 3cm = 4 x length of the side in □ ABCD So, perimeter of □ ABCD = 4 x length of the side iii)In quadrilateral PQRS, . Given PQ = 2cm, QR = 2cm, RS = 2cm and PS = 2cm Perimeter of PQRS = PQ + QR + RS + PS = 2 + 2 + 2 + 2 = 8cm Perimeter of PQRS = 8cm. By observing that perimeter of PQRS = 8cm = 4×2 cm = 4 x length of the side in PQRS So, perimeter of PQRS = 4 x length of the side. Question 2. Measurements of two rectangular fields are 50m x 30m and 60m x 40m. Find their perimeters. Check whether the perimeters are 2 x length + 2 x breadth. Solution: Given, in rectangle ABCD, AB = DC = 50 m and BC = DA = 30 m Perimeter of rectangle ABCD = AB + BC + CD + DA = 50 + 30 + 50 + 30 = 2 x 50 + 2x 30 = 160 m ∴ Perimeter of rectangle ABCD = 2 x length + 2 x breadth Given in rectangle PQRS, PQ = RS = 60 m and QR = SP = 40 m Perimeter of rectangle PQRS = PQ + QR + RS + SP = 60 + 40 + 60 + 40 = 2 x 60 + 2 x 40 = 200 m ∴ Perimeter of rectangle PQRS = 2 x length + 2 x breadth Question 3. Find the perimeter of a) An equilateral triangle whose side is 3.5cm. b) A square whose side is 4.8cm. Solution: a) Given side of an equilateral triangle is 3.5 cm. We know that, perimeter of an equilateral triangle = 3 x length of the side = 3 x 3.5 = 10.5 cm . ∴ Perimeter = 10.5 cm b) Given side of a square = 4.8 cm We know that, perimeter of a square = 4 x length of the side 4 x 4.8 ∴ Perimeter = 19.2 cm Question 4. Length and breadth of top of one table is 160cm and 90cm respectively. Find how . much length of beading is required for each table. Solution: Given the length of top of table =160 cm breadth of top of table = 90 cm g To find the length of beading we have to find the ® perimeter of the table top. . So, perimeter of table top = 160 + 90 + 160 + 90 = 500 cm 160 cm Required length of beading of table top = 500 cm. Question 5. Manasa has 24cm of metallic wire with her. She wanted to make some polygons with equal sides whose sides are integral without milling into pieces values. Find how many such polygons she can make with the length of 24cm metallic wire? Solution: Given length of metallic wire = 24 cm We know that 24 = 1 x 24 = 2 x 12 = 3 x 8 = 4 x 6 ∴ 24 can be divided into 1,2,3,4,6,8,12 & 24 But we can’t form polygons with sides 1 & 2. ∴ The polygons with sides 3, 4, 6, 8, 12 and 24 can be formed. The length of the sides are equal so and units. They are 8 cm, 6cm, 4cm, 3cm, 2cm and 1cm respectively Question 6. Find the perimeter of the following figures. (i) and (ii) i) Perimeter of the given polygon is sum of the lengths of its all sides. Perimeter = 5cm + 3cm+ 1cm + 2cm + 1 cm + 1 cm + 1 cm + 1 cm + 1cm + 2cm + 1 cm + 3cm Perimeter = 22 cm ii) Perimeter of the given polygon is sum of the lengths of its all sides. Perimeter = 1cm + 5cm + 1 cm + lcm+ 2cm + 1cm + 1cm + 1cm + 1cm + 1cm + 2cm + 1cm Perimeter = 18cm Question 7. Statement P : So many rectangles exists with the same perimeter. Statement Q : So many squares exists with the same perimeter. Which option is correct? A) P wrong Q correct B) P correct Q wrong C) P and Q are correct D) P and Q are wrong Solution: B) P correct Q wrong. Lengths and breadths of a rectangle can change for the same perimeter. But, the side of a square cannot change for the same perimeter.
# How do you know if a number is a power of two? ## How do you know if a number is a power of two? Method-2: Keep dividing by 2 Keep dividing the number by two, i.e, do n = n/2 iteratively until n becomes 1. In any iteration, if n%2 becomes non-zero and n is not 1 then n is not a power of 2. If n becomes 1 then it is a power of 2. ## Is power of 2 A LeetCode? Power of Two – LeetCode. Given an integer n , return true if it is a power of two. Otherwise, return false . An integer n is a power of two, if there exists an integer x such that n == 2x . ## Is power of 4 A LeetCode? Power of Four – LeetCode. Given an integer n , return true if it is a power of four. An integer n is a power of four, if there exists an integer x such that n == 4x . ## What is the power of 2 C++? Explanation. A simple method for this is to simply take the log of the number on base 2 and if you get an integer then the number is the power of 2. ## How do you know if a number is a power of 4? A simple method is to take a log of the given number on base 4, and if we get an integer then the number is the power of 4. 2. Another solution is to keep dividing the number by 4, i.e, do n = n/4 iteratively. ## How do you check whether a number is power of 3 or not? Suppose we have a number n. We have to check whether the number is the power of 3 or not. So if the number is like n = 27, that is the power of 3, the result will be true, if n = 15, it will be false. ## How do you know if a number is a power of N? Following are detailed step. 1) Initialize pow = x, i = 1 2) while (pow < y) { pow = powpow i = 2 } 3) If pow == y return true; 4) Else construct an array of powers from x^i to x^(i/2) 5) Binary Search for y in array constructed in step 4. ## How do you check if a number is a power of 5? First check if last digit is 5. If last digit is 5; divide it by 5. If result of division is 1, then number is power of 5. ## How do you find the power of a number without a calculator? So, for example, this is how you would solve 6^3 without a calculator, from start to finish. Write: 6 6 6, because the base number is 6 and the exponent is 3. Then write: 6 x 6 x 6, to place multiplication signs between each of the base numbers. After that, multiply out the first multiplication sign, or 6 x 6 = 36. ## How do you tell if a number is a power of 10? Number is a power of 10 if it’s equal to 10, 100, 1000 etc. 1 is also 0-th power of 10. Other numbers like 2, 3, 11, 12 etc. are not powers of 10. ## How do I find the power of a number? The power of a number is shown by the exponent. The exponent indicates how many times the base number is going to be multiplied by itself. Any number that is raised to a power is going to grow very fast! Begin typing your search term above and press enter to search. Press ESC to cancel.
# Quadratic Equation – Exercise 9.6 – Class X So far we have learnt to find the roots of given quadratic equations by different methods. We see that the roots are all real numbers. Is it possible to determine the nature of roots of a given quadratic equation, without actually finding them? Now, let us learn about this. Study the following examples: 1. ### Consider the quadratic equation x2 – 2x + 1 = 0 This is in the form of ax2 + bx + c = 0; a = 1 , b = -2 and c = 1 1. ### Consider the equation x2 – 2x – 3 = 0 This is in the form ax2 + bx + c = 0, where a = 1 , b = -2 , c = -3 1. ### Consider the quadratic equation x2 – 2x + 3 = 0 This is in the form ax2 + bx + c = 0, where a = 1 , b = -2 , c = 3 From the above examples, it is evident that the roots of a quadratic equation can be real and equal, real and distinct or imaginary. Also, observe that the value of b2 – 4ac determines the nature of the roots. We say the nature of roots depends on the values of b2 – 4ac. The value of the expression b2 – 4ac discriminates the nature of the roots of ax2 + bx + c = 0 and so it is called the discriminant of the quadratic equation. It is denoted by symbol ∆ and real as delta. In general, the roots of the quadratic equation ax2 + bx + c = 0 are The above results are presented in the table given below: Discriminant Nature of roots ∆ = 0 real and equal ∆ > 0 real and distinct ∆ < 0 no real roots(imaginary roots) Example 1: Determine the nature of the roots of the equation 2x2 – 5x – 1 = 0 Solution: This is in the form ax2 + bx + c = 0. The coefficient are a = 2, b = -5, c = -1 ∆ = b2 – 4ac = (-5)2 – 4(2)(-1) = 25 + 8 = 33 Example 2: Determine the nature of the roots of the eqation 4x2 – 4x + 1 = 0 Solution: Consider the equation 4x2 – 4x + 1 = 0 This is in the form ax2 + bx + c = 0. The coefficient are a = 4, b = -4, c = 1 ∆ = b2 – 4ac = (-4)2 – 4(4)(1) = 16 – 16 = 0 Therefore, the roots of are real and equal. Example 3: For what positive values of m, roots of the equation x2 + mx + 4 = 0 are (i) equal (ii) distinct Solution: Consider the equation x2 + mx + 4 = 0 This is in the form ax2 + bx + c = 0. The coefficient are a = 1, b = m, c = 4 ∆ = b2 – 4ac ∆ = m2 – 4(1)(4) ∆ = m2 – 16 (i) If roots are equal then ∆ = 0 m2 – 16 = 0 m2 = 16 m = ±4 (ii) If roots are distinct, then ∆ > 0 m2 – 16 >0 m2 > 16 m > √16 m > 4 ## Quadratic equation – Exercise 9.6 – Class X 1. Discuss the nature of roots of the following equations: (i) y2 – 7y + 2 = 0 (ii) x2 – 2x + 3 = 0 (iii) 2n2 + 5n – 1 = 0 (iv) a2 + 4a + 4 = 0 (v) x2 + 3x – 4 = 0 (vi) 3d2 – 2d + 1 = 0 2. For what positive values of m roots of the following equation are a) Equal b) Distinct c) Imaginary (i) a2 – ma + 1 = 0 (ii) x2 – mx + 9 = 0 (iii) r2 – (m + 1)r + 4 = 0 (iv) mk2 – 3k + 1 = 0 3. Find the value of ‘p’ for which the quadratic equations have equal roots. (i) x2 – px + 9 = 0 (ii) 2a2 + 3a + p = 0 (iii) pk2 – 12k + 9 = 0 (iv) 2y2 – py + 1 = 0 (v) (p + 1)n2 + 2(p + 3)n + (p + 8) = 0 (vi) (3p + 1)c2 + 2(p + 1)c + p = 0 ## Quadratic equation – Exercise 9.6 – Class X 1. Discuss the nature of roots of the following equations: (i) y2 – 7y + 2 = 0 Solution: Consider the equation y2 – 7y + 2 = 0 This is in the form ax2 + bx + c = 0. The coefficient are a = 1, b = -7, c = 2 ∆ = b2 – 4ac ∆ = (-7)2 – 4(1)(2) ∆ = 49 – 8 ∆ = 41 Hence, ∆ = 41 , therefore, ∆ > 0 Therefore, the roots of the equation y2 – 7y + 2 = 0 are real and distinct. (ii) x2 – 2x + 3 = 0 Solution: Consider the equation x2 – 2x + 3 = 0 This is in the form ax2 + bx + c = 0. The coefficient are a = 1, b = -2, c = 3 ∆ = b2 – 4ac ∆ = (-2)2 – 4(1)(3) ∆ = 4 – 12 ∆ = -8 Hence, ∆ = -8 , therefore, ∆ < 0 Thus, the roots of the equation x2 – 2x + 3 = 0 are imaginary. (iii) 2n2 + 5n – 1 = 0 Solution: Consider the equation 2n2 + 5n – 1 = 0 This is in the form ax2 + bx + c = 0. The coefficient are a = 2, b = 5, c = -1 ∆ = b2 – 4ac ∆ = (5)2 – 4(2)(-1) ∆ = 25 + 8 ∆ = 33 Hence, ∆ = 33 , therefore, ∆ > 0 Therefore, the roots of the equation 2n2 + 5n – 1 = 0 are real and distinct. (iv) a2 + 4a + 4 = 0 Solution: Consider the equation a2 + 4a + 4 = 0 This is in the form ax2 + bx + c = 0. The coefficient are a = 1, b = 4, c = 4 ∆ = b2 – 4ac ∆ = (4)2 – 4(1)(4) ∆ = 16 – 16 ∆ = 0 Hence, ∆ = 0 Therefore, the roots of the equation a2 + 4a + 4 = 0 are real and equal (v) x2 + 3x – 4 = 0 Solution: Consider the equation x2 + 3x – 4 = 0 This is in the form ax2 + bx + c = 0. The coefficient are a = 1, b = 3, c = -4 ∆ = b2 – 4ac ∆ = (3)2 – 4(1)(-4) ∆ = 9 + 16 ∆ = 25 Hence, ∆ = 25 , therefore, ∆ > 0 Therefore, the roots of the equation x2 + 3x – 4 = 0 are real and distinct. (vi) 3d2 – 2d + 1 = 0 Solution: Consider the equation 3d2 – 2d + 1 = 0 This is in the form ax2 + bx + c = 0. The coefficient are a = 3, b = -2, c = 1 ∆ = b2 – 4ac ∆ = (-2)2 – 4(3)(1) ∆ = 4 – 12 ∆ = -8 Hence, ∆ = -8 , therefore, ∆ < 0 Therefore, the roots of the equation 3d2 – 2d + 1 = 0 are imaginary. 