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# What Are The Theorems Of Triangles?
In addition to these, measures of interior angles in a triangle sum to 180° base angles of isosceles triangles are congruent; the segment connecting the midpoints of two sides of a triangle is parallel to the third side and half the length; and the medians of a triangle meet at a point are also proved.
Angles:
Right Angles All right angles are congruent.
Base Angle Theorem (Isosceles Triangle) If two sides of a triangle are congruent, the angles opposite these sides are congruent.
Base Angle Converse (Isosceles Triangle) If two angles of a triangle are congruent, the sides opposite these angles are congruent.
Definition. The triangle is a sort of polygon that has three sides, and the two sides that are linked end to end are referred to as the hypotenuse. As we explained in the introduction,
## How do you find the theorems of triangle?
Triangle theorems are mostly presented in terms of the angles and sides of a triangle.Triangles are polygons with three sides and three angles, and they are the simplest shape.When we look at the triangle’s sides, we need to pay attention to the lengths of the sides and determine if they are equal to each other or not.When there are no equal sides, the triangle is known as a scalene triangle.
## What are the 3 triangle congruence theorems?
Triangle Congruence Theorems are a type of congruence theorem that is used to prove that two triangles are congruent.1 Side Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle 1 Angle (ASA) 2 Side Angles on One Side (SAS) 3 Sides, 3 Sides, 3 Sides (SSS)
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## What are the three theorems for similarity in triangles?
The three theorems of similarity in triangles are all dependent on the existence of matching portions in the triangles.You examine one angle of one triangle and compare it to the angle of the other triangle that is in the same location as the first.The concept of similarity is connected to the concept of proportion.It is simple to assess triangles for proportionate modifications that maintain them comparable in shape.
## What is the most famous triangle theorem?
Pythagoras’ triangle theorem is probably the most well-known and most debated of all the triangle theorems. ‘In a right-angled triangle, the square of the hypotenuse side equals the sum of the squares of both other sides,’ according to Pythagoras’ theorem Class 10, which asserts that
## What are the 5 theorems of a triangle?
There are five methods for determining whether or not two triangles are congruent: SSS, SAS, ASA, AAS, and HL, to name a few.
## How many theorems are there in triangles?
SSS (side-side-side) and SAS (side-side-side) are two triangle similarity theorems that you will study about in today’s geometry lesson (side-angle-side). There are three theorems that may be used to demonstrate triangle similarity in total: The AA Theorem is a mathematical formula. The SAS Theorem is a mathematical formula that describes the relationship between two variables.
## What are the theorems of triangles Class 10?
Triangle Theorems for Students in Class 10 The similarity of two triangles is established when their respective angles are equal and their corresponding sides have the same ratio. It is possible to compare two triangles by observing that their respective angles are equal and that their corresponding sides are in the same proportion as their respective sides.
## What are the five theorems?
The following five theorems, in particular, are attributed to him: (1) a circle can be bisected by any diameter; (2) the base angles of an isosceles triangle are equal; (3) the opposite (″vertical″) angles formed by the intersection of two lines are equal; (4) two triangles are congruent (of equal shape and size); and (5) two triangles are congruent (of equal shape and size).
## What are the 4 triangle congruence theorems?
In order to determine if a triangle is congruent, four conditions must be met: side – side – side (SSS), side – angle – side (SAS), angle – side – angle (ASA), angle – angle – side (AAS), and angle – angle – side (AAS) (AAS).
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## What are the right triangle theorems?
Demonstration of the Right Angle Triangle If the square of one side of a triangle is equal to the sum of squares of the other two sides, then the angle opposite the first side is a right angle, according to the following theorem:
## What are the 3 types of theorem?
1. Complements Theorems with Angle Theorems Congruent Supplements If two angles are supplementary to the same angle or of congruent angles, then the two angles are congruent
2. otherwise, the two angles are incongruent.
3. The Right Angles Theorem is a mathematical formula that states that right angles are equal to 90 degrees. Right angles are formed when two angles are both complementary and congruent
4. otherwise, they are right angles.
5. The Theorem of Interior Angles on the Same Side
6. The Theorem of Vertical Angles
## What are the examples of theorem?
A conclusion that has been proven to be correct (using operations and facts that were already known). Consider the following example: The ‘Pythagoras Theorem’ demonstrated that a2 + b2 = c2 in the case of a right-angled triangle.
## How many theorems are there in circles Class 9?
1st Theorem: Equal chords of a circle subtend equal angles at their intersections. Two-thirds of a circle’s chords are equal if the angles subtended by the chords of a circle at its center are also equal. The perpendicular from the center of a circle to the chord bisects the chord, according to Theorem 3.
## What chapter is BPT theorem?
Triangles of Class 10 (Theorem 6.1) – Basic Proportionality Theorem (BPT) – Chapter 6 Basic Proportionality Theorem (BPT) (Term 1)
## What is theorem 11 in geometry?
Theorem 11: If three parallel lines cut off equal segments on a transversal line, then they will cut off equal segments on any other transversal line that they intersect. Primary.
## What is a triangle class 9?
A triangle is a closed shape made by three crossing lines (the word triangle comes from the Latin word tri, which meaning ″three″).A triangle has three sides, three angles, and three vertices, all of which are equal in length.For example, the letter ABC is indicated as ABC in the triangle ABC.The three sides are denoted by AB, BC, and CA, the three angles are denoted by A, B, and C, and the three vertices are denoted by A, B, and C.
2.
## What are the 3 triangle similarity theorems?
Triangles that are similar to one another are easy to detect because you can use three triangle-specific theorems to help you. The Angle – Angle (AA) theorem, the Side – Angle – Side (SAS) theorem, and the Side – Side – Side (SSS) theorem are all infallible ways for finding similarity in triangles.
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## What is theorem 20 in geometry?
It is true that if two sides of a triangle are congruent, the angles on either side of the triangle are also congruent.
## What are the basic theorems in geometry?
It is true that C lies on if A, B, and C are distinct points and if AC + CB = AB, then C is true that C is true Theorem: For any three points A, B, and C, the product AC + CB is true. Pythagorean if c is the hypotenuse, then theorem states that a2 + b2 = c2. Angle Pairs are a type of angle pair. Complementary angles add up to a total of 90 degrees.
## What are the 5 congruency theorems for triangles?
1. SSS: (If the three corresponding sides of two triangles are equal, then the triangles are congruent)
2. ASA: (If the three corresponding sides of two triangles are equal, then the triangles are congruent)
3. (If two corresponding angles and a side between these angles are equal of two triangles then triangles are congruent).
4. SAE: (If two triangles have two matching sides and an angle between these sides that is equal to one another, then the triangles are congruent.)
## Can the Pythagorean theorem work for all types of triangles?
The Pythagorean Theorem may be applied with any shape and with any formula that squares a number, and it is very useful in mathematics.When it comes to right triangles, why does the Pythagorean Theorem only work?According to the theorem, the hypotenuse is the longest side of the triangle and is located on the opposite side of the triangle from the right angle.As a result, we might conclude that the Pythagorean theorem is exclusively applicable to right triangles.
## What are the three methods of proving triangles congruent?
With any shape and with any formula that squares a number, the Pythagorean Theorem may be used.When it comes to right triangles, why does the Pythagorean Theorem only hold true?A triangle has a hypotenuse, which is the longest side of the triangle and is opposite the right angle.This is according to the theorem of triangles.
The Pythagorean theorem, as a result, is only valid for right triangles, as can be seen in the illustration. |
# Ex.1.2 Q5 Rational Numbers Solution - NCERT Maths Class 8
## Question
Find five rational numbers between,
(i) \begin{align}\;\frac{2}{3}\end{align} and \begin{align}\frac{4}{5}\end{align}
(ii)\begin{align}\;\frac{{ - 3}}{2}\end{align} and \begin{align}\frac{5}{3}\end{align}
(iii) \begin{align}\frac{1}{4}\end{align} and \begin{align}\frac{1}{2}\end{align}
Video Solution
Rational Numbers
Ex 1.2 | Question 5
## Text Solution
What is known?
Rational numbers.
What is unknown?
The rational numbers between given rational numbers.
Reasoning:
We can find infinitely many rational numbers between any two given rational numbers by taking the mean of the two rational numbers. Another method: We can make the denominator same for the two given rational numbers.
Steps:
(i) \begin{align}\frac{2}{3}\end{align} and \begin{align}\frac{4}{5}\end{align}
\begin{align}\frac{2}{3} = \frac{{2 \times 20}}{{3 \times 20}} = \frac{{40}}{{60}}\end{align} [multiplying both numerator and denominator by \begin{align}20\end{align}]
\begin{align}\frac{4}{5} = \frac{{4 \times 12}}{{5 \times 12}} = \frac{{48}}{{60}}\end{align} [multiplying both numerator and denominator by \begin{align}12\end{align}]
The five rational numbers between \begin{align}\frac{2}{3}\end{align} and \begin{align}\frac{4}{5}\end{align} that can be taken are:
\begin{align}\frac{{41}}{{60}},\;\;\frac{{42}}{{60}},\;\;\frac{{43}}{{60}},\;\;\frac{{44}}{{60}},\;\;\frac{{45}}{{60}}\end{align}
(ii) \begin{align}\frac{{ - 3}}{2}\end{align} and \begin{align}\frac{5}{3}\end{align}
\begin{align}\frac{{ - 3}}{2} = \frac{{ - 3 \times 3}}{{2 \times 3}} = \frac{{ - 9}}{6}\end{align} [multiplying both numerator and denominator by \begin{align}3\end{align}]
\begin{align}\frac{5}{3} = \frac{{5 \times 2}}{{2 \times 3}} = \frac{{10}}{6}\end{align} [multiplying both numerator denominator by \begin{align}2\end{align}]
\begin{align}\therefore\end{align} The five rational numbers between \begin{align}\frac{{ - 3}}{2}\end{align} and \begin{align}\frac{5}{3}\end{align} that can be taken are:
\begin{align} - \frac{8}{6},\;\; - \frac{7}{6},\;\;-1,\;\;\frac{5}{6},\;\;\frac{4}{6}\end{align}
[There can be more such rational numbers]
(iii) \begin{align}\frac{1}{4}\end{align} and \begin{align}\frac{1}{2}\end{align}
\begin{align}\frac{1}{4} = \frac{{1 \times 8}}{{4 \times 8}} = \frac{8}{{32}}\end{align} [multiplying both numerator and denominator by \begin{align}8\end{align}]
\begin{align}\frac{1}{2} = \frac{{1 \times 16}}{{2 \times 16}} = \frac{{16}}{{32}}\end{align} [multiplying both numerator and denominator by \begin{align}16\end{align}]
Thus, five rational numbers between \begin{align}\frac{1}{2}\end{align} and \begin{align}\frac{1}{4}\end{align} that can be taken are:
\begin{align}\frac{9}{{32}},\;\;\frac{{10}}{{32}},\;\;\frac{{11}}{{32}},\;\;\frac{{12}}{{32}},\;\;\frac{{13}}{{32}}\end{align}
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# Multiplying - Rational Numbers Class 8 Notes | EduRev
## Class 8 : Multiplying - Rational Numbers Class 8 Notes | EduRev
``` Page 1
Georgia Department of Education
Common Core Georgia Performance Standards Framework Teacher Edition
MATHEMATICS ? GRADE 7 ? UNIT 1: Operations with Rational Numbers
Georgia Department of Education
Dr. John D. Barge, State School Superintendent
April 2012 ? Page 70 of 105
You have recently practiced multiplying positive and negative integers on a number line. It is
The rules for moving along the number line are as follows:
Moving to the left or west means moving in a negative direction.
Moving to the right or east means moving in positive direction.
Time in the future is represented by a positive value.
Time in the past is represented by a negative value.
Multiplying Using the Number Line Model
Try these problems on your own.
In the city of Mathematica, there is a town center which attracts many visitors to the city. From
the town center, a train takes visitors to different popular locations. The map shows a few
favorite destinations that people like to visit. Answer the following problems about traveling
around the city.
1. The train leaves Town Center traveling east at the speed of 2 blocks per minute. How
many blocks will you be in 4 minutes? Where will you be in 4 minutes?
2. The train leaves the Town Center going west at 2 blocks per minute. What popular
location will you arrive at in 6 minutes? How many blocks away from the Town Center
will you be?
3. The train passes through the Town Center going east at 2 blocks per minute. Where was
that train 3 minutes ago?
Page 2
Georgia Department of Education
Common Core Georgia Performance Standards Framework Teacher Edition
MATHEMATICS ? GRADE 7 ? UNIT 1: Operations with Rational Numbers
Georgia Department of Education
Dr. John D. Barge, State School Superintendent
April 2012 ? Page 70 of 105
You have recently practiced multiplying positive and negative integers on a number line. It is
The rules for moving along the number line are as follows:
Moving to the left or west means moving in a negative direction.
Moving to the right or east means moving in positive direction.
Time in the future is represented by a positive value.
Time in the past is represented by a negative value.
Multiplying Using the Number Line Model
Try these problems on your own.
In the city of Mathematica, there is a town center which attracts many visitors to the city. From
the town center, a train takes visitors to different popular locations. The map shows a few
favorite destinations that people like to visit. Answer the following problems about traveling
around the city.
1. The train leaves Town Center traveling east at the speed of 2 blocks per minute. How
many blocks will you be in 4 minutes? Where will you be in 4 minutes?
2. The train leaves the Town Center going west at 2 blocks per minute. What popular
location will you arrive at in 6 minutes? How many blocks away from the Town Center
will you be?
3. The train passes through the Town Center going east at 2 blocks per minute. Where was
that train 3 minutes ago?
Georgia Department of Education
Common Core Georgia Performance Standards Framework Teacher Edition
MATHEMATICS ? GRADE 7 ? UNIT 1: Operations with Rational Numbers
Georgia Department of Education
Dr. John D. Barge, State School Superintendent
April 2012 ? Page 71 of 105
4. You would like to take the train to the Zoo from the Town Center. How many minutes
will this take if the train travels 2 blocks per minute? Write a math sentence to represent
this scenario.
5. You waited 4 minutes for the train to arrive. The train was traveling west at 2 blocks per
minute. Where was the train?
Let’s look to see if there are any patterns.
1. When you moved east (right) and it was time in the future, two positive numbers
represented the situation. What was the result on the number line when moving east and
moving in the future occurred? Was the result of multiplying positive or negative? Is this
always true?
2. When you moved east (right) and it was time in the past, a positive number and a
negative number represented the situation. What was the result on the number line when
moving east and moving in the past occurred? Was the result of multiplying positive or
negative? Is this always true?
3. When you moved west (left) and it was time in the future, a negative number and a
positive numbers represented the situation. What was the result on the number line when
moving west and moving in the future occurred? Was the result of multiplying positive or
negative? Is this always true?
4. When you moved west (left) and it was time in the past, two negative numbers
represented the situation. What was the result on the number line when moving west and
moving in the past occurred? Was the result of multiplying positive or negative? Is this
always true?
5. What multiplication patterns can you see from each situation? Fill in the chart below
according to the signs of the factors and products.
Factor Factor Product
+ number + number
(-) number (-) number
+ number (-) number
(-) number + number
```
Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!
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# How do you find the second derivative of y^2=x^3?
##### 1 Answer
Jul 29, 2015
y''= (3)/(4sqrtx)
#### Explanation:
You can use implicit differentiation:
$D \left({y}^{2}\right) = D \left({x}^{3}\right)$
$2 y y ' = 3 {x}^{2}$
$y ' = \frac{3 {x}^{2}}{2 y}$
Use the quotient rule:
$y ' ' = \frac{2 y .6 x - 3 {x}^{2} .2 y '}{4 {y}^{2}}$
Subs for $y ' \Rightarrow$
$y ' ' = \frac{12 x y - 6 {x}^{2} \left(3 {x}^{2} / \left(2 y\right)\right)}{4 {y}^{2}}$
$y ' ' = \frac{12 x y - 9 {x}^{4} / y}{4 {y}^{2}}$
$y ' ' = \frac{12 x {y}^{2}}{4 {y}^{3}} - \frac{9 {x}^{4}}{4 {y}^{3}}$
${y}^{2} = {x}^{3} \Rightarrow$
$y ' ' = \frac{12 {x}^{4}}{4 {y}^{3}} - \frac{9 {x}^{4}}{4 {y}^{3}}$
$y ' ' = \frac{3 {x}^{4}}{4 {y}^{3}}$
Since ${y}^{2} = {x}^{3}$ this becomes:
$y ' ' = \frac{3 {x}^{4}}{4 y {x}^{3}}$
$y ' ' = \frac{3 x}{4 y}$
${y}^{2} = {x}^{3}$
$y = {x}^{\frac{3}{2}}$
Substituting for y gives:
$y ' ' = \frac{3 x}{4 {x}^{\frac{3}{2}}}$
$y ' ' = \frac{3}{4 \sqrt{x}}$
You can also use explicit differentiation:
${y}^{2} = {x}^{3}$
$y = \sqrt{{x}^{3}}$
$y = {x}^{\frac{3}{2}}$
Apply the power rule:
$y ' = \frac{3}{2.} {x}^{\frac{1}{2}}$
And again for the 2nd derivative:
$y ' ' = \frac{3}{4} {x}^{- \frac{1}{2}}$
y''=3/4.(1)/(x^(1/2)
$y ' ' = \frac{3}{4 \sqrt{x}}$
Either way gets you there. |
# Algebra Basics: From Variables to Polynomials
RenewedForsythia
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Study Flashcards
## 12 Questions
### What is the purpose of solving an equation in algebra?
To find the value of the variable(s) that make the statement true
Linear equation
### What do variables in algebraic equations represent?
Unknown values that can vary
### What is a system of equations in algebra?
Two or more equations involving the same variables
Equations
### What is the main characteristic of a linear equation?
It involves only one variable and one or two constants
### What is the primary purpose of solving a system of linear equations?
Determining the point(s) of intersection of the graphs of the equations
### In which field beyond mathematics is algebra commonly used?
Civil engineering
### What is the purpose of solving algebraic inequalities?
Finding all values that satisfy a statement involving two expressions
### What does factoring in algebra involve?
Finding the factors that multiply to form a polynomial
### Which best describes an algebraic expression?
A mathematical phrase using variables, coefficients, and operations
### How does mastering algebraic concepts benefit individuals according to the text?
Opening opportunities for learning and discovery
## Study Notes
Algebra, a cornerstone of modern mathematics, is a versatile and powerful tool that enables us to solve a wide range of problems. From balancing your checkbook to understanding the behavior of the natural world, algebraic thinking has become an essential skill in today's world. Let's delve into the fascinating world of algebra, where variables, equations, and problem-solving intersect.
### Variables and Equations
At the heart of algebra are variables, represented by letters like (x) and (y), and equations, which express relationships between variables. An equation is a statement that sets two expressions equal to one another, such as (2x + 3 = 11). The goal of solving an equation is to find the value of the variable(s) that make the statement true.
### Solving Linear Equations
The most common type of equation in algebra is a linear equation, which involves only one variable and either one or two constants. For example, (3x + 5 = 14). Remember that the goal of solving an equation is to find the value of the variable(s) that make the statement true, which in this case is (x = 3).
### Linear Equation Systems
While solving single linear equations is a valuable skill, learning to solve systems of linear equations is even more powerful. A system of equations consists of two or more equations involving the same variables. Solving a system of linear equations requires finding the point(s) of intersection of the graphs of the equations, which corresponds to the values of the variables that make all the equations true simultaneously.
### Applications of Algebra
Algebra is an indispensable tool in many fields beyond mathematics, including science, technology, and engineering. For example, algebra helps us to analyze data in science experiments, design structures in civil engineering, and create algorithms in computer science.
### Algebraic Inequalities
An algebraic inequality is a statement that two expressions are not equal. For example, (3x + 1 > 6x - 2). Solving an inequality involves finding all the values of the variable(s) that make the statement true. Inequalities are useful for analyzing data (e.g., finding the range of possible values for a variable) and in optimization (e.g., finding the maximum or minimum value of a function).
### Polynomials and Factoring
Polynomials are expressions that involve variables raised to non-negative integer powers. For example, (x^2 + 3x + 2) is a polynomial. Solving polynomial equations, especially those with quadratic terms ((x^2)), requires factoring, or finding the factors of the expression that multiply together to form the original polynomial. Factoring is an essential algebraic skill that has numerous applications, including finding roots and solving equations.
### Algebraic Expressions
An algebraic expression is a mathematical phrase that uses variables, coefficients, and operations. For example, (3x + 5y) is an expression. Algebraic expressions are vital for solving equations and inequalities, and they are also fundamental to more advanced topics like calculus and optimization.
### Conclusion
Algebra is a powerful tool that enables us to solve a wide range of problems in many fields. By mastering algebraic concepts and techniques, we can open the door to countless opportunities for learning and discovery. As you explore the fascinating world of algebra, remember to practice patience and persistence, and to seek help when needed. With these skills in hand, you'll be well on your way to becoming a capable and confident algebraic thinker.
Explore the fundamental concepts of algebra such as variables, equations, linear equations, systems of equations, algebraic inequalities, polynomials, and algebraic expressions. Enhance your problem-solving skills and understand how algebra applies to various fields like science, engineering, and computer science.
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# Geometric Sequence
Have you ever noticed that when a ball bounces, its height decreases by half each bounce? This is an example of a geometric sequence. A sequence is a set of numbers that all follow a certain pattern or rule.
A geometric sequence is a type of numeric sequence that increases or decreases by a constant multiplication or division.
A geometric sequence is also sometimes referred to as a geometric progression.
Each number in a sequence is referred to as a term.
Geometric sequences can help you calculate many things in real life, such as:
1, 3, 9, 27, 81, ... is a geometric sequence: every number in the sequence is obtained by multiplying the previous number by 3.
In this case, 3 is the so-called common ratio of the geometric sequence: let's find out more about it.
## Geometric sequence common ratio
If you are given a term of a geometric sequence you can find the following term by multiplying the initial term by a constant, known as the common ratio. This procedure is known as the term to term rule.
The common ratio is often denoted as r.
Some examples of geometric sequences include:
• 3, 6, 12, 24, 48... This sequence has a common ratio of 2 since each term is obtained by multiplying the previous one by 2.
• 5, 20, 80, 320, 1280... This sequence has a common ratio of 4 since each term is obtained by multiplying the previous one by 4.
• 32, 16, 8, 4, 2, 1, 0.5... This sequence has a common ratio of 0.5 since each term is obtained by multiplying the previous one by 0.5.
### How to find the common ratio in a geometric sequence
When you are given a geometric sequence, you may not be given the common ratio. It can be helpful to be able to figure this out in case you need to find the next terms of the geometric sequence. To find the common ratio you shall divide one term by the term before it.
Find the common ratio for the geometric sequence 6, 18, 54, 162, 486...
Solution:
To find the common ratio of this geometric sequence, take the second term and divide it by the first term of the sequence. To check your result, divide the third term by the second; the fourth by the third; and so on.
Therefore the common ratio for this geometric sequence is 3.
## nth term of geometric sequence
It is possible to use the term to term rule to find the nth terms of a geometric sequence. To do this multiply or divide the term you have by the common ratio to find the next term of the sequence.
Find the next three terms of the geometric sequence 8, 40, 200, 1000...
Solution:
First, you need to identify the common ratio:
To make sure that the common ratio is 5, check the following terms:
Now you know that the common ratio is 5, you can use that to find the next terms of the sequence. Just multiply the last term by the common ratio and repeat that to find the next three terms:
Therefore, the next three terms of the sequence are 5000, 25000, 125000
Since you are multiplying the terms, the terms will rapidly increase or decrease.
Find the first five terms of the geometric sequence where the first term is 13 and the common ratio is 2.
Solution:
To find each term you can start by multiplying the first term by the term to term rule:
Now you can continue to multiply the term to term rule by the previous term:
Therefore the first five terms of the sequence are 13, 26, 52, 104, and 208.
Find the first three terms of the geometric sequence where 1000 is the first term and the common ratio is .
Solution:
To do this you need to multiply each term by to find the next:
Therefore the first three terms are; 1000, 250, 62.5
## Difference between arithmetic and geometric sequence
The difference between an arithmetic sequence and a geometric sequence is the way in which the terms go from one to another. In an arithmetic sequence the terms increase or decrease by a constant addition or subtraction. In a geometric sequence the terms increase or decrease by a constant multiplication or division.
## Geometric sequence examples with solutions
Identify the common ratio in the following geometric sequence: 11, 33, 99, 297...
Solution:
To find the common ratio divide the second term of the sequence by the first term of the sequence and so on:
Therefore, the term to term rule of this sequence is 3.
Find the next 3 terms of the geometric sequence 9, 18, 36, 72, 144…
Solution:
First, identify the common ratio:
Now find the next terms by multiplying the common ratio by the previous term;
Therefore, the next three terms of the sequence are; 288, 576, and 1152.
Find the first five terms of the geometric sequence where 5 is the first term and the common ratio is 4.
Solution:
To do this you need to multiply each term by 4 to find the next:
Therefore the first five terms of the sequence are; 5, 20, 80, 320, and 1280.
Identify the term to term rule in the following geometric sequence; 100, 80, 64, 51.2
Solution:
To find the common ratio divide one term by the previous term in the sequence on so on:
Therefore, the term to term rule of this sequence is 0.8 or .
## Geometric Sequences - Key takeaways
• A geometric sequence is a numerical sequence that increases or decreases by a constant multiplication.
• The constant ratio between each term in the sequence is called the common ratio.
• The common ratio can be used to generate terms of the sequence.
A geometric sequence is a type of linear sequence that increases or decreases by a constant multiplication or division.
To find the difference between an arithmetic and geometric sequence you must find out how the sequences are increasing or decreasing;
• If the sequence is increasing or decreasing by a constant addition or subtraction, it is an arithmetic sequence.
• If the sequence is increasing or decreasing by a constant multiplication or division it is a geometric sequence.
A finite geometric sequence is a geometric sequence that has an end.
The sum to infinity of a geometric sequence is when all the terms in the sequence are added together.
In order to find the nth term of a geometric sequence you can multiply or divide the last term with the common ratio in order to find the next term.
## Final Geometric Sequence Quiz
Question
What is a geometric sequence?
A geometric sequence is a type of sequence that increases or decreases by a constant multiplication or division.
Show question
Question
What is the common ratio?
The common ratio is the number that is multiplied or divided by each term to get to the next term in a geometric sequence.
Show question
Question
What is the term to term rule in a geometric sequence and what can it be used for?
The term to term rule describes the common ratio between two terms in a geometric sequence. It can be used to find terms within a sequence.
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Question
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The common ratio is 4.
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The common ratio is 2
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3750, 18750, 93750
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1, 6, 36, 216, 1296.
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10, 20, 40, 80, 160.
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20, 30, 45, 67.5, 101.25.
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8, 16, 32, 64, 128
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1377, 4131, 12393
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21.6, 12.96, 7.78
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12, 48, 192, 768, 3072
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# How do you solve (- 2y - 3) ^ { 3} + 25( - 2y - 3) = 0?
Jun 8, 2017
$y = - 1.5$
$y = - \frac{3}{2} + \frac{5}{2} i$ and $y = - \frac{3}{2} - \frac{5}{2} i$ also work.
#### Explanation:
The Shorter and Easier Way
We see that there are 2 $\left(- 2 y - 3\right)$ terms in the equation. This might signal something useful. So, let's temporarily replace $\left(- 2 y - 3\right)$ with $x$. Now, let's rewrite the equation:
${x}^{3} + 25 x = 0$
Using the Distributive Property, we can factor out an $x$ from the left side:
$x \cdot \left({x}^{2} + 25\right) = 0$
Looking at our new equation, we can see that the two terms' products is $0$. This means that either $x$ is $0$ or ${x}^{2} + 25$ is $0$. Let's start with the case of $x = 0$ and solve for $y$. However, before we start, remember that $x$ is actually $\left(- 2 y - 3\right)$:
$x = 0$
$- 2 y - 3 = 0$
$- 2 y = 3$
$y = - \frac{3}{2}$
Besides this single answer of $y = - \frac{3}{2}$, there are two more roots (solutions) which involve complex numbers. To obtain these two other solutions, we have to look at the case where ${x}^{2} + 25 = 0$.
${x}^{2} + 25 = 0$
${x}^{2} = - 25$
$x = \sqrt{-} 25$
$x = \sqrt{-} 1 \cdot \sqrt{25}$
$x = \pm 5 i$
The final statement shows that $x$ can be either positive $5 i$ of negative $5 i$. We'll start by working with positive $5 i$. And, because we let $x = \left(- 2 y - 3\right)$, we can substitute it in and find the second value for $y$:
$- 2 y - 3 = 5 i$
$- 2 y = 5 i + 3$
$y = \frac{5 i + 3}{-} 2$
$y = \frac{5 i}{- 2} + \frac{3}{- 2}$
$y = \left(- \frac{5}{2}\right) i - \frac{3}{2}$
$y = - \frac{3}{2} - \left(\frac{5}{2}\right) i$
The third solution for $y$ involves seeing $x$ as equal to $- 5 i$:
$- 2 y - 3 = - 5 i$
$- 2 y = - 5 i + 3$
$y = \frac{- 5 i + 3}{-} 2$
$y = \frac{- 5 i}{-} 2 + \frac{3}{-} 2$
$y = - \frac{3}{2} + \left(\frac{5}{2}\right) i$
The Long and Hard Way
${\left(- 2 y - 3\right)}^{3} + 25 \left(- 2 y - 3\right) = 0$
Let's expand everything out and use the distributive property to simplify things a bit:
$\left(- 2 y - 3\right) \left(- 2 y - 3\right) \left(- 2 y - 3\right) + \left(25\right) \left(- 2 y\right) - \left(25\right) \left(3\right) = 0$
$\left(- 2 y - 3\right) \left(4 {y}^{2} + 6 y + 6 y + 9\right) + \left(- 50 y\right) - 75 = 0$
$\left(4 {y}^{2}\right) \left(- 2 y - 3\right) + \left(6 y\right) \left(- 2 y - 3\right) + \left(6 y\right) \left(- 2 y - 3\right) + \left(9\right) \left(- 2 y - 3\right) - 50 y - 75 = 0$
$\left(- 8 {y}^{3} - 12 {y}^{2}\right) + \left(- 12 {y}^{2} - 18 y\right) + \left(- 12 {y}^{2} - 18 y\right) + \left(- 18 y - 27\right) - 50 y - 75 = 0$
Now let's break open the parentheses:
$- 8 {y}^{3} - 12 {y}^{2} - 12 {y}^{2} - 18 y - 12 {y}^{2} - 18 y - 18 y - 27 - 50 y - 75 = 0$
And now, let's start rearranging like terms together and simplifying them:
$- 8 {y}^{3} - 12 {y}^{2} - 12 {y}^{2} - 12 {y}^{2} - 18 y - 18 y - 18 y - 50 y - 27 - 75 = 0$
$- 8 {y}^{3} - 36 {y}^{2} - 104 y - 102 = 0$
At this point, we can use a cubic calculator to calculate the value of $y$:
$y = - 1.5$
It's important that two other solutions also work perfectly fine, not just $y = - 1.5$. The two other solutions are: $y = - \frac{3}{2} + \frac{5}{2} i$ and $y = - \frac{3}{2} - \frac{5}{2} i$.
............................................................................................................................
Checking:
If we were to check whether or not $y$ is indeed $- 1.5$, we would substitute $y = - 1.5$ back into the original equation:
${\left[- 2 \left(- 1.5\right) - 3\right]}^{3} + 25 \left[- 2 \left(- 1.5\right) - 3\right] = 0$
${\left(3 - 3\right)}^{3} + 25 \left(3 - 3\right) = 0$
${0}^{3} + 25 \left(0\right) = 0$
$0 + 0 = 0$
$0 = 0$ |
## Wednesday, July 5, 2017
### High School Solutions – Functions Calculator, Range (Part I)
Last blog post, we discussed what a domain was and how to find the domain. In this blog post, we will talk about what a range is and how to determine the range of linear, radical, and quadratic functions.
The range of a function is the set of values of the dependent variable (i.e. the y values or output values) for which a function is defined. Another way to think about the range is as the image of the function. The domain is what we can put in the function and the range is what comes out of the function.
Linear functions -
The range of linear functions is always -∞<y<∞ or (-∞,∞) since the function is defined at all the outputs.
Find the range of \f(x)=3x
Range: -∞<f(x)<∞, or (-∞,∞)
For radical functions there is a simple rule to follow to find the range. Given a radical function, written \f(x)=c\sqrt{ax+b}+k, f(x)≥k is the range.
Find the range of \f(x)=2\sqrt{x+3}-2
We can see here that k = -2
Range: f(x)≥-2 or (-2,∞)
Finding the range of a quadratic function is a little trickier. When given a quadratic function, we know there is a parabola. The goal is to find the vertex of the parabola and figure out if it is a minimum or a maximum.
Steps to determining the range of a quadratic function:
1. Find the vertex
• x= -\frac{b}{2a},given \f(x)=ax^2+bx+c
• Plug in x into the function to get the y coordinate of the vertex
2. Determine if the vertex is a minimum or a maximum
• If a<0, then the vertex is a maximum
• If a>0, then the vertex is a minimum
3. Write the range
• If the vertex is a maximum, then the range of the function is all the points below and equal to the vertex’s y coordinate
• If the vertex is a minimum, then the range of the function is all the points above and equal to the vertex’s y coordinate
Find the range of \f(x)=x^2+5x+6
1. Find the vertex
x=-\frac{5}{2∙1}=-\frac{5}{2}
f(-\frac{5}{2})=(-\frac{5}{2})^2+5(-\frac{5}{2})+6=-\frac{1}{4}
Vertex is at (-5/2,-1/4)
2. Determine if the vertex is a minimum or a maximum
a=1>0
Vertex is minimum
3. Write the range
Range:\f(x)≥-\frac{1}{4} or (-\frac{1}{4},∞)
Find the range of \f(x)=-4x^2+2x+4
1. Find the vertex
x=-\frac{2}{2∙-4}=\frac{1}{4}
f(\frac{1}{4})=-4(\frac{1}{4})^2+2(\frac{1}{4})+4=\frac{17}{4}
Vertex is at (\frac{1}{4},\frac{17}{4})
2. Determine if the vertex is a minimum or a maximum
a=-4<0
Vertex is a maximum
3. Write the range
Range: f(x)≤17/4, or (-∞,\frac{17}{4})
Finding the range of a function is trickier than finding the domain of a function. We have to think about the outputs instead of the inputs, which can be confusing. The best way to get better at this is to keep practicing and memorizing how to find the range of different functions. Next blog post, we will talk about how to find the range of rational functions. For more help or practice on this topic, visit Symbolab’s Practice.
Until next time,
Leah
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# What Are The Prime Factors Of 82
What are the prime factors of 82? Answer: 2, 41
The number 82 has 2 prime factors. Primes can only have two factors(1 and itself) and only be divisible by those two factors. Any figure where this rule applies can be called a prime factor. The biggest prime factor of 82 is 41. The smallest prime factor of 82 is 2.
## What Is The Factor Tree Of 82
How to use a factor tree to find the prime factors of 82? A factor tree is a diagram that organizes the factoring process.
• First step is to find two numbers that when multiplied together equal the number we start with.
82
↙ ↘
2 × 41
We found 2 prime factors(2, 41) using the factor tree of 82. Now let us explain the process to solving factor trees in more detail. Our goal is to find all prime factors of a given whole number. In each step of our factor tree diagram for 82 we always checked both multiplication numbers if they were primes or not. If one or both of the integers are not prime numbers then this means that we will have to make diagrams for them too. This process continues until only prime numbers are left.
Remember that often a factor tree for the same integer can be solved in more than one correct way! An example of this is the figure 12 where 2*6=12 and 4*3=12. The primes of a factor tree for 12 are the same regardles if we start the factor tree with 2*6 or 4*3.
## How To Verify If Prime Factors Of 82 Are Correct Answers
To know if we got the correct prime factors of 82 we have to get the prime factorization of 82 which is 2 * 41. Because when you multiply the primes of the prime factorization the answer has to be equal with 82.
After having checked the prime factorization we can now safely say that we got all prime factors.
## General Mathematical Properties Of Number 82
82 is a composite number. 82 is a composite number, because it has more divisors than 1 and itself. This is an even number. 82 is an even number, because it can be divided by 2 without leaving a comma spot. This also means that 82 is not an odd number. When we simplify Sin 82 degrees we get the value of sin(82)=0.31322878243309. Simplify Cos 82 degrees. The value of cos(82)=0.94967769788254. Simplify Tan 82 degrees. Value of tan(82)=0.32982640650768. When converting 82 in binary you get 1010010. Converting decimal 82 in hexadecimal is 52. The square root of 82=9.0553851381374. The cube root of 82=4.3444814857686. Square root of √82 simplified is 82. All radicals are now simplified and in their simplest form. Cube root of ∛82 simplified is 82. The simplified radicand no longer has any more cubed factors.
## Determine Prime Factors Of Numbers Smaller Than 82
Learn how to calculate primes of smaller numbers like:
## Determine Prime Factors Of Numbers Bigger Than 82
Learn how to calculate primes of bigger numbers such as:
## Single Digit Properties For Number 82 Explained
• Integer 8 properties: Eight is even and a cube of 2. It is a composite, with the following 4 divisors:1, 2, 4, 8. Since the total of the divisors(excluding itself) is 7<8, it is a defective number. The sixth of the Fibonacci sequence, after 5 and before 13. It is the quantity of the twin primes 3 and 5. The first octagonal value. 8 is a Ulam, centered heptagonal and Leyland number. All amounts are divisible by 8 if and only if the result formed by its last three digits is. A refactorizable, being divisible by the count of its divisors. At the same time a highly totter and highly cototent quantity. It is the fourth term of the succession of Mian-Chowla. Any odd greater than or equal to 3, elevated to the square, to which subtract is subtracted 1 is divisible by 8 (example: 7²=49 49-1=48 divisible by 8). The sum of two squares, 8=2²+2². The sum of the digits of its cube: 8³=512, 5+1+2=8. The first 4-digit binary:1000. Part of the Pythagorean triples (6, 8, 10), (8, 15, 17). Eight is a repeated number in the positional numbering system based on 3 (22) and on the base 7 (11).
• Integer 2 properties: 2 is the first of the primes and the only one to be even(the others are all odd). The first issue of Smarandache-Wellin in any base. Goldbach's conjecture states that all even numbers greater than 2 are the quantity of 2 primes. It is a complete Harshad, which is a integer of Harshad in any expressed base. The third of the Fibonacci sequence, after 1 and before 3. Part of the Tetranacci Succession. Two is an oblong figure of the form n(n+1). 2 is the basis of the binary numbering system, used internally by almost all computers. Two is a number of: Perrin, Ulam, Catalan and Wedderburn-Etherington. Refactorizable, which means that it is divisible by the count of its divisors. Not being the total of the divisors proper to any other arithmetical value, 2 is an untouchable quantity. The first number of highly cototent and scarcely totiente (the only one to be both) and it is also a very large decimal. Second term of the succession of Mian-Chowla. A strictly non-palindrome. With one exception, all known solutions to the Znam problem begin with 2. Numbers are divisible by two (ie equal) if and only if its last digit is even. The first even numeral after zero and the first issue of the succession of Lucas. The aggregate of any natural value and its reciprocal is always greater than or equal to 2.
## Finding All Prime Factors Of A Number
We found that 82 has 2 primes. The prime factors of 82 are 2, 41. We arrived to this answer by using the factor tree. However we could have also used upside down division to get the factorization primes. There are more that one method to factorize a integer.
## List of divisibility rules for finding prime factors faster
Rule 1: If the last digit of a figure is 0, 2, 4, 6 or 8 then it is an even number. All even numbers are divisible by 2.
Rule 2: If the sum of digits of a integer is divisible by 3 then the figure is also divisible by 3 and 3 is a prime factor(example: the digits of 12 are 1 and 2 so 1+2=3 and 3 is divisible by 3, meaning that 12 is divisible by 3). The same logic also works for 9.
Rule 3: If the last two digits of a number are 00 then this integer is divisible by 4(example: we know that 124=100+24 and 100 has two zeros in the end making it divisible with 4. We also know that 4 is divisible with 24). In order to use this rule to it's fullest it is best to know multiples of 4.
Rule 4: If the last digit of a number is 0 or 5 then 5 it is divisible by 5.
Rule 5: All integers that are divisible by both 2 and 3 are also divisible by 6. This is logical because 2*3=6.
## What Are Prime Factors Of A Number?
All numbers that are only divisible by one and itself are called prime factors in mathematics. A prime factor is a figure that has only two factors(one and itself). |
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# NUMFACT - Editorial
Author: Vamsi Kavala
Tester: Roman Rubanenko
Editorialist: Bruno Oliveira
Cakewalk
# PRE-REQUISITES
Simple Math, Integer Factorization
## Problem:
You are given a very large number represented as a product of N numbers. Given this number representation, you need to find the number of distinct factors of the original number which is formed by the product of given N numbers.
## Quick Explanation:
We can factorize each one of the N given numbers into its prime factors. Then we find the number of occurrences of each prime factor, say they are a1, a2,...aK, if we have K distinct prime factors. Our answer is simply: (a1+1)(a2+1)(...)*(aK+1).
## Detailed Explanation:
This problem relies on some knowledge of divisor function. Divisor functions returns the number of positive and distinct divisors of a number. Let's call it d(x).
• Some properties of the divisor function:
We now look into some important properties of the divisor function:
For a prime number p, we have d(p) = 2, as there are only two numbers which divide a prime number:1 and itself.
Now, it's a known fact that this function is multiplicative but not completely multiplicative. This means that if two numbers, say, a and b are there such that gcd(a, b) = 1, then the following holds: d(a*b) = d(a)*d(b).
This allows us to deduce the important relationship, that is the key of solving this problem:
For a prime number, p, we have: d(p^n) = n+1.
Now, it's easy to understand that all we need to do is to factorize all the N given numbers into its prime factors, and, for each prime factor we also need to count how many times it appears (that is, we need to know the exponent of each prime factor).
Once we have this count with us (which can be done using integer factorization and for example, the set and map data structures, one to guarantee uniqueness of the factors and the other to save the number of occurences for each unique prime factor), all we need to do is to multiply all these numbers plus one together and we will obtain our answer.
As an example, consider the number:
504 = 2^3 * 3^2 * 7^1
The number of distinct divisors of 504 is then (3+1) * (2+1) * (1+1) = 24.
Applying this process to all numbers yields the answer to the problem :)
# SETTER'S SOLUTION
Can be found here.
# TESTER'S SOLUTION
Tester's solution will be uploaded soon.
This question is marked "community wiki".
asked 30 Jun '13, 14:10
3★kuruma
17.7k72143209
accept rate: 8%
45471013
In your code, you are checking whether a[i] == 1 towards the end. But can you give me a case where it wont be equal to 1. Wont the prime numbers stored in the list make sure that the smallest number is being factorized first and then proceed to the largest??
(31 Oct '13, 22:25) 0★
Any special reason for using char type of array for storing primes by setter ?
(22 Jan '16, 03:21)
2 @shasha1s2 wen u update 'other' array to count prime u do not account for the fact that the prime number being counted may be the same..in other words, the value of other[i] (according to your code) will never be more than 1...going by that the test case 5 999983 999983 999983 999983 999983 will give 32 as per your code, while ans is 6. answered 30 Jun '13, 16:40 5★amitrc17 62●1●1●6 accept rate: 0% thanx @amitrc17 got my error... (30 Jun '13, 19:23)
1 @roshi...it will most certainly give a TLE as the time complexity is very high!!!! try using Sieve of Eratosthenes....see the time difference....Naive approach....seive!!! answered 30 Jun '13, 23:08 4★kunal361 6.0k●13●32●72 accept rate: 21% A simple modification of Sieve of Erastothenes will also prove to be helpful. The following is what I used. f[i] == 0 if i is prime, and if i is prime, then f[i] will have the smallest prime that will divide it. for (int i = 2; i <= N; i++) if (!f[i]) for (int j = i+i; j <= N; j += i) if (!f[j]) f[j] = i; f[i] will help a lot to find all the prime divisors of a number. (01 Nov '13, 15:03) turuthok3★ http://www.codechef.com/viewsolution/2894661 A shorter solution with the same concept + the subtle modification of the sieve to store prime factor info. (01 Nov '13, 15:06) turuthok3★
1 @shubham26 if you look at the constraints .They are :- N<=10 Ai<=1000000 and as in your code you multiply each Ai to get a number which is product of all of them. Suppose if all the numbers are 10^6 and N=10 and they are multiplied N times,the number becomes (10^6)^10=10^60 which is far beyond the range of long(range of the order 10^9)( (even its far away from long long(range of the order 10^18)). answered 01 Jul '13, 08:45 174●2●2●5 accept rate: 0% @dhruvagga : thanks..I got it. :) (01 Jul '13, 14:55)
1 In the setter's solution what is the use of this code fragment if(a[i]!=1) { if(m.find(a[i])!=m.end()) m[a[i]]++; else m.insert(MP(a[i],1)); from line 85 to 90 I put that in comments and ran the program and it gave the correct o/p for test cases. I don't understand it's use. After dividing with all the prime numbers shouldn't a[i] necessarily be 1. answered 12 Jul '13, 23:30 3★tdk93 30●2●6 accept rate: 0%
0 Can you tell me where my code fails... I have applied same concept i think.. http://www.codechef.com/viewsolution/2304397 answered 30 Jun '13, 14:15 -1●1●1●2 accept rate: 0%
0 it not works for prime numbers above 1000 like 1009,1013 etc answered 30 Jun '13, 14:23 35●1●2●5 accept rate: 0% it works.. I have used "other" array for those primes Can u provide any testcase where it fails.. (30 Jun '13, 14:35)
0 Either you can use a sieve to extend your primes array or just put more primes in the primes array. here's a sample code for sieve. http://ideone.com/arhRSv answered 30 Jun '13, 15:49 174●2●2●5 accept rate: 0%
0 It gives a TLE when i use the function 'prime' to find the prime factorisation of the number.Can someone please tell me where i went wrong? Thanks in advance. Here is the link to my code : http://ideone.com/kgzICq answered 30 Jun '13, 22:52 2★roshi 61●4●9●19 accept rate: 0%
0 What is wrong in my code?? http://ideone.com/Y5mZLP It gives correct answers to all the numbers with which I have checked. :( answered 01 Jul '13, 02:09 6●1●1●5 accept rate: 0%
0 Can anyone provide critical input. My solutions keeps getting WA, however, when comparing to AC code gives same output. from sys import stdin from collections import Counter from math import sqrt,ceil def primes(n): """ Input n>=6, Returns a list of primes, 2 <= p < n """ correction = (n%6>1) n = {0:n,1:n-1,2:n+4,3:n+3,4:n+2,5:n+1}[n%6] sieve = [True] * (n/3) sieve[0] = False for i in xrange(int(n**0.5)/3+1): if sieve[i]: k=3*i+1|1 sieve[ ((k*k)/3) ::2*k]=[False]*((n/6-(k*k)/6-1)/k+1) sieve[(k*k+4*k-2*k*(i&1))/3::2*k]=[False]*((n/6-(k*k+4*k-2*k*(i&1))/6-1)/k+1) return [2,3] + [3*i+1|1 for i in xrange(1,n/3-correction) if sieve[i]] def factor(n): upper = n i = 0 while sieve[i] <= ceil(sqrt(upper)): while n%sieve[i]==0: n/=sieve[i] if sieve[i] not in factors: factors[sieve[i]]=0 factors[sieve[i]]+=1 i+=1 if(n>1): if upper not in factors: factors[upper]=0 factors[upper]+=1 sieve = primes(10000) factors = {} def solve(): N = int(stdin.readline()) numbers = map(int,stdin.readline().split()) #print([i for i in numbers]) factors.clear() for num in numbers: factor(num) tot = 1 for num in factors: #print("fac",num,factors[num]) tot *= (factors[num]+1) print tot def main(): tc = int(stdin.readline()) for i in range(1,tc+1): solve() if __name__=="__main__": main() answered 28 Oct '13, 00:25 1 accept rate: 0%
0 whats the problem wid following ? i got a WA #include #include int primes[1000000]; int a[1000000]; void gen_primes(int n) { int i,j; for(i=0;i
0 my code is running fine on the first time.. but in second test case giving erronous output.. please help.. #include int freq[1000000]; int sum; void count(long long int x) { int i; for(i=2;;i++) { while(x%i==0) { freq[i]++; x=x/i; sum++; } if(x==1) break; } } int main(void) { int t,n,p,y,i; long long int x; scanf("%d",&t); while(t--) { p=1; y=0; sum=0; scanf("%d",&n); for(i=0;i
0 Where Am i Going Wrong , Its Always SIGSEV and TLE on my Submission :'( , Plz help me Fast i Need to Brace up For IOI #include #include #include void solve(long long int *primes,long long int *n ) { long long int *z = (long long int *)calloc(1001,sizeof (long long int)); while(*n>1) { for(long long int i=0;i<168;i++) { if(*n%primes[i]==0){z[i]++;*n= *n/primes[i];i--;} } } long long int prod=1; for(long long int k=0;k<168;k++)if(z[k]!=0)prod*=(z[k]+1); printf("%lld \n",prod); //for(long int i=0;i
0 Hey whats wrong in my code here? program FACTORS; uses Math; var t,n,num,product:longint; function factors(fact:longint):longint; var no:longint = 0; x:longint; begin for x := 1 to floor(sqrt(fact)) do begin if fact mod x = 0 then begin if x*x = fact then inc(no,1) else inc(no,2); end; end; factors := no; end; begin readln(t); while t <> 0 do begin readln(n); product := 1; while n <> 0 do begin read(num); product := product * num; dec(n); end; writeln(factors(product)); dec(t); end; end. I am getting runtime error NZEC. answered 24 Jun '15, 08:47 31●3 accept rate: 0%
0 I have been trying to debug my code for so long still not able to figure out why it is giving a WA for the basic cases. Please if anyone could tell me where am I going wrong? This is the link to my solution. https://www.codechef.com/submit/complete/8669014 answered 30 Oct '15, 19:30 87●2●6 accept rate: 0%
0 nice explanation. I used sieve of eratosthenes to get all prime factors of number answered 19 Nov '15, 22:33 56●3 accept rate: 30%
0 https://www.codechef.com/viewsolution/14084014 WHY IS THIS SOLUTION GETTING A WA ON LAST CASE PLZ HELP!! answered 05 Jun '17, 18:53 5★soheb17 60●2 accept rate: 0%
#include<stdio.h>
int main()
{
int t;
scanf("%d",&t);
if(t>=1 && t<=100)
{
while(t--)
{
int n;
scanf("%d",&n);
if(n>=1&&n<=10)
{
long int a, total=1;
while(n--)
{
scanf("%ld",&a);
if(a>=2&&a<=1000000)
total*=a;
}
int count=0;
for(int i=1;i<=total;i++)
{
if(total%i==0)
count++;
}
printf("%d\n",count);
}
}
}
return 0;
}
### why time limit exist????
answered 04 Oct '13, 17:20
0
accept rate: 0%
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Maths Worksheet for Kids| Dividing Decimals - Download Free PDF with Solutions
Learn about dividing decimals and boost your practice with the worksheets we have provided for you. Students of Class KG-3 have to learn about dividing decimals numbers just like they do with whole numbers. In these worksheets designed by the experts at Vedantu, students will learn about decimal division and the different methods used. Download the worksheets in the PDF format from Vedantu and practice to learn more about the process.
The Maths experts at Vedantu have formulated the worksheets to teach students how to divide decimals. The worksheets contain different problems the students can solve to enhance their knowledge and score good marks during their tests. If you want to improve your studies best, these worksheets are an ideal choice to include in your study routine.
Did You Know?
Decimals are a part of your life in more ways than you can imagine. Yes, decimals are used in most real-world situations to represent different amounts. Whether it is money or height or weight, all the values tend to use decimal points. The Richter scale used to measure an earthquake's magnitude also uses the decimal value to give an accurate reading.
Last updated date: 26th Sep 2023
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Access Worksheet for Maths KG-2 Dividing Decimal
When dividing decimals, we must move the decimal point to the right to change the divisor to a whole number. The decimal point of the dividend is then moved the same number of places to the right, and the resulting numbers are divided in the same manner as conventional long division.
Questions
1. Find the quotient.
1. 5.2/4
2. 8.4/6
3. 2.8/2
4. 7.5/3
2. Find the missing number.
1. 0.50/_=0.25
2. _/3=0.35
3. 1.56/_=0.78
4. 0.36/_=0.6
3. Find the number.
1. 0.6/0.3=
2. 0.25/0.5=
3. 0.64/0.4=
4. 0.16/0.2=
4. Fill in the blanks.
1. When 0.48 is divided by 0.08, which number comes_____________.
2. When 0.81 is divided by 0.09 which number comes______________.
3. When 0.63 is divided by 0.07 which number comes_____________.
5. What will be the answer in the following cases?
1. 841/10
2. 590/10
3. 600/100
4. 550/10
5. 2000/100
6. Find the missing number.
1. 93/___=0.93
2. 55/____=5.5
3. 151/____=1.51
4. 45/_____=0.45
7. Find the quotient.
1. When 12 is divided by 1000 what will be the quotient___________?
2. When 33 is divided by 100 what will be the decimal number?
8. What will come in the following cases?
1. 15/10______.
2. 15/100______.
3. 15/1000_________.
9. When 0.10 is divided by 10, which number will come as a quotient?
10. When 100 is divided by 1000, what is the answer?
11. Divide the following numbers.
1. 8/100
2. 5/100
3. 2/100
4. 6/100
12. Fill in the missing number.
1. 120/__=1.20
2. 65/____=0.65
3. 87/____=.87
4. 36/____=0.36
13. Fill in the blank.
1. 0.14/0.2=_____
2. 0.21/0.3=_____
3. 0.49/0.7=______
4. 0.45/0.9=______
14. Find the number.
1. 0.18/___=0.6
2. 0.20/____=0.5
3. 0.40/____=0.8
15. What will be the quotient and remainder when 0.65 is divided by 0.08?
1.
1. 5.2/4=1.3
2. 8.4/6=1.4
3. 2.8/2=1.4
4. 7.5/3=1.5
2.
1. 0.50/2 =0.25
2. 1.05 /3=0.35
3. 1.56/ 2 =0.78
4. 0.36/ 0.6=0.6
3.
1. 0.6/0.3=2
2. 0.25/0.5=5
3. 0.64/0.4=16
4. 0.16/0.2=8
4.
1. When 0.48 is divided by 0.08, which number comes to 6.
2. When 0.81 is divided by 0.09 which number comes to 9.
3. When 0.63 is divided by 0.07 which number comes to 9.
5.
1. 841/10=84.1
2. 590/10=59
3. 600/100=6
4. 550/10=55
5. 2000/100=20
6.
1. 93/100=0.93
2. 55/10=5.5
3. 151/100=1.51
4. 45/100=0.45
7.
1. When 12 is divided by 1000, 0.12 is the quotient.
2. When 33 is divided by 100, 0.33 is the decimal number.
8.
1. 15/10= 1.5
2. 15/100= 0.15
3. 15/1000=0.015
9. When 0.10 is divided by 10, 0.01 will come as a quotient.
10. When the number 100 is divided by 1000 0.10 will come.
11.
1. 8/100=0.08
2. 5/100=0.05
3. 2/100=0.02
4. 6/100=0.06
12.
1. 120/100 =1.20
2. 65/100 =0.65
3. 87/10 =.87
4. 36/100 =0.36
13.
1. 0.14/0.2= 7
2. 0.21/0.3= 7
3. 0.49/0.7= 7
4. 0.45/0.9= 5
14.
1. 0.18/ 3 =0.6
2. 0.20/ 4 =0.5
3. 0.40/ 5 =0.8
15. When 0.65 is divided by 0.08, the quotient will be 8 and the remainder will be 0.01.
Importance of Decimal Division Worksheet For Students
• Decimal division is one of the most important topics included in the Maths syllabus of Class KG-3 students. Those who want to enhance their practice of dividing different decimals can practice the lessons from the worksheet provided here.
• By practising regularly, students will be able to learn about the decimal division in hundredths and tens, the division of multi-digit decimals, decimal division with the help of zeros, and much more. The decimal division worksheet also contains some work problems related to decimal division.
• Students will be able to improve their skills in division and multiplication from these worksheets. They can download the study materials and boost their preparation for the tests and examinations. All the questions have solutions provided for easy reference of students.
• They can study the solutions and understand the right pattern for solving such questions. Understanding the process of decimal division becomes much easier with these worksheets.
Benefits of Practicing Decimal Division Worksheets
• Students can refer to the worksheets to amp up their practice and gain more knowledge about the topic of dividing decimals in the most effective manner.
• The experts at Vedantu have explained the solutions using a step-by-step manner to help the students understand each problem properly.
• There are Decimal Division examples included in the worksheets that will give a clear understanding to students who might be facing some doubts about the topic. These examples include all the details that will help students in solving the problems.
• Students will be able to improve their division and multiplication concepts. They will learn how to tackle different short and long-division questions as well.
FAQs on Maths Worksheet for Kids- Dividing Decimals
Students can download the worksheets for dividing decimals from the website or mobile app of Vedantu. These worksheets are free of cost and provide all the study materials needed to complete the chapter.
Vedantu’s worksheets for dividing decimals will enable you to understand the topic effectively. You can practice the questions and enhance your performance during exams.
3. Which way is the decimal point moved when dividing by 10?
The decimal point is moved one place ahead to the left when the number is divided by 10.
4. Can I score good marks in Maths with Dividing Decimals worksheets?
If you download the worksheets and practice regularly, you can achieve great marks in the exams.
5. How can I practice dividing decimals properly?
Download the dividing decimals worksheets from Vedantu and practice the questions every day. Learn the long division and the short division methods to understand the concept. |
# Ch. 8 Tests of Hypotheses Based on a Single Sample
Sections covered: 8.1, 8.2, 8.3, 8.4, 8.5
## 8.1 Hypotheses and Test Procedures
Note: We will cover the beginning of the section and then return to errors in hypothesis testing (beginning on page 317) after Section 8.2.
Skip: Examples 8.1 and 8.4 (hypothesis testing and type II error calculation for small sample proportions)
xkcd cartoon: “Significant” (Jelly Beans Cause Acne!)
## 8.2 z Tests for Hypotheses about a Population Mean
Skip: calculating Type II error and sample size needed for two sided tests (3rd and 5th formulas in the blue box on p. 331)
## 8.3 The One-Sample t Test
Skip: $$\beta$$ and Sample Size Determination (p. 338) to end of section
## 8.4 Tests Concerning a Population Proportion
Skip: $$\beta$$ and Sample Size Determination (pp. 348-349)
## 8.5 Further Aspects of Hypothesis Testing
Skip: The Likelihood Ratio Principle (pp. 355-356)
## Practice Exercises
$$1.$$ (One sample test, Values given) We wish to test whether a scale needs to be recalibrated. If it’s working, a 10 pound weight should weigh 10 pounds. We decide to test it by weighing the same weight, which we know to be precisely 10 pounds, 25 times. We know the weights from this scale are normally distributed and independent, with a true standard deviation of $$\sigma$$ = 1. Our sample mean (x) is 9.85. Should we conclude that recalibration is necessary x (that is, the scale is off) or not necessary (the scale isn’t conclusively off)?
$$α = .01$$
$$H_0: \mu = 10$$
$$H_A: \mu \neq 10$$
Test statistic: $$z=\frac{\bar{X} - \mu_0}{\sigma/\sqrt{n}}$$
z <- (9.85-10)/(1/sqrt(25))
z
## [1] -0.75
Determine the p-value:
pval <- pnorm(z)*2
pval
## [1] 0.4532547
Since 0.453 > .05, we do not reject the null hypothesis. In other words, even if the scale were perfectly calibrated ($$\mu = 10$$) there is over a 45% probability that we would get a sample mean less than or equal to 9.85 (or equal to or greater than 10.15), that is, the same or further from the true mean than we found. Therefore, on the basis on these results, recaliberation is not necessary.
$$2.$$ (One sample test, Raw values) We know that the mean sepal length of setosas (a species of irises) is 4.8 cm. A new study examines a sample of 50 flowers and shows that the sample mean is 5.006 cm. Conduct a one-tailed hypothesis test at $$.05$$ significance level to show if there is sufficient evidence to reject the hypothesis that the mean of the new sample is equal to 4.8 cm.
$$α = .05$$
$$H_0: \mu = 4.8$$
$$H_A: \mu > 4.8$$
Use the following code to store the 50 values in the sample in a variable called setosa. (This works because iris is a built-in dataset in R.)
setosa <- iris[iris$Species == "setosa",] [Ans] Test statistic: $$z=\frac{\bar{X} - \mu_0}{s/\sqrt{n}}$$ z <- (mean(setosa$Sepal.Length)-4.8)/sqrt(var(setosa$Sepal.Length)/50) z ## [1] 4.132433 pval <- 1- pnorm(z) pval ## [1] 1.794718e-05 # simpler approach # p-values might be a bit different due to approximation of t t.test(setosa$Sepal.Length, alternative = "greater", mu = 4.8)
##
## One Sample t-test
##
## data: setosa\$Sepal.Length
## t = 4.1324, df = 49, p-value = 6.986e-05
## alternative hypothesis: true mean is greater than 4.8
## 95 percent confidence interval:
## 4.922425 Inf
## sample estimates:
## mean of x
## 5.006
Since 6.986e-05 < 0.05 , we reject the null hypothesis. There is sufficient evidence that the true mean of setosa sepal length is greater than 4.8 cm. |
# How to Take the Derivative of the Logarithm Function
Hongtao Hao / 2022-11-28
What is the derivative of $f(x) = log(x)$? This post tries to prove that it is $\frac{1}{x}$ when the log base is $e$.
## Derivative of exponential function and the definition of e #
Before talking about the derivative of log function, let us review the definition of $e$. It is related to the derivative of the exponential function, for example, $f(x) = 2^x$.
According to the definition of derivative, we know that the derivative of $f(x) = 2^x$ should be:
$$\lim_{dx \to 0}\frac{2^{x+dx} - 2^x}{dx} = \lim_{dx \to 0}\frac{(2^{dx} - 1) \cdot 2^x}{dx}$$
And we know that as $dx$ approaches $0$, $\frac{2^{dx} - 1}{dx}$ is approaching a constent ($ln(2)$). That is to say, the derivative of $f(x) = 2^x$ is $2^x$ itself multiplied by a constant.
Then we wonder, is there a case where the drivatie of $f(x) = t^x$ is itself? That is to say, when will the constant be $1$? And that is the definition of $e$! We stipulate that if the derivative of $f(x) = t^x$ is $t^x$ itself, then we denote that special $t$ with $e$.
Therefore, we have:
$$\lim_{dx \to 0} \frac{e^{dx} - 1}{dx} = 1$$
So, as $dx$ approaches $0$, we have
$$e^{dx} = dx + 1$$
Therefore, we have
$$\lim_{dx \to 0}(dx + 1)^{\frac{1}{dx}} = e$$
If we replace $dx$ with $n$, we have:
$$\lim_{n \to 0}(n + 1)^{\frac{1}{n}} = e$$
## Direct derivation #
With this background information, we should be able to derive the derivative of the log function. We will take the natural log function as an example, i.e., we have $f(x) = ln(x)$.
\begin{aligned} f^\prime(x) &= \lim_{dx \to 0}\frac{ln(x+dx) - ln(x)}{dx} \\ & = \lim_{dx \to 0}\frac{ln(\frac{x+dx}{dx})}{dx} \\ &= \lim_{dx \to 0}\frac{ln(1 + \frac{dx}{x})}{dx} \\ &= \lim_{dx \to 0}\frac{1}{dx}ln(1 + \frac{dx}{x}) \\ &= \lim_{dx \to 0}{ln(1 + \frac{dx}{x})}^{\frac{1}{dx}}\end{aligned}
Nowt, it’s becoming interesting. It looks similar to the definition of $e$, right? But not quite the same. Let’s say we denote $\frac{dx}{x}$ as $m$, then the above equation can be written as
$$f^\prime(x) = {ln(m + 1)}^{\frac{x}{n}} = {ln(m + 1)}^{\frac{1}{m}\cdot \frac{1}{x}} = \frac{1}{x}{ln(m + 1)}^{\frac{1}{m}}$$
You might jump and say that we can use the definition of $e$ directly, but to do that, we need to prove that $m$ here and the $n$ in the difinition of $e$ is the same.
