1. The point of inflection of the function y = x3 − 3x2 + 6x + 2000 is x = 1. At this point, the slope is 3.
2. The optimal length for Karen to climb is when x = √3/3, which is the solution to the equation 1 − 3x2 = 0 derived from finding the critical points of her change in position, x − x3.
3. By applying the Fundamental Theorem of Calculus, the constant C in the solution to the given integral 1/4π ∫0 sin x + C dx is equal to 2√2 − 4/π.
This document contains 10 math problems with solutions. The problems cover topics like probability, combinatorics, geometry, and complex analysis. The solutions provide detailed step-by-step workings to arrive at the final answers.
The document contains 10 geometry problems with solutions. Problem 1 asks about the total number of sides of seven polygons given the sum of their interior angles. Problem 2 asks about the shaded area of an infinitely repeating pattern of equilateral triangles. Problem 3 asks about the ratio of the diagonal to the side of a regular pentagon.
This document contains 10 math problems with solutions. Problem 1 asks the reader to calculate the product of the smallest and largest prime factors of the number of words in the problem statement. The answer is 1681. Problem 2 asks what percentage of gold King Midas would need to earn today to end up with as much gold as he started with after spending a certain percentage yesterday. The answer is 100x-1%. Problem 3 asks the reader to find all integer pairs (a,b) such that ab + a - 3b = 5. The answer lists the four pairs.
1. The document contains 19 math problems from various topics including calculus, algebra, geometry, probability, and number theory. The problems range in complexity from computing simple expressions to word problems involving multiple steps.
This document provides examples of solutions to problems involving number theory, algebra, geometry, and probability. It contains the following types of problems:
- Number theory word problems involving finding missing terms in sequences
- Solving quadratic equations and determining equations with transformed roots
- Calculating areas and perimeters of geometric shapes like octagons and hexagons
- Finding equations of lines transformed by parallel shifts
- Probability problems involving counting outcomes, finding probabilities of events, and calculating combinations and permutations
The document aims to demonstrate smart solutions to a variety of mathematical problems across different topics through clear explanations and step-by-step working.
1. The document contains 10 math problems from an advanced topics test involving probabilities of dice rolls, digits in square numbers, gear rotations, team formations in a Pokémon-style game, drawing cards from a deck, complex number solutions, evaluating expressions involving square roots and trigonometric functions.
2. The problems cover a wide range of mathematical topics including probability, number theory, combinatorics, algebra, and trigonometry.
3. The level of difficulty ranges from straightforward calculations to more complex problems requiring multiple steps and conceptual understanding.
This document provides the answers to a mathematics exam for 10th grade students. It includes multiple choice questions with explanations and word problems with step-by-step solutions. The topics covered include algebra, logarithms, quadratic equations, and inequalities.
1. The document contains 21 multiple choice math questions covering topics like area, perimeter, volume, coordinate geometry, factoring, and algebraic equations.
2. For each question, the question stem and possible multiple choice answers are provided, along with the correct answer and an explanation of the mathematical steps taken to arrive at the solution.
3. The questions progress from easier concepts involving basic formulas to more complex problems requiring multiple steps of algebraic manipulation or geometric reasoning.
This document contains 10 math problems with solutions. The problems cover topics like probability, combinatorics, geometry, and complex analysis. The solutions provide detailed step-by-step workings to arrive at the final answers.
The document contains 10 geometry problems with solutions. Problem 1 asks about the total number of sides of seven polygons given the sum of their interior angles. Problem 2 asks about the shaded area of an infinitely repeating pattern of equilateral triangles. Problem 3 asks about the ratio of the diagonal to the side of a regular pentagon.
This document contains 10 math problems with solutions. Problem 1 asks the reader to calculate the product of the smallest and largest prime factors of the number of words in the problem statement. The answer is 1681. Problem 2 asks what percentage of gold King Midas would need to earn today to end up with as much gold as he started with after spending a certain percentage yesterday. The answer is 100x-1%. Problem 3 asks the reader to find all integer pairs (a,b) such that ab + a - 3b = 5. The answer lists the four pairs.
1. The document contains 19 math problems from various topics including calculus, algebra, geometry, probability, and number theory. The problems range in complexity from computing simple expressions to word problems involving multiple steps.
This document provides examples of solutions to problems involving number theory, algebra, geometry, and probability. It contains the following types of problems:
- Number theory word problems involving finding missing terms in sequences
- Solving quadratic equations and determining equations with transformed roots
- Calculating areas and perimeters of geometric shapes like octagons and hexagons
- Finding equations of lines transformed by parallel shifts
- Probability problems involving counting outcomes, finding probabilities of events, and calculating combinations and permutations
The document aims to demonstrate smart solutions to a variety of mathematical problems across different topics through clear explanations and step-by-step working.
1. The document contains 10 math problems from an advanced topics test involving probabilities of dice rolls, digits in square numbers, gear rotations, team formations in a Pokémon-style game, drawing cards from a deck, complex number solutions, evaluating expressions involving square roots and trigonometric functions.
2. The problems cover a wide range of mathematical topics including probability, number theory, combinatorics, algebra, and trigonometry.
3. The level of difficulty ranges from straightforward calculations to more complex problems requiring multiple steps and conceptual understanding.
This document provides the answers to a mathematics exam for 10th grade students. It includes multiple choice questions with explanations and word problems with step-by-step solutions. The topics covered include algebra, logarithms, quadratic equations, and inequalities.
1. The document contains 21 multiple choice math questions covering topics like area, perimeter, volume, coordinate geometry, factoring, and algebraic equations.
2. For each question, the question stem and possible multiple choice answers are provided, along with the correct answer and an explanation of the mathematical steps taken to arrive at the solution.
3. The questions progress from easier concepts involving basic formulas to more complex problems requiring multiple steps of algebraic manipulation or geometric reasoning.
The document contains multiple choice questions testing various math concepts. The questions cover topics like: properties of angles, algebraic expressions, sets, integers, functions, rates of change, equations, and word problems involving percentages.
This document provides problems in four areas: number theory, algebra, geometry, and probability. It includes:
1) Five number theory exercises involving divisibility, remainders, and properties of numbers.
2) Five algebra exercises involving solving equations, finding roots, and relationships between coefficients and roots.
3) Five geometry exercises involving properties of shapes like triangles, cylinders, and trapezoids.
4) Five probability exercises calculating chances of outcomes and applying distributions to real-world scenarios.
The document provides the problems in each area along with the full worked out solutions and explanations. It covers a range of fundamental mathematical concepts across multiple domains.
The document contains questions related to CAT, MAT, GMAT entrance exams. It discusses various topics like probability, permutations and combinations, averages, ratios etc. and provides solutions to sample questions in 3-4 sentences each. The overall document aims to help exam preparation by providing practice questions on common quantitative topics.
1. The document presents problems involving number theory, algebra, geometry, and probability. For number theory, it provides exercises and solutions involving sums of powers and finding the nearest integer of a difference. For algebra, it solves systems of equations and determines values based on given equations. For geometry, it calculates areas and volumes. For probability, it finds probabilities of arrangements and outcomes of dice rolls and ball draws.
1. The document presents problems involving number theory, algebra, geometry, and probability. For number theory, it provides exercises and solutions involving sums of powers and nearest integers. For algebra, it solves systems of equations and determines values based on relationships between variables. For geometry, it calculates areas and volumes. For probability, it determines probabilities of events occurring based on arrangements and selections from sets.
The document discusses various types of numbers including natural numbers, whole numbers, and integers. It provides examples and explanations related to properties of these numbers. Some key points include:
- Natural numbers start from 1 and do not include 0, negative numbers, or decimals.
- Whole numbers include all natural numbers and 0.
- Integers include whole numbers and their negatives.
- Examples are provided to illustrate properties like divisibility, perfect squares, and solving word problems involving sums and products of numbers.
- The last part discusses Donkey's stable number based on his true and false answers to questions about divisibility, being a square, and the first digit. It is determined his number must
This document discusses solving multi-step linear equations in 3 sentences or less:
The document provides examples and steps for solving multi-step linear equations, which may require using the distributive property, combining like terms, or having variables on both sides of the equation. It also discusses setting up and solving equations related to consecutive integers and word problems involving linear equations with more than one step.
The document discusses rules for indices and factorizing algebraic expressions. It provides examples of:
- Multiplying and dividing terms with the same base using index rules like ab × ac = ab+c and ab ÷ ac = ab-c.
- Expanding single and double brackets by distributing terms.
- Finding common factors to group like terms.
- Factorizing quadratics and using differences of squares.
- Solving equations set equal to zero by factorizing.
