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# Einstein Riddle/Zebra Puzzle in Prolog
In this post we solve another logic puzzle deriving rules from the puzzle and stating them in Prolog. And this time it is a rather famous puzzle one: the Zebra puzzle (also called Einstein riddle/puzzle).
# Solving Logic Puzzles in Prolog: Puzzle 3 of 3
There are four researchers: Ainsley, Madeline, Sophie and Theodore. The goal is to find out their sports competition discipline, birth year and research interests (while knowing that each of the mentioned attributes is different amongst them). In order so solve the puzzle, a couple of hints is provided from which the solution can be derived.
# Solving Logic Puzzles in Prolog: Puzzle 2 of 3
Eilen, Ada, Verena and Jenny participated in a painting competition. Find out who painted which subject and who took which place in the competition, using the hints provided.
# Solving Logic Puzzles in Prolog: Puzzle 1 of 3
There are 4 students: Carrie, Erma, Ora und Tracy. Each has one scholarship and one major subject they study. The goal is to find out which student has which scholarship and studies which subject (with all scholarships and majors being different from each other) from the hints provided.
# Jodici solver: Python vs Prolog
Jodici is a fun and intuitive number placement puzzle. It consists of a circle which a) contains 3 nested rings and b) is divided into 6 cake-piece-like sectors. As with Sudoku, the goal is to fill in all numbers, while satisfying certain rules: each field must contain an integer [1,9], with each such integer being used twice in total. Further, each sector sums up to 15 and each ring to 30.
# Flower disk rotation puzzle solver: Python vs Prolog
The flower disk rotation puzzle consists of 4 wooden, stacked disks. The disks are connected at their center via a pole, so that they can be rotated. Each disk contains holes that are arranged around the disk center in the form of a flower. The holes are uniformly spread, so that there is space for 12 holes - but each disk only has 9 of these 12 possible holes (the position of holes differ per disk). The remaining 3 areas are instead made of solid wood. The goal is to rotate the disks so that all holes are covered by at least one of the disks (as we have a total amount of 4*3=12 solid areas for a total of 12 holes, each solid area must cover exactly one hole).
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## Topic outline
• ### UNIT 1: SET OF NUMBERS
Key Unit competence: Classify numbers into naturals, integers, rational and irrationals.
1.0. Introductory Activity 1
1.1. Natural numbers
1.0.1. Definition
Activity 1.1.1
For any school or organization, they register their different assert for good
management.
For example
1. The number of desks may be 152
2. The number of Mathematics textbooks may be 2000.
3. The number of classrooms is 15
4. The number of kitchen may be 1
5. The number of car may be 0
6. And so on
All of these numbers are elements of a set. Which set do you think, they can
belong to?
Can you give other example of elements of that set?
CONTENT SUMMARY
Usually, when counting, people begin by one, followed by two, then three and so
on. The numbers we use in counting including zero, are called Natural numbers.
The set of natural numbers is denoted by = … 0, 1, 2, 3, 4, . { }
On a number line, natural numbers are represented as follows:
Application activity 1.1.1
1. Write down, first ten elements of natural numbers starting from zero.
2. Apart from recording assets, give two examples of where natural
numbers can be used in daily life
2 Mathematics ECLPE | TTC Year 1
1.1.2. Sub sets of natural numbers
Activity 1.1.2
(a) Use a dictionary or internet to define the terms: even, odd and prime numbers.
(b) You are given the set of natural numbers between 0 and 20,
(i) make a set of odd numbers.
(ii) make a set of even numbers.
(iii) make a set of prime numbers.
(iv) identify even numbers which are prime numbers?
(v) How many odd numbers are prime numbers?
(c) Represent the information from (b) in a Venn diagram.
CONTENT SUMMARY
There are several subsets of natural numbers:
(a)Even numbers
Even numbers are numbers which are divisible by 2 or numbers which are multiples of 2.
Even numbers from 0 to 20 are 0, 2, 4, 6, 8, 10,12, 14, 16, 18 and 20.
The set of even numbers is E = {2,4,6,8,} and it is a subset of natural numbers.
(b) Odd numbers
Odd numbers are numbers which leave a remainder of 1 when divided by 2. Odd
numbers between 0 and 20 are 1,3,5,7,9,11,13,15,17 and 19. The set of odd numbers
is O = {1,3,5,7,} and it is a subset of natural numbers.
(c) Prime numbers
Prime number is a number that has only two divisors 1 and itself. Prime numbers
between 0 and 20 are 2, 3, 5, 7, 11, 13, 17 and 19. The set of prime numbers is
P = {2,3,5,7,11,13,19,} and it is a subset of natural numbers.
Application activity 1.1.2
Given E = {1, 4, 8, 11, 16, 25, 49, 53, 75}, list the elements of the following subsets
(a) Even numbers
(b) Odd numbers
(c) Prime numbers
Represent the above information on a Venn diagram.
1.1.3 Operations and properties on natural numbers
Activity 1.1.3
1. From the given any three natural numbers a ,b and c , investigate the following operations
a) is a b + always a natural number? Use example to justify your answer.
b) is a b + and b a + always giving the same answer? Use example to justify your
c) is a bc + + and (ab c + +) always giving the same answer? Use example to
2. Given any three natural numbers a ,b and c , investigate the following operations:
a b − and b a − , a bc − − and ab c − −
What do you notice? Is always the answer an element of natural numbers?
3. Given any three natural numbers a ,b and c , investigate the following
operations : a b × , a b × and b a × , a bc × × and ab c × × , a bc × + and
ab ac + , a bc × − and ab ac −
4. Given any two natural numbers a ,b different from zero, investigate the
following operations a b ÷ , a b ÷ and b a ÷ ,
CONTENT SUMMARY
(i) Closure property: The sum of any two natural numbers is always a natural
number. This is called ‘Closure property of addition’ of natural numbers. Thus, is
closed under addition. If a and b are any two natural numbers, then (a + b) is also a
natural number.
Example: 2 + 4 = 6 is a natural number.
(ii) Commutative property: If a and b are any two natural numbers, then, a+b=b+a.
Addition of two natural numbers is commutative.
Example: 2+ 4 = 6 and 4 + 2 = 6. Hence, 2 + 4 = 4 + 2
(iii) Associative property:
If a, b and c are any three natural numbers, then a + (b + c) = (a + b) + c. Addition
of natural numbers is associative.
Example: 2 + (4 + 1) = 2 + (5) = 7 and (2 + 4) + 1 = (6) + 1 = 7. Hence, 2+(4+1)=(2+4)+1
Subtraction
(i) Closure property: The difference between any two natural numbers need not be
a natural number. Hence is not closed under subtraction.
Example: 2 - 5 = -3 is a not natural number.
(ii) Commutative property: If a and b are any two natural numbers, then
(a-b) ≠ (b-a). Subtraction of two natural numbers is not commutative.
Example: 5 - 2 = 3 and 2 - 5 = -3. Hence, 5 - 2 ≠ 2 – 5. Therefore, Commutative
property is not true for subtraction.
(iii) Associative property: If a, b, c and d are any three natural numbers, then
a - (b - d) ≠ (a - b) – d. Subtraction of natural numbers is not associative.
Example: 2 - (4 - 1) = 2 - 3 = -1 and (2 - 4) - 1 = -2 - 1 = -3
Hence, 2 - (4 - 1) ≠ (2 - 4) – 1. Therefore, Associative property is not true for
subtraction.
Multiplication
(i) Closure property: If a and b are any two natural numbers, then a x b = ab is also
a natural number. The product of two natural numbers is always a natural number.
Hence is closed under multiplication.
Example: 5 x 2 = 10 is a natural number.
(ii) Commutative property: If a and b are any two natural numbers, then
a x b = b x a. Multiplication of natural numbers is commutative.
Example: 5 x 9 = 45 and 9 x 5 = 45. Hence, 5 x 9 = 9 x 5. Therefore, Commutative
property is true for multiplication.
(iii) Associative property: If a, b and c are any three natural numbers,
then a x (b x c) = (a x b) x c. Multiplication of natural numbers is associative.
Example: 2 x (4 x 5) = 2 x 20 = 40 and (2 x 4) x 5 = 8 x 5 = 40 , Hence,
2 x (4 x 5) = (2 x 4) x 5. Therefore, Associative property is true for multiplication.
(iv) Multiplicative identity: a is any natural number, then a x 1 = 1 x a = a. The
product of any natural number and 1 is the whole number itself. ‘One’ is the
multiplicative identity for natural numbers.
Example: 5 x 1 = 1 x 5 = 5
Division
(i) Closure property: When we divide of a natural number by another natural number,
the result does not need to be a natural number. Hence, is not closed under
multiplication.
Example: When we divide the natural number 3 by another natural number 2, we
get 1.5 which is not a natural number.
(ii) Commutative property: If a and b are two natural then a ÷ b ≠ b ÷ a. Division
of natural numbers is not commutative.
Example: 2 ÷ 1 = 2 and 1 ÷ 2 = 1.5. Hence, 2 ÷ 1 ≠ 1 ÷ 2
Therefore, Commutative property is not true for division.
(iii) Associative property: If a, b and c are any three natural numbers,
then a ÷ (b ÷ c) ≠ (a ÷ b) ÷ c. Division of natural numbers is not associative.
Example : 3 ÷ (4 ÷ 2) = 3 ÷ 2 = 1.5 and (3 ÷ 4) ÷ 2 = 0.75 ÷ 2 = 0.375
Hence, 3 ÷ (4 ÷ 2) ≠ (3 ÷ 4) ÷ 2. Therefore, Associative property is not true for
division.
Distributive Property
(i) Distributive property of multiplication over addition :
If a, b and c are any three natural numbers, then a x (b + c) = ab + ac. Multiplication
of natural numbers is distributive over addition.
Example : 2 x (3 + 4) = 2 x 3 + 2 x 4 = 6 + 8 = 14
2 x (3 + 4) = 2 x (7) = 14. Hence, 2 x (3 + 4) = 2 x3 + 2 x 4
Therefore, Multiplication is distributive over addition.
(ii) Distributive property of multiplication over subtraction:
If a, b and c are any three natural numbers, then a x (b - c) = ab – ac. Multiplication
of natural numbers is distributive over subtraction.
Example: 2 x (4 - 1) = (2x4) - (2x1) = 8 - 2 = 6
2 x (4 - 1) = 2 x (3) = 6. Hence, 2 x (4 - 1) = (2x4) - (2x1)
Therefore, multiplication is distributive over subtraction.
Application activity 1.1.3
1.2 Integers
1.2.1 Definition
Activity 1.2.1
Carry out the following activities
1. Using a Maths dictionary, define what an integer is.
2. What integer would you give to each of the following situation?
(a) A fish which is 50 m below the water level.
(b) Temperature of the room which is 42ºC.
(c) A boy who is 2 m below the ground level in a hole.
(d) A bird which is 3 m high on a tree
CONTENT SUMMARY
Integers are whole numbers which have either negative or positive sign and include
zero. The set of integers is represented by . Iintegers can be negative {-1,-2,-3,-4,-5,... },
positive {1, 2, 3, 4, 5, ... }, or zero {0}
The set of integers is represented using Carly brackets as follow :
{ ..., -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, ... }
Example 1:
Taking 4 steps forward can be considered as +4 (positive 4). This means it is 4 steps
Taking 4 steps backward is considered as -4 (negative 4). This means it is 4 steps
behind the starting point.
Example 2:
Application activity 1.2.1
John borrowed $3 to pay for his lunch. Alex borrowed$5 to pay for her lunch.
Virginia had enough money for lunch and has $3 left over. Place these people on the number line to find who is poorest and who is richest given that they will pay the same amount. 1.2.2 Sub sets of integers Activity 1.2.2 Discuss the following: From integers between 12 and +20, form small sets which contain; a) odd numbers b) even numbers c) factors of 6 d) multiples of 3. CONTENT SUMMARY Integers have several subsets such as set of natural numbers, set of even numbers, set of odd numbers, set of prime numbers, set of negative numbers and so on. Some of special subset of integers are the following: Application activity 1.2.2 Among them, • which are negative integers? • Which are positive integers? • Which are neither positive nor negative integers? • Which are not integers? 1.2.3 Operations and properties on integers Activity 1.2.3 1. Work out the following on a number line: (a) (+3) + (+2) (b) –(5) + - (3) (c) (+4) + (-3) (d) Which side of the number line did you move when adding a negative number to a positive number? (e) Which side of the number line did you move when adding the given numbers which are all negative? 2. Work out the following and show your solutions on a number line. (a) (-4)-(+3) (b) (+5) – (+3) (c) (–6) – (-6) (d) On your number line, which direction do you move when subtracting two negative numbers? (e) In case you have two positive numbers that you are finding the difference, which side of the number line would you move? 3. Work out the following: (a) (+5) × (-6) (b) (+5) × (+6) (c) (-5) × (+6) (d) (-5) × (-6) (e) Did you obtain the same results in all the four tasks? 4. Work out the following and show your solutions on a number line. (a) (-4) ÷(+4) (b) (+4) ÷ (+4) (c) (–4) ÷ (-4) (d) (+4) ÷ (-4) (e) Did you obtain the same results in all the four questions above? CONTENT SUMMARY The table below shows how addition, subtraction, multiplication and division of integers are performed. Application activity 1.2.3 A common example of negative integer usage is the thermometer. Thermometers are similar to number lines, but vertical. They have positive integers above zero and negative integers below zero. Commonly, people recognize a temperature of -25°C as cold. People use this number system to measure and represent the temperature of the air. Also, if it is -23°C outside, and the temperature drops 3 degrees, what is temperature now? If we picture the thermometer, we know that as the temperature drops, we look downwards on the thermometer 1.3 Rational and irrational numbers 1.3.1 Definition of rational numbers Activity 1.3.1 CONTENT SUMMARY From any two integers a and b , we deduce fractions expressed in the form a/b , where b is a non-zero integer. A rational number is a number that can be expressed as a fraction where both the numerator and the denominator in the fraction are integers; the denominator in a rational number cannot be zero. As fractions are rational numbers, thus set of fractions is known as a set of rational numbers. Application activity 1.3.1 From research on internet or using reference books, identify different type of fractions and give example for each type. 1.3.2 Sub sets of rational numbers Activity 1.3.2 Knowing that a rational number is a number that can be in the form of p q, where p and q are integers and q is not zero. Tell whether the given statements are true or false. Explain your choice. 1. All integers are rational numbers. 2. No rational numbers are whole numbers. 3. All rational numbers are integers. 4. All whole numbers are rational numbers CONTENT SUMMARY From the concept of subset and definition of a set of rational numbers we can establish some subsets of rational numbers; among them we have: • Integers and its subsets: 0 , , , ,... +−+ Z Z;Z Z.......... • Natural numbers and its subsets: , , + N N prime numbers, odd numbers, even numbers,… • Counting numbers and its subsets: ,+ N square numbers, prime numbers, odd numbers, even numbers … Application activity 1.3.2 Using Venn diagram, establish the relationship between the following sets: Natural numbers; Integers and rational numbers. 1.3.3 Operations and properties on rational numbers Activity 1.3.3 CONTENT SUMMARY The following table shows how addition, subtraction, multiplication and division of rational numbers are performed Application activity 1.3.3 Read carefully the following text and make a research to find out two more examples where fractions or rational numbers are used in real life and then present your findings in written and oral forms.“Imagine you are shopping with your$100 in birthday money. You really want a few items you have had your eye on for a while, but they are all very expensive. You are waiting for the items to go on sale, and when they do, you rush down to the store. Instead of being marked with a new price, though, the store has a large sign that reads: All items are currently 75% off. This sounds like great news, but without doing some Math, there is no way to know if you have enough money. Knowing that 75% is ¾ off the cost of each item is the best way to get started”.
1.3.4 Definition of irrational numbers
Activity 1.3.4
CONTENT SUMMARY
Application activity 1.3.4
Using internet or reference books, from your own choice of appropriate numbers, verify and discuss if the given statement is always true, sometimes true or never true.
1. The sum of rational number and irrational number is irrational.
2. The product of rational number and irrational number is irrational.
3. The sum of two irrational numbers is irrational.
4. The product of two irrational numbers is irrational.
5. Between two rational numbers, there is an irrational number.
6. If you divide an irrational number by another, the result is always an irrational number
1.4 Real numbers
1.4.1 Definition
Activity 1.4.1
From the following Venn diagram, representing the set of counting numbers, natural numbers, integers, decimals, rational numbers and irrational numbers, verify and discuss if the given statement is whether true or false.
1. The set of counting numbers is a subset of natural numbers.
2. The intersection of set of integers and counting numbers is the set of natural numbers.
3. The intersection of set of integers and natural numbers is the set of counting numbers.
4. The union of set of natural numbers and counting numbers is the set of natural numbers.
5. The intersection of set of rational numbers and irrational numbers is the set of irrational numbers.
6. The union of set of rational numbers and irrational numbers is a set of irrational numbers.
CONTENT SUMMARY
The set of rational numbers and the set of irrational numbers combined together, form the set of real numbers. The set of real numbers is denoted by R . Real numbers are represented on a number line as infinite points or they are set of decimal numbers found on a number line. This is illustrated on the number line below
Application activity 1.4.1
Some examples of applications of real numbers in our daily life are identified below. From research activity; find out at least 3 examples of other applications of real numbers in our daily life.
Real numbers help us to count and to measure out quantities of different items. For instance, in catering you may have to ask the client how many sandwiches they need for the event. Certainly, those working in accounts and other financial related jobs may use real numbers mostly. Even when relaxing at the end of the day in front of the television flicking from one channel to the next you are using real numbers.
1.4.2 Subsets of real numbers and intervals
Activity 1.4.2
Carry out research on sets of real numbers to determine its subsets and present your finds using a number line.
Application activity 1.4.2
Note that, in writing intervals, it is also possible to include only one endpoint in an interval. Included point is geometrically represented by the solid dot. From this notice, complete the following table.
1.4.3 Operations and properties on real numbers
CONTENT SUMMARY
The following table shows how addition, subtraction, multiplication and division of real numbers are performed
Application activity 1.4.3
1. Discuss whether Closure property under division for real numbers is satisfied.
2. Ngoma District wants to sell two fields on the same price. One has width of 60m out 160m of length. Another one has the width of 100m as it is its length. Among these fields, what is biggest? Interpret your result.
1.4. END UNIT ASSESSMENT 1
1. List three rational numbers between 0 and 1
2. Identify the sets to which each of the following numbers belong by ticking (˅) in the appropriate boxes (cells).
• ### UNIT 2: SET THEORY
Key Unit competence: Solve problems that involve Sets operations, using Venn diagram.
2.0. Introductory Activity 2
At a TTC school of 500 students-teachers, there are 125 students enrolled in Mathematics club, 257 students who play sports and 52 students that are enrolled in Mathematics club and play sports.
If M stands for the set of students in Mathematics club, S stands for the set of students in Sports and U stands for all students at the school or universal set,
1. Complete the following table
2.1 Sets and Venn diagrams
Activity 2.1
Given set A = { 1, 3, 5, 7, 9 } and B = { 1, 2, 3, 4, 5 },
• Identify common elements for both sets.
• Determine elements of set A which are not in set B
• Determine elements of set B which are not in set A
• Determine all elements in set A and set B
• Represent elements of set A in a Venn diagram
• Represent elements of set B in a Venn diagram
• Represent elements of sets A and B using one Venn diagram
CONTENT SUMMARY
A well-defined collection of objects is called a set. Each member of a set is called an element. All elements of a set follow a certain rule and share a common property amongst them.
A set that contains all the elements and sets in a given scenario is called a Universal Set (U).
Venn Diagrams consist of closed shapes, generally circles, which represent sets. The capital letter outside the circle denotes the name of the set while the small letters inside the circle denote the elements of the set.
The various operations of sets are represented by partial or complete overlap of these closed figures. Regions of overlap represent elements that are shared by sets.
In practice, sets are generally represented by circles. The universal set is represented by a rectangle that encloses all other sets.
The given figure is a representation of a Venn diagram. Here each of the
circles A, B and C represents a set of elements.
• Set A has the elements a, d, e and g.
• Set B has the elements b, d, g and f.
• Set C has the elements e, g, f and c.
• Both A and B have the elements d and g.
• Both B and C have the elements g and f.
• Both C and A have the elements e and g.
• A, B and C all have the element g.
The circular pattern used to represent a set and its elements is called a Venn diagram.Venn diagrams are an efficient way of representing and analysing sets and performing set operations. As such, the usage of Venn diagrams is just the elaboration of a solving technique. Problems that are solved using Venn diagrams are essentially problems based on sets and set operations.
Example
1. Given set A= {2, 4, 5, 7, 8}, represent set A on a Venn diagram.
Solution
First, express the data in terms of set notation and then fill the data in the Venn diagram for easy solution.
When drawing Venn diagrams, some important facts like “intersection”, “union” and “complement” should be well considered and represented.
Example: Consider the Venn diagram below.
List the elements of set M and N.
Solution
Application activity 2.1
Consider these two sets A = {2,4,6,8,10 ] and B = { 2,3,5,7} . Represent them in a Venn diagram
2.2 Operations of Sets
Activity 2.2
Consider a class of students that form the universal set. Set A is the set of all students who were present in the English class, while Set B is the set of all the students who were present in the History class. It is obvious that there were students who were present in both classes as well as those who were not present in either of the two classes. The shaded part shows the elements which are considered in the diagram.
Observe the diagrams and identify which one to represent the following:
• All students who were absent in the English class
• All students who were present in at least one of the two classes.
• All the students who were present for both English as well as History classes.
• All the students who have attended only the English class and not the History class
• All the students who have attended just the History class and not the English class.
CONTENT SUMMARY
From the activity, we realize that two or more sets can be represented using one Venn diagram and from the representations, different sets can be determined. To determine those sets, one may perform different operations on sets such as:
• Intersection of sets
• Union of sets
• Universal sets
• Simple difference of sets
• Symmetric difference of sets
• Complement of sets
a. The intersection of sets
The common elements which appear in two or more sets form the intersection of sets. The symbol used to denote the intersection of sets is∩ . The intersection of sets A and B is denoted by A ∩ B and consist of those elements
a) Universal set
A set that contains all the subsets under consideration is known as a universal set. A Universal set is denoted by the symbolU .
Example
Consider a school in which one can find various categories of people such as pupils, teachers and other workers or staff.
1. Use S to represent the set of people in a school.
2. Write all the subsets of S.
Solution
1. We can present the set of all people in the school with sets as follows:
Set S pupils teachers workers = { ,, . }
Thus, set S contains all the subsets of the various categories of people in the school.
Let us use sets P, T and W to represent the subsets of set S.
2.3 Analysis, interpretation and presentation of a problem using Venn diagram
Activity 2.3
1. A survey was carried out in a shop to find the number of customers who bought bread or milk or both or neither. Out of a total of 79 customers for the day, 52 bought milk, 32 bought bread and 15 bought neither.
a. Without using a Venn diagram, find the number of customers who:
iii. bought milk only
b. With the aid of a Venn diagram, work out questions (i), (ii) and (iii) in (a) above.
c. Which of the methods in (a) and (b) above is easier to work with? Give reasons for your answer.
2. The Venn diagram below shows the number of senior one students in a school who like Mathematics (M), Physics (P) and Kinyarwanda (K). Some like more than one subject in total 55 students like Mathematics.
CONTENT SUMMARY
1. Venn diagrams are great for comparing things in a visual manner and to quickly identify overlaps. They are diagrams containing circles that show the logical relations between a collection of sets or groups
Venn diagrams are used in many areas of life where people need to categorize or group items, as well as compare and contrast different items. Although Venn diagrams are primarily a thinking tool, they can also be used for assessment. However, students must already be familiar with them.
Example: In a room, there are 5 people a, b, c, d, e. Out of them, a, b and c are Males while d and e are Females. Also, a and e study science while b, c and d study English. The set of males is M = {a, b, c} and the set of females is F = {d, e}
2.4. Modelling and solving problems involve Set operations using Venn diagram
Activity 2.4
A survey was carried out in Kigali. 50 people were asked about their preferred hotel for taking lunch among Hilltop, Serena and Lemigo hotels. It was found out that 15 people ate at Hilltop, 30 people ate at Serena, 19 people ate at Lemigo, 8 people ate at Hilltop and Serena, 12 people ate at Hilltop and Lemigo, 7 people ate at Serena and Lemigo. 5 people ate at Hilltop, Serena, and Lemigo.
a. Model the problem using variables and represent the information on a Venn diagram.
b. How many people ate at Hilltop?
c. How many ate at Hilltop and Serena but not at Lemigo?
d. How many people did not eat from any of these three hotels?
CONTENT SUMMARY
Pictorial representations of sets represented by closed figures are called set diagrams or Venn diagrams and they are used to illustrate various operations like union and intersection. A Mathematician John Venn introduced the concept of representing the sets
pictorially by means of closed geometrical figures called Venn diagrams. In Venn diagrams, the Universal Set U is represented by a rectangle and all other sets under consideration by circles within the rectangle. Venn diagrams are useful in solving simple logical problems.
A Venn Diagram is an illustration that shows logical relationships between two or more sets (grouping items). Venn diagram uses circles (both overlapping and non-overlapping) or other shapes. Commonly, Venn diagrams show how given items are similar and different.
Despite Venn diagram with 2 or 3 circles are the most common type, there are also many diagrams with a larger number of circles (5,6,7,8,10…)theoretically, they can have unlimited circles.
Venn Diagram in case of two elements
Where w is the number of elements that belong to none of the sets A, B or C
Tip: Always start filling values in the Venn diagram from the innermost value or intersection part
Example:
150 TTC student-teachers were interviewed. 85 were were registered for a Math class, 70 were were registered for an English class while 50 were registered for both Math and English.
Model this problem using variables and Venn diagram and find out the following:
• How many student-teachers signed up only for Math class?
• How many student-teachers signed up only for English class?
• How many student-teachers signed up for Math or English?
• How many student-teachers signed up for neither Math or English?
Solution
• Let x be the number of student- teachers who signed up for both Math and English.
• The number of student- teachers who signed up only for Math class is 85 – x. Knowing that x = 50, student- teachers who signed up only for Math is 35.
• The number of student- teachers who signed up only for English class is 70 – x. Knowing that x = 50, student- teachers who signed up only for English is 20.
• The number of student- teachers who signed up for Math or English is given by the total number of all student teachers in both sets. This is 35 +50 + 20 = 105
• The number of student- teachers who signed up for neither Math or English is given by the total number of all TTC student- teachers who were interviewed minus the total number of all student teachers in both sets.
This is 150 - 105 = 45
Application activity 2.4
The Venn diagram below shows the number of year two student-teachers in SME who like Mathematics (M), Physics (P) and Kinyarwanda (K). Some like more than one subject in total 55 students like Mathematics.
a. How many student- teachers who like the three subjects?
b. Find the total number of year two student- teachers in SME.
c. How many student-teachers who like Physics and Kinyarwanda only?
2.5. END UNIT ASSESSMENT 2
1. In a class, 15 students play Volleyball, 11play Basketball, 6 play both games and everyone plays at least one of the games. Find the total number of students in the class.
2. Out of 17 teachers in a school, 10 teach Economics and 9 teach Mathematics. The number of teachers who teach both subjects is twice that of those who teach none of the subjects. With the aid of a Venn diagram, find the number that teach:
(a) Both subjects
(b) None of the subject
(c) Only one subject
3. In a class the students are required to take part in at least two sports chosen from football, gymnastics and tennis. 9 students play football and gymnastics; 19 play football and tennis; 6 play all the three sports. If there were 30 students in the class, draw a Venn diagram to show this information. With the help of a Venn diagram, find out how many students did not participate in any of the sports.
4. At a certain school, 100 students were interviewed about the subject they like. 28 students took Physical Education (PE), 31 took Biology (BIO), 42 took English (ENG), 9 took PE and BIO, 10 took PE and ENG, 6 took BIO and ENG, while 4 students took all three subjects.
• Represent the situation in the following Venn diagram.
• Model the problem using variables and find out the following:
a. How many students took none of the three subjects?
b. How many students took PE, but not BIO or ENG?
c. How many students took BIO and PE but not ENG?
• ### UNIT 3: PROBLEMS ON RATIOS AND PROPORTIONS
Key Unit competence: Apply ratios, proportions and multiplier proportion change to solve real life related problems
3.0. Introductory Activity 3
In daily life people compare quantities, share proportionally things or objects for different reasons. Do you ever wonder why ratio and proportion are necessary in daily life situations?
For example, look at your classmates:
a. Find out the number of boys, then the number of girls. Can you express the number of boys or girls in terms of fraction or ratio?
b. Can you equally share a certain number of Mathematics textbooks to different groups in your classroom and then figure out the ratio of Mathematics textbooks per learner?
c. Give examples where the concept of ratio and proportion are used in daily life situations
3.1 Equal and unequal share, Ratio, Direct and indirect proportions.
3.1.1 Equal and unequal share, Ratio and proportion
Activity 3.1.1
1. Suppose that two brothers from your village received 7 000 Frw from their child who lives in the city. The condition to share this amount of money is that for every 2 Frw that the young brother gets, the other one gets 3 Frw. The two brothers have come to you for help after they disagreed on how to share such money.
(i) In what ratio would you share the money between them?
(ii) Tell your partner how you would share the money and how much each would get.
2. Read carefully the following word problem and express the given data into fraction or ratio.
a. John and Lucy partnered to save money for x time and later buy a taxi. For every 800 Frw that John saved, Lucy saved 120 Frw. In what fraction or ratio were their contributions? What other simplest fraction or ratio is same as this?
b. Jane and David sold milk to a vendor in the morning. Jane sold 4 500 ml while David sold 7.5 litres. In what fraction or ratio are their milk sales?
CONTENT SUMMARY
• Equal and unequal share
There are many cases in real life where people or organizations need to share items or resources equally or unequally. For example, a father may want to share 24 acre of land among his two sons. One of them who is disabled gets double of what the other son gets. In such case, the land is unequally shared.
Ratios:
The mathematical term ‘ratio’ defines the relationship between two numbers of the same kind. The relationship between these numbers is expressed in the form “a to b” or more commonly in the form a b
A ratio is used to represent how much of one object or value there is in relation to another object or value.
Example: If there are 10 apples and 5 oranges in a bowl, then the ratio of apples to oranges would be 10 to 5 or 10: 5. This is equivalent to 2:1. In contrast, the ratio of oranges to apples would be 1:2.
Ratios occur in many situations such as in business where people compare profit to loss, in sports where compare wins to losses etc.
Application activity 3.1.1
Ingabire, Mugenzi and Gahima have jointly invested in buying and selling of shares in the Rwanda stock exchange market. In one sale as they invested different amount of money, they realised a gain of 1 080 000 Frw and intend to uniquely share it in the ratio 2:3:4 respectively. How much did Mugenzi get?
3.1.2 Direct and indirect proportions
CONTENT SUMMARY
• Direct proportion
The mathematical term ‘proportional’ describes two quantities which always have the same relative size or ‘ratio’.
Example: An object weighs 2kg on the 1st day. If weight is proportional to age, then the object will weigh 4kg on the 2nd day, 6kg on the 3rd day, 8kg on the 4th day, 10kg on the 5th day and so on.
There are many phenomena in real life that involve a squared relationship.
For example: • The area (A) of a circle varies directly with the square of the radius r
• The value (V) of a diamond varies directly with the square of its mass (M) make the following comparison:
3.2 Calculation of proportional and compound proportional change
Activity 3.2
1. Discuss with your classmate what you understand by the word multiplier.
• Consider a shirt that is sold at a 20% discount.
• What is the percentage of the selling price?
• Convert this percentage you have gotten into fraction. What do you notice?
2. Consider a shirt with a marked price of 500 Frw. After negotiating with the customer, the shirt is sold at a 10% lower. Discuss with your classmate the change in price and the new price (selling price) of the shirt in Frw.
3. Consider that 3 people working at the same rate can cultivate 2 acres of land in 3 days. What do you think will happen if the working days are increased to five and people are still working at the same rate? Discuss.
2. A farmer gets 80 litres of milk from his cow. The amount of milk from the cow reduced by 5% after illness. What is the new amount of milk produced by the cow?
Repeated proportional change is an extremely useful mathematical process because it can be used to calculate real world financial problems such as compound interest.
‘Compound interest’ refers to the interest added to a deposit or loan. The added interest will also earn interest as time passes. To calculate the compound interest of a loan, we may use repeated proportional change.
Example
(a) £500 is borrowed for 6 years at 5 % compound interest. Calculate the amount of compound interest which will be paid
Solution
(a) From the question, we know that 5% compound interest is added each year. This means that there will be 105% of the original amount borrowed at the end of the first year or 1.05.
With 1.05 as your multiplier, you can calculate the total amount of money borrowed after 6 years. The money was borrowed for 6 years, so you must raise 1.05 to the power of 6. Therefore:
Total amount of money borrowed is 500 x (1.05) 6 = 670.047= £670
The question has asked you to calculate the amount of compound interest. To do so, you must subtract the original amount borrowed (£500) from the value you have just generated:
670 - 500 = 170. As a result, the amount of compound interest which will be paid is £170
To calculate compound proportionality problems, one can use the unitary method or compound rule of three
Example: 9 men working in a factory produce a certain number of pans in 6 working days. How long will it take 12 men to produce the same number of pans if they work at the same rate?
Application activity 3.2
1. What is the multiplier of 45% decrease?
2. Deborah’s salary last year was 15 000 Frw. This year it was increased by 20%. What is her salary this year?
3. In 2004 a company processed 800 tonnes of maize. In 2005, the company decreased production by 30 %. How many tonnes did the company process in 2005?
4. Four men working at the same rate can dig a piece of land in ten days. How long would it take five men to do the same job?
3.3 Problems involving direct and indirect proportions
Activity 3.3
Read carefully the given problems and discuss how can you solve problems involving direct and indirect proportions.
1. F is directly proportional to x. When F is 6, x is 4. Find the value of F when x is 5.
2. A is directly proportional to the square of B. When A is 10, B is 2. Find the value of A when B is 3.
3. A is inversely proportional to B. When A is 10, B is 2. Find the value of A when B is 8
4. F is inversely proportional to the square of x. When F is 20, X is 3. Find the value of F when x is 5.
CONTENT SUMMARY
Two values x and y are directly proportional to each other when the ratio x : y or x y ∝ is a constant (i.e. always remains the same). This would mean that x and y will either increase together or decrease together by an amount that would not change the ratio.
Knowing that the ratio does not change allows you to form an equation to find the value of an unknown variable.
Example:
If two pencils cost $1.50, how many pencils can you buy with$9.00?
Problem solving plan
1. Understand the problem: Think what information is given and what information is required
2. Decide on a strategy: List the strategies with which you think the solution can be found
3. Apply the strategy: Find the solution using the strategy you have chosen
4. Look back:
• Have you verified your solution?
• Are there other solutions?
• Can you solve a simpler problem?
• Have you answered the question as it was initially stated?
Example
1. The voltage V(volts) across an electrical circuit is directly proportional to the current I (Amperes) flowing through the circuit. When I=1.2 A, V=78V
a) Express V in terms of I
b) Find V when I=2A
c) Find I when V=162.5V
• ### UNIT 4 :PROPOSITIONAL AND PREDICATE LOGIC
Key Unit competence: Use Mathematical logic as a tool of reasoning and argumentation in daily situation
4.1 Definition
4.1.1 Simple statement and compound statements
true or false, but never both, and to always have the same truth value. The two truth values of proposition are true and false and are denoted by the symbols T and F respectively. Occasionally they are also denoted by the symbols 1 and 0 respectively.
Application activity 4.1.1
1. Find out which of the following sentences are statements and which are not. Justify your answer.
a) Uganda is a member of East African Community.
b) The sun is shining.
c) Come to class!
d) The sum of two prime numbers is even.
e) It is not true that China is in Europe.
f) May God bless you!
2. Write down the truth value (T or F) of the following statements
a) Paris is in Italy.
b) 13 is a prime number.
c) Kigeri IV Rwabugiri was the king of the Kingdom of Rwanda
d) Lesotho is a state of South Africa.
e) Tanzania is in east of Rwanda and is in SADC (Southern African Development Community)
4.1.2 Truth tables
Activity 4.1.2
Suppose we are given two simple statements, named p and q to get a compound statement C pq ( , ).
a) What are the possibilities for the truth-values of p and of q?
b) Using a table,
i) How many possibilities are there, for their pairs of truth-values?
ii) How many possibilities are there, for the triples of truth-values of three statements p, q and r?
The way we will define compound statements is to list all the possible combinations of the truth-values (abbreviated T and F) of the simple statements (that are being combined into a compound statement) in a table, called a truth table. The name of each statement is at the top of a column of the table. The truth values used to define the truth-value of the compound statement appear in the last column.
Application activity 4.1.2
Write down the truth table for three propositions p, q and r
4.2 Logical connectives
4.2.1 Negation
Activity 4.2.1
Let p, q, r, s be the propositions
1. It is raining
2. Uganda is African country
3. London is in France
Give a verbal sentence which describes the opposite proposition
Application activity 4.2.1
1. Write the negation of each of the following statements.
a) Today is raining.
b) Sky is blue
c) My native country is Rwanda.
d) Benimana is smart and healthy.
4.2.2 Conjunction
4.2.3 Disjunction
Determine the truth value of each of the following statements
1. Paris is in France or 4 + 4= 8
2. The sun is a planet or the Jupiter is a star.
3. Paris is in Japan or 3 +4 =7
4. The first president of United States of America is George Washington or was inaugurated in 1879.
5. Nairobi is the capital of Tanzania or1+1 =2
6. The French revolution started in 1789 or ended in 1799
4.2.4 Conditional statement
Activity 4.2.4
Rephrase the following statements without changing the meaning
1. If you buy me a pen, I will go to school
2. If the earth is flat, then mars is flat
3. If you are tall, then you will be a member of our volleyball team
4. If you do not buy these shoes, then I will not go with you
5. If you do not pay school fees, then you will not get you school report
Application activity 4.2.4
1. Using the statements p :Mico is fat and p :Mico is happy
Assuming that “not fat” is thin, write the following statements in symbolic form
a) If Mico is fat then she is happy
b) Mico is unhappy implies that Mico is thin
2. Write the following statements in symbolic form and their truth table
a) If n is prime, then n is odd or n is 2.
b) If x is nonnegative, then x is positive or x is 0.
c) If Tom is Ann’s father, then Jim is her uncle and Sue is her aunt.
4.2.5 Bi-conditional statements
CONTENT SUMMARY
A tautology is a compound statement that is always true regardless of the truth values of the individual statements substituted for its statement variables.
Example
• The statement “The main idea behind data compression is to compress data” is a tautology since it is always true.
• The statement “I will either get paid or not get paid” is a tautology since it is always true
If you are given a statement and want to determine if it is a tautology, then all you need to do is construct a truth table for the statement and look at the truth values in the final column. If all of the values are T (for true), then the statement is a tautology.
The statement “I will either get paid or not get paid” is a tautology since it is always true. We can use p to represent the statement “I will get paid” and not p (written ¬p) to represent “I will not get paid.”
4.3 Quantifiers and their negations: Existential and Universal quantifiers
4.3.1 Predicates
4.3.2 Quantifiers
4.3.3 Negation of quantifiers
Activity 4.3.3
Negate the following statements
1. All grapefruit are pink.
2. Some celebrities are modest.
3. No one weighs more than two thousand pounds.
4. Some people are more than ten feet tall.
5. All snakes are poisonous.
6. Some whales can stay under water for two days without surfacing for air.
7. All birds can fly
Application activity 4.3.3
Negate each of the following statements and write the answer in symbolic form
1. Some students are math majors
2. Every real number is positive, negative or zero
3. Every good boy does fine
4. There is a broken desk in our classroom
5. Lockers must be turned in by the last day of class
6. Haste makes waste
4.4 END UNIT ASSESSMENT 4
• ### UNIT 5: OPERATION ON POLYNOMIALS
Key Unit competence: Perform operations, on polynomials and solve related problems.
5.1 Defining and comparing polynomials
5.2. Operations on polynomials
5.3 Factorization of polynomials
5.4 Expansion of polynomials
• ### UNIT 6:LINEAR AND QUADRATIC EQUATION AND INEQUALITIES
Key Unit competence: Solve algebraically or graphically linear, quadratic equations or inequalities
6.1.1. linear equations
When completing the square to solve quadratic equation, remember that you must preserve the equality. When you add a constant to one side of the equation, be sure to add the same constant to the other side of equation.
6.3.1. Linear inequalities
6.4 Solving word problems involving linear or quadratic equations
Activity 6.4
1. Make a research on internet and by means of examples prepare a short presentation on how linear equations and quadratic equations can be used in daily life.
2. Read carefully the following word problems and model or rewrite them using a mathematical statement.
• Six less than two times a number is equal to nine
• Jane paid 22100 Frw for shoes and clothes. She paid 2100 Frw more for clothes than she did for shoes. How much did Jane pay for shoes?
3. A manufacturer develops a formula to determine the demand for its product depending on the price in Rwandan franc. The formula is , where is the price per unit, and is the number of units in demand. At what price will the demand drop to 1000 units?
CONTENT SUMMARY
It is not always easy to tell what kind of equation a word problem involves, until you start translating it to math symbols.
Words like “together,” “altogether,” or “combined” often indicate that the problem involves addition. The word “left,” as in “he had xx amount left,» often indicates subtraction.
The following table explains the keywords used when writing equations from verbal models.
6.5 Solving and discussing parametric equations
Application activity 6.5
• ### UNIT 7:PROBLEMS ON POWERS, INDICES, RADICALS AND LOGALITHMS
Key Unit competence: Solve problems related to powers, indices, radical and common logarithms
7.0. Introductory Activity
7.1.1 Definition of powers/ indices and radicals
7.1.2 Properties of indices and radicals
7.2 Operations on indices and radicals
7.3. Decimal logarithm
7.3.1 Definition
Activity 7.3.1
What is the real number for which 10 must be raised to obtain:
1) 1 2) 10
3) 100 4) 1000
5) 10000 6) 100000
d. Logarithms are used to determine growth decay
Example:
7.4. END UNIT ASSESSMENT 7
• ### UNIT: 8 :PARAMETERS OF CENTRAL TENDENCES AND DISPERSION
Key Unit competence: Extend understanding, analysis and interpretation of data arising from problems and questions in daily life to the standard deviation
8.0. Introductory Activity 8
1. During 6 consecutive days, a fruit-seller has recorded the number of fruits sold per type
Which type of fruits had the highest number of fruits sold?
b) Which type of fruits had the least number of fruits sold?
c) What was the total number of fruits sold that week?
d) Find out the average number of fruits sold per day.
2. During the welcome test of Mathematics out of 10 , 10 student-teachers of year one ECLPE scored the following marks: 3, 5,6,3,8,7,8,4,8 and 6.
a) Determine the mean mark of the class.
b) What is the mark that was obtained by many students?
c) Compare and discuss about the mean mark of the class and the mark for every student-teacher. What advice could you give to the Mathematics tutor?
d) organize all marks in a table and try to present them in an X-Y Cartesian plane by making x-axis the number of marks and y-axis the number of student-teachers.
8.1 Collection and presentation of grouped and ungrouped data
8.1.1. Collection and presentation of ungrouped data
Activity 8.1.1
1. Observe the information provided by the graph and complete the table by corresponding the number of learners and their age
CONTENT SUMMARY
Every day, people come across a wide variety of information in form of facts or categorical data, numerical data in form of tables.
For example, information related to profit/ loss of the school, attendance of students and tutors, used materials, school expenditure in term or year, student-teachers’ results. These categorical or numerical data which is numerical or otherwise, collected with a definite purpose is called data.
Statistics is the branch of mathematics that deals with the collection, presentation, interpretation and analysis of data.
A sequence of observations, made on a set of objects included in the sample drawn from population, is known as statistical data.
Statistical data can be organized and presented in different forms such as raw tables, frequency distribution tables, graphs, etc.
Qualitative data
Qualitative data is a categorical measurement expressed not in terms of numbers, but rather by means of a natural language description. In statistics, it is often used interchangeably with “categorical” data. Categorical data represent characteristics such as a person’s gender, marital status, hometown, or the types of movies they like. Categorical data can take on numerical values (such as “1” indicating male and “2” indicating female), but those numbers don’t have mathematical meaning. It couldn’t add them together, for example.
Quantitative data
Quantitative data is a numerical measurement expressed not by means of a natural language description, but rather in terms of numbers. These data have meaning as a measurement, such as a person’s height, weight, IQ, or blood pressure; or they’re a count, such as the number of stock shares a person owns, or how many pages you can read of your favorite book before you fall asleep.
Numerical data can be further broken into two types: discrete and continuous.
• Discrete data represent items that can be counted; they take on possible values that can be listed out. The list of possible values may be fixed (also called finite); or it may go from 0, 1, 2, on to infinity (making it countably infinite). For example, the number of heads in 100 coin flips takes on values from 0 through 100 (finite case), but the number of flips needed to get 100 heads takes on values from 100 (the fastest scenario) up to infinity (if you never get to that 100th heads). Its possible values are listed as 100, 101, 102, 103, . . . (representing the countably infinite case).
• Continuous data represent measurements; their possible values cannot be counted and can only be described using intervals on the real number line. For example, the exact amount of gas purchased at the pump for cars with 20-gallon tanks would be continuous data from 0 gallons to 20 gallons, represented by the interval [0, 20], inclusive. You might pump 8.40 gallons, or 8.41, or 8.414863 gallons, or any possible number from 0 to 20. In this way, continuous data can be thought of as being uncountably infinite.
After the collection of data, tally is used to organize and present in a frequency distribution table. Tally means to count by grouping the number of times an item has occurred. When the data are arranged in this way we say that we have obtained
the frequency distribution.
Raw data
Data which have been arranged in a systematic order are called raw data or ungrouped data.
For Example, the following are marks out of 20 for 12 student-teachers.
13 10 15 17 17 18
17 17 11 10 17 10
Frequency distribution
A frequency distribution is a table showing how often each value (or set of values) of the collected data occurs in a data set. A frequency table is used to summarize categorical or numerical data. Data presented in the form of a frequency distribution are called grouped data.
Example
The following data of marks, out of 20, obtained by 12 student-teachers can be presented in a frequency distribution table.
13 10 15 17 17 18
17 17 11 10 17 10
The set of outcomes is displayed in a frequency table, as illustrated below:
Cumulative frequency
The cumulative frequency corresponding to a particular value is the sum of all frequencies up to the last value including the first value. Cumulative frequency can also defined as the sum of all previous frequencies up to the current point.
Example:
The set of data below shows marks obtained by student-teachers in Mathematics. Draw a cumulative table for the data.
Solution:
The cumulative frequency at a certain point is found by adding the frequency at the present point to the cumulative frequency of the previous point.
STEM AND LEAF DISPLAYS
Is a plot where each data value is split into a leaf usually the last digit and a stem the other digit. The stem values are listed down, and the leaf values are listed next to them. This way the stem groups the scores and each leaf indicates a score within that group
Example: The mathematical competence scores of 10 student-teachers participating in mathematics competition are as follows:15,16,21,23,23,26,26,30,32,41. Construct a stem and leaf display for these data by using1, 2, 3, and 4 as your stems.
This means that data are concentrated in twenties.
Example: The following are results obtained by student-teachers in French out of 50.
37, 33, 33, 32, 29, 28, 28, 23, 22, 22, 22, 21, 21, 21, 20, 20, 19, 19, 18, 18, 18, 18, 16, 15, 14, 14, 14, 12, 12, 9, 6
Use stem and leaf to display data
Solution:
Numbers 3, 2, 1, and 0, arranged as a stems to the left of the bars. The other numbers come in the leaf part.
3|2337
2|001112223889
1|2244456888899
0|69
From this, data are concentrated in ones
Application activity 8.1.1
1. At the beginning of the school year, a Mathematics test was administered in year 1 to 50 student-teachers to test their level of understanding. Their results out of 20 were recorded as follows:
Construct a frequency distribution table to help the tutor and the school administration to easily recognize the level of understanding for the Year 1 student-teachers.
2. Differentiate qualitative and quantitative data from the list below: Product rating, basketball team classification, number of student-teachers in the classroom, weight, age, Number of rooms in a house, number of tutors in school.
8.1.2. Collection and presentation of grouped data
Activity 8.1.2
The mass of 50 tomatoes (measured to the nearest g) were measured and recorded in the table below
Construct a frequency distribution table, using equal class intervals of width 5g and taking the lower class boundary of the first interval as 84.5g.
CONTENT SUMMARY
When the range of data is large, the data must be grouped into classes that are more than one unit in width. In this case a grouped frequency distribution is used. Data in this case are grouped in a frequency distribution using groups or classes.
• Class limits: The class limits are the lower and upper values of the class
• Lower class limit: Lower class limit represents the smallest data value that can be included in the class.
• Upper class limit: Upper class limit represents the largest data value that can be included in the class.
• Class boundaries: Class boundaries are the midpoints between the upper class limit of a class and the lower class limit of the next class. Therefore, each class has a lower and an upper class boundary.
Example: The following data represent the marks obtained by 40 students in Mathematics test. Organize the data in the frequency table; grouping the values into classes, starting from 41-50
54 83 67 71 80 65 70 73 45 60 72 82 79 78 65 54 67 64 54 76 45 63 49 52
60 70 81 67 45 58 69 53 65 43 55 68 49 61 75 52
Solution:
Example
Using the frequency table above, determine the class boundaries of the three classes.
Solution
For the first class, 41-50
The lower class boundary is the midpoint between 40 and 41, that is 40.5
The upper class boundary is the midpoint between 50 and 51, that is 50.5
For the second class, 51-60
The lower class boundary is the midpoint between 50 and 51, that is 50.5
The upper class boundary is the midpoint between 60 and 61, that is 60.5
For the third class, 61-70
The lower class boundary is the midpoint between 60 and 61, that is 60.5
The upper class boundary is the midpoint between 70 and 71, that is 70.5
Class width
Class width is the difference between the upper class boundary and lower class boundary
Application activity 8.1.2
Suppose a researcher wished to do a study on the distance (in kilometres) that the employees of a certain department store travelled to work each day. The researcher collected the data by asking each employee the approximate distance is the store from his or her home. Data are collected in original form called raw data as follow:
Since little information can be obtained from looking at raw data, help the researcher to make a frequency distribution table (use 1-3 km and 16-18 km as class limits) so that some general observations can easily be done by the researcher.
8.2 Measure of central tendencies: Mean, median and mode
Activity 8.2
1. Observe the following marks of 10 students in mathematics test: 5, 4, 10, 3, 3, 4, 7, 4, 6, 5 and complete the table below
Calculate the mean, median, mode and range of this set
2. In three classes of Year three in TTC, during the quiz of mathematics out of 5 marks, 100 student-teachers obtain marks as shown in the table below:
• What is the marks obtained by most of the students?
• You are asked to calculate the mean mark for the class. Explain how you should find it.
• What is the marks obtained by most of the students?
• You are asked to calculate the mean mark for the class. Explain how you should find it.
CONTENT SUMMARY
For the comparison of one frequency distribution with another, it requires to summarize it in such a manner that its essence is expressed in few figures only. For this purpose, we find most representative value of data. This representative value of the data is known as the measure of central tendency or averages.
a) Mean, mode , median and range of ungrouped data
Thus for any particular set of ungrouped data, it is possible to select some typical values or average to represent or describe the entire set such a descriptive value is referred to as a measure of central tendency or location such as mean, mode, median
The median:
The data is arranged in order from the smallest to the largest, the middle number is then selected. This really the central number of the range and is called the median.
Mean, mode, median and range of grouped data
For any particular set of grouped data, it is possible to select some typical values or average to represent or describe the entire set such a descriptive value is referred to as a measure of central tendency or location such as mean, mode, median and range.
Range
In the case of grouped data the range is defined as the difference between the upper limit of the highest class and the lower limit of the smallest class.
Application activity 8.2
1. A group of student-teachers from SME were asked how many books they had read in previous year; the results are shown in the frequency table below. Calculate the mean, median and mode number of books read.
2. During oral presentation of internship report for year three studentteachers the first 10 student-teachers scored the following marks out of 10:
8, 7, 9, 10, 8, 9, 8, 6, 7 and 10
a) Calculate the mean of the group
b) Calculate the median and Mode
8.3 Graphical representation of grouped and ungrouped data
Activity 8.3
1. In a class of year 1, student-teachers are requested to provide their sizes in order to be given their sweaters. The graph below shows the sizes of sweaters for 30 students. Observe and interpret it by finding out the number of students with small, medium, large, extra-large. Guess the name of the graph.
2. The mass of 50 tomatoes (measured to the nearest g) were measured and recorded in the table below.
a) Determine cumulative frequency
b) Try to draw a histogram, the frequency polygon and the cumulative frequency polygon to illustrate the data.
3. A pie chart shows the mass (in tonnes) of each type of food eaten in a certain month. Observe the pie chart and make a corresponding frequency distribution table by considering the number of tones for each type of food as frequency.
CONTENT SUMMARY
After the data have been organized into a frequency distribution, they can be presented in a graphical form. The purpose of graphs in statistics is to convey the data to the viewers in a pictorial form. It is easier for most people to comprehend the
meaning of data presented graphically than data presented numerically in tables or frequency distributions.
The most commonly used graphs are: Bar graph, Pie chart, Histogram, Frequency polygon, Cumulative frequency graph or Ogive.
A bar graph
A bar chart or bar graph is a chart or graph that presents numerical data with rectangular bars with heights or lengths proportional to the values that they represent.
The bars can be plotted vertically or horizontally. A vertical bar chart is sometimes called a line graph
Pie chart
A pie chart is used to display a set of categorical data. It is a circle, which is divided into segments. Each segment represents a particular category. The area of each segment is proportional to the number of cases in that category.
Example:
A TTC tutor conducted a survey to check the level of how student-teachers like some subjects (English, Mathematics, French, Entrepreneurship, Kinyarwanda) taught in SME option. The survey was done on 60 student-teachers and the table below summarizes the results.
In order to clearly present his/ her findings, the tutor presented the data on a bar graph and then on a pie chart as follows:
1. Bar graph
Histogram
Histogram is a statistical graph showing frequency of data. The horizontal axis details the class boundaries, and the vertical axis represents the frequency. Blocks are drawn such that their areas (rather than their height, as in a bar chart) are proportional to the frequencies within a class or across several class boundaries. There are no spaces between blocks.
Example: Draw a histogram for the frequency distribution given in the table below.
Step 2: Draw Histogram (frequency against class boundaries)
Frequency polygon
In a frequency polygon, a line is drawn by joining all the midpoints of the top of the bars of a histogram.
A frequency polygon gives the idea about the shape of the data distribution. The two end points of a frequency polygon always lie on the x-axis.
Example
Use the histogram above to draw a frequency polygon.
Solution
Example:
Construct a histogram, a frequency polygon, a cumulative frequency graph or ogive to represent the data shown below for the record high temperatures for each of the 50 states
Application activity 8.3
1. At the beginning of the school year, a Mathematics test was administered in year 1 for 50 student-teachers to test their levels in this subject.. Their results out of 20 were recorded as follow:
Present the data using a bar graph to help the tutor and the school administration to easily recognize the level of understanding for the Year 1 student-teachers.
2. The cumulative frequency distribution table below shows distances (in km) covered by 20 runners during the week of competition.
a) Construct a histogram
b) Construct a frequency polygon
c) Construct an ogive.
8.4 Measure of dispersion: Range, variance, Standard Deviation and coefficient of variation
Activity 8.4
CONTENT SUMMARY
The word dispersion has a technical meaning in statistics. The average measures the center of the data. It is one aspect of observations. Another feature of the observations is how the observations are spread about the center. The observation may be close to the center or they may be spread away from the center. If the observations are close to the center (usually the arithmetic mean or median), we say that dispersion, scatter or variation is small. If the observations are spread away from the center, we say that dispersion is large.
The study of dispersion is very important in statistical data. If in a certain factory there is consistence in the wages of workers, the workers will be satisfied. But if some workers have high wages and some have low wages, there will be unrest among the low paid workers and they might go on strikes and arrange demonstrations. If in a certain country some people are very poor and some are very high rich, we say there is economic disparity. It means that dispersion is large.
The extent or degree in which data tend to spread around an average is also called the dispersion or variation. Measures of dispersion help us in studying the extent to which observations are scattered around the average or central value. Such measures are helpful in comparing two or more sets of data with regard to their variability.
Properties of a good measure of dispersion
i. It should be simple to calculate and easy to understand
ii. It should be rigidly defined
iii. Its computation be based on all the observations
iv. It should be amenable to further algebraic treatment
Some measures of dispersion are Quartiles, variance, Range, standard deviation, coefficient of variation
Variance
Variance measures how far a set of numbers is spread out. A variance of zero indicates that all the values are identical. Variance is always non-negative: a small variance indicates that the data points tend to be very close to the mean and hence to each other, while a high variance indicates that the data points are very spread out around the mean and from each other.
The variance is denoted and defined by
Example 1
Calculate the variance of the following distribution: 9, 3, 8, 8, 9, 8, 9, 18
STANDARD DEVIATION
The standard deviation has the same dimension as the data, and hence is comparable to deviations from the mean. We define the standard deviation to be the square root of the variance. thus, the standard deviation is denoted and defined by
Example 1
The six runners in a 200 meter race clocked times (in seconds) of 24.2, 23.7, 25.0, 23.7, 24.0,4.6
a) Find the mean and standard deviation of these times.
b) These readings were found to be 10% too low due to faulty timekeeping.
Write down the new mean and standard deviation.
Example 2
The heights (in meters) of six children are 1.42, 1.35, 1.37, 1.50, 1.38 and 1.30.
Calculate the mean height and the standard deviation of the heights
Coefficient of variation
The coefficient of variation measures variability in relation to the mean (or average) and is used to compare the relative dispersion in one type of data with the relative dispersion in another type of data. It allows us to compare the dispersions of two different distributions if their means are positive. The greater dispersion corresponds to the value of the coefficient of greater variation.
The coefficient of variation is a calculation built on other calculations: the standard deviation and the mean as follows:
Example:
One data series has a mean of 140 and standard deviation 28.28. The second data series has a mean of 150 and standard deviation 24. Which of the two has a greater dispersion?
Solution:
Range
In the case of ungrouped data, the range of a set of observations is the difference in values between the largest and the smallest observations in the set. In the case of grouped data the range is defined as the difference between the upper limit of the highest class and the lower limit of the smallest class.
Example
Calculate the range of the following set of the data set: 1, 3, 4, 5, 5, 6, 9, 14 and 21
Solution: From the given series the lowest data is 1 and the highest data is 21
The Range = highest value −lowest value
Range = −= 21 1 20
Application activity 8.4
1. Out of 4 observations done by tutor of English, arranged in descending order, the 5th, 7th, 8th and 10th observations are respectively 89, 64, 60 and 49.
Calculate the median of all the 4 observations.
2. In the following statistical series, calculate the standard deviation of the following set of data 56,54,55,59,58,57,55
3. In the classroom of SME the first five student-teachers scored the following marks out of 10 in a quiz of Mathematics
5, 6, 5, 2, 4, 7, 8, 9, 7, 5
a) Calculate the mean, median and the modal mark
b) Determine the quartiles and interquartile range
c) Calculate the variance and the standard deviation
d) Determine the coefficient of variation
8.5 Application of statistics in daily life
Activity 8.5
Using internet or reference books from the school library, make a research to provide in written form at least one example of where the following statistical terminologies are needed and used in real life situations:
• Frequency distribution
• Statistical graphs like Bar graph, Histogram, Frequency polygon Mean,
• Variance
Make a presentation of your findings to the whole clas
CONTENT SUMMARY
Statistics is concerned with scientific method for collecting and presenting, organizing and summarizing and analyzing data as well as deriving valid conclusions and making reasonable decisions on the basis of this analysis
Today, statistics is widely employed in government, business and natural and social sciences. Statistical methods are applied in all field that involve decision making.
For example, in agriculture, statistics can help to ensure the amount of crops are grown this year in comparison to previous year or in comparison to required amount of crop for the country. It may also be helpful to analyze the quantity and size of grains grown due to use of different fertilizer.
In education, statistics can be used to analyze and decide on the money spend on girls education in comparison to boys education.
Nowadays, graphs are used almost everywhere. In stock market, graphs are used to determine the profit margins of stock. There is always a graph showing how prices have changed over time, food price, budget forecasts, exchange rates…
Mean is used as one of comparing properties of statistics. It is defined as the average of all the clarifications. Mean helps Teachers to see the average marks of the students.
A large standard deviation indicates that the data points can spread far from the mean and a small standard deviation indicates that they are clustered closely around the mean. Standard deviation is often used to compare real-world data against a model to test the model
In finance, standard deviation is often used as a measure of the risk associated with price-fluctuations of a given asset (stocks, bonds, property, etc.), or the risk of a portfolio of assets, Standard deviation provides a quantified estimate of the uncertainty of future returns.
Application activity 8.5
1. Using internet or reference books from the school library, make a research to provide in written form at least one example of where the following statistical terminologies are needed and used in real life situations:
• Frequency distribution
• Statistical graphs like Pie chart, cumulative frequency polygon or ogive
• Mode and median
• Standard deviation
Make a presentation of your findings to the whole class
2. Collect data on 20 students’ heights in your class and organize them in a frequency distribution. Calculate the mean of 20 students’ heights, variance as well as standard deviation. What conclusion can you make based on the calculated mean, variance and standard deviation?
8.6. END UNIT ASSESSMENT 8
1. Use the graph below to answer the questions that follows
a) Use the graph to estimate the mode.
b) State the range of the distribution.
c) Draw a frequency distribution table from the graph
2. In test of mathematics, 10 student-teachers got the following marks:
6, 7, 8, 5, 7, 6, 6, 9, 4, 8
a) Calculate the mean, mode, quartiles and interquartile range
b) Calculate the variance and standard deviation
c) Calculate the coefficient of variation
3. A survey taken in a restaurant shows the following number of cups of coffee consumed with each meal. Construct an ungrouped frequency distribution. 0 2 2 1 1 2 3 5 3 2 2 2 1 0 1 2 4 4 0 1 0 1 4 4 2 2 0 1 1 5
4. The amount of protein (in grams) for a variety of fast-food sandwiches is reported here. Construct a frequency distribution using six classes. Draw a histogram, frequency polygon, and ogive for the data, using relative frequencies. Describe the shape of the histogram.
REFERENCES
1. J. Sadler, D. W. S. Thorning: Understanding Pure Mathematics, Oxford University Press 1987.
2. Arthur Adam Freddy Goossens: Francis Lousberg. Mathématisons 65. DeBoeck,3e edition 1991.
3. Charles D. Hodgman, M.S., Samuel M. Selby, Robert C.Weast. Standard Mathematical Table. Chemical Rubber Publishing Company, Cleveland, Ohio 1959.
4. David Rayner, Higher GCSE Mathematics, Oxford University Press 2000
5. Direction des Progammes de l’Enseignement Secondaire. Géometrie de
l’Espace 1er Fascule. Kigali, October1988
6. Direction des Progammes de l’Enseignement Secondaire. Géometrie de
l’Espace 2ème Fascule. Kigali, October1988
7. Frank Ebos, Dennis Hamaguchi, Barbana Morrison & John Klassen, Mathematics Principles & Process, Nelson Canada A Division of International Thomson Limited 1990
8. George B. Thomas, Maurice D. Weir & Joel R. Hass, Thomas’ Calculus Twelfth Edition, Pearson Education, Inc. 2010
9. Geoff Mannall & Michael Kenwood, Pure Mathematics 2, Heinemann Educational Publishers 1995
10. H.K. DASS...Engineering Mathematics. New Delhi, S. CHAND&COMPANY LTD, thirteenth revised edition 2007.
11. Hubert Carnec, Genevieve Haye, Monique Nouet, ReneSeroux, Jacqueline
Venard. Mathématiques TS Enseignement obligatoire. Bordas Paris 1994.
12. James T. McClave, P.George Benson. Statistics for Business and Economics.
USA, Dellen Publishing Company, a division of Macmillan, Inc 1988.
13. J CRAWSHAW, J CHAMBERS: A concise course in A-Level statistics with
worked examples, Stanley Thornes (Publishers) LTD, 1984.
14. Jean Paul Beltramonde, VincentBrun, ClaudeFelloneau, LydiaMisset, Claude Talamoni. Declic 1re S Mathématiques. Hachette-education, Paris 2005.
15. JF Talber & HH Heing, Additional Mathematics 6th Edition Pure & Applied, Pearson Education South Asia Pte Ltd 1995
16. J.K. Backhouse, SPTHouldsworth B.E.D. Copper and P.J.F. Horril. Pure Mathematics 2. Longman, third edition 1985, fifteenth impression 1998.
17. Mukasonga Solange. Mathématiques 12, AnalyseNumérique. KIE, Kigali 2006
18. N. PISKOUNOV, Calcul Différential et Integral tom II 9ème édition.
Editions MIR. Moscou, 1980.
19. Paule Faure- Benjamin Bouchon, Mathématiques Terminales F. Editions
Nathan, Paris 1992.
20. Peter Smythe: Mathematics HL & SL with HL options, Revised Edition,
Mathematics Publishing Pty. Limited, 2005.
21. Robert A. Adms & Christopher Essex, Calculus A complete course Seventh
Edition, Pearson Canada Inc., Toronto, Ontario 2010
22. Seymour Lipschutz. Schaum’s outline of Theory and Problems of Finite Mathematics. New York, Schaum Publisher, 1966
23. Seymour Lipschutz. Schaum’s outline of Theory and Problems of linear algebra. McGraw-Hill 1968.
24. Shampiyona Aimable : Mathématiques 6. Kigali, Juin 2005.
25. Yves Noirot, Jean–Paul Parisot, Nathalie Brouillet. Cours de Physique Mathématiques pour la Physique. Paris, DUNOD, 1997.
26. Swokowski, E.W. (1994). Pre-calculus: Functions and graphs, Seventh
edition. PWS Publishing Company, USA.
27. Allan G. B. (2007). Elementary statistics: a step by step approach, seventh
edition, Von Hoffmann Press, New York.
28. David R. (2000). Higher GCSE Mathematics, revision and Practice. Oxford
University Press, UK.
29. Ngezahayo E.(2016). Subsidiary Mathematics for Rwanda secondary
Schools, Learners’ book 4, Fountain publishers, Kigali.
30. REB. (2015). Subsidiary Mathematics Syllabus, MINEDUC, Kigali, Rwanda.
31. REB. (2019). Mathematics Syllabus for TTC-Option of LE, MINEDUC,
Kigali Rwanda.
32. Peter S. (2005). Mathematics HL&SL with HL options, Revised edition. Mathematics Publishing PTY. Limited.
33. Elliot M. (1998). Schaum’s outline series of Calculus. MCGraw-Hill Companies, Inc. USA.
34. Frank E. et All. (1990). Mathematics. Nelson Canada, Scarborough, Ontario (Canada)
35. Gilbert J.C. et all. (2006). Glencoe Advanced mathematical concepts, MCGraw-Hill Companies, Inc. USA.
36. Robert A. A. (2006). Calculus, a complete course, sixth edition. Pearson Education Canada, Toronto, Ontario (Canada).
37. Sadler A. J & Thorning D.W. (1997). Understanding Pure mathematics, Oxford university press, UK
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# 2.02 Operations with fractions
Lesson
We've already learned how to add, subtract, multiply and divide fractions. Similarly, we've looked at each of these operations with negative numbers.
The process is just the same when we have questions with negative fractions - we'll just combine these two skills and their rules to complete the operation.
Review: Fraction operations
• Adding and subtracting: be sure there is a common denominator, and then add or subtract the numerators and keep the denominator
• Multiplying: Multiply the numerators, multiply the denominators, and simplify.
• Dividing: Take the reciprocal of the divisor and multiply. Then, follow the rules for multiplication.
• Mixed numbers: Change the mixed number to an improper fraction, and proceed as if this were a normal fraction operation. You may be expected to change the solution back to a mixed number.
Remember! Operations with integers
• Subtracting a negative number is the same as adding a positive number, e.g. $5-\left(-3\right)=5+3$5(3)=5+3
• Adding a negative number is the same as subtracting a positive number, e.g. $2+\left(-7\right)=2-7$2+(7)=27.
• The product or quotient of two negative terms is a positive answer, e.g. $-6\times\left(-9\right)=54$6×(9)=54 or $\left(-144\right)\div\left(-12\right)=12$(144)÷(12)=12
• The product or quotient of a positive and a negative term is a negative answer, e.g. $-5\times10=-50$5×10=50 or $20\div\left(-4\right)=-5$20÷(4)=5
Using the number line or zero pairs may be helpful with working positive and negative fractions.
#### Worked examples
##### Question 1
Calculate: $3+4-\left(-\frac{4}{5}\right)$3+4(45).
Think: We will evaluate the addition and subtraction, working from left to right.
Do:
$3+4-\left(-\frac{4}{5}\right)$3+4−(−45) $=$= $7-\left(-\frac{4}{5}\right)$7−(−45) $=$= $7+\frac{4}{5}$7+45 $=$= $7\frac{4}{5}$745
Reflect: What if this problem were written as a decimal? Is $3+4-\left(-0.8\right)$3+4(0.8) easier to compute? Do you still end up with the same solution when you convert the decimal back to a mixed number?
#### Practice questions
##### Question 2
Evaluate $-10\times\left(-2\frac{1}{4}\right)$10×(214), giving your answer as a mixed number in simplest form.
##### Question 3
Evaluate $-8\frac{4}{7}+3\frac{3}{7}$847+337, writing your answer as a mixed number in its simplest form.
##### Question 4
Evaluate $4\frac{2}{3}\div\left(-1\frac{2}{5}\right)$423÷(125), giving your answer as a mixed number in simplest form.
### Outcomes
#### 7.NS.1
Apply and extend previous understandings of addition and subtraction to add and subtract integers and other rational numbers; represent addition and subtraction on a horizontal or vertical number line diagram.
#### 7.NS.1d
d. Apply properties of operations as strategies to add and subtract rational numbers.
#### 7.NS.2
Apply and extend previous understandings of multiplication and division and of fractions to multiply and divide integers and other rational numbers.
#### 7.NS.2c
c. Apply properties of operations as strategies to multiply and divide rational numbers.
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Math Calculators, Lessons and Formulas
It is time to solve your math problem
« Basic Concepts
Analytic Geometry: (lesson 2 of 4)
## Lines
Only two pieces of information are needed to completely describe a given line. However, we have some flexibility on which two pieces of information we use
1. specifying the slope and the "y intercept", b, of the line (slope - intercept form).
2. Specifying the slope of the line and one point on the line (point slope form).
3. Specifying two points through which the line passes (two point form).
### Slope-Intercept Form
The most useful form of straight-line equations is the "slope-intercept" form:
y = mx + b
This is called the slope-intercept form because "m" is the slope and "b" gives the y-intercept. (That means the point (0,b) is where the line cross the y-axis.)
The Slope-Intercept Form of the equation of a straight line introduces a new concept, that of the y-intercept. The y-intercept describes the point where the line crosses the y-axis. (At this set of coordinates, the 'y' value is zero, and the 'x' value is the y-intercept.)
Example 1:
1. y = 5x + 7
2. y = -3x + 23
3. y = 2x (or y = 2x + 0)
### Point-Slope Form
The other format for straight-line equations is called the "point-slope" form. Suppose that we want to find the equation of a straight line that passes through a known point and has a known slope. For this one, they give you a point (x1, y1) and a slope m, and have you plug it into this formula:
y - y1 = m(x - x1)
Example 2:
1. y - 4 = -2(x - 1)
2. y - 8 = 3(x - 2)
3. y - 12 = 4(x - 3)
Example 3:
Find the equation of a line passing through the point (4, 2) and having a slope of 3.