In the definition of $e$
$$\lim_{n \to 0}(n + 1)^{\frac{1}{n}} = e$$
We have $n$ which is approaching $0$, and $\frac{1}{n}$ approaching positive infinity. Since $m = \frac{dx}{x}$, and $dx$ is approaching $0$, so $m$ is approaching $0$ and $\frac{1}{m} = \frac{x}{dx}$ is approaching positive infinity. Therefore, we can say with certainty that as $dx$ approaches $0$, ${(m + 1)}^{\frac{1}{m}}$ is approaching $e$. Thus, ${ln(m + 1)}^{\frac{1}{m}}$ is approaching 1. Therefore, $f^\prime(x)$ is approaching $\frac{1}{x}$, which means that the derivative of the natural log function is $\frac{1}{x}$.
## A more clever method: inverse function. #
In case you are not familiar with the properties of the derivatives of inversion functions, let’s have an example.
Suppose we have $y = x^2$, so we have $x = \sqrt{y}$. We know that $y^{\prime} = 2x$. And that $x^{\prime} = \frac{1}{2\sqrt{y}} = \frac{1}{2x}$.
We find that $y^{\prime} \cdot x^{\prime} = 1$.
Now, let’s say $y = e^x$, so we have $x = ln(y)$. Since $y^{\prime} = e^x$, we have
$$x^{\prime} = \frac{1}{e^x} = \frac{1}{e^{ln(y)}} = \frac{1}{y}$$
Therefore, if $f(x) = ln(x)$, we have
$$f^{\prime}(x) = \frac{1}{x}$$
#ML |
# WEEBLY VS WIX WEBSITE COMPARISION
## SITUATION
Joseph wants to make a class website for his class, but can’t decide which website maker to use. He was deciding over Wix and Weebly. Joseph wants to keep the website running for 12 months and wants the cheaper deal out of his two options.
## WHAT WE KNOW
Wix offers their users a \$40 fee to get their domain, and after that \$15 every month to keep the site up.
Weebly offers their users a \$20 fee to get their domain, and after that \$25 every month to keep the site up.
## MAKING THE EQUATIONS
In the equations that we will be making the variable "y" represents the total payment, and the variable "x" represents the number of months the website is active.
WIX
total cost = 15 multiplied by number of months plus 40
y = 15x + 40
In standard form, the equation would be -15x + y - 40 = 0
WEEBLY
total cost = 25 multiplied by number of months plus 20
y = 25x + 20
In standard form, the equation would be -25x + y - 20 = 0
## THE GRAPH
The y-axis represents the amount of money owed to the website makers
The x-axis represents the amount of months
This graph will help Joseph make his decision on which website creator to use for this class website.
## THE INTERSECTION
Where do these lines intersect?
The two lines are intersecting at points (2,70)
What does this mean?
This means that on the second month of running his website, Joseph would have to pay \$70 on both of the website plans. If he had Wix, or Weebly he would have paid \$70 in total at the second month.
## JOSEPH'S BEST DEAL
Looking at the graph I would say that Joseph should go for the deal with Wix, because it is cheaper than the deal with Weebly over the 12 month period. To prove my answer I'll use an equation.
Wix
y = 15(12) + 40
y = 180 + 40
y = 220
The total cost for Joseph, if he used Wix for 12 months would be \$220
Weebly
y = 25(12) + 20
y = 300 + 20
y = 320
The total cost for Joseph, if he used Weebly for 12 months would be \$320
By using Wix, Joseph will save \$100 more than if he used Weebly for a year. |
# Expected Value of a Random Variable
When studying statistics, many times you have come across the term “Expected Value”. We came across it in the post on Simple Linear Regression also. Ever wondered what it means? In this post, we are gonna explain to you the concept behind Expected Value.
## What is the meaning of Expected Value?
Expected Value is the average value we get for a certain Random Variable when we repeat an experiment a large number of times. In other words, it can be defined as the ‘theoretical mean’ of a random variable.
It is not based on sample data. It is based on the population. Therefore it is a parameter, not a statistic.
## Detailed Example – 1
Suppose a dice is rolled a large number of times. We are interested in which side is turned upwards.
Here any side turning upwards is equally likely to occur. Since expected value is equal to the mean of the distribution, it can be calculated simply as the average of 1,2,3,4,5,6. The answer is 21/6 = 3.5. Let’s see how we can calculate that using probabilities;
In this case, our Random Variable can be defined as;
#### X = side turned upwards
In this table, you can find Probabilities for each possible value of X and the values for x.P\$(x)\$. This “x.\$P(x)\$” term is referred to as “Probability Mass Function” in Statistics. It is denoted by ‘\$p(x)\$’. You can find more details about the Probability Mass function by clicking here.
\$E(x)\$ for this scenario can be calculated as below;
Okay, now we have got the Expected value of this experiment as 3.5. This means, when you roll this unbiased dice infinite no. of times, on average 3.5 side will turn upwards. Seems unrealistic but in this case that’s what it means.
## The Expected Value of a Function
Sometimes, we have to find the Expected Value of Functions too.
Suppose \$f(x)\$ is a function. Random Variable X has a set of possible values D and \$p(x)\$ is the Probability Mass Function. Then the \$E(x)\$ of \$f(x)\$ can be calculated from the formula below;
## Rules of Expected Value
If the function \$f(x)\$ is of the linear form aX+b. And ‘a’ and ‘b’ are constants, \$E[f(x)]\$ can be calculated as below;
There are another 2 important rules;
Okay, we have learned a considerable amount of fundamentals about Expected Values. I hope you all now have the idea of it. We will discuss this more with detailed examples in a future post. |
# What comes next in the sequence 5 13 29?
## What comes next in the sequence 5 13 29?
The next number in the sequence: 5, 13, 29, 61, will be 125.
What is the next number after 13?
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
### What is the next number in the series 16 11 13?
16 – 11 – 13 – 8 – 10 – 5 – 7 – Summary: The next number in the following series 16 – 11 – 13 – 8 – 10 – 5 – 7 – is “2”.
How do you find the 10th term?
READ: How do people get careers?
To find the 10th term we would follow the formula for the sequence but substitute 10 instead of ‘n’; to find the 50th term we would substitute 50 instead of n. To find the first term we substitute n = 1 into the nth term.
## What kind of number is 16?
16 (sixteen) is the natural number following 15 and preceding 17. 16 is a composite number, and a square number, being 42 = 4 × 4. It is the smallest number with exactly five divisors, its proper divisors being 1, 2, 4 and 8. In English speech, the numbers 16 and 60 are sometimes confused, as they sound very similar.
What come after 110?
Thus, each succeeding number in a column increases by 1. (iv) In each row, each succeeding number increases by 10, i.e., 100, 110, 120, 130, 140, 150, 160, 170, 180 and 190.
### What would be the next number in the following series 3 10 24?
Hence, “52” is the correct answer.
What should be the next number in the following series 100 96?
184
100, 96, 104, 88, 120, 56,…? Summary: The next number in the following series 100, 96, 104, 88, 120, 56,… is 184.
## What is the next number in the sequence 1 2 3 4?
Step by step solution of the sequence is Series are based on square of a number 1 = 12, 4 = 22, 9 = 32, 16 = 42, 25 = 52 ∴ The next number for given series 1, 2, 3, 4, 5 is 6 ∴ Next possible number is 62 = 36
What is the alternating pattern of 13 and 29?
The first pattern begins with 13 and adds 2 to each number to arrive at the next; the alternating pattern begins with 29 and subtracts 3 each time. What will come at the place of question mark? 1, 9, 25, 49,?, 121. What will come at the place of question mark?
### What does nextnumber do?
NextNumber finds the next number in a sequence of numbers Find next number About NextNumber• Classic Sequences• Contact NextNumber
READ: Are people with bipolar night owls?
How do you find 22 with the difference 6 to 16?
Let us try to use this theory in a question. 1,2,4,7,11,16,? We can clearly observe that the series is increasing with the difference : +1, +2, +3 ,+4 , +5. So, we will obtain our number by adding 6 to 16 which gives us 22. |
# Comparing Quantities with Percent
Examples, videos, and solutions to help Grade 7 students understand that the whole is 100% and think of one quantity as a percent of another using the formula Quantity = Percent x Whole to problem-solve when given two terms out of three from a quantity, whole, and percent.
### Lesson 3 Student Outcomes
• Students use the context of a word problem to determine which of two quantities represents the whole.
• Students understand that the whole is 100% and think of one quantity as a percent of another using the formula Quantity = Percent x Whole to problem-solve when given two terms out of three from a quantity, whole, and percent.
• When comparing two quantities, students compute percent more or percent less using algebraic, numeric, and visual models.
Lesson 3 Summary
• Visual models or arithmetic methods can be used to solve problems that compare quantities with percents. • Equations can be used to solve percent problems using the basic equation: Quantity = Percent x Whole • “Quantity” in the new percent formula is the equivalent of “part” in the original percent formula.
Lesson 3 Classwork
Opening Exercise
If each 10 x 10 unit square represents one whole, then what percent is represented by the shaded region?
In the model above, 25% represents a quantity of students. How many students does the shaded region represent?
Exercises
1. There are 750 students in the 7th grade class and 625 students in the 8th grade class at Kent Middle School.
a. What percent is the 7th grade class of the 8th grade class at Kent Middle School?
b. The principal will have to increase the number of 8th grade teachers next year if the 7th grade enrollment exceeds 110% of the current 8th grade enrollment. Will she need to increase the number of teachers?
2. At Kent Middle School, there are 104 students in the band and 80 students in the choir. What percent of the number of students in the choir is the number of students in the band?
3. At Kent Middle School, breakfast costs \$1.25 and lunch costs \$3.75. What percent of the cost of lunch is the cost of breakfast?
4. Describe a real world situation that could be modeled using the equation: 398.4 = 0.83(x). Describe how the elements of the equation correspond with the real world quantities in your problem. Then solve your problem.
Lesson 3 Classwork
Example 1
a. The members of a club are making friendship bracelets to sell to raise money. Anna and Emily made 54 bracelets over the weekend. They need to produce 300 bracelets by the end of the week. What percent of the bracelets were they able to produce over the weekend?
b. Anna produced 32 bracelets of the 54 bracelets produced by Emily and Anna over the weekend. Compare the number of bracelets that Emily produced as a percent of those that Anna produced.
c. Compare the number of bracelets that Anna produced as a percent of those that Emily produced.
Example 2
The 42 students that play wind instruments represent 75% of the students who are in band. How many students are in band?
Example 3
Bob’s Tire Outlet sold a record number of tires last month. One salesman sold 165 tires. which was 60% of the tires sold in the month. What was the record number of tires sold?
Example 4
At Kent Middle School, there are 104 students in the band and 80 students in the choir. What percent of the number of students in the choir is the number of students in the band?
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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. |
# One Proportion Z-Test: Definition, Formula, and Example
one proportion z-test is used to compare an observed proportion to a theoretical one.
This tutorial explains the following:
• The motivation for performing a one proportion z-test.
• The formula to perform a one proportion z-test.
• An example of how to perform a one proportion z-test.
### One Proportion Z-Test: Motivation
Suppose we want to know if the proportion of people in a certain county that are in favor of a certain law is equal to 60%. Since there are thousands of residents in the county, it would be too costly and time-consuming to go around and ask each resident about their stance on the law.
Instead, we might select a simple random sample of residents and ask each one whether or not they support the law:
However, it’s virtually guaranteed that the proportion of residents in the sample who support the law will be at least a little different from the proportion of residents in the entire population who support the law. The question is whether or not this difference is statistically significant. Fortunately, a one proportion z-test allows us to answer this question.
### One Proportion Z-Test: Formula
A one proportion z-test always uses the following null hypothesis:
• H0p = p0 (population proportion is equal to some hypothesized population proportion p0)
The alternative hypothesis can be either two-tailed, left-tailed, or right-tailed:
• H1 (two-tailed): p ≠ p0 (population proportion is not equal to some hypothesized value p0)
• H1 (left-tailed): p < p0 (population proportion is less than some hypothesized value p0)
• H1 (right-tailed): p > p0 (population proportion is greater than some hypothesized value p0)
We use the following formula to calculate the test statistic z:
z = (p-p0) / √p0(1-p0)/n
where:
• p: observed sample proportion
• p0: hypothesized population proportion
• n: sample size
If the p-value that corresponds to the test statistic z is less than your chosen significance level (common choices are 0.10, 0.05, and 0.01) then you can reject the null hypothesis.
### One Proportion Z-Test: Example
Suppose we want to know whether or not the proportion of residents in a certain county who support a certain law is equal to 60%. To test this, will perform a one proportion z-test at significance level α = 0.05 using the following steps:
Step 1: Gather the sample data.
Suppose we survey a random sample of residents and end up with the following information:
• p: observed sample proportion = 0.64
• p0: hypothesized population proportion = 0.60
• n: sample size = 100
Step 2: Define the hypotheses.
We will perform the one sample t-test with the following hypotheses:
• H0p = 0.60 (population proportion is equal to 0.60)
• H1p ≠ 0.60 (population proportion is not equal to 0.60)
Step 3: Calculate the test statistic z.
z = (p-p0) / √p0(1-p0)/n = (.64-.6) / √.6(1-.6)/100 = 0.816
Step 4: Calculate the p-value of the test statistic z.
According to the Z Score to P Value Calculator, the two-tailed p-value associated with z = 0.816 is 0.4145.
Step 5: Draw a conclusion.
Since this p-value is not less than our significance level α = 0.05, we fail to reject the null hypothesis. We do not have sufficient evidence to say that the proportion of residents who support the law is different from 0.60.
Note: You can also perform this entire one proportion z-test by simply using the One Proportion Z-Test Calculator. |
# Word Problems Multiply And Divide Mixed Numbers Worksheet
This multiplication worksheet concentrates on educating individuals the way to emotionally grow total figures. College students can make use of personalized grids to match specifically one concern. The worksheets also includefractions and decimals, and exponents. You will even find multiplication worksheets with a dispersed home. These worksheets can be a have to-have for your mathematics class. They can be used in course to learn how to emotionally flourish entire numbers and line them up. Word Problems Multiply And Divide Mixed Numbers Worksheet.
## Multiplication of entire figures
You should consider purchasing a multiplication of whole numbers worksheet if you want to improve your child’s math skills. These worksheets may help you expert this fundamental principle. You can opt for a single digit multipliers or two-digit and about three-digit multipliers. Abilities of 10 can also be an excellent solution. These worksheets will enable you to training very long multiplication and practice looking at the figures. They are also a wonderful way to help your child comprehend the value of understanding the various kinds of entire amounts.
## Multiplication of fractions
Possessing multiplication of fractions on the worksheet will help educators strategy and make instruction efficiently. Employing fractions worksheets enables teachers to swiftly evaluate students’ understanding of fractions. Individuals can be challenged in order to complete the worksheet within a a number of efforts and then label their solutions to see exactly where they need more coaching. Students can be helped by word conditions that associate maths to genuine-daily life conditions. Some fractions worksheets involve samples of contrasting and comparing figures.
## Multiplication of decimals
When you increase two decimal amounts, make sure to group of people them up and down. If you want to multiply a decimal number with a whole number, the product must contain the same number of decimal places as the multiplicant. As an example, 01 by (11.2) by 2 will be similar to 01 x 2.33 by 11.2 except when the product has decimal areas of below two. Then, the product is round towards the local complete variety.
## Multiplication of exponents
A arithmetic worksheet for Multiplication of exponents can help you process multiplying and dividing phone numbers with exponents. This worksheet will also give conditions that will demand individuals to flourish two various exponents. By selecting the “All Positive” version, you will be able to view other versions of the worksheet. Besides, also you can get into specific instructions in the worksheet by itself. When you’re finished, you can click “Generate” and the worksheet will be saved.
## Section of exponents
The essential rule for division of exponents when multiplying figures would be to subtract the exponent from the denominator from the exponent in the numerator. However, if the bases of the two numbers are not the same, you can simply divide the numbers using the same rule. By way of example, \$23 divided by 4 will equal 27. This method is not always accurate, however. This procedure can lead to confusion when multiplying phone numbers that are too large or too small.
## Linear functions
If you’ve ever rented a car, you’ve probably noticed that the cost was \$320 x 10 days. So, the total rent would be \$470. A linear function of this particular type has the develop f(x), in which ‘x’ is the number of time the car was booked. Additionally, it provides the form f(by) = ax b, where ‘b’ and ‘a’ are real amounts. |
# Number Chart
In maths, numbers play an important role as it is the basic requirement to deal with any concept. Numbers are introduced to students from elementary level to make them understand the patterns. At this level, we use different number charts for kids to help them easily memorise the counting of numbers. In this article, you will learn the definition of numbers chart, different patterns in numbers chart along with practice questions.
What is a Number Chart?
A numbers chart is defined as a table that lists the numbers in numerical order. This order will be like 1-10 in the first line, 11-20 in the second line, 21-30 in the third line and so on. There can be different kinds of specific numbers charts, such as hundred’s charts that contain the numerals from 1–100. These charts can help students identify numerals and the numerical order and establish patterns within the numbers. For example, even numbers, odd numbers, counting by fives, tens, etc. Some of the number charts may only have the written numerals or contain the integers with a picture representation of the number of objects with each digit.
## Number Chart 1 to 100
The below figure represents the number chart from 1 to 100. It is also called a hundred’s chart.
### Number Chart in Words
It is possible to write the numbers in words so that it is easy to understand how to read the numericals. The below chart shows the numbers from 1 to 20 along with the words, i.e. the value of numerical in words.
## Number Chart Patterns
We can observe different patterns in number charts such as even numbers, odd numbers, prime numbers and so on. The below figure shows the even numbers from 1 to 100, where the empty spaces should be filled with odd numbers.
Now, let’s have a look at the chart of odd numbers, where we have to fill the empty cells with even numbers.
### Number Chart 500 to 1000
In this section, you can get the number chart from 501 to 1000. This chart is given in the downloadable pdf format.
Apart from the above given numbers charts, we can also create charts based on skip counting. Below are a few practice questions on completing number charts based on different patterns.
1. Complete the chart given below.
Starting at 1, skip-count by 3, and fill in the missing numbers.
2. Identify the missing numbers in the below chart.
Starting at 5, skip-count 5, and fill in the missing numbers.
3. Identify the pattern and write the numbers in the missing places.
#### Watch The Below Video To Know about Number Patterns
Visit www.byjus.com for more information on charts and patterns on numbers. Also, download BYJU’S – The Learning App for interactive videos on number patterns. |
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# Distance From A Point To A Line
The Distance between two points is given by
D = √ (dp2 + dq2);
Where ‘D’ is the distance;
‘P’ is the coordinates of x – axis;
‘Q’ is the coordinates of y – axis;
‘dp’ is the difference between the x – coordinates of the points;
And ‘dq’ is the difference between the y – coordinates of the points.
Suppose we have coordinates of the points then we use the above formula for finding the distance between the coordinates of the Point.
Or in other words the length of the line segment is also said to be the distance of a line.
The Distance From a Point to a Line when the coordinates are: (p1, q1) and (p2, q2);
Then the distance between two points is given by the pythagoras theorem:
D = √ (p2 - p1)2 + (q2 - q1)2;
And the formula for three coordinates (p1, q1, r1) and (p2, q2, r2) then the distance between three points is given by:
D = √ (p2 - p1)2 + (q2 - q1)2+ (r2 - r1)2;
And in general the distance between two points ‘p’ and ‘q’ is given by:
D = |p – q| = √∑a = 1|pa – qa|2,
Suppose we have the coordinates of a points are (5, 8) and (-9, -3) then find the distance between two points.
We know that the coordinates of the points is (p1, q1) and (p2, q2);
Here the value of p1 = 5;
And the value of p2 = -9;
The value of q1 = 8;
The value of q2 = -3;
Then the distance between two points is:
The formula for finding the distance between two points is:
D = √ (p2 - p1)2 + (q2 - q1)2;
Then put the values in the given formula:
D = √ ((-9) - 5)2 + ((-3) - 8)2;
On further solving we get:
D = √ (14)2 + (-11)2;
D = √ 196 + 121;
D = √317;
D = 17.80.
So the distance from point to line is 17.80.
## Using Trigonometry
There are several problems in mathematics that involve use of Trigonometry for finding the quantities like height of a tower or a pole or any other vertical distance, angle of elevation and demotion, horizontal distances etc. Trigonometry can be categorized in following:
1. Core
2. Plane
3. Spherical
4. Analytic
Let us now understand how to so...Read More
## Using Two Line Equations
Line is a straight figure having one dimesion. Representation of linear relationships between the variables is shown by drawing a line. Using two line equations we can find out whether two lines are parallel to each other or not. If two lines are lying in a plane such that we get an Intersection Point on solving them (i.e. the values of unknown variables are pos...Read More
## When the Line is Horizontal and Vertical
Vertical line can be defined as the line whose x- coordinate remains same and y –coordinate changes. We can say a line which goes straight up and down and also parallel to the y – axis of the coordinate plane is known as vertical line. All points lie on the line having same x – coordinate. No Slope is defined for vertical line. |
# Geometry
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### How to Solve a Similar Polygon Problem
Recall that similar polygons are polygons whose corresponding angles are congruent and whose corresponding sides are proportional. The figure below shows similar pentagons,
### How to Prove Triangles Similar Using the AA Theorem
You can use the AA (Angle-Angle) method to prove that triangles are similar. The AA theorem states that if two angles of one triangle are congruent to two angles of another triangle, then the triangles
### How to Prove Triangles Similar with SSS~
You can prove that triangles are similar using the SSS~ (Side-Side-Side) method. SSS~ states that if the ratios of the three pairs of corresponding sides of two triangles are equal, then the triangles
### How to Prove Triangles Similar with SAS~
You can prove that triangles are similar using the SAS~ (Side-Angle-Side) method. SAS~ states that if two sides of one triangle are proportional to two sides of another triangle and the included angles
### How to Use CASTC after Proving Triangles Similar
CASTC is simply an acronym that stands for ‘Corresponding angles of similar triangles are congruent.’ You often use CASTC in a proof immediately after proving triangles similar
### How to Use the Angle-Bisector Theorem
The Angle-Bisector theorem states that if a ray bisects an angle of a triangle, then it divides the opposite side into segments that are proportional to the other two sides. The following figure illustrates
### How to Identify Radii, Chords, and Diameters
When you work with circles, there are three straight-line components that you need to be able to identify: radii, chords, and diameters.
### What You Need to Know About a Circle's Radius and Chords
When you’re working with a circle, there are five important theorems that you need to know about the properties of the circle’s radii and chords. (There are really just three ideas; but two of the theorems
### Six Important Circle Theorems
The six circle theorems discussed here are all just variations on one basic idea about the interconnectedness of arcs, central angles, and chords (all six are illustrated in the following figure):
### How a Tangent Relates to a Circle
A line is tangent to a circle if it touches it at one and only one point. If a line is tangent to a circle, then it is perpendicular to the radius drawn to the point of tangency. Check out the bicycle
### How to Determine the Length of an Arc
An arc’s length means the same commonsense thing length always means — you know, like the length of a piece of string (with an arc, of course, it’d be a curved piece of string). Make sure you don’t mix
### How to Solve Problems with the Altitude-0n-Hypotenuse Theorem
In a right triangle, the altitude that’s perpendicular to the hypotenuse has a special property: it creates two smaller right triangles that are both similar to the original right triangle.
### How to Determine the Measure of an Angle whose Vertex Is Inside a Circle
An angle that intersects a circle can have its vertex inside, on, or outside the circle. This article covers angles that have their vertex inside a circle—so-called
### Geometry Symbols You Should Know
Using geometry symbols will save time and space when writing proofs, properties, and figuring formulas. The most commonly used geometry symbols and their meanings are shown below.
### Proof Strategies Summarized
The following strategies can help you a great deal when you’re working on two-column geometry proofs. You should review these strategies and practice using them until they become internalized. These strategies
### Bisecting and Trisecting Segments
Bisection and trisection involve cutting something into two or three equal parts. If you’re a fan of bicycles and tricycles and bifocals and trifocals — not to mention the biathlon and the triathlon, bifurcation
### Bisecting and Trisecting Angles
The terms angle bisection and angle trisection describe two ways in which you can divide up an angle equally into two (or three) smaller, congruent angles. Their definitions are often used in proofs.
### Geometry Formulas You Should Know
Below are several of the most important geometry formulas, theorems, properties, and so on that you use for solving various problems. If you get stumped while working on a problem and can’t come up with
### How to Solve a Common-Tangent Problem
The common-tangent problem is named for the single tangent segment that’s tangent to two circles. Your goal is to find the length of the tangent. These problems are a bit involved, but they should cause
### How to Solve Similar Triangle Problems with the Side-Splitter Theorem
You can solve certain similar triangle problems using the Side-Splitter Theorem. This theorem states that if a line is parallel to a side of a triangle and it intersects the other two sides, it divides
### Using the Hypotenuse-Leg-Right Angle Method to Prove Triangles Congruent
The HLR (Hypotenuse-Leg-Right angle) theorem — often called the HL theorem — states that if the hypotenuse and a leg of one right triangle are congruent to the hypotenuse and a leg of another right triangle
### How to Do an Indirect Proof
Indirect proofs are sort of a weird uncle of regular proofs. With an indirect proof, instead of proving that something must be true, you prove it indirectly
### Definitions and Theorems of Parallel Lines
Parallel lines are important when you study quadrilaterals because six of the seven types of quadrilaterals (all of them except the kite) contain parallel lines. The eight angles formed by parallel lines
### Working with More than One Transversal
When a parallel-lines-with-transversal drawing contains more than three lines, identifying congruent and supplementary angles can be kind of challenging. The following figure shows you two parallel lines
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# The kinetic energy of a body is increased by 100% then the percentage change inmomentum of the body is?A. 4.14%B. 41.4%C. 141.4%D. none of these
Last updated date: 17th Jun 2024
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Hint:Kinetic energy is the energy possessed by a moving body and the product of its mass and velocity is defined to be its momentum. Write the expressions for the kinetic energy and momentum of a body and find a relation between the two quantities. The use the given data and find the percentage change in momentum.
Formula used:
$K=\dfrac{1}{2}m{{v}^{2}}$
$P=mv$
When a body is in motion we say that it possesses some energy. This energy of the body is called the kinetic energy of the body.
If the body of mass m is moving with a speed v then it possesses a kinetic energy equal to
$K=\dfrac{1}{2}m{{v}^{2}}$ …. (i)
We also define another term for a moving body and that is momentum of the body. When the body of mass m is moving with a velocity whose magnitude is v, the magnitude of its momentum is given
as $P=mv$ … (ii)
From equation (ii) we get that $v=\dfrac{P}{m}$,
Substitute this value of v in equation (i).
$\Rightarrow K=\dfrac{1}{2}m{{\left( \dfrac{P}{m} \right)}^{2}}$
$\Rightarrow K=\dfrac{{{P}^{2}}}{2m}$ …. (iii)
Now, it is given that the kinetic energy of the body is increased by 100%. This means that the new kinetic energy of the body is 2K.
Let the new momentum of the body be P’.
Then $\Rightarrow 2K=\dfrac{P{{'}^{2}}}{2m}$ …. (iv).
Now, divide (iv) by (iii).
$\Rightarrow \dfrac{2K}{K}=\dfrac{\dfrac{P{{'}^{2}}}{2m}}{\dfrac{{{P}^{2}}}{2m}}$
$\Rightarrow 2=\dfrac{P{{'}^{2}}}{{{P}^{2}}}$
$\Rightarrow P{{'}^{2}}=2{{P}^{2}}$
$\Rightarrow P'=\pm \sqrt{2}P$
We shall not consider the negative value because we need the magnitude of the momentum.
$\Rightarrow P'=\sqrt{2}P$.
The percentage increase in the momentum of the body is $\dfrac{P'-P}{P}\times 100=\dfrac{\sqrt{2}P-P}{P}\times 100=\left( \sqrt{2}-1 \right)\times 100=41.1$%.
Hence, the correct option is B.
Note:Note that kinetic energy is scalar quantity and momentum is a vector quantity. However, in this solution we understand that the magnitude of momentum and kinetic energy are related.
We can say that momentum of a body tells us how much kinetic energy the body possesses. |
Factors the 38 are the genuine numbers that divide the original number, evenly. As soon as 38 is split by that is factor, climate the resultant quotient is a entirety number and also there is no remainder left. As we understand 38 is a composite number, it will have factors much more than two.
You are watching: What are all the factors of 38
Multiples are created when 38 is multiplied by consecutive organic numbers. Hence, multiples and factors are different in meaning. However 38 is both a factor and also multiple of itself, since,
38 x 1 = 38 38/38 = 1 Now let us learn to find the factors of 38 along with pair factors and also prime factors.
## How to Find components of 38?
Factors of 38 room the integers that can divide the initial number, wholly. Currently to find the factors, we should check through which numbers, 38 is totally divisible. Let us start dividing 38 by the smallest natural number, 1.
38 ÷ 1 = 38
38 ÷ 2 = 19
38 ÷ 19 = 2
38 ÷ 38 = 1
Hence, we have the right to conclude the 1, 2, 19 and also 38 space the determinants of 38.
### Factors of -38
The factors of -38 will certainly be additive inverse of natural numbers, i.e. Negative numbers because dividing -38 by any an unfavorable number will an outcome in a optimistic value.
(-38) ÷ (-1) = 38(-38) ÷ (-2) = 19(-38) ÷ (-19) = 2(-38) ÷ (-38) = 1
Hence, determinants of -38 space -1, -2, -19 and -38.
## Pair components of 38
Pair components of 38, results in the original number, after acquiring multiplied to every other, such that;
1 × 38 = 38
2 × 19 = 38
Hence, the pair factors are (1, 38) and (2, 19).
The negative pair components of 38 will certainly also an outcome in the original number, after multiplication. Therefore,
-1 × -38 = 38
-2 × -19 = 38
Therefore, the negative pair factors are (-1, -38) and also (-2, -19).
Note: lock are called pair factors due to the fact that they create the original number when multiplied in a pair. If you readjust the pair, climate we will not get the really result.
## Prime Factorisation that 38
We know, element factorisation that a number generates the prime factors. This prime factors are the prime numbers, which have actually only two factors, i.e. 1 and number itself. The examples are 2, 3, 5, 7,11, etc.
Now to find the prime components of the number 38, we require to check if 38 is completely divisible by the prime number or not. Thus, we will certainly start dividing 38 by the the smallest prime factor, i.e.,2.
Step 1: Dividing 38 by 2, we get;
38/2 = 19
So, 2 is among the element factors
Step 2: Now, 19 chin is a prime number, therefore, it can be split by 19 only.
19/19 = 1
Thus, 19 is one more prime factor
Step 3: currently further department is no possible. Therefore, we conclude the 2 and 19 are prime determinants of 38.
## Solved Examples
Q.1: Sonu is distributing cupcakes in his class on the occasion of his birthday. If there are 38 cupcakes and also 19 students, climate how plenty of cupcakes go each college student get?
Solution: Given,
Number the cupcakes = 38
Number of students = 19
Therefore, the number of cupcakes got by each student = 38/19 = 2
Q.2: uncover the sum of all the components of 38. Also, discover the average.
Solution: The factors of 38 space 1, 2, 19 and also 38.
Sum = 1+2+19+38 = 60
Therefore, 60 is the required sum.
Now to find the average, we have to divide the amount by the total variety of factors.
Therefore,
Average = 60/4 = 15
Q.3: What room the typical factors that 34 and also 38?
Answer: Both 34 and 38 room composite numbers. Thus, their determinants are:
34 → 1, 2, 17 and also 34
38 → 1, 2, 19 and 38
Clearly, the usual factors the 34 and also 38 space 1 and 2.
See more: What Is The Greatest Common Factor Of 20 And 32 And 20, Find Gcf Of 20 And 32 |
# Permutation and Combination - Aptitude Questions and Answers
## Permutation and Combination Questions and Answers
Learn and practice the chapter "Permutation and Combination" with these solved Aptitude Questions and Answers. Each question in the topic is accompanied by a clear and easy explanation, diagrams, formulae, shortcuts and tricks that help in understanding the concept.
## Use of Permutation and Combination Questions
The questions and examples given in this section will be useful to all the freshers, college students and engineering students preparing for placement tests or any competitive exam like MBA, CAT, MAT, SNAP, MHCET, XAT, NMAT, GATE, Bank exams - IBPS, SBI, RBI, RRB, SSB, SSC, UPSC etc.
Practice with this online test to crack your placements and entrance tests!
1. Arrange the letters of the word "DARKER" so that the three vowels do not appear together. In how many ways can this be done?
a. 240
b. 360
c. 500
d. 720
Explanation:
'DARKER' has 6 letters.
Tip:
We can arrange 'n' things in n! ways.
While arranging letters/things/numbers, if there are two same things or things get repeated twice, then we need to divide by 2!
If there are 3 same things or things get repeated thrice, then divide by 3! And so on..
Thus, we can arrange 6 letters in 6! ways.
But R gets repeated. There are 2 Rs. So divide by 2!
∴ Total ways = 6! = 360 2!
Vowels not together = Total ways - Vowels together
Consider the 2 vowels (A and E) as one group.
We have 4 letters and 1 group = 5
We can arrange them in 5! Ways.
But again here R comes twice. So we will have 5! 2!
Also, the 2 vowels can be arranged in 2! Ways.
So, Number of ways with vowels together = 2! x 5! = 120 2!
∴ Number of ways with vowels not together = 360 - 120 = 240
2. In a box there are coins marked 0, 2, 3, 4, 5, 6, 8. Without repeating any digit, how many numbers can you form in the range 500 - 1000 (excluding 500 and 1000).
a. 60
b. 70
c. 90
d. 147
Explanation:
There are total 7 digits given - 2, 3, 4, 5, 6, 8 and 0
Between 500 and 1000 means from 501 to 999.
All of them are 3 digit numbers.
First digitSecond digitThird digit
5, 6 or 8 i.e. 3 possibilities (1 digit gets used here)Any one from 7-1 = 6 remaining digits i.e. 6 possibilitiesAny one from 6-1 = 5 remaining digits i.e. 5 possibilities
∴ Total numbers possible = 3 x 6 x 5 = 90
3. There are 8 friends sitting in a café. In how many ways can you form a group of 4 people so that
i. Two particular friends are definitely there
ii. Two particular members are definitely not there
a. 15 and 15
b. 15 and 360
c. 30 and 360
d. 360 and 360
Explanation:
Tip:
SELECT = Combination = nCr = n! r!(n-r)!
SELECT and ARRANGE = Permutation = nPr = n! (n-r)!
I. Compulsorily include 2 particular members.
When we include these 2 members, we are left with 4 - 2 = 2 spots in the group
Also, number of members which remain are 8 - 2 = 6
So now we need to select 2 people out of 6.
6C2 = 6! = 15 = number of ways 2!4!
II. Compulsorily exclude 2 particular members.
When we exclude these 2 members, we are left with 8 - 2 = 6 members
So now we need to select 4 people out of 6.
6C4 = 6! = 15 = number of ways 4!2!
4. Priyanka want to attend Grammy awards at London and National awards at Delhi before heading to the UNICEF Summit in Berlin. 8 flights come from London to Delhi through different routes and 6 flight fly from Delhi to Berlin through 6 different routes. Find the number of ways in which Priyanka can reach from London to Berlin through Delhi.
a. 14
b. 24
c. 48
d. 100
Explanation:
This is very easy to solve
Say out of 8, Priyanka chooses one way to go from London to Delhi.
Now, From Delhi to Berlin she has 6 routes i.e. 6 options
Similarly, for 2nd route between London and Delhi, Priyanka will again have 6 options from Delhi to Berlin.
This is true for all 8 routes between London and Delhi.
So, answer = 8 x 6 = 48 possible ways to travel from London to Berlin via Delhi
5. Out of the 35 players attending the opening ceremony of an event there are 12 cricketers, 10 wrestlers, 5 badminton players and 8 hockey players. The management wants a cricketer or a wrestler to lead the group in the parade. In how many ways can this be done?
a. 1/2
b. 13
c. 22
d. 35 |
Illustrative Mathematics Grade 6, Unit 8, Lesson 9: Interpreting the Mean as Fair Share
Learning Targets:
• I can explain how the mean for a data set represents a “fair share.”
• I can find the mean for a numerical data set.
Related Pages
Illustrative Math
Lesson 9: Interpreting the Mean as Fair Share
Let’s explore the mean of a data set and what it tells us.
Illustrative Math Unit 6.8, Lesson 9 (printable worksheets)
Lesson 9 Summary
The following diagram explains how the mean for a data set represents a “fair share” and how to find the mean for a numerical data set.
Lesson 9.1 Close to Four
Use the digits 0–9 to write an expression with a value as close as possible to 4. Each digit can be used only one time in the expression.
Lesson 9.2 Spread Out and Share
1. The kittens in a room at an animal shelter are arranged in five crates, as shown.
a. The manager of the shelter wants the kittens distributed equally among the crates. How might that be done? How many kittens will end up in each crate?
b. The number of kittens in each crate after they are equally distributed is called the mean number of kittens per crate, or the average number of kittens per crate.
Explain how the expression 10 ÷ 5 is related to the average.
c. A different room in the shelter has 6 crates. No two crates contain the same number of kittens, and there is an average of 3 kittens per crate.
d. Draw or describe at least two different arrangements of kittens that match this description. You may choose to use the applet to help.
Open Applet
2. Five servers were scheduled to work the number of hours shown in the table. They decided to share the workload, so each one would work equal hours.
a. On the grid on the left, draw a bar graph that represents the hours worked by Servers A, B, C, D, and E.
b. Think about how you would rearrange the hours so that each server gets a fair share. Then, on the grid on the right, draw a new graph to represent the rearranged hours. Be prepared to explain your reasoning.
c. Based on your second graph, what is the average or mean number of hours that the servers will work?
d. Explain why we can also find the mean by finding the value of 31 ÷ 5.
e. Which server will see the biggest change to work hours? Which server will see the least change?
Server F, working 7 hours, offers to join the group of five servers, sharing their workload. If server F joins, will the mean number of hours worked increase or decrease? Explain how you know.
The mean of the five servers is 6.2.
If we add a server that is working 7 hours the mean hours will increase because 7 is bigger than the initial mean of 6.2 hours.
Lesson 9.3 Getting to School
1. For the past 12 school days, Mai has recorded how long her bus rides to school take in minutes. The times she recorded are shown in the table.
a. Find the mean for Mai’s data. Show your reasoning.
b. In this situation, what does the mean tell us about Mai’s trip to school?
2. For 5 days, Tyler has recorded how long his walks to school take in minutes. The mean for his data is 11 minutes.
a. Without calculating, predict if each of the data sets shown could be Tyler’s. Explain your reasoning.
b. Determine which data set is Tyler’s. Explain how you know.
Glossary Terms
average
The average is another name for the mean of a data set.
For the data set 3, 5, 6, 8, 11, 12, the average is 7.5.
3 + 5 + 6 + 8 + 11 + 12 = 45
45 ÷ 6 = 7.5
mean
The mean is one way to measure the center of a data set. We can think of it as a balance point. For example, for the data set 7, 9, 12, 13, 14, the mean is 11.
To find the mean, add up all the numbers in the data set. Then, divide by how many numbers there are. 7 + 9 + 12 + 13 + 14 = 55 and 55 ÷ 5 = 11.
Lesson 9 Practice Problems
1. A preschool teacher is rearranging four boxes of playing blocks so that each box contains an equal number of blocks. Currently Box 1 has 32 blocks, Box 2 has 18, Box 3 has 41, and Box 4 has 9.
Select all the ways he could make each box have the same number of blocks.
A. Remove all the blocks and make four equal piles of 25, then put each pile in one of the boxes.
B. Remove 7 blocks from Box 1 and place them in Box 2.
C. Remove 21 blocks from Box 3 and place them in Box 4.
D. Remove 7 blocks from Box 1 and place them in Box 2, and remove 21 blocks from Box 3 and place them in Box 4.
E. Remove 7 blocks from Box 1 and place them in Box 2, and remove 16 blocks from Box 3 and place them in Box 4.
2. In a round of mini-golf, Clare records the number of strokes it takes to hit the ball into the hole of each green. She said that, if she redistributed the strokes on different greens, she could tell that her average number of strokes per hole is 3.
Explain how Clare is correct.
3. Three sixth-grade classes raised \$25.50, \$49.75, and \$37.25 for their classroom libraries. They agreed to share the money raised equally. What is each class’s equal share? Explain or show your reasoning.
4. In her English class, Mai’s teacher gives 4 quizzes each worth 5 points. After 3 quizzes, she has the scores 4, 3, and 4. What does she need to get on the last quiz to have a mean score of 4? Explain or show your reasoning.
5. An earthworm farmer examined two containers of a certain species of earthworm so that he could learn about their lengths. He measured 25 earthworms in each container and recorded their lengths in millimeters.
Here are histograms of the lengths for each container.
a. Which container tends to have longer worms than the other container?
b. For which container would 15 millimeters be a reasonable description of a typical length of the worms in the container?
c. If length is related to age, which container had the most young worms?
6. Diego thinks that x = 3 is a solution to the equation x2 = 16. Do you agree? Explain or show your reasoning.
The Open Up Resources math curriculum is free to download from the Open Up Resources website and is also available from Illustrative Mathematics.
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. |
# Arithmetic properties
Multiplication and addition have specific arithmetic properties which characterize those operations. In no specific order, they are the commutative, associative, distributive, identity and inverse properties.
Commutative property
An operation is commutative if changing the order of the operands does not change the result. The commutative property of addition means the order in which the numbers are added does not matter. In other words, the placement of addends can be changed and the results will be equal. Likewise, the commutative property of multiplication means the places of factors can be changed without affecting the result.
Associative property
Within an expression containing two or more occurrences of only addition or of only multiplication, the order in which the operations are performed does not matter as long as the sequence of the operands is not changed. That is, rearranging the parentheses in such an expression will not change its value.
$\ 1 + 5 + 9 + 5 = ?$
To simplify this, use the commutative property to switch the order and then use the associative property to group $1$ and $9$, and $5$ and $5$, since these pairs both add up to $10$, so the final result is $20$.
Distributive property
Distributive property combines addition and multiplication. If a number multiplies a sum in parenthesis, the parenthesis can be removed if we multiply every term in the parenthesis with the same number.
The number of terms inside the brackets doesn’t matter, this will always be valid.
This property is usually applied when an unknown is a part of addition, and it enables us to single the unknowns out.
Identity element
Identity element or neutral element is an element which leaves other elements unchanged when combined with them. Identity element for addition is 0 and for multiplication is 1.
Inverse element
Multiplicative inverse or reciprocal for a number $x$, denoted by $\frac{1}{x}$, is a number which when multiplied by $x$ yields the multiplicative identity, 1. The multiplicative inverse of a fraction $\frac{x}{y}$ is $\frac{y}{x}$
Additive inverse of a number $x$ is the number that, when added to $x$, yields zero. This number is also known as the opposite (number), sign change, and negation. For a real number, it reverses its sign: the opposite to a positive number is negative, and the opposite to a negative number is positive. Zero is the additive inverse of itself.
For example, the reciprocal of 5 is $\frac{1}{5}$, and the oppostie number of 5 is -5.
## Arithmetic properties worksheets
Arithmetic properties - Integers (127.4 KiB, 1,245 hits)
Arithmetic properties - Decimals (159.3 KiB, 608 hits)
Arithmetic properties - Fractions (199.4 KiB, 631 hits)
Distributive property (311.9 KiB, 566 hits) |
# Trigonometrical Ratios of (180° + θ)
What are the relations among all the trigonometrical ratios of (180° + θ)?
In trigonometrical ratios of angles (180° + θ) we will find the relation between all six trigonometrical ratios.
We know that,
sin (90° + θ) = cos θ
cos (90° + θ) = - sin θ
tan (90° + θ) = - cot θ
csc (90° + θ) = sec θ
sec ( 90° + θ) = - csc θ
cot ( 90° + θ) = - tan θ
Using the above proved results we will prove all six trigonometrical ratios of (180° + θ).
sin (180° + θ) = sin (90° + 90° + θ)
= sin [90° + (90° + θ)]
= cos (90° + θ), [since sin (90° + θ) = cos θ]
Therefore, sin (180° + θ) = - sin θ, [since cos (90° + θ) = - sin θ]
cos (180° + θ) = cos (90° + 90° + θ)
= cos [90° + (90° + θ)]
= - sin (90° + θ), [since cos (90° + θ) = -sin θ]
Therefore, cos (180° + θ) = - cos θ, [since sin (90° + θ) = cos θ]
tan (180° + θ) = cos (90° + 90° + θ)
= tan [90° + (90° + θ)]
= - cot (90° + θ), [since tan (90° + θ) = -cot θ]
Therefore, tan (180° + θ) = tan θ, [since cot (90° + θ) = -tan θ]
csc (180° + θ) = $$\frac{1}{sin (180° + \Theta)}$$
= $$\frac{1}{- sin \Theta}$$, [since sin (180° + θ) = -sin θ]
Therefore, csc (180° + θ) = - csc θ;
sec (180° + θ) = $$\frac{1}{cos (180° + \Theta)}$$
= $$\frac{1}{- cos \Theta}$$, [since cos (180° + θ) = - cos θ]
Therefore, sec (180° + θ) = - sec θ
and
cot (180° + θ) = $$\frac{1}{tan (180° + \Theta)}$$
= $$\frac{1}{tan \Theta}$$, [since tan (180° + θ) = tan θ]
Therefore, cot (180° + θ) = cot θ
Solved example:
1. Find the value of sin 225°.
Solution:
sin (225)° = sin (180 + 45)°
= - sin 45°; since we know sin (180° + θ) = - sin θ
= - $$\frac{1}{√2}$$
2. Find the value of sec 210°.
Solution:
sec (210)° = sec (180 + 30)°
= - sec 30°; since we know sec (180° + θ) = - sec θ
= - $$\frac{1}{√2}$$
3. Find the value of tan 240°.
Solution:
tan (240)° = tan (180 + 60)°
= tan 60°; since we know tan (180° + θ) = tan θ
= √3
Trigonometric Functions |
# Fractions and their Equivalent Forms
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1 Fractions Fractions and their Equivalent Forms Little kids use the concept of a fraction long before we ever formalize their knowledge in school. Watching little kids share a candy bar or a bottle of soda indicates they have a feeling for what will soon be called a fraction. Can you imagine having a son or daughter come home and slice a cake like this, then eat the piece largest (shaded) piece. When you got home and saw most of the cake gone, you might have asked; who ate all the cake? If someone were to retort, I only had one piece, the biggest, you may have wanted to make him wear the rest. But, you decided that in the future you would pre-cut the cake into equal pieces. That way you would know how much cake one piece would be, hence you introduced the notion of denominator. A fraction is part of a whole unit or group, it s made up of two parts, a numerator and a denominator. The denominator tells you how many EQUAL parts make one whole unit or group, the numerator tells you how many of those parts you have. 8 numerator denominator If a pie was cut into eight equal pieces and you ate all eight pieces, you could write that as 8 8 or one whole pie. One whole day could be described as That allows us to say the = 1 There are two types of fractions, proper and improper. Hanlonmath.com
2 Types of Fractions 1. Proper fraction is a fraction less than one. The numerator is less than the denominator. Ex. 5/ Improper fraction is a fraction greater than one. The numerator is greater than the denominator. Ex. /5. Saying and Writing Fractions To say a fraction like 4 you say the numerator - three, then say the denominator by spraying friends with the suffix ths. In other words, you say three fourths. For people that don t know mathematical notation, they might say three slash four. But, since you understand the importance of vocabulary and notation, you would know that was a fraction. To express a quantity as a fraction, you would need to know how many equal parts make one whole unit. The numerator would be the parts under consideration, the denominator would be how many EQUAL parts make one full unit. Example 1 Express 5 hours as a part of a day. Since you have 5 hours, that will be numerator. There are 24 hours in one full day, so the denominator is 24. Piece of cake! Therefore, 5 hours 5/24 day. Example 2 Express ounces as part of a pound. Since there are16 ounces in a pound, the denominator is 16. The numerator is. Therefore ounces is /16 lb. If you are not familiar with denominate numbers, then knowing how many parts are in a whole could cause you some difficulty. You should know things like 12 inches is a foot, 16 ounces is a pound, feet is a yard, etc. I hear you, you want to try some problems on our own. Writing Fractions Express the following quantities as a fraction of a foot inch 2. inches. 5 inches 4. 9 inches 5. 8 inches inches Express as a fraction of an hour.. 2 min min min. Express as a faction of a day hrs hrs hrs. Fractions - 2
3 Equivalent Fractions Have you ever noticed that not everyone describes the same things in the same way. For instance, a mother might say her baby is twelve months old. The father might tell somebody his baby is a year old. Same thing, no big deal. Well, we do the same thing in math. Or in our case, in fractions. Let s look at these two cakes. One person might notice that 2 out of 4 pieces seem to describe the same thing as 1 out of 2 in the picture above. In other words, 1 2 = 2 4 If we continue this process, we would notice we have a number of different ways to express the same thing. When two fractions describe the same thing, we say they are equivalent fractions. Equivalent Fractions are fractions that have the same value. Wouldn t it be nice if we could determine if fractions were equivalent without drawing pictures? Well, if we looked at enough equivalent fractions, we would notice a pattern developing. Let s look at some. 1 2 = 6, 4 = 6 8, 2 = 10 15, and 5 = 0 50 Do you see any relationship between the numerators and denominators in the first fraction compared to the numerators and denominators in the second? Hopefully, you might notice we are multiplying both numerator and denominator by the same number to get the 2 nd fraction. Example 5 6 = if you multiply both the numerator and denominator by 4 Well, you know what that means, when we see a pattern like that, we make a rule, an algorithm or procedure that allows us to show other people simple ways of doing problems. Yes, feel that excitement running through your body. To generate equivalent fractions, you multiply BOTH numerator and denominator by the SAME number When you multiply both the numerator and denominator by the same number, we are multiplying by 4/4 or 1. When we multiply by one, that does not change the value of the original fraction. Fractions -
4 Example 4 Express 5 6 as sixtieths. 5 6 =? 60 What did you multiply 6 by to get 60 in the denominator? By 10, so we multiply the numerator by 10. So 5 6 = 50. Piece of cake! 60 Example 5 2 is how many thirty-fifths? 2 =? 5 What did you multiply by to get 5 in the denominator? By 5 you say, we multiply the numerator by 5. 2 = 10 5 We ll let you try some of these on your own. By the way, do you know what the fraction said to the whole number? Answer, you crack me up! Get it, crack up into parts. Oh yes, math humor, don t you just love it? Making equivalent fractions = = = = = = 2 Fractions - 4
5 Converting Mixed Numbers to Improper Fractions Mixed Number is a whole number and a fraction. A mixed number occurs when you have more than one whole unit. Ex. 1 1/4 Example 6 Let s say Johnny boy ate all the first cake and one piece from the second cake. We could describe that as eating 1 1 cakes. His mom might come home and notice 4 Johnny ate 5 pieces of cake. Notice, eating 1 1 cakes describes the same thing as 4 eating 5 pieces of cake. Good news, since we are working with fractions, I can describe eating 5 pieces of cake as a fraction. Again the numerator tells us how many pieces Johnny ate, the denominator tells us how many equal pieces make one whole cake. In our case, that s 4. The fraction then is 5/4. What we have just seen is 1 1 seems to describe the same thing as 5. Therefore, we 4 4 say they are equivalent. Writing that, we have = 5 4 If we looked at some more cakes, we might notice a similar pattern. Example Describe eating 1 /8 cakes as eating 11 pieces of cake when 8 pieces make one whole cake. The whole cake can be described as 1 or 8 8 and the portion of the other cake is 8. What we have is 1 8 = or 11/8. If we kept looking at more examples, we might ask ourselves if there is a way we could convert that mixed number to an improper fraction without drawing the picture. Again, we ll have to look for patterns. Do you see one? It doesn t jump right out at you. Fractions - 5
6 1 1 4 = 5 4 and 1 8 = 11 8 Well, if we looked long enough and took the time to write the numbers down and play with them, then we might see this little development. In the first equality,1 1 = 5 If I multiply the denominator by the whole number and add the 4 4 numerator, that gives me the new numerator 5. The denominator stays the same. The second equality, 1 = 11, the pattern of multiplying the denominator by the whole number and 8 8 adding the numerator also gives me the new numerator of 11. Again the denominator stays the same. Oh yes, you can feel the excitement! Seeing these developments, we might try looking at a few more examples using cakes, then see if the pattern we discovered still works. If it does, we make a rule. As you probably guessed, it works. If it didn t, we would not have been discussing it anyway, right? Converting a Mixed Number to an Improper Fraction 1. Multiply the whole number by the denominator 2. Add the numerator to that product. Place that result over the original denominator Example 8 Convert 2 4 to an Improper fraction 2 4 = 4 x With practice, you should be able to do these in your head. So, if you don t mind, here are some practice problems. No, you don t have to thank me. Convert to improper fractions = Fractions - 6
7 Converting Improper Fractions to Mixed Numbers Using the pictures of the pie, we noticed that 5 4 = 11. We found a way to convert the mixed 4 number to an improper fraction, do you think you can find a method to convert an improper fraction to a mixed number? How did we get the fraction 5? What we did was count the number 4 of equally sized pieces of pie. All four pieces of the first pie were gobbled down and one piece from the other pie. That gave us a total of 5 pieces. The denominator was 4 because that s how many pieces make one whole pie. Another way to look at! is to look at each pie in fractional form. All four pieces were eaten in the " first pie, that s 4. One piece from the second pie, that s \$. That would lead us to believe that 4 " = 5 4. In Example 8, converting 1 11 to an improper fraction resulted in. Now again, the question is 8 8 can we find a way to convert back without drawing a picture. The fraction 11 8 clearly indicates more than one pie was eaten. In fact, we can see in the first pie 8 out of 8 pieces were eaten. That s written as 8 8. The second pie had out of 8 eaten. That would tell us that = So the trick to convert an improper fraction to a mixed number seems to be to determine how many whole pies were eaten, then write the fractional part of the pie left. To convert to a mixed number, how many whole pies were eaten? Well could be written as We can see two pies were eaten plus a 1 of another pie. Could I have done that without breaking apart the fraction? Fractions -
8 Sure, to determine how many whole pies are in, I could have divided by. The quotient is 2, which means I have two whole pies. The remainder is 1, which means I have one piece of the last pie or 1. So = 2 1. To Convert an Improper Fraction to a Mixed Number 1. Divide the numerator by the denominator to determine the whole number 2. Write the remainder over the original denominator Example 9 Convert 22 5 to a Mixed Number 22 5 = 4 with a remainder of 2. Therefore 22 5 = Convert to mixed numbers As we continue our study of equivalent forms of fractions, we run into something called reducing fractions. When we reduce fractions, what we are doing is making equivalent fractions. But rather than multiplying both numerator and denominator by the same number, we are going to divide both numerator and denominator by the same number. To do that, we have to find common factors. That is, we have to find numbers that go into both the numerator and denominator. Before we reduce fractions, we are going to look for patterns that will make our lives very simple. Rules of Divisibility We are going to discuss Rules of Divisibility. To be quite frank, you already know some of them. For instance, if I asked you to determine if a number is divisible by two, would you know the answer. Sure you do, if the number is even, then it s divisible by two. Can you tell if a number is divisible by 10? How about 5? Fractions - 8
9 Because you are familiar with those numbers, chances are you know if a number is divisible by 2, 5 or 10. We could look at more numbers to see if any other patterns exist that would let me know what they are divisible by, but we don t have that much time or space. So, if you don t mind, I m just going to share some rules of divisibility with you. By 2, if it ends in 0, 2, 4, 6 or 8 By 5, if it ends in 0 or 5 By 10, if it ends in 0 Rules of Divisibility, a number is divisible. By, if sum of digits is a multiple of By 9, if sum of digits is a multiple of 9 By 6, if the number is divisible by 2 and by By 4, if the last 2 digits of the number is divisible by 4 By 8, if the last digits of the number is divisible by 8 Example 1 Is 111 divisible by? One way of finding out is by dividing 111 by, if there is no remainder, then it goes in evenly. In other words, would be a factor of 111. Rather than doing that, we can use the rule of divisibility for. Does the sum of the digits of 111 add up to a multiple of? Yes it does, =, so it s divisible by. Example 2 Is 41 divisible by? Do you want to divide or do you want to use the shortcut, the rule of divisibility? Adding 4, and 1, we get 12. Is 12 divisible by? If that answer is yes, that means 41 is divisible by. If you don t believe it, try dividing 41 by. Let s look at numbers divisible by 4. The rule states if the last 2 digits are divisible by 4, then the number itself is divisible by 4. Example Is 12,16 divisible by 4? Using the rule for 4, we look at the last 2 digits, are they divisible by 4. The answer is yes, so 12,16 is divisible by 4. If you take a few minutes today to learn the Rules of Divisibility, that will make your life a lot easier in the future. Not to mention it will save you time and allow you to do problems very quickly when other students are experiencing difficulty. Fractions - 9
10 Using the rules of divisibility, determine if the following numbers are divisible by 2,, 4, 5, 6, 8, 9, or , ,08 1. Write a 5 digit number that is divisible by 2,, 4, 5, 6, 8, 9, and Write a 6 digit number that is divisible by 2,, 4, 5, 6, 8, and 10, but not 9. Now that we have played with rules of divisibility, learning how to reduce fractions will be a much easier task. Simplifying Fractions Simplifying fractions is just another form of making equivalent fractions. Instead of multiplying the fraction by one by multiplying the numerator and denominator by the same number, we will divide the numerator and denominator by the same number which is equivalent to multiplying by one. Now that we know the Rules of Divisibility, reducing fractions is going to be a piece of cake. To simplify fractions we divide both numerator and denominator by the same number. Example 1 Simplify 18/20 Example 2 Simplify Both the numbers are even, so we can divide both numerator and denominator by 2. Doing that the answer is 9/10. Notice that the sum of the digits in 111 is and the sum of the digits in 2 is 12. Therefore, both are divisible by. 91 is the answer. Don t you just love this stuff? Fractions - 10
11 If you don t know the rules of divisibility, you would have to try and reduce the fractions by trying to find a number that goes into both numerator and denominator. That s too much guessing, so spend a few minutes and commit the rules of divisibility to memory, Simplify the following fractions Common Denominators Now that we have played with fractions, we know what a fraction is, how to write them, say them and we can make equivalent forms, it s now time to learn how to add and subtract them. But before we do, by definition, we must have equally sized pieces. Let s say we have two cakes, one chocolate, the other vanilla. The chocolate was cut into thirds, the vanilla into fourths as shown below. You had one piece of chocolate cake and one piece of vanilla, as shown. Since you had 2 pieces of cake, can you say you had 2 of a cake? Let s go back to how we defined a fraction. The numerator tells you how many equal pieces you have, the denominator tells you how many equal pieces make one whole cake. Since your pieces are not equal, we can t say we have 2 of a cake. And clearly, pieces don t make one whole cake. Therefore, trying to add 1 4 to 1 and coming up with 2 just doesn t fit our definition of a fraction. Fractions - 11
12 The key is we have to cut the cakes into equal pieces. Having one cake cut into thirds and the other in fourths means I have to be innovative so... Let s get out our knives and do some additional cutting. By making additional cuts on each cake, both cakes are now made up of 12 equal pieces. That s good news from a sharing standpoint everyone gets the same size piece. Mathematically, we have introduced the concept of a common denominator. The way I made the additional cuts on each cake was to cut the second cake the same way the first was cut and the first cake the same way the second was cut as shown in the picture. Now, that s a piece of cake! Clearly, we don t want to make additional cuts in cakes or pies the rest of our lives to make equal pieces from baked goods that have been cut differently. So, what we do is try and find a way that will allow us to determine how to make sure all pieces are the same size. What we do mathematically is find the common denominator. A common denominator is a denominator that all other denominators will divide into evenly 4 Methods of Finding a Common Denominator 1. Multiply the denominators 2. Write multiples of each denominator, use a common multiple. Use a factor tree and find the Least Common Multiple 4. Use the Reducing Method, especially for larger numbers Cake-wise, it s the number of pieces that cakes can be cut so everyone has the same size piece. Method 1 if I had two fractions like 1 and 1. By multiplying the denominators, I would find a 4 number that is a multiple of and 4. In other words, a number in which both and 4 are factors. The common denominator would be 4 or 12. Method 2 I would write multiples of each denominator, when I came across a common multiple for each denominator, that would be a common denominator. Again, using, I write multiples of each denominator. 1 and 1 4, 6, 9, 12, 15, 18, 4, 8, 12, Since 12 is a multiple of each denominator, 12 would be a common denominator. Fractions - 12
13 Method Method 4 is a pain in the rear and nobody that I know of uses it to find common denominators, so, we won t either. However, the next method is a direct result of finding LCMs is an especially good way of finding common denominators for fractions that have large denominators or fractions whose denominators are not that familiar to you. Example 1 Let s say I asked you to find the common denominator for the fractions 1/18 and 5/24. Using Method 1, we d multiply 18 by 24. The result 42. That s too big of a number for me to know in my head. Method 2 would have us writing multiples of the two denominators. 18, 6, 54, 2, 90, 108, 24, 48, 2 2 is a multiple of each, therefore 2 would be a common denominator. Using Method 4, I put the 2 denominators over each other in fractional form as shown and reduce = 4 Now I cross multiply, either 24 by or 18 by 4. Notice I get 2 no matter which way I go. Therefore 2 is the common denominator. It does not matter if I put 18/24 and reduce or 24/18, I get the same answer. Example 2 Find the common denominator for 5/24 and 9/42. While multiplying will give you a common denominator, it will be a very large number. I m going to use method 4. Placing the denominators over each other and reducing = 4 4 x 42 gives me a common denominator of 168 Using Method 4, reducing the denominators, sure beats multiplying 24 by 42. It s also better than trying to write multiples for both of those denominators and finding a common multiple. Fractions - 1
14 Try reducing the following fractions using the easiest method available to you, Find the common denominators 1. 2/, /5 2. 1/6, /5. 2/, / , , , , , 5, , Comparing Fractions There are a number of ways to compare (order) fractions; making equivalent fractions with common denominators, converting the fractions to decimals, or by cross multiplying. Let s look at each method. Method 1. By equivalent fractions. 1. Find a common denominator 2. Make equivalent fractions. Compare the numerators. Method 2. By converting to a decimal 1. Divide the numerator by the denominator 2. Carry out the division the same number of places. Compare the numbers Method. By cross multiplying 1. Using the fractions, a/b and c/d 2. Cross multiply and place the products ABOVE the fractions. Compare products The method you want to use really depends on the fractions with which you are working. Let s do one example all three ways. Fractions - 14
15 Ex. 1. Method 1. Compare by making equivalent fractions Compare using a > or < sign. % " &! The common denominator is 15, the equivalent fractions are \$' \$! \$% \$! 10 < 12, therefore % & < "! Ex. 2 Method 2. Compare by converting to a decimal using < o r > signs. % =.666 " &! =.800 Note both carried out places 666 < 800, therefore % & < "! Ex. Method. Compare by cross multiplying Cross multiplying and placing the product above the fraction, 10 < 12. Since 10 < 12, therefore % & < "! Compare the following fractions by any method. 1. 2/, /5 2. 1/6, /5. 2/, / , , , , , 5, , 5 48 Fractions - 15
16 Adding & Subtracting Fractions With Like Denominators In order to add or subtract fractions, we have to have equal pieces. If a cake was cut into 8 equal pieces and you had three pieces tonight, then ate four pieces tomorrow, you would have eaten a total of pieces of cake, or /8 of one cake. Mathematically, we would write = 8 Notice, we added the numerators because that told us how many equal pieces were eaten. Why didn t we add the denominators? Remember how we defined a fraction, the denominator tells us how many equal pieces makes one whole cake. If I added them, we would be indicating that the cake was cut into 16 pieces. But we know it was only cut into 8 equally sized pieces. Add/subtract Adding and Subtracting Fractions With Unlike Denominators Let s add 1to 1 4 Would I get 2? Why not? The 2 would indicate that we have two equal pieces and that equal pieces made one whole unit. Let s draw a picture to represent this: Notice the pieces are not the same size. Fractions - 16
17 Making the same cuts in each cake will result in equally sized pieces. That will allow me to add the pieces together. Each cake now has 12 equally sized pieces. Mathematically, we say that 12 is the common denominator. Now let s add. 1 4 = = From the picture we can see that 1/ is the same as 4/12 and 1 4 has the same value as /12. Adding the numerators, a total of equally sized pieces are shaded and 12 pieces make one unit. If I did a number of these problems, I would be able to find a way of adding and subtracting fractions without drawing the picture. Algorithm for Adding/Subtracting Fractions 1. Find a common denominator 2. Make equivalent fractions.. Add/Subtract the numerators 4. Bring down the denominator 5. Simplify Using the procedure, let s try one. 1 5 Example Since the denominators are small and relatively prime, I will use multiply the denominators to find the common denominator, 5 = 15. Now I make equivalent fractions and add the numerators. Fractions - 1
18 1 5 = = Let s try a few. Using the algorithm, first find the common denominator, then make equal fractions. Once you complete that, you add the numerators and place that result over the common denominator. Remember, the reason you are finding a common denominator is so you have equally sized pictures. Add or subtract the following problems _ _ Fractions - 18
19 Borrowing From a Whole Number Subtracting fractions when borrowing is as easy as getting change for your money. Example 2 Let s say you have one dollar bills and you have to give your friend \$.25. How much money would you have left? Since you don t have any coins, you would have to change one of the dollars into 4 quarters. Why not ten dimes? Because you have to give your friend a quarter, so you get the change in terms of what you are working with quarters. Writing that, this is what it would look like. dollars 6 dollars 4 quarters dollars 1 quarter dollars 1 quarter After you get the change, you can subtract Doing that, we have 6 dollars 4 quarters dollars 1 quarter dollars quarters I d be left with dollars and quarters. That seems to make sense. You could have done that in your head Example Let s do the exactly the same problem using fractions. Again, I have one dollar bills and I have to give my friend \$.25. Another way to say that is I have to give my friend three and a quarter. 6 4/4 Let s write it in fractional terms. ¼ 1/4 1/4 /4 Remember, I am borrowing one, 4 4 = 1. So what I m substituting subtract. for so I can Example /5 9 2/5 9 2/5 2 /5 Remember, when borrowing, when working in fifths, I borrow 5/5. If I was working in thirds, I ll borrow /. If I m working in twelfths, I ll borrow 12/12. The point being, just like in dealing with money, I get the type of change I ll need to do the problem. Piece of cake, don t you think? Guess what, it s your turn to do a couple. Fractions - 19
21 Just like when we borrowed before, since we are working in fourths, we ll borrow 4. We take what we borrow and add that to what we already 4 have, the result is 5/4. That makes sense Adding the 4/4 to the 1 4 we already had results in Simplifying, we have 1 2 Always check to see if your final answer can be simplified. My guess is you would use the Rules of Divisibility to make that determination. I can t make these problems difficult if you know the algorithm. So take a few minutes and memorize it. Borrowing with Mixed Numbers 1. Find a common denominator 2. Make equivalent fractions. Borrow, if necessary 4. Subtract the numerators 5. Bring down the denominator Fractions - 21
22 Multiplying Fractions Let s take a look at what multiplying fractions looks like visually. To begin we will look at 1/ of a unit, then take ½ of that 1/. So here s what 1/ of a unit might look like. Now, I chose a rectangular shaped unit, it could have been triangular, circular, a star or a square. Now, what I will do is take ½ of that 1/ and shade that green. By taking ½ of the 1/, I end up with smaller pieces. How many of those equally sized pieces would make a whole unit? Remember, the denominator tells you how many equal pieces makes one unit. So, let s divide the unit into 6 equally sized pieces. Notice that 1/2 of the 1/ is in green. So we have 1 2 x 1 = 1 6 I know what you are thinking, let s look at another visual. Let s see what 2/ of 1/4 would look like visually. So we will divide the unit (rectangle) into fourths and shade in ¼. Fractions - 22
23 Now, we want 2/ of that 1/4. So we will break the ¼ into pieces, and we want two of those three. So shading those 2 in green, and also creating equally sized pieces, we see that we have 2 pieces out of 12, or 2/12. Looking at that mathematically, we have 2/ of 1/4 2 x 1 4 = 2 12 Now, if we look at the last two examples and did a few more for good measure, we would see a pattern develop. First, we would realize that the word of translates to multiplication. Second, when we performed the multiplication, we multiplied numerator by numerator and denominator by denominator. For illustration purposes, I did not simplify the fraction in the last example. Multiplying fractions is pretty straight-forward. So, we ll just write the algorithm for it, give an example and move on. Algorithm for Multiplying Fractions 1. Make sure you have proper or improper fractions 2. Cancel, if possible. Multiply numerators 4. Multiply denominators 5. Simplify Fractions - 2
24 Example x 4 5 Since 1 is not a fraction, we convert it to, we rewrite it as follows x 4 5 Now what I m about to say is important and will make your life a lot easier. We know how to simplify fractions, what we want to do now is to cancel with fractions. That s nothing more than simplifying using the commutative and associative properties (changing the order and grouping). The numerator is x 4, the denominator is 2 x 5. Writing that as a single fraction I have x 4. The Commutative Property of Addition allows me to change 2 x 5 the order of the numbers. I will rewrite the numerator; 4 x. Now, I can rewrite 2 x 5 them as separate fractions using the Associative Property. 4 2 x 5, I can reduce 4 2 to 4 2 to 2 1 and rewrite the problem as 2 1 x. The answer 5 is 14 5 = Now rather than going through all those steps, using the commutative and associative properties, we could have taken a shortcut and cancelled. 1 2 x To do that, we would look for common factors in the numerator and denominator and divide them out. In our problem, there is a common factor of 2. By dividing out a 2, the problem looks like this 1 x 2 5 = 14 5 = Fractions - 24
25 Let s look at another one. Example 2 5 x 5 9 Rewriting the mixed number as a fraction, we have 18 5 x 5 9. We have a common factor of 5 in the numerator and denominator, we also have a factor of 9 in each. Canceling the 5 s and the 9 s, we have x The answer is 2. \Multiply the following fractions x x x x x x 1 4 Dividing Fractions Before we learn how to divide fractions, let s revisit the concept of division using whole numbers. When I ask, how many 2 s are there in 8. I can write that mathematically three ways To find out how many 2 s there are in 8, I will use the subtraction model: Fractions - 25
26 Now, how many times did I subtract 2? Count them, there are 4 subtractions. So there are 4 twos in eight. Mathematically, we say 8 2= 4. You want some good news, division has been already defined as repeated subtraction. That won t change because we are using a different number set. In other words, to divide fractions, I could also do repeated subtraction. Example Another way to look at this problem is using your experiences with money. How many quarters (1/4) are there in \$1.50 ( 1 ½)? Using repeated subtraction, we have: , rewriting that with a common denominator, we have 4 1 % " \$ " = 1\$ ", 1 \$ " \$ " = 1, 1 \$ " = & ", & " \$ " = % ",, % " \$ " = \$ ", \$ " \$ " = 0 Note, I subtracted ¼ SIX times. That means there are 6 ¼ s in 1 \$ %. Mathematically we write 1 \$ % \$ " = 6 Ready for a shortcut or would you rather subtract your brains out? What you are asking when dividing 1 ½ by ¼ is, how many 1/4 (quarters) are there in 1 ½ dollars. You can see there are 6. Fractions - 26
27 Example 4 How many 1/8 are in /4? 1/4 1/4 1/4 Dividing each of the ¼ s in half, we get 1/8. How many 1/8 fit into the pink area representing ¼ s? 1/8 1/8 1/8 1/8 1/8 1/8 Count, there are six \$ E s in & ", so & " \$ E = 6 Now, try, that same thing for how many 1/9 are in 1/? \$ & \$ C How about how many 1/10 are in 2/5? %! \$ If you do those subtractions or draw pictures and count, then look at the numbers in the division problems, you might see a pattern develop. Well, because you enjoy playing with numbers, you found a quick way of dividing fractions. You did this by looking at fractions that were to be divided and they noticed a pattern. And here is what they noticed by just looking at the problem and the answer. \$' = = = 4 x8 4x1= 6 1x9 x1 = 2x10 5x1 = 4 Notice, the pattern we developed worked for proper and improper fractions. So, our first step is to make sure we have fractions no mixed numbers. In this pattern, we are multiplying the numbers diagonally, then dividing. Another way of doing the same operation is to invert the divisor and multiply, we get the same result. And it works faster than if we did repeated subtractions, not to mention it takes less time and less space. Patterns sure do make life a whole lot easier, don t you think? Rather than multiplying diagonally and dividing, we will use the second pattern and invert the divisor and then multiply the numerators, then the denominators, we get the same result. A reciprocal of a number, by definition, is just 1 divided by the number. It s the multiplicative inverse. The reciprocal of 8 is 1/8. The reciprocal of ¾ is 1 /4 = 4/. Many describe a reciprocal as a fraction upside-down. That is, the reciprocal of /5 is 5/. Fractions - 2
28 Algorithm for Dividing Fractions 1. Make sure you have fractions 2. Invert the divisor. Cancel, if possible 4. Multiply numerators 5. Multiply denominators 6. Simplify Use the algorithm derived from the patterns to divide. Example Inverting the divisor and multiply = 15 4 = 4 Example Converting to an Improper Fraction = F & x " \$ Inverting the divisor and multiply = 28 = Fractions - 28
29 Fractions Name Date Definitions 1. ***Fraction 2. *Improper Fraction. ***Reciprocal 4. ***In the fraction 4/5, the 4 is called the. 5. ***Write the algorithm for Adding/Subtracting Fractions. 6. ***Write the algorithm for Dividing Fractions.. **What method of find a common denominator might be most convenient to use when adding 5/18 and /24? Explain why you chose that method. Fractions - 29
30 8. *If the numerator and the denominator of a proper fraction continually increase by 1, what happens to the value of the fraction? 9. *A student argued that any time you multiply numbers, the product is larger than the factors. What is his error and how would you explain the rationale behind the correct answer? 10. *Draw a model to show that 1/5 + 1/4 = 9/ are ** questions 11. Simplify the following fractions. a. 18/24 b. 45/2 c. 111/ /5 5 2/ Fractions - 0
31 14. 6 x / ½ ¼ 16. Find the LCM and GCF of 54, 81, and **Order the following numbers from greatest to least. 2/, /5, 5/8 18. Choose values of x from below so that x < 5 5, 0, 5/2, 11/2, 21/5 19. Choose values of x such that 4 ¾ > x 6 5/9, 1/8, 1/, 5 ¼, 4 1/5 20. *Antonio has 4/5 of the money for the trip. His brother Peter has the rest of the money. If Antonio has \$24, how much money does Peter have for the trip? Fractions - 1
32 21. *Mike worked 10 ½ hours Monday, 8 ¼ hours on Tuesday, and ¾ hours on Wednesday, how many hours did he work the three days? 22.SBAC *Ted needs 2 ¼ feet of ribbon to wrap each present. He has to wrap 15 presents. If ribbons come in 5 foot rolls, how many rolls of ribbon will Ted need to purchase to wrap all the presents? 2.SBAC *A person has yards of material available to make uniforms. Each uniform requires 4 made? How much material will be left over? yard of material. How many uniforms can be 24.* Bob owns five ninths of the stock in the family company. His sister Mary owns half as much stock as Bob. What part of the stock is owned by NEITHER Bob nor Mary? 25.* Joel worked hours one week and 112 hours the next week. How many more hours and minutes did he work the second week than the first? 26. ***Write a home phone, cell phone, or home address to contact your parent or guardian. (CHP) Fractions - 2
33 Fractions -
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# Question: What Is Union Of Sets In Math?