2. For what positive values of m roots of the following equation are a) Equal b)Distinct c)Imaginary (i) a2 – ma + 1 = 0 Solution: Consider the equation a2 – ma + 1 = 0 This is in the form ax2 + bx + c = 0. The coefficient are a = 1, b = -m, c = 1 ∆ = b2 – 4ac ∆ = (-m)2 – 4(1)(1) ∆ = m2 – 4 (i) If roots are equal then ∆ = 0 m2 – 4 = 0 m2 = 4 Therefore, at m = 2 the roots of the quadratic equation a2 – ma + 1 = 0 are equal. (ii) If roots are distinct, then ∆ > 0 m2 – 4 >0 m2 > 4 m > √4 ⸫ m > +2 (iii) If roots are imaginary, then ∆ < 0 m2 – 4 < 0 m2 < 4 m < √4 ⸫ m < -2 (ii) x2 – mx + 9 = 0 Solution: Consider the equation x2 – mx + 9 = 0 This is in the form ax2 + bx + c = 0. The coefficient are a = 1, b = -m, c = 9 ∆ = b2 – 4ac ∆ = (-m)2 – 4(1)(9) ∆ = m2 – 36 (i) If roots are equal then ∆ = 0 m2 – 36 = 0 m2 = 36 m = 6 (ii) If roots are distinct, then ∆ > 0 m2 – 36 >0 m2 > 36 m > √36 ⸫ m > +6 (iii) If roots are imaginary, then ∆ < 0 m2 – 36 < 0 m2 < 36 m < √36 ⸫ m < -6 (iii) r2 – (m + 1)r + 4 = 0 Solution: Consider the equation r2 – (m + 1)r + 4 = 0 This is in the form ax2 + bx + c = 0. The coefficient are a = 1, b = -(m+1), c = 4 ∆ = b2 – 4ac ∆ = [-(m+1)]2 – 4(1)(4) ∆ = (m + 1)2 – 16 ∆ = m2 + 2m + 1 – 16 ∆ = m2 + 2m – 15 (i) If roots are equal then ∆ = 0 m2 + 2m – 15 =0 m2 + 5m – 3m – 15 = 0 m(m + 5)-3(m + 5) = 0 (m – 3)(m + 5) = 0 m – 3 = 0 or m + 5 = 0 m = 3 or m = -5 Therefore, at no value m can be the quadratic equation mk2 – 3k + 1 = 0 which has equal roots. (ii) If roots are distinct, then ∆ > 0 m2 + 2m – 15 >0 m2 – 3m + 5m – 15 > 0 m(m – 3)-1(m +3) > 0 (m + 3)(m – 1) > 0 m + 3 > 0 or m – 1 > 0 ⸫ m > 1 (iii) If roots are imaginary, then ∆ < 0 m2 + 2m – 3 < 0 m2 +3m – m – 3 < 0 m(m + 3)-1(m +3) < 0 (m + 3)(m – 1) < 0 m + 3 < 0 or m – 1 < 0 ⸫ m < -3 (iv) mk2 – 3k + 1 = 0 Solution: Consider the equation mk2 – 3k + 1 = 0 This is in the form ax2 + bx + c = 0. The coefficient are a = m, b = -3, c = 1 ∆ = b2 – 4ac ∆ = [-3]2 – 4(m)(1) ∆ = 9 – 4m (i) If roots are equal then ∆ = 0 9 – 4m =0 -4m = – 9 m = 9/4 Therefore, at m = 9/4 the quadratic equation mk2 – 3k + 1 = 0 has equal roots (ii) If roots are distinct, then ∆ > 0 9 – 4m > 0 -4m > – 9 ⸫ m > 9/4 (iii) If roots are imaginary, then ∆ < 0 9 – 4m < 0 -4m < – 9 ⸫ m < 9/4 3.. Find the value of ‘p’ for which the quadratic equations have equal roots. (i) x2 – px + 9 = 0 Solution: x2 – px + 9 = 0 This is in the form ax2 + bx + c = 0. The coefficient are a = 1, b = -p, c = 9 ∆ = (-p)2 – 4x1x9 ∆ = [p]2 – 36 ∆ = p2 – 36 If roots of quadratic equation x2 – px + 9 = 0 then ∆ = 0 p2 – 36 = 0 p2 = 36 p = √36 = 6 (ii) 2a2 + 3a + p = 0 Solution: 2a2 + 3a + p = 0 This is in the form ax2 + bx + c = 0. The coefficient are a = 2, b = 3, c = p ∆ = (3)2 – 4x2xp ∆ = [3]2 – 8p ∆ = 9 – 8p If roots of quadratic equation 2a2 + 3a + p = 0 then ∆ = 0 9 – 8p = 0 9 = 8p p = 9/8 (iii) pk2 – 12k + 9 = 0 Solution: pk2 – 12k + 9 = 0 This is in the form ax2 + bx + c = 0. The coefficient are a = p, b = -12, c = 9 ∆ = b2 – 4ac ∆ = [-12]2 – 4(p)(9) ∆ = 144 – 36p If roots of quadratic equation pk2 -12k + 9 = 0 then ∆ = 0 144 – 36p = 0 36p = 144 p = 144/36 = 24/6 = 4 Therefore, p = 4 . (iv) 2y2 – py + 1 = 0 Solution: 2y2 – py + 1 = 0 This is in the form ax2 + bx + c = 0. The coefficient are a = 2, b = -p, c = 1 ∆ = b2 – 4ac ∆ = [-p]2 – 4(2)(1) ∆ = p2 – 8 If roots of quadratic equation 2y2 – py + 1 = 0 then ∆ = 0 p2 – 8 = 0 p2 = 8 p = √8 = 2√2 Therefore, p = 2√2 (v) (p + 1)n2 + 2(p + 3)n + (p + 8) = 0 Solution: Consider the quadratic equation (p + 1)n2 + 2(p + 3)n + (p + 8) = 0 This is in the form ax2 + bx + c = 0. The coefficient are a = (p + 1), b = 2(p + 3), c = (p + 8) ∆ = b2 – 4ac ∆ = [2(p+3)]2 – 4(p+1)(p+8) ∆ = 4(p + 3)2 – 4[p2 + 8p + p + 8] ∆ = p2 + 6p + 9 – p2 – 9p – 8 ∆ =-3p + 1 If roots of quadratic equation (p + 1)n2 + 2(p + 3)n + (p + 8) = 0 then ∆ = 0 -3p + 1 = 0 -3p = – 1 p = 1/3 (vi) (3p + 1)c2 + 2(p + 1)c + p = 0 Solution: Consider the quadratic equation (3p + 1)c2 + 2(p + 1)c + p = 0 This is in the form ax2 + bx + c = 0. The coefficient are a = (3p + 1), b = 2(p + 1), c = p ∆ = b2 – 4ac ∆ = [2(p + 1)]2 – 4(3p+1)p ∆ = 4[p2 + 2p + 1] – 12p2 – 4p ∆ = 4p2 + 8p + 4 – 12p2 – 4p ∆ = -8p2 + 4p + 4 ∆ = 8p2 – 4p – 4 If roots of quadratic equation (3p + 1)c2 + 2(p + 1)c + p = 0 then ∆ = 0 8p2 – 4p – 4 = 0 8p2 – 8p + 4p – 4 = 0 8p(p – 1)+4(p – 1) = 0 (8p + 4)(p – 1) = 0 8p + 4 = 0 or p – 1 = 0 8p = -4 or p = 1 p = –4/8 or p = 1 Therefore, p = –1/2 or p = 1 Quadratic Equations – Exercise 9.1 – Class X Quadratic Equations – Exercise 9.2 – Class X Quadratic Equations – Exercise 9.3 – Class X Quadratic Equations – Exercise 9.4 – Class X Quadratic Equations – Exercise 9.5 – Class X
# How do you solve the following system?: 4x-y=-1 , 4x-3y=15 Dec 23, 2015 The solution for the system of equations is color(blue)(x=-9/4,y=-8 #### Explanation: $\textcolor{b l u e}{4 x} - y = - 1$.......equation $\left(1\right)$ $\textcolor{b l u e}{4 x} - 3 y = 15$.........equation $\left(2\right)$ Solving by elimination Subtracting equation $2$ from $1$ $\cancel{\textcolor{b l u e}{4 x}} - y = - 1$ $\cancel{- \textcolor{b l u e}{4 x}} + 3 y = - 15$ $2 y = - 16$ $y = \frac{- 16}{2}$ color(blue)(y=-8 Finding $x$ by substituting $y$ in equation $1$: $4 x - y = - 1$ $4 x = - 1 + y$ $4 x = - 1 - 8$ $4 x = - 9$ color(blue)(x=-9/4
# How to find median of discrete frequency distriburion and continuous frequency distribution andFurther how to find m.d.(m) in discreate frquency and in continuous frequency distributiotion Example- Class - - - - - Frequency 0-10 - - - - - - - -10 10-40 - - - - - - -15 40-50 - - - - - - -5 50-70 - - - - - - -5 Solution- Solution: Median of continuous frequecy : Given : In a continuous frequency distribution, the median of the data is 21. And If each observation is increased by 5 . Then the new median is also increased by 5 , so new median = 21 + 5 = 26 ( Ans ) We can understand it As : We have observations of number of toy for every children and total number of children = 20 3, 5, 6, 6, 8, 8, 9, 9, 10, 11, 11, 12, 12, 12, 14, 15, 15, 18, 18, 20 First we form a continues frequency table , As : Number of toys 1 - 5 6 - 10 11 - 15 16 - 20 Frequency 2 7 8 3 Here frequency is number of children that have toys in between ( 1 -5 , 6 - 10 , 11- 15 and 16 - 20 ) Here ${\sum }_{}$Fx = 222 SO mean of that data is $\frac{222}{20}$ = 11.1 , So our median class in 11 - 15 , And median = L + Here 1 ) L is the lower class boundary of the group containing the median . ( here L = 11 ) 2 ) n is the total number of data . ( here n = 20 ) 3 ) cfb is the cumulative frequency of the groups before the median group.( Here cfb = 9 ) 4 ) fm is the frequency of the median group. ( here fm = 8 ) 5 ) w is the group width. ( Here w = 5 ) So, Median = 11 + = 11.625 Now we observe the change in median after increase its observation by 5 , now our new observations is 8, 10, 11, 11, 13, 13, 14, 14, 15, 16, 16, 17, 17, 17, 19, 20, 20, 23, 23, 25 First we form a continues frequency table , As : Number of toys 1 - 5 6 - 10 11 - 15 16 - 20 \| 20 - 25 Frequency 0 2 7 8 \| 3 Here ${\sum }_{}$Fx = 320 SO mean of that data is $\frac{320}{20}$ = 16 , So our median class in 16 - 20 , And median = L + Here 1 ) L is the lower class boundary of the group containing the median . ( here L = 16 ) 2 ) n is the total number of data . ( here n = 20 ) 3 ) cfb is the cumulative frequency of the groups before the median group.( Here cfb = 9 ) 4 ) fm is the frequency of the median group. ( here fm = 8 ) 5 ) w is the group width. ( Here w = 5 ) So, Median = 16 + = 16.625 So , we can see that after increment of 5 in observations ,we get our new median increased by 5 . Example of mean deviation:- Calculate the mean deviation about median age for the age distribution of 100 persons given below: Age Number 16-20 5 21-25 6 26-30 12 31-35 14 36-40 26 41-45 12 46-50 16 51-55 9 Solution:- The given data is not continuous. Therefore, it has to be converted into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval. The table is formed as follows. Age Number fi Cumulative frequency (c.f.) Mid-point xi \|xi – Med.\| fi \|xi – Med.\| 15.5-20.5 5 5 18 20 100 20.5-25.5 6 11 23 15 90 25.5-30.5 12 23 28 10 120 30.5-35.5 14 37 33 5 70 35.5-40.5 26 63 38 0 0 40.5-45.5 12 75 43 5 60 45.5-50.5 16 91 48 10 160 50.5-55.5 9 100 53 15 135 100 735 The class interval containing theor 50th item is 35.5 – 40.5. Therefore, 35.5 – 40.5 is the median class. It is known that, Here, l = 35.5, C = 37, f = 26, h = 5, and N = 100 Thus, mean deviation about the median is given by, To calculate mean deviation from median, the following formula is used When Mean deviation are taken from Median (M) For Individual Series For Discrete and Continuous Series The calculation process of Mean Deviation from median can be better understood with the help of the following example. Example: For the following data calculate Mean Deviation from median. X f 0 – 10 5 10 – 20 9 20 – 30 6 30 – 40 3 40 – 50 2 X f C.f Mid-Point \|D\| From median f\|D\| 0 – 10 5 5 5 13.88 69.4 10 – 20 9 14 15 3.88 34.92 20 – 30 6 20 25 6.12 36.72 30 – 40 3 23 35 16.12 48.36 40 – 50 2 25 45 26.12 52.24 N = 25 66.12 179.18 13th item whose class interval is 10 – 20 Thus, the value of Mean Deviation calculated from median is 7.16 • 0 What are you looking for?