Solution:
### Two Point Form
If two points are available we will use the two point form equation for a line,
The slope formula is
Example 4:
Find the equation of the line that passes through the points (2, 4) and (1, 2).
Solution:
Example 5:
Find the equation of a line through the points (1,2) and (3,1). What is its slope? What is its y intercept? Solution: We first find the slope of the line by finding the ratio of the change in y over the change is x. Thus
Now we can use the point - slope form to obtain:
### Standard Form
In the Standard Form of the equation of a straight line, the equation is expressed as:
Ax + By = C
where A and B are not both equal to zero.
1. 7x + 4y = 6
2. 2x - 2y = -2
3. -4x + 17y = -432
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# 3.3: Circles and Arrows 1
Difficulty Level: At Grade Created by: CK-12
Can you figure out the value of each letter in the diagram below? In this concept, we will use problem solving steps to help us find the value of letters in circles and arrows diagrams.
### Guidance
In order to solve the problem above, use the problem solving steps.
• Start by describing what you see in the diagram.
• Next, figure out what your job is in the problem. In all of these problems your job will be to figure out the value of the two letters in the diagram.
• Then, make a plan for how you will solve. There is usually more than one way to solve the problem. You will want to figure out one letter first and then the next letter.
• Next, solve the problem.
• Finally, check to make sure that the values you found work for all four of the arrows.
#### Example A
Figure out the value of each letter.
Solution:
We can use the problem solving steps to help us solve this problem.
\begin{align*}& \mathbf{Describe:} && \text{Two rows and two columns with numbers and letters.}\\ &&& \text{Arrows point to sums.}\\ & \mathbf{My \ Job:} && \text{Figure out the values of letters} \ C \ \text{and} \ D.\\ & \mathbf{Plan:} && \text{Start with the first column. Solve for} \ C.\\ &&& \text{Then figure out} \ D.\\ & \mathbf{Solve:} && C+C=6, \ \text{so} \ C=3,\\ &&& \text{In the first row,} \ C+D=4\\ &&& \text{Replace} \ C \ \text{with} \ 3.\\ &&& \text{Then} \ 3+D=4. \ D=4-3, \ \text{or} \ 1.\\ & \mathbf{Check:} && \text{Replace all} \ C\text{s with} \ 3.\\ &&& \text{Replace all} \ D\text{s with} \ 1.\\ &&& \text{Add rows and columns. Check the sums.}\\ &&& \text{Row} \ 1: 3+1=4 \qquad \text{Column} \ 1: 3+3=6\\ &&& \text{Row} \ 2: 3+2=5 \qquad \text{Column} \ 2: 1+2=3\end{align*}
#### Example B
Figure out the value of each letter.
Solution:
We can use the problem solving steps to help us solve this problem.
\begin{align*}& \mathbf{Describe:} && \text{Two rows and two columns with numbers and letters.}\\ &&& \text{Arrows point to sums.}\\ & \mathbf{My \ Job:} && \text{Figure out the values of letters} \ E \ \text{and} \ F.\\ & \mathbf{Plan:} && \text{Start with the second row. Solve for} \ F.\\ &&& \text{Then figure out} \ E.\\ & \mathbf{Solve:} && F+F=4, \ \text{so} \ F=2,\\ &&& \text{In the first column,} \ E+F=7\\ &&& \text{Replace} \ F \ \text{with} \ 2.\\ &&& \text{Then} \ E+2=7. \ E=7-2, \ \text{or} \ 5.\\ & \mathbf{Check:} && \text{Replace all} \ E\text{s with} \ 5.\\ &&& \text{Replace all} \ F\text{s with} \ 2.\\ &&& \text{Add rows and columns. Check the sums.}\\ &&& \text{Row} \ 1: 5+1=6 \qquad \text{Column} \ 1: 5+2=7\\ &&& \text{Row} \ 2: 2+2=4 \qquad \text{Column} \ 2: 1+2=3\end{align*}
#### Example C
Figure out the value of each letter.
Solution:
We can use the problem solving steps to help us solve this problem.
\begin{align*}& \mathbf{Describe:} && \text{Two rows and two columns with numbers and letters.}\\ &&& \text{Arrows point to sums.}\\ & \mathbf{My \ Job:} && \text{Figure out the values of letters} \ G \ \text{and} \ H.\\ & \mathbf{Plan:} && \text{Start with the second column. Solve for} \ G.\\ &&& \text{Then figure out} \ H.\\ & \mathbf{Solve:} && G+G=8, \ \text{so} \ G=4,\\ &&& \text{In the first column,} \ 3+H=4\\ &&& \text{Then} \ H=4-3, \ \text{or} \ 1.\\ & \mathbf{Check:} && \text{Replace all} \ H\text{s with} \ 1.\\ &&& \text{Replace all} \ G\text{s with} \ 4.\\ &&& \text{Add rows and columns. Check the sums.}\\ &&& \text{Row} \ 1: 3+4=7 \qquad \text{Column} \ 1: 3+1=4\\ &&& \text{Row} \ 2: 1+4=5 \qquad \text{Column} \ 2: 4+4=8\end{align*}
#### Concept Problem Revisited
We can use the problem solving steps to help us solve this problem.
\begin{align*}& \mathbf{Describe:} && \text{Two rows and two columns with numbers and letters.}\\ &&& \text{Arrows point to sums.}\\ & \mathbf{My \ Job:} && \text{Figure out the values of letters} \ A \ \text{and} \ B.\\ & \mathbf{Plan:} && \text{Start with the first row. Solve for} \ A.\\ &&& \text{Then figure out} \ B.\\ & \mathbf{Solve:} && A+A=4, \ \text{so} \ A=2,\\ &&& \text{In the first column,} \ A+B=5\\ &&& \text{Replace} \ A \ \text{with} \ 2.\\ &&& \text{Then} \ 2+B=5. \ B=5-2, \ \text{or} \ 3.\\ & \mathbf{Check:} && \text{Replace all} \ A\text{s with} \ 2.\\ &&& \text{Replace all} \ B\text{s with} \ 3.\\ &&& \text{Add rows and columns. Check the sums.}\\ &&& \text{Row} \ 1: 2+2=4 \qquad \text{Column} \ 1: 2+3=5\\ &&& \text{Row} \ 2: 3+6=9 \qquad \text{Column} \ 2: 2+6=8\end{align*}
### Vocabulary
In math, an unknown is a letter that stands for a number that we do not yet know the value of. In this concept, when you figured out the value of the letters in the circles and arrows diagrams you were solving for unknowns.
### Guided Practice
Figure out the value of each letter in each circles and arrows diagram.
1.
2.
3.
Answers:
1. \begin{align*}J = 1, K = 5\end{align*}
2. \begin{align*}L = 2, M = 3\end{align*}
3. \begin{align*}N = 5, P = 3\end{align*}
### Practice
Figure out the value of each letter in each circles and arrows diagram.
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# Mathematical properties of establishment in integers
Perhaps the best thing, before moving forward on a definition of each of the mathematical properties that can be found in the Establishment of integers, is to briefly review some concepts, which will allow you to understand these laws in your mathematical context Precise.
## Fundamental definitions
In this way, it may also be useful to focus this theoretical review on two specific notions: Integers and Integer Radiation, because they are directly related to the nature of the numerical elements and the operation mathematics on the basis of which each of these properties arises. Here’s each one:
## The Whole Numbers
In this sense, it will begin by saying then that mathematics has defined integers as those numerical elements with which you can account or represent whole or exact amounts, that is, that in them don´t take place neither the amounts fractional, or those with decimal expressions.
Similarly, this discipline has pointed out that integers can also be identified as the elements that make up the eponymous numeric set, which is also known as the Z numerical set, a collection where the integers grouped as follows:
Positive integers: First, positive integers can be found as a subset of Z, a grouping that will also be known as set N, or set of Natural Numbers. These elements will be characterized by being located in the number line to the right of zero, from where they will extend from 1 to the . They can count the elements of a grouping, or express an accounting quantity.
Negative integers: on the other hand, within the Z set you will also find the negative integers, which will be understood as the inverses of the positive numbers, which is why they will be found in the number line to the left of the zero, from where it is range from -1 to -O. Thanks to negative integers, the Z set can be used when accounting for debt or lack of a specific amount.
Zero: Finally, zero will also be part of the natural numbers set. However, this element will not be considered a number as such, but as the total absence of quantity. Therefore, it isn´t assumed as negative and positive, while being conceived as inverse of itself.
## Integer establishment
In another order of ideas, it will also be necessary to make a brief revision to the definition of Radication in whole numbers, which has been explained by Mathematics as an operation in which two integers try to determine a third, which complies with the ownership that being elevated to one of them, results in the other, hence the Establishment of integers is also seen by some authors as an inverse form of Empowering.
Thus, this discipline states that the Establishment of integers will be established between three numerical elements, each defined in turn as follows:
Index: this will be one of the numbers on which the Radication operation will be established. Your mission will be to point out to the root how many times you must multiply yourself, in order to result in the Radicando.
Radicando: for its part, this element will be the second on which the Radication operation will be established. It will serve to point out to the root what the correct result should be, as long as it is raised to the index.
Root: Finally, the Root will be interpreted as the final result of the operation, that is, as the number that will have the quality that once raised to the Index, results in the Radicando.
## Properties of the Establishment of Integers
With these definitions in mind, it may then be much easier to address an explanation of each of the two mathematical laws that can be found in reference to the Establishment of whole numbers, and which have been explained in turn from the next Way:
## The two signs of square roots
In this sense, the first mathematical property that will be found in the Establishment of integers will be that always and without exception the square roots of integers will have two signs, since whether it is negative, or positive, when raised to the square will result in a positive establishment. An example of this property may be the following:
√25 = 5 → 52 = 25
√25 = -5 → -5 = -5 . -5= 25
## The positive character of the square root establishment
Likewise, the Mathematics notes that in the operation of Radication of whole numbers the property indicating that always and in any case the establishment of square roots must be a positive integer, as well as non-zero, since it is considers it impossible, mathematically speaking a file operation in which the establishment is negative, that is, that there is no or no viable solution, since there is no number that raised squared can result in a negative whole number .
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## Finding Volume of Right Rectangular Prism: A 6th Grade Lesson Plan
written by: Donna Ventura • edited by: Carly Stockwell • updated: 2/3/2014
Your 6th grade class will investigate the concept of volume of right rectangular prisms.
• slide 1 of 5
Introduction: Find volumes of right rectangular prisms by filling prisms with unit cubes with fractional edge lengths. Students explore the connection between filling a prims with unit cubes and the volume formula.
Lesson Objective: The lesson is aligned to the Common Core State Standards for Mathematics – 6.G.2 Geometry – Find the volume of a right rectangular prism with fractional edge lengths by packing it with unit cubes of the appropriate unit fraction edge lengths, and show that the volume is the same as would be found by multiplying the edge lengths of the prism.
Materials Required: Unit Cubes, Calculator
• slide 2 of 5
### Lesson Procedure: Finding the Volume of a Right Rectangular Prism
Look at the right rectangular prism on the right. The measurement of the length is 2¼ inches. The measurement of the width is 1/2 inch. The measurement of the height is 1 inch.
Change the dimensions of the right rectangular prism to improper-like fractions. The dimensions are 9/4, 2/4, and 4/4 inches.
To find the volume of this right rectangular prism, fill it with ¼ inch cubes. It takes 64 – ¼ inch cubes to fill one unit cube.
The right rectangular prism can be filled with ¼ inch cubes. It takes 9 x 2 x 4 cubes to fill this right rectangular prism. It takes 72 – ¼ inch cubes to fill this right rectangular prism.
The volume of the right rectangular prism is 9/4 x 2/4 x 4/4 = 72/64 or 1 8/64 or 1 1/8 cubic inches.
• slide 3 of 5
### Individual or Group Work:
Look at the right rectangular prisms below. Fill the prisms with unit cubes of the appropriate unit fraction edge lengths.
1. The measurement of the length is 2 inches. The measurement of the width is 1 inch. The measurement of the height is 3¾ inches. How many cubes with side lengths of ¼ would be needed to fill the prism? What is the volume of the prism? Show how you found your answers.
2. The measurement of the length is 1½ inches. The measurement of the width is ½ inch. The measurement of the height is 4 inches. How many cubes with side lengths of 1/2 inch would be needed to fill the prism? What is the volume of the prism? Show how you found your answers.
3. The measurement of the length is 4 1/2 inches. The measurement of the width is 1 ½ inch. The measurement of the height is 2 inches. How many cubes with side lengths of 1/2 inch would be needed to fill the prism? What is the volume of the prism? Show how you found your answers.
4. The measurement of the length is 3/4 inches. The measurement of the width is ½ inch. The measurement of the height is 2 inches. How many cubes with side lengths of 1/4 inch would be needed to fill the prism? What is the volume of the prism? Show how you found your answers.
5. The measurement of the length is 2½ inches. The measurement of the width is 3 ½ inch. The measurement of the height is 4 inches. How many cubes with side lengths of 1/2 inch would be needed to fill the prism? What is the volume of the prism? Show how you found your answers.
• slide 4 of 5
1. It takes 8 x 4 x 15 cubes, or 480 cubes with side lengths of ¼ inch. The volume of the right rectangular prism is 480/64 = 7 ½ cubic inches or 2 x 1 x 3 ¾ = 7 ½ cubic inches.
2. It takes 3 x 1 x 8 cubes, or 24 cubes with side lengths of ½ inch. The volume of the right rectangular prism is 24/8 = 3 cubic inches or 1 ½ x ½ x 4 = 3 cubic inches.
3. It takes 9 x 3 x 4 cubes, or 108 cubes with side lengths of ½ inch. The volume of the right rectangular prism is 108/8 = 13 ½ cubic inches or 4 ½ x 1 1/2 x 2 = 13 1/2 cubic inches.
4. It takes 3 x 2 x 8 cubes, or 48 cubes with side lengths of ¼ inch. The volume of the right rectangular prism is 48/64 = ¾ cubic inches or ¾ x ½ x 2 = ¾ cubic inches.
5. It takes 5 x 7 x 8 cubes, or 280 cubes with side lengths of ½ inch. The volume of the right rectnagular prism is 280/8 = 35 cubic inches or 2 ½ x 3 ½ x 4 = 35 cubic inches.
• slide 5 of 5
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### Mean Geometrically
A and B are two points on a circle centre O. Tangents at A and B cut at C. CO cuts the circle at D. What is the relationship between areas of ADBO, ABO and ACBO?
### Golden Triangle
Three triangles ABC, CBD and ABD (where D is a point on AC) are all isosceles. Find all the angles. Prove that the ratio of AB to BC is equal to the golden ratio.
### Sangaku
The square ABCD is split into three triangles by the lines BP and CP. Find the radii of the three inscribed circles to these triangles as P moves on AD.
# So Big
### Why do this problem?
At first sight there might not seem to be enough information given in this question to find a solution. The first step is to interpret what is known from the geometry given the inscribed circle, and then to evaluate that information in terms of how it might be used to express what is known in terms of the three variables $a$, $b$ and $r$ and how that might be used to give an expression for the area of the triangle.
One of the methods of solving this problem is an application of the formula for tan$(A + B)$ combined with the formula for the area of a triangle.
### Possible approach
Set this as an exercise in applying the formula for $tan(A + B)$ or better still, if you want the students to revise their trigonometry, you can make it a more challenging problem solving activity and leave it to the students to decide what they have to use.
### Key questions
Can we make a list of everything that is known from the information given?
Can we see a way of using the information given to find the area of the triangle?
Can we add lines to make pairs of congruent triangles?
Can we find the area of the whole triangle in terms of these smaller congruent triangles?
What can we say about the angles at the centre of the circle?
If three of the angles at the centre add up to $180$ degrees what can we say about the tangents of these angles?
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MATHEMATICS Learner’s Study and Revision Guide for Grade 12 LOGARITHMS
Revision notes and exercises by Roseinnes Phahle
Preparation for the Mathematics examination brought to you by Kagiso Trust
Contents Unit 3 Logarithms: meaning and definition
3
Numbers that are not powers of 10
4
Relationship between logs and index laws
4
Laws of logarithms
5
More examples
5
Exerc ise 3.1
8
Applications of logarithms: Exercise 3.2
9
10
About this guide 1. This guide gives you a complete revision of logarithms: their meaning, how to evaluate them using an electronic calculator, simplifying expressions involving logarithms, solving equations requiring the use of logarithms and real life examples of the applications of logarithms. 2. You can work out your solutions to the examples in the blank spaces provided. 3. Don’t work alone. Work through this guide in a team with your classmates.
Logarithms
REVISION UNIT 3: LOGARITHMS The logarithm of a number is the index (power) to which a base must be raised to give that number. As an example, 16 = 2 4 , this means that 4 is the logarithm of 16 to base 2. In practice, the name Logarithm is abbreviated to Log:
16 = 2 4 which is in index form ↔ Log 2 16 = 4 which is in logarithmic form Here are further examples in which the interchange between index and logarithmic forms is shown:
1000 = 10 3
Log10 1000 = 3
100 = 10 2
Log10 100 = 2
10 = 101
Log10 10 = 1
When the base is 10 we need not show it in writing so that we simply write:
Log1000 = 3 Log100 = 2 Log10 = 1 Evaluating logs using a calculator You can verify all of the above results using the Log key on a calculator. For example, to find the log of 1000 press the following keys in the sequence shown:
What do you get?
log 1000 ) = The log key on the calculator works only when the base is 10.
We need not use a calculator to find the logs of numbers that are powers of 10 as in the above examples. The logs of such numbers are simply the powers to which these numbers are raised when written to base 10. Thus log 1 000 000 = 6 because 1 000 000 = 10 6 . Check this on your calculator by keying in log 1 000 000 = 3
Preparation for the Mathematics examination brought to you by Kagiso Trust
Numbers that are not a power of 10 What happens though when we have to find the log of a number that is not a simple power of 10? In this case you have no choice but to use a calculator. Example:
log 5342 = 3,7277 to 4 places of decimals. Check this on your calculator.
Example:
Thus in index form, the above is 5342 = 10 3, 7277
The antilog You can check the last example by using what is known as the anti-log key given by the sequence: SHIFT
10 x Example:
Evaluate the anti-log of 3,7277
Solution:
On your calculator enter the following keys in the sequence shown: SHIFT
10 x 3,7277 = What do you get? Relationship between logs and index laws As we have seen, a logarithm is simple the power or index to which a number is raised given a base. There is therefore a relationship between the laws governing logarithms and indices. Example:
368 × 0,583 = 10 2,5658 × 10 −0, 2343 writing each number in index form = 10 2,5658+ (−0, 2343 ) adding indices because we are multiplying = 10 2,5658−0, 2343 = 10 2,3315
What this implies is that
log(368 × 0,583) = log 368 + log 0,583 log(214,544 ) = 2,5658 + (− 0,2343) 2,3315 = 2,3315
Logarithms
To finish the multiplication, do the antilog of 2,3315 SHIFT
10 x 2,3315 = What do you get? First law of logarithms From the foregoing, we can deduce that log AB = log A + log B Second law of logarithms In a similar way we can illustrate the second law which is
log
A = log A - log B B
Third law of logarithms The third law is logA x = xlogA
log12 3 = 3 log 12
Example:
= 3× 1,07918 = 3,2375 Check this answer by working out log12 3 = log 12 × 12 × 12 = log 1728 = What do you get? More examples 1. Log ABCD = log A + log B Log C + log D 2.
log
AB = log A + log B - log C C
3.
log
1 1 = log 3 = log 10 −3 = −3 log 10 = −3 × 1 = −3 1000 10
5
Preparation for the Mathematics examination brought to you by Kagiso Trust
RECALL: Only when the base is 10 that we need not write it in. Otherwise when the base is not 10 we can convert to base 10 by applying the change of base formula which is at the top of the next page. What you should know Convert from exponential (index) to logarithmic form
Convert from logarithmic to exponential (index) form
Example
Work out the solutions in the boxes below
5 3 = 125
Log 1 81 = −4 3
log A A = 1
Simplify log 8 8 + 2log 3 3
log A 1 = 0
Simplify 5log16 1
log n A p = plog n A
Simplify log 5 25
Simplify logarithms
log10 100 + log 6 216 + log 3
1 27
Logarithms
What you should know Change of base:
logA log B A = logB
log B A =
1 log A B
log 1 A = −log B A B
Example
Work out the solutions in the boxes below
Use a calculator to evaluate
log 9 17
Write in an alternative form:
log 2 9
Write in an alternative form and simplify: log 1 32 2
Solve equations
Solve inequalities
Solve 6 x = 12 (Take logs on both sides and use a calculator)
log 1 x ≥ 4 3
(Note: logx is only defined for x > 0)
Inverse function NOTE: The inverse function is covered in UNIT 6 but can you at this stage answer this question? NOTE: The three forms in which the inverse function can be written.
Write the inverse of
f ( x ) = log a x in the forms:
y= 1.
f
−1
(x ) =
−1
:x→
and
f
7
Preparation for the Mathematics examination brought to you by Kagiso Trust
The following exercise involves expressions and equations requiring a knowledge of logarithmic laws such as you have practiced above. EXERCISE 3.1 QUESTION 1 Simplify the following expressions: 1.1
log 3 3 − log 3 81 + log 3 1
1.2
1 log 1 243 + log 2 64 4 3
1.3
log k k 5 − 2log k k
1.4
log x
1 1 + log 1 x + log m m + log x x + log 1 x x x x
QUESTION 2
QUESTION 3
Solve the following equations:
Solve the following to 4 decimal places:
2.1
x = log 2 128
3.1
2x = 7
2.2
log 3 x = 4
3.2
43x = 9
2.3
1 log 2 = 3 x 64
3.3
5 2 x +1 = 17
2.4
log x 8 = 2
3.4
2.(3 x ) = 2,38
2.5
logx + log5 = log( x + 16 )
3.5
3.(2 2 x ) = 0,5
2.6
log 3 x + log 3 (x + 6 ) = 3
3.6
x = log 5 8
2.7
log 2 x( x + 2 ) = 3
3.7
25(1 + 0,025) x = 100 x
2.8
log( x + 7 ) = 2log( x + 1)
3.8
1 1 3 = 3 3 3
2.9
log(2 x + 1) − log(x − 1) = 1
3.9
2 x .7 x = log14
Logarithms
APPLICATIONS OF LOGARITHMS EXERCISE 3.2 QUESTION 1 The formula for compound interest is A = P(1 + i )n , where A is the amount after n interest periods, P the initial amount or principal, and i the interest per period. How many years will it take for money to treble when it is compounded annually at 5,5% per year? Give your answer in years and correct to the nearest month. (Hint: Let A = 3P .) QUESTION 2 In the above problem after how many years will it take for the money to treble if it is compounded monthly at the same rate of interest per year? Give your answer in years and correct to the nearest month. (Hint: i = interest rate divided by number of months in the year and n will then be number of months.) QUESTION 3 The animal population of a certain game reserve is estimated to be falling at a yearly rate of 9% of its total population at the beginning of each year. Determine after how long it will take for the population to decrease to 1500 if the population is estimated at 35 000 at the beginning of a particular year. Give your answer in years and correct to the nearest month. QUESTION 4 A steel bar whose temperature is To =1500 C will obviously undergo a fall in temperature if it is immersed in a liquid whose temperature is Tl = 18 C . Given that the temperature T of a hot object placed into a cooling liquid is calculated by using the formula T = Tl + (To − Tl )10 −0, 08t where t is the time in minutes, how long will it take for the bar to cool to 35 C ? Give your answer in minutes and correct to the nearest second. QUESTION 5 In a chrome electroplating process, the mass m in grams of the chrome plating increases according to the formula m = 200 − 2 t 2 , where t is the time in minutes. How many minutes does it take to form 120 g of the plating? Give your answer in minutes and correct to the nearest second.
9
Preparation for the Mathematics examination brought to you by Kagiso Trust
ANSWERS EXERCISE 3. 1 QUESTION 1 1.1 -2 1.2 2 1.3 3 1.4 1
QUESTION 2 2.1 x = 7 2.2 x = 243 2.3 x = −2 2.4 x = 3 2.5 x = 4 2.6 x = 3 (reject x = −9 because negative numbers do not have logarithms) 2.7 x = 2 (similarly reject x = −4 ) 2.8 x = 2 (similarly reject x = −3 ) 2.9 x =
11 8
QUESTION 3 3.1 x = 2,8074 3.2 x = 0,5283 3.3 x = 0,3802 3.4 x = 0,1583 3.5 x = −1,2925 3.6 x = 1,2920 3.7 x = 56,1421 3.8 x = −0,0959 3.9 x = 0,0517
EXERCISE 3.2 1. n = 20years 6 months 2. n = 20 years 0 months 3. n = 33 years 5 months 4. t = 24 minutes 15 seconds 5. t = 12 minutes 39 seconds
KT Classroom Unit 3: Logarithms
Kagiso Trust's KT Classroom: Unit 3_Logarithms
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# 7.4Finite Population Correction Factor
Introductory Business Statistics 2e7.4 Finite Population Correction Factor
We saw that the sample size has an important effect on the variance and thus the standard deviation of the sampling distribution. Also of interest is the proportion of the total population that has been sampled. We have assumed that the population is extremely large and that we have sampled a small part of the population. As the population becomes smaller and we sample a larger number of observations the sample observations are not independent of each other. To correct for the impact of this, the Finite Correction Factor can be used to adjust the variance of the sampling distribution. It is appropriate when more than 5% of the population is being sampled and the population has a known population size. There are cases when the population is known, and therefore the correction factor must be applied. The issue arises for both the sampling distribution of the means and the sampling distribution of proportions. The Finite Population Correction Factor for the variance of the means shown in the standardizing formula is:
$Z= x¯ − µ σ n · N−n N−1 Z= x¯ − µ σ n · N−n N−1$
and for the variance of proportions is:
$σp' = p(1−p) n × N−n N−1 σp'= p(1−p) n × N−n N−1$
The following examples show how to apply the factor. Sampling variances get adjusted using the above formula.
## Example 7.1
It is learned that the population of White German Shepherds in the USA is 4,000 dogs, and the mean weight for German Shepherds is 75.45 pounds. It is also learned that the population standard deviation is 10.37 pounds.
### Problem
If the sample size is 100 dogs, then find the probability that a sample will have a mean that differs from the true probability mean by less than 2 pounds.
## Try It 7.1
For the 5,000 students in University A, the average height of the students is 170 cm. The population standard deviation is 5 cm.
If the sample size is 100 students, find the probability that the sample mean is within 1 cm of the true mean.
## Example 7.2
When a customer places an order with Rudy's On-Line Office Supplies, a computerized accounting information system (AIS) automatically checks to see if the customer has exceeded their credit limit. Past records indicate that the probability of customers exceeding their credit limit is .06.
### Problem
Suppose that on a given day, 3,000 orders are placed in total. If we randomly select 360 orders, what is the probability that between 10 and 20 customers will exceed their credit limit?
## Try It 7.2
There is a 0.01 probability that a student gets food poisoning eating food in a cafeteria. Suppose there are 2,000 orders in a day in the cafeteria. If we select 200 orders randomly, what is the probability that between 5 and 10 students will get food poisoning?
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# The Angle Measure of an Arc
### New York State Common Core Math Geometry, Module 5, Lesson 7
Student Outcomes
• Define the angle measure of arcs, and understand that arcs of equal angle measure are similar.
• Restate and understand the inscribed angle theorem in terms of arcs: The measure of an inscribed angle is half the angle measure of its intercepted arc.
• Explain and understand that all circles are similar.
The Angle Measure of an Arc
Classwork
Opening Exercise
If the measure of β πΊπ΅πΉ is 17Β°, name three other angles that have the same measure and explain why.
What is the measure of β πΊπ΄πΉ? Explain.
Can you find the measure of β π΅π΄π·? Explain.
Example
What if we started with an angle inscribed in the minor arc between π΄ and πΆ?
Exercises
1. In circle π΄, ππ΅πΆ : ππΆπΈ : ππΈπ· : ππ·π΅ = 1: 2: 3: 4. Find the following angles of measure.
a. πβ π΅π΄πΆ
b. πβ π·π΄πΈ
c. ππ·π΅
d. ππΆπΈπ·
2. In circle π΅, π΄π΅ = πΆπ·. Find the following angles of measure.
a. ππΆπ·
b. ππΆπ΄π·
c. ππ΄πΆπ·
3. In circle π΄, π΅πΆ is a diameter and πβ π·π΄πΆ = 100Β°. If ππΈπΆ = 2ππ΅π· , find the following angles of measure.
a. πβ π΅π΄πΈ
b. ππΈπΆ
c. ππ·πΈπΆ
4. Given circle π΄ with πβ πΆπ΄π· = 37Β°, find the following angles of measure.
a. ππΆπ΅π·
b. πβ πΆπ΅π·
c. πβ πΆπΈπ·
Lesson Summary
Theorems:
• INSCRIBED ANGLE THEOREM: The measure of an inscribed angle is half the measure of its intercepted arc.
• Two arcs (of possibly different circles) are similar if they have the same angle measure. Two arcs in the same or congruent circles are congruent if they have the same angle measure.
• All circles are similar
Relevant Vocabulary
• ARC: An arc is a portion of the circumference of a circle.
• MINOR AND MAJOR ARC: Let πΆ be a circle with center π, and let π΄ and π΅ be different points that lie on πΆ but are not the endpoints of the same diameter. The minor arc is the set containing π΄, π΅, and all points of πΆ that are in the interior of β π΄ππ΅. The major arc is the set containing π΄, π΅, and all points of πΆ that lie in the exterior of β π΄ππ΅.
• SEMICIRCLE: In a circle, let π΄ and π΅ be the endpoints of a diameter. A semicircle is the set containing π΄, π΅, and all points of the circle that lie in a given half-plane of the line determined by the diameter.
• INSCRIBED ANGLE: An inscribed angle is an angle whose vertex is on a circle and each side of the angle intersects the circle in another point.
• CENTRAL ANGLE: A central angle of a circle is an angle whose vertex is the center of a circle.
• INTERCEPTED ARC OF AN ANGLE: An angle intercepts an arc if the endpoints of the arc lie on the angle, all other points of the arc are in the interior of the angle, and each side of the angle contains an endpoint of the arc.
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
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# If $f\left( x \right) = \left\{ {\dfrac{{{x^2} + \propto }}{{2\sqrt {{x^2} + 1 + B} }}} \right\}$ ${\text{for}}\;x \geqslant 0 \\ {\text{for}}\;x < 0 \\$And $f\left( {\dfrac{1}{2}} \right) = 2$ is continuous at x =0, value of $\left( { \propto ,P} \right)$ is :A) $\left(\dfrac{7}{4}, -\dfrac{1}{4}\right)$B) $\left(4-\sqrt{5},2-\sqrt{5}\right)$C) $\left(0, -1\right)$D) $\left(\dfrac{7}{4}, \dfrac{1}{4}\right)$
Last updated date: 13th Jun 2024
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Hint: The function to be continuous at a particular point like say a, then we have a condition for a, where $f\left( x \right)$ function needs to have its left hand limit, Right hand limit and value equal to each other. Like $\Rightarrow \dfrac{{\lim }}{{x \to {a^ - }}}\;f\left( x \right) = f\left( a \right) = \;\dfrac{{\lim }}{{x \to {a^ + }}}f\left( x \right)$
Complete step by step solution: let’s begin with the given function which is represented as
$\begin{gathered} f\left( x \right) = \left\{ {\dfrac{{{x^2} + \propto }}{{2\sqrt {{x^2} + 1 + B} }}} \right\}\;\;\;\;\;\;\;\; \\ \\ \end{gathered}$ $\begin{gathered} x \geqslant 0 \\ x < 0 \\ \end{gathered}$
As we know that from given data,$f\left( {{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}} \right) = 2,$ so for this we will use the function $f\left( x \right) = {x^2} + \propto$ because ${\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}$ is greater than 0
So, $f\left( {\dfrac{1}{2}} \right) = \dfrac{1}{4} + \propto = 2$
$\Rightarrow \; \propto \; = 2 - \dfrac{1}{4} = \dfrac{7}{4}$
Now we need to find the value of B, and the other information given is they are continuous at x = 0 for being continuous of a function the left hand limit, Right hand limit should be equal.
$so\dfrac{{\;\lim }}{{x \to {0^ - }}}f\left( x \right) = \dfrac{{\lim }}{{x \to {0^ + }}}f\left( x \right)$
$\Rightarrow \dfrac{{\lim }}{{x \to {0^ - }}}\;\left( {{x^2} + \propto } \right) = \dfrac{{\lim }}{{x \to {0^ + }}}2\sqrt {{x^2} + 1} + B)$
$\Rightarrow \lim {\left( 0 \right)^2} + \propto \; = 2\sqrt {{0^2} + } 1 + B$
$\propto - 2 + B$
So we get,$B = \; \propto - 2\;and \propto = {\raise0.5ex\hbox{$\scriptstyle 7$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 4$}}$
$\Rightarrow$ we get $B = \; \propto - 2 = \dfrac{7}{4} - 2 = - \dfrac{1}{4}$
Hence we get the value of $\left( { \propto ,B} \right) = \left( {{\raise0.5ex\hbox{\scriptstyle 7} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle {4,}}}{\raise0.5ex\hbox{\scriptstyle { - 1}} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 4}}} \right)$ option A is the correct answer.
Note: we know that function to get continuous left-hand limit, Right hand limit and function value should be equal.
$\dfrac{{\lim }}{{x \to {a^ - }}}f\left( x \right) = f\left( a \right) = \dfrac{{\lim }}{{x \to {a^ + }}}f\left( x \right)$ Similarly, for differentiability of a function at point a is checked by
$\dfrac{{\lim }}{{x \to {a^ - }}}f\left( x \right) = f'\left( a \right) = \dfrac{{\lim }}{{x \to {a^ + }}}f'\left( x \right)$
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# How to Use thePower Rule of Exponents
## Summary
The formalized definition of the power rule of exponents is...
$$(x^{a})^{b} = x^{ab}$$
Basically, it tells us that if we have a power raised to another exponent, we can multiply the two exponents to simplify the expression. You can also use this rule in reverse.
The power rule of exponents is also known as the "power property of exponents" or the "powers of powers" rule.
## How to Use the Power Rule of Exponents
1. Identify the exponent that is being raised to another exponent.
2. Multiply the exponents.
3. Write the simplified power with the result of Step 2 as the exponent of the original base.
## Examples
$$(2^{3})^{4}=2^{12}$$
$$(7^{5})^{3}=7^{15}$$
$$(y^{9})^{2}=y^{18}$$
## Why Does It Work?
To understand why the power rule of exponents works, let's calculate $$(2^{3})^{4}$$ with order of operations instead of using the power rule.
$$(2^{3})^{4}$$
Following the order of operations, I'll simplify the stuff inside parentheses first. The exponent (3) tells me that I need to multiply 2 by itself 3 times.
$$(2 \times 2 \times 2)^{4}$$
The exponent outside the parentheses (4) tells me that I need to multiply all the stuff inside the parentheses by itself 4 times.
$$(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times2 \times 2) \times (2 \times2 \times 2)$$
So, when $$(2^{3})^{4}$$ is fully expanded, we have twelve 2's multiplied by each other.
We can rewrite that as $$2^{12}$$.
So, that is why...
$$(2^{3})^{4}= 2^{12}$$
The reason you multiply the exponents is because the outside exponent multiplies the effect of the inside exponent. This is also true when variables are involved.
$$(y^{9})^{2}$$
Following the order of operations, I'll simplify the stuff inside parentheses first. The exponent (9) tells me that I need to multiply y by itself 9 times.
$$(yyyyyyyyy)^{2}$$
The exponent outside the parentheses (2) tells me that I need to multiply the stuff inside the parentheses by itself 2 times.
$$(yyyyyyyyy)(yyyyyyyyy)$$
There are 18 y's multiplied together, which is $$y^{18}$$.
So, that is why...
$$(y^{9})^{2}= y^{18}$$
## Reverse Power Rule of Exponents
Sometimes, it can be helpful to use the reverse power rule of exponents to write a power with one exponent as a power of powers.
$$x^{ab}=(x^{a})^{b}$$
For example, if you were trying to factor perfect squares or perfect cubes, you may want to write $$x^{30}$$ like this...
$$x^{30}=(x^{15})^{2}$$
OR
$$x^{30}=(x^{10})^{3}$$
Or if you were trying to rewrite rational exponents as roots, you may want to write $$x^{\frac{3}{2}}$$ like this...
$$x^{\frac{3}{2}}=$$$$(x^{\frac{1}{2}})^{3}$$
OR
$$x^{\frac{3}{2}}=$$$$(x^{3})^{\frac{1}{2}}$$
This would be helpful because an exponent of $$\frac{1}{2}$$ is the same thing as a square root
### What if I have multiple exponents?
#### Show/Hide
If you have multiple exponents (as powers within powers), you should multiply all of the exponents.
$$[(3^{4})^{7}]^{2}=3^{56}$$
$$[(x^{2})^{5}]^{6}=x^{60}$$
$$[(y^{3})^{7}]^{4}=y^{84}$$
I'll simplify an example using order of operations, so you can see WHY we multiply all of the exponents...
$$[(8^{2})^{4}]^{3}$$
I'll start by simplifying the stuff inside the inner-most set of parentheses. The exponent (2) tells me to multiply 8 by itself two times.
$$[(8\times8)^{4}]^{3}$$
The next exponent (4) tells me to multiply the stuff inside the inner parentheses by itself 4 times.
$$[(8\times8)\times(8\times8)\times(8\times8)\times(8\times8)]^{3}$$
The last exponent tells me to multiply the stuff inside the outermost parentheses by itself 3 times.
$$[(8\times8)\times(8\times8)\times(8\times8)\times(8\times8)]\times$$
$$[(8\times8)\times(8\times8)\times(8\times8)\times(8\times8)]\times$$
$$[(8\times8)\times(8\times8)\times(8\times8)\times(8\times8)]$$
There are twenty-four 8's multiplied by each other, so…
$$[(8^{2})^{4}]^{3}=8^{24}$$
### What if I have multiple numbers in the base?
#### Show/Hide
If you have multiple numbers inside the parentheses, you can use the distributive property of exponents to distribute the outer exponent to each factor within the parentheses.
If any of the bases don't have an exponent written, remember that they have an "invisible" exponent of 1
$$(a^{2}b^{5})^{3}=a^{6}b^{15}$$
$$(2x^{7})^{3}=8x^{21}$$
$$(x^{2}y^{3}z^{4})^{5}=x^{10}y^{15}z^{20}$$
I'll simplify an example with order of operations, so you can see WHY we distribute the exponent to each factor inside the parentheses...
$$(x^{3}y^{4}z)^{2}$$
I'll simplify the inside of the parentheses first. The exponent (3) tells me that I need to multiply x by itself 3 times. The exponent (4) tells me that I need to multiply y by itself 4 times. The z has an "invisible" exponent of 1, so it is multiplied once.
$$(xxxyyyyz)^{2}$$
The outer exponent tells me that I need to multiply the stuff inside parentheses by itself 2 times.
$$(xxxyyyyz)(xxxyyyyz)$$
There are 6 x’s, 8 y’s, and 2 z's multiplied together, so…
$$(x^{3}y^{4}z)^{2}=x^{6}y^{8}z^{2}$$
### What if the base is a fraction?
#### Show/Hide
If you have a fraction inside the parentheses, you can use the division version of the distributive property of exponents to distribute the outside exponent to the numerator and denominator of the fraction.
$$(\frac{7^2}{5^3})^4 = \frac{7^{8}}{5^{12}}$$
$$(\frac{a^6}{9^4})^2 = \frac{a^{12}}{9^{8}}$$
$$(\frac{x^{12}}{y^3})^5 = \frac{x^{60}}{y^{15}}$$
In this example, you can see WHY we distribute the exponent to the numerator and denominator...
$$(\frac{x^2}{y^5})^4$$
I'll simplify the inside of the parentheses first. The exponent (2) tells me that I need to multiply x by itself 2 times. The exponent (5) tells me that I need to multiply y by itself 5 times.
$$(\frac{xx}{yyyyy})^{4}$$
The next exponent tells me I need to multiply the stuff inside the parentheses by itself 4 times.
$$(\frac{xx}{yyyyy})(\frac{xx}{yyyyy})(\frac{xx}{yyyyy})(\frac{xx}{yyyyy})$$
There are 8 x’s in the numerator and 20 y’s in the denominator, so...
$$(\frac{x^2}{y^5})^4 = \frac{x^{8}}{y^{20}}$$
### What if the exponent is negative?
#### Show/Hide
If one (or more) of the exponents is negative, multiply the exponents together using the rules for multiplying negative numbers
$$(2^{-4})^{3}= 2^{-12}$$
$$(x^{4})^{-7}= x^{-28}$$
$$(y^{-3})^{-6}= y^{18}$$
In this example, you can see WHY we can multiply the exponents as negative numbers...
$$(5^{-3})^{2}$$
I'll start by simplifying the inside of the parentheses first. The (-3), tells me that I need to multiply 5 by itself 3 times in the denominator of a fraction.
$$(\frac{1}{5\times5\times5})^{2}$$
The next exponent (2) tells me that I need to multiply the stuff inside the parentheses by itself two times.
$$(\frac{1}{5\times5\times5})(\frac{1}{5\times5\times5})$$
There are six 5's multiplied in the denominator of a fraction, so...
$$(5^{-3})^{2}= 5^{-6}$$
## Free Online Practice Problems
Khan Academy - Powers of Powers
Khan Academy - Powers of Products & Quotients (Structured Practice)
Khan Academy - Powers of Products & Quotients (Integer Exponents)
Khan Academy - Powers of Products & Quotients
## Free Printable Worksheets
MathWorksheets4Kids - Power Rule of Exponents
Math Aids - Products to a Power Worksheet Generator
Math Aids - Quotients to a Power Worksheet Generator
To use the worksheet generator from Math Aids, select the options you want for your worksheet and then click the "Create It" button.
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# 2010 AMC 12A Problems/Problem 20
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
Arithmetic sequences $\left(a_n\right)$ and $\left(b_n\right)$ have integer terms with $a_1=b_1=1 and $a_n b_n = 2010$ for some $n$. What is the largest possible value of $n$?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 288 \qquad \textbf{(E)}\ 2009$
## Solution
### Solution 1
Since $\left(a_n\right)$ and $\left(b_n\right)$ have integer terms with $a_1=b_1=1$, we can write the terms of each sequence as
\begin{align*}&\left(a_n\right) \Rightarrow \{1, x+1, 2x+1, 3x+1, ...\}\\ &\left(b_n\right) \Rightarrow \{1, y+1, 2y+1, 3y+1, ...\}\end{align*}
where $x$ and $y$ ($x\leq y$) are the common differences of each, respectively.
Since
\begin{align*}a_n &= (n-1)x+1\\ b_n &= (n-1)y+1\end{align*}
it is easy to see that
$a_n \equiv b_n \equiv 1 \mod{(n-1)}$.
Hence, we have to find the largest $n$ such that $\frac{a_n-1}{n-1}$ and $\frac{b_n-1}{n-1}$ are both integers; equivalently, we want to maximize $\gcd(a_n-1, b_n-1)$.
The prime factorization of $2010$ is $2\cdot{3}\cdot{5}\cdot{67}$. We list out all the possible pairs that have a product of $2010$, noting that these are the possible values of $(a_n, b_n)$ and we need $a_n \leq b_n$:
$$(2,1005), (3, 670), (5,402), (6,335), (10,201),(15,134),(30,67)$$
and soon find that the largest $n-1$ value is $7$ for the pair $(15, 134)$, and so the largest $n$ value is $\boxed{8\ \textbf{(C)}}$.
### Solution 2
As above, let $a_n=(n-1)x+1$ and $b_n=(n-1)y+1$ for some $1\leq x\leq y$.
Now we get $2010 = a_n b_n = (n-1)^2xy + (n-1)(x+y) + 1$, hence $2009 = (n-1)( (n-1)xy + x + y )$. Therefore $n-1$ divides $2009 = 7^2 \cdot 41$. And as the second term is greater than the first one, we only have to consider the options $n-1\in\{1,7,41\}$.
For $n=42$ we easily see that for $x=y=1$ the right side is less than $49$ and for any other $(x,y)$ it is way too large.
For $n=8$ we are looking for $(x,y)$ such that $7xy + x + y = 2009/7 = 7\cdot 41$. Note that $x+y$ must be divisible by $7$. We can start looking for the solution by trying the possible values for $x+y$, and we easily discover that for $x+y=21$ we get $xy + 3 = 41$, which has a suitable solution $(x,y)=(2,19)$.
Hence $n=8$ is the largest possible $n$. (There is no need to check $n=2$ anymore.)
## Alternative Thinking
Since
$a_n*b_n = 2010,$
and
$a_n \le b_n$,
it follows that
$a_n \le \sqrt{2010} \Rightarrow a_n \le 44$.
But $a_n$ and $b_n$ are also integers, so $a_n$ must be a factor of $2010$ smaller than $44$. Notice that $2010 = 2*3*5*67$. Therefore $a_n = 2, 3, 5, 6, 10, 15,$ or $30$ and $b_n = 1005, 670, 402, 335, 201, 134,$ or $67$; respectively.
Notice that the term $a_m$ is equivalent to the first term $a_1 = 1$ plus $(m-1)$ times the common difference for that particular arithmetic sequence. Let the common difference of $(a_n)$ be $k$ and the common difference of $(b_n)$ be $i$ (not $\sqrt{-1}$). Then
$a_n$ (the $n$th term, not the sequence itself) $=1 + k(n-1)$
and
$b_n = 1 + i(n-1)$
Subtracting one from all the possible values listed above for $a_n$ and $b_n$, we get
$k(n-1) = 1, 2, 4, 5, 9, 14, 29$
and
$i(n-1) = 1004, 669, 401, 334, 200, 133, 66$
In order to maximize $n$, we must maximize $n-1$. Therefore $k$ and $i$ are coprime and $n-1$ is the GCF of any corresponding pair. Inspecting all of the pairs, we see that the GCF is always $1$ except for the pair $(14, 133),$ which has a GCF of $7$. Therefore the maximum value of $n$ is $8 \Rightarrow \boxed{\text{C}}$.
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## 1. Median of a triangle
Triangle
A triangle is a closed figure made by three line segments. It has three vertices, three sides and three angles.
A triangle ABC is given having three vertices A, B, C and three sides AB, BC, CA and three angles are ABC, BCA, CAB.
Classification of triangles:
1. With respect to angles are:
(a) Acute-angled triangles
(b) Obtuse-angled triangles
(c) Right-angled traingles
(a) Acute-angled triangles
A triangle in which measurement of each of the three angles are less than 90°.
(b) Obtuse-angled triangle :
A triangle in which one angle is obtuse angle i.e. greater than 90° but less then 180°.
(c) Right-angled triangle:
A triangle in which one angle is exactly 90°.
• The side opposite to the right angle is called the hypotenuse.
• The other two sides are called as the legs of the right-angled triangle (or base and perpendicular).
• In the given figure AC is the hypotenuse and AB and BC are the legs of DABC.
## 1. Median of a triangle
Chapter 6
Triangle and its properties
Median of a triangle
Median of a triangle is a line segment joining a vertex to the midpoint of the opposing side, bisecting it.
A median connects a vertex of a triangle to the mid-point of the opposite side.
In the ∆ ABC, the line segment AD joining the mid-point of BC to its opposite vertex A is called a median of the triangle.
Properties of Median of a Triangle
Every triangle has exactly three medians one from each vertex and they all intersect each other at the triangle's centroid.
• The 3 medians always meet at a single point, no matter what the shape of the triangle is.
• The point where the 3 medians meet is called the centroid of the triangle. Point O is the centroid of the triangle ABC.
• Each median of a triangle divides the triangle into two smaller triangles which have equal area.
• In fact, the 3 medians divide the triangle into 6 smaller triangles of equal area.
In ∆ ABC, three medians are AD, CE and BF and they are intersecting the point O which is centroid of the triangle.
## 2. Altitude of a triangle
Medians of a Triangle and Centroid
Medians :-
AB, BC, CA are the sides of and AD is a line segment which intersects side BC of at point D such that AD bisects the side BC. This line segment AD is called the median of .
A median is a line segment which connects a vertex of a triangle to the mid point of the opposite side.
Centroid :-
• Centroid is the point of intersection of all three medians of a triangle.
• In the given figure G is the centroid of
• A centroid of a triangle cuts the median in the ratio 2 : 1
AG : GD = 2 : 1
## 2. Altitude of a triangle
Altitude of a triangle
An altitude of a triangle is a line segment through a vertex and perpendicular to (i.e., forming a right angle with) a line containing the base (the side opposite the vertex).
In ∆ ABC, AD is the altitude of triangle ABC.
Through each vertex, an altitude can be drawn. So, there are at most three altitudes in a triangle.
Properties of Altitudes of a Triangle
• Every triangle has 3 altitudes, one from each vertex. AE, BF and CD are the 3 altitudes of the triangle ABC.
• The altitude is the shortest distance from the vertex to its opposite side.
• The 3 altitudes always meet at a single point, no matter what the shape of the triangle is.
• The point where the 3 altitudes meet is called the ortho-centre of the triangle. Point O is the ortho-centre of the triangle ABC.
• The altitude of a triangle may lie inside or outside the triangle.
## 3. Exterior angles of a triangle
ALTITUDES OF A TRIANGLE AND ORTHOCENTRE
AD is a line segment which is perpendicular to the side BC of ABC .This line segment AD is called as altitude of
Altitude is also called as height of a triangle. In the given figure AD is the height of ∆ ABC from vertex A to side BC.
• Orthocentre is the meeting point of all three altitudes of a triangle.
• In the given figure H is the orthocentre of ∆ ABC.
• In obtuse-angled triangle the orthocentre lies outside the triangle.
• In Acute-angled triangle the orthocentre lies inside the triangle.
• In a Right-angled triangle, the right angle lies at the vertex at which the right angle is formed.
## 3. Exterior angles of a triangle
Exterior angles of a triangle
An exterior angle of a triangle is equal to the sum of the opposite interior angles.
In the above figure, ACD is the exterior angle of the Δ ABC.
So, ACD = CAB + CBA
At each vertex of a triangle, an exterior angle of the triangle may be formed by extending one side of the triangle.
## 4. Apply angle sum property of a triangle
Exterior Angle Property of a Triangle -
Illustration 1
Is it possible to draw a triangle whose sides are 3 cm, 4 cm and 7 cm ?
Solution
Sides of a triangle are 3 cm, 4cm, 7 cm.
Here 3 + 4 = 7
and we know that sum of two sides of a triangles is always greater than the third side so it is not possible to draw a triangle whose sides are 3 cm, 4 cm and 7 cm.
Illustration 2
In each of the following there are three positive numbers. State if these numbers could possibly be the lengths of the sides of a triangle.
(i) 2, 3, 4 (ii) 2.5, 1.5, 4
Solution
(i) We have,
2 + 3 > 4, 2 + 4 > 3 and 3 + 4 > 2
That is, the sum of any two of the given numbers is greater than the third number.
So, 2 cm, 3 cm and 4 cm can be the lengths of the sides of a triangle.
(ii) We have,
2.5 + 1.5 4.
So, the given numbers cannot be the lengths of the sides of a triangle.
Illustration 3
The length of two sides of a triangle are 12 cm and 15 cm. Between what two measure should the length of the third side fall ?
Solution
Let x cm be the length of the third side.
Then 12 + x > 15; 15 + x > 12 and 12 + 15 > x.
x > 15 – 12; x > 12 –15 and 27 > x.
x > 3 ; x > – 3 and 27 > x.
A number greater than 3 is obviously greater than – 3.
x > 3 and 27 > x.
Hence, x lies between 3 cm and 27 cm.
Property : Angles opposite to equal sides of a triangle are equal.
Property : Sides opposite to equal angles of a triangle are equal.
Property : In a right triangle, if a, b are the lengths of the sides and c that of the hypotenuse, then c2 = a2 + b2.
(Hypotenuse)2 = (Base)2 + (Perpendicular)2
Property : If the sides of a triangle are of lengths a, b and c such that c2 = a2 + b2, then the triangle is right angled and the side of length c is the hypotenuse.
Note : Three positive numbers a, b, c in this order are said to form a pythagorean triplet, if
c2 = a2 + b2. Triplets (3, 4, 5) (5, 12, 13), (8, 15, 17), (7, 24, 25) and (12, 35, 37) are some pythagorean triplets.
Illustration 1
The sides of certain triangles are given below. Determine which of them are right triangles :
(i) a = 6 cm, b = 8 cm and c = 10 cm
(ii) a = 5 cm, b = 8 cm and c = 11 cm.
Solution
(i) Here the larger side is c = 10 cm.
We have : a2 + b2 = 62 + 82 = 36 + 64 = 100 = c2.
So, the triangle with the given sides is a right triangle.
(ii) Here, the larger side is c = 11 cm
Clearly, a2 + b2 = 25 + 64 = 89 c2.
So, the triangle with the given sides is not a right triangle.
Illustration 2
A ladder 25 m long reaches a window of a building 20 m above the ground. Determine the distance of the foot of the ladder from the building.
Solution
Suppose that AB is the ladder, B is the window and CB is the building. Then, triangle ABC is a right triangle, with right angle at C.
AB2 = AC2 + BC2
252 = AC2 + 202
AC2 = 625 – 400 = 225
Illustration 3
A ladder 17 m long reaches a window which is 8 m above the ground on one side of the street. Keeping its foot at the same point, the ladder is turned to the other side of the street to reach a window at a height of 15 m. Find the width of the street.
Solution
Let AB be the street and C be the foot of the ladder. Let D and E be the windows at the height of 8 m and 15m respectively from the ground.
Then, CD and CE are the two positions of the ladder.
In right triangle, DDAC, we have
= {(17)2 – (8)2} m2 = 289 – 64 = 225 m
In right triangle, DCBE, we have
CB2 + BE2 = CE2
CB2 = (CE2 – BE2) = {(17)2 – (15)2} m2 = 64 m2
Hence, width of the street = AB = (AC + CB) = (15 + 8) m = 23 m.
Illustration 4
A tree has broken at a height of 5 m from the ground and its top touches the ground at a distance of 12m from the base of the tree. Find the original height of tree.
Solution
Let AB be the tree and Let C be the point at which it broke.
Then CB takes the position CD.
Original height of tree i.e., AB
i.e. AC + BC AC + CD (
i.e. ACD, using pythagoras theorm, we
CD2 = (5)2 + (12)2 = 25 + 144 = 169
CD2 = 132
CD = 13 m
So. height of tree = AC + BC
= AC + CD
= (5+13) m
= 18 m
Hence, height of the tree = 18 m.
## 4. Apply angle sum property of a triangle
Apply angle sum property of a triangle
The sum of the measures of the three angles of a triangle is 180°
In Δ ABC,
A + B + C = 180°
## 5. Two special triangles:
Angle Sum Property of a Triangle:
The sum of the measures of three angles of a triangle is 180°.
Proof of Angle Sum Property of a Triangle
Given : A triangle ABC.
To prove : ∠ A +∠B +∠C = 180°
Construction : Draw a line segment PQ through A and parallel to BC.
Proof: Mark the angles as indicated in the figure.
Hence, ∠ A +∠B +∠C = 180°
or sum of the angles of a triangle is 180°
ANGLE BISECTORS OF A TRIANGLE AND IN-CENTRE
Angle Bisectors :-
Angle bisector of a triangle is a line segment which bisect the angle and whose end points lies on the vertex and its opposite side.
• In the given figure, AD is the bisector of
Incentre
• In-centre is the meeting point of all three angle bisectors of a triangle.
• In the given figure, I is the Incentre of
• In-centre always lies inside the triangle.
• The perpendicular distance from incentre to the side of triangle is always same.
IP = IQ = IR
• A circle can be drawn inside the triangle by assuming I as centre and IP as radius.
PERPENDICULAR BISECTOR AND CIRCUMCENTRE
Perpendicular Bisector
It is a line segment which is perpendicular to the side of a triangle and passing through the mid point of the side.
In the given figure MN is the perpendicular bisector for
## 5. Special triangles
2. With respect to sides are:
(a) Scalene triangles (b) Isosceles triangles (c) Equilateral triangles
(a) Scalene Triangle: A triangle in which no two sides are equal in length is called a scalene triangle.
• Length of all sides and measurement of all angles is different.
(b) Isosceles Triangle
(i) Two sides have same length
(ii) The angles opposite to the equal sides are equal.
If in DABC, AB = AC then B = C this is an isoceles triangle.
Converse is also true
If in a triangle two angles are equal then sides opposite to equal angles are equal in length.
In , if B = C then AB = AC.
(c) Equilateral Triangle
In an equilateral triangle
(i) All sides are equal in length.
(ii) Measurement of each angle is 60°.
In , if AB = BC = AC then ∠A = ∠B =∠C = 60°
or if then ∠A = ∠B = ∠C= 60° AB = BC = AC
## 5. Special triangles
Special triangles
Two Special Triangles: Equilateral and Isosceles
A triangle in which all the three sides are of equal lengths and each angle has measure 60o is called an equilateral triangle.
In an equilateral triangle,
AB = BC = CA
And ÐA = ÐB = ÐC = 600
A triangle in which two sides are of equal lengths is called an isosceles triangle
In triangle XYZ,
XY = XZ
Sum of the Lengths of Two Sides of a Triangle
The sum of the lengths of any two sides of a triangle is greater than the length of the third side.
In the above figure,
AB + BC > AC
Also, the difference between the lengths of any two sides of a triangle is smaller than the length of the third side.
## 6. Sides of a triangle
Perimeter of a Triangle
The sum of the lengths of the sides of a triangle is called its Perimeter.
In AC = b, BC = a, AB = c then perimeter is BC + CA + AB = a + b + c
TRIANGULAR INEQUALITY PROPERTY
The sum of the lengths of any two sides of a triangle is always greater than the third side.
AB + BC > AC
AB + AC > BC
AC + BC > AB
The difference between the length of any two sides of a triangle is smaller than the length of the third side.
AB – BC < CA
BC – CA < AB
CA – AB < BC
Example
XY + YZ > XZ
3 + 4 > 5
similary YZ + XZ > XY
4 + 5 > 3
similary XY + XZ > YZ
3 + 5 > 4
## 6. Sides of a triangle
Sides of a triangle
Right-Angled Triangles and Pythagoras Property
In a right angle triangle, the side opposite to the right angle is called the hypotenuse and the other two sides are known as the base and perpendicular of the right-angled triangle.
In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of base and perpendicular.
(Hypotenuse)2 = (Base)2 + (Perpendicular)2]
If a triangle holds Pythagoras property, then the triangle must be right-angled.
Problem 1: PQR is a triangle, right angled at P. If PQ = 10 cm and PR = 24 cm, find QR.
Solution:
Given: PQ = 10 cm, PR = 24 cm
Let QR be x cm.
In right angled triangle QPR,
(Hypotenuse)2 = (Base)2 + (Perpendicular)2 [By Pythagoras theorem]
=> (QR)2 = (PQ)2 + (PR)
=> x = 102 + 242
=> x = 100 + 576
=> x= 100 + 576 = 676
=> x = 676
=> x = √676
=> x = 26 cm
Thus, the length of QR is 26 cm.
Problem 2: A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.
Solution:
Let AC be the ladder and A be the window.
Given: AC = 15 m, AB = 12 m, CB = a m
In right angled triangle ACB,
(Hypotenuse)2 = (Base)2 + (Perpendicular)2 [By Pythagoras theorem]
=> (AC)2 = (CB)2 + (AB)2
=> 152 = a2 + 122
=> 225 = a2 + 144
=> a2 = 225 – 144
=> a2 = 81
=> a = √81
=> a = 9 cm
Thus, the distance of the foot of the ladder from the wall is 9 m.
## 7.Right-angles triangles and Pythagoras property
3. Pythagoras Property of Right-Angled Triangle
In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the legs.
Converse of Pythagoras Theorem
In a triangle if the square of one side is equal to the sum of the squares of the remaining two sides then the angle opposite to the first side is a right angle.
It is clear that:-
AC2 > AB2 and AC2 > BC2
AC > AB and AC > BC.
Hence, we can say that in right triangle the hypotenuse is the largest side.
A ABC is such that AB = 4 cm, AC = 3cm, BC = 5 cm and BC2 = AB2 + AC2 52 = 42 + 32
then measure of A should be 90°.
Note.: If the pythagoras property holds, the triangle must be right angled.
Pythagorian Triplets
It is a set of three natural numbers a, b, c such that a2 + b2 = c2. Such a set of three number it is called as Pythagorean Triplet. For example- (3, 4, 5), (5, 12, 13), (6, 8, 10) etc.
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#### explain solution RD Sharma class 12 chapter Derivative As a Rate Measure exercise 12.2 question 8 maths
Answer$\frac{d s}{d t}=\frac{5}{2}=2.5\; km/hr$
Hint: The rate at which the length of the man’s shadow increase will be $\frac{ds}{dt}$ .
Given: A man 2 meters high walk at a uniform speed of 5 km/hr away from lamp post 6 meters high.
Solution:
Suppose AB the lamp post and let MN be of the man of height 2 m.
Suppose AM = l meter and MS be the shadow of the man
Suppose length of the shadow MS=s
Given as man walk at the speed of 5 km/hr
$\frac{d I}{d t}=5\; km/hr$
So considering $\Delta ASB,$
$\tan \theta=\frac{A B}{A S}=\frac{6}{I+s}$................(i)
Then considering $\Delta MNS,$
$\tan \theta=\frac{M N}{M S}=\frac{2}{s}$...................(ii)
From equation (i) and (ii)
\begin{aligned} &\frac{6}{l+s}=\frac{2}{s} \\\\ &6 s=2(I+s) \\\\ &6 s=(2 I+2 s) \end{aligned}
\begin{aligned} &6 s-2 s=2 l \\\\ &4 s=2 l \\\\ &2 s=1 \end{aligned}..............(iii)
By applying derivative, With respect to time on both side
$\frac{d l}{d t}=\frac{d(2 s)}{d t}$ $I=2 s\left(\text { from } e q^{n} \text { iii }\right)$
\begin{aligned} &\frac{d l}{d t}=2 \frac{d s}{d t} \\\\ &5=2 \frac{d s}{d t} \end{aligned} $\left(\frac{d l}{d t}=5\right)(\text { given })$
\begin{aligned} &\frac{5}{2}=\frac{d s}{d t} \\\\ &2.5=\frac{d s}{d t} \end{aligned}
Thus, the rate at which the length of his shadow increases by 2.5 km/hr
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# Shapes in a lattice
### Problem
Imagine a rectangular grid (lattice in the math parlance) with the distance between the nodes equal to 1. Also, let there be given a shape with the area less than 1 but otherwise arbitrary. Show that it's possible to place this shape onto the plane in such a manner that no grid nodes will fall inside the shape.
Solution
Imagine a rectangular grid (lattice in the math parlance) with the distance between the nodes equal to 1. Also, let there be given a shape with the area less than 1 but otherwise arbitrary. Show that it's possible to place this shape onto the plane in such a manner that no grid nodes will fall inside the shape.
### Solution
There are two ways to look at the problem. As the problem was actually formulated, one starts with a lattice and fits a given shape trying to avoid the nodes. The second way is to place the shape onto the plane and then draw a grid around it so that no nodes will fall inside the shape. We may also combine the two: place the shape onto the given lattice, and then, if necessary, shift the lattice a little to avoid intersections.
I'll pursue the latter course. After placing the shape on top of the grid, cut the plane into squares along the grid lines. (We already used such a trick while Shredding the torus.) Stack the squares (without rotating them) on top of each other. Let the plane be transparent and the shape opaque. Take now a look from the top of the stack down. What you'll see is an outline of a single square with a blot formed by the intersection of several pieces of the given shape that fell into different squares. The area of this intersection is clearly less than 1. Therefore, the blot can't cover the whole square. There must be a point that does not belong to the blot. Pierce the whole stack with a vertical line through this point. Now disassemble the stack and place the squares back into their original positions on the plane. Each of the squares will contain a point (the trace of the vertical line) none of which belongs to the shape. The collection of these points forms a new grid of side 1. This may be looked at as the new position of the old grid. Slide it there.
Logic and math logic, in particular, might be flawless, and yet lead to a wrong result. This may happen if one starts with a faulty premise. I do not say that what we obtained is wrong. However, the whole formulation has relied on an intuitive notion of a planar shape and its area. I just want to draw your attention to the fact that our intuition may not be sufficient basis for mathematical derivations.
Let's ask ourselves a few questions?
Is a shape a set? Yes, probably. In so far as everything may be declared a set.
A set of what? It's a set of points, a point set.
Is every point set a shape?
It's a hard question and a source of our quandary. Let the lattice be associated with the integer grid lines of a coordinate system with the origin at one of the nodes. Consider a set of all points (x, y) with rational coordinates. Does this set have a shape? There is only a countable number of such pairs. Sometimes (in the theory of Generalized Functions) it's convenient to ascribe a positive measure to a single point. Measure is a rigorous math notion that formalizes intuitive numerical properties of geometric objects, like length, area, and volume. (I've mentioned the Hausdorff measures when talking about Fractal Dimension. However, talking of areas, having a point with a positive area is completely counterintuitive. So that we should assume that area as a measure, vanishes at every point.
Now, a measure may be expected to have some additive properties. We would expect, for example, that the area of two nonintersecting pieces is the sum of the component areas. In the Measure Theory, we are looking into infinite (but only countably infinite) sums (also known as series.) Thus the area of a countable set of points is also 0. Note that we can't apply the same reasoning to an uncountable set. If we did, then all and every set would have a zero area. Not very interesting.
If you go ahead and apply the stacking procedure to the above set of points with rational coordinates then the blot will actually appear to cover the whole of the square. We, of course, know that it's only an appearance. The set is countable while the square is not. Therefore, we are assured of existence of a point through which we would be able to pierce the stack.
What if we take the set of points (x, y) inside a single square with at least one coordinate irrational. By the additive property, we know (or, at least, expect) the area of such a set to be 1. There are still holes.
As the third example, let's consider a set that consists of two parts. The set of points with rational coordinates in one square plus the set of points with at least one irrational coordinate in another square. The stack in this case will have no holes.
May we be sure that it's impossible to remove a subset from the second part and augment the first part a little bit so that at the end we would have a weird set of area less than 1 with the union of two (shifted) halves covering the entire square? The answer to this question is a firm "No" - no, it's impossible to do that with just two parts. But who will vouch if we allow an infinite number of small parts?
Well, you should not worry about this either. If the set of measure less than 1 is spread over a number (perhaps, infinite) of squares then piling them on top of each other will create a set of measure less than 1 inside a single square so that it won't be able to cover a whole square whose measure is 1. However, the musings above might give you an idea why it is at least desirable to develop rigorous definitions for such simple and mundane notions as shape and area.
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# At a Glance - Continuity at a Point via Pictures
## The Pencil Rule of Continuity
A continuous function is one that we can draw without lifting our pencil, pen, or Crayola crayon.
Here are some examples of continuous functions:
If a function is continuous at xc we can start with our pencil a little to the left of x = c and trace the graph until our pencil is a little to the right of xc, without lifting our pencil along the way.
We will now return to those functions that are continuous at x = c. We can trace each function with a pencil, from one side of x = to the other, without lifting the pencil.
If a function isn't continuous at x = c, we say it's discontinuous at x = c.
### Sample Problem
This function is not continuous at x = c, since the function isn't even defined at x = c. We can't compare the value of f(c) to , since neither exists!
### Sample Problem
This function "jumps" at x = c. To draw the graph we would have to draw one line, stop at x = c and lift the pencil, then draw another line. As far as the limit definition goes, doesn't even exist (the one-sided limits disagree). Therefore f can't possibly be continuous at c.
### Sample Problem
This function also jumps at x = c. To draw the graph we would have to draw a line, lift the pencil and draw a dot at x = c, then lift the pencil again to draw the remaining line. In this graph both f(c) and exist, but the function value disagrees with the limit.
If a function f is discontinuous at x = c, then at least one of three things need to go wrong. Either
• f(c) is undefined (therefore we can't draw it at all),
• we need to move the pencil either just before or just after we reach x = c ( doesn't exist), or
• we need to move the pencil either just before or just after we reach x = c and we need to draw a separate little dot for f(c).
In other words: for a function f(x) to be continuous at x = c, three things need to happen:
• The function must be defined at x = c (that is, f(c) must exist)
• The limit must exist (both one-sided limits must exist and agree)
• The value f(c) must agree with the limit .
#### Example 1
Look at the graph of the function f(x).The following three statements must all hold for f to be continuous at c.The function f is defined at x = c. The limit exists. The value f(c) agrees with the limit .For each given value of c, determine whether each of the three statements holds. Use this to determine whether f is continuous at the given value of c.c = -2 c = -1 c = 0 c = 2 c = 5
#### Example 2
Look at the graph of the function f(x).The following three statements must all hold for f to be continuous at c.The function f is defined at x = c. The limit exists. The value f(c) agrees with the limit .For each given value of c, determine whether each of the three statements hold. Use this to determine whether f is continuous at the given value of c.c = -2 c = -1 c = 0 c = 1 c = 2
#### Example 3
Look at the graph of the function f(x).Determine whether the function f is continuous at each given value. If not, explain.x = -2 x = 0 x = 2 x = 4 x = 5
#### Example 4
Look at the graph of the function g(x).Determine whether the function g is continuous at each given value. If not, explain.x = -20 x = -10 x = 0 x = 5 x = 10
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12-4. The Most General Applications of Bernoulli’s Equation
Learning Objectives
• Calculate using Torricelli’s theorem.
• Calculate power in fluid flow.
Torricelli’s Theorem
Figure 1 shows water gushing from a large tube through a dam. What is its speed as it emerges? Interestingly, if resistance is negligible, the speed is just what it would be if the water fell a distance $h$ from the surface of the reservoir; the water’s speed is independent of the size of the opening. Let us check this out. Bernoulli’s equation must be used since the depth is not constant. We consider water flowing from the surface (point 1) to the tube’s outlet (point 2). Bernoulli’s equation as stated in previously is
$P1+12ρv12+ρgh1=P2+12ρv22+ρgh2.$
Both ${P}_{1}$ and ${P}_{2}$ equal atmospheric pressure (${P}_{1}$ is atmospheric pressure because it is the pressure at the top of the reservoir. ${P}_{2}$ must be atmospheric pressure, since the emerging water is surrounded by the atmosphere and cannot have a pressure different from atmospheric pressure.) and subtract out of the equation, leaving
$12ρv12+ρgh1=12ρv22+ρgh2.$
Solving this equation for ${v}_{2}^{2}$, noting that the density $\rho$ cancels (because the fluid is incompressible), yields
$v22=v12+2g(h1−h2).$
We let $h={h}_{1}-{h}_{2}$; the equation then becomes
$v22=v12+2gh$
where $h$ is the height dropped by the water. This is simply a kinematic equation for any object falling a distance $h$ with negligible resistance. In fluids, this last equation is called Torricelli’s theorem. Note that the result is independent of the velocity’s direction, just as we found when applying conservation of energy to falling objects.
Figure 1: (a) Water gushes from the base of the Studen Kladenetz dam in Bulgaria. (credit: Kiril Kapustin; http://www.ImagesFromBulgaria.com) (b) In the absence of significant resistance, water flows from the reservoir with the same speed it would have if it fell the distance $h$ without friction. This is an example of Torricelli’s theorem.