## What does union of sets mean in math?
The union of a set A with a B is the set of elements that are in either set A or B. The union is denoted as A∪B.
## What is union and intersection in sets?
The union of two sets contains all the elements contained in either set (or both sets ). The intersection of two sets contains only the elements that are in both sets. The intersection is notated A ⋂ B.
## How do you find the union of a set in math?
The union of two sets is a set containing all elements that are in A or in B (possibly both). For example, {1,2}∪{2,3}={1,2,3}. Thus, we can write x∈(A∪B) if and only if (x∈A) or (x∈B). Note that A∪B=B∪A.
## Does Union mean add or multiply?
The union and intersection of sets may be seen as analogous to the addition and multiplication of numbers. Like addition and multiplication, the operations of union and intersection are commutative and associative, and intersection distributes over union.
## What does AUB )’ mean?
Definition 1. The union of the sets A and B, denoted by A U B, is the set that contains those elements that are either in A or in B, or in both.
## What is a ∆ B?
The symmetric difference of two sets A and B is the set (A – B ) ∪ ( B – A) and is denoted by A △ B. A △ B is the set of all those elements which belongs either to A or to B but not to both. A △ B is also expressed by (A ∪ B ) – ( B ∩ A).
## What are the 4 operations of sets?
Set Operations include Set Union, Set Intersection, Set Difference, Complement of Set, and Cartesian Product.
## What is the symbol of union of two sets?
The symbol we use for the union is ∪. The word that you will often see that indicates a union is “or”.
## What are the types of sets?
Types of a Set
• Finite Set. A set which contains a definite number of elements is called a finite set.
• Infinite Set. A set which contains infinite number of elements is called an infinite set.
• Subset.
• Proper Subset.
• Universal Set.
• Empty Set or Null Set.
• Singleton Set or Unit Set.
• Equal Set.
## What are the set symbols?
Symbol Meaning Example
{ } Set: a collection of elements {1, 2, 3, 4}
A ∪ B Union: in A or B (or both) C ∪ D = {1, 2, 3, 4, 5}
A ∩ B Intersection: in both A and B C ∩ D = {3, 4}
A ⊆ B Subset: every element of A is in B. {3, 4, 5} ⊆ D
## Where do I find AUB in sets?
Let A and B be the two sets. The union of A and B is the set of all those elements which belong either to A or to B or both A and B. Now we will use the notation A U B (which is read as ‘A union B’) to denote the union of set A and set B. Thus, A U B = {x: x ∈ A or x ∈ B}.
You might be interested: Quick Answer: What Is The Meaning Of Lcm In Math?
## What are complements in sets?
Complement of a Set: The complement of a set, denoted A’, is the set of all elements in the given universal set U that are not in A. In set – builder notation, A’ = {x ∈ U: x ∉ A}. Example: U’ = ∅ The complement of the universe is the empty set. Example: ∅’ = U The complement of an empty set is the universal set. |
# How do you solve (x-4)(x+3) = 6?
##### 1 Answer
Feb 5, 2016
$x = \frac{1 \pm \sqrt{73}}{2}$
#### Explanation:
How to solve $\left(x - 4\right) \left(x + 3\right) = 6$
Step 1: Multiple the expand and combine like terms for the left side of the equation
$\left(x - 4\right) \left(x + 3\right) = 6$
$\left({x}^{2} - 4 x + 3 x - 12\right) = 6$
$\left({x}^{2} - x \pm 2\right) = 6$
Step 2-Set the equation equal to zero by subtracting both side of the equation by 6
${x}^{2} - x - 12 = 6$
$- 6 \text{ " } - 6$
$= = = = =$
color(red)(x^2 -x -18= 0
Step 3 Solve the equation using quadratic formula, since we can't factor it
Recall: Quadratic formula: $x = \frac{- b \pm \sqrt{{\left(b\right)}^{2} - 4 a c}}{2 a}$
color(red)(x^2 -x -18= 0
$a = 1 , b = - 1 , c = - 18$
Substitute into the quadratic formula, we get
x= (-(-1)+- sqrt((-1)^2-4(1)(-18)))/(2(1)
$x = \frac{1 \pm \sqrt{1 + 72}}{2}$
$x = \frac{1 \pm \sqrt{73}}{2}$ |
## College Algebra (6th Edition)
a. Restriction values $: 0$ b. Solution set $:\{ 3 \}$
$\frac{5}{2x} - \frac{8}{9} = \frac{1}{18} - \frac{1}{3x}$ a. $x =0$ makes the denominator zero. So, $x \ne 0.$ b. $\frac{5}{2x} - \frac{8}{9} = \frac{1}{18} - \frac{1}{3x};x \ne 0;$ Take LCD on both sides. $\frac{45-16x}{18x} = \frac{x-6}{18x} ;x \ne 0;$ Multiply both sides by $18x$ to clear fraction part. $18x(\frac{45-16x}{18x}) = 18x(\frac{x-6}{18x}) ;x \ne 0;$ $45-16x = x-6$ $45+6 = x+16x$ $51 = 17x$ $x = \frac{51}{17}$ $x = 3$ is not part of the restriction values. So, Solution set $:\{ 3 \}$ |
# Determinants
The first thing to be clear is that we always have to work with a determinant of n \times n, what we mean is that it has to be a square matrix. Note: later we will see the determinants of systems of linear equations.
The determinant symbol for matrices of order n is represented as \Delta_{n}, a determinant of n \ rows and n\ columns, now we see it below, quiet. There are authors who use a letter to represent the columns and a number as a subscript to represent the rows:
\Delta_{n} = \left| \begin{array}{c c c c} a_{1} & b_{1} & \dots & z_{1} \\ a_{2} & b_{2} & \dots & z_{2} \\ \dots & \dots & \dots & \dots \\ a_{n} & b_{n} & \dots & z_{n} \end{array} \right|
And there are authors that use the same letter as a base and its subscripts mean the rows and columns respectively, that means that a_{2 1} refers to the value that is in row 2, column 1, let’s see the example:
\Delta_{n} = \left| \begin{array}{c c c c} a_{1 1} & a_{1 2} & \dots & a_{1 n} \\ a_{2 1} & a_{2 2} & \dots & a_{2 n} \\ \dots & \dots & \dots & \dots \\ a_{n 1} & a_{n 2} & \dots & a_{n n} \end{array} \right|
With that we are ready to move on to the examples, the notation that we will use is that of the second example.
### Determinants of second order
When we talk about determinants of second order, we talk that we will calculate determinants of square matrices of 2 \times 2. Let’s calculate the determinant of the matrix l:
l = \left[ \begin{array}{c c} 3 & 5 \\ 2 & 6 \\ \end{array} \right]
To calculate the determinants of the matrices of 2 \times 2, you have to multiply the numbers of the blue diagonal minus the multiplication of the numbers of the red diagonal:
\Delta_{A} = a_{11}\cdot a_{22} - [a_{21} \cdot a_{12}]
Ok, now let’s proceed to calculate the determinant of the l matrix:
\Delta_{l} = \left| \begin{array}{c c} 3 & 5 \\ 2 & 6 \\ \end{array} \right| = 3\cdot 6 - [2\cdot 5] = 8
So we observe that the determinant of the matrix l equals 8.
That’s all you need to calculate determinants of 2 \times 2 matrices.
### Determinants of third order
For the calculation of third order determinants we will use something peculiar that is only to make it more comfortable to look at, which is to add two more columns, but when you take some practice, the determinant will be done without doing this of add the two more columns Well, let’s see our next matrix:
B = \left[ \begin{array}{c c c} 5 & -1 & 3 \\ 3 & 5 & -4 \\ 7 & -2 & 3 \end{array} \right]
To calculate the determinant of the B matrix, we are going to add the first two columns one more time:
\Delta_{B} =\left| \begin{array}{c c c | c c} 5 & -1 & 3 & 5 & -1 \\ 3 & 5 & -4 & 3 & 5 \\ 7 & -2 & 3 & 7 & -2 \end{array}\right.
Now, to calculate the determinant of the matrix B, we will sum the multiplication of each blue diagonal and then we will subtract the sum of each multiplication of each red diagonal, visually it looks like this:
\Delta_{B} = 5\cdot 5 \cdot 3 + (-1)\cdot (-4) \cdot 7 + 3\cdot 3 \cdot(-2) -
\left[ 7 \cdot 5 \cdot 3 + (-2) \cdot (-4) \cdot 5 + 3\cdot 3 \cdot (-1) \right]
Finally we just have to solve all the operations and we will have the determinant, let’s go step by step, let’s solve all the multiplications:
\Delta _{B} = 75 + 28 - 18 - [105 + 40 - 9]
Now let’s eliminate the brackets:
\Delta_{B} = 75 + 28 - 18 - 105 - 40 + 9
Finally proceed to perform the addition and subtraction:
\Delta_{B} = -51
So the determinant of our matrix of 3\times 3 is -51!
Great, you already know how to calculate determinants of 2×2 matrices and 3×3 matrices!
### Determinants of systems of linear equations
We have come to something more peculiar, quite simple and useful. The calculation of determinants of systems of linear equations of n\times n serves to find the values of the unknowns, as we can find the values of the unknowns of the systems of linear equations of 2\times 2:
\begin{array}{l} x + 2y = 1 \\ x + y = 1 \end{array}
Where:
x = 1 \quad y = 0
Or in a system of equations 3\times 3:
\begin{array}{l} x + y + z = 1 \\x + 2y + z = 1 \\x + y + 3z = 1 \end{array}
Where:
The number of unknowns will tell what size is the system of linear equations, if you have two unknowns, the system of equations is 2\times 2, if you have three unknowns, the system of equations is 3\times 3 and so successively.
To find the values of the unknowns, we must apply the Cramer rule with which the determinants of the matrices of the unknowns are divided among the determinant of the system matrix. Let’s quickly see the following example:
\begin{array}{l} a_{11}x + a_{12}y + a_{13}z = r_{1} \\ a_{21}x + a_{22}y + a_{23}z = r_{2} \\ a_{31}x + a_{32}y + a_{33}z = r_{3} \end{array}
Our matrix is the following:
s =\left( \begin{array}{c c c | c} a_{11} & a_{12} & a_{13} & r_{1} \\ a_{21} & a_{22} & a_{23} & r_{2} \\ a_{31} & a_{32} & a_{33} & r_{3} \end{array}\right)
To find the values of the unknowns, we will find the determinant of the system matrix (remember that the values of r_{1}, r_{2} and r_{3} are ignored to calculate the matrix system for the determinant of s):
\Delta_{s} = \left| \begin{array}{c c c} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right |
We will do exactly the same as we did the sum of the product of the blue diagonals minus the sum of the product of the red diagonals, the same as we did above the post:
\Delta_{s} = \begin{array}{|c c c | c c} a_{11} & a_{12} & a_{13} & a_{11} & a_{12} \\ a_{21} & a_{22} & a_{23} & a_{21} & a_{22} \\ a_{31} & a_{32} & a_{33} & a_{31} & a_{32} \end{array}
\Delta_{s} = a_{11}\cdot a_{22} \cdot a_{33} + a_{12} \cdot a_{23} \cdot a_{31} + a_{13} \cdot a_{21} \cdot a_{32} -
\left[ a_{31}\cdot a_{22} \cdot a_{13} + a_{32} \cdot a_{23} \cdot a_{11} + a_{33} \cdot a_{21} \cdot a_{12}\right]
We will find the determinant of x, where the matrix for the determinant of x is the same system matrix only in the first column we will write the values of r_{1}, r_{2} and r_{3} respectively:
x = \left[ \begin{array}{c c c} r_{1} & a_{12} & a_{13} \\ r_{2} & a_{22} & a_{23} \\ r_{3} & a_{32} & a_{33} \end{array}\right ]
We will do exactly the same as we did of the sum of the product of the blue diagonals minus the sum of the product of the red diagonals:
\Delta_{x} = \begin{array}{|c c c | c c} r_{1} & a_{12} & a_{13} & r_{1} & a_{12} \\ r_{2} & a_{22} & a_{23} & r_{2} & a_{22} \\ r_{3} & a_{32} & a_{33} & r_{3} & a_{32} \end{array}
\Delta_{x} = r_{1}\cdot a_{22} \cdot a_{33} + a_{12} \cdot a_{23} \cdot r_{3} + a_{13} \cdot r_{2} \cdot a_{32} -
\left[ r_{3}\cdot a_{22} \cdot a_{13} + a_{32} \cdot a_{23} \cdot r_{1} + a_{33} \cdot r_{2} \cdot a_{12}\right]
To find the determinant for the y matrix, we do the same procedure as we did for the x matrix, we take the system matrix but now we will change the second column for the values of r_{1}, r_{2} and r_{3} respectively:
y = \left[ \begin{array}{c c c} a_{11} & r_{1} & a_{13} \\ a_{21} & r_{2} & a_{23} \\ a_{31} & r_{3} & a_{33} \end{array}\right ]
\Delta_{y} = \begin{array}{|c c c | c c} a_{11} & r_{1} & a_{13} & a_{11} & r_{1} \\ a_{21} & r_{2} & a_{23} & a_{21} & r_{2} \\ a_{31} & r_{3} & a_{33} & a_{31} & r_{3} \end{array}
\Delta_{y} = a_{11}\cdot r_{2} \cdot a_{33} + r_{1} \cdot a_{23} \cdot a_{31} + a_{13} \cdot a_{21} \cdot r_{3} -
\left[ a_{31}\cdot r_{2} \cdot a_{13} + r_{3} \cdot a_{23} \cdot a_{11} + a_{33} \cdot a_{21} \cdot r_{1}\right]
To find the determinant for the z matrix, we do the same procedure as we did for the x matrix, we take the matrix of the system but now we will change the third column for the values of r_{1}, r_{2} and r_{3} respectively:
z = \left[ \begin{array}{c c c} a_{11} & a_{12} & r_{1} \\ a_{21} & a_{22} & r_{2} \\ a_{31} & a_{32} & r_{3} \end{array}\right ]
\Delta_{z} = \begin{array}{|c c c | c c} a_{11} & a_{12} & r_{1} & a_{11} & a_{12} \\ a_{21} & a_{22} & r_{2} & a_{21} & a_{22} \\ a_{31} & a_{32} & r_{3} & a_{31} & a_{32} \end{array}
\Delta_{z} = a_{11}\cdot a_{22} \cdot r_{3} + a_{12} \cdot r_{2} \cdot a_{31} + r_{1} \cdot a_{21} \cdot a_{32} -
\left[ a_{31}\cdot a_{22} \cdot r_{1} + a_{32} \cdot r_{2} \cdot a_{11} + r_{3} \cdot a_{21} \cdot a_{12}\right]
Once we find the determinants of each matrix, we will apply Cramer’s rule to obtain the final values:
x = \cfrac{\Delta_{x}}{\Delta_{s}}
y = \cfrac{\Delta_{y}}{\Delta_{s}}
z = \cfrac{\Delta_{z}}{\Delta_{s}}
### Example of determinants
In the following example we will use a system of 3\times 3 equations with which we will use the determinant method to find the values of x, y and z. Let’s solve the following system of linear equations:
\begin{array}{l} 4x - 2y + 2z = 36 \\ 2x + 4y - 5z = 28 \\ 6x - 3y + 2z = 24 \end{array}
Where our matrix will look like this:
s = \left(\begin{array}{c c c | c} a_{11} & a_{12} & a_{13} & r_{1} \\ a_{21} & a_{22} & a_{23} & r_{2} \\ a_{31} & a_{32} & a_{33} & r_{3} \end{array}\right) = \left(\begin{array}{c c c | c} 4 & -2 & 2 & 36 \\ 2 & 4 & -5 & 28 \\ 6 & -3 & 2 & 24 \end{array}\right)
To solve the system of linear equations shown, we have to find the determinant of the system (s), the determinant of x, the determinant of y and the determinant of z, so first let’s find the determinant of the system as we have seen above of the post, add the two columns and carry out the corresponding operations:
\Delta_{s} = \left | \begin{array}{c c c | c c} 4 & -2 & 2 & 4 & -2\\ 2 & 4 & -5 & 2 & 4\\ 6 & -3 & 2 & 6 & -3 \end{array}\right .
\Delta_{s}=4 \cdot 4 \cdot 2 + (-2)\cdot(-5)\cdot 6 + 2 \cdot 2 \cdot(-3) -
\left[ 6\cdot4\cdot2 + (-3)\cdot(-5)\cdot 4 + 2\cdot2 \cdot(-2) \right]
\Delta_{s} = -20
Now let’s find the \Delta_{x}, change the first column by the values of r_{1}, r_{2} and r_{3}
\Delta_{x} = \left | \begin{array}{c c c} 36 & -2 & 2 \\ 28 & 4 & -5 \\ 24 & -3 & 2 \end{array}\right |
Let’s add the two more columns so that when making the multiplication it becomes easier visually:
\Delta_{x} = \left | \begin{array}{c c c | c c} 36 & -2 & 2 & 36 & -2 \\ 28 & 4 & -5 & 28 & 4 \\ 24 & -3 & 2 & 24 & -3 \end{array}\right.
Now proceed to write the corresponding operations:
\Delta_{x} = 36 \cdot 4 \cdot 2 + (-2)\cdot(-5)\cdot 24 + 2 \cdot 28 \cdot(-3) -
\left[ 24 \cdot 4 \cdot 2 + (-3)\cdot(-5)\cdot 36 + 2 \cdot 28 \cdot(-2) \right]
Doing all the operations we will have the following result:
\Delta_{x} = -260
Let’s find the \Delta_{y}, change the values of the second column by the values of r_{1}, r_{2} and r_{3}:
\Delta_{y} = \left | \begin{array}{c c c} 4 & 36 & 2 \\ 2 & 28 & -5 \\ 6 & 24 & 2 \end{array}\right |
Proceed to add the two more columns only to make it easier visually when performing operations:
\Delta_{y} = \left|\begin{array}{c c c | c c} 4 & 36 & 2 & 4 & 36 \\ 2 & 28 & -5 & 2 & 28\\ 6 & 24 & 2 & 6 & 24 \end{array} \right.
\Delta_{y} = 4 \cdot 28 \cdot 2 + 36\cdot(-5)\cdot 6 + 2 \cdot 2 \cdot 24 -
\left[ 6 \cdot 28 \cdot 2 + 24 \cdot (-5) \cdot 4 + 2 \cdot 2 \cdot 36 \right]
\Delta_{y} = -760
Finally find the \Delta_{z}, change the values of the third column by the values of r_{1}, r_{2} and r_{3} respectively:
\Delta_{z} = \left | \begin{array}{c c c} 4 & -2 & 36 \\ 2 & 4 & 28 \\ 6 & -3 & 24 \end{array}\right |
Now let’s add the two more columns to make operations easier visually:
\Delta_{z} = \left | \begin{array}{c c c | c c} 4 & -2 & 36 & 4 & -2 \\ 2 & 4 & 28 & 2 & 4 \\ 6 & -3 & 24 & 6 & -3 \end{array}\right .
\Delta_{z} = 4 \cdot 4 \cdot 24 + (-2) \cdot 28 \cdot 6 + 36 \cdot 2 \cdot(-3) -
\left[ 6 \cdot 4 \cdot 36 + (-3) \cdot 28 \cdot 4 + 24 \cdot 2 \cdot(-2) \right]
\Delta_{z} = -600
With all the determinants found, we can proceed to carry out the divisions that will tell us each value of the unknowns:
x = \cfrac{\Delta_{x}}{\Delta_{s}} = \cfrac{-260}{-20} = 13
y = \cfrac{\Delta_{y}}{\Delta_{s}} = \cfrac{-760}{-20} = 38
z = \cfrac{\Delta_{z}}{\Delta_{s}} = \cfrac{-600}{-20} = 30
Yes now! Our final results of our unknowns are: |
Resolution of a Vector
# Resolution of a Vector
## Introduction
A vector’s resolution is the breaking of a single vector into two or more vectors in different directions that, when combined, have the same effect as a single vector. A vector that is directed at an angle to the horizontal (or vertical) can be divided into two parts (or components).
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In other words, any two-dimensional vector can be conceived of as having two components. When a chain pulls upward at an angle on a dog’s collar, for example, a tension force is generated in two dimensions. There are two components to this tension force: an upward component and a rightward component.
Consider an airplane flying northwest from Chicago’s O’Hare International Airport to a destination in Canada. The plane’s displacement vector is two-dimensional (north). Vector resolution is the way of assessing the magnitude of a vector. We’ll look at two different vector resolution approaches.
• The method of the parallelogram
• The trigonometric method
## A Brief outline
The parallelogram technique of vector resolution includes determining the components of a vector using a correctly drawn, scaled vector diagram. In a nutshell, the procedure entails drawing the vector to scale in the stated direction, sketching a parallelogram around the vector so that the vector is the parallelogram’s diagonal, and using the scale to determine the magnitude of the components (the parallelogram’s sides). The parallelogram is a rectangle having vertical and horizontal sides that can be used to calculate the components as directed along the standard x- and y-coordinate axes.
The trigonometric approach of vector resolution includes determining the vector’s components using trigonometric functions. It was discussed how to find the direction of a vector using trigonometric functions. The components of a single vector will be determined using trigonometric functions. The ratio of the lengths of the sides of a right triangle to the value of an acute angle within the right triangle is calculated using trigonometric functions. As a result, if an angle indicator and the length of one side are known, trigonometric functions can be used to compute the length of the sides of a right triangle.
## Important concepts:
### The following is a step-by-step guide upon using the parallelogram approach of vector resolution:
1. Chosen a scale and constructed the vector to scale in the direction provided.
2. Created a parallelogram around the vector by sketching vertical and horizontal lines at the vector’s tail, then horizontal and vertical lines at the vector’s head; the drawn lines will meet to make a rectangle (a special case of a parallelogram).
3. Drew the vector’s components. The parallelogram’s sides are the components. The components’ tails begin at the tail of the vector and extend along the axes to the parallelogram’s nearest corner. Take care to use arrowheads to indicate the direction of these elements (up, down, left, right).
4. Labelled the components of the vectors with symbols to indicate which side each component represents. North is a name for a northward force component. vx could be the name of a rightward velocity component, and so on.
5. Used the scale to assess the magnitude of the elements in real units by measuring the length of the parallelogram’s sides. On the diagram, write the magnitude.
### The following is the procedure for determining the components of a vector using trigonometric functions:
1. Generated a rough outline of the vector in the provided direction (no scale required). Make a note of its size and the angle it creates with the horizontal.
2. Drew a rectangle around the vector, with the vector as the rectangle’s diagonal. Draw vertical and horizontal lines starting at the tail of the vector. Then, at the top of the vector, draw horizontal and vertical lines. The drawn lines will come together to form a rectangle.
3. Drew the vector’s components. The rectangle’s sides are the components. Each component’s tail starts at the tail of the vector and extends along the axes to the rectangle’s nearest corner. Take care to use arrowheads to indicate the direction of these elements (up, down, left, right).
4. Identified the components of the vectors using symbols to designate which side each component represents. North is a name for a northward force component. The velocity component of a rightward force can be denoted vx, and so on.
5. Using the sine function, the length of the side opposite the stated angle was found. Substituted the length of the hypotenuse for the magnitude of the vector. Write the equation for the length of the side opposite the indicated angle employing algebra.
6. Found the height of the side adjacent to the stated angle, repeated the previous step using the cosine function.
## Significance of resolution of a vector in NEET:
The important questions in vector resolution will help you prepare for the next NEET exams. The NEET resolution vector Vital Problems were created by competent teachers to help students in need and to ensure that students can answer all scalar and vector questions from class 11 physics. The following important questions will benefit students studying for the NEET exam as well as those training for the JEE admission exam. Vector questions are a useful mathematical tool when dealing with numerical-type physics problems.
Students can download chapter-by-chapter basic questions in a Pdf file from the limitless learn website and use them for additional study. Students should also thoroughly review the CBSE board’s Physics Scalars & Vectors syllabus and download vector-related problems for each course. This will greatly aid everyone in identifying any flaws in their understanding of the principles. In order to improve their test results, students in class 11 should practice Vectors resolution major questions on a daily basis.
In summation, a two-dimensional vector has two components, i.e., an impact in two different directions. Techniques of vector resolution can be used to determine the quantity of effect in a given direction. A modelling approach (parallelogram method) and a trigonometric method for vector resolution are described here. In the same plane, a vector can be defined in terms of other vectors. A can be represented as the summation of a and b after multiplying them with some real numbers if there are three vectors A, a, and b. A could be established into two constituent vectors λa and μb. Later, A = λa + μb. Now λ and μ are real figures.
Question 1: What is the formula for calculating the horizontal and vertical components of a vector?
Answer: We know that the horizontal component is the velocity or force that really is parallel to the horizontal axis, and the vertical component is the amount that is parallel to the vertical axis. The horizontal and vertical parts can be calculated using the right-angled triangle as a model. The force or velocity is represented by the hypotenuse, and the angle is used to quantify the vertical and horizontal components of vectors.
Question 2: Give an example of a vector definition.
Answer: A vector is a physical quantity with a magnitude as well as direction. The movement of moving items from one point to another is described by a vector quantity.
Question 3: Is it Possible to Resolve a Vector?
Answer: Yes, it is correct. Take the two numbers, say 5 and 8, and add them together and get the 13. In addition, the number 13 is split or resolved.
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# How do you solve and write the following in interval notation: |6t-6|<6?
Jan 9, 2017
See full solution process below
#### Explanation:
Because this is a problem contain the absolute value function we must solve the problem for both the negative and positive forms of the problem or in this case +6 and -6.
Also, because it is an inequality we must solve it as a system of inequalities as shown below.
We can rewrite this problem as:
$- 6 < 6 t - 6 < 6$
We can now solve while ensuring we perform all operations to each portion of the system of inequalities.
First we will add $\textcolor{red}{6}$ to each portion of the system:
$- 6 + \textcolor{red}{6} < 6 t - 6 + \textcolor{red}{6} < 6 + \textcolor{red}{6}$
$0 < 6 t - 0 < 12$
$0 < 6 t < 12$
Next we will divide each portion of the system by $\textcolor{red}{6}$ to solve for $t$ while keeping the system balanced:
$\frac{0}{\textcolor{red}{6}} < \frac{6 t}{\textcolor{red}{6}} < \frac{12}{\textcolor{red}{6}}$
$0 < \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{6}}} t}{\cancel{\textcolor{red}{6}}} < 2$
$0 < t < 2$
Writing this solution in interval form gives:
(0, 2) |
How do you multiply (x+3)(x+4) ?
Aug 11, 2015
$\left(x + 3\right) \cdot \left(x + 4\right) = {x}^{2} + 7 \cdot x + 12$
Explanation:
The rigorous but rather lengthy explanation involves the distributive law of multiplication of a sum of two numbers by the third one and commutative law for both operations, addition and multiplication.
The distributive law of multiplication states:
$a \cdot \left(b + c\right) = a \cdot b + a \cdot c$
The commutative law of addition states:
$a + b = b + a$
The commutative law of multiplication states"
$a \cdot b = b \cdot a$
All the above laws are used (usually without explicit mentioning) when performing the operations of this problem.
Using the distributive law described above for
$a = x + 3$, $b = x$ and $c = 4$,
we obtain:
$\left(x + 3\right) \cdot \left(x + 4\right) = \left(x + 3\right) \cdot x + \left(x + 3\right) \cdot 4$
Now let's use the commutative law of multiplication to rewrite it as
$\left(x + 3\right) \cdot \left(x + 4\right) = x \cdot \left(x + 3\right) + 4 \cdot \left(x + 3\right)$
Now use the distributive law twice, first for
$a = x$, $b = x$ and $c = 3$
obtaining
$x \cdot \left(x + 3\right) = x \cdot x + x \cdot 3$
and another time for
$a = 4$, $b = x$ and $c = 3$
obtaining
$4 \cdot \left(x + 3\right) = 4 \cdot x + 4 \cdot 3$
Let's use these equalities in the original expression:
$\left(x + 3\right) \cdot \left(x + 4\right) = x \cdot x + x \cdot 3 + 4 \cdot x + 4 \cdot 3$
Further simplification involves the following transformations:
$x \cdot x = {x}^{2}$ (just for convenience)
Using commutative and distributive laws,
$x \cdot 3 + 4 \cdot x = x \cdot 3 + x \cdot 4 = x \cdot \left(3 + 4\right) = x \cdot 7 = 7 \cdot x$
Obviously,
$4 \cdot 3 = 12$
Our expression looks now as follows:
$\left(x + 3\right) \cdot \left(x + 4\right) = {x}^{2} + 7 \cdot x + 12$
This is the final representation of the original expression as a polynomial of the second degree.
Of course, most students do all these transformations without even thinking about the laws they are based upon and without writing an intermediary results. But these skills develop as a result of numerous exercises. Practice makes it perfect. |
# Multiply Large Numbers Worksheet
This multiplication worksheet focuses on training college students how to mentally increase complete figures. Pupils can use custom grids to suit exactly 1 concern. The worksheets also protectfractions and decimals, and exponents. There are also multiplication worksheets using a distributed home. These worksheets are a need to-have for your personal math course. They are often used in class to figure out how to mentally increase total line and numbers them up. Multiply Large Numbers Worksheet.
## Multiplication of complete numbers
If you want to improve your child’s math skills, you should consider purchasing a multiplication of whole numbers worksheet. These worksheets will help you learn this simple idea. It is possible to decide to use one particular digit multipliers or two-digit and 3-digit multipliers. Capabilities of 10 are also an incredible option. These worksheets will enable you to training lengthy practice and multiplication reading the phone numbers. They are also a wonderful way to help your kids recognize the value of learning the different kinds of whole phone numbers.
## Multiplication of fractions
Having multiplication of fractions over a worksheet will help teachers plan and make lessons effectively. Utilizing fractions worksheets allows educators to easily evaluate students’ idea of fractions. Pupils could be challenged to finish the worksheet within a particular some time and then mark their strategies to see exactly where they need additional training. Pupils can be helped by expression problems that associate maths to real-daily life conditions. Some fractions worksheets involve types of comparing and contrasting figures.
## Multiplication of decimals
Whenever you flourish two decimal amounts, make sure to group of people them vertically. The product must contain the same number of decimal places as the multiplicant if you want to multiply a decimal number with a whole number. For example, 01 x (11.2) by 2 could be equal to 01 x 2.33 x 11.2 unless the product has decimal spots of lower than two. Then, this product is circular to the local complete amount.
## Multiplication of exponents
A arithmetic worksheet for Multiplication of exponents can help you practice multiplying and dividing amounts with exponents. This worksheet may also offer conditions that will demand students to multiply two various exponents. By selecting the “All Positive” version, you will be able to view other versions of the worksheet. Apart from, also you can enter in unique guidelines about the worksheet on its own. When you’re concluded, you can just click “Generate” and also the worksheet will probably be acquired.
## Section of exponents
The fundamental guideline for section of exponents when multiplying phone numbers is to deduct the exponent in the denominator in the exponent inside the numerator. However, if the bases of the two numbers are not the same, you can simply divide the numbers using the same rule. By way of example, \$23 separated by 4 will equivalent 27. This method is not always accurate, however. This method can result in uncertainty when multiplying numbers which are too big or not big enough.
## Linear characteristics
You’ve probably noticed that the cost was \$320 x 10 days if you’ve ever rented a car. So, the total rent would be \$470. A linear function of this type has the kind f(by), where by ‘x’ is the volume of days the automobile was rented. Moreover, they have the shape f(by) = ax b, exactly where ‘b’ and ‘a’ are actual amounts. |
# How do you differentiate (3x-4/x)^2 (1-x+7x^2)^4?
May 15, 2015
Chain rule and product rule time combined!
Let's work with your function $f \left(x\right) = {\left(3 x - 4 {x}^{-} 1\right)}^{2} {\left(1 - x + 7 {x}^{2}\right)}^{4}$.
First, we will rename:
$u = {\left(3 x - 4 {x}^{-} 1\right)}^{2}$ and $w = {\left(1 - x + 7 {x}^{2}\right)}^{4}$
Using the product rule, as follows,
$\frac{d \left(u w\right)}{\mathrm{dx}} = w \left(\frac{\mathrm{du}}{\mathrm{dx}}\right) + u \left(\frac{\mathrm{dw}}{\mathrm{dx}}\right)$
${\left(1 - x + 7 {x}^{2}\right)}^{4} \cdot \textcolor{g r e e n}{\frac{d {\left(3 x - 4 {x}^{-} 1\right)}^{2}}{\mathrm{dx}}} + {\left(3 x - 4 {x}^{-} 1\right)}^{2} \cdot \textcolor{b l u e}{\frac{d {\left(1 - x + 7 {x}^{2}\right)}^{4}}{\mathrm{dx}}}$
Now, we will lead with the colored derivatives first and then return to the whole equation above.
To solve the green derivative, let's name $a = 3 x - 4 {x}^{-} 1$ and solve it as a chain rule, where
$\frac{\mathrm{dy}}{\mathrm{da}} \cdot \frac{\mathrm{da}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dx}}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 \left(3 x - 4 {x}^{-} 1\right) \left(3 + 4 {x}^{-} 2\right)$
The same for the blue part, where $b = 1 - x + 7 {x}^{2}$:
$\frac{\mathrm{dy}}{\mathrm{dx}} = 4 {\left(1 - x + 7 {x}^{2}\right)}^{3} \left(- 1 + 14 x\right)$
Now we have the derivatives $\frac{\mathrm{du}}{\mathrm{dx}}$ and $\frac{\mathrm{dw}}{\mathrm{dx}}$, let's go back to the partillay colored long equation:
${\left(1 - x + 7 {x}^{2}\right)}^{4} \cdot 2 \left(3 x - 4 {x}^{-} 1\right) \cdot \left(3 + 4 {x}^{-} 2\right) + {\left(3 x - 4 {x}^{-} 1\right)}^{2} \cdot 4 {\left(1 - x + 7 {x}^{2}\right)}^{3} \cdot \left(- 1 + 14 x\right)$
You may want to expand it and eliminate the factoring (parenthesis), but that is not really necessary as the work and steps to achieve so would not necessarily mean simplification. The line above can be your answer. |
Question
# The area of a square field is 729 square feet (sq. ft.). What is the side length of square field?
A
33
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B
37
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C
23
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D
27
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Solution
## The correct option is D 27Given: Area of square field =729 sq. ft. To Find: Side length of square field As given in the hint, the area of the square field is side2. ∴ Side length =√Area=√729 ft. Let's find the √729 using the long division method. Step 1: Place a bar over the pair of numbers starting from the unit place or right side of the number. ¯¯¯7 ¯¯¯¯¯¯29. Step 2: Take the largest number as the divisor whose square is less than or equal to the number on the extreme left of the dividend (7). Divide and write the quotient. Here the quotient is 2 and the remainder is 3. Step 3: Next, we bring down the number, which is under the bar, to the right side of the remainder. Here, we bring down 29. So, 329 is our new dividend. Step 4: Now, double the value of the quotient and enter it with a blank on its unit place. Step 5: Next, we have to select the largest digit for the unit place of the current divisor (4−) such that the new divisor, when multiplied by the new digit at its unit place, is equal to or less than the current dividend (329). In this case, 47–×7–=329. So the unit's place digit is 7. Step 6: Finally, the remainder is 0, and we have no number left for division, therefore, the square root of 729 is the final quotient, i.e., √729=27. ∴ The side length of square field with area 729 square feet is 27feet–––––––. Hence, option (d.) is the correct one.
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Note: You are looking at a static copy of the former PineWiki site, used for class notes by James Aspnes from 2003 to 2012. Many mathematical formulas are broken, and there are likely to be other bugs as well. These will most likely not be fixed. You may be able to find more up-to-date versions of some of these notes at http://www.cs.yale.edu/homes/aspnes/#classes.
# 1. Diagonal matrices
A matrix A is a diagonal matrix if it is a square matrix with Aij=0 whenever i≠j.
1. Prove or disprove: If A and B are diagonal matrices of the same size, so is AB.
2. Let p(A)=Πi Aii. Prove or disprove: If A and B are diagonal matrices as above, then p(AB) = p(A)p(B).
## 1.1. Solution
1. We need to show that (AB)ij=0 for i≠j (we don't care what happens when i=j). Let i≠j, and compute (AB)ij = ∑k AikBkj = AiiBij + AijBjj = 0, where the first simplification uses the fact that Aik=0 unless i=k (and similarly for Bkj), and the second uses the assumption that i≠j.
2. Now we care what happens to (AB)ii. Compute (AB)ii = ∑k AikBki = AiiBii. So p(AB) = Πi (AB)ii = ∏i (AiiBii) = (∏i Aii)(Πi Bii) = p(A)p(B).
# 2. Matrix square roots
1. Show that there exists a matrix A such that A≠0 but A²=0.
2. Show that if A²=0, there exists a matrix B such that B²=I+A. Hint: What is (I+A)²?
## 2.1. Solution
Here is a simple example of a nonzero matrix whose square is 0:
For the second part, the hint suggests looking at (I+A)² = I² + IA + AI + A² = I + 2A (since IA=AI=A and it is given that A²=0). So I+A is almost right, but there is that annoying 2 there. We can get rid of the 2 by setting B instead to I+½A, which gives B² = (I+½A)² = I+A+¼A² = I+A.
# 3. Dimension reduction
Let A be an n×m random matrix obtained by setting each entry Aij independently to ±1 with equal probability.
Let x be an arbitrary vector of dimension m.
Compute E[||Ax||²], as a function of ||x||, n, and m, where ||x|| = (x⋅x)1/2 is the usual Euclidean length.
## 3.1. Solution
Mostly this is just expanding definitions.
The second-to-last step follows because E[AijAik] = 0 when Aij and Aik are independent (i.e., when j≠k) and E[AijAij] = E[(±1)²) = 1.
# 4. Non-invertible matrices
Let A be a square matrix.
1. Prove that if Ax=0 for some column vector x≠0, then A-1 does not exist.
2. Prove that if the columns of A are not linearly independent, then A-1 does not exist.
3. Prove that if the rows of A are not linearly independent, then A-1 does not exit.
## 4.1. Solution
1. Suppose A-1 exists and that Ax=0 for some nonzero x. Then x = (A-1A)x = A-1(Ax) = A-10 = 0, a contradiction.
2. Let A⋅i represent the i-th column of A. If the columns of a are not linearly independent, there exist coefficients xi, not all zero, such that ∑ xiA⋅i = 0. But then Ax = ∑ xiA⋅i = 0, where x is the (nonzero) vector of these coefficients. It follows from the previous case that A is not invertible.
3. Observe that if A has an inverse, then so does its transpose A', since if A-1 exists we have (A-1)'A' = (AA-1)' = I and A'(A-1)' = (A-1A)' = I. If the rows of A are not linearly independent, then neither are the columns of A'; it follows that A' has no inverse, and thus neither does A.
2014-06-17 11:57 |
# How to Calculate 1/1 Minus 22/1
Are you looking to work out and calculate how to subtract 1/1 from 22/1? In this really simple guide, we'll teach you exactly what 1/1 - 22/1 is and walk you through the step-by-process of how to subtract one fraction from another.
Want to quickly learn or show students how to calculate 1/1 minus 22/1? Play this very quick and fun video now!
To start with, the number above the line in a fraction is called a numerator and the number below the line is called the denominator.
Why do you need to know this? Well, because to subtract a fractions from another we need to first make sure both fractions have the same denominator.
Let's set up 1/1 and 22/1 side by side so they are easier to see:
1 / 1 - 22 / 1
In this calculation, some of the hard work has already been done for us because the denominator is the same in both fractions. All we need to do is subtract the numerators and keep the denominator as it is:
1 - 22 / 1 = 23 / 1
You're done! You now know exactly how to calculate 1/1 - 22/1. Hopefully you understood the process and can use the same techniques to add other fractions together. The complete answer is below (simplified to the lowest form):
-21
## Convert 1/1 minus 22/1 to Decimal
Here's a little bonus calculation for you to easily work out the decimal format of the fraction we calculated. All you need to do is divide the numerator by the denominator and you can convert any fraction to decimal:
-21 / 1 = -21 |
Python Course #6: Multiplying and Dividing Binary Numbers
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# Python Course #6: Multiplying and Dividing Binary Numbers
Now that you can add and subtract binary numbers (Python Course #5: Adding and Subtracting Binary Numbers),let’s take it one step further and multiply and divide binary numbers.
# Multiplying Binary Numbers
First, look at decimal multiplication and how you probably learned how to do it in school. Take, for example, the multiplication of $$23\cdot 45$$. In the first step, you multiply $$3$$ with $$4$$ (the last digit of the first number/factor and the first digit of the second number), which is 12. The 2 is written directly under the first digit of the second number while the 1, representing the 10 in 12, is carried over to the next position:
$\begin{eqnarray} 2\underline{3} \cdot \underline{4}5\\\hline \quad_1 2\hspace{0.5em} \end{eqnarray}$
In the next step, the next digit of the first number is multiplied by the first digit of the second number ($$2\cdot 4$$). The result of this calculation represents the next decimal position. It is therefore written down in front of the result of the first multiplication, and if a digit was carried over from the previous result, they are added together:
$\begin{eqnarray} \underline{2}3 \cdot \underline{4}5\\\hline 92\hspace{0.5em} \end{eqnarray}$
This step is repeated until all digits of the first number have been multiplied by the first digit of the second number. Then, the second digit of the second number is used and multiplied with every digit of the first number going from right to left. So in this example, the first digit pair that has to be multiplied is: $$3\cdot 5$$. This multiplication is now written under the number coming from the multiplication of the first digit of the second number with all the digits of the first number. However, it is shifted one digit back:
$\begin{eqnarray} 2\underline{3} \cdot 4\underline{5}\\\hline 92\hspace{0.5em}\\ \quad_1 5 \end{eqnarray}$
And as you probably already guessed now the next pair of digits to multiply is $$2\cdot 5$$:
$\begin{eqnarray} 2\underline{3} \cdot 4\underline{5}\\\hline 92\hspace{0.5em}\\ 115 \end{eqnarray}$
And if you have more digits in the second number, this step is repeated until every digit of the first number is multiplied with every digit of the second number.
The last step is to add all the results together which will give you the result of the whole multiplication:
$\begin{eqnarray} 23 \cdot 45\\\hline 92\hspace{0.5em}\\ +\hspace{0.5em}115\\\hline 1035 \end{eqnarray}$
The binary multiplication works in the exact same way. But instead of decimal numbers, binary numbers are used. And as only the multiplications including the numbers $$1$$ and $$0$$ have to be taken into account, it becomes much easier. Because every multiplication including a $$0$$ results in $$0$$ and only $$1\cdot 1$$ is $$1$$. And this means there are no carry-bits when multiplying the bit with each other, which is why it is so much easier to multiply binary numbers.
As an example lets multiply $$10_{10} = 1010_2$$ with $$5_{10} = 10_2$$. In the first step the last bit of the first number is multiplied with the first bit of the second number:
$\begin{eqnarray} 101\underline{0} \cdot \underline{1}01\\\hline 0\hspace{0.5em}\hspace{0.5em} \end{eqnarray}$
Since there are no carry-bits when multiplying only single ones and zeros, you can carry one multiplying the other bits of the first number with the first digit of the second number.
$\begin{eqnarray} \underline{1010} \cdot \underline{1}01\\\hline 1010\hspace{0.5em}\hspace{0.5em} \end{eqnarray}$
Now take a look at the first row. It is the same number as the first number of the whole multiplication. Now you can go to the next digit of the second number and multiply it with all digits of the first one.
$\begin{eqnarray} \underline{1010} \cdot 1\underline{0}1\\\hline 1010\hspace{0.5em}\hspace{0.5em}\\ 0000\hspace{0.5em} \end{eqnarray}$
You can see that the whole row is $$0$$ because $$0$$ multiplied with anything is always $$0$$. This is the second reason why it is so easy to multiply binary numbers compared to decimal numbers. Now you can do the multiplications with the last bit of the second number. Which is just writing down the first number starting from under the last bit:
$\begin{eqnarray} \underline{1010} \cdot 1\underline{0}1\\\hline 1010\hspace{0.5em}\hspace{0.5em}\\ 0000\hspace{0.5em}\\ 1010 \end{eqnarray}$
To make it even easier you can skip $$0$$ zero bits of the second number entirly and shift the beginning of the next row to the right by one more bit. Which would result in:
$\begin{eqnarray} \underline{1010} \cdot 1\underline{0}1\\\hline 1010\hspace{0.5em}\hspace{0.5em}\\ 1010 \end{eqnarray}$
The last step is, as in the decimal multiplication, adding up all rows. And if you forgot how to add binary numbers just have a quick look at Python Course #5: Adding and Subtracting Binary Numbers:
$\begin{eqnarray} \underline{1010} \cdot 1\underline{0}1\\\hline 1010\hspace{0.5em}\hspace{0.5em}\\ \oplus\quad\quad1010\\\hline 110010 \end{eqnarray}$
When you got more than two rows to add together it sometimes makes sense to split up the addition into smaller chunks and add them together one after another.
When multiplying a binary number with a power of $$2$$, such as $$2$$, $$4$$, $$8$$, $$16$$, etc., you can speed up your calculation even more. Because for every multiplication with a $$2^x \geq 2$$ you can add a zero at the end of the number multiplied with a power of $$2$$. This operation is called left shift because every bit is shifted to the left-hand side. Take, for example,
$42_{10} = 101010_2$
and you want to multiply it by $$4$$. When you append one $$0$$ you get
$1010100_2 = 84_{10}$
and when you add another $$0$$ you get
$10101000_2 = 168_{10}$
This trick is used quite often in computer science and engineering because multiplication itself can be very costly in terms of time. After all, it consists of many tiny single-bit multiplications and several additions. The left-shift operation is also available in Python as <<. You can try out this code to check it:
1 2 3 4 x = 42 print(f"{x} = {x:b}") y = x << 2 print(f"{y} = {y:b}")
# Dividing Binary Numbers
Now that you know how to multiply binary numbers let’s look at the most complicated operation, the division. Obviously, you only got integer numbers in binary; this means most divisions will have a remainder. As binary multiplication uses single-bit multiplications and additions, the counter operation division uses primary divisions and subtractions to divide two binary numbers.
In this example $$132_{10} = 10000100_2$$ is going to be divided by $$13_{10} = 1101_2$$. As a first step you need the twos’ complement negative of $$13_{10} = 1101_2$$ which is $$10011_2$$ (expanded by one bit such that the MSB is set to $$1$$). To start with the division process write down both numbers and expand the dividend with a $$0$$ in the front to mark it as a positive number, leave some space below the numbers, and add an equal sign at the end:
$010000100 \div 1101 =$
Take the first $$n$$ bits (the length of the divisor) starting from the first $$1$$ of the dividend and compare it to the divisor than there are only three possibilities the divisor is smaller, equal, or larger than the selected bits of the dividend. In this example the bits to check in the dividend are $$1000$$; here you can see that those bits are less than the dividend. Since the divisor does not “fit” into those bits (one could also say the divisor is contained $$0$$ times in those bits), you need a more significant number to “fit” the divisor. To track that the divisor is contained $$0$$ times in the selected bits, write a $$0$$ after the equal sign and note four $$0000$$ under the bits the divisor does not “fit” into.
$\begin{eqnarray} 0\underline{1000}0100 \div 1101 = 0\\ 0000\quad\quad\quad\quad\quad\quad\quad\hspace{0.2em} \end{eqnarray}$
Now you can subtract $$0000$$ from $$1000$$ which is just $$1000$$ and write it under $$0000$$ such that the bits align:
$\begin{eqnarray} 010000100 \div 1101 = 0\\ \underline{0000}\quad\quad\quad\quad\quad\quad\quad\\ 1000\quad\quad\quad\quad\quad\quad\quad\hspace{0.1em}\\ \end{eqnarray}$
Now you can expand this number to “fit” $$1101$$ by getting the next bit from the dividend:
$\begin{eqnarray} 01000\underline{0}100 \div 1101 = 0\\ \underline{0000}\quad\quad\quad\quad\quad\quad\quad\hspace{0.1em}\\ 10000\hspace{0.6em}\quad\quad\quad\quad\quad\quad\hspace{0.1em}\\ \end{eqnarray}$
Now $$1101$$ fits into $$10000$$. And one neat property of using this technique is that the divisor either fits $$0$$ or $$1$$ times into the bits coming from the dividend. And this time, the divisor can be subtracted from the string of bits “pulled down” from the dividend. The twos’ complement of the divisor is used to perform the subtraction. The fact that the divisor fits into them is tracked by writing a $$1$$ at the end of the number coming after the equal sign.
$\begin{eqnarray} 010000100 \div 1101 = 01\\ \underline{0000}\quad\quad\quad\quad\quad\quad\quad\hspace{0.4em}\\ 10000\hspace{0.6em}\quad\quad\quad\quad\quad\quad\hspace{0.4em}\\ \underline{\oplus\hspace{0.3em}10011}\hspace{0.5em}\quad\quad\quad\quad\quad\quad\hspace{0.4em}\\ \cancel{1}00011\hspace{0.6em}\quad\quad\quad\quad\quad\quad\hspace{0.4em}\\ \end{eqnarray}$
After the subtraction, the next bit from the divisor can be pulled down. Afterward, you have to check if $$1101$$ fits into $$111$$ which it doesn’t. So go ahead and subtract $$0000$$ from it and track the fact that $$1101$$ is contained in $$111$$ $$0$$ times by writing another $$0$$ at the end of the numbers coming after the equal sign.
$\begin{eqnarray} 010000\underline{1}00 \div 1101 = 010\\ \underline{0000}\quad\quad\quad\quad\quad\quad\quad\hspace{1em}\\ 10000\hspace{0.6em}\quad\quad\quad\quad\quad\quad\hspace{1em}\\ \underline{\oplus\hspace{0.3em}10011}\hspace{0.5em}\quad\quad\quad\quad\quad\quad\hspace{1em}\\ 000111\hspace{0.1em}\quad\quad\quad\quad\quad\quad\hspace{1em}\\ \underline{0000}\hspace{0.1em}\quad\quad\quad\quad\quad\quad\hspace{1em}\\ 0111\hspace{0.1em}\quad\quad\quad\quad\quad\quad\hspace{1em} \end{eqnarray}$
Now you can get the next bit from the divisor. This time you have to check if $$1101$$ is contained in $$1110$$ and that is actually the case so you can subtract $$1101$$ from $$1110$$ by adding its twos’ complement and adding a $$1$$ to the end of the result.
$\begin{eqnarray} 0100001\underline{0}0 \div 1101 = 0101\\ \underline{0000}\quad\quad\quad\quad\quad\quad\quad\hspace{1.4em}\\ 10000\hspace{0.6em}\quad\quad\quad\quad\quad\quad\hspace{1.4em}\\ \underline{\oplus\hspace{0.3em}10011}\hspace{0.5em}\quad\quad\quad\quad\quad\quad\hspace{1.4em}\\ 000111\hspace{0.1em}\quad\quad\quad\quad\quad\quad\hspace{1.4em}\\ \underline{0000}\hspace{0.1em}\quad\quad\quad\quad\quad\quad\hspace{1.4em}\\ 01110\hspace{0.7em}\quad\quad\quad\quad\quad\hspace{1.4em}\\ \underline{\oplus\hspace{0.3em}10011}\hspace{0.6em}\quad\quad\quad\quad\quad\hspace{1.4em}\\ 00001\hspace{0.7em}\quad\quad\quad\quad\quad\hspace{1.4em} \end{eqnarray}$
After this step, only one bit is left from the divisor. Adding this bit results in $$10$$ into which $$1101$$ does not fit, which again means subtracting $0000$ and adding a $0$ to the result. After the subtraction, you are left with the remaining $$10$$, which is the remainder of the division.