Ncert solutions for class 7 maths chapter 1 integers exercise 1.2jobazine India
This document provides solutions to exercises involving integer addition and subtraction from NCERT Class 7 Maths Chapter 1. It gives examples of finding integer pairs whose sum or difference equals a given number. It also demonstrates that the order of adding integers does not matter, and integers have additive inverses and identities. For one question, it shows that teams A and B scored the same total (-30) despite the order of their individual scores. It fills in blanks to complete statements demonstrating the commutative, associative, additive inverse and identity properties for integer addition.
This document contains a quiz with math problems of varying difficulty levels: easy, average, and difficult. The easy problems are worth 1 point each and cover topics like the intersection of lines, factoring, algebraic expressions, evaluating expressions, and identifying quadrants and graphs. The average problems are worth 3 points each and involve factoring quadratics, simplifying expressions, and operations with monomials. The difficult problems are worth 5 points each and require solving systems of equations, finding areas of rectangles, identifying mathematicians, and identifying perfect square trinomials. An answer key is provided with the solutions.
The document summarizes key concepts about sequences and simultaneous equations in algebra 2. It provides examples and explanations of linear sequences, writing the nth term formula, and solving simultaneous equations using substitution and elimination methods. Sample examination questions are also included to assess understanding of finding missing terms in a sequence, writing the nth term formula, and solving simultaneous equations word problems involving two unknown variables.
The document is a mathematics exam paper with 25 questions from sections A, B, and C worth varying marks. Section A has 7 questions worth 2 marks each about topics like finding the greatest common divisor of polynomials, solving systems of equations, and probability. Section B has 12 questions worth 3 marks each involving topics like rationalizing expressions, finding terms of arithmetic progressions, and properties of cyclic quadrilaterals. Section C has 6 questions worth 5 marks each. The summary provides an overview of the document's content and structure without copying significant portions.
The document provides important facts and formulae related to numbers. It discusses the following key points:
1. The Hindu-Arabic numeral system uses 10 digits (0-9) to represent any number. A group of digits forming a number is called a numeral.
2. Types of numbers include natural numbers, whole numbers, integers, even/odd numbers, prime/composite numbers. Tests for divisibility by various numbers are outlined.
3. Shortcut methods for multiplication like distributive law are described. Basic formulae for exponents, progressions, and the division algorithm are listed.
This document contains shortlisted problems for the Junior Balkan Mathematics Olympiad in 2016 in Romania. It lists the contributing countries that proposed problems, the problem selection committee members, and 4 algebra problems, 1 combinatorics problem, and 1 geometry problem. The combinatorics problem asks the student to find the least positive integer k such that k! multiplied by the sum of reciprocals of digits in numbers up to 2016 is an integer. It provides the solution by calculating this sum incrementally up to 999, then 1999, then 2016.
This document provides examples of recurrence relations and their solutions. It begins by defining convergence of sequences and limits. It then provides examples of recurrence relations, solving them using algebraic and graphical methods. One example finds the 6th term of a sequence defined by a recurrence relation to be 2.3009. Another example solves a recurrence relation algebraically to express the general term un in terms of n. The document emphasizes using graphical methods like sketching graphs to prove properties of sequences defined by recurrence relations.
This document contains a 10 question geometry test covering topics like:
- Finding the total number of sides of polygons given the sum of their interior angles.
- Calculating the shaded area of an infinitely repeating pattern of triangles.
- Finding the ratio of the diagonal to the side of a regular pentagon.
- Calculating the area of a rhombus given lengths of its sides and diagonals.
- Finding the distance between the starting positions of two bikers racing on inner and outer lanes of a circular track.
- Calculating the area of a quadrilateral formed by lines intersecting sides of an equilateral triangle.
- Finding the ratio of radii of mutually tangent disks of different sizes.
This document contains 10 math problems for an algebra test, ranging from finding integer pairs that satisfy an equation to evaluating infinite sums involving binary expansions of numbers. The problems cover a variety of algebra topics including functions, inequalities, number bases, and series.
Tes OSN matematika berisi 10 soal yang mencakup materi persamaan, ketidakteraturan, fungsi polinomial, dan bilangan. Soal-soal tersebut harus dikerjakan lengkap dan benar dalam waktu 120 menit.
Tugas hari Ahad memberikan 4 soal matematika yang harus diselesaikan untuk pembinaan OSN dan dikumpulkan pada hari Kamis berikutnya. Soal pertama meminta menentukan nilai dari produk x dan y berdasarkan dua persamaan yang menghubungkan x dan y dengan akar-akar bilangan. Soal kedua meminta menentukan angka terakhir dari suatu ekspresi rumit. Soal ketiga meminta menentukan nilai terkecil n pada suatu barisan agar m
The document contains multiple choice questions testing various math concepts. The questions cover topics like: properties of angles, algebraic expressions, sets, integers, functions, rates of change, equations, and word problems involving percentages.
This document provides problems in four areas: number theory, algebra, geometry, and probability. It includes:
1) Five number theory exercises involving divisibility, remainders, and properties of numbers.
2) Five algebra exercises involving solving equations, finding roots, and relationships between coefficients and roots.
3) Five geometry exercises involving properties of shapes like triangles, cylinders, and trapezoids.
4) Five probability exercises calculating chances of outcomes and applying distributions to real-world scenarios.
The document provides the problems in each area along with the full worked out solutions and explanations. It covers a range of fundamental mathematical concepts across multiple domains.
The document contains questions related to CAT, MAT, GMAT entrance exams. It discusses various topics like probability, permutations and combinations, averages, ratios etc. and provides solutions to sample questions in 3-4 sentences each. The overall document aims to help exam preparation by providing practice questions on common quantitative topics.
1. The document presents problems involving number theory, algebra, geometry, and probability. For number theory, it provides exercises and solutions involving sums of powers and finding the nearest integer of a difference. For algebra, it solves systems of equations and determines values based on given equations. For geometry, it calculates areas and volumes. For probability, it finds probabilities of arrangements and outcomes of dice rolls and ball draws.
1. The document presents problems involving number theory, algebra, geometry, and probability. For number theory, it provides exercises and solutions involving sums of powers and nearest integers. For algebra, it solves systems of equations and determines values based on relationships between variables. For geometry, it calculates areas and volumes. For probability, it determines probabilities of events occurring based on arrangements and selections from sets.
The document discusses various types of numbers including natural numbers, whole numbers, and integers. It provides examples and explanations related to properties of these numbers. Some key points include:
- Natural numbers start from 1 and do not include 0, negative numbers, or decimals.
- Whole numbers include all natural numbers and 0.
- Integers include whole numbers and their negatives.
- Examples are provided to illustrate properties like divisibility, perfect squares, and solving word problems involving sums and products of numbers.
- The last part discusses Donkey's stable number based on his true and false answers to questions about divisibility, being a square, and the first digit. It is determined his number must
This document discusses solving multi-step linear equations in 3 sentences or less:
The document provides examples and steps for solving multi-step linear equations, which may require using the distributive property, combining like terms, or having variables on both sides of the equation. It also discusses setting up and solving equations related to consecutive integers and word problems involving linear equations with more than one step.
The document discusses rules for indices and factorizing algebraic expressions. It provides examples of:
- Multiplying and dividing terms with the same base using index rules like ab × ac = ab+c and ab ÷ ac = ab-c.
- Expanding single and double brackets by distributing terms.
- Finding common factors to group like terms.
- Factorizing quadratics and using differences of squares.
- Solving equations set equal to zero by factorizing.
Ncert solutions for class 7 maths chapter 1 integers exercise 1.2jobazine India
This document provides solutions to exercises involving integer addition and subtraction from NCERT Class 7 Maths Chapter 1. It gives examples of finding integer pairs whose sum or difference equals a given number. It also demonstrates that the order of adding integers does not matter, and integers have additive inverses and identities. For one question, it shows that teams A and B scored the same total (-30) despite the order of their individual scores. It fills in blanks to complete statements demonstrating the commutative, associative, additive inverse and identity properties for integer addition.
This document contains a quiz with math problems of varying difficulty levels: easy, average, and difficult. The easy problems are worth 1 point each and cover topics like the intersection of lines, factoring, algebraic expressions, evaluating expressions, and identifying quadrants and graphs. The average problems are worth 3 points each and involve factoring quadratics, simplifying expressions, and operations with monomials. The difficult problems are worth 5 points each and require solving systems of equations, finding areas of rectangles, identifying mathematicians, and identifying perfect square trinomials. An answer key is provided with the solutions.
The document summarizes key concepts about sequences and simultaneous equations in algebra 2. It provides examples and explanations of linear sequences, writing the nth term formula, and solving simultaneous equations using substitution and elimination methods. Sample examination questions are also included to assess understanding of finding missing terms in a sequence, writing the nth term formula, and solving simultaneous equations word problems involving two unknown variables.