Accelerate your child’s learning Smartick is a fun way to learn math! Jul11 # Geometric Figures and Straight Lines Straight lines are key to recognizing shapes that we often see in the real world: geometric figures. In today’s post, we will classify geometric figures formed from straight lines and think about where we can find them. We’ll start with the simplest: ### Two Straight LinesCan you make a figure with two straight lines? Well, of course, you can! However, we do not call this figure a geometrical figure. In order to be a geometrical figure, it must be “closed.” That is to say, if we put a ball inside the figure, it could not escape. In the figure we have made, the ball could escape! If you try to move the two lines, you will see that in no way can it be closed. You could take two straight objects, for example, two chopsticks, and try. ### Basic Geometric Figures Let’s now try it with three straight lines… That’s it! Now you can make a figure that the ball cannot escape. Each of the straight lines we used now act as a part of the figure. We will call each straight line a side of the figure. So, this figure has three sides and is called a triangle. When we say “triangle” we are either referring to how to arrange the three lines or the enclosing space. Can you imagine where we can find triangles around us? I found some in Smartick, let’s see if you recognize them! ### 4-Sided Geometric Figures Let’s see what we can do with 4 straight lines… We could make, for example, We call these figures quadrilaterals. There are some very special quadrilaterals that have pairs of parallel sides. That is, there are two sides which are parallel to each other, and the other two are also. In addition, the sides that are parallel are equal. How could we make these? We get two equal sides and put them parallel to each other: How can we place the other two sides so that they are also parallel? That’s it! Simply by joining them together. At this point we have three possibilities: • All the sides are equal, creating a diamond. Where can you find diamonds? I found two! • If the four angles of the figure are right angles, then we have created a rectangle. We can find many rectangles around us … I’m sure that you can think of many! • If we create a geometric shape that meets both of the above conditions and also has all equal sides and four right angles it is a square. ### Geometric Figures With More Than 4 Sides We can keep adding all the lines we want to create different shapes with different numbers of sides. They are especially important when they have equal sides. For example, the following would be the pentagon (5-sided). We can find them, for example, in the drawing of soccer balls. And then we have the hexagon (6-sided). We can find them … in beehives! How do bees know how to make such perfect geometric shapes? And then heptagons, octagons, enneagon, nonagons … any missing? Of course! What can we do with curved lines? With curved lines, it does not make sense to classify them because they vary so much more … but I can tell you that there is a very, very important curved figure that I am sure you already know of… the circle! We get a circle when we maintain the distance from a point. For example, with a compass, we set a point with a needle. Then, keeping the opening, we draw the circle with a pencil. You can find a lot of circles around you, on the wheels of your car or a tennis ball. If you want, you can continue learning elementary math with Smartick! Learn More: Fun is our brain’s favorite way of learning Diane Ackerman Smartick is a fun way to learn math • 15 fun minutes a day • Adapts to your child’s level • Millions of students since 2009 1 Comment; • Khatki shahanvaZ mohmmad rafikNov 04 2022, 1:22 AM What are these figures called? Reply
## Monday, March 5, 2007 ### 9.2 Arithmetic Sequences and Partial Sums 9.2 Arithmetic Sequences and Partial Sums Arithmetic Sequences - a sequence whose consecutive terms have a common difference A. Definition of Arithmetic Sequences: A sequence is arithmetic if the differences between consecutive terms are the same. So, the sequence a1, a2, a3, a4, ... , an, ... is arithmetic if there is a number d such that a2 - a1 = a3 - a2 = a4 - a3 = ... = d. The number d is the common difference of the arithmetic sequence. Example 1: Given the sequence an = 1 + (n - 1)4, find the first five terms of the sequence an the common difference: a2 = 1 + (2 - 1)4 = 1 + (1)(4) = 1 + 4 = 5 a3 = 1 + (3 - 1)4 = 1 + (2)(4) = 1 + 8 = 9 a4 = 1 + (4 - 1)4 = 1 + (3)(4) = 1 + 12 = 13 a5 = 1 + (5 - 1)4 = 1 + (5)(4) = 1 + 20 = 21 a2 - a1 = a3 - a2 = a4 - a3 = ... = d. 9 - 5 = 13 - 9 = 21 - 13 = 4 = d B. The nth term of an Arithmetic Sequence: last year you learned it as an = a1 + (n - 1) d distributing d we get: an = a1 + dn - d = dn + (a1 - d) therefore let’s let c = (a1 - d) so we have an = dn + c where d is the common difference between consecutive terms of the sequence and c = a1 - d Example 2: write the first five terms of the arithmetic sequence, find the common difference and write the nth term of the sequence as a function of n. a1 = 200, ak+1 = ak - 20 a2 = a1+1 = a1 - 20 = 200 - 20 = 180 a3 = a2+1 = a2 - 20 = 180 - 20 = 160 a4 = a3+1 = a3 - 20 = 160 - 20 = 140 a5 = a4+1 = a4 - 20 = 140 - 20 = 120 so d = -20 and c = a1 - d = 200 - -20 = 220 therefore an = dn + c = -20n + 220 Example 3: The first two terms of the arithmetic sequences are given. Find the missing term. a1 = 3, a2 = 13 , a9 = ____ d = 13 - 3 = 10 c = a1 - d = 3 - 10 = -7 so therefore an = dn + c = 10n - 7 a9 = 10n - 7 = 10 (9) - 7 = 90 - 7 = 83 Example 4: Find a formula for an for the arithmetic sequence: a1 = 15, d = 4 c = a1 - d = 15 - 4= 11 so therefore an = dn + c = 4n +11 Example 5: Find a formula for an for the arithmetic sequence: a1 = -4, a5 = 16 You may have noticed by now that an = dn + c is a linear function so therefore d is the slope of the line so using this concept of the slope formula ▵y/▵x We have: (16 - -4) / (5 - 1) = (20)/ (4) = 5 = d c = a1 - d = - 4 - 5 = -9 so therefore an = dn + c = 5n - 9 C. The Sum of a Finite Arithmetic Sequence Carl Friedrich Gauss (1777 - 1855) was ten years old, his teacher asked him to add all the integers from 1 to 100. Gauss was able to answer the teacher within a few moments, the teacher was amazed. Here is how he did it: 1 + 2 + 3 + 4 + ... + 97 + 98 + 99 + 100 He took the 1 + 100 = 101, 2 + 99 = 101, 3 + 98 = 101 so taking 100 / 2 = 50, then 50 * 101 = 5050. The Sum of a Finite Arithmetic Sequence: The sum of a finite arithmetic sequence with n terms is Sn = (n/2)(a1 + an) Example 6: find the indicated nth partial sum of the arithmetic sequence: -6, -2, 2, 6, ... , n = 50 the common difference is -2 - -6 = 2 - -2 = 6 - 2 = 4 c = a1 - d = - 6 - 4 = -10 so therefore an = dn + c = 4n - 10 a1 = -6 and a50 = 4(50) - 10 = 200 - 10 = 190 Sn = (n/2)(a1 + an) = (n/2)(a1 + a50) = (50 / 2) (-6 + 190) = (25)(184) = 4600 9.2 Homework #53 page 635; # 9 - 19 odd, 35 - 47 odd, 59 - 79 odd
# A convex mirror used as a rear-view mirror in a car has a radius of curvature of 3 m. If a bus is located at a distance of 5 m from this mirror, find the position of image. What is the nature of the image? Given: The mirror is convex. Distance of the object from the mirror, $u$ = $-$5 m Radius of curvature of the mirror, $R$ = 3 m Focal length of the mirror, $f$ = 1.5 m $(\because f=\frac{R}{2})$ To find: Distance or position of the image, $v$, and its magnification $m$. Solution: From the mirror formula, we know that- $\frac {1}{f}=\frac {1}{v}+\frac {1}{u}$ Substituting the given values we get- $\frac {1}{1.5}=\frac {1}{v}+\frac {1}{(-5)}$ $\frac {1}{1.5}=\frac {1}{v}-\frac {1}{5}$ $\frac {1}{5}+\frac {1}{1.5}=\frac {1}{v}$ $\frac {1}{5}+\frac {10}{15}=\frac {1}{v}$ $\frac {1}{v}=\frac {3+10}{15}$ $\frac {1}{v}=\frac {13}{15}$ $v=\frac {15}{13}$ $v=+1.15m$ Thus, the distance of the image $v$ is 1.15 m from the mirror, and the positive sign implies that the image forms behind the mirror (on the right). Now, from the magnification formula, we know that- $m=-\frac {v}{u}$ Substituting the given values we get- $m=-\frac {1.15}{(-5)}$ $m=\frac {115}{500}$ $m=+0.22$ Thus, the magnification is $0.22$ which is less than 1, which means the image is small in size, and the positive sign implies that the image is virtual and erect. Hence, the image is virtual, erect, and small in size. Tutorialspoint Simply Easy Learning Updated on: 10-Oct-2022 102 Views
# Question #b8d66 Feb 6, 2018 $y = 4$ $x = 2$ #### Explanation: If $4 x - 2 y = 0$ and $4 x + y = 12$, then: $4 x - 4 x + y - \left(- 2 y\right) = 12 - 0$ $\implies 0 x + y + 2 y = 12$ $\implies 3 y = 12$ $\implies y = 4$ Since $4 x + y = 12$ we get that $4 x + 4 = 12$ $\implies 4 x = 12 - 4$ $\implies 4 x = 8$ $\implies x = 2$ So $y = 4$ and $x = 2$. Feb 6, 2018 $x = 2 \mathmr{and} y = 4$ #### Explanation: Notice that the $x$ terms are exact;y the same in both equations. If you subtract the equations, the $x$ terms will be eliminated and you will be able to solve for $y$ $\text{ "4x +y =12" } \rightarrow A$ $\text{ "4x -2y =0" } \rightarrow B$ $A - B : \text{ " 0x+3y =12" } \rightarrow C$ $\textcolor{w h i t e}{w w w w w w w w w w w} y = 4$ Once you know the value for $y$ you can use it to find $x$ Substitute $4 \text{ for " y " in } A$ $\text{ "4x +y =12" } \rightarrow A$ $\text{ } 4 x + \left(4\right) = 12$ $\text{ } 4 x = 8$ $\text{ } x = 2$ Check in $B$ $\text{ "4x -2y =0" } \rightarrow B$ $\text{ } 4 \left(2\right) - 2 \left(4\right) = 0$ $\text{ } 8 - 8 = 0$
# RD SHARMA Solutions for Class 10 Maths Chapter 8 - Circles Page / Exercise ## Chapter 8 - Circles Exercise Ex. 8.2 Question 24 A is a point at a distance 13 cm from the centre O of a circle of radius 5 cm. AP and AQ are the tangents to the circle at P and Q. If a tangent BC is drawn at a point R lying on the minor arc PQ to intersect AP at B and AQ at C, find the perimeter of the ∆ABC. Solution 24 AP = AQ , BP = PR and CR = CQ (tangents from an external point) Perimeter of ∆ABC = AB + BR + RC + CA = AB + BP + CQ + CA = AP + AQ = 2AP ∆APO is a right-angled triangle. AO2 = AP2 + PO2 132 = AP2 + 52 AP2 = 144 AP = 12 Perimeter of ∆ABC = 24 cm Question 46 In Fig., BC is a tangent to the circle with centre O. OE bisects AP. Prove that ∆AEO ~ ∆ABC Solution 46 In ∆AOP, OA = OP (radii) ∆AOP is an isosceles triangle. OE is a median. In an isosceles triangle,the median drawnOEA = 90o In ∆AOE and ∆ABC, ABC = OEA = 90o A is common. ∆AEO ~ ∆ABC…(AA test) Question 50 In Fig., there are two concentric circles with centre O. PRT and PQS are tangents to the inner circle from a point P lying on the outer circle. If PR = 5 cm, find the length of PS. Solution 50 PR = PQ…(tangents fromexternal points) PQ = 5 cm Also, OQ is perpendicular to PS …(tangent is perpendicular to the radius) Now, in a circle,a perpendicular drawn from the centre of a circle bisects the chord. So, OQ bisects PS. PQ = QS QS = 5 cm PS = 10 cm Question 51 In Fig., PQ is a tangent from an external point P to a circle with centre O and OP cuts the circle at T and QOR is a diameter. If POR = 130˚ and S is a point on the circle, find 1 + 2. Solution 51 In DPQR, POR is an external angle. So, POR = PQO + OPQ Now, PQ is tangent to the circle with radius OQ. PQO = 90o 130˚ = 90˚ + OPQ OPQ = 40o 1 = 40o Now, Minor arc RT subtends a 130˚ angle at the centre. So, it will subtend a 65˚angle at any other