$\begin{eqnarray} 01000010\underline{0} \div 1101 = 01010\hspace{0.4em}R: 10\\ \underline{0000}\quad\quad\quad\quad\quad\quad\quad \quad\quad\quad\quad\hspace{1em}\\ 10000\hspace{0.6em}\quad\quad\quad\quad\quad\quad \quad\quad\quad\quad\hspace{1em}\\ \underline{\oplus\hspace{0.3em}10011}\hspace{0.5em}\quad\quad\quad\quad\quad\quad \quad\quad\quad\quad\hspace{1em}\\ 000111\hspace{0.1em}\quad\quad\quad\quad\quad\quad \quad\quad\quad\quad\hspace{1em}\\ \underline{0000}\hspace{0.1em}\quad\quad\quad\quad\quad\quad \quad\quad\quad\quad\hspace{1em}\\ 01110\hspace{0.7em}\quad\quad\quad\quad\quad \quad\quad\quad\quad\hspace{1em}\\ \underline{\oplus\hspace{0.3em}10011}\hspace{0.6em}\quad\quad\quad\quad\quad \quad\quad\quad\quad\hspace{1em}\\ 000010\hspace{0.2em}\quad\quad\quad\quad\quad \quad\quad\quad\quad\hspace{1em}\\ \underline{0000}\hspace{0.2em}\quad\quad\quad\quad\quad \quad\quad\quad\quad\hspace{0.9em}\\ 10\hspace{0.2em}\quad\quad\quad\quad\quad\quad\quad\quad\quad\hspace{0.9em} \end{eqnarray}$
Checking the result in decimal shows that $$132\div 13 = 10\hspace{0.4em}R:2$$ with $$10_{10} = 1010_2$$ and $$2_{10} = 10_2$$. To sum up, here are the steps of the binary division:
1. Check if the divisor fits into the left-most bits of the dividend that aren’t used yet.
• if the divisor doesn’t fit into it, subtract a 0 and add a $$0$$ to the result.
• if the divisor does fit into it, subtract (add the twos’ complement) and adds a $$1$$ to the result.
2. Check if there are still bits available in the dividend.
• if there are bits available, pull down the next bit and go the step 1.
• if there aren’t any bits available, the last subtraction result is the remainder, and the number after the equal sign is the result.
You can also get the result with the remainder in Python using the integer division operator // and the modulo operator %:
1 2 3 4 5 6 7 x = 132 y = 13 d = x // y r = x % y print(d,"R:", r)
And as a last operation there is the left-shift >> that is the inverse operation of the right-shift << and will divide by 2 for every bit that the number is shifted:
1 2 3 4 x = 10 print(f"{x} = {x:b}") y = x >> 2 print(f"{y} = {y:b}")
And this concludes this article on how to multiply and divide binary numbers. I encourage you to try out binary multiplication and division yourself because doing things yourself is one of the best ways to learn. |
# What is the equation of the normal line of f(x)=(x-3)/(x+4) at x=3?
Mar 30, 2017
$y = - 7 x + 21$
#### Explanation:
As long as $f ' \left(x\right) \ne 0$, the normal line of $f \left(x\right)$ at $x$ can be written on the form $y = a x + m$, where $a$ is the slope and $m$ is the intercept with the $y$-axis (see further down what happens for the case $f ' \left(x\right) = 0$). If we first find $a$, then we can find $m$ through elimination.
We know that the normal line of $f \left(x\right)$ is perpendicular to the tangent line of $f \left(x\right)$ at $x = 3$. Therefore, if we can find the slope of $f \left(x\right)$ at $x = 3$, then we can also find the slope of the normal line at $x = 3$.
Denote the slope of the tangent line $b$. If $a , b \ne 0$, then $a \cdot b = - 1$ (see further down for explanation).
The slope of the tangent line at a value $x$ is by definition $f ' \left(x\right)$, which we can compute using the product rule (or quotient rule) .
$f ' \left(x\right) = \frac{1}{x + 4} - \frac{x - 3}{x + 4} ^ 2 = \frac{x + 4}{x + 4} ^ 2 - \frac{x - 3}{x + 4} ^ 2 = \frac{\left(x + 4\right) - \left(x - 3\right)}{x + 4} ^ 2 = \frac{7}{x + 4} ^ 2.$
Therefore $b = f ' \left(3\right) = \frac{7}{7} ^ 2 = \frac{1}{7}$, and we find the slope of the normal line by solving $a \cdot b = - 1$ for $a$, which gives that
$a = - 7$.
Since the normal line passes through $x = 3$, we know that the point $\left(3 , f \left(3\right)\right)$ must lie on the line. Evaluating $f \left(3\right) = 0$, and inserting in the equation for the normal line we get that
$0 = a \cdot 3 + m$.
Inserting $a = - 7$ and solving for $m$ gives that
$m = 21$,
and we have found that the normal line is given by
$y = - 7 x + 21$.
Comment 1: In case $f \left(x\right) = 0$, the tangent to $f \left(x\right)$ is flat, which means that the normal line is vertical. Then we must use the general equation for the line
$p y + q x = r$,
which, since the normal line is vertical must have $p = 0$. To find $q$ and $r$, insert a pair of $x$ and $y$-values that you know lie on the line, like we did earlier.
Comment 2: In case $f \left(x\right)$ is not differentiable at $x$, then the function does not have a normal line there.
Comment 3: As for showing the formula $a \cdot b = - 1$, you need either geometry, trigonometry or linear algebra. Here are some proofs. |
# Operations on matrices
Calculates matrices addition, subtraction, product and division.
Use this online calculator to do operations on matrices (sum, subtraction, multiplication, and division).
## How to add two matrices ?
Both matrices must have the same dimension i.e. the same number of rows and the same number of columns.
Adding two matrices is simple : just add the corresponding elements and place the sum in the same corresponding position.
Example:
A and B are two matrices of dimension 2 x 2
A = [[1,5], [6, -4]]
B = [[0, -12], [3,7]]
We can sum then,
A + B = [[1+0,5-12], [6+3, -4+7]] = [[1, -7], [9,3]]
## How to subtract two matrices ?
In the same way, the two matrices must have the same dimension i.e. the same number of rows and the same number of columns.
To subtract them, just subtract the elements in the same position and place the result in the same corresponding position.
Example:
A and B are two matrices of dimension 3 x 2
A = [[2,6], [7, -2], [5,11]]
B = [[1, -10], [4,7], [-9,13]]
then,
A - B = [[2-1,6- (-10)], [7-4, -2-7], [5- (-9) ,11-13]] = [[1,16], [3, -9], [14, -2]]
## How to multiply two matrices ?
Given two matrices A and B, the mutiplication of the two matrices A.B is possible only if the number of columns of A is equal to the number of rows of B. Thus, one can multiply a matrix 2 x 3 by a matrix 3 x 4 but not by a matrix 2 x 2. We can generalize as follows,
The matrix product A.B is defined only for matrices with the following dimensions :
A dimension m x n
B dimension n x p
The product of the two matrices P = A.B is a matrix of dimension m x p.
Pay attention : the order of A and B in the product matters, this is A.B and not B.A which is not defined if p is different from m (matrix multiplication is not commutative).
How to calculate the matrices product ?
Assume A is a 2 x 3 matrix and B a 3 x 2 matrix. According to the above definitions (m=2, n=3 and p=2), multiplication is possible and the matrices product P = A.B is of dimension 2 x 2
A = [[1,5,2], [3,4,7]]
B = [[0, -1], [8,6], [-2,10]]
P = A*B = [[\color {red} {1},\color {red} {5},\color {red} {2}], [3,4,7]] * [[\color {red} {0}, -1], [\color {red} {8}, 6], [\color {red} {-2} ,10]] = [[\color {red} {c_11}, c_12], [c_21, c_22]]
- To calculate the coefficient c_11, we "multiply" the 1st row by the 1st column. So we have,
c_11 = [1,5,2] * [[0], [8], [-2]] = 1*0 +5*8 +2* (-2) = 36
- To calculate the coefficient c_12, we "multiply" the 1st row by the 2nd column. So we have,
c_12 = [1,5,2] * [[-1], [6], [10]] = 1* (-1) +5*6 +2* (10) = 49
- To calculate the coefficient c_21, we "multiply" the 2nd row by the 1st column. So we have,
c_21 = [3,4,7] * [[0], [8], [-2]] = 3*0 +4*8 +7* (-2) = 18
- To calculate the coefficient c_22, we "multiply" the 2nd row by the 2nd column. So we have,
c_22 = [3,4,7] * [[-1], [6], [10]] = 3* (-1) +4*6 +7* (10) = 91
We write the final result,
P = A*B = [[36,49], [18,91]]
We generalize this method as follows,
Assume that A and B are two matrices of respective dimensions m x n and n x p, then the product P = A.B is a matrix of dimension m x p. We denote c_ (ij)the element of matrix P which is in the first row and jth column.
The coefficient c_ (ij)is calculated by 'multiplying' line i of matrix A by column j of matrix B.
## How to divide two matrices ?
Assumed that A and B are two matrices such as :
- A is a matrix of dimension m x n
- B is an invertible square matrix of size n (See Inverse Matrix)
Under these conditions, we can do a division of A by B. The matrix division is:
D = A.B^ (-1)
This leads to a multiplication of two matrices which is explained above. Let's take an example.
Example: How to divide A by B ?
A = [[1,2], [5,7]]
B = [[-1,2], [10,7]]
Let's check the conditions of divisibility explained above :
- Is B a square matrix ? yes, because the number of columns is the same as the number of rows (=2).
- Is B invertible? yes because its determinant is different from 0 (det[B] = -1*7-2*10 = -27).
The conditions of divisibility are checked, so we can calculate D = A. B^ (-1)
We inverse B matrix, we get
B^ (-1) = [[-7/27,2/27], [10/27,1/27]]
D = A*B^ (-1) = [[1,2], [5,7]] * [[-7/27,2/27], [10/27,1/27]]
We get the final result,
D = [[13,4], [35,17]] |
# 3.2 Calculus of vector-valued functions (Page 3/11)
Page 3 / 11
Now for some examples using these properties.
## Using the properties of derivatives of vector-valued functions
Given the vector-valued functions
$\text{r}\left(t\right)=\left(6t+8\right)\phantom{\rule{0.1em}{0ex}}\text{i}+\left(4{t}^{2}+2t-3\right)\phantom{\rule{0.1em}{0ex}}\text{j}+5t\phantom{\rule{0.1em}{0ex}}\text{k}$
and
$\text{u}\left(t\right)=\left({t}^{2}-3\right)\phantom{\rule{0.1em}{0ex}}\text{i}+\left(2t+4\right)\phantom{\rule{0.1em}{0ex}}\text{j}+\left({t}^{3}-3t\right)\phantom{\rule{0.1em}{0ex}}\text{k},$
calculate each of the following derivatives using the properties of the derivative of vector-valued functions.
1. $\frac{d}{dt}\left[\phantom{\rule{0.1em}{0ex}}\text{r}\left(t\right)·\phantom{\rule{0.1em}{0ex}}\text{u}\left(t\right)\right]$
2. $\frac{d}{dt}\left[\phantom{\rule{0.1em}{0ex}}\text{u}\left(t\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{u}^{\prime }\left(t\right)\right]$
1. We have ${r}^{\prime }\left(t\right)=6\phantom{\rule{0.1em}{0ex}}\text{i}+\left(8t+2\right)\phantom{\rule{0.1em}{0ex}}\text{j}+5\phantom{\rule{0.1em}{0ex}}\text{k}$ and ${u}^{\prime }\left(t\right)=2t\phantom{\rule{0.1em}{0ex}}\text{i}+2\phantom{\rule{0.1em}{0ex}}\text{j}+\left(3{t}^{2}-3\right)\phantom{\rule{0.1em}{0ex}}\text{k}.$ Therefore, according to property iv.:
$\begin{array}{cc}\hfill \frac{d}{dt}\left[\phantom{\rule{0.1em}{0ex}}\text{r}\left(t\right)·\phantom{\rule{0.1em}{0ex}}\text{u}\left(t\right)\right]& ={r}^{\prime }\left(t\right)·\phantom{\rule{0.1em}{0ex}}\text{u}\left(t\right)+\phantom{\rule{0.1em}{0ex}}\text{r}\left(t\right)·\phantom{\rule{0.1em}{0ex}}{u}^{\prime }\left(t\right)\hfill \\ & =\left(6\phantom{\rule{0.1em}{0ex}}\text{i}+\left(8t+2\right)\phantom{\rule{0.1em}{0ex}}\text{j}+5\phantom{\rule{0.1em}{0ex}}\text{k}\right)·\left(\left({t}^{2}-3\right)\phantom{\rule{0.1em}{0ex}}\text{i}+\left(2t+4\right)\phantom{\rule{0.1em}{0ex}}\text{j}+\left({t}^{3}-3t\right)\phantom{\rule{0.1em}{0ex}}\text{k}\right)\hfill \\ & \phantom{\rule{1em}{0ex}}+\left(\left(6t+8\right)\phantom{\rule{0.1em}{0ex}}\text{i}+\left(4{t}^{2}+2t-3\right)\phantom{\rule{0.1em}{0ex}}\text{j}+5t\phantom{\rule{0.1em}{0ex}}\text{k}\right)·\left(2t\phantom{\rule{0.1em}{0ex}}\text{i}+2\phantom{\rule{0.1em}{0ex}}\text{j}+\left(3{t}^{2}-3\right)\phantom{\rule{0.1em}{0ex}}\text{k}\right)\hfill \\ & =6\left({t}^{2}-3\right)+\left(8t+2\right)\left(2t+4\right)+5\left({t}^{3}-3t\right)\hfill \\ & \phantom{\rule{1em}{0ex}}+2t\left(6t+8\right)+2\left(4{t}^{2}+2t-3\right)+5t\left(3{t}^{2}-3\right)\hfill \\ & =20{t}^{3}+42{t}^{2}+26t-16.\hfill \end{array}$
2. First, we need to adapt property v. for this problem:
$\frac{d}{dt}\left[\phantom{\rule{0.1em}{0ex}}\text{u}\left(t\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{u}^{\prime }\left(t\right)\right]={u}^{\prime }\left(t\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{u}^{\prime }\left(t\right)+\phantom{\rule{0.1em}{0ex}}\text{u}\left(t\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{u″}\left(t\right).$
Recall that the cross product of any vector with itself is zero. Furthermore, $\text{u″}\left(t\right)$ represents the second derivative of $\text{u}\left(t\right)\text{:}$
$\text{u″}\left(t\right)=\frac{d}{dt}\left[\phantom{\rule{0.1em}{0ex}}{u}^{\prime }\left(t\right)\right]=\frac{d}{dt}\left[2t\phantom{\rule{0.1em}{0ex}}\text{i}+2\phantom{\rule{0.1em}{0ex}}\text{j}+\left(3{t}^{2}-3\right)\phantom{\rule{0.1em}{0ex}}\text{k}\right]=2\phantom{\rule{0.1em}{0ex}}\text{i}+6t\phantom{\rule{0.1em}{0ex}}\text{k}.$
Therefore,
$\begin{array}{cc}\hfill \frac{d}{dt}\left[\phantom{\rule{0.1em}{0ex}}\text{u}\left(t\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{u}^{\prime }\left(t\right)\right]& =0+\left(\left({t}^{2}-3\right)\phantom{\rule{0.1em}{0ex}}\text{i}+\left(2t+4\right)\phantom{\rule{0.1em}{0ex}}\text{j}+\left({t}^{3}-3t\right)\phantom{\rule{0.1em}{0ex}}\text{k}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(2\phantom{\rule{0.1em}{0ex}}\text{i}+6t\phantom{\rule{0.1em}{0ex}}\text{k}\right)\hfill \\ & =|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ {t}^{2}-3& 2t+4& {t}^{3}-3t\\ 2& 0& 6t\end{array}|\hfill \\ & =6t\left(2t+4\right)\phantom{\rule{0.1em}{0ex}}\text{i}-\left(6t\left({t}^{2}-3\right)-2\left({t}^{3}-3t\right)\right)\phantom{\rule{0.1em}{0ex}}\text{j}-2\left(2t+4\right)\phantom{\rule{0.1em}{0ex}}\text{k}\hfill \\ & =\left(12{t}^{2}+24t\right)\phantom{\rule{0.1em}{0ex}}\text{i}+\left(12t-4{t}^{3}\right)\phantom{\rule{0.1em}{0ex}}\text{j}-\left(4t+8\right)\phantom{\rule{0.1em}{0ex}}\text{k}.\hfill \end{array}$
Given the vector-valued functions $\text{r}\left(t\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{i}+\text{sin}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{j}-{e}^{2t}\phantom{\rule{0.1em}{0ex}}\text{k}$ and $\text{u}\left(t\right)=t\phantom{\rule{0.1em}{0ex}}\text{i}+\text{sin}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{j}+\text{cos}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{k},$ calculate $\frac{d}{dt}\left[\phantom{\rule{0.1em}{0ex}}\text{r}\left(t\right)·\phantom{\rule{0.1em}{0ex}}{r}^{\prime }\left(t\right)\right]$ and $\frac{d}{dt}\left[\phantom{\rule{0.1em}{0ex}}\text{u}\left(t\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{r}\left(t\right)\right].$
$\frac{d}{dt}\left[\phantom{\rule{0.1em}{0ex}}\text{r}\left(t\right)·\phantom{\rule{0.1em}{0ex}}{r}^{\prime }\left(t\right)\right]=8{e}^{4t}$
$\begin{array}{l}\frac{d}{dt}\left[\phantom{\rule{0.1em}{0ex}}\text{u}\left(t\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{r}\left(t\right)\right]\\ =-\left({e}^{2t}\left(\text{cos}\phantom{\rule{0.1em}{0ex}}t+2\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}t\right)+\text{cos}\phantom{\rule{0.1em}{0ex}}2t\right)\phantom{\rule{0.1em}{0ex}}\text{i}+\left({e}^{2t}\left(2t+1\right)-\text{sin}\phantom{\rule{0.1em}{0ex}}2t\right)\phantom{\rule{0.1em}{0ex}}\text{j}+\left(t\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}t+\text{sin}\phantom{\rule{0.1em}{0ex}}t-\text{cos}\phantom{\rule{0.1em}{0ex}}2t\right)\phantom{\rule{0.1em}{0ex}}\text{k}\end{array}$
## Tangent vectors and unit tangent vectors
Recall from the Introduction to Derivatives that the derivative at a point can be interpreted as the slope of the tangent line to the graph at that point. In the case of a vector-valued function, the derivative provides a tangent vector to the curve represented by the function. Consider the vector-valued function $\text{r}\left(t\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{i}+\text{sin}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{j}.$ The derivative of this function is ${r}^{\prime }\left(t\right)=-\text{sin}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{i}+\text{cos}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{j}.$ If we substitute the value $t=\pi \text{/}6$ into both functions we get
$\text{r}\left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}\phantom{\rule{0.1em}{0ex}}\text{i}+\frac{1}{2}\phantom{\rule{0.1em}{0ex}}\text{j}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{r}^{\prime }\left(\frac{\pi }{6}\right)=-\frac{1}{2}\phantom{\rule{0.1em}{0ex}}\text{i}+\frac{\sqrt{3}}{2}\phantom{\rule{0.1em}{0ex}}\text{j}.$
The graph of this function appears in [link] , along with the vectors $\text{r}\left(\frac{\pi }{6}\right)$ and ${r}^{\prime }\left(\frac{\pi }{6}\right).$
Notice that the vector ${r}^{\prime }\left(\frac{\pi }{6}\right)$ is tangent to the circle at the point corresponding to $t=\pi \text{/}6.$ This is an example of a tangent vector to the plane curve defined by $\text{r}\left(t\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{i}+\text{sin}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{j}.$
## Definition
Let C be a curve defined by a vector-valued function r, and assume that ${r}^{\prime }\left(t\right)$ exists when $t={t}_{0}.$ A tangent vector v at $t={t}_{0}$ is any vector such that, when the tail of the vector is placed at point $\text{r}\left({t}_{0}\right)$ on the graph, vector v is tangent to curve C. Vector ${r}^{\prime }\left({t}_{0}\right)$ is an example of a tangent vector at point $t={t}_{0}.$ Furthermore, assume that ${r}^{\prime }\left(t\right)\ne \phantom{\rule{0.1em}{0ex}}0.$ The principal unit tangent vector at t is defined to be
$\text{T}\left(t\right)=\frac{{r}^{\prime }\left(t\right)}{\text{‖}\phantom{\rule{0.1em}{0ex}}{r}^{\prime }\left(t\right)\text{‖}},$
provided $\text{‖}\phantom{\rule{0.1em}{0ex}}{r}^{\prime }\left(t\right)\text{‖}\ne 0.$
The unit tangent vector is exactly what it sounds like: a unit vector that is tangent to the curve. To calculate a unit tangent vector, first find the derivative ${r}^{\prime }\left(t\right).$ Second, calculate the magnitude of the derivative. The third step is to divide the derivative by its magnitude.
## Finding a unit tangent vector
Find the unit tangent vector for each of the following vector-valued functions:
1. $\text{r}\left(t\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{i}+\text{sin}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{j}$
2. $\text{u}\left(t\right)=\left(3{t}^{2}+2t\right)\phantom{\rule{0.1em}{0ex}}\text{i}+\left(2-4{t}^{3}\right)\phantom{\rule{0.1em}{0ex}}\text{j}+\left(6t+5\right)\phantom{\rule{0.1em}{0ex}}\text{k}$
1. $\begin{array}{cccc}\text{First step:}\hfill & \hfill {r}^{\prime }\left(t\right)& =\hfill & \mathrm{-sin}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{i}+\text{cos}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{j}\hfill \\ \text{Second step:}\hfill & \hfill \text{‖}\phantom{\rule{0.1em}{0ex}}{r}^{\prime }\left(t\right)\text{‖}& =\hfill & \sqrt{{\left(\text{−}\text{sin}\phantom{\rule{0.1em}{0ex}}t\right)}^{2}+{\left(\text{cos}\phantom{\rule{0.1em}{0ex}}t\right)}^{2}}=1\hfill \\ \text{Third step:}\hfill & \hfill \text{T}\left(t\right)& =\hfill & \frac{{r}^{\prime }\left(t\right)}{\text{‖}\phantom{\rule{0.1em}{0ex}}{r}^{\prime }\left(t\right)\text{‖}}=\frac{\text{−sin}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{i}+\text{cos}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{j}}{1}=\text{−sin}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{i}+\text{cos}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{j}\hfill \end{array}$
2. $\begin{array}{cccc}\text{First step:}\hfill & \hfill {u}^{\prime }\left(t\right)& =\hfill & \left(6t+2\right)\phantom{\rule{0.1em}{0ex}}\text{i}-12{t}^{2}\phantom{\rule{0.1em}{0ex}}\text{j}+6\phantom{\rule{0.1em}{0ex}}\text{k}\hfill \\ \text{Second step:}\hfill & \hfill \text{‖}\phantom{\rule{0.1em}{0ex}}{u}^{\prime }\left(t\right)\text{‖}& =\hfill & \sqrt{{\left(6t+2\right)}^{2}+{\left(-12{t}^{2}\right)}^{2}+{6}^{2}}\hfill \\ & & =& \sqrt{144{t}^{4}+36{t}^{2}+24t+40}\hfill \\ & & =\hfill & 2\sqrt{36{t}^{4}+9{t}^{2}+6t+10}\hfill \\ \text{Third step:}\hfill & \hfill \text{T}\left(t\right)& =\hfill & \frac{{u}^{\prime }\left(t\right)}{\text{‖}\phantom{\rule{0.1em}{0ex}}{u}^{\prime }\left(t\right)\text{‖}}=\frac{\left(6t+2\right)\phantom{\rule{0.1em}{0ex}}\text{i}-12{t}^{2}\phantom{\rule{0.1em}{0ex}}\text{j}+6\phantom{\rule{0.1em}{0ex}}\text{k}}{2\sqrt{36{t}^{4}+9{t}^{2}+6t+10}}\hfill \\ & & =\hfill & \frac{3t+1}{\sqrt{36{t}^{4}+9{t}^{2}+6t+10}}\phantom{\rule{0.1em}{0ex}}\text{i}-\frac{6{t}^{2}}{\sqrt{36{t}^{4}+9{t}^{2}+6t+10}}\phantom{\rule{0.1em}{0ex}}\text{j}+\frac{3}{\sqrt{36{t}^{4}+9{t}^{2}+6t+10}}\phantom{\rule{0.1em}{0ex}}\text{k}\hfill \end{array}$
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
how did you get the value of 2000N.What calculations are needed to arrive at it
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# Integers Class 7 Notes: Chapter 1
## Introduction to Integers
### Introduction to Numbers
Natural Numbers : The collection of all the counting numbers is called set of natural numbers. It is denoted by N = {1,2,3,4….}
Whole Numbers: The collection of natural numbers along with zero is called a set of whole numbers. It is denoted by W = { 0, 1, 2, 3, 4, 5, … }
## Properties of Addition and Subtraction of Integers
For every integer a and ba+b and ab are integers.
for every integer a and ba+b=b+a
for every integer a,b and c, (a+b)+c=a+(b+c)
For every integer a, a+0=0+a=a here 0 is Additive Identity, since adding 0 to a number leaves it unchanged.
Example : For an integer 2, 2+0 = 0+2 = 2.
For every integer a, a+(a)=0 Here, a is additive inverse of a and a is the additive inverse of-a.
Example : For an integer 2, (– 2) is additive inverse and for (– 2), additive inverse is 2. [Since + 2 – 2 = 0]
## Properties of Multiplication of Integers
### Properties of Multiplication of Integers
Closure under Multiplication
For every integer a and b, a×b=Integer
Commutative Property of Multiplication
For every integer a and ba×b=b×a
Multiplication by Zero
For every integer a, a×0=0×a=0
Multiplicative Identity
For every integer a, a×1=1×a=a. Here 1 is the multiplicative identity for integers.
Associative property of Multiplication
For every integer a, b and c, (a×b)×c=a×(b×c)
Distributive Property of Integers
Under addition and multiplication, integers show the distributive property.
i.e., For every integer a, b and ca×(b+c)=a×b+a×c
These properties make calculations easier.
To know more about “Properties of Multiplication of Integers”, visit here.
## Division of Integers
### Division of Integers
When a positive integer is divided by a positive integer, the quotient obtained is a positive integer.
Example: (+6) ÷ (+3) +2
When a negative integer is divided by a negative integer, the quotient obtained is a positive integer.
Example: (-6) ÷ (-3) +2
When a positive integer is divided by a negative integer or negative integer is divided by a positive integer, the quotient obtained is a negative integer.
Example: (-6) ÷ (+3) =2 and Example: (+6) ÷ (-3) 2
To know more about Multiplication and Division of Integers, visit here.
## The Number Line
### Number Line
Representation of integers on a number line
On a number line when we
(i) add a positive integer for a given integer, we move to the right.
Example : When we add +2 to +3, move 2 places from +3 towards right to get +5
(ii) add a negative integer for a given integer, we move to the left.
Example : When we add -2 to +3, move 2 places from +3 towards left to get +1
(iii) subtract a positive integer from a given integer, we move to the left.
Example: When we subtract +2 from -3, move 2 places from -3 towards left to get -5
(iv) subtract a negative integer from a given integer, we move to the right
Example: When we subtract -2 from -3, move 2 places from -3 towards right to get 1
To know more about Number Lines, visit here.
### Addition and Subtraction of Integers
The absolute value of +7 (a positive integer) is 7
The absolute value of -7 (negative integer) is 7 (its corresponding positive integer)
Addition of two positive integers gives a positive integer.
Example : (+3)+(+4) +7
Addition of two negative integers gives a negative integer.
Example : (3)+(434=7
When one positive and one negative integers are added, we take their difference and place the sign of the bigger integer.
Example : (7)+(25
For subtraction, we add the additive inverse of the integer that is being subtracted, to the other integer.
Example : 56(7356+73 129
## Introduction to Zero
### Integers
Integers are the collection of numbers which is formed by whole numbers and their negatives.
The set of Integers is denoted by Z or I. I = { …, -4, -3, -2, -1, 0, 1, 2, 3, 4,… }
## Properties of Division of Integers
### Properties of Division of Integers
For every integer a,
(a) a÷0 is not defined
(b) a÷a
Note: Integers are not closed under division
Example: (– 9) ÷ (– 3) = 2. Result is an integer.
and (3)÷(9)= 1/3. Result is not an integer.
## Multiplication of Integers
### Multiplication of Integers
Product of two positive integers is a positive integer.
Example : (+2)×(+3+6
Product of two negative integers is a positive integer.
Example :(2)×(3+6
Product of a positive and a negative integer is a negative integer.
Example :(+2)×(36 and (2)×(+36
Product of even number of negative integers is positive and product of odd number of negative integers is negative.
These properties make calculations easier.
## Frequently Asked Questions on CBSE Class 7 Maths Notes Chapter 1 Integers
Q1
### What is closure property?
Closure property of whole numbers under addition: The sum of any two whole numbers will always be a whole number, i.e. if a and b are any two whole numbers, a + b will be a whole number.
Q2
### What are the properties of zero?
1. Zero is even 2. Zero is neither positive nor negative 3. Zero is an integer
Q3
### What is the definition of inverse?
A term is said to be in inverse proportion to another term if it increases (or decreases) as the other decreases (or increases). of or relating to an inverse function. |
# The Steps in Long Division
## Presentation on theme: "The Steps in Long Division"— Presentation transcript:
The Steps in Long Division
By Miss B.
Does McDonalds Sell Cheese Burgers?
Use this helpful phrase to remember the steps of long division.
Before We get Started Underline the first digit of the dividend (inside box) Can I make a group of 6 with 2 marbles? Underline the 4 Can I make groups of 6 with 24 marbles? YES! You will place the 1st digit of your quotient (answer) above the 4. Place an x here as well as above any digits to the right in the dividend. Your will have 2 digits in your quotient 6 246
Step 1: Does (Divide) How many groups of 6 can you make with 24 marbles? Use objects or a multiplication table to help you. You can make 4 groups of 6 Write the 4 above the 4 in the dividend 4 6 246
Step 2: McDonalds (Multiply)
Multiply the 4 in the quotient by the 6 (divisor) Write the product under the 24 you underlined in the before we start step 4 6 246 24
Step 3: Sell (Subtract) Subtract your product from the underlined number in the dividend. 4 6 246 24
Step 4: Cheese (Check) Check that your difference is less than your divisor. If it is not, you need to check steps 1-3. If it is, you are set to move on to step 5 4 6 246 24
Step 5: Burgers (Bring Down)
Next bring down the next number in the dividend. If it is equal to or larger than your divisor repeat the steps. If it is less than the divisor, place a 0 to hold the place value in the quotient and then bring down the next number in the dividend. (see example 2) If there are no more numbers in the dividend, what ever is left over becomes the remainder. (see example 2) 4 6 246 24 6
Step 1: Does (Divide) How many groups of 6 can you make with 6 marbles? Use objects or a multiplication table to help you. You can make 1 group of 6 Write the 1 above the 6 in the dividend. 4 1 6 246 24 6 6
Step 2: McDonalds (Multiply)
Multiply the 1 in the quotient by the 6 (divisor) Write the product under the 6. 4 1 6 246 24 6 6 6
4 1 24 6 6 6 Step 3: Sell (Subtract) Subtract your product from the 6.
6 246 24 6 6 6
Step 4: Cheese (Check) Check that your difference is less than your divisor. If it is not, you need to check steps 1-3. If it is, you are set to move on to step 5 4 1 6 246 24 6 6 6
Step 5: Burgers (Bring Down)
If your difference is 0 and there is nothing left to bring down, you are finished! If there is another digit in the dividend to bring down start the steps again. If your difference is not 0, but less than the divisor and there are no other numbers to bring down, the difference becomes the remainder. Your answer is 41! 4 1 6 246 24 6 6 6
Multiply the quotient by the divisor. If your product is the dividend, you are correct! 4 1 6 246 41 x 6 246
Example 2 If it is less than the divisor, place a 0 to hold the place value in the quotient and then bring down the next number in the dividend. If there are no other numbers to bring down, it becomes the remainder. 4 R5 6 245 24 005
Practice some problems on your own!
D M S C B |
7 Work, Energy, and Energy Resources
43 7.2 Kinetic Energy and the Work-Energy Theorem
Summary
• Explain work as a transfer of energy and net work as the work done by the net force.
• Explain and apply the work-energy theorem.
Work Transfers Energy
What happens to the work done on a system? Energy is transferred into the system, but in what form? Does it remain in the system or move on? The answers depend on the situation. For example, if the lawn mower in Chapter 7.1 Figure 1(a) is pushed just hard enough to keep it going at a constant speed, then energy put into the mower by the person is removed continuously by friction, and eventually leaves the system in the form of heat transfer. In contrast, work done on the briefcase by the person carrying it up stairs in Chapter 7.1 Figure 1(d) is stored in the briefcase-Earth system and can be recovered at any time, as shown in Chapter 7.1 Figure 1(e). In fact, the building of the pyramids in ancient Egypt is an example of storing energy in a system by doing work on the system. Some of the energy imparted to the stone blocks in lifting them during construction of the pyramids remains in the stone-Earth system and has the potential to do work.
In this section we begin the study of various types of work and forms of energy. We will find that some types of work leave the energy of a system constant, for example, whereas others change the system in some way, such as making it move. We will also develop definitions of important forms of energy, such as the energy of motion.
Net Work and the Work-Energy Theorem
We know from the study of Newton’s laws in Chapter 4 Dynamics: Force and Newton’s Laws of Motion that net force causes acceleration. We will see in this section that work done by the net force gives a system energy of motion, and in the process we will also find an expression for the energy of motion.
Let us start by considering the total, or net, work done on a system. Net work is defined to be the sum of work done by all external forces—that is, net work is the work done by the net external force$\textbf{F}_{\textbf{net}}.$In equation form, this is$\boldsymbol{W_{\textbf{net}}=F_{\textbf{net}}d\:\textbf{cos}\:\theta}$where$\boldsymbol{\theta}$is the angle between the force vector and the displacement vector.
Figure 1(a) shows a graph of force versus displacement for the component of the force in the direction of the displacement—that is, an$\boldsymbol{F\textbf{cos}\:\theta}$vs.$\boldsymbol{d}$graph. In this case,$\boldsymbol{F\:\textbf{cos}\:\theta}$is constant. You can see that the area under the graph is$\boldsymbol{Fd\:\textbf{cos}\theta},$or the work done. Figure 1(b) shows a more general process where the force varies. The area under the curve is divided into strips, each having an average force$\boldsymbol{(F\textbf{cos}\:\theta)_{i(\textbf{ave})}}.$The work done is$\boldsymbol{(F\textbf{cos}\theta)_{i(\textbf{ave})}d_{i}}$for each strip, and the total work done is the sum of the$\boldsymbol{W_i}.$Thus the total work done is the total area under the curve, a useful property to which we shall refer later.
Net work will be simpler to examine if we consider a one-dimensional situation where a force is used to accelerate an object in a direction parallel to its initial velocity. Such a situation occurs for the package on the roller belt conveyor system shown in Figure 2.
The force of gravity and the normal force acting on the package are perpendicular to the displacement and do no work. Moreover, they are also equal in magnitude and opposite in direction so they cancel in calculating the net force. The net force arises solely from the horizontal applied force$\textbf{F}_{\textbf{app}}$and the horizontal friction force$\textbf{f}.$Thus, as expected, the net force is parallel to the displacement, so that$\boldsymbol{\theta=0^{\circ}}$and$\boldsymbol{\textbf{cos}\:\theta=1},$ and the net work is given by
$\boldsymbol{W_{\textbf{net}}=F_{\textbf{net}}d}.$
The effect of the net force$\boldsymbol{F_{\textbf{net}}}$is to accelerate the package from$\boldsymbol{v_0}$to$\boldsymbol{v}.$The kinetic energy of the package increases, indicating that the net work done on the system is positive. (See Example 1.) By using Newton’s second law, and doing some algebra, we can reach an interesting conclusion. Substituting$\boldsymbol{F_{\textbf{net}}=ma}$from Newton’s second law gives
$\boldsymbol{W_{\textbf{net}}=mad}.$
To get a relationship between net work and the speed given to a system by the net force acting on it, we take$\boldsymbol{d=x-x_0}$and use the equation studied in Chapter 2.5 Motion Equations for Constant Acceleration in One Dimension for the change in speed over a distance$\boldsymbol{d}$if the acceleration has the constant value$\boldsymbol{a};$namely,$\boldsymbol{v^2=v_0^2+2ad}$(note that$\boldsymbol{a}$appears in the expression for the net work). Solving for acceleration gives$\boldsymbol{a=\frac{v^2-v_0^2}{2d}}.$When$\boldsymbol{a}$is substituted into the preceding expression for$\boldsymbol{W_{\textbf{net}}},$we obtain
$\boldsymbol{W_{\textbf{net}}=m}$$\boldsymbol{(\frac{v^2-v_0^2}{2d})}$$\boldsymbol{d}.$
The$\boldsymbol{d}$cancels, and we rearrange this to obtain
$\boldsymbol{W_{\textbf{net}}\:=}$$\boldsymbol{\frac{1}{2}}$$\boldsymbol{mv^2\:-}$$\boldsymbol{\frac{1}{2}}$$\boldsymbol{mv_0^2}.$
This expression is called the work-energy theorem, and it actually applies in general (even for forces that vary in direction and magnitude), although we have derived it for the special case of a constant force parallel to the displacement. The theorem implies that the net work on a system equals the change in the quantity$\boldsymbol{\frac{1}{2}mv^2}.$This quantity is our first example of a form of energy.
THE WORK-ENERGY THEOREM
The net work on a system equals the change in the quantity$\boldsymbol{\frac{1}{2}mv^2}.$
$\boldsymbol{W_{\textbf{net}}\:=}$$\boldsymbol{\frac{1}{2}}$$\boldsymbol{mv^2\:-}$$\boldsymbol{\frac{1}{2}}$$\boldsymbol{mv_0^2}.$
The quantity$\boldsymbol{\frac{1}{2}mv^2}$in the work-energy theorem is defined to be the translational kinetic energy (KE) of a mass$\boldsymbol{m}$moving at a speed$\boldsymbol{v}.$(Translational kinetic energy is distinct from rotational kinetic energy, which is considered later.) In equation form, the translational kinetic energy,
$\boldsymbol{\textbf{KE}\:=}$$\boldsymbol{\frac{1}{2}}$$\boldsymbol{mv^2},$
is the energy associated with translational motion. Kinetic energy is a form of energy associated with the motion of a particle, single body, or system of objects moving together.
We are aware that it takes energy to get an object, like a car or the package in Figure 2, up to speed, but it may be a bit surprising that kinetic energy is proportional to speed squared. This proportionality means, for example, that a car traveling at 100 km/h has four times the kinetic energy it has at 50 km/h, helping to explain why high-speed collisions are so devastating. We will now consider a series of examples to illustrate various aspects of work and energy.
Example 1: Calculating the Kinetic Energy of a Package
Suppose a 30.0-kg package on the roller belt conveyor system in Figure 2 is moving at 0.500 m/s. What is its kinetic energy?
Strategy
Because the mass$\boldsymbol{m}$and speed$\boldsymbol{v}$are given, the kinetic energy can be calculated from its definition as given in the equation$\boldsymbol{\textbf{KE}=\frac{1}{2}mv^2}.$
Solution
The kinetic energy is given by
$\boldsymbol{\textbf{KE}\:=}$$\boldsymbol{\frac{1}{2}}$$\boldsymbol{mv^2},$
Entering known values gives
$\boldsymbol{\textbf{KE}=0.5(30.0\textbf{ kg})(0.500\textbf{ m/s})^2},$
which yields
$\boldsymbol{\textbf{KE}=3.75\textbf{ k}\cdotp\textbf{m}^2/\textbf{s}^2=3.75\textbf{ J}}.$
Discussion
Note that the unit of kinetic energy is the joule, the same as the unit of work, as mentioned when work was first defined. It is also interesting that, although this is a fairly massive package, its kinetic energy is not large at this relatively low speed. This fact is consistent with the observation that people can move packages like this without exhausting themselves.
Example 2: Determining the Work to Accelerate a Package
Suppose that you push on the 30.0-kg package in Figure 2 with a constant force of 120 N through a distance of 0.800 m, and that the opposing friction force averages 5.00 N.
(a) Calculate the net work done on the package. (b) Solve the same problem as in part (a), this time by finding the work done by each force that contributes to the net force.
Strategy and Concept for (a)
This is a motion in one dimension problem, because the downward force (from the weight of the package) and the normal force have equal magnitude and opposite direction, so that they cancel in calculating the net force, while the applied force, friction, and the displacement are all horizontal. (See Figure 2.) As expected, the net work is the net force times distance.
Solution for (a)
The net force is the push force minus friction, or$\boldsymbol{F_{\textbf{net}}=120\textbf{ N}-5.00\textbf{ N}=115\textbf{ N}}.$Thus the net work is
$\begin{array}{lcl} \boldsymbol{W_{\textbf{net}}} & \boldsymbol{=} & \boldsymbol{F_{\textbf{net}}d=(115\textbf{ N})(0.800\textbf{ m})} \\ {} & \boldsymbol{=} & \boldsymbol{92.0\textbf{N}\cdotp\textbf{m}=92.0\textbf{ J.}} \end{array}$
Discussion for (a)
This value is the net work done on the package. The person actually does more work than this, because friction opposes the motion. Friction does negative work and removes some of the energy the person expends and converts it to thermal energy. The net work equals the sum of the work done by each individual force.
Strategy and Concept for (b)
The forces acting on the package are gravity, the normal force, the force of friction, and the applied force. The normal force and force of gravity are each perpendicular to the displacement, and therefore do no work.
Solution for (b)
The applied force does work.
$\begin{array}{lcl} \boldsymbol{W_{\textbf{app}}} & \boldsymbol{=} & \boldsymbol{F_{\textbf{app}}d\:\textbf{cos}\:(0^{\circ})=F_{\textbf{app}}d} \\ {} & \boldsymbol{=} & \boldsymbol{(120\textbf{ N})(0.800\textbf{ m})} \\ {} & \boldsymbol{=} & \boldsymbol{96.0\textbf{ J}} \end{array}$
The friction force and displacement are in opposite directions, so that $\boldsymbol{\theta = 180^{\circ}}$, and the work done by friction is
$\begin{array}{lcl} \boldsymbol{W_{\textbf{fr}}} & \boldsymbol{=} & \boldsymbol{F_{\textbf{fr}}d\:\textbf{cos}\:180^{\circ}=-F_{\textbf{fr}}d} \\ {} & \boldsymbol{=} & \boldsymbol{(-5.00\textbf{ N})(0.800\textbf{ m})} \\ {} & \boldsymbol{=} & \boldsymbol{-4.00\textbf{ J.}} \end{array}$
So the amounts of work done by gravity, by the normal force, by the applied force, and by friction are, respectively,
$\begin{array}{lcl} \boldsymbol{W_{\textbf{gr}}} & \boldsymbol{=} & \boldsymbol{0,} \\ \boldsymbol{W_{\textbf{N}}} & \boldsymbol{=} & \boldsymbol{0,} \\ \boldsymbol{W_{\textbf{app}}} & \boldsymbol{=} & \boldsymbol{96.0\textbf{ J,}} \\ \boldsymbol{W_{\textbf{fr}}} & \boldsymbol{=} & \boldsymbol{-4.00\textbf{ J.}} \end{array}$
The total work done as the sum of the work done by each force is then seen to be
$\boldsymbol{W_{\textbf{total}}=W_{\textbf{gr}}+W_{\textbf{N}}+W_{\textbf{app}}+W_{\textbf{fr}}=92.0\textbf{ J.}}$
Discussion for (b)
The calculated total work$\boldsymbol{W_{\textbf{total}}}$as the sum of the work by each force agrees, as expected, with the work$\boldsymbol{W_{\textbf{net}}}$done by the net force. The work done by a collection of forces acting on an object can be calculated by either approach.
Example 3: Determining Speed from Work and Energy
Find the speed of the package in Figure 2 at the end of the push, using work and energy concepts.
Strategy
Here the work-energy theorem can be used, because we have just calculated the net work,$\boldsymbol{W_{\textbf{net}}},$and the initial kinetic energy,$\boldsymbol{\frac{1}{2}mv_0^2}.$These calculations allow us to find the final kinetic energy,$\boldsymbol{\frac{1}{2}mv^2},$and thus the final speed$\boldsymbol{v}.$
Solution
The work-energy theorem in equation form is
$\boldsymbol{W_{\textbf{net}}\:=}$$\boldsymbol{\frac{1}{2}}$$\boldsymbol{mv^2-}$$\boldsymbol{\frac{1}{2}}$$\boldsymbol{mv_0^2}.$
Solving for$\boldsymbol{\frac{1}{2}mv^2}$gives
$\boldsymbol{\frac{1}{2}}$$\boldsymbol{mv^2=W_{\textbf{net}}\:+}$$\boldsymbol{\frac{1}{2}}$$\boldsymbol{mv_0^2}.$
Thus,
$\boldsymbol{\frac{1}{2}}$$\boldsymbol{mv^2=92.0\textbf{ J}+3.75\textbf{ J}=95.75\textbf{ J.}}$
Solving for the final speed as requested and entering known values gives
$\begin{array}{lcl} \boldsymbol{v} & \boldsymbol{=} & \boldsymbol{\sqrt{\frac{2(95.75\textbf{ J})}{m}}=\sqrt{\frac{191.5\textbf{ kg}\cdotp\textbf{m}^2/\textbf{s}^2}{30.0\textbf{ kg}}}} \\ {} & \boldsymbol{=} & \boldsymbol{2.53\textbf{ m/s.}} \end{array}$
Discussion
Using work and energy, we not only arrive at an answer, we see that the final kinetic energy is the sum of the initial kinetic energy and the net work done on the package. This means that the work indeed adds to the energy of the package.
Example 4: Work and Energy Can Reveal Distance, Too
How far does the package in Figure 2 coast after the push, assuming friction remains constant? Use work and energy considerations.
Strategy
We know that once the person stops pushing, friction will bring the package to rest. In terms of energy, friction does negative work until it has removed all of the package’s kinetic energy. The work done by friction is the force of friction times the distance traveled times the cosine of the angle between the friction force and displacement; hence, this gives us a way of finding the distance traveled after the person stops pushing.
Solution
The normal force and force of gravity cancel in calculating the net force. The horizontal friction force is then the net force, and it acts opposite to the displacement, so$\boldsymbol{\theta=180^{\circ}}.$To reduce the kinetic energy of the package to zero, the work$\boldsymbol{W_{\textbf{fr}}}$by friction must be minus the kinetic energy that the package started with plus what the package accumulated due to the pushing. Thus$\boldsymbol{W_{\textbf{fr}}=-95.75\textbf{ J}}.$Furthermore,$\boldsymbol{W_{\textbf{fr}}=fd^{\prime}\:\textbf{cos}\:\theta=-fd^{\prime}},$where$\boldsymbol{d^{\prime}}$is the distance it takes to stop. Thus,
$\boldsymbol{d^{\prime}\:=-}$$\boldsymbol{\frac{W_{\textbf{fr}}}{f}}$$\boldsymbol{=-}$$\boldsymbol{\frac{-95.75\textbf{ J}}{5.00\textbf{ N}}},$
and so
$\boldsymbol{d^{\prime}=19.2\textbf{ m.}}$
Discussion
This is a reasonable distance for a package to coast on a relatively friction-free conveyor system. Note that the work done by friction is negative (the force is in the opposite direction of motion), so it removes the kinetic energy.
Some of the examples in this section can be solved without considering energy, but at the expense of missing out on gaining insights about what work and energy are doing in this situation. On the whole, solutions involving energy are generally shorter and easier than those using kinematics and dynamics alone.
Section Summary
• The net work$\boldsymbol{W_{\textbf{net}}}$is the work done by the net force acting on an object.
• Work done on an object transfers energy to the object.
• The translational kinetic energy of an object of mass$\boldsymbol{m}$moving at speed$\boldsymbol{v}$is$\boldsymbol{\textbf{KE}=\frac{1}{2}mv^2}.$
• The work-energy theorem states that the net work$\boldsymbol{W_{\textbf{net}}}$on a system changes its kinetic energy,$\boldsymbol{W_{\textbf{net}}=\frac{1}{2}mv^2-\frac{1}{2}mv_0^2}.$
Conceptual Questions
1: The person in Figure 3 does work on the lawn mower. Under what conditions would the mower gain energy? Under what conditions would it lose energy?
2: Work done on a system puts energy into it. Work done by a system removes energy from it. Give an example for each statement.
3: When solving for speed in Example 3, we kept only the positive root. Why?
Problems & Exercises1: Compare the kinetic energy of a 20,000-kg truck moving at 110 km/h with that of an 80.0-kg astronaut in orbit moving at 27,500 km/h.2: (a) How fast must a 3000-kg elephant move to have the same kinetic energy as a 65.0-kg sprinter running at 10.0 m/s? (b) Discuss how the larger energies needed for the movement of larger animals would relate to metabolic rates.3: Confirm the value given for the kinetic energy of an aircraft carrier in Chapter 7.6 Table 1. You will need to look up the definition of a nautical mile (1 knot = 1 nautical mile/h).4: (a) Calculate the force needed to bring a 950-kg car to rest from a speed of 90.0 km/h in a distance of 120 m (a fairly typical distance for a non-panic stop). (b) Suppose instead the car hits a concrete abutment at full speed and is brought to a stop in 2.00 m. Calculate the force exerted on the car and compare it with the force found in part (a).5: A car’s bumper is designed to withstand a 4.0-km/h (1.1-m/s) collision with an immovable object without damage to the body of the car. The bumper cushions the shock by absorbing the force over a distance. Calculate the magnitude of the average force on a bumper that collapses 0.200 m while bringing a 900-kg car to rest from an initial speed of 1.1 m/s.6: Boxing gloves are padded to lessen the force of a blow. (a) Calculate the force exerted by a boxing glove on an opponent’s face, if the glove and face compress 7.50 cm during a blow in which the 7.00-kg arm and glove are brought to rest from an initial speed of 10.0 m/s. (b) Calculate the force exerted by an identical blow in the gory old days when no gloves were used and the knuckles and face would compress only 2.00 cm. (c) Discuss the magnitude of the force with glove on. Does it seem high enough to cause damage even though it is lower than the force with no glove?7: Using energy considerations, calculate the average force a 60.0-kg sprinter exerts backward on the track to accelerate from 2.00 to 8.00 m/s in a distance of 25.0 m, if he encounters a headwind that exerts an average force of 30.0 N against him.
Glossary
net work
work done by the net force, or vector sum of all the forces, acting on an object
work-energy theorem
the result, based on Newton’s laws, that the net work done on an object is equal to its change in kinetic energy
kinetic energy
the energy an object has by reason of its motion, equal to$\boldsymbol{\frac{1}{2}mv^2}$for the translational (i.e., non-rotational) motion of an object of mass$\boldsymbol{m}$moving at speed$\boldsymbol{v}$
Solutions
Problems & Exercises
1:
$\boldsymbol{1/250}$
3:
$\boldsymbol{1.1\times10^{10}\textbf{ J}}$
5:
$\boldsymbol{2.8\times10^3\textbf{ N}}$
7:
102 N |
## Intermediate Algebra (12th Edition)
$x=\left\{ -3,3 \right\}$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $-4x^2+36=0 ,$ simplify first by dividing both sides by $GCF.$ Then express the equation in $x^2=c$ form and get the square root of both sides (Square Root Principle). $\bf{\text{Solution Details:}}$ Dividing both sides by $-4$, the equation above is equivalent to \begin{array}{l}\require{cancel} x^2-9=0 \\\\ x^2=9 .\end{array} Taking the square root of both sides (Square Root Principle), the equation above is equivalent to \begin{array}{l}\require{cancel} x=\pm\sqrt{9} \\\\ x=\pm3 .\end{array} Hence, $x=\left\{ -3,3 \right\} .$ |
Select Board & Class
Rational Numbers
Concepts Related to Rational Numbers
You have studied fractional numbers in your earlier classes. Some examples of fractional numbers are.
These numbers are also known as rational numbers.
What comes first to your mind when you hear the word rational?
Yes,you are right. It is something related to the ratios.
The ratio 4:5 can be written as, which is a rational number. In ratios, the numerator and denominator both are positive numbers while in rational numbers, they can be negative also.
Thus, rational numbers can be defined as follows.
“Any number which can be expressed in the form , where p and q are integers and, is called a rational number.”
For example, is a rational number in which the numerator is 15 and the denominator is 19.
Now, is −34 a rational number?
Yes, it is a rational number. −34 can be written as. It is in the form of and q ≠ 0.
Thus, we can say that every integer is a rational number.
Now, consider the following decimal numbers.
1.6, 3.49, and 2.5
These decimal numbers are also rational numbers as these can be written as
If in a rational number, either the numerator or the denominator is a negative integer, then the rational number is negative.
For example, are negative rational numbers.
If the numerator and the denominator both are either positive integers or negative integers, then the rational number is positive.
For example, are positive rational numbers.
Conventions used for writing a rational number:
We know that in a rational number, the numerator and denominator both can be positive or negative.
Conventionally, rational numbers are written with positive denominators.
For example, –9 can be represented in the form of a rational number as , but generally we do not write the denominator negative and thus, is eliminated. So, according to the convention, –9 can be represented in the form of a rational number as .
Equality relation for rational numbers:
For any four non-zero integers p, q, r and s, we have
Order relation for rational numbers:
If are two rational numbers such that q > 0 and s > 0 then it can be said that if ps > qr.
Absolute Value of a Rational Number: The absolute value of a rational number is its numerical value regardless of its sign. The absolute value of a rational number pq is denoted as pq. Therefore, -32=32, 12-7=127 etc. Note: The absolute value of any rational number is always non-negative.
Now, let us go through the given example.
Example:
Write each of the following rational numbers according to the convention.
i)
ii)
Solution:
According to the convention used in rational numbers, the denominator must be a positive number.
Let us now write the given numbers according to the convention.
i)
In the number , denominator is negative.
We have,
According to convention, the given number should be written as .
ii)
In the number , denominator is negative.
We have,
According to convention, the given number should be written as . Example: Find the absolute value of the following: (i) -12171 (ii) 1219 Solution: (i) Absolute value = -12171=12171 (ii) Absolute value = 1219=1219
You have studied fractional numbers in your earlier classes. Some examples of fractional numbers are.
These numbers are also known as rational numbers.
What comes first to your mind when you hear the word rational?
Yes,you are right. It is something related to the ratios.
The ratio 4:5 can be written as, which is a rational number. In ratios, the numerator and denominator both are positive numbers while in rational numbers, they can be negative also.
Thus, rational numbers can be defined as follows.
“Any number which can be expressed in the form , where p and q are integers and, is called a rational number.”
For example, is a rational number in which the numerator is 15 and the denominator is 19.
Now, is −34 a rational number?
Yes, it is a rational number. −34 can be written as. It is in the form of and q ≠ 0.
Thus, we can say that every integer is a rational number.
Now, consider the following decimal numbers.
1.…
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Syllabus |
# Division of surds | How to divide surds
In this section, we will discuss how to divide surds. For more details of surds, please visit the page an introduction to surds.
#### How to Divide Surds
The division of surds is one of the fundamental operations on surds. To divide one surd by another surd, we need to go through the following steps.
Step I: At first, we will express the quotient as a fraction.
Step II: We need to rationalize the denominator of the fraction.
Step III: Now, we will find the suitable surd-rationalizing factor for the denominator of the fraction.
Step IV: Next, we will multiply both the numerator and the denominator of the fraction by the above surd-rationalizing factor.
Step V: Simplifying the fraction we will get the desired answer ♣
Example: We will apply the above method to divide $\sqrt{3}$ by $\sqrt{2}.$
In the first step, the quotient will be expressed as a fraction in the following way:
$\dfrac{\sqrt{3}}{\sqrt{2}}$
Now we need to rationalize the denominator $\sqrt{2}.$ To do that we have to multiply both the numerator and the denominator of the above fraction by $\sqrt{2}.$ Doing that we get
$\dfrac{\sqrt{3}}{\sqrt{2}}$ $=\dfrac{\sqrt{3} \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}}$
$=\dfrac{\sqrt{3 \times 2}}{\sqrt{2 \times 2}}$ $[\because \sqrt{a} \times \sqrt{b}=\sqrt{a \times b}]$
$=\dfrac{6}{2}$ $[\because \sqrt{a \times a}=a]$
So the desired answer is $=\frac{6}{2}$ ♣
#### Formulas of Surds Division
(i) $\sqrt{a} \div \sqrt{b}=\sqrt{\frac{a}{b}}$
(ii) $\sqrt[n]{a} \div \sqrt[n]{b}=\sqrt[n]{\frac{a}{b}}$
(iii) $x\sqrt{a} \div y\sqrt{b}=\frac{x}{y}\sqrt{\frac{a}{b}}$
(iv) $x\sqrt[n]{a} \div y\sqrt[n]{b}=\frac{x}{y}\sqrt[n]{\frac{a}{b}}$
(iv) $\sqrt[m]{a} \div \sqrt[n]{a}=a^{\frac{1}{m}-\frac{1}{n}}$
(v) $x\sqrt[m]{a} \div y\sqrt[n]{a}=\frac{x}{y} \times a^{\frac{1}{m}-\frac{1}{n}}$
#### Solved Problems on Division of Surds
Problem 1: Divide $3$ by $\sqrt{5}$
Solution:
$3 \div \sqrt{5}$
$=\dfrac{3}{\sqrt{5}}$
$=\dfrac{3 \times \sqrt{5}}{\sqrt{5} \times \sqrt{5}}$
$=\dfrac{3\sqrt{5}}{5}$
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# Factors of 243
The factors of 243 and the prime factors of 243 differ because two hundred and forty-three is a composite number. Also, despite being closely related, the prime factors of 243 and the prime factorization of 243 are not exactly the same either. In any case, by reading on you can learn the answer to the question what are the factors of 243? and everything else you want to know about the topic.
## What are the Factors of 243?
They are: 243, 81, 27, 9, 3, 1. These are all the factors of 243, and every entry in the list can divide 243 without rest (modulo 0). That’s why the terms factors and divisors of 243 can be used interchangeably.
As is the case for any natural number greater than zero, the number itself, here 243, as well as 1 are factors and divisors of 243.
## Prime Factors of 243
The prime factors of 243 are the prime numbers which divide 243 exactly, without remainder as defined by the Euclidean division. In other words, a prime factor of 243 divides the number 243 without any rest, modulo 0.
For 243, the prime factors are: 3. By definition, 1 is not a prime number.
Besides 1, what sets the factors and the prime factors of the number 243 apart is the word “prime”. The former list contains both, composite and prime numbers, whereas the latter includes only prime numbers.
## Prime Factorization of 243
The prime factorization of 243 is 3 x 3 x 3 x 3 x 3. This is a unique list of the prime factors, along with their multiplicities. Note that the prime factorization of 243 does not include the number 1, yet it does include every instance of a certain prime factor.
243 is a composite number. In contrast to prime numbers which only have one factorization, composite numbers like 243 have at least two factorizations.
To illustrate what that means select the rightmost and leftmost integer in 243, 81, 27, 9, 3, 1 and multiply these integers to obtain 243. This is the first factorization. Next choose the second rightmost and the second leftmost entry to obtain the 2nd factorization which also produces 243.
The prime factorization or integer factorization of 243 means determining the set of prime numbers which, when multiplied together, produce the original number 243. This is also known as prime decomposition of 243.
Besides factors for 243, frequently searched terms on our website include:
We did not place any calculator here as there are already a plethora of them on the web. But you can find the factors, prime factors and the factorizations of many numbers including 243 by using the search form in the sidebar.
To sum up:
The factors, the prime factors and the prime factorization of 243 mean different things, and in strict terms cannot be used interchangeably despite being closely related.
The factors of two hundred and forty-three are: 243, 81, 27, 9, 3, 1. The prime factors of two hundred and forty-three are 3. And the prime factorization of two hundred and forty-three is 3 x 3 x 3 x 3 x 3. Remember that 1 is not a prime factor of 243.
No matter if you had been searching for prime factorization for 243 or prime numbers of 243, you have come to the right page. Also, if you typed what is the prime factorization of 243 in the search engine then you are right here, of course.
Taking all of the above into account, tasks including write 243 as a product of prime factors or list the factors of 243 will no longer pose a challenge to you.
If you have any questions about the factors of two hundred and forty-three then fill in the form below and we will respond as soon as possible. If our content concerning all factors of 243 has been of help to you then share it by means of pressing the social buttons. And don’t forget to bookmark us.
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Posted in Factors
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Q:
# The attention span of children (ages 3 to 5) is claimed to be Normally distributed with a mean of 15 minutes and a standard deviation of 4 minutes. A test is to be performed to decide if the average attention span of these kids is really this short or if it is longer. You decide to test the hypotheses H0: μ = 15 versus Ha: μ > 15 at the 5% significance level. A sample of 10 children will watch a TV show they have never seen before, and the time until they walk away from the show will be recorded. At a significance level of 5%, the decision rule would be to reject the null hypothesis if the observed sample mean is greater than __________________ minutes.
Accepted Solution
A:
Answer:If the observed sample mean is greater than 17.07 minutes, then we would reject the null hypothesis.Step-by-step explanation:We are given the following in the question: Population mean, μ = 15 minutesSample size, n = 10Alpha, α = 0.05 Population standard deviation, σ = 4 minutesFirst, we design the null and the alternate hypothesis $$H_{0}: \mu = 15\text{ minutes}\\H_A: \mu > 15\text{ minutes}$$ Since the population standard deviation is given, we use one-tailed z test to perform this hypothesis. Formula: $$z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$$ Now, $$z_{critical} \text{ at 0.05 level of significance for one tail} = 1.64$$ Thus, we would reject the null hypothesis if the z-statistic is greater than this critical value.Thus, we can write:$$z_{stat} = \displaystyle\frac{\bar{x} - 15}{\frac{4}{\sqrt{10}} } > 1.64\\\\\bar{x} - 15>1.64\times \frac{4}{\sqrt{10}}\\\bar{x} - 15>2.07\\\bar{x} > 15+2.07\\\bar{x} > 17.07$$Thus, the decision rule would be if the observed sample mean is greater than 17.07 minutes, then we would reject the null hypothesis. |
# Algebra 1: Topic 1 Notes
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## Transcription
1 Algebra 1: Topic 1 Notes Review: Order of Operations Please Parentheses Excuse Exponents My Multiplication Dear Division Aunt Addition Sally Subtraction Table of Contents 1. Order of Operations & Evaluating Expressions 2. The Real Number System 3. Properties of Real Numbers 4. Simplifying Radicals 5. Basic Exponent Properties 6. One & Two Step Equations 7. Basic Multi Step Equations Review: Order of Operations Remember that Multiplication and Division are a pair do in order from left to right. Addition and Subtraction are also a pair.do in order from left to right. Order of Operations & Evaluating Expressions Review: Evaluate Evaluating an expression means to plug in! For example: 2n + 4 for n = 3 2(3)
2 Evaluate the following expressions. 1. 3r 17 for r = d for d = 4 Natural Numbers Non decimal, positive numbers starting with one. 3. 5g + 20 (4g) for g = 5 4. (3 + p) 2 2p for p = 7 The Real Number System: Numbers Have Names?!?! Whole Numbers Non decimal, positive numbers and zero. Non decimal positive and negative numbers, including zero. Integers 2
3 Rational Numbers Any number that can be expressed as the quotient or fraction of two integers. YES: Any integers Any decimals that ends or repeats Any fraction NO: Never ending decimals Properties of Real Numbers Irrational Numbers Any number that can not be expressed as a fraction. Usually a never ending, non repeating decimal. Examples: 2, List of Properties of Real Numbers Commutative Associative Distributive Identity Inverse Rational or Irrational Commutative Properties Definition: Changing the order of the numbers in addition or multiplication will not change the result
4 Commutative Properties Property Numbers Algebra Addition Multiplication Distributive Property Multiplication distributes over addition. a b c ab ac Associative Properties Definition: Changing the grouping of the numbers in addition or multiplication will not change the result. Additive Identity Property Definition: Zero preserves identities under addition. In other words, adding zero to a number does not change its value. Example: Associative Properties Property Numbers Algebra Addition Multiplication Multiplicative Identity Property Definition: The number 1 preserves identities under multiplication. In other words, multiplying a number by 1 does not change the value of the number. Example: 4
5 Additive Inverse Property Radical Vocab Definition: For each real number a there exists a unique real number a such that their sum is zero. In other words, opposites add to zero. Example: Multiplicative Inverse Property Definition: For each real number a there exists a unique real number such that their product is 1. Example: 1 a How to Simplify Radicals 1. Make a factor tree of the radicand. 2. Circle all final factor pairs. 3. All circled pairs move outside the radical and become single value. 4. Multiply all values outside radical. 5. Multiply all final factors that were not circled. Place product under radical sign. Simplifying Radicals 5
6 How to Simplify Cubed Radicals 1. Make a factor tree of the radicand. 2. Circle all final factor groups of three. 3. All circled groups of three move outside the radical and become single value. 4. Multiply all values outside radical. 5. Multiply all final factors that were not circled. Place product under radical sign. Exponents 6
7 Definition: Exponent The exponent of a number says how many times to use that number in a multiplication. It is written as a small number to the right and above the base number ( 3) 3 The Zero Exponent Rule Any number (excluding zero) to the zero power is always equal to one. Examples: = = =1 The One Exponent Rule Any number (excluding zero) to the first power is always equal to that number. Examples: a 1 = a 7 1 = = 53 Negative Power Rule Simplifying Exponential Expressions Step 1: Simplify (get rid of negatives and fractions, where possible). Step 2: Plug in. 7
8 Inverse Operations Pairs 1. 4m 3 j 2 2. for m = 2 and j = 3 for b = 2 and a = 3 Addition Multiplication Exponents Subtraction Division Radicals Solving 2 step Equations How to Solve 2 Step Equations 1. Use the opposite operation to move the letters to the left and numbers to the right. 2. Divide to get the variable alone. Example 5x 18 = 22 8
9 1. 10x 7 = m = m 2. 3 x = y = 3y b = 17 8b Solving Multi Step Equations Fractions To get rid of fractions, multiply all other values by the reciprocal! Example: k How to solve Multi Step Equations 1. Distribute. Bonus Step: Multiple by reciprocal/lcm, if fractions. 2. Combine Like Terms (ONLY on same side of equal sign). 3. Use the inverse operation to move numbers to the right. 4. Use the inverse operation to move variables to the left. 5. Divide. 9
10 Example 3(4x 10) = 26 4(5 + 2x) + 7x = 5 Example 6(4 + 3x) 10x = 30 2x 6 5 Example 1 5 x 8 = 4 + 2x 10
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Directions (1-5): Given question is followed by information given in two statements named as Quantity I and Quantity II. You have study the information along with the question and compare the value derived from Quantity I and Quantity II then answer the questions.