The document is a mathematics exam paper with 25 questions from sections A, B, and C worth varying marks. Section A has 7 questions worth 2 marks each about topics like finding the greatest common divisor of polynomials, solving systems of equations, and probability. Section B has 12 questions worth 3 marks each involving topics like rationalizing expressions, finding terms of arithmetic progressions, and properties of cyclic quadrilaterals. Section C has 6 questions worth 5 marks each. The summary provides an overview of the document's content and structure without copying significant portions.
The document provides important facts and formulae related to numbers. It discusses the following key points:
1. The Hindu-Arabic numeral system uses 10 digits (0-9) to represent any number. A group of digits forming a number is called a numeral.
2. Types of numbers include natural numbers, whole numbers, integers, even/odd numbers, prime/composite numbers. Tests for divisibility by various numbers are outlined.
3. Shortcut methods for multiplication like distributive law are described. Basic formulae for exponents, progressions, and the division algorithm are listed.
This document contains shortlisted problems for the Junior Balkan Mathematics Olympiad in 2016 in Romania. It lists the contributing countries that proposed problems, the problem selection committee members, and 4 algebra problems, 1 combinatorics problem, and 1 geometry problem. The combinatorics problem asks the student to find the least positive integer k such that k! multiplied by the sum of reciprocals of digits in numbers up to 2016 is an integer. It provides the solution by calculating this sum incrementally up to 999, then 1999, then 2016.
This document provides examples of recurrence relations and their solutions. It begins by defining convergence of sequences and limits. It then provides examples of recurrence relations, solving them using algebraic and graphical methods. One example finds the 6th term of a sequence defined by a recurrence relation to be 2.3009. Another example solves a recurrence relation algebraically to express the general term un in terms of n. The document emphasizes using graphical methods like sketching graphs to prove properties of sequences defined by recurrence relations.
This document contains a 10 question geometry test covering topics like:
- Finding the total number of sides of polygons given the sum of their interior angles.
- Calculating the shaded area of an infinitely repeating pattern of triangles.
- Finding the ratio of the diagonal to the side of a regular pentagon.
- Calculating the area of a rhombus given lengths of its sides and diagonals.
- Finding the distance between the starting positions of two bikers racing on inner and outer lanes of a circular track.
- Calculating the area of a quadrilateral formed by lines intersecting sides of an equilateral triangle.
- Finding the ratio of radii of mutually tangent disks of different sizes.
This document contains 10 math problems for an algebra test, ranging from finding integer pairs that satisfy an equation to evaluating infinite sums involving binary expansions of numbers. The problems cover a variety of algebra topics including functions, inequalities, number bases, and series.
Tes OSN matematika berisi 10 soal yang mencakup materi persamaan, ketidakteraturan, fungsi polinomial, dan bilangan. Soal-soal tersebut harus dikerjakan lengkap dan benar dalam waktu 120 menit.
Tugas hari Ahad memberikan 4 soal matematika yang harus diselesaikan untuk pembinaan OSN dan dikumpulkan pada hari Kamis berikutnya. Soal pertama meminta menentukan nilai dari produk x dan y berdasarkan dua persamaan yang menghubungkan x dan y dengan akar-akar bilangan. Soal kedua meminta menentukan angka terakhir dari suatu ekspresi rumit. Soal ketiga meminta menentukan nilai terkecil n pada suatu barisan agar m
Tes ini berisi 18 soal tipe pilihan ganda dan 2 soal uraian yang mencakup materi geometri dan kombinatorik. Soal-soal tersebut meliputi konsep segitiga, lingkaran, bilangan bulat, dan permutasi yang diujikan untuk mengukur kemampuan peserta olimpiade sains nasional bidang matematika.
Dokumen tersebut merupakan kurikulum mata pelajaran Pendidikan Agama Islam (PAI) di Pondok Pesantren Darul 'Ulum yang mencakup 6 mata pelajaran yaitu: 1) Al-Qur'an, 2) Hadits, 3) Aqidah-Akhlaq, 4) Fiqih, 5) Bahasa Arab, dan 6) Sejarah Keislaman. Tujuan utamanya adalah agar siswa dapat memahami dan menerapkan ajaran agama Islam dalam kehidupan seh
Dokumen tersebut berisi ringkasan pembahasan soal-soal latihan logaritma untuk persiapan olimpiade sains nasional. Terdapat 13 soal latihan yang mencakup sifat-sifat dasar logaritma seperti perubahan basis, logaritma persamaan, dan grafik fungsi logaritma.
Dokumen tersebut berisi tugas matematika untuk persiapan Olimpiade Sains Nasional (OSN) yang terdiri dari 4 soal dan harus dikumpulkan pada 10 Januari 2016.
The document contains 18 math word problems with their step-by-step solutions. The problems cover a range of topics including arithmetic sequences, geometric sequences, percentages, factorials, trigonometry, and more. The final problem asks to find the 12th term of a sequence where the first two terms are 3 and 2, and subsequent terms are the sum of all preceding terms. The solution shows this forms a geometric sequence and calculates the 12th term as 2,560.
1. The document shows methods for calculating the area of rectangles by splitting them into smaller rectangles.
2. It demonstrates that the area of the original rectangle equals the sum of the areas of the smaller rectangles.
3. Algebraic formulas are developed to represent splitting rectangles and multiplying sums and differences.
The document defines and describes different types of real numbers including natural numbers, whole numbers, integers, rational numbers, and irrational numbers. It provides examples of each type of number. Real numbers consist of all rational and irrational numbers. A Venn diagram shows the relationships between the different subsets of real numbers. Euclid's division algorithm and its application to find the highest common factor of two numbers is also explained in the document.
The document discusses different types of real numbers including rational and irrational numbers. It provides examples and definitions of natural numbers, whole numbers, integers, rational numbers, and irrational numbers. It also includes information on Euclid's division algorithm and its application in finding the highest common factor of two numbers. Examples are provided to illustrate the algorithm.
The document discusses different types of real numbers including rational and irrational numbers. It provides examples and definitions of natural numbers, whole numbers, integers, rational numbers, and irrational numbers. It also includes information on Euclid's division algorithm and its application in finding highest common factors and lowest common multiples. Examples of proving the irrationality of square roots like √5 are given.
The document discusses different types of real numbers including rational and irrational numbers. It provides examples and definitions of natural numbers, whole numbers, integers, rational numbers, and irrational numbers. It also includes information on Euclid's division algorithm and its application in finding the highest common factor of two numbers. Examples are provided to illustrate the algorithm.
This document contains a mock CAT exam with multiple choice questions and explanations. It consists of two pages. The first page lists 60 multiple choice questions with answer options A-D. The second page provides explanations for the questions and solutions to problems. It discusses topics like probability, ratios, geometry, time/speed/distance word problems, and data interpretation from graphs.
This document contains 6 problems and their solutions from a Regional Mathematical Olympiad competition in 2010. Problem 1 involves proving that the area of one triangle is the geometric mean of the areas of two other triangles in a hexagon with concurrent diagonals. Problem 2 proves that if three quadratic polynomials have a common root, then their coefficients must be equal. Problem 3 counts the number of 4-digit numbers divisible by 4 but not 8 having non-zero digits. Problem 4 finds the smallest positive integers whose reciprocals satisfy certain relationships. Problem 5 proves that the reflection of a point in a line lies on a side of a triangle. Problem 6 determines the number of values of n where an is greater than an+1 for a defined sequence
The document outlines 9 multiplication shortcuts or tricks using properties of numbers. These include multiplying numbers by 11 by adding the digits, squaring numbers ending in 9 by placing 9s and appending other digits, squaring numbers ending in 5 by omitting the 5 and multiplying the remaining number by the next higher number and appending 25, and multiplying numbers where the ones digits sum to 10 by multiplying the tens and ones places separately and placing the products successively.
This section introduces differential equations and their use in mathematical modeling. It provides examples of verifying solutions to differential equations by direct substitution. Typical problems show finding an integrating constant to satisfy an initial condition. Differential equations are derived from descriptions of real-world phenomena involving rates of change. The section establishes foundational knowledge of differential equations and their solution methods.
Triangle ABC is given, with altitudes CD and BE from vertices C and B to opposite sides AB and AC respectively being equal. It is proved that triangle ABC must be isosceles by showing that triangles CBD and BCE are congruent by the right angle-hypotenuse (RHS) criterion, implying corresponding angles are equal, and then using corresponding parts of congruent triangles to show sides AB and AC are equal, making triangle ABC isosceles.
Factoring polynomials involves finding common factors that can be divided out of terms, similar to factoring numbers but with variables; this is done by looking for a single variable or number that is a common factor of all terms that can be pulled out in front of parentheses. The document provides examples of different types of factoring polynomials including using the greatest common factor, difference of squares, grouping, and perfect squares and cubes.
1. The document contains 10 math problems with solutions. The problems cover topics like arithmetic progressions, rates of change, probability, and geometry.