point on the circle.RST = 65˚ 2 = 65˚ 1 + 2 =105˚ Question 52 In Fig., PA and PB are tangents to the circle from an external point P. CD is another tangent touching the circle at Q. If PA = 12 cm, QC = QD = 3 cm, then find PC + PD. Solution 52 AP = PB = 12 cm, AC = CQ = 3 cm and QD = DB = 3 cm …(tangent from external point) PA = 12 cm, PC + CA = 12 PC + 3 = 12 PC= 9 …(i) Now, PB = 12 PD + DB = 12 PD + 3 = 12 PD = 9 …(ii) PC + PD = 18 cm…from (i) and (ii) Question 1 Solution 1 Question 2 Solution 2 Question 3 Solution 3 Question 4 If from any point on the common chord of two intersecting circles, tangents be drawn to the circles, prove that they are equal. Solution 4 Question 5 Solution 5 Question 6 Out of the two concentric circles, the radius of the outer circle is 5 cm and the chord AC of length 8 cm is a tangent to the inner circle. Find the radius of the inner circle. Solution 6 Question 7 A chord PQ of a circle is parallel to the tangent drawn at a point R of the circle. Prove that R bisects the arc PRQ. Solution 7 Question 8 Prove that a diameter AB of a circle bisects all those chords which are parallel to the tangent at the point A. Solution 8 Question 9 If AB, AC, PQ are tangents in Fig. 8.56 and AB = 5 cm, find the perimeter of ΔAPQ. Solution 9 Question 10 Solution 10 Question 11 In Fig., PQ is tangent at a point R of the circle with centre O. If ∠TRQ = 30°, find m∠PRS. Solution 11 Question 12 Solution 12 Question 13 In a right triangle ABC in which ∠B =90°, a circle is drawn with AB as diameter intersecting the hypotenuse AC at P. Prove that the tangent to the circle at P bisects BC. Solution 13 Question 14 From an external point P, tangents PA and PB are drawn to a circle with centre O. At one point E on the circle tangent is drawn, which intersects PA and PB at C and D respectively. If PA = 14 cm, find the perimeter of Δ PCD. Solution 14 Question 15 In fig., ABC is a right triangle right-angled at B such that BC = 6 cm and AB = 8 cm. Find the radius of its incirde. Solution 15 Question 16 Prove that the tangent drawn at the mid-point an arc of a circle is parallel to the chord joining the end points of the arc. Solution 16 Question 17 Solution 17 Question 18 Two tangent segments PA and PB are drawn to a circle with centre O such that APB = 120o. Prove that OP = 2 AP. Solution 18 Question 19 Solution 19 Question 20 AB is a diameter and AC is a chord of a circle with centre O such that BAC = 30°. The tangent at C intersects AB at a point D. Prove that BC =BD Solution 20 Question 21 In fig., a circle touches all the four sides of a quadrilateral ABCD with AB = 6 cm, BC = 7 cm and CD = 4 cm. Find AD. Solution 21 Question 22 Solution 22 Question 23 Solution 23 Question 25 In Fig., a circle is inscribed in a quadrilateral ABCD in which B= 90°. If AD=23 cm, AB =29 cm and DS =5 cm, find the radius r of the circle. Solution 25 Question 26 In Fig., there are two concentric circles with centre O of radii 5 cm and 3 cm. From an external point P, tangents PA and PB are drawn to these circles. If AP = 12 cm, find the length of BP. Solution 26 Question 27 In Fig., AB is a chord of length 16 cm of a circle of radius 10 cm. The tangents at A and B intersect at a point P. Find the length of PA. Solution 27 Question 28 In Fig., PA and PB are tangents from an external point P to a circle with centre O. LN touches the circle at M. Prove that PL + LM = PN + MN. Solution 28 Question 29 In Fig., BDC is a tangent to the given circle at point D such that BD = 30 cm and CD = 7 cm. The other tangents BE and CF are drawn respectively from B and C to the Calculate (i) AF (ii) radius of the circle. Solution 29 Question 30 Solution 30 Question 31 In the given figure, tangents PQ and PR are drawn from an external point P to a circle with centre O, such that RPQ = 30°. A chord RS is drawn parallel to the tangent PQ. Find RQS. Solution 31 Since RS is drawn parallel to the tangent PQ, SRQ = PQR Also, PQ = PR PQR = PRQ In ∆PQR, PQR + PRQ + QPR = 180° PQR + PQR + 30° = 180° 2PQR = 150° PQR = 75° SRQ =PQR = 75° (alternate angles) Also, RSQ =RQP = 75° (the angle between a tangent and a chord through the point of contact is equal to an angle in the alternate segment.) In ∆RSQ, RSQ + SRQ + RQS = 180° 75° + 75° + RQS = 180° RQS =30° Question 32 From an external point P, tangents PA = PB are drawn to a circle with centre O. If PAB = 50°, then find AOB. Solution 32 Question 33 Solution 33 Question 34 Solution 34 Question 35 The common tangents AB and CD to two circles with centres O and O' intersect at E between their centres. Prove that the points O, E and O' are collinear. Solution 35 Question 36 In Fig, common tangents PQ and RS to two circles intersects at A. Prove that PQ = RS. Solution 36 Question 37 Two concentric circles are of diameters 30 cm and 18 cm. Find the length of the chord of the larger circle which touches the smaller circle. Solution 37 Since AC is the tangent to the circle with radius 9 cm, we have OB AC. Hence, by applying the Pythagoras Theorem, we have, OA2 = OB2 + AB2 152 = 92 + AB2 AB2 = 152 - 92 AB2 = 225 - 81 = 144 AB = 12 cm We know that the perpendicular drawn from the centre of a circle to any chord of the circle bisects the chord. Here, OB is the perpendicular and AC is the length of the chord of the circle with radius 15 cm. So, AC = 2 × AB = 2 × 12 = 24 cm Length of the chord of the larger circle which touches the smaller circle = 24 cm. Question 38 AB and CD are common tangents to two circles of equal radii. Prove that AB =CD. Solution 38 Question 39 A triangle PQR is drawn to circumscribe a circle of radius 8 cm such that the segments QT and TR, into which QR is divided by the point of contact T, are of lengths 14 cm and 16 cm respectively. If area of ∆ PQR is 336 cm2, find the sides PQ and PR. Solution 39 Let PA = PB = x Tangents drawn from an external point are equal in length. QB = QT = 14 cm , RA = RT = 16 cm PR = (x + 16) cm, PQ = (x + 14)cm, QR = 30 cm = x + 30 Area of PQR Area of PQR = 336 cm2 Side PR = (12 + 16) = 28 cm Side PQ = (12 + 14) = 26 cm Question 40 In Fig., the tangent at a point C of a circle and a diameter AB when extended intersect at P. If PCA =110°, find CBA. Solution 40 Question 41 AB is a chord of a circle with centre O, AOC is a diameter and AT is the tangent at A as shown in figure. Prove that BAT = ACB. Solution 41 Question 42 In the given figure, a ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC are of lengths 8 cm and 6 cm respectively. Find the lengths of sides AB and AC, when area of ABC is 84 cm2. Solution 42 Let M and N be the points where AB and AC touch the circle respectively. Tangents drawn from an external point to a circle are equal AM=AN BD=BM=8 cm and DC=NC=6 cm Question 43 In the given figure, AB is a diameter of a circle with centre O and AT is a tangent. If AOQ = 58°, find ATQ. Solution 43 AOQ=58° (given) In right BAT, ABT + BAT + ATB=180° 29° + 90° + ATB=180° ATB = 61° that is, ATQ = 61° Question 44 In Fig., OQ : PQ = 3 : 4 and perimeter of ΔPOQ = 60 cm. Determine PQ, QR and OP. Solution 44 Question 45 Solution 45 Question 47 In fig., PO ⊥ QO . The tangents to the circle at P and Q intersect at a point T. Prove that PQ and OT are right bisectors of each other. Solution 47 Question 48 In fig., O is the centre of the circle and BCD is tangent to it at C. Prove that BAC + ACD = 90o. Solution 48 Question 49 Prove that the centre of a circle touching two intersecting lines lies on the angle bisector of the lines. Solution 49 ## Chapter 8 - Circles Exercise Ex. 8.1 Question 1 Fill in the blanks: (i) The common point of a tangent and the circle is called ....... . (ii) A circle may have ....... parallel tangents. (iii) A tangent to a circle intersects it in ....... point(s). (iv) A line intesecting a circle in two points is called a ....... . (v) The angle between tangent at a point on a circle and the radius through the point is ....... . Solution 1 Fill in the blanks: (i) The common point of a tangent and the circle is called point of contact . (ii) A circle may have two parallel tangents. (iii) A tangent to a circle intersects it in one point(s). (iv) A line intesecting a circle in two points is called a secant . (v) The angle between tangent at a point on a circle and the radius through the point is 90o . Question 2 Solution 2 Question 3 Solution 3 Question 4 Solution 4 ## Chapter 8 - Circles Exercise 8.48 Question 1 A tangent PQ at a point P of a circle of radius 5 cm meets line through the centre O at a point Q such that OQ = 12 cm. Length PQ is (a) 12 cm (b) 13 cm (c) 8.5 cm (d) Solution 1 So, OP = 5 cm OQ = 12 cm So, the correct option is (d). Question 2 From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is (a) 7 cm (b) 12 cm (c) 15 cm (d) 24.5 cm Solution 2 PQ is a tangent to the circle So, OP+ PQ2 = OQ2 OP= OQ- PQ2 = (25)- (24)2 = 49 OP = 7 So, the correct option is (a). Question 3 The length of tangent from point A at a circle, of radius 3 cm, is 4 cm. The distance of A from the centre of the circle is (a) (b) 7 cm (c) 5 cm (d) 25 cm Solution 3 Given OP = 3 cm PA = 4 cm Hence, OA= OP2 + PA2 OA2 = 3+ 42 = 25 OA = 5 cm So, the correct option is (c). Question 4 Solution 4 ## Chapter 8 - Circles Exercise 8.49 Question 5 Solution 5 Question 6 Solution 6 Question 7 Solution 7 Question 8 Solution 8 Tangents from same point to circle have equal length. Hence Bb = Ba bC = Cc Ac = Aa Let Ba = x then Bb = x bc = 6 - x and Aa = 8 - x and Cc = 6 - x and Ac = 8 - x So AC = AC + cC = 6 - x + 8 - x AC = 14 - 2x ......(1) Also AC= AB+ BC2 = 82 + 6 = 100 AC = 10 .....(2) from (1) & (2) 14 - 2x = 10 4 = 2x x = 2 also aB = Ob = radius = 2 cm So, the correct option is (b). Question 9 Solution 9 Question 10 If four sides of a quadrilateral ABCD are tangential to a circle, then (a) AC + AD = BD + CD (b) AB + CD = BC + AD (c) AB + CD = AC + BC (d) AC + AD = BC + DB Solution 10 Tangents from same point are of equal length. AP = AS, PB = BQ QC = CR, RD = DS AB = AP + PB .....(1) BC = BQ + QC ......(2) CD = CR + RD .....(3) AD = AS + DS .....(4) AB + CD = AP + BP + CR + RD = AS + BQ + CQ + DS = (AS + DS) + (BQ + CQ) from (2) & (4) AB + CD = AD + BC So, the correct option is (b). Question 11 The length of the tangent drawn from a point 8 cm away from the centre of a circle of radius 6 cm is (a) (b) (c) 10 cm (d) 5 cm Solution 11 Given OQ = 8 cm OP = 6 cm OP+ PQ2 = OQ2 6+ PQ2 = 82 PQ= 64 - 36 = 28 PQ = So, the correct option is (b). Question 12 AB and CD are two common tangents to circles which touch each other at C. If D lies on AB such that CD = 4 cm, then AB is equal to (a) 4 cm (b) 6 cm (c) 8 cm (d) 12 cm Solution 12 DA and DC are tangents to circle from same point so, DA = DC ......(1) similarly DB = DC ......(2) (1) + (2) 2DC = DA + DB 2DC = AB AB = 2 × 4 = 8 cm So, the correct option is (c). Question 13 In Figure, if AD, AE and BC are tangents to the circle at D, E and F respectively. Then, (a) AD = AB + BC + CA (b) 2AD = AB + BC + CA (c) 3AD = AB + BC + CA (d) 4AD = AB + BC + CA Solution 13 CD = CF ......(2) BF = BE .....(3) from (1) = (AB + BE) + (AC + CD) = AB + BF + AC + CF = AB + AC + BC So, the correct option is (b). Question 14 In figure, RQ is a tangent to the circle with centre O. If SQ = 6 cm and QR = 4 cm, then OR = (a) 8 cm (b) 3 cm (c) 2.5 cm (d) 5 cm Solution 14 ## Chapter 8 - Circles Exercise 8.50 Question 15 Solution 15 Question 16 Solution 16 Question 17 AP and PQ are tangents drawn from a point A to a circle with centre O and radius 9 cm. If OA = 15 cm, then AP + AQ = (a) 12 cm (b) 18 cm (c) 24 cm (d) 36 cm Solution 17 AP = PQ ....