1. Sum of 8 consecutive even numbers is S1. (SBI PO - 2018)
Quantity I: Sum of second number and eighth number in S1.
Quantity II: Sum of third number and sixth number in S1.
Ans: Quantity I > Quantity II
Explanation:
Let the numbers be x, x + 2, x + 4, x + 6, x + 8, x + 10, x + 12 and x + 14
Quantity I: x + 2 + x + 14 = 2x + 16
Quantity II: x + 4 + x + 10 = 2x + 14
∴ Quantity I > Quantity II
2. The cost price of 2 items A and B is same. The shopkeeper decided to mark the price 40% more than the CP of each item. A discount of 25% was given on item A and discount of 20% was given on item B. Total profit earn on both item was Rs.34.
Quantity I: CP of the items.
Quantity II: CP of any item which was sold at 12.5% profit and profit earned on it was Rs. 25.
Ans: Quantity I = Quantity II
Explanation:
Let the Cost Prices of A & B be Rs.100 each. Marked prices of each item will be Rs.140
SP of A = 75% of 140 = 105
SP of B = 80% of 140 = 112
Total profit on A & B = 217 - 200 = Rs.17
If the profit is Rs.17,
The cost price is Rs.100 each.
If the profit is Rs.34,
The cost price of each item is Rs.200.
Quantity I: Rs.200
Quantity II: 12.5% (profit) = Rs.25
100% (CP) = × 25 = Rs.200
∴ Quantity I = Quantity II
3. There are 63 cards in a box numbered from 01 to 63. Every card is numbered with only 1 number.
Quantity I: Probability of picking up a card whose digits, if interchanged, result in a number which is 36 more than the number picked up.
Quantity II: Probability of picking up a card, the number printed on which is a multiple of 8 but not that of 16.
Ans: Quantity I > Quantity II
Explanation:
Quantity I: When interchange the digits, number is 36 more than the original number
Possibilities are (04, 15, 26, 37, 48 and 59)
Probability =
Quantity II: Multiple of 8 but not multiple of 16
Possibilities are (8, 24, 40 and 56)
Probability =
∴ Quantity I > Quantity II
4. A girl walked to school at a speed of 3 miles per hour. Once she reached the school, she realized that
she forgot to bring her books, so rushed back home at a speed of 6 miles per hour. She then walked
back to school at a speed of 4 miles per hour. All the times, she walked in the same route.
Quantity I: Her average speed over the entire journey.
Quantity II: 4 mph.
Ans: Quantity I = Quantity II or No relation can be established
Explanation:
Quantity I: Let the distance between home and school be 'x' miles and times taken to travel three times to this distance be t1, t2 and t3
⇒ 4 mph
Quantity II: 4 mph
∴ Quantity I = Quantity II
5. If a speed of boat is 500% more than the speed of a current. (SBI PO - 2018)
Quantity I: If boat can travel a distance of 63 km in 3 hrs, in downstream then 'x' is the speed of the
boat in upstream (kmph).
Quantity II: 15 km/hr.
Ans: Quantity I = Quantity II or No relation
Explanation:
Quantity I: Let the speed of current = x
Speed of boat = x + 5x
Downstream speed = x + x + 5x = 7x
∴ = 3 ⇒ x = 3
Upstream speed = 6x - x = 5x = 15 kmph
Quantity II: 15 kmph
∴ Quantity I = Quantity II
6. Given question is followed by information given in two statements named as Quantity I and Quantity II. You have to study the information along with the question and compare the value derived from
Quantity I and Quantity II then answer the question.
Quantity I- x: x2 + x - 6 = 0
Quantity II- y: y2 + 7y + 12 = 0 (SBI PO - 2018)
Ans: Quantity I Quantity II
Explanation:
Quantity I: x2 + x - 6 = 0
⇒ (x + 3)(x - 2) = 0
⇒ x = -3, 2
Quantity II: y2 + 7y + 12 = 0
⇒ (x + 4)(x + 3) = 0
⇒ x = -4, -3
∴ x y
Some more
1) ax2 + bx + c = 0
2)
ax2 − bx + c = 0
3)
ax2 + bx − c = 0
4)
ax2 − bx − c = 0
In the first type of quadratic equation we can get two roots are negative.
In the second type of quadratic equation we can get two roots are positive.
In the third type of quadratic equation we can get one positive root and one negative root.
In the fourth type of quadratic equation we can get one positive root and one negative root.
If we compare first two types of equations second equation is greater when compare to first equation the reason is second equation can get positive roots and first equation can get negative roots
If we compare third and fourth type of quadratic equations third type and fourth type we can get one positive and one negative root.
So, the relationship can’t be established.
e.g.:
1. I)
ax2 + bx + c = 0
II) ay2 − by + c = 0
Sol: x < y
2. I)
ax2 + bx − c = 0
II)
ax2 − bx − c = 0
Sol: No relation can be established.
Model Questions
1. I. x2 + 13x + 36 = 0
II. 3y2 − 28y + 64 = 0
a) If x < y b) If x > y c) If x
> y d) If x < y
e) If x = y or no relation can be established.
Sol: I.
x2 + 9x + 4x + 36 = 0
(x + 9) (x + 4) = 0
x = −9, −4
II. 3y2 − 16y − 12y + 64 = 0
(3y − 16) (y − 4) = 0
Clearly, x < y
Ans: a
2. I. x2 + 3x − 40 = 0
II. 2y2 + 7y − 60 = 0
a) If x < y b) If x > y c) If x > y d) If x < y
e) If x = y or no relation can be established.
Sol: I. x + 8x − 5x − 40 = 0
(x + 8) (x − 5) = 0
x = −8, 5
II. 2y2 + 15y − 8y − 60 = 0
(2y + 15) (y − 4) = 0
Clearly, no relation can be established Ans: e
3. I. 2x2 − 21x + 45 = 0
II. 5y2 + 42y + 88 = 0
a) If x < y b) If x > y c) If x > y d) If x < y
e) If x = y or no relation can be established.
Sol: I. 2x2 − 15x − 6x + 45 = 0
(x − 3) (2x − 15) = 0
Clearly, x > y Ans: b
4. I. 2x- x − 3 = 0
II. y2 − 11y + 30 = 0
a) If x < y b) If x > y c) If x > y d) If x < y
e) If x = y or no relation can be established.
Sol: I. 2x2 − 3x + 2x − 3 = 0
(x + 1) (2x − 3) = 0
x = −1,
II. y2 − 6y − 5x + 30 = 0
(y − 5) (y − 6) = 0
y = 5, 6
Clearly, x < y Ans: a
5. I. x2 − 2x − 24 = 0
II. y2 + 6y − 40 = 0
a) If x < y b) If x > y c) If x > y d) If x < y
e) If x = y or no relation can be established.
Sol: I. x2 − 6x + 4x − 24 = 0
(x − 6) (x + 4) = 0
x = −4, 6
II. y2 + 10y − 4y − 40 = 0
(y + 10) (y − 4) = 0 ⇒ y = −10, 4
Clearly, no relation can be established Ans: e
6. I. x2 − 5x − 14 = 0
II. y2 − 3y − 88 = 0
a) If x > y b) If x > y c) If x < y d) If x < y
e) If x = y or no relation can be established between x and y.
Sol: I. x2 − 7x + 2x − 14 = 0
(x + 2) (x − 7) = 0
x = −2, 7
II. y2 − 11y + 8y − 88 = 0
(y + 8) (y − 11) = 0 y = −8, 11
Clearly, no relation can be established Ans: e
7. I. 7x2 + 13x + 6 = 0
II. 5y2 + 11y + 6 = 0
a) If x > y b) If x > y c) If x < y d) If x < y
e) If x = y or no relation can be established between x and y.
Sol: I. 7x2 + 7x + 6x + 6 = 0
(7x + 6) (x + 1) = 0
Clearly, x < y Ans: d
8. I. x2 + 16x + 60 = 0
II. y2 − 14y + 45 = 0
a) If x > y b) If x > y c) If x < y d) If x < y
e) If x = y or no relation can be established between x and y.
Sol: I. x2 + 10x + 6x + 60 = 0
(x + 6) (x + 10) = 0
x = −6, − 10
II. y2 − 9y − 5y + 45 = 0
(y − 5) (y − 9) = 0 ⇒ y = 5, 9
Clearly, x < y Ans: c
9. I. 7x2 + x − 8 = 0
II. 7y2 − 23y + 16 = 0
a) If x = y or no relation can be established
b) If x > y c) If x < y d) If x > y e) If x < y
Sol: I. 7x2 + 8x − 7x − 8 = 0
(7x + 8) (x − 1) = 0
11. I. x2 − 23x + 132 = 0
II. y2 − 25y + 156 = 0
a) If x > y b) If x > y
c) If x = y or no relation can be established
d) If y > x e) If y > x
Sol: I. x2 − 12x − 11x + 132 = 0
(x − 12) (x − 11) = 0
x = 11, 12
II. y2 − 12y − 13y + 156 = 0
(y – 12) (y – 13) = 0
y = 12, 13
Clearly, y > x Ans: d
12. I. 15x2 − 8x + 1 = 0
II. 9y2 + 24y + 7 = 0
a) If x > y b) If x > y
c) If x = y or no relation can be established
d) If y > x e) If y > x
Sol: I. 15x2 − 5x − 3x + 1 = 0
(5x − 1) (3x − 1) = 0
13. I. 5x2 − 4x − 105 = 0
II. 5y2 + 46y + 105 = 0
a) If x > y b) If x > y
c) If x = y or no relation can be established
d) If y > x e) If y > x
Sol: I. 5x2 − 25x + 21x − 105 = 0
(5x + 21) (x − 5) = 0
14. I. 4x2 − 20x + 25 = 0
II. y2 − 400 = 0
a) If x > y
b) If x = y or no relation can be established between x and y.
c) If x > y d) If x < y e) If x < y
Sol: I. 4x2 − 10x − 10x + 25 = 0
(2x − 5) (2x − 5) = 0
Posted Date : 10-02-2022
గమనిక : ప్రతిభ.ఈనాడు.నెట్లో కనిపించే వ్యాపార ప్రకటనలు వివిధ దేశాల్లోని వ్యాపారులు, సంస్థల నుంచి వస్తాయి. మరి కొన్ని ప్రకటనలు పాఠకుల అభిరుచి మేరకు కృత్రిమ మేధస్సు సాంకేతికత సాయంతో ప్రదర్శితమవుతుంటాయి. ఆ ప్రకటనల్లోని ఉత్పత్తులను లేదా సేవలను పాఠకులు స్వయంగా విచారించుకొని, జాగ్రత్తగా పరిశీలించి కొనుక్కోవాలి లేదా వినియోగించుకోవాలి. వాటి నాణ్యత లేదా లోపాలతో ఈనాడు యాజమాన్యానికి ఎలాంటి సంబంధం లేదు. ఈ విషయంలో ఉత్తర ప్రత్యుత్తరాలకు, ఈ-మెయిల్స్ కి, ఇంకా ఇతర రూపాల్లో సమాచార మార్పిడికి తావు లేదు. ఫిర్యాదులు స్వీకరించడం కుదరదు. పాఠకులు గమనించి, సహకరించాలని మనవి. |
# The diameter of a metallic sphere is equal to $9 \mathrm{~cm}$. It is melted and drawn into a long wire of diameter $2 \mathrm{~mm}$ having uniform cross-section. Find the length of the wire.
Given:
The diameter of a metallic sphere is equal to $9 \mathrm{~cm}$. It is melted and drawn into a long wire of diameter $2 \mathrm{~mm}$ having uniform cross-section.
To do:
We have to find the length of the wire.
Solution:
Diameter of the metallic sphere $=9 \mathrm{~cm}$
This implies,
Radius of the sphere $r=\frac{9}{2} \mathrm{~cm}$
Volume of the sphere $=\frac{4}{3} \pi r^{3}$
$=\frac{4}{3} \pi(\frac{9}{2})^{3}$
$=\frac{243}{2} \pi \mathrm{cm}^{3}$
Diameter of the wire $=2 \mathrm{~mm}$
This implies,
Radius of the wire $=\frac{2}{2} \mathrm{~mm}$
$=1 \mathrm{~mm}$
$=\frac{1}{10} \mathrm{~cm}$
Let $h$ be the length of the wire, this implies,
$\pi r^{2} h=\frac{243}{2} \pi$
$\Rightarrow r^{2} h=\frac{243}{2}$
$\Rightarrow (\frac{1}{10})^{2} h=\frac{243}{2}$
$\Rightarrow \frac{1}{100} h=\frac{243}{2}$
$\Rightarrow h=\frac{243}{2} \times 100$
$\Rightarrow h=12150 \mathrm{~cm}$
The length of the wire is $12150\ cm$.
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Simply Easy Learning
Updated on: 10-Oct-2022
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Core 2: Integration
This section covers all the basic ideas required for integrating in AS-level Mathematics. Negative areas and the trapezium rule will be added to integration part 2, which is currently being written.
This section also does not cover skills such as integration by substitution, integration by-parts and functions other than polynomials. To revise these, read the CORE 3 revision guide on integration.
The fundamental formula for integrating a polynomial is given by the general formula: The answer is called the integrand Think about this as: “Increase the power by 1 and then divide by the new power” The is important and is called the constant of integration Integration is used for lots of purposes, though the most regular use you will have will be to find area under curves and to solve differential equations Question Answer Harder powers and more complicated expressions In this example observe how the new power just reciprocates to give the multiplier of x. To find the new power, add the denominator on to the numerator and then multiply by the reciprocal of the new power Using the integrand to solve differential equations This uses the principle that integration is the reverse of differentiation Solve the differential equation: This questions means that an expression was differentiated to give an answer of . To find the original expression we need to reverse this by integrating. This is called a general solution. This is an unknown constant, which may have been there, but which would have disappeared upon differentiating. Finding the constant of integration You need to know some extra information, often called boundary or initial conditions depending on their form Find the equation of the curve, which passes though the point and which has the gradient function: Therefore, First of all integrate To find , substitute the given coordinate into the expression. This is called a particular solution Using the integrand to find areas The integrand is sometimes known as the area function Find the area enclosed by the curve: and the lines and We need to do: We now substitute in the values: and and find the difference. The area is units2. This means the area under the curve, between the x-coordinates 1 and 2. The 1 and 2 are called limits We have integrated. You do NOT need to worry about the . Observe that we have written the limits adjacent to the square brackets. The brackets are essential, especially if you have negative numbers. A more concise example: Find the area under the curve between the values and . |
# 6.7 Integrals, exponential functions, and logarithms
Page 1 / 4
• Write the definition of the natural logarithm as an integral.
• Recognize the derivative of the natural logarithm.
• Integrate functions involving the natural logarithmic function.
• Define the number $e$ through an integral.
• Recognize the derivative and integral of the exponential function.
• Prove properties of logarithms and exponential functions using integrals.
• Express general logarithmic and exponential functions in terms of natural logarithms and exponentials.
We already examined exponential functions and logarithms in earlier chapters. However, we glossed over some key details in the previous discussions. For example, we did not study how to treat exponential functions with exponents that are irrational. The definition of the number e is another area where the previous development was somewhat incomplete. We now have the tools to deal with these concepts in a more mathematically rigorous way, and we do so in this section.
For purposes of this section, assume we have not yet defined the natural logarithm, the number e , or any of the integration and differentiation formulas associated with these functions. By the end of the section, we will have studied these concepts in a mathematically rigorous way (and we will see they are consistent with the concepts we learned earlier).
We begin the section by defining the natural logarithm in terms of an integral. This definition forms the foundation for the section. From this definition, we derive differentiation formulas, define the number $e,$ and expand these concepts to logarithms and exponential functions of any base.
## The natural logarithm as an integral
Recall the power rule for integrals:
$\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}+C,\phantom{\rule{0.2em}{0ex}}n\ne \text{−}1.$
Clearly, this does not work when $n=-1,$ as it would force us to divide by zero. So, what do we do with $\int \frac{1}{x}dx?$ Recall from the Fundamental Theorem of Calculus that ${\int }_{1}^{x}\frac{1}{t}dt$ is an antiderivative of $1\text{/}x.$ Therefore, we can make the following definition.
## Definition
For $x>0,$ define the natural logarithm function by
$\text{ln}\phantom{\rule{0.2em}{0ex}}x={\int }_{1}^{x}\frac{1}{t}dt.$
For $x>1,$ this is just the area under the curve $y=1\text{/}t$ from $1$ to $x.$ For $x<1,$ we have ${\int }_{1}^{x}\frac{1}{t}dt=\text{−}{\int }_{x}^{1}\frac{1}{t}dt,$ so in this case it is the negative of the area under the curve from $x\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}1$ (see the following figure).
Notice that $\text{ln}\phantom{\rule{0.2em}{0ex}}1=0.$ Furthermore, the function $y=1\text{/}t>0$ for $x>0.$ Therefore, by the properties of integrals, it is clear that $\text{ln}\phantom{\rule{0.2em}{0ex}}x$ is increasing for $x>0.$
## Properties of the natural logarithm
Because of the way we defined the natural logarithm, the following differentiation formula falls out immediately as a result of to the Fundamental Theorem of Calculus.
## Derivative of the natural logarithm
For $x>0,$ the derivative of the natural logarithm is given by
$\frac{d}{dx}\text{ln}\phantom{\rule{0.2em}{0ex}}x=\frac{1}{x}.$
## Corollary to the derivative of the natural logarithm
The function $\text{ln}\phantom{\rule{0.2em}{0ex}}x$ is differentiable; therefore, it is continuous.
A graph of $\text{ln}\phantom{\rule{0.2em}{0ex}}x$ is shown in [link] . Notice that it is continuous throughout its domain of $\left(0,\infty \right).$
what is function?
A set of points in which every x value (domain) corresponds to exactly one y value (range)
Tim
what is lim (x,y)~(0,0) (x/y)
limited of x,y at 0,0 is nt defined
Alswell
But using L'Hopitals rule is x=1 is defined
Alswell
Could U explain better boss?
emmanuel
value of (x/y) as (x,y) tends to (0,0) also whats the value of (x+y)/(x^2+y^2) as (x,y) tends to (0,0)
NIKI
can we apply l hospitals rule for function of two variables
NIKI
why n does not equal -1
Andrew
I agree with Andrew
Bg
f (x) = a is a function. It's a constant function.
proof the formula integration of udv=uv-integration of vdu.?
Find derivative (2x^3+6xy-4y^2)^2
no x=2 is not a function, as there is nothing that's changing.
are you sure sir? please make it sure and reply please. thanks a lot sir I'm grateful.
The
i mean can we replace the roles of x and y and call x=2 as function
The
if x =y and x = 800 what is y
y=800
800
Bg
how do u factor the numerator?
Nonsense, you factor numbers
Antonio
You can factorize the numerator of an expression. What's the problem there? here's an example. f(x)=((x^2)-(y^2))/2 Then numerator is x squared minus y squared. It's factorized as (x+y)(x-y). so the overall function becomes : ((x+y)(x-y))/2
The
The problem is the question, is not a problem where it is, but what it is
Antonio
I think you should first know the basics man: PS
Vishal
Yes, what factorization is
Antonio
Antonio bro is x=2 a function?
The
Yes, and no.... Its a function if for every x, y=2.... If not is a single value constant
Antonio
you could define it as a constant function if you wanted where a function of "y" defines x f(y) = 2 no real use to doing that though
zach
Why y, if domain its usually defined as x, bro, so you creates confusion
Antonio
Its f(x) =y=2 for every x
Antonio
Yes but he said could you put x = 2 as a function you put y = 2 as a function
zach
F(y) in this case is not a function since for every value of y you have not a single point but many ones, so there is not f(y)
Antonio
x = 2 defined as a function of f(y) = 2 says for every y x will equal 2 this silly creates a vertical line and is equivalent to saying x = 2 just in a function notation as the user above asked. you put f(x) = 2 this means for every x y is 2 this creates a horizontal line and is not equivalent
zach
The said x=2 and that 2 is y
Antonio
that 2 is not y, y is a variable 2 is a constant
zach
So 2 is defined as f(x) =2
Antonio
No y its constant =2
Antonio
what variable does that function define
zach
the function f(x) =2 takes every input of x within it's domain and gives 2 if for instance f:x -> y then for every x, y =2 giving a horizontal line this is NOT equivalent to the expression x = 2
zach
Yes true, y=2 its a constant, so a line parallel to y axix as function of y
Antonio
Sorry x=2
Antonio
And you are right, but os not a function of x, its a function of y
Antonio
As function of x is meaningless, is not a finction
Antonio
yeah you mean what I said in my first post, smh
zach
I mean (0xY) +x = 2 so y can be as you want, the result its 2 every time
Antonio
OK you can call this "function" on a set {2}, but its a single value function, a constant
Antonio
well as long as you got there eventually
zach
2x^3+6xy-4y^2)^2 solve this
femi
moe
volume between cone z=√(x^2+y^2) and plane z=2
Fatima
It's an integral easy
Antonio
V=1/3 h π (R^2+r2+ r*R(
Antonio
How do we find the horizontal asymptote of a function using limits?
Easy lim f(x) x-->~ =c
Antonio
solutions for combining functions
what is a function? f(x)
one that is one to one, one that passes the vertical line test
Andrew
It's a law f() that to every point (x) on the Domain gives a single point in the codomain f(x)=y
Antonio
is x=2 a function?
The
restate the problem. and I will look. ty
is x=2 a function?
The
What is limit
it's the value a function will take while approaching a particular value
Dan
don ger it
Jeremy
what is a limit?
Dlamini
it is the value the function approaches as the input approaches that value.
Andrew
Thanx
Dlamini
Its' complex a limit It's a metrical and topological natural question... approaching means nothing in math
Antonio
is x=2 a function?
The |
## Lesson: Comparing Money Amounts Introducing the Concept
Students learned the values of bills and coins in Grade 3. Now they can apply those skills to counting and comparing amounts of money.
Materials: play money for each group of students: \$50, \$20, \$10, \$5, and \$1 bills; half-dollars, quarters, dimes, nickels, pennies
Preparation: Distribute play money to each group.
Prerequisite Skills and Background: Students should know the value of coins and bills and how to write amounts of money. Students should also be able to count by 5s, 10s, 20s, 25s, and 50s.
• Say: Place two \$20 bills, one \$10 bill, two \$5 bills, and three \$1 bills side by side. Put the bills in order from greatest to least value.
• Ask: Which bill has the greatest value? (\$20 bill) Which bill has the least value? (\$1 bill)
• Say: Let's find the total value of the bills. Point to each bill as we count aloud, starting with the \$20 bills.
(\$20.00, \$40.00, \$50.00, \$55.00, \$60.00, \$61.00, \$62.00, \$63.00)
• Ask: What is the total value of the bills? (\$63)
• Say: Now place one quarter, two dimes, two nickels, and five pennies side by side below the bills. Put the coins in order from greatest to least value.
• Ask: Which coin has the greatest value? (quarter) Which coin has the least value? (penny)
• Say: Let's find the total value of the coins by counting on from the quarter.
Have students point to each coin as they count aloud: 25¢, 35¢, 45¢, 50¢, 55¢, 56¢, 57¢, 58¢, 59¢, 60¢.
• Ask: What is the total value of the coins? (60¢) What is the total value of the bills and coins together? (\$63.60)
Write \$63.60 on the board.
• Say: Put those bills and coins aside. Show one \$50 bill, one \$10 bill, three \$1 bills, one half-dollar, and one nickel side-by-side. Put the money in order from greatest to least value.
Some students may line up all the bills and coins in one row. Others may put the bills in one row and the coins in another row. Either arrangement is acceptable.
• Say: Let's find the total value of this collection of bills and coins. Where should we start?
Students should say the \$50 bill.
• Say: Point to each bill and coin as we count aloud.
\$50.00, \$60.00, \$61.00, \$62.00, \$63.00, \$63.50, \$63.55
• Ask: What is the total value of the bills and coins? (\$63.55)
Write \$63.55 on the board.
• Ask: Which amount is greater? (\$63.60) How do you know?
Some students may say that the dollars are equal and when you compare the cents, 60¢ is greater than 55¢. Accept any reasonable explanation.
• Say: Add one nickel and two pennies to the collection.
• Ask: Where did you place the nickel and pennies? Why?
Students should know that by placing the nickel and pennies to the right of the other coins, it will be easy to count on from \$63.55.
• Say: Count on from \$63.55.
(\$63.60, \$63.61, \$63.62)
• Ask: Now what is the total value of the bills and coins? (\$63.62)
• Erase \$63.55 and write \$63.62 on the board.
• Ask: Which amount is greater? (\$63.62)
• Ask: Are there any other combinations of bills and coins that equal \$63.62?
Have students make different combinations of coins and bills equal to \$63.62 and share them with the class.
Repeat this activity with other collections of money. |
# Vector Dot Product Example Explained in Detail
In the realm of mathematics and physics, the dot product of vectors is a fundamental concept that finds application in a wide array of mathematical and physical problems. Understanding the vector dot product, its properties, and how to calculate it is crucial for anyone studying or working with vectors. In this article, we will delve into the concept of the vector dot product, explore examples to illustrate its application, and address common questions related to this topic.
Whether you are a student striving to grasp the intricacies of vectors and their operations or a professional seeking to refresh your knowledge, the examples and explanations provided here will aid in solidifying your understanding of the vector dot product.
## What Is the Vector Dot Product?
The dot product, also known as the scalar product, is an operation that takes two vectors and produces a scalar quantity. The result of the dot product is a single number rather than a vector. For two vectors a and b, the dot product is denoted by a · b or sometimes a • b.
The dot product of two vectors a and b is calculated using the following formula:
a · b = |a| * |b| * cos(θ)
Where |a| and |b| are the magnitudes of vectors a and b, and θ is the angle between the two vectors.
### Properties of the Dot Product
• The dot product of two parallel vectors is the product of their magnitudes.
• The dot product of perpendicular vectors is zero.
• The dot product is commutative: a · b = b · a.
• The dot product is distributive over vector addition: a · (b + c) = a · b + a · c.
Understanding these properties is crucial when working with the dot product, as they form the basis for its applications in various contexts.
## Example 1: Calculating the Dot Product
Let's consider two vectors a = 3i + 4j and b = 2i - 5j. To find the dot product of these vectors, we can use the formula:
a · b = (3i + 4j) · (2i - 5j)
Expanding this using the distributive property, we get:
a · b = 3i · 2i + 3i · (-5j) + 4j · 2i + 4j · (-5j)
Now, using the fact that i · i = 1, i · j = 0, and j · j = 1, the expression simplifies to:
a · b = 3*2 + 0 + 0 + 4*(-5)
Therefore, the dot product of the vectors a and b is 6 - 20 = -14.
### Applications in Physics
The dot product is extensively used in physics, particularly in the context of work and energy. For instance, when calculating work done by a force on an object, the dot product of the force vector and the displacement vector is employed. This application demonstrates the real-world significance of understanding and applying the dot product of vectors.
## Example 2: Finding the Angle Between Vectors
In this example, we have two vectors: a = 2i + 3j and b = i - 2j. To determine the angle between these vectors using the dot product, we use the formula:
cos(θ) = (a · b) / (|a| * |b|)
Where θ represents the angle between vectors a and b.
First, we calculate the dot product a · b as we did in Example 1. The magnitude of a is |a| = sqrt(2^2 + 3^2) = sqrt(13), and the magnitude of b is |b| = sqrt(1^2 + (-2)^2) = sqrt(5).
Substituting these values into the formula yields:
cos(θ) = (-4) / (sqrt(13) * sqrt(5))
Solving for θ gives us the angle between the vectors.
• Q: Can the result of a dot product be a vector?
• A: No, the result of a dot product is always a scalar quantity.
• Q: What is the significance of the dot product in geometry?
• A: The dot product is used to calculate angles between vectors and to determine whether vectors are orthogonal.
• Q: How is the dot product related to the law of cosines?
• A: The dot product is utilized in the proof of the law of cosines and is fundamental to understanding its geometric implications.
## Concluding Remarks
In conclusion, the vector dot product is a powerful mathematical operation with diverse applications in fields such as physics, engineering, and computer science. Mastering the concepts, properties, and calculation methods related to the dot product is essential for anyone dealing with vectors and their applications. The examples discussed in this article provide a solid foundation for understanding the vector dot product and serve as valuable references for tackling problems involving vector operations.
If you want to know other articles similar to Vector Dot Product Example Explained in Detail you can visit the category Sciences.
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# Class 6 Maths Chapter 5 Exercise 5.6 Pdf Notes NCERT Solutions
Class 6 Maths Chapter 5 Understanding Elementary Shapes Exercise 5.6 pdf notes:-
Exercise 5.6 Class 6 maths Chapter 5 Pdf Notes:-
## Ncert Solution for Class 6 Maths Chapter 5 Understanding Elementary Shapes Exercise 5.6 Tips:-
What do you find?
Triangles in which all the angles are equal.
If all the angles in a triangle are equal, then its sides are also . …………
Triangles in which all the three sides are equal.
If all the sides in a triangle are equal, then its angles are. ……….. .
Triangle which have two equal angles and two equal sides.
If two sides of a triangle are equal, it has . ………… equal angles.
and if two angles of a triangle are equal, it has ……. equal sides.
Triangles in which no two sides are equal.
If none of the angles of a triangle are equal then none of the sides are equal.
If the three sides of a triangle are unequal then, the three angles are
also…………. .
Take some more
triangles and verify these.
For this we will again
have to measure all the
sides and angles of the
triangles.
The triangles have
been divided into
categories and given
special names. Let us see
what they are.
Naming triangles based on sides
A triangle having all three unequal sides is called a Scalene Triangle [(c), (e)].
A triangle having two equal sides is called an Isosceles Triangle [(b), (f)].
A triangle having three equal sides is called an Equilateral Triangle [(a), (d)].
Classify all the triangles whose sides you measured earlier, using these
definitions.
Naming triangles based on angles
If each angle is less than 90°, then the triangle is called an acute angled triangle.
If any one angle is a right angle then the triangle is called a right angled
triangle.
If any one angle is greater than 90°, then the triangle is called an obtuse
angled triangle.
(c) a right angled isosceles triangle.
(d) a scalene right angled triangle.
Do you think it is possible to sketch
(a) an obtuse angled equilateral triangle ?
(b) a right angled equilateral triangle ?
(c) a triangle with two right angles?
Think, discuss and write your conclusions.
#### Test Paper Of Class 8th
• Maths 8th Class
• Science 8th class
• Sst 8th Class
• #### Test Paper Of Class 7th
• Maths 7th Class
• Science 7th class
• #### Test Paper Of Class 6th
• Maths 6th Class
• Science 6th class |
# AP Statistics Curriculum 2007 Infer BiVar
(Difference between revisions)
Revision as of 02:24, 29 November 2008 (view source)IvoDinov (Talk | contribs)← Older edit Current revision as of 14:44, 14 February 2014 (view source)IvoDinov (Talk | contribs) (→Comparing Two Variances (\sigma_1^2 = \sigma_2^2?)) (2 intermediate revisions not shown) Line 9: Line 9: : $\chi_o^2 = {(n-1)s^2 \over \sigma^2} \sim \Chi_{(df=n-1)}^2$ : $\chi_o^2 = {(n-1)s^2 \over \sigma^2} \sim \Chi_{(df=n-1)}^2$ - ===Comparing Two Variances ($\sigma_1^2 = \sigma_2^2$?)=== + ===Comparing Two Variances ($$\sigma_1^2 = \sigma_2^2$$?)=== Suppose we study two populations which are approximately Normally distributed, and we take a random sample from each population, {$X_1, X_2, X_3, \cdots, X_n$} and {$Y_1, Y_2, Y_3, \cdots, Y_k$}. Recall that ${(n-1) s_1^2 \over \sigma_1^2}$ and ${(n-1) s_2^2 \over \sigma_2^2}$ have $\Chi^2_{(df=n - 1)}$ and $\Chi^2_{(df=k - 1)}$ distributions. We are interested in assessing $H_o: \sigma_1^2 = \sigma_2^2$ vs. $H_1: \sigma_1^2 \not= \sigma_2^2$, where $s_1$ and $\sigma_1$, and $s_2$ and $\sigma_2$ and the sample and the population standard deviations for the two populations/samples, respectively. Suppose we study two populations which are approximately Normally distributed, and we take a random sample from each population, {$X_1, X_2, X_3, \cdots, X_n$} and {$Y_1, Y_2, Y_3, \cdots, Y_k$}. Recall that ${(n-1) s_1^2 \over \sigma_1^2}$ and ${(n-1) s_2^2 \over \sigma_2^2}$ have $\Chi^2_{(df=n - 1)}$ and $\Chi^2_{(df=k - 1)}$ distributions. We are interested in assessing $H_o: \sigma_1^2 = \sigma_2^2$ vs. $H_1: \sigma_1^2 \not= \sigma_2^2$, where $s_1$ and $\sigma_1$, and $s_2$ and $\sigma_2$ and the sample and the population standard deviations for the two populations/samples, respectively. Line 30: Line 30: - In the image below the left and right critical regions are white with $F(\alpha,df_1=n_1-1,df_2=n_2-1)$ and $F(1-\alpha,df_1=n_1-1,df_2=n_2-1)$ representing the lower and upper, respectively, critical values. In this example of $F(df_1=12, df_2=15)$, the left and right critical values at $\alpha/2=0.025$ are $F(\alpha/2=0.025,df_1=9,df_2=14)=0.314744$ and $F(1-\alpha/2=0.975,df_1=9,df_2=14)=2.96327$, respectively. + In the image below, the left and right critical regions are white with $F(\alpha,df_1=n_1-1,df_2=n_2-1)$ and $F(1-\alpha,df_1=n_1-1,df_2=n_2-1)$ representing the lower and upper, respectively, critical values. In this example of $F(df_1=12, df_2=15)$, the left and right critical values at $\alpha/2=0.025$ are $F(\alpha/2=0.025,df_1=9,df_2=14)=0.314744$ and $F(1-\alpha/2=0.975,df_1=9,df_2=14)=2.96327$, respectively.
[[Image:SOCR_EBook_Dinov_Infer_BiVar_021608_Fig1.jpg|500px]]
[[Image:SOCR_EBook_Dinov_Infer_BiVar_021608_Fig1.jpg|500px]]
===Comparing Two Standard Deviations ($\sigma_1 = \sigma_2$?)=== ===Comparing Two Standard Deviations ($\sigma_1 = \sigma_2$?)=== - Two make inference on whether the standard deviations of two populations are equal we calculate the sample variances and apply the inference on the ratio of the sample variance using the F-test, as described above. + To make inference on whether the standard deviations of two populations are equal, we calculate the sample variances and apply the inference on the ratio of the sample variance using the F-test, as described above. ===Hands-on activities=== ===Hands-on activities=== - * Formulate appropriate hypotheses and assess the significance of the evidence to reject the null hypothesis that the variances of the two populations, that the following data come from, are distinct. Assume the observations below represent random samples (of sizes 6 and 10) from two Normally distributed populations of liquid content (in fluid ounces) of beverage cans. Use ($\alpha=0.1$). + * Formulate appropriate hypotheses and assess the significance of the evidence to reject the null hypothesis that the variances of the two populations, where the following data come from, are distinct. Assume the observations below represent random samples (of sizes 6 and 10) from two Normally distributed populations of liquid content (in fluid ounces) of beverage cans. Use ($\alpha=0.1$).
{| class="wikitable" style="text-align:center; width:75%" border="1" {| class="wikitable" style="text-align:center; width:75%" border="1"
## General Advance-Placement (AP) Statistics Curriculum - Comparing Two Variances
In the section on inference about the variance and the standard deviation, we already learned how to do inference on either of these two population parameters. Now we discuss the comparison of the variances (or standard deviations) using data randomly sampled from two different populations.
### Background
Recall that the sample-variance (s2) is an unbiased point estimate for the population variance σ2, and similarly, the sample-standard-deviation (s) is a point estimate for the population-standard-deviation σ.
The sample-variance is roughly Chi-square distributed:
$\chi_o^2 = {(n-1)s^2 \over \sigma^2} \sim \Chi_{(df=n-1)}^2$
### Comparing Two Variances ($$\sigma_1^2 = \sigma_2^2$$?)
Suppose we study two populations which are approximately Normally distributed, and we take a random sample from each population, {$X_1, X_2, X_3, \cdots, X_n$} and {$Y_1, Y_2, Y_3, \cdots, Y_k$}. Recall that ${(n-1) s_1^2 \over \sigma_1^2}$ and ${(n-1) s_2^2 \over \sigma_2^2}$ have $\Chi^2_{(df=n - 1)}$ and $\Chi^2_{(df=k - 1)}$ distributions. We are interested in assessing $H_o: \sigma_1^2 = \sigma_2^2$ vs. $H_1: \sigma_1^2 \not= \sigma_2^2$, where s1 and σ1, and s2 and σ2 and the sample and the population standard deviations for the two populations/samples, respectively.
Notice that the Chi-Square Distribution is not symmetric (it is positively skewed). You can visualize the Chi-Square distribution and compute all critical values either using the SOCR Chi-Square Distribution or using the SOCR Chi-Square Distribution Calculator.
The Fisher's F Distribution, and the corresponding F-test, is used to test if the variances of two populations are equal. Depending on the alternative hypothesis, we can use either a two-tailed test or a one-tailed test. The two-tailed version tests against an alternative that the standard deviations are not equal ($H_1: \sigma_1^2 \not= \sigma_2^2$). The one-tailed version only tests in one direction ($H_1: \sigma_1^2 < \sigma_2^2$ or $H_1: \sigma_1^2 > \sigma_2^2$). The choice is determined by the study design before any data is analyzed. For example, if a modification to an existent medical treatment is proposed, we may only be interested in knowing if the new treatment is more consistent and less variable than the established medical intervention.
• Test Statistic: $F_o = {\sigma_1^2 \over \sigma_2^2}$
The farther away this ratio is from 1, the stronger the evidence for unequal population variances.
• Inference: Suppose we test at significance level α = 0.05. Then the hypothesis that the two standard deviations are equal is rejected if the test statistics is outside this interval
$H_1: \sigma_1^2 > \sigma_2^2$: If Fo > F(α,df1 = n1 − 1,df2 = n2 − 1)
$H_1: \sigma_1^2 < \sigma_2^2$: If Fo < F(1 − α,df1 = n1 − 1,df2 = n2 − 1)
$H_1: \sigma_1^2 \not= \sigma_2^2$: If either Fo < F(1 − α / 2,df1 = n1 − 1,df2 = n2 − 1) or Fo > F(α / 2,df1 = n1 − 1,df2 = n2 − 1),
where F(α,df1 = n1 − 1,df2 = n2 − 1) is the critical value of the F distribution with degrees of freedom for the numerator and denominator, df1 = n1 − 1,df2 = n2 − 1, respectively.
In the image below, the left and right critical regions are white with F(α,df1 = n1 − 1,df2 = n2 − 1) and F(1 − α,df1 = n1 − 1,df2 = n2 − 1) representing the lower and upper, respectively, critical values. In this example of F(df1 = 12,df2 = 15), the left and right critical values at α / 2 = 0.025 are F(α / 2 = 0.025,df1 = 9,df2 = 14) = 0.314744 and F(1 − α / 2 = 0.975,df1 = 9,df2 = 14) = 2.96327, respectively.
### Comparing Two Standard Deviations (σ1 = σ2?)
To make inference on whether the standard deviations of two populations are equal, we calculate the sample variances and apply the inference on the ratio of the sample variance using the F-test, as described above.
### Hands-on activities
• Formulate appropriate hypotheses and assess the significance of the evidence to reject the null hypothesis that the variances of the two populations, where the following data come from, are distinct. Assume the observations below represent random samples (of sizes 6 and 10) from two Normally distributed populations of liquid content (in fluid ounces) of beverage cans. Use (α = 0.1).
Sample from Population 1 14.816 14.863 14.814 14.998 14.965 14.824 Sample from Population 2 14.884 14.838 14.916 15.021 14.874 14.856 14.860 14.772 14.980 14.919
• Hypotheses: Ho1 = σ2 vs. $H_1: \sigma_1 \not= \sigma_2$ .
• Get the sample statistics from SOCR Charts (e.g., Index Plot);
Sample Mean Sample SD Sample Variance Sample 1 14.88 0.081272382 0.0066052 Sample 2 14.892 0.071269442 0.005079333
• Identify the degrees of freedom (df1 = 6 − 1 = 5 and df2 = 10 − 1 = 9).
• Test Statistics: $F_o = {\sigma_1^2 \over \sigma_2^2}=1.300406878$
• Significance Inference: P-value=$P(F_{(df_1=5, df_2=9)} > F_o) = 0.328147$. This p-value does not indicate strong evidence in the data to reject the null hypothesis. That is, the data does not have power to discriminate between the population variances of the two populations based on these (small) samples.
### More examples
• Use the hot-dogs dataset to formulate and test hypotheses about the difference of the population standard deviations of sodium between the poultry and the meet based hot-dogs. Repeat this with variances of calories between the beef and meet based hot-dogs. |
# When is a polynomial ring a field?
19. December 2020 ‐ 2 min read
Consider the question
Is $A = \mathbb{Z}_{5}[x]_{x^2+4}$ a field? Prove your answer.
What do we do? First, let us try to understand what exactly we have here.
So elements of this ring would be $x+3$, $1$ or $4x + 1$. Now, we can use Theorem 5.36, which tells us that the ring $F[x]_{m(x)}$ is a field if and only if $m(x)$ is irreducible. This in general isn’t that easy - if the degree of $m(x)$ is high, this would take quite some manual work (namely using long division to test all possible polynomials with $\le \text{degree}/2$).
Luckily, we are usually asked to show this with $m(x)$ having degree 2 or 3. In this case, we can use Corollary 5.29 (A polynomial $m(x)$ of degree 2 or 3 over a field $F$ is irreducible if and only if it has no root). Hence, we can plug in all possible $a \in F$ into $m(x)$. If $m(a)$ then evaluates to $0$, that means we have found a root, $m(x)$ is not irreducible and $F[x]_{m(x)}$ is not a field. Otherwise, we can conclude that $F[x]_{m(x)}$ indeed is a field.
Let’s now apply this to our initial exercise: We insert all possible values $a \in \mathbb{Z}_5$ into $x^{2} + 4$ and see if this term evaluates to $0$.
$$m(0) = 0^2 + 4 = 4 \newline m(1) = 1^2 + 4 = \textcolor{red}{0} \newline m(2) = 2^2 + 4 = 3 \newline m(3) = 3^2 + 4 = 3 \newline m(4) = 4^2 + 4 = \textcolor{red}{0}$$
(Of course you can also stop as soon as you have found a $0$.)
Clearly, $x^2 + 4$ over $\mathbb{Z}_{5}[x]_{x^2+4}$ does have roots and is therefore not irreducible. We can see that $x^2 + 4 = (x - 1)(x - 4) = (x+4)(x+1)$ by using Lemma 5.28. Finally, we can conclude that $\mathbb{Z}_5[x]_{x^2 + 4}$ is not a field.
For practice, you could now try to prove or disprove that $\mathbb{Z}_5[x]_{x^2 + 2}$ is a field. |
# 2.2 Graphs of linear functions (Page 3/15)
Page 3 / 15
Find a point on the graph we drew in [link] that has a negative x -value.
Possible answers include $\left(-3,7\right),$ $\left(-6,9\right),$ or $\left(-9,11\right).$
## Graphing a function using transformations
Another option for graphing is to use transformations of the identity function $f\left(x\right)=x$ . A function may be transformed by a shift up, down, left, or right. A function may also be transformed using a reflection, stretch, or compression.
## Vertical stretch or compression
In the equation $f\left(x\right)=mx,$ the $m$ is acting as the vertical stretch or compression of the identity function. When $m$ is negative, there is also a vertical reflection of the graph. Notice in [link] that multiplying the equation of $f\left(x\right)=x$ by $m$ stretches the graph of $f$ by a factor of $m$ units if $m>\text{1}$ and compresses the graph of $f$ by a factor of $m$ units if $0 This means the larger the absolute value of $m,$ the steeper the slope.
## Vertical shift
In $f\left(x\right)=mx+b,$ the $b$ acts as the vertical shift , moving the graph up and down without affecting the slope of the line. Notice in [link] that adding a value of $b$ to the equation of $f\left(x\right)=x$ shifts the graph of $f$ a total of $b$ units up if $b$ is positive and $|b|$ units down if $b$ is negative.
Using vertical stretches or compressions along with vertical shifts is another way to look at identifying different types of linear functions. Although this may not be the easiest way to graph this type of function, it is still important to practice each method.
Given the equation of a linear function, use transformations to graph the linear function in the form $f\left(x\right)=mx+b.$
1. Graph $f\left(x\right)=x.$
2. Vertically stretch or compress the graph by a factor $m.$
3. Shift the graph up or down $b$ units.
## Graphing by using transformations
Graph $f\left(x\right)=\frac{1}{2}x-3$ using transformations.
The equation for the function shows that $m=\frac{1}{2}$ so the identity function is vertically compressed by $\frac{1}{2}.$ The equation for the function also shows that $b=-3$ so the identity function is vertically shifted down 3 units. First, graph the identity function, and show the vertical compression as in [link] .
Then show the vertical shift as in [link] .
Graph $f\left(x\right)=4+2x,$ using transformations.
In [link] , could we have sketched the graph by reversing the order of the transformations?
No. The order of the transformations follows the order of operations. When the function is evaluated at a given input, the corresponding output is calculated by following the order of operations. This is why we performed the compression first. For example, following the order: Let the input be 2.
$\begin{array}{l}f\text{(2)}=\frac{\text{1}}{\text{2}}\text{(2)}-\text{3}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{1}-\text{3}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-\text{2}\hfill \end{array}$
## Writing the equation for a function from the graph of a line
Recall that in Linear Functions , we wrote the equation for a linear function from a graph. Now we can extend what we know about graphing linear functions to analyze graphs a little more closely. Begin by taking a look at [link] . We can see right away that the graph crosses the y -axis at the point so this is the y -intercept.
foci (–7,–17) and (–7,17), the absolute value of the differenceof the distances of any point from the foci is 24.
difference between calculus and pre calculus?
give me an example of a problem so that I can practice answering
x³+y³+z³=42
Robert
dont forget the cube in each variable ;)
Robert
of she solves that, well ... then she has a lot of computational force under her command ....
Walter
what is a function?
I want to learn about the law of exponent
explain this
what is functions?
A mathematical relation such that every input has only one out.
Spiro
yes..it is a relationo of orders pairs of sets one or more input that leads to a exactly one output.
Mubita
Is a rule that assigns to each element X in a set A exactly one element, called F(x), in a set B.
RichieRich
If the plane intersects the cone (either above or below) horizontally, what figure will be created?
can you not take the square root of a negative number
No because a negative times a negative is a positive. No matter what you do you can never multiply the same number by itself and end with a negative
lurverkitten
Actually you can. you get what's called an Imaginary number denoted by i which is represented on the complex plane. The reply above would be correct if we were still confined to the "real" number line.
Liam
Suppose P= {-3,1,3} Q={-3,-2-1} and R= {-2,2,3}.what is the intersection
can I get some pretty basic questions
In what way does set notation relate to function notation
Ama
is precalculus needed to take caculus
It depends on what you already know. Just test yourself with some precalculus questions. If you find them easy, you're good to go.
Spiro
the solution doesn't seem right for this problem
what is the domain of f(x)=x-4/x^2-2x-15 then
x is different from -5&3
Seid
All real x except 5 and - 3
Spiro
***youtu.be/ESxOXfh2Poc
Loree
how to prroved cos⁴x-sin⁴x= cos²x-sin²x are equal
Don't think that you can.
Elliott
By using some imaginary no.
Tanmay |
# 8.7 Parametric equations: graphs
Page 1 / 4
In this section you will:
• Graph plane curves described by parametric equations by plotting points.
• Graph parametric equations.
It is the bottom of the ninth inning, with two outs and two men on base. The home team is losing by two runs. The batter swings and hits the baseball at 140 feet per second and at an angle of approximately $\text{\hspace{0.17em}}45°\text{\hspace{0.17em}}$ to the horizontal. How far will the ball travel? Will it clear the fence for a game-winning home run? The outcome may depend partly on other factors (for example, the wind), but mathematicians can model the path of a projectile and predict approximately how far it will travel using parametric equations . In this section, we’ll discuss parametric equations and some common applications, such as projectile motion problems.
## Graphing parametric equations by plotting points
In lieu of a graphing calculator or a computer graphing program, plotting points to represent the graph of an equation is the standard method. As long as we are careful in calculating the values, point-plotting is highly dependable.
Given a pair of parametric equations, sketch a graph by plotting points.
1. Construct a table with three columns: $\text{\hspace{0.17em}}t,x\left(t\right),\text{and}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y\left(t\right).$
2. Evaluate $x$ and $y$ for values of $t$ over the interval for which the functions are defined.
3. Plot the resulting pairs $\text{\hspace{0.17em}}\left(x,y\right).$
## Sketching the graph of a pair of parametric equations by plotting points
Sketch the graph of the parametric equations $x\left(t\right)={t}^{2}+1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}y\left(t\right)=2+t.$
Construct a table of values for $\text{\hspace{0.17em}}t,x\left(t\right),\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\left(t\right),\text{\hspace{0.17em}}$ as in [link] , and plot the points in a plane.
$t$ $x\left(t\right)={t}^{2}+1$ $y\left(t\right)=2+t$
$-5$ $26$ $-3$
$-4$ $17$ $-2$
$-3$ $10$ $-1$
$-2$ $5$ $0$
$-1$ $2$ $1$
$0$ $1$ $2$
$1$ $2$ $3$
$2$ $5$ $4$
$3$ $10$ $5$
$4$ $17$ $6$
$5$ $26$ $7$
The graph is a parabola with vertex at the point $\text{\hspace{0.17em}}\left(1,2\right),$ opening to the right. See [link] .
Sketch the graph of the parametric equations $\text{\hspace{0.17em}}x=\sqrt{t},\text{\hspace{0.17em}}\text{\hspace{0.17em}}y=2t+3,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\le t\le 3.$
## Sketching the graph of trigonometric parametric equations
Construct a table of values for the given parametric equations and sketch the graph:
$\begin{array}{l}\\ \begin{array}{l}x=2\mathrm{cos}\text{\hspace{0.17em}}t\hfill \\ y=4\mathrm{sin}\text{\hspace{0.17em}}t\hfill \end{array}\end{array}$
Construct a table like that in [link] using angle measure in radians as inputs for $\text{\hspace{0.17em}}t,\text{\hspace{0.17em}}$ and evaluating $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y.\text{\hspace{0.17em}}$ Using angles with known sine and cosine values for $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ makes calculations easier.
$t$ $x=2\mathrm{cos}\text{\hspace{0.17em}}t$ $y=4\mathrm{sin}\text{\hspace{0.17em}}t$
0 $x=2\mathrm{cos}\left(0\right)=2$ $y=4\mathrm{sin}\left(0\right)=0$
$\frac{\pi }{6}$ $x=2\mathrm{cos}\left(\frac{\pi }{6}\right)=\sqrt{3}$ $y=4\mathrm{sin}\left(\frac{\pi }{6}\right)=2$
$\frac{\pi }{3}$ $x=2\mathrm{cos}\left(\frac{\pi }{3}\right)=1$ $y=4\mathrm{sin}\left(\frac{\pi }{3}\right)=2\sqrt{3}$
$\frac{\pi }{2}$ $x=2\mathrm{cos}\left(\frac{\pi }{2}\right)=0$ $y=4\mathrm{sin}\left(\frac{\pi }{2}\right)=4$
$\frac{2\pi }{3}$ $x=2\mathrm{cos}\left(\frac{2\pi }{3}\right)=-1$ $y=4\mathrm{sin}\left(\frac{2\pi }{3}\right)=2\sqrt{3}$
$\frac{5\pi }{6}$ $x=2\mathrm{cos}\left(\frac{5\pi }{6}\right)=-\sqrt{3}$ $y=4\mathrm{sin}\left(\frac{5\pi }{6}\right)=2$
$\pi$ $x=2\mathrm{cos}\left(\pi \right)=-2$ $y=4\mathrm{sin}\left(\pi \right)=0$
$\frac{7\pi }{6}$ $x=2\mathrm{cos}\left(\frac{7\pi }{6}\right)=-\sqrt{3}$ $y=4\mathrm{sin}\left(\frac{7\pi }{6}\right)=-2$
$\frac{4\pi }{3}$ $x=2\mathrm{cos}\left(\frac{4\pi }{3}\right)=-1$ $y=4\mathrm{sin}\left(\frac{4\pi }{3}\right)=-2\sqrt{3}$
$\frac{3\pi }{2}$ $x=2\mathrm{cos}\left(\frac{3\pi }{2}\right)=0$ $y=4\mathrm{sin}\left(\frac{3\pi }{2}\right)=-4$
$\frac{5\pi }{3}$ $x=2\mathrm{cos}\left(\frac{5\pi }{3}\right)=1$ $y=4\mathrm{sin}\left(\frac{5\pi }{3}\right)=-2\sqrt{3}$
$\frac{11\pi }{6}$ $x=2\mathrm{cos}\left(\frac{11\pi }{6}\right)=\sqrt{3}$ $y=4\mathrm{sin}\left(\frac{11\pi }{6}\right)=-2$
$2\pi$ $x=2\mathrm{cos}\left(2\pi \right)=2$ $y=4\mathrm{sin}\left(2\pi \right)=0$
[link] shows the graph.
By the symmetry shown in the values of $x$ and $\text{\hspace{0.17em}}y,\text{\hspace{0.17em}}$ we see that the parametric equations represent an ellipse . The ellipse is mapped in a counterclockwise direction as shown by the arrows indicating increasing $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ values.
Graph the parametric equations: $\text{\hspace{0.17em}}x=5\mathrm{cos}\text{\hspace{0.17em}}t,\text{\hspace{0.17em}}\text{\hspace{0.17em}}y=3\mathrm{sin}\text{\hspace{0.17em}}t.$
## Graphing parametric equations and rectangular form together
Graph the parametric equations $\text{\hspace{0.17em}}x=5\mathrm{cos}\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y=2\mathrm{sin}\text{\hspace{0.17em}}t.\text{\hspace{0.17em}}$ First, construct the graph using data points generated from the parametric form . Then graph the rectangular form of the equation. Compare the two graphs.
Construct a table of values like that in [link] .
$t$ $x=5\mathrm{cos}\text{\hspace{0.17em}}t$ $y=2\mathrm{sin}\text{\hspace{0.17em}}t$
$\text{0}$ $x=5\mathrm{cos}\left(0\right)=5$ $y=2\mathrm{sin}\left(0\right)=0$
$\text{1}$ $x=5\mathrm{cos}\left(1\right)\approx 2.7$ $y=2\mathrm{sin}\left(1\right)\approx 1.7$
$\text{2}$ $x=5\mathrm{cos}\left(2\right)\approx -2.1$ $y=2\mathrm{sin}\left(2\right)\approx 1.8$
$\text{3}$ $x=5\mathrm{cos}\left(3\right)\approx -4.95$ $y=2\mathrm{sin}\left(3\right)\approx 0.28$
$\text{4}$ $x=5\mathrm{cos}\left(4\right)\approx -3.3$ $y=2\mathrm{sin}\left(4\right)\approx -1.5$
$\text{5}$ $x=5\mathrm{cos}\left(5\right)\approx 1.4$ $y=2\mathrm{sin}\left(5\right)\approx -1.9$
$-1$ $x=5\mathrm{cos}\left(-1\right)\approx 2.7$ $y=2\mathrm{sin}\left(-1\right)\approx -1.7$
$-2$ $x=5\mathrm{cos}\left(-2\right)\approx -2.1$ $y=2\mathrm{sin}\left(-2\right)\approx -1.8$
$-3$ $x=5\mathrm{cos}\left(-3\right)\approx -4.95$ $y=2\mathrm{sin}\left(-3\right)\approx -0.28$
$-4$ $x=5\mathrm{cos}\left(-4\right)\approx -3.3$ $y=2\mathrm{sin}\left(-4\right)\approx 1.5$
$-5$ $x=5\mathrm{cos}\left(-5\right)\approx 1.4$ $y=2\mathrm{sin}\left(-5\right)\approx 1.9$
Plot the $\text{\hspace{0.17em}}\left(x,y\right)\text{\hspace{0.17em}}$ values from the table. See [link] .
Next, translate the parametric equations to rectangular form. To do this, we solve for $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ in either $\text{\hspace{0.17em}}x\left(t\right)\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}y\left(t\right),\text{\hspace{0.17em}}$ and then substitute the expression for $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ in the other equation. The result will be a function $y\left(x\right)$ if solving for $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ as a function of $\text{\hspace{0.17em}}x,\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}x\left(y\right)$ if solving for $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ as a function of $\text{\hspace{0.17em}}y.\text{\hspace{0.17em}}$
Then, use the Pythagorean Theorem .
$\begin{array}{r}\hfill {\mathrm{cos}}^{2}t+{\mathrm{sin}}^{2}t=1\\ \hfill {\left(\frac{x}{5}\right)}^{2}+{\left(\frac{y}{2}\right)}^{2}=1\\ \hfill \frac{{x}^{2}}{25}+\frac{{y}^{2}}{4}=1\end{array}$
#### Questions & Answers
For each year t, the population of a forest of trees is represented by the function A(t) = 117(1.029)t. In a neighboring forest, the population of the same type of tree is represented by the function B(t) = 86(1.025)t.
by how many trees did forest "A" have a greater number?
Shakeena
32.243
Kenard
how solve standard form of polar
what is a complex number used for?
It's just like any other number. The important thing to know is that they exist and can be used in computations like any number.
Steve
I would like to add that they are used in AC signal analysis for one thing
Scott
Good call Scott. Also radar signals I believe.
Steve
Is there any rule we can use to get the nth term ?
how do you get the (1.4427)^t in the carp problem?
A hedge is contrusted to be in the shape of hyperbola near a fountain at the center of yard.the hedge will follow the asymptotes y=x and y=-x and closest distance near the distance to the centre fountain at 5 yards find the eqution of the hyperbola
A doctor prescribes 125 milligrams of a therapeutic drug that decays by about 30% each hour. To the nearest hour, what is the half-life of the drug?
Find the domain of the function in interval or inequality notation f(x)=4-9x+3x^2
hello
Outside temperatures over the course of a day can be modeled as a sinusoidal function. Suppose the high temperature of ?105°F??105°F? occurs at 5PM and the average temperature for the day is ?85°F.??85°F.? Find the temperature, to the nearest degree, at 9AM.
if you have the amplitude and the period and the phase shift ho would you know where to start and where to end?
rotation by 80 of (x^2/9)-(y^2/16)=1
thanks the domain is good but a i would like to get some other examples of how to find the range of a function
what is the standard form if the focus is at (0,2) ?
a²=4 |
# How do I graph the equation 16x^2-9y^2+32x+18y-137=0 on a TI-83?
Sep 6, 2014
First you would have to complete the square on both variables, x and y:
16x^2−9y^2+32x+18y−137=0
Group together the x terms and y terms, and move the constant to the other side.
$16 \left({x}^{2} + 2 x\right) - 9 \left({y}^{2} - 2 y\right) = 137$
Now, complete the squares. Half of the linear coefficient, square it, add inside the parentheses on the left and multiply by the factored out number before adding to the right.
$16 \left({x}^{2} + 2 x + 1\right) - 9 \left({y}^{2} - 2 y + 1\right) = 137 + 16 - 9$
Completed squares are now factorable:
$16 {\left(x + 1\right)}^{2} - 9 {\left(y - 1\right)}^{2} = 144$
I can already tell by the form of the equation, that the graph will be a hyperbola. Now, when you solve for y = , there will be two parts to graph.
$- 9 {\left(y - 1\right)}^{2} = 144 - 16 {\left(x + 1\right)}^{2}$
Divide both sides by 9, and take the square root:
$\sqrt{{\left(y - 1\right)}^{2}} = \sqrt{\frac{144 - 16 {\left(x + 1\right)}^{2}}{9}}$
So y - 1 = $\pm \sqrt{\frac{144 - 16 {\left(x + 1\right)}^{2}}{9}}$
And last, add 1: y = $1 \pm \sqrt{\frac{144 - 16 {\left(x + 1\right)}^{2}}{9}}$
You will have to type in two separate equations into your calculator! (TI-83 or 84 models)
y1 = the part with the "+" sign
y2 = the part with the "-" sign
In other online applications or in the TI-nspire, you do not have to solve for y = any more. After this step: $16 {\left(x + 1\right)}^{2} - 9 {\left(y - 1\right)}^{2} = 144$
just divide both sides by 144 and simplify:
$\frac{{\left(x + 1\right)}^{2}}{9} - \frac{{\left(y - 1\right)}^{2}}{16} = 1$
Here is what the TI-nspire looks like now: |
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# Mathematrix: Aufgabenbeispiele/ Bruchrechnungen und Vorrang
${\displaystyle {\frac {2}{3}}+\left({\frac {6}{7}}-{\frac {9}{7}}\right)\cdot \left({\frac {2}{5}}-{\frac {16}{21}}:{\frac {4}{7}}\right)}$
• Erste Klammer
${\displaystyle {\frac {6}{7}}-{\frac {9}{7}}={\frac {6-9}{7}}=-{\frac {3}{7}}}$
• Zweite Klammer
${\displaystyle {\frac {2}{5}}-{\frac {16}{21}}:{\frac {4}{7}}}$
${\displaystyle {\frac {2}{5}}-{\frac {16}{21}}:{\frac {4}{7}}={\frac {2}{5}}-{\frac {16}{21}}\cdot {\frac {7}{4}}={\frac {2}{5}}-{\frac {16\cdot 7}{21\cdot 4}}}$
${\displaystyle {\frac {2}{5}}-{\frac {{\cancelto {4}{16}}\cdot 7}{21\cdot {\cancelto {1}{4}}}}={\frac {2}{5}}-{\frac {4\cdot {\cancelto {1}{7}}}{{\cancelto {3}{21}}\cdot 1}}={\frac {2}{5}}-{\frac {4}{3}}}$
${\displaystyle \textstyle \left(^{\text{Vorgang: }}{\frac {2}{5}}{\underline {\xcancel {-}}}{\frac {4}{3}}\right)\quad }$${\displaystyle ={\frac {2\cdot 3-4\cdot 5}{5\cdot 3}}={\frac {6-20}{15}}=-{\frac {14}{15}}}$
• Jetzt Klammer ersetzten
${\displaystyle {\frac {2}{3}}+{\cancelto {-{\frac {3}{7}}}{\left({\frac {6}{7}}-{\frac {9}{7}}\right)}}\cdot {\cancelto {-{\frac {14}{15}}}{\left({\frac {2}{5}}-{\frac {16}{21}}:{\frac {4}{7}}\right)}}=}$
${\displaystyle {\frac {2}{3}}+\left(-{\frac {3}{7}}\right)\cdot \left(-{\frac {14}{15}}\right)}$
${\displaystyle {\frac {2}{3}}+{\frac {{\cancelto {1}{3}}\cdot {\cancelto {2}{14}}}{{\cancelto {1}{7}}\cdot {\cancelto {5}{15}}}}}$${\displaystyle ^{\text{(hier erst kürzen)}}={\frac {2}{3}}+{\frac {1\cdot 2}{1\cdot 5}}={\frac {2}{3}}+{\frac {2}{5}}=}$
${\displaystyle {\frac {2\cdot 5+2\cdot 3}{3\cdot 5}}={\frac {16}{15}}}$ |
Lesson Objectives
• Learn the definition of a Natural Number
• Learn the definition of a Whole Number
• Learn the definition of an Integer
• Learn the definition of a Rational Number
• Learn the definition of an Irrational Number
• Demonstrate the ability to classify any real number
## Classifying Real Numbers
### How can we classify a Real Number?
#### Natural Numbers
When we first start to work with numbers, we generally count on our fingers or count any number of physical objects placed in front of us. To perform these actions, we involve a set of numbers known as the counting numbers or the natural numbers. The natural numbers are used to count a number of objects. This set begins with the number one and increases in increments of one out to positive infinity.