2. One problem involves finding the value of n given that the sum of even numbers between 1 and n is a specific value. The solution uses the formula for the sum of an arithmetic progression.
3. Another problem asks what fraction of a solution must be replaced if the original solution was 40% and replaced with 25% solution to get a final concentration of 35%. The solution sets up an equation to solve for the fraction replaced.
The document contains solutions to 18 math and probability problems. Some key details:
- Problem 1 involves finding an odd number n such that the sum of even numbers between 1 and n equals 79*80.
- Problem 2 calculates the price at which a bushel of corn costs the same as a peck of wheat, given changing prices.
- Problem 3 determines the minimum number of people needed to have over a 50% chance that one was born in a leap year.
This document contains 5 math problems involving factorizing expressions, solving equations, evaluating expressions for given values, expanding expressions, and finding the highest common factor. It also provides context on working with straight line graphs, including finding the gradient and y-intercept of a line from its equation, finding the gradient between two points, finding the midpoint and a point that divides a line segment in a given ratio, and finding the x- and y-intercepts of a line.
This document provides an overview of quadratic equations, including definitions, methods for solving quadratic equations such as factoring, completing the square, and using the quadratic formula, and applications of quadratic equations. Key topics covered include defining linear and quadratic equations, solving quadratics by factoring when possible and using completing the square or the quadratic formula when not factorable, deriving the quadratic formula, interpreting the discriminant, and modeling real-world situations with quadratic equations.
1) The value of the expression 44 + 4 × 4 − 4 is 268.
2) Two of the four statements about sequences and patterns are true - that it follows the Fibonacci sequence and is an arithmetic sequence with a common difference of 4.
3) The area of the gray region overlapping two rectangles is 32 cm^2.
The document provides solutions to 15 problems from a CAT exam. Here are summaries of 3 of the problems:
1) The problem involves finding the remainder when a sum of terms of the form (163 + 173 + ...) is divided by 70. The solution shows the sum can be written as 70k, so the remainder is 0.
2) The problem involves 4 tanks losing chemical at different rates per minute. It is found that tank D loses chemical the fastest, at 50 units per minute, and will empty first after 20 minutes, when its initial 1000 units are completely lost.
3) The problem involves two circles intersecting at two points to form a square. It is noted that the area common to
1. The document contains solutions to 20 math problems. It provides step-by-step working and explanations for finding values, solving equations, calculating probabilities and averages.
2. Questions include solving for values of n in an equation, finding sums of logarithmic and arithmetic sequences, determining factors of polynomials, calculating probabilities from experiments, and solving systems of equations.
3. The solutions demonstrate various mathematical techniques such as using formulas, factoring, eliminating terms, and setting up proportional relationships to systematically arrive at final numeric answers.
Tes matrikulasi mata pelajaran fisika untuk siswa SMA Darul Ulum 2 Unggulan BPPT Jombang tahun pelajaran 2016-2017 terdiri dari 30-35 soal pilihan ganda dengan empat pilihan jawaban untuk setiap soal. Tes ini diselenggarakan untuk mengukur kemampuan siswa dalam mata pelajaran fisika dan dilaksanakan dalam waktu 40 menit.
Surat ini meminta guru-guru untuk menyusun modul dan soal ujian matrikulasi untuk kelas 10 yang mencakup mata pelajaran sejarah, ekonomi, matematika, fisika dan biologi. Modul berisi soal latihan pilihan ganda untuk 4 pertemuan dan ujian berisi soal pilihan ganda untuk masing-masing mata pelajaran. Semua dokumen diminta dikirim dalam format softcopy melalui email pada tanggal 14 Juli 2016.
Ringkasan dokumen tersebut adalah:
Dokumen tersebut merupakan petunjuk untuk mengerjakan tes awal kemampuan di SMA Darul Ulum 2 Unggulan BPPT Jombang yang mencakup instruksi untuk memeriksa lembar soal dan jawaban, mengisi identitas di lembar jawaban, waktu pengerjaan soal, dan larangan yang berlaku selama mengerjakan tes.
Ringkasan dokumen tersebut adalah sebagai berikut:
Dokumen tersebut merupakan soal ujian awal kemampuan (TAKUP) mata pelajaran Fiqih untuk siswa program studi MIPA/IPS di SMA Darul Ulum 2 Unggulan BPPT Jombang tahun 2016 yang terdiri dari 30 soal pilihan ganda dan petunjuk pengerjaan soal.
Tes awal kemampuan untuk siswa SMA diadakan pada bulan Juli 2016 untuk mata pelajaran matematika. Soal tes terdiri dari 30 pertanyaan pilihan ganda dengan lima pilihan jawaban untuk setiap pertanyaan. Siswa diberi waktu 90 menit untuk mengerjakan soal.
Tes awal kemampuan ujian nasional (TAKUN) mata pelajaran matematika untuk siswa program studi MIPA/IPS SMA tahun pelajaran 2016/2017 yang diselenggarakan oleh Yayasan Darul Ulum SMA Darul Ulum 2 Unggulan BPPT Jombang. Soal tes berjumlah 30 butir pilihan ganda dengan waktu pengerjaan 90 menit.
Surat ini memberitahukan tentang Tes Awal Kemampuan Ujian Nasional & Ujian Pondok (TAKUN-TAKUP) 2016/2017, diantaranya penyusun soal berasal dari guru mata pelajaran, rujukan kisi-kisi ujian, dua tipe soal dengan tingkat kesulitan yang seimbang beserta kunci jawaban, daftar penyusun soal dan mata pelajaran, alokasi waktu mengerjakan soal, serta batas akhir pengumpulan nask
Soal seleksi olimpiade sains tingkat sekolah bidang matematika tahun 2016 terdiri dari 15 soal teori bilangan dan kombinatorik yang harus diselesaikan peserta dalam waktu 60 menit. Soal-soal meliputi konsep bilangan bulat, kuadrat sempurna, bilangan asli, kombinasi, dan permutasi.
বাংলাদেশের অর্থনৈতিক সমীক্ষা ২০২৪ [Bangladesh Economic Review 2024 Bangla.pdf] কম্পিউটার , ট্যাব ও স্মার্ট ফোন ভার্সন সহ সম্পূর্ণ বাংলা ই-বুক বা pdf বই " সুচিপত্র ...বুকমার্ক মেনু 🔖 ও হাইপার লিংক মেনু 📝👆 যুক্ত ..
আমাদের সবার জন্য খুব খুব গুরুত্বপূর্ণ একটি বই ..বিসিএস, ব্যাংক, ইউনিভার্সিটি ভর্তি ও যে কোন প্রতিযোগিতা মূলক পরীক্ষার জন্য এর খুব ইম্পরট্যান্ট একটি বিষয় ...তাছাড়া বাংলাদেশের সাম্প্রতিক যে কোন ডাটা বা তথ্য এই বইতে পাবেন ...
তাই একজন নাগরিক হিসাবে এই তথ্য গুলো আপনার জানা প্রয়োজন ...।
বিসিএস ও ব্যাংক এর লিখিত পরীক্ষা ...+এছাড়া মাধ্যমিক ও উচ্চমাধ্যমিকের স্টুডেন্টদের জন্য অনেক কাজে আসবে ...
How to Make a Field Mandatory in Odoo 17Celine George
In Odoo, making a field required can be done through both Python code and XML views. When you set the required attribute to True in Python code, it makes the field required across all views where it's used. Conversely, when you set the required attribute in XML views, it makes the field required only in the context of that particular view.
How to Setup Warehouse & Location in Odoo 17 InventoryCeline George
In this slide, we'll explore how to set up warehouses and locations in Odoo 17 Inventory. This will help us manage our stock effectively, track inventory levels, and streamline warehouse operations.
How to Fix the Import Error in the Odoo 17Celine George
An import error occurs when a program fails to import a module or library, disrupting its execution. In languages like Python, this issue arises when the specified module cannot be found or accessed, hindering the program's functionality. Resolving import errors is crucial for maintaining smooth software operation and uninterrupted development processes.
it describes the bony anatomy including the femoral head , acetabulum, labrum . also discusses the capsule , ligaments . muscle that act on the hip joint and the range of motion are outlined. factors affecting hip joint stability and weight transmission through the joint are summarized.
Chapter wise All Notes of First year Basic Civil Engineering.pptxDenish Jangid
Chapter wise All Notes of First year Basic Civil Engineering
Syllabus
Chapter-1
Introduction to objective, scope and outcome the subject
Chapter 2
Introduction: Scope and Specialization of Civil Engineering, Role of civil Engineer in Society, Impact of infrastructural development on economy of country.
Chapter 3
Surveying: Object Principles & Types of Surveying; Site Plans, Plans & Maps; Scales & Unit of different Measurements.