(1) and OA= OP + PA2 (15)= (9)+ AP2 AP= 225 - 81 = 144 AP = 12 AP + AQ = 2AP = 24 cm So, the correct option is (c). Question 18 At one end of a diameter PQ of a circle of radius 5 cm, tangent XPY is drawn to the circle. The length of chord AB parallel to XY and at a distance of 8 cm from P is (a) 5 cm (b) 6 cm (c) 7 cm (d) 8 cm Solution 18 Question 19 Solution 19 ## Chapter 8 - Circles Exercise 8.51 Question 20 In the adjacent figure, if AB = 12 cm, BC = 8 cm and AC = 10 cm, then AD = (a) 5 cm (b) 4 cm (c) 6 cm (d) 7 cm Solution 20 AB = 12 cm BC = 8 cm AC = 10 cm AF = x BD = 12 - x and BE = BD = 12 - x CE = BC - BE = 8 - (12 - x) = x - 4 and CE = CF = x - 4 AC = AF + FC = x + x - 4 AC = 2x - 4 Given, AC = 10 cm so 2x - 4 = 10 2x = 14 x = 7 cm So, the correct option is (d). Question 21 In figure, if AP = PB, then (a) AC = AB (b) AC = BC (c) AQ = QC (d) AB = BC Solution 21 AP = BP given and AP = AQ also BP = BR from this, we conclude that AQ = BR .....(1) We know CR = CQ .....(2) from (1) & (2) AQ + CR = BR + CR AQ + CQ = BR + CR AC = BC So, the correct option is (b). Question 22 In figure, if AP = 10 cm, then BP = Solution 22 AP = 10 cm AO = 6 cm OB = 3 cm AP2 + OA= OP2 OP= 102 + 62 OP= 136 Also OB+ BP2 = OP 32 + BP= 136 BP2 = 136 - 9 So, the correct option is (b). Question 23 Solution 23 ## Chapter 8 - Circles Exercise 8.52 Question 24 In Figure, if quadrilateral PQRS circumscribes a circle, then PD + QB = (a) PQ (b) QR (c) PR (d) PS Solution 24 PA = PD AQ = QB and PQ = PA + AQ PQ = PD + QB Hence PD + QB = PQ So, the correct option is (a). Question 25 In figure, two equal circles touch each other at T, if QP = 4.5 cm, then QR = (a) 9 cm (b) 18 cm (c) 15 cm (d) 13.5 cm Solution 25 PQ = PT .....(1) and PT = PR .....(2) so from (1) & (2) PQ = PR PQ = PR = 4.5 cm QR = PQ + PR = 4.5 + 4.5 = 9 cm So, the correct option is (a). Question 26 Solution 26 Question 27 Solution 27 ## Chapter 8 - Circles Exercise 8.53 Question 28 In figure, PR = (a) 20 cm (b) 26 cm (c) 24 cm (d) 28 cm Solution 28 radius of circle 1 = 3 cm radius of circle 2 = 5 cm OP= OQ+ QP and O'S+ SR2 = O'R OP= 4+ 32 O'R= 5+ 122 = 16 + 9 O'R2 = 169 = 25 O'R' = 13 cm OP = 5 cm OO' = OK + KO' = 3 + 5 = 8 cm PR = PO + OK + KO' + O'R = 5 + 3 + 5 + 13 = 26 cm So, the correct option is (b). Question 29 Solution 29 Question 30 Two concentric circles of radii 3 cm and 5 cm are given. Then length of chord BC which touches the inner circle at P is equal to (a) 4 cm (b) 6 cm (c) 8 cm (d) 10 cm Solution 30 OB = OC = OA = 5 cm OQ = OP = 3 cm OB= OQ+ BQ2 BQ2 = OB- OQ2 = 5- 32 = 16 BQ = 4 cm also BQ = BP BP = 4 cm In ΔOPC, OP2 + PC2 = OC2 PC = OC2 - OP2 = 5- 32 = 16 PC = 4 cm BC = BP + PC = 4 + 4 = 8 cm So, the correct option is (c). Question 31 In figure, there are two concentric circles with centre O. PR and PQS are tangents to the inner circle from point plying on the outer circle. If PR = 7.5 cm, then PS is equal to (a) 10 cm (b) 12 cm (c) 15 cm (d) 18 cm Solution 31 Given PR = 7.5 cm so PR = PQ PQ = 7.5 cm PS is the chord to the larger circle. We know that, perpendicular drawn from centre bisect the chords. Hence PQ = QS PS = PQ + QS = 2PQ = 2 × 7.5 = 15 cm So, the correct option is (c). ## Chapter 8 - Circles Exercise 8.54 Question 32 In figure, if AB = 8 cm and PE = 3 cm, then AE = (a) 11 cm (b) 7 cm (c) 5 cm (d) 3 cm Solution 32 AC = AB .....(1) BD = DP ......(2) PE = EC ......(3) AB = 8 so AC = 8 cm PE = 3 so EC = 3 cm AE = AC - EC = 8 - 3 = 5 cm So, the correct option is (c). Question 33 Solution 33 Question 34 Solution 34 ## Chapter 8 - Circles Exercise 8.55 Question 35 In figure, the sides AB, BC and CA of a triangle ABC, touch a circle at P,Q and R respectively. If PA = 4 cm, BP = 3 cm and AC = 11 cm, then length of BC is (a) 11 cm (b) 10 cm (c) 14 cm (d) 15 cm Solution 35 PA = AR AR = 4 cm BP = BQ and QC = RC BQ = 3 cm Given AC = 11 AR + RC = 11 4 + RC = 11 RC = 7 so QC = 7 cm BC = BQ + QC = 3 + 7 = 10 cm So, the correct option is (b). Question 36 Solution 36 EK = 9 cm and EK = EM Hence EM = 9 cm .....(1) Also EK = ED + DK and DK = DH EK = ED + HD .......(2) EM = EF + FM and FM = FH EM = EF + FH ......(3) (2) + (3) EK + EM = ED + EF + DH + HF 18 = ED + DF + EF perimeter = 18 cm So, the correct option is (a). Question 37 Solution 37 ## Chapter 8 - Circles Exercise 8.56 Question 38 Solution 38 AB = 29 cm DS = 5 cm DS = DR so DR = 5 cm = 23 - 5 = 18 cm AR = AQ AQ = 18 cm BQ = AB - AQ = 29 - 18 BQ = 11 cm As OP \|\| BQ and OQ \|\| PB Hence, OP = BQ OP = 11 cm So, the correct option is (a). Question 39 In a right triangle ABC, right angled at B, BC = 12 cm and AB = 5 cm. The radius of the circle inscribed in the triangle (in cm) is (a) 4 (b) 3 (c) 2 (d) 1 Solution 39 AB = 5 cm BC = 12 cm AB + BC2 = AC2 AC= 52 + 122 = 169 AC = 13 cm Let BQ = x AQ = AR = 5 - x CR = AC - AR = 13 - (5 - x) = x + 8 And CP = CR = x + 8 so BP = BC - PC = 12 - (x + 8) = 4 - x But BP = BQ = x 4 - x = x x = 2 and BQ \|\| OP and OQ \|\| PB so BQ = PO PO = 2 cm So, the correct option is (c). Question 40 Solution 40 Question 41 Solution 41 Question 42 In figure, QR is a common tangent to the given circles touching externally at the point T. The tangent at T meets QR at P. If PT = 3.8 cm, then the length of QR (in cm) is (a) 3.8 (b) 7.6 (c) 5.7 (d) 1.9 Solution 42 PT = 3.8 cm We know PQ = PT and PT = PR Hence PQ = 3.8 cm and PR = 3.8 cm Now, QR = QP + PR = 3.8 + 3.8 QR = 7.6 cm So, the correct option is (b). ## Chapter 8 - Circles Exercise 8.57 Question 43 In figure, a quadrilateral ABCD is drawn to circumscribe a circle such that its sides AB, BC, CD and AD touch the circle at P, Q, R and S respectively. If AB = x cm, BC = 7 cm, CR = 3 cm and AS = 5 cm, then x = (a) 10 (b) 9 (c) 8 (d) 7 Solution 43 AB = x cm BC = 7 cm CR = 3 cm AS = 5 cm CR = CQ CQ = 3 cm given BC = 7 cm BQ = BC - QC = 7 - 3 = 4 cm And BQ = BP so BP = 4 cm Also AS = AP Hence AP = 5 cm AB = AP + BP = 5 + 4 = 9 cm x = 9 cm So, the correct option is (b). Question 44 If angle between two radii of a circle is 130°, the angle between the tangents at the ends of radii is a. 90° b. 50° c. 70° d. 40° Solution 44 Question 45 If two tangents inclined at an angle of 60° are drawn to a circle of radius 3 cm, then length of each tangent is equal to Solution 45 Question 46 If radii of two concentric circles are 4 cm and 5 cm, then the length of each chord of one circle which is tangent to the other circle is a. 3 cm b. 6 cm c. 9 cm d. 1 cm Solution 46 Question 47 At one end A of a diameter AB of a circle of radius 5 cm, Tangent XAY is drawn to the circle. The length of the chord CD parallel to XY and at a distance 8 cm from A is a. 4 cm b. 5 cm c. 6 cm d. 8 cm Solution 47 Question 48 From a point P which is at a distance 13 cm from the centre O of a circle of radius 5 cm, the pair of tangents PQ and PR to the circle are drawn. Then the area of the quadrilateral PQOR is a. 60 cm2 b. 65 cm2 c. 30 cm2 d. 32.5 cm2 Solution 48 Question 49 If PA, PB are tangents to the circle with centre O such that APB = 50°, then OAB is equal to a. 25° b. 30° c. 40° d. 50° Solution 49 Question 50 The pair of tangents AP and AQ drawn from an external point to a circle with centre O are perpendicular to each other and length of each tangent is 5 cm. The radius of the circle is a. 10 cm b. 7.5 cm c. 5 cm d. 2.5 cm Solution 50 ## Chapter 8 - Circles Exercise 8.58 Question 51 In figure, if AOB = 125° , then COD is equal to a. 45° b. 35° c. 55° d. 62° Solution 51 Question 52 In figure, if PQR is tangent to a circle at Q whole centre is O , AB is a chord parallel to PR and BQR = 70°, thenAQB is equal to a. 20° b. 40° c. 35° d. 45° Solution 52
# How to Graph ln(x) An error occurred trying to load this video. Try refreshing the page, or contact customer support. Coming up next: Defining and Graphing Ellipses in Algebra ### You're on a roll. Keep up the good work! Replay Your next lesson will play in 10 seconds • 0:03 The Steps to Solve ln(x) • 2:51 Logarithmic vs… Save Save Want to watch this again later? Timeline Autoplay Autoplay Speed Speed #### Recommended Lessons and Courses for You Lesson Transcript Instructor: Laura Pennington Laura received her Master's degree in Pure Mathematics from Michigan State University. She has 15 years of experience teaching collegiate mathematics at various institutions. This lesson will explore the properties of the function ln(x). We will use these properties to graph ln(x), and we will take it a bit further and look at how we can use the relationship between ln(x) and e^x to graph ln(x). ## Steps to Solve ln(x) We are going to use the properties of logarithms to graph f(x) = ln(x). A logarithmic function has the form f(x) = log a (x), and log a (x) represents the number we raise a to in order to get x. We call a the base of the logarithmic function. The function f(x) = ln(x) is a logarithmic function with base e, where e is an irrational number with value e = 2.71828 (rounded to 5 decimal places). Instead of writing the natural logarithm as log e (x), we use the notation ln(x). We are going to use the following properties of the graph of f(x) = log a (x) to graph f(x) = ln(x). • The x-intercept, or where the graph crosses the x-axis, of the graph is (1, 0). • The y-axis is a vertical asymptote of the graph. In other words, the graph approaches the y-axis, but does not touch it. • The domain of the function is all real numbers strictly greater than 0. • The range of the function is all real numbers. • If the base of the function is greater than 1, then the function is increasing, or rising from left to right, and takes on the following general shape. If the base of the function is greater than 0 and less than 1, then the function is decreasing, or falling from left to right, and takes on the following general shape. Let's examine these properties for f(x) = ln(x). We know the base is e, and e > 1. Therefore, the function is increasing and takes on the general shape shown above when the base is greater than 1. Furthermore, we know that the graph passes through the point (1, 0) and that it approaches the y-axis, but never actually touches it. Lastly, because the domain is all real numbers strictly greater than 0, and the range is all real numbers, we know the entire graph will fall to the right of the y-axis. These facts give us an idea of what the graph will look like. To be more exact, we can plot a few strategic points, so we know the graph is accurate. To find points, we choose some strategic values of x plug them into y = ln(x) and find the corresponding y-value. x-value y = ln(x) e = 2.7 y = ln(e) = 1 e 2 = 7.4 y = ln(e 2 ) = 2 e 3 = 20.1 y = ln(e 3 ) = 3 Thus, we have the points (2.7, 1), (7.4, 2), (20.1, 3). We plot these along with our other point (1, 0), and connect the dots with a smooth curve that takes on the shape described above. Here's the solution to the problem: The graph of f(x) = ln(x) is shown. To unlock this lesson you must be a Study.com Member. ### Register to view this lesson Are you a student or a teacher? #### See for yourself why 30 million people use Study.com ##### Become a Study.com member and start learning now. Back What teachers are saying about Study.com ### Earning College Credit Did you know… We have over 200 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level.