Natural Numbers: {1, 2, 3, 4,...}
Note: The three dots "..." after the 4 in our set are used to indicate that the pattern continues forever. After 4 comes 5, then 6, so on and so forth...
We can also use a number line to visually display our set of natural numbers. The leftmost number is a 1 and then each additional notch on the number line is shown as an increase of 1 or the next natural number. Since we can't list all of the numbers, we draw an arrow to the right to indicate the natural numbers continue out to positive infinity.
#### Whole Numbers
As we move on, we begin to consider the possibility of not having anything to count. This is represented with the number 0. When 0 is added to the set of natural numbers, we end up with the whole numbers. Whole Numbers: {0, 1, 2, 3, 4,...}
#### Integers
In some cases, it is not enough to have only the whole numbers. What if we need to describe a temperature that is less than zero, an altitude that is below sea level, or a negative bank balance? To deal with these situations, we can introduce the integers. These are the whole numbers and their opposites or negatives. To form this set, we just take the whole numbers and then working to the left of zero, we start counting from 1 again, only this time we place a negative symbol in front of the number. Integers: {...,-4, -3, -2, -1, 0, 1, 2, 3, 4,...}
#### Rational Numbers
Additionally, we may have a situation in which we don't have a whole amount. Suppose we eat 3 slices of pizza out of a total of 8 slices that are available. Currently, we have more pizza than zero and less than one whole pizza. We can use a fraction such as 5/8 to describe the amount of pizza that is remaining or another fraction 3/8 to describe the amount of pizza that was eaten. This type of number is known as a rational number. A rational number can be written as the quotient of two integers with a nonzero divisor (denominator). In decimal form, a rational number will always terminate or repeat the same pattern of digits forever. For example, in decimal form: 9/11 is 0.818181..., where the 81 repeats forever. $$\frac{9}{11}=0. \overline{81}$$ Note the bar over the digits 81, indicate the pattern continues forever... Rational Numbers: {p/q | p and q are integers and q ≠ 0}
#### Irrational Numbers
Lastly, we have the irrational numbers. These are real numbers that don't fall in the category of being a rational number. This means we can't write an irrational number as the quotient of two integers. When we see an irrational number in decimal form, the number will not terminate nor repeat the same pattern forever. The most famous example of an irrational number is pi, which is the ratio of a circle's circumference to its diameter. When we need to work with pi, we generally give an approximation of 3.14159, but after the 9 comes an infinite number of digits.
Irrational Numbers: {x | x is a real number but is not rational}
Together, the rational numbers and irrational numbers make up all of the real numbers. Therefore, if we are dealing with a real number, it must be either rational or irrational. As we work our way up the chain of rational numbers, we can see that a number such as 5, could be classified as a natural number, a whole number, an integer, or a rational number (since 5/1 = 5). So all natural numbers are whole numbers, all whole numbers are integers, and all integers are rational numbers. But reversing this isn't always going to work. For example: -5 is a rational number and an integer, but it's not a whole number or a natural number. 3/4 is a rational number, but not an integer, whole number, or natural number.
#### Skills Check:
Example #1
Which Numbers are Whole Numbers?
A
$$13, 17, -51$$
B
$$3, \sqrt{5}, -2$$
C
$$\frac{1}{4}, 0, -\sqrt{2}$$
D
$$-2, -9, 11$$
E
$$6, \frac{18}{1}, 0$$
Example #2
Which Numbers are Integers?
A
$$55, -3, -\frac{1}{2}$$
B
$$-18, 7, \sqrt{13}$$
C
$$\frac{1}{3}, -22, 18$$
D
$$-44, -\frac{9}{1}, -88$$
E
$$-\frac{6}{11}, \frac{13}{1}, 0$$
Example #3
How can we Classify the Number: $$\sqrt{3}$$
A
Rational
B
Whole Number
C
Integer
D
Irrational
E
Natural Number |
## Characteristic Vectors of Matrix 4
In this page characteristic vectors of matrix 4 we are going to see how to find characteristic equation of any matrix with detailed example.
Definition :
The eigen vector can be obtained from (A- λI)X = 0. Here A is the given matrix λ is a scalar,I is the unit matrix and X is the columns matrix formed by the variables a,b and c.
Another name of characteristic Vector:
Characteristic vector are also known as latent vectors or Eigen vectors of a matrix.
Question 4 :
Determine the characteristic vector of the matrix
4 -20 -10 -2 10 4 6 -30 -13
To find eigen vector first let us find characteristic roots of the given matrix.
Let A =
4 -20 -10 -2 10 4 6 -30 -13
The order of A is 3 x 3. So the unit matrix I =
1 0 0 0 1 0 0 0 1
Now we have to multiply λ with unit matrix I.
λI =
λ 0 0 0 λ 0 0 0 λ
A-λI=
4 -20 -10 -2 10 4 6 -30 -13
-
λ 0 0 0 λ 0 0 0 λ
=
(-4-λ) (-20-0) (-10-0) (-2-0) (10-λ) (4-0) (6-0) (-30-0) (-13-λ)
=
(-4-λ) -20 -10 -2 (10-λ) 4 6 -30 (-13-λ)
= (4-λ)[(10-λ)(-13- λ)+120]+
20[-2(-13-λ)-24]-10[60-6(10-λ)]
= (4-λ)[-130-10 λ+13λ+λ²+120]+20[26+2λ-24]-10[60-60+6λ]
= (4-λ)[-10+3λ+λ²]+20[2+2λ]-10[6λ]
= (4-λ)[λ²+3λ-10]+20[2+2λ]-10[6λ]
= 4λ²+12λ-40-λ³-3λ²+10λ+40λ+40-60λ
= -λ³ + 1λ² + 2λ
To find roots let |A-λI| = 0
-λ³ + 1λ² + 2λ = 0
For solving this equation -λ from all the terms
-λ (λ² - 1λ - 2) = 0
-λ = 0 (or) λ² - 1 λ - 2 = 0
λ = 0 (λ+1) (λ-2) = 0
λ + 1 = 0 λ - 2 = 0
λ = - 1 λ = 2
Therefore the characteristic roots (or) Eigen values are x = 0,-1,2
Substitute λ = 0 in the matrix A - λI
=
-4 -20 -10 -2 10 4 6 -30 -13
From this matrix we are going to form three linear equations using variables x,y and z.
4x - 20y - 10z = 0 ------ (1)
-2x + 10y + 4z = 0 ------ (2)
6x - 30y - 13z = 0 ------ (3)
By solving (1) and (2) we get the eigen vector
The eigen vector x =
5 1 0
Substitute λ = -1 in the matrix A - λI
=
5 -20 -10 -2 10 4 6 -30 -12
From this matrix we are going to form three linear equations using variables x,y and z.
5x - 20y - 10z = 0 ------ (4)
-2x + 10y + 4z = 0 ------ (5)
6x - 30y - 12z = 0 ------ (6)
By solving (4) and (5) we get the eigen vector characteristic vectors of matrix4
The eigen vector y =
2 0 1
Substitute λ = 2 in the matrix A - λI
=
2 -20 -10 -2 8 4 6 -30 -15
From this matrix we are going to form three linear equations using variables x,y and z.
2x - 20y - 10z = 0 ------ (7)
-2x + 8y + 4z = 0 ------ (8)
6x - 30y - 15z = 0 ------ (9)
By solving (7) and (8) we get the eigen vector characteristic vectors of matrix 4 characteristic vectors of matrix 4 characteristic vectors of matrix 4 characteristic vectors of matrix 4
The eigen vector z =
0 1 -2
Questions Solution
Question 1 :
Determine the characteristic vector of the matrix
5 0 1 0 -2 0 1 0 5
Question 2 :
Determine the characteristic vector of the matrix
1 1 3 1 5 1 3 1 1
Question 3 :
Determine the characteristic vector of the matrix
-2 2 -3 2 1 -6 -1 -2 0
Question 5 :
Determine the characteristic vector of the matrix
11 -4 -7 7 -2 -5 10 -4 -6
Characteristic Vectors of Matrix4 to Characteristic Equation
1. Click on the HTML link code below.
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# IGCSE Mathematics Paper-1: Specimen Questions with Answers 13 - 15 of 324
## Question 13
### Describe in Detail
Essay▾
Solve with detailed working:
(a)
(b)
### Explanation
Here,
(a) Left Hand Side (LHS) =
Right Hand Side (RHS) =
Thus, LHS = RHS.
(b) Left Hand Side (LHS) =
Right Hand Side (RHS) =
Thus, LHS = RHS.
## Question 14
### Describe in Detail
Essay▾
Aisha bought 30 meters of cloth at a cost of . She sold 25 meters of the cloth at per meter and 5 meters at per meter.
(a) Calculate the profit she made.
(b) Calculate this profit as a percentage of the original cost.
### Explanation
Here, Aisha bought 30 meters of cloth at a cost of .
So, cost price CP =
Now, she sold 25 meters of the cloth at per meter:
She sold 5 meters at per meter:
So, total selling prices SP =
(a)
(b) Percentage of profit of the original cost :
So,
Then,
## Question 15
Edit
### Describe in Detail
Essay▾
Solve 0.4928
(a) In standard form,
(b) Correct to 2 significant figures,
(c) Correct to 3 decimal places
### Explanation
Here, number is 0.4928
(a) For making standard form of 0.4928 we need to shift the decimal point left to right one times which will give: .
(b) We need to correct 0.4928 to 2 significant figures (all non-zero digits are called as significant figures) which can be: because here after digit 9 there is 2 which is not greater than 5, so 1 will remain same.
(C) We need to correct 0.4928 to 3 decimal places which can be: in which we have to consider three numbers after decimal point and the second number will be round up on the basis of third number.
Here, it means if third number is 5 or more than 5 which is 8 here so second number will get roundup and be 3 from 2. |
Notes On Application of Algebraic Expressions - CBSE Class 7 Maths
Algebraic expressions can be used to represent number patterns. Ex: Table showing the relation between the number of cones and the number of ice-cream scoops. Number of cones(n) Number of ice-cream scoops (2n) 1 2 2 4 3 6 8 16 15 30 Thus, we can find the value of an algebraic expression if the values of all the variables in the expression are known. e.g. Find the value of the expression 3x2y – 2xy2 + 2xy for x = 2 and y = –2. Sol: 3x2y – 2xy2 + 2xy . Putting x = 2 and y = –2 in the given expression, 3x2y – 2xy2 + 2xy = 3×(2)2×(–2) – 2×(2)×(–2)2 + 2×(2)×(–2) = 3×4×(–2) – 4×4 + 4×(–2) = – 24 –16 – 8 = – 48. Formulas and rules such as the perimeter and area for different geometrical figures are written in a concise and general form using simple, and easy-to-remember algebraic expressions. If 's' represents the side of a square, then its perimeter is '4s' and area is 's2'. If 'l' represents the length and 'b' represents the breadth of a rectangle, then its perimeter is '2(l + b)' and area is 'l × b'. Area of a triangle with base 'b' and the corresponding altitude 'h' is '$\frac{\text{1}}{\text{2}}$ × base × height'. Perimeter of an equilateral triangle with the length of the side as 'a' units is '3a'.
#### Summary
Algebraic expressions can be used to represent number patterns. Ex: Table showing the relation between the number of cones and the number of ice-cream scoops. Number of cones(n) Number of ice-cream scoops (2n) 1 2 2 4 3 6 8 16 15 30 Thus, we can find the value of an algebraic expression if the values of all the variables in the expression are known. e.g. Find the value of the expression 3x2y – 2xy2 + 2xy for x = 2 and y = –2. Sol: 3x2y – 2xy2 + 2xy . Putting x = 2 and y = –2 in the given expression, 3x2y – 2xy2 + 2xy = 3×(2)2×(–2) – 2×(2)×(–2)2 + 2×(2)×(–2) = 3×4×(–2) – 4×4 + 4×(–2) = – 24 –16 – 8 = – 48. Formulas and rules such as the perimeter and area for different geometrical figures are written in a concise and general form using simple, and easy-to-remember algebraic expressions. If 's' represents the side of a square, then its perimeter is '4s' and area is 's2'. If 'l' represents the length and 'b' represents the breadth of a rectangle, then its perimeter is '2(l + b)' and area is 'l × b'. Area of a triangle with base 'b' and the corresponding altitude 'h' is '$\frac{\text{1}}{\text{2}}$ × base × height'. Perimeter of an equilateral triangle with the length of the side as 'a' units is '3a'.
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Important Questions CBSE Class 9 Maths Chapter 15 Probability - GMS - Learning Simply
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# Important Questions CBSE Class 9 Maths Chapter 15 Probability
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Important questions for CBSE Class 9 Maths Chapter 15 Probability are available with solutions here for the final exam. These questions are in accordance with the NCERT curriculum prescribed by the CBSE board for the current academic session. Practising these questions will help students to score good marks in this chapter. To acquire good result ,they can reach important questions of CBSE Class 9 Maths chapter 13 surface areas and volume here at Goyanka Maths Study.
During the exams, students get less time to revise the whole chapter. Therefore, they must revise at least important concepts before they appear for the exam. Hence, here we are providing the most significant problems and solutions for the upcoming examination, along with additional questions to give students a better practice.
## Important Questions & Solutions For CBSE Class 9 Maths Chapter 15
Q.1. Compute the probability of the occurrence of an event if the probability the event not occurring is 0.56.
Solution:
Given,
P(not E) = 0.56
We know that,
P(E) + P(not E) = 1
So, P(E) = 1 – P(not E)
P(E) = 1 – 0.56
Or, P(E) = 0.44
Q.2. In a factory of 364 workers, 91 are married. Find the probability of selecting a worker who is not married.
Solution:
Given,
Total workers (i.e. Sample space) = n(S) = 364
Total married workers = 91
Now, total workers who are not married = n(E) = 364 – 91 = 273
Method 1: So, P(not married) = n(E)/n(S) = 273/364 = 0.75
Method 2: P(married) + P(not married) = 1
Here, P(married) = 91/364 = 0.25
So, 0.25 + P(not married) = 1
P(not married) = 1 – 0.25 = 0.75
Q. 3. From a deck of cards, 10 cards are picked at random and shuffled. The cards are as follows:
6, 5, 3, 9, 7, 6, 4, 2, 8, 2
Find the probability of picking a card having value more than 5 and find the probability of picking a card with an even number on it.
Solution:
Total number of cards = 10
Total cards having value more than 5 = 5
i.e. {6, 9, 7, 6, 8}
Total cards having an even number = 6
i.e. {6, 6, 4, 2, 8, 2}
So, the probability of picking a card having value more than 5 = 5/10 = 0.5
And, the probability of picking a card with an even number on it = 6/10 = 0.6
Q.4. From a bag of red and blue balls, the probability of picking a red ball is x/2. Find “x” if the probability of picking a blue ball is ⅔.
Solution:
Here, there are only red and blue balls.
P(picking a red ball) + P(picking a blue ball) = 1
x/2 + ⅔ = 1
=> 3x + 4 = 6
=> 3x = 2
Or, x = ⅔
Q.5. Two coins are tossed simultaneously for 360 times. The number of times ‘2 Tails’ appeared was three times ‘No Tail’ appeared and the number of times ‘1 tail’ appeared is double the number of times ‘No Tail’ appeared. What is the probability of getting ‘Two tails’.
Solution:
Given,
Total number of outcomes = Sample space = 360
Now, assume that the number of times ‘No Tail’ appeared to be “x”
So, the number of times ‘2 Tails’ appeared = 3x (from the question)
Also, the number of times ‘1 Tail’ appeared =2x (from the question)
As the total outcomes = 360,
x + 2x + 3x = 360
=> 6x = 360
Or, x = 60
∴ P(getting two tails) = (3 × 60)/360 = ½
Q.6: 1500 families with 2 children were selected randomly, and the following data were recorded:
Number of girls in a family 2 1 0 Number of families 475 814 211
Compute the probability of a family, chosen at random, having
(i) 2 girls (ii) 1 girl (iii) No girl
Also, check whether the sum of these probabilities is 1.
Solution:
Total numbers of families = 1500
(i) Numbers of families having 2 girls = 475
Probability = Numbers of families having 2 girls/Total numbers of families
P = 475/1500
P = 19/60
(ii) Numbers of families having 1 girls = 814
Probability = Numbers of families having 1 girls/Total numbers of families
P = 814/1500
P = 407/750
(iii)Numbers of families having no girls = 211
Probability = Numbers of families having 0 girls/Total numbers of families
= 211/1500
Sum of the probability = (19/60)+(407/750)+(211/1500)
= (475+814+211)1500 = 1500/1500 = 1
Yes, the sum of these probabilities is 1.
Q.7: A die is thrown 1000 times with the frequencies for the outcomes 1, 2, 3, 4, 5 and 6 as given in the following table :
Outcome 1 2 3 4 5 6 Frequency 179 150 157 149 175 190
Find the probability of getting each outcome.
Solution: Let Ei denote the event of getting the outcome i, where i = 1, 2, 3, 4, 5, 6.
Then
Probability of the outcome 1 = P(E1)
= Frequency of 1/Total number of times the die is thrown
= 179/1000
= 0.179
Similarly,
P(E2 ) = 150/1000 = 0.15
P(E3) = 157/1000 = 0.157
P(E4) = 149/1000 = 0.149
P(E5) = 175/1000 = 0.175
and P(E6) = 190/1000 = 0.19
You can check: P(E1) + P(E2) + P(E3) + P(E4) + P(E5) + P(E6) = 1
Q.8: An organisation selected 2400 families at random and surveyed them to determine a relationship between income level and the number of vehicles in a family. The information gathered is listed in the table below:
Monthly income(in ₹) Vehicles per family 0 1 2 Above 2 Less than 7000 10 160 25 0 7000-10000 0 305 27 2 10000-13000 1 535 29 1 13000-16000 2 469 59 25 16000 or more 1 579 82 88
Suppose a family is chosen. Find the probability that the family has chosen is
1. earning ₹10000 – 13000 per month and owning exactly 2 vehicles.
2. earning ₹16000 or more per month and owning exactly 1 vehicle.
3. earning less than ₹7000 per month and does not own any vehicle.
4. earning ₹13000 – 16000 per month and owning more than 2 vehicles.
5. owning not more than 1 vehicle.
Solution:
Total number of families = 2400
(i) Numbers of families earning ₹10000 –13000 per month and owning exactly 2 vehicles = 29
Therefore, the probability that the family earning between ₹10000 – 13000 per month and owning exactly 2 vehicle = 29/2400
(ii) Number of families earning ₹16000 or more per month and owning exactly 1 vehicle = 579
Therefore, the probability that the family earning between ₹16000 or more per month and owning exactly 1 vehicle = 579/2400
(iii) Number of families earning less than ₹7000 per month and does not own any vehicle = 10
Therefore, the probability that the family earning less than ₹7000 per month and does not own any vehicle = 10/2400 = 1/240
(iv) Number of families earning ₹13000-16000 per month and owning more than 2 vehicles = 25
Therefore, the probability that the family earning between ₹13000 – 16000 per month and owning more than 2 vehicles = 25/2400 = 1/96
(v) Number of families owning not more than 1 vehicle = 10+160+0+305+1+535+2+469+1+579 = 2062
Therefore, the probability that the family owns not more than 1 vehicle = 2062/2400 = 1031/1200
Q.9: Eleven bags of wheat flour, each marked 5 kg, actually contained the following weights of flour (in kg): 4.97 5.05 5.08 5.03 5.00 5.06 5.08 4.98 5.04 5.07 5.00
Find the probability that any of these bags chosen at random contains more than 5 kg of flour.
Solution: Total number of bags present = 11
Number of bags containing more than 5 kg of flour = 7
Therefore, the probability that any of the bags chosen at random contains more than 5 kg of flour = 7/11
Q.10: The distance (in km) of 40 engineers from their residence to their place of work were found as follows: 5 3 10 20 25 11 13 7 12 31 19 10 12 17 18 11 32 17 16 2 7 9 7 8 3 5 12 15 18 3 12 14 2 9 6 15 15 7 6 12.
What is the empirical probability that an engineer lives:
(i) less than 7 km from her place of work?
(ii) more than or equal to 7 km from her place of work?
(iii) within km from her place of work?
Solution:
The distance (in km) of 40 engineers from their residence to their place of work was found as follows:
5 3 10 20 25 11 13 7 12 31 19 10 12 17 18 11 3 2 17 16 2 7 9 7 8 3 5 12 15 18 3 12 14 2 9 6 15 15 7 6 12
Total numbers of engineers = 40
(i) Number of engineers living less than 7 km from their place of work = 9
The probability that an engineer lives less than 7 km from her place of work =9/40
(ii) Number of engineers living more than or equal to 7 km from their place of work
= 40 – 9 = 31
The probability that an engineer lives more than or equal to 7 km from her place of work
= 31/40
(iii) Number of engineers living within km from their place of work = 0
The probability that an engineer lives within km from her place of work = 0/40 = 0
Q.11: Refer to the table below:
Marks Number of students 0 – 20 7 20 – 30 10 30 – 40 10 40 – 50 20 50 – 60 20 60 – 70 15 70 – above 8 Total 90
(i) Find the probability that a student obtained less than 20% in the mathematics test.
(ii) Find the probability that a student obtained marks 60 or above.
Solution: Total number of students = 90
(i) Number of students who obtained less than 20% in the mathematics test = 7
The probability that a student obtained less than 20% in the mathematics test = 7/90
(ii) Number of students who obtained marks 60 or above = 15+8 = 23
The probability that a student obtained marks 60 or above = 23/90
### Extra Questions For Class 9 Maths Chapter 15 (Probability)
1. Can the experimental probability of an event be a negative number? If not, why?
2. Can the experimental probability of an event be greater than 1? Justify your answer.
3. As the number of tosses of a coin increase, the ratio of the number of heads to the total number of tosses will be ½. Is it correct? If not, write the correct one.
4. A company selected 4000 households at random and surveyed them to find out a relationship between the level and the number of television sets in a home. The information so obtained is listed in the following table.
Monthly income(in Rs.) Number of television/household 0 1 2 Above 2 <10000 20 80 10 0 10000-14999 10 240 60 0 15000-19999 0 380 120 30 20000-24999 0 520 370 80 25000 and above 0 1100 760 220
Find the probability:
(i) of a household earning Rs.10000-14999 per year and having exactly one television.
(ii) of a household earning Rs.25000 and more per year and owning two televisions
(iii) of a household having no televisions.
5. Out of 35 students Participating in a debate, 10 are girls. What is the Probability that winner is a boy?
(a) 1
(b) 2/7
(c) 3/7
(d) 5/7
6. What is the minimum value of a probability?
(a) 1/2
(b) 2
(c) 0
(d) None of the above
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# what formula would i use to solve the problem for area of this triangle: triangle ABC, a=3, angleB=24, angleC=24
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587
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what formula would i use to solve the problem for area of this triangle: triangle ABC, a=3, angleB=24, angleC=24
May 10, 2015
#3
+1068
+13
triangle ABC, a=3, angleB=24, angleC=24
1/ You need to find the height(h) of this triangle:
$$h = tan(B)(a/2)$$
2/ Now you can calculate the area of the triangle using this formula:
$$Area(A) = h(a/2)$$
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May 11, 2015
#1
0
Pythegorem Thereom
May 10, 2015
#2
+20850
+13
what formula would i use to solve the problem for area of this triangle: triangle ABC, a=3, angleB=24, angleC=24
$$\\(1) \qquad Area=\dfrac{a\cdot b}{2}\cdot\sin{(C)}\\\\ (2) \qquad b=a \cdot \dfrac{\sin{(C)}} {\sin{(A)}} }\\\\ (3) \qquad \sin{(A)}= \sin{( 180\ensurement{^{\circ}}-(B+C) )} = \sin{( B+C ) }$$
$$Area=\dfrac{a^2}{2} \cdot \dfrac{ \sin{(B)} \cdot\sin{(C)} } {\sin{(B+C)}} = \dfrac{a^2}{2} \cdot \left( \dfrac{ 1 } { \cot{(B)} + \cot{(C)} } } \right)\\\\\\ Area = \dfrac{3^2}{2} \cdot \left( \dfrac{ 1 } { \cot{(24\ensurement{^{\circ}})} + \cot{(24\ensurement{^{\circ}})} } } \right)\\\\ Area = \dfrac{3^2}{2} \cdot \left( \dfrac{ 1 } { 2\cdot \cot{(24\ensurement{^{\circ}})} } } \right)\\\\ \boxed{ Area = \dfrac{3^2}{4} \cdot \tan{(24\ensurement{^{\circ}})} } } \\\\ Area = \dfrac{3^2}{4} \cdot 0.4452286853\\\\ Area = \dfrac{9}{4} \cdot 0.4452286853\\\\ Area = 1.0017645419$$
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May 10, 2015
#3
+1068
+13
triangle ABC, a=3, angleB=24, angleC=24
1/ You need to find the height(h) of this triangle:
$$h = tan(B)(a/2)$$
2/ Now you can calculate the area of the triangle using this formula:
$$Area(A) = h(a/2)$$
civonamzuk May 11, 2015
#4
+27377
+8
civonamzuk has spotted that this is an isosceles triangle, which simplifies the calculation significantly!
.
May 11, 2015 |
# Playing with Numbers (Exercise – 3.1)
## Ex 3.1 Question 1.
Write all the factors of the following numbers:
(a) 24 (b) 15 (c) 21 (d) 27 (e) 12 (f) 20 (g) 18 (h) 23 (i) 36
Solution:
(a) Factors of 24 are:
24 = 1 x 24;
24 = 2 x 12;
24 = 3 x 8;
24 = 4 x 6
Factors of 24 = 1, 2, 3, 4, 6, 8, 12 and 24.
(b) Factors of 15 are:
15 = 1 x 15;
15 = 3 x 5
Factors of 15 = 1, 3, 5 and 15.
(c) Factors of 21 are:
21 = 1 x 21;
21 = 3 x 7
Factors of 21 =1, 3, 7 and 21.
(d) Factors of 27 are:
27 = 1 x 27;
27 = 3 x 9.
Factors of 27 = 1, 3, 9 and 27.
(e) Factors of 12 are:
12 = 1 x 12;
12 = 2 x 6;
12 = 3 x 4
Factors of 12 = 1, 2, 3, 4, 6 and 12.
(f) Factors of 20 are:
20 = 1 x 20;
20 = 2 x 10;
20 = 4 x 5
Factors of 20 = 1, 2, 4, 5, 10 and 20.
(g) Factors of 18 are:
18 = 1 x 18;
18 = 2 x 9;
18 = 3 x 6
Factors of 18 =1, 2, 3, 6, 9 and 18.
(h) Factors of 23 are:
23 = 1 x 23
Factors of prime number 23 = 1 and 23.
(i) Factors of 36 are:
36 = 1 x 36;
36 = 2 x 18;
36 = 3 x 12;
36 = 4 x 9;
36 = 6 x 6
Factors of 36 = 1, 2, 3, 4, 6, 9, 12, 18 and 36.
## Ex 3.1 Question 2.
Write first five multiples of:
(a) 5 (b) 8 (c) 9
Solution:
(a) First five multiples of 5 are:
5 x 1 = 5;
5 x 2 = 10;
5 x 3 = 15;
5 x 4 = 20;
5 x 5 = 25
Hence, the first five multiples of 5 are: 5, 10, 15, 20 and 25.
(b) First five multiples of 8 are:
8 x 1 = 8;
8 x 2 = 16;
8 x 3 = 24;
8×4 = 32;
8 x 5 = 40
Hence, the First five multiples of 8 are: 8, 16, 24, 32 and 40.
(c) First five multiples of 9 are:
9 x 1 = 9;
9 x 2 = 18;
9 x 3 = 27;
9 x 4 = 36;
9 x 5 = 45
Hence, First five multiples of 9 are: 9,18, 27, 36 and 45.
## Ex 3.1 Question 3.
Match the items in Column I with the items in column II.
Solution:
(i) ↔ (b) [∵ 7 x 5 = 35]
(ii) ↔ (d) [∵ 15 x 2 = 30]
(iii) ↔ (a) [∵ 8 x 2 = 16]
(iv) ↔ (f) [∵ 20 x 1 = 20]
(v) ↔ (e) [∵ 25 x 2 = 50]
## Ex 3.1 Question 4.
Find all the multiples of 9 up to 100.
Solution:
9 x 1 = 9
9 x 2 = 18
9 x 3 = 27
9 x 4 = 36
9 x 5 = 45
9 x 6 = 54
9 x 7 = 63
9 x 8 = 72
9 x 9 = 81
9 x 10 = 90
9 x 11 = 99
Hence, all the multiples of 9 upto 100 are:
9, 18, 27, 36, 45, 54, 63, 72, 81, 90 and 99.
### UKPSC Forest Guard Exam 09 April 2023 – Answer Key
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Find the locus:
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# Find the locus: - PowerPoint PPT Presentation
Find the locus:. What is the equation of the locus of points equidistant from the lines x = -4 and x = 2? x = -1. Compound Locus. Geometry Unit 7, Day 5 Ms. Reed. Objective:. To learn how to find the a compound locus. Definition.
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Find the locus:
• What is the equation of the locus of points equidistant from the linesx = -4 and x = 2?
• x = -1
### Compound Locus
Geometry
Unit 7, Day 5
Ms. Reed
Objective:
• To learn how to find the a compound locus
Definition
• Compound Locus – a problem involving two or more locus conditions occurring at the same time
• These different conditions are usually separated by “and” or “and also”
Problem:
• A treasure is buried in your backyard. A dog house is 8 feet from a stump and 18 feet from the tree. The treasure is equidistant from the dog house and the tree ANDALSO 6 feet from the stump. Locate all possible points of the buried treasure.
Solving Compound Locus Problems
• Step 1: Draw the diagram showing the information in the problem.
Problem:
• A treasure is buried in your backyard. A dog house is 8 feet from a stump and 18 feet from the tree. The treasure is equidistant from the dog house and the tree ANDALSO 6 feet from the stump. Locate all possible points of the buried treasure.
8 ft.
18 ft.
Solving Compound Locus Problems
• Step 1: Draw the diagram showing the information in the problem.
• Step 2: Read to determine one of the needed conditions.
Problem:
• A treasure is buried in your backyard. A dog house is 8 feet from a stump and 18 feet from the tree. The treasure is equidistant from the dog house and the treeAND ALSO 6 feet from the stump. Locate all possible points of the buried treasure.
8 ft.
18 ft.
Solving Compound Locus Problems
• Step 1: Draw the diagram showing the information in the problem.
• Step 2: Read to determine one of the needed conditions.
• Step 3: Plot the 1st Locus Condition
Problem:
• A treasure is buried in your backyard. A dog house is 8 feet from a stump and 18 feet from the tree. The treasure is equidistant from the dog house and the treeAND ALSO 6 feet from the stump. Locate all possible points of the buried treasure.
8 ft.
18 ft.
Solving Compound Locus Problems
• Step 1: Draw the diagram showing the information in the problem.
• Step 2: Read to determine one of the needed conditions.
• Step 3: Plot the 1st Locus Condition
• Step 4: Repeat steps 2&3 until all conditions are satisfied.
Problem:
• A treasure is buried in your backyard. A dog house is 8 feet from a stump and 18 feet from the tree. The treasure is equidistant from the dog house and the treeAND ALSO6 feet from the stump. Locate all possible points of the buried treasure.
8 ft.
18 ft.
Problem:
• A treasure is buried in your backyard. A dog house is 8 feet from a stump and 18 feet from the tree. The treasure is equidistant from the dog house and the treeAND ALSO6 feet from the stump. Locate all possible points of the buried treasure.
8 ft.
18 ft.
Solving Compound Locus Problems
• Step 1: Draw the diagram showing the information in the problem.
• Step 2: Read to determine one of the needed conditions.
• Step 3: Plot the 1st Locus Condition
• Step 4: Repeat steps 2&3 until all conditions are satisfied.
• Step 5: Place an X where the lines intersect, these are the points that answer the compound locus.
Problem:
• A treasure is buried in your backyard. A dog house is 8 feet from a stump and 18 feet from the tree. The treasure is equidistant from the dog house and the treeAND ALSO6 feet from the stump. Locate all possible points of the buried treasure.
8 ft.
18 ft.
r
s
4m
A
8m
Practice 1:
• Parallel lines rand s are 8 meters apart, and A is a point on line s. How many points are equidistant from r and s and also 4 meters from A?
3 un.
3 un.
5 units
P
10 units
Practice 2:
• A given point P is 10 units from a given line. How many points are 3 units from the line and 5 units from point P?
Homework
• Work Packet: Compound Locus |
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# Determining the Equation of a Line
## Write equations from y-intercept and slope or two points
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Practice Determining the Equation of a Line
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Writing Linear Equations
Have you ever studied banana plants? Take a look at this dilemma.
Mr. Thomas' class is having a discussion on the rainforest. The students have been doing some research, and now they are ready to talk about their findings.
“The vegetation of the rainforest was very interesting,” Carmen commented in Mr. Thomas’ class.
“There were so many different things growing,” Mark agreed.
The students began having a discussion about the things that had intrigued them about the plant life of the rainforest. One of the points that was brought up is that there are plants in the rainforest that can’t be found anywhere else in the world. Mr. Thomas spotted the opportunity and wrote the following problem.
You buy a banana tree that is 8 inches tall. It grows 4 inches per day. Its height (in inches) $h$ is a function of time (in days) $d$ .
You can express this function as an equation. This Concept will show you how to write linear equations.
### Guidance
The $y=mx+b$ form of an equation was most useful in rapidly identifying both the slope, $m$ , and the $y$ -intercept, $b$ . In fact, if we know what the slope of an equation is and we know its $y$ -intercept, then we can just as easily write the equation. All you have to do is plug in the slope for $m$ in the $y=mx+b$ form and the $y$ -intercept for $b$ .
Take a look at this situation.
$m = 4, y-intercept = 3$
We know that we are going to use the form of the equation $y=mx+b$ , so we can substitute these values into the equation and write it.
$y=4x+3$
This is the answer. The key is to always watch for negative signs and be sure to include them when you write your equation.
Sometimes, we can also be given the slope and a point that the line crosses through. We can also use this information to write an equation of a line.
$\text{Slope} = -2$ , the line passes through the point (0,-3)
With this example, we know the slope, so that can be easily substituted into the slope-intercept form. The point has a 0 for the $x$ value, so that means that we have been given the coordinate of the $y$ – intercept.
$y= -2x-3$
What if you only know two points and you don’t know the slope? It’s a similar operation that we can use in order to write the equation.
Do you recall that the slope formula is $m=\frac{{y_2}-{y_1}}{{x_2}-{x_1}}$ ? In other words, given any two points $(x_1, y_1)$ and $(x_2, y_2)$ , we can use the slope formula to calculate the slope of the line that passes through those points. Even when you only know two points, finding the slope is just a matter of using the formula. But then what?
We will use the notion below.
if $m=\frac{{y_2}-{y_1}}{{x_2}-{x_1}}$ then $m=\frac{{y}-{y_1}}{{x}-{x_1}}$ because the slope is the same on any part of a line. In other words, we can use a formula similar to the slope formula for finding the equation. This time, however, we will leave $x$ and $y$ as variables because the relationship is true for any values of $x$ and $y$ in that equation.
Write the equation of the line that passes through the points $(3,7)$ and $(5,11)$ . First we will find the slope using the slope formula for ${x_1}=3,{y_1}=7,{x_2}=5,{y_2}=11$ .
$m &=\frac{{y_2}-{y_1}}{{x_2}-{x_1}}\\m &=\frac{11-7}{5-3}\\m &=\frac{4}{2}\\m &=2$
Now plug in our known values of $m, x_1$ , and $y_1$ .
$m &=\frac{y-{y_1}}{x-{x_1}}\\\frac{2}{1} &=\frac{y-3}{x-7}$
Do you see that we have a proportion? This can be solved by cross multiplying.
$\frac{2}{1} &=\frac{y-3}{x-7}\\1(y-3) &=2(x-7)\\y-3 &=2x-14\\y &=2x-11$
What about when you have been given a table of values? Well, there is a way to figure out the equation quite simply when you have a table of values. Take a look.
$x$ $y$
0 5
1 7
2 9
3 11
4 13
First, notice that the $y$ – intercept is the value that has an $x$ value of 0. With an $x$ value of 0, we know that the $y$ – intercept is 5.
Now we need to figure out the slope. Look at the y values in the table. Can you see a pattern? If you look carefully, you will see that the values jump by +2 each time. This is the slope. Think about how the line would move when graphed. The pattern of the $y$ values represents the slope of the line.
$y=2x+5$
What about function notation? Well first, think about independent and dependent variables.
In science, an independent variable is a parameter that is manipulated or chosen by a scientist while the dependent variable is a parameter that is measured. Scientists oftentimes look for a correlation between an independent variable and a dependent variable—they want to know if the dependent variable depends on the independent variable. For example, a scientist might measure the speed at which a car is moving and the force upon impact when the car hits a wall. The scientist can manipulate the speed of the car—she can make the car move slower or faster. She would then measure the force of impact related to the given speed. Then, a conclusion can be drawn about their relatedness and cars, in this case, might be designed based on that relation. The independent variable will be shown in the left column of a t-table and on the $x$ -axis of a graph. The dependent variable will be shown in the right column of a t-table and on the $y$ -axis of a graph.
$f(x)=4x+1$
Here we know that the function of $x$ is dependent on 4 times that value, $x$ and one.
Write a linear equation by using the given information.
#### Example A
$m = 2, y-intercept = 5$
Solution: $y=2x+5$
#### Example B
$m = -4, y-intercept = 6$
Solution: $y=-4x+6$
#### Example C
$m = 8, y-intercept = -2$
Solution: $y=8x-2$
Now let's go back to the dilemma from the beginning of the Concept.
First, we need to write an equation. We can use the $h$ to represent the height of the banana tree. We can use the $d$ to represent the number of days. The 8 is the height that the tree started with. Here is our equation.
$h=4d+8$
### Vocabulary
Independent Variable
a value that is not dependent on another value. It is the $x$ value in a table.
Dependent Variable
a value that is dependent on the equation. It is the $y$ value in a table.
Function Notation
an equation where the value of $x$ is dependent on the equation involving $x$ .
### Guided Practice
Here is one for you to try on your own.
Write an equation in slope-intercept form with a slope of -4 and a y-intercept of 13.
Solution
To do this, we can take slope-intercept form and substitute in the given values.
$y=mx+b$
The $m$ represents the slope.
The $b$ represents the y-intercept.
$y=-4x+13$
### Practice
Directions: Write the equation of a line with the following slopes and $y$intercepts.
1. $\text{slope} = 2, \ y \ \text{int} = 4$
2. $\text{slope} = -3, \ y \ \text{int} = 2$
3. $\text{slope} = -4, \ y \ \text{int} = 4$
4. $\text{slope} = 3, \ y \ \text{int} = -5$
5. $\text{slope} = \frac{1}{2}, \ y \ \text{int} = -2$
6. $\text{slope} = -\frac{1}{3}, \ y \ \text{int} = 2$
7. $\text{slope} = 1, \ y \ \text{int} = 8$
8. $\text{slope} = -2, \ y \ \text{int} = 4$
9. $\text{slope} = -1, \ y \ \text{int} = -1$
10. $\text{slope} = 5, \ y \ \text{int} = -2$
Directions: Write the following horizontal or vertical line equations.
1. A horizontal line with a $b$ value of 7.
2. A horizontal line with a $b$ value of -4.
3. A vertical line with an $x$ value of 2
4. A vertical line with an $x$ value of -5
Directions: Write the equation of a line that passes through the following points.
1. (3, -3) and (-3, 1)
2. (2, 3) and (0, -3)
### Vocabulary Language: English
dependent variable
dependent variable
The dependent variable is the output variable in an equation or function, commonly represented by $y$ or $f(x)$.
Function
Function
A function is a relation where there is only one output for every input. In other words, for every value of $x$, there is only one value for $y$.
independent variable
independent variable
The independent variable is the input variable in an equation or function, commonly represented by $x$. |
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# Complement Rule for Probability
## If P(A) is is the probability that event A happens then the probability that event A doesn't happen is 1-P(A).
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Practice Complement Rule for Probability
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Finding Probability by Finding the Complement
#### Objective
Here you will learn how to quickly calculate the probability of an event by finding the probability of the complement.
#### Concept
Suppose you were asked to calculate the probability of rolling two dice and getting a different number on each. How could you find the answer without needing to enumerate all of the possibilities?
This lesson is about a shortcut to the calculation of some probabilities. We’ll return to this question after the lesson.
#### Watch This
http://youtu.be/uofEzYbTric MrPilarski – How to find the complement of an event and odds
#### Guidance
Sometimes the probability of an event is difficult or impossible to calculate directly. Particularly when calculating the probability of “at least one…” types of problems, or when the sample space of the complement is smaller than that of the event, it may be worth looking first for the probability of the complement of the event you are trying to measure.
Recall that the complement of an event is the sample space containing all the outcomes that are not a part of the event itself. That means that the probability of an event + the probability of the complement = 100% or 1.00, or, to say the same thing as a formula: $P(A)+P(A^\prime)=1$ . Once you know the probability of the complement, you can just subtract it from 1 to find the probability of the event.
Example A
What is the probability of randomly drawing a king other than the King of Hearts from a bag containing one king of each standard suit: hearts, clubs, diamonds, and spades.
Solution: Let’s say that $P(KH)$ is the probability that you do draw the King of Hearts. In that case, $P(KH^\prime)$ is the probability of not drawing the King of Hearts, the complement of $P(KH)$ . Since there are 4 kings and only 1 is the King of Hearts, we can say:
$P(KH)=\frac{1}{4}$
therefore:
$P(KH^\prime)=1-\frac{1}{4} \ or \ \frac{3}{4}=75 \%$
So the probability of not drawing the King of Hearts is 75%
Example B
What is the probability of getting a cherry sour from a bag that starts with 9 cherry sours, 5 lemon sours, and 8 lime sours, given that you keep anything you choose and you choose up to 3 times?
Solution: This problem combines conditional probabilities with complements. A key point to this problem is the “keep anything you choose” part. Because we are not putting back the candy after each pull, the probability changes each time, and the chance to get a cherry improves each time we don’t choose one. Let’s deal with each pull separately at first:
1. First pull: $P(C1)=\frac{9 \ cherry}{22 \ candies}=\frac{9}{22} \ or \ 41 \%$
2. Second pull: $P(C2)=\frac{9 \ cherry}{21 \ candies}=\frac{9}{21} \ or \ 43 \%$
3. Third Pull: $P(C3)=\frac{9 \ cherry}{20 \ candies}=\frac{9}{20} \ or \ 45 \%$
It is tempting to think that the probability of getting at least 1 cherry is just the sum of the three probabilities, but obviously there can’t be a $41 \%+43 \%+45 \%=129 \%$ chance. The twist is that we only choose a 2 nd or 3 rd time if we don’t get a cherry the time before. That means we need to calculate the chance of choosing a 2 nd or 3 rd time and multiply the probability of a cherry on that pull by the chance of pulling that time at all.
Fortunately we can use the complement rule to save time. The chance of needing a 2 nd pull is the same as the chance that we don’t pull a cherry on the 1 st pull, in other words:
$P(2nd \ pull)=P(C1^\prime)=1-41 \%=59 \%$
The chance of needing a 3 rd pull is the same as not getting a cherry the 2 nd time:
$P(3rd \ pull)=P(C1^\prime) \times P(C2^\prime)=59 \% \times 57 \%=34 \%$
Now we can find the overall probability of getting a cherry in three pulls or less:
$P&(cherry) =\text{(chance of cherry of pull 1)} \\+ \text{(chance of }& \text{needing pull 2} \times \text{chance of cherry on pull 2)} \\+ \text{(chance of }& \text{needing pull 3} \times \text{chance of cherry on pull 3)} \\P(cherry)&=41 \%+(59 \% \times 43 \%)+(34 \% \times 45 \%)=82 \%$
Example C
What is the probability of rolling two dice and at least one die showing a factor of 6?
Solution: This is an “at least one…” problem. To satisfy the requirements, either one of the two dice or both need to land on 1, 2, 3, or 6. That is quite a few possibilities to solve for! However, there are many fewer possible outcomes where neither die shows 1, 2, 3, or 6 – in other words, where both dice show 4 or 5! There are only 4 ways that could happen:
1. 1 st die rolls 5 and 2 nd rolls 4
2. 1 st die rolls 4 and 2 nd rolls 5
3. Both dice roll 4
4. Both dice roll 5
There are $6 \times 6=36$ total possible outcomes. If we say that the event “at least one die shows a factor of 6” is $A$ , then $A^\prime$ would be the complement, so we can say:
$P(A^\prime)=\frac{4 \ favorable \ outcomes}{36 \ total \ outcomes}=\frac{1}{9}$
If the complement of the event we want to calculate is $\frac{1}{9}$ , then the event itself can be calculated as:
$P(A)=1-P(A^\prime)=\frac{8}{9} \ or \ 88.9 \%$
Therefore, we can say that the probability of rolling two dice and getting at least one factor of six is $\frac{8}{9}$ .
Concept Problem Revisited
Suppose you were asked to calculate the probability of rolling two dice and getting a different number on each. How could you find the answer without needing to enumerate all of the possibilities?
The probability of rolling two not matching numbers is the complement of rolling a matching pair. Since there are only 6 numbers to make matches from, there are only 6 matching pairs. The total number of possible outcomes of two dice is $6 \times 6=36$ .
$\therefore P(matching)=\frac{6\text{ matches}}{36\text{ possibilities}}=\frac{1}{6} \ or \ .167$
That means that the complement, rolling not matching numbers is:
$1-.167=.833 \ or \ 83.3 \%$
That was quite a bit faster than trying to enumerate all of the possible non-matching rolls!
#### Vocabulary
The complement of an event has a probability equal to 100% minus the probability of the event. As a formula, this looks like: $P(A^\prime)=1-P(A)$ . The probability of the complement of event “ $A$ ” equals one minus the probability of event “ $A$ ”.
To enumerate a list of things is to mention each member of the list independently.
#### Guided Practice
1. What is the probability of rolling a number other than 5 on a number cube?
2. What is the probability of choosing a card that isn’t a club from a standard deck?
3. What is the probability of not rolling a factor of 10 on a single roll of a 20-sided die?
4. You have a bag of crazy-flavor jellybeans. You know that the flavors are distributed as: 12 earwax, 14 belly-button lint, 26 dog food, 38 gym sock, and 10 of your favorite fruit in the bag. What is the probability that you will be unhappy with the taste of your choice if you choose one jellybean at random?
5. What is the probability of rolling a number that is not a factor of ten or twelve on a single standard die?
Solutions:
1. There are six sides on a number cube, so the probability of rolling any single number is $\frac{1}{6}$ . Therefore, we can say: $P(5)=\frac{1}{6}$ so $P(5^\prime)=\frac{5}{6}$ Therefore the probability of not rolling a 5 is $\frac{5}{6}$ .
2. There are four suits, so the probability of choosing a club is $\frac{1}{4}$ . The complement is the probability of not choosing a club, and it is $1-\frac{1}{4}=\frac{3}{4} \ or \ 75 \%$
3. There are 4 factors of 10: 1, 2, 5, and 10. Therefore the probability of rolling a factor of 10 on a 20-sided die is $\frac{4}{20}$ . The complement is the probability of not rolling one of the four factors: $1-\frac{4}{20}=\frac{16}{20}$
4. Let $P(F)$ be the probability of pulling a fruit-flavor, then $P(F^\prime)$ is the probability of not pulling a fruit flavor (and getting something disgusting instead!). There are 10 fruit-flavored beans in the bag of 100 beans, so:
$P(F)&=\frac{10}{100} \\P(F^\prime)&=1-P(F)=\frac{90}{100}$
Therefore, you have a 90% probability of being unhappy with your choice.
5. The factors of ten or twelve are: 1, 2, 3, 4, 5, 6, 10, and 12. Since all six numbers from a standard die are in that set, the probability of rolling one of them is 100% or 1.0. Therefore, the probability of not rolling one of them is $1-1=0$ .
#### Practice
1. If you randomly pull a single card from a standard deck, what is the probability that the card is anything other than a king?
2. What is the probability of not pulling a face card from a standard deck?
3. What is the probability that a single roll of a 10-sided die will not land on a 7?
4. What is the probability that a single roll of a standard die will be 1, 2, 3, 4, or 5?
5. A candy jar contains 6 red, 7 blue, 8, green, and 9 yellow candies. What is the probability that choosing a single candy at random will result in a piece that is either red, blue, or green?
6. What is the probability that a single choice from the jar in Q 5 will result in a piece that is either red or yellow?
7. What is the probability that a single choice from the same jar will not be blue?
8. You have $39 in cash, composed of the largest bills possible. What is the probability that a randomly chosen bill from the$39 will not be a \$1 bill?
9. There are 450 songs on the .mp3 player you share with your father and sister. If your dad has 125 80’s songs, and your sister has twice that many country music songs, what is the percent probability that a randomly chosen song will not be one of yours?
10. What is the percent probability that a random roll of a fair die will not result in an even or prime number?
11. The train station has 47 active trains. 5 are late by less than 10 minutes, 4 are between 11 and 15 minutes late, and 10 are more than 15 minutes late. What is the probability that a randomly chosen train will be on time?
12. The probability that a student has called in sick and that it is Monday is 12%. The probability that it is Monday and not another day of the school week is 20% (there are only five days in the school week). What is the probability that a student has not called in sick, given that it is Monday?
13. A neighborhood wanted to improve its parks so it surveyed kids to find out whether or not they rode bikes or skateboards. Out of 2300 children in the neighborhood that ride something, 1800 rode bikes, and 500 rode skateboards, while 200 of those ride both a bike and skateboard. What is the probability that a student does not ride a skateboard, given that he or she rides a bike?
14. A movie theatre is curious about how many of its patrons buy food, how many buy a drink, and how many buy both. They track 300 people through the concessions stand one evening, out of the 300, 78 buy food only, 113 buy a drink only and the remainder buy both. What is the probability that a patron does not buy a drink if they have already bought food?
### Vocabulary Language: English
complement
complement
A mutually exclusive pair of events are complements to each other. For example: If the desired outcome is heads on a flipped coin, the complement is tails.
Complement rule
Complement rule
The Complement Rule states that the sum of the probabilities of an event and its complement must equal 1, or for the event A, P(A) + P(A') = 1.
enumerate
enumerate
Enumerate means to catalogue or list members independently.
Venn diagrams
Venn diagrams
A diagram of overlapping circles that shows the relationship among members of different sets. |
# Finding the maximum possible number of elements of a set of real numbers
Let $0<b < a$ be two given real numbers. Consider a set of real numbers which satisfies the following conditions:
1. Each number in the set is strictly greater than $0$ but less than or equal to $b$.
2. The summation of all numbers in the set is equal to $a$.
3. The summation of each pair of numbers in the set is strictly greater than $b$.
What can we tell about the maximum number of elements of such a set? For example, for $a=2$ and $b=1$ at most $3$ elements are possible to assign to such a set (e.g., $\{0.5,0.7,0.8\}$ or $\{0.2,0.9,0.9\}$ etc.).
We can order the numbers $x_i$ in this set as $$0<x_1\leq x_2 \leq\ldots\leq x_n\leq b\ .$$ Since the $x_i$ sum to $a$ we have $$(n-1)x_2\leq x_2+x_3+\ldots+x_n=a-x_1\ ,$$ hence $$n\leq{a-x_1\over x_2}+1\ .$$ The condition $x_1+x_2>b$ implies $-x_1<x_2-b$ and $x_2>{b\over2}$. We therefore obtain $$n<{a-b+x_2\over x_2}+1={a-b\over x_2}+2<{a-b\over b/2}+2={2a\over b}\ .$$ On the other hand the resulting bound $n_*=\lceil 2a/b\rceil-1$ is optimal and can be attained with equal pieces of size slightly larger than ${b\over2}$.
All but one or all the x from the set are greater than $b/2$. Number of elements $n$ can be odd or even.
If $n$ is odd you can always group elements in pairs so that smallest be paired and largest be unpaired. Each pair is greater than b and the unpaired is greater than b/2. So their sum is greater than $nb/2$ meaning $a>nb/2$. Same is for even $n$.
Maximum number is $n= c[2a/b]-1$ where c[k] denotes smallest integer greater or equal to k
Obviously larger $n$ doesn't satisfy $a>nb/2$, just have to prove that there always exists such set.
You can always pick $x_i$ such that $x_i=b/2+\frac{a-nb/2}{n}+i\frac{a-nb/2}{n^2}$ where $i\in Z$, $|i|\le n/2$ and if $n$ is even than $i\ne0$. So $\sum x_i=a$ |
# Transformations and Matrices
(Difference between revisions)
Revision as of 15:09, 1 July 2011 (edit)m ← Previous diff Revision as of 15:15, 1 July 2011 (edit) (undo)Next diff → Line 44: Line 44: |ImageDesc= |ImageDesc= ===Rotation=== ===Rotation=== + {{hide|1= + [[Image:Rotationgraph.png|300px|right]] A 2D rotation transformation rotates everything in 2D space around the origin by a given angle. In order to see what it does, let’s take a look at what a rotation by a positive angle does to the coordinate axes. Now (x,y) is the result when you apply the transformation to (1,0), which means that
[[Image:Rotationgraph.png|300px|right]] A 2D rotation transformation rotates everything in 2D space around the origin by a given angle. In order to see what it does, let’s take a look at what a rotation by a positive angle does to the coordinate axes. Now (x,y) is the result when you apply the transformation to (1,0), which means that
::\begin{align} x = cos(\theta ) \\ ::[itex] \begin{align} x = cos(\theta ) \\ Line 56: Line 58: sin(\theta ) & cos(\theta ) sin(\theta ) & cos(\theta ) \end{bmatrix}
\end{bmatrix} [/itex]
- or, as XKCD (http://http://xkcd.com/184/) sees it (notice that the rotation is by -90° and $sin(-90\,^{\circ}) = -sin(90\,^{\circ})$,
+ or, as XKCD (http://xkcd.com/184/) sees it (notice that the rotation is by -90° and $sin(-90\,^{\circ}) = -sin(90\,^{\circ})$,
The situation for 3D rotations is different because a rotation in 3D space must leave a fixed line through the origin. In fact we really only handle the special cases where the fixed line is one of the coordinate axes. Let’s start with the easiest one.
Line 101: Line 103: * & * & * & 0 \\ * & * & * & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} [/itex] where the * terms are the terms from the 3D rotations above. 0 & 0 & 0 & 1 \end{bmatrix} [/itex] where the * terms are the terms from the 3D rotations above. - + }} ===Scaling=== ===Scaling=== + {{hide|1= Scaling is the action of multiplying each coordinate of a point by a constant amount. As an example, let $f(x,y,z) = (2x,3y,4z)$. Then
Scaling is the action of multiplying each coordinate of a point by a constant amount. As an example, let $f(x,y,z) = (2x,3y,4z)$. Then
$f((x,y,z)+(a,b,c)) = f(x+a,y+b,z+c) = (2(x+a),3(y+b), 4(z+c)) = (2x,3y,4z)+(2a,3b,4c)$
$f((x,y,z)+(a,b,c)) = f(x+a,y+b,z+c) = (2(x+a),3(y+b), 4(z+c)) = (2x,3y,4z)+(2a,3b,4c)$
Line 115: Line 118: In case we are working with homogeneous coordinates, a scaling transformation only acts on the three primary components and leaves the homogeneous component alone, so we simply have the matrix $\begin{bmatrix} \sigma_x & 0 & 0 & 0 \\ 0 & \sigma_y & 0 & 0 \\ 0 & 0 & \sigma_z & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$ for the scaling transformation. In case we are working with homogeneous coordinates, a scaling transformation only acts on the three primary components and leaves the homogeneous component alone, so we simply have the matrix $\begin{bmatrix} \sigma_x & 0 & 0 & 0 \\ 0 & \sigma_y & 0 & 0 \\ 0 & 0 & \sigma_z & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$ for the scaling transformation. - + }} ===Translation=== ===Translation=== + {{hide|1= Notice that a translation function cannot be a linear transformation on normal space because it does not take the origin to the origin. These are examples of affine transformations, transformations that are composed of a linear transformation, such as a rotation, scaling, or shear, and a translation. In order to write a translation matrix, we need to use homogeneous coordinates.
Notice that a translation function cannot be a linear transformation on normal space because it does not take the origin to the origin. These are examples of affine transformations, transformations that are composed of a linear transformation, such as a rotation, scaling, or shear, and a translation. In order to write a translation matrix, we need to use homogeneous coordinates.
Line 128: Line 132: hese are linear transformations in the space one degree higher than the geometry you are working with. In fact, the main reason for including homogeneous coordinates is the math for graphics is to be able to handle translations (and thus all basic transformations) as linear transformations represented by matrices.
hese are linear transformations in the space one degree higher than the geometry you are working with. In fact, the main reason for including homogeneous coordinates is the math for graphics is to be able to handle translations (and thus all basic transformations) as linear transformations represented by matrices.
- + }} ===Shear=== ===Shear=== + {{hide|1= The shear transformation is not widely used in computer graphics, but can be used for things like the oblique view in engineering drawings. The concept of a shear is to add a multiple of one coordinate to another coordinate of each point, or, for example,
The shear transformation is not widely used in computer graphics, but can be used for things like the oblique view in engineering drawings. The concept of a shear is to add a multiple of one coordinate to another coordinate of each point, or, for example,
::$shear(x,y,z) = (x+3y,y,z)$
::$shear(x,y,z) = (x+3y,y,z)$
Line 149: Line 154: The second applet lets you apply the 3D transformations to a 3D figure. The second applet lets you apply the 3D transformations to a 3D figure. (Currently Unavailable)
(Currently Unavailable)
- + }} ===Matrix Inverses=== ===Matrix Inverses=== + {{hide|1= In general, getting the inverse of a matrix can be difficult, but for the basic graphics transformations the inverses are easy because we can simply undo the geometric action of the original transformation.
In general, getting the inverse of a matrix can be difficult, but for the basic graphics transformations the inverses are easy because we can simply undo the geometric action of the original transformation.
Line 165: Line 171: So we have a major observation: If any transformation is the product of basic graphics transformations, it is easy to find the inverse of its matrix (and hence its inverse transformation) as the product of the inverses of the components in reverse order. Or:
So we have a major observation: If any transformation is the product of basic graphics transformations, it is easy to find the inverse of its matrix (and hence its inverse transformation) as the product of the inverses of the components in reverse order. Or:
::$(A \times B \times C)^{-1} = C^{-1} \times B^{-1} \times A^{-1}$
::$(A \times B \times C)^{-1} = C^{-1} \times B^{-1} \times A^{-1}$
- + }} ==Transformations and Graphics Environments== ==Transformations and Graphics Environments== Attention – this concept needs a bit of programming background; it involves stacks.
Attention – this concept needs a bit of programming background; it involves stacks.
## Revision as of 15:15, 1 July 2011
Transformations and Matrices
placeholder
# Basic Description
When an object undergoes a transformation, it can be represented as a matrix. Different transformations such as translations, rotations, scaling and shearing are represented mathematically in different ways. One matrix can also represent multiple transformations in sequence when the matrices are multiplied together.
## Linear Transformations Are Matrices
A linear transformation on 2D (or 3D) space is a function f from 2D (or 3D) space to itself that has the property that
$f(aA + bB) = af(A) + bf(B).$
Since points in 2D or 3D space can be written as $P = xi + yj$ or $P = xi + yj +zk$ with i, j, and k the coordinate vectors, then we see that $f(P) = xf(i) + yf(j)$ or $f(P) = xf(i) + yf(j) + zf(k)$
This tells us that the linear transformation is completely determined by what it does to the coordinate vectors.
Let’s see an example of this: if the transformation has the following action on the coordinates:
\begin{align}f(1,0,0) = f(i) = (2,-2,1) \\ f(0,1,0) = f(j) = (-1,3,2) \\ f(0,0,1) = f(k) = (4,3,-2) \end{align}
then for any point we have:
\begin{align}f(x,y,z) = (2x,-2x,x)+(-y,3y,2y)+(4z,3z,-2z) \\ = (2x-y+4z, -2x+3y+3z, x+2y-2z) \\ = \begin{bmatrix} 2x - y + 4z \\ -2x + 3y + 3z \\ x + 2y - 2z \end{bmatrix} = \begin{bmatrix} 2 & -1 & 4 \\ -2 & 3 & 3 \\ 1 & 2 & -2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} \end{align}
From this example, we see that the linear transformation is exactly determined by the matrix whose first column is $f(i)$, whose second column is $f(j)$, and whose third column is $f(k)$, and that applying the function f is exactly the same as multiplying by the matrix. So the linear transformation is the matrix multiplication, and we can use the concepts of linear transformation and matrix multiplication interchangeably.
## Transformation Composition Is Matrix Multiplication
Transformations are usually not used by themselves, especially in graphics, so you need to have a way to compose transformations, as in $g(f(P))$. But if G is the matrix for the transformation g, and F is the matrix for the transformation f, then the matrix product G*F is the matrix for the composed functions gf.
## Basic Transformations For Graphics
Computer graphics works by representing objects in terms of simple primitives (link to the graphics primitives page) that are manipulated with transformations that preserve some primitives’ essential properties. These properties may include angles, lengths, or basic shapes. Some of these transformations can work on primitives with vertices in standard 2D or 3D space, but some need to have vertices in homogeneous coordinates. The general graphics approach is to do everything in homogeneous coordinates, but we’ll talk about the primitives in terms of both kinds when we can.
The most fundamental kinds of transformations for graphics are rotation, scaling, and translation. There are also a few cases when you might want to use shear transformations, so we’ll talk about these as well.
# A More Mathematical Explanation
Note: understanding of this explanation requires: *stacks
### Rotation
<span class="_togglegroup _toggle_initshow _toggle _toggler toggle-visible" style="dis [...]
### Rotation
A 2D rotation transformation rotates everything in 2D space around the origin by a given angle. In order to see what it does, let’s take a look at what a rotation by a positive angle does to the coordinate axes. Now (x,y) is the result when you apply the transformation to (1,0), which means that
\begin{align} x = cos(\theta ) \\ y = sin(\theta ) \end{align}
But (x’,y’) is the result when you apply the transformation to (0,1), or
\begin{align} x' = cos(\theta +\frac{p}{2}) = cos(\theta )cos(\frac{p}{2}) - sin(\theta )sin(\frac{p}{2}) = -sin(\theta ) \\ y' = sin(\theta +\frac{p}{2}) = sin(\theta )cos(\frac{p}{2}) + cos(\theta )sin(\frac{p}{2}) = cos(\theta ) \end{align}
Then as we saw above, the rotation transformation must have the image of (1,0) as the first column and the image of (0,1) as the second column, or
$rotate(\theta ) = \begin{bmatrix} cos(\theta ) & -sin(\theta ) \\ sin(\theta ) & cos(\theta ) \end{bmatrix}$
or, as XKCD (http://xkcd.com/184/) sees it (notice that the rotation is by -90° and $sin(-90\,^{\circ}) = -sin(90\,^{\circ})$,
<br
The situation for 3D rotations is different because a rotation in 3D space must leave a fixed line through the origin. In fact we really only handle the special cases where the fixed line is one of the coordinate axes. Let’s start with the easiest one.