Linear Measurements: Instruments used. Linear Measurement by Tape, Ranging out Survey Lines and overcoming Obstructions; Measurements on sloping ground; Tape corrections, conventional symbols. Angular Measurements: Instruments used; Introduction to Compass Surveying, Bearings and Longitude & Latitude of a Line, Introduction to total station.
Levelling: Instrument used Object of levelling, Methods of levelling in brief, and Contour maps.
Chapter 4
Buildings: Selection of site for Buildings, Layout of Building Plan, Types of buildings, Plinth area, carpet area, floor space index, Introduction to building byelaws, concept of sun light & ventilation. Components of Buildings & their functions, Basic concept of R.C.C., Introduction to types of foundation
Chapter 5
Transportation: Introduction to Transportation Engineering; Traffic and Road Safety: Types and Characteristics of Various Modes of Transportation; Various Road Traffic Signs, Causes of Accidents and Road Safety Measures.
Chapter 6
Environmental Engineering: Environmental Pollution, Environmental Acts and Regulations, Functional Concepts of Ecology, Basics of Species, Biodiversity, Ecosystem, Hydrological Cycle; Chemical Cycles: Carbon, Nitrogen & Phosphorus; Energy Flow in Ecosystems.
Water Pollution: Water Quality standards, Introduction to Treatment & Disposal of Waste Water. Reuse and Saving of Water, Rain Water Harvesting. Solid Waste Management: Classification of Solid Waste, Collection, Transportation and Disposal of Solid. Recycling of Solid Waste: Energy Recovery, Sanitary Landfill, On-Site Sanitation. Air & Noise Pollution: Primary and Secondary air pollutants, Harmful effects of Air Pollution, Control of Air Pollution. . Noise Pollution Harmful Effects of noise pollution, control of noise pollution, Global warming & Climate Change, Ozone depletion, Greenhouse effect
Text Books:
1. Palancharmy, Basic Civil Engineering, McGraw Hill publishers.
2. Satheesh Gopi, Basic Civil Engineering, Pearson Publishers.
3. Ketki Rangwala Dalal, Essentials of Civil Engineering, Charotar Publishing House.
4. BCP, Surveying volume 1
Beyond Degrees - Empowering the Workforce in the Context of Skills-First.pptxEduSkills OECD
Iván Bornacelly, Policy Analyst at the OECD Centre for Skills, OECD, presents at the webinar 'Tackling job market gaps with a skills-first approach' on 12 June 2024
Main Java[All of the Base Concepts}.docxadhitya5119
This is part 1 of my Java Learning Journey. This Contains Custom methods, classes, constructors, packages, multithreading , try- catch block, finally block and more.
Film vocab for eal 3 students: Australia the movie
Sols
1. Advanced Topics Solutions
Stanford Mathematics Tournament 2000
1. Assume we have 4 colors - 1, 2, 3, and 4. Fix the bottom as color 1. On the remaining sides you can
have colors 2, 3, 4 (in that order), or 2, 4, 3, which are not rotationally identical. So, there are 2 ways
to color it.
2. Since cos 2π
3 = −1
2 and sin 2π
3 =
√
3
2 , we can write the first term as (cos 2π
3 + i sin 2π
3 )
6
. Since cos 4π
3 =
−1
2 and sin 4π
3 = −
√
3
2 , we can write the second term as (cos 4π
3 + i sin 4π
3 )
6
. Now, we apply DeMoivre’s
Theorem to simplify the first expression to (cos 6 · 2π
3 + sin 6 · 2π
3 ) = (cos 4π + sin 4π) = 1 + 0 = 1.
Similarly, we simplify the second expression to (cos 6 · 4π
3 + sin 6 · 4π
3 ) = (cos 8π + sin 8π) = 1 + 0 = 1.
Thus, the total sum is 1 + 1 = 2.
3. We know that 1
n2+2n = 1
n(n+2) =
1
n − 1
n+2
2 . So, if we sum this from 1 to ∞, all terms except for
1
1
2 +
1
2
2
will cancel out (a ”telescoping” series). Therefore, the sum will be 3
4
.
4. The possibilities for the numbers are:
• all five are divisible by 3
• three are divisible by 3, one is ≡ 1 (mod 3) and one is ≡ 2 (mod 3)
• two are divisible by 3, and the other three are either ≡ 1 (mod 3) or ≡ 2 (mod 3)
• one is divisible by 3, two are ≡ 1 (mod 3) and two are ≡ 2 (mod 3)
• four are ≡ 1 (mod 3) and one is ≡ 2 (mod 3)
• four are ≡ 2 (mod 3) and one is ≡ 1 (mod 3)
This gives us 1001 possible combinations out of 15
5 or 3003. So, the probability is 1001
3003 =1
3
.
5. 153,370,371,407
6. There are 6 people that could get their hat back, so we must multiply 6 by the number of ways that
the other 5 people can arrange their hats such that no one gets his/her hat back. So, the number of
ways this will happen is (6 · derangement of 5), or 6 ∗ 44 = 264. Since there are 6! = 720 possible
arrangements of hats, the probability of exactly one person getting their hat back is 264
720 =11
30
.
7. We can view these conditions as a geometry diagram as seen below. So, we know that e
f = 3
4 (since
e = a − b = 3
4 c − 3
4 d = 3
4 f and we know that e2 + f2 = 15 (since this is
√
a2 + c2 −
√
b2 + d2). Also,
note that ac + bd − ad − bc = (a − b)(c − d) = ef. So, solving for e and f, we find that e2
+ f2
= 225,
so 16e2
+ 16f2
= 3600, so (4e)
2
+ (4f)
2
= 3600, so (3f)
2
+ (4f)
2
= 3600, so f2
(32
+ 42
) = 3600, so
25f2
= 3600, so f2
= 144 and f = 12. Thus, e = 3
4 · 12 = 9. Therefore, ef = 9 ∗ 12 = 108.
b
a
d
c
e
f
2. 8. It suffices to consider the complements of the graphs, so we are looking for graphs with 9 vertices,
where each vertex is connected to 2 others. There are 4 different graphs - see below.
9. The probability of the Reals hitting 0 singles is (2
3 )
3
. The probability of the Reals hitting exactly 1
single is 3
2 ·(2
3 )
3
· 1
3 , since there are 3 spots to put the two outs (the last spot must be an out, since the
inning has to end on an out). The probability of the Reals hitting exactly 2 singles is 4
2 ·(2
3 )
3
· (1
3 )
3
.
The probability of the Reals hitting exactly 3 singles is 5
2 ·(2
3 )
3
· (1
3 )
3
. If any of these happen, the
Alphas win right away. Adding these gives us a 656
729 chance of this happening. If exactly 4 singles occur
(with probability 6
2 ·(2
3 )
3
·(1
3 )
4
), then there is a 2
5 chance that the Alphas win. The probability of this
happening is 2
5 · 40
729 . , the total probability of the Alphas winning is the sum of these two probabilities,
or 656
729 + 16
729 =224
243
.
10. A will say yes when B says no to n − 1 or n, as A will then know B’s number is one greater than A’s
number. Thus, A responds first, after n−1
2
“no” responses if n is odd, after n
2
“no” responses
if n is even.
3. Algebra Solutions
Stanford Mathematics Tournament 2000
1. The only integers that satisfy |x| + 5 < 7 are the ones that satisfy |x| < 2 - namely, −1, 0, 1. The
integers that satisfy |x − 3| > 2 are 6, 7, 8, . . . and 0, −1, −2, . . .. So, the integers that satisfy both are
0, −1, and there are 2 of them.
2. 20003
−1999·20002
−19992
·2000+19993
can be factored into (2000 − 1999)20002
+19992
(−2000 + 1999),
which reduces to 20002
− 19992
. This factors into (2000 + 1999)(2000 − 1999), which is equal to 3999.
3. Let the scores be a,b,c,d,e, where 0 ≤ a ≤ b ≤ c ≤ d ≤ e ≤ 100. So, the mean is 1
5 (a + b + c + d + e),
and the median is c. So, we want to maximize 1
5 (a + b + c + d + e) − c. To do this, we must maximize
d and e and minimize or maximize c. One way to do this is to let a = b = c = 0 and d = e = 100, so
the difference between the mean and the median is 1
5 (0 + 0 + 0 + 100 + 100) − 0) = 200
5 = 40. If we
maximize c, then c = d = e = 100, and then the mean is 1
5 (0 + 0 + 100 + 100 + 100) = 60, and the
median is 60, with a difference of 40 as well.
4. A shortest path is x → x2
→ x4
→ x8
→ x12
→ x24
→ x25
→ x50
→ x100
→ x200
→ x400
→ x800
→
x1600
→ x2000
, using 13 multiplications.