Figure 2: Pressure in the nozzle of this fire hose is less than at ground level for two reasons: the water has to go uphill to get to the nozzle, and speed increases in the nozzle. In spite of its lowered pressure, the water can exert a large force on anything it strikes, by virtue of its kinetic energy. Pressure in the water stream becomes equal to atmospheric pressure once it emerges into the air.
All preceding applications of Bernoulli’s equation involved simplifying conditions, such as constant height or constant pressure. The next example is a more general application of Bernoulli’s equation in which pressure, velocity, and height all change. (See Figure 2.)
Example 1: Calculating Pressure: A Fire Hose Nozzle
Fire hoses used in major structure fires have inside diameters of 6.40 cm. Suppose such a hose carries a flow of 40.0 L/s starting at a gauge pressure of $1\text{.}\text{62}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$. The hose goes 10.0 m up a ladder to a nozzle having an inside diameter of 3.00 cm. Assuming negligible resistance, what is the pressure in the nozzle?
Strategy
Here we must use Bernoulli’s equation to solve for the pressure, since depth is not constant.
Solution
Bernoulli’s equation states
$P 1 + 1 2 ρv 1 2 +ρ gh1 =P2+ 12 ρv22 +ρ gh2 ,$
where the subscripts 1 and 2 refer to the initial conditions at ground level and the final conditions inside the nozzle, respectively. We must first find the speeds ${v}_{1}$ and ${v}_{2}$. Since $Q={A}_{1}{v}_{1}$ , we get
$v1=QA1=40.0×10−3m3/sπ(3.20×10−2m)2=12.4m/s.$
Similarly, we find
$v2=56.6 m/s.$
(This rather large speed is helpful in reaching the fire.) Now, taking ${h}_{1}$ to be zero, we solve Bernoulli’s equation for ${P}_{2}$:
$P2=P1+12ρ v 1 2 − v 2 2 −ρgh2.$
Substituting known values yields
$P2=1.62×106N/m2+12(1000kg/m3)(12.4m/s)2−(56.6m/s)2−(1000kg/m3)(9.80m/s2)(10.0m)=0.$
Discussion
This value is a gauge pressure, since the initial pressure was given as a gauge pressure. Thus the nozzle pressure equals atmospheric pressure, as it must because the water exits into the atmosphere without changes in its conditions.
Power in Fluid Flow
Power is the rate at which work is done or energy in any form is used or supplied. To see the relationship of power to fluid flow, consider Bernoulli’s equation:
$P+12ρv2+ρgh=constant.$
All three terms have units of energy per unit volume, as discussed in the previous section. Now, considering units, if we multiply energy per unit volume by flow rate (volume per unit time), we get units of power. That is, $\left(E/V\right)\left(V/t\right)=E/t$. This means that if we multiply Bernoulli’s equation by flow rate $Q$, we get power. In equation form, this is
$P+12ρv2+ρghQ=power.$
Each term has a clear physical meaning. For example, $\text{PQ}$ is the power supplied to a fluid, perhaps by a pump, to give it its pressure $P$. Similarly, $\frac{1}{2}{\mathrm{\rho v}}^{2}Q$ is the power supplied to a fluid to give it its kinetic energy. And $\rho \text{ghQ}$ is the power going to gravitational potential energy.
Making Connections: Power:
Power is defined as the rate of energy transferred, or $E/t$. Fluid flow involves several types of power. Each type of power is identified with a specific type of energy being expended or changed in form.
Example 2: Calculating Power in a Moving Fluid
Suppose the fire hose in the previous example is fed by a pump that receives water through a hose with a 6.40-cm diameter coming from a hydrant with a pressure of $0\text{.}\text{700}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$. What power does the pump supply to the water?
Strategy
Here we must consider energy forms as well as how they relate to fluid flow. Since the input and output hoses have the same diameters and are at the same height, the pump does not change the speed of the water nor its height, and so the water’s kinetic energy and gravitational potential energy are unchanged. That means the pump only supplies power to increase water pressure by $0\text{.}\text{92}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$ (from $0.700×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$ to $1.62×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$).
Solution
As discussed above, the power associated with pressure is
$power = PQ = 0.920×106N/m2 40.0×10−3m3/s. = 3.68×104W=36.8kW$
Discussion
Such a substantial amount of power requires a large pump, such as is found on some fire trucks. (This kilowatt value converts to about 50 hp.) The pump in this example increases only the water’s pressure. If a pump—such as the heart—directly increases velocity and height as well as pressure, we would have to calculate all three terms to find the power it supplies.
Summary
• Power in fluid flow is given by the equation $\left({P}_{1}+\frac{1}{2}{\mathrm{\rho v}}^{2}+\rho \text{gh}\right)Q=\text{power}\text{,}$ where the first term is power associated with pressure, the second is power associated with velocity, and the third is power associated with height.
Conceptual Questions
Exercise 1
Based on Bernoulli’s equation, what are three forms of energy in a fluid? (Note that these forms are conservative, unlike heat transfer and other dissipative forms not included in Bernoulli’s equation.)
Exercise 2
Water that has emerged from a hose into the atmosphere has a gauge pressure of zero. Why? When you put your hand in front of the emerging stream you feel a force, yet the water’s gauge pressure is zero. Explain where the force comes from in terms of energy.
Exercise 3
The old rubber boot shown in Figure 3> has two leaks. To what maximum height can the water squirt from Leak 1? How does the velocity of water emerging from Leak 2 differ from that of leak 1? Explain your responses in terms of energy.
Figure 3: Water emerges from two leaks in an old boot.
Exercise 4
Water pressure inside a hose nozzle can be less than atmospheric pressure due to the Bernoulli effect. Explain in terms of energy how the water can emerge from the nozzle against the opposing atmospheric pressure.
Problems & Exercises
Exercise 1
Hoover Dam on the Colorado River is the highest dam in the United States at 221 m, with an output of 1300 MW. The dam generates electricity with water taken from a depth of 150 m and an average flow rate of $\text{650}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{3}\text{/s}$. (a) Calculate the power in this flow. (b) What is the ratio of this power to the facility’s average of 680 MW?
Show/Hide Solution
Solution
(a) $\text{9.56}×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{W}$
(b) 1.4
Exercise 2
A frequently quoted rule of thumb in aircraft design is that wings should produce about 1000 N of lift per square meter of wing. (The fact that a wing has a top and bottom surface does not double its area.) (a) At takeoff, an aircraft travels at 60.0 m/s, so that the air speed relative to the bottom of the wing is 60.0 m/s. Given the sea level density of air to be $1\text{.}\text{29}\phantom{\rule{0.25em}{0ex}}{\text{kg/m}}^{3}$, how fast must it move over the upper surface to create the ideal lift? (b) How fast must air move over the upper surface at a cruising speed of 245 m/s and at an altitude where air density is one-fourth that at sea level? (Note that this is not all of the aircraft’s lift—some comes from the body of the plane, some from engine thrust, and so on. Furthermore, Bernoulli’s principle gives an approximate answer because flow over the wing creates turbulence.)
Exercise 3
The left ventricle of a resting adult’s heart pumps blood at a flow rate of $\text{83}\text{.}0\phantom{\rule{0.25em}{0ex}}{\text{cm}}^{3}\text{/s}$, increasing its pressure by 110 mm Hg, its speed from zero to 30.0 cm/s, and its height by 5.00 cm. (All numbers are averaged over the entire heartbeat.) Calculate the total power output of the left ventricle. Note that most of the power is used to increase blood pressure.
Show/Hide Solution
Solution
1.26 W
Exercise 4
A sump pump (used to drain water from the basement of houses built below the water table) is draining a flooded basement at the rate of 0.750 L/s, with an output pressure of $3.00×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$. (a) The water enters a hose with a 3.00-cm inside diameter and rises 2.50 m above the pump. What is its pressure at this point? (b) The hose goes over the foundation wall, losing 0.500 m in height, and widens to 4.00 cm in diameter. What is the pressure now? You may neglect frictional losses in both parts of the problem.
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Question 1: Find the amount and the compound interest on Rs. $\displaystyle 5000$ for $\displaystyle 2$ years at $\displaystyle 8\%$ per annum, compounded annually.
$\displaystyle \text{ Principal for the first year } = 5000 \text{ Rs. }$
$\displaystyle \text{ Interest for the first year } = \Big( \frac{8}{100} \Big) \times{}5000=400 \text{ Rs. }$
$\displaystyle \text{ Principal for the second year } = 5400 \text{ Rs. }$
$\displaystyle \text{ Amount at the end of second year } = (1+ \frac{8}{100} ) \times 5400 =5832 \text{ Rs. }$
$\displaystyle \text{Interest for the second year } = \frac{8}{100} \time 5400 = 432 \text{ Rs. }$
$\displaystyle \text{Compound interest } = 400+432=832 \text{ Rs. }$
$\displaystyle \\$
Question 2: Find the amount and the compound interest on Rs. $\displaystyle 8000$ for $\displaystyle 2$ years at $\displaystyle 6\%$ per annum, compounded annually.
$\displaystyle \text{ Principal for the first year } = 8000 \text{ Rs. }$
$\displaystyle \text{ Interest for the first year } = \frac{6}{100} \times 8000 =480 \text{ Rs. }$
$\displaystyle \text{Principal for second year } = 8000+480=8480 \text{ Rs. }$
$\displaystyle \text{Interest for second year } = \frac{6}{100} \times 8480=508.80 \text{ Rs. }$
$\displaystyle \text{Principal at the end of second year } = 8480+508.80=8988.80 \text{ Rs. }$
$\displaystyle \text{Total compound interest } = 480+508.80=988.80 \text{ Rs. }$
$\displaystyle \\$
Question 3: Find the amount and the compound interest on Rs. $\displaystyle 2500$ for $\displaystyle 2$ years, compounded annually, the rate of interest being $\displaystyle 6\%$ during the first year and $\displaystyle 8\%$ during second year.
$\displaystyle \text{ Principal for the first year } = 2500 \text{ Rs. }$
$\displaystyle \text{Interest earned by end of first year } = \frac{6}{100} \times 2500 = 150 \text{ Rs. }$
$\displaystyle \text{ Principal for the second year } = 2500+150=2650 \text{ Rs. }$
$\displaystyle \text{Interest earned for second year } = \frac{8}{100} \times 2650=212 \text{ Rs. }$
$\displaystyle \text{Total interest } = 150+212=362 \text{ Rs. }$
$\displaystyle \text{Amount at the end of the second year } = 2650+212=2862 \text{ Rs. }$
$\displaystyle \\$
Question 4: Find the amount and the compound interest on Rs. $\displaystyle 2500$ for $\displaystyle 3$ years at $\displaystyle 6\%$ per annum, compounded annually.
$\displaystyle \text{ Principal for the first year } = 25000 \text{ Rs. }$
$\displaystyle \text{Interest for first year } = \frac{6}{100} \times 25000 =1500 \text{ Rs. }$
$\displaystyle \text{Principal for 2nd year } = Rs.25000+1500=26500 \text{ Rs. }$
$\displaystyle \text{Interest for 2nd year } = \frac{6}{100} \times 26500=1590 \text{ Rs. }$
$\displaystyle \text{Principal for 3rd year } = 26500+1590=28090 \text{ Rs. }$
$\displaystyle \text{Interest for 3rd year } = \frac{6}{100} \times 28090=1685.40 \text{ Rs. }$
$\displaystyle \text{Amount at the end of the 3rd year } =28090+1685.4 =29775.4 \text{ Rs. }$
$\displaystyle \\$
Question 5: Find the amount and the compound interest on Rs. $\displaystyle 10000$ for $\displaystyle 3$ years at $\displaystyle 10 \%$ per annum, compounded annually.
$\displaystyle \text{Principal for the 1st year } = 10000 \text{ Rs. }$
$\displaystyle \text{Interest for 1st year } = \frac{10}{100} \times 10000=1000 \text{ Rs. }$
$\displaystyle \text{Principal for 2nd year } = 10000+100=11000 \text{ Rs. }$
$\displaystyle \text{Interest for 2nd year } = \frac{10}{100} \times 11000=1100 \text{ Rs. }$
$\displaystyle \text{Principal for 3rd year } = 11000+1100=12100 \text{ Rs. }$
$\displaystyle \text{Interest for 3rd year } = \frac{10}{100} \times 121000=1210 \text{ Rs. }$
$\displaystyle \text{Amount at the end of the 3rd year } = 12100+1210=13310 \text{ Rs. }$
$\displaystyle \text{Total compounded interest } = 1000+1100+1210=3210 Rs.$
$\displaystyle \\$
Question 6: ‘A’ took a loan of Rs. $\displaystyle 25000$ from corporate bank at $\displaystyle 12\%$ per annum, compounded annually. How much amount he will have to pay at the end of $\displaystyle 3$ years?
$\displaystyle \text{Principal for the 1st year } = 25000 \text{ Rs. }$
$\displaystyle \text{Interest for the 1st year } = \frac{12}{100} \times 25000=3000 \text{ Rs. }$
$\displaystyle \text{Principal for 2nd year } = 28000 \text{ Rs. }$
$\displaystyle \text{Interest on 2nd the year } = \frac{12}{100} \times 28000=3360 \text{ Rs. }$
$\displaystyle \text{Principal for the 3rd year } = 31600 \text{ Rs. }$
$\displaystyle \text{Interest for the 3rd year } = \frac{12}{100} \times 31360=3763.2 \text{ Rs. }$
$\displaystyle \text{Amount at the end of 3rd year } = 35123.20 \text{ Rs. }$
$\displaystyle \text{ Total compounded interest } = 10152 \text{ Rs. }$
$\displaystyle \\$
Question 7: ‘A’ deposited Rs. $\displaystyle 15625$ in a bank at $\displaystyle 8\%$ per annum, compounded annually. How much amount will he get after $\displaystyle 3$ years?
$\displaystyle \text{Principal for 1st year } = 15625 \text{ Rs. }$
$\displaystyle \text{Interest for 1st year } = \frac{8}{100} \times 15625=1250 \text{ Rs. }$
$\displaystyle \text{Principal for 2nd year } = 16875 \text{ Rs. }$
$\displaystyle \text{Interest for 2nd year } = \frac{8}{100} \times 16875=1350 \text{ Rs. }$
$\displaystyle \text{Principal for the 3rd year } = 18225 \text{ Rs. }$
$\displaystyle \text{ Interest for the 3rd year } = \frac{8}{100} \times 18225=1458 \text{ Rs. }$
$\displaystyle \text{Amount at the end of the 3rd year } = 19683 \text{ Rs. }$
$\displaystyle \\$
Question 8: A person lent out Rs. $\displaystyle 16000$ on simple interest and the same sum on compound interest for $\displaystyle 2$ years at $\displaystyle 12.5\%$ per annum. Find the ratio of the amounts received by him as interest after $\displaystyle 2$ years.
Simple Interest
Principal for the 1st year $\displaystyle = 16000 \text{ Rs. }$
Interest for the 1st year $\displaystyle = \frac{12.5}{100} \times 16000=2000 \text{ Rs. }$
Interest for 2nd year $\displaystyle = 2000 \text{ Rs. }$
Total interest $\displaystyle = 4000 \text{ Rs. }$
Compound Interest
Principal for 1st year $\displaystyle = 16000 \text{ Rs. }$
Interest at the end of the 1st year $\displaystyle = \frac{12.5}{100} \times 16000=2000 \text{ Rs. }$
Principal for 2nd year $\displaystyle = 18000 \text{ Rs. }$
Interest at the end of the 2nd year $\displaystyle = \frac{12.5}{100} \times 18000=2250 \text{ Rs. }$
Total compound interest $\displaystyle =2000+2250=4250 \text{ Rs. }$
Ratio of the interest $\displaystyle = 4000 \colon 4250 \ or \ 16 \colon 17$
$\displaystyle \\$
$\displaystyle \text{ Note: We could also solve by using the formula } A=P \Big\{ \Big( 1+ \frac{r}{100} \Big) \Big\}^n$
$\displaystyle \text{Answer 1: } A=5000 \Big\{ \Big( 1+ \frac{8}{100} \Big) \Big\}^2 = 5832 \text{ Rs. }$
$\displaystyle \text{Answer 2: } A=8000 \Big\{ \Big( 1+ \frac{6}{100} \Big) \Big\}^2 = 8988.80 \text{ Rs. }$
$\displaystyle \text{Answer 3: } A=25000 \Big\{ \Big( 1+ \frac{6}{100} \Big) \Big\}^2 =29775.4 \text{ Rs. }$
$\displaystyle \text{Answer 4: } A=10000 \Big\{ \Big( 1+ \frac{10}{100} \Big) \Big\}^2 =13310 \text{ Rs. }$
$\displaystyle \text{Answer 5: } A=25000 \Big\{ \Big( 1+ \frac{12}{100} \Big) \Big\}^2 =35123.2 \text{ Rs. }$
$\displaystyle \text{Answer 6: } A=15625 \Big\{ \Big( 1+ \frac{8}{100} \Big) \Big\}^2 =19683 \text{ Rs. }$
$\displaystyle \text{Answer 7: } A=16000 \Big\{ \Big( 1+ \frac{12.5}{100} \Big) \Big\}^2 = 20250 \text{ Rs. }$
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# Common Core: 6th Grade Math : Draw Polygons in the Coordinate Plane and Solve for Side Lengths: CCSS.Math.Content.6.G.A.3
## Example Questions
### Example Question #94 : Coordinate Geometry
Select the graph that displays the polygon created using the following coordinates:
Explanation:
When we are given coordinate points, it's important to know the difference between the x-axis and the y-axis, and which order these points are given. The x-axis is the axis that runs left to right and the y-axis is the axis the runs up and down. When coordinate points are written, the x value goes first, followed by the y value .
Knowing this information, we can plot the points and use straight lines to connect them in a counter-clockwise or clockwise direction. The provided coordinate points should create the following graph:
### Example Question #95 : Coordinate Geometry
Select the graph that displays the polygon created using the following coordinates:
Explanation:
When we are given coordinate points, it's important to know the difference between the x-axis and the y-axis, and which order these points are given. The x-axis is the axis that runs left to right and the y-axis is the axis the runs up and down. When coordinate points are written, the x value goes first, followed by the y value .
Knowing this information, we can plot the points and use straight lines to connect them in a counter-clockwise or clockwise direction. The provided coordinate points should create the following graph:
### Example Question #96 : Coordinate Geometry
Select the graph that displays the polygon created using the following coordinates:
Explanation:
When we are given coordinate points, it's important to know the difference between the x-axis and the y-axis, and which order these points are given. The x-axis is the axis that runs left to right and the y-axis is the axis the runs up and down. When coordinate points are written, the x value goes first, followed by the y value .
Knowing this information, we can plot the points and use straight lines to connect them in a counter-clockwise or clockwise direction. The provided coordinate points should create the following graph:
### Example Question #97 : Coordinate Geometry
Select the graph that displays the polygon created using the following coordinates:
Explanation:
When we are given coordinate points, it's important to know the difference between the x-axis and the y-axis, and which order these points are given. The x-axis is the axis that runs left to right and the y-axis is the axis the runs up and down. When coordinate points are written, the x value goes first, followed by the y value .
Knowing this information, we can plot the points and use straight lines to connect them in a counter-clockwise or clockwise direction. The provided coordinate points should create the following graph:
### Example Question #98 : Coordinate Geometry
Select the graph that displays the polygon created using the following coordinates:
Explanation:
When we are given coordinate points, it's important to know the difference between the x-axis and the y-axis, and which order these points are given. The x-axis is the axis that runs left to right and the y-axis is the axis the runs up and down. When coordinate points are written, the x value goes first, followed by the y value .
Knowing this information, we can plot the points and use straight lines to connect them in a counter-clockwise or clockwise direction. The provided coordinate points should create the following graph:
### Example Question #99 : Coordinate Geometry
Select the graph that displays the polygon created using the following coordinates:
Explanation:
When we are given coordinate points, it's important to know the difference between the x-axis and the y-axis, and which order these points are given. The x-axis is the axis that runs left to right and the y-axis is the axis the runs up and down. When coordinate points are written, the x value goes first, followed by the y value .
Knowing this information, we can plot the points and use straight lines to connect them in a counter-clockwise or clockwise direction. The provided coordinate points should create the following graph:
### Example Question #100 : Coordinate Geometry
Select the graph that displays the polygon created using the following coordinates:
Explanation:
When we are given coordinate points, it's important to know the difference between the x-axis and the y-axis, and which order these points are given. The x-axis is the axis that runs left to right and the y-axis is the axis the runs up and down. When coordinate points are written, the x value goes first, followed by the y value .
Knowing this information, we can plot the points and use straight lines to connect them in a counter-clockwise or clockwise direction. The provided coordinate points should create the following graph:
### Example Question #81 : Draw Polygons In The Coordinate Plane And Solve For Side Lengths: Ccss.Math.Content.6.G.A.3
Select the graph that displays the polygon created using the following coordinates:
Explanation:
When we are given coordinate points, it's important to know the difference between the x-axis and the y-axis, and which order these points are given. The x-axis is the axis that runs left to right and the y-axis is the axis the runs up and down. When coordinate points are written, the x value goes first, followed by the y value .
Knowing this information, we can plot the points and use straight lines to connect them in a counter-clockwise or clockwise direction. The provided coordinate points should create the following graph:
### Example Question #81 : Draw Polygons In The Coordinate Plane And Solve For Side Lengths: Ccss.Math.Content.6.G.A.3
Select the graph that displays the polygon created using the following coordinates:
Explanation:
When we are given coordinate points, it's important to know the difference between the x-axis and the y-axis, and which order these points are given. The x-axis is the axis that runs left to right and the y-axis is the axis the runs up and down. When coordinate points are written, the x value goes first, followed by the y value .
Knowing this information, we can plot the points and use straight lines to connect them in a counter-clockwise or clockwise direction. The provided coordinate points should create the following graph:
### Example Question #83 : Draw Polygons In The Coordinate Plane And Solve For Side Lengths: Ccss.Math.Content.6.G.A.3
Select the graph that displays the polygon created using the following coordinates:
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Next: Sinusoids as geometric series Up: Time shifts Previous: Time shifts Contents Index
# Complex numbers
Complex numbers are written as:
where and are real numbers and . (In this book we'll use capital letters to denote complex numbers and lowercase for real numbers.) Since a complex number has two real components, we use a Cartesian plane (in place of a number line) to graph it, as shown in Figure 7.1. The quantities and are called the real and imaginary parts of , written as:
If is a complex number, its magnitude, written as , is just the distance in the plane from the origin to the point :
and its argument, written as , is the angle from the positive axis to the point :
If we know the magnitude and argument of a complex number (say they are and , for instance) we can reconstruct the real and imaginary parts:
A complex number may be written in terms of its real and imaginary parts and (this is called rectangular form), or alternatively in polar form, in terms of and :
The rectangular and polar formulations are equivalent, and the equations above show how to compute and from and and vice versa.
The main reason we use complex numbers in electronic music is because they magically encode sums of angles. We frequently have to add angles together in order to talk about the changing phase of an audio signal as time progresses (or as it is shifted in time, as in this chapter). It turns out that, if you multiply two complex numbers, the argument of the product is the sum of the arguments of the two factors. To see how this happens, we'll multiply two numbers and , written in polar form:
giving:
Here the minus sign in front of the term comes from multiplying by itself, which gives . We can spot the cosine and sine summation formulas in the above expression, and so it simplifies to:
And so, by inspection, it follows that the product has magnitude and argument .
We can use this property of complex numbers to add and subtract angles (by multiplying and dividing complex numbers with the appropriate arguments) and then to take the cosine and sine of the result by extracting the real and imaginary parts of the result.
Subsections
Next: Sinusoids as geometric series Up: Time shifts Previous: Time shifts Contents Index
Miller Puckette 2006-03-03
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# How to Calculate Expected Value
#### Table of Contents
Any random variable has multiple outcomes, which represents by a certain value of possibility. Expect value gives us expected value on a particular event over the mean value of a large set of independent outcomes of the random variable. As we understand that the measures of a particular event over a specific instance tell us about the probability, whereas expected values tell us about the average outcome of a random variable.
## What is Expected Value?
We know that the probability distribution is a set of random variables, and the mean of random variables is called the expected value. The expected value naturally refers to what anyone desired or expect if an experiment is repeated infinite times. Mathematically, it is the average of all the possible outcomes of a random experiment; it doesn’t mean the event with the highest probability is the expected value.
How to calculate the expected value: Expected value of a random variable can be calculated in the following ways with certain considerations:
Let x is a discrete random variable and p(x) is the probability of random variable, mean is denoted by μ.
1. Fundamental formula for expected value
It is the sum of the probability of the random variable and the number of times the experiment happens:
Expected value, (μ)=xP(x)
Example: Considering a dice is rolled, so the probability of which number comes on its face?
Expected value,
() = 1*(1/6) + 2*(1/6) + 3*(1/6) + 4*(1/6) + 5*(1/6) + 6*(1/6) = 3.5
1. Calculation of expected value for binomial random variables
It is the multiplication of the number of trials and probability of success event.
Example: A coin is tossed 5 times and the probability of getting a tail in each trial is 0.5.
So, Number of trials (X) = 5, and Probability of success event = 0.5.
Expected value = X*P(X) = 5 * 0.5 = 2.5
1. Calculation of expected value for multiple events:
Consider a situation; you are an investor for 2 different companies (event). Both companies handle two projects simultaneously with different probabilities so that achieving success by different success values.
Company 1 has two different strategies of probability 0.3 and 0.7 to achieve project value worth \$3000 and \$4000 respectively.
Company 2 has two different strategies of probability 0.25 and 0.75 to achieve project value worth \$3000 and \$4000 respectively.
For such scenario, the expected values are as follows:
Expected value for company 1 = 0.3*\$3000 + 0.7*\$4000 = \$3700;
Expected value for company 2 = 0.25*\$3000 + 0.75*\$4000 = \$3750;
Since, EV (Company 2) > EV (Company 1)
\$3750 > \$3700
With the expected value, we can say that Company B has a better strategy than company A.
1. Calculation of expected value for a continuous random variable: If possible values consist of either a particular interval on the number line or a combination of disjoint intervals. The expected value of the continuous random variable is the average of a random variable. Where probability density function is defines the area under the function or curve.
Mathematically: A continuous function f(a) is varies from minus infinity to plus infinity and a is the random variable.
Expected value, Ea=-∞+∞a.fa.da
1. Calculation of expected value for an Arbitrary Function: Expected value for an arbitrary function is quite similar to the continuous random variable with certain consideration that is the event is represented by a function f(a) of a random variable g(a) instead of only a.
Simply: a random variable a is replaced by random variable g(a).
Expected value, E[g(a)] = -∞+∞a.fa.da
The expected value has an essential role, especially in statistics and decision theory. An unbiased estimated value of the calculation is exactly equal to the desired parameter value. On the other hand, in decision theory, it is the exact choice of uncertainty. Simply, we can say the expected value of an independent random variable, is the precise value of the arithmetic mean of the result. Also, we have calculated the expected value in the different statistics scenario probability distribution function.
## Daily life problem
The price of shooting a target is \$5, there are 50 different targets are in the wall. If you successfully hit the target you will get the price of \$100 for a particular target out of 50. If you miss the target you will lose the \$5 for a chance. What is the expected value for such a scenario?
Expected Value = Price money * Probability (Hit the target)
+ Price money * Probability (Miss the target);
Expected Value = \$100(1/50) – \$5(49/50) = \$2 – \$4.9 = -\$2.9
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# How do you find the fraction notation and simplify $16.6\%$?
Last updated date: 14th Jul 2024
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Hint: In this type of problem where we need to find the fraction from the percent we must know what percent means. If the person scores $n\%$ it simply means that he scored $n$ out of $100$ and we can write it in the form of the fraction as $\dfrac{n}{{100}}$ and then we can also remove the decimal by multiplying in the denominator of the fraction with $10$.
Complete step by step solution:
Here we are given to find the fraction of term which is given as $16.6\%$
We need to find the fraction from the percent we must know what percent means. If the person scores $n\%$ it simply means that he scored $n$ out of $100$ and we can write it in the form of the fraction as $\dfrac{n}{{100}}$.
So we can write $16.6\% = \dfrac{{16.6}}{{100}}$
So we know that whenever we are given the term in the form of decimal we can convert it into the fraction by taking that term in the form of the numerator and denominator. In the denominator comes the ${10^n}$ where $n$ is the number of digits that are present after the decimal in the numerator.
For example: If we have the term $100.12$ so we can write it as $\dfrac{{10012}}{{100}}$
Here we have two terms after the decimal in the numerator, so we have put two zeroes with $1$ in the denominator in order to remove the decimal point from the numerator.
So in a similar way, we know that in $16.6$ we have one digit after the decimal in the given decimal. So we will write ${10^1} = 10$ in the denominator and the decimal will be removed from the numerator.
So we can write:
$16.6 = \dfrac{{166}}{{10}}$
So we can say that:
$16.6\% = \dfrac{{16.6}}{{100}} = \dfrac{{166}}{{1000}}$
Now we need to convert it into the simplest form. Here we know that both ${\text{166 and 1000}}$ are divisible by $2$ and hence we can cancel them and we will be getting:
$\dfrac{{166}}{{1000}} = \dfrac{{83}}{{500}}$
Now there is no number that can divide both ${\text{83 and 500}}$ completely.
Hence it is the simplest form of the fraction which is $\dfrac{{83}}{{500}}$.
Note:
Here in this kind of problems, a student can make mistake by leaving the fraction of $16.6$ as $\dfrac{{166}}{{10}}$ which is also one of the answers but to get the proper result we must convert it into the simplest form where both are not divisible by any common factor.
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# A body starts from the origin and moves along the X-axis such that the velocity at any instant is given by (4t3 – 2t), where t is in sec and velocity in m/s. What is the acceleration of the particle, when it is 2 m from the origin?
By BYJU'S Exam Prep
Updated on: September 25th, 2023
(a) 28 m/s2
(b) 22 m/s2
(c) 12 m/s2
(d) 10 m/s2
The acceleration of the particle, when it is 2 m from the origin is 22 m/s2. Steps to find the acceleration of the particle, when it is 2 m from the origin:
Step 1: Given that:
Velocity, v = 4t3 – 2t
Step 2: Now we have to find the equation:
We know that Acceleration a = dv/dt
Substituting the values we get:
So, a = d (4t3 – 2t)/ dt
a = dv/dt = 12t2 – 2
We also know that:
v = dx/dt
Or ∫dx = ∫vdt
The above equation can be written as:
and x = ∫vdt = ∫(4t3 – 2t)dt = t4 – t2
When a particle is 2 meters from its source, the question states,
We can write:
t4 – t2 = 2
Or
t4 – t2 – 2 = 0
t4 + t2 – 2t2 – 2 = 0
t2 (t2 + 1) – 2 (t2 + 1) = 0
After solving the above polynomial we get:
(t2 – 2) (t2 + 1) = 0
t = √2 sec (only possible physical root)
The equation for acceleration will now substitute this number for t.
Acceleration at t = √2 sec given by:
a = 12t2 – 2 = 12 x 2 – 2 = 22 m/s2
therefore the acceleration is 22 m/s2.
Summary:
## A body starts from the origin and moves along the X-axis such that the velocity at any instant is given by (4t3 – 2t), where t is in sec and velocity in m/s. What is the acceleration of the particle, when it is 2 m from the origin? (a) 28 m/s2 (b) 22 m/s2 (c) 12 m/s2 (d) 10 m/s2
A body starts from the origin and moves along the X-axis such that the velocity at any instant is given by (4t3 – 2t), where t is in sec and velocity in m/s. The acceleration of the particle, when it is 2 m from the origin is 22 m/s2.
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### A-SSE: Seeing Structure in Expressions
#### A-SSE.A: Interpret the structure of expressions
A-SSE.A.1: Interpret expressions that represent a quantity in terms of its context.
A-SSE.A.1a: Interpret parts of an expression, such as terms, factors, and coefficients.
A-SSE.A.1b: Interpret complicated expressions by viewing one or more of their parts as a single entity.
A-SSE.A.2: Use the structure of an expression to identify ways to rewrite it. For example, see x^4 – y^4 as (x²)² – (y²)², thus recognizing it as a difference of squares that can be factored as (x² – y²)(x² + y²).
#### A-SSE.B: Write expressions in equivalent forms to solve problems
A-SSE.B.3: Choose and produce an equivalent form of an expression to reveal and explain properties of the quantity represented by the expression.
A-SSE.B.3a: Factor a quadratic expression to reveal the zeros of the function it defines.
A-SSE.B.3b: Complete the square in a quadratic expression to reveal the maximum or minimum value of the function it defines.
A-SSE.B.3c: Use the properties of exponents to transform expressions for exponential functions.
### A-APR: Arithmetic with Polynomials and Rational Expressions
#### A-APR.A: Perform arithmetic operations on polynomials
A-APR.A.1: Understand that polynomials form a system analogous to the integers, namely, they are closed under the operations of addition, subtraction, and multiplication; add, subtract, and multiply polynomials.
#### A-APR.B: Understand the relationship between zeros and factors of polynomials
A-APR.B.2: Know and apply the Remainder Theorem: For a polynomial p(x) and a number a, the remainder on division by x – a is p(a), so p(a) = 0 if and only if (x – a) is a factor of p(x).
A-APR.B.3: Identify zeros of polynomials when suitable factorizations are available, and use the zeros to construct a rough graph of the function defined by the polynomial.
#### A-APR.C: Use polynomial identities to solve problems
A-APR.C.4: Prove polynomial identities and use them to describe numerical relationships.
A-APR.C.5: Know and apply the Binomial Theorem for the expansion of (x + y)^n in powers of x and y for a positive integer n, where x and y are any numbers, with coefficients determined for example by Pascal’s Triangle.
### A-CED: Creating Equations
#### A-CED.A: Create equations that describe numbers or relationships
A-CED.A.1: Create equations and inequalities in one variable and use them to solve problems.
A-CED.A.2: Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales.
A-CED.A.3: Represent constraints by equations or inequalities, and by systems of equations and/or inequalities, and interpret solutions as viable or nonviable options in a modeling context.
A-CED.A.4: Rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations.
### A-REI: Reasoning with Equations and Inequalities
#### A-REI.A: Understand solving equations as a process of reasoning and explain the reasoning
A-REI.A.1: Explain each step in solving a simple equation as following from the equality of numbers asserted at the previous step, starting from the assumption that the original equation has a solution. Construct a viable argument to justify a solution method.
A-REI.A.2: Solve simple rational and radical equations in one variable, and give examples showing how extraneous solutions may arise.
#### A-REI.B: Solve equations and inequalities in one variable
A-REI.B.3: Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.
A-REI.B.4: Solve quadratic equations in one variable.
A-REI.B.4a: Use the method of completing the square to transform any quadratic equation in x into an equation of the form (x - p)² = q that has the same solutions. Derive the quadratic formula from this form.
A-REI.B.4b: Solve quadratic equations by inspection (e.g., for x² = 49), taking square roots, completing the square, the quadratic formula and factoring, as appropriate to the initial form of the equation. Recognize when the quadratic formula gives complex solutions and write them as a ± bi for real numbers a and b.
#### A-REI.C: Solve systems of equations
A-REI.C.5: Prove that, given a system of two equations in two variables, replacing one equation by the sum of that equation and a multiple of the other produces a system with the same solutions.
A-REI.C.6: Solve systems of linear equations exactly and approximately (e.g., with graphs), focusing on pairs of linear equations in two variables.
A-REI.C.8: Represent a system of linear equations as a single matrix equation in a vector variable.
A-REI.C.9: Find the inverse of a matrix if it exists and use it to solve systems of linear equations (using technology for matrices of dimension 3 × 3 or greater).
#### A-REI.D: Represent and solve equations and inequalities graphically
A-REI.D.10: Understand that the graph of an equation in two variables is the set of all its solutions plotted in the coordinate plane, often forming a curve (which could be a line).
A-REI.D.11: Explain why the x-coordinates of the points where the graphs of the equations y = f(x) and y = g(x) intersect are the solutions of the equation f(x) = g(x); find the solutions approximately, e.g., using technology to graph the functions, make tables of values, or find successive approximations. Include cases where f(x) and/or g(x) are linear, polynomial, rational, absolute value, exponential, and logarithmic functions.
A-REI.D.12: Graph the solutions to a linear inequality in two variables as a half plane (excluding the boundary in the case of a strict inequality), and graph the solution set to a system of linear inequalities in two variables as the intersection of the corresponding half-planes.
Correlation last revised: 9/15/2020
This correlation lists the recommended Gizmos for this state's curriculum standards. Click any Gizmo title below for more information.
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# Rounding Off to The Nearest 1000
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## What is Rounding Off of a Number?
Rounding off numbers is an interesting concept. It rounds off a number to the nearest tens, hundreds, thousands, or so on. This concept is used in our daily life. For example, there are 495 students in a school, so we round off the number and say that there are around 500 students in the school. Let us understand the full concept of what is the nearest thousand and nearest hundred and Rounding off to the Nearest 1000 and 100 with a few rounds to the nearest thousand examples.
## Number Line Representation
Let us understand the concept of rounding to the nearest thousand examples through number line representation.
Round to Nearest Thousand
In this example, we can understand that below 3500 numbers would round off to 3000, and above 3500 would round off to 4000. So 3450 would round off to 3000 and 3890 would round off to 4000. This will be followed in all the numbers which need to get rounded off to the nearest thousand.
## Rounding Numbers to The Nearest 1000
To round to the nearest thousand examples we write the estimated value which is a simpler representation of the number.
Round to Nearest Thousand
We have to understand the place value of the numbers in rounding off to the nearest concept. To round off to the nearest thousand, we need to write the multiple of 1000 which is closest to that specific number. For example, 1120 students are studying in a university, in this case, we usually say that there are around 1000 students in round figures.
## Solved Examples
Q 1. Round 3240 to the Nearest thousand
Ans: To solve this question we have to see whether 3240 is closer to 3000 or 4000 in the counting series. In this case, 3240 is closer to 3000, hence the round-off figure of 3240 is 3000.
Q 2. Round 7980 to the Nearest thousand
Ans: To solve this question we have to check whether 7980 is closer to 7000 or 8000 in the counting. In this case, 7980 is closer to 8000, hence 8000 is the round figure of 7980.
Q 3. Round 8039 to the Nearest Thousand
Ans: To solve this question we have to check whether 8039 is closer to 8000 or 9000 in the counting. In this case, 8039 is closer to 8000, hence 8000 is the round figure of 8039.
Q 4. Round 16796 to the Nearest Thousand
Ans: To solve this question we have to check whether 16796 is closer to 16000 or 17000 in the counting. In this case, 16796 is closer to 16000, hence 17000 is the round figure of16796.
## Practice Questions
Q 1. What is the nearest thousand of 5321?
Ans: 5000 is the round-off number.
Q 2. What is the nearest thousand of 8999?
Ans: 9000 is the nearest thousand.
Q 3. What is the nearest thousand of 6294?
Ans: 6000 is the nearest thousand.
Q 4. What is the nearest thousand 1964?
Ans: 2000 is the nearest thousand.
Q 5. What is the nearest thousand of 9839
Ans: 10000 is the nearest thousand.
## Summary
Rounding Numbers to the nearest thousand is a similar concept to rounding numbers to the nearest tens or hundred. We use this concept in our daily lives while counting money or buying things. Whenever we go to the market for shopping, we keep money with us in round figures. Rounding off the numbers to the nearest tens, hundred, and thousand is done by keeping the place value of digits in mind. Kids must learn and understand this topic in greater detail as it helps in various other concepts and real life.
Last updated date: 26th Sep 2023
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Total views: 112.5k
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## FAQs on Rounding Off to The Nearest 1000
1. Why do we round off?
Rounding numbers make them simpler and easier to use. Although they're slightly less accurate, their values are still relatively close to what they originally were. People round numbers in many different situations, including many real-world situations you'll find yourself in on a regular basis.
2. Is 5 rounded up or down?
If the digit is less than 5, round the previous digit down; if it's 5 or greater, round the previous digit up.
3. Which numbers cannot be rounded off?
0 and negative numbers cannot be rounded off.
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# Differential Equations. Definition A differential equation is an equation involving derivatives of an unknown function and possibly the function itself.
## Presentation on theme: "Differential Equations. Definition A differential equation is an equation involving derivatives of an unknown function and possibly the function itself."— Presentation transcript:
Differential Equations
Definition A differential equation is an equation involving derivatives of an unknown function and possibly the function itself as well as the independent variable. Example Definition The order of a differential equation is the highest order of the derivatives of the unknown function appearing in the equation 1 st order equations2 nd order equation Examples In the simplest cases, equations may be solved by direct integration. Observe that the set of solutions to the above 1 st order equation has 1 parameter, while the solutions to the above 2 nd order equation depend on two parameters.
Linear or non-linear Differential equations are said to be non-linear if any products exist between the dependent variable and its derivatives, or between the derivatives themselves. Product between two derivatives ---- non-linear Product between the dependent variable themselves ---- non-linear
4 An object is dropped from a height h at t = 0. Determine velocity v and displacement y. Solution by immediate integration FREE FALL
Separable Differential Equations A separable differential equation can be expressed as the product of a function of x and a function of y. Multiply both sides by dx and divide both sides by y 2 to separate the variables. (Assume y 2 is never zero.) Example
A separable differential equation can be expressed as the product of a function of x and a function of y. Combined constants of integration Separable Differential Equations Example
Family of solutions (general solution) of a differential equation The picture on the right shows some solutions to the above differential equation. The straight lines y = x and y = -x are special solutions. A unique solution curve goes through any point of the plane different from the origin. The special solutions y = x and y = -x go both through the origin. Example
Initial conditions In many physical problems we need to find the particular solution that satisfies a condition of the form y(x 0 )=y 0. This is called an initial condition, and the problem of finding a solution of the differential equation that satisfies the initial condition is called an initial-value problem. Find a solution to y 2 = x 2 + C satisfying the initial condition y(0) = 2. 2 2 = 0 2 + C C = 4 y 2 = x 2 + 4 Example
Separable differential equation Combined constants of integration Example
We now have y as an implicit function of x. We can find y as an explicit function of x by taking the tangent of both sides.
A population of living creatures normally increases at a rate that is proportional to the current level of the population. Other things that increase or decrease at a rate proportional to the amount present include radioactive material and money in an interest-bearing account. If the rate of change is proportional to the amount present, the change can be modeled by: Law of natural growth or decay
Rate of change is proportional to the amount present. Divide both sides by y. Integrate both sides. Exponentiate both sides.
Example of the first order linear ODE – RC circuit Find the time dependence of the electric current i(t) in the given circuit. Now we take the first derivative of the last equation with respect to time The general solution is: Constant K can be calculated from initial conditions. We know that particular solution is
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## Links Forward
### Areas and determinants
The determinant of a $2 \times 2$ matrix is given by
$\det \begin{bmatrix} a & b \\ c & d \end{bmatrix} = ad - bc.$
The determinant has a nice geometric interpretation in terms of areas, and this interpretation can also be applied to linear transformations.
We discussed earlier (in "The effect of a linear transformation") how two vectors ${\bf v}$ and ${\bf w}$ can be used to construct a parallelogram, the parallelogram spanned by ${\bf v}$ and ${\bf w}$. We discussed how, if the matrix $M = \begin{bmatrix} {\bf v} & {\bf w} \end{bmatrix}$ has ${\bf v}$ and ${\bf w}$ as its columns, then the linear transformation $T_M$ sends the unit square to this parallelogram. We also discussed the orientation of this parallelogram and how $T_M$ can preserve or reverse orientation.
We now ask: what is the area of this parallelogram? It is closely related to the determinant of the matrix $M$.
For instance, if ${\bf v} = (1,0)$ and ${\bf w} = (0,1)$, then ${\bf v}$ and ${\bf w}$ span the unit square, and these two vectors form the matrix
$M = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \quad \text{which has determinant 1.}$
Similarly, if ${\bf v} = (2,0)$ and ${\bf w} = (3,-5)$, then we obtain a parallelogram with base length $2$ and height $5$, hence with area $10$; and the corresponding matrix is
$M = \begin{bmatrix} 2 & 3 \\ 0 & -5 \end{bmatrix}, \quad \text{which has determinant -10.}$
The determinant equals the negative area. But in this case the parallelogram is negatively oriented and the linear transformation reverses orientation.
In fact, the determinant of $M = \begin{bmatrix} {\bf v} & {\bf w} \end{bmatrix}$ gives the signed area of the parallelogram spanned by ${\bf v}$ and ${\bf w}$, i.e. positive or negative area according to orientation.
The linear transformation $T_M$ sends the unit square to this parallelogram, and in fact takes the tessellation of the plane by unit squares by a tessellation by these parallelograms. So unit squares of area $1$ are taken to parallelograms of signed area $\det M$. Thus $T_M$ expands areas by a factor of $|\det M|$, and preserves or reverses orientation accordingly as $\det M$ is positive or negative. For any region $R$ in the plane, its image $T_M (R)$ under $T_M$ has area $|\det M|$ times the area of $R$.
Exercise 38
In this exercise we prove that, if $M = \begin{bmatrix} {\bf v} & {\bf w} \end{bmatrix}$, then $\det M$ is signed area of the parallelogram spanned by ${\bf v}$ and ${\bf w}$. Let $A({\bf v}, {\bf w})$ be this signed area.
1. Show that for any real number $k$, $A({\bf v}, {\bf w}) = A({\bf v}, {\bf w} + k{\bf v})$ and $A({\bf v}, {\bf w}) = A({\bf v} + k {\bf w}, {\bf w})$.
2. Show that $A({\bf v}, {\bf w}) = -A({\bf w}, {\bf v})$.
3. Show that $A \left( \begin{bmatrix} a \\ 0 \end{bmatrix}, \begin{bmatrix} b \\ d \end{bmatrix} \right) = ad$ and $A \left( \begin{bmatrix} a \\ c \end{bmatrix}, \begin{bmatrix} b \\ 0 \end{bmatrix} \right) = -bc$.
4. Show that, if $d \neq 0$, then $A \left( \begin{bmatrix} a \\ c \end{bmatrix}, \begin{bmatrix} b \\ d \end{bmatrix} \right) = A \left( \begin{bmatrix} a - \frac{bc}{d} \\ 0 \end{bmatrix}, \begin{bmatrix} b \\ d \end{bmatrix} \right) = ad -bc$.
5. Conclude that $A({\bf v}, {\bf w}) = \det M$, where $M = \begin{bmatrix} {\bf v} & {\bf w} \end{bmatrix}$.
#### Example
Find the determinant of the matrix corresponding to the rotation $\text{Rot}_{{\bf 0}, \frac{\pi}{3}}$ of $\pi/3$ about the origin, and explain what this means for areas and orientations.
Solution
The corresponding rotation matrix (with $\phi = \pi/3$) is
$M = \begin{bmatrix} \cos \phi & -\sin \phi \\ \sin \phi & \cos \phi \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \\ - \frac{\sqrt{3}}{2} & \frac{1}{2} \end{bmatrix}, \quad \text{which has} \quad \det M = %\frac{1}{2} \cdot \frac{1}{2} - \frac{\sqrt{3}}{2} \cdot \left( - \frac{\sqrt{3}}{2} \right) = \frac{1}{4} + \frac{3}{4} = 1.$
Thus the linear transformation $\text{Rot}_{{\bf 0}, \pi/3}$ preserves areas and orientations.
In fact, any rotation preserves areas, so any rotation matrix has determinant 1. Similarly, reflections preserve area, but reverse orientation, so reflection matrices have determinant $-1$. Projections collapse all areas down to zero, so have determinant zero.
Exercise 39
Prove these facts. Show any rotation matrix has determinant $1$, any projection matrix has determinant $0$, and any reflection matrix has determinant $-1$.
What about dilation? Dilation from a line $L$ by factor $k$ expands lengths by a factor of $k$ in the direction perpendicular to $L$, but leaves lengths unchanged in the direction of $L$. Thus, areas expand by a factor of $k$.
#### Example
Let $L$ be the line $y=x$. Find the determinant of the matrix corresponding to dilation from $L$ by a factor of $3$. Explain what this means for areas and orientations.
Solution
Using the formula for a dilation matrix, with $k=3$ and $\phi = \pi/4$, $\text{Dil}_{L,3}$ has matrix
$\begin{bmatrix} 1 + (k-1) \sin^2 \phi & (1-k) \sin \phi \cos \phi \\ (1-k) \sin \phi \cos \phi & 1 + (k-1) \cos^2 \phi \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \quad \text{which has determinant 3.}$
Hence $\text{Dil}_{L,k}$ expands areas by a factor of $3$, and preserves orientation.
Exercise 40
Show that the determinant of a general dilation matrix is $k$.
Next page - Links Forward - Isometries
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# Can you take the inverse of both sides?
## Can you take the inverse of both sides?
While it’s true in general that you can take the reciprocal of both sides, unfortunately, you can only take the reciprocal of a single number or a single fraction, NOT a sum or difference of fractions.
### What is it called when you do something to both sides?
Reciprocal describes something that’s the same on both sides. The word mutual is a near synonym in most uses: reciprocal/mutual friendship, describing, a relationship in which two people feel the same way about each other, or do or give similar things to each other.
Why do you apply the inverse of the given operation to both sides?
To isolate the variable, we must reverse the operations acting on the variable. We do this by performing the inverse of each operation on both sides of the equation. Performing the same operation on both sides of an equation does not change the validity of the equation, or the value of the variable that satisfies it.
What is it called when both sides of an equation are equal?
In an equation, the quantities on both sides of the equal sign are equal. The “equa” at the beginning of equation will be familiar from other words such as “equal,” “equality,” and “equate.” All of these words have to do with making things balance out.
## How do you reverse inverse?
UNDOING A ONE-TO-ONE FUNCTION; INVERSE FUNCTIONS
2. ‘cube’ is undone by ‘take the cube root’
3. ‘multiply by 3 ‘ is undone by ‘divide by 3 ‘
### What is the inverse of equation?
The inverse function returns the original value for which a function gave the output. If you consider functions, f and g are inverse, f(g(x)) = g(f(x)) = x. A function that consists of its inverse fetches the original value. Then, g(y) = (y-5)/2 = x is the inverse of f(x).
What does it mean if something is symmetrical?
: having sides or halves that are the same : having or showing symmetry.
Why do we do inverse operation?
Mathematically, inverse operations are opposite operations. Addition is the opposite of subtraction; division is the opposite of multiplication, and so on. Inverse operations are used to solve simple algebraic equations to more difficult equations that involve exponents, logarithms, and trigonometry.
## What are the two golden rules of solving equations?
Do unto one side of the equation, what you do to the other! An equation is like a balance scale. If we put something on, or take something off of one side, the scale (or equation) is unbalanced. When solving math equations, we must always keep the ‘scale’ (or equation) balanced so that both sides are ALWAYS equal.
### What are the 4 properties of equality?
Following are the properties of equality:
• Reflexive property of equality: a = a.
• Symmetric property of equality:
• Transitive property of equality:
• Subtraction property of equality:
• Multiplication property of equality:
• Division property of equality;
• Substitution property of equality:
How do you find the inverse of an angle?
Inverse Tangent: If you know the opposite side and adjacent side of an angle in a right triangle, you can use inverse tangent to find the measure of the angle. Inverse tangent is also called arctangent and is labeled or arctan. The “-1” indicates inverse.
How are inverse operations used to solve an equation?
The goal in solving an equation is to get the variable by itself on one side of the equation and a number on the other side of the equation. To isolate the variable, we must reverse the operations acting on the variable. We do this by performing the inverse of each operation on both sides of the equation.
## Which is an example of an additive inverse?
The additive inverse. The first type of opposite is the one you might be most familiar with: positive numbers and negative numbers. For example, the opposite of 4 is -4, or negative four. On a number line, 4 and -4 are both the same distance from 0, but they’re on opposite sides. This type of opposite is also called the additive inverse.
### What do you call a multiplicative inverse number?
It’s called the multiplicative inverse, but it’s more commonly called a reciprocal. To understand the reciprocal, you must first understand that every whole number can be written as a fraction equal to that number divided by 1. For example, 6 can also be written as 6/1. Variables can be written this way too. For instance, x = x/1.
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## Wednesday, 18 May 2011
### Non-Vedic Maths: One more here, one more there, do I care?
Multiplying two numbers whose last digits add up to powers of 10
This is my first post in the series of Non-Vedic maths, where I will be explaining speed calculation techniques. These techniques are NOT based on the ancient Indian Vedas but can be derived using elementary maths.
In this first instalment of the series I will be looking at how to multiply two numbers whose last digits add up to powers of 10 and whose leading digits are identical. To explain the method let us first look at two digit numbers. As an example, we multiply 63 with 67. For this method to work the first digit has to be the same for both numbers and the last digits have to add up to 10. Then we can write our to numbers as
10a + b and 10a + c
where b + c = 10 and a is the leading digit. In our example we have a = 6, b = 3 and c = 7. Multiplying the two numbers gives
\begin{alignat}{1} (10a+b)(10a+c) &= 100a^2 + 10(ab + ac) + bc\\ &= 100a^2 + 10a(b+c) + bc\\ &= 100a^2 + 100a + bc\\ &= 100a(a+1)+bc \end{alignat}
In the third step we used that b+c=10. The product bc will always be less than 100 so the digits will not interfere with the first term, which is a multiple of 100. In our example we have
63 * 67 = 100 * 6 * 7 + 3 * 7 = 100 * 42 + 21 = 4221
The rule that can be extracted from this is:
Take the first digit and multiply with one more than itself. Multiply the last two digits. The final answer is made up of the two results by joining the results together.
We can represent this graphically in the following way.
The method clearly also works if a is not a single digit number, but it can have any number of digits. If we wanted to multiply 116 with 114 we will get
100 * 11 * 12 + 6 * 4 = 13200 + 24 = 13224.
The method can be generalised if the trailing two digits add up to 100 or the trailing three digits add up to 1000 and so on. The maths is similar to the previous case.
\begin{alignat}{1} (100a+b)(100a+c) &= 100a^2 + 100(ab + ac) + bc\\ &= 10000a^2 + 100a(b+c) + bc\\ &= 10000a^2 + 10000a + bc\\ &= 10000a(a+1)+bc\end{alignat}
Again, bc will always be less than 10000 because each of them is less than 100. To see an example, lLet's multiply the numbers 436 and 464. Here the last two digits add up to 100 and the leading digit is the same in both numbers.
\begin{alignat}{1} 436 * 464 &= 10000*4*5 + 36*64\\ &= 10000*20 + 2304\\ &= 202304 \end{alignat}
Of course one has to multiply two digit numbers, which can be a little more complicated. In another post I will present methods that can make this multiplication easier.
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# Squares Tricks
Here are some of the Squares Tricks which you might find useful next time you go for an exam.
### Shortcut tricks to find square of numbers which is nearer to 10x
Trick is X2 = (X+Y)(X-Y) + Y2
Here X is the actual number whose square we want to calculate.
To apply this trick we have to find out the value of ‘Y’. Let’s understand with the help of examples.
• Calculate square of 98
Let’s apply the above trick. We know that by adding 2 in 98, we get 100. Therefore value of Y is 2.
Thus,
982 = (98+2)(98-2) + 22 = 100 x 96 + 4 = 9604
• Calculate square of 103
Let’s apply the above trick. We know that by subtracting 3 from 103, we get 100. Therefore value of Y is 3.
Thus,
1032=(103-3)(103+3)+32=10600+9=10609
### Tricks to find out the value of a perfect square
This is a four step trick. Let’s understand it with the help of an example.
### Trick to find square of numbers between 40 and 50
Let’s understand this trick with the help of examples below:
• Calculate square of 46
• Step 1: Subtract 46 from 50 i.e. 50-46 = 4
• Step 2: Now subtract result of step 1 from 25 i.e. 25-4 = 21
21 will be the first two digits of our answer.
• Step 3: Now square the number obtained in step 1 i.e.
42 = 16 (This will be last two digits of our answer)
Therefore, answer is 462 = 2116.
### Trick to find square of numbers between 51 and 59
There are two methods to calculate the square of numbers between 51 and 59. Let’s apply these methods to calculate the square of 57.
• Method 1:
• Step 1: Subtract 50 from 57 i.e. 57-50 = 7
• Step 2: Now add result of step 1 to 25 i.e. 25+7 = 32
32 will be the first two digits of our answer.
• Step 3: Now square the number obtained in step 1 i.e.
72 = 49 (This will be last two digits of our answer)Therefore, answer is 572 = 3249.
• Method 2:
Here trick is 5A2=25+A/A2
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# Class 8 Maths MCQ – Data Handling – Equally Likely Outcomes
This set of Class 8 Maths Chapter 5 Multiple Choice Questions & Answers (MCQs) focuses on “Data Handling – Equally Likely Outcomes”.
1. What is the probability of getting head when a coin is tossed?
a) $$\frac{1}{2}$$
b) $$\frac{1}{3}$$
c) $$\frac{1}{4}$$
d) $$\frac{3}{2}$$
Explanation: A coin has two sides i.e. head and tail. When a coin is tossed, we can get either head or tail. The likelihood of getting either head or tail is equal. Hence, the probability of getting head when a coin is tossed is $$\frac{1}{2}$$.
2. If 2 coins are tossed, what is the probability of getting heads on both coins?
a) $$\frac{1}{2}$$
b) $$\frac{1}{3}$$
c) $$\frac{1}{4}$$
d) $$\frac{3}{2}$$
Explanation: A coin has two sides i.e. head and tail. When a coin is tossed, we can get either head or tail. The likelihood of getting either head or tail is equal. Hence, the probability of getting head when a coin is tossed is $$\frac{1}{2}$$. The probability of getting head on both the coins is $$\frac{1}{4}$$.
3. If 2 coins are tossed, what is the probability of getting different outcomes on both coins?
a) $$\frac{1}{2}$$
b) $$\frac{1}{3}$$
c) $$\frac{1}{4}$$
d) $$\frac{3}{2}$$
Explanation: A coin has two sides i.e. head and tail. When a coin is tossed, we can get either head or tail. The likelihood of getting either head or tail is equal. We want to head in one coin and tail in other coin. The four possible outcomes are HH, TH, HT, TT. Hence the probability of getting different outcomes on both coins is $$\frac{1}{2}$$.
4. What is the probability of getting 6 when a die is rolled?
a) $$\frac{1}{2}$$
b) $$\frac{1}{3}$$
c) $$\frac{1}{4}$$
d) $$\frac{1}{6}$$
Explanation: A die has 6 faces. The faces are numbered 1, 2, 3, 4, 5 and 6. When a die is rolled, there are 6 outcomes possible. Probability of getting 6 when a die is rolled is $$\frac{1}{6}$$.
5. What is the probability of getting an even number when a die is rolled?
a) $$\frac{1}{2}$$
b) $$\frac{1}{3}$$
c) $$\frac{1}{4}$$
d) $$\frac{1}{6}$$
Explanation: A die has 6 faces. The faces are numbered 1, 2, 3, 4, 5 and 6. When a die is rolled, there are 6 outcomes possible. Outcome of getting even number when a die is rolled is 3. Hence, probability of getting even number when a die is rolled is $$\frac{3}{6} = \frac{1}{2}$$.
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6. A bag has 6 balls; two of each color red, blue, green. What is the probability of picking a red ball from bag?
a) $$\frac{1}{2}$$
b) $$\frac{1}{3}$$
c) $$\frac{1}{4}$$
d) $$\frac{1}{6}$$
Explanation: A bag has 6 balls; two of each color red, blue, green. When a ball is picked from bag, it can be of red, blue or green. So possible outcome of picking ball from bag is 3. Hence, the probability of picking a red ball from the bag is $$\frac{1}{3}$$.
7. A card is picked from well shuffled deck of 52 cards. What is the probability of picking a red card?
a) $$\frac{1}{2}$$
b) $$\frac{1}{13}$$
c) $$\frac{1}{4}$$
d) $$\frac{1}{26}$$
Explanation: Deck of cards contains 52 cards. There are 13 cards of hearts, spades, clubs and diamonds. The 13 cards are A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K. Heart and diamond cards are red cards and clubs and spades are black card. Hence, we have 26 red cards and 26 black cards in a deck. The probability of picking a red card is $$\frac{1}{2}$$.
8. What is the probability of drawing a queen card from deck of cards?
a) $$\frac{1}{2}$$
b) $$\frac{1}{13}$$
c) $$\frac{1}{4}$$
d) $$\frac{1}{26}$$
Explanation: Deck of cards contains 52 cards. There are 13 cards of hearts, spades, clubs and diamonds. The 13 cards are A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K. Queen card can belong to heart, spade, club or diamond. Hence, 4 outcomes are possible when drawing a queen card from deck of cards. Probability of picking a queen from deck of cards is $$\frac{1}{13}$$.
9. A bag of fruits contains apple, orange, lemons and tangerines. What is the probability of picking up an apple?
a) $$\frac{1}{2}$$
b) $$\frac{1}{13}$$
c) $$\frac{1}{4}$$
d) $$\frac{1}{3}$$
Explanation: A bag contains four types of fruits apple, orange, lemons and tangerines. The outcome of picking a fruit at random from the bag is picking either of the four fruits. Hence, 4 is the correct answer. Probability of picking an apple from the bag is $$\frac{1}{4}$$.
10. What is the probability of picking a spade card from deck of cards?
a) $$\frac{1}{2}$$
b) $$\frac{1}{13}$$
c) $$\frac{1}{4}$$
d) $$\frac{1}{3}$$
Explanation: Deck of cards contains 52 cards. There are 13 cards of hearts, spades, clubs and diamonds. The 13 cards are A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K. The probability of picking a spade card from deck of cards is $$\frac{1}{4}$$.
Sanfoundry Global Education & Learning Series – Mathematics – Class 8.
To practice all chapters and topics of class 8 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]
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# Two More Math Problems: Continued Fractions, Nested Square Roots, Digits of Pi
These problems are for college undergrads after a first course in calculus. They are provided with solutions, and could be used by college professors as exercises or exam questions.
1. Digits of Pi/4
Prove that in base b, if b is an even integer, n > 3, and x = Pi/4, then the n-th digit of x, denoted as a(n), is given by the formula below. We start with n = 1 after the decimal point, for the first digit. Also show that the formula below is not valid if the base b is an odd integer, or if x is different from Pi/4.
where the brackets represent the integer part (also called floor) function.
Solution
Regardless of the number x in [0, 1] and the base b, the n-th digit a(n) of x can be computed as follows:
See here for details. Thus we have
Using the angle difference formula for sinus, the fact that n > 3, b is an even integer, and x = Pi/4, it simplifies to
The result for a(n) follows immediately.
2. Continued Fractions and Nested Square Roots
Let us consider the two following expressions, assuming a is a strictly positive real number:
Prove that x is an integer if and only if a is the product of two consecutive integers. Prove that the same is true for y
Solution
Let's focus on the first case. The second case is almost identical. The strictly positive number x must satisfy x^2 = a + x, thus x = (1 + SQRT(1 + 4a)) / 2. In order for x to be an integer, 1 + 4a must be a perfect odd square, which is possible only if a is the product of two consecutive integers. For instance,
• If a = 1 * 2, then x = 2
• If a = 2 * 3, then x = 3
• If a = 3 * 4, then x = 4
• If a = 4 * 5, then x = 5
• and so on.
Note that the expansion of the number x = 2 in the nested square root numeration system, when x tends to 2, has all its "digits" equal to a = 1 * 2. See this spreadsheet for details. More on this here
For related articles from the same author, click here or visit www.VincentGranville.com. Follow me on on LinkedIn.
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# The value of $\cos 10\cos 20\cos 30...\cos 900...\cos 1790$ is?A.$\dfrac{1}{{\sqrt 2 }}$B.0C.1D.None of these
Last updated date: 13th Jun 2024
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Hint: First, we will rewrite the given values and then use the cosine value $\cos 90 = 0$, in the obtained value. Then we will simplify the obtained value to find the required value.
We are given that $\cos 10\cos 20\cos 30...\cos 900...\cos 1790$.
$\Rightarrow \cos 10\cos 20\cos 30\cos 40\cos 50\cos 60\cos 70\cos 80\cos 90...\cos 900...\cos 1790$
Using the cosine value $\cos 90 = 0$, in the above equation, we get
$\Rightarrow \cos 10\cos 20\cos 30\cos 40\cos 50\cos 60\cos 70\cos 80 \times 0 \times ...\cos 900...\cos 1790$
$\Rightarrow 0$
Thus, the value of $\cos 10\cos 20\cos 30...\cos 900...\cos 1790$ is 0.
Note: The trick part of this question is to rewrite the given value to reach the $\cos 90 = 0$, in the given problem. In solving these types of questions, the key concept is to have a good understanding of the basic trigonometric values and learn how to use the values from trigonometric tables. Students should have a grasp of trigonometric values, for simplifying the given equation. Avoid calculation mistakes.
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Worksheet on Finding the Quartiles and the Interquartile Range of Raw Data
In worksheet on finding the quartiles and the interquartile range of raw and arrayed data we will solve various types of practice questions on measures of central tendency. Here you will get 5 different types of questions on finding the quartiles and the interquartile range of raw and arrayed data.
1. The number of problems worked out by a student on seven days of a week were following.
5, 9, 15, 11, 13, 17, 7
Find the (i) lower quartile, (ii) upper quartile, (iii) interquartile range, (iv) semi-interquartile range, and (v) range for the distribution.
2. Find the (i) lower quartile, (ii) upper quartile, and (iii) interquartile range for the following data.
2, 1, 0, 3, 1, 2, 3, 4, 3, 5
3. Find for the given distribution.
(i) the lower quartile, (ii) the upper quartile, and (iii) the interquartile range.
Variate
1
2
3
4
5
6
7
8
Frequency
8
1
7
15
11
6
10
5
4. Find for the given distribution.
(i) the lower quartile, (ii) the upper quartile, and (iii) the interquartile range.
Hint: Arrange the variates in ascending order.
Variate
30
40
10
20
50
60
Frequency
11
30
15
8
12
9
5. Find for the given distribution.
(i) the lower quartile, (ii) the upper quartile, and (iii) the interquartile range.
Variate
5
10
20
30
50
60
80
Cumulative Frequency
7
12
21
35
42
50
56
Hint: Here $$\frac{3N}{4}$$ = $$\frac{3 × 56}{4}$$ = 42 = cumulative frequency of the variate 50. So, Q3 = $$\frac{50 + 60}{2}$$.
Answers on Worksheet on Finding the quartiles and the interquartile range of raw and arrayed data are given below to check the exact answers of the questions.
1. (i) 7
(ii) 15
(iii) 8
(iv) 4
(v) 12
2. (i) 1
(ii) 3
(iii) 2
3. (i) 3
(ii) 7
(iii) 4
4. (i) 20
(ii) 40
(iii) 20
5. (i) 20
(ii) 55
(iii) 35
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# Day Problems Evaluate each expression for 1. a – 2b2. b ÷ c 3. a ÷ c4. -2abc.
## Presentation on theme: "Day Problems Evaluate each expression for 1. a – 2b2. b ÷ c 3. a ÷ c4. -2abc."— Presentation transcript:
Day Problems Evaluate each expression for 1. a – 2b2. b ÷ c 3. a ÷ c4. -2abc
1.7 The Distributive Property Distributive Property –For every real number a, b, and c, a (b + c) = ab + ac(b + c) a = ba + ca a (b – c) = ab – ac(b – c) a = ba – ca –Examples: 5 (20 + 6) = 5 (20) + 5 (6) (20 + 6) 5 = 20 (5) + 6 (5) 9 ( 30 – 2) = 9 (30) – 9 (2) (30 – 2) 9 = 30 (9) – 2 (9)
Simplifying a Numerical Expression Use the distributive property to simplify 34 (102). 34 (102) = 34 (100 + 2) = 34 (100) + 34 (2) = 3400 + 68 = 3468
Simplifying an Expression Simplify each expression. a. 2 (5x + 3) = 2 (5x) + 2 (3) = 10x + 6 b.
Using the Multiplication Property of -1 Simplify –(6m + 4). -(6m + 4) = -1 (6m + 4) = -1 (6m) + (-1)(4) = -6m – 4
Algebraic Expressions 9/22/10 Term – a number, a variable, or the product of a number and one or more variables. Ex. 6a 2 – 5ab + 3b – 12 Constant – a term that has no variable. Ex. -12 Coefficient – a numerical factor of a term. Ex. 6, -5, and 3
Like Terms Like terms – have exactly the same variable factors. Like TermsNot Like Terms 3x and -2x8x and 7y -5x 2 and 9x 2 5y and 2y 2 xy and –xy4y and 5xy -7x 2 y 3 and 15x 2 y 3 x 2 y and xy 2 An algebraic expression in simplest form has NO like terms.
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Three Squares
What is the greatest number of squares you can make by overlapping three squares?
Two Dice
Find all the numbers that can be made by adding the dots on two dice.
Biscuit Decorations
Andrew decorated 20 biscuits to take to a party. He lined them up and put icing on every second biscuit and different decorations on other biscuits. How many biscuits weren't decorated?
Robot Monsters
Age 5 to 7Challenge Level
We had a couple of very full solutions to this problem. Jack and Alex from Woodfall Junior School wrote:
The biggest robot you can make is 22cm using 6cm head 8cm body and 8cm legs
The smallest robot you can make is 14cm using 3cm head 5cm body and 6cm legs
The bits that are left over using 4cm head 7cm body and 7cm legs that equals 18cm
They went on to say:
There are 27 ways to make different monsters but only 9 different heights of monster altogether:
Head Body Legs Total height 3 5 6 14 3 5 7 15 3 5 8 16 3 7 6 16 3 7 7 17 3 7 8 18 3 8 6 17 3 8 7 18 3 8 8 19 6 5 6 17 6 5 7 18 6 5 8 19 6 7 6 19 6 7 7 20 6 7 8 21 6 8 6 20 6 8 7 21 6 8 8 22 4 8 8 20 4 8 7 19 4 8 6 18 4 7 8 19 4 7 6 17 4 7 7 18 4 5 8 17 4 5 7 16 4 5 6 15
We tried to use a systematic order to make sure we found all 27 monsters. We found all the monsters with 3cm heads first, then 6cm, then 4cm. We did the same with the bodies and legs.
Ruth from Swanborne House School commented:
There are 9 possible monster heights between 14cm and 22cm (14/15/16/17/18/19/20/21/22). To prove that you could make all of them we drew a diagram of all the possible combinations, starting from the three heads. We found that once we had drawn the first head + body and leg combination for the 3cm head, we could work out the other combinations for the 3cm head quite easily, because only the body measurement varied. Then we found it was easy to adjust from this to the other two heads, and we didn't need to do the full diagram.
Here is the diagram that Ruth sent:
Children from Seven Mills Primary School in the London borough of Tower Hamlets created robots with the body part cards but then used Numicon pieces to represent the height of each part. This meant that they were easily able to compare the heights of their robots and spot which ones were missing. Here are some photos to show what they did:
Three very good ways of solving this problem - well done!
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# EQUATION OF THE LINE ALONG THE ALTITUDE OR MEDIAN OF TRIANGLE
Equation of the Line Along the Altitude or Median of Triangle :
Here we are going to see how to find the equation of the line along the altitude or median of the triangle.
## Equation of the Line Along the Altitude or Median of Triangle - Examples
Question 14 :
Vertices of triangle ABC are A (2,-4) ,B(3,3) and C (-1,5). Find the equation of the straight line along the altitude from the vertex B.
Solution :
A line drawn from the vertex B is perpendicular to the line AC.So the product of their slopes will be equal to -1.