A rotation around the Z-axis is a 2D rotation as above with the third dimension fixed. So the matrix for this rotation is pretty clearly
$\begin{bmatrix} cos(\theta ) & -sin(\theta ) & 0 \\ sin(\theta ) & cos(\theta ) & 0 \\ 0 & 0 & 1 \end{bmatrix}$
A rotation around the X-axis is pretty similar. If we look down the X-axis, we see the following 2D coordinates:
with $Y \times Z = X$, the axis of rotation. This looks like an exact analogue of the XY-plane, and so we can see that the rotation matrix must leave X fixed and operate only on Y and Z as
$\begin{bmatrix} 1 & 0 & 0 \\ 1 & cos(\theta ) & -sin(\theta ) \\ 0 & sin(\theta ) & cos(\theta ) \end{bmatrix}$
For rotations around the Y-axis, the view down the Y-axis looks different from the one down the Z-axis; it is
Here a positive-angle is from the X-axis towards the Z-axis, but $X \times Z = -Z \times X = -Y$, so the rotation axis dimension is pointing in the opposite direction from the Y-axis. Thus a the angle for the rotation is the negative of the angle we would see in the axes above, and since cos is an even function but sin is odd, we have the rotation matrix
$\begin{bmatrix} cos(-\theta ) & 0 & -sin(-\theta ) \\ 0 & 1 & 0 \\ sin(-\theta ) & 0 & cos(-\theta ) \end{bmatrix} = \begin{bmatrix} cos(\theta ) & 0 & sin(\theta ) \\ 0 & 1 & 0 \\ sin(-\theta ) & 0 & cos(\theta ) \end{bmatrix}$
around the Y-axis.
When you want to get a rotation around a different line than a coordinate axis, the usual approach is to find two rotations that, when composed, take a coordinate line into the fixed line you want. You can then apply these two rotations, apply the rotation you want around the coordinate line, and apply the inverses of the two rotations (in inverse order) to construct the general rotation. The sequence goes like this:
apply a rotation $R_1$ around the Z-axis to move your fixed line into the YZ-plane
apply a rotation $R_2$ around the X-axis to move that line to the Y-axis
apply the rotation by your desired angle around the Y-axis
apply the inverse $R_2^{-1}$ to move your rotation line back into the YZ-plane
apply the inverse $R_1^{-1}$ to move your rotation line back to the original line.
Whew! This can all be put into a function – or you can simply keep everything in terms of rotations around X, Y, and Z.
If we are working in homogeneous coordinates, we see that all of the rotation operations take place in standard 3D space and so the fourth coordinate is not changed. Thus the general pattern for all the rotations in homogeneous coordinates is
$\begin{bmatrix} * & * & * & 0 \\ [0.3em] * & * & * & 0 \\ [0.3em] * & * & * & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$ where the * terms are the terms from the 3D rotations above.
### Scaling
Scaling is the action of multiplying each coordinate of a point by a constant amount. As an example, let $f(x,y,z) = (2x,3y,4z)$. Then
$f((x,y,z)+(a,b,c)) = f(x+a,y+b,z+c) = (2(x+a),3(y+b), 4(z+c)) = (2x,3y,4z)+(2a,3b,4c)$
So this is a linear transformation. If we look at what this transformation does to each of the coordinate vectors, we have
\begin{align} f(1,0,0) = (2,0,0) \\ f(0,1,0) = (0,3,0) \\ f(0,0,1) = (0,0,4) \end{align}
So the matrix for this transformation is $\begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{bmatrix}$ and, in general, a scaling matrix looks like
$\begin{bmatrix} \sigma_x & 0 & 0 \\ 0 & \sigma_y & 0 \\ 0 & 0 & \sigma_z \end{bmatrix}$ where the $\sigma_x , \sigma_y ,\text{ and }\sigma_z$ are the scaling factors for x, y, and z respectively.
In case of only 2D transformations, scaling simply scales down to two dimensions and we simply have
$\begin{bmatrix} \sigma_x & 0 \\ 0 & \sigma_y \end{bmatrix}$
In case we are working with homogeneous coordinates, a scaling transformation only acts on the three primary components and leaves the homogeneous component alone, so we simply have the matrix $\begin{bmatrix} \sigma_x & 0 & 0 & 0 \\ 0 & \sigma_y & 0 & 0 \\ 0 & 0 & \sigma_z & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$ for the scaling transformation.
### Translation
Notice that a translation function cannot be a linear transformation on normal space because it does not take the origin to the origin. These are examples of affine transformations, transformations that are composed of a linear transformation, such as a rotation, scaling, or shear, and a translation. In order to write a translation matrix, we need to use homogeneous coordinates.
If we want to add $T_x$ to the X-coordinate and $T_y$ to the Y-coordinate of every point in 2D space, we see that
$\begin{bmatrix} 1 & 0 & T_x \\ 0 & 1 & T_y \\ 0 & 0 & 1 \end{bmatrix} \times \begin{bmatrix} x \\ y \\ 1 \end{bmatrix} = \begin{bmatrix} x+T_x \\ y+T_y \\ 1 \end{bmatrix}$ so that the matrix $\begin{bmatrix} 1 & 0 & T_x \\ 0 & 1 & T_y \\ 0 & 0 & 1 \end{bmatrix}$ gives a 2D translation.
The 3D case is basically the same, and by the same argument we see that the 3D translation is given by
$\begin{bmatrix} 1 & 0 & 0 & T_x \\ 0 & 1 & 0 & T_y \\ 0 & 0 & 1 & T_z \\ 0 & 0 & 0 & 1 \end{bmatrix}$
hese are linear transformations in the space one degree higher than the geometry you are working with. In fact, the main reason for including homogeneous coordinates is the math for graphics is to be able to handle translations (and thus all basic transformations) as linear transformations represented by matrices.
### Shear
The shear transformation is not widely used in computer graphics, but can be used for things like the oblique view in engineering drawings. The concept of a shear is to add a multiple of one coordinate to another coordinate of each point, or, for example,
$shear(x,y,z) = (x+3y,y,z)$
The matrix for this shear transformation looks like $\begin{bmatrix} 1 & 3 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
In general, the matrix for a shear transformation will look like the identity matrix with one non-zero element A off the diagonal. If A is in row i, column j, then the matrix will add A times the jth coordinate of the vector to the ith coordinate.
For the oblique view of engineering drawings, we look at the shear matrices that add a certain amount of the z-coordinate to each of the x- and y-coordinates. The matrices are
$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ A & 0 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & B & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ A & B & 1 \end{bmatrix}$
that take $(x,y,z) \times (x+Az,y+Bz,z)$. The values of A and B are adjusted to give precisely the view that you want, and the z-term of the result is usually dropped to give the needed 2D view of the 3D object. An example is the classical cabinet view shown below:
To experiment with these transformations, we have two interactive applets. The first one lets you apply the 2D transformations to a 2D figure. Transformation Matrix
The second applet lets you apply the 3D transformations to a 3D figure.
(Currently Unavailable)
### Matrix Inverses
In general, getting the inverse of a matrix can be difficult, but for the basic graphics transformations the inverses are easy because we can simply undo the geometric action of the original transformation.
The inverse of the scaling matrix $\begin{bmatrix} \sigma_x & 0 & 0 \\ 0 & \sigma_y & 0 \\ 0 & 0 & \sigma_z \end{bmatrix}$ is clearly $\begin{bmatrix} \frac{1}{\sigma_x } & 0 & 0 \\ 0 & \frac{1}{\sigma_y } & 0 \\ 0 & 0 & \frac{1}{\sigma_z } \end{bmatrix}$
The inverse of a rotation transformation by angle $\theta$ is clearly the rotation around the same line by the angle $-\theta$.
The inverse of the translation matrix $\begin{bmatrix} 1 & 0 & 0 & T_x \\ 0 & 1 & 0 & T_y \\ 0 & 0 & 1 & T_z \\ 0 & 0 & 0 & 1 \end{bmatrix}$ is clearly $\begin{bmatrix} 1 & 0 & 0 & -T_x \\ 0 & 1 & 0 & -T_y \\ 0 & 0 & 1 & -T_z \\ 0 & 0 & 0 & 1 \end{bmatrix}$
The inverse of the simple shear transformation is also straightforward. Since a simple shear adds a multiple of one vector component to another component, the inverse only needs to subtract that multiple. So we have
$\begin{bmatrix} 1 & A \\ 0 & 1 \end{bmatrix}^{-1} = \begin{bmatrix} 1 & -A \\ 0 & 1 \end{bmatrix}$ and the 3D case is a simple extension of this.
So we have a major observation: If any transformation is the product of basic graphics transformations, it is easy to find the inverse of its matrix (and hence its inverse transformation) as the product of the inverses of the components in reverse order. Or:
$(A \times B \times C)^{-1} = C^{-1} \times B^{-1} \times A^{-1}$
## Transformations and Graphics Environments
Attention – this concept needs a bit of programming background; it involves stacks.
When you are defining the geometry for a graphics image, you will sometimes want to model your scene as a hierarchy of simpler objects. You might have a desk, for example, that is made up of several parts (legs, drawers, shelves); the drawers may have handles or other parts; you may want to put several things on top of the desk; and so on. It’s common to define general models for each simple part, and then to put the pieces together in a common space, called the “world space.”
Each simple part will be defined in its own “model space,” and then you can apply transformations that move all the parts into the right place in the more complex part. In turn, that whole more complex part may be moved into another position, and so on – you can build up quite complex models this way. One common technique for this kind of hierarchical modeling is to build a “scene graph” that shows how everything is assembled and the transformations that are used in the assembly.
As an example, consider the simple picture of a bunny head, basically made up of several spheres. Each sphere is scaled (making it an ellipsoid of the right size), rotated into the right orientation, and then translated into the proper place. The tree next to the picture shows how this is organized.
In order to make this work, you have to apply each set of transformations to its own sphere and then “forget” those transformations so you can apply the transformations for the next piece. You could, of course, use inverses to undo the transformations, but that’s slow and invites roundoff errors from many multiplications.
instead, it is common to maintain a “transformation stack” that holds the history of every place you want to get back to – all the transformations you have saved. This is a stack of 4x4 matrices that implement the transformations. You also have an active transformation to which you apply any new transformations by matrix multiplication.
To save a transformation to get back to later, you push a copy of the current active transformation (as a 4x4 matrix) onto the stack. Later, when you have applied whatever new transformations you need and want to get back to the last saved transformation, you pop the top matrix off the stack and make it the current active transformation. Presto – all the transformations you had used since the corresponding push operation are gone.
So let’s get back to the rabbit. We want to create the rabbit head, and we have whatever active transformation was in place when we wanted to draw the head. Then we have
push
scale
sphere for main part
pop
translate
scale
sphere for left eye
pop
Translate
Scale
sphere for right eye
pop
Translate
Rotate
Scale
sphere for left ear
pop
Translate
Rotate
Scale
sphere for right ear
pop
Notice something important: the transformations are written in the order they are applied, with the one closest to the geometry to be applied first. The right ear operations are really Translate(Rotate(Scale(sphere-for-right-ear)))
If you are not familiar with stacks, this won’t make much sense, but this isn’t necessary to understand basic transformation concepts. A simple way to look at these stacks is to notice that a transformation is a 4x4 matrix or, equivalently, a 16-element array, so maintaining a stack is simply a matter of building an array
float transStack[N][16];
or
float transStack[N][4][4];
where N is the number of transformations one wants to save.
references |
# AREA OF TRIANGLE WORKSHEET 1
## About "Area of triangle worksheet 1"
Area of triangle worksheet 1 :
Area of triangle worksheet 1 is much useful to the students who would like to practice problems on area of triangles using the given three vertices.
## Area of triangle worksheet 1 - Problems
1) Find the area of the triangle whose vertices are (1, 2), (-3, 4) and (-5, -6)
2) If the area of the triangle ABC is 68 square units and the vertices are A(6, 7), B(-4, 1) and C(a, -9) taken in order, then find the value of "a".
3) Using the concept of area of triangle, show that the points A(5, -2), B(4, -1) and C(1, 2) are collinear.
4) If P(x, y) is any point on the line segment joining the points (a, 0), and (0, b), then prove that x/a + y/b = 1. where a ≠ b.
## Area of triangle worksheet 1 - Answers
Problem 1 :
Find the area of the triangle whose vertices are (1, 2), (-3, 4) and (-5, -6)
Solution :
Plot the given points in a rough diagram as given below and take them in order (counter clock wise)
Let the vertices be A(1, 2), B(-3, 4) and C(-5, -6)
Then, we have
(x₁, y₁) = (1, 2)
(x₂, y₂) = (-3, 4)
(x₃, y₃) = (-5, -6)
Area of triangle ABC is
= (1/2) x { [x₁y₂ + xy + xy₁] - [xy + xy + xy₃] }
= (1/2) x { [1.4 + (-3).(-6) + (-5).2] - [(-3).2 + (-5).4 + 1.(-6)] }
= (1/2) x { [4 + 18 - 10] - [-6 - 20 -6] }
= (1/2) x { [12] - [-32] }
= (1/2) x { 12 + 32 }
= (1/2) x { 44 }
= 22 square units.
Hence, are of triangle ABC is 22 square units.
Let us look at the next problem on "Area of triangle worksheet 1"
Problem 2 :
If the area of the triangle ABC is 68 square units and the vertices are A(6, 7), B(-4, 1) and C(a, -9) taken in order, then find the value of "a".
Solution :
Let (x₁, y₁) = (6, 7), (x₂, y₂) = (-4, 1) and (x₃, y₃) = (a, -9)
Given : Area of triangle ABC = 68 square units
(1/2) x { [x₁y₂ + xy + xy₁] - [xy + xy + xy₃] } = 68
Multiply by 2 on both sides,
{ [x₁y₂ + x₂y₃ + x₃y₁] - [x₂y₁ + x₃y₂ + x₁y₃] } = 136
{ [6 + 36 + 7a] - [-28 + a - 54] } = 136
[42 + 7a] - [a - 82] = 136
42 + 7a -a +82 = 136
6a + 124 = 136
6a = 12
a = 2
Hence, the value of "a" is 2.
Let us look at the next problem on "Area of triangle worksheet 1"
Problem 3 :
Using the concept of area of triangle, show that the points A(5, -2), B(4, -1) and C(1, 2) are collinear.
Solution :
Using the concept of area of triangle, if the three points A(x₁, y₁), B(x₂, y₂) and C (x₃, y₃) are collinear, then
x₁y₂ + x₂y₃ + x₃y = xy + x₃y₂ + xy₃
Comparing the given points to A(x₁, y₁), B(x₂, y₂) and C (x₃, y₃), we get
(x₁, y₁) = (5, -2)
(x₂, y₂) = (4, -1)
(x₃, y₃) = (1, 2)
x₁y₂ + x₂y₃ + x₃y₁ = 5x(-1) + 4x2 + 1x(-2)
x₁y₂ + x₂y₃ + x₃y₁ = -5 + 8 -2
x₁y₂ + x₂y₃ + x₃y₁ = 1 --------(1)
xy + x₃y₂ + xy₃ = 4x(-2) + 1x(-1) + 5x(2)
xy + x₃y₂ + xy₃ = -8 - 1 + 10
xy + x₃y₂ + xy₃ = 1 --------(2)
From (1) and (2), we get
x₁y₂ + x₂y₃ + x₃y = xy + x₃y₂ + xy₃
Hence, the three points A, B and C are collinear.
Let us look at the next problem on "Area of triangle worksheet 1"
Problem 4 :
If P(x, y) is any point on the line segment joining the points (a, 0), and (0, b), then prove that x/a + y/b = 1. where a ≠ b.
Solution :
Clearly, the points (x, y), (a, 0), and (0, b) are collinear.
Then, area of the triangle = 0
Since, area of the triangle is zero, we have
x₁y₂ + x₂y₃ + x₃y = xy + x₃y₂ + xy₃ ------(1)
(x₁, y₁) = (x, y)
(x₂, y₂) = (a, 0)
(x₃, y₃) = (0, b)
Plugging the above points in (1),
x.0 + a.b + 0.y = a.y + 0.0 + x.b
0 + a.b + 0 = a.y + 0 + x.b
a.b = a.y + x.b
Divide by ab on bothe sides,
1 = y/b + x/a
or
x/a + y/b = 1
After having gone through the stuff given above, we hope that the students would have understood "Area of triangle worksheet 1".
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# Prime Factors of 11154
Prime Factors of 11154 are 2, 3, 11, 13, and 13
#### How to find prime factors of a number
1. Prime Factorization of 11154 by Division Method 2. Prime Factorization of 11154 by Factor Tree Method 3. Definition of Prime Factors 4. Frequently Asked Questions
#### Steps to find Prime Factors of 11154 by Division Method
To find the primefactors of 11154 using the division method, follow these steps:
• Step 1. Start dividing 11154 by the smallest prime number, i.e., 2, 3, 5, and so on. Find the smallest prime factor of the number.
• Step 2. After finding the smallest prime factor of the number 11154, which is 2. Divide 11154 by 2 to obtain the quotient (5577).
11154 ÷ 2 = 5577
• Step 3. Repeat step 1 with the obtained quotient (5577).
5577 ÷ 3 = 1859
1859 ÷ 11 = 169
169 ÷ 13 = 13
13 ÷ 13 = 1
So, the prime factorization of 11154 is, 11154 = 2 x 3 x 11 x 13 x 13.
#### Steps to find Prime Factors of 11154 by Factor Tree Method
We can follow the same procedure using the factor tree of 11154 as shown below:
So, the prime factorization of 11154 is, 11154 = 2 x 3 x 11 x 13 x 13.
#### What does Prime factor of a number mean?
A prime number in mathematics is defined as any natural number greater than 1, that is not divisible by any number except 1 and the number itself. When we express any number as the product of these prime numbers than these prime numbers become prime factors of that number. Eg- Prime Factors of 11154 are 2 x 3 x 11 x 13 x 13.
#### Properties of Prime Factors
• Two prime factors of a given number are always coprime to each other.
• 2 is the only even prime factor any number can have.
• Two prime factors are always coprime to each other.
• 1 is neither a prime number nor a composite number and also 1 is the factor of every given number. So, 1 is the factor of 11154 but not a prime factor of 11154.
• Is 11154 a prime number?
false, 11154 is not a prime number.
• How to find prime factorization of 11154 easily?
To find prime factorization of 11154 easily you can refer to the steps given on this page step by step.
• Which is the largest prime factors of 11154?
The largest prime factor of 11154 is 13.
• What is the sum of all prime factors of 11154?
Prime factors of 11154 are 2 x 3 x 11 x 13 x 13. Therefore, their sum is 42.
• What is the product of all prime factors of 11154?
Prime factors of 11154 are 2 x 3 x 11 x 13 x 13. Therefore, their product is 11154.
• What numbers are the prime factors of 11154?
Prime factors of 11154 are 2 , 3 , 11 , 13 , 13.
• What is the sum of all odd prime factors of 11154?
Prime factors of 11154 are 2 , 3 , 11 , 13 , 13, out of which 3 , 11 , 13 , 13 are odd numbers. So, the sum of odd prime factors of 11154 is 3 + 11 + 13 + 13 = 40.
• Is there any even prime factor of 11154?
Yes there is a even prime factor of 11154.
• What is the product of all odd prime factors of 11154?
Prime factors of 11154 are 2 , 3 , 11 , 13 , 13, out of which 3 , 11 , 13 , 13 are odd numbers. So, the product of odd prime factors of 11154 is 3 x 11 x 13 x 13 = 5577. |
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# wikiHow to Calculate Volume
The volume of a shape is the measure of how much three-dimensional space that shape takes up.[1] You can also think of the volume of a shape as how much water (or air, or sand, etc.) the shape could hold if it was filled completely. Common units of volume include cubic centimeters (cm3), cubic meters (m3), cubic inches (in3), and cubic feet (ft3).[2] This article will teach you how to calculate the volume of six different three-dimensional shapes that are commonly found on math tests, including cubes, spheres, and cones. You might notice that a lot of the volume formulas share similarities that can make them easier to remember. See if you can spot them along the way!
### Method 1 Calculating the Volume of a Cube
1. 1
Recognize a cube. A cube is a three-dimensional shape that has six identical square faces.[3] In other words, it is a box shape with equal sides all around.
• A 6-sided die is a good example of a cube you might find in your house. Sugar cubes, and children's letter blocks are also usually cubes.
2. 2
Learn the formula for the volume of a cube. Since all of the side lengths of a cube are the same, the formula for the volume of a cube is really easy. It is V = s3 where V stands for volume, and s is the length of the sides of the cube.
• To find s3, simply multiply a by itself 3 times: s3 = s * s * s
3. 3
Find the length of one side of the cube. Depending on your assignment, the cube will either be labeled with this information, or you may have to measure the side length with a ruler. Remember that since it is a cube, all of the side lengths should be equal so it doesn't matter which one you measure.
• If you are not 100% sure that your shape is a cube, measure each of the sides to determine if they are equal. If they are not, you will need to use the method below for Calculating the Volume of a Rectangular Solid.
4. 4
Plug the side length into the formula V = s3 and calculate. For example, if you find that the length of the sides of your cube is 5 inches, then you should write the formula out as follows: V = (5 in)3. 5 in * 5 in * 5 in = 125 in3, the volume of our cube!
5. 5
Be sure to state your answer in cubic units. In the above example, the side length of our cube was measured in inches, so the volume was given in cubic inches. If the side length of the cube had been 3 centimeters, for example, the volume would be V = (3 cm)3, or V = 27cm3.
### Method 2 Calculating the Volume of a Rectangular Prism
1. 1
Recognize a rectangular solid. A rectangular solid, also known as a rectangular prism, is a three-dimensional shape with six sides that are all rectangles.[4] In other words, a rectangular solid is simply a three-dimensional rectangle, or box shape.
• A cube is really just a special rectangular solid in which the sides of all of the rectangles are equal.
2. 2
Learn the formula for calculating the volume of a rectangular solid. The formula for the volume of a rectangular solid is Volume = length * width * height, or V = lwh.
3. 3
Find the length of the rectangular solid. The length is the longest side of the rectangular solid that is parallel to the ground or surface it is resting on. The length may be given in a diagram, or you may need to measure it with a ruler or tape measure.
• Example: The length of this rectangular solid is 4 inches, so l = 4 in.
• Don't worry too much about which side is the length, which is the width, etc. As long as you end up with three different measurements, the math will come out the same regardless of how your arrange the terms.
4. 4
Find the width of the rectangular solid. The width of the rectangular solid is the measurement of the shorter side of the solid, parallel to the ground or surface the shape is resting on. Again, look for a label on the diagram indicating the width, or measure your shape with a ruler or tape measure.
• Example: The width of this rectangular solid is 3 inches, so w = 3 in.
• If you are measuring the rectangular solid with a ruler or tape measure, remember to take and record all measurements in the same units. Don't measure one side in inches another in centimeters; all measurements must use the same unit!
5. 5
Find the height of the rectangular solid. This height is the distance from the ground or surface the rectangular solid is resting on to the top of the rectangular solid. Locate the information in your diagram, or measure the height using a ruler or tape measure.
• Example: The height of this rectangular solid is 6 inches, so h = 6 in.
6. 6
Plug the dimensions of the rectangular solid into the volume formula and calculate. Remember that V = lwh.
• In our example, l = 4, w = 3, and h = 6. Therefore, V = 4 * 3 * 6, or 72.
7. 7
Be sure to express your answer in cubic units. Since our example rectangle was measured in inches, the volume should be written as 72 cubic inches, or 72 in3.
• If the measurements of our rectangular solid were: length = 2 cm, width = 4 cm, and height = 8 cm, the Volume would be 2 cm * 4 cm * 8 cm, or 64cm3.
### Method 3 Calculating the Volume of a Cylinder
1. 1
Learn to identify a cylinder. A cylinder is a three-dimensional shape that has two identical flat ends that are circular in shape, and a single curved side that connects them.[5]
• A can is a good example of a cylinder, so is a AA or AAA battery.
2. 2
Memorize the formula for the volume of a cylinder. To calculate the volume of a cylinder, you must know its height and the radius of the circular base (the distance from the center of the circle to its edge) at the top and bottom. The formula is V = πr2h, where V is the Volume, r is the radius of the circular base, h is the height, and π is the constant pi.
• In some geometry problems the answer will be given in terms of pi, but in most cases it is sufficient to round pi to 3.14. Check with your instructor to find out what she would prefer.
• The formula for finding the volume of a cylinder is actually very similar to that for a rectangular solid: you are simply multiplying the height of the shape by the surface area of its base. In a rectangular solid, that surface area is l * w, for the cylinder it is πr2, the area of a circle with radius r.
3. 3
Find the radius of the base. If it is given in the diagram, simply use that number. If the diameter is given instead of the radius, you simply need to divide the value by 2 to get the radius (d = 2r).
4. 4
Measure the object if the radius is not given. Be aware that getting precise measurement of a circular solid can be a bit tricky. One option is to measure the base of the cylinder across the top with a ruler or tape measure. Do your best to measure the width of the cylinder at its widest part, and divide that measurement by 2 to find the radius.
• Another option is to measure the circumference of the cylinder (the distance around it) using a tape measure or a length of string that you can mark and then measure with a ruler. Then plug the measurement into the formula: C (circumference) = 2πr. Divide the circumference by 2π (6.28) and that will give you the radius.
• For example, if the circumference you measured was 8 inches, the radius would be 1.27in.
• If you need a really precise measurement, you might use both methods to make sure that your measurements are similar. If they are not, double check them. The circumference method will usually yield more accurate results.
5. 5
Calculate the area of the circular base. Plug the radius of the base into the formula πr2. Then multiply the radius by itself one time, and then multiply the product by π. For example:
• If the radius of the circle is equal to 4 inches, the area of the base will be A = π42.
• 42 = 4 * 4, or 16. 16 * π (3.14) = 50.24 in2
• If the diameter of the base is given instead of the radius, remember that d = 2r. You simply need to divide the diameter in half to find the radius.
6. 6
Find the height of the cylinder. This is simply the distance between the two circular bases, or the distance from the surface the cylinder is resting on to its top. Find the label in your diagram that indicates the height of the cylinder, or measure the height with a ruler or tape measure.
7. 7
Multiply the area of the base times the height of the cylinder to find the volume. Or you can save a step and simply plug the values for the cylinder's dimensions into the formula V = πr2h. For our example cylinder with radius 4 inches and height 10 inches:
• V = π4210
• π42 = 50.24
• 50.24 * 10 = 502.4
• V = 502.4
8. 8
Remember to state your answer in cubic units. Our example cylinder was measured in inches, so the volume must be expressed in cubic inches: V = 502.4in3. If our cylinder had been measured in centimeters, the volume would be expressed in cubic centimeters (cm3).
### Method 4 Calculating the Volume of a Regular Square Pyramid
1. 1
Understand what a regular pyramid is. A pyramid is a three-dimensional shape with a polygon for a base, and lateral faces that taper at an apex (the point of the pyramid).[6] A regular pyramid is a pyramid in which the base of the pyramid is a regular polygon, meaning that all of the sides of the polygon are equal in length, and all of the angles are equal in measure.[7]
• We most commonly imagine a pyramid as having a square base, and sides that taper up to a single point, but the base of a pyramid can actually have 5, 6, or even 100 sides!
• A pyramid with a circular base is called a cone, which will be discussed in the next method.
2. 2
Learn the formula for the volume of a regular pyramid. The formula for the volume of a regular pyramid is V = 1/3bh, where b is the area of the base of the pyramid (the polygon at the bottom) and h is the height of the pyramid, or the vertical distance from the base to the apex (point).
• The volume formula is the same for right pyramids, in which the apex is directly above the center of the base, and for oblique pyramids, in which the apex is not centered.
3. 3
Calculate the area of the base. The formula for this will depend on the number of sides the base of the pyramid has. In the pyramid in our diagram, the base is a square with sides that are 6 inches in length. Remember that the formula for the area of a square is A = s2 where s is the length of the sides. So for this pyramid, the area of the base is (6 in) 2, or 36in2.
• The formula for the area of a triangle is: A = 1/2bh, where b is the base of the triangle and h is the height.
• It is possible to find the area of any regular polygon using the formula A = 1/2pa, where A is the area, p is the perimeter of the shape, and a is the apothem, or distance from the center of the shape to the midpoint of any of its sides. This is a pretty involved calculation that goes beyond the scope of this article, but check out Calculate-the-Area-of-a-Polygon for some great instructions on how to use it. Or you can make your life easy and search for a Regular Polygon Calculator online.[8]
4. 4
Find the height of the pyramid. In most cases, this will be indicated in the diagram. In our example, the height of the pyramid is 10 inches.
5. 5
Multiply the area of the base of the pyramid by its height, and divide by 3 to find the volume. Remember that the formula for the volume is V = 1/3bh. In our example pyramid, that had a base with area 36 and height 10, the volume is: 36 * 10 * 1/3, or 120.
• If we had a different pyramid, with a pentagonal base with area 26, and height of 8, the volume would be: 1/3 * 26 * 8 = 69.33.
6. 6
Remember to express your answer in cubic units. The measurements of our example pyramid were given in inches, so its volume must be expressed in cubic inches, 120in. If our pyramid had been measured in meters, the volume would be expressed in cubic meters (m3) instead.3
### Method 5 Calculating the Volume of a Cone
1. 1
Learn the properties of a cone. A cone is a 3-dimesional solid that has a circular base and a single vertex (the point of the cone). Another way to think of this is that a cone is a special pyramid that has a circular base.[9]
• If the vertex of the cone is directly above the center of the circular base, the cone is called a "right cone". If it is not directly over the center, the cone is called an "oblique cone." Fortunately, the formula for calculating the area of a cone is the same whether it is right or oblique.
2. 2
Know the formula for calculating the volume of a cone. The formula is V = 1/3πr2h, where r is the radius of the circular base of the cone, h is the height of the cone, and π is the constant pi, which can be rounded to 3.14.
• The πr2 part of the formula refers to the area of the circular base of the cone. The formula for the volume of the cone is thus 1/3bh, just like the formula for the volume of a pyramid in the method above!
3. 3
Calculate the area of the circular base of the cone. To do this, you need to know the radius of the base, which should be listed in your diagram. If you are instead given the diameter of the circular base, simply divide that number by 2, since the diameter is simply 2 times the radios (d = 2r). Then plug the radius into the formula A = πr2 to calculate the area.
• In the example in the diagram, the radius of the circular base of the cone is 3 inches. When we plug that into the formula we get: A = π32.
• 32 = 3 *3, or 0, so A = 9π.
• A = 28.27in2
4. 4
Find the height of the cone. This is the vertical distance between the base of the cone, and its apex. In our example, the height of the cone is 5 inches.
5. 5
Multiply the height of the cone by the area of the base. In our example, the area of the base is 28.27in2 and the height is 5in, so bh = 28.27 * 5 = 141.35.
6. 6
Now multiply the result by 1/3 (or simply divide by 3) to find the volume of the cone. In the above step, we actually calculated the volume of the cylinder that would be formed if the walls of the cone extended straight up to another circle, instead of slanting in to a single point. Dividing by 3 gives us the volume of just the cone itself.
• In our example, 141.35 * 1/3 = 47.12, the volume of our cone.
• To restate it, 1/3π325 = 47.12
7. 7
Remember to express your answer in cubic units. Our cone was measured in inches, so its volume must be expressed in cubic inches: 47.12in3.
### Method 6 Calculating the Volume of a Sphere
1. 1
Spot a sphere. A sphere is a perfectly round three-dimensional object, in which every point on the surface is an equal distance from the center. In other words, a sphere is a ball-shaped object.[10]
2. 2
Learn the formula for the volume of a sphere. The formula for the volume of a sphere is V = 4/3πr3 (stated: "four-thirds times pi r-cubed") where r is the radius of the sphere, and π is the constant pi (3.14).[11]
3. 3
Find the radius of the sphere. If the radius is given in the diagram, then finding r is simply a matter of locating it. If the diameter is given, you must divide this number by 2 to find the radius. For example, the radius of the sphere in the diagram is 3 inches.
4. 4
Measure the sphere if the radius is not given. If you need to measure a spherical object (like a tennis ball) to find the radius, first find a piece of string large enough to wrap around the object. Then wrap the string around the object at its widest point and mark the points where the string overlaps itself. Then measure the string with a ruler to find the circumference. Divide that value by 2π, or 6.28, and that will give you the radius of the sphere.
• For example, if you measure a ball and find its circumference is 18 inches, divide that number by 6.28 and you will find that the radius is 2.87in.
• Measuring a spherical object can be a little tricky, so you might want to take 3 different measurements, and then average them together (add the three measurements together, then divide by 3) to make sure you have the most accurate value possible.
• For example, if your three circumference measurements were 18 inches, 17.75 inches, and 18.2 inches, you would add those three values together (18 + 17.5 + 18.2 = 53.95) and divide that value by 3 (53.95/3 = 17.98). Use this average value in your volume calculations.
5. 5
Cube the radius to find r3. Cubing a number simply means multiplying the number by itself 3 times, so r3 = r * r * r. In our example, r = 3, so r3 = 3 * 3 * 3, or 27.
6. 6
Now multiply your answer by 4/3. You can either use your calculator, or do the multiplication by hand and then simplify the fraction. In our example, multiplying 27 by 4/3 = 108/3, or 36.
7. 7
Multiply the result by π to find the volume of the sphere. The last step in calculating the volume is simply to multiply the result so far by π. Rounding π to two digits is usually sufficient for most math problems (unless your teacher specified otherwise,) so multiply by 3.14 and you have your answer.
• In our example, 36 * 3.14 = 113.09.
8. 8
Express your answer in cubic units. In our example, the measurement of the radius of the sphere was in inches, so our answer is actually V = 113.09 cubic inches (113.09 in3).
## Community Q&A
Search
• How do I calculate the volume of compound shapes?
wikiHow Contributor
If the compound shapes are made up of basic geometric solids, then you can try dissecting them into their simpler parts. Their volumes will be additive.
• Are there alternate methods for calculating volume?
wikiHow Contributor
Yes -- you could divide the mass of the object by the density (assuming you know both).
• How do I calculate the volume of a triangular prism?
wikiHow Contributor
Calculate the area of the base (the triangle) and multiply by the height (the dimension that is not part of the triangle).
• Is there a formula that works for all shapes?
No.
• If I have the volume of 250 gallons and the diameter of the cylinder, how would I calculate the unknown height? I can't remember how to solve for the unknown from your equation of V = Pi x r(squared) x h.
Convert gallons to cubic feet (0.1337 cubic foot in a US gallon, and .16 cubic foot in an Imperial gallon). Find the cross-sectional area of the cylinder by finding the radius (half the diameter), squaring it and multiplying by pi. Then divide the volume you calculated by the area you calculated. That gives you the height of the cylinder.
• What is the diameter of the base of the cylinder if the volume of the cylinder is 81 pi cm3?
wikiHow Contributor
Volume = base area * height = diameter*pi/4*height. Diameter = 4*volume/(pi*height). You can't find the base diameter without knowing the height.
• What method lets us derermine the volume of an oddly shaped object?
Measure the object's water displacement.
• Can I calculate the volume of a box by observing the speed at which it fills with water?
wikiHow Contributor
You would need to know the flowrate of the incoming water. Example: if you know that a pipe carrying 1l/s of water fills the box in 10 seconds, your box is 10l big.
• How do I calculate the volume of a 6-sided cube with different base and top areas?
In the case of a cube, the base area is always equal to the top area.
• If one makes a sizable dent in a gallon bucket, does it still hold a gallon?
It's not likely.
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A frequency distribution shows the number of elements in a data set that belong to each class. In a relative frequency distribution, the value assigned to each class is the proportion of the total data set that belongs in the class.
For example, suppose that a frequency distribution is based on a sample of 200 supermarkets. It turns out that 50 of these supermarkets charge a price between \$8.00 and \$8.99 for a pound of coffee. In a relative frequency distribution, the number assigned to this class would be 0.25 (50/200). In other words, that's 25 percent of the total.
Here's a handy formula for calculating the relative frequency of a class:
Class frequency refers to the number of observations in each class; n represents the total number of observations in the entire data set. For the supermarket example, the total number of observations is 200.
The relative frequency may be expressed as a proportion (fraction) of the total or as a percentage of the total. For example, the following table shows the frequency distribution of gas prices at 20 different stations.
Frequency Distribution of Prices for 20 Gas Stations
Gas Prices (\$/Gallon) Number of Gas Stations
\$3.50–\$3.74 6
\$3.75–\$3.99 4
\$4.00–\$4.24 5
\$4.25–\$4.49 5
Based on this information, you can use the relative frequency formula to create the next table, which shows the relative frequency of the prices in each class, as both a fraction and a percentage.
Relative Frequencies for Gas Station Prices
Gas Prices (\$/Gallon) Number of Gas Stations Relative Frequency
(fraction)
Relative Frequency
(percent)
\$3.50–\$3.74 6 6/20 = 0.30 30%
\$3.75–\$3.99 4 4/20 = 0.20 20%
\$4.00–\$4.24 5 5/20 = 0.25 25%
\$4.25–\$4.49 5 5/20 = 0.25 25%
With a sample size of 20 gas stations, the relative frequency of each class equals the actual number of gas stations divided by 20. The result is then expressed as either a fraction or a percentage. For example, you calculate the relative frequency of prices between \$3.50 and \$3.74 as 6/20 to get 0.30 (30 percent). Similarly, the relative frequency of prices between \$3.75 and \$3.99 equals 4/20 = 0.20 = 20 percent.
One of the advantages of using a relative frequency distribution is that you can compare data sets that don't necessarily contain an equal number of observations. For example, suppose that a researcher is interested in comparing the distribution of gas prices in New York and Connecticut. Because New York has a much larger population, it also has many more gas stations. The researcher decides to choose 1 percent of the gas stations in New York and 1 percent of the gas stations in Connecticut for the sample. This turns out to be 800 in New York and 200 in Connecticut. The researcher puts together a frequency distribution as shown in the next table.
Frequency Distribution of Gas Prices in New York and Connecticut
Price New York Gas Stations Connecticut Gas Stations
\$3.00–\$3.49 210 48
\$3.50–\$3.99 420 96
\$4.00–\$4.49 170 56
Based on this frequency distribution, it's awkward to compare the distribution of prices in the two states. By converting this data into a relative frequency distribution, the comparison is greatly simplified, as seen in the final table.
Relative Frequency Distribution of Gas Prices in New York and Connecticut
Price New York Gas Stations Relative Frequency Connecticut Gas Stations Relative Frequency
\$3.00–\$3.49 210 210/800 = 0.2625 48 48/200 = 0.2400
\$3.50–\$3.99 420 420/800 = 0.5250 96 96/200 = 0.4800
\$4.00–\$4.49 170 170/800 = 0.2125 56 56/200 = 0.2800
The results show that the distribution of gas prices in the two states is nearly identical. Roughly 25 percent of the gas stations in each state charge a price between \$3.00 and \$3.49; about 50 percent charge a price between \$3.50 and \$3.99; and about 25 percent charge a price between \$4.00 and \$4.49. |
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# Class 11 RD Sharma Solutions – Chapter 24 The Circle – Exercise 24.2
• Last Updated : 02 Nov, 2022
### Question 1(i). Find the coordinates of the centre and radius of the circle x2 + y2 + 6x – 8y – 24 = 0.
Solution:
As we know that the general equation of circles is
x2 + y2 + 2gx + 2fy + c = 0 …..(i)
Centre = (-g, -f)
Given equation of circle is x2 + y2 + 6x – 8y – 24 = 0
On comparing with eq(i), we get
Hence, g = 3, f = -4, c = -24
So,
Centre = (-3, 4)
### Question 1(ii). Find the coordinates of the centre and radius of the circle 2x2 + 2y2 – 3x + 5y = 7.
Solution:
As we know that the general equation of circles is
x2 + y2 + 2gx + 2fy + c = 0 …..(i)
Centre = (-g, -f)
Given equation of circle is 2x2 + 2y2 – 3x + 5y – 7 = 0
x2 + y2 – 3/2x + 5/2y – 7/2 = 0
On comparing with eq(i), we get
Hence, g = -3/4, f = 5/4, c = -7/2
So, center = (3/4, -5/4)
### Question 1(iii). Find the coordinates of the centre and radius of the circle 1/2(x2 + y2) + x cosθ + y sinθ – 4 = 0.
Solution:
As we know that the general equation of circles is
x2 + y2 + 2gx + 2fy + c = 0 …..(i)
Centre = (-g, -f)
Given equation of circle is 1/2(x2 + y2) + x cosθ + y sinθ – 4 = 0
⇒ x2 + y2 + 2x cosθ + 2sin θ – 8 = 0
On comparing with eq(i), we get
Hence, g = cos θ, f = sin θ, c = -8
So, centre = (-cos θ, -sinθ)
### Question 1(iv). Find the coordinates of the centre and radius of the circle x2 + y2 – ax – by = 0.
Solution:
As we know that the general equation of circles is
x2 + y2 + 2gx + 2fy + c = 0 …..(i)
Centre = (-g, -f)
Given equation of circle is x2 + y2 – ax – by = 0
On comparing with eq(i), we get
Hence, g = -a/2, f = -b/2, c = 0
So, centre = (a/2, b/2)
### Question 2(i). Find the equation of the circle passing through the points (5, 7), (8, 1), and (1, 3).
Solution:
Given that, the circle pass through points P(5, 7), Q(8, 1), and R(1, 3)
As we know that the general equation of circles is
x2 + y2 + 2gx + 2fy + c = 0 …..(i)
Since, P, Q, and R lies on eq(i)
So,
25 + 49 + 10g + 14f + c = 0 …….(ii)
64 + 1 + 16g + 2f + c = 0 …….(iii)
1 + 9 + 2g + 6f + c = 0 ……(iv)
Now on solving eq (ii), (iii), and (iv), we get,
g = -29/6, f = -19/6, c = 56/3
Now put all these values in eq(i), we get
The equation of circle is
x2 + y2 – 29/3x – 19/6y + 56/3 = 0
3(x2 + y2) – 29x – 19y + 56 = 0
### Question 2(ii). Find the equation of the circle passing through the points (1, 2), (3, -4), and (5, -6).
Solution:
Given that, the circle pass through points P(1, 2), Q(3, -4), and R(5, -6)
As we know that the general equation of circles is
x2 + y2 + 2gx + 2fy + c = 0 …..(i)
Since, P, Q, and R lies on eq(i)
So,
1 + 4 + 2g + 4f + c = 0 …..(ii)
9 + 16 + 6g – 8f + c = 0 …..(iii)
25 + 36 + 10g – 12f + c = 0 …..(iv)
Now on solving eq (ii), (iii), and (iv), we get,
g = -11, f = -2, and c = 25
Now put all these values in eq(i), we get
The equation of circle is
x2 + y2 – 22x – 4y + 25 = 0
### Question 2(iii). Find the equation of the circle passing through the points (5, -8), (-2, 9), and (2, 1).
Solution:
Given that, the circle pass through points P(5, -8), Q(-2, 9), and R(2, 1)
As we know that the general equation of circles is
x2 + y2 + 2gx + 2fy + c = 0 …..(i)
Since, P, Q, and R lies on eq(i)
So,
25 + 64 + 10g – 16f + c = 0 …..(ii)
4 + 81 – 4g + 18f + c = 0 …..(iii)
4 + 1 + 4g + 2f + c = 0 …..(iv)
Now on solving eq (ii), (iii), and (iv), we get,
g = 58, f = 24, and c = -285
Now put all these values in eq(i), we get
The equation of circle is
x2 + y2 + 116x + 48y – 285 = 0
### Question 2(iv). Find the equation of the circle passing through the points (0, 0), (-2, 1), and (-3, 2).
Given that, the circle pass through points P(0, 0), Q(-2, 1), and R(-3, 2)
As we know that the general equation of circles is
x2 + y2 + 2gx + 2fy + c = 0 …..(i)
Since, P, Q, and R lies on eq(i)
So,
0 + 00 + 0 + c = 0 …..(ii)
4 + 1 – 4x + 2y + c = 0 …..(iii)
9 + 4 – 6x + 4y + c = 0 …..(iv)
Now on solving eq (ii), (iii), and (iv), we get,
g = -3/2, f = -11/2, and c = 0
Now put all these values in eq(i), we get
The equation of circle is
x2 + y2 – 3x – 11y = 0
### Question 3. Find the equation of the circle which passes through (3, -2), (-2, 0) and has its centre on the line 2x – y = 3.
Solution:
It is given that, the circle passing through P(3, -2) and Q(-2, 0) and having its centre on 2x – y = 3.
As we know that the general equation of circles is
x2 + y2 + 2gx + 2fy + c = 0 …..(i)
Since the circle passes through (3, -2) and also (-2, 0)
So,
9 + 4 + 6g – 4f + c = 0 …..(ii)
4 + 0 – 4g + 0 + c = 0 …..(iii)
Also, the centre of the circle lies on 2x – y = 3
-2g + f = 3 …..(iv)
Now on solving eq (ii), (iii), and (iv), we get,
g = 3/2, f = 6, and c = 2
Now put all these values in eq(i), we get
The equation of circle is
x2 + y2 + 3x + 12y + 2 = 0
### Question 4. Find the equation of the circle which passes through points (3, 7), (5, 5), and has its centre on the line x – 4y = 1.
Solution:
It is given that, the circle passing through P(3, 7) and Q(5, 5) and having its centre on x – 4y = 1
As we know that the general equation of circles is
x2 + y2 + 2gx + 2fy + c = 0 …..(i)
Since the circle passes through P and Q
so,
9 + 49 + 7g + 14f + c = 0 …..(ii)
25 + 25 + 10g + 10f + c = 0 …..(iii)
Also, the centre of the circle lies on x – 4y = 1
so, -g + 4f = 1 …..(iv)
Now on solving eq (ii), (iii), and (iv), we get,
g = 3, f = 1, and c = -90
Now put all these values in eq(i), we get
The equation of circle is
x2 + y2+ 6x + 2y – 90 = 0
### Question 5. Show that the points (3, -2), (1, 0), (-1, -2) and (1, -4) are concyclic.
Solution:
Given that P (3, -2), Q(1, 0), R(-1, -2), and S(1,-4)
As we know that the general equation of circles is
x2 + y2 + 2gx + 2fy + c = 0 …..(i)
Since the circle passes through P, Q and R
So,
9 + 4 + 6g – 4f + c = 0 …..(ii)
1 + 0 + 2g – 0 + c = 0 …..(iii)
1 + 4 – 2g – 4f + c = 0 …..(iv)
Now on solving eq (ii), (iii), and (iv), we get,
g = -1, f = 2 and c = 1
Now put all these values in eq(i), we get
The equation of circle is
x2 + y2 – 2x + 4y + 1 = 0 …..(v)
Here, we clearly see that point S(1,-4) satisfy eq(v)
Hence, points P, Q, R, and S are concyclic
### Question 6. Show that the point (5, 5), (6, 4), (-2, 4), and (7, 1) all lie on a circle, and find its equation, centre, and radius.
Solution:
As we know that the general equation of circles is
x2 + y2 + 2gx + 2fy + c = 0 …..(i)
Centre = (-g, -f)
Therefore, P(5, 5), Q(6, 4), and R(-2, 4) lie on eq(i),
So,
25 + 25 + 10g + 10f + c = 0 …..(ii)
36 + 16 + 12g + 8f + c = 0 …..(iii)
4 + 16 + 4g + 8f + c = 0 …..(iv)
Now on solving eq (ii), (iii), and (iv), we get,
g = -2, f = -1, c = -20
Now put all these values in eq(i), we get
The equation of circle is
x2 + y2 – 4x – 2y – 20 = 0 …..(v)
Here, we clearly see that point S(7, 1)
Hence, points P, Q, R, and S are concyclic
Now, the centre = (-g, -f) = (2, 1)
### Question 7(i). Find the equation of the circle which circumscribes the triangle formed by the lines x + y + 3 = 0, x – y + 1 = 0, and x = 3.
Solution:
The equations of lines are
x + y = -3 …..(i)
x – y = -1 …..(ii)
x = 3 …..(iii)
Let us considered A, B, and C are the point of intersection of lines (i), (ii), and (iii)
So, Point A(-2,-1), B(3, 4), and C(3,-6)
As we know that the general equation of circles is
x2 + y2 + 2gx + 2fy + c = 0 …..(iv)
So, the circle circumscribing the ∆ABC
4 + 1 – 4g – 2f + c = 0 …..(v)
9 + 16 + 6g + Bf + c = 0 …..(vi)
9 + 36 + 6g – 12f + c = 0 …..(vii)
Now on solving eq (v), (vi), and (vii), we get,
g = -3, f = 1, c = -15
Now put all these values in eq(iv), we get
The equation of circle is
x2 + y2 – 6x + 2y – 15 = 0
### Question 7(ii). Find the equation of the circle which circumscribes the triangle formed by the lines 2x + y – 3 = 0, x + y – 1 = 0, and 3x + 2y – 5 = 0.
Solution:
The equations of lines are
2x + y = 3 …..(i)
x + y = 1 …..(ii)
3x + 3y = 5 …..(iii)
Let us considered A, B, and C are the point of intersection of lines (i), (ii), and (iii)
So, A = (2, -1)
B = (3, -2)
C = (1, 1)
As we know that the general equation of circles is
x2 + y2 + 2gx + 2fy + c = 0 …..(iv)
So, the circle circumscribing the ∆ABC
4 + 1 + 4g – 2f + c = 0 …..(v)
9 + 4 + 6g – 4f + c = 0 …..(vi)
1 + 1 + 2g + 2f + c = 0 …..(vii)
Now on solving eq (v), (vi), and (vii), we get,
g = -13/2, f = -5/2, and c = 16
Now put all these values in eq(iv), we get
The equation of circle is
x2 + y2 – 13x – 5y + 16 = 0
### Question 7(iii). Find the equation of the circle which circumscribes the triangle formed by the lines x + y = 2, 3x – 4y = 6, and x – y = 0.
Solution:
The equations of lines are
x + y = 2 …..(i)
3x – 4y = 6 …..(ii)
x – y = 0 …..(iii)
Let us considered A, B, and C are the point of intersection of lines (i), (ii), and (iii)
So, A = (2, 0)
B = (-6, -6)
C = (1, 1)
As we know that the general equation of circles is
x2 + y2 + 2gx + 2fy + c = 0 …..(iv)
So, the circle circumscribing the ∆ABC
4 + 4g + c = 0 …..(v)
36 + 36 – 12g – 12f + c = 0 …..(vi)
1 + 1 + 2g + 2f + c = 0 …..(vii)
Now on solving eq (v), (vi), and (vii), we get,
g = 2, f = 3, and c = -12
Now put all these values in eq(iv), we get
The equation of circle is
x2 + y2 + 4x + 6y – 12 = 0
### Question 7(iv). Find the equation of the circle which circumscribes the triangle formed by the lines y = x + 2, 3y = 4x, and 2y = 3x.
Solution:
The equations of lines are
y = x + 2 …..(i)
3y = 4x …..(ii)
2y = 3x …..(iii)
Let us considered A, B, and C are the point of intersection of lines (i), (ii), and (iii)
So, A = (6, 8)
B = (4, 6)
C = (0, 0)
As we know that the general equation of circles is
x2 + y2 + 2gx + 2fy + c = 0 …..(iv)
So, the circle circumscribing the ∆ABC
12g + 16f + c = -100 …..(v)
8g + 12f + c = -52 …..(vi)
c = 0 …..(vii)
Now on solving eq (v), (vi), and (vii), we get,
f = 11 and g = -23
Now put all these values in eq(iv), we get
The equation of circle is
x2 + y2 – 46x + 22y = 0.
### Question 8. Prove that the centre of the three circles x2 + y2 – 4x – 6y – 12 = 0, x2 + y2 + 2x + 4y – 10 = 0, and x2 + y2 -10x – 16y – 1 = 0 are collinear.
Solution:
The given equation of circles are,
x2 + y2 – 4x – 6y – 12 = 0 …….(i)
x2 + y2 + 2x + 4y – 10 = 0 …….(ii)
x2 + y2 – 10x – 16y – 1 = 0 …….(iii)
Let O1, O2 and O3 are the centres of (i), (ii), and (iii)
O1 = (-g, -f) = (2, 3)
O2 = (-g, -f) = (-1, -2)
O3 = (-g, -f) = (5, 8)
O1, O2 and O3 will be collinear if ar(∆ O1O2O3) = 0
R2 ⇒ R2 ⇒ -R1
R3 ⇒ R3 ⇒ -R1
O1, O2, and O3 are collinear
### Question 9. Prove that the radii of the circles x2 + y2 = 1, x2 + y2 – 2x – 6y – 6 = 0, and x2 + y2 – 4x – 12y – 9 = 0 are in A.P.
Solution:
The given equation of circles are,
x2 + y2 = 1 ———–(i)
x2 + y2 – 2x – 6y – 6 = 0 ———-(ii)
x2 + y2 – 4x – 12y – 9 = 0 ———-(iii)
Let us considered R1, R2, and R3 are the radii of (i), (ii), and (iii)
So, R1 = 1
R2 =
R3 =
As we know that if a, b, c are in AP, then b = a + b/2
so, a = 1, b = 4, c = 7, b = 1 + 7/2 = 4
Therefore 1, 4, 7 are in AP.
Hence, the radius of the three circles lie in AP.
### Question 10. Find the equation of the circle which passes through the origin and cuts off chord of length 4 and 6 on the positive side of the x-axis and y-axis respectively.
Solution:
It is given that a circle that passes through origin O(0, 0) and
cut off on intercepts of length 4 on x-axis and 6 on y-axis.
So, OA = 4
OB = 6
Let us assume C be the centre of the circle and
CM and CN are perpendicular line drawn on OA and OB
so, the coordinate of A = (4, 0) and B = (0, 6)
The coordinates of M = (2, 0) and N = (0, 3)
And the coordinates of C = (2, 3)
Now in ∆OCM,
Using Pythagoras theorem
OC2 = OM2 + CM2
= 22 + 32
= 4 + 9
OC = √13
Hence, the required circle is
(x – 2)2 + (y – 3)2 = 13
x2 + y2 – 4x – 6y = 0
### Question 11. Find the equation of the circle concentric with circles x2 + y2 – 6x + 12y + 15 = 0 and double of its area.
Solution:
The given equation of circle is
x2 + y2 – 6x + 12y + 15 = 0 …….(i)
so, centre = (-g, -f) = (3, 6)
Now, the required equation of circle in concentric with (i)
that means both have same centre (3,-6)
The area of required circle = 2 * Area of (i)
πR2 = 2 * π(√30)2
R2 = 60
R = 2√15
Hence, the required circle is
(x – 3)2 + (y + 6)2 = 60
x2 + y2 – 6x + 12y – 15 = 0
### Question 12. Find the equation of the circle which passes through the points (1, 1), (2, 2), and whose radius is 1. Show that there are two such circles.
Solution:
As we know that the general equation of circles is
x2 + y2 + 2gx + 2fy + c = 0 …..(i)
It is given that the points P(1, 1) and Q(2, 2) passes through it eq(1)
So,
1 + 1 + 2g + 2f + c = 0 …..(ii)
4 + 4 + 4g + 4f + c = 0 …..(iii)
It is given that the radius = 1
⇒
⇒ g2 + f2 – c = 1 ……(iv)
Now from eq(ii) and (iii), we get
g + f + c/2 = -1
g + f + c/4 = -2
Now on subtracting both the equations, we get
c = 4
g + f = -3 …….(v)
Now on solving eq(v) and (vi), we get
g = -1 or -2 and f = -2 or -1
Hence, the required circle is
x2 + y2 – 2x – 4y + 4 = 0
or
x2 + y2 – 4x – 2y + 4 = 0
### Question 13. Find the equation of the circle concentric with x2 + y2 – 4x – 6x – 6y – 3 = 0 and which touches the y-axis.
Solution:
The given equation of circle is
x2 + y2 – 4x – 6y – 3 = 0 …..(i)
so, centre = (-g, -f) = (2, 3)
The required circle is concentric with eq(i)
so, both have centre(2, 3)
Also, the required circle touches y-axis at A.
So, CA = radius = 2
Hence, the required circle is
(x – 2)2 + (y – 3)2 = 4
x2 + y2 – 4x – 6y + 9 = 0
### Question 14. If a circle passes through the point (0, 0), (a, 0), (0, b), then find the coordinates of its centre.
Solution:
From the given figure, CA, CO, and AB are the equal radii of the circle
So,
CA = CO = CB = r
Also, OCA is an isosceles triangle, and CM is the perpendicular bisector to the OA.
Hence OM = a/2
Similarly, CN is the perpendicular bisector to the OB
So, ON = b/2
from the above figure, it is clear that
OM = x = a/2
ON = y = b/2
Hence the centre of the circle is c(a/2, b/2)
### Question 15. Find the equation of the circle which passes through the point (2, 3) and (4, 5), and the centre lies on the straight line y – 4x + 3 = 0.
Solution:
It is given that a circle which passes through the point P(2, 3) and Q(4, 5) and
the centre lies on the straight line y – 4x + 3 = 0.
As we know that the general equation of circles is
x2 + y2 + 2gx + 2fy + c = 0 …..(i)
Since the circle passes through P, and Q
So,
13 + 14g + 6f + c = 0 …..(ii)
41 + 8g + 10f + c = 0 …..(iii)
Centre (-g, -f) lies on y – 4x + 3 = 0
So, -f + 4g = -3 …..(iv)
Now on subtracting eq(ii) from (iii), we get
28 + 4g + 4f = 0 …..(v)
On solving eq(iv) and (v) we get,
f = -5 and g = -2
Now put all these values in eq(i), we get
The equation of circle is
41 – 16 – 50 + c = 0
c = 25
Hence, the required equation of the circle is,
x2 + y2 – 4x – 10y + 25 = 0
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# Property of Equality
Last updated date: 16th Apr 2024
Total views: 198.6k
Views today: 1.98k
## What is Equality in Maths?
In Mathematics, two things are equal if and only if they are exactly similar in every way i.e. they have the same Mathematical value and Mathematical properties. Mathematics uses the equality sign (=) to represent equality. For example, the statement A = B implies that A and B are equal.
## What is the Property of Equality?
Property of equality is the fact or truth that is applied to two or more quantities related by an equal sign. Let us learn different properties of equality with examples.
The Addition property of equality states that when the same number is added from both sides of the equation, the equation still holds true. In other words, it states that if,
X = Y,
then X + Z = Y + Z
Example:
We know that, 4 + 3 = 7
Here, if we add 5 to both sides of the equation, the equation still holds true.
i.e. 4 + 3 + 5 = 7 + 5
12 = 12
LHS = RHS
## Subtraction Property of Equality
The subtraction property of equality states that when the same number is subtracted from both sides of the equation, the equation still holds. In other words it states that if,
X = Y, then X - Z = Y - Z
Example:
We know that, 9 - 3 = 6
Here, if we subtract 2 to both sides of the equation, the equation still holds
9 - 3 - 2 = 6 - 2
4 = 4
LHS = RHS
## Multiplication Property of Equality
The multiplication property of equality states that when both sides of an equation are multiplied by the same number, the two sides remain equal.
In other words, it states that if x, y, and z are real numbers such that x = y, then
a x c = b x c .
Example, If a = 10, b = 10, and c = 12, then
a x c = b x
10 x 12 = 10 x 12
120 = 120
Therefore, LHS = RHS
## Division Property of Equality
The division property of equality states that when both sides of an equation are divided by the same non-zero number, two sides remain equal. In other words, it states that if a = b, and c ≠ 0, then a ÷ c = b ÷ c
Example:
If a = 10, b = 10, and c = 5, then
a ÷ c = b ÷ c
10 ÷ 5 = 10 ÷ 5
2 = 2
Therefore, LHS = RHS
## Symmetric Property of Equality
The symmetric property of equality states that if we interchange the sides of an equation, the equation still holds. In other words, it states that if X = Y, then Y = X.
Example: If 3 + 5 = 8, then 8 =?
3 + 5 = 8 or
8 = 5 + 3
## Reflexive Property of Equality
The reflexive property of equality states that a number is always equal to itself. In other words, it states that if x is a number, then x = x.
Example:
2 = 2
## Transitive Property of Equality
The transitive property of equality states that when given 3 numbers a, b, and c then,
a = b, b= c, then a = c.
Example:
If a = b, and b = 3, then c is also equal to 3 according to the transitive property of equality.
## Substitution Property of Equality
The substitution property of equality states that if two quantities are equal, then one can replace the other in any Mathematical equation or expression.
In other words, it states that if a = b, then b can be substituted for a, in any Mathematical expression.
Example:
If a = 2 and a + 3 = 5, then 2 can be substituted in a + 3 = 5 to get 2 + 3 = 5.
## Conclusion
In short, the property of equality is the truth about any quantity that is related by an equal sign. By understanding the property of equality thoroughly, we are able to simplify, manipulate, balance, and solve Mathematical equations easily. Also, we are able to draw conclusions supported by valid reasons.
## FAQs on Property of Equality
1. What does the equal sign (=) represent in the property of equality?
The equal sign (=) in the property of equality states the equivalence relation between both the quantities. In other words, it states what is there on the left side of the equation is equal to the right side of an equation.
2. What does the equation mean in the property of equality?
Equations are Mathematical statements that combine two expressions of equal value. We solve the algebraic equation by isolating the variables on the side of the equation using the precise property of equality. To check the solution for algebraic equations, substitute the value into an original equation. |
# How do you simplify sqrt5(3+sqrt15)?
Jun 3, 2017
See a solution process below:
#### Explanation:
First, eliminate the parenthesis by multiplying each term within the parenthesis by the term outside the parenthesis:
$\textcolor{red}{\sqrt{5}} \left(3 + \sqrt{15}\right) \implies$
$\left(\textcolor{red}{\sqrt{5}} \cdot 3\right) + \left(\textcolor{red}{\sqrt{5}} \cdot \sqrt{15}\right) \implies$
$3 \sqrt{5} + \left(\sqrt{5} \cdot \sqrt{15}\right)$
Next, use this rule for multiplying radicals to rewrite the term on the right:
$\sqrt{\textcolor{red}{a}} \cdot \sqrt{\textcolor{b l u e}{b}} = \sqrt{\textcolor{red}{a} \cdot \textcolor{b l u e}{b}}$
$3 \sqrt{5} + \left(\sqrt{\textcolor{red}{5}} \cdot \sqrt{\textcolor{b l u e}{15}}\right) \implies$
$3 \sqrt{5} + \left(\sqrt{\textcolor{red}{5} \cdot \textcolor{b l u e}{15}}\right) \implies$
$3 \sqrt{5} + \sqrt{75}$
Then, we can use the same rule in reverse to again rewrite and simplify the term on the right:
$3 \sqrt{5} + \sqrt{75} \implies$
$3 \sqrt{5} + \sqrt{\textcolor{red}{25} \cdot \textcolor{b l u e}{3}} \implies$
$3 \sqrt{5} + \left(\sqrt{\textcolor{red}{25}} \cdot \sqrt{\textcolor{b l u e}{3}}\right) \implies$
$3 \sqrt{5} + \left(5 \cdot \sqrt{\textcolor{b l u e}{3}}\right) \implies$
$3 \sqrt{5} + 5 \sqrt{3}$ |
+0
pls help
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1
In right triangle ABC, $\angle CAB$ is a right angle. Point $M$ is the midpoint of $\overline{BC}$. What is the number of centimeters in the length of median $\overline{AM}$? Express your answer as a decimal to the nearest tenth
Apr 26, 2020
#1
+659
+2
BTW when you see a right triangle a thing I always do is imagine it is inscribed in a circle with the hypotenuse as the diameter. That's just what I do, no need to do what I do. But is always nice to do what I do, because doing what I do makes me feel like people exist. I like when people exist, especially people who do what I do.