5. The price starts at $100. Clearly, the order of price changes does not matter. It is reduced by 10%
three times ( $100 → $90 → $81 → $72.90), and the new price is $72.90. It is increased by 10% four
times ($72.90 → $80.19 → $88.209 → $97.0299 → $106.73289) , and the new price is $106.73289.
Rounded to the nearest cent, this is $106.73.
6. Every day Edward works, he gets 1
9 of the test done. Similarly, every day Barbara works, she gets 1
10
of the test done, every day Abhinav works, he gets 1
11 of the test done, and every day Alex works, he
gets 1
12 of the test done. So, after 4 days (after everyone has worked on the test one day, they have
completed 1
9 + 1
10 + 1
11 + 1
12 = 38.535% of the test. After 8 days, they have completed twice that,
or 77.0707% of the test. After Edward, Barbara, and Abhinav each work one more day, the test will
be complete in the minimum amount of time, so the test will take 11 days to complete. If the least
efficient workers work after the 8th day, the test still takes 11 days to complete.
7. All multiplicatively perfect numbers have exactly 4 distinct positive divisors, or 1. So, we must look
for numbers that are either
• 1
• a product of two distinct primes
• a cube of a prime
Numbers satisfying one of these conditions less than 100 are: 1, 6, 8, 10, 14, 15, 21, 22, 26, 27, 33, 34,
35, 38, 39, 46, 51, 55, 57, 58, 62, 65, 69, 74, 77, 82, 85, 86, 87, 91, 93, 94, 95. There are 33 of these.
8. 168 = 23
· 3 · 7. There are only 2 combinations of these whose sums allow indistinguishability of the
ages. If there are 27 trees, 2, 4, 21 and 1, 12, 14 years are possible. If there are 21 trees, 2, 7, 12 and
3, 4, 14 are possible. So, the possible ages of the oldest daughter are 12, 14, 21.
4. 9. We can view these conditions as a geometry diagram as seen below. So, we know that e
f = 3
4 (since
e = a − b = 3
4 c − 3
4 d = 3
4 f and we know that e2 + f2 = 15 (since this is
√
a2 + c2 −
√
b2 + d2). Also,
note that ac + bd − ad − bc = (a − b)(c − d) = ef. So, solving for e and f, we find that e2
+ f2
= 225,
so 16e2
+ 16f2
= 3600, so (4e)
2
+ (4f)
2
= 3600, so (3f)
2
+ (4f)
2
= 3600, so f2
(32
+ 42
) = 3600, so
25f2
= 3600, so f2
= 144 and f = 12. Thus, e = 3
4 12 = 9. Therefore, ef = 9 ∗ 12 = 108.
b
a
d
c
e
f
10. a = 1 clearly does not work, since if x = 1, then x4
+ a2
= 2, which is prime. a = 2 clearly does not
work, since if x = 1, then x4
+ a2
= 5, which is also prime. Here is a table for a’s and values of x that
show they do not work:
a x a4
+ x2
3 10 10009
4 1 17
5 2 41
6 1 37
7 20 160049
So, consider a = 8; i.e. the sum x4
+64. This is the same as (x2
+8)2
−16x2
=(x2
+ 4x + 8)(x2
− 4x + 8)
by the difference of squares. This is clearly not prime for any integer x. So, the answer is a = 8.
5. Calculus Solutions
Stanford Mathematics Tournament 2000
1. y = x3
−3x2
+6x+2000, so y = 3x2
−6x+6 and y = 6x−6, so the point of inflection is the solution
to 6x − 6 = 0, or x = 1. At x = 1, the slope is f |x=1 = 3(1)
2
− 6(1) + 6 = 3.
2. The change in Karen’s position is x − x3
. The optimal length to climb is at a critical point. The only
realistic critical point is at the solution to 1 − 3x2
= 0 or x =
√
3
3
.
3.
1
4 π
0 sin x+C = 0, from the statement of the problem. So, [− cosx + Cx]|
π
4
0 = 0. Thus, cos π
4 + π
4 C+1 =
0. So, −
√
2
2 + π
4 C + 1 = 0, and solving for C, we find that C = 2
√
2−4
π
.
4. Let y = tan x. So, we want to find the minimum of y + 1
y , where 0 ≤ y ≤ ∞. Taking the derivative and
minimizing, we find that the minimum occurs at y = 1, so the minimum of the given function occurs
at arctan 1 = π
4
.
5. f(x) =
∞
i=1
xi
i . So, f (x) =
∞
i=0 xi
= 1
1−x . Thus, f(x) = − ln(1 − x).
6. Assume the pipe barely fits around the corner (i.e. it is in contact with the corner). The lower corner
is at (0, 0) and the upper corner is at (6, 6
√
5). Call x0 the point on the lower wall it hits at the tightest
spot. Given an x0, the longest a pipe could be with one end at x0 and leaning against the (6, 6
√
5)
corner is x2
0 + (6
√
5 + 36
√
5
x0−6 )
2
. We want the minimum of all of these ”longest pipes”, because the
pipe needs to fit at all angles around the corner. Taking the derivative (without the square root for
simplicity) and setting it equal to 0, we need to solve x3
0 − 6x2
0 + 36x0 − 1296 = 0. We can quickly find
that x0 = 12 is the only good solution, so the maximum length is 12
√
6.
7. For each value of x, we want to find the minimum (maximum) of y for the range of x0. Therefore, take
dy
dx0
, treating x as a constant. Set this equal to 0, and solve for x0 relative to x. Plug this in for x0 in
the given family to obtain the envelope y = x + 1
x
, x = 0.
8. Let I denote the given integral. Under the transformation θ → π
2 −θ, I transforms to
π
2
0
ln(cos(θ))dθ.
So,
2I =
π
2
0 ln(sin θ cos θ)dθ
=
π
0
(ln(sin 2θ) − ln 2)d(2θ)/2
= − π
2 ln 2 + 1
2
π
0
sin(α)dα
= −π
2 · ln 2 +
π
2
0 sin(α)dα
= −π
2 · ln 2 + I
giving I = −π
2
ln 2.
9. Note first that ([f(x)]2
− x)
2
= f(x), so if f(a) = 4, then (16 − a)
2
= 4, so a = 14. Now, f (x) =
1+
f (x)
2([f(x)]2−x)
2f(x) , so f (14) = 4
31
.
10. Solution: Throughout this solution we will use the fact that when light bounces off a mirror, the angle
of incidence is equal to the angle of reflection. First the beam hits the point (8,-1), then (6,1), (4,-1),
(2,1), and then is travelling along the line y = x − 1. Thus the beam hits the parabola at the point
(1+ 1−
√
5
2 , 1−
√
5
2 ). To estimate
√
5, notice that 222
= 484 and 232
= 529, so
√
5 =
√
500
10 = 2.2 . . .. Thus
1−
√
5
2 = −.6 . . ., so the light hits the parabola at approximately (.4,-.6). The slope of the tangent to
the parabola at this point is −1
2 (.4)−1/2
, which is about -.8, so we need to find the slope of the beam
after it reflects off of this tangent. For purposes of finding this slope, change coordinates so that the
point of intersection is the origin. The beam is coming in along y = x, and y = 1.2x is perpendicular
to the tangent. The diagram below should clarify the setup.
We will find the new path of the light by finding the reflection about the line y = 1.2x of a point on
its incoming path. We know the point (1,1.2) is on the line y = 1.2x, so a perpendicular through this
point is y − 1.2 = −.8(x − 1), which intersects y = x at the point (1.1,1.1). Thus the new path goes
through the point (.9,1.3), so it has slope 1.4 (all values rounded to one decimal place). Going back
6. reflected
beam
y=x
perpendicular to
tangent
tangent
to our original coordinate system, the light is now travelling along the line y + .6 = 1.4(x − .4), so it
next hits the mirror at (1.5,1). After that the x coordinate increases by 2/1.4 = 1.4 between bounces,
so it hits (2.9,-1), (4.3,1), (5.7,-1), (7.1,1), (8.5,-1), and finally (9.9,1). A closer examination of the
approximations made (e.g. by refining them to two decimal places) reveals that the last bounce is
actually further to the left (at (9.21,1), to be more precise), so indeed the light does bounce 12 times.
7. General Solutions
Stanford Mathematics Tournament 2000
1. Since d = a − c, substitute into the equation b = 2c + d to get b = 2c + a − c = a + c. Also, substitute
into 2c = d+a−1 to get 2c = a−c+a−1, or 3c = 2a−1. Now, since b = a+c, we can substitute into
a = 2b + c to get a = 2a + 2c + c, or a = −3c. Since we know 3c = 2a − 1 from above, we substitute in
to get 3c = −6c − 1, or c = −1
9 . Thus, we find that a = 1
3 , d = 4
9 , and b = 2
9
.
