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# Quick Lesson on Area of a Sector
To recall, an industry is a portion of a circle that is confined between its two radii and the arc adjoining them. Let’s take a look at the concept of the Area of a Sector.
For example, a pizza slice is an instance of a field that stands for a portion of the pizza. There are two sorts of markets, small as well as a significant call. A minor field is less than a semi-circle market, whereas a substantial industry is higher than a semi-circle.
In this post, you will indeed find out:
What is the area of a sector?
How to locate the location of a field &
The formula for the area of a sector.
What is the area of an Industry?
Industry in the area is confined by the two spans of a circle and the arc. In short words, the area of a field is a fraction of the area of the circle.
#### How to Find the Area of a Sector?
To determine the area of a market, you need to understand the following two specifications:
The size of the circle’s radius.
The procedure of the main angle or the size of the arc. Hence, the central angle subtended by an arc of a sector at the center of a circle. Can give the central angle in levels or radians.
With the above two parameters, locating the area of a circle is as simple as ABCD. It is just an issue of connecting in the values in the area of the sector formula provided below.
### The formula for the area of a sector
There are three formulas for determining the location of a market. Each of these formulas is used relying on the type of info given regarding the market.
Area of an industry when the central angle is provided in degrees
If the angle of the sector is given up levels, then the formula for the area of industry is,
Area of a sector = (θ/ 360) πr2
A = (θ/ 360) πr2
Where θ = the central angle in degrees
Pi (π) = 3.14 as well as r = the radius of a field.
Location of a sector given the central angle in radians
If the central angle is given up radians, after that the formula for computing the area of industry is;
Area of a field = (θr2)/ 2.
Where, θ = the procedure of the central angle given up radians.
Location of a field given the arc length.
Given the size of the arc, the area of a market is provided by.
Area of a sector = rL/2.
Where r = distance of the circle.
L = arc size.
Let’s exercise several example troubles involving the area of a field.
#### Example
Calculate the area of the field revealed listed below.
Solution.
Location of a market = (θ/ 360) πr2.
Now, it implies (130/360) x 3.14 x 28 x 28.
Hence, it is equal to 888.97 cm2. |
Cube and Cuboid Questions and Answers
Questions of Cube and Cuboid.
Definition of Cube and Cuboid
A cube is a shape having three dimensions and 6 faces, 12 edges, and 8 corners. In a cube all the edges are of similar kind and include square-shaped faces. A Cube, which is a three-dimensional dense image have the same measurements of all of its dimensions. In total, there are 6 faces, 8 vertices & 12 edges in a cube. Vertex denotes the corners and an edge denotes the side.
These dimensions are known as length, breadth, and height. Conversely if a cube has six rectangular faces then, it is known as a cuboid. Each face of the cuboid is in rectangular shape, and at least four rectangles among them are identical.
Important Facts
The most important thing to consider while solving the questions of such types is to visualizing the cube in your mind. By looking at the cube you can clearly identify the basic terminologies such as the face, vertex, and edge of a cube Each line segment in the cuboid is edge, and the points where the edges meet is vertex. All edges are the sides, and vertices are the corners of the cuboid, and the opposite faces are parallel rectangles, which include similar dimensions.
Rules
• When a cube have its side measuring unit as ‘a’ and is painted on every face, and then it is reduced into smaller parts with measuring unit of sides as ‘b. Then you are expected to answer the quantity of cubes with ‘n’ faces painted. Reasonably if you see one specific edge of any big cube and visualize the smaller parts of that by making $\frac{a}{b}$ then the number of smaller cubes will be calculated by $(\frac{a}{b})^3$
• As we know that all the reduced cube parts will always have at least one face in the inner side, which means not on the exterior side; therefore, all the reduced cubes will have faces that are not painted. Also as the larger cubes meet at the corner points, i.e. 3, hence, the lesser cubes will be having a limit of 3 painted faces. Therefore, the smaller cubes with 3 faces painted = Number of large cube’s corners = every time 8 cubes. The only condition is that all the faces of the larger cubes are painted.
• To get the total number of smaller cubes having 2 faces painted only, we need to check the cube edges points. These are the points where 2 faces of the larger cubes meet. To solve these questions, you always need to include the corner cubes also, hence if you will remove 2 cubes from the total number of cubes on each edge, then you will easily get the answer.