Update all PDFs # Equivalent Expressions? Alignments to Content Standards: 7.EE.A If we multiply $\frac{x}{2} + \frac34$ by 4, we get $2x+3$. Is $2x+3$ an equivalent expression to $\frac{x}{2} + \frac34$? ## IM Commentary The purpose of this task is to directly address a common misconception held by many students who are learning to solve equations. Because a frequent strategy for solving an equation with fractions is to multiply both sides by a common denominator (so all the coefficients are integers), students often forget why this is an "allowable" move in an equation and try to apply the same strategy when they see an expression. Two expressions are equivalent if they have the same value no matter what the value of the variables in them. After learning to transform expressions and equations into equivalent expressions and equations, it is easy to forget the original definition of equivalent expressions and mix up which transformations are allowed for expressions and which are allowed for equations. ## Solution No, $2x + 3$ and $\frac{x}{2} + \frac34$ are not equivalent expressions because they do not yield the same result for most values of $x$. For example, when $x = 1$, we get $$2(1) + 3 = 5$$ and $$\frac{(1)}{2}+\frac34=\frac54 \neq 5$$ Therefore, they are not equivalent. In fact, the expression $2x + 3$ will be 4 times as big as $\frac{x}{2} + \frac34$ for all values of $x$, since we obtained it by multiplying $\frac{x}{2} + \frac34$ by 4. #### Cam says: almost 3 years The two expressions $\frac{x}{2}+\frac34$ and $2x+3$ are equal when $x=-3/2$, but the question of equivalence is a question about whether the two expressions are equal for all values of $x$. Since there is at least one real number $x$ for which the two expressions are not equal (e.g., $x=1$), the two expressions are not equivalent. #### James says: almost 3 years If we restrict this to being in the first quadrant then the conclusion is right for all x's from zero to infinity but the fact that the slopes vary implies that the two lines must intersect i.e. there must be a point that the two solutions are equivalent (but only one since the two are linear expressions). #### James says: almost 3 years Except when x = -3/2. At that value only are they equivalent.
Home Â» What is a Probability Distribution Table? (Definition & Example) # What is a Probability Distribution Table? (Definition & Example) A probability distribution table is a table that displays the probability that a random variable takes on certain values. For example, the following probability distribution table tells us the probability that a certain soccer team scores a certain number of goals in a given game: The left-hand column shows the number of goals and the right-hand column tells us the probability that the team will score this number of goals. For example: • The probability that the team scores exactly 0 goals is 0.18. • The probability that the team scores exactly 1 goal is 0.34. • The probability that the team scores exactly 2 goals is 0.35. And so on. ### Properties of a Probability Distribution Table A probability distribution table has the following properties: 1. All probabilities must add up to 1.Â For a probability distribution table to be valid, all of the individual probabilities must add up to 1. We can verify that the previous probability distribution table is valid: Sum of probabilities = 0.18 + 0.34 + 0.35 + 0.11 + 0.02 = 1. 2. The mean can be calculated. The formula to calculate the mean of a given probability distribution table is: Î¼ = Î£x * P(x) where: • x: Data value • P(x): Probability of value For example, consider our probability distribution table for the soccer team: The mean number of goals for the soccer team would be calculated as: Î¼ = 00.18Â +Â 10.34Â +Â 20.35Â +Â 30.11Â +Â 40.02Â =Â Â 1.45Â goals. 3. The standard deviation can be calculated. The formula to calculate the standard deviation of a given probability distribution table is: Ïƒ = âˆšÎ£(xi-Î¼)2Â P(xi) where: • xi:Â The ithÂ value • Î¼:Â The mean of the distribution • P(xi):Â The probability of the ithÂ value For example, hereâ€™s how to calculate the standard deviation of goals scored by the soccer team: The standard deviation is the square root of the sum of the values in the third column: Standard deviation = âˆš(.3785 + .0689 + .1059 + .2643 + .1301)Â =Â 0.9734 ### How to Visualize a Probability Distribution Table The easiest way to visualize the values in a probability distribution table is by using a histogram, which displays the values of the random variable along the x-axis and the probability of those values along the y-axis: This lets us quickly visualize the probability values from the table. In particular, we can see that there is a high probability that the team scores 2 goals or less while there is a tiny probability that the team scores as many as 4 goals.
# If the sum of first four terms of an A.P. is 40 and that of first 14 terms is 280. Find the sum of its first n terms. Given: The sum of first four terms of an A.P. is 40 and that of first 14 terms is 280. To do: We have to find the sum of its first $n$ terms. Solution: Let the first term be $a$ and the common differnce be $d$. We know that, Sum of $n$ terms$S_{n} =\frac{n}{2}(2a+(n-1)d)$ $S_{4}=\frac{4}{2}[2(a)+(4-1)d]$ $40=2(2a+3d)$ $20=2a+3d$ $2a=20-3d$......(i) $S_{14}=\frac{14}{2}[2(a)+(14-1)d]$ $280=7(2a+13d)$ $40=2a+13d$ $20-3d+13d=40$ (From (i)) $10d=40-20$ $d=\frac{20}{10}$ $d=2$ This implies, $2a=20-3(2)$ $2a=20-6$ $a=\frac{14}{2}$ $a=7$ The sum of $n$ terms $S_n=\frac{n}{2}(2a+(n-1)d)$ $S_n=\frac{n}{2}[2(7)+(n-1)2]$ $=n(7+n-1)$ $=n(n+6)$ $=n^2+6n$ Hence, the sum of $n$ terms is $n^2+6n$. Tutorialspoint Simply Easy Learning Updated on: 10-Oct-2022 43 Views
# Year Five Mathematics Worksheets Place value is a fundamental concept in mathematics and is essential for understanding numbers and how they work. Place value refers to the value of a digit in a number based on its position. In this article, we will focus on place value up to thousandths. Ones: The first place value is ones. This refers to the value of the rightmost digit in a number. For example, in the number 12, the 2 is in the ones place and represents 2. Tenths: The next place value is tenths. This refers to the value of the digit one place to the right of the ones place. It represents a fraction of one and is expressed as a decimal. For example, in the number 12.5, the 5 is in the tenths place and represents 5/10 or 0.5. Hundredths: The next place value is hundredths. This refers to the value of the digit two places to the right of the ones place. It represents a fraction of one and is expressed as a decimal. For example, in the number 12.56, the 6 is in the hundredths place and represents 6/100 or 0.06. Thousandths: The final place value we will cover is thousandths. This refers to the value of the digit three places to the right of the ones place. It represents a fraction of one and is expressed as a decimal. For example, in the number 12.567, the 7 is in the thousandths place and represents 7/1000 or 0.007. It’s important for kids to understand place value as it is used in many areas of mathematics, including addition, subtraction, multiplication, and division. Understanding place value also helps kids understand the value of numbers and how to compare and round them. In conclusion, place value is a critical concept in mathematics that helps kids understand numbers and how they work. By understanding place value up to thousandths, kids will have a strong foundation for future success in math and will be able to work with decimals with ease. # Year Five Math Worksheet for Kids – Understanding Place Value Up To Thousandths Taking too long? \| Open in new tab ## Applied Machine Learning & Data Science Projects and Coding Recipes for Beginners A list of FREE programming examples together with eTutorials & eBooks @ SETScholars # Projects and Coding Recipes, eTutorials and eBooks: The best All-in-One resources for Data Analyst, Data Scientist, Machine Learning Engineer and Software Developer Topics included: Classification, Clustering, Regression, Forecasting, Algorithms, Data Structures, Data Analytics & Data Science, Deep Learning, Machine Learning, Programming Languages and Software Tools & Packages. (Discount is valid for limited time only) `Disclaimer: The information and code presented within this recipe/tutorial is only for educational and coaching purposes for beginners and developers. Anyone can practice and apply the recipe/tutorial presented here, but the reader is taking full responsibility for his/her actions. The author (content curator) of this recipe (code / program) has made every effort to ensure the accuracy of the information was correct at time of publication. The author (content curator) does not assume and hereby disclaims any liability to any party for any loss, damage, or disruption caused by errors or omissions, whether such errors or omissions result from accident, negligence, or any other cause. The information presented here could also be found in public knowledge domains.`
# A-level Mathematics/OCR/C2/Sequences and Series ## Definitions A sequence is simply a list of numbers in a particular order. We call these numbers the terms of the sequence. For instance, 2,4,6,8 are the first four terms in the sequence of even positive integers. When we take the sum of the terms in a sequence, we get a series. For example, 2+4+6+8+... is a series. We denote the terms in a sequence by ${\displaystyle T_{n}}$ where ${\displaystyle n}$ is the number of the term in question. For example, we have ${\displaystyle T_{1}=2}$, ${\displaystyle T_{2}=4}$, ${\displaystyle T_{3}=6}$, and so on, in the sequence described above. A definition is a rule that tells us how to compute each term in a sequence. For example, a rule for the sequence above is ${\displaystyle T_{n}=2n}$. A relation describes how each term is related to other terms. For instance, a relation for the above sequence is ${\displaystyle T_{n+1}=T_{n}+2}$. ## Sigma (${\displaystyle \Sigma }$) Notation As you might have suspected, describing a series with the help of some of its terms isn't always a good idea --- if too few terms are used, the series can be ambiguous to your reader; on the other hand, you risk insulting your reader by writing out too many terms! To express a series succinctly, we use the sigma ${\displaystyle (\Sigma )}$ notation instead. In general, a series may be written as ${\displaystyle \sum _{k=k_{0}}^{N}f(k)}$, which means "sum of all terms beginning with ${\displaystyle f(k_{0})}$ up to and including ${\displaystyle f(N)}$". Hence, ${\displaystyle \sum _{k=k_{0}}^{N}f(k)=f(k_{0})+f(k_{0}+1)+f(k_{0}+2)+\cdots +f(N)}$. As an example, the series 2+4+6+8+... may be written as ${\displaystyle \sum _{k=1}^{\infty }2k}$. ## Recognising Simple Progressions A progression is just another word for a sequence. In this module, you are expected to be well-acquainted with two very common types of progressions --- the arithmetic progression and the geometric progression. Briefly, an arithmetic progression or AP is a sequence in which each successive term is the sum of the previous term and a fixed value. An example of an AP is 1,4,7,10,..., where the difference between successive terms is 3. A geometric progression or GP is a sequence in which each successive term is the product of the previous term and a fixed value. An example of a GP is 2,4,8,16,32,..., where each term is twice the value of the previous term. ## Arithmetic Progression (AP) An arithmetic progression (AP) is a sequence that can be written in the following way: ${\displaystyle a,a+d,a+2d,a+3d,a+4d,\ldots }$, where ${\displaystyle a,d}$ are constants. The first term ${\displaystyle T_{1}}$ in the AP is denoted by ${\displaystyle a}$, and the common difference between subsequent terms is denoted by ${\displaystyle d}$. Thus, the series 1,4,7,10,..., is an arithmetic progression with ${\displaystyle a=1}$ and ${\displaystyle d=3}$. ### Rules The common difference ${\displaystyle d}$ can be calculated by ${\displaystyle d=T_{n}-T_{n-1}}$, where ${\displaystyle n=2,3,4,...