2. Let x be the temperature we are looking for, so x = 9
5 + 32. So, −4
5 x = 32, so x = −5
4 · 32 =−40.
3. Let x be the length of Henry’s shadow in feet. Using similar triangles, we find that 5.5
12 = x
5 , so
x = 5.5
12 · 5 = 11
24 · 5 =55
24
.
4. Let x be the number of students and y be the number of non-students. We then have the equations
x + y = 3000 and 10x + 15y = 36250. Substituting, we find that 10(3000 − y) + 15y = 36250, or
30000 − 10y + 15y = 36250, so 30000 + 5y = 36250. So, 5y = 6250, and y = 1250, so x =1750.
5. The total number of degrees in an octagon is (8 − 2) · 180 = 1080. Since the degrees are evenly
distributed among the angles, the measure of one interior angle is 1080
8 =135◦
.
6. Pick any card first, then pick the other face-down card.
(a) 1
3
(b) 2
3
7.
√
19992000 ≈ 4471.241, and [4471.241] =4471.
8. The time that Bobo can juggle is the number of cows times seconds. So, we get the following table:
cows started juggling 1 2 3 4 5 6 7 8 9 10 11
total time 64 110 141 160 165 162 154 144 126 130 132
cows started juggling 12 13 14 15 16 17 18 19 20 21 22
total time 132 130 126 120 112 102 90 76 60 42 22
Thus, we see that the maximum occurs with 5 cows, and the total time is 165 seconds = 23
4
minutes.
9. Let p be the price of HMMT. So p = k · x
y , where k is a constant to be determined. We know that
when x = 8 and y = 4 that p = 12, so, solving for k, we find that k = 6. So, when x = 4 and y = 8,
we find that p = 6 · 4
8 =3.
10. On the 12 foot sides, he needs 7 posts, and on the 20 foot sides, he needs 9 posts, so he needs
7 + 9 + 7 + 9 = 32 total posts. Let x be the number of normal fenceposts and y be the number of
strong fenceposts, so x + y = 32. To spend $70, we have the equation 2x + 3y = 70. Substituting, we
find that 2(32 − y) + 3y = 70, so 64 − 2y + 3y = 70, and 64 + y = 70, so y =6.
11. Substituting, we find that 3+n
3−n = 3, so 3 + n = 9 − 3n, thus 4n = 6. So, n =3
2
.
12. Yes - one possible path is 4 → 1 → 2 → 4 → 5 → 2 → 3 → 5, so the difference between the start island
and end island is 1.
13. The number of rearrangements keeping 1 number in its spot and rearranging the other 5 such that
none are in the right spot is 44. There are 6 numbers to fix, this gives us an answer of 44 · 6 =264.
14. Feburary 26, 2000 is a Saturday. April 24, 2000 is 23 · 365 + 1 + 1 + 1 + 1 + 1 + 1 = 8401 days away
from April 24, 1977 (including the leap years). So, Feburary 26, 2000 is 8401−3 −31 −24 = 8343 days
after April 24, 1977. Now, 8343
7 = 1191 with a remainder of 6. So, 6 days before Saturday is Sunday.
15. Notice that 36
= 729 while 54
= 625, and since ln 5 > ln 3, it follows that 54
ln 3 < 36
ln 5, so
53
∗ 5 ln 3 < 35
∗ 3 ln 5, and so by laws of logarithms, 53
ln 35
< 35
ln 53
. Again applying laws of
logarithms, it follows that ln (35
)(53
)
< ln (53
)(35
)
. So, since ln x is an increasing function, it follows
that (35
)
(53
)
< (53
)
(35
)
, and so it follows that (53
)
(35
)
is greater. One can also use the laws of
exponents to reduce the values to 3(54
)
and 5(36
)
. The second is clearly larger.
8. 16. Expressing the total time Joe has biked in hours leads to the equation x
20 + x
20 + x+2
14 = 1. So, x = 5.
Thus, we can construct the diagram below, and find the total time that it takes to get back: 13∗ 1
78 = 1
6
hours, or 10 minutes.
5 mi. Friend’s house
shore 7 mi. Gma’s house
13 mi.
5 mi.
12 mi.
17. There are 7! ways to arrange those letters. However, for every distinct arrangement, there are 2!∗2! = 4
total arrangements of the 2 T’s and 2 N’s. Therefore, the total number of distinct ways to arrange the
letters is 7!
2!2!
= 1260.
18. The only digits possible are 4, 6, 8, and 9. The only groups of numbers allowed keeping the average at
5 are 8444 are 6464. There are 4 ways to arrange 8444 and 6 ways to arrange 6464 so there are only
10 combinations to try.
19. Let y be the number of coins in the chest. From the problem, we know that y ≡ 5 (mod 11), y ≡ 3
(mod 10), and y ≡ 0 (mod 9). Combining these gives us that y ≡ 423 (mod 990), so the answer is
423.
20. The area of the big circle is (5
2 )
2
π = 25
4 π. The area of the circle with diameter AB is π, and the
area of the circle with diameter BC is 9
4 π. Thus, the percentage of the big circle that is shaded is
25
4 π−π− 9
4 π
25
4 π
= 25−4−9
25 = 12
25
= 48%.
21. It is clear that we can split the figure into 12 equilateral triangles, all of which have side length s. So,
since the area of one of these triangles is s2
√
3
4 , the total area is 3s2
√
3.
22. Aiso =
Aeq
2 so ∆A = Aeq − Aiso = 1
2 Aeq. Also, Aeq = 42
·
√
3
4 = 4
√
3, so ∆A = 2
√
3. Thus, the area is
2
√
3.
23. Notice that 72k+1
≡ 3 (mod 4) for k ∈ Z (k is an integer). Also, 71
≡ 7 (mod 100), 72
≡ 49
(mod 100), 73
≡ 43 (mod 100), 74
≡ 1 (mod 100), with the cycle repeating afterwards. So, clearly
7777
≡ 3 (mod 4), and since the cycle has period 4 (mod 100), we can conclude that 7777
≡ 43
(mod 100).
24. If there are no empty boxes, there are 126 ways of distributing the identical candy. If there is one
empty box, there are 84 ways of distributing the candy and 5 ways of choosing the empty box, so there
are 84 · 5 = 420 ways. If there are two empty boxes, there are 36 ways of distributing the candy and
6 combinations of assigning the empty boxes so that they are not adjacent, so there are 36 · 6 = 216
ways. If there are three empty boxes, there are 9 ways of distributing the candy, and only 1 way of
arranging the empty boxes. Having four or five empty boxes is impossible, since some two would have
to be adjacent. So, the total number of ways is 126 + 420 + 216 + 9 = 771.
25. We can place 9 points as shown, all at least 1
2 unit apart, but the next point must be less than 1
2 , so
10 points must be placed. There is no arrangement of 10 points with distance at least 1
2 . The proof
of this is a simple application of the Pigeonhole Principle.
9. 26. Janet is at (5, −5) and Tim is at (7, 4). They are
√
85 miles apart.
10. Geometry Solutions
Stanford Mathematics Tournament 2000
1. Consider the board labeled as below, with labels for columns and rows. To choose any rectangle on
the board, it is sufficient to choose some number (1-8) of adjacent columns, and some number (1-8) of
adjacent rows, since the rectangle can be created by forming the intersection of the columns and rows.
For instance, the intersection of columns 2,3 and rows 3,4,5 is the rectangle shaded below. So, there
are 8 ways to choose 1 adjacent column, 7 ways to choose 2 adjacent columns, . . ., 1 way to choose 8
adjacent columns, so there are 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 36 total ways to choose the columns,
and 36 ways to choose the rows. Thus, the total number of ways to choose a rectangle (i.e. the total
number of rectangles) is 362
= 1296.
1 2 3 4 5 6 7 8
1
2
3
4
5
7
6
8
2. The maximum occurs in an equilateral triangle, in which case the sides a = b − c are given by a2
+
b2
+ c2
= 3a2
= 96, so a = 4
√
2. Thus, the medians are 3 · (a
√
3
2 ) =6
√
6.
3. Since ABCD is a square, we can write (PA)
2
+ (PC)
2
= (PB)
2
+ (PC)
2
. So, we can substitute in for
PA,PC, and PD to get that PB =24.
4. Notice that in general, when there is a rectangle of side length x and y, the area of the non-triangle
regions (created by drawing a line connecting the midpoint of two opposite lines and a line connecting
two opposite corners - see diagram for examples) is simply 3
4 of the original area of the box, since the
area of the excluded triangles are 1
2 · y
2 · x
2 + 1
2 · y
2 · x
2 = 2 · xy
2 = xy
4 , so the desired area is simply
ab − ab
4 = 3
4 ab, as desired. So, if we consider the four rectangles making up the large rectangle that
are divided this way (the one with dimensions a
4 xb
3 , etc.), we can say that the total shaded area is 3
4
of the total area of the rectangle - that is, 3
4
ab.