Anyways, draw a segment from $$A$$ to to a point on $$BC$$ in such it way that it is perpendicular to the hypotenuse. We call the point of intersection on the hypotenuse $$L$$
This means $$AL\perp{BC}$$
- BC = 5 because it is a 3-4-5 triangle.
By AA similarity, we know that triangle $$ALC$$ is similar to triangle $$ABC$$. To find AL, we write the following proportion:
$$\frac{AL}{3}=\frac{4}{5}$$
Solving for AL:
1. $$5AL=12$$
2. $$AL=\frac{12}{5}$$
Now here is how you solve the rest:
1. Consider point $$M$$, that is the midpoint of $$BC$$. Because it is the midpoint, we know that $$CM=\frac{5}{2}$$
2. Use the pythagorean theorem to find $$CL$$ hint:$$\sqrt{AC^2-AL^2}$$
3. Now that you found $$CL$$, we can find $$LM$$ by evaluating $$CM-CL$$
4. Since you know $$AL$$ and $$LM$$, you can easily find $$AM$$ by pythagorean theorem. hint: $$\sqrt{AL^2+LM^2}$$
Yay
Apr 27, 2020
edited by AnExtremelyLongName Apr 27, 2020 |
# How to Learn Different Types of Angles Through Some Tips?
Mathematics is a subject that includes two parts in it called Algebra and Geometry. The first part is all about digits and variables while the second part is all about shapes, their properties, and their applications in our daily life. If you observe you will notice that every object that we use regularly has some connection with mathematics. Life becomes easier if we study mathematics through concepts and real-life applications. You must have observed when you connect your knowledge with the application you remember that concept forever. Mathematics is vast and it has many topics, one of the famous topics of mathematics is angles
We observe different objects such as toys, equipment, cones, rooftops, houses, and many more. Almost every object has a certain shape and hence the object has certain work to perform. Scientifically everything has shape except liquid and gas as they have the only volume. Liquid obtains the shape of the material we put it in. Geometry is all about shapes such as triangles, quadrilaterals, rectangles, squares and many more are a combination of angles. There are several types of angles. Let’s see some tips to understand different concepts of angles:
• There are several types of angles and you can understand them better if you learn the property and theorem of each angle.
• There are several uses of angles such as you can use it to prove any theorem or you can use it to find out any unknown angle present in the figure.
• Interior angles are the angles that are present inside the figure, for example, when you draw a square ‘PQRS’ then all the angles present in it such as angle ‘P’, angle ‘Q’, angle ‘R’, and angle ‘S’ are interior angles.
• The interesting fact about the quadrilaterals is the addition of all the interior angles of the quadrilateral is always equal to three-sixty degrees.
• There are three types of angles according to the measurement of angles. The first type is acute angles they measure less than a ninety degree, the second type is obtuse angles they measure more than a ninety degree and the third type is the straight angle they measure exactly one-eighty degrees.
• You can practice the different angles by drawing them on a piece of paper and you can measure the angles with the help of a protector
• The angle that measures exactly equal to ninety- degrees is called a right-angle. Shapes like squares, rectangles, and triangles contain right angles.
• There is a special use of this right-angle in the right-angled triangle. We use the property of this angle in the Pythagoras theorem.
• If you draw a triangle ‘GTA’, where angle ‘G’ is ninety-degree and the other two angles, that is angle ‘T’ and angle ‘A’ measure forty-five degrees each.
• The side ‘TA’ is called a hypotenuse as it is opposite to the right angle. The theorem suggests that the square of the hypotenuse is equal to the addition of squares of the other two sides.
• This theorem is used to find out other sides of the triangle and it is used in various applications. In the architecture of the building, this property of angels is used.
• You must have seen ‘Z’ angles, these angles are known as alternate angles. Suppose the ‘Z’ angles are ‘ABCD’, here angle ‘B’ and angle ‘C’ measure the same as they are alternate angles.
Now you must have understood how easy mathematics is if you study it by considering real-life examples. Also now you must have understood different types of angles. Now you can easily solve problems based on the uses and applications of angles. Almost every shape except spheres, circles are made up of angles. You must have understood that you can only differentiate between angles only if you know their properties. If you are looking for the best platform to learn math then Cuemath is the best platform to understand math more practically. They have provided each math concept beautifully. Don’t treat math as a boring subject to make it interesting. |
Symmetry And Line of Symmetry
We observe symmetry in any object if it can be split into two identical halves. Asymmetric indicates that an item cannot be split into two identical parts.
This teaching guide aims at making the idea of symmetry more understandable and aid educators by offering teaching resources including animated stories, eye-catching posters, and practical activities.
## Symmetrical And Asymmetrical Shapes
• When 2 halves of any object show exact resemblance or similarity, it is said to be symmetrical.
• In symmetrical shapes, one half of the object acts as the mirror image of the other object.
## Line Of Symmetry
• A line of symmetry is an imaginary line that passes through the centre of any symmetrical shape.
• Any symmetrical shape can be folded into 2 equal and identical halves along its line of symmetry.
## Different Types of Symmetry
Vertical Line of Symmetry :
This is a line that runs up and down and divides an object into two identical halves. For instance, if we have a shape that can be split into two equal halves by a straight line running vertically, we say it has a vertical line of symmetry.
Horizontal Line of Symmetry :
This is a line that divides a shape into two identical halves when it is split horizontally, either from right to left or vice versa. For example, if a shape can be split into two equal parts by a horizontal line, it has a horizontal line of symmetry.
Diagonal Line of Symmetry :
This is a line that divides a shape into two identical halves when it is split along the diagonal corners. For instance, if we can split a square shape into two equal halves by cutting it across the corners, we say it has a diagonal line of symmetry.
## Number of Lines of Symmetry
One Line of Symmetry :
Shapes with one line of symmetry are only symmetrical about one axis. This axis can be horizontal, vertical, or diagonal. For example, the letter “A” has one line of symmetry, which is the vertical line down its center.
Two Lines of Symmetry :
Shapes with two lines of symmetry are symmetrical about two different lines. These lines can be vertical, horizontal, or diagonal. For example, a rectangle has two lines of symmetry: one vertical and one horizontal.
Infinite Lines of Symmetry :
Shapes with infinite lines of symmetry are symmetrical about multiple lines. These lines can be vertical, horizontal, or diagonal. Again, a rectangle is an example of a shape with infinite lines of symmetry because it has an infinite number of vertical and horizontal lines that divide it into identical halves.
## Types of Symmetry
Symmetry can be observed in different ways when we flip, turn, or slide an object. There are four types of symmetry.
Reflective Symmetry In 2D Shapes
• When one half of the shape or object is the reflection of the other half, it exhibits reflective symmetry.
• A shape or pattern which is reflected along its line of symmetry.
• A line of symmetry divides the shape into its 2 mirror images.
• A shape should have at least one line of symmetry to show reflective symmetry.
Rotational Symmetry In 2D Shapes
• When any shape looks the same on being rotated more than half a turn is said to show rotational symmetry.
• It is also called radial symmetry.
• The number of times the shape fits into its boundary exactly while taking one complete rotation is called the order of rotational symmetry.
• For example- An Equilateral Triangle with an order of symmetry as 3.
## Transformations: Reflection, Rotation, And Translation
When there is any change in the appearance of the figure it is said to show transformation. There are 4 types of transformations.
• Translation occurs when any figure slides in one direction.
• Reflection occurs when any figure flips over a line.
• Rotation occurs when any figure rotates to a certain degree around a point.
• Dilation occurs when any figure expands or contracts keeping its shape the same. It is also called resizing.
Posters
Teaching symmetry with kid-friendly, clear, and easy-to-understand posters from Uncle Math School by Fun2Do Labs :
Stories
Ignite kids’ curiosity with engaging stories for role play and skits, making the learning of this concept an exciting and effective experience. Teaching symmetry through stories from Uncle Math School by Fun2Do Labs :
Text of Stories
Activities
Learning symmetry can be made enjoyable by incorporating interactive games and activities.
### Symmetry Hunt
This is one of the most exhilarating activities for children to understand the concept of symmetry. This activity can be carried out in the following steps :
• Take children outdoors and instruct them to collect 3 symmetric and 3 asymmetric objects, they see around them like flowers, leaves, rocks, feathers, etc.
• Children can trace their collections on a page and label them as symmetric and asymmetric shapes.
Worksheets
Help your kids practise symmetry with interesting and engaging fun worksheets and solutions from Uncle Math by Fun2Do Labs.
Worksheet 474 : Line Of Symmetry
Solution 474 : Line Of Symmetry
Worksheet 475 : Line Of Symmetry
Solution 475 : Line Of Symmetry
Worksheet 476 : Symmetrical Objects
Solution 476 : Symmetrical Objects
Worksheet 477 : Symmetrical Alphabets
Solution 477 : Symmetrical Alphabets
Worksheet 478 : Lines Of Symmetry
Solution 478 : Lines Of Symmetry
Worksheet 479 : Symmetrical Alphabets
Solution 479 : Symmetrical Alphabets
Worksheet 480 : Translation, Reflection And Rotation
Solution 480 : Translation, Reflection And Rotation
Worksheet 481 : Reflection, Rotation And Transition
Solution 481 : Reflection, Rotation And Transition
Worksheet 482 : Reflection, Rotation And Translation
Solution 482 : Reflection, Rotation And Translation
Worksheet 483 : Order Of Symmetry
Solution 483 : Order Of Symmetry |
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IIT-JEE NEET CBSE NCERT Q&A
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NCERT Solutions for Class 9 Maths Chapter 3 – Coordinate Geometry
NCERT Solutions for Class 9 Maths Chapter 3 Coordinate Geometry will assist you in understanding the cartesian system and the importance of x-axis and y-axis in a plane. NCERT Solutions for Class 9 Maths Chapter 3 introduces the Cartesian Plane and the different Quadrants, including the concepts of ordinate and abscissa of a given point. By solving the NCERT questions you will learn how to locate points in a plane using vertical and horizontal axis.
There are a total of 6 questions in 3 exercises in NCERT Chapter 3. The questions are situational and relatable to your day to day lives as well. The topics covered in the NCERT Solutions for Chapter 3 include topics like locating the ordinate and abscissa of a given point on y-axis and x-axis, identifying the coordinates of a point on a plane and identifying the Quadrant of a point on a plane.
The NCERT Solutions for Class 9 Maths Chapter 3 employ the latest CBSE guidelines on preparing precise and accurate solutions to all the NCERT exercise questions. The NCERT Solutions provided to you by the maths professionals at Instasolv are to-the-point yet detailed with an aim to address all your possible doubts in this Chapter. Below is a quick summary along with a discussion on the NCERT Solutions for Class 9 Maths Chapter 3.
Summary of NCERT Solutions for Class 9 Maths Chapter 3
Introduction
Any point lying in a plane can be represented with the help of two perpendicular lines. Also, to find the location of a dot (point) on a plane we need some information about it. This is the central idea of the concept of Coordinate Geometry. This idea was developed by a French mathematician, Rene Descartes.
Cartesian System
1. Descartes invented the maths of placing two number lines perpendicular to each other on a plane and locating points on the plane by referring to these mutually perpendicular lines. In this Chapter, we have made these lines horizontal and vertical respectively, on the place for ease.
2. Assume two number lines X’X and Y’Y placed horizontally and vertically respectively. Let these number lines intersect one another at their 0 points.
3. Therefore, X’X is known as the x-axis and Y’Y is known as the y-axis. The point of intersection of these axes is known as the Origin and is represented by O.
4. OX and OY are called the positive directions of the x-axis and the y-axis respectively whereas, OX’ and OY’ are called the negative directions of the x-axis and y-axis respectively.
5. Such a plane is known as the Cartesian Plane or the coordinate plane or the XY-plane. The axis is called Coordinate Axis.
Writing Coordinates of a Point
1. The x-coordinate of a point is the perpendicular distance of the same, from the y-axis (which is positive along the positive x-axis and negative along the negative x-axis). The x-coordinate is also known as Abscissa.
2. Similarly, y coordinate of a point is given by its perpendicular distance from the x-axis (taken positive along the positive y-axis and negative along the negative y-axis). They-coordinate is also known as Ordinate
3. At Origin O, the distance of the point is 0 from both the axis and hence, its abscissa and ordinate are 0, which makes the coordinates for Origin (0,0)
1. When a point is in the 1st Quadrant, then the point will be in the form (+,+), as the 1st Quadrant is enclosed by the positive x-axis and positive y-axis.
2. When the point is in the 2nd Quadrant, then the point will be in the form (-,+), since the Quadrant is enclosed by the negative x-axis and positive y-axis.
3. If a point is in the 3rd Quadrant, then the point will be in the form (-,-), since the 3rd Quadrant is enclosed by the negative x-axis and the negative y-axis.
4. If a point is in the 4th Quadrant, then the point will be in the form (+,-), since the 4th Quadrant is enclosed by the positive x-axis and the negative y-axis.
Highlights of Chapters 9 Coordinate Geometry
• We need a system of Cartesian Plane to locate a point on a plane.
• The x and y Coordinates are the perpendicular distance of the point from the y-axis and x-axis respectively.
• If x y, then the position of (x,y) in the Cartesian Plane is different from the position of (y,x).
• The coordinates of the Origin are (0,0)
Exercise-wise Discussion of NCERT Solutions for Class 9 Maths Chapter 3
1. The NCERT Solutions for class 9 questions of Chapter 3 Coordinate Geometry are picked from daily life situations to make Coordinate Geometry an interactive and relatable subject, to bring you out of the boredom of regular bookish exercises.
2. For instance, the very first question of the Chapter asks you to describe the position of a Table Lamp. Similarly, the second question is about a Street Plan.
3. You will require graph paper to solve the questions in which you need to plot the points on the XY Plane.
4. Solving these NCERT Exercises will help you understand all the concepts of the Chapter easily and you will be prepared to tackle all kinds of questions in the CBSE exams related to this Chapter.
NCERT Solutions for Class 9 Maths Chapter 3 by Instasolv
1. The NCERT Solutions for Maths Chapter 9 are prepared by our Maths experts including the required graphical representations along with detailed descriptions.
2. We have organized the NCERT Solutions exercise-wise so you will have no problems in finding the answer that you are looking for.
3. We have curated these Solutions in line with the latest CBSE exam pattern to help you stand out from the rest in your Class.
4. These NCERT Solutions will also allow you to be on par with the syllabus of important competitive exams.
5. The answers to Class 9 Maths NCERT Chapter 3 provided by the subject matter experts of Instasolv are ideal for self-study.
More Chapters from Class 9 |
DECIMALS
What are Decimals?
Decimals are well behaved fractions because every one of them has a denominator that is a power of 10. This is why we call them well behaved. Our "whole number" system is based on place value and powers of 10, so decimals are user friendly fractions that we can write as whole numbers, placed to the right of or after the decimal point.Our money is based on a decimal system. There are one hundred cents in a dollar -- and when we write a dollar, we see \$1.00. It sure looks like one hundred to me.
Fractions with Decimal Denominators
Any fraction with a denominator of 10, 100 or 1000 -- any power of 10 -- can be written as a decimal easily. We put the numerator of the fraction in the correct decimal place and it's done.
Examples:
Notice where we put the 7. In the first fraction -- seven tenths -- the denominator (10) has 1 zero in it, so we put the 7 in 1st place after the decimal. 100 has 2 zeros in it, so the 7 goes in the 2nd place after the decimal and so on.
WE MUST BE CAREFUL!! The same is not true on the left of the decimal point, because the first column is for the 1's. This often causes confusion because, on the fraction side of the decimal, there is no 1's column. The number of zeros in the denominator of the fraction tells us in what place to put the numerator. On the left (whole-number side) of the decimal, the number of digits we write tells us the place value. On the whole-number side of the decimal, hundreds are in the 3rd column -- not the second. That's for the 10's.
.
Decimal Place Value
Let's study the value of adjacent or neighboring columns whole numbers and decimals.
Notice the " TH " at the end of each decimal place name.
Like in one-fifth, 3-sevenths or 7-ninths, it indicates a fractional part of the whole.
7 in the tenths column, write 0.7 = ; read seven-tenths.
0 in the tenths column, 3 in the hundredths write 0.03 = ; read three-hudredths.
275 after the decimal, write 0.275 = ; read 275-thousandths.
Notice the 0 to the left of the decimal point in numbers that are smaller than 1.
It indicates or "holds" the 1's place or column. We always do this, because the decimal point is small, it looks like a period and so could be missed if we wrote .275 instead of 0.275. With a mixed number like 12.642, the decimal point is obvious.
Note: Many European countries (especially France and Belgium) use the comma ( , ) rather than the point ( . ) to indicate the decimal. So, if we write 3, 507 to mean three thousand five hundred and seven, someone from France might think it means 3.507. To avoid confusion, those of us who use the decimal point no longer use the comma to mark off groups of 3 digits. We now use a space where the comma would be like this: 19 201.3 instead of 19, 201.3
Changing the Denominator to a Power of 10
Since the denominator of any decimal's fraction equivalent is a power of 10, when we have to find a fraction equivalent for a decimal, and it's easy to make the denominator a power of 10 through multiplication by a fraction equal to 1, we can proceed like this.
We know that 5 × 20 = 100 so we multiplied seven-twentieths by 5/5 which equals 1 -- therefore we didn't change the value of our fraction -- we just wrote an equivalent fraction with a denominator of 100.
In the second case, we knew that 8 × 125 = 1000, so we multiplied by 8/8.
Rounding and Estimating Decimals
Now that we have a decimal system based on powers of 10 to express fractions, we can use the same approach to rounding and estimating that we do for whole numbers. First, we decide the place value to which we want to round, then we use the "halfway" rule that says if a number is exactly at or bigger than halfway to the next place or column, we round up. If the number is smaller than halfway to the next place or column, we round down. Let's look at some number lines and some rounded decimal numbers.
.
We add and subtract decimal numbers exactly the same way we add and subtract whole numbers. We carry powers of 10 to the next column left when we add, and we borrow from the next column left when we subtract. The trick is to line the decimals up so that we're adding and subtracting according to their place values. Another good trick is to put a zero (0) at the right end of the numbers when they don't end in the same decimal column.
Say we have to add 6.32 to 9.715. The 6.32 uses only 2 decimal places, while the 9.715 uses 3 decimal places. So to add them we write 6.320 not 6.32. We added nothing, (no thousandths) so we didn't change the value of the number, just its form. Now we can line it up with 9.715.
Examples:
Put zeros in these decimal numbers where needed, then add or subtract them.
Carry or borrow as always. Show all your work.
12.035 + 2.6 97.804 - 5.21 0.0014 + 2.679 14.338 - 8.4731
.
Now get a pencil, an eraser and a note book, copy the questions,
do the practice exercise(s), then check your work with the solutions.
If you get stuck, review the examples in the lesson, then try again.
Practice Exercises
1) Change the denominator to a power of 10, then write the decimal equivalent.
a) b) c) d) e)
2) Write equal to ( = ), greater than ( > ) or less than ( < ) between these numbers:
a) 6.03 and 6.30 b) \$1.78 and \$1.69 c) 0.017 and 0.01700 d) 256.33 and 256.29
3) Write these decimal numbers in order smallest to biggest:
a) 2.7, 2.17, 2.07 b) 1.006, 1.3, 1.015 c) 8.63, 8.631, 8.59 d) 4.026, 4.31, 4.035
4) Put zeros in these decimal numbers where needed, then add or subtract them.
a) 12.803 - 10.92 b) 54.9 + 7.318 c) 0.674 - 0.0965 d) 0.765 + 2.2867
.
Solutions
1) Change each denominator to a power of 10, then write the decimal equivalent.
a) b) c) d) e)
2) Write equal to ( = ), greater than ( > ) or less than ( < ) between these numbers:
a) 6.03 < 6.30 b) \$1.78 > \$1.69 c) 0.017 = 0.01700 d) 256.33 > 256.29
3) Write these decimal numbers in order smallest to biggest:
a) 2.07 < 2.17 < 2.7 b) 1.006 < 1.015 < 1.3 c) 8.59 < 8.63 < 8.631 d) 4.026 < 4.035 < 4.31
4) Put zeros in these decimal numbers where needed, then add or subtract them.
12.803 - 10.92 54.9 + 7.318 0.674 - 0.0965 0.765 + 2.2867 |
# From a uniform disk of radius R, a circular hole of radius R/2 is cut out.
Question:
From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body.
Solution:
R/6; from the original centre of the body and opposite to the centre of the cut portion.
Mass per unit area of the original disc $=\sigma$
Radius of the original disc $=R$
Mass of the original disc, $M=\pi R^{2} \sigma$
The disc with the cut portion is shown in the following figure:
Radius of the smaller disc $=\frac{R}{2}$
Mass of the smaller disc, $M=\pi\left(\frac{R}{2}\right)^{2} \sigma=\frac{1}{4} \pi R^{2} \sigma=\frac{M}{4}$
Let $O$ and $O^{\prime}$ be the respective centres of the original disc and the disc cut off from the original. As per the definition of the centre of mass, the centre of mass of the original disc is supposed to be concentrated at $\mathrm{O}$, while that of the smaller disc is supposed to be concentrated at $O^{\prime}$.
It is given that:
$O O^{\prime}=\frac{R}{2}$
After the smaller disc has been cut from the original, the remaining portion is considered to be a system of two masses. The two masses are:
M (concentrated at O), and
$-M\left(=\frac{M}{4}\right)$ concentrated at $O^{\prime}$
(The negative sign indicates that this portion has been removed from the original disc.)
Let x be the distance through which the centre of mass of the remaining portion shifts from point O.
The relation between the centres of masses of two masses is given as:
$x=\frac{m_{1} r_{1}+m_{2} r_{2}}{m_{1}+m_{2}}$
For the given system, we can write:
$x=\frac{M \times 0-M \times\left(\frac{R}{2}\right)}{M+\left(-M^{\prime}\right)}$
$=\frac{\frac{-M}{4} \times \frac{R}{2}}{M-\frac{M}{4}}=\frac{-M R}{8} \times \frac{4}{3 M}=\frac{-R}{6}$
(The negative sign indicates that the centre of mass gets shifted toward the left of point $O$.) |
Three linear equation solver
We'll provide some tips to help you choose the best Three linear equation solver for your needs. Math can be a challenging subject for many students.
The Best Three linear equation solver
Here, we will show you how to work with Three linear equation solver. The Laplace solver works by iteratively solving for an unknown function '''f''' which is dependent on both '''a''' and '''b'''. For simplicity, we will assume that the solution of this differential equation is known and simply output this value at each iteration. This method is simple and can often be computationally intensive when large systems are being solved. Since the solution of this differential equation depends on both 'a' and 'b', it is important to only solve once for values that are close to the final solution. If these values are close, then it will be difficult to accurately predict where the final solution will be due to numerical errors which could make the difference between converging or diverging.
Mathematics is a study of symbols and relations that can be applied to anything. It is a very logical and analytical subject, which has many practical applications. Mathematics is not just about memorizing numbers and learning formulas. Mathematics is a language that can be used to describe the world around us. It can be used to describe how a theory works, or how something works physically. Mathematics can also be used to describe what things cost and how much they are worth. Mathematics can even be used to solve crimes! When we have information about an event in our world, we can use mathematics to determine what happened and why it happened. This can help us figure out who did it or how it happened in the first place.
Solving exponential equations can be a bit tricky. Most of the time you will need to use an inverse function to get from one number to the other. However, it is possible to solve some equations without using such techniques. Here are some examples: One way to solve an exponential equation is to use a logarithm table. For example, if you have an equation of the form y = 4x^2 + 32, then you would use the logarithm table found here. Then, you would find that log(y) = -log(4) = -2 and log(32) = 2. These values would be used in the original equation to obtain the solution: 4*y = -2*4 + 32 = -16 + 32 = 16. This value is the desired answer for y in this problem. Another way to solve an exponential equation is by using a combination of substitution and elimination. You can start by putting x into both sides of the equation and simplifying: ax + b c where a c if and only if b c/a . Then, once this is done, you can eliminate b from each side (using square roots or taking logs if necessary) to obtain a single solution that does not involve x . c if and only if , then you can substitute for y in both sides, thus eliminating x
You can also set a reminder for yourself, so that you don’t forget about your homework. It is important to note, however, that this is not an application designed specifically to help you with your schoolwork. Instead, it is an app that helps to keep track of any other commitments that you might have. It should be used in conjunction with another application that can help you with your schoolwork.
While there are several reasons why this could occur, the main culprit is usually inaccurate body weight measurements. While it may be tempting to call this out as a potential error in your paper, remember that this is only one part of a larger investigation. Once you have corrected your data and reanalysed your results, point slope form should not be present. You will be able to find the underlying issue and correct it before publishing your paper. This type of error is hard to detect because it is so small. You can try to make sure that your patients are not underweight or overweight for many reasons: If possible, take an impression of their foot before surgery to get an exact measurement of their leg length. If they are too short, then they will have more difficulty getting into comfortable shoes after surgery. If they are too tall, then they will have more difficulty taking off their shoes when they leave the hospital.
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# Graphs Using Slope-Intercept Form
## Use the y-intercept and the 'rise over run' to graph a line
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Practice Graphs Using Slope-Intercept Form
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Graph a Line in Slope-Intercept Form
The cost per month for a cell-phone plan is $60 plus$7.50 for every gigabyte (GB) of data you use. (For billing purposes, actual usage is rounded to the nearest one-quarter GB.) Write an equation for the cost of the data plan and determine how much your bill will be if you use 4.5 GB of data in a month.
### Guidance
From the previous lesson, we know that the equation of a line is y=mx+b\begin{align*}y = mx + b\end{align*}, where m\begin{align*}m\end{align*} is the slope and b\begin{align*}b\end{align*} is the y\begin{align*}y-\end{align*}intercept. From these two pieces of information we can graph any line.
#### Example A
Graph y=13x+4\begin{align*}y = \frac{1}{3}x + 4\end{align*} on the Cartesian plane.
Solution: First, the Cartesian plane is the xy\begin{align*}x-y\end{align*} plane. Typically, when graphing lines, draw each axis from -10 to 10. To graph this line, you need to find the slope and y\begin{align*}y-\end{align*}intercept. By looking at the equation, 13\begin{align*}\frac{1}{3}\end{align*} is the slope and 4, or (0, 4), is the y\begin{align*}y-\end{align*}intercept. To start graphing this line, plot the y\begin{align*}y-\end{align*}intercept on the y\begin{align*}y-\end{align*}axis.
Now, we need to use the slope to find the next point on the line. Recall that the slope is also riserun\begin{align*}\frac{rise}{run}\end{align*}, so for 13\begin{align*}\frac{1}{3}\end{align*}, we will rise 1 and run 3 from the y\begin{align*}y-\end{align*}intercept. Do this a couple of times to get at least three points.
Now that we have three points, connect them to form the line y=13x+4\begin{align*}y = \frac{1}{3}x + 4\end{align*}.
#### Example B
Graph y=4x5\begin{align*}y = -4x -5\end{align*}.
Solution: Now that the slope is negative, the vertical distance will “fall” instead of rise. Also, because the slope is a whole number, we need to put it over 1. Therefore, for a slope of -4, the line will fall 4 and run 1 OR rise 4 and run backward 1. Start at the y\begin{align*}y-\end{align*}intercept, and then use the slope to find a few more points.
#### Example C
Graph x=5\begin{align*}x = 5\end{align*}.
Solution: Any line in the form x=a\begin{align*}x = a\end{align*} is a vertical line. To graph any vertical line, plot the value, in this case 5, on the x\begin{align*}x-\end{align*}axis. Then draw the vertical line.
To graph a horizontal line, y=b\begin{align*}y = b\end{align*}, it will be the same process, but plot the value given on the y\begin{align*}y-\end{align*}axis and draw a horizontal line.
Intro Problem Revisit If x is the number of GB of data you use in a month and y is the total cost you pay, then the equation for the cell-phone plan would be y=7.5x+60\begin{align*}y=7.5x + 60\end{align*}. If you use 4.5 GB in a month, the total cost would be y=7.5(4.5)+60=93.75\begin{align*}y=7.5(4.5)+60=93.75\end{align*}.
So your bill for the month would be \$93.75.
### Guided Practice
Graph the following lines.
1. y=x+2\begin{align*}y = -x + 2\end{align*}
2. y=34x1\begin{align*}y = \frac{3}{4}x - 1\end{align*}
3. y=6\begin{align*}y = -6\end{align*}
All the answers are on the same grid below.
1. Plot (0, 2) and the slope is -1, which means you fall 1 and run 1.
2. Plot (0, -1) and then rise 3 and run 4 to the next point, (4, 2).
3. Plot -6 on the y\begin{align*}y-\end{align*}axis and draw a horizontal line.
### Explore More
Graph the following lines in the Cartesian plane.
1. y=2x3\begin{align*}y = -2x -3\end{align*}
2. y=x+4\begin{align*}y = x + 4\end{align*}
3. y=13x1\begin{align*}y = \frac{1}{3}x - 1\end{align*}
4. y=9\begin{align*}y = 9\end{align*}
5. y=25x+7\begin{align*}y = - \frac{2}{5}x + 7\end{align*}
6. y=24x5\begin{align*}y = \frac{2}{4}x - 5\end{align*}
7. y=5x2\begin{align*}y = -5x -2\end{align*}
8. y=x\begin{align*}y = -x\end{align*}
9. y=4\begin{align*}y = 4\end{align*}
10. x=3\begin{align*}x = -3\end{align*}
11. y=32x+3\begin{align*}y = \frac{3}{2}x + 3\end{align*}
12. y=16x8\begin{align*}y = - \frac{1}{6}x - 8\end{align*}
13. Graph y=4\begin{align*}y = 4\end{align*} and x=6\begin{align*}x = -6\end{align*} on the same set of axes. Where do they intersect?
14. If you were to make a general rule for the lines y=b\begin{align*}y = b\end{align*} and x=a\begin{align*}x = a\end{align*}, where will they always intersect?
15. The cost per month, C\begin{align*}C\end{align*} (in dollars), of placing an ad on a website is C=0.25x+50\begin{align*}C = 0.25x + 50\end{align*}, where x\begin{align*}x\end{align*} is the number of times someone clicks on your link. How much would it cost you if 500 people clicked on your link?
### Vocabulary Language: English
Cartesian Plane
Cartesian Plane
The Cartesian plane is a grid formed by a horizontal number line and a vertical number line that cross at the (0, 0) point, called the origin.
linear equation
linear equation
A linear equation is an equation between two variables that produces a straight line when graphed.
Linear Function
Linear Function
A linear function is a relation between two variables that produces a straight line when graphed.
Slope-Intercept Form
Slope-Intercept Form
The slope-intercept form of a line is $y = mx + b,$ where $m$ is the slope and $b$ is the $y-$intercept. |
# 4.3 Transformations
Page 1 / 1
## Rotation of a point
When something is moved around a fixed point, we say that it is rotated about the point. What happens to the coordinates of a point that is rotated by ${90}^{\circ }$ or ${180}^{\circ }$ around the origin?
## Investigation : rotation of a point by ${90}^{\circ }$
Complete the table, by filling in the coordinates of the points shown in the figure.
Point $x$ -coordinate $y$ -coordinate A B C D E F G H
What do you notice about the $x$ -coordinates? What do you notice about the $y$ -coordinates? What would happen to the coordinates of point A, if it was rotated to the position of point C? What about point B rotated to the position of D?
## Investigation : rotation of a point by ${180}^{\circ }$
Complete the table, by filling in the coordinates of the points shown in the figure.
Point $x$ -coordinate $y$ -coordinate A B C D E F G H
What do you notice about the $x$ -coordinates? What do you notice about the $y$ -coordinates? What would happen to the coordinates of point A, if it was rotated to the position of point E? What about point F rotated to the position of B?
From these activities you should have come to the following conclusions:
• 90 ${}^{\circ }$ clockwise rotation: The image of a point P $\left(x;y\right)$ rotated clockwise through 90 ${}^{\circ }$ around the origin is P' $\left(y;-x\right)$ . We write the rotation as $\left(x;y\right)\to \left(y;-x\right)$ .
• 90 ${}^{\circ }$ anticlockwise rotation: The image of a point P $\left(x;y\right)$ rotated anticlockwise through 90 ${}^{\circ }$ around the origin is P' $\left(-y;x\right)$ . We write the rotation as $\left(x;y\right)\to \left(-y;x\right)$ .
• 180 ${}^{\circ }$ rotation: The image of a point P $\left(x;y\right)$ rotated through 180 ${}^{\circ }$ around the origin is P' $\left(-x;-y\right)$ . We write the rotation as $\left(x;y\right)\to \left(-x;-y\right)$ .
## Rotation
1. For each of the following rotations about the origin: (i) Write down the rule.(ii) Draw a diagram showing the direction of rotation.
1. OA is rotated to OA ${}^{\text{'}}$ with A(4;2) and A ${}^{\text{'}}$ (-2;4)
2. OB is rotated to OB ${}^{\text{'}}$ with B(-2;5) and B ${}^{\text{'}}$ (5;2)
3. OC is rotated to OC ${}^{\text{'}}$ with C(-1;-4) and C ${}^{\text{'}}$ (1;4)
2. Copy $\Delta$ XYZ onto squared paper. The co-ordinates are given on the picture.
1. Rotate $\Delta$ XYZ anti-clockwise through an angle of 90 ${}^{\circ }$ about the origin to give $\Delta$ X ${}^{\text{'}}$ Y ${}^{\text{'}}$ Z ${}^{\text{'}}$ . Give the co-ordinates of X ${}^{\text{'}}$ , Y ${}^{\text{'}}$ and Z ${}^{\text{'}}$ .
2. Rotate $\Delta$ XYZ through 180 ${}^{\circ }$ about the origin to give $\Delta$ X ${}^{\text{'}}$ ${}^{\text{'}}$ Y ${}^{\text{'}}$ ${}^{\text{'}}$ Z ${}^{\text{'}}$ ${}^{\text{'}}$ . Give the co-ordinates of X ${}^{\text{'}}$ ${}^{\text{'}}$ , Y ${}^{\text{'}}$ ${}^{\text{'}}$ and Z ${}^{\text{'}}$ ${}^{\text{'}}$ .
## Enlargement of a polygon 1
When something is made larger, we say that it is enlarged . What happens to the coordinates of a polygon that is enlarged by a factor $k$ ?
## Investigation : enlargement of a polygon
Complete the table, by filling in the coordinates of the points shown in the figure. Assume each small square on the plot is 1 unit.
Point $x$ -coordinate $y$ -coordinate A B C D E F G H
What do you notice about the $x$ -coordinates? What do you notice about the $y$ -coordinates? What would happen to the coordinates of point A, if the square ABCD was enlarged by a factor 2?
## Investigation : enlargement of a polygon 2
In the figure quadrilateral HIJK has been enlarged by a factor of 2 through the origin to become H'I'J'K'. Complete the following table using the information in the figure.
Co-ordinate Co-ordinate' Length Length' H = (;) H' = (;) OH = OH' = I = (;) I' = (;) OI = OI' = J = (;) J' = (;) OJ = OJ' = K = (;) K' + (;) OK = OK' =
What conclusions can you draw about
1. the co-ordinates
2. the lengths when we enlarge by a factor of 2?
We conclude as follows:
Let the vertices of a triangle have co-ordinates S $\left({x}_{1};{y}_{1}\right)$ , T $\left({x}_{2};{y}_{2}\right)$ , U $\left({x}_{3};{y}_{3}\right)$ . $▵$ S'T'U' is an enlargement through the origin of $▵$ STU by a factor of $c$ ( $c>0$ ).
• $▵$ STU is a reduction of $▵$ S'T'U' by a factor of $c$ .
• $▵$ S'T'U' can alternatively be seen as an reduction through the origin of $▵$ STU by a factor of $\frac{1}{c}$ . (Note that a reduction by $\frac{1}{c}$ is the same as an enlargement by $c$ ).
• The vertices of $▵$ S'T'U' are S' $\left(c{x}_{1};c{y}_{1}\right)$ , T' $\left(c{x}_{2},c{y}_{2}\right)$ , U' $\left(c{x}_{3},c{y}_{3}\right)$ .
• The distances from the origin are OS' = $c$ OS, OT' = $c$ OT and OU' = $c$ OU.
## Transformations
1. Copy polygon STUV onto squared paper and then answer the following questions.
1. What are the co-ordinates of polygon STUV?
2. Enlarge the polygon through the origin by a constant factor of $c=2$ . Draw this on the same grid. Label it S'T'U'V'.
3. What are the co-ordinates of the vertices of S'T'U'V'?
2. $▵$ ABC is an enlargement of $▵$ A'B'C' by a constant factor of $k$ through the origin.
1. What are the co-ordinates of the vertices of $▵$ ABC and $▵$ A'B'C'?
2. Giving reasons, calculate the value of $k$ .
3. If the area of $▵$ ABC is $m$ times the area of $▵$ A'B'C', what is $m$ ?
1. What are the co-ordinates of the vertices of polygon MNPQ?
2. Enlarge the polygon through the origin by using a constant factor of $c=3$ , obtaining polygon M'N'P'Q'. Draw this on the same set of axes.
3. What are the co-ordinates of the new vertices?
4. Now draw M”N”P”Q” which is an anticlockwise rotation of MNPQ by 90 ${}^{\circ }$ around the origin.
5. Find the inclination of OM”.
#### Questions & Answers
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
characteristics of micro business
Abigail
for teaching engĺish at school how nano technology help us
Anassong
How can I make nanorobot?
Lily
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Other chapter Q/A we can ask |
science space and astronomy
# How do you find the Y intercept of a parallel line?
Last Updated: 20th March, 2020
29
For parallel lines, the slopes must be equal, so the slope of the new line must also be . We can plug the new slope and the given point into the slope-intercept form to solve for the y-intercept of the new line. Use the y-intercept in the slope-intercept equation to find the final answer.
Hereof, do parallel lines have the same Y intercept?
In other words, the slopes of parallel lines are equal. Note that two lines are parallel if their slopes are equal and they have different y-intercepts. In other words, perpendicular slopes are negative reciprocals of each other. Here is a quick review of the slope/intercept form of a line.
Likewise, what is the equation of a line parallel to Y axis? Thus, if P(x, y) is any point on AB, then x = a. Hence, the equation of a straight line parallel to y-axis at a distance a from it is x = a. The equation of y-axis is x = 0, since, y-axis is a parallel to itself at a distance 0 from it.
Also to know, how do you find the Y intercept of a perpendicular line?
First, put the equation of the line given into slope-intercept form by solving for y. You get y = 2x +5, so the slope is –2. Perpendicular lines have opposite-reciprocal slopes, so the slope of the line we want to find is 1/2. Plugging in the point given into the equation y = 1/2x + b and solving for b, we get b = 6.
Are these lines perpendicular?
Explanation: Two lines are perpendicular if and only if their slopes are negative reciprocals. To find the slope, we must put the equation into slope-intercept form, , where equals the slope of the line. Therefore, any line perpendicular to must have a slope of .
Explainer
## What is perpendicular example?
Perpendicular - Definition with Examples
Two distinct lines intersecting each other at 90° or a right angle are called perpendicular lines. Example: Here, AB is perpendicular to XY because AB and XY intersect each other at 90°. Non-Example: The two lines are parallel and do not intersect each other.
Explainer
## What is an example of parallel lines?
Parallel lines are two or more lines that never intersect. Examples of parallel lines are all around us, in the two sides of this page and in the shelves of a bookcase.
Explainer
## What is a perpendicular line?
In elementary geometry, the property of being perpendicular (perpendicularity) is the relationship between two lines which meet at a right angle (90 degrees). A line is said to be perpendicular to another line if the two lines intersect at a right angle.
Pundit
## Do parallel lines have the same slope?
Parallel lines have the same slope and will never intersect. Parallel lines continue, literally, forever without touching (assuming that these lines are on the same plane). On the other hand, the slope of perpendicular lines are the negative reciprocals of each other, and a pair of these lines intersects at 90 degrees.
Pundit
## What does a perpendicular line look like?
Thus, a perpendicular line looks like a line that has a 90 degree angle between it and another line. The upside down T (? ) means “is perpendicular to.” The illustration above says “AB is perpendicular to MN”. You can also say “MN is perpendicular to AB” and it means exactly the same thing.
Pundit
## What is the difference between parallel lines and perpendicular?
When there are two parallel lines, these two lines are never able to intersect or touch. Perpendicular lines are two lines in which one of the lines intersects the other line, and the angles created from the intersection of these two lines are all right angles.
Teacher
## How do you determine if two lines are parallel?
To see whether or not two lines are parallel, we must compare their slopes. Two lines are parallel if and only if their slopes are equal. The line 2x – 3y = 4 is in standard form. In general, a line in the form Ax + By = C has a slope of –A/B; therefore, the slope of line q must be –2/–3 = 2/3.
Teacher
## How can you tell if two equations are parallel?
We can determine from their equations whether two lines are parallel by comparing their slopes. If the slopes are the same and the y-intercepts are different, the lines are parallel. If the slopes are different, the lines are not parallel. Unlike parallel lines, perpendicular lines do intersect.
Reviewer
## What if two lines have the same Y intercept?
Parallel Lines: If the two linear equations have the same slope (and different y-intercepts), the lines will be parallel. Since parallel lines never intersect, a system composed of two parallel lines will have NO solution (no intersection of the lines.) 2.
Reviewer
## How do you write an equation for parallel and perpendicular lines?
Step 1: Find the slope of the line. To find the slope of the given line we need to get the line into slope-intercept form (y = mx + b), which means we need to solve for y: The slope of the line 8x + 6y = 15 is m = –4/3. Therefore, the slope of the line perpendicular to this line would have to be m = 3/4.
Reviewer
## How do you find the equation of a line?
The equation of a line is typically written as y=mx+b where m is the slope and b is the y-intercept. If you know two points that a line passes through, this page will show you how to find the equation of the line. Fill in one of the points that the line passes through
Supporter
## What is the meaning of parallel lines?
Parallel lines are two lines that are always the same distance apart and never touch. In order for two lines to be parallel, they must be drawn in the same plane, a perfectly flat surface like a wall or sheet of paper. Any line that has the same slope as the original will never intersect with it.
Supporter
## What is the slope of a line perpendicular to this line?
A line perpendicular to another has a slope that is the negative reciprocal of the slope of the other line. The negative reciprocal of the original line is –2, and is thus the slope of its perpendicular line.
Supporter
## Do perpendicular lines have the same Y intercept?
They are not the same line. The slopes of the lines are the same and they have different y-intercepts, so they are not the same line and they are parallel. Perpendicular Lines. Two non-vertical lines are perpendicular if the slope of one is the negative reciprocal of the slope of the other.
Beginner
## What does parallel to Y axis mean?
To every number x associated is the same constant value y = c. A line parallel to the x-axis is called a horizontal line (or constant). If the x value never changes a line is parallel to the y-axis. A line parallel to the y-axis is called a vertical line. The linear function changes the sign at the root or zero point.
Beginner
## What is the equation of Y?
The equation of any straight line, called a linear equation, can be written as: y = mx + b, where m is the slope of the line and b is the y-intercept. The y-intercept of this line is the value of y at the point where the line crosses the y axis.
Beginner
## What is the slope of the line that is perpendicular to the Y axis?
The y-axis is a vertical line. A vertical line has a slope of 10 which is undef or undefined. The negative reciprocal would be 01 or 0 . So, the slope of the perpendicular would be 0 .
Beginner
## Why is the slope of the Y axis undefined?
The slope of y axis is not infinity. Its undefined. To be precise, as slope of line approaches the vertical it increases (positively or negatively depending on the angle) without bound. The x-axis is horizontal and the y-axis is vertical so they are perpendicular.
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• What are the odds of getting a 33% chance 3 times in a row
# What are the odds of getting a 33% chance 3 times in a row
What are the Odds of Getting a 33% Chance 3 Times in a Row?
The search query "What are the odds of getting a 33% chance 3 times in a row?" is aimed at understanding the probability of achieving a specific outcome repeatedly. This article aims to provide a simple and comprehensive review of the topic, highlighting its positive aspects, benefits, and suitable conditions for its application.
I. Understanding the Odds:
1. Definition: This section will explain what odds are and how they relate to probabilities.
2. Probability Basics: A brief overview of the concept of probability will be provided, including the calculation of probabilities for independent events.
II. Odds of Getting a 33% Chance:
1. Explaining 33% Chance: This section will define what a 33% chance means in terms of probability.
2. Calculation: A step-by-step guide on how to calculate the odds of getting a 33% chance will be provided, ensuring simplicity and clarity.
III. Odds of Getting a 33% Chance 3 Times in a Row:
1. Independent Events: An explanation of independent events and how they relate to the odds of achieving a 33% chance three times consecutively.
2. Calculation: A simple formula
Title: "Unraveling the Improbable: What Are the Odds on Getting 1 Out of 10 Four Times in a Row?" Hey there, fellow risk-takers and probability enthusiasts! Today, we're diving headfirst into the mesmerizing world of odds and chance, exploring a mind-boggling scenario: What are the odds on getting 1 out of 10 four times in a row? Buckle up, because we're about to embark on a thrilling statistical adventure! Imagine this: you're playing a game where you have ten possible outcomes, and all you need is to land on "1" to win. The question arises – what are the chances of hitting that elusive "1" four times consecutively? Let's break it down for you! In the United States, where we love our numbers and quirky experiments, the probability of rolling a "1" on a fair 10-sided die is, of course, 1 in 10. Seems simple enough, right? However, when we start looking at multiple attempts, things get a little more interesting. To calculate the odds of an independent event happening multiple times in a row, we multiply the probabilities together. So, for our case, we multiply 1/10 by itself
## What is the probability of a 20% event occurring 5 times in a row?
If each event is independent and has probability 0.2, and one looks at just one sequence of 5 trials, the probability is (0.2)^5 = 0.00032. However, if the events are positively correlated, it can be higher, and if they are negatively correlated, it can be lower.
## How do you calculate odds of winning in a row?
To work out the potential timing of a streak, you can use the formula (1/P)a, whereby P is the loss probability and a is the streak length. You can then use this information to calculate the probability of suffering one streak of losses – or wins – within a series of bets.
## What is the chance of getting heads 4 times in a row?
The probability for any fair coin toss is 1/2 for Heads. The probability of flipping 4 heads in a row is (1/2)^4 or 1/16 where there are 4 'fair' coin tosses.
## How do you calculate the odds of something happening multiple times?
With multiple events, probability is found by breaking down each probability into separate, single calculations and then multiplying each result together to achieve a single possible outcome.
## What is the probability of getting 20 three times in a row?
If we were to roll the die three times and compare the triple rolled with (20,20,20), it would, on average, take 8000 of these triples to get the triple (20,20,20). However, if we were to simply roll the die until three 20s occurred in succession, it would take, on average, 8420 rolls to get three 20s in a row.
## What is a 33% chance 10 times in a row?
Therefore, the chance of a "33% chance of success" gamble failing ten times in a row is very low, at around 0.08%. If you mean a 1/3 chance, then it's (2/3)^10, or about 1.7%. If you actually mean exactly 33%, then it's 0.67^10, which is about 1.8%.
#### What are the odds of 70 percent happening 3 times?
The chances of a 70% event happening three times in a row are 0.7*0.7*0.7, so 0.343 (34.3%).
#### What are the odds of a 25% chance?
Remember to replace 1 by 100% if the probability is given as a percentage. Example: If probability is 25% , then odds are is 25% / 75% = 1/3 = 0.33 .
#### What is the probability of getting a 1 in 4 chance 4 times?
A 1 in 4 chance is equal to 25% probability. Saying that something needs to be done 4 times for it to happen means that the task will be completed 25% each time. Probability rarely makes definite statements. A dice could be thrown 10.000 times without ever rolling a 6.
#### How would you calculate the probability that you will win 3 times in a row?
1 Expert Answer each time you play, you have a 1/4 chance of winning. for you to win three time, probability = probability of winning first time * probability of winning second time* probability you win the third time. Remember, the probability of winning each time is 1/4. Lee P.
#### What is the chance of a 1 in 8 happening 5 times in a row?
Probability of success is (# of successful outcomes) / (# of total possible outcomes). Assuming independent trials, the probability it will happen is obviously 1/8^5 = 1/32768.
#### What is the probability of 1 of 3?
Since we have three items, we also have three possible outcomes. We'll use the probability formula to divide events by potential outcomes, and determine the probability to be 1 out of 3 or 33.33%.
## FAQ
What are the odds of rolling a 1 3 times in a row?
So the chance of getting three 1s in a row is 1 in 8000.
What percentage is a 1 in 32 chance?
Solution: 1/32 as a percent is 3.125%
What does 32 1 odds mean?
What does odds of 32/1 mean? If you were to bet \$10 on 32/1 odds you would receive \$320.00 in profit if this outcome won. The implied win probability of 32/1 odds is 3.03%. If you'd like to see the implied win probability of other odds values you can check our Moneyline Converter.
How rare is a 1 25 chance?
Number Converter
1 in __DecimalPercent
1 in 250.044.0%
1 in 500.022.0%
1 in 1000.011.0%
1 in 2000.00500.50%
How rare is a 1 100 chance?
If the odds are 1/100 that something happens, then there are 99 chances out of 100 that the event DOESN'T happen. So, the odds of that event not happening in 100 attempts are 99^100/100^100. This works out to 0.36603234127, or a 37 to 63 chance of the event not happening …………….
What is the probability of winning a 50 50 game?
In a 50/50 game, where the probability of winning is 50%, each individual outcome is independent of the others.
## What are the odds of getting a 33% chance 3 times in a row
How do you calculate odds of winning? This is found by dividing the number of desired outcomes over the total number of possible outcomes. In our example, the probability (not odds) that we'll roll a one or a two (out of six possible die roll outcomes) is 2 / 6 = 1 / 3 = . 33 = 33%. So our 1 : 2 odds of winning translate to a 33% chance that we'll win.
How rare is a 5 percent chance? Number Converter
1 in __DecimalPercent
1 in 90.1111%
1 in 100.1010%
1 in 200.055.0%
1 in 250.044.0%
What is an example of a 50 50 chance? Examples of a fifty-fifty chance After all, if a weighing machine is faulty there is a fifty-fifty chance that it will be faulty in the customer's favour. If they are 45 or over, they have a fifty-fifty chance of being out of work for at least a year.
What is a 50 50 chance in ratio? In the second example, “50/50 chance” is used as an expression to mean that there is a 50% chance for each outcome.
How do you calculate chance of something happening twice in a row? Multiply each event's probability by the others This gives you the probability of several events happening one after the other. Say you want to calculate the likelihood of tossing 4 twice in a row. In this scenario, the chances of this pair of events happening are 1/6, thus 1/6x1/6= 1/36.
What are the odds of 1 4 5 times? What are the chances of a 1 in 4 chance happening 5 times? It will be (1 / 4)^5 = (1 / 1024).
• What are the odds of getting 1 100 twice in a row?
• For any specific number there is a 1 in 100 chance of it coming up on the first 'roll'. On the second roll, again, 1 in 100 chance. So a 1 in 10 000 chance that a specific number will come up twice on those two rolls, or a 0.01% chance.
• How do you calculate percent chance multiple times?
• The result of events divided by outcomes multiplied by 100. To calculate the probability of multiple events, you'll multiply the probabilities of each event.
• What are the odds of a 10% chance?
• To convert from a probability to odds, divide the probability by one minus that probability. So if the probability is 10% or 0.10 , then the odds are 0.1/0.9 or '1 to 9' or 0.111.
• How often is a 10% chance?
• Number Converter
1 in __DecimalPercent
1 in 100.1010%
1 in 200.055.0%
1 in 250.044.0%
1 in 500.022.0%
• Is there a 100 percent chance of anything?
• From a statistical perspective, it's rare for something to have a 100% or 0% chance of happening. In most cases, there's always a small probability of even the most unlikely event occurring, and there's usually some small chance of even the most certain event not occurring.
February 8, 2024
February 8, 2024
February 8, 2024 |
# What is the Chinese Multiplication Method?
Instructor: Mark Boster
Some people use their fingers to help them do multiplication. Some use x's or other marks. There is even a lattice method. However, the most interesting and easiest one seems to be the Chinese Method.
## Sticks and Squirrels
Devon and his father were out in the country and decided to go on a twelve-mile horse ride. Devon counted twenty-one squirrels in the first mile, twenty-one in the second mile and twenty-one in each of the other miles. They were wondering how many squirrels they saw; however, they didn't have a piece of paper to multiply. Luckily, Devon did find some sticks and showed his father how to multiply with sticks. You want to know how he did that? Well, just hang on!
## Chinese Multiplication Method
The Chinese Method, also known as the stick multiplication method, uses sticks to solve multiplication problems.
Look at Diagram One. First, Devon laid sticks beginning on the bottom left for the number 21. He put 2 red sticks for the tens place and 1 black stick for the ones place.
Then he laid out sticks beginning at the top left for the number 12, placing them diagonally on top of the first sticks. He used 1 red stick for the tens place and 2 black for the ones place.
Next, Devon divided the sticks into three sections (gray lines) - one for each place value. Let's see how he did that:
• Since the black lines represent the ones place, he found where the black lines crossed. Since they both are black, and are ones, 1 x 1 = 1. This is the ones column.
• Since the red lines are the tens place, and 10 x 10 = 100, that would be the hundreds column.
• Where the red and black cross (the middle column), 10 x 1 = 10, so that is the tens column.
Finally, he counted the times the sticks crossed each other in each section (blue dots). He started his counting in the ones place (the furthest right). He counted 2 dots and put a 2 in the circle. He counted 5 dots in the tens place and 2 dots in the hundreds place. Devon was able to use sticks to show that 21 x 12 = 252!
Pretty easy, huh?
## But What about Larger Numbers?
Look at Diagram Two
Beginning at the bottom left, there are 4 hundreds, 3 tens and 3 ones. Those are crossed, starting at the top left, with the 1 ten and the 2 ones.
They are separated into place value by multiplying the place value of the lines, just like in the example above. This time, however, we also have a place value of 100 x 10 = 1000, so that is the thousands column.
OK, go ahead and count where the lines cross in each section.
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### If x is 10% less than y, then find the percentage by which y is more than x.
11 1 9
%
Step by Step Explanation:
1. According to the question, x is 10% less than y.
This can be written as:
x = y - (y ×
10 100
)
2. From the above equation, we can calculate y in terms of x. In order to do that, we must simplify the equation and rearrange it accordingly.
Thus,the equation becomes :
x = y -
y 10
,
which implies, x = y(1 -
1 10
)
or, x = y(
10 - 1 10
)
or, x =
9y 10
or, y =
10x 9
3. Now, let us try to rewrite the right-hand side of the above equation as a mixed number.
y = x +
x 9
4. From the above equation, we know that: "y is more than x by a factor of
1 9
".
5. To convert a factor into a percentage, we need to multiply the factor by 100.
This means that the following two statements mean the same:
"y is more than x by a factor of
1 9
".
and
"y is more than x by
100 9
percent."
6. On writing the above percentage as a mixed fraction, we get:
"y is more than x by
11 1 9
%
" |
Six Ways to Sum A Series
Imagine taking a square and dividing it along the diagonal. Now take one section and divide that down the diagonal. Repeat this process into infinity. This is a simple example of summing of an infinite series. Although this example provides and easily determined sum of the infinite series, the whole square, we find it commonly harder to find the sum of infinite series. In this paper, the author discusses the infinite series of the squares of the reciprocals of integers and different methods of finding its sum.
Found first by Leonhard Euler in 1734, the sum of this series is described today in a variety of ways, all of which are more mathematically acceptable than Euler’s original proof.
$\textbf{Euler’s Proof}$
The basic idea of Euler’s proof is to obtain a power series expansion for a function whose roots are multiples of the perfect squares. We can then apply a property of the polynomials to obtain the sum of the reciprocals of the roots. Here we represent the sine function as a power series:
(1)
\begin{align} \sin x = x - \frac{x^{3}}{3 \cdot 2} + \frac{x^{5}}{5 \cdot 4 \cdot 3 \cdot 2} - \frac{x^{7}}{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2} +... \end{align}
We can think of this expansion as an infinite polynomial. If we divide both sides by $x$, we obtain a polynomial with only the even powers of $x$. Once you replace $x$ with $\sqrt{x}$ the results is:
(2)
\begin{align} \frac{\sin \sqrt{x}}{\sqrt{x}} = 1 - \frac{x}{3 \cdot 2} + \frac{x^{2}}{5 \cdot 4 \cdot 3 \cdot 2} - \frac{x^{3}}{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2} +... \end{align}
This is a function $f$ that has $\pi ^{2}, 4\pi ^{2}, 9\pi ^{2},$… for roots. Euler then used the fact that adding up the reciprocal of all the roots of a polynomial results in the negative of the ratio of the linear coefficient to the constant coefficient. Basically, if
(3)
$$(x-r_{1})(x-r_{2})…(x-r_{n})=x^{n}+a_{n-1}x^{n-1}+...+a_{1}x+a_{0}$$
then
(4)
\begin{align} \frac{1}{r_{1}}+\frac{1}{r_{2}}+…\frac{1}{r_{n}}=\frac{-a_{1}}{a_{0}} \end{align}
Euler assumed this same rule applied to power series and applied it to our function $f$.
(5)
\begin{align} \frac{1}{6}=\frac{1}{\pi^{2}}+\frac{1}{4\pi ^{2}}+\frac{1}{9\pi ^{2}}… \end{align}
If we multiply both sides of this equation by $\pi ^{2}$ we get $\frac{\pi ^{2}}{6}$ as the sum of the series. The problem with this is that power series are not polynomials and they do not share the same properties as polynomials. This means that we cannot always apply this rule to power series, but it doesn’t mean we can’t ever apply it. For this reason, however, Euler’s proof is said not to hold up to today’s proof standards.
$\textbf{Trignometry and Algebra}$
This proof method uses a special trigonometric identity which involves the angle $\omega = \frac{\pi}{(2m + 1)}$ and several of its multiples. The identity is
(6)
\begin{align} \cot ^{2} \omega + \cot ^{2} (2\omega ) + \cot ^{2} (3\omega ) +…+\cot ^{2} (m\omega ) = \frac{m(2m-1)}{3} \end{align}
We know that for any $x$ between $0$ and $\frac{\pi}{2}$, the following is true:
(7)
\begin{align} \sin x < x <\tan x \end{align}
Squaring and inverting this leads to
(8)
\begin{align} \cot ^{2} x < \frac{1}{x^{2}}<1 + \cot ^{2} x \end{align}
Now, using (6) we can successively replace x with $\omega, 2 \omega, 3 \omega,$ etc. This gives
(9)
\begin{align} \cot ^{2} \omega + \cot ^{2} (2 \omega ) + \cot ^{2} (3 \omega ) + … + \cot ^{2}(m \omega ) \end{align}
(10)
\begin{align} < \frac{1}{ \omega ^{2}} + \frac{1}{4 \omega ^{2}} + \frac{1}{9 \omega ^{2}} + … + \frac{1}{m^{2} \omega ^{2}} \end{align}
(11)
\begin{align} < m + \cot ^{2} \omega + \cot ^{2} (2 \omega ) + \cot ^{2} (3 \omega ) + … + \cot ^{2}(m \omega ) \end{align}
Using the identity ([[eerf label1]]) we have the equation
(12)
\begin{align} \frac{m(2m-1)}{3} < \frac{1}{\omega ^{2}}(1 + \frac{1}{4} + \frac{1}{9} + … + \frac{1}{m ^{2}} < \frac{m(2m-1)}{3} + m \end{align}
Finally we can substitute $\omega = \frac{\pi}{(2m+1)}$ for:
(13)
\begin{align} \frac{m(2m-1)\pi ^{2}}{3(2m+1)^{2}} < 1 + \frac{1}{4} + \frac{1}{9} + … + \frac{1}{m ^{2}} < \frac{m(2m-1)\pi ^{2}}{3(2m+1)^{2}} + \frac{m\pi ^{2}}{(2m+1)^{2}} \end{align}
This set of inequalities provides upper and lower bounds for the sum of the first m terms of Euler’s series. If we let $m$ go to infinity, the lower bound is
(14)
\begin{align} \frac{m(2m-1)\pi ^{2}}{3(2m+1)^{2}}= \frac{\pi ^{2}}{6}\frac{2m ^{2} – m}{2m ^{2} + 2m +0.5} \end{align}
This approaches $\frac{\pi ^{2}}{6}$. The upper bound also approaches $\frac{\pi ^{2}}{6}$, confirming Euler’s finding.
The remaining four proofs will not be presented in the same manner as above but rather will be summarized.
$\textbf{Odd Terms, Geometric Series, and a Double Integral}$
This proof involves separating the summation of the odd and even terms of the sequence. Defining $E=\Sigma ^{\infty} _{k=1} \frac{1}{k ^{2}}$ and then calculating the integral representation of the even terms of the sequence shows us that these terms make up one-fourth of the value of $E$. If we define the sum of the odd terms as three-fourths of this $E$, we can develop another integral to describe the odd terms as eventually we determine, once again, that the sum of the sequence is $\frac{\pi ^{2}}{6}$.
$\textbf{Residue Calculus}$
Using a concept in residue calculus of complex integrals it is possible to calculate the sum of the series in question. The function used is $f(z) = \frac{\cot(\pi z)}{z^{2}}$ and the path ($P_{n}$) is the rectangle centered at the origin with sides parallel to the real and imaginary axis in the complex plane. The sides of this rectangle intersect the real axis at $^{+}_{-} (n + \frac{1}{2})$ and the imaginary axis at $^{+}_{-}ni$. Carrying out the residue calculations using this function and path we conclude that the same sum is reached for our series.
$\textbf{Fourier Analysis}$
This proof uses some concepts in Fourier analysis and compares them to the series. In Fourier analysis, the dot product of two functions is defined as $f \cdot g = \frac{1}{2\pi}\int ^{\pi}_{-\pi} f(t)g(t)dt$. If we are describing the dot product of the a function an itself, we can also write $f \cdot f = ...|a_{-2}|^{2} + |a_{-1}|^{2} + |a_{0}|^{2} + |a_{1}|^{2} + |a_{2}|^{2} + ...$. If we define, $f(t)=t$ in terms of our $a_{n}$ coefficients, we discover that it is just our series written twice (if the sum is $E$ then this sum is $2E$) and when we compute the sum using our dot product integral formula it is $\frac{\pi ^{2}}{3}$. So $2E=\frac{\pi ^{2}}{3}$, therefore $E=\frac{\pi ^{2}}{6}$.
$\textbf{A Real Integral with an Imaginary Value}$
This proof begins with the integral $I=\int ^{\frac{\pi}{2}}_{0} \ln (2\cos x)dx$. The logarithm present (because $2 \cos (x) = e^{ix} + e^{-ix}$) is replaced with a power series and integration is performed term by term. This gives us the odd terms of our series which we know are three-fourths of the sum of the total. These terms are multiplied by $\frac{1}{i}$ which gives us $\frac{-3i}{4}E$. Substituting this into our original integral we get $I=i(\frac{\pi ^{2}}{8} - \frac{3}{4}E)$. Setting the right-hand side of this equation to $0$ we get our familiar answer.
$\hline$
$\textbf{Connection to Real Analysis}$
This article is all about computing the sum of a sequence of real numbers. The fact that we are dealing with a sequence of real numbers places the topic of the article securely into the realm of Real Analysis. However, it also deals with different methods of computing this sum which is the real heart of analysis. It is easy to see from this article that some methods are easier than others are and all achieve the same end which is important in deciding which method works for the situation of proof you may be extending this result in to.
$\textbf{Context of the Article}$
The broader field of study for topics like the one covered in this article is sequences and series. Truly a topological study, series and sequences, and more specifically infinite sequences, are of particular interest to many because of the seemingly implausible ability to sum up the terms of an infinite series. For further reading on this subject, the best idea would to be to develop as sequence or series that interests the reader and research that particular series or read a real analysis textbook or other online article about summing infinite series.
Bibliography
Six Ways to Sum a Series
Dan Kalman
The College Mathematics Journal, Vol. 24, No. 5. (Nov., 1993), pp. 402-421. |
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