}$. The ${\displaystyle n}$th term is given by ${\displaystyle T_{n}=a+(n-1)d}$. The sum of the first ${\displaystyle n}$ terms of an AP (with ${\displaystyle T_{1}}$ as its first term and ${\displaystyle T_{n}}$ as its last term) is given by ${\displaystyle S_{n}={\frac {n}{2}}(T_{1}+T_{n})={\frac {n}{2}}(2a+(n-1)d)}$. In fact, more generally, the sum of ${\displaystyle n}$ consecutive terms in an AP is given by ${\displaystyle {\frac {\mathrm {number\,of\,terms} }{2}}(\mathrm {first\,term} +\mathrm {last\,term} )}$. ### Example What is the sum of the even numbers 2, 4, 6, 8, ..., 100? The given sequence can be expressed as an AP with ${\displaystyle T_{1}=a=2}$, and ${\displaystyle d=2}$. We want the sum of the first 50 terms of the AP: ${\displaystyle S_{50}={\frac {50}{2}}(2+100)={\frac {50}{2}}\left(2(2)+(50-1)2\right)=2550}$. ## Geometric Progression (GP) A geometric progression (GP) is a sequence that can be written in the following way: ${\displaystyle a,ar,ar^{2},ar^{3},ar^{4},\ldots }$, where ${\displaystyle a,r}$ are constants. The first term ${\displaystyle T_{1}}$ in the GP is denoted by ${\displaystyle a}$, and the common ratio between subsequent terms is denoted by ${\displaystyle r}$. ### Rules The common ratio ${\displaystyle r}$ can be calculated by ${\displaystyle r={\frac {T_{n}}{T_{n-1}}}}$, where ${\displaystyle n=2,3,4,...}$. The ${\displaystyle n}$th term is given by ${\displaystyle T_{n}=ar^{n-1}}$. The sum of the first ${\displaystyle n}$ terms of an GP is given by ${\displaystyle S_{n}={\frac {a(1-r^{n})}{1-r}}}$. Proof of this is given by: ${\displaystyle S_{n}=a+ar+ar^{2}+\ldots +ar^{n-1}}$ ${\displaystyle rS_{n}=ar+ar^{2}+ar^{3}+\ldots +ar^{n}}$ ${\displaystyle S_{n}-rS_{n}=a-ar^{n}}$ ${\displaystyle S_{n}(1-r)=a(1-r^{n})}$ ${\displaystyle S_{n}={\frac {a(1-r^{n})}{1-r}}}$ ### Sum Of An Infinite Geometric Series We say that the geometric series ${\displaystyle \sum _{k=0}^{\infty }ar^{k}=a+ar+ar^{2}+ar^{3}+ar^{4}+\ldots }$ is convergent if the sum to infinity ${\displaystyle S_{\infty }}$ approaches some limit. This occurs when ${\displaystyle \|r\|<1}$. Hence, if ${\displaystyle \|r\|<1}$, then ${\displaystyle S_{\infty }=\lim _{n\rightarrow \infty }{\frac {a(1-r^{n})}{1-r}}={\frac {a}{1-r}}}$. Proof of this is given by: ${\displaystyle S_{\infty }=a+ar+ar^{2}+ar^{3}+...}$ ${\displaystyle rS_{\infty }=ar+ar^{2}+ar^{3}+ar^{4}+...}$ ${\displaystyle S_{\infty }-rS_{\infty }=a}$ ${\displaystyle S_{\infty }(1-r)=a}$ ${\displaystyle S_{\infty }={\frac {a}{1-r}}}$ ## Binomial expressions A binomial is a polynomial with two parts in the form ${\displaystyle (a+b)}$, such as ${\displaystyle (x+1)}$. When a binomial is raised to a power, you could simplify it by multiplying out the brackets several times. The expanded polynomial is called a binomial expansion, and all binomial expansions follow a pattern that can be used to expand binomials quicker than multiplying out several brackets. For now, we will only look at binomial expressions which are raised to positive integers. ### Expansions of ${\displaystyle (x+1)}$ Here are the expansions of ${\displaystyle x+1}$ raised to different powers. ${\displaystyle (x+1)^{1}=}$ ${\displaystyle 1x+1}$ ${\displaystyle (x+1)^{2}=}$ ${\displaystyle 1x^{2}+2x+1}$ ${\displaystyle (x+1)^{3}=}$ ${\displaystyle 1x^{3}+3x^{2}+3x+1}$ ${\displaystyle (x+1)^{4}=}$ ${\displaystyle 1x^{4}+4x^{3}+6x^{2}+4x+1}$ ${\displaystyle (x+1)^{5}=}$ ${\displaystyle 1x^{5}+5x^{4}+10x^{3}+10x^{2}+5x+1}$ If you look at the coefficient of each term, you may notice a pattern. These numbers are called binomial coefficients and are found by adding the two numbers above it. ### Pascal's triangle Binomial coefficients are more commonly known as Pascal's triangle, named after Blaise Pascal. The first 10 lines of Pascal's triangle are: (1) 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 1 9 36 84 126 126 84 36 9 1 1 10 45 120 210 252 210 120 45 10 1 1 11 55 165 330 462 462 330 165 55 11 1 Since each number is found by adding the two numbers above it, it is possible to find a few lines of the triangle to help you expand binomials. For binomials raised to powers greater than 10, you should use the binomial coefficient formula. ### Binomial coefficient formula When a binomial is raised to a large power, it may be too time consuming to find the binomial coefficients by writing out Pascal's triangle. Fortunately, there is a formula that can find any line of Pascal's triangle. If ${\displaystyle n}$ is the power of the expansion, and ${\displaystyle r}$ is the number of the term in a single row, the binomial coefficient formula is: ${\displaystyle {n \choose r}={\frac {n!}{r!(n-r)!}}}$ The ${\displaystyle !}$ means factorial and multiplies ${\displaystyle n}$ by every integer less than itself, down to 1. So ${\displaystyle 3!=3\times 2\times 1=6}$. To find the binomial coefficients, you use the formula with the required value of ${\displaystyle n}$, and ${\displaystyle r=0}$, ${\displaystyle r=1}$, ${\displaystyle r=2}$, and so on, until ${\displaystyle r=n}$. Most scientific calculators will have two buttons that will be useful in this process, one is the factorial button, usually labelled n! and the other will actually find ${\displaystyle {n \choose r}}$ and is often labelled nCr or ${\displaystyle C_{r}^{n}}$. (The C stands for "choose" or "combination" which is based on the formula's use in probability.) You should be aware that Pascal's triangle is symmetrical, so once a coefficient is repeated, you can write down the rest of the coefficients with ease. ### Expanding binomials Now that you know how to find the coefficients in a binomial expansion, you can easily expand any binomial that is raised to a positve integer by following these simple steps: For a binomial in the form ${\displaystyle (a+b)^{n}}$, 1. Write down ${\displaystyle a}$ in descending powers, from ${\displaystyle n}$ to ${\displaystyle 0}$ 2. Write down ${\displaystyle b}$ in ascending powers, from ${\displaystyle 0}$ to ${\displaystyle n}$, making sure that you place the terms so that the powers add up to ${\displaystyle n}$ 3. Add the binomial coefficients to each term, either from line ${\displaystyle n}$ in Pascal's triangle (ignoring the 1 at the top), or by using the binomial coefficient formula. You then simplify where necessary. For example, for the expansion of ${\displaystyle (x+2)^{4}}$: ${\displaystyle x}$ in descending powers: ${\displaystyle x^{4}}$ ${\displaystyle +}$ ${\displaystyle x^{3}}$ ${\displaystyle +}$ ${\displaystyle x^{2}}$ ${\displaystyle +}$ ${\displaystyle x^{1}}$ ${\displaystyle +}$ ${\displaystyle x^{0}}$ ${\displaystyle 2}$ in ascending powers: ${\displaystyle 2^{0}}$ ${\displaystyle +}$ ${\displaystyle 2^{1}}$ ${\displaystyle +}$ ${\displaystyle 2^{2}}$ ${\displaystyle +}$ ${\displaystyle 2^{3}}$ ${\displaystyle +}$ ${\displaystyle 2^{4}}$ Grouping everything together we now have: ${\displaystyle x^{4}}$${\displaystyle 2^{0}}$ ${\displaystyle +}$ ${\displaystyle x^{3}}$${\displaystyle 2^{1}}$ ${\displaystyle +}$ ${\displaystyle x^{2}}$${\displaystyle 2^{2}}$ ${\displaystyle +}$ ${\displaystyle x^{1}}$${\displaystyle 2^{3}}$ ${\displaystyle +}$ ${\displaystyle x^{0}}$${\displaystyle 2^{4}}$ ${\displaystyle 1x^{4}}$${\displaystyle 2^{0}}$ ${\displaystyle +}$ ${\displaystyle 4x^{3}}$${\displaystyle 2^{1}}$ ${\displaystyle +}$ ${\displaystyle 6x^{2}}$${\displaystyle 2^{2}}$ ${\displaystyle +}$ ${\displaystyle 4x^{1}}$${\displaystyle 2^{3}}$ ${\displaystyle +}$ ${\displaystyle 1x^{0}}$${\displaystyle 2^{4}}$ Finally simplifying will now give us: ${\displaystyle x^{4}}$ ${\displaystyle +}$ ${\displaystyle 8x^{3}}$ ${\displaystyle +}$ ${\displaystyle 24x^{2}}$ ${\displaystyle +}$ ${\displaystyle 32x^{1}}$ ${\displaystyle +}$ $16$ This process is summarised in the equation known as the binomial theorem: ${\displaystyle (x+y)^{n}=\sum _{r=0}^{n}{n \choose r}x^{r}y^{n-r}}$ In case you are not familiar with sigma notion this means: ${\displaystyle (x+y)^{n}={n \choose 0}x^{0}y^{n}+{n \choose 1}x^{1}y^{n-1}+{n \choose 2}x^{2}y^{n-2}+\dots +{n \choose {n-2}}x^{n-2}y^{2}+{n \choose {n-1}}x^{n-1}y^{1}+{n \choose n}x^{n}y^{0}}$ Several simplifications can be made but they aren't worth memorising as you will pick them up automatically: ${\displaystyle (x+y)^{n}=y^{n}+nxy^{n-1}+{n \choose 2}x^{2}y^{n-2}+\dots +{n \choose {n-2}}x^{n-2}y^{2}+nx^{n-1}y+x^{n}}$ This is part of the C2 (Core Mathematics 2) module of the A-level Mathematics text. Appendix A: Formulae
# Difference between revisions of "2007 AMC 10A Problems/Problem 15" Three circles are inscribed in a rectangle of width w and height h as shown. Two of the circles are congruent and are each tangent to two adjacent sides of the rectangle and to each other. The other circle is larger and is tangent to three sides of the rectangle and to the two smaller circles. What the ratio of h to w? Express your answer as a decimal to the nearest hundredth. ## Problem Four circles of radius $1$ are each tangent to two sides of a square and externally tangent to a circle of radius $2$, as shown. What is the area of the square? $[asy] unitsize(5mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); real h=3sqrt(2)/2; pair O0=(0,0), O1=(h,h), O2=(-h,h), O3=(-h,-h), O4=(h,-h); pair X=O0+2dir(30), Y=O2+dir(45); draw((-h-1,-h-1)--(-h-1,h+1)--(h+1,h+1)--(h+1,-h-1)--cycle); draw(Circle(O0,2)); draw(Circle(O1,1)); draw(Circle(O2,1)); draw(Circle(O3,1)); draw(Circle(O4,1)); draw(O0--X); draw(O2--Y); label("2",midpoint(O0--X),NW); label("1",midpoint(O2--Y),SE); [/asy]$ $\text{(A)}\ 32 \qquad \text{(B)}\ 22 + 12\sqrt {2}\qquad \text{(C)}\ 16 + 16\sqrt {3}\qquad \text{(D)}\ 48 \qquad \text{(E)}\ 36 + 16\sqrt {2}$ ## Solution Draw a square connecting the centers of the four small circles of radius $1$. This square has a diagonal of length $6$, as it includes the diameter of the big circle of radius $2$ and two radii of the small circles of radius $1$. Therefore, the side length of this square is $$\frac{6}{\sqrt{2}} = 3\sqrt{2}.$$ The radius of the large square has a side length $2$ units larger than the one found by connecting the midpoints, so its side length is $$2 + 3\sqrt{2}.$$ The area of this square is $(2+3\sqrt{2})^2 = 22 + 12\sqrt{2}$ $(B).$ ## Solution 2 We draw the long diagonal of the square. This diagonal yields $2\sqrt{2}+1+1+2+2=2\sqrt{2}+6$. We know that the side length $s$ in terms of the diagonal $d$ is $s=\frac{d}{\sqrt{2}}$, so our side length is $\frac{\sqrt{2}+6}{\sqrt{2}}$. However, we are trying to look for the area, so squaring $\frac{\sqrt{2}+6}{\sqrt{2}}$ yields $\frac{44+24\sqrt{2}}{2}=\boxed{\text{B}22+12\sqrt{2}$ (Error compiling LaTeX. ! File ended while scanning use of \boxed.)