5. The area of ABF = 1
2 bh = 1
2 · 3h = 3
2 (3 sin 60◦
) = 9
2 sin 60◦
. The area of FCDE = area of ABF
- area of FGC = 9
2 sin 60◦
− 1
2 sin 60◦
= 4 sin 60◦
. Therefore, area of ABCDE = area of ABF+
area of FCDE = 9
2 sin 60◦
+ 4 sin 60◦
= 17
2 sin 60◦
= 17
√
3
4
.
E
F
G
D
A
BC
h1
2
h2
b
11. 6. The area of an equilateral triangle with side length s is
√
3
4 s2
. Therefore, the side length is 4
√
2 and
height h = 2
√
6. Now, if r is the radius of the inscribed circle, then r = h
3 , since we have an equilateral
triangle. Thus the area is πr2
=8π
3
.
7. Let’s “resize” the coordinates to be x = x, y = 2y
3 . This keeps the origin at (0,0), but turns our ellipse
into a circle of radius 2. Thus now, the triangle is equilateral, and we can see it now has area 3
√
3.
Once we expand back, we can see we are just multiplying the area by 3
2 and so the answer is 9
√
3
2
.
8. By standard formula, we have that the radius of the inscibed circle, r, is r = ab sin A
2a+b (isoceles triangle
that is formed by cutting the pyramid vertically in half (cuts the base into 2 equal rectangles)).
h2
+ (b
2 )
2
= a2
gives a = h2 + b2
4 . Also sin A = h
a . Therefore, r = bh
2 h2+ b2
4 +b
. Note that the
diameter of the cube is the diameter of the sphere. Let l be the length of the side of the cube, so the
diameter of the cube is l
√
3 = 2r, so l = 2r√
3
. So, the volume of the pyramid is 1
3 b2
h and the cube
volume is l3
. So, the ratio is 25
√
3
6
.
r
a
C
h
A
b=base
A
a
9. A hexagon can be formed by removing any vertex, removing all vertices connected to that vertex, and
then removing any edges connected to any of the removed vertices, and these are the only hexagons in
the diagram. Thus, since there are 10 vertices, there are 10 hexagons in the figure.
10. Consider the flattened version of the situation. Then let O be the center fo the spheres, A be the first
reflection point, B be the point of C1 such that OB is perpendicular to OP. Then since OB = 1,OP =√
3, PBO = 60◦
and P, B, A are collinear, BAO = 60◦
implies that POA = 30◦
. Therefore, each
reflection takes the ray 1
12 around the circle, so there are 11 reflections.
12. Team Test Solutions
Stanford Mathematics Tournament 2000
1. .6445 rounds to .645 to .65 to .7. Otherwise .6444... rounds to .644. So the smallest number is .6445.
2. Let c =price, p=purity, d =diameter, h=depth of gold mine, ki=constant. We are given c = k1p2
d3
,
p = k2
c2
h2 , and d = k3
3
√
ch. So, c = k1k2
2
c4
h4 k3
3ch3
= k4c5 1
h . Thus, k4c4
= h, and c = k5h
1
4 . Thus, p
varies as h
1
4 .
3. The sum of the numbers from 700 to 799 is 799·800
2 − 699·700
2 = 74950. The sum of the numbers from 70 to
79 is 79·80
2 − 69·70
2 = 745. So, all numbers that end from 70 to 79 (excluding those starting with 7, since
we counted those from 700 to 799) is 745·9+10(100 + 200 + . . . + 600 + 800 + 900) = 44705. The sum
of all numbers ending in 7 is 9(7+17+27+37+47+57+67+87+97)+9(100+200+. . .+600+800+900) =
38187. So, the total sum of numbers containing a 7 is 74950 + 44705 + 38186 = 157842.
4. 738,826. This can be arrived at by stepping down, starting with finding how many combinations are
there that begin with a letter other than V or W, and so forth. The answer is 8·9!
2·2 + 4·7!
2 + 4 · 6! + 4 ·
4! + 3! + 2! + 2! = 738826.
5. 0 = cos (α + β) + sin (α − β) = cos α · cos β − sin α · sin β + sin α · cos β − sin β cos α = (cos α + sin α) ·
(cos β − sin β). So cos α + sin α = 0 or cos β − sin β = 0. Then tan α = −1 or tan β = 1. Since tan β is
given as 1
2000 , tan α =−1.
6. Since α3
−α−1 = 0, then α10
= α8
+α7
. So, we can reduce our expression to 3α8
−3α6
−3α5
+4α4
+
2α3
−4α2
−6α−17. Also, 3α8
−3α6
−3α5
= 0, so our expression reduces to 4α4
+2α3
−4α2
−6α−17.
Also, 4α4
− 4α2
− 4α0
, so our expression reduces to 2α3
− 2α − 17. Now, 2α3
− 2α − 2 = 0, so our
expression reduces to -15, which is our answer.
7. Another 4-digit number that satisfies this property is 9801, since 9801=9*1089.
8. If she has a silver dollar, then she would have too many other coins, as 0 half dollars, 2 quarters, 3
dimes, etc. would be greater than the total. So she has no silver dollars, and at least one of every
other denomination. Continuing, it turns out the only feasible solution is 0 silver dollars, 1 half dollar,
2 quarters, 3 dimes, 4 nickels, 8 pennies, for a total of 18 coins.
9. It suffices to consider x ≥ 1, since 4(−x)
4
+ 1 = 4(x)
4
+ 1, and 4(0) + 1 = 1 is not prime. So,
4x4
+ 1 = (4x4
+ 4x2
+ 1) − 4x2
= (2x2
+ 1)
2
− (2x)
2
= (2x2
+ 1 − 2x)(2x2
+ 1 + 2x). For integers x,
both 2x2
− 2x + 1 and 2x2
+ 2x + 1 are integers, so this factors 4x4
+ 1 unless 2x2
− 2x + 1 = ±1
or 2x2
+ 2x + 1 = ±1. Since x > 0, then 2x2
+ 2x + 1 > 1, so we must have 2x2
− 2x + 1 = ±1.
2x2
− 2x + 1 = −1 is absurd (4x4
+ 1, 2x2
+ 2x + 1 > 0, so 2x2
− 2x + 1 = 4x4
+1
2x2+2x+1 > 0), so we solve
2x2
− 2x + 1 = 1, or 2x2
− 2x = 0, so x(x − 1) = 0, and x = 0 or x = 1. We have already rejected
x = 0, so the only case left is x = 1, or 4(1)
4
+ 1 = 5.
10. The second hand crosses the minute hand 59 times an hour. The second hand crosses the hour hand
60 times an hour, except for 2 of the hours, due to the movement of the hour hand. The minute hand
and the hour hand cross 22 times total, because the hour hand completes 2 rotations in a day, and
the minute hand completes 24. The second, hour, and minute hand all coincide only at noon and
midnight, but we’ve counted each of these 12:00’s 3 times instead of once. Therefore, the answer is
59 · 60 + 60 · 24 − 2 + 22 − 2 · 2, giving us 2872 crossings.
11. f(x) is either 0 or something of the form ±xm
, where m ≥ 0.
12. A’s position is (a − Vat, 0) and P’s position is (0, b − Vbt). So, at time t, the distance between them
is (a − Vat)2
+ (b − Vbt)2
. Notice that this distance is the same as the distance between the point
(a, b) and the line (Vat, Vbt), which is the same as the line Vbx − Vay = 0. The distance from a line
Ax + By + C = 0 and (x0, y0) is |Ax0+By0+C|
√
A2+B2
, so the answer is |bVa−aVb|
√
V 2
a +V 2
b
13. Given any 4 vertices, there is exactly one intersection of all the diagonals connecting them. So, the
answer is n
4
.
13. 14. x0 if n ≡ 0 (mod 4),1+x0
1−x0
if n ≡ 1 (mod 4), − 1
x0
if n ≡ 2 (mod 4), x0−1
x0+1 if n ≡ 3 (mod 4).
15. Consider a regular n-gon with radius r. Let x be the side length of the n-gon. So, since the central
angle is 2π
n (see diagram below), use the Law of Cosines to find that x2
= r2
+ r2
− 2r ∗ r cos 2π
n ,
so x2
= 2r2
(1 − cos 2π
n ). Thus, x = r
√
2 1 − cos 2π
n . So, the total perimeter of the n-gon is nx =
nr
√
2 1 − cos 2π
n . Now, if we take limn→∞ of the perimeter, the result will be 2πn, since the n-gon
approaches a cirle, so limn→∞ nr
√
2 1 − cos 2π
n = 2πr, and so limn→∞ nr 1 − cos 2π
n
= πr
√
2.
rr
x
2π
n