Slope of AC = (y₂ - y₁)/(x₂ - x₁)
x₁ = 2, y₁ = -4, x₂ = -1 , y₂ = 5
= (5 - (-4))/(-1 - 2)
= (5 + 4)/(-3)
= -9/3 = -3
slope of line drawn from the vertex B = -1/m
= -1/(-3) = 1/3
Equation of the line drawn from the vertex B:
(y - y₁) = m (x - x₁)
B (3,3) slope = 1/3
x₁ = 3, y₁ = 3
(y - 3) = (1/3) (x - 3)
3 (y - 3) = 1 (x - 3)
3 y - 9 = x - 3
x - 3y - 3 + 9 = 0
x - 3y + 6 = 0
Question 15 :
If the vertices of triangle ABC are A (-4,4), B(8,4) and C(8,10). Find the equation of the along the median from the vertex A.
Solution :
Since AD is median,it passes through the midpoint of the side BC.
x₁ = 8, y₁ = 4, x₂ = 8 , y₂ = 10
= [(x + x)/2 , (y₁ + y₂)/2]
= [(8 + 8)/2 , (4 + 10)/2]
= 16/2 , 14/2
= (8,7)
(y - y₁)/(y₂ - y₁) = (x - x₁)/(x₂ - x₁)
(-4,4) (8,7)
x₁ = -4, y₁ = 4, x₂ = 8 , y₂ = 7
(y -4)/(7-4) = (x-(-4))/(8-(-4))
(y -4)/3 = (x+4)/(8+4)
(y -4)/3 = (x+4)/12
12(y - 4) = 3(x + 4)
12 y - 48 = 3 x + 12
3 x - 12 y + 12 + 48 = 0
3 x - 12y + 60 = 0
Dividing the whole equation by 3
x - 4 y + 20 = 0
After having gone through the stuff given above, we hope that the students would have understood, how to find the equation of the line along the altitude or median of triangle
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# 2002 AMC 12A Problems/Problem 18
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
Let $C_1$ and $C_2$ be circles defined by $(x-10)^2 + y^2 = 36$ and $(x+15)^2 + y^2 = 81$ respectively. What is the length of the shortest line segment $PQ$ that is tangent to $C_1$ at $P$ and to $C_2$ at $Q$?
$\text{(A) }15 \qquad \text{(B) }18 \qquad \text{(C) }20 \qquad \text{(D) }21 \qquad \text{(E) }24$
## Solution 1
First examine the formula $(x-10)^2+y^2=36$, for the circle $C_1$. Its center, $D_1$, is located at (10,0) and it has a radius of $\sqrt{36}$ = 6. The next circle, using the same pattern, has its center, $D_2$, at (-15,0) and has a radius of $\sqrt{81}$ = 9. So we can construct this diagram: $[asy] unitsize(0.3cm); defaultpen(0.8); path C1=circle((10,0),6); path C2=circle((-15,0),9); draw(C1); draw(C2); draw( (-25,0) -- (17,0) ); dot((10,0)); dot((-15,0)); pair[] p1 = intersectionpoints(C1, circle((5,0),5) ); pair[] p2 = intersectionpoints(C2, circle((-7.5,0),7.5) ); dot(p1[1]); dot(p2[0]); draw((10,0)--p1[1]--p2[0]--(-15,0)); label("C_1",(10,0) + 6*dir(-45), SE ); label("C_2",(-15,0) + 9*dir(225), SW ); label("D_1",(10,0), SE ); label("D_2",(-15,0), SW ); label("Q", p2[0], NE ); label("P", p1[1], SW ); label("O", (0,0), SW ); [/asy]$ Line PQ is tangent to both circles, so it forms a right angle with the radii (6 and 9). This, as well as the two vertical angles near O, prove triangles S$_2$QO and S$_1$PO similar by AA, with a scale factor of 6:9, or 2:3. Next, we must subdivide the line D$_2$D$_1$ in a 2:3 ratio to get the length of the segments D$_2$O and D$_1$O. The total length is 10 - (-15), or 25, so applying the ratio, D$_2$O = 15 and D$_1$O = 10. These are the hypotenuses of the triangles. We already know the length of D$_2$Q and D$_1$P, 9 and 6 (they're radii). So in order to find PQ, we must find the length of the longer legs of the two triangles and add them.
$15^2 - 9^2 = (15-9)(15+9) = 6 \times 24 = 144$
$\sqrt{144} = 12$
$10^2-6^2 = (10-6)(10+6) = 4 \times 16 = 64$
$\sqrt{64} = 8$
Finally, the length of PQ is $12+8=\boxed{20}$, or (C).
## Solution 2
Using the above diagram, imagine that segment $\overline{QS_2}$ is shifted to the right to match up with $\overline{PS_1}$. Then shift $\overline{QP}$ downwards to make a right triangle. We know $\overline{S_2S_1} = 25$ from the given information and the newly created leg has length $\overline{QS_2} + \overline{PS_1} = 9 + 6 = 15$. Hence by Pythagorean theorem $15^2 + {\overline{QP}}^2 = 25^2$.
$\overline{QP} = \boxed{20}$, or C.
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## Algebra: A Combined Approach (4th Edition)
$x=4$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $\sqrt{x}+2=x ,$ isolate first the radical. Then raise both sides to the exponent equal to the index of the radical. Use concepts of solving quadratic equations to find the values of the variable. Finally, do checking of the solutions with the original equation. $\bf{\text{Solution Details:}}$ Using the properties of equality, the given equation is equivalent to \begin{array}{l}\require{cancel} \sqrt{x}=x-2 .\end{array} Get rid of the radical symbol by raising both sides to the exponent equal to $2 ,$ the same index as the radical. Hence, the equation above becomes \begin{array}{l}\require{cancel} (\sqrt{x})^2=(x-2)^2 \\\\ x=(x-2)^2 .\end{array} Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to \begin{array}{l}\require{cancel} x=(x)^2-2(x)(2)+(2)^2 \\\\ x=x^2-4x+4 .\end{array} In the form $ax^2+bx+c=0,$ the equation above is equivalent to \begin{array}{l}\require{cancel} -x^2+(x+4x)-4=0 \\\\ -x^2+5x-4=0 \\\\ -1(-x^2+5x-4)=-1(0) \\\\ x^2-5x+4=0 .\end{array} Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the factored form of the equation above is \begin{array}{l}\require{cancel} (x-4)(x-1)=0 .\end{array} Equating each factor to $0$ (Zero Product Property), the solutions are \begin{array}{l}\require{cancel} x-4=0 \\\\\text{OR}\\\\ x-1=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x-4=0 \\\\ x=4 \\\\\text{OR}\\\\ x-1=0 \\\\ x=1 .\end{array} Upon checking, only $x=4$ satisfies the original equation.
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# What can 12/32 be simplified?
## What can 12/32 be simplified?
Therefore, 12/32 simplified to lowest terms is 3/8.
What is to the simplest form?
Thus, finding the simplest form of a fraction means reducing the top and bottom of the fraction to the smallest whole number possible. The simplest form is the smallest possible equivalent fraction of the number.
### How can I reduce 24 and 32?
Reduce 24/32 to lowest terms
• Find the GCD (or HCF) of numerator and denominator. GCD of 24 and 32 is 8.
• 24 ÷ 832 ÷ 8.
• Reduced fraction: 34. Therefore, 24/32 simplified to lowest terms is 3/4.
What is 12 32 as a decimal?
12/32 as a decimal is 0.375.
## What is 12 32 as a percent?
Now we can see that our fraction is 37.5/100, which means that 12/32 as a percentage is 37.5%.
What is an example of simplest form?
A fraction is said to be in its simplest form if 1 is the only common factor of its numerator and denominator. For example,89,because 1 is the only common factor of 8 and 9 in this fraction.
### What is the simplest form of 3 4?
34 is already in the simplest form. It can be written as 0.75 in decimal form (rounded to 6 decimal places)….Steps to simplifying fractions
• Find the GCD (or HCF) of numerator and denominator. GCD of 3 and 4 is 1.
• 3 ÷ 14 ÷ 1.
• Reduced fraction: 34. Therefore, 3/4 simplified to lowest terms is 3/4.
What is the simplest form of 32 and 24?
Reduce 32/24 to lowest terms
• Find the GCD (or HCF) of numerator and denominator. GCD of 32 and 24 is 8.
• 32 ÷ 824 ÷ 8.
• Reduced fraction: 43. Therefore, 32/24 simplified to lowest terms is 4/3.
## What is the GCF 24 and 32?
8
The GCF of 24 and 32 is 8. To calculate the greatest common factor of 24 and 32, we need to factor each number (factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24; factors of 32 = 1, 2, 4, 8, 16, 32) and choose the greatest factor that exactly divides both 24 and 32, i.e., 8.
What is 13 32 as a decimal?
13/32 as a decimal is 0.40625.
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# How do you divide using synthetic division: (y^4 - 4y^2 + y + 4)/(y + 2)?
Nov 26, 2015
(y^4-4y^2+y+4)/(y+2) = (y^3-2y^2+1 with remainder $\left(+ 2\right)$
#### Explanation:
Synthetic Division
{: (,,,color(brown)(y^4),color(brown)(y^3),color(brown)(y^2),color(brown)(y^1),color(brown)(y^0)), (color(brown)("[1] dividend coefficients"),,,1,(+0),-4,+1,+1), (color(brown)("[2]"),,,,-2,+4,+0,-2), (,,,"-----","-----","-----","-----","-----"), (color(brown)("[3] negative of constant from divisor"),xx(-2),color(blue)("||"),1,-2,0,1,color(red)(+2)), (,,,color(brown)(y^3),color(brown)(y^2),color(brown)(y^1),color(brown)(y^0),color(brown)("Remainder")) :}
The test in $\textcolor{b r o w n}{\text{brown}}$ would not normally be written. It is there to help with the explanation only.
Line $\textcolor{b r o w n}{\text{[3]}}$
- the value to the left of $\textcolor{b l u e}{\text{||}}$ (when dividing by a monic binomial) is the negative of the constant term. In this case the constant term of the monic binomial ($x - 1$) is $\left(- 1\right)$ so its negativeis $\left(+ 1\right)$
-the values to the right of color(blue)("||") are the sum of the numbers in the column above each location. These values become the coefficients of the reduced polynomial terms, except for the final number which is the remainder.
Line $\textcolor{b r o w n}{\text{[2]}}$
-is the product of the negative divisor coefficient (to the left of the $\textcolor{b l u e}{\text{||}}$ in line $\textcolor{b r o w n}{\text{[3]}}$ and the sum from the previous column in line $\textcolor{b r o w n}{\text{[3]}}$ (to the right of the $\textcolor{b l u e}{\text{||}}$
:}
The test in $\textcolor{b r o w n}{\text{brown}}$ would not normally be written. It is there to help with the explanation only.
Line $\textcolor{b r o w n}{\text{[3]}}$
- the value to the left of $\textcolor{b l u e}{\text{||}}$ (when dividing by a monic binomial) is the negative of the constant term. In this case the constant term of the monic binomial ($x - 1$) is $\left(- 1\right)$ so its negative is $\left(+ 1\right)$
-the values to the right of #color(blue)("||") are the sum of the numbers in the column above each location. These values become the coefficients of the reduced polynomial terms, except for the final number which is the remainder.
Line $\textcolor{b r o w n}{\text{[2]}}$
-is the product of the negative divisor coefficient (to the left of the $\textcolor{b l u e}{\text{||}}$ in line $\textcolor{b r o w n}{\text{[3]}}$ and the sum from the previous column in line $\textcolor{b r o w n}{\text{[3]}}$ (to the right of the $\textcolor{b l u e}{\text{||}}$
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How do I find the extraneous solution of sqrt(x+4)=x-2?
Mar 8, 2018
$x = 0 \text{ is an extraneous solution}$
Explanation:
$\text{square both sides to 'undo' the radical}$
${\left(\sqrt{x + 4}\right)}^{2} = {\left(x - 2\right)}^{2}$
$\Rightarrow x + 4 = {x}^{2} - 4 x + 4$
$\text{rearrange into standard form}$
$\Rightarrow {x}^{2} - 5 x = 0 \leftarrow \textcolor{b l u e}{\text{in standard form}}$
$\Rightarrow x \left(x - 5\right) = 0$
$\Rightarrow x = 0 \text{ or } x = 5$
$\textcolor{b l u e}{\text{As a check}}$
$\text{Substitute these values into the original equation}$
$x = 0 \to \sqrt{4} = 2 \text{ and } 0 - 2 = - 2$
$2 \ne - 2 \Rightarrow x = 0 \textcolor{red}{\text{ is an extraneous solution}}$
$x = 5 \to \sqrt{9} = 3 \text{ and "5-2=3larr" True}$
$\Rightarrow x = 5 \textcolor{red}{\text{ is the solution}}$
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# Differentiation of Inverse Functions: Example 2
by Batool Akmal
My Notes
• Required.
Learning Material 2
• PDF
DLM Differentiation of Inverse Functions Calculus Akmal.pdf
• PDF
Report mistake
Transcript
00:01 Now that we’ve done that and we’ve practiced a little bit of changing dx/dy to dy/dx, let’s now move to the more fun parts. We’re going to learn how to differentiate sin inverse of x, cos inverse of x and tan inverse of x. Now, these are standard rules. So, we have set results for these already which you can use but we’ll alternate between deriving the actual result and also just by using the result in some of the questions. But let’s just have a go at firstly deriving the actual differential result by working through this. We’ll learn a little bit about inverse differentiation as we go along. If we look at this question here, we have y equals to sin inverse of x. Now, so far we know that the differential of sin of x is cos of x but we haven’t yet done sin inverse of x. Now, the first step that you can do here is you can obviously change this inverse back to its original. So remember where the sin inverse comes from, it must have come from this side of the equation. So, you would have had sin of y which you can take to the other side to make it into a sin inverse of x. So if I take it back and rewrite this as sin y = x and then I write the x first, so I’m just rewriting this equation. So I write it as x = sin of y. Remember we’re trying to find the gradient, so we’re trying to find dy/dx. The first thing that we could do is rather rearranging it because this is the equation rearranged. We’ve taken it back to basics or to what it should look like.
01:39 So we’ve taken the sine back. Instead of doing anything else fiddly with this, we can just differentiate here. So we can say let’s differentiate dx with respect to dy. That gives you cos of y.
01:51 So you’re just differentiating sin y with respect to y which gives you cos y. This doesn’t mean anything, dx/dy at the moment. But you can change it to the gradient by just flipping it. So we want dy/dx, so you’re trying to just change places. You’re swapping it. If you do that here, imagine this is over 1, you’re going to have to do that on the other side as well. So you’re flipping the whole equation.
02:18 We have dy/dx equals to 1 over cos of y. Now, we’re almost done but the standard result that you usually see in textbooks or in formula books gives this answer in terms of x's and we have cos y.
02:35 We have dy/dx or the gradient of sin inverse of x here is 1 over cos y. Now, if we want to change this to an answer in terms of x, we’re going to have to use some identities. The identity that we use is this, so sin²y + cos²y = 1. We’ve learned that sin²x + cos²x = 1 or sin²θ + cos²θ = 1. It really doesn’t matter what the angle is. So we’re going to use this identity, rearrange this so we’ll get cos²y = 1 – sin²y.
03:14 Just to get cos y by itself, this is what we want, you can square root the other side.
03:20 So, we’ve got cos y = √(1 – sin²y). Let’s replace this now. So we’re going to substitute this in here.
03:30 We can now say that dy/dx equals to 1 over √(1 – sin²y). Still in terms of y, not in terms of x, if you just look at the very first thing you wrote, so we said that sin y or sin²y in this case is the same as this bit here. So we said that sin y = x therefore this value must be x².
04:02 Let me just repeat that again. At the start, you said that x was equal to sin y.
04:07 Here, you can see your sin y but you’ve got two of them. So you got sin² of y. That must be equal to x².
04:14 You can now change this to 1 over √(1 – x²). If you look at your textbooks or if you look at your formula books, you will often be given y equals to sin inverse of x. Differentiate.
04:30 So the derivative of that dy/dx is 1 over √(1 –x²). It really depends on the question, what they ask, whether they want you to just go straight to the result which is here or whether they want you to derive the result which is what we’ve done here. But it’s nice to know that just by using inverse differentiation and by using the fact that you can change dx/dy to dy/dx, we can actually derive the actual answer ourselves. So you don’t really need your formula books or textbooks when you can do it all by yourself.
### About the Lecture
The lecture Differentiation of Inverse Functions: Example 2 by Batool Akmal is from the course Differentiation of Inverse Functions.
### Included Quiz Questions
1. dy/dx = 1 / √(1 - x²)
2. dy/dx = 1 / √(1 + x²)
3. dy/dx = -1 / √(1 - x²)
4. dy/dx = -1 / √(1 + x²)
5. dy/dx = √(1 - x²)
1. x = cos(y) , dx/dy = -sin(y)
2. x = sin(y) , dx/dy = cos(y)
3. x = 1/cos(y) , dx/dy = sin(y)/cos²(y)
4. x = cos(y) , dx/dy = sin(y)
5. x = cos⁻¹(y) , dx/dy = -1/√(1-y²)
1. cosΘ = √(1 - sin²Θ)
2. cosΘ = √(1 + sin²Θ)
3. cosΘ = 1 - sin²Θ
4. cosΘ = 1 + sin²Θ
5. cosΘ = √(1 - sinΘ)
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Technology
# How to Find the Radius: A Comprehensive Guide
## Different Methods to Find the Radius of a Circle
When it comes to finding the radius of a circle, there are several methods you can use depending on the information you have available. Here are some of the most common methods:
1. Measure the distance from the center to the edge: This is the simplest method, requiring only a ruler or tape measure. Place the ruler or tape measure at the center of the circle and extend it to the edge. The distance from the center to the edge is the radius.
2. Use the Pythagorean Theorem: If you know the length of a chord (a line segment connecting two points on the circle) and the distance from the midpoint of the chord to the center of the circle (also known as the sagitta), you can use the Pythagorean Theorem to find the radius.
3. Use trigonometry: If you know the length of a chord and the angle it subtends (the angle between the two radii connecting the center of the circle to the endpoints of the chord), you can use trigonometry to find the radius.
4. Use the circumference: If you know the circumference of the circle, you can use the formula C = 2Ï€r to solve for the radius.
5. Use the diameter: If you know the diameter of the circle, you can use the formula d = 2r to solve for the radius.
## How to Find the Radius with Circumference or Diameter
If you know the circumference or diameter of a circle, you can use a simple formula to find the radius. Here’s how:
1. Finding the Radius with Circumference:
If you know the circumference (C) of a circle, you can use the formula C = 2Ï€r to solve for the radius (r). Simply divide both sides of the equation by 2Ï€ to isolate the radius. The formula becomes r = C / 2Ï€.
2. Finding the Radius with Diameter:
If you know the diameter (d) of a circle, you can use the formula d = 2r to solve for the radius (r). Simply divide both sides of the equation by 2 to isolate the radius. The formula becomes r = d / 2.
It’s important to note that if you only know the circumference or diameter, you won’t have the complete picture of the circle’s size and shape. If possible, it’s best to measure or calculate the radius directly using other methods as well.
## Solving for Radius in Real-World Scenarios
Knowing how to find the radius of a circle is a useful skill in many real-world scenarios. Here are some examples:
1. Finding the Radius of a Tire: To find the radius of a tire, measure the distance from the center of the tire to the edge. This will give you the radius, which is important for calculating the tire’s circumference, speed, and other properties.
2. Calculating the Area of a Circle: To calculate the area of a circle, you need to know the radius. Use one of the methods mentioned earlier to find the radius, and then use the formula A = πr² to calculate the area.
3. Estimating the Size of a Circle: In some situations, you may need to estimate the size of a circle without measuring it directly. For example, if you’re trying to plan the layout of a garden or park, you might want to estimate the size of circular areas for landscaping or other features. In this case, you can use your knowledge of radius, diameter, and circumference to make an educated guess about the circle’s size.
Whether you’re working with tires, circles in a park, or anything in between, being able to find the radius of a circle is a valuable skill.
## Tips and Tricks for Finding the Radius Quickly and Accurately
While there are several methods to find the radius of a circle, some are more efficient and accurate than others. Here are some tips and tricks to help you find the radius quickly and accurately:
1. Use a compass: If you have a compass, it can be a very accurate tool for measuring the radius of a circle. Simply place the point of the compass at the center of the circle and extend the other end to the edge. Then, measure the distance between the two points to find the radius.
2. Use a ruler or tape measure: If you don’t have a compass, a ruler or tape measure can also be used to measure the radius. Just make sure to measure from the center of the circle to the edge.
3. Memorize common circle measurements: Knowing the circumference and diameter of common circle sizes (such as a full-size basketball or a dinner plate) can help you estimate the radius quickly without needing to measure.
4. Practice using formulas: The formulas for finding the radius with circumference or diameter are simple and easy to memorize. Practicing using these formulas can help you solve for the radius quickly and accurately.
5. Check your work: After finding the radius, double-check your work using a different method to ensure accuracy. If your measurements don’t match up, check for errors in your calculations or measurements.
By using these tips and tricks, you can find the radius of a circle quickly and accurately, making it easier to solve problems and complete tasks that require this knowledge.
## Common Mistakes to Avoid When Finding the Radius
While finding the radius of a circle may seem straightforward, there are some common mistakes that can lead to inaccurate results. Here are some mistakes to watch out for:
1. Measuring from the edge of the circle instead of the center: It’s important to measure from the center of the circle to the edge, not from one edge to the other.
2. Confusing the diameter with the radius: The diameter is the distance across the circle, while the radius is the distance from the center to the edge. Make sure you know which measurement you are working with.
3. Using the wrong formula: Make sure you use the correct formula for the information you have. For example, using the formula for finding the radius with diameter when you only have the circumference will not give you the correct answer.
4. Rounding too early: When using formulas, make sure to carry out all calculations before rounding to avoid losing accuracy.
5. Forgetting to double-check: Always double-check your work using a different method to ensure accuracy. This is especially important when working on important projects or tasks where accuracy is crucial.
By avoiding these common mistakes and being diligent in your measurements and calculations, you can find the radius of a circle accurately and confidently.
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###### Brian McCall
Univ. of Wisconsin
J.D. Univ. of Wisconsin Law school
Brian was a geometry teacher through the Teach for America program and started the geometry program at his school
##### Thank you for watching the video.
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# Parallel Planes and Lines - Concept
Brian McCall
###### Brian McCall
Univ. of Wisconsin
J.D. Univ. of Wisconsin Law school
Brian was a geometry teacher through the Teach for America program and started the geometry program at his school
Share
In Geometry, a plane is any flat, two-dimensional surface. Two planes that do not intersect are said to be parallel. Parallel planes are found in shapes like cubes, which actually has three sets of parallel planes. The two planes on opposite sides of a cube are parallel to one another.
In Geometry when we talk about this concept of two things being parallel, we aren't just talking about two parallel lines. We could be talking about well, the obvious the two coplanar lines that's what we're going to see the most. But a line and plane can be parallel to each other and two planes can be parallel to each other.
So let's start off by identifying two coplanar lines in this cube right here. So this cube we'll assume that we have 6 congruent faces and that opposite faces are parallel. So if we start off by saying well two coplanar lines, if I look at this front face, so that's going to be one plane. That's going to be place a, b, c, d. I could say that this segment a, b so I'm going to write segment a, b is parallel to segment c, d. So those will be 2 that are in the same plane that will never intersect. Now what about a line and a plane? How can those be parallel? Well taking that same plane a, b, c, d if I took one edge let's say a, b so I'm going to say line segment a, b line segment a, b intersects this plane a, b, c, d it also intersects this plane a, b, e, f. Which means it could be parallel to this bottom face, so the bottom face is c, d, h and g. So we can say that line segment a, b is parallel to plane c, d, h, g so that line will never intersect that plane they're considered parallel.
And last what about two planes? Well since we have a cube we have 3 pairs of parallel planes, so we could start off with front plane a, b, c, d. So I'm going to say a, b, c, d is parallel to the face that is opposite to it e, f, g, h, so I'm going to say e, f, g, h but we could also consider the other 2 pairs, so we could say this side face a, e, h, d. a, e, h, d is parallel to this other side face b, f, g, c and last we could say our two bottom faces or the top and the bottom face. So we have a, b, f, e is parallel to this bottom face which is c, d, h, g.
So two coplanar lines if we look at our cube, there's lots of them but I only named one pair and that was a, b, and c, d. We said that we could have a line parallel to a plane again there's many and I just chose one. And we can say that two planes can be parallel if they never intersect. And because there's only 3 pairs I decided to write them all up, so don't just think that parallelism applies only to two coplanar lines. It could also apply to a line and a plane and two planes.
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# Eureka Math Precalculus Module 4 Lesson 14 Answer Key
## Engage NY Eureka Math Precalculus Module 4 Lesson 14 Answer Key
### Eureka Math Precalculus Module 4 Lesson 14 Example Answer Key
Example 1.
A designer wants to test the safety of a wheelchair ramp she has designed for a building before constructing it, so she creates a scale model. To meet the city’s safety requirements, an object that starts at a standstill from the top of the ramp and rolls down should not experience an acceleration exceeding 2.4 $$\frac{\mathrm{m}}{\mathrm{s}^{2}}$$ .
a. A ball of mass 0.1 kg is used to represent an object that rolls down the ramp. As it is placed at the top of the ramp, the ball experiences a downward force due to gravity, which causes it to accelerate down the ramp. Knowing that the force applied to an object is the product of its mass and acceleration, create a sketch to model the ball as it accelerates down the ramp.
b. If the ball rolls at the maximum allowable acceleration of 2.4 $$\frac{\mathrm{m}}{\mathrm{s}^{2}}$$, what is the angle of elevation for the ramp?
Framp = mball×aball = 0.1 kg×2.4 $$\frac{\mathrm{m}}{\mathrm{s}^{2}}$$ = 0.24 N parallel to the ramp and directed down the ramp
F = mball×agravity = 0.1 g×9.8 $$\frac{\mathrm{m}}{\mathrm{s}^{2}}$$ = 0.98 N directed down toward the ground
sin (θ) = $$\frac{F_{\text {ramp }}}{F}$$ = $$\frac{0.24 \mathrm{~N}}{0.98 \mathrm{~N}}$$
sin-1 (sin (θ)) = sin-1($$\frac{0.24 \mathrm{~N}}{0.98 \mathrm{~N}}$$) ≈ 14.2°
The angle of elevation for the ramp is approximately 14.2°.
c. If the designer wants to exceed the safety standards by ensuring the acceleration of the object does not exceed 2.0 $$\frac{\mathrm{m}}{\mathrm{s}^{2}}$$ , by how much will the maximum angle of elevation decrease?
Framp = mball×aball = 0.1 kg×2.0 $$\frac{\mathrm{m}}{\mathrm{s}^{2}}$$ = 0.2 N parallel to the ramp and directed down the ramp
F = mball×agravity = 0.1 g×9.8 $$\frac{\mathrm{m}}{\mathrm{s}^{2}}$$ = 0.98 N directed down toward the ground
sin (θ) = $$\frac{F_{\text {ramp }}}{F}$$ = $$\frac{0.24 \mathrm{~N}}{0.98 \mathrm{~N}}$$
sin-1 (sin (θ)) = sin-1 ($$\frac{0.24 \mathrm{~N}}{0.98 \mathrm{~N}}$$) ≈ 11.8°
The maximum angle of elevation would be approximately 11.8°, which is 2.4 degrees less than the maximum angle permitted to meet the safety specification.
d. How does the mass of the ball used in the scale model affect the value of θ? Explain your response.
It doesn’t. The mass of the ball is a common factor in the numerator and denominator in the ratio used to calculate θ.
Example 2.
The declination of the sun is the path the sun takes overhead the earth throughout the year. When the sun passes directly overhead, the declination is defined as 0°, while a positive declination angle represents a northward deviation and a negative declination angle represents a southward deviation. Solar declination is periodic and can be roughly estimated using the equation
δ = -23.44°(cos(($$\frac{360}{365}$$)(N + 10))), where N represents a calendar date (e.g., N = 1 is January 1, and δ is the declination angle of the sun measured in degrees).
a. Describe the domain and range of the function.
D: 1 ≤ N ≤ 365 where N is a counting number
R: -23.44° ≤ δ ≤ 23.44°
b. Write an equation that represents N as a function of δ.
δ = -23.44°(cos(($$\frac{360}{365}$$)(N + 10)))
$$\frac{\delta}{-23.44^{\circ}}$$ = (cos(($$\frac{360}{365}$$)(N + 10)))
cos-1 ($$\frac{\delta}{-23.44^{\circ}}$$) = ($$\frac{360}{365}$$)(N + 10)
-10+(365/360) cos-1 ($$\frac{\delta}{-23.44^{\circ}}$$) = N
c. Determine the calendar date(s) for the given angles of declination:
i. 10°
N = -10+($$\frac{360}{365}$$) cos-1 ($$\frac{10^{\circ}}{-23.44^{\circ}}$$) ≈ 107 and 238
N = 107 corresponds to a calendar date of April 17, and N = 238 corresponds to August 26.
ii. -5.2°
N = -10+($$\frac{360}{365}$$) cos-1 ($$\frac{-5.2^{\circ}}{-23.44^{\circ}}$$) ≈ 68 and 277
N = 68 corresponds to a calendar date of March 9, and N = 277 corresponds to October 4.
iii. 25°
No date will correspond to this angle because it lies outside of the domain of the function.
d. When will the sun trace a direct path above the equator?
When the sun passes directly overhead, the declination is 0°. This means that
N = -10+($$\frac{360}{365}$$) cos-1 ($$\frac{0^{\circ}}{-23.44^{\circ}}$$) ≈ 81 and 264, which correspond to the calendar dates March 22 and September 21.
### Eureka Math Precalculus Module 4 Lesson 14 Exercise Answer Key
Exercise 1.
A vehicle with a mass of 1,000 kg rolls down a slanted road with an acceleration of 0.07 $$\frac{0.24 \mathrm{~N}}{0.98 \mathrm{~N}}$$ . The frictional force between the wheels of the vehicle and the wet concrete road is 2,800 newtons.
a. Sketch the situation.
b. What is the angle of elevation of the road?
Froad = mvehicle×avehicle+Ffr = 1000 kg×0.07 $$\frac{0.24 \mathrm{~N}}{0.98 \mathrm{~N}}$$ = 70 N + 2800 N = 2870 N
Fgravity = mvehicle×agravity = 1000 kg×9.8 $$\frac{0.24 \mathrm{~N}}{0.98 \mathrm{~N}}$$ = 9800 N
sin (θ) = $$\frac{F_{\text {road }}}{F_{\text {gravity }}}$$ = $$\frac{2870 \mathrm{~N}}{9800 \mathrm{~N}}$$
sin-1 (sin (θ)) = sin-1 ($$\frac{2870 \mathrm{~N}}{9800 \mathrm{~N}}$$) ≈ 17°
The angle of elevation for the road is approximately 17°.
c. What is the maximum angle of elevation the road could have so that the vehicle described would not slide down the road?
If the vehicle does not slide, then the frictional force must be greater than or equal to the downward force parallel to the road. So, for the maximum angle θ, Froad = Ffriction = 2800 N.
sin (θ) = $$\frac{F_{\text {road }}}{F_{\text {gravity }}}$$ = $$\frac{2870 \mathrm{~N}}{9800 \mathrm{~N}}$$
sin-1 (sin (θ)) = sin-1 ($$\frac{2870 \mathrm{~N}}{9800 \mathrm{~N}}$$) ≈ 16.6°
The angle of elevation for the road that would prohibit sliding is approximately 16.6°.
Exercises 2–3
Exercise 2.
The average monthly temperature in a coastal city in the United States is periodic and can be modeled with the equation y = -8 cos((x-1)(π/6))+17.5, where y represents the average temperature in degrees Celsius and
x represents the month, with x = 1 representing January.
a. Write an equation that represents x as a function of y.
x = 1+$$\frac{6}{\pi}$$ cos-1 ($$\frac{y-17.5}{-8}$$)
b. A tourist wants to visit the city when the average temperature is closest to 25° Celsius. What recommendations would you make regarding when the tourist should travel? Justify your response.
x = 1+$$\frac{6}{\pi}$$ cos-1 ($$\frac{25-17.5}{-8}$$)) ≈ 6.32
If the tourist wants to visit when the temperature is closest to 25° Celsius, she should travel about the second week in June.
Exercise 3.
The estimated size for a population of rabbits and a population of coyotes in a desert habitat are shown in the table. The estimated population sizes were recorded as part of a long-term study related to the effect of commercial development on native animal species.
a. Describe the relationship between sizes of the rabbit and coyote populations throughout the study.
The rabbit population started at approximately 15,000 rabbits and then decreased while the coyote population increased (perhaps because of the abundance of prey for the coyotes). Over time, both the rabbit and coyote populations declined until the coyote population was about 1,800, when the rabbit population increased again. Both species’ population numbers appear to cycle, with the coyotes’ values shifted about
3 years from the rabbit’s values.
b. Plot the relationship between the number of years since the initial count and the number of rabbits. Fit a curve to the data.
Equation of curve: r = 5000 cos($$\frac{\pi n}{6}$$)+10000
c. Repeat the procedure described in part (b) for the estimated number of coyotes over the course of the study.
Equation of curve: c = 200 sin($$\frac{\pi n}{6}$$)+2000
d. During the study, how many times was the rabbit population approximately 12,000? When were these times?
r = 5000 cos($$\frac{\pi n}{6}$$)+10000
n = $$\frac{6}{\pi}$$ cos-1 ($$\frac{r-10000}{5000}$$)
n = $$\frac{6}{\pi}$$ cos-1 ($$\frac{12000-10000}{5000}$$) ≈ 2.2
Given that the function cycles every 12 years, n = 0 + 2.2 = 2.2; n = (12-2.2) = 9.8; n = (12 + 2.2) = 14.2; and n = (24-2.2) = 21.8. The rabbit population was approximately 12,000 four times at 2.2, 9.8, 14.2, and 21.8 years.
e. During the study, when was the coyote population estimate below 2,100?
c = 200 sin($$\frac{\pi n}{6}$$)+2000
n = $$\frac{6}{\pi}$$ sin-1 ($$\frac{c-2000}{200}$$)
n = $$\frac{6}{\pi}$$ sin-1 ($$\frac{2100-2000}{200}$$) = 1,5,13,17
By analyzing the graph of the coyote population estimates, the population of coyotes was less than 2,100 prior to n = 1, between n = 5 and n = 13, and from n = 17 to n = 24.
### Eureka Math Precalculus Module 4 Lesson 14 Problem Set Answer Key
Question 1.
A particle is moving along a line at a velocity of y = 3 sin($$\frac{2 \pi x}{5}$$)+2 $$\frac{m}{s}$$ at location x meters from the starting point on the line for 0 ≤ x ≤ 20.
a. Find a formula that represents the location of the particle given its velocity.
y = $$\frac{5}{2 \pi}$$⋅sin-1($$\frac{x-2}{3}$$)
b. What is the domain and range of the function you found in part (a)?
The domain is -1 ≤ x ≤ 5, and the range is –$$\frac{5}{4}$$ ≤ y ≤ $$\frac{5}{4}$$.
c. Use your answer to part (a) to find where the particle is when it is traveling 5 m/s for the first time.
The particle will be located at x = 1.2 meters from the starting point on the line.
d. How can you find the other locations the particle is traveling at this speed?
In this case, the velocity is a maximum, so it will only occur once every period. All other values can be found by adding multiples of 5 to the location. If it was not a maximum, we could subtract the location from 5/2 to find another value within the same period and then add multiples of 5 to find analogous values in other periods.
Question 2.
In general, since the cosine function is merely the sine function under a phase shift, mathematicians and scientists regularly choose to use the sine function to model periodic phenomena instead of a mixture of the two. What behavior in data would prompt a scientist to use a tangent function instead of a sine function?
The tangent function has infinitely many vertical asymptotes and rapidly takes on extreme values. Since the tangent function is the ratio between the sine and cosine functions, it will probably show up when comparing the ratio of two sets of periodic data. Otherwise, the extreme values would be reasons to use the tangent function.
Question 3.
A vehicle with a mass of 500 kg rolls down a slanted road with an acceleration of 0.04 $$\frac{\mathrm{m}}{s^{2}}$$ . The frictional force between the wheels of the vehicle and the road is 1,800 newtons.
a. Sketch the situation.
b. What is the angle of elevation of the road?
Froad = ma + Ffr = 500 kg×0.04 $$\frac{\mathrm{m}}{s^{2}}$$ + 1800 N = 20 N + 1800 N = 1820 N
Fg = mg = 500 kg×9.8 $$\frac{\mathrm{m}}{s^{2}}$$ = 4900 N
sin(θ) = $$\frac{F_{\text {road }}}{F_{\mathrm{g}}}$$ = $$\frac{1820 \mathrm{~N}}{4900 \mathrm{~N}}$$
sin-1 (sin(θ)) = sin-1 ($$\frac{1820 \mathrm{~N}}{4900 \mathrm{~N}}$$) ≈ 21.8°
The angle of elevation for the road is approximately 21.8°.
c. The steepness of a road is frequently measured as grade, which expresses the slope of a hill as a ratio of the change in height to the change in the horizontal distance. What is the grade of the hill described in this problem?
We need to find the perpendicular force. We get
cos(21.8) = $$\frac{F_{\text {perp }}}{4900}$$
Fperp ≈ 4549.46.
So the grade of the hill is $$\frac{1820}{4549.46}$$ ≈ 40%.
Question 4.
Canton Avenue in Pittsburgh, PA is considered to be one of the steepest roads in the world with a grade of 37%.
a. Assuming no friction on a particularly icy day, what would be the acceleration of a 1,000 kg car with only gravity acting on it?
tan(θ) = 0.37
θ ≈ 20.30
Since the acceleration due to gravity is the only acceleration on the car, the acceleration due to gravity is being transferred into an acceleration as the car goes down the hill. We can envision the force due to gravity as two separate forces, one that is parallel to the road and the second that is perpendicular to the road. The force parallel to the road is F = m⋅g⋅sin(θ), and the acceleration parallel to the road is
a = g⋅sin(θ)
a = 9.8 sin(20.3)
a ≈ 3.4
The acceleration is 3.4 $$\frac{\mathrm{m}}{s^{2}}$$ .
b. The force due to friction is equal to the product of the force perpendicular to the road and the coefficient of friction μ. For icy roads and a non-moving vehicle, assume the coefficient of friction is μ = 0.3. Find the force due to friction for the car above. If the car is in park, will it begin sliding down Canton Avenue if the road is this icy?
The force perpendicular to the road is 9191.3 N. F = m⋅g⋅cos(θ) = 1000⋅9.8⋅cos(20.3) ≈ 9191.3
Thus, the force due to friction is approximately 2757.4 N. μF ≈ 2757.4.
The force parallel to the road is 3400.0 N. F = 1000⋅9.8⋅sin(20.3) ≈ 3400.0
Because the force parallel to the road is greater than the force due to friction, the car will slide down the road once Canton Avenue gets this icy.
c. Assume the coefficient of friction for moving cars on icy roads is μ = 0.2. What is the maximum angle of road that the car will be able to stop on?
The car will be able to slow (and eventually stop) when the force due to friction is greater than the force parallel to the road, so we need to solve,
9800⋅sin(θ) = 0.2⋅9800 cos(θ)
$$\frac{\sin (\theta)}{\cos (\theta)}$$ = 0.2
tan(θ) = 0.2.
So the car will be able to slow on any hill with less than a 20% grade, which corresponds to an angle of 11.3°.
Question 5.
Talladega Superspeedway has some of the steepest turns in all of NASCAR. The main turns have a radius of about 305 m and are pitched at 33°. Let N be the perpendicular force on the car and N_v and N_h be the vertical and horizontal components of this force, respectively. See the diagram below.
a. Let μ represent the coefficient of friction; recall that μN gives the force due to friction. To maintain the position of a vehicle traveling around the bank, the centripetal force must equal the horizontal force in the direction of the center of the track. Add the horizontal component of friction to the horizontal component of the perpendicular force on the car to find the centripetal force. Set your expression equal to (mv2)/r, the centripetal force.
The force due to friction is μN, and the horizontal component then is μN cos(θ).
The horizontal perpendicular force would be N sin(θ).
We get, $$\frac{m v^{2}}{r}$$ = N sin(θ)+μN cos(θ).
b. Add the vertical component of friction to the force due to gravity, and set this equal to the vertical component of the perpendicular force.
N cos(θ) is the vertical component of the perpendicular force.
The force due to gravity is mg, and the vertical force of friction is μN sin(θ).
We get, N cos(θ) = mg + μN sin(θ).
c. Solve one of your equations in part (a) or part (b) for m, and use this with the other equation to solve for v.
m = $$\frac{N \cos (\theta)-\mu N \sin (\theta)}{g}$$
$$\frac{N \cos (\theta)-\mu N \sin (\theta)}{g} \cdot \frac{v^{2}}{r}$$ = N sin(θ)+μN cos(θ)
$$\frac{v^{2}}{g r}$$ = $$\frac{N \sin (\theta)+\mu N \cos (\theta)}{N \cos (\theta)-\mu N \sin (\theta)}$$
We can factor out the N and cancel the common factor from here onward.
v2 = gr⋅$$g r \cdot \frac{\sin (\theta)+\mu \cos (\theta)}{\cos (\theta)-\mu \sin (\theta)}$$
v = $$\sqrt{g r \cdot \frac{\sin (\theta)+\mu \cos (\theta)}{\cos (\theta)-\mu \sin (\theta)}}$$
d. Assume μ = 0.75, the standard coefficient of friction for rubber on asphalt. For the Talladega Superspeedway, what is the maximum velocity on the main turns? Is this about how fast you might expect NASCAR stock cars to travel? Explain why you think NASCAR takes steps to limit the maximum speeds of the stock cars.
v = $$\sqrt{g r \cdot \frac{\sin (\theta)+\mu \cos (\theta)}{\cos (\theta)-\mu \sin (\theta)}}$$
= $$\sqrt{9.8 \cdot 305 \cdot \frac{\sin (33)+0.75 \cos (33)}{\cos (33)-0.75 \sin (33)}}$$
≈ 90.3
90.3 m/s is about 202 mph, which is the speed most people associate with NASCAR stock cars (although they usually go slower than this). This means that at Talladega, the race cars are able to go at their maximum speeds through the main turns.
If the cars go any faster than this, then they would drift up toward the wall, which may prompt them to oversteer and possibly spin out of control. NASCAR may limit the speeds of the cars because the racetracks themselves are not designed for cars that can go faster.
e. Does the friction component allow the cars to travel faster on the curve or force them to drive slower? What is the maximum velocity if the friction coefficient is zero on the Talladega roadway?
The friction component is in the direction toward the center of the racetrack (away from the walls), so it allows them to travel faster more safely. If μ = 0, the equation becomes v = $$\sqrt{g r \tan (\theta)}$$. Therefore, the maximum velocity is about 44.1 $$\frac{\mathrm{m}}{\mathrm{s}}$$ or 99 mph.
f. Do cars need to travel slower on a flat roadway making a turn than on a banked roadway? What is the maximum velocity of a car traveling on a 305 m turn with no bank?
They need to travel much slower on a flat roadway making a turn than when traveling on a banked roadway. The normal force does not prevent the car from spinning out of control the way it does on a banked turn.
If θ = 0, then the equation becomes v = . Therefore, the maximum velocity is about 47.3$$\frac{\mathrm{m}}{\mathrm{s}}$$ or 106 mph, which is much slower than the velocity that cars can travel on the banked turns (202 mph).
Question 6.
At a particular harbor over the course of 24 hours, the following data on peak water levels was collected (measurements are in feet above the MLLW):
a. What appears to be the average period of the water level?
It takes 13 hours to get from the first low-point to the second, and 13 hours to get from the first high-point to the second, so the average period is $$\frac{13 + 13}{2}$$ = 13.
b. What appears to be the average amplitude of the water level?
There are three areas we can examine to get amplitudes, from 1:30 to 7:30, 7:30 to 14:30, and 14:30 to 20:30. We get amplitudes of $$\frac{8.21-(-0.211)}{2}$$ = 4.2105, $$\frac{8.21-(-0.619)}{2}$$ = 4.4145, and $$\frac{7.518-(-0.619)}{2}$$ = 4.0685. We get 4.231 as the average amplitude.
c. What appears to be the average midline for the water level?
3.748
d. Fit a curve of the form y = A sin(ω(x-h))+k or y = A cos(ω(x-h))+k modeling the water level in feet as a function of the time.
Since it would make our curve more inaccurate to guess at what point the water levels will cross the midline, we can either start at 1:30 or 7:30 and use the cosine function. For 7:30, we get
y = 4.231 cos($$\frac{2 \pi}{13}$$ (x-7.5))+3.748.
e. According to your function, how many times per day will the water level reach its maximum?
It should reach its maximum levels twice a day usually, but there is the rare possibility that it will reach its maximum only once.
f. How can you find other time values for a particular water level after finding one value from your function?
The values repeat every 13 hours, so immediately once a value is found, adding any multiple of 13 will give other values that work. Additionally, if you have the inverse cosine value for a particular water level (but have not solved for x yet), then take the opposite, solve normally, and you will have another time to which you can add multiples of 13.
g. Find the inverse function associated with the function in part (d). What is the domain and range of this function? What type of values does this function output?
y = $$\frac{13}{2 \pi}$$ cos-1($$\frac{x-3.748}{4.231}$$)+7.5
The domain is all real numbers x, such that -0.483 ≤ x ≤ 7.979, and the range is all real numbers y such that 7.5 ≤ y ≤ 14.
### Eureka Math Precalculus Module 4 Lesson 14 Exit Ticket Answer Key
Question 1.
The minimum radius of the turn r needed for an aircraft traveling at true airspeed v is given by the following formula
r = $$\frac{v^{2}}{g \tan (\theta)}$$
where r is the radius in meters, g is the acceleration due to gravity, and θ is the banking angle of the aircraft. Use
g = 9.78 $$\frac{\mathrm{m}}{s^{2}}$$ instead of 9.81 $$\frac{\mathrm{m}}{s^{2}}$$ to model the acceleration of the airplane accurately at 30,000 ft.
a. If an aircraft is traveling at 03 $$\frac{m}{s}$$, what banking angle is needed to successfully turn within 1 km?
1000 = $$\frac{103^{2}}{9.78 \cdot \tan (\theta)}$$
θ = tan-1($$\frac{103^{2}}{1000 \cdot 9.78}$$)
≈ 47.328
θ = tan-1($$\frac{v^{2}}{r \cdot g}$$)
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# How to Solve Subtraction Equations
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Learning how to solve subtraction equations is an important skill for all students. This skill can be used to solve fractional problems, add and subtract fractions, and more. These problems involve comparing values. It is also important to understand when two numbers are equal. Here are some examples. If x is greater than three, then it is equal to five. If x is smaller than three, it is equal to -b.
## Problems involving fractions
Solving subtraction equations can be difficult if you are working with fractions. These types of problems often require factoring and grouping. Factoring fractions is easier if the factors are the same and you don’t need to worry about the sign.
Fractions are a very common concept in mathematics and they appear in many practical situations. There are two types of fractions: arithmetic fractions, which are numbers, and algebraic fractions, which contain algebraic expressions. Each type has a different way of solving the problem.
For example, let’s say that Mary needs to use 3/4 cup of flour for her recipe, but she only has 2/4 cup. This means that she needs to buy 3/4 cup of flour and has only two-fourths of a cup in her pantry. If she uses the common denominator method, she will be able to get 8 out of 10 correct answers for two of the three activities.
The common denominator is the expression of all the factors of both the numerator and denominator. The least common denominator, on the other hand, contains the least amount of factors. For complicated fractions, the fraction will be in the numerator and denominator, or both. Whenever possible, it is easier to solve these problems using a mathematical calculator.
The common denominator method can be used to simplify equations with fractions. This method requires multiple operations. It is best to choose the method that is easiest for you to follow. Then, check your solution by substituting it into the original equation. You can also clear fractions by multiplying the fractions with the least common denominator.
Sometimes, the equation will be too complex to be solved using one operation. In such cases, you need to simplify the equation by removing the parentheses and multiplying both sides with the same denominator. Sometimes, you will have to use both methods, or a combination of both.
## Substitution method
The substitution method is a simple method to solve equations that have two variables. It involves solving the first equation for the first variable and then substituting the other variable into the equation. Finally, we write the solution as an ordered pair, and check to make sure that the solution is valid in both equations.
The substitution method is best used when the coefficient of a term is the same. For example, Ax+By+C=0. This method also works well if the other variable has the same coefficient, such as Ax+By+C. Using this method, you can solve other equations by combining like terms.
Using the substitution method is also a good option for solving simultaneous equations. However, you have to be careful to avoid introducing fractions because it can lead to miscalculations. If you have a fraction that needs to be solved, it is best to remove it from the equation.
In addition to solving equations using this method, you can also multiply two equations. You should make sure that both equations have the same coefficients. This will give you a solution that will satisfy both equations. And, it will also be faster. With practice, you can apply the substitution method to solve equations.
The substitution method is an easier way to solve equations. It involves setting the coefficient of or to one. This way, you will know the value of the other variable without having to perform any additional steps. When solving subtraction equations, you can use the substitution method to solve the equations. This method is easier than using the addition method, and it also tends to make fewer mistakes.
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# DECIMAL REPRESENTATION OF RATIONAL NUMBERS
Let us consider the rational number p/q, where both p and q are integers and ≠ 0.
We can get the decimal representation of the rational number p/q by long division division.
When we divide p by q using long division method either the remainder becomes zero or the remainder never becomes zero and we get a repeating string of remainders.
Case 1 (Remainder = 0) :
Let us express 7/16 in decimal form. Then 7/16 = 0.4375.
In this example, we observe that the remainder becomes zero after a few steps.
Also the decimal expansion of 7/16 terminates.
Similarly, using long division method we can express the following rational numbers in decimal form as
1/2 = 0.5
7/5 = 1.5
-8/25 = -0.32
In the above examples, the decimal expansion terminates or ends after a finite number of steps.
Key Concept (Terminating Decimal) :
When the decimal expansion of p/q, q ≠ 0 terminates (i,e., comes to an end), the decimal expansion is called terminating.
Case 2 ( Remainder ≠ 0) :
Does every rational number has a terminating decimal expansion?
Before answering the question, let us express 5/11 and 7/6 in decimal form.
Thus, the decimal expansion of a rational number need not terminate.
In the above examples, we observe that the remainders never become zero. Also we note that the remainders repeat after some steps. So, we have a repeating (recurring) block of digits in the quotient.
Key Concept (Non-terminating and Recurring) :
In the decimal expansion of p/q, q ≠ 0 when the remainder never becomes zero, we have a repeating (recurring) block of digits in the quotient. In this case, the decimal expansion is called non-terminating and recurring.
To simplify the notation, we place a bar over the first block of the repeating (recurring) part and omit the remaining blocks.
So, we can write the expansion of 5/11 and 7/6 as follows..
The following table shows decimal representation of the reciprocals of the first ten natural numbers. We know that the reciprocal of a number n is 1/n. Obviously, the reciprocals of natural numbers are rational numbers.
Number
Reciprocal
Type of Decimal
1
1.0
Terminating
2
0.5
Terminating
3
0.333.......
Non-terminating and recurring
4
0.25
Terminating
5
0.2
Terminating
6
0.1616.......
Non-terminating and recurring
7
0.142857142857.......
Non-terminating and recurring
8
0.125
Terminating
9
0.111.......
Non-terminating and recurring
10
0.1
Terminating
Thus we see that,
A rational number can be expressed by either a terminating or a non-terminating and recurring (repeating) decimal expansion.
The converse of this statement is also true.
That is, if the decimal expansion of a number is terminating or non-terminating and recurring (repeating), then the number is a rational number.
## Practice Questions
Express the following rational numbers as decimal numbers.
1) 3/4
2) 5/8
3) 9/16
4) 7/25
5) 47/99
6) 1/999
7) 26/45
8) 27/110
9) 2/3
10) 14/9
1) 3/4 = 0.75
2) 5/8 = 0.625
3) 9/16 = 0.5625
4) 7/25 = 0.0.28
5) 47/99 = 0.474747.........
6) 1/999 = 0.001001001........
7) 26/45 = 0.577777........
8) 27/110 = 0.2454545.........
9) 2/3 = 0.6666........
10) 14/9 = 1.5555..........
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## Recent Articles
1. ### Exponential vs Linear Growth Worksheet
May 23, 22 01:59 AM
Exponential vs Linear Growth Worksheet
2. ### Linear vs Exponential Growth
May 23, 22 01:59 AM
Linear vs Exponential Growth - Concept - Examples
|
# How do you differentiate f(x)=tan(e^(1/x)) using the chain rule?
Nov 6, 2016
$\frac{d}{\mathrm{dx}} \left[f \left(g \left(x\right)\right)\right] = f ' \left(g \left(x\right)\right) g ' \left(x\right)$
#### Explanation:
The chain rule allows us to take the derivative of a composition of two or more functions. For two functions:
$\frac{d}{\mathrm{dx}} \left[f \left(g \left(x\right)\right)\right] = f ' \left(g \left(x\right)\right) g ' \left(x\right)$
We can see that the function $f \left(x\right)$ is composed of three functions, each nestled inside the last.
By the chain rule, we first need to take the derivative of the outermost function and work our way in. It's like taking the function apart layer by layer until we get all the way through.
First, we take the derivative of the tangent function, which is ${\sec}^{2}$. This gives us:
${\sec}^{2} \left({e}^{\frac{1}{x}}\right)$
Next, we take the derivative of the $e$ term. This, as always, is just itself. We multiply this by the first portion of the derivative as we determined above. Thus, we have so far:
${\sec}^{2} \left({e}^{\frac{1}{x}}\right) \cdot {e}^{\frac{1}{x}}$
Lastly, we take the derivative of $\frac{1}{x}$. Recognizing that this is equivalent to ${x}^{-} 1$, we have $- {x}^{- 2}$. And so, our final answer becomes:
$- {\sec}^{2} \left({e}^{\frac{1}{x}}\right) {e}^{\frac{1}{x}} {x}^{-} 2$
This can also be written as:
$\frac{- {\sec}^{2} \left({e}^{\frac{1}{x}}\right) {e}^{\frac{1}{x}}}{x} ^ 2$
|
Lesson Objectives
• Demonstrate an understanding of decimals
• Demonstrate an understanding of fractions
• Learn how to convert between a decimal and a percentage
• Learn how to convert between a fraction and a percentage
• Learn how to multiply with percentages
## How to Convert between Decimals, Fractions, and Percentages
### What is a Percent?
We previously learned that decimals and fractions are used to describe parts of a whole. A percent is another commonly used method to describe part of a whole. A percent is a fraction whose denominator is 100. The literal meaning of percent is "parts per hundred" or parts (number in the numerator) per (division or fraction bar) 100 (denominator). When we encounter a percent, we see the "%" symbol placed behind the number. As an example, suppose we had 100 boxes with 30 shaded yellow and 70 shaded blue:
In our above example, we can see that 30 boxes are shaded yellow out of a total of 100. This can be written with a fraction as: $$\frac{30}{100}$$ To write this as a percentage we take the top number of the fraction (30) followed by the "%" symbol: $$30\%$$ Similarly, we know there are 70 blue boxes. We can show this with a fraction as: $$\frac{70}{100}$$ To write this as a percentage we take the top number of the fraction (70) followed by the "%" symbol: $$70\%$$ We can say that 30 percent (30%) of the boxes are yellow and 70 percent (70%) of the boxes are blue.
### Converting a Fraction with a Denominator of 100 to a Percent
When converting to a percentage, the easy scenario occurs when we have a fraction with a denominator of 100. In this case, we can write the number in the numerator, followed by the "%" symbol. Let's try a few examples.
Example 1: Convert each into a percentage $$\frac{9}{100}$$ We write the numerator (9) followed by the percentage symbol "%":
$$\frac{9}{100} = 9\%$$ Example 2: Convert each into a percentage $$\frac{84}{100}$$ We write the numerator (84) followed by the percentage symbol "%":
$$\frac{84}{100} = 84\%$$ We don't have to start with a denominator of 100 to use a percentage. We previously learned that proportions were equivalent ratios/fractions. In this case, we are looking to convert our fraction into an equivalent fraction where 100 is the denominator. Let's take a look at a few examples.
Example 3: Convert each into a percentage $$\frac{17}{25}$$ The denominator is 25, what can we multiply by 25 that would result in a product of 100?
? x 25 = 100
Use a related division statement:
100 ÷ 25 = 4
We will create an equivalent fraction where 100 is the denominator by multiplying the numerator and denominator by 4: $$\frac{17}{25} \cdot \frac{4}{4} = \frac{68}{100}$$ We can then convert this into a percentage: $$\frac{68}{100} = 68\%$$ We can place this in terms of our original fraction: $$\frac{17}{25} = 68\%$$ Example 4: Convert each into a percentage $$\frac{15}{50}$$ The denominator is 50, what can we multiply by 50 that would result in a product of 100?
? x 50 = 100
Use a related division statement:
100 ÷ 50 = 2
We will create an equivalent fraction where 100 is the denominator by multiplying the numerator and denominator by 2: $$\frac{15}{50} \cdot \frac{2}{2} = \frac{30}{100}$$ We can then convert this into a percentage: $$\frac{30}{100} = 30\%$$ We can place this in terms of our original fraction: $$\frac{15}{50} = 30\%$$
### Converting Decimals into Percents
In most cases, it will not be practical to convert a fraction into a percentage using an equivalent fraction where the denominator is 100. Let's suppose we saw: $$\frac{6}{13}$$ We would need to multiply 6 and 13 by the fraction 100/13 in order to get a denominator of 100. This procedure would be extremely tedious and fortunately, there is an easier way. Let's begin by talking about how to convert a decimal into a percent:
• Move the decimal point two places to the right
• Add the percentage "%" symbol to the end of the number
We can reverse this process to go from a percent back to a decimal:
• Delete the percentage symbol "%"
• Move the decimal point two places to the left
Let's look at a few examples.
Example 5: Convert each decimal into a percent
0.005
We move our decimal point two places to the right:
0.005 » 0.5
Place a percentage symbol "%" at the end of the number:
0.5 » 0.5%
0.005 = 0.5%
Example 6: Convert each decimal into a percent
0.98
We move our decimal point two places to the right:
0.98 » 98.0
Place a percentage symbol "%" at the end of the number:
98 » 98%
0.98 = 98%
Example 7: Convert each percent into a decimal
34%
We delete our percentage symbol:
34% » 34
We move our decimal point two places to the left:
34.0 » 0.34
34% = 0.34
Now let's return to the situation where we have a fraction that does not easily transform into an equivalent fraction where 100 is the denominator. In this case, we will divide the numerator by the denominator and obtain a decimal form for the number. We can then transform our decimal into a percentage. Suppose we ran into 9/15 and wanted this as a percentage. We could divide 9 by 15 and get our decimal form:
9 ÷ 15 = 0.6
We can then transform this decimal into a percentage:
0.6 » 60%
When we compare this to our other method, we can see how tedious the process would be:
$$\require{cancel}\frac{9 \cdot \frac{100}{15}}{15 \cdot \frac{100}{15}} = \frac{\cancel{9}3 \cdot \frac{\cancel{100}20}{\cancel{15}}}{\cancel{15} \cdot \frac{100}{\cancel{15}}} = \frac{60}{100}$$ $$\frac{60}{100} = 60\%$$ Let's try another example.
Example 8: Convert each fraction into a percentage $$\frac{5}{8}$$ Divide 5 by 8:
5 ÷ 8 = 0.625
Convert to a percentage:
0.625 » 62.5%
Our answer: $$\frac{5}{8} = 62.5\%$$
### Multiplying with Percents
In some cases, we may need to find a certain percentage of something. For example, we may have 500 containers and want to ship 10% of the containers. We can find the number of containers to ship by multiplying the number 500 by 10%. In most cases, the easiest way to do this is by converting to a decimal and then performing the multiplication:
500 x 0.10 = 50
10% of 500 is 50
We would ship 50 containers.
Let's take a look at a few examples.
Example 9: Find 20% of 300 and 1425:
300
Multiply 300 by .2:
300 x 0.20 = 60
20% of 300 is 60
1425
Multiply 1425 by .2:
1425 x 0.2 = 285
20% of 1425 is 285
Example 10: Solve the following word problem
A government department spends $11,000 each year on supplies. Due to budget cuts, they are asked to decrease their spending by 22%. Given this decrease, what is the new budget for supplies? First let's find 22% of 11,000: 0.22 x 11,000 = 2420 This means the department will need to decrease its spending by$2420. We can subtract the initial spending ($11,000) minus the decrease ($2420):
11,000 - 2420 = 8580
The department's new budget for supplies will be \$8580.
### Percents More than 100
In some cases, we may see percents that are larger than 100. We may see statements such as: sales are up 120% or XYZ company's stock is up 340%. Having a percentage larger than 100 depends on the context of what's being discussed. Suppose we all sit for a math exam with a total of 100 questions. In this particular case, we cannot say we answered 130% of the questions since at most we can answer 100 questions or 100% of the questions. As another example, suppose we heard that a particular toothpaste was favored by 190% of the people surveyed. This would be another example where a percentage over 100 does not make logical sense. If every person surveyed favored the toothpaste, the result would be 100%. We generally will see percentages over 100 when we talk about increases. We may see a stock index start with a value of 6000. If the value doubles, we could say it increased by 100%. This means it went from 6000 to 12,000. Now, what if our stock index increased from 6000 to 18,000? In this case, we could say we had an increase of 200%.
|
# The value of the determinant
Question:
The value of the determinant $\left|\begin{array}{ccc}x & x+y & x+2 y \\ x+2 y & x & x+y \\ x+y & x+2 y & x\end{array}\right|$ is
(a) $9 x^{2}(x+y)$
(b) $9 y^{2}(x+y)$
(c) $3 y^{2}(x+y)$
(d) $7 x^{2}(x+y)$
Solution:
$\left|\begin{array}{ccc}x & x+y & x+2 y \\ x+2 y & x & x+y \\ x+y & x+2 y & x\end{array}\right|$
$=\left|\begin{array}{ccc}-2 y & y & y \\ x+2 y & x & x+y \\ -y & 2 y & -y\end{array}\right|$ [Applying $R_{1} \rightarrow R_{1}-R_{2}$ and $R_{3} \rightarrow R_{3}-R_{2}$ ]
$=y^{2}\left|\begin{array}{ccc}-2 & 1 & 1 \\ x+2 y & x & x+y \\ -1 & 2 & -1\end{array}\right|$ [Taking $(y)$ common from $R_{1}$ and from $R_{3}$ ]
$=y^{2}\left|\begin{array}{ccc}-2 & -3 & 3 \\ x+2 y & 3 x+4 y & -y \\ -1 & 0 & 0\end{array}\right|$ [Applying $C_{2} \rightarrow C_{2}+2 C_{1}$ and $C_{3} \rightarrow C_{3}-C_{1}$ ]
$=y^{2}[-1(3 y-9 x-12 y)]$
$=y^{2}[9 y+9 x]$
$=9 y^{2}(y+x)$
Hence, the correct option is (b).
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# Two years ago, Dilip was three times as old as his son and two years hence, twice his age will be equal to five times his son. Find their present ages.
Last updated date: 21st Jul 2024
Total views: 383.4k
Views today: 6.83k
Answer
Verified
383.4k+ views
Hint: Notice that we get an equality from the information of present age and age after two years. We can use this information to formulate an equation which on solving will give us Dilip and his son’s age.
Complete step by step answer:
1) Assign variables for father and son’s age. One can assign any variable.
Let Dilip’s age be x and son’s age be y.
Therefore, 2 years ago, the age of the father and son is:
Dilip’s age $=x$
Son’s age $= y$
Therefore, Dilip’s age $= 3y$
At present, their age is;
Son’s age $= y + 2$
Dilip’s age$= x + 2$$= 3y +2$ ------(1)
And, after two years, their ages will be:
Son’s age $= y + 4$
Dilip’s age $= x + 4 = 3y + 4$
2) We have that 2 years later, two times (twice) Dilip’s age will be 5 times his son after 2 years,
Therefore, we get the equation,
$= 5 ( y + 4 ) = 3 ( y + 4 )$
Which on solving, would give us
$\begin{array}{l}5y + 20 = 6y + 8\\5y + 20 - 6y - 8 = 0\\ - y + 12 = 0\\\therefore y = 12\end{array}$
Hence, the age of Dilip’s son is 12. On substituting this in the equation (1), we get,
$3(12) + 4 = 36 + 4 = 40$
Thus, Dilip’s age is 40 at present.
Note:
1) In questions like this, writing down the information helps clear the picture.
2) Students are prone to rushing it in the end when it comes to such problems. Make sure not to make any silly mistakes.
3) Number the equations which would be required later in the problem.
4) These kinds of problems often seem easy and in a hurry students are prone to make mistakes. Rather than treating it like a puzzle, treat it like a mathematics problem and don’t solve it in your head.
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# How do you solve 3x + \frac { 1} { 2} = \frac { 2} { 3} x - 6?
Mar 25, 2018
$x = - \frac{39}{14}$
#### Explanation:
$3 x + \frac{1}{2} = \frac{2}{3} x - 6$
$3 x - \frac{2}{3} x = - \frac{1}{2} - 6$
$\frac{3 \left(3\right) - 2}{3} x = \frac{2 - 2 \left(12\right)}{2}$ $\Leftarrow$ finding the LCD
$\frac{7}{3} x = - \frac{13}{2}$ $\Leftarrow$ bring the $\frac{7}{3}$ to the other side to isolate for $x$
$x = - \frac{13}{2} \div \frac{7}{3}$
$x = - \frac{13}{2} \cdot \frac{3}{7}$
$x = - \frac{39}{14}$
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Congruent Shapes
Start Practice
## What Are Congruent Shapes?
Congruent shapes have the exact same shape and size.
Of the shapes below, only the triangles are congruent.
If you put two congruent shapes on top of each other, the top one completely covers the bottom one.
Tip: congruent shapes can have different colors.
To test if two shapes are congruent, they need to have:
✅ the same shape
✅ the same size
### Are These Shapes Congruent?
Let's check.
(1) Are they the same shape?
✅ Yes, they're the same shape, a circle.
(2) Are they the same size?
One circle is big. The other one is smaller than it.
❎ No, they're not the same size.
👉 These shapes are not congruent.
### Example 2
Are these congruent?
Let's check!
(1) Are they the same shape?
One shape has 5 sides. The other shape has 8 sides.
❎ No, they're not the same shape.
(2) Are they the same size?
❎ No, they're not the same size.
This pair does not check out on the two tests.
👉 These shapes are not congruent.
### Example 3
(1) Are these triangles the same shape?
✅ Yes, they're the same shape, a triangle.
(2) Are they the same size?
✅ Yes, they're the same size.
👉 These two shapes are congruent! 😃
Their colors are different.
Are they still congruent?
Yes, because color doesn't matter when looking for congruence.
The triangles are congruent because they have the same shape and size.
### Example 4
Are these congruent? 🤔
It might seem like they're not.
But let's check.
(1) Are they the same shape?
✅ Yes, they're both triangles. They are the same kind of triangle.
Their sides and angles are the same.
(2) Are they the same size?
It's tough to tell.
What if we rotate it to the right?
Like this:
Do their sizes look the same now?
✅ Yes, they have the same size!
This means that these triangles are congruent.
Tip: Congruent shapes match even if you flip, slide, or turn them.
Great job learning about congruent shapes!
Now, ace the practice. 💪 It'll help you remember what congruent shapes are for longer.
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# Slope-Intercept Form Of A Line
## The Definition, Formula, and Problem Example of the Slope-Intercept Form
Slope-Intercept Form Of A Line – One of the numerous forms used to represent a linear equation one that is commonly found is the slope intercept form. It is possible to use the formula of the slope-intercept identify a line equation when you have the straight line’s slope , and the y-intercept. It is the point’s y-coordinate where the y-axis intersects the line. Learn more about this particular linear equation form below.
## What Is The Slope Intercept Form?
There are three fundamental forms of linear equations: the traditional, slope-intercept, and point-slope. Even though they can provide identical results when utilized in conjunction, you can obtain the information line that is produced more quickly using this slope-intercept form. It is a form that, as the name suggests, this form employs a sloped line in which you can determine the “steepness” of the line reflects its value.
This formula can be used to calculate the slope of straight lines, the y-intercept or x-intercept where you can apply different formulas that are available. The equation for this line in this specific formula is y = mx + b. The straight line’s slope is signified through “m”, while its y-intercept is signified through “b”. Every point on the straight line can be represented using an (x, y). Note that in the y = mx + b equation formula the “x” and the “y” are treated as variables.
## An Example of Applied Slope Intercept Form in Problems
The real-world In the real world, the “slope intercept” form is commonly used to show how an item or problem evolves over its course. The value provided by the vertical axis represents how the equation tackles the degree of change over the amount of time indicated via the horizontal axis (typically time).
A basic example of this formula’s utilization is to determine how much population growth occurs within a specific region as the years pass by. Based on the assumption that the population in the area grows each year by a certain amount, the point value of the horizontal axis will grow by one point with each passing year and the point value of the vertical axis is increased to represent the growing population by the amount fixed.
You may also notice the starting point of a problem. The beginning value is at the y’s value within the y’intercept. The Y-intercept marks the point where x is zero. By using the example of a problem above the beginning point could be the time when the reading of population starts or when the time tracking begins , along with the related changes.
This is the point in the population where the population starts to be documented in the research. Let’s suppose that the researcher began to do the calculation or measure in 1995. This year will be the “base” year, and the x = 0 points will occur in 1995. So, it is possible to say that the 1995 population represents the “y”-intercept.
Linear equation problems that utilize straight-line equations are typically solved in this manner. The beginning value is depicted by the y-intercept and the rate of change is expressed in the form of the slope. The primary complication of the slope intercept form is usually in the horizontal interpretation of the variable especially if the variable is attributed to a specific year (or any kind in any kind of measurement). The first step to solve them is to make sure you are aware of the variables’ meanings in detail.
|
# Evaluate Integrals
Evaluate integrals: Tutorials with examples and detailed solutions. Also exercises with answers are presented at the end of the page. In what follows, C is the constant of integration.
## Examples
### Example 1
Evaluate the integral
6 cos x sinx dx
Solution to Example 1:
We first use the trigonometric identity 2sin x cos x = sin (2x) to rewrite the integral as follows
6 cos x sinx dx = 3 sin 2x dx
Substitution: Let u = 2x which leads to du / dx = 2 or du = 2 dx or dx = du / 2, the above integral becomes
6 cos x sinx dx = 3 (1/2) sin u du
We now use integral formulas for sine function to obtain
6 cos x sinx dx = - (3/2) cos u + c
We now substitute u by 2x into the above result to obtain the final result as follows
6 cos x sinx dx = - (3/2) cos 2x + c
As an exercise, differentiate - (3/2) cos 2x + c to obtain 6 sin x cos x which is the integrand in the given integral. This is a way to check the answer to integrals evaluation.
### Example 2
Evaluate the integral
x √ (x + 1) dx
Solution to Example 2:
Substitution: Let u = x + 1 which leads to du = dx. We also have x = u - 1. The given integral becomes
x √(x + 1) dx = (u - 1) u 1/2 du = (u 3/2 - u 1/2) du
We now use property for integral of sum of functions and the formula for integration of power function
= (2 / 5) u
5/2 - (2 / 3) u 3/2 + c
We now substitute u by x + 1 into the above result to obtain the final result as follows
= (2 / 5) (x + 1)
5/2 - (2 / 3) (x + 1) 3/2 + c + c
To check the final answer, differentiate the indefinite integral obtained to obtain the integrand x √(x + 1) in the given integral.