• The cubes with one face painted can only be at the cubes at the face of the bigger cubes.
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Question 1
Time: 00:00:00
The following questions (1-5) are based on the information given below:
A cuboid shaped wooden block has 8 cm length, 4 cm breadth and 1 cm height.
Two faces measuring 4 cm x 1 cm are coloured in black.
Two faces measuring 8 cm x 1 cm are coloured in yellow.
Two faces measuring 8 cm x 4 cm are coloured in purple.
The block is divided into 8 equal cubes of side 1 cm (from 8 cm side), 4 equal cubes of side 1 cm(from 4 cm side).
How many cubes having yellow, purple and black colors on at least one side of the cube will be formed?
16
16
12
12
10
10
4
4
Once you attempt the question then PrepInsta explanation will be displayed.
Start
Question 2
Time: 00:00:00
A cuboid shaped wooden block has 8 cm length, 4 cm breadth and 1 cm height.
Two faces measuring 4 cm x 1 cm are coloured in black.
Two faces measuring 8 cm x 1 cm are coloured in yellow.
Two faces measuring 8 cm x 4 cm are coloured in purple.
The block is divided into 8 equal cubes of side 1 cm (from 8 cm side), 4 equal cubes of side 1 cm(from 4 cm side).
How many small cubes will be formed?
8
8
12
12
16
16
32
32
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Start
Question 3
Time: 00:00:00
A cuboid shaped wooden block has 8 cm length, 4 cm breadth and 1 cm height.
Two faces measuring 4 cm x 1 cm are coloured in black.
Two faces measuring 8 cm x 1 cm are coloured in yellow.
Two faces measuring 8 cm x 4 cm are coloured in purple.
The block is divided into 8 equal cubes of side 1 cm (from 8 cm side), 4 equal cubes of side 1 cm(from 4 cm side).
How many cubes will have 4 colored sides and two non-colored sides?
8
8
4
4
16
16
10
10
Once you attempt the question then PrepInsta explanation will be displayed.
Start
Question 4
Time: 00:00:00
A cuboid shaped wooden block has 8 cm length, 4 cm breadth and 1 cm height.
Two faces measuring 4 cm x 1 cm are coloured in black.
Two faces measuring 8 cm x 1 cm are coloured in yellow.
Two faces measuring 8 cm x 4 cm are coloured in purple.
The block is divided into 8 equal cubes of side 1 cm (from 8 cm side), 4 equal cubes of side 1 cm(from 4 cm side).
How many cubes will have Purple color on two sides and rest of the four sides having no color?
12
12
10
10
8
8
4
4
Once you attempt the question then PrepInsta explanation will be displayed.
Start
Question 5
Time: 00:00:00
A cuboid shaped wooden block has 8 cm length, 4 cm breadth and 1 cm height.
Two faces measuring 4 cm x 1 cm are coloured in black.
Two faces measuring 8 cm x 1 cm are coloured in yellow.
Two faces measuring 8 cm x 4 cm are coloured in purple.
The block is divided into 8 equal cubes of side 1 cm (from 8 cm side), 4 equal cubes of side 1 cm(from 4 cm side).
How many cubes will remain if the cubes having black and purple colored are removed?
4
4
8
8
16
16
24
24
Once you attempt the question then PrepInsta explanation will be displayed.
Start
Question 6
Time: 00:00:00
The following questions (6-10) are based on the information given below:
There is a cuboid whose dimensions are 6 x 3 x 3 cm.
The opposite faces of dimensions 6 x 3 are colored yellow.
The opposite faces of other dimensions 6 x 3 are colored red.
The opposite faces of dimensions 3 x 3 are colored green.
Now the cuboid is cut into small cubes of side 1 cm.
How many small cubes will have only two faces colored?
12
12
24
24
16
16
12
12
Once you attempt the question then PrepInsta explanation will be displayed.
Start
Question 7
Time: 00:00:00
There is a cuboid whose dimensions are 6 x 3 x 3 cm.
The opposite faces of dimensions 6 x 3 are colored yellow.
The opposite faces of other dimensions 6 x 3 are colored red.
The opposite faces of dimensions 3 x 3 are colored green.
Now the cuboid is cut into small cubes of side 1 cm.
How many small cubes have three faces colored?