Boxing Pythagoras On the Pythagorean Theorem In right-angled triangles, the square on the side subtending the right-angle is equal to the (sum of the) squares on the sides containing the right-angle. Euclid’s Elements, Book 1, Proposition 47 (R. Fitzpatrick, trans.) Figure 1: A right triangle with squares on its sides The Pythagorean Theorem is my favorite math problem of all time. I feel so strongly about this particular bit of geometry that I have the theorem tattooed on my chest. Over my heart. In the original Greek. Yeah, I’m that kind of nerd. Most people have some vague recollection from their high school math classes that the Pythagorean Theorem is $a^2+b^2=c^2$; and a few even remember that the in that equation refers to the hypotenuse of a right triangle, while the a and b refer to the other two legs. However, most of the time, people were just taught to memorize this theorem– they weren’t taught how to prove that it was actually true. Now, the Internet is full of all kinds of really clever visual proofs involving rearranging copies of the triangle in order to form the different squares, but I’m not really a huge fan of these. They make it very easy to see that the Pythagorean Theorem is true, but they don’t really make it easy to see why the Pythagorean Theorem is true. So, today, I wanted to discuss my favorite proof for the Pythagorean Theorem, which comes to us by way of Euclid’s Elements, which was the standard textbook for math in the West for around 2000 years. In Figure 1, you’ll see that we have our triangle $\triangle ABC$, where $\angle ABC$ is a right angle. The figures ABDE, BCFG, and ACHJ are squares. Figure 2: Dividing the square on the hypotenuse My first step is to draw a line from point B which is perpendicular to $\overline{HJ}$. You’ll notice that this divides the big square, ACHJ, into two rectangles: AJKL and CHKL. Now, it’s fairly obvious that adding these two rectangles together gives you the big square, so I’m going to attempt to prove that square ABDE has the same area as rectangle AJKL, and that square BCFG has the same area as rectangle CHKL. Figure 3: Proving ABDE = AJKL Let’s start with square ABDE. I’m going to draw two more line segments, now, $\overline{EC}$ and $\overline{BJ}$. You might notice that this forms two new triangles, $\triangle ACE$ and $\triangle ABJ$. These two triangles are congruent, since $\overline{EA}=\overline{AB}$, $\overline{AJ}=\overline{AC}$, and $\angle EAC=\angle BAJ$. Now, if you remember your geometry, you’ll recall that the area of a triangle is equal to its base times its height, divided by two, while the area of a rectangle is just its base times its height. Looking at $\triangle ACE$, we can use $\overline{AE}$ as its base, and we can see easily see that its height must be equal to $\overline{AB}$. Since square ABDE and $\triangle ACE$ share their base and height, we can conclude that square ABDE has double the area of $\triangle ACE$. By that same token, $\triangle ABJ$ can be seen to have a base $\overline{AJ}$ and a height $\overline{AL}$, which means that rectangle AJKL must be double the area of $\triangle ABJ$. Since $\triangle ACE$ and $\triangle ABJ$ are equal, that means the area of square ABDE and rectangle AJKL must also be equal. Figure 4: Proving BCFG = CHKL Now we can repeat the exact same process with square BCFG and rectangle CHKL. First I draw $\overline{BH}$ and $\overline{AF}$. This gives me $\triangle BCH$ and $\triangle ACF$, which are congruent. Square BCFG shares a base and height with $\triangle ACF$, and is therefore double $\triangle ACF$. Rectangle CHKL shares a base and height with $\triangle BCH$, and is therefore double $\triangle BCH$. Since $\triangle BCH$ and $\triangle ACF$ are equal, we know that square BCFG must equal rectangle CHKL. Figure 5: The Theorem proved And there you have it! We proved that the squares adjacent to the right angle are equal in area to two rectangles, and we know that those rectangles add up to the area of the square opposite to the right angle. Therefore, the sum of the squares adjacent to the right angle is equal to the square opposite to the right angle.
UK USIndia Every Question Helps You Learn Jason asked his friends how many times they had visited the cinema this year. # Handling Data 3 (Difficult) Calculating the average value from a set of data is an important part of data handling, and you need to be familiar with it if you want to do well in your Eleven Plus maths exam. If you have played the first two quizzes in this section, then it shouldn’t be a problem. This third quiz gives you another opportunity to revise! Here are some terms you need to know about averages: • Range is the difference between the highest and lowest values in a set of data • Mean is the average when you add all the values together, then divide by the number of values • Median is the average when you line up the data in ascending order and choose the middle value • Mode is the average worked out by choosing the value which appears most often in a set of data That may sound a little complicated, but it’s quite easy really – once you know what you are doing, that is! Do you? Well, it’s time to find out by playing the quiz. Take your time and be sure to work out your answers properly. Good luck! 1. This question is based on the following data set which represents the number of times that Jason’s friends had visited the cinema this year: 12, 1, 4, 6, 3, 5, 8. What is the range of the given data? 11 8 12 2 Range = highest value - lowest value = 12 - 1 = 11 2. This question is based on the following data set which represents the number of times that Jason’s friends had visited the cinema this year: 12, 1, 4, 6, 3, 5, 8. What is the mean of the given data? 5 5.6 6 5.7 Mean = (the sum of the data values) ÷ the number of data values. The sum of the data values = 12 + 1 + 4 + 6 + 3 + 5 + 8 = 39. The number of data values = 7 ? mean = 39 ÷ 7 = 5.571 = 5.6 (1 decimal place) 3. This question is based on the following data set which represents the number of times that Jason’s friends had visited the cinema this year: 12, 1, 4, 6, 3, 5, 8. What is the mode of the given data? 5 5.5 6 There isn't a mode The mode is the data value that occurs most often: in this set of data, there isn't a mode as each value occurs only once 4. This question is based on the following data set which represents the number of times that Jason’s friends had visited the cinema this year: 12, 1, 4, 6, 3, 5, 8. What is the median of the given data? 3 4 5 6 The median is the middle value of the data: you have to arrange the data in order BEFORE you can find it: 1, 3, 4, 5, 6, 8, 12. You have seven values, so the median is the 4th value = 5. It is usually best to arrange the data in ascending order. DON'T make the mistake of giving the median the value of its position: in this case 4 because it is the 4th value 5. This question is based on the following data set which represents the number of times that Jason’s friends had visited the cinema this year: 12, 1, 4, 6, 3, 5, 8. How many of Jason’s friends took part in the survey? 7 39 12 24 You have seven data values: each value corresponds to one person. DON'T make the mistake of calling the sum of the data values the number of people: it is NOT 39 6. This question is based on the following data set which represents the number of times that Jason’s friends had visited the cinema this year: 12, 1, 4, 6, 3, 5, 8. Jason decided to include one more friend in his survey. If that friend hadn't been to the cinema this year, how would that affect the value of the 'mean'? It would increase the value of the 'mean' It would lower the value of the 'mean' It wouldn't affect the value of the 'mean' There isn't enough information to answer this question Just because something is zero, it doesn't mean that it isn't important. Mean = (the sum of the data values) ÷ the number of data values. The sum of the data values = 0 + 12 + 1 + 4 + 6 + 3 + 5 + 8 = 39. So '0' won't affect the sum BUT the number of data values now = 8 because you have an extra person ? mean = 39 ÷ 8 = 4.875 = 4.9 (1 decimal place) 7. This question is based on the following data set which represents the number of times that Jason’s friends had visited the cinema this year: 12, 1, 4, 6, 3, 5, 8. How many people went to the cinema the same number of times? 12 8 6 0 None of the numbers occurs more than once, so nobody went to the cinema the same number of times 8. This question is based on the following data set which represents the number of times that Jason’s friends had visited the cinema this year: 12, 1, 4, 6, 3, 5, 8. How many people went to the cinema more times than the mean number of times? 5 4 3 2 The mean is 5.6: only three values > 5.6. You can't include the person who went to the cinema 5 times because 5 < 5.6. D'oh! 9. Which of the following statements is correct? The mode is always one of the data values The mode is never one of the data values The mode is sometimes one of the data values The mode is the middle number in a set of data The mode is the number which appears most often, so it is always one of the data values 10. Why does data have to be ordered to calculate the mode? So that you know how many data values you have So that you can find the'middle' value To make calculating the answer easier It doesn't have to be ordered The mode is the value which appears most often - you can usually spot this at a glance, without orderinfg the data - but you do have to order the data if you want to find the median value Author: Frank Evans
# Quantitative Aptitude Questions for IBPS Exam Set – 27 Hello Aspirants. Welcome to Online Quantitative Aptitude section in AffairsCloud.com. Here we are creating question sample From all topics , which are Important for upcoming IBPS and RRB exams. We have included Some questions that are repeatedly asked in exams. 1. In what proportion must water be mixed with alcohol to gain 13 1/3% by selling it at cost price? A. 1 : 5 B. 2 : 15 C. 3: 10 D. 10 : 3 B. 2 : 15 Explanation: Ratio: 13 1/3 : 100 40/3 : 100 40 : 300 2: 15 2. If 2 men or 4 boys can do a piece of work in 126 days, how long will 6 men and 9 boys take to do the same work? A. 12 days B. 16 days C. 20 days D. 24 days D. 24 days Explanation: 12624 / 29 + 46 = 24 3. At simple interest, Rs 64 amounts to Rs 83.2 in 2 yrs, what will rs 86 amount to in 4 yrs? A. Rs 137.6 B. Rs 130 C. Rs 135.6 D. Rs 135.7 A. Rs 137.6 Explanation: Rs 64 in 2 yrs give SI = 83.2–64 = Rs 19.2 Interest on Rs 86 for 4 yrs = 19.2 * [86/64] [4/2] = Rs 51.6 Amount on Rs 86 = 86+51.6 =137.6 4. 8 men and 16 women do a work in 8 days, then in how many days 6 men and 12 women do the same work? A. 6 days B. 10 days C. 10 2/3 days D. Cannot be determined C. 10 2/3 days Explanation: In this 8 : 16 is same as 6 : 12 So we can use as: 8m + 16w = 8 days 8(1m + 2w) = 8 days 1m + 2w = 88 days 6(1m + 2w) = 88/ 6 days 6m + 12w = 10 2/3 days 5. A is twice good as workman as B. A and B together do a piece of work in 18 days, then in how many days A does the work alone? A. 15 days B. 18 days C. 25 days D. 27 days D. 27 days Explanation: A+B do work in 18 days, this implies twice good + once good workman do in 18 days. 3 times good in 18 days 1 time good in 183 days A(twice good) work in 183 / 2 = 27 days 6. A man purchases apples at 11 for Rs 10 and sells them at 10 for Rs 11. Find the profit/loss percent? A. 21% B. 22 2/3% C. 25% D. 27% A. 21% Explanation: Profit% = (1111 – 1010 / 1010) 100 = 21% 7. The incomes of A and B are in the ratio 2 : 5 and their expenditures are in the ratio 4 : 9. If A saves Rs 2000 and B saves Rs 6000, what is B’s income? A. Rs 15,000 B. Rs 12,000 C. Rs 14,000 D. Rs 13,000 A. Rs 15,000 Explanation: 2x – 4y = 2000 5x – 9y = 6000 Solve the equations, x = 3000 So A’s income = 5x = 53000 = 15,000 8. A man bought a shirt at 5/6th of its list price and sold it at 20% above its list price. What is his percentage gain in the transaction? A. 26% B. 34% C. 40% D. 44% D. 44% Explanation: 6/5 [100+20] – 100 = 44% 9. In going 30 km upstream and come back, speed of boat in still water is 12 km/hr and speed of stream is 4 km/hr. If it takes 2 hrs to go upstream, then find the time taken for downstream journey? A. 1 hr B. 2 hr C. 3 hr D. 4 hr A. 1 hr Explanation: Use: B = [tu + td] / [tu – td] R 12 = [2+ td] / [2- td] * 3 td = 1 hr 10. A man can row 9 km/hr in still water. When the river is running at 6 km/hr, it takes him 6 hrs to row to a place and come back. How far is the place? A. 15 km B. 18 km C. 20 km D. 22 km A. 15 km Explanation: Time is total time to go upstream and downstream B is speed of boat in still water, R is speed of river/stream Distance = time * [B^2 – R^2] / 2B = 3 [9^2 – 6^2] / 2*9 AffairsCloud Recommends Oliveboard Mock Test AffairsCloud Ebook - Support Us to Grow
# An isosceles triangle has sides A, B, and C, such that sides A and B have the same length. Side C has a length of 8 and the triangle has an area of 64 . What are the lengths of sides A and B? Mar 8, 2018 Side color(brown)(a = b = 16.49 units #### Explanation: A median if triangle divides it in 2 triangles of equal areas. And if it is isosceles triangle the median is perpendicular bisector. $c = 8$ $a = b , h e i g h t = h$ Half base = c / 2 = 8 / 2 = 4 Area of triangle = 64 sq units Area of half triangle = 32 sq units 1/2 * 4 * h = 32 $h e i g h t$ $h = 16$ units By using Pythagoras theorem a = b = sqrt(4^2 + 16^2) = sqrt 272 = color(brown)(16.49 units Mar 8, 2018 $A = B = 4 \sqrt{17} \text{units}$ #### Explanation: Let $h$ the height $\text{area of triangle} = \frac{1}{2} \times h \times C$ $64 = \frac{1}{2} \times h \times 8$ $h = 16 \text{ units}$ ∆ MHN: # $M {N}^{2} = {h}^{2} + N {H}^{2}$ ${A}^{2} = {16}^{2} + {\left(\frac{8}{2}\right)}^{2}$ $A = \sqrt{272}$ $A = 4 \sqrt{17}$ $A = B = 4 \sqrt{17} \text{units}$

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