### Example 3
Evaluate the integral
cos 2 dx
Solution to Example 3:
Use the trigonometric identity cos 2 = (1 + cos(2x)) / 2 to rewrite the given integral as
cos 2 dx = (1 + cos(2x)) / 2 dx
Substitute: u = 2x so that du = 2 dx and dx = du / 2, and the given integral can be written as
=
(1 / 4) (1 + cos(u)) du
Integrate to obtain
= (1 / 4) u + (1 / 4) sin (u) + c
Substitute u by 2x and simplify
= x / 2 + (1 / 4) sin (2x) + c = x / 2 + (1/2) sin x cos x + c
As an exercise, check the final answer by differentiation.
### Example 4
Evaluate the integral
x 3 e x 4 dx
Solution to Example 4:
Substitution: Let u = x 4 so that du / dx = 4 x 3 which leads to (1 / 4) du = x 3 dx, so that the given integral can be written as
=
(1 / 4) e u du
We now use formula for integral of exponential function to obtain
= (1 / 4) e
u + c
Substitute u by u = x 4
= (1 / 4) e
x 4 + c
## Exercises
Use the table of integrals and the properties above to evaluate the following integrals. [Note that you may need to use more than one of the above properties for one integral].
1.
√(x + 1) dx
2.
sin 2 x dx
3.
x cos(x 2) dx
4.
x e x 2 dx
### Answers to Above Exercises
1. (2 / 3) (x+1)
3/2
2. x / 2 - (1/2) sin x cos x
3. (1 / 2) sin(x
2)
4. (1 / 2) e
x 2
More references on
integrals and their applications in calculus.
|
File
```NS.09:
Fractions, Decimals,
Percents, and Ratios
Objective: I can convert between
decimals, fractions, percents, and
ratios.
.01 = 1/100 = 1% = 1:100
Mathematical Practices:
MP.2: I can use reasoning habits to help
me think about numbers in many ways.
MP.8: I can solve problems by looking for
rules and repeated patterns.
Notes: Key Vocabulary:
Positive Rational Numbers: Numbers
that are greater than zero and can be
expressed as a ratio. 6:10
Fraction: part of a whole group. It has a
numerator and denominator. ¾
Decimal: a number containing a decimal
point, written as base ten. 0.345
Percent: a special ratio that compares a
number to 100 using the symbol %
Ratio: A comparison of two numbers or
measures using division. 2 to 4, 2:4, or
2/4
Convert: to change
Notes: Essential Questions: (Skip 4
lines between questions.)
1.
What are different ways to express the
same value? How do you convert
between each value?
2.
What is a conversion? Why do you need
to convert between values?
Note:
Fractions, decimals, ratios and
percents are different ways to represent
the same value.
Each conversion requires a specific
procedure.
Conversion Foldable:
1. Get a sheet of white paper and fold in
half horizontally. (like a hamburger)
2. Fold in half again from the folded
position.
3. From this position, fold into thirds so
you have a small square.
To convert (change) a fraction to a
decimal, divide the numerator by the
denominator.
0.75
3 = 4 3.00
4
Fraction
Decimal
Change these fractions to decimals:
5
8
2
3
Practice time
To convert a decimal to a fraction, use the
place value of the last digit to determine
the denominator and put the numbers to
the right of the decimal point in the
numerator. Reduce to lowest terms.
0.75 =
75
100
= 3
4
The 5 is in the hundredths place. Then,
reduce to lowest terms.
Decimal
Fraction
Convert the decimals into a fraction:
0.92
2.15
0.37
Let’s try a couple.
To convert a decimal to a percent, move
the decimal point two places to the right
0.75 =
75%
Decimal
Percent
Change decimals to percent:
0.67
0.125
1.73
Try these
To convert a percent to a decimal, move
the decimal point two places to the left
and remove the % sign.
75.% = 0.75
Percent
Decimal
Change the percents to decimals:
44%
59.2%
315%
You can do it!!!
What are the three ways to write a ratio?
Apply what you have learned to convert a
decimal and a percent to a ratio.
Not too tough, right?
Convert these numbers into a ratio:
0.4
76%
0.37
Don’t forget to reduce.
```
|
## Definition
Golomb's sequence, named after Solomon Golomb, is a curious sequence of whole numbers that describes itself. It is defined in the following way: it is a non-decreasing sequence of whole numbers where the nth term gives the number of times n occurs in the sequence, and the first term is 1. From this we can begin constructing it: The second element must be greater than 1 as there is only one 1. It must be 2, and so must be the third element. Given this, there must be 2 threes, and from here on we may merely refer to the terms in the sequence and continue from there. The first several terms of the sequence are: $1, 2, 2, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 9, 9, 9, 9, \\ 9, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12,...$
## Recurrence Relation
The sequence can be given an explicit recurrence relation by stating it in the following way, using the self-describing property: To determine the next term in the sequence, go back the number of times that the previous term occurred (this will put you at the next-smallest value), then add one. For example, to determine the 12 term (6), count the number of times that the value of the 11th term (5) occurs (3 times). Step back that many terms (to the 9th term: 5) then add one to that value (6). This then gives the recurrence relation: $a(n+1)=1+a\left ( n+1-a(a(n)) \right )$ Where $a(1)=1$.
## Asymptotic Behavior
The recurrence relation allows us to give an asymptotic expression for the value of the sequence. Let us suppose the sequence grows like $a(n)=A n^\alpha$ Let us put this into the recurrence relation: $A(n+1)^\alpha=1+A\left ( n+1-A(A n^\alpha)^\alpha \right )^\alpha$ Simplifying and rearranging, we obtain: $1=\frac{1}{A(n+1)^\alpha}+\left (1-A^{1+\alpha}\frac{n^{\alpha^2}}{n+1} \right )^\alpha$ As $\alpha<1$, $\frac{n^{\alpha^2}}{n+1}$ goes to zero. For small x, $(1+x)^b\rightarrow 1+bx$. Thus, asymptotically: $1\approx\frac{1}{A(n+1)^\alpha}+1-\alpha A^{1+\alpha}\frac{n^{\alpha^2}}{n+1}$ $\alpha A^{2+\alpha}n^{\alpha^2}(n+1)^{\alpha-1} \approx 1$ Thus it must be the case that $\alpha^2+\alpha-1=0$ $A=\alpha^{-\frac{1}{2+\alpha}}$ The solution to the first equation is $\alpha=\left \{\varphi-1,-\varphi \right \}$ Where $\varphi$ is the golden ratio. As the exponent is clearly positive, we find the sequence is asymptotic to: $a(n)\rightarrow \varphi^{2-\varphi}n^{\varphi-1}$ Below we plot the ratio of these two expressions:
## Definition and Background
A continued fraction is a representation of a number $x$ in the form $x=a_0+\cfrac{b_0}{a_1+\cfrac{b_1}{a_2+\cfrac{b_2}{a_3+\cfrac{b_3}{\ddots}}}}$ Often, the b's are taken to be all 1's and the a's are integers. This is called the canonical or simple form. There are numerous ways of representing continued fractions. For instance, $x=a_0+\cfrac{1}{a_1+\cfrac{1}{a_2+\cfrac{1}{a_3+\cfrac{1}{\ddots}}}}$ can be represented as $x=a_0+\overset{\infty}{\underset{k=1}{\mathrm{K}}}\frac{1}{a_k}$ Or as $\left [ a_0;a_1,a_2,a_3,... \right]$
## Construction Algorithm
The continued fraction terms can be determined as follows: Given $x$, set $x_0=x$. Then $a_k=\left \lfloor x_k \right \rfloor$ $x_{k+1}=\frac{1}{x_k-a_k}$ Continue until $x_k=a_k$.
## Convergents
The convergents of a continued fraction are the rational numbers resulting from taking the first n terms of the continued fraction. Let $P_n$ and $Q_n$ be the numerators and deominators respectively of the nth convergent (the one that includes $a_n$). It is not difficult to show that $P_n=a_nP_{n-1}+P_{n-2}$ $Q_n=a_nQ_{n-1}+Q_{n-2}$ An alternate way of saying this is that $\begin{bmatrix} a_n & 1\\ 1 & 0 \end{bmatrix} \begin{bmatrix} P_{n-1} & Q_{n-1}\\ P_{n-2} & Q_{n-2} \end{bmatrix} = \begin{bmatrix} P_{n} & Q_{n}\\ P_{n-1} & Q_{n-1} \end{bmatrix}$ Where $\begin{bmatrix} P_{-1} & Q_{-1}\\ P_{-2} & Q_{-2} \end{bmatrix}= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$ And therefore ${}^L\prod^n_{k=0} \begin{bmatrix} a_k & 1\\ 1 & 0 \end{bmatrix} = \begin{bmatrix} P_{n} & Q_{n}\\ P_{n-1} & Q_{n-1} \end{bmatrix}$ Let $p_n=\frac{P_n}{P_{n-1}}$ $q_n=\frac{Q_n}{Q_{n-1}}$ Then $p_n=a_n+\frac{1}{p_{n-1}}$ $q_n=a_n+\frac{1}{q_{n-1}}$ We find that $\frac{P_{n+1}}{Q_{n+1}}=a_0+\sum_{k=0}^{n}\frac{(-1)^k}{Q_kQ_{k+1}}$ And thus $\left | x- \frac{P_{n}}{Q_{n}}\right |<\frac{1}{Q_nQ_{n+1}}$ As $a_n \geq 1$, $Q_n \geq F_n$ i.e. the nth Fibonacci number. This, then, implies Hurwitz's theorem: For any irrational number x, there exist infinitely many ratios $P/Q$ such that $\left | x-\frac{P}{Q} \right |<\frac{k}{Q^2}$ Only if $k \geq 1/\sqrt{5}$.
## Periodic Continued Fractions
Suppose that for $k \geq N$, $a_{k+M}=a_k$. Let $[a_0;a_1,a_2,...a_{N-2}]=\frac{P_{Y1}}{Q_{Y1}}$ $[a_0;a_1,a_2,...a_{N-1}]=\frac{P_{Y2}}{Q_{Y2}}$ $\left [a_N;a_1,a_2,...a_{N+M-2} \right ]=\frac{P_{Z1}}{Q_{Z1}}$ $\left [a_N;a_1,a_2,...a_{N+M-1} \right ]=\frac{P_{Z2}}{Q_{Z2}}$ Then x satisfies the formula $x=\frac{P_{Y2}\cdot y+P_{Y1}}{Q_{Y2} \cdot y+Q_{Y1}}$ Where y satisfies $y=\frac{P_{Z2}\cdot y+P_{Z1}}{Q_{Z2} \cdot y+Q_{Z1}}$ Thus a continued fraction will be eventually periodic if and only if it is the solution of some quadratic polynomial.
## Generic Continued Fractions
Let x be uniformly chosen between 0 and 1. We define a sequence of random variables as follows $\xi_0=x$ $\xi_{n+1}=\frac{1}{\xi_n}-\left \lfloor \frac{1}{\xi_n} \right \rfloor$ Clearly, if $x=[0;a_1,a_2,a_3,...]$ Then $\xi_n=[0;a_{n+1},a_{n+2},a_{n+3},...]$ Let us assume that, asymptotically, the $\xi$'s approach a single distribution. Based on our definitions, this would imply that $P(\xi_{n+1} < z)=\sum_{k=1}^{\infty} P \left (\frac{1}{k} < \xi_n < \frac{1}{k+z} \right )$ Differentiating both sides gives the required relationship: $f_\xi(z)=\sum_{k=1}^{\infty}\frac{f_\xi\left ( \tfrac{1}{k+z} \right )}{(k+z)^2}$ Let us test the function $f_\xi(z)=\frac{A}{1+z}$ $\sum_{k=1}^{\infty}\frac{A}{1+\tfrac{1}{k+z}}\frac{1}{(k+z)^2}=\sum_{k=1}^{\infty}\frac{1}{(1+k+z)(k+z)}$ $\sum_{k=1}^{\infty}\frac{1}{(1+k+z)(k+z)}=\sum_{k=1}^{\infty}\frac{1}{k+z}-\frac{1}{k+z+1}=\frac{A}{1+z}$ It can be proved more rigorously that this is indeed the asymptotic probability density function, with $A=1/\ln(2)$. Thus $P(\xi_{n} < z)=\log_2(1+z)$ From this we can easily find the asymptotic density function for the continued fraction terms. The probability that $a_{n+1}=k$ is the same as the probability that $\left \lfloor \tfrac{1}{\xi_n} \right \rfloor=k$. This is then $P(a_{n+1}=k)=P\left ( \frac{1}{k+1} < \xi_n \leq \frac{1}{k} \right )=\log_2(1+\tfrac{1}{k})-\log_2(1+\tfrac{1}{k+1})$ $P(a_{n+1}=k)=\log_2\left ( \frac{(k+1)^2}{k(k+2)} \right )=\log_2\left ( 1+\frac{1}{k(k+2)} \right )$ This is called the Gauss-Kuzmin Distribution.
From this we can then easily find the asymptotic geometric mean of the terms $\underset{n \to \infty}{\lim}\sqrt[n]{\prod_{k=1}^{n}a_k}=\exp\left (\underset{n \to \infty}{\lim} \frac{1}{n}\sum_{k=1}^{\infty} \ln(a_k)\right )=\exp\left ( E(\ln(a_k)) \right )$$\underset{n \to \infty}{\lim}\sqrt[n]{\prod_{k=1}^{n}a_k}= \exp\left (\sum_{j=1}^{\infty}P(a_k=j)\ln(j) \right )$$\underset{n \to \infty}{\lim}\sqrt[n]{\prod_{k=1}^{n}a_k}= \prod_{j=1}^\infty \left ( 1+\frac{1}{j(j+2)} \right )^{\log_2(j)}=2.685452001...=K_0$ This value is called Khinchin's Constant.
Let us now look at the asymptotic behavior of the convergents. Namely, we wish to examine the asymptotic behavior of the denominators. First note that $\xi_n=\frac{1}{\xi_{n-1}}-a_n$ If we let $y_n=1/\xi_n$, we then have $y_{n-1}=a_n+\frac{1}{y_n}$ From above we have that $q_n=a_n+\frac{1}{q_{n-1}}$ As, asymptotically, $\xi_n \sim \xi_{n-1}$, this implies that, asymptotically, $y_n \sim y_{n-1} \sim 1/\xi_n$ and therefore $q_n \sim q_{n-1} \sim 1/\xi_n$. Thus $f_q(z)=\left\{\begin{matrix} \frac{1}{z^2}\frac{1}{\ln(2)}\frac{1}{1+1/z} \\ 0 \end{matrix}\right.\; \; \begin{matrix} z > 1 \\ z \leq 1 \end{matrix}$ As $Q_n=\prod_{k=1}^{n}q_k$ We have $\underset{n \to \infty}{\lim}\sqrt[n]{Q_n}=\underset{n \to \infty}{\lim}\sqrt[n]{\prod_{k=1}^{n}q_k}=\exp\left (\underset{n \to \infty}{\lim}\frac{1}{n}\sum_{k=1}^{\infty}\ln(q_k) \right )$$\underset{n \to \infty}{\lim}\sqrt[n]{Q_n}=\exp\left (E(\ln(q_n)) \right )= \exp\left (\int_{1}^{\infty}\ln(z)\frac{1}{z^2}\frac{1}{\ln(2)}\frac{1}{1+1/z}dz \right )$$\underset{n \to \infty}{\lim}\sqrt[n]{Q_n}= \exp\left (-\frac{1}{\ln(2)}\int_{0}^{1}\frac{\ln(z)}{1+z}dz \right )$$\underset{n \to \infty}{\lim}\sqrt[n]{Q_n}=\exp\left ( \frac{\pi^2}{12\ln(2)} \right )=3.275823...$ This value (or sometimes its natural log) is called Levy's constant.
We want to know how efficient continued fractions are for representing numbers relative to place-value expansions. Suppose we are working in base b. We want to find how many terms in the continued fraction expansion are required to obtain an approximation good to m base-b digits. We will have obtained such an approximation when the error is less than $b^{-m}$ but greater than $b^{-(m+1)}$. From above we have $\left | x- \frac{P_{n}}{Q_{n}}\right |<\frac{1}{Q_nQ_{n+1}}$ Thus $b^{-(m+1)} < \left | x- \frac{P_{n}}{Q_{n}}\right | < \frac{1}{Q_nQ_{n+1}} < \frac{1}{Q_n^2} \leq b^{-m}$ Rearranging, we have $b^m \leq Q_n^2 < b^{m+1}$ $b^{\frac{m}{2n}} \leq \sqrt[n]{Q_n} < b^{\frac{m+1}{2n}}$ Thus, as the center expression approaches a limit for large n, it follows that $m/n$ does as well. Namely, by rearranging, we find that for n the number of continued fraction terms needed to express x in base b up to m decimal places, $\underset{m,n \to \infty}{\lim}\frac{m}{n}=\frac{\pi^2}{6\ln(2)\ln(b)}$ This is known as Loch's Theorem. In particular, for base 10, this implies that each continued fraction term provides on average 1.03064... digits of precision. In fact, base 10 is the largest integral base for which the continued fraction is more efficient.
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# How do you identify 19th term of a geometric sequence where a1 = 14 and a9 = 358.80?
Dec 29, 2015
An explanation is given below.
#### Explanation:
We are to find $19$th term of Geometric Sequence
Given ${a}_{1} = 14$ and ${a}_{9} = 358.80$
The general term of a Geometric Sequence is given by
${a}_{n} = a \cdot {r}^{n - 1}$
Where $a$ is the first term also known as ${a}_{1}$ and $r$ is the common ratio.
We have ${a}_{1}$ if we get $r$ we can easily find ${a}_{19}$ by using $19$ for $n$
Let us start by writing the given term using $r$
${a}_{9} = a \cdot {r}^{8}$
If we divide ${a}_{9}$ by ${a}_{1}$ we would get an equation in $r$
$\frac{a {r}^{8}}{a} = \frac{358.80}{14}$
${r}^{8} = 25.628571428571428571428571428571$
Taking $8$th root.
$r = \sqrt[8]{25.628571428571428571428571428571}$
r=1.4999975504465127405341330547934"
$r \approx 1.5$
${a}_{19} = 14 {\left(1.5\right)}^{19}$
a_19 = 31,035.729480743408203125"#
${a}_{19} \approx 31035.73$
|
# If y=ce^(2x)+De^(-2x) then show that y'' -4y = 0 ?
Aug 15, 2017
$y ' ' - 4 y = 0$
#### Explanation:
This is an inverse problem. Given a solution, find a candidate differential equation which has is as solution.
Assuming the differential equation is second-order linear homogeneous such as
${c}_{1} y ' ' + {c}_{2} y ' + {c}_{3} y = 0$
after substituting the solution we have
$\left(4 {c}_{1} - 2 {c}_{2} + {c}_{3}\right) {D}_{0} {e}^{- 2 x} + {C}_{0} \left(4 {c}_{1} + 2 {c}_{2} + {c}_{3}\right) {e}^{2 x} = 0$
This relationship must be true for all $x$ so
$\left\{\begin{matrix}4 {c}_{1} - 2 {c}_{2} + {c}_{3} = 0 \\ 4 {c}_{1} + 2 {c}_{2} + {c}_{3} = 0\end{matrix}\right.$
now solving for ${c}_{1} , {c}_{2}$ we obtain
${c}_{1} = - {c}_{3} / 4 , {c}_{2} = 0$ so the differential equation is
$- {c}_{3} / 4 y ' ' + {c}_{3} y = 0$ or
$y ' ' - 4 y = 0$
Aug 15, 2017
Refer to the Explanation.
#### Explanation:
We note that, the given eqn. $y = C {e}^{2 x} + D {e}^{- 2 x} \ldots \ldots \left(1\right) ,$ contains
2 arbitrary constants.
Therefore, the reqd. Diff. Eqn. must be of Second Order.
To find it, we diif. $\left(1\right)$ twice.
$y = C {e}^{2 x} + D {e}^{- 2 x} \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \left(C {e}^{2 x}\right) 2 + \left(D {e}^{- 2 x}\right) \left(- 2\right) , \mathmr{and} ,$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 \left\{C {e}^{2 x} - D {e}^{- 2 x}\right\} ,$
$\therefore \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{d}{\mathrm{dx}} \left[2 \left\{C {e}^{2 x} - D {e}^{- 2 x}\right\}\right] ,$
$= 2 \left[\left(C {e}^{2 x}\right) \left(2\right) - \left(D {e}^{- 2 x}\right) \left(- 2\right)\right] ,$
$\Rightarrow \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = 4 \left[C {e}^{2 x} + D {e}^{- 2 x}\right] = 4 y ,$
$\Rightarrow \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 - 4 y = 0 ,$ is the Desired Diff. Eqn.
Aug 15, 2017
$y ' ' - 4 y = 0$
#### Explanation:
We have:
$y = c {e}^{2 x} + D {e}^{- 2 x}$ .... [A}
As others have indicated we are not solving a second order Differentiation Equation with constant coefficients, but rather forming one given the solution.
Recognizing the solution is that of a second order Differentiation Equation with constant coefficients we can instantly write down the appropriate DE.
The Auxiliary Equation that produced this solution would require two distinct real solution, $m = 2$ and m=-2)
Hence the associated Auxiliary Equation would be:
$\left(m - 2\right) \left(m + 2\right) = 0 \implies {m}^{2} - 4 = 0$
Hence the DE associated with this Auxiliary Equation is:
$y ' ' + 0 y ' - 4 y = 0$
$y ' ' - 4 y = 0$
|
# Convert Decimal to Binary and Vice Versa
14,680
23
9
## Introduction: Convert Decimal to Binary and Vice Versa
This is a technique that helped me out a lot when learning about AND OR NOR gates, creating subnet masks and helped me pass time seeing how high I could get with the binary sequence (not very far). I was able to teach my son this method when he was 7 or 8.
I plan to eventually make this a Computer Science Collection of Instructables to help teach kids. If anyone would like to help out, provide suggestions and/or write Instructables for this collection let me know.
I keep seeing my Assembly Language Quick Reference Guide on my desk.... There's nothing quick about Assembly at least not for me.
### Teacher Notes
Teachers! Did you use this instructable in your classroom?
Add a Teacher Note to share how you incorporated it into your lesson.
## Step 1: Binary Sequence
The key to this method is laying out the numbers binary sequence which is powers of 2 since binary is base 2. This that much more difficult than counting by 2's. Note that we use base 10 considering we have 10 digits and computers at their lowest level use base 2. People argue that some computers use base 16 but remember that a switch is either on or off, 1 or 0. The numbers will need to be from right to left the lowest number to the highest number. This is illustrated by the second picture in the collection.
## Step 2: Binary to Base 10
Here we will use the sequence table (2nd Image) to find the base 10 number. Use the visual (1st image) to get a better idea of how this works.
So here you have a given binary number.
1. You will fill in the table from right to left with the 1's & 0's from the binary number.
2. You will find the decimal numbers that have a 1 and not a zero. Remember that 1 is on and 0 is off.
3. Add the decimal numbers you found in the previous step together to get the decimal value of the binary number that you were given.
## Step 3: Decimal to Binary
Here we will use the sequence table (2nd Image) to find the base 2 number. Use the visual (1st image) to get a better idea of how this works.
So here you have a given decimal (base 10) number.
1. Find the highest number that will go into your given decimal number.
2. Place a 1 in the table for each number that you were able to subtract.
3. Subtract that number from your decimal number.
4. Find the next highest number that will go into the number from number 3.
5. Repeat steps 1-5 until you have a zero remainder.
6. Your binary number will not start with zero but everywhere else that you have a blank fill in a zero (see example)
23 3.5K
13 3.5K
56 8.2K
## 9 Discussions
Oh so many memories from my Computer Architecture class. Writing in assembly isn't the most fun or happiest of things.
But this is definitely a good way of doing this!
At one point I felt that if I loaded one more binary digit into a register I would surely go mad.
It's good for kids to do this kind of exercise a few times so they understand the logic behind number system relationships, but for actual coding I'd want them to use a calculator for conversions.
I don't disagree. My motive is for kids to have a good solid foundation.
Same basic idea with any form of math. Kids should not do their work on a calculator until they know how to solve problems "manually", which requires them to understand the logic of problem solving.
What no hex? Where's the hex. Like tomato says, "Oh so many memories..."
Haha I have to review it before I finish up the hex instructions. The binary was easy because I had committed it to memory.
|
College Algebra with Corequisite Support
5.6Rational Functions
College Algebra with Corequisite Support5.6 Rational Functions
Learning Objectives
In this section, you will:
• Use arrow notation.
• Solve applied problems involving rational functions.
• Find the domains of rational functions.
• Identify vertical asymptotes.
• Identify horizontal asymptotes.
• Graph rational functions.
Corequisite Skills
Learning Objectives
• Determine the values for which a rational expression is undefined (IA 7.1.1)
• Find x- and y-intercepts (IA 3.1.4)
Rational Expression
A rational expression is an expression of the form $pq,pq,$ where p and q are polynomials and $q≠0.q≠0.$
Here are some examples of rational expressions:
$−24565x12y4x+1x2−94x2+3x−12x−8 −24565x12y4x+1x2−94x2+3x−12x−8$
Practice Makes Perfect
Evaluate the following expression for the given values $5x-10x+35x-10x+3$
1.
$x=3x=3$
2.
$x=2x=2$
3.
$x=-3x=-3$
4.
Why do we have a problem with evaluating this expression for $x=−3x=−3$ but not for $x=2x=2$ ?
We say that this rational expression is undefined because its denominator equals 0.
How To
Determine the values for which a rational expression is undefined.
1. Step 1. Set the denominator equal to zero.
2. Step 2. Solve the equation.
Example 1
Determine the value for which the rational expression $x+32x2+9x-5x+32x2+9x-5$ is undefined
Practice Makes Perfect
Determine the value for which each rational expression is undefined.
5.
$3y25x3y25x$
6.
$3x+52x-33x+52x-3$
7.
$m-5m2+m-6m-5m2+m-6$
8.
$2aa2-92aa2-9$
9.
Determine the value for which the function $f(x)=3x-2f(x)=3x-2$ is undefined.
What is the connection between the value you found and the graph of this function?
Because function is not defined at $x=2x=2$ , we say that the Domain of this function is
The domain of a rational function includes all real numbers except those that cause the denominator to equal zero.
10.
Determine the domain of the function and express using interval notation.
1. $f(x)=x3x-1f(x)=x3x-1$
2. $f(x)=3x2-5x+6f(x)=3x2-5x+6$
Vocabulary
The $xx$-intercept is a point where the graph intersects the ________ axis. The $yy$-coordinate at this point is always ________.
The $yy$-intercept is a point where the graph intersects the ________ axis. The $xx$-coordinate at this point is always ________.
How To
Find the x-intercept and y-intercept of a line.
Example: $3x−4y=93x−4y=9$
Find the $xx$ -intercept and $yy$ -intercept of a line.
To find the $xx$ -intercept, replace $yy$ with zero and find $xx$ $xx$ -int: $3x−4(0)=93x−4(0)=9$ , $3x=93x=9$ , $x=3x=3$ (3, 0) To find the $yy$ -intercept, replace $xx$ with zero and find $yy$ $yy$ -int: $3(0)–4y=93(0)–4y=9$ , $–4y=9–4y=9$ , $y=–94y=–94$ (0, $–94–94$ )
Practice Makes Perfect
Find $xx$ - and $yy$ -intercept of each of the following functions. Express each as an ordered pair.
11.
12.
$y=5x-1y=5x-1$
13.
$y=1(x-2)+3y=1(x-2)+3$
Suppose we know that the cost of making a product is dependent on the number of items, $x, x,$ produced. This is given by the equation $C(x)=15,000x−0.1 x 2 +1000. C(x)=15,000x−0.1 x 2 +1000.$ If we want to know the average cost for producing $x x$ items, we would divide the cost function by the number of items, $x. x.$
The average cost function, which yields the average cost per item for $x x$ items produced, is
$f(x)= 15,000x−0.1 x 2 +1000 x f(x)= 15,000x−0.1 x 2 +1000 x$
Many other application problems require finding an average value in a similar way, giving us variables in the denominator. Written without a variable in the denominator, this function will contain a negative integer power.
In the last few sections, we have worked with polynomial functions, which are functions with non-negative integers for exponents. In this section, we explore rational functions, which have variables in the denominator.
Using Arrow Notation
We have seen the graphs of the basic reciprocal function and the squared reciprocal function from our study of toolkit functions. Examine these graphs, as shown in Figure 1, and notice some of their features.
Figure 1
Several things are apparent if we examine the graph of $f(x)= 1 x . f(x)= 1 x .$
1. On the left branch of the graph, the curve approaches the x-axis
2. As the graph approaches $x=0 x=0$ from the left, the curve drops, but as we approach zero from the right, the curve rises.
3. Finally, on the right branch of the graph, the curves approaches the x-axis
To summarize, we use arrow notation to show that $x x$ or $f(x) f(x)$ is approaching a particular value. See Table 1.
Symbol Meaning
$x→ a − x→ a −$ $x x$ approaches $a a$ from the left ( $x but close to $a a$ )
$x→ a + x→ a +$ $x x$ approaches $a a$ from the right ( $x>a x>a$ but close to $a a$ )
$x→∞ x→∞$ $x x$ approaches infinity ( $x x$ increases without bound)
$x→−∞ x→−∞$ $x x$ approaches negative infinity ( $x x$ decreases without bound)
$f(x)→∞ f(x)→∞$ the output approaches infinity (the output increases without bound)
$f(x)→−∞ f(x)→−∞$ the output approaches negative infinity (the output decreases without bound)
$f(x)→a f(x)→a$ the output approaches $a a$
Table 1
Local Behavior of $f(x)= 1 x f(x)= 1 x$
Let’s begin by looking at the reciprocal function, $f(x)= 1 x . f(x)= 1 x .$ We cannot divide by zero, which means the function is undefined at $x=0; x=0;$ so zero is not in the domain. As the input values approach zero from the left side (becoming very small, negative values), the function values decrease without bound (in other words, they approach negative infinity). We can see this behavior in Table 2.
$x x$ –0.1 –0.01 –0.001 –0.0001 $f(x)= 1 x f(x)= 1 x$ –10 –100 –1000 –10,000
Table 2
We write in arrow notation
As the input values approach zero from the right side (becoming very small, positive values), the function values increase without bound (approaching infinity). We can see this behavior in Table 3.
$x x$ 0.1 0.01 0.001 0.0001 $f(x)= 1 x f(x)= 1 x$ 10 100 1000 10,000
Table 3
We write in arrow notation
See Figure 2.
Figure 2
This behavior creates a vertical asymptote, which is a vertical line that the graph approaches but never crosses. In this case, the graph is approaching the vertical line $x=0 x=0$ as the input becomes close to zero. See Figure 3.
Figure 3
Vertical Asymptote
A vertical asymptote of a graph is a vertical line $x=a x=a$ where the graph tends toward positive or negative infinity as the inputs approach $a. a.$ We write
End Behavior of $f(x)= 1 x f(x)= 1 x$
As the values of $x x$ approach infinity, the function values approach 0. As the values of $x x$ approach negative infinity, the function values approach 0. See Figure 4. Symbolically, using arrow notation
Figure 4
Based on this overall behavior and the graph, we can see that the function approaches 0 but never actually reaches 0; it seems to level off as the inputs become large. This behavior creates a horizontal asymptote, a horizontal line that the graph approaches as the input increases or decreases without bound. In this case, the graph is approaching the horizontal line $y=0. y=0.$ See Figure 5.
Figure 5
Horizontal Asymptote
A horizontal asymptote of a graph is a horizontal line $y=b y=b$ where the graph approaches the line as the inputs increase or decrease without bound. We write
Example 1
Using Arrow Notation
Use arrow notation to describe the end behavior and local behavior of the function graphed in Figure 6.
Figure 6
Try It #1
Use arrow notation to describe the end behavior and local behavior for the reciprocal squared function.
Example 2
Using Transformations to Graph a Rational Function
Sketch a graph of the reciprocal function shifted two units to the left and up three units. Identify the horizontal and vertical asymptotes of the graph, if any.
Analysis
Notice that horizontal and vertical asymptotes are shifted left 2 and up 3 along with the function.
Try It #2
Sketch the graph, and find the horizontal and vertical asymptotes of the reciprocal squared function that has been shifted right 3 units and down 4 units.
Solving Applied Problems Involving Rational Functions
In Example 2, we shifted a toolkit function in a way that resulted in the function $f(x)= 3x+7 x+2 . f(x)= 3x+7 x+2 .$ This is an example of a rational function. A rational function is a function that can be written as the quotient of two polynomial functions. Many real-world problems require us to find the ratio of two polynomial functions. Problems involving rates and concentrations often involve rational functions.
Rational Function
A rational function is a function that can be written as the quotient of two polynomial functions
$f(x)= P(x) Q(x) = a p x p + a p−1 x p−1 +...+ a 1 x+ a 0 b q x q + b q−1 x q−1 +...+ b 1 x+ b 0 ,Q(x)≠0 f(x)= P(x) Q(x) = a p x p + a p−1 x p−1 +...+ a 1 x+ a 0 b q x q + b q−1 x q−1 +...+ b 1 x+ b 0 ,Q(x)≠0$
Example 3
Solving an Applied Problem Involving a Rational Function
A large mixing tank currently contains 100 gallons of water into which 5 pounds of sugar have been mixed. A tap will open pouring 10 gallons per minute of water into the tank at the same time sugar is poured into the tank at a rate of 1 pound per minute. Find the ratio of sugar to water, in pounds per gallon in the tank after 12 minutes. Is that a greater ratio of sugar to water, in pounds per gallon than at the beginning?
Try It #3
There are 1,200 freshmen and 1,500 sophomores at a prep rally at noon. After 12 p.m., 20 freshmen arrive at the rally every five minutes while 15 sophomores leave the rally. Find the ratio of freshmen to sophomores at 1 p.m.
Finding the Domains of Rational Functions
A vertical asymptote represents a value at which a rational function is undefined, so that value is not in the domain of the function. A reciprocal function cannot have values in its domain that cause the denominator to equal zero. In general, to find the domain of a rational function, we need to determine which inputs would cause division by zero.
Domain of a Rational Function
The domain of a rational function includes all real numbers except those that cause the denominator to equal zero.
How To
Given a rational function, find the domain.
1. Set the denominator equal to zero.
2. Solve to find the x-values that cause the denominator to equal zero.
3. The domain is all real numbers except those found in Step 2.
Example 4
Finding the Domain of a Rational Function
Find the domain of $f(x)= x+3 x 2 −9 . f(x)= x+3 x 2 −9 .$
Analysis
A graph of this function, as shown in Figure 8, confirms that the function is not defined when $x=±3. x=±3.$
Figure 8
There is a vertical asymptote at $x=3 x=3$ and a hole in the graph at $x=−3. x=−3.$ We will discuss these types of holes in greater detail later in this section.
Try It #4
Find the domain of $f(x)= 4x 5(x−1)(x−5) . f(x)= 4x 5(x−1)(x−5) .$
Identifying Vertical Asymptotes of Rational Functions
By looking at the graph of a rational function, we can investigate its local behavior and easily see whether there are asymptotes. We may even be able to approximate their location. Even without the graph, however, we can still determine whether a given rational function has any asymptotes, and calculate their location.
Vertical Asymptotes
The vertical asymptotes of a rational function may be found by examining the factors of the denominator that are not common to the factors in the numerator. Vertical asymptotes occur at the zeros of such factors.
How To
Given a rational function, identify any vertical asymptotes of its graph.
1. Factor the numerator and denominator.
2. Note any restrictions in the domain of the function.
3. Reduce the expression by canceling common factors in the numerator and the denominator.
4. Note any values that cause the denominator to be zero in this simplified version. These are where the vertical asymptotes occur.
5. Note any restrictions in the domain where asymptotes do not occur. These are removable discontinuities, or “holes.”
Example 5
Identifying Vertical Asymptotes
Find the vertical asymptotes of the graph of $k(x)= 5+2 x 2 2−x− x 2 . k(x)= 5+2 x 2 2−x− x 2 .$
Removable Discontinuities
Occasionally, a graph will contain a hole: a single point where the graph is not defined, indicated by an open circle. We call such a hole a removable discontinuity.
For example, the function $f(x)= x 2 −1 x 2 −2x−3 f(x)= x 2 −1 x 2 −2x−3$ may be re-written by factoring the numerator and the denominator.
$f(x)= ( x+1 )( x−1 ) ( x+1 )( x−3 ) f(x)= ( x+1 )( x−1 ) ( x+1 )( x−3 )$
Notice that $x+1 x+1$ is a common factor to the numerator and the denominator. The zero of this factor, $x=−1, x=−1,$ is the location of the removable discontinuity. Notice also that $x–3 x–3$ is not a factor in both the numerator and denominator. The zero of this factor, $x=3, x=3,$ is the vertical asymptote. See Figure 10. [Note that removable discontinuities may not be visible when we use a graphing calculator, depending upon the window selected.]
Figure 10
Removable Discontinuities of Rational Functions
A removable discontinuity occurs in the graph of a rational function at $x=a x=a$ if $a a$ is a zero for a factor in the denominator that is common with a factor in the numerator. We factor the numerator and denominator and check for common factors. If we find any, we set the common factor equal to 0 and solve. This is the location of the removable discontinuity. This is true if the multiplicity of this factor is greater than or equal to that in the denominator. If the multiplicity of this factor is greater in the denominator, then there is still an asymptote at that value.
Example 6
Identifying Vertical Asymptotes and Removable Discontinuities for a Graph
Find the vertical asymptotes and removable discontinuities of the graph of $k(x)= x−2 x 2 −4 . k(x)= x−2 x 2 −4 .$
Try It #5
Find the vertical asymptotes and removable discontinuities of the graph of $f(x)= x 2 −25 x 3 −6 x 2 +5x . f(x)= x 2 −25 x 3 −6 x 2 +5x .$
Identifying Horizontal Asymptotes of Rational Functions
While vertical asymptotes describe the behavior of a graph as the output gets very large or very small, horizontal asymptotes help describe the behavior of a graph as the input gets very large or very small. Recall that a polynomial’s end behavior will mirror that of the leading term. Likewise, a rational function’s end behavior will mirror that of the ratio of the function that is the ratio of the leading terms.
There are three distinct outcomes when checking for horizontal asymptotes:
Case 1: If the degree of the denominator > degree of the numerator, there is a horizontal asymptote at $y=0. y=0.$
In this case, the end behavior is $f(x)≈ 4x x 2 = 4 x . f(x)≈ 4x x 2 = 4 x .$ This tells us that, as the inputs increase or decrease without bound, this function will behave similarly to the function $g(x)= 4 x , g(x)= 4 x ,$ and the outputs will approach zero, resulting in a horizontal asymptote at $y=0. y=0.$ See Figure 12. Note that this graph crosses the horizontal asymptote.
Figure 12 Horizontal asymptote $y=0 y=0$ when
Case 2: If the degree of the denominator < degree of the numerator by one, we get a slant asymptote.
In this case, the end behavior is $f(x)≈ 3 x 2 x =3x. f(x)≈ 3 x 2 x =3x.$ This tells us that as the inputs increase or decrease without bound, this function will behave similarly to the function $g(x)=3x. g(x)=3x.$ As the inputs grow large, the outputs will grow and not level off, so this graph has no horizontal asymptote. However, the graph of $g(x)=3x g(x)=3x$ looks like a diagonal line, and since $f f$ will behave similarly to $g, g,$ it will approach a line close to $y=3x. y=3x.$ This line is a slant asymptote.
To find the equation of the slant asymptote, divide $3 x 2 −2x+1 x−1 . 3 x 2 −2x+1 x−1 .$ The quotient is $3x+1, 3x+1,$ and the remainder is 2. The slant asymptote is the graph of the line $g(x)=3x+1. g(x)=3x+1.$ See Figure 13.
Figure 13 Slant asymptote when $f(x)= p(x) q(x) ,q(x)≠0 f(x)= p(x) q(x) ,q(x)≠0$ where degree of
Case 3: If the degree of the denominator = degree of the numerator, there is a horizontal asymptote at $y= a n b n , y= a n b n ,$ where $a n a n$ and $b n b n$ are the leading coefficients of $p( x ) p( x )$ and $q( x ) q( x )$ for $f(x)= p(x) q(x) ,q(x)≠0. f(x)= p(x) q(x) ,q(x)≠0.$
In this case, the end behavior is $f(x)≈ 3 x 2 x 2 =3. f(x)≈ 3 x 2 x 2 =3.$ This tells us that as the inputs grow large, this function will behave like the function $g(x)=3, g(x)=3,$ which is a horizontal line. As $x→±∞,f(x)→3, x→±∞,f(x)→3,$ resulting in a horizontal asymptote at $y=3. y=3.$ See Figure 14. Note that this graph crosses the horizontal asymptote.
Figure 14 Horizontal asymptote when
Notice that, while the graph of a rational function will never cross a vertical asymptote, the graph may or may not cross a horizontal or slant asymptote. Also, although the graph of a rational function may have many vertical asymptotes, the graph will have at most one horizontal (or slant) asymptote.
It should be noted that, if the degree of the numerator is larger than the degree of the denominator by more than one, the end behavior of the graph will mimic the behavior of the reduced end behavior fraction. For instance, if we had the function
$f(x)= 3 x 5 − x 2 x+3 f(x)= 3 x 5 − x 2 x+3$
with end behavior
$f(x)≈ 3 x 5 x =3 x 4 , f(x)≈ 3 x 5 x =3 x 4 ,$
the end behavior of the graph would look similar to that of an even polynomial with a positive leading coefficient.
Horizontal Asymptotes of Rational Functions
The horizontal asymptote of a rational function can be determined by looking at the degrees of the numerator and denominator.
• Degree of numerator is less than degree of denominator: horizontal asymptote at $y=0. y=0.$
• Degree of numerator is greater than degree of denominator by one: no horizontal asymptote; slant asymptote.
• Degree of numerator is equal to degree of denominator: horizontal asymptote at ratio of leading coefficients.
Example 7
Identifying Horizontal and Slant Asymptotes
For the functions listed, identify the horizontal or slant asymptote.
1. $g(x)= 6 x 3 −10x 2 x 3 +5 x 2 g(x)= 6 x 3 −10x 2 x 3 +5 x 2$
2. $h(x)= x 2 −4x+1 x+2 h(x)= x 2 −4x+1 x+2$
3. $k(x)= x 2 +4x x 3 −8 k(x)= x 2 +4x x 3 −8$
Example 8
Identifying Horizontal Asymptotes
In the sugar concentration problem earlier, we created the equation $C(t)= 5+t 100+10t . C(t)= 5+t 100+10t .$
Find the horizontal asymptote and interpret it in context of the problem.
Example 9
Identifying Horizontal and Vertical Asymptotes
Find the horizontal and vertical asymptotes of the function
$f(x)= (x−2)(x+3) (x−1)(x+2)(x−5) f(x)= (x−2)(x+3) (x−1)(x+2)(x−5)$
Try It #6
Find the vertical and horizontal asymptotes of the function:
$f(x)= (2x−1)(2x+1) (x−2)(x+3) f(x)= (2x−1)(2x+1) (x−2)(x+3)$
Intercepts of Rational Functions
A rational function will have a y-intercept at $f(0) f(0)$ , if the function is defined at zero. A rational function will not have a y-intercept if the function is not defined at zero.
Likewise, a rational function will have x-intercepts at the inputs that cause the output to be zero. Since a fraction is only equal to zero when the numerator is zero, x-intercepts can only occur when the numerator of the rational function is equal to zero.
Example 10
Finding the Intercepts of a Rational Function
Find the intercepts of $f(x)= (x−2)(x+3) (x−1)(x+2)(x−5) . f(x)= (x−2)(x+3) (x−1)(x+2)(x−5) .$
Try It #7
Given the reciprocal squared function that is shifted right 3 units and down 4 units, write this as a rational function. Then, find the x- and y-intercepts and the horizontal and vertical asymptotes.
Graphing Rational Functions
In Example 9, we see that the numerator of a rational function reveals the x-intercepts of the graph, whereas the denominator reveals the vertical asymptotes of the graph. As with polynomials, factors of the numerator may have integer powers greater than one. Fortunately, the effect on the shape of the graph at those intercepts is the same as we saw with polynomials.
The vertical asymptotes associated with the factors of the denominator will mirror one of the two toolkit reciprocal functions. When the degree of the factor in the denominator is odd, the distinguishing characteristic is that on one side of the vertical asymptote the graph heads towards positive infinity, and on the other side the graph heads towards negative infinity. See Figure 17.
Figure 17
When the degree of the factor in the denominator is even, the distinguishing characteristic is that the graph either heads toward positive infinity on both sides of the vertical asymptote or heads toward negative infinity on both sides. See Figure 18.
Figure 18
For example, the graph of $f(x)= (x+1) 2 (x−3) (x+3) 2 (x−2) f(x)= (x+1) 2 (x−3) (x+3) 2 (x−2)$ is shown in Figure 19.
Figure 19
• At the x-intercept $x=−1 x=−1$ corresponding to the $(x+1) 2 (x+1) 2$ factor of the numerator, the graph "bounces", consistent with the quadratic nature of the factor.
• At the x-intercept $x=3 x=3$ corresponding to the $(x−3) (x−3)$ factor of the numerator, the graph passes through the axis as we would expect from a linear factor.
• At the vertical asymptote $x=−3 x=−3$ corresponding to the $(x+3) 2 (x+3) 2$ factor of the denominator, the graph heads towards positive infinity on both sides of the asymptote, consistent with the behavior of the function $f(x)= 1 x 2 . f(x)= 1 x 2 .$
• At the vertical asymptote $x=2, x=2,$ corresponding to the $(x−2) (x−2)$ factor of the denominator, the graph heads towards positive infinity on the left side of the asymptote and towards negative infinity on the right side.
How To
Given a rational function, sketch a graph.
1. Evaluate the function at 0 to find the y-intercept.
2. Factor the numerator and denominator.
3. For factors in the numerator not common to the denominator, determine where each factor of the numerator is zero to find the x-intercepts.
4. Find the multiplicities of the x-intercepts to determine the behavior of the graph at those points.
5. For factors in the denominator, note the multiplicities of the zeros to determine the local behavior. For those factors not common to the numerator, find the vertical asymptotes by setting those factors equal to zero and then solve.
6. For factors in the denominator common to factors in the numerator, find the removable discontinuities by setting those factors equal to 0 and then solve.
7. Compare the degrees of the numerator and the denominator to determine the horizontal or slant asymptotes.
8. Sketch the graph.
Example 11
Graphing a Rational Function
Sketch a graph of $f(x)= (x+2)(x−3) (x+1) 2 (x−2) . f(x)= (x+2)(x−3) (x+1) 2 (x−2) .$
Try It #8
Given the function $f(x)= (x+2) 2 (x−2) 2 (x−1) 2 (x−3) , f(x)= (x+2) 2 (x−2) 2 (x−1) 2 (x−3) ,$ use the characteristics of polynomials and rational functions to describe its behavior and sketch the function.
Writing Rational Functions
Now that we have analyzed the equations for rational functions and how they relate to a graph of the function, we can use information given by a graph to write the function. A rational function written in factored form will have an x-intercept where each factor of the numerator is equal to zero. (An exception occurs in the case of a removable discontinuity.) As a result, we can form a numerator of a function whose graph will pass through a set of x-intercepts by introducing a corresponding set of factors. Likewise, because the function will have a vertical asymptote where each factor of the denominator is equal to zero, we can form a denominator that will produce the vertical asymptotes by introducing a corresponding set of factors.
Writing Rational Functions from Intercepts and Asymptotes
If a rational function has x-intercepts at $x= x 1 , x 2 ,..., x n , x= x 1 , x 2 ,..., x n ,$ vertical asymptotes at $x= v 1 , v 2 ,…, v m , x= v 1 , v 2 ,…, v m ,$ and no then the function can be written in the form:
$f(x)=a (x− x 1 ) p 1 (x− x 2 ) p 2 ⋯ (x− x n ) p n (x− v 1 ) q 1 (x− v 2 ) q 2 ⋯ (x− v m ) q n f(x)=a (x− x 1 ) p 1 (x− x 2 ) p 2 ⋯ (x− x n ) p n (x− v 1 ) q 1 (x− v 2 ) q 2 ⋯ (x− v m ) q n$
where the powers $p i p i$ or $q i q i$ on each factor can be determined by the behavior of the graph at the corresponding intercept or asymptote, and the stretch factor $a a$ can be determined given a value of the function other than the x-intercept or by the horizontal asymptote if it is nonzero.
How To
Given a graph of a rational function, write the function.
1. Determine the factors of the numerator. Examine the behavior of the graph at the x-intercepts to determine the zeroes and their multiplicities. (This is easy to do when finding the “simplest” function with small multiplicities—such as 1 or 3—but may be difficult for larger multiplicities—such as 5 or 7, for example.)
2. Determine the factors of the denominator. Examine the behavior on both sides of each vertical asymptote to determine the factors and their powers.
3. Use any clear point on the graph to find the stretch factor.
Example 12
Writing a Rational Function from Intercepts and Asymptotes
Write an equation for the rational function shown in Figure 22.
Figure 22
Media
Access these online resources for additional instruction and practice with rational functions.
5.6 Section Exercises
Verbal
1.
What is the fundamental difference in the algebraic representation of a polynomial function and a rational function?
2.
What is the fundamental difference in the graphs of polynomial functions and rational functions?
3.
If the graph of a rational function has a removable discontinuity, what must be true of the functional rule?
4.
Can a graph of a rational function have no vertical asymptote? If so, how?
5.
Can a graph of a rational function have no x-intercepts? If so, how?
Algebraic
For the following exercises, find the domain of the rational functions.
6.
$f(x)= x−1 x+2 f(x)= x−1 x+2$
7.
$f(x)= x+1 x 2 −1 f(x)= x+1 x 2 −1$
8.
$f(x)= x 2 +4 x 2 −2x−8 f(x)= x 2 +4 x 2 −2x−8$
9.
$f(x)= x 2 +4x−3 x 4 −5 x 2 +4 f(x)= x 2 +4x−3 x 4 −5 x 2 +4$
For the following exercises, find the domain, vertical asymptotes, and horizontal asymptotes of the functions.
10.
$f(x)= 4 x−1 f(x)= 4 x−1$
11.
$f( x )= 2 5x+2 f( x )= 2 5x+2$
12.
$f(x)= x x 2 −9 f(x)= x x 2 −9$
13.
$f(x)= x x 2 +5x−36 f(x)= x x 2 +5x−36$
14.
$f( x )= 3+x x 3 −27 f( x )= 3+x x 3 −27$
15.
$f(x)= 3x−4 x 3 −16x f(x)= 3x−4 x 3 −16x$
16.
$f(x)= x 2 −1 x 3 +9 x 2 +14x f(x)= x 2 −1 x 3 +9 x 2 +14x$
17.
$f(x)= x+5 x 2 −25 f(x)= x+5 x 2 −25$
18.
$f(x)= x−4 x−6 f(x)= x−4 x−6$
19.
$f( x )= 4−2x 3x−1 f( x )= 4−2x 3x−1$
For the following exercises, find the x- and y-intercepts for the functions.
20.
$f(x)= x+5 x 2 +4 f(x)= x+5 x 2 +4$
21.
$f(x)= x x 2 −x f(x)= x x 2 −x$
22.
$f(x)= x 2 +8x+7 x 2 +11x+30 f(x)= x 2 +8x+7 x 2 +11x+30$
23.
$f(x)= x 2 +x+6 x 2 −10x+24 f(x)= x 2 +x+6 x 2 −10x+24$
24.
$f(x)= 94−2 x 2 3 x 2 −12 f(x)= 94−2 x 2 3 x 2 −12$
For the following exercises, describe the local and end behavior of the functions.
25.
$f( x )= x 2x+1 f( x )= x 2x+1$
26.
$f( x )= 2x x−6 f( x )= 2x x−6$
27.
$f( x )= −2x x−6 f( x )= −2x x−6$
28.
$f( x )= x 2 −4x+3 x 2 −4x−5 f( x )= x 2 −4x+3 x 2 −4x−5$
29.
$f( x )= 2 x 2 −32 6 x 2 +13x−5 f( x )= 2 x 2 −32 6 x 2 +13x−5$
For the following exercises, find the slant asymptote of the functions.
30.
$f(x)= 24 x 2 +6x 2x+1 f(x)= 24 x 2 +6x 2x+1$
31.
$f(x)= 4 x 2 −10 2x−4 f(x)= 4 x 2 −10 2x−4$
32.
$f(x)= 81 x 2 −18 3x−2 f(x)= 81 x 2 −18 3x−2$
33.
$f(x)= 6 x 3 −5x 3 x 2 +4 f(x)= 6 x 3 −5x 3 x 2 +4$
34.
$f(x)= x 2 +5x+4 x−1 f(x)= x 2 +5x+4 x−1$
Graphical
For the following exercises, use the given transformation to graph the function. Note the vertical and horizontal asymptotes.
35.
The reciprocal function shifted up two units.
36.
The reciprocal function shifted down one unit and left three units.
37.
The reciprocal squared function shifted to the right 2 units.
38.
The reciprocal squared function shifted down 2 units and right 1 unit.
For the following exercises, find the horizontal intercepts, the vertical intercept, the vertical asymptotes, and the horizontal or slant asymptote of the functions. Use that information to sketch a graph.
39.
$p( x )= 2x−3 x+4 p( x )= 2x−3 x+4$
40.
$q( x )= x−5 3x−1 q( x )= x−5 3x−1$
41.
$s( x )= 4 ( x−2 ) 2 s( x )= 4 ( x−2 ) 2$
42.
$r( x )= 5 ( x+1 ) 2 r( x )= 5 ( x+1 ) 2$
43.
$f( x )= 3 x 2 −14x−5 3 x 2 +8x−16 f( x )= 3 x 2 −14x−5 3 x 2 +8x−16$
44.
$g( x )= 2 x 2 +7x−15 3 x 2 −14x+15 g( x )= 2 x 2 +7x−15 3 x 2 −14x+15$
45.
$a( x )= x 2 +2x−3 x 2 −1 a( x )= x 2 +2x−3 x 2 −1$
46.
$b( x )= x 2 −x−6 x 2 −4 b( x )= x 2 −x−6 x 2 −4$
47.
48.
$k( x )= 2 x 2 −3x−20 x−5 k( x )= 2 x 2 −3x−20 x−5$
49.
$w( x )= ( x−1 )( x+3 )( x−5 ) ( x+2 ) 2 (x−4) w( x )= ( x−1 )( x+3 )( x−5 ) ( x+2 ) 2 (x−4)$
50.
$z( x )= ( x+2 ) 2 ( x−5 ) ( x−3 )( x+1 )( x+4 ) z( x )= ( x+2 ) 2 ( x−5 ) ( x−3 )( x+1 )( x+4 )$
For the following exercises, write an equation for a rational function with the given characteristics.
51.
Vertical asymptotes at $x=5 x=5$ and $x=−5, x=−5,$ x-intercepts at $(2,0) (2,0)$ and $(−1,0), (−1,0),$ y-intercept at $( 0,4 ) ( 0,4 )$
52.
Vertical asymptotes at $x=−4 x=−4$ and $x=−1, x=−1,$ x-intercepts at $( 1,0 ) ( 1,0 )$ and $( 5,0 ), ( 5,0 ),$ y-intercept at $(0,7) (0,7)$
53.
Vertical asymptotes at $x=−4 x=−4$ and $x=−5, x=−5,$ x-intercepts at $( 4,0 ) ( 4,0 )$ and $( −6,0 ), ( −6,0 ),$ Horizontal asymptote at $y=7 y=7$
54.
Vertical asymptotes at $x=−3 x=−3$ and $x=6, x=6,$ x-intercepts at $( −2,0 ) ( −2,0 )$ and $( 1,0 ), ( 1,0 ),$ Horizontal asymptote at $y=−2 y=−2$
55.
Vertical asymptote at $x=−1, x=−1,$ Double zero at $x=2, x=2,$ y-intercept at $(0,2) (0,2)$
56.
Vertical asymptote at $x=3, x=3,$ Double zero at $x=1, x=1,$ y-intercept at $(0,4) (0,4)$
For the following exercises, use the graphs to write an equation for the function.
57.
58.
59.
60.
61.
62.
63.
64.
Numeric
For the following exercises, make tables to show the behavior of the function near the vertical asymptote and reflecting the horizontal asymptote
65.
$f(x)= 1 x−2 f(x)= 1 x−2$
66.
$f(x)= x x−3 f(x)= x x−3$
67.
$f(x)= 2x x+4 f(x)= 2x x+4$
68.
$f(x)= 2x (x−3) 2 f(x)= 2x (x−3) 2$
69.
$f(x)= x 2 x 2 +2x+1 f(x)= x 2 x 2 +2x+1$
Technology
For the following exercises, use a calculator to graph $f( x ). f( x ).$ Use the graph to solve $f( x )>0. f( x )>0.$
70.
$f(x)= 2 x+1 f(x)= 2 x+1$
71.
$f(x)= 4 2x−3 f(x)= 4 2x−3$
72.
$f(x)= 2 ( x−1 )( x+2 ) f(x)= 2 ( x−1 )( x+2 )$
73.
$f(x)= x+2 ( x−1 )( x−4 ) f(x)= x+2 ( x−1 )( x−4 )$
74.
$f(x)= (x+3) 2 ( x−1 ) 2 ( x+1 ) f(x)= (x+3) 2 ( x−1 ) 2 ( x+1 )$
Extensions
For the following exercises, identify the removable discontinuity.
75.
$f(x)= x 2 −4 x−2 f(x)= x 2 −4 x−2$
76.
$f(x)= x 3 +1 x+1 f(x)= x 3 +1 x+1$
77.
$f(x)= x 2 +x−6 x−2 f(x)= x 2 +x−6 x−2$
78.
$f(x)= 2 x 2 +5x−3 x+3 f(x)= 2 x 2 +5x−3 x+3$
79.
$f(x)= x 3 + x 2 x+1 f(x)= x 3 + x 2 x+1$
Real-World Applications
For the following exercises, express a rational function that describes the situation.
80.
A large mixing tank currently contains 200 gallons of water, into which 10 pounds of sugar have been mixed. A tap will open, pouring 10 gallons of water per minute into the tank at the same time sugar is poured into the tank at a rate of 3 pounds per minute. Find the concentration (pounds per gallon) of sugar in the tank after $t t$ minutes.
81.
A large mixing tank currently contains 300 gallons of water, into which 8 pounds of sugar have been mixed. A tap will open, pouring 20 gallons of water per minute into the tank at the same time sugar is poured into the tank at a rate of 2 pounds per minute. Find the concentration (pounds per gallon) of sugar in the tank after $t t$ minutes.
For the following exercises, use the given rational function to answer the question.
82.
The concentration $C C$ of a drug in a patient’s bloodstream $t t$ hours after injection is given by $C(t)= 2t 3+ t 2 . C(t)= 2t 3+ t 2 .$ What happens to the concentration of the drug as $t t$ increases?
83.
The concentration $C C$ of a drug in a patient’s bloodstream $t t$ hours after injection is given by $C(t)= 100t 2 t 2 +75 . C(t)= 100t 2 t 2 +75 .$ Use a calculator to approximate the time when the concentration is highest.
For the following exercises, construct a rational function that will help solve the problem. Then, use a calculator to answer the question.
84.
An open box with a square base is to have a volume of 108 cubic inches. Find the dimensions of the box that will have minimum surface area. Let $x x$ = length of the side of the base.
85.
A rectangular box with a square base is to have a volume of 20 cubic feet. The material for the base costs 30 cents/ square foot. The material for the sides costs 10 cents/square foot. The material for the top costs 20 cents/square foot. Determine the dimensions that will yield minimum cost. Let $x x$ = length of the side of the base.
86.
A right circular cylinder has volume of 100 cubic inches. Find the radius and height that will yield minimum surface area. Let $x x$ = radius.
87.
A right circular cylinder with no top has a volume of 50 cubic meters. Find the radius that will yield minimum surface area. Let $x x$ = radius.
88.
A right circular cylinder is to have a volume of 40 cubic inches. It costs 4 cents/square inch to construct the top and bottom and 1 cent/square inch to construct the rest of the cylinder. Find the radius to yield minimum cost. Let $x x$ = radius.
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## USING METHOD OF INTEGRATION ,HOW TO FIND AREA OF TRIANGLE BOUNDED BY THREE LINES
Using method of integration find the area of triangle,using the method of integration find the area of the region bounded by the lines,area of triangle by integration method
## Using Method of Integration , How to find the area of triangle bounded by three lines
2x + y = 0 , 3x - 2y = 6 and x - 3y + 5 = 0
## Solution
Given lines are
2x + y = 4 -------- (1)
3x - 2y = 6 --------- (2) and
x - 3y = -5 --------- (3)
If these lines are intersecting then we have to find their coordinates of points of intersection .
### To Find Coordinate of Point A
Multiply (1) by 2 and adding to (2) , we get
4x + 2y + 3x - 2y = 8 + 6
7x = 14 ⇒ x =2
Putting x = 2 in (1) , we get
Area under Curve
2(2) + y = 4
4 + y = 4 ⇒ y = 0
∴ (1) and (2) meets at point A(2,0).
### To Find Coordinate of Point B
To find point of intersection (2) and (3);
Multiply (3) by -3 and adding to (2) , we get
3x - 2y -3x +9y = 6 +15
7y = 21 ⇒ y = 3
Putting y = 3 in (3) , we get
x-3(3) = -5
⇒ x = -5 +9 ⇒ x = 4
∴ (2) and (3) meets at point B(4,3).
### To Find Coordinate of Point C
To find point of intersection (1) and (3);
Multiply (1) by 3 and adding to (3) , we get
6x + 3y + x - 3y = 12 - 5
7x = 7 ⇒ x =1
Putting x = 1 in (1) , we get
2(1) + y = 4
⇒ y =4 - 2 ⇒ y = 2
∴ (1) and (3) meets at point C(1,2).
we get points of intersection of (1) and (2) A(2,0), points of intersection of (2) and (3) B(4,3) and points of intersection of (1) and (3) C(1,2).
## Required Area = Shaded Area = Area DCBED - Area DCAD -Area ABEA
Also read my previous post How to find area bounded by two circles
For better understanding watch this video
## Conclusion
Thanks for visiting this website and spending your precious time to read how to find area of triangle Using method of integration find the area of triangle,using the method of integration find the area of the region bounded by the lines,area of triangle by integration method.
If you are a mathematician Don't forget to visit my Mathematics You tube channel ,Mathematics Website and Mathematics Facebook Page , whose links are given below
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#### 1 comment:
1. Thanks for sharing this article here about the calculation of the Perimeter Of A Triangle. Your article is very informative and I will share it with my other friends as the information is really very useful. Keep sharing your excellent work.
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The limit inferior and the limit superior of a sequence
Prerequirements:
In this post we’ll see two concepts of mathematical analysis which will be useful in number theory: the limit inferior ($\lim \inf$) and the limit superior ($\lim \sup$) of a sequence.
Like for the post about asymptotic analysis, we don’t claim to cover the topic in depth, but out intent is rather to give an intuitive idea of what the limit superior and the limit inferior of a sequence are, without neglecting the mathematical rigour.
Sequences and intervals
Let’s consider a real sequence $(a_n)_{n \geq 0} = a_0, a_1, a_2, \ldots, a_n, \ldots$, that is a sequence where all the $a_n$ are real numbers. Indeed it’s a function from $\mathbb{N}$ to $\mathbb{R}$ that maps $n$ to $a_n$, so we can represent it in a Cartesian plane where $n$ is on the horizontal axis and $a_n$ on the vertical axis, like in Figure 1.
The sequence shown in Figure 1 is given by the following mathematical definition:
$$a_n = \begin{cases} 5 & \text{if n = 4} \\ -\frac{7}{2} & \text{if n = 7} \\ |n \mathrm{\ mod\ } 6 - 3| - 1 & \text{otherwise} \end{cases}$$
Without discussing the details of the algebraic expression of the last case, that in this context has little importance, it’s clear that the cases $n = 4$ and $n = 7$ are treated as exceptions; hence we can imagine that, continuing on the right towards infinity, the graph will always have the same behaviour, oscillating between the values -1 and 2. So we can identify two intervals on the vertical axis, which correspond to two horizontal bands in the graph (Figure 2):
• The interval $[-\frac{7}{2}, 5]$, that contains all the terms of the sequence
• The interval $[-1, 2]$, that contains all the terms of the sequence, except two
Using the statistical language, we can say that the terms $a_4$ and $a_7$ are outliers, i.e. some values that clearly differ from the others, and that for this reason are negligible. In descriptive statistics, two possible analyses that can be made on a data set are just the identification of the outliers and of some intervals that contain most data, a bit like we have done till now with our sequence. But the comparison with statistics ends here, because there is a big difference between the study of a statistical sample and the study of a sequence: the latter is an infinite sequence of data, whereas a statistical sample is finite. This changes many things. For example we could ask ourselves:
• If the value 5 was repeated infinite times, but only every 1000 terms of the sequence, would it still be negligible?
• Does it make sense to ask what is an interval containing more than tha 99% of the sequence terms?
The answers to both questions are negative:
• If the value 5 was repeated every 1000 terms, though it may be considered “rare”, it still would repeat itself infinite times! One thing is to neglect a single abnormal value, another thing is to neglect infinite ones.
• The 99%, or any fixed percentage greater than 0, of an infinite amount is still an infinite amount. There is no numeric difference between the 99%, the 50% or the 100% of infinite: they are all infinite in the same way.
So let’s try to understand what makes sense to ask ourselves when we study a sequence, containing infinite terms. An important concept, on which we’ll focus hereinafter, is the distinction between “all” and “almost all”. We’ll apply it to the terms of the sequence, but it’s more generally defined on any set of infinite elements:
• “All” means… just all the set! No element excluded.
• “Almost all” means all the set, except a finite number of elements.
Applying this concept to the set of the terms of a sequence, we can say that it’s meaningful to ask:
• What is an interval that contains all the terms of a sequence?
• What is an interval that contains almost all the terms of a sequence?
As we saw, in the case of the sequence shown in Figure 1, an interval that contains all its terms is $[-\frac{7}{2}, 5]$, whereas one that contains almost all of them is $[-1, 2]$; in fact the latter interval contains all the terms except two, that is a finite number. These intervals are particular, because they are also the smallest possible ones, each for the respective category. In fact:
• Also $[-4, 6]$, $[-5, 10]$, $[-10, 10]$, … are intervals that contain all the terms of the sequence, but they aren’t the smallest intervals with this property. The smallest is $[-\frac{7}{2}, 5]$ because, if we made it smaller on the right, it would not include the term $a_4 = 5$; if we made it smaller on the left, it would not include the term $a_7 = -\frac{7}{2}$.
• Also $[-2, 5]$, $[4, 4]$ and $[-3, 4]$ are intervals that contain almost all the terms of the sequence, but they aren’t the smallest intervals with this property. The smallest is $[-1, 2]$ because, if we made it smaller on the right, it would not include all the sequence terms equal to 2, that are infinite; similarly, if we made it smaller on the left, it would not include all the terms equal to -1, that are also infinite.
We already used several times the word “infinite”, and we’ll continue often to use it in this post. It’s important to point out that we always refer to an infinity of countable kind, that is the same kind of infinity of the natural numbers. There are other kinds of infinity that are said to be uncountable, but we’ll not consider them in this post.
Infimum, supremum, limit inferior and limit superior of a real sequence
Now we could ask ourselves: does the smallest interval containing all, or almost all, the terms of a sequence always exist for all sequences?
The answer is anything but trivial. For the first question it’s affirmative, considering closed intervals:
Existence of the smallest closed interval that contains all the terms of a sequence
For every real sequence $(a_n)$, there exists one and only one closed interval such that:
• It contains all the terms of the sequence
• It’s the smallest closed interval with the previous property
What does “smallest interval” exactly mean?
When in mathematics one person speaks about an object that is smaller, or greater, than another one, he or she always should clarify in what sense. When the objects are integer or real numbers, the question is rather trivial, but if we have two intervals $I$ and $J$, how can we establish which one is the smallest? There are at least two ways:
• The smallest interval is the one with the smallest width (where the width is given by the difference between the two endpoints)
• The smallest interval is the one that is contained in the other one
Both these ways to interpret the concept of smallest interval have their own problems:
• In the first way, unlimited intervals cannot be adequately compared. For example the intervals $[-1, +\infty]$, $[-\infty, 1000]$ e $[-\infty, +\infty]$ have all the same width (infinite), but they are very different from each other.
• In the second way, some unlimited intervals can be easily compared (for example $[-1, +\infty]$ is greater than $[3, +\infty]$, because it contains it). On the other hand, many couples of intervals are not comparable, in the sense that we cannot establish which is the smallest (for example the intervals $[0, 2]$ and $[1, 3]$ are not comparable, because neither of the two contains the other one).
In this post we’ll adopt the second definition, the one based on containment. This definition is also the most general, because it can be applied to any set, not only to intervals. The adoption of one definition rather that the other one, however, has relative importance in this post, because all that we’ll say, with small formal modifications, would still be valid if intervals were compared to themselves by width.
The following property, of which we’ll omit the proof, lets us define the infimum ($\inf$) and the supremum ($\sup$) of a real sequence, that are simply the endpoints of the interval given by Property A.1:
$\inf$ and $\sup$ of a real sequence
Given a real sequence $(a_n)$, let $I = [i, s]$ be the smallest closed interval that contains all the sequence terms. Then we define the following:
$$\begin{gathered}\inf a_n := i \\ \sup a_n := s \end{gathered}$$
For example, in our case:
$$[\inf a_n, \sup a_n] = \left[-\frac{7}{2}, 5\right]$$
We can note that in Definition A.5 it’s not specified if the $\inf$ and the $\sup$ are two terms of the sequence. In the sequence $a_n$ of Figure 5 it’s so, but generally the $\inf$ and the $\sup$ may be only some limit values that indeed do not correspond to any particular term. For example, the $\inf$ of the sequence $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots$ is 0, but no term of the sequence is equal to zero.
In the case of unbounded sequences, the $\inf$ or the $\sup$ can be infinite. In this case the interval $I$ of Definition A.5 can be of the kind $[-\infty, s]$, $[i, +\infty]$ or $[-\infty, +\infty]$. For example, the Fibonacci sequence $1, 2, 3, 5, \ldots$ (in which each term is the sum of the two previous ones) has 1 as $\inf$ and $+\infty$ as $\sup$. Note that the intervals of the kind $[-\infty, s]$, $[i, +\infty]$ or $[-\infty, +\infty]$ are closed, as required by Definition A.5. In fact $+\infty$ and/or $-\infty$, as appropriate, belong to such intervals, as if they were real numbers. The intervals $[-\infty, s]$, $[i, +\infty]$ and $[-\infty, +\infty]$ so are not intervals in $\mathbb{R}$, but in the set $\mathbb{R} \cup \{-\infty, +\infty\}$.
Definition A.5 defines simultaneously the $\inf$ and the $\sup$ of a sequence, but there is also a way to define them separately. Let’s see this alternative definition, which is the canonical one:
$\inf$ and $\sup$ of a real sequence, canonical definition
Given a real sequence $(a_n)$, we define the following:
$$\inf a_n := \begin{cases} \begin{gathered}\text{the greatest real number} \\ \text{less than or equal to all} \\ \text{the sequence terms} \end{gathered} & \text{if (a_n) is bounded below} \\ -\infty & \text{otherwise} \end{cases}$$
$$\sup a_n := \begin{cases} \begin{gathered}\text{the smallest real number} \\ \text{greater or equal to all} \\ \text{the sequence terms} \end{gathered} & \text{if (a_n) is bounded above} \\ +\infty & \text{otherwise} \end{cases}$$
The two definitions are equivalent, meaning that, if Definition A.5 is adopted as definition of the $\inf$ and the $\sup$, then Definition A.6 can be proved as a property; conversely, if Definition A.6 is adopted, then Definition A.5 can be proved as a property. In this post we chose Definition A.5 because we consider it simpler and more intuitive.
On the basis of the previous example, we could be tempted to say that a property similar to A.1 is valid for the intervals that contain almost all the terms of the sequence, i.e. that there is always one of them which is the smallest. We’ll see that this is not true, taking as an example the sequence shown in Figure 3.
This sequence, that we’ll call $(b_n)_{n \geq 0}$, is defined starting from $(a_n)$, as follows:
$$b_n := \begin{cases} \frac{a_n + 1}{n + 1} + 1 & \text{if a_n \gt 0} \\ \frac{a_n}{n} & \text{otherwise} \end{cases} \tag{1}$$
As it’s clear from the picture, the positive terms of the sequence tend to 1, whereas the negative ones tend to zero. The sequence on the whole does not have a limit, but it can be divided into two subsequences having different limits. But this is not the true difference with the previous sequence $a_n$, because also in $a_n$ some subsequences with a limit can be found: for example the terms $a_0, a_6, a_{12}, a_{18}, \ldots, a_{6k}, \ldots$ are all equal to 2, so all the more true they have limit 2.
The most important difference between the sequences $(a_n)$ and $(b_n)$ is that in the first one there exists the smallest interval containing almost all the terms of the sequence, in the second one it does not exist. Let’s see why. First of all in the sequence $(b_n)$ no term is equal to 1 or to 0, even if the positive terms tend to 1 and the negative ones tend to 0: this is clear from the graph.
Why is no term of the sequence $b_n$ equal to 1 or 0?
The reason is very simple. Any term of the sequence $b_n$ is given by the formula $\frac{a_n + 1}{n + 1} + 1$ or by the formula $\frac{a_n}{n}$. Let’s suppose that $b_n = 0$. If $b_n$ was obtained by the formula $\frac{a_n}{n} = 0$, $a_n$ should be 0, but no term of the sequence $(a_n)$ is 0; if instead it was obtained by the formula $\frac{a_n + 1}{n + 1} + 1 = 0$, $a_n$ should be $-n - 2$, but it cannot be because, according to (1), the formula $b_n = \frac{a_n + 1}{n + 1} + 1 = 0$ is valid only when $a_n \gt 0$. We can reason in the same way for proving also that no term of the sequence is equal to 1.
The values 0 and 1 are so theoretical limits that are never actually reached: all the terms of the sequence are either smaller than 0, or greater than 1. It follows that the interval $[0, 1]$ does not contain any term. Then, if we want to find the smallest interval that contains almost all the terms of the sequence, we have to extend the interval $[0, 1]$ both on the right and on the left, in order to include both the terms greater than 1 and those ones smaller than zero (both infinite).
Actually, it can be proved that any interval greater than $[0, 1]$ contains almost all the terms of the sequence, and that there is no one which is smaller than all the others.
Why does any interval greater than $[0, 1]$ contain almost all the terms of the sequence, and there is no one that is smaller than all the others?
Let’s suppose for simplicity to extend the width of the interval of the same amount both on the right and on the left; for example, if the amount is $\frac{1}{5}$ we’ll obtain the interval $I := [0 -\frac{1}{5}, 1 + \frac{1}{5}] = [-\frac{1}{5}, \frac{6}{5}]$. How many terms of the sequence does this interval contain?
Let’s start from the negative terms. Since they tend to zero, becoming smaller and smaller in absolute value, from a certain point of the sequence on (that is “definitively”, using the terminology of asymptotic analysis) they will always be greater or equal to $-\frac{1}{5}$. So there exists a certain constant $k$ such that, for every $n \geq k$, all the negative terms $b_n$ are greater or equal to $-\frac{1}{5}$. Therefore the only sequence terms that can be negative and not belong to the interval $I$ are $b_0, b_1, b_2, \ldots, b_{k-1}$: they are at most $k$ terms (actually they are less than $k$, because the listed terms include also the positive ones; but in the worst hypothesis, that is if no positive term exists, they would be $k$).
About the positive terms we can argument in a similar way. Since they tend to 1, becoming smaller and smaller, they will be definitively smaller or equal to $\frac{6}{5}$. So there exists a certain constant $h$ such that, for every $n \geq h$, all the positive terms $b_n$ are less than or equal to $\frac{6}{5}$. So the only sequence terms that can be positive and not belong to the interval $I$ are $b_0, b_1, b_2, \ldots, b_{h-1}$: they are at most $h$ terms.
Summarizing, at most $k + h$ sequence terms, that is a finite number of terms, can be outside of the interval $I = [-\frac{1}{5}, \frac{6}{5}]$. So this interval contains almost all the terms of the sequence (see Figure 4).
But then, if the interval $[0, 1]$ does not contain any term $b_n$, and the interval $[-\frac{1}{5}, \frac{6}{5}]$ contains almost all of them, certainly the smallest interval that contains almost all the terms must be intermediate between the two. However, no matter how long we can try to look for this interval, we’ll never find it… bacause it does not exist! In fact, like for constructing the interval $I$ we started from the interval $[0, 1]$ widening it by $\frac{1}{5}$ both on the right and on the left, if instead of $\frac{1}{5}$ we took another positive constant $\epsilon$, obtaining the interval $I_{\epsilon} := [-\epsilon, 1 + \epsilon]$, nothing would change, because also this interval would contain almost all the terms, for any $\epsilon$. In fact, even supposing the constant $\epsilon$ to be very small, there will always exist some indexes $k_{\epsilon}$ and $h_{\epsilon}$ such that all the negative sequence terms with index greater or equal to $k_{\epsilon}$ are greater or equal to $-\epsilon$, an all the positive terms with index greater or equal than $h_{\epsilon}$ are less than or equal to $1 + \epsilon$. The smaller $\epsilon$ is, the bigger $k_{\epsilon}$ and $h_{\epsilon}$ will be, but they will still be finite. So the interval $I_{\epsilon}$ does not include at most $k_{\epsilon} + h_{\epsilon}$ sequence terms, that is a finite number, so it contains almost all of them.
From that we can deduce that there is not one smallest interval that contains almost all the sequence terms. This comes from the fact that there is no real positive number that is the smallest (in contrast to integer numbers). In fact, we saw that any constant $\epsilon \gt 0$ corresponds to an interval $I_{\epsilon} = [-\epsilon, 1 + \epsilon]$ which contains almost all the sequence terms; making $\epsilon$ smaller, this interval becomes smaller and smaller, but there not exists a smallest interval, because there not exists a smallest positive $\epsilon$.
Summarizing, we saw that the interval $[0, 1]$, though not containing any term of the sequence $(b_n)$, has a particular property: any interval properly containing it, contains almost all the sequence terms. In other words, we pass from zero to almost all the terms with even a very small enlargement of the interval, no middle ground. But another interesting thing is that the interval $[0, 1]$ is the smallest interval with this property. In fact, we can prove that the property is valid also for all the intervals greater than $[0, 1]$, while it is not for all the smaller ones:
• We said that, if $A$ is an interval greater than $[0, 1]$, it contains almost all the sequence terms. So the property holds for $A$ because, as $A$ itself contains almost all the terms, this will be all the more true also for any interval greater than $A$.
• If $B$ in an interval smaller than $[0, 1]$, the property does not hold for $B$, because it’s not true that all the intervals greater than $B$ contains almost all the sequence terms; in fact $[0, 1]$ in greater than $B$ but does not contain any term.
So $[0, 1]$ is not the smallest interval that contains almost all the sequence terms (that does not exist), but the smallest interval such that any interval greater than it contains almost all the sequence terms. It can be proved (but we don’t do it) that such an interval exists for any real sequence:
Existence of the smallest interval such that any interval greater than it contains almost all the terms of a real sequence
For any real sequence $(a_n)$, there exists an interval such that:
• Any interval greater than it contains almost all the terms of the sequence
• It’s the smallest interval with the previous property
We can note that the interval that satisfies Property A.2 does not necessarily contain almost all the sequence terms: the intervals greater than it do. This is just what happens in our example: the interval $[0, 1]$ does not contain almost all the sequence terms, actually it does not contain any one!
As well as Property A.1 lets us define $\inf$ and $\sup$ of a sequence, Property A.2 lets us define $\lim \inf$ and $\lim \sup$, that are the endpoints of the interval that satisfies it:
$\lim \inf$ and $\lim \sup$ of a real sequence
Given a real sequence $(a_n)$, let $I = [i, s]$ be the interval that satisfies Property A.2. Then we define the following:
$$\begin{gathered} \lim \inf a_n := i \\ \lim \sup a_n := s \end{gathered}$$
In the case of the sequence $b_n$, as we saw:
$$[\lim \inf b_n, \lim \sup b_n] = [0, 1]$$
Alternative definition of $\lim \inf$ and $\lim \sup$
Like we pointed out before about $\inf$ and $\sup$, similarly the canonical definition of the $\lim \inf$ and the $\lim \sup$ of a sequence is not Definition A.7, but the following that defines them separately:
$\lim \inf$ and $\lim \sup$ of a real sequence, canonical definition
Let $(a_n)$ be a real sequence.
We define $\lim \sup a_n$ as the real number $s$ such that:
• If $(a_n)$ is bounded above:
• Almost all the sequence terms are less than or equal to $s$
• It’s the smallest number with the previous property
• Otherwise, $s = +\infty$
We define $\lim \inf a_n$ as the real number $i$ such that:
• If $(a_n)$ is bounded below:
• Almost all the sequence terms are greater or equal to $i$
• It’s the greatest number with the previous property
• Otherwise, $i = -\infty$
As well as Definitions A.5 and A.6, Definitions A.7 and A.8 are equivalent: having chosen one of the two, the other can be proved as a property.
Relationship between the limit superior, the limit inferior and the limit of a sequence
We conclude by stating two important properties that link the $\lim \inf$ and the $\lim \sup$ of a sequence, on one side, and its limit, on the other side.
We saw that, in the sequence $(b_n)$ shown in Figure 3, the $\lim \inf$ and the $\lim \sup$ are different, and the sequence doesn’t have a limit, but it oscillates continuously between the $\lim \inf$ (0) and the $\lim \sup$ (1), approximating now to one and now to the other. This fact is not a characteristic of our example, but it’s true in general. In fact the following Proposition can be proved:
Any real sequence with $\lim \inf$ different from $\lim \sup$ does not have a limit
Any real sequence $(a_n)$ such that $\lim \inf a_n \neq \lim \sup a_n$ does not have a limit.
Let’s see instead what happens when $\lim \inf$ and $\lim \sup$ are equal.
Let’s consider the sequence $(c_n)$ defines as follows:
$$c_n := \frac{a_n}{n + 1}$$
By drawing the graph we can see clearly that this sequence tends to zero:
Let’s ask ourselves now: according to Definition A.7, what are the $\lim \inf$ and the $\lim \sup$ of this sequence? We’ll start, like before, from the analysis of the intervals that contain almost all the sequence terms. Intuitively, since the sequence terms tend to concentrate around the zero, both above and below it, an interval that contains almost all the terms necessarily must contain the zero.
Indeed, like before, it can be proved that any interval that contains zero in its interior, contains almost all the terms of the sequence.
We can consider by simplicity an interval with the same width both on the right and on the left of the zero, that is of the kind $[-\epsilon, \epsilon]$, for some constant $\epsilon \gt 0$. Indeed, it can be proved (using the definition of limit) that such an interval contains, for any $\epsilon$, almost all the terms of the sequence, in particular all the terms with an index greater or equal to a certain constant $k_{\epsilon} \gt 0$ (see Figure 6).
So we get back to the same argument as before: by making $\epsilon$ smaller we can find infinite intervals of the kind $[-\epsilon, \epsilon]$, all of which contain almost all the terms of the sequence.
But we want to find a closed interval which satisfies Definition A.7. So this interval must be smaller than all the intervals that contain 0 in their interior… Then there is only one possibility, and it’s a very particular interval: it’s the interval $[0, 0]$, having 0 as both the endpoints, so not as an interior point. It’s indeed the set reduced to just the number 0, but it’s still a closed interval, and it satisfies Definition A.7.
The interval [0, 0] satisfies Definition A.7 because it satisfies both points of it:
• All the intervals of the kind $[-\epsilon, \epsilon]$, that are greater than $[0,0]$, contain almost all the sequence terms (the same thing is true for any greater interval, also with endpoints different in absolute value, but for simplicity we don’t consider this more general case).
• About the fact that it’s the smallest interval with the previous property, the problem does not arise, as it is reduced to a single point, so there is no smaller interval (the empty set is not an interval).
So, according to Definition A.7:
$$[\lim \inf c_n, \lim \sup c_n] = [0, 0]$$
We can note a strong analogy between this sequence and the previous one: for example, as well as the interval $[0,1]$ does not contain any term of the sequence $(b_n)$, the interval $[0,0]$ does not contain any term of $(c_n)$.
But there is a substantial difference between the two sequences: the sequence $(c_n)$ has a limit, and its $\lim \inf$ coincides with its $\lim \sup$, instead the sequence $(b_n)$ has the $\lim \inf$ different from the $\lim \sup$ and, by Proposition A.1, has not a limit. Also for the case of $(c_n)$, the relationship we noted among the limit inferior, the limit superior and the limit is true generally for any real sequence:
Any real sequence with $\lim \inf$ equal to $\lim \sup$ has a limit
Any real sequence $(a_n)$ such that $\lim \inf a_n = \lim \sup a_n$ has a limit, and $\lim a_n = \lim \inf a_n = \lim \sup a_n$.
Joining Propositions A.1 and A.2, we can immediately obtain the following Property:
A real sequence has a limit if and only if its $\lim \inf$ and its $\lim \sup$ are equal.
A real sequence $(a_n)$ has a limit if and only if $\lim \inf a_n = \lim \sup a_n$, and in this case $\lim a_n = \lim \inf a_n = \lim \sup a_n$.
In theory the previous Property may be used as the definition of the limit of a sequence: this way this limit would not be defined as a separate entity, but on the basis of the limit inferior and the limit superior.
In the next post we’ll apply this Property to the function $\frac{\pi(x)}{x}$.
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# Chapter 3 - Part A Descriptive Statistics: Numerical Methods
## Presentation on theme: "Chapter 3 - Part A Descriptive Statistics: Numerical Methods"— Presentation transcript:
Chapter 3 - Part A Descriptive Statistics: Numerical Methods
Measures of Location Measures of Variability % x
Measures of Location Mean Median Mode Percentiles Quartiles
Example: Apartment Rents
Given below is a sample of monthly rent values (\$) for one-bedroom apartments. The data is a sample of 70 apartments in a particular city. The data are presented in ascending order.
Mean The mean of a data set is the average of all the data values.
If the data are from a sample, the mean is denoted by . If the data are from a population, the mean is denoted by (mu).
Example: Apartment Rents
Mean
Median The median is the measure of location most often reported for annual income and property value data. A few extremely large incomes or property values can inflate the mean.
Median The median of a data set is the value in the middle when the data items are arranged in ascending order. For an odd number of observations, the median is the middle value. For an even number of observations, the median is the average of the two middle values.
Example: Apartment Rents
Median Median = 50th percentile i = (p/100)n = (50/100)70 = Averaging the 35th and 36th data values: Median = ( )/2 = 475
Mode The mode of a data set is the value that occurs with greatest frequency. The greatest frequency can occur at two or more different values. If the data have exactly two modes, the data are bimodal. If the data have more than two modes, the data are multimodal.
Example: Apartment Rents
Mode 450 occurred most frequently (7 times) Mode = 450
Percentiles A percentile provides information about how the data are spread over the interval from the smallest value to the largest value. Admission test scores for colleges and universities are frequently reported in terms of percentiles.
Percentiles The pth percentile of a data set is a value such that at least p percent of the items take on this value or less and at least (100 - p) percent of the items take on this value or more. Arrange the data in ascending order. Compute index i, the position of the pth percentile. i = (p/100)n If i is not an integer, round up. The p th percentile is the value in the i th position. If i is an integer, the p th percentile is the average of the values in positions i and i +1.
Example: Apartment Rents
90th Percentile i = (p/100)n = (90/100)70 = 63 Averaging the 63rd and 64th data values: 90th Percentile = ( )/2 = 585
Quartiles Quartiles are specific percentiles
First Quartile = 25th Percentile Second Quartile = 50th Percentile = Median Third Quartile = 75th Percentile
Example: Apartment Rents
Third Quartile Third quartile = 75th percentile i = (p/100)n = (75/100)70 = 52.5 = 53 Third quartile = 525
Measures of Variability
It is often desirable to consider measures of variability (dispersion), as well as measures of location. For example, in choosing supplier A or supplier B we might consider not only the average delivery time for each, but also the variability in delivery time for each.
Measures of Variability
Range Interquartile Range Variance Standard Deviation Coefficient of Variation
Range The range of a data set is the difference between the largest and smallest data values. It is the simplest measure of variability. It is very sensitive to the smallest and largest data values.
Example: Apartment Rents
Range Range = largest value - smallest value Range = = 190
Interquartile Range The interquartile range of a data set is the difference between the third quartile and the first quartile. It is the range for the middle 50% of the data. It overcomes the sensitivity to extreme data values.
Example: Apartment Rents
Interquartile Range 3rd Quartile (Q3) = 525 1st Quartile (Q1) = 445 Interquartile Range = Q3 - Q1 = = 80
Variance The variance is a measure of variability that utilizes all the data. It is based on the difference between the value of each observation (xi) and the mean (x for a sample, m for a population).
Variance The variance is the average of the squared differences between each data value and the mean. If the data set is a sample, the variance is denoted by s2. If the data set is a population, the variance is denoted by 2. If the original data is measured in terms of “unit”, then the variance will be measured in terms of “unit-square” or “unit2”.
Standard Deviation The standard deviation of a data set is the positive square root of the variance. It is measured in the same units as the data, making it more easily comparable, than the variance, to the mean. If the data set is a sample, the standard deviation is denoted s. If the data set is a population, the standard deviation is denoted (sigma).
Coefficient of Variation
The coefficient of variation indicates how large the standard deviation is in relation to the mean. If the data set is a sample, the coefficient of variation is computed as follows: If the data set is a population, the coefficient of variation is computed as follows:
Example: Apartment Rents
Variance Standard Deviation Coefficient of Variation
End of Chapter 3, Part A
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# What Is The HCF Of 2 And 3?
## What is the highest common factor of 2 and 3?
Multiply the circled numbers together.
This is the greatest common factor.
In the example, gcf(18,36,90)=2⋅3⋅3=18 gcf ( 18 , 36 , 90 ) = 2 ⋅ 3 ⋅ 3 = 18 ..
## What is a factor of 2 and 3?
For example, -2 and 3 AND 2 and -3 are both factor pairs of -6. Factors are whole numbers that are multiplied together to produce another number.
## What is the HCF of 2?
HCF (Highest Common Factor) or GCD (Greatest Common Divisor) of two numbers is the largest number that divides both of them. For example GCD of 20 and 28 is 4 and GCD of 98 and 56 is 14.
## How is HCF calculated?
The highest common factor is found by multiplying all the factors which appear in both lists: So the HCF of 60 and 72 is 2 × 2 × 3 which is 12. The lowest common multiple is found by multiplying all the factors which appear in either list: So the LCM of 60 and 72 is 2 × 2 × 2 × 3 × 3 × 5 which is 360.
## What is the GCF of 3 and 4?
We found the factors and prime factorization of 3 and 4. The biggest common factor number is the GCF number. So the greatest common factor 3 and 4 is 1.
## What is HCF of three numbers?
To find out HCF of three given numbers using division method, Step 1: Find out HCF of any two numbers. Step 2: Find out the HCF of the third number and the HCF obtained in step 1. Step 3: HCF obtained in step 2 will be the HCF of the three numbers.
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ML Aggarwal Solutions for Chapter 10 Reflection Class 10 Maths ICSE
Here, we are providing the solutions for Chapter 10 Reflection from ML Aggarwal Textbook for Class 10 ICSE Mathematics. Solutions of the tenth chapter has been provided in detail. This will help the students in understanding the chapter more clearly. Class 10 Chapter 10 Reflection ML Aggarwal Solutions for ICSE is one of the most important chapter for the board exams which is based on plotting the points on Cartesian plane, finding coordinates of reflection of given points, mirror line and finding coordinates of triangle. We have also added chapter test and multiple choice questions.
Exercise 10.1
1. Find the co-ordinates of the images of the following points under reflection in the x-axis:
(i) (2, -5)
(ii) (–3/2, -1/2)
(iii) (-7, 0)
Co-ordinates of the images of the points under reflection in the x-axis will be
(i) Image of (2, -5) will be (2, 5)
(ii) Image of (-3/2, -1/2) will be (-3/2, 1/2)
(iii) Image of (-7, 0) will be (-7, 0)
2. Find the co-ordinates of the images of the following points under reflection in the y-axis:
(i) (2, -5)
(ii) – 3/2, 1/2
(iii) (0, - 7)
Co – ordinates of the image of the points under reflection in the y- axis
(i) Image of (2, - 5) will be (-2, 5)
(ii) Image of -3/2, 1/2 will be 3/2, 1/2
(iii) Image of (0, -7) will be (0, -7)
3. Find co-ordinates of the images of the following points under reflection in the origin.
(i) (2, -5)
(ii) (-3/2, -1/2)
(iii) (0, 0)
Co-ordinates of the image of the points under reflection in the y-axis
(i) Image of (2, -5) will be (-2, 5)
(ii) Image of -3/2, -1/2 will be 3/2, 1/2
(iii) Image of (0, 0) will be (0, 0)
4. The image of a point P under reflection in the x-axis is (5, -2).
Write down the co-ordinates of P.
As the image of a point (5, -2) under x-axis is P
∴ Co-ordinates of P will be (5, 2)
5. A point P is reflected in the x-axis. Co-ordinates of its image are (8, -6).
(i) Find the co-ordinates of P.
(iii) Find the co-ordinates of the image of P under reflection in the y-axis.
The co-ordinates of image of P which is reflected in x-axis are (8, -6), then
(i) Co-ordinates of P will be (8, 6)
(ii) Co-ordinates of image P under reflection in the y-axis will be (-8, 6)
6. A point P is reflected in the origin. Co-ordinates of its image are (2, -5). Find
(i) the co-ordinates of P.
(ii) The co-ordinates of the image of P in the x- axis.
The co-ordinates of image of a point P which is reflected in origin are (2, -5), then
(i) Co-ordinates of P will be (-2, 5)
(ii) Co-ordinates of the image of P in the x-axis will be (-2, - 5)
7. (i) The point P (2, 3) is reflected in the line x = 4 to the point P’. Find the co-ordinates of the point P’.
(ii) Find the image of the point P (1, - 2) in the line x = - 1.
(i) (a) Draw axis XOX’ and YOY’ and take 1 cm = 1 unit
(b) Plot point P (2, 3) on it.
(c) Draw a line x = 4 which is parallel to y-axis.
(d) From P, draw a perpendicular on x = 4, which intersects x = 4 at Q.
(e) Produce PQ to P’ such that QP’ = QP.
∴ P’ is the reflection of P in the x = 4
Co-ordinates of P’ are (6, 3)
(ii) (a) Draw axis XOX’ and YOY’ and take 1 cm = 1 unit.
(b) Plot the point P(1, -2) on it.
(c) Draw a line x = - 1 which is parallel to y-axis.
(d) From p, draw a perpendicular on the line x = -1, which meets it at Q.
(e) Produce PQ to P’ such that PQ = QP’
P’ is the image or reflection of P in the line x = - 1
Co-ordinates of P’ are (-3, -2)
Co –ordinates of P’ are (-3, - 2)
8.(i) The point P (2, 4) on reflection in the line y = 1 is mapped onto P’. Find the co-ordinates of P’.
(ii) Find the image of the point P (-3, -5) in the line y = - 2.
(i) (a) Draw axis XOX’ and YOY’ and take 1 cm = 1 unit.
(b) Plot point P (2, 4) on it.
(c) Draw a line y = 1 which is parallel to the x-axis.
(d) From P, draw a perpendicular on y = 1 which meets it at Q.
(e) Produce PQ to P’ such that QP’ = PQ.
P’ is the image of P, whose co-ordinates are (2, -2)
(ii) (a) Draw axis XOX’ and YOY’ and take 1 cm = 1 unit.
(b) Plot point P (-3, -5) on it.
(c) Draw a line y = - 2 which is parallel to the x-axis.
(d) From P, draw a perpendicular on y = - 2 which meets it at Q.
(e) Produce PQ to P’ such that QP’ = PQ.
Then P’ is the image of P, whose co-ordinates are (-3, 1)
9. The point (-4, -5) on reflection in y-axis is mapped on P’. The point P’ on reflection in the origin is mapped on P’’. Find the co-ordinates of P’ and P’’. Write down a single transformation that maps P onto P’’.
P’ is the image of point P (-4, -5) in y-axis
∴ Co-ordinates of P’ will be (4, -5)
Again P’’ is the image of P’ under reflection in origin will be (-4, 5).
The single transformation that maps P onto P’’ is the x-axis.
10. Write down the co-ordinates of the image of the point (3, -2) when :
(i) reflected in the x-axis
(ii) reflected in the y-axis
(iii) reflected in the x-axis followed by reflection in the y-axis.
(iv) reflected in the origin.
Co-ordinates of the given points are (3, -2).
(i) Co-ordinates of the image reflected in y – axis will be (3, 2)
(ii) Co-ordinates of the image reflected in y-axis will be (-3, -2)
(iii) Co-ordinates of the point reflected in x- axis followed by reflection in the y-axis will be (-3, 2).
(iv) Co-ordinates of the point reflected in the origin will be (-3, 2)
11. Find the co-ordinates of the image of (3, 1) under reflection in x-axis followed by a reflection in the line x = - 1.
(i) Draw axis XOX’ and YOY’ taking 1 cm = 1 unit.
(ii) Plot a point P (3, 1)
(iii) Draw a line x = 1, which is parallel to y-axis.
(iv) From P, draw a perpendicular on x-axis meeting it at Q.
(v) Produce PQ to P’ such that QP’ = PQ, then
P’ is the image of P is x-axis. Then co-ordinates of P’ will be (3, - 1)
(vi) From P’, draw a perpendicular on x = 1 meeting it at R.
(vii) Produce P’R to P’’ such that RP’’ = P’R
∴ P’’ is the image of P’ in the line x = 1
Co- ordinates of P’’ are (-1, -1)
12. If P’ (-4, - 3) is the image of a point P under refection in the origin, find
(i) the co-ordinates of P.
(ii) the co-ordinates of the image of P under reflection in the line y = - 2.
(i) Reflection of P is P’ (- 4, -3) in the origin
∴ Co-ordinates of P will be (4, 3)
Draw a line y = - 2, which is parallel to x- axis.
(ii) From P, draw a perpendicular on y = - 2 meetings it at Q
Produce PQ to P’’ such that QP’’ = PQ
∴ P’’ will the image of P in the line y = - 2
∴ Co-ordinates of P’’ will be (4, -7)
13. A point P (a, b) is reflected in the x- axis to P’ (2, -3), write down the values of a and b. P’’ is the image of P, when reflected in the y-axis. Write down the co-ordinates of P’’. Find the co-ordinates of P’’, when P is reflected in the line parallel to y-axis such that x = 4.
P’ (2, -3) is the reflection of P (a, b) in the x-axis
∴ Co –ordinates of P’ will be (a, - b) but P’ is (2, -3)
Comparing a = 2, b = 3
∴ Co –ordinates of P will be (2, 3)
P’’ is the image of P when reflected in y-axis
∴ Co-ordinate of P’’ will be (-2, 3)
Draw a line x = 4, which is parallel to y-axis
And P’’ is the image of P when it is reflected in the line x = 4,
Then P’’ is its reflection Co-ordinates of P’’ will be (6, 3).
14. (i) Point P (a, b) is reflected in the x-axis to P’ (5, -2). Write down the values of and b.
(ii) P’’ is the image of P when reflected in the y-axis. Write down the co-ordinates of P’’.
(iii) Name a single transformation that maps P’ to P’’.
(i) Image of P (a, b) reflected in the x-axis to P’ (5, -2)
∴ a = 5 and b = 2
(ii) P’’ is the image of P when reflected in the y-axis
∴ Its co-ordinates will be (-5, -2).
(iii) The single transformation that maps P’ to P’’ is the origin.
15. Points A and B have co-ordinates (2, 5) and (0, 3). Find
(i) the image A’ of A under reflection in the x-axis.
(ii) the image B’ of B under reflection in the line AA’.
Co-ordinates of A are (2, 5) and of B are (0, 3)
(i) Co-ordinates of A’, the image of A reflected in the x- axis will be (2, -5)
(ii) Co-ordinates of B’, the image of B under reflection in the line AA’ will be (4, 3).
16. Plot the points A(2, -3), B (- 1, 2) and C (0, -2) on the graph paper. Draw the triangle formed by reflecting these points in the x-axis. Are the two triangles congruent ?
The points A (2, -3), B (-1, 2) and C (0, -2) has been plotted on the graph paper as shown and are joined to form a triangle ABC. The co-ordinates of the images of A, B and C reflected in x-axis will be A’ (2, 3), B’ (-1, -2), C’ (0, 2) respectively and are joined to from another ∆A’B’C’
Yes, these two triangles are congruent.
17. The points (6, 2), (3, -1) and (-2, 4) are the vertices of a right angled triangle. Check whether it remains a right angled triangle after reflection in the y-axis.
Let A (6, 2), B (3, -1) and C (-2, 4) be the points of a right-angled triangle then the co-ordinates of the images of A, B, C reflected in y-axis be
A’ (-6, 2), B’ (- 3, -1) and C’ (2, 4)
By joining these points, we find that ∆A’B’C’ is also a right angled triangle.
18. The triangle ABC where A (1, 2), B (4, 8), C (6, 8) is reflected in the x-axis to triangle A’ B’ C’. The triangle A’ B’ C’ is then reflected in the origin to triangle A” B” C”. Write down the co-ordinates of A”, B”, C”. Write down a single transformation that maps ABC onto A” B’’ C’’.
The co-ordinates of ∆ABC are A (1, 2) B (4, 8), C (6, 8) which are reflected in x-axis as A’ B’ and C’.
∴ The co-ordinates of A’ (1, -2), B (4, -8) and C (6, - 8).
A’, B’ and C’ are again reflected in origins to form an ∆A’’B’’C’’.
∴ The co-ordinates of A’’ will be (-1, 2), B’’(-4, 8) and C’’(-6, 8)
The single transformation that maps ABC onto A’’ B’’ C’’ is y-axis.
19. The image of a point P on reflection in a line l is point P’. Describe the location of the line l.
The line will be the right bisector of the line segment joining P and P’.
20. Given two points P and Q, and that (1) the image of P on reflection in y-axis is the point Q and (2) the mid-point of PQ is invariant on reflection in x-axis. Locate
(i) the x-axis
(ii) the y- axis and
(iii) the origin.
Q is the image of P on reflection in y- axis.
And mid point of PQ is invariant on reflection in x-axis.
(i) x- axis will be the line joining the points P and Q.
(ii) The line perpendicular bisector of line segment PQ is the y-axis.
(iii) The origin will be the mid point of line segment PQ.
21. The point (-3, 0) on reflection in a line is mapped as (3, 0) and the point (2, -3) on reflection in the same line is mapped as (-2, -3).
(i) Name the mirror line.
(ii) Write the co-ordinates of the image of (-3, - 4) in the mirror line.
The point (-3, 0) is the image of point (3, 0)
And point (2, -3) is image of point (-2, -3) reflected on the same line.
(i) It is clear that the mirror line will be y- axis.
(ii) The co-ordinates of the image of the point (-3, -4)
Reflected in the same line i.e., y-axis will be (3, -4).
22. Use graph paper for this (take 2 cm = 1 unit long both x and y axis). ABCD is a quadrilateral whose vertices are A (2, 2), B (2, -2), C (0, -1) and D (0, 1).
(i) Reflect quadrilateral ABCD on the y-axis and name it as A’B’CD.
(ii) Write down the coordinates of A’ and B’.
(iii) Name two points which are invariant under the above reflection.
(iv) Name the polygon A’B’CD.
(i) Quadrilateral ABCD is reflected on the y-axis and named as A’B’CD.
(ii) As A’ is the refection of A (2, 2) about the line x = 0 (y-axis)
Thus, the coordinate of A’ are (-2, 2).
And, as B’ is the reflection of B are (-2, -2).
(iii) Points C (0, 1) and D (0, 1) are invariant under the above reflection.
(iv) The polygon A’B’CD is a trapezium since A’B’ || CD.
23. Use graph paper for this question.
(i) The point P (2, -4) is reflected about the line x = 0 to get the image Q. Find the co-ordinates of Q.
(ii) Point Q is reflected about the line y = 0 to get the image R. Find the co-ordinates of R.
(iii) Name the figure PQR.
(iv) Find the area of figure PQR.
(i) Since the point Q is the reflection of the point P (2, -4) in the line x = 0, the co-ordinates of Q are (2, 4).
(ii) Since R is the reflection of Q (2, 4) about the line y = 0, the co-ordinates of R are (-2, 4).
(iii) Figure PQR is the right angled triangle PQR.
(iv) Area of ∆PQR = 1/2 × QR × PQ
= 1/2 × 4 × 8
= 16 sq. units.
24. Use graph paper for this question. The point P (5, 3) was reflected in the origin to get the image P’.
(i) Write down the co-ordinates of P’.
(ii) If M is the foot of perpendicular from P to the x-axis, find the co-ordinates of M.
(iii) If N is the foot of the perpendicular from P’ to the x-axis, find the co-ordinates of N.
(iv) Name the figure PMP’N.
(v) Find the area of the figure PMP’N.
P’ is the image of point P (5, 3) reflected in the origin.
(i) Co-ordinates of P’ will be (-5, -3).
(ii) M is the foot of the perpendicular from P to the x-axis. Co-ordinates of M will be (5, 0)
(iii) N is the foot of the perpendicular from P’ to x-axis. Co-ordinates of N will be (-5, 0).
(iv) By joining the points, the figure PMP’N is a parallelogram.
(v) Area of the parallelogram = 2 × Area of ∆ MPN
= 2 × 1/2 × MN × PM
= MN × PM
= 10 × 3
= 30 sq. units.
25. Using a graph paper, plot the points A (6, 4) and B (0, 4).
(i) Reflect A and B in the origin to get the images A’ and B’.
(ii) Write the co-ordinates of A’ and B’.
(iii) State the geometrical name for the figure ABA’B’.
(iv) Find its perimeter.
(i) A (6, 4), B (0, 4)
(ii)
(iii) ABA’B’ is a parallelogram
(iv)
Perimeter = Sum of all sides = 6 + 10 + 6 + 10 = 32 units
26. Use graph paper to answer this question.
(i) Plot the points A (4, 6) and B (1, 2)
(ii) If A’ is the image of A when reflected in x- axis, write co-ordinates of A’.
(iii) If B’ is the image of B when B is reflected in the line AA’. Write the co-ordinates of B’.
(iv) Give the geometrical name for the figure ABA’B’.
(i) Plotting the points A (4, 6) and B (1, 2) on the given graph.
(ii) A’ = (4, -6)
(iii) B’ = (7, 2)
(iv) In the quadrilateral ABA’B’, we have AB = AB’ and A’B = A’B’
Hence, ABA’B’ is a kite.
27. The points A (2, 3), B (4, 5) and C (7, 2) are the vertices of ABC.
(i) Write down the co-ordinates of A1, B1, C1 if A1B1C1 is the image of ABC when reflected in the origin.
(ii) Write down the co-ordinates of A2, B2, C2 if A2B2C2 is the image of ABC when reflected in the x-axis.
(iii) Assign the special name to the quadrilateral BCC2B2 and find its area.
Points A (2, 3), B (4, 5) and C (7, 2) are the vertices of ∆ABC.
A1, B1 and C1 are the images of A, B and C reflected in the origin.
(i) Co-ordinates of A1 = (-2, -3) of B1 (-4, -5) and of C1 (-7, -2).
(ii) Co-ordinates of A2, B2 and C2 the images of A, B and C when reflected in x-axis are A2 (2, -3), B2 (4, -5), C2 (7, -2)
(iii) The quadrilateral formed by joining the points, BCC2B2 is an isosceles trapezium and its area
= 1/2(BB2 + CC2) ×3
= 1/2(10 + 4) × 3
= 1/2 × 14 × 3
= 21 sq. units.
28. The point P (3, 4) is reflected to P’ in the x-axis and O’ is the image of O (origin) in the line PP’. Find :
(i) the co-ordinates of P’ and O’,
(ii) the length of segments PP’ and OO’.
(iii) the perimeter of the quadrilateral POP’O’.
P’ is the image of P (3, 4) reflected in x- axis. And O’ is the image of O the origin in the line P’P.
(i) Co-ordinates of P’ are (3, -4) and co-ordinates of O’ reflected in PP’ are (6, 0)
(ii) Length of PP’ = 8 units and OO’ = 6 units
(iii) Perimeter of POP’O’ is
4 × OP
29. Use a graph for this question. (Take 10 small divisions = 1 unit on both axes). P and Q have co-ordinates (0, 5) and (-2, 4).
(i) P is invariant when reflected in an axis. Name the axis.
(ii) Find the image of Q on reflection in the axis found in (i).
(iii) (0, k) on reflection in the origin is invariant. Write the value of k.
(iv) Write the co-ordinates of the image of Q, obtained by reflecting it in the origin followed by reflection in x-axis.
(i) Two points P (0, 5) and Q (-2, 4) are given as the abscissa of P is 0. It is invariant when is reflected in y-axis.
(ii) Let Q’ be the image of Q on reflection in y-axis. Co-ordinate of Q’ will be (2, 4)
(iii) (0, k) on reflection in the origin is invariant. Co-ordinates of image will be (0, 0). k = 0
(iv) The reflection of Q in the origin is the point Q’’ and its co-ordinates will be (2, -4) and reflection of Q’’ (2, -4) in x-axis is (2, 4) which is the point Q’.
Multiple Choice Questions
Choose the correct answer from the given four options (1 to 7):
1. The reflection of the point P (-2, 3) in the x-axis is
(a) (2, 3)
(b) (2, -3)
(c) (-2, -3)
(d) ( -2, 3)
Reflection of the point P (-2, 3) in x-axis is (-2, -3) (c)
2. The reflection of the point P (-2, 3) in the y-axis is
(a) (2, 3)
(b) (2, -3)
(c) (-2, -3)
(d) (0, 3)
The reflection of the point P (-2, 3) under reflection in y-axis (2, 3) (a)
3. If the image of the point P under reflection in the x-axis is (-3, 2), then the coordinates of the point P are
(a) (3, 2)
(b) (-3, -2)
(c) (3, -2)
(d) (-3, 0)
The image of the point P under reflection in the x-axis is (-3, 2), then the co-ordinates of the point P will be (-3, -2) (b)
4. The reflection of the point P(1, -2) in the line y = -1 is
(a) (-3, -2)
(b) (1, -4)
(c) (1, 4)
(d) (1, 0)
The reflection of the point P (1, -2) in the line y = - 1 is (1, 0) (d)
5. The reflection of the point A (4, -1) in the line x = 2 is
(a) (0, - 1)
(b) (8, - 1)
(c) (0, 1)
(d) none of these
The reflection of A (4, -1) in the line x = 2 will be A’ (0, -1) (a)
Question 6. The reflection of the point (-3, 0) in the origin is the point
(a) (0, -3)
(b) (0, 3)
(c) (3, 0)
(d) none of these
Reflection of the point (-3, 0) in origin will be (3, 0) (c)
7. Which of the following points is invariant with respect to the line y = -2 ?
(a) (3, 2)
(b) (3, -2)
(c) (2, 3)
(d) (-2, 3)
The variant points are (3, -2) (b)
Chapter Test
1. The point P (4, -7) on reflection in x-axis is mapped onto P’ . Then P’ on reflection in the y-axis is mapped onto P’’. Find the co-ordinate of P’ and P’’. Write down a single transformation that maps P onto P’’.
P’ is the image of P (4, -7) reflected in x-axis
∴ Co-ordinates of P’ are (4, 7)
Again P’’ is the image of P’ reflected in y-axis
∴ Co-ordinates of P’’ are (-4, 7)
∴ Single transformation that maps P and P’’ is in the origin.
2. The point P (a, b) is first reflected in the origin and then reflected in the y-axis to P’ . If P’ has co-ordinates (3, -4), evaluate a, b.
The co-ordinates of image of P (a, b) reflected in origin are (-a, -b).
Again the co-ordinates of P’, image of the above point (-a, -b) reflected in the y-axis are (a, -b).
But co-ordinates of P’ are (3, -4)
∴ a = 3 and –b = - 4
b = 4
Hence a = 3, b = 4.
3. A point P (a, b) becomes (-2, c) after reflection in the x-axis, and P becomes (d, 5) after reflection in the origin. Find the value of a, b, c and d.
If the image of P (a, b) after reflected in the x-axis be (a, -b) but it is given (-2, c).
a = -2, c = -b.
if P is reflected in the origin, then its co-ordinates will be (-a, -b), but it is given (d, 5)
∴ -b = 5
⇒ b = - 5
d = - a = - (-2) = 2,
C = - b = -(-5) = 5
Hence a = -2, b = -5, c = 5, d = 2
4. A (4, -1), B (0, 7) and C (-2, 5) are the vertices of a triangle. ∆ ABC is reflected in y-axis and then reflected in the origin. Find the co-ordinates of the final images of the vertices.
A (4, -1), B (0, 7) and C (-2, 5) are the vertices of ∆ABC.
After reflecting in y-axis, the co-ordinates of points will be A’ (-4, -1), B’ (0, 7), C’ (2, 5) Again reflecting in origin, the co-ordinates of images of the vertices will be A’’ (4, 1), B’’ (0, -7), C’’ (-2, -5)
5. The point A (4, - 11), B (5, 3), C (2, 15), and D (1, 1) are the vertices of a parallelogram. If the parallelogram is reflected in the y-axis and then in the origin, find the co-ordinates of the final images. Check whether it remains a parallelogram. Write down a single transformation that brings the above change.
The points A (4, -11), B (5, 3), C (2, 15) and D (1, 1) are the vertices of a parallelogram.
After reflecting in-axis, the images of these points will be A’ (-4, 11), B’ (-5, 3), C (-2, 15) and D’ (-1, 1).
Again reflecting these points in origin, the image of these points will be A’’ (4, -11), B’’ (5, -3), C’’ (2, -15), D’’(0, -1)
Yes, the reflection of a single transformation is in the x-axis.
6. Use a graph paper for this question (take 2cm = 1 unit on both x and y axes).
(i) Plot the following points: A (0, 4), B (2, 3), C (1, 1) and D (2, 0)
(ii) Reflect points B, C, D on 7-axis and write down their co-ordinates. Name the images as B’, C’, D’ respectively.
(iii) Join points A, B, C, D, D’, C’, B’ and A in order, so as to form a closed figure. Write down the equation of line of symmetry of the figure formed.
(i) On graph A (0, 4), B (2, 3), C (1, 1) and D (2, 0)
(ii) B’ = (-2, 3), C’ = (-1, 1), D’ (-2, 0)
(iii) The equation of the line of symmetry is x = 0
7. The triangle OAB is reflected in the origin O to triangle OA’B’. A’ and B’ have coordinates (-3, -4) and (0, -5) respectively.
(i) Find the co-ordinates of A and B.
(ii) Draw a diagram to represent the given information.
(iii) What kind of figure is the quadrilateral ABA’B’ ?
(iv) Find the coordinates of A’’, the reflection of A in the origin followed by reflection in the y-axis.
(v) Find the co-ordinates of B’’, the reflection of B in the x-axis followed by reflection in the origin.
(i) ∆OAB is reflected in the origin O to ∆OA’B’.
Co-ordinates of A’ = (-3, -4), B’ (0, -5).
∴ Co-ordinates of A’ will be (3, 4) and of B will be (0, 5).
(ii) The diagram representing the given information has been drawn at the end.
(iii) The figure in the diagram is a rectangle.
(iv) The co-ordinate of B’, the reflection of B is the x-axis is (0, - 5) and co-ordinates of B’’, the reflection in origin of the point (0, - 5) will be (0, 5).
(v) The co-ordinates of the points, the reflection of A in the origin are (-3, -4)
and coordinates of A’’, the reflected in y-axis of the point (-3, -4) are (3, -4)
The solutions provided for Chapter 10 Reflection of ML Aggarwal Textbook. This solutions of ML Aggarwal Textbook of Chapter 10 Reflection contains answers to all the exercises given in the chapter. These solutions are very important if you are a student of ICSE boards studying in Class 10. While preparing the solutions, we kept this in our mind that these should based on the latest syllabus given by ICSE Board.
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## Solution to Puzzle #112: Weights and Balance Revisited
As I mentioned in the question, this is an age old problem, but I was still intrigued by the twist provided to this puzzle by a colleague’s daughter. Many people gave a correct answer to the first two parts of the puzzle, only Pratik Poddar came back with the right answer for the 3rd part.
Here are the answers (and I will add explanations later):
Part 1: Weights need to be 1, 2, 4, 8, ….essentially all the powers of 2
Part 2: Weights need to be 1, 3, 9, 27, 81….essentially powers of 3
Part 3: Weights need to be twice that of powers of 3, i.e. 2, 6, 18, 54, ….
The basic principle to be adopted is that you choose weights in a manner, so that no weight can be represented in two different ways. For example, if you had two weights of 5 kg each, then you can make 10 kg with it (in Part 1). However, there is a bit of “wastage” here as well, as 5 kg itself can be formed by two different weights.
In Part 1, if you begin with a weight of 1 kg, you can only weigh 1 kg with it. For 2 kg, you need to add a weight of 2 kg. With the two of them, you can weigh 3 kg also, but not 4 kg, so you add a weight of 4 kg. Now with these three weights, you can go up to 7 kg, but not 8 kg…you can continue this way and realise that if you have weights that are a power of 2, you can create any weight with these combinations, by effectively using the binary notation.
In a balance, you now have a special property, that you can put weights on both sides and therefore not only add weights, but also subtract weights. So lets begin again:
– you start with a weight of 1 kg
– Add a weight of 2 kg. Now you can measure 2 kg with it. However, you can now create 1 kg in two different ways – 1 kg by itself and also by putting 2 kg on one side and 1 kg on the other side (difference). That is “wastage”
– So instead of 2 kg, we try 3 kg. With 3 kg, you can measure 2 kg through a difference of 3 kg and 1 kg. You can also measure 3 kg, and also 4 kg. But you cannot measure 5 kg.
– Through trial and error, you will realise that there is wastage by trying 5 kg, 6 kg, 7 kg and 8 kg. Once you try 9 kg, you can measure 5-8 kg by subtracting 1-4 kg from 9 kg.
– A better way to think about this is – since you can measure all the weights up to 4 kg, the next weight you need to chose is 4 x 2 + 1 (9 kg), because all the intermediate weights can be measured by subtracting from 9. With these weights, you can measure all the way up to 13, and hence the next weight needs to be 2 x 13 + 1 = 27 kg…and so on.
Part 3: The trick here is that once you know that whatever you are trying to measure is heavier than lets say 15 kg and lighter than 17 kg, and you know that it can only be a whole number, then you know that it is 16 kg, without actually having a combination of weights that gives you 16. This is what Arushi Gupta pointed out.
When you think about this further, you can conclude that if you can measure all alternate numbers, then the ones inbetween can be inferred. The easiest way to do this is from Part 2, you know that you can measure all #s, so by multiplying the weights by 2, you can measure 2, 4, 6, 8, 10, ….and infer the odd ones!
Hope you enjoyed the puzzle!
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### Direct Proof
#### Definition
A direct proof is one of the most familiar forms of proof. It is used to prove statements of the form “if p then q” or “p implies q”. The method of the proof is to take an original statement p, which we assume to be true, and use it to show directly that another statement q is true.
#### Worked Examples
###### Example 1
Directly prove that if $n$ is an odd integer then $n^2$ is also an odd integer.
###### Solution
Let $p$ be the statement that $n$ is an odd integer and $q$ be a statement that $n^2$ is an odd integer.
Assume that $n$ is an odd integer, then by definition $n=2k+1$ for some integer $k$. We will now use this to show that $n^2$ is also an odd integer.
\begin{align} n^2 &= (2k+1)^2\\ &=(2k+1)(2k+1)\\ &=4k^2+2k+2k+1\\ &=4k^2+4k+1\\ &=2(2k^2+2k)+1 \end{align}
Hence, we have shown that $n^2$ has the form of an odd integer since $2k^2+2k$ is an integer. Therefore we have shown that $p\Rightarrow q$ and so we have completed the proof.
###### Example 2
Directly prove that if $m$ and $n$ are odd integers then $mn$ is also an odd integer.
###### Solution
Assume than $m$ and $n$ are odd integers. Then by definition $m=2k+1$ and $n=2l+1$ for some integers $k$ and $l$.
Note: that different integers $k$ and $l$ are used in the definitions of $m$ and $n$.
Now we will use this to show that $mn$ is also an odd integer.
\begin{align} mn &= (2k+1)(2l+1)\\ &=4kl+2k+2l+1\\ &=2(2kl+k+l)+1 \end{align}
Hence we have shown that $mn$ has the form of an odd integer since $2kl +k+l$ is an integer.
#### More Support
You can get one-to-one support from Maths-Aid.
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# Tangents, Perpendiculars and Geometric Mean: What Is This About? A Mathematical Droodle
### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.
What if applet does not run?
Explanation
The applet illustrates a problem that exploits relationships between angles and arcs in a circle. Hidden below the surface, there is a segment that serves as the geometric mean of other two:
Let AB be a chord in a circle and P a point on the circle. Let Q be the feet of the perpendicular from P to AB, and R and S the feet of the perpendiculars from P to the tangents to the circle at A and B. Prove that PQ is the geometric mean of PR and PS: PQ2 = PR·PS.
The problem does not require additional construction beyond the observation that quadrilaterals ARPQ and BSPQ are cyclic. This is because each contains a pair of right angles opposite each other.
### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.
What if applet does not run?
We conclude that, in ARPQ,
∠RAP = ∠RQP and also ∠PAQ = ∠PRQ
as inscribed and subtended by the same chord. Similarly, in BSPQ,
∠SBP = ∠SQP and also ∠PBQ = ∠PSQ.
But in the given circle the inscribed ∠ABP is subtended by chord AP that forms with the tangent at A an equal ∠RAP. It follows that
∠RQP = ∠RAP = ∠ABP = ∠PBQ = ∠PSQ,
so that
∠RQP = ∠PSQ.
Similarly,
∠SQP = ∠PRQ.
Thus triangles PRQ and PQS are similar and therefore PR/PQ = PQ/PS.
A different solution was found by Vo Duc Dien.
### References
1. T. Andreescu, R. Gelca, Mathematical Olympiad Challenges, Birkhäuser, 2004, pp. 6-7.
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### Frequently Asked Questions on Chapter 1 Real Numbers
Euclid’s division algorithm to find the HCF of 135 and 225?
135 and 225
As you can see, from question 225 is greater than 135. Therefore, by Euclid’s division algorithm, we have, 225 = 135 × 1 + 90 Now, remainder 90 ≠ 0, thus again using division lemma for 90, we get, 135 = 90 × 1 + 45 Again, 45 ≠ 0, repeating the above step for 45, we get, 90 = 45 × 2 + 0 The remainder is now zero, so our method stops here. Since, in the last step, the divisor is 45, therefore, HCF (225,135) = HCF (135, 90) = HCF (90, 45) = 45. Hence, the HCF of 225 and 135 is 45.
Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Let a be any positive integer and b = 6. Then, by Euclid’s algorithm, a = 6q + r, for some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5, because 0≤r<6. Now substituting the value of r, we get, If r = 0, then a = 6q Similarly, for r= 1, 2, 3, 4 and 5, the value of a is 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5, respectively. If a = 6q, 6q+2, 6q+4, then a is an even number and divisible by 2. A positive integer can be either even or odd Therefore, any positive odd integer is of the form of 6q+1, 6q+3 and 6q+5, where q is some integer.
Express each number as a product of its prime factors 5005?
5005 By Taking the LCM of 5005, we will get the product of its prime factors. Hence, 5005 = 5 × 7 × 11 × 13 × 1 = 5 × 7 × 11 × 13
Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers 26 and 91.
6 and 91 Expressing 26 and 91 as product of its prime factors, we get, 26 = 2 × 13 × 1 91 = 7 × 13 × 1 Therefore, LCM (26, 91) = 2 × 7 × 13 × 1 = 182 And HCF (26, 91) = 13
Verification
Now, product of 26 and 91 = 26 × 91 = 2366 And Product of LCM and HCF = 182 × 13 = 2366 Hence, LCM × HCF = product of the 26 and 91.
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 what can you say about the prime factors of q 43.123456789?
43.123456789 Since it has a terminating decimal expansion, it is a rational number in the form of p/q and q has factors of 2 and 5 only.
Find the HCF of 135 and 225 using Euclid’s division
By using Euclid’s division algorithm, the HCF of 135 and 225 can be calculated as:
We know that 225 > 135
So, 225 = 135 × 1 + 90
Since, remainder 90 ≠ 0
135 = 90 × 1 + 45
Remainder 45 ≠ 0
So, 90 = 45 × 2 + 0
Hence the remainder is = 0, so the method stops here.
Therefore HCF of (135, 225) is 45.
Find the maximum number of columns in which they can march? When an army contingent of 616 members is to march behind an army band of 32 members in a parade. Consider the two groups marching in the same number of columns.
Given:
Number of army contingent members = 616
Number of army band members = 32
If the two groups have to march in the same column, we have to find out the highest common factor between the two groups.
HCF(616, 32), gives the maximum number of columns in which they can march.
By Using Euclid’s algorithm, we get
We know that 616 > 32
616 = 32 × 19 + 8
Since, 8 ≠ 0
32 = 8 × 4 + 0
Now we have got remainder as 0.
Therefore, HCF (616, 32) = 8.
Hence, the maximum number of columns in which they can march is 8.
Express 5005 as a product of its prime factors:
To find the product of its prime factors, firstly we need to take LCM of 5005
So, 5005 = 5 × 7 × 11 × 13 × 1
Hence, 5005 = 5 × 7 × 11 × 13
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are viewing an older version of this Concept. Go to the latest version.
# Area of Composite Shapes
## Shapes in the real world come in all sizes. Learn how to break down and calculate the area of composite shapes using the sum of areas of each part.
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Practice Area of Composite Shapes
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Area of Composite Shapes
What if you wanted to find the area of a shape that was made up of other shapes? How could you use your knowledge of the area of rectangles, parallelograms, and triangles to help you? After completing this Concept, you'll be able to answer questions like these.
### Guidance
Perimeter is the distance around a shape. The perimeter of any figure must have a unit of measurement attached to it. If no specific units are given (feet, inches, centimeters, etc), write “units.” Area is the amount of space inside a figure. If two figures are congruent, they have the same area. This is the congruent areas postulate. This postulate needs no proof because congruent figures have the same amount of space inside them. Keep in mind that two figures with the same area are not necessarily congruent.
A composite shape is a shape made up of other shapes. To find the area of such a shape, simply find the area of each part and add them up. The area addition postulate states that if a figure is composed of two or more parts that do not overlap each other, then the area of the figure is the sum of the areas of the parts.
#### Example A
Find the area of the figure below. You may assume all sides are perpendicular.
Split the shape into two rectangles and find the area of each.
Atop rectangleAbottom square=62=12 ft2=33=9 ft2\begin{align*}A_{top \ rectangle} &= 6 \cdot 2=12 \ ft^2\\ A_{bottom \ square} &= 3 \cdot 3=9 \ ft^2\end{align*}
The total area is 12+9=21 ft2\begin{align*}12 + 9 = 21 \ ft^2\end{align*}.
#### Example B
• Divide the shape into two triangles and one rectangle.
• Find the area of the two triangles and rectangle.
• Find the area of the entire shape.
Solution:
• One triangle on the top and one on the right. Rectangle is the rest.
• Area of triangle on top is 8(5)2=20 units2\begin{align*}\frac{8(5)}{2}=20 \ units^2\end{align*}. Area of triangle on right is 5(5)2=12.5 units2\begin{align*}\frac{5(5)}{2}=12.5 \ units^2\end{align*}. Area of rectangle is 375 units2\begin{align*}375 \ units^2\end{align*}.
• Total area is 407.5 units2\begin{align*}407.5 \ units^2\end{align*}.
#### Example C
Find the area of the figure below.
Divide the figure into a triangle and a rectangle with a small rectangle cut out of the lower right-hand corner.
\begin{align*}A &= A_{top \ triangle}+A_{rectangle}-A_{small \ triangle}\\ A &= \left(\frac{1}{2} \cdot 6 \cdot 9\right)+(9 \cdot 15)\left) - (\frac{1}{2} \cdot 3 \cdot 6\right)\\ A &= 27+135-9\\ A &= 153 \ units^2\end{align*}
Watch this video for help with the Examples above.
### Vocabulary
Perimeter is the distance around a shape. The perimeter of any figure must have a unit of measurement attached to it. If no specific units are given (feet, inches, centimeters, etc), write “units.” Area is the amount of space inside a figure and is measured in square units. A composite shape is a shape made up of other shapes.
### Guided Practice
1. Find the area of the rectangles and triangle.
2. Find the area of the whole shape.
1. Rectangle #1: \begin{align*}\text{Area }= 24(9+12)=504 \ units^2\end{align*}. Rectangle #2: \begin{align*}\text{Area }=15(9+12)=315 \ units^2\end{align*}. Triangle: \begin{align*}\text{Area }=\frac{15(9)}{2}=67.5 \ units^2\end{align*}.
2. You need to subtract the area of the triangle from the bottom right corner. \begin{align*}\text{Total Area }=504+315+67.5-\frac{15(12)}{2}=796.5 \ units^2\end{align*}
### Practice
Use the picture below for questions 1-2. Both figures are squares.
1. Find the area of the unshaded region.
2. Find the area of the shaded region.
Find the area of the figure below. You may assume all sides are perpendicular.
Find the areas of the composite figures.
Use the figure to answer the questions.
1. What is the area of the square?
2. What is the area of the triangle on the left?
3. What is the area of the composite figure?
Find the area of the following figures.
1. Find the area of the unshaded region.
2. Lin bought a tract of land for a new apartment complex. The drawing below shows the measurements of the sides of the tract. Approximately how many acres of land did Lin buy? You may assume any angles that look like right angles are \begin{align*}90^\circ\end{align*}. (1 acre \begin{align*}\approx\end{align*} 40,000 square feet)
3. Linus has 100 ft of fencing to use in order to enclose a 1200 square foot rectangular pig pen. The pig pen is adjacent to the barn so he only needs to form three sides of the rectangular area as shown below. What dimensions should the pen be?
### Vocabulary Language: English
Composite
A number that has more than two factors.
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# Linear Approximations and Differentials
Linear Approximation
Let $y=f(x)$ be a differentiable function. The function $f(x)$ can be approximated by the tangent line to $y=f(x)$ at $a$ if $x$ is near $a$. Such an approximation is called a linear approximation.
If $x\approx a$ then $\Delta x=x-a\approx 0$, so we have
\begin{align*}
\frac{\Delta y}{\Delta x}&\approx \frac{dy}{dx}\\
&=f'(a).
\end{align*}
This means that
$$\frac{f(x)-f(a)}{x-a}\approx f'(a),$$
i.e.
\label{eq:lineapprox}
f(x)\approx f(a)+f'(a)(x-a).
The equation \eqref{eq:lineapprox} is called the linear approximation or tangent line approximation of $f$ at $a$. The linear function
L(x):=f(a)+f'(a)(x-a)
is called the linearization of $f$ at $a$. Notice that $L(x)$ is the equation of tangent line to $f$ at $a$.
Example. Find the linearlization of $f(x)=\sqrt{x+3}$ at $a=1$ and use it to approximate $\sqrt{3.98}$ and $\sqrt{4.05}$.
Solution.
Linear approximation of f(x)=sqrt(x+3) at a=1
$f'(x)=\frac{1}{2\sqrt{x+3}}$, so
\begin{align*}
L(x)&=f(1)+f'(1)(x-1)\\
&=2+\frac{1}{4}(x-1)\\
&=\frac{x}{4}+\frac{7}{4}.
\end{align*}
When $x\approx 1$, we have the approximation
$$\sqrt{x+3}\approx \frac{x}{4}+\frac{7}{4}.$$
Setting $x+3=3.98$ we find $x=1.98$. Hence,
\begin{align*}
\sqrt{3.98}&\approx \frac{0.98}{4}+\frac{7}{4}\\
&=1.995.
\end{align*}
Setting $x+3=4.05$ we find $x=1.05$. Hence,
\begin{align*}
\sqrt{4.05}&\approx \frac{1.05}{4}+\frac{7}{4}\\
&=2.0125.
\end{align*}
Example. Use linear approximation to estimate $\sqrt{99.8}$.
Solution. In order to use linear approximation we need to choose $f(x)$, $x$ and $a$. First clearly from the given quantity we see that $f(x)=\sqrt{x}$ and thereby $x=99.8$. Since $f'(x)=\frac{1}{2\sqrt{x}}$, the linear approximation of $\sqrt{99.8}$ at $a$ is $$\sqrt{99.8}\approx \sqrt{a}+\frac{1}{2\sqrt{a}}(99.8-a)$$ How do we choose a suitable $a$? There are two criteria you have to have in mind. One is $a$ has to be close to $x$ for the linear approximation to be useful. Second $a$ needs to be chosen so that $f(a)$ and $f'(a)$ can be calculated easily (meaning by hand without aid of a calculator). Why is this important? You have to understand that the use of linear approximation is not assuming any use of a calculator. (If you can use a calculator, what is the point of doing this approximation?) This is a method that was developed when there were no calculators available so people could calculate values like $\sqrt{99.8}$ by hand. Considering the two criteria, we find that $a=100$ is the one. Hence, $$\sqrt{99.8}\approx \sqrt{100}+\frac{1}{2\sqrt{100}}(99.8-100)=10+\frac{1}{20}(-0.2)=9.99$$
Example. Use linear approximation to estimate $\cos 29^\circ$.
Solution. $f(x)=\cos x$ and $x=29^\circ=\frac{29\pi}{180}$ ($29^\circ$ is not a number but $\frac{29\pi}{180}$ is). Since $f'(x)=-\sin x$, the linear approximation of $\cos 29^\circ$ at $a$ is $$\cos 29^\circ\approx \cos a-\sin a \left(\frac{29\pi}{180}-a\right)$$ The suitable $a$ is $=\frac{30\pi}{180}=\frac{\pi}{6}$ in the spirit of the two criteria we discussed in the example above. Therefore, we have $$\cos 29^\circ\approx \cos\frac{\pi}{6}-\sin\frac{\pi}{6}\left(-\frac{\pi}{180}\right)=\frac{\sqrt{3}}{2}+\frac{\pi}{360}$$
Differentials
Differentials
As seen in the above figure, when $\Delta x\approx 0$, $\Delta x=dx$ and $\Delta y\approx dy$. On the other hand, $\frac{dy}{dx}=f'(x)$. Hence, we obtain
\label{eq:differential}
\Delta y\approx f'(x)\Delta x.
Example. The radius of a sphere was measured and found to be 21 cm with a possible error in measurement of at most 0.05 cm. What is the maximum error in using this value of the radius to compute the volume of the sphere?
Solution. Let $V$ denote the volume of a sphere of radius $r$. Then $V=\frac{4}{3}\pi r^3$. What we are trying to find is $\Delta V$ with $\Delta r=0.05$ cm. As seen in \eqref{eq:differential}, $\Delta V\approx dV$, so we find $dV$ instead because finding $dV$ is easier than findingthe exact error $\Delta V$. Differentiating $V$ with respect to $r$, we obtain
\begin{align*}
dV&=4\pi r^2 dr\\
&=4\pi r^2\Delta r\\
&=4\pi\cdot(21)^2\cdot 0.05\\
&=277.
\end{align*}
So the maximum error in the calculated volume is about 277 $\mbox{cm}^3$.
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# Advanced Algorithms Analysis and Design
By
Dr Nazir A. Zafar Advanced Algorithms Analysis and Design
Lecture No. 8
Recurrence Relations
(Algorithms Design and Analysis Techniques)
Dr Nazir A. Zafar
## Advanced Algorithms Analysis and Design
Today Covered
Solution: General Homogenous Recurrences when Roots are distinct Roots are repeated multiplicity of a root is k many roots with different multiplicities Non-homogenous Recurrence Relations Characteristics of various type of non-homogenous recurrence relations Solution to various non-homogenous recurrence relations Conclusion
Dr Nazir A. Zafar
## Solution to Generalized Recurrence with Distinct Roots
Dr Nazir A. Zafar
## Solution to Gen. Recurrence with Distinct Roots
Statement: Find the general solution of the kth order recurrence given below assuming that all roots are distinct a0 tn + a1 tn-1 + ..+ak tn-k = 0 Solution: Let Tn = xn, x is a constant as yet unknown. If we assume that Tn is a solution of equation a0 tn + a1 tn-1 + ..+ak tn-k = 0 Then, a0xn + a1xn-1 +.+ akxn-k = 0 Equation satisfied if x = 0 ,trivial solution, no interest. Otherwise, the equation is satisfied if and only if a0 + a1x1 +.+ akxk = 0
Dr Nazir A. Zafar Advanced Algorithms Analysis and Design
## Solution to Recurrence with Distinct Roots
a0 + a1x1 +.+ akxk = 0
This equation of degree k in x is called the characteristic equation of above recurrence and P(x) = a0 + a1x1 +.+ akxk is called its characteristic polynomial Fundamental theorem of algebra states that polynomial P(x) of degree k has k roots, not necessarily distinct It means that it can be factorized as a product of k terms P(x) = (x-r1) (x-r2) (x-r3). . .(x-rk) where ri may be complex numbers. Moreover, these ri are only solutions of equation P(x) = 0
Dr Nazir A. Zafar Advanced Algorithms Analysis and Design
## Solution to Recurrence with Distinct Roots
Consider any root ri of the characteristic polynomial
## P(x) = a0 + a1x1 +.+ akxk ( x ri )
i 1
Since, p(r1) = 0, p(r2) = 0, p(r3) = 0, . . ., p(rk) = 0 Hence all x = ri , for i {1, 2, . . .,k} are solutions to above characteristic polynomial. Therefore, r1n, r2n, . . ., rkn are solution to our original recurrence relation. Since linear combination of solutions is also a solution to recurrence, therefore below is a solution.
T (n) ci ri
i 1
Dr Nazir A. Zafar
## Solution to Recurrence with Distinct Roots
T (n) ci ri
i 1 k n
where, c1, c2, . . ., ck are all constants to be determined finding particular solution The remarkable fact is that this equation has only solutions of this form provided all ri are distinct. Constants can be determined from k initial conditions by solving system of k linear equations in k unknowns
Dr Nazir A. Zafar
## Solving Recurrence Problem with Distinct Roots
Problem: Consider the recurrence tn = n if n = 0, 1, 2 tn = 7.tn-2 + 6.tn-3 otherwise Find the general solution of the recurrence above.
Solution: First we rewrite the recurrence. tn - 7.tn-2 - 6tn-3 = 0 The characteristic equation become. x3 7x + 6 = (x + 1) (x + 2)(x - 3) The roots are: r1 = -1, r2 = - 2 and r3 = + 3 tn = c1 (-1)n + c2 (-2)n + c3 (3n)
Dr Nazir A. Zafar Advanced Algorithms Analysis and Design
## Solving Recurrence Problem with Distinct Roots
tn = c1 (-1)n + c2 (-2)n + c3 (3n)
The initial conditions give c1 + c 2 + + c 3 = 0 - c1 - 2c2 + 3c3 = 1 c1 + 4c2 + 9c3 = 2 for n = 0 for n = 1 for n = 2
Solving these equations, we obtain c1 = -1/4, c2 = 0 and c3 = 1/4 Therefore, tn = c1 (-1/4)(-1)n + (1/4).(3)n
Dr Nazir A. Zafar Advanced Algorithms Analysis and Design
## Solution to Generalized Recurrence with One Repeated Root
Dr Nazir A. Zafar
## Solution to Recurrence with one Repeated Root
Statement: If the characteristic polynomial P(x) = a0 + a1x1 +.+ akxk then conclude that if r is a double root then tn = rn
## and tn = n rn are both solutions to recurrence.
Solution: It can be found as in case of second order linear homogenous recurrence relation. Since r is a multiple root. By definition of multiple roots, there exists a polynomial q(x) of degree k-2 such that the following holds p(x) = (x r)2q(x), for every n k
Dr Nazir A. Zafar Advanced Algorithms Analysis and Design
## Solution to Recurrence with one Repeated Root
Consider the kth degree polynomials un (x) = a0 xn + a1 xn-1 + . . . + ak xn-k and vn (x) = a0 n xn + a1 (n-1)xn-1 + . . . + ak (n-k)xn -k It is to be noted that vn(x) = x.un (x), where un (x) denotes the derivative of un (x) with respect to x . But un (x) can be written as un (x) = xn-k p(x) = xn-k (x-r)2q(x) = (x-r)2(xn-k q(x)) Using rule for computing the derivative of a product of functions, we obtain that derivative of un(x) with respect to x is given by
Dr Nazir A. Zafar Advanced Algorithms Analysis and Design
## Solution to Recurrence with one Repeated Root
un (x) = 2 (x-r) xn-k q(x) +(x-r)2(xn-k q(x)) therefore un (r) = 0, which implies that vn (r) = r . un (r) = 0 for all n k. It means: a0 n rn + a1 (n-1) rn-1 +. . .+ ak (n-k) rn-k = 0 Hence, tn = n rn is also solution to the recurrence. Now we conclude that if r is a double root then tn = rn and tn = n rn are both solutions to the recurrence. Rest of k-2 are all distinct roots hence general solution
## tn c1r c2nr b r b2r2 ... bk 2rk 2
n n n 11 n
Dr Nazir A. Zafar
where c1, c2, b1, b2,. . ., and bk-2 are all real constants
## Higher Order Homogenous Recurrence with k-multiplicity of a Root
Dr Nazir A. Zafar
## Solution to Recurrence with Root of Multiplicity k
Now if we have to solve recurrence order k, then to solve polynomial, degree k, given below is sufficient Statement: If the characteristic polynomial P(x) = a0 + a1x1 +.+ akxk has r as double root then it can be written as p(x) = (x r)2q(x), for every n k and solutions are: rn and n rn has r as triple root then it can be written as p(x) = (x r)3q1(x), for every n k and solutions are: rn, n.rn and n2.rn
Dr Nazir A. Zafar Advanced Algorithms Analysis and Design
## K-Multiplicity of a Root: General Result
r has multiplicity k then it can be written as p(x) = (x r)k, for every n k and solutions are: rn, n.rn, n2.rn,. . ., nk-1.rn then general solution is
t n c1r c2 nr ... ck n r
n n
k 1 n
tn c j n r
j 1
j 1 n
## where b1, b2,. . ., bk are all real constants
Dr Nazir A. Zafar Advanced Algorithms Analysis and Design
## Multiplicity of Roots: More General Result
If there are l roots, r1, r2, . . ., rl with multiplicities m1, m2, . . ., ml respectively, of the polynomial:
P(x) = a0 + a1x1 +. . .+ akxk s.t. m1 + m2 + . . .+ ml = k then the general solution to the recurrence is
## t n c11r1 c12 nr1 c13n r1 ... c1m1 n
n n 2 n n n 2 n
m1 1 n 1 m2 1
r r2
n
c21r2 c22 nr2 c23n r2 ... c2 m2 n ... cl1rl cl 2 nrl cl 3n rl ... clml n
n n 2 n
Dr Nazir A. Zafar
ml 1 n l
## Advanced Algorithms Analysis and Design
Contd..
t n c11r1 c12 nr1 c13n r1 ... c1m1 n
n n 2 n n n 2 n m1 1 n 1 m2 1
r r2
n
c21r2 c22 nr2 c23n r2 ... c2 m2 n ... cl1rl cl 2 nrl cl 3n rl ... clml n
n n 2 n
ml 1 n l
tn c1 j n r c2 j n r2 ... clj n r
j 1 j 1 n 1 j 1 n j 1 j 1
m1
m2
ml
j 1 n l
tn cij n r
i 1 j 1
Dr Nazir A. Zafar
mi
j 1 n i
## where all ci, j are constants
Problem
Statement: Consider the recurrence tn = n if n = 0, 1, 2 tn = 5tn-1 - 8tn-2 + 4tn-3 otherwise Find general solution of the recurrence above.
Solution: First we rewrite the recurrence. tn - 5tn-1 + 8tn-2 - 4tn-3 = 0 The characteristic equation become. x3 5x2 + 8x -4 = (x-1) (x-2)2 The roots are: r1 = 1 of multiplicity m1 = 1 and r2 = 2 of multiplicity m2 = 2, and hence the general solution is tn = c 1 1 n + c 2 2 n + c 3 n 2 n
Dr Nazir A. Zafar Advanced Algorithms Analysis and Design
Example
The initial conditions give c1 + c2 = 0 c1 + 2c2 + 2c3 = 1 c1 + 4c2 + 8c3 = 2
## Solving these equations, we obtain
c1 = -2, c2 = 2 and c3 = -1/2
## Therefore, tn = c 1 1 n + c 2 2 n + c 3 n 2 n = -2 + 2.2n .n.2n = 2n+1 n.2n-1 - 2
Dr Nazir A. Zafar Advanced Algorithms Analysis and Design
## Non-homogeneous Recurrence of Higher Order
Dr Nazir A. Zafar
## Advanced Algorithms Analysis and Design
Non-homogenous Recurrence
Solution of a linear recurrence with constant coefficients becomes more difficult when the recurrence is not homogeneous That is when linear combination is not equal to zero In particular, it is no longer true that any linear combination of solutions is a solution. Consider the following recurrence
a0 tn + a1 tn-1 + . . .+ ak tn-k = bn p(n) The left-hand side is the same as before, but on the right-hand side we have bnp(n) where b is a constant and p(n) is a polynomial in n of degree d.
Dr Nazir A. Zafar Advanced Algorithms Analysis and Design
## Generalization: Non-homogeneous Recurrences
If a recurrence is of the form
n 1
## Then the characteristics polynomial is
(a0 x k a1 x k 1 . . . ak )( x b1d 1 )
Which contains one factor for the left hand side And other factor corresponding to the right hand side, where d is degree of polynomial Characteristics polynomial can be used to solve the above recurrence.
Dr Nazir A. Zafar Advanced Algorithms Analysis and Design
## Problem 1: Non-homogeneous Recurrences
Problem: Consider the recurrence below. Find its solution tn 2tn-1 = 3n Solution: Compare the above recurrence with a0 tn + a1 tn-1 + . . .+ ak tn-k = bn p(n) Here: b = 3, p(n) = 1, a polynomial of degree 0. Reducing to homogeneous case, we are given tn 2tn-1 = 3n (1) Replace n by n - 1 and multiply by 3, we obtain tn-1 2tn-2 = 3n-1 3tn-1 6tn-2 = 3n (2)
Dr Nazir A. Zafar Advanced Algorithms Analysis and Design
## Problem 1: Non-homogeneous Recurrences
From (1) and (2) tn 2tn-1 = 3n (1) + 3tn-1 6tn-2 = 3n (2) Subtracting 2 from equation 1, we get tn - 5tn-1 + 6tn-2 = 0 The characteristic equation is x2 - 5x + 6 = 0 Roots of this equation are: x = 2 and x = 3 And therefore general solution is tn = c 1 2 n + c 2 3 n
Dr Nazir A. Zafar
## Problem 1: Non-homogeneous Recurrences
It is not always true that an arbitrary choice of c1 and c2 produces a solution to the recurrence even when initial conditions are not taken into account. Note: It is to be noted that solutions: tn = 2n and tn = 3 n which are solutions to reduced recurrence, are not solution to original one What is the reason?
Dr Nazir A. Zafar Advanced Algorithms Analysis and Design
## Problem 2: Non-homogeneous Recurrences
Problem Find general solution of the following recurrence. tn - 2tn-1 = (n + 5) 3n n 1 Solution: The manipulation needed to transform this into a homogeneous recurrence is slightly more complicated than with first example. tn - 2tn-1 = (n + 5) 3n n 1 (1) replace n by n-1, n-2, we get tn-1 - 2tn-2 = (n + 4) 3n-1 n 2 (2) tn-2 - 2tn-3 = (n + 3) 3n-2 n 3 (3)
Dr Nazir A. Zafar Advanced Algorithms Analysis and Design
## Problem 2: Non-homogeneous Recurrences
Above equations can be written as tn - 2tn-1 = 9(n + 5) 3n-2 n 1 (4) tn-1 - 2tn-2 = 3(n + 4) 3n-2 n 2 (5) tn-2 - 2tn-3 = (n + 3) 3n-2 n 3 (6) Our objective is to eliminate the right hand side of the above equations to make it homogenous. Multiply (5) by -6, and (6) by 9 we get tn - 2tn-1 - 6tn-1 + 12tn-2 + 9tn-2 - 18tn-3
Dr Nazir A. Zafar
## Problem 2: Non-homogeneous Recurrences
After simplification, the above equations can be written as tn - 2tn-1 - 6tn-1 + 12tn-2 + 9tn-2 - 18tn-3 = (9n + 45) 3n-2 = (-18n 72) 3n-2 = (9n + 27) 3n-2
Adding these equation, we get homogenous equation, which can be solved easily tn - 8tn-1 + 21tn-2 - 18tn-3 = 0
Dr Nazir A. Zafar Advanced Algorithms Analysis and Design
## Problem 2: Non-homogeneous Recurrences
tn 8tn-1 + 21tn-2 - 18tn-3 = 0
The characteristics equation of the above homogenous equation is: x3 8x2 +21x -18 = 0 ( x-2) (x-3)2 = 0 and hence, x = 2, 3, 3 General solution is: tn = c1 2n + c2 3n + c3 n 3n For n = 0, 1, 2 We can find values of c1, c2, c3 and then tn = (t0 - 9) 2n + (n + 3)3n+1
Dr Nazir A. Zafar
## Tower of Hanoi: Problem 3
Tower of Hanoi is a mathematical game or puzzle. It consists of three towers, a number of disks of different sizes which can slide onto any tower. The puzzle starts with disks stacked in order of size on one tower, smallest at top, making a conical shape. Objective is to move entire stack to another tower, obeying following rules: Only one disk may be moved at a time. Each move consists of taking upper disk from one of towers and sliding it onto another tower You can put on top of other disks already present No disk may be placed on top of a smaller disk.
Dr Nazir A. Zafar
## More Generalized Non-homogeneous Recurrences
If a recurrence is of the form
n 1 n 2
## Then the characteristics polynomial is
d (a0 x k a1 x k 1 . . . ak )( x b1d1 1 )( x b2 2 1 )...,
Which contains one factor for the left hand side And other factor corresponding to the each term on right hand side. Once the characteristics polynomial is obtained the recurrence can be solved as before.
Dr Nazir A. Zafar
## Problem 6 : Non-homogeneous Recurrences
Consider the recurrence tn = 2tn-1 + n + 2n otherwise
Solution: Compare the recurrence: tn - 2tn-1 = n + 2n with
n 1 n 2
## Here, b1 = 1, p1(n) = n, b2 = 2, and p2(n) = 1. Degree of p1(n) = d1 = 1, Degree of p2(n) = d2 = 0.
Dr Nazir A. Zafar
## Problem 6: Non-homogeneous Recurrences
The characteristic polynomial: (x-2) (x-1)2 (x-2) The roots are, x = 1, 2, both of multiplicity 2. All solutions of recurrence therefore have form tn = c1 1n + c2 n1n + c3 2n + c4 n 2n n + 2n = (2c2 - c1 ) c2 n + c4 2n For n = 0, 1, 2, 3 c1, c2, c3 and c4 can be solved and hence solution is tn= n.2n +2n+1 n -2
Dr Nazir A. Zafar
## Advanced Algorithms Analysis and Design
Conclusion
Recursive relations are important because used in divide and conquer, dynamic programming and in many other e.g. optimization problems Analysis of such algorithms is required to compute complexity Recursive algorithms may not follow homogenous behavior, it means there must be some cost in addition to recursive cost Solution to homogenous recursive relations is comparatively easy, but not so easy in case of non-homogenous recursive relations
Dr Nazir A. Zafar Advanced Algorithms Analysis and Design
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# 6a.4 - Hypothesis Test for One-Sample Proportion
## Overview Section
In this section, we will demonstrate how we use the sampling distribution of the sample proportion to perform the hypothesis test for one proportion.
Recall that if $$np$$ and $$n(1-p)$$ are both greater than five, then the sample proportion, $$\hat{p}$$, will have an approximate normal distribution with mean $$p$$, standard error $$\sqrt{\frac{p(1-p)}{n}}$$, and the estimated standard error $$\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$.
In hypothesis testing, we assume the null hypothesis is true. Remember, we set up the null hypothesis as $$H_0\colon p=p_0$$. This is very important! This statement says that we are assuming the unknown population proportion, $$p$$, is equal to the value $$p_0$$.
Since this is true, then we can follow the same logic above. Therefore, if $$np_0$$ and $$n(1-p_0)$$ are both greater than five, then the sampling distribution of the sample proportion will be approximately normal with mean $$p_0$$ and standard error $$\sqrt{\frac{p_0(1-p_0)}{n}}$$.
We can find probabilities associated with values of $$\hat{p}$$ by using:
$$z^*=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}$$
## Example 6-4 Section
Referring back to a previous example, say we take a random sample of 500 Penn State students and find that 278 are from Pennsylvania. Can we conclude that the proportion is larger than 0.5?
Is 0.556(=278/500) much bigger than 0.5? What is much bigger?
This depends on the standard deviation of $$\hat{p}$$ under the null hypothesis.
$$\hat{p}-p_0=0.556-0.5=0.056$$
The standard deviation of $$\hat{p}$$, if the null hypothesis is true (e.g. when $$p_0=0.5$$) is:
$$\sqrt{\dfrac{p_0(1-p_0)}{n}}=\sqrt{\dfrac{0.5(1-0.5)}{500}}=0.0224$$
We can compare them by taking the ratio.
$$z^*=\dfrac{\hat{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}=\dfrac{0.556-0.5}{\sqrt{\frac{0.5(1-0.5)}{500}}}=2.504$$
Therefore, assuming the true population proportion is 0.5, a sample proportion of 0.556 is 2.504 standard deviations above the mean.
The $$z^*$$ value we found in the above example is referred to as the test statistic.
Test statistic
The sample statistic one uses to either reject $$H_0$$ (and conclude $$H_a$$ ) or fail to reject $$H_0$$.
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# How do I find the upper bound of a polynomial?
May 17, 2016
See explanation...
#### Explanation:
If the term of highest degree is of odd degree or has a positive coefficient, then there is no upper bound (unless you are being asked for the upper bound over an interval).
Otherwise, in the general case of:
$f \left(x\right) = {a}_{n} {x}^{n} + {a}_{n - 1} {x}^{n - 1} + \ldots + {a}_{1} x + {a}_{0}$
you need to find the zeros of the derivative:
$f ' \left(x\right) = n {a}_{n} {x}^{n - 1} + \left(n - 1\right) {a}_{n - 1} {x}^{n - 2} + \ldots + {a}_{1}$
and evaluate $f \left(x\right)$ at those zeros.
The highest of these values will be the upper bound of the polynomial.
Example
What is the upper bound of the following polynomial?
$f \left(x\right) = - {x}^{4} + 4 {x}^{3} - 2 {x}^{2} - 4 x + 1$
Note that this polynomial has even degree and a negative leading coefficient, so does have an upper bound.
We find:
$f ' \left(x\right) = - 4 {x}^{3} + 12 {x}^{2} - 4 x - 4$
$= - 4 \left({x}^{3} - 3 {x}^{2} + x + 1\right)$
$= - 4 \left(x - 1\right) \left({x}^{2} - 2 x - 1\right)$
$= - 4 \left(x - 1\right) \left(x - 1 - \sqrt{2}\right) \left(x - 1 + \sqrt{2}\right)$
So evaluate $f \left(x\right)$ for each of the three zeros of $f ' \left(x\right)$.
$f \left(x\right) = - {x}^{4} + 4 {x}^{3} - 2 {x}^{2} - 4 x + 1 = 1 - x \left(x \left(\left(x - 4\right) x + 2\right) + 4\right)$
So:
$f \left(1\right) = - 1 + 4 - 2 - 4 + 1 = - 2$
$f \left(1 + \sqrt{2}\right) = 1 - \left(1 + \sqrt{2}\right) \left(\left(1 + \sqrt{2}\right) \left(\left(\left(1 + \sqrt{2}\right) - 4\right) \left(1 + \sqrt{2}\right) + 2\right) + 4\right) = 2$
$f \left(1 - \sqrt{2}\right) = 1 - \left(1 - \sqrt{2}\right) \left(\left(1 - \sqrt{2}\right) \left(\left(\left(1 - \sqrt{2}\right) - 4\right) \left(1 - \sqrt{2}\right) + 2\right) + 4\right) = 2$
So the upper bound of $f \left(x\right)$ is $2$
graph{-x^4+4x^3-2x^2-4x+1 [-3.79, 6.21, -2.48, 2.52]}
|
## Linear Algebra and Its Applications, exercise 1.2.13
Exercise 1.2.13. Given the equation x + 4y = 7 for a line in the x-y plane, find the equation for a line that is parallel to the first line and passes through the point (0, 0). The first line passes through the point (1, 3) (3, 1); find another line that also passes through that point.
Answer: Solving for y in the first equation, we have
$x + 4y = 7 \Rightarrow 4y = -x + 7 \Rightarrow y = - \frac{1}{4}x + \frac{7}{4}$
This line has slope 1/4 and intersects the y axis at the point (0, 7/4). The parallel line (also of slope 1/4) passing through the origin has the equation $y = - \frac{1}{4}x$.
If we assume a slope of 1 then the corresponding line has an equation of the form y = x + a for some a. If we further assume that the line goes through the point (1, 3) (3, 1) then a = 2 a = -2. So the line with equation y = x + 2 y = x – 2 satisfies the condition for the second part of the exercise.
UPDATE: Corrected the point of intersection. (I had transcribed the exercise incorrectly.) Thanks to freaky4you for the fix!
NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.
If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.
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### 2 Responses to Linear Algebra and Its Applications, exercise 1.2.13
1. freaky4you says:
Meeting Point is (3,1) not (1,3) so the answer will be y=x-2
• hecker says:
Fixed. Thanks for finding and reporting the error!
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P Pankaj Sanodiya
Let the speed of the train and bus be u and v respectively Now According to the question, And Let, Now, our equation becomes And By Cross Multiplication method, And Hence, Hence the speed of the train and bus are 60 km/hour and 80 km/hour respectively.
P Pankaj Sanodiya
Let the number of days taken by woman and man be x and y respectively, The proportion of Work done by a woman in a single day The proportion of Work done by a man in a single day Now, According to the question, Also, Let, Now, our equation becomes And By Cross Multiplication method, So,
P Pankaj Sanodiya
Given Equations, As we can see by adding and subtracting both equations we can make our equations simple to solve. So, Adding (1) and )2) we get, Subtracting (2) from (1) we get, Now, Adding (3) and (4) we get, Putting this value in (3) Hence, .
P Pankaj Sanodiya
Given, And Now, Subtracting (1) from (2), we get Substituting this in (1), we get, . Hence,
P Pankaj Sanodiya
Given equation, Now By Cross multiplication method,
P Pankaj Sanodiya
Given Two equations, Now By Cross multiplication method,
P Pankaj Sanodiya
Given Equations, Now By Cross multiplication method,
P Pankaj Sanodiya
As we know that in a quadrilateral the sum of opposite angles is 180 degree. So, From Here, Also, Multiplying (1) by 3 we get, Now, Subtracting, (2) from (3) we get, Substituting this value in (1) we get, Hence four angles of a quadrilateral are :
P Pankaj Sanodiya
Given two equations, And Points(x,y) which satisfies equation (1) are: X 0 1 5 Y -5 0 20 Points(x,y) which satisfies equation (1) are: X 0 1 2 Y -3 0 3 GRAPH: As we can see from the graph, the three points of the triangle are, (0,-3),(0,-5) and (1,0).
P Pankaj Sanodiya
Given, Also, As we know that the sum of angles of a triangle is 180, so Now From (1) we have Putting this value in (2) we have Putting this in (3) And Hence three angles of triangles
P Pankaj Sanodiya
Let the speed of the train be v km/h and the time taken by train to travel the given distance be t hours and the distance to travel be d km. Now As we Know, Now, According to the question, Now, Using equation (1), we have Also, Adding equations (2) and (3), we obtain: Substituting the value of x in equation (2), we obtain: Putting this value in (1) we get, Hence...
P Pankaj Sanodiya
Let the number of rows be x and number of students in a row be y. Total number of students in the class = Number of rows * Number of students in a row Now, According to the question, Also, Subtracting equation (2) from (1), we get: Substituting the value of y in equation (1), we obtain: Hence, The number of rows is 4...
P Pankaj Sanodiya
Let the amount of money the first person and the second person having is x and y respectively
Noe, According to the question.
$x + 100 = 2(y - 100)$
$\Rightarrow x - 2y =-300...........(1)$
Also
$y + 10 = 6(x - 10)$
$\Rightarrow y - 6x =-70..........(2)$
Multiplying (2) by 2 we get,
$2y - 12x =-140..........(3)$
Now adding (1) and (3), we get
$-11x=-140-300$
$\Rightarrow 11x=440$
$\Rightarrow x=40$
Putting this value in (1)
$40-2y=-300$
$\Rightarrow 2y=340$
$\Rightarrow y=170$
Thus two friends had 40 Rs and 170 Rs respectively.
View More
P Pankaj Sanodiya
Let the age of Ani be , age of Biju be , Case 1: when Ani is older than Biju age of Ani's father Dharam: and age of his sister Cathy : Now According to the question, Also, Now subtracting (1) from (2), we get, putting this in (1) Hence the age of Ani and Biju is 19 years and 16 years respectively. Case 2: And Now Adding (3) and (4), we get, putting it in (3) Hence the age of...
P Pankaj Sanodiya
Let the speed of Ritu in still water be x and speed of current be y, Let's solve this problem by using relative motion concept, the relative speed when they are going in the same direction (downstream)= x +y the relative speed when they are going in the opposite direction (upstream)= x - y Now, As we know, Relative distance = Relative speed * time . So, According to the question, And, Now,...
P Pankaj Sanodiya
Given Equations, Let, Now, our equation becomes And Now, Adding (1) and (2), we get Putting this value in (1) Now, And Now, Adding (3) and (4), we get Putting this value in (3), Hence,
P Pankaj Sanodiya
Given Equations, Let, Now, our equation becomes And By Cross Multiplication method, Now, And, Adding (3) and (4) we get, Putting this value in (3) we get, And Hence,
P Pankaj Sanodiya
Given Equations, Let, Now, our equation becomes And By Cross Multiplication method, And Hence,
|
# What number can be subtracted from both the numerator
## Subtracting Fractions
That is NOT the same as the number that can be subtracted from "34/53" Multiplying both sides by the common denominator of 2(n) gives.
can you re welcome from moana an old cloth you might use to clean with
Working with fractions is a basic mathematical principle needed for understanding further math topics and real world applications. Adding and subtracting fractions work on the same principle. Simplifying fractions before completing any other operations makes the process easier and lets you see if you need to complete any further steps. The simplest form of a fraction is the standard form of the fraction used for both common fractions and mixed numbers. Determine if the two fractions have a common denominator. Set both fractions to have a lowest common denominator.
When subtracting fractions, the first thing to check is if the denominators are the same. Fractions with the same denominators are called like fractions. To subtract fractions with like denominators, subtract the numerators , and write the difference over the denominator. You may get an answer which is not in lowest terms , even if the fractions you were adding and subtracting both were. In this case, you have to reduce the fraction.
This calculator subtracts two fractions. It accepts proper, improper, mixed fractions and whole number inputs. If they exist, the solutions and answers are provided in simplified, mixed and whole formats. There general steps to subtract fractions are described below. If the inputs are mixed fractions or whole numbers, convert them to improper fractions. Multiply the left and right fractions by a factor so each of the fractions have the LCM as the denominator. Subtract the left and right numerators.
Mathematically, when we want to refer to only a part of something whole, we use fractions. For example, when someone says they drank half a glass of milk, mathematically they mean that they drank glass of milk and is a fraction. Mike invited his friends over to play some board games and have fun on the weekend. With 3 slices gone, only pizza was left and Mike offered his little sister to take as much as she want. She took 3 slices i.
Go there now. Throughout this post, assume that I am talking about positive fractions with a positive numerator and positive denominator. - Chat or rant, adult content, spam, insulting other members, show more. Harm to minors, violence or threats, harassment or privacy invasion, impersonation or misrepresentation, fraud or phishing, show more.
How to Add or Subtract Mixed Numbers
What number must be subtracted from both the numerator and the What are the biggest tracker networks and what can I do about them?.
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.
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# Is f(x)=(-2x^3-x^2-5x+2)/(x+1) increasing or decreasing at x=0?
Mar 19, 2016
decreasing at x = 0
#### Explanation:
To test if a function is increasing / decreasing at x = a , require to check the sign of f'(a)
• If f'(a) > 0 then f(x) is increasing at x = a
• If f'(a) < 0 then f(x) is decreasing at x = a
Require to find f'(x)
differentiate using the $\textcolor{b l u e}{\text{ Quotient rule }}$
If $f \left(x\right) = g \frac{x}{h \left(x\right)} \text{ then } f ' \left(x\right) = \frac{h \left(x\right) . g ' \left(x\right) - g \left(x\right) . h ' \left(x\right)}{h \left(x\right)} ^ 2$
$\text{-------------------------------------------------------------------------}$
$g \left(x\right) = - 2 {x}^{3} - {x}^{2} - 5 x + 2 \Rightarrow g ' \left(x\right) = - 6 {x}^{2} - 2 x - 5$
h(x) = x+1 $\Rightarrow h ' \left(x\right) = 1$
$\text{------------------------------------------------------------------------}$
substitute these values into f'(x)
$\Rightarrow f ' \left(x\right) = \frac{\left(x + 1\right) . \left(- 6 {x}^{2} - 2 x - 5\right) - \left(- 2 {x}^{3} - {x}^{2} - 5 x + 2\right) .1}{x + 1} ^ 2$
and f'(0) = $\frac{1. \left(- 5\right) - 2.1}{1} = - 7$
since f'(0) < 0 , f(x) is decreasing at x = 0
Here is the graph of f(x)
graph{(-2x^3-x^2-5x+2)/(x+1) [-10, 10, -5, 5]}
|
# 5.8 Modeling using variation
Page 1 / 14
In this section, you will:
• Solve direct variation problems.
• Solve inverse variation problems.
• Solve problems involving joint variation.
A used-car company has just offered their best candidate, Nicole, a position in sales. The position offers 16% commission on her sales. Her earnings depend on the amount of her sales. For instance, if she sells a vehicle for $4,600, she will earn$736. She wants to evaluate the offer, but she is not sure how. In this section, we will look at relationships, such as this one, between earnings, sales, and commission rate.
## Solving direct variation problems
In the example above, Nicole’s earnings can be found by multiplying her sales by her commission. The formula $\text{\hspace{0.17em}}e=0.16s\text{\hspace{0.17em}}$ tells us her earnings, $\text{\hspace{0.17em}}e,\text{\hspace{0.17em}}$ come from the product of 0.16, her commission, and the sale price of the vehicle. If we create a table, we observe that as the sales price increases, the earnings increase as well, which should be intuitive. See [link] .
$\text{\hspace{0.17em}}s\text{\hspace{0.17em}}$ , sales price $e=0.16s$ Interpretation
$4,600 $e=0.16\left(4,600\right)=736$ A sale of a$4,600 vehicle results in $736 earnings.$9,200 $e=0.16\left(9,200\right)=1,472$ A sale of a $9,200 vehicle results in$1472 earnings.
$18,400 $e=0.16\left(18,400\right)=2,944$ A sale of a$18,400 vehicle results in $2944 earnings. Notice that earnings are a multiple of sales. As sales increase, earnings increase in a predictable way. Double the sales of the vehicle from$4,600 to $9,200, and we double the earnings from$736 to $1,472. As the input increases, the output increases as a multiple of the input. A relationship in which one quantity is a constant multiplied by another quantity is called direct variation . Each variable in this type of relationship varies directly with the other. [link] represents the data for Nicole’s potential earnings. We say that earnings vary directly with the sales price of the car. The formula $\text{\hspace{0.17em}}y=k{x}^{n}\text{\hspace{0.17em}}$ is used for direct variation. The value $\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$ is a nonzero constant greater than zero and is called the constant of variation . In this case, $\text{\hspace{0.17em}}k=0.16\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}n=1.\text{\hspace{0.17em}}$ We saw functions like this one when we discussed power functions. ## Direct variation If $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ are related by an equation of the form $\text{\hspace{0.17em}}y=k{x}^{n}\text{\hspace{0.17em}}$ then we say that the relationship is direct variation and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ varies directly with, or is proportional to, the $\text{\hspace{0.17em}}n\text{th}\text{\hspace{0.17em}}$ power of $\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$ In direct variation relationships, there is a nonzero constant ratio $\text{\hspace{0.17em}}k=\frac{y}{{x}^{n}},\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$ is called the constant of variation , which help defines the relationship between the variables. Given a description of a direct variation problem, solve for an unknown. 1. Identify the input, $\text{\hspace{0.17em}}x,$ and the output, $\text{\hspace{0.17em}}y.\text{\hspace{0.17em}}$ 2. Determine the constant of variation. You may need to divide $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ by the specified power of $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ to determine the constant of variation. 3. Use the constant of variation to write an equation for the relationship. 4. Substitute known values into the equation to find the unknown. ## Solving a direct variation problem The quantity $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ varies directly with the cube of $\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$ If $\text{\hspace{0.17em}}y=25\text{\hspace{0.17em}}$ when $\text{\hspace{0.17em}}x=2,\text{\hspace{0.17em}}$ find $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ when $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is 6. The general formula for direct variation with a cube is $\text{\hspace{0.17em}}y=k{x}^{3}.\text{\hspace{0.17em}}$ The constant can be found by dividing $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ by the cube of $\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$ $\begin{array}{ccc}\hfill k& =& \frac{y}{{x}^{3}}\hfill \\ & =& \frac{25}{{2}^{3}}\hfill \\ & =& \frac{25}{8}\hfill \end{array}$ Now use the constant to write an equation that represents this relationship. $y=\frac{25}{8}{x}^{3}$ Substitute $\text{\hspace{0.17em}}x=6\text{\hspace{0.17em}}$ and solve for $\text{\hspace{0.17em}}y.$ $\begin{array}{ccc}\hfill y& =& \frac{25}{8}{\left(6\right)}^{3}\hfill \\ & =& 675\hfill \end{array}$ #### Questions & Answers sin^4+sin^2=1, prove that tan^2-tan^4+1=0 SAYANTANI Reply what is the formula used for this question? "Jamal wants to save$54,000 for a down payment on a home. How much will he need to invest in an account with 8.2% APR, compounding daily, in order to reach his goal in 5 years?"
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A = P(1 + r/n) ^rt
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its a very simple
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What is the value of log-1
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Log(-1)
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What is the value of i^i
Masum
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No
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tan20°×tan30°×tan45°×tan50°×tan60°×tan70°
jaldi batao
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Find the value of x between 0degree and 360 degree which satisfy the equation 3sinx =tanx
what is sine?
what is the standard form of 1
1×10^0
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Evalute exponential functions
30
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The sides of a triangle are three consecutive natural number numbers and it's largest angle is twice the smallest one. determine the sides of a triangle
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x exposant 4 + 4 x exposant 3 + 8 exposant 2 + 4 x + 1 = 0
x exposent4+4x exposent3+8x exposent2+4x+1=0
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ranges
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factoring polynomial
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# Trigonometric ratios homework
Apps can be a great way to help learners with their math. Let's try the best Trigonometric ratios homework. Math can be a challenging subject for many students.
## The Best Trigonometric ratios homework
First determine the y intercept. The y intercept is the value where the line crosses the Y axis. It is sometimes referred to as the "zero" point, or reference point, along the line. The y intercept of an equation can be determined by drawing a vertical line down through the origin of each graph and placing a dot at the intersection of the two lines (Figure 1). When graphing a parabola, the y intercept is placed at the origin. When graphing a line with a slope 1, then both y-intercepts are placed at 0. When graphing a line with a slope >1, then both y-intercepts are moved to positive infinity. In order to solve for x intercept on an equation, first use substitution to solve for one of the variables in terms of another variable. Next substitute back into original equation to find x-intercept. In order to solve for y intercept on an equation, first use substitution to solve for one of the variables in terms of another variable. Next substitute back into original equation to find y-intercept. Example: Solve for x-intercept of y = 4x + 10 Solution: Substitute 4x + 5 = 0 into original problem: y = 4x + 10 => y = 4(x + 5) => y =
Solving for angle in a right triangle is actually quite easy, but it’s important to remember a few things. Firstly, you can never solve for the "length" of the side unless that side is a right triangle (which means all three sides are equal). Secondly, when solving for angle in a right triangle, you always need to have an initial guess as to what angle you’re looking for. Lastly, the values for angles must always be expressed in degrees. One of the most common problems with solving for angle in a right triangle is when you try to find the "perpendicular bisector" of one of the sides. When this happens, it's usually because you're trying to find *the* perpendicular bisector of the side instead of finding its length (which would give you a third angle). The easiest way to avoid this problem is to always think about which side you're looking at first and make sure that angle is always used as your starting point.
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# Bihar Board Class 10th Maths Solutions 11 Constructions Ex 11.2
Bihar Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.2 Textbook Questions and Answers.
## BSEB Bihar Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.2
Question 1.
Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.
Solution:
Steps of Construction :
1. Take a point 0 as centre and draw a circle of radius 6 cm.
2. Mark a point P at a distance of 10 cm from the centre O.
3. Join OP and bisect it. Let M be its mid-point of OP.
4. With M as centre and MP as radius, draw a circle to intersect the previous circle at Q and R.
5. Join PQ and PR. Then, PQ and PR are the required tangents.
On measurement, we find that PQ = PR = 8 cm.
Justification :
On joining OQ, we find that ∠PQO = 90°, as ∠PQO is the angle in a semi-circle.
∴ PQ ⊥ OQ.
Since OQ is the radius of the given circle, so PQ has to be a tangent to the circle. Similarly, PR is also a tangent to the circle.
Question 2.
Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also, verify the measurement by actual calculation.
Solution:
Steps of Construction :
1. Take a point O and draw two concentric circles of radii 4 cm and 6 cm respectively.
2. Mark a point P on the bigger circle.
3. Join OP and bisect it. Let M be its mid-point.
4. With M as centre and MP as radius, draw a circle to intersect the smaller circle at Q and R.
5. Join PQ and PR. Then, PQ and PR are the required tangents.
On measuring, we find that PQ = PR = 4.5 cm (approx.).
Calculation : From A OQP,
OP² = OQ² + PQ²
or 6² = 4² + PQ²
or PQ² = 36 – 16 = 20
So, PQ = $$\sqrt{20}$$cm = 4-47 cm (approx.)
Similarly, PR = 4.47 cm (approx.).
Justification :
On joining OQ, we find that ∠PQO = 90°, as ∠PQO is the angle in a semi-circle.
∴ PQ ⊥ OQ.
Since OQ is the radius of the given circle, so PQ has to be a tangent to the circle. Similarly, PR is also a tangent to the circle.
Question 3.
Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.
Solution:
Steps of Construction :
1. Take a point O, draw a circle of radius 3 cm with this point as centre.
2. Take two points P and Q on one of its extended diameter such that OP = OQ = 7 cm.
3. Bisect OP and OQ. Let their respective mid-points be M1 and M2.
4. With M1 as centre and M1P as radius, draw a circle to intersect the circle at T1 and T2.
5. Join PT1 and PT2. Then, PT1 and PT2 are the required tangents.
Similarly, the tangents QTS and QT4 can be obtained.
Justification :
On joining OT4, we find ∠PT10 = 90°, as an angle in a semi-circle.
∴ PT1 ⊥ OT1
Since OT1 is a radius of the given circle, so PT1 has to be a tangent to the circle.
Similarly, PT2, QT3 and QT4 are also tangents to the circle.
Question 4.
Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°.
Solution:
Steps of Construction :
1. With O as centre and radius = 5 cm, draw a circle.
2. Draw any diameter AOC.
3. Draw a radius OL such that ∠COL = 60° (i.e., the given angle).
4. At L, draw LM ⊥ OL.
5. At A, draw AN ⊥ OA.
6. These two perpendiculars intersect each other at P. Then, PA and PL are the required tangents.
Justification :
Since OA is the radius, so PA has to be a tangent to the circle.
Similarly, PL is the tangent to the circle.
∠APL = 360° – ∠OAP – ∠OLP – ∠AOL
= 360° – 90° – 90° – (180° – 60°)
= 360° – 360° + 60° = 60°
Thus, tangents PA and PL are inclined to each other at an angle of 60°.
Question 5.
Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.
Solution:
Steps of Construction :
1. Draw a line segment AB = 8 cm.
2. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw a circle of radius 3 cm.
3. Draw perpendicular bisector of the line segment AB. It intersects the line segment AB at O. Clearly, O is the mid-point of AB. With O as centre, draw a circle of radius OA or OB, intersecting circle with centre B at T4 and T4, circle with centre A at T3 and T4.
4. Join AT1( AT2, BT3 and BT4). Then, these are the required tangents.
Justification :
On joining BT1 we find that ∠BT1A = 90°, as ∠BT1A is the angle in a semi-circle.
∴ AT1 ⊥ BT1
Since BT1 is the radius of the given circle, so AT1 has to be a tangent to the circle.
Similarly, AT2, BT3 and BT4 are the tangents.
Question 6.
Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.
Solution:
Steps of Construction :
1. With the given data, draw a A ABC, in which AB = 6 cm, BC = 8 cm and ∠B = 90°.
2. Draw BD ⊥ AC.
3. Draw perpendicular bisectors of BC and BD. They meet at a point O.
4. Taking O as centre and OD as radius, draw a circle. This circle passes through B, C and D.
5. Join OA.
6. Draw perpendicular bisector of OA. This perpendicular bisector meets OA at M.
7. Taking M as centre and MA as radius, draw a circle which intersects the circle drawn in step 4 at point B and P.
8. Join AB and AP.
These are the required tangents from A.
Question 7.
Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle.
Solution:
Steps of Construction :
1. Draw a circle with the help of a bangle.
2. Draw a secant ARS from an external point A. Produce RA to C such that AR = AC.
3. With CS as diameter, draw a semi-circle.
4. At the point A, draw AB ⊥ AS, cutting the semi-circle at B.
5. With A as centre and AB as radius, draw an arc to intersect the given circle, in T and T’. Join AT and AT’. Then, AT and AT’ are the required pair of tangents.
Aliter:
1. Mark any three points P, Q and R on the circle.
2. Draw the perpendicular bisectors of PQ and QR to meet at O, the centre of the circle.
3. Join OA and locate its mid-point M.
4. With M as centre and OM as radius, draw a circle to intersect the previous circle at T1 and T2.
5. Join AT1 and AT2, which are the required tangents.
|
Refer to our Texas Go Math Grade 5 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 5 Module 10 Assessment Answer Key.
Vocabulary
Choose the best term from the box.
Vocabulary
input/output table
pattern
pattern rule
Question 1.
You can use a graph or an __________ to show a pattern. (p. 375)
pattern rule,
Explanation:
We will use a graph or pattern rule to show a pattern,
Pattern rules mean numerical pattern is a sequence of numbers
that has been created based on a formula or rule called a pattern rule.
Pattern rules can use one or more mathematical operations to describe
the relationship between consecutive numbers in the pattern.
A __________ is an ordered set of numbers or objects. (p. 375)
Pattern,
Explanation:
Pattern means an ordered set of numbers or objects in which the
order helps you predict what will come next.
Use the rule to complete the table. (TEKS 5.4.C)Concepts and Skills
Question 3.
Rule: m = n + 10.
Explanation:
Given Rule m = n + 10,
if n = 1 then m = 1 + 10 = 11, (1,11),
if n = 2 then m = 2 + 10 = 12, (2,12),
if n= 3 then m = 3 + 10 = 13, (3,13) as shown
above in the table.
Question 4.
Rule: p = s × 4.
Explanation:
Given p = s XÂ 4,
if s = 2 then p = 2 X 4 = 8, (2,8),
if s = 4 then p = 4 X 4 = 16, (4,16),
if s = 6 then p = 6 X 4 = 24, (6, 24),
if s = 8 then p = 8 X 4 = 32,(8,32) as shown
above in the table.
Decide if the pattern shown in the table is additive or multiplicative.
Write a rule to describe the pattern. (TEKS 5.4.D)
The pattern is ____________.
Rule: ____________
Rule is t = r + 2,
Explanation:
As we see if input r = 5 then output t = 5 + 2 = 7,
if input r = 6 then output t = 6 + 2 = 8 and
if input r = 7 then output t = 7 + 2 = 9 therefore
the pattern is additive and rule is t = r + 2.
Question 6.
The pattern is ____________.
Rule: ____________
The pattern is multiplicative,
Rule: p = a X 7,
Explanation:
As we see if input a = 1 then output p = 1 X 7 = 7,
if input a = 2 then output p = 2 X 7= 14
if input a = 3 then output p = 3 X 7 = 21 and
if input a = 4 then output p = 4 X 7 = 28 therefore
the pattern is multiplicative and rule is p = a x 7.
Use the graph for 7-8.
Question 7.
The graph shows the relationship between the number of
work days and the number of weeks. Which rule best describes the
the pattern in the graph? (TEKS 5.4.D)
(A) d = w + 5
(B) w = 5d
(C) d = 5w
(D) w = d + 5
(B) w = 5d,
Explanation:
The graph shows the relationship between the number of
work days and the number of weeks. the rule best describes the
pattern in the graph is w = 5d as if d = 1 then w = 5 XÂ 1 = 5,
d = 2 then w = 5 X 2 = 10,
if d = 3 then w = 5 X 3 = 15,
if d = 4 then w = 5 X 4 = 20 so the rule is w = 5d
which matches with (B).
Which of the following number pairs extends the pattern on the graph? (TEKS 5.4.C)
(A) (5, 25)
(B) (6, 25)
(C) (25, 5)
(D) (5, 20)
(A) (5, 25),
Explanation:
Number pairs extend the pattern on the graph is as w = 5d after d = 4
the next one is d = 5 so w = 5 x 5 = 25 therefore next number pair
extends the pattern on the graph is (5,25) matches with (A).
Question 9.
Party favors cost $1, plus$5 for shipping. The rule is t = 5 + p,
where t is the total cost in dollars and p is the number of party favors bought.
Which point on the graph shows the total cost of 4 party favors? (TEKS 5.4.C)
(A) Point A
(B) Point B
(C) Point C
(D) Point D
(D) Point D
Explanation:
Given Party favors cost $1, plus$5 for shipping. The rule is t = 5 + p,
where t is the total cost in dollars and p is the number of party favors bought.
the point on the graph shows the total cost of 4 party favors if p = 4
then t = 5 + 4 = 9 so point D, (4,9) matches with (D).
Texas Go Math Grade 5 Answer Key Pdf Module 10 Question 10.
The rule for a pattern is e = 6w. What is the output, e, when the input, w, is 6? (TEKS 5.4.C)
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Common Core: 2nd Grade Math : Add and Subtract Within 100: CCSS.Math.Content.2.NBT.B.5
Example Questions
← Previous 1 3 4 5
Example Question #1 : Add And Subtract Within 100: Ccss.Math.Content.2.Nbt.B.5
Solve:
Explanation:
When we add with multi-digit numbers, we start with the numbers in the ones place, and then move to the left.
Let's add the digits in the ones place:
Because the sum is greater than , we have to carry the to the tens place. So far, your work should look like the following:
Next, we add the numbers in the tens position, including the that was carried over:
Let's add the digits in the tens place:
Example Question #2 : Add And Subtract Within 100: Ccss.Math.Content.2.Nbt.B.5
Solve:
Explanation:
When we add with multi-digit numbers, we start with the numbers in the ones place, and then move to the left.
Let's add the digits in the ones place:
Next, we add the numbers in the tens position.
Let's add the digits in the tens place:
Example Question #3 : Add And Subtract Within 100: Ccss.Math.Content.2.Nbt.B.5
Solve:
Explanation:
When we add with multi-digit numbers, we start with the numbers in the ones place, and then move to the left.
Let's add the digits in the ones place:
Next, we add the numbers in the tens position.
Let's add the digits in the tens place:
Example Question #4 : Add And Subtract Within 100: Ccss.Math.Content.2.Nbt.B.5
Solve:
Explanation:
When we add with multi-digit numbers, we start with the numbers in the ones place, and then move to the left.
Let's add the digits in the ones place:
Because the sum is greater than , we have to carry the to the tens place. So far, your work should look like the following:
Next, we add the numbers in the tens position, including the that was carried over:
Let's add the digits in the tens place:
Example Question #5 : Add And Subtract Within 100: Ccss.Math.Content.2.Nbt.B.5
Solve:
Explanation:
When we add with multi-digit numbers, we start with the numbers in the ones place, and then move to the left.
Let's add the digits in the ones place:
Next, we add the numbers in the tens position.
Let's add the digits in the tens place:
Example Question #6 : Add And Subtract Within 100: Ccss.Math.Content.2.Nbt.B.5
Solve:
Explanation:
When we add with multi-digit numbers, we start with the numbers in the ones place, and then move to the left.
Let's add the digits in the ones place:
Because the sum is greater than , we have to carry the to the tens place. So far, your work should look like the following:
Next, we add the numbers in the tens position, including the that was carried over:
Let's add the digits in the tens place:
Example Question #7 : Add And Subtract Within 100: Ccss.Math.Content.2.Nbt.B.5
Solve:
Explanation:
When we add with multi-digit numbers, we start with the numbers in the ones place, and then move to the left.
Let's add the digits in the ones place:
Because the sum is greater than , we have to carry the to the tens place. So far, your work should look like the following:
Next, we add the numbers in the tens position, including the that was carried over:
Let's add the digits in the tens place:
Example Question #8 : Add And Subtract Within 100: Ccss.Math.Content.2.Nbt.B.5
Solve:
Explanation:
When we add with multi-digit numbers, we start with the numbers in the ones place, and then move to the left.
Let's add the digits in the ones place:
Because the sum is greater than , we have to carry the to the tens place. So far, your work should look like the following:
Next, we add the numbers in the tens position, including the that was carried over:
Let's add the digits in the tens place:
Example Question #9 : Add And Subtract Within 100: Ccss.Math.Content.2.Nbt.B.5
Solve:
Explanation:
When we add with multi-digit numbers, we start with the numbers in the ones place, and then move to the left.
Let's add the digits in the ones place:
Next, we add the numbers in the tens position.
Let's add the digits in the tens place:
Example Question #10 : Add And Subtract Within 100: Ccss.Math.Content.2.Nbt.B.5
Solve:
Explanation:
When we add with multi-digit numbers, we start with the numbers in the ones place, and then move to the left.
Let's add the digits in the ones place:
Because the sum is greater than , we have to carry the to the tens place. So far, your work should look like the following:
Next, we add the numbers in the tens position, including the that was carried over:
Let's add the digits in the tens place:
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# 2.3 Differentiabilityap Calculus
Example 1:
2,3 69, 3 xx fx xx is (A) undefined. (B) continuous but not differentiable. (C) differentiable but not continuous. (D) neither continuous nor differentiable. (E) both continuous and differentiable. (calculator not allowed) f(x) x 2if x 3 4x 7if x 3 Let f be the function given above. Math AP®︎/College Calculus AB Differentiation: definition and basic derivative rules Connecting differentiability and continuity: determining when derivatives do and do not exist Differentiability and continuity.
Show that the function has a solution between 2 and 3.
Solution
:
Plugging in 2 and 3 into f(x), we see that f(2) = ln(2) 1 -0.307 and f(3) = ln(3) 1 0.099. Since the first number is negative and the second number is positive and f(x) is a continuous function on the interval [2, 3], by the Intermediate Value Theorem, f(x) must have a solution between 2 and 3.
Example 2:
Show that the function has only one real solution.
Solution:
### 2.3 Differentiabilityap Calculus Test
First we use the Intermediate Value Theorem to show that there is at least one solution. We can use the theorem since f(x) is a continuous function everywhere. Notice that f(1) = -1 and f(2) = 45. That means that somewhere between 1 and 2, f(x) = 0.
Well, we have shown that there is at least one solution to the equation. Now we have to show that it is the only solution. To do that, we shall show that it is not possible for f(x) to have a second solution.
We do this by taking the derivative. . Notice that . That means that f(x) is a strictly increasing function. A strictly increasing function will only have one x-intercept (solution), and thus f(x) has only one real solution.
Problems For You To Solve
Multiple Choices:
1. In what situation f'(x) might NOT fail to exist？
A. cuspe
B. conner
C. jump
D. continuous
E. IDK
2. Let f be a continuous function on the closed interval [-3.6]. If f (-3) =- 1 and f (6) = 3, then the Intermediate Value Theorem guarantees that
A. f(0)=0
B. for at least one c between -3 and 6
C. for all x between -3 and 6
D. f(c)=1 for at least one c between -3 and 6
E. f(c)=0 for at least one c between -1 and 3
3. The function f is continuous on the closed interval [0, 2] and has values that are given in the table above. The equation must have at least two solutions in the interval [0, 2] if k=
A. 0
B. 1
C. 2
D. 3
E. 4
4. Let f be a function that is differentiable on the open interval (1, 10). If f(2) = -5, f(5) = 5, and f(9) = -5, which of the following must be true?
I. f has at least 2 zeros.
II. The graph of f has at least one horizontal tangent.
III. For some c, 2 < c < 5 , f(c)=3.
A. None
B. I only
C. I and II only
D. I and III only
E. I, II and III
5. The function f is continuous for and differentiable for -2 < x < 1. If f(-2) =-5 and f(l) = 4, which of the following statements could be false?
A. There exists c, where -2 < c < 1, such that f (c) = 0.
B. There exists c, where -2 < c < 1, such that f '(c) = 0.
C. There exists c, where -2 < c < 1, such that f(c) =3.
D. There exists c, where -2 < c < 1, such that f '(c) = 3.
6. Let f(x) = x x and g(x) =sin x. Assertion : gof is differentiable at x = 0 and its derivative is continuous at that point Reason : gof is twice differentiable at x = 0.
C. Both assertion and reason are true but reason is not the correct explanation of assertion
7. Function f(x) =Ixl + x-1 is not differentiable at
B. x=0,1
D. x=1,2
8. If f(x) = x , then f'(0) =
B. X
D. None of these
9.
10.
11. Let f be a differentiable odd function defined on R. (That is f(-x) = -f(x) for all x in R.) Let a be a positive number. How many solution(s) could the equation af'(x) = f(a) have?
A. 0
C. 2
12. Let f be a quadratic function defined on the interval [a,b] with 0 < a < b. Which one of the followings is the value of c in (a,b) such that f(b)-f(a) = f'(c)(b-a) ?
A. (a+b)/2
13. The Mean Value Theorem is applied to the function f(x) = x3 + qx2 + 5x - 6 on the interval [0,2]. Suppose that the number c determined by the theorem is equal to 2. Which one of the followings is the value of q ?
### 2.3 Differentiabilityap Calculus Notes
D. -4
14. Which of the following functions satisfy the conditions of the Mean value Theorem on their domains ?
(I) f(x) = x¾ for all x in [-1,1]. (II) g(x) = x-1 for all x in [-1,1]. (III) h(x) = x/(1-x2) for all x in (-1,1). (IV) k(x) = 1- x 3 for all x in [-1,½].
B. I and IV
D. II and IV
15. Let g(x) be a differentiable function defined on R and f(x) = g(x)sinx. How many real solutions does the equation g(x)cosx + g'(x)sinx = 0 have ?
Free Response:
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# What Is 4/75 as a Decimal + Solution With Free Steps
The fraction 4/75 as a decimal is equal to 0.053.
The decimal form and the fractional form are interchangeable. We know that Division is one of the four primary operators of mathematics therefore by applying the long division method the fraction can be converted to its equivalent decimal. When the long division is performed on the fraction 4/75 it results in its equivalent recurring decimal.
Here, we are more interested in the division types that result in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers.
Now, we introduce the method used to solve said fraction to decimal conversion, called Long Division, which we will discuss in detail moving forward. So, let’s go through the Solution of fraction 4/75.
## Solution
First, we convert the fraction components, i.e., the numerator and the denominator, and transform them into the division constituents, i.e., the Dividend and the Divisor, respectively.
This can be done as follows:
Dividend = 4
Divisor = 75
Now, we introduce the most important quantity in our division process: the Quotient. The value represents the Solution to our division and can be expressed as having the following relationship with the Division constituents:
Quotient = Dividend $\div$ Divisor = 4 $\div$ 75
This is when we go through the Long Division solution to our problem. The solution for fraction 4/75 is given in the figure below.
Figure 1
## 4/75 Long Division Method
We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 4 and 75, we can see how 4 is Smaller than 75, and to solve this division, we require that 4 be Bigger than 75.
This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If so, we calculate the Multiple of the divisor closest to the dividend and subtract it from the Dividend. This produces the Remainder, which we then use as the dividend later.
Since 4 when multiplied by 10 becomes 40 which is still smaller than 75. Therefore we will multiply 40 by 10 again and add a zero in the quotient after the decimal point. By doing this the dividend will become 400 which is bigger than 75 and hence divisible by 75.
Now, we begin solving for our dividend 400.
We take this 400 and divide it by 75; this can be done as follows:
400 $\div$ 75 $\approx$ 5
Where:
75 x 5 = 375
This will lead to the generation of a Remainder equal to 400 – 375 = 25. Now this means we have to repeat the process by Converting the 25 into 250 and solving for that:
250 $\div$ 75 $\approx$ 3
Where:
75 x 3 = 225
Finally, we have a Quotient generated after combining the three pieces of it as 0.053, with a Remainder equal to 25.
Images/mathematical drawings are created with GeoGebra.
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# Kerala Syllabus 9th Standard Maths Solutions Chapter 3 Pairs of Equations
## Kerala State Syllabus 9th Standard Maths Solutions Chapter 3 Pairs of Equations
### Kerala Syllabus 9th Standard Maths Pairs of Equations Text Book Questions and Answers
Textbook Page No. 36
Do each problem below either in your head, or using an equation with one letter, or two equations with two letters:
Pairs Of Equations Class 9 Questions And Answers Kerala Syllabus Question 1.
In a rectangle of perimeter one metre, one side is five centimetres longer than the other. What are the lengths of the sides?
Shortest side = x
Longest side = x + 5
Perimeter = 1 m = 100 cm
2(x + x + 5) = 100
2x + 5 = 50; 2x = 45;
x = 22.5
∴ Shortest side = 22.5
Longest side = 22.5 + 5 = 27.5
Pairs Of Equations Questions And Answers Kerala Syllabus Question 2.
A class has 4 more girls than boys. On a day when only 8 boys were absent, the number of girls was twice that of the boys. How many girls and boys are there in the class?
Number of boys = x
Number of girls = x + 4
2(x – 8) = x + 4
2x – 16 = x + 4
2x – x = 4 +16; x = 20
∴ Number of boys = 20
Number of girls = 24
Pairs Of Equations Problems Kerala Syllabus Question 3.
A man invested 10000 rupees, split into two schemes, at annual rates of interest 8% and 9%. After one year he got 875 rupees as interest from both. How much did he invest in each?
If one part is x then
the remaining part is 10000 – x
$$x\times \frac { 8 }{ 100 } +\left( 10000-x \right) \times \frac { 9 }{ 100 } =100$$
8x + 90000 – 9x = 87500
90000 – 87500 = x
2500 = x
one part = 2500 and
remaining part = 7500
Kerala Syllabus 9th Standard Maths Chapter 3 Question 4.
A three and a half metre long rod is to be cut into two pieces, one piece is to be bent into a square and the other into an equilateral triangle. The length of their sides must be the same. How should it be cut?
Total length = 3½ m
Since the sides of a square and equilateral triangle are equal, all these 7 sides are equal.
∴ Length of one side
$$3\frac { 1 }{ 2 } \div 7=\frac { 7 }{ 2 } \div 7=\frac { 1 }{ 2 }$$m
Length of the rod for the square
$$= 4\times \frac { 1 }{ 2 }$$ = 2m
Length of the rod for the equilateral triangle = $$3\times \frac { 1 }{ 2 }$$ = $$1\frac { 1 }{ 2 }$$m
Class 9 Maths Chapter 3 Kerala Syllabus Question 5.
The distance travelled in t seconds by an object starting with a speed of u metres/second and moving along a straight line with speed increasing at the rate of a metres/second every second is given by ut + $$\frac { 1 }{ 2 }$$ at² metres. An object moving in this manner travels 10 metres in 2 seconds and 28 metres in 4 seconds. With what speed did it start? At what rate does its speed change?
If t = 2
ut + $$\frac { 1 }{ 2 }$$at²= 10
2u + 2a= 10
u + a = 5 — (1)
If t = 4
4u + 8a = 28
u + 2a = 7 — (2)
from (1) and (2)
a = 2
∴ u = 3
Textbook Page No. 40
Pairs Of Equations Class 9 Questions And Answers Pdf Question 1.
Raju bought seven notebooks of two hundred pages and five of hundred pages, for 107 rupees. Joseph bought five notebooks of two hundred pages and seven of hundred pages, for 97 rupees. What is the price of each kind of notebook?
Cost of 200 page note book = x
Cost of 100 page note book = y
7x + 5y= 107 …………(1)
5x + 7y = 97 …………(2)
(1) × 5 $$\Rightarrow$$ 35x + 25y = 535 …………(3)
(2) × 7 $$\Rightarrow$$ 35x + 49y = 679 …………(4)
(4) – (3) $$\Rightarrow$$ 24y = 144
y = $$\frac{144}{24}$$ = 6
Substitute y = 6 in equation (1)
7x + 30 = 107; 7x = 77
x = $$\frac{77}{7}$$ = 11
Price of the 200 pages notebook = Rs. 11
Price of the 100 pages notebook = Rs. 6
Pairs Of Equations Class 9 Extra Questions And Answers Question 2.
Four times a number and three times number added together make 43. Two times the second number, subtracted from three times the first give 11. What are the numbers?
Let the first number = x and
the second number = y
4x + 3y = 43 …………(1)
3x – 2y = 11 …………(2)
(1) × 3 $$\Rightarrow$$ 12x + 9y= 129 …………(3)
(2) × 4 $$\Rightarrow$$ 12x – 8y = 44 …………(4)
(3) -(4) $$\Rightarrow$$ 17y = 85; y = $$\frac{85}{17}$$ = 5
Substitute y = 5 in equation (1)
4x + 3y = 43
4x + 15 = 43
4x = 43 – 15 = 28
∴ x = $$\frac{28}{4}$$ = 7, y = 5
First number = 7
Second number = 5
9th Standard Maths Chapter 3 Kerala Syllabus Question 3.
The sum of the digits of two – digit number is 11. The number got by interchanging the digits is 27 more than the original number. What is the number?
If the numbers are x and y
x + y = 11 …………(1)
10x + y + 27 = 10y + x
9x – 9y = -27
X – y = -3 …………(2)
(1) + (2) 2x = 8; x = 4
x + y = 11
4 + y = 11
y = 7
∴ Required number is 47.
Kerala Syllabus 9th Standard Maths Notes Question 4.
Four years ago, Rahim’s age was three times Ramu’s age. After two years, it would just be double. What are their ages now?
Ramu’s present age = x
Rahim’s present age = y
4 years back,
Ramu’s age = x – 4
Rahim’s age = y – 4
3(x – 4) = y – 4
3x – 12 = y – 4
3x – y = 8 ……….(1)
After 2 years,
Ramu’s age = x + 2
Rahim’s age = y + 2
2(x + 2) = y + 2
2x + 4 = y + 2
2x – y = -2 ……….(2)
(1) – (2) $$\Rightarrow$$ x = 10
3x – y = 8; 30 – y = 8; y = 22
x = 10, y = 22
Ramu’s present age = 10
Rahim’s present age = 22
Pair Of Equations Class 9 Kerala Syllabus Question 5.
If the length of a rectangle is in-creased by 5 metres and breadth decreased by 3 metres, the area would decrease by 5 square metres. If the length is increased by 3 metres and breadth increased by 2 metres, the area would increase by 50 square metres. What are the length and breadth?
length = x; breadth = y
(x + 5)(y – 3) = xy – 5
xy – 3x + 5y – 15 = xy – 5
– 3x + 5y = + 10
3x – 5y = -10 ………..(1)
(x + 3)(y + 2) = xy + 50
xy + 2x + 3y + 6 = xy + 50
2x + 3y = 44 ………..(2)
(2) × 1 $$\Rightarrow$$ 6x-10y = -20 ……….(3)
(3) × 2 $$\Rightarrow$$ 6x + 9y = 132 …………(4)
(3)- (4) $$\Rightarrow$$ -19y = -152
y = $$\frac{-152}{-19}$$ = 8, 2x + 3y = 44
2x + 24 = 44; 2x = 20; x = 10
∴ x = 10, y = 8
Length of the rectangle = 10 m
Breadth of the rectangle = 8m
Textbook Page No. 42
Hsslive Guru Maths 9th Kerala Syllabus Question 1.
A 10 metre long rope is to be cut into two pieces and a square is to be made using each. The difference in the areas enclosed must be 1$$\frac{1}{4}$$ square metres. How should it be cut?
Length of one piece = x m
Length of other piece = (10 – x) m
∴ Rope is divided into 6 m and 4 m.
9th Maths Notes Kerala Syllabus Question 2.
The length of a rectangle is 1 metre more than its breadth. Its area is 3$$\frac{3}{4}$$ square metres. What are its length and breadth?
Length = x
x = y + 1; x – y = 1
$$x y = 3 \frac{3}{4} =\frac{15}{4}$$
(x + y)² = (x – y)² + 4xy
1² + 4 × $$\frac{15}{4}$$ = 1 + 15 = 16
x – y = 1; x + y = 4
2x = 5; x = 5/2 = 2.5
y=1.5
∴ Length = 2.5 m
Hsslive Maths Class 9 Kerala Syllabus Question 3.
The hypotenuse of a right triangle is 6$$\frac{1}{2}$$ centimetres and its area is 7$$\frac{1}{2}$$ square centimetres. Calculate the lengths of its perpendicular sides.
The perpendicular sides are x and y. Given that,
From (3) and (4)
2x = 24/2 = 12
∴ x = 6
6 – y = 7/2
∴ y = 5/2 = 2.5
∴ Perpendicular sides = 6 and 2.5
### Kerala Syllabus 9th Standard Maths Pairs of Equations Exam Oriented Text Book Questions and Answers
Kerala Syllabus 9th Standard Notes Maths Question 1.
There are some oranges in a bag. When 10 oranges more added in the bag; the numbers become 3 times of the oranges initially had. Then how many oranges were there in the bag initially.
Let the number of oranges initially taken = x
X + …….. = 3X
3X – X = ……..
2X = ……..
X = ……… /2 = ……..
x + 10 = 3x; 3x – x = 10 ; 2x = 10;
x = $$\frac{10}{2}$$ = 5
Kerala Syllabus 9th Standard Maths Notes Pdf Question 2.
A box contains some white balls and some black balls. The number of black balls is 8 more than the number of white balls. The total number of balls is 4 times the number of white balls. Find out the number of white balls and the number of black balls.
Number of white balls = x
Number of black balls = ……… + 8
Total number of balls = ……. ‘ x;
i. e. (x) + (x + 8) = ………. x;
2x + 8 = ……. x;
8 = …… x – 2x
= …… x; x = 8/ …….
white balls = ……….
black balls = ……… + 8 = ……..
Number of white balls = x
Number of black balls = x + 8
Total number of balls = 4 ‘ x;
(x) + (x + 8) = 4x ; 2x + 8 = 4x;
8 = 4x – 2x = 2x ; x = $$\frac{8}{2}$$ = 4
white balls = 4
black balls = 4 + 8 = 12
Pairs Of Equations Questions Kerala Syllabus Question 3.
The sum of two numbers is 36 and the difference is 8. Find the numbers.
Let x, y be the numbers
x + y = 36
x – y = 8
(x + y) + (x – y) = ……. + ……
2x = …..
x = …… /2 = ……..
x – y = 8
……. – y = 8
…….. – 8 = y
x + y = 36; x – y = 8
(x + y) + (x – y) = 36 + 8
2x = 44; x = $$\frac{44}{2}$$ = 22
x – y = 8; 22 – y = 8;
22 – 8 = y; y = 14
numbers 22, 14
Kerala Syllabus 9th Std Maths Notes Question 4.
The cost of 2 pencils and 5 pens is Rs 17, two pencils and 3 pens of the same rate is Rs 11. Find out the prices of a pencil and a pen.
Let the price of pencil = x;
Price of pen = y;
∴ 2x + 5y = …….. 2x +….. y = …..
2x + …… y = 11
(2x + 5y) (-2x + ….. y) = -11
…… y = …….
y = $$\frac{…..}{……}$$
2x + 5x ….. = 17;
2x = 17 – ……..; x = ……. /2; = ……..
2x + 5y = 17; 2x + 3y = 11;
(2x + 5y) – (2x + 3y) = 17 – 11; 2y = 6
y = $$\frac{6}{2}$$ = 3; 2x + 5 × 3 = 17;
2x = 17 – 15;
x = $$\frac{2}{2}$$ = 1
Price of pencil = Rs 1
Price of pen = Rs 3
Question 5.
Twice of a number added with thrice of another number gives 23. Four times the first number and 5 times the second number when added gives 41. Find out the numbers.
First number = x
Second number = y
∴ 2x + 3y = ……
4x + 5y = ……
2(2x + 3y) = 2 …….
4x + 6y = …….
(4x + 6y) – (4x + 5y) = ( …… ) – ( ….. )
y = …….; 2x + 3x ……. = …….
2x = ( …… ) – ( ……. ); x = ………./2
2x + 3 y = 23
4x + 5y = 41
2(2x + 3y) = 2 × 23; 4x + 6y = 46
(4x + 6y) – (4x + 5y) = 46 – 41;
y = 5
2x + 3 × 5 = 23
2x = 23 – 15;
x = $$\frac{8}{2}$$ = 4
Price of pencil = Rs 4
Price of pen = Rs 5
Question 6.
Rama spends Rs 97 to buy 4 two hundred page note books and 5 hundred page note books. Geetha spends Rs 101 to buy 5 two hundred page note book and 4 one hundred page note books. What is the prices of two types of note books?
Let the cost of two hundred page note books = x
The cost of one hundred page note books = y
(1) 4x + 5y = 97
(1) × 5 $$\Rightarrow$$ 20x + ……..y = ……..
(2) 5x + 4y = 101
(2) × 4 $$\Rightarrow$$ 20x + …….y = …….
……y = ( ….. ) – ( ……. );
y = …….. /……..
4x + 5x( ……. ) = 97
4x = 97 – ( ……. )
x = ……../4
(1) 4x + 5y = 97
(2) 5x + 4y = 101
(1) × 5 → 20x + 25y = 485
(2) × 4 → 20x + 16y = 404
9y = 485 – 404
y = $$\frac{81}{9}$$ = 9
4x + 5 × 9 = 97
4x = 97 – 45 = 52
Cost of two hundred page note book = Rs 13
Cost of one hundred page note book = Rs 9
Question 7.
6 years back the age of Muneer was 3 times the age of Mujeeb. After 4 years the age of Muneer becomes twice the age of Mujeeb. Find the age of two of them now.
Age of Mujeeb 6 years back = x
Age of Muneer 6 years back = 3x
After 4 years
3x + 4 = 2(x + 4)
3x + 4 = 2x + 8
3x – 2x = 8 – 4;
x = 4
Age of Mujeeb 6 years back = 4 + 6 = 10
Age of Muneer 6 years back = 3(4 + 6) = 18 years.
Question 8.
The cost of 4 chairs and 5 tables is Rs 6600 and the cost of 5 chairs and 3 tables is Rs 5000 at the same prices. What are the prices of a table and a chair?
Cost of a chair = Rs a
Cost of a table = Rs b
4a + 5b = 6600 ¾ …………(1)
5a + 3b = 5000 ¾ …………(2)
(1) × 5 → 20a + 25b = 33000
(2) × 4 → 20a + 12b = 20000
(20a + 25b) – (20a + 12b) = 33000 – 20000
(1) $$\Rightarrow$$ 13b = 13000
b = $$\frac{13000}{13}$$ = Rs 1000
4a + 5b = 6600;
4a + 5 × 1000 = 6600
4a = 6600 – 5000 = 1600
a = $$\frac{1600}{4}$$ = Rs 400
Cost of a table = Rs 1000
Cost of a chair = Rs 400
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# How do you express sin(pi/ 8 ) * cos(( ( 2 pi) / 3 ) without using products of trigonometric functions?
Oct 31, 2016
$\frac{\sqrt{2 - \sqrt{2}}}{2}$
#### Explanation:
$P = \sin \left(\frac{\pi}{8}\right) . \cos \left(\frac{2 \pi}{3}\right)$
Trig table --> $\cos \left(\frac{2 \pi}{3}\right) = - \frac{1}{2}$, then P can be expressed as:
$P = - \left(\frac{1}{2}\right) \sin \left(\frac{\pi}{8}\right)$
We can evaluate the value of sin (pi)/8 by the trig identity:
$2 {\sin}^{2} \left(\frac{\pi}{8}\right) = 1 - \cos \left(\frac{2 \pi}{8}\right) = 1 - \cos \left(\frac{\pi}{4}\right) = 1 - \frac{\sqrt{2}}{2}$
${\sin}^{2} \left(\frac{\pi}{8}\right) = \frac{2 - \sqrt{2}}{4}$
$\sin \left(\frac{\pi}{8}\right) = \frac{\sqrt{2 - \sqrt{2}}}{2}$
Only the positive value is accepted because $\sin \left(\frac{\pi}{8}\right)$ is positive
Finally,
$P = - \left(\frac{1}{2}\right) \sin \left(\frac{\pi}{8}\right) = - \frac{\sqrt{2 - \sqrt{2}}}{4}$
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## DEV Community is a community of 867,901 amazing developers
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Posted on • Updated on
# Recursion Explained (with Examples)
“To understand recursion, one must first understand recursion” - Unknown
Recursion is a method of solving problems where you solve smaller portions of the problem until you solve the original, larger problem. A method or function is recursive if it can call itself.
``````function understandRecursion(doIUnderstandRecursion) {
const recursionAnswer = confirm('Do you understand recursion?');
if(recursionAnswer === true) { // base case
return true;
}
}
``````
For the example above, notice the base case and recursive call which make this a recursive algorithm. Recursive functions must have a base case, or a condition in which no recursive call is made. I think the best way to understand recursion is to look at examples so let’s walk through two common recursive problems.
## Example 1: Calculating the Factorial of a Number
Calculating the factorial of a number is a common problem that can be solved recursively. As a reminder, a factorial of a number, n, is defined by n! and is the result of multiplying the numbers 1 to n. So, `5!` is equal to `5*4*3*2*1`, resulting in `120`.
Let’s first take a look at an iterative solution:
``````function factorial(num) {
let total = 1;
for(let n = num; n > 1; n--) {
total *= n;
}
}
``````
The iterative solution above is fine but let’s try rewriting it using recursion. When we think about solving this problem recursively, we need to figure out what our subproblems will be. Let’s break it down:
1. We know `factorial(5) = 5 * factorial(4)` aka `5! = 5 * 4!`.
2. To continue, `factorial(5) = 5 * (4 * factorial(3))` which equals `5 * (4 * (3 * factorial(2))` and so on…
3. ...Until you get `5 * 4 * 3 * 2 * 1` and the only remaining subproblem is `1!`.
4. `factorial(1)` and `factorial(0)` always equals 1 so this will be our base case.
Using this line of thinking, we can write a recursive solution to our factorial problem:
``````function factorial(n) {
if(n === 1 || n === 0) { // base case
return 1;
}
return n * factorial(n - 1); // recursive call
}
``````
## Example 2: Fibonacci Sequence
Another fun problem that can be solved using recursion is the Fibonacci sequence problem. As a reminder, the Fibonacci sequence is a series of numbers: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, and so on. The pattern involves totaling the two previous numbers so 0 + 1 = 1, 1 + 1 = 2, 1 + 2 = 3, 2 + 3 = 5, etc. In other words, the Fibonacci number at position `n` (for `n > 2`) is the Fibonacci of `(n - 1)` plus the Fibonacci of `(n - 2)`.
Again, I think it’s helpful to see an iterative solution first:
``````function fibonacci(n) {
if(n === 0) return 0;
if(n === 1) return 1;
let fibNMinus2 = 0;
let finNMinus1 = 1;
let fibN = n;
for(let i = 2; i <= n; i++) { // n >= 2
fibN = fibNMinus1 + fibNMinus2; // f(n-1) + f(n-2)
fibNMinus2 = fibNMinus1;
fibNMinus1 = fibN;
}
return fibN;
}
``````
As you’ll see, the recursive solution looks much simpler:
``````function fibonacci(n) {
if(n === 0) return 0; // base case 1
if(n === 1) return 1; // base case 2
return fibonacci(n - 1) + fibonacci(n - 2); // recursive call
}
``````
If you were to call fibonacci(5), the following represents the calls that would be made:
### Fibonacci with Memoization
I wanted to take this opportunity to mention another approach to this problem, called memoization. Memoization consists of an optimization technique that stores the values of the previous results, similar to a cache, making our recursive solution faster. If you look back at the calls made to compute `fibonacci(5)` in the image above, you can see that `fibonacci(3)` was computed twice, so we can store its result so that when we compute it again, we already have it.
Take a look at how our `fibonacci` solution changes when we add memoization:
``````function fibonacci(n) {
const memo = [0, 1]; // cache all computed results here
const fib = (n) => {
if(memo[n] != null) return memo[n]; // base case
return memo[n] = fib(n - 1, memo) + fib(n - 2, memo); // recursive call
};
return fib(n);
}
``````
## Why Use Recursion?
To be completely frank, a recursive solution is almost always slower than an iterative one. That being said, if you look back at our Fibonacci solutions, the recursive solution is much easier to read plus memoization can help bridge the gap in speed. Recursion is generally easier to understand and usually requires less code.
## Conclusion
Now that we’ve gone over some examples, I hope recursion is a little easier for you to grasp and that you can see why we would use it. In a future post, I plan to take a look at the tree data structure which uses recursion in many of its methods so stay tuned! This article only scratches the surface of recursion’s potential so here are a few resources you might find helpful if you want to continue your studies.
## Discussion (19)
Aaron McCollum
Thanks Christina! This helps my understanding of recursion. I'm going through a JS course right now and just came across this for the first time and was like "what?"
Christina
It makes me so glad to hear that this helped, best of luck with your studies! I'm curious which JS course you're taking...
Aaron McCollum
The JS intro course on freeCodeCamp. I just discovered CodeWars as well, which I'll throw in there to break things up. Near the end of the JavaScript Introduction part of the course and just got into recursion. I have not figured out the solution yet but your article has helped.
Christina
Hasn't heard of CodeWars before but I started programming on freeCodeCamp so I'm a big fan of their courses! Good luck!
Aaron McCollum
Thanks! Have you made it through their entire course offering? (Before Python's update I mean)
Christina
Not yet, only the responsive web design and JS algorithms ones so far. I've taken several other courses through Coursera, Scrimba, and Codecademy though. There are so many excellent, free resources or there!
Aaron McCollum
True that! I've done free versions of Codecademy before - that was my first HTML course almost a year ago today. Ah the memories.
David • Edited on
Welcome to the world of programming, Aaron -- I used to hate recursion because it was not intuitive to me at all no matter who explained it. When I learned it there weren't such abundant resources. It took me years to come back around to learning it (all the while hoping no one around me would find out that I couldn't do it)
Here are a few things that may help with some lingering questions (e.g. when to use it?). Whenever you've a problem and the only thing that comes to mind as the answer the problem is to "enumerate" or "try all possibilities" or an "exhaustive search" is usually a signal for recursion, because the recursive parts are probably repetitive...
One thing that I was aware of, but wish someone had really forced me to understand was thinking about how to do recursion in terms of a decision tree aka recursive tree -- "in my current position, what are my options?"
With Fibonacci, we have two branches (i.e. two decisions) at EVERY (node) call with one exception (base case) when the value that comes into our function is `0` or `1`
Drawing this out by hand is an exercise that has made it click for me. While I can mostly do this in my head now or in a comment in the code, if I can't determine what my decisions and exceptions are, then I cannot write the code
Aaron McCollum
Thanks David! I appreciate this. 😊
Joshua Baker
Also, for some algorithms, an iterative solution may not be an option.
Actually, that's false. Anything that can be implemented with recursion can also be implemented with iteration. After all, the CPU doesn't know recursion -- it only knows jump instructions. So every recursive function in a high-level language eventually gets translated into an iterative subroutine.
The following booklet contains a few problems solved with both recursion and also iterative approach:
On the same site you can also explore the following two playgrounds with problems solved with both recursion and iterative approach:
Flood fill
codeguppy.com/code.html?t=flood_fill
Find element in a hierarchical structure
codeguppy.com/code.html?t=find_max
Christina
Thanks for pointing that out and for the explanation, totally makes sense! I've gone ahead and taken it out :)
David • Edited on
In fact, what Josh notes is actually what's done in practice most of the time (when possible) and what you've done turning your recursive solution into a "memoized" version. For big company interviews, these types of questions are becoming common-place -- where recursion (already a challenge for some) is the brute force method, but with a large enough input would result in an overflow of the call stack. This optimization that you've done is often known as "dynamic programming" and while Fib is the classic example, there are certainly some problems that require some exceptional thought to convert from standard recursion (much more elegant IMO) to one that's iterative (DP) and you can certainly find tons on Leetcode. I wouldn't worry too much about it unless you're super curious -- took me forever to learn.
There's also one "in-between"/hybrid memoized version -- in "dynamic programming" there are typically two approaches as well: 1) top-down, 2) bottom-up. I wrote a gist (in JS) for a friend trying to understand it a couple months back: the `fibIterative` is the top-down approach, and the `fibOptimal` is similar to yours but instead of two well-named variables that you have, I've stuck them in a two-element array/tuple: gist.github.com/wulymammoth/bdb5e3...
James Carter
It's good to know you can do all recusive algorithms iteratively also, I had always wondered.
Your example above does not iterate unless you change n > 1 not n < 1, it had me going for a bit you and should always test your code ;-)
function factorial(num) {
let total = 1;
for(let n = num; n > 1; n--) {
total *= n;
}
}
Christina
Oops, thanks for pointing that typo out, I've made the change in the post.
Shannon Crabill
I was dreading learning recursion, but this write up helped for it to click.
Thank you!
Christina
That makes me so happy to hear since I've been in your shoes too!
Athul Muralidhar
i absolutely love recursion! it could be a bit tricky to wrap your head around, but once you get the gist of it, things just are more simpler and elegant
Christina
Yes! I love how recursion appears in so many walks of life: programming, math, art, nature, etc. MC Escher is one of my favorite artists due to his use of recursion!
|
# 2016 AMC 12A Problems/Problem 23
## Problem
Three numbers in the interval $\left[0,1\right]$ are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?
$\textbf{(A) }\frac16\qquad\textbf{(B) }\frac13\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac23\qquad\textbf{(E) }\frac56$
## Solution 1: Super WLOG
Because we can let the the sides of the triangle be any variable we want, to make it easier for us when solving, let’s let the side lengths be $x,y,$ and $a$. WLOG assume $a$ is the largest. Then, $x+y>a$, meaning the solution is $\boxed{\textbf{(C)}\;1/2}$, as shown in the graph below. $[asy] pair A = (0,0); pair B = (1,0); pair C = (1,1); pair D = (0,1); pair E = (0,0); draw(A--B--C--D--cycle); draw(B--D,dashed); fill(B--D--C--cycle,gray); label("0",A,SW); label("a",B,S); label("a",D,W); label("y",(0,.5),W); label("x",(.5,0),S); label("x+y>a",(5/7,5/7)); [/asy]$
## Solution 2: Conditional Probability
WLOG, let the largest of the three numbers drawn be $a>0$. Then the other two numbers are drawn uniformly and independently from the interval $[0,a]$. The probability that their sum is greater than $a$ is $\boxed{\textbf{(C)}\;1/2.}$
## Solution 3: Calculus
When $a>b$, consider two cases:
1) $0, then $\int_{0}^{\frac{1}{2}} \int_{0}^{a}2b \,\text{d}b\,\text{d}a=\frac{1}{24}$
2)$\frac{1}{2}, then $\int_{\frac{1}{2}}^{1} \left(\int_{0}^{1-a}2b \,\text{d}b + \int_{1-a}^{a}1+b-a \,db\right)\text{d}a=\frac{5}{24}$
$a is the same. Thus the answer is $\frac{1}{2}$.
## Solution 4: Geometry
The probability of this occurring is the volume of the corresponding region within a $1 \times 1 \times 1$ cube, where each point $(x,y,z)$ corresponds to a choice of values for each of $x, y,$ and $z$. The region where, WLOG, side $z$ is too long, $z\geq x+y$, is a pyramid with a base of area $\frac{1}{2}$ and height $1$, so its volume is $\frac{\frac{1}{2}\cdot 1}{3}=\frac{1}{6}$. Accounting for the corresponding cases in $x$ and $y$ multiplies our answer by $3$, so we have excluded a total volume of $\frac{1}{2}$ from the space of possible probabilities. Subtracting this from $1$ leaves us with a final answer of $\frac{1}{2}$.
## Solution 5: More Calculus
The probability of this occurring is the volume of the corresponding region within a $1 \times 1 \times 1$ cube, where each point $(x,y,z)$ corresponds to a choice of values for each of $x, y,$ and $z$. We take a horizontal cross section of the cube, essentially picking a value for z. The area where the triangle inequality will not hold is when $x + y < z$, which has area $\frac{z^2}{2}$ or when $x+z or $y+z, which have an area of $\frac{(1-z)^2}{2}+\frac{(1-z)^2}{2} = (1-z)^2.$ Integrating this expression from 0 to 1 in the form
$\int_0^1 \left(\frac{z^2}{2} + (1-z)^2\right) dz = \bigg[\frac{z^3}{2} - z^2 + z \biggr |_0^1 = \frac{1}{2} -1 + 1 = \frac{1}{2}$
### Solution 5.1 (better explanation)
This problem is going to require some geometric probability, so let's dive right in. Take three integers $x$, $y$, and $z$. Then for the triangle inequality to hold, the following $3$ inequalities must be true
$$x + y > z$$ $$y + z > x$$ $$x + z > y$$
Now, it would be really easy if these equations only had two variables instead of $3$, because then we could graph it in a $2$-dimensional plane instead of a $3$-dimensional cube. So, we assume $z$ is a constant. We will deal with it later.
Now, since we are graphing, we should probably write these equations in terms of $y$ so they are in slope-intercept form and are easier to graph.
$$y > -x + z$$ $$y > x - z$$ $$y < x + z$$
Now, note that all solutions $(x,y)$ are in a $1$ $x$ $1$ square in the $xy$-plane because $x$, $y \in [0, 1]$.
I recommend drawing the following figure to get an idea of what is going on. The first line is a line with a negative slope that cuts off a $45-45-90$ triangle with side length $z$ of the bottom left corner of the square. The second line is a line with a positive slope that cuts off a $45-45-90$ triangle with side length $1-z$ off the top left corner of the square. The third line also has a positive slope and cuts a $45-45-90$ triangle with side length $1 - z$ off the bottom right corner of the square.
Note: All triangles are $45-45-90$ because the lines have slopes of $1$ or $-1$.
Using the $>$ and $<$ signs in the lines, we see that the area that satisfies all three inequalities is the area not enclosed in the three triangles. So, our plan of attack will be to
Find the area of the triangles -> Subtract that from the area of the square -> Use probability to get the answer.
Except, now, we have one problem. $z$ is still a variable. But, we want $z$ to be a constant. Well, what if we just took the area over every possible value of $z$? Well, that would be a bit hard, if not impossible to do by hand, but there is a handy math tool that will let us do that: the integral!
To find the area of the triangles, our plan of attack will be
Find the area in terms of $z$ -> take the integral from $0$ to $1$ of the expression for the area (this will cover every possible value of $z$
The area of the triangles is
$$\frac{z^2}{2} + (1-z)^2 = \frac{1}{2}\left(-3x^2 + 4x\right)$$.
The integral from $0$ to $1$ is
$$\frac{1}{2}\int_0^1\left(-3x^2 + 4x\right)dx = \frac{1}{2}$$
The total area of all the possible unit squares is quite obviously
$$\int_0^11dx = 1$$
Thus, the area not enclosed by the triangles is $1 - \dfrac{1}{2} = \dfrac{1}{2}$, and the total area of the square is $1$. Thus, the desired probability is
$$\frac{\frac{1}{2}}{1} = \boxed{\textbf{(C) }\frac12}$$
~Extremelysupercooldude
## Solution 6: Geometry in 2-D
WLOG assume that $z$ is the largest number and hence the largest side. Then $x,y \leq z$. We can set up a square that is $z$ by $z$ in the $xy$ plane. We are wanting all the points within this square that satisfy $x+y > z$. This happens to be a line dividing the square into 2 equal regions. Thus the answer is $\frac{1}{2}$.
[][] diagram for this problem goes here (z by z square)
## Solution 7: More WLOG, Complementary Probability
The triangle inequality simplifies to considering only one case: $\text{the smallest side+ the second smallest side} > \text{the largest side}$. Consider the complement (the same statement, except with a less than or equal to). Assume (WLOG) $a$ is the largest, so on average $a=1/2$ (now equal to becomes a degenerate case with probability $0$, so we no longer need to consider it). We now want $b+c<1/2$, so imagine choosing $b+c$ at once rather than independently. But we know that $b+c$ is between $0$ and $2$. The complement is thus: $(1/2-0)/2=1/4$. But keep in mind that we choose each $b$ and $c$ randomly and independently, so if there are $k$ ways to choose $b+c$ together, there are $2k$ ways to choose them separately, and therefore the complement actually doubles to match each case (a good example of this is to restrict b and c to integers such that if $b+c=3$, then we only count this once, but in reality: we have two cases $1+2$, and $2+1$; similar reasoning also generalizes to non-integral values). The complement is then actually $2(1/4)=1/2$. Therefore, our desired probability is given by $1-\text{complement}=1/2, C$
## Solution 8: 3D geometry
We can draw a 3D space where each coordinate is in the range [0,1]. Drawing the lines $x+y>z, x+z>y,$ and $y+z>x,$ We have a 3D space that consists of two tetrahedrons. One is a regular tetrahedron with side length $\sqrt2$ and the other has 3 sides of length $\sqrt2$ and 3 sides of length $1.$ The volume of this region is $\frac 1 2$. Hence our solution is $C.$
## Solution 9(Fastest solution if you have no time): Stick Solution
Consider a stick of length $1$. Cutting the stick at two random points gives a triangle from the three new segments. These two random points must be on opposite sides of the halfway mark. Thus, after the first cut is made, there is $\boxed{\textbf{(C) }\frac12}$ probability that the second cut is on the opposite side.
-this isn't necessarily true: what if the cuts are very very close to the edges? -happypi31415
-alanisawesome2018
- AMBRIGGS
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## 5.6 Vector Triple Products
The volume of a parallelepiped with sides A, B and C is the area of its base (say the parallelogram with area |BC| ) multiplied by its altitude, the component of A in the direction of BC. This is the magnitude of ABC; but it is also the magnitude of the determinant of the matrix with columns A, B and C, so these linear functions of the vectors here are the same up to sign. The usual sign convention gives
A(BC) = det(A, B, C)
This product is not changed by cyclically permuting the vectors (for example to B, C, A) or by reversing the order of the factors in the dot product.
We can deduce then that ABC = CAB = ABC. In words, we can switch the dot and cross product without changing anything in this entity. (In either formula of course you must take the cross product first.) This product, like the determinant, changes sign if you just reverse the vectors in the cross product.
The vector triple product, A(BC) is a vector, is normal to A and normal to BC which means it is in the plane of B and C. And it is linear in all three vectors.
We can deduce it is a multiple of B that is linear in A and C plus a multiple of C that is linear in A and B, with the condition that it is normal to A.
Any multiple of B(AC) - C(AB) will obey all these conditions.
What multiple is A(BC)?
Suppose A(BC) = q(B(AC) - C(AB)) holds.
Earlier we saw that the square of the area of a parallelogram with sides A and B can be written either as (AA)( BB) - (AB)( AB) or (BA)(BA). By interchanging the dot and first cross product on the right here you can rewrite this equality as
(BA)(BA) = B(A(BA)) =(AA)( BB) ) - (AB)( AB)
If we identify A with C in A(BC) and take the dot product of A(BA) with B we find q = 1, and we get
A(BC) = B(AC) - C(AB)
This is sometimes called the back cab rule to make it easier to remember the appropriate signs. When using this name remember that the parentheses here are all as far back as possible in this expression The easiest way to get the signs right here without remembering anything is to guess a sign and then check it on the case A = i = C, B = j.
Exercise 5.13 Suppose we have a vector A in three dimensions and an unknown vector v, but we do know Av and Av. Can we find v? YES! How?
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What is a Linear Function?
Linear functions are algebraic equations whose graphs are straight lines with unique values for their slope and y-intercepts.
Learning Objectives
Describe the parts and characteristics of a linear function
Key Takeaways
Key Points
• A linear function is an algebraic equation in which each term is either a constant or the product of a constant and (the first power of) a single variable.
• A function is a relation with the property that each input is related to exactly one output.
• A relation is a set of ordered pairs.
• The graph of a linear function is a straight line, but a vertical line is not the graph of a function.
• All linear functions are written as equations and are characterized by their slope and $y$-intercept.
Key Terms
• relation: A collection of ordered pairs.
• variable: A symbol that represents a quantity in a mathematical expression, as used in many sciences.
• linear function: An algebraic equation in which each term is either a constant or the product of a constant and (the first power of) a single variable.
• function: A relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output.
What is a Linear Function?
A linear function is an algebraic equation in which each term is either a constant or the product of a constant and (the first power of) a single variable. For example, a common equation, $y=mx+b$, (namely the slope-intercept form, which we will learn more about later) is a linear function because it meets both criteria with $x$ and $y$ as variables and $m$ and $b$ as constants. It is linear: the exponent of the $x$ term is a one (first power), and it follows the definition of a function: for each input ($x$) there is exactly one output ($y$). Also, its graph is a straight line.
Graphs of Linear Functions
The origin of the name “linear” comes from the fact that the set of solutions of such an equation forms a straight line in the plane. In the linear function graphs below, the constant, $m$, determines the slope or gradient of that line, and the constant term, $b$, determines the point at which the line crosses the $y$-axis, otherwise known as the $y$-intercept.
Graphs of linear functions: The blue line, $y=\frac{1}{2}x-3$ and the red line, $y=-x+5$ are both linear functions. The blue line has a positive slope of $\frac{1}{2}$ and a $y$-intercept of $-3$; the red line has a negative slope of $-1$ and a $y$-intercept of $5$.
Vertical and Horizontal Lines
Vertical lines have an undefined slope, and cannot be represented in the form $y=mx+b$, but instead as an equation of the form $x=c$ for a constant $c$, because the vertical line intersects a value on the $x$-axis, $c$. For example, the graph of the equation $x=4$ includes the same input value of $4$ for all points on the line, but would have different output values, such as $(4,-2),(4,0),(4,1),(4,5),$ etcetera. Vertical lines are NOT functions, however, since each input is related to more than one output.
Horizontal lines have a slope of zero and is represented by the form, $y=b$, where $b$ is the $y$-intercept. A graph of the equation $y=6$ includes the same output value of 6 for all input values on the line, such as $(-2,6),(0,6),(2,6),(6,6)$, etcetera. Horizontal lines ARE functions because the relation (set of points) has the characteristic that each input is related to exactly one output.
Slope
Slope describes the direction and steepness of a line, and can be calculated given two points on the line.
Learning Objectives
Calculate the slope of a line using “rise over run” and identify the role of slope in a linear equation
Key Takeaways
Key Points
• The slope of a line is a number that describes both the direction and the steepness of the line; its sign indicates the direction, while its magnitude indicates the steepness.
• The ratio of the rise to the run is the slope of a line, $m = \frac{rise}{run}$.
• The slope of a line can be calculated with the formula $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$, where $(x_1, y_1)$ and $(x_2, y_2)$ are points on the line.
Key Terms
• steepness: The rate at which a function is deviating from a reference.
• direction: Increasing, decreasing, horizontal or vertical.
Slope
In mathematics, the slope of a line is a number that describes both the direction and the steepness of the line. Slope is often denoted by the letter $m$. Recall the slop-intercept form of a line, $y = mx + b$. Putting the equation of a line into this form gives you the slope ($m$) of a line, and its $y$-intercept ($b$). We will now discuss the interpretation of $m$, and how to calculate $m$ for a given line.
The direction of a line is either increasing, decreasing, horizontal or vertical. A line is increasing if it goes up from left to right which implies that the slope is positive ($m > 0$). A line is decreasing if it goes down from left to right and the slope is negative ($m < 0$). If a line is horizontal the slope is zero and is a constant function ($y=c$). If a line is vertical the slope is undefined.
Slopes of Lines: The slope of a line can be positive, negative, zero, or undefined.
The steepness, or incline, of a line is measured by the absolute value of the slope. A slope with a greater absolute value indicates a steeper line. In other words, a line with a slope of $-9$ is steeper than a line with a slope of $7$.
Calculating Slope
Slope is calculated by finding the ratio of the “vertical change” to the “horizontal change” between any two distinct points on a line. This ratio is represented by a quotient (“rise over run”), and gives the same number for any two distinct points on the same line. It is represented by $m = \frac{rise}{run}$.
Visualization of Slope: The slope of a line is calculated as “rise over run.”
Mathematically, the slope m of the line is:
$\displaystyle m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$
Two points on the line are required to find $m$. Given two points $(x_1, y_1)$ and $(x_2, y_2)$, take a look at the graph below and note how the “rise” of slope is given by the difference in the $y$-values of the two points, and the “run” is given by the difference in the $x$-values.
Slope Represented Graphically: The slope $m =\frac{y_{2} – y_{1}}{x_{2} – x_{1}}$ is calculated from the two points $\left( x_1,y_1 \right)$ and $\left( x_2,y_2 \right)$.
Now we’ll look at some graphs on a coordinate grid to find their slopes. In many cases, we can find slope by simply counting out the rise and the run. We start by locating two points on the line. If possible, we try to choose points with coordinates that are integers to make our calculations easier.
Example
Find the slope of the line shown on the coordinate plane below.
Find the slope of the line: Notice the line is increasing so make sure to look for a slope that is positive.
Locate two points on the graph, choosing points whose coordinates are integers. We will use $(0, -3)$ and $(5, 1)$. Starting with the point on the left, $(0, -3)$, sketch a right triangle, going from the first point to the second point, $(5, 1)$.
Identify points on the line: Draw a triangle to help identify the rise and run.
Count the rise on the vertical leg of the triangle: $4$ units.
Count the run on the horizontal leg of the triangle: $5$ units.
Use the slope formula to take the ratio of rise over run:
\displaystyle \begin{align} m &= \frac{rise}{run} \\ &= \frac{4}{5} \end{align}
The slope of the line is $\frac{4}{5}$. Notice that the slope is positive since the line slants upward from left to right.
Example
Find the slope of the line shown on the coordinate plane below.
Find the slope of the line: We can see the slope is decreasing, so be sure to look for a negative slope.
Locate two points on the graph. Look for points with coordinates that are integers. We can choose any points, but we will use $(0, 5)$ and $(3, 3)$.
Identify two points on the line: The points $(0, 5)$ and $(3, 3)$ are on the line.
$\displaystyle m =\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$
Let $(x_1, y_1)$ be the point $(0, 5)$, and $(x_2, y_2)$ be the point $(3, 3)$.
Plugging the corresponding values into the slope formula, we get:
\displaystyle \begin{align} m &= \frac{3-5}{3-0} \\ &= \frac{-2}{3} \end{align}
The slope of the line is $- \frac{2}{3}$. Notice that the slope is negative since the line slants downward from left to right.
Direct and Inverse Variation
Two variables in direct variation have a linear relationship, while variables in inverse variation do not.
Learning Objectives
Recognize examples of functions that vary directly and inversely
Key Takeaways
Key Points
• Two variables that change proportionally to one another are said to be in direct variation.
• The relationship between two directly proportionate variable can be represented by a linear equation in slope -intercept form, and is easily modeled using a linear graph.
• Inverse variation is the opposite of direct variation; two variables are said to be inversely proportional when a change is performed on one variable and the opposite happens to the
other.
• The relationship between two inversely proportionate variables cannot be represented by a linear equation, and its graphical representation is not a line, but a hyperbola.
Key Terms
• hyperbola: A conic section formed by the intersection of a cone with a plane that intersects the base of the cone and is not tangent to the cone.
• proportional: At a constant ratio. Two magnitudes (numbers) are said to be proportional if the second varies in a direct relation arithmetically to the first.
Direct Variation
Simply put, two variables are in direct variation when the same thing that happens to one variable happens to the other. If $x$ and $y$ are in direct variation, and $x$ is doubled, then $y$ would also be doubled. The two variables may be considered directly proportional.
For example, a toothbrush costs $2$ dollars. Purchasing $5$ toothbrushes would cost $10$ dollars, and purchasing $10$ toothbrushes would $20$ cost dollars. Thus we can say that the cost varies directly as the value of toothbrushes.
Direct variation is represented by a linear equation, and can be modeled by graphing a line. Since we know that the relationship between two values is constant, we can give their relationship with:
$\displaystyle \frac{y}{x} = k$
Where $k$ is a constant.
Rewriting this equation by multiplying both sides by $x$ yields:
$\displaystyle y = kx$
Notice that this is a linear equation in slope-intercept form, where the $y$-intercept $b$ is equal to $0$.
Thus, any line passing through the origin represents a direct variation between $x$ and $y$:
Directly Proportional Variables: The graph of $y = kx$ demonstrates an example of direct variation between two variables.
Revisiting the example with toothbrushes and dollars, we can define the $x$-axis as number of toothbrushes and the $y$-axis as number of dollars. Doing so, the variables would abide by the relationship:
$\displaystyle \frac{y}{x} = 2$
Any augmentation of one variable would lead to an equal augmentation of the other. For example, doubling $y$ would result in the doubling of $x$.
Inverse Variation
Inverse variation is the opposite of direct variation. In the case of inverse variation, the increase of one variable leads to the decrease of another. In fact, two variables are said to be inversely proportional when an operation of change is performed on one variable and the opposite happens to the other. For example, if $x$ and $y$ are inversely proportional, if $x$ is doubled, then $y$ is halved.
As an example, the time taken for a journey is inversely proportional to the speed of travel. If your car travels at a greater speed, the journey to your destination will be shorter.
Knowing that the relationship between the two variables is constant, we can show that their relationship is:
$\displaystyle yx = k$
Where $k$ is a constant known as the constant of proportionality. Note that as long as $k$ is not equal to $0$, neither $x$ nor $y$ can ever equal $0$ either. We can rearrange the above equation to place the variables on opposite sides:
$\displaystyle y=\frac{k}{x}$
Notice that this is not a linear equation. It is impossible to put it in slope-intercept form. Thus, an inverse relationship cannot be represented by a line with constant slope. Inverse variation can be illustrated with a graph in the shape of a hyperbola, pictured below.
Inversely Proportional Function: An inversely proportional relationship between two variables is represented graphically by a hyperbola.
Zeroes of Linear Functions
A zero, or $x$-intercept, is the point at which a linear function’s value will equal zero.
Learning Objectives
Practice finding the zeros of linear functions
Key Takeaways
Key Points
• A zero is a point at which a function ‘s value will be equal to zero. Its coordinates are $(x,0)$, where $x$ is equal to the zero of the graph.
• Zeros can be observed graphically or solved for algebraically.
• A linear function can have none, one, or infinitely many zeros. If the function is a horizontal line ( slope = $0$), it will have no zeros unless its equation is $y=0$, in which case it will have infinitely many. If the line is non-horizontal, it will have one zero.
Key Terms
• zero: Also known as a root; an $x$ value at which the function of $x$ is equal to $0$.
• linear function: An algebraic equation in which each term is either a constant or the product of a constant and (the first power of) a single variable.
• y-intercept: A point at which a line crosses the $y$-axis of a Cartesian grid.
The graph of a linear function is a straight line. Graphically, where the line crosses the $x$-axis, is called a zero, or root. Algebraically, a zero is an $x$ value at which the function of $x$ is equal to $0$. Linear functions can have none, one, or infinitely many zeros. If there is a horizontal line through any point on the $y$-axis, other than at zero, there are no zeros, since the line will never cross the $x$-axis. If the horizontal line overlaps the $x$-axis, (goes through the $y$-axis at zero) then there are infinitely many zeros, since the line intersects the $x$-axis multiple times. Finally, if the line is vertical or has a slope, then there will be only one zero.
Finding the Zeros of Linear Functions Graphically
Zeros can be observed graphically. An $x$-intercept, or zero, is a property of many functions. Because the $x$-intercept (zero) is a point at which the function crosses the $x$-axis, it will have the value $(x,0)$, where $x$ is the zero.
All lines, with a value for the slope, will have one zero. To find the zero of a linear function, simply find the point where the line crosses the $x$-axis.
Zeros of linear functions: The blue line, $y=\frac{1}{2}x+2$, has a zero at $(-4,0)$; the red line, $y=-x+5$, has a zero at $(5,0)$. Since each line has a value for the slope, each line has exactly one zero.
Finding the Zeros of Linear Functions Algebraically
To find the zero of a linear function algebraically, set $y=0$ and solve for $x$.
The zero from solving the linear function above graphically must match solving the same function algebraically.
Example: Find the zero of $y=\frac{1}{2}x+2$ algebraically
First, substitute $0$ for $y$:
$\displaystyle 0=\frac{1}{2}x+2$
Next, solve for $x$. Subtract $2$ and then multiply by $2$, to obtain:
\displaystyle \begin{align} \frac{1}{2}x&=-2\\ x&=-4 \end{align}
The zero is $(-4,0)$. This is the same zero that was found using the graphing method.
Slope-Intercept Equations
The slope-intercept form of a line summarizes the information necessary to quickly construct its graph.
Learning Objectives
Convert linear equations to slope-intercept form and explain why it is useful
Key Takeaways
Key Points
• The slope -intercept form of a line is given by $y = mx + b$ where $m$ is the slope of the line and $b$ is the $y$-intercept.
• The constant $b$ is known as the $y$-intercept. From slope- intercept form, when $x=0$, $y=b$, and the point $(0,b)$ is the unique point on the line also on the $y$-axis.
• To graph a line in slope-intercept form, first plot the $y$-intercept, then use the value of the slope to locate a second point on the line. If the value of the slope is an integer, use a $1$ for the denominator.
• Use algebra to solve for $y$ if the equation is not written in slope-intercept form. Only then can the value of the slope and $y$-intercept be located from the equation accurately.
Key Terms
• slope: The ratio of the vertical and horizontal distances between two points on a line; zero if the line is horizontal, undefined if it is vertical.
• y-intercept: A point at which a line crosses the $y$-axis of a Cartesian grid.
Slope-Intercept Form
One of the most common representations for a line is with the slope-intercept form. Such an equation is given by $y=mx+b$, where $x$ and $y$ are variables and $m$ and $b$ are constants. When written in this form, the constant $m$ is the value of the slope and $b$ is the $y$-intercept. Note that if $m$ is $0$, then $y=b$ represents a horizontal line. Note that this equation does not allow for vertical lines, since that would require that $m$ be infinite (undefined). However, a vertical line is defined by the equation $x=c$ for some constant $c$.
Converting an Equation to Slope-Intercept Form
Writing an equation in slope-intercept form is valuable since from the form it is easy to identify the slope and $y$-intercept. This assists in finding solutions to various problems, such as graphing, comparing two lines to determine if they are parallel or perpendicular and solving a system of equations.
Example
Let’s write an equation in slope-intercept form with $m=-\frac{2}{3}$, and $b=3$. Simply substitute the values into the slope-intercept form to obtain:
$\displaystyle y=-\frac{2}{3}x+3$
If an equation is not in slope-intercept form, solve for $y$ and rewrite the equation.
Example
Let’s write the equation $3x+2y=-4$ in slope-intercept form and identify the slope and $y$-intercept. To solve the equation for $y$, first subtract $3x$ from both sides of the equation to get:
$\displaystyle 2y=-3x-4$
Then divide both sides of the equation by $2$ to obtain:
$\displaystyle y=\frac{1}{2}(-3x-4)$
Which simplifies to $y=-\frac{3}{2}x-2$. Now that the equation is in slope-intercept form, we see that the slope $m=-\frac{3}{2}$, and the $y$-intercept $b=-2$.
Graphing an Equation in Slope-Intercept Form
We begin by constructing the graph of the equation in the previous example.
Example
We construct the graph the line $y=-\frac{3}{2}x-2$ using the slope-intercept method. We begin by plotting the $y$-intercept $b=-2$, whose coordinates are $(0,-2)$. The value of the slope dictates where to place the next point.
Since the value of the slope is $\frac{-3}{2}$, the rise is $-3$ and the run is $2$. This means that from the $y$-intercept, $(0,-2)$, move $3$ units down, and move $2$ units right. Thus we arrive at the point $(2,-5)$ on the line. If the negative sign is placed with the denominator instead the slope would be written as $\frac{3}{-2}$, we can instead move up $3$ units and left $2$ units from the $y$-intercept to arrive at the point $(-2,1)$, also on the line.
Slope-intercept graph: Graph of the line $y=-\frac{3}{2}x-2$.
Example
Let’s graph the equation $12x-6y-6=0$. First we solve the equation for $y$ by subtracting $12x$ to obtain:
$\displaystyle -6y-6=-12x$
Next, add $6$ to get:
$\displaystyle -6y=-12x+6$
Finally, divide all terms by $-6$ to get the slope-intercept form:
$\displaystyle y=2x-1$
The slope is $2$, and the $y$-intercept is $-1$. Using this information, graphing is easy. Start by plotting the $y$-intercept $(0,-1)$, then use the value of the slope, $\frac{2}{1}$, to move up $2$ units and right $1$ unit.
Slope-intercept graph: Graph of the line $y=2x-1$.
Point-Slope Equations
The point-slope equation is another way to represent a line; only the slope and a single point are needed.
Learning Objectives
Use point-slope form to find the equation of a line passing through two points and verify that it is equivalent to the slope-intercept form of the equation
Key Takeaways
Key Points
• The point – slope equation is given by $y-y_{1}=m(x-x_{1})$, where $(x_{1}, y_{1})$ is any point on the line, and $m$ is the slope of the line.
• The point-slope equation requires that there is at least one point and the slope. If there are two points and no slope, the slope can be calculated from the two points first and then choose one of the two points to write the equation.
• The point-slope equation and slope-intercept equations are equivalent. It can be shown that given a point $(x_{1}, y_{1})$ and slope $m$, the $y$-intercept ($b$) in the slope-intercept equation is $y_{1}-mx_{1}$.
Key Terms
• point-slope equation: An equation of a line given a point $(x_{1}, y_{1})$ and a slope $m$: $y-y_{1}=m(x-x_{1})$.
Point-Slope Equation
The point-slope equation is a way of describing the equation of a line. The point-slope form is ideal if you are given the slope and only one point, or if you are given two points and do not know what the $y$-intercept is. Given a slope, $m$, and a point $(x_{1}, y_{1})$, the point-slope equation is:
$\displaystyle y-y_{1}=m(x-x_{1})$
Verify Point-Slope Form is Equivalent to Slope-Intercept Form
To show that these two equations are equivalent, choose a generic point $(x_{1}, y_{1})$. Plug in the generic point into the equation $y=mx+b$. The equation is now, $y_{1}=mx_{1}+b$, giving us the ordered pair,$(x_{1}, mx_{1}+b)$. Then plug this point into the point-slope equation and solve for $y$ to get:
$\displaystyle y-(mx_{1}+b)=m(x-x_{1})$
Distribute the negative sign through and distribute $m$ through $(x-x_{1})$:
$\displaystyle y-mx_{1}-b=mx-mx_{1}$
Add $mx_{1}$ to both sides:
$\displaystyle y-mx_{1}+mx_{1}-b=mx-mx_{1}+mx_{1}$
Combine like terms:
$\displaystyle y-b=mx$
Add $b$ to both sides:
$\displaystyle y-b+b=mx+b$
Combine like terms:
$\displaystyle y=mx+b$
Therefore, the two equations are equivalent and either one can express an equation of a line depending on what information is given in the problem or what type of equation is requested in the problem.
Example: Write the equation of a line in point-slope form, given a point $(2,1)$ and slope $-4$, and convert to slope-intercept form
Write the equation of the line in point-slope form:
$\displaystyle y-1=-4(x-2)$
To switch this equation into slope-intercept form, solve the equation for $y$:
$\displaystyle y-1=-4(x-2)$
Distribute $-4$:
$\displaystyle y-1=-4x+8$
Add $1$ to both sides:
$\displaystyle y=-4x+9$
The equation has the same meaning whichever form it is in, and produces the same graph.
Line graph: Graph of the line $y-1=-4(x-2)$, through the point $(2,1)$ with slope of $-4$, as well as the slope-intercept form, $y=-4x+9$.
Example: Write the equation of a line in point-slope form, given point $(-3,6)$ and point $(1,2)$, and convert to slope-intercept form
Since we have two points, but no slope, we must first find the slope:
$\displaystyle m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
Substituting the values of the points:
\displaystyle \begin{align} m&=\frac{-2-6}{1-(-3)}\\&=\frac{-8}{4}\\&=-2 \end{align}
Now choose either of the two points, such as $(-3,6)$. Plug this point and the calculated slope into the point-slope equation to get:
$\displaystyle y-6=-2[x-(-3)]$
Be careful if one of the coordinates is a negative. Distributing the negative sign through the parentheses, the final equation is:
$\displaystyle y-6=-2(x+3)$
If you chose the other point, the equation would be: $y+2=-2(x-1)$ and either answer is correct.
Next distribute $-2$:
$\displaystyle y-6=-2x-6$
Add $6$ to both sides:
$\displaystyle y=-2x$
Again, the two forms of the equations are equivalent to each other and produce the same line. The only difference is the form that they are written in.
Linear Equations in Standard Form
A linear equation written in standard form makes it easy to calculate the zero, or $x$-intercept, of the equation.
Learning Objectives
Explain the process and usefulness of converting linear equations to standard form
Key Takeaways
Key Points
• The standard form of a linear equation is written as: $Ax + By = C$.
• The standard form is useful in calculating the zero of an equation. For a linear equation in standard form, if $A$ is nonzero, then the $x$-intercept occurs at $x = \frac{C}{A}$.
Key Terms
• zero: Also known as a root, a zero is an $x$-value at which the function of $x$ is equal to 0.
• slope-intercept form: A linear equation written in the form $y = mx + b$.
• y-intercept: A point at which a line crosses the y-axis of a Cartesian grid.
Standard Form
Standard form is another way of arranging a linear equation. In the standard form, a linear equation is written as:
$\displaystyle Ax + By = C$
where $A$ and $B$ are both not equal to zero. The equation is usually written so that $A \geq 0$, by convention. The graph of the equation is a straight line, and every straight line can be represented by an equation in the standard form.
For example, consider an equation in slope -intercept form: $y = -12x +5$. In order to write this in standard form, note that we must move the term containing $x$ to the left side of the equation. We add $12x$ to both sides:
$\displaystyle y + 12x = 5$
The equation is now in standard form.
Using Standard Form to Find Zeroes
Recall that a zero is a point at which a function ‘s value will be equal to zero ($y=0$), and is the $x$-intercept of the function. We know that the y-intercept of a linear equation can easily be found by putting the equation in slope-intercept form. However, the zero of the equation is not immediately obvious when the linear equation is in this form. However, the zero, or $x$-intercept of a linear equation can easily be found by putting it into standard form.
For a linear equation in standard form, if $A$ is nonzero, then the $x$-intercept occurs at $x = \frac{C}{A}$.
For example, consider the equation $y + 12x = 5$.
In this equation, the value of $A$ is 1, and the value of $C$ is 5. Therefore, the zero of the equation occurs at $x = \frac{5}{1} = 5$. The zero is the point $(5, 0)$.
Note that the $y$-intercept and slope can also be calculated using the coefficients and constant of the standard form equation. If $B$ is non-zero, then the y-intercept, that is the y-coordinate of the point where the graph crosses the y-axis (where $x$ is zero), is $\frac{C}{B}$, and the slope of the line is $-\frac{A}{B}$.
Example: Find the zero of the equation $3(y - 2) = \frac{1}{4}x +3$
We must write the equation in standard form, $Ax + By = C$, which means getting the $x$ and $y$ terms on the left side, and the constants on the right side of the equation.
Distribute the 3 on the left side:
$\displaystyle 3y - 6 = \frac{1}{4}x +3$
$\displaystyle 3y = \frac{1}{4}x + 9$
Subtract $\frac{1}{4}x$ from both sides:
$\displaystyle 3y - \frac{1}{4}x = 9$
Rearrange to $Ax + By = C$:
$\displaystyle - \frac{1}{4}x+3y = 9$
The equation is in standard form, and we can substitute the values for $A$ and $C$ into the formula for the zero:
\displaystyle \begin{align} x &= \frac{C}{A} \\&= \frac{9}{-\frac{1}{4}} \\&= -36 \end{align}
The zero is $(-36, 0)$.
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CHAPTER 3: The Normal Distributions
Presentation on theme: "CHAPTER 3: The Normal Distributions"— Presentation transcript:
CHAPTER 3: The Normal Distributions
Basic Practice of Statistics - 3rd Edition CHAPTER 3: The Normal Distributions Basic Practice of Statistics 7th Edition Lecture PowerPoint Slides Chapter 5
In Chapter 3, We Cover … Density curves Describing density curves
Normal distributions The rule The standard Normal distribution Finding Normal proportions Using the standard Normal table Finding a value given a proportion
Density Curves We now have a toolbox of graphical and numerical methods for describing distributions. What is more, we have a clear strategy for exploring data on a single quantitative variable. Exploring a distribution Always plot your data: Make a graph, usually a histogram or a stemplot. Look for the overall pattern (shape, center, and variability) and for striking deviations such as outliers. Calculate a numerical summary to briefly describe center and variability. Now we add one more step to this strategy: Sometimes, the overall pattern of a large number of observations is so regular that we can describe it by a smooth curve.
Density Curves Example: Here is a histogram of vocabulary scores of 947 seventh graders. The smooth curve drawn over the histogram is a mathematical model for the distribution.
Density Curves The areas of the shaded bars in this histogram represent the proportion of scores in the observed data that are less than or equal to This proportion is equal to Now the area under the smooth curve to the left of 6.0 is shaded. If the scale is adjusted so that the total area under the curve is exactly 1, then this curve is called a density curve. The proportion of the area to the left of 6.0 is now equal to
Density Curves Density curve A density curve is a curve that:
is always on or above the horizontal axis. has an area of exactly 1 underneath it. A density curve describes the overall pattern of a distribution. The area under the curve and above any range of values on the horizontal axis is the proportion of all observations that fall in that range. Caution: No set of real data is exactly described by a density curve!
Describing Density Curves
7 Median and Mean of a Density Curve The median of a density curve is the equal-areas point, the point that divides the area under the curve in half. The mean of a density curve is the balance point, at which the curve would balance if made of solid material. The median and the mean are the same for a symmetric density curve. They both lie at the center of the curve. The mean of a skewed curve is pulled away from the median in the direction of the long tail.
Density Curves The mean and standard deviation computed from actual observations (data) are denoted by x and s, respectively. The mean and standard deviation of the actual distribution represented by the density curve are denoted by µ (“mu”) and σ (“sigma”), respectively.
Normal Distributions One particularly important class of density curves are the Normal curves, which describe Normal distributions. All Normal curves are symmetric, single-peaked, and bell-shaped Any specific Normal curve is described by giving its mean µ and standard deviation σ.
Normal Distributions Unlike most distributions, any particular Normal distribution is completely specified by two numbers: its mean µ and standard deviation σ: The mean is located at the center of the symmetric curve and is the same as the median. Changing µ without changing σ moves the Normal curve along the horizontal axis without changing its variability. The standard deviation σ controls the variability of a Normal curve. When the standard deviation is larger, the area under the normal curve is less concentrated about the mean. The standard deviation is the distance from the center to the change-of-curvature points on either side.
Normal Distributions Normal Distributions
A Normal distribution is described by a Normal density curve. Any particular Normal distribution is completely specified by two numbers: its mean µ and standard deviation σ. The mean of a Normal distribution is at the center of the symmetric Normal curve. The standard deviation is the distance from the center to the change-of-curvature points on either side.
The Rule The Rule In the Normal distribution with mean µ and standard deviation σ: Approximately 68% of the observations fall within σ of µ. Approximately 95% of the observations fall within 2σ of µ. Approximately 99.7% of the observations fall within 3σ of µ.
Normal Distributions The distribution of Iowa Test of Basic Skills (ITBS) vocabulary scores for seventh-grade students in Gary, Indiana, is close to Normal. Suppose the distribution is N(6.84, 1.55). Sketch the Normal density curve for this distribution. What percent of ITBS vocabulary scores is between 3.74 and 9.94? What percent of the scores is above 5.29?
Standardizing All Normal distributions are the same if we measure in units of size σ from the mean µ as center. Changing to these units is called standardizing. STANDARDIZING AND z-SCORES If x is an observation from a distribution that has mean and standard deviation σ, the standardized value of x is z = x − µ/σ A standardized value is often called a z-score.
Standardizing Example:
The heights of women aged 20 to 29 in the United States are approximately Normal with µ = 64.2 and σ = 2.8 inches. The standardized height is z = height − 64.2/2.8 A woman's standardized height is the number of standard deviations by which her height differs from the mean height of all women aged 20 to 29. A woman 70 inches tall, for example, has standardized height z = 70 − 64.2/2.8 = 2.07 or 2.07 standard deviations above the mean. Similarly, a woman 5 feet (60 inches) tall has standardized height z = 60 − 64.2/2.8 = −1.50 or 1.5 standard deviations less than the mean height.
The Standard Normal Distribution
The standard Normal distribution is the Normal distribution with mean 0 and standard deviation 1. If a variable x has any Normal distribution N(µ,σ) with mean µ and standard deviation σ, then the standardized variable z = x − µ/σ has the standard Normal distribution, N(0,1).
Cumulative Proportions
The cumulative proportion for a value x in a distribution is the proportion of observations in the distribution that are less than or equal to x.
Cumulative Proportions—Example
Example: Who qualifies for college sports? The combined scores of the almost 1.7 million high school seniors taking the SAT in 2013 were approximately Normal with mean 1011 and standard deviation 216. What percent of high school seniors meet this SAT requirement of a combined score of 820 or better? Here is the calculation in a picture: So about 81% qualified for college sports.
Cumulative Proportions—Example
Example: Normal calculation using software To find the numerical value of the cumulative proportion in our college sports example using software, plug in mean 1010 and standard deviation 214 and ask for the cumulative proportion for 820. Software often uses terms such as “cumulative distribution” or “cumulative probability.” Minitab’s output:
The Standard Normal Table
Because all Normal distributions are the same when we standardize, we can find areas under any Normal curve from a single table. The Standard Normal Table Table A is a table of areas under the standard Normal curve. The table entry for each value z is the area under the curve to the left of z. Suppose we want to find the proportion of observations from the standard Normal distribution that are less than 0.81. We can use Table A: P(z < 0.81) = .7910 Z .00 .01 .02 0.7 .7580 .7611 .7642 0.8 .7881 .7910 .7939 0.9 .8159 .8186 .8212
Normal Calculations Find the proportion of observations from the standard Normal distribution that are between –1.25 and 0.81. Can you find the same proportion using a different approach? 1 – ( ) = 1 – =
Normal Calculations Using Table A to find Normal proportions
Step 1. State the problem in terms of the observed variable x. Draw a picture that shows the proportion you want in terms of cumulative proportions. Step 2. Standardize x to restate the problem in terms of a standard Normal variable z. Step 3. Use Table A and the fact that the total area under the curve is 1 to find the required area under the standard Normal curve.
Basic Practice of Statistics - 3rd Edition
Finding a Value Given a Proportion SAT reading scores for a recent year are distributed according to a N(504, 111) distribution. How high must a student score in order to be in the top 10% of the distribution? We’ll look at the same two approaches we did for finding Normal proportions—using software, and then using table A. Chapter 5
Basic Practice of Statistics - 3rd Edition
Finding a Value Given a Proportion Example: using software How high must a student score in order to be in the top 10% of the distribution? We want to find the SAT score x with area 0.1 to its right under the Normal curve with mean µ = 504 and standard deviation σ = 111. That’s the same as finding the SAT score x with area 0.9 to its left. Most software will tell you x when you plug in mean 504, standard deviation 111, and cumulative proportion 0.9. Here is Minitab’s output: Chapter 5
Basic Practice of Statistics - 3rd Edition
Normal Calculations SAT reading scores for a recent year are distributed according to a N(504, 111) distribution. How high must a student score in order to be in the top 10% of the distribution? In order to use table A, equivalently, what score has cumulative proportion 0.90 below it? .10 ? N(504, 111) .90 Chapter 5
Basic Practice of Statistics - 3rd Edition
Normal Calculations .10 ? How high must a student score in order to be in the top 10% of the distribution? Look up the closest probability (closest to 0.10) in the table. Find the corresponding standardized score. The value you seek is that many standard deviations from the mean. z .07 .08 .09 1.1 .8790 .8810 .8830 1.2 .8980 .8997 .8015 1.3 .8147 .8162 .8177 z = 1.28 Chapter 5
Basic Practice of Statistics - 3rd Edition
Normal Calculations .10 ? How high must a student score in order to be in the top 10% of the distribution? z = 1.28 We need to “unstandardize” the z-score to find the observed value (x): x = z(111) = [(1.28 ) (111)] = (142.08) = A student would have to score at least to be in the top 10% of the distribution of SAT reading scores for this particular year. Chapter 5
“Backward” Normal Calculations
Using Table A Given a Normal proportion Step 1. State the problem in terms of the given proportion. Draw a picture that shows the Normal value, x, you want in relation to the cumulative proportion. Step 2. Use Table A, the fact that the total area under the curve is 1, and the given area under the standard Normal curve to find the corresponding z-value. Step 3. Unstandardize z to solve the problem in terms of a non-standard Normal variable x.
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# 5.3: Angle Bisectors in Triangles
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Apply the Angle Bisector Theorem and its converse.
• Understand concurrency for angle bisectors.
## Review Queue
1. Construct the angle bisector of an \begin{align*}80^\circ\end{align*} angle (Investigation 1-4).
2. Draw the following: \begin{align*}M\end{align*} is on the interior of \begin{align*}\angle LNO\end{align*}. \begin{align*}O\end{align*} is on the interior of \begin{align*}\angle MNP\end{align*}. If \begin{align*}\overleftrightarrow{NM}\end{align*} and \begin{align*}\overleftrightarrow{NO}\end{align*} are the angle bisectors of \begin{align*}\angle LNO\end{align*} and \begin{align*}\angle MNP\end{align*} respectively, write all the congruent angles.
3. Find the value of \begin{align*}x\end{align*}.
Know What? The cities of Verticville, Triopolis, and Angletown are joining their city budgets together to build a centrally located airport. There are freeways between the three cities and they want to have the freeway on the interior of these freeways. Where is the best location to put the airport so that they have to build the least amount of road?
In the picture to the right, the blue roads are proposed.
## Angle Bisectors
In Chapter 1, you learned that an angle bisector cuts an angle exactly in half. In #1 in the Review Queue above, you constructed an angle bisector of an \begin{align*}80^\circ\end{align*} angle. Let’s analyze this figure.
\begin{align*}\overrightarrow{BD}\end{align*} is the angle bisector of \begin{align*}\angle ABC\end{align*}. Looking at point \begin{align*}D\end{align*}, if we were to draw \begin{align*}\overline{ED}\end{align*} and \begin{align*}\overline{DF}\end{align*}, we would find that they are equal. Recall from Chapter 3 that the shortest distance from a point to a line is the perpendicular length between them. \begin{align*}ED\end{align*} and \begin{align*}DF\end{align*} are the shortest lengths between \begin{align*}D\end{align*}, which is on the angle bisector, and each side of the angle.
Angle Bisector Theorem: If a point is on the bisector of an angle, then the point is equidistant from the sides of the angle.
In other words, if \begin{align*}\overleftrightarrow{BD}\end{align*} bisects \begin{align*}\angle ABC, \overrightarrow{BE} \bot \overline{ED}\end{align*}, and \begin{align*}\overrightarrow{BF} \bot \overline{DF}\end{align*}, then \begin{align*}ED = DF\end{align*}.
Proof of the Angle Bisector Theorem
Given: \begin{align*}\overrightarrow{BD} \end{align*} bisects \begin{align*}\angle ABC, \overrightarrow{BA} \bot \overline{AD}\end{align*}, and \begin{align*}\overrightarrow{BC} \bot \overline{DC}\end{align*}
Prove: \begin{align*}\overline{AD} \cong \overline{DC}\end{align*}
Statement Reason
1. \begin{align*}\overrightarrow{BD}\end{align*} bisects \begin{align*}\angle ABC, \overrightarrow{BA} \bot \overline{AD}, \overrightarrow{BC} \bot \overline{DC}\end{align*} Given
2. \begin{align*}\angle ABD \cong \angle DBC\end{align*} Definition of an angle bisector
3. \begin{align*}\angle DAB\end{align*} and \begin{align*}\angle DCB\end{align*} are right angles Definition of perpendicular lines
4. \begin{align*}\angle DAB \cong \angle DCB\end{align*} All right angles are congruent
5. \begin{align*}\overline{BD} \cong \overline{BD}\end{align*} Reflexive PoC
6. \begin{align*}\triangle ABD \cong \triangle CBD\end{align*} AAS
7. \begin{align*}\overline{AD} \cong \overline{DC}\end{align*} CPCTC
The converse of this theorem is also true. The proof is in the review questions.
Angle Bisector Theorem Converse: If a point is in the interior of an angle and equidistant from the sides, then it lies on the bisector of the angle.
Because the Angle Bisector Theorem and its converse are both true we have a biconditional statement. We can put the two conditional statements together using if and only if. A point is on the angle bisector of an angle if and only if it is equidistant from the sides of the triangle.
Example 1: Is \begin{align*}Y\end{align*} on the angle bisector of \begin{align*}\angle XWZ\end{align*}?
Solution: In order for \begin{align*}Y\end{align*} to be on the angle bisector \begin{align*}XY\end{align*} needs to be equal to \begin{align*}YZ\end{align*} and they both need to be perpendicular to the sides of the angle. From the markings we know \begin{align*}\overline{XY} \bot \overrightarrow{WX}\end{align*} and \begin{align*}\overline{ZY} \bot \overrightarrow{WZ}\end{align*}. Second, \begin{align*}XY = YZ = 6\end{align*}. From this we can conclude that \begin{align*}Y\end{align*} is on the angle bisector.
Example 2: \begin{align*}\overrightarrow{MO}\end{align*} is the angle bisector of \begin{align*}\angle LMN\end{align*}. Find the measure of \begin{align*}x\end{align*}.
Solution: \begin{align*}LO = ON\end{align*} by the Angle Bisector Theorem Converse.
\begin{align*}4x-5 &= 23\\ 4x &= 28\\ x &= 7\end{align*}
## Angle Bisectors in a Triangle
Like perpendicular bisectors, the point of concurrency for angle bisectors has interesting properties.
Investigation 5-2: Constructing Angle Bisectors in Triangles
Tools Needed: compass, ruler, pencil, paper
1. Draw a scalene triangle. Construct the angle bisector of each angle. Use Investigation 1-4 and #1 from the Review Queue to help you.
Incenter: The point of concurrency for the angle bisectors of a triangle.
2. Erase the arc marks and the angle bisectors after the incenter. Draw or construct the perpendicular lines to each side, through the incenter.
3. Erase the arc marks from #2 and the perpendicular lines beyond the sides of the triangle. Place the pointer of the compass on the incenter. Open the compass to intersect one of the three perpendicular lines drawn in #2. Draw a circle.
Notice that the circle touches all three sides of the triangle. We say that this circle is inscribed in the triangle because it touches all three sides. The incenter is on all three angle bisectors, so the incenter is equidistant from all three sides of the triangle.
Concurrency of Angle Bisectors Theorem: The angle bisectors of a triangle intersect in a point that is equidistant from the three sides of the triangle.
If \begin{align*}\overline{AG}, \overline{BG}\end{align*}, and \begin{align*}\overline{GC}\end{align*} are the angle bisectors of the angles in the triangle, then \begin{align*}EG = GF = GD\end{align*}.
In other words, \begin{align*}\overline{EG}, \overline{FG}\end{align*}, and \begin{align*}\overline{DG}\end{align*} are the radii of the inscribed circle.
Example 3: If \begin{align*}J, E\end{align*}, and \begin{align*}G\end{align*} are midpoints and \begin{align*}KA = AD = AH\end{align*} what are points \begin{align*}A\end{align*} and \begin{align*}B\end{align*} called?
Solution: \begin{align*}A\end{align*} is the incenter because \begin{align*}KA = AD = AH\end{align*}, which means that it is equidistant to the sides. \begin{align*}B\end{align*} is the circumcenter because \begin{align*}\overline{JB}, \overline{BE}\end{align*}, and \begin{align*}\overline{BG}\end{align*} are the perpendicular bisectors to the sides.
Know What? Revisited The airport needs to be equidistant to the three highways between the three cities. Therefore, the roads are all perpendicular to each side and congruent. The airport should be located at the incenter of the triangle.
## Review Questions
Construction Construct the incenter for the following triangles by tracing each triangle onto a piece of paper and using Investigation 5-2. Draw the inscribed circle too.
1. Is the incenter always going to be inside of the triangle? Why?
2. For an equilateral triangle, what can you conclude about the circumcenter and the incenter?
For questions 6-11, \begin{align*}\overrightarrow{AB}\end{align*} is the angle bisector of \begin{align*}\angle CAD\end{align*}. Solve for the missing variable.
Is there enough information to determine if \begin{align*}\overrightarrow{AB}\end{align*} is the angle bisectorof \begin{align*}\angle CAD\end{align*}? Why or why not?
What are points \begin{align*}A\end{align*} and \begin{align*}B\end{align*}? How do you know?
1. The blue lines are congruent The green lines are angle bisectors
2. Both sets of lines are congruent The green lines are perpendicular to the sides
3. Fill in the blanks in the Angle Bisector Theorem Converse. Given: \begin{align*}\overline{AD} \cong \overline{DC}\end{align*}, such that \begin{align*}AD\end{align*} and \begin{align*}DC\end{align*} are the shortest distances to \begin{align*}\overrightarrow{BA}\end{align*} and \begin{align*}\overrightarrow{BC}\end{align*} Prove: \begin{align*}\overrightarrow{BD}\end{align*} bisects \begin{align*}\angle ABC\end{align*}
Statement Reason
1.
2. The shortest distance from a point to a line is perpendicular.
3. \begin{align*}\angle DAB\end{align*} and \begin{align*}\angle DCB\end{align*} are right angles
4. \begin{align*}\angle DAB \cong \angle DCB\end{align*}
5. \begin{align*}\overline{BD} \cong \overline{BD}\end{align*}
6. \begin{align*}\triangle ABD \cong \triangle CBD\end{align*}
7. CPCTC
8. \begin{align*}\overrightarrow{BD}\end{align*} bisects \begin{align*}\angle ABC\end{align*}
Determine if the following descriptions refer to the incenter or circumcenter of the triangle.
1. A lighthouse on a triangular island is equidistant to the three coastlines.
2. A hospital is equidistant to three cities.
3. A circular walking path passes through three historical landmarks.
4. A circular walking path connects three other straight paths.
Constructions
1. Construct an equilateral triangle.
2. Construct the angle bisectors of two of the angles to locate the incenter.
3. Construct the perpendicular bisectors of two sides to locate the circumcenter.
4. What do you notice? Use these points to construct an inscribed circle inside the triangle and a circumscribed circle about the triangle.
Multi- Step Problem
1. Draw \begin{align*}\angle ABC\end{align*} through \begin{align*}A(1, 3), B(3, -1)\end{align*} and \begin{align*}C(7, 1)\end{align*}.
2. Use slopes to show that \begin{align*}\angle ABC\end{align*} is a right angle.
3. Use the distance formula to find \begin{align*}AB\end{align*} and \begin{align*}BC\end{align*}.
4. Construct a line perpendicular to \begin{align*}AB\end{align*} through \begin{align*}A\end{align*}.
5. Construct a line perpendicular to \begin{align*}BC\end{align*} through \begin{align*}C\end{align*}.
6. These lines intersect in the interior of \begin{align*}\angle ABC\end{align*}. Label this point \begin{align*}D\end{align*} and draw \begin{align*}\overrightarrow{BD}\end{align*}.
7. Is \begin{align*}\overrightarrow{BD}\end{align*} the angle bisector of \begin{align*}\angle ABC\end{align*}? Justify your answer.
1. \begin{align*}\angle LNM \cong \angle MNO \cong \angle ONP\!\\ \angle LNO \cong \angle MNP\end{align*}
1. \begin{align*}5x+11=26\!\\ {\;} \ \ \ \ \ 5x=15\!\\ {\;}\qquad \ x=3\end{align*}
2. \begin{align*}9x-1=2(4x+5)\!\\ 9x-1=8x+10\!\\ {\;}\quad \quad x=11^\circ \end{align*}
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## Basic College Mathematics (10th Edition)
(a) $\displaystyle 8\frac{3}{8}$ (b) $\displaystyle 3\frac{21}{40}$
We add or subtract the fractions using both methods: (a) Method 1: $\displaystyle 4\frac{5}{8}+3\frac{3}{4}$ $=\displaystyle 4\frac{5}{8}+3\frac{6}{8}$ $=\displaystyle 7\frac{11}{8}$ $=\displaystyle 8\frac{3}{8}$ Method 2: $\displaystyle 4\frac{5}{8}+3\frac{3}{4}$ $\displaystyle =\frac{37}{8}+\frac{15}{4}$ $\displaystyle =\frac{37}{8}+\frac{30}{8}$ $\displaystyle =\frac{67}{8}$ $\displaystyle =8\frac{3}{8}$ We confirm that we obtained the same answer with both methods. (b) Method 1: $\displaystyle 12\frac{2}{5}-8\frac{7}{8}$ $=\displaystyle 12\frac{16}{40}-8\frac{35}{40}$ $=\displaystyle 11\frac{56}{40}-8\frac{35}{40}$ $=\displaystyle 3\frac{21}{40}$ Method 2: $\displaystyle 12\frac{2}{5}-8\frac{7}{8}$ $=\displaystyle \frac{62}{5}-\frac{71}{8}$ $=\displaystyle \frac{496}{40}-\frac{355}{40}$ $=\displaystyle \frac{141}{40}$ $=\displaystyle 3\frac{21}{40}$ We confirm that we obtained the same answer with both methods. Both methods have their advantages and disadvantages. In this case, it seems that working with whole numbers is a bit simpler because the numbers are smaller.
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# What does it mean to pivot (linear algebra)?
So I'm told that if the matrix is symmetric positive definitive (as the one below is), pivoting is not required when using Gaussian elimination.
$A = \begin{bmatrix} 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 2 \end{bmatrix}$
So I've been googling around trying to actually get a definition of what pivoting is and I can't find a straightforward answer. I've read that it means to do a row swap. I've read that it means to "make an element above or below leading one into a zero."
I guess what I've always just done is: make $a_{1,1} = 1$, make zeros below. Make $a_{2,2} = 1$, make zeros below. etc.
• interestingly, I'm in that class right now – Collin Feb 27 '14 at 4:44
• But it's dated 2011... – IAmNoOne Feb 27 '14 at 4:56
• and the lecture slides haven't changed a bit lol – Collin Feb 27 '14 at 5:04
Think of it as follows.
If you have the following equations $$2x+2y=6\\-x+y=-1$$ The solution is $$x=2\\y=1$$ We get it by(dividing the first equation by 2, add to the second, divide the result by 2, ..)
Now the first two equations could be represented as a matrix $$\pmatrix{2&2&6\\-1&1&-1}$$ and the second two equations could be represented as a matrix also $$\pmatrix{1&0&2\\0&1&1}$$ Pivoting means to take the first matrix to the second matrix using row operations as you do with equations.
For your matrix notice the following
$$A = \begin{bmatrix} 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 2 \end{bmatrix}\implies \begin{bmatrix} -1 & 2 & -1 \\ 2 & -1 & 0\\ 0 & -1 & 2 \end{bmatrix} \\ \implies \begin{bmatrix} 1 & -2 & 1 \\ 2 & -1 & 0\\ 0 & -1 & 2 \end{bmatrix} \implies \begin{bmatrix} 1 & -2 & 1 \\ 0 & 3 & -2\\ 0 & -1 & 2 \end{bmatrix}$$
• So how could you do Gaussian elimination on my original matrix without using those types of row operations? – Collin Feb 27 '14 at 6:37
• I have added some new@user2079802 – Semsem Feb 27 '14 at 6:45
Pivoting in the word sense means turning or rotating. In the Gauß algorithm it means rotating the rows so that they have a numerically more favorable make-up.
The straight-forward implementation of the LU decomposition has no pivoting. However, it may encounter zeros or near zeros on the diagonal while entries below in the same column have an appreciable size.
So the natural idea is to pick the largest of the remaining entries, call it the pivot (turning axis) and use that row as the basis for the elimination step. To keep constructing the echelon form, rows are swapped or rotated (most efficiently using a row index array), adding permutation steps to the elementary row transformations.
The result of the pivoted Gauß algorithm is a PLU decomposition, where P is a permutation matrix that has in each row and column exactly one entry 1, all other 0.
As to the original matrix, the discretization of minus the second derivative is indeed positive definite. To show that requires an eigenvalue analysis.
For positive definite matrices $A$, the naked LU decomposition without pivoting works, since the diagonal entries that are encountered are quotients of main diagonal minors, and all of them are positive. Symmetry results in $U=D\,L^\top$, so that the $A=LDL^\top$ can be cheaply obtained. If wanted, the square root of $D$ may be distributed to the factors to obtain the Cholesky decomposition.
For an invertible matrix $A,$ gaussian elimination is equivalent to constructing an $LU$ factorization of $A$ where $L$ is unit lower triangular and $U$ is upper triangular. There are many algorithms for doing this but if you aren't careful you can get runaway cancellation errors or swamping. By partial pivoting you can keep a lid of sorts on some runaway errors by making sure that at each step of the algorithm you re-order the rows/columns so that any scaling that is done of subsequent rows/columns is by a number of absolute value less than 1. This is not the same idea as reducing a matrix to row echelon form as that is already implicit in performing the $LU$ factorization. Rather partial pivoting refers to a numerical technique in the implementation of an $LU$ (or many other) factorization. This is unnecessary and indeed numerically dubious for a symmetric positive definite matrix since the cholesky factorization can be employed instead.
Pivoting is a more general technique than partial pivoting but the underlying concerns are the same. I would suggest going over your notes or googling the relevant terms and talking with your instructor for more clarification.
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Last updated on 6/23/22
Assess Linearity
Now that you have a better understanding of what linear regression is, and how it applies to companies across industries and departments, let's dive into the first important concept behind it: linearity.
Linearity is, of course, at the heart of linear regression and its variants. It is a simple and elegant property widely used in forecastingneural networks, and many other complex machine learning algorithms. As you will see, it is a very robust element of data analysis.
You probably already have a good idea of what linearity is. When asking to define it, the answer usually involves the idea of a straight line that best fits the samples. For instance, plotting the weight versus the height of a sample of school children gives us the figure below on the left. The goal of linear interpolation is to find the straight line that is closest to all the points in the graph.
And to refresh your memory, the general equation of a line is:
is the classic equation for a line in the two-dimensional (2D) plane. is the slope, is the constant, and for different values of and we obtain different lines as illustrated by the figure below.
Let's go back to the mathematical definition of linearity of a function. Consider a function where is a real number (also called a scalar). The exact mathematical definition of a linear function is the following:
is a linear function if, and only if, it verifies these two conditions:
The first condition is called the multiplicative property:
• for whatever number
The second condition is called the additive property:
•
That's it! It is quite a simple but powerful definition.
You can also verify that a line corresponds to a linear function. Let's start with the equation of a line that goes through the origin point (0,0):
When the slope , the line goes up, and when , the line goes down. And since , all the lines generated by go through the origin point with coordinates .
You can verify the multiplicative property:
The Catch
What about lines that do not go through the point of origin? Lines whose equation is:
with
In that case, you can verify that the function is not linear, at least not according to the strictest definition of linearity. You have:
You can also verify that as an exercise.
The presence of the constant breaks the two conditions required to make the function linear.
Linear functions that respect the additive and multiplicative properties are called homogeneous linear functions.
And now for some nonlinear functions!
What is a nonlinear function?
Any function that does not rely only on multiplication and addition is nonlinear
Examples of nonlinear functions include:
•
This function does not verify the additive or the multiplicative property. Let's verify the additive property:
• is nonlinear since
• similarly is also nonlinear for the same reason:
You get the drift.
The next function is also nonlinear, although it is linear on each segment defined by being negative or positive.
The function spanning negative and positive values of x is nonlinear. This function is called a rectified linear unit (ReLU) is it an activation function widely used in deep learning to introduce nonlinearities in the neural network.
So far, we have only worked with real numbers. The linearity property of a function can extend to vectors.
A vector of dimension N is simply a set of N numbers taken together as a single object.
In the 2D plane, a vector has 2 numbers . In a 3D volume, a vector has 3 numbers and so forth. Generalizing to N dimensions, a vector has N numbers
To multiply two vectors with , simply write:
The function defined as where and are both dimensional vectors is a linear function in the strictest sense of the definition.
Consider the relations between , our target variable, and two predictors, and .
In this example, is the weight of a middle school student, is the age, and the height. You can imagine that there's a linear relationship between the weight of the student and his/her age and height and write:
Where are coefficients we would like to estimate.
You can rewrite the above equation as a vector equation:
With the vector
Here, you have a linear function with two variables.
If there were more than two predictors of the student's weight but, instead predictors, the equation of the weight as a linear function of all the N predictors would still take the form where would now be a vector of coefficients and a vector of predictors.
What's interesting about linear function is that the linear feature is conserved by linear transformations:
• The sum of two linear functions is still a linear function.
• The multiplication of a linear function is also a linear function.
• The composition of two linear functions is a linear function.
The last point means that if two linear functions are defined by and , the composition of g and f : is still linear! Linearity is very meta.
This means that if you apply linear regression to a dataset, you can transform your variables and still be in the same context of linear functions. Even when your dataset shows some strong nonlinearities.
Linear Modeling of Nonlinearities
The linear model can also extend to model nonlinearities in the data.
Let's say you want to include the height as a predictor of the school children's weight as well as the square of the height. Although the square of the height is not a linear function of the height, you can still build a linear model as such:
This is a linear model with respect to two variables: and . We have simply created a new predictor and used it in the linear model.
How do you know if the relation between your variables is linear? The question is whether there is a linear relationship between one or several of your predictors and the target variable.
One way to check the linearity is to plot the target versus the predictors for each of the predictors in the dataset.
If the plot shows a distinct trend, you can conclude that there is some amount of linearity between the two variables. When the plot shows a different pattern, the relation is not linear.
The following plot shows the relation between the weight and the height in school children. The straight line is the linear interpolation of the data.
As expected, the weight increases with the height. Although not perfect, the relationship between the two variables shows some degree of linearity.
If you look at the height versus the age, the linearity of the relationship is less obvious as shown by the following figure:
The auto-mpg dataset is another dataset we're going to work with throughout the course. The original dataset comes from UCI Machine Learning repository. The auto-mpg dataset lists the fuel consumption in miles per gallon for over 300 cars from the 80s. Each sample has information on acceleration, displacement, weight, and some categorical variables. We'll come back to that dataset later in the course.
For now, let's look at the variables weight and acceleration with respect to the fuel efficiency/ miles per gallon (mpg) target variable.
The plot below on the left shows the mpg versus weight and on the right the mpg versus acceleration. As expected, the heavier the car, the worse its fuel consumption (lower number of miles per gallon).
• The plot on the left shows a distinct decreasing relation between mpg and weight. Some degree of linearity exists between the two variables.
• The plot on the right is hardly interpretable as a linear relation. The points form a sort of central blob, and it would be unwise to infer anything from it.
Looking at plots to assess the linearity of the relation between two variables is fine, but it does not give you enough information on the degree of that linearity. How close is the relation to a single straight line?
At this point several important questions are left unanswered:
• How can you effectively measure the degree of linearity?
• How can you be certain that a linear model properly captures the information contained in the dataset?
To illustrate that last point, let's look at the Height vs. Age plot above. If you trace a linear interpolation and a quadratic one (the curved line) on the same graph, how do you know which interpolation is better? Furthermore, the point on the right seems to have a big impact on the quadratic line, but by how much? What would happen if we removed it?
Linearity is central to this course as well as a wide range of techniques with many diverse application domains, from forecasting and time series to neural networkssignal processing, and modeling of natural phenomena.
In this chapter, we covered the following key elements:
• A strict and mathematical definition of linearity based on additive and multiplicative properties.
• That linear functions deal with scalars as well as vectors with the same simple expression
• That assessing the linearity between two variables can be done visually, but the conclusion is not always obvious or unique.
Looking back at the plots in this chapter, two types of relations between variables emerge:
• One where the plot shows a swarm of points, with no distinguishable patterns; a blob of point. This was the case for the Mpg vs. Acceleration plot.
• Other plots where some patterns can be distinctively observed, such as the Weight vs. Height in schoolchildren plot.
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