24
24
20
20
16
16
8
8
Once you attempt the question then PrepInsta explanation will be displayed.
Start
Question 8
Time: 00:00:00
There is a cuboid whose dimensions are 6 x 3 x 3 cm.
The opposite faces of dimensions 6 x 3 are colored yellow.
The opposite faces of other dimensions 6 x 3 are colored red.
The opposite faces of dimensions 3 x 3 are colored green.
Now the cuboid is cut into small cubes of side 1 cm.
How many small cubes will have no face colored?
1
1
2
2
4
4
8
8
Once you attempt the question then PrepInsta explanation will be displayed.
Start
Question 9
Time: 00:00:00
There is a cuboid whose dimensions are 6 x 3 x 3 cm.
The opposite faces of dimensions 6 x 3 are colored yellow.
The opposite faces of other dimensions 6 x 3 are colored red.
The opposite faces of dimensions 3 x 3 are colored green.
Now the cuboid is cut into small cubes of side 1 cm.
How many small cubes will have only one face colored?
10
10
12
12
14
14
18
18
Once you attempt the question then PrepInsta explanation will be displayed.
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Question 10
Time: 00:00:00
There is a cuboid whose dimensions are 6 x 3 x 3 cm.
The opposite faces of dimensions 6 x 3 are colored yellow.
The opposite faces of other dimensions 6 x 3 are colored red.
The opposite faces of dimensions 3 x 3 are colored green.
Now the cuboid is cut into small cubes of side 1 cm.
Two cubes, each of edge 20 cm, are joined to form single cuboid. What is the surface area of the new cuboid so formed?
2000 $cm^{2}$
2000 $cm^{2}$
4000 $cm^{2}$
4000 $cm^{2}$
2400 $cm^{2}$
2400 $cm^{2}$
6000 $cm^{2}$
6000 $cm^{2}$
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Question 11
Time: 00:00:00
A cube of 2.7 liters volume will have each edge closest to
18 cm
18 cm
24 cm
24 cm
14 cm
14 cm
36 cm
36 cm
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Question 12
Time: 00:00:00
A cube having sides measuring 14 cms is cut into smaller cubes with side measuring 2cms. Calculate the total no of smaller cubes.
21952
21952
243
243
343
343
22542
22542
Once you attempt the question then PrepInsta explanation will be displayed. |
HomeTren&dThe Face Value of a Number: Understanding its Significance and Applications
# The Face Value of a Number: Understanding its Significance and Applications
Author
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Numbers are an integral part of our everyday lives. From counting objects to solving complex mathematical equations, numbers play a crucial role in various aspects of our society. However, beyond their numerical value, numbers also possess a face value that holds significant meaning and applications. In this article, we will explore the concept of the face value of a number, its importance, and how it is utilized in different fields.
## What is the Face Value of a Number?
The face value of a number refers to the numerical value that a digit holds in a given number. It is the value that the digit represents on its own, without considering its position within the number. For example, in the number 456, the face value of the digit 4 is 4, the face value of the digit 5 is 5, and the face value of the digit 6 is 6.
The face value of a number is determined by its place value, which is the value of the digit based on its position within the number. Place value is a fundamental concept in mathematics that helps us understand the significance of each digit in a number. By understanding the face value and place value of a number, we can perform various mathematical operations and solve problems efficiently.
## The Importance of Face Value in Mathematics
The face value of a number is essential in mathematics as it helps us understand the numerical value of each digit and its contribution to the overall value of the number. It allows us to perform arithmetic operations such as addition, subtraction, multiplication, and division accurately.
For example, when adding two numbers, we align the digits based on their place value and add the corresponding digits together. The face value of each digit determines the sum of the digits in the result. Similarly, when multiplying numbers, the face value of each digit determines the product of the digits in the result.
Understanding the face value of a number also helps in solving equations and inequalities. By considering the face value of each digit, we can determine the value of the unknown variable and find solutions to mathematical problems.
## Applications of Face Value in Different Fields
The concept of face value extends beyond mathematics and finds applications in various fields. Let’s explore some of these applications:
### 1. Finance and Accounting
In finance and accounting, face value is commonly used in the context of bonds and stocks. The face value of a bond represents the amount that the issuer promises to repay to the bondholder at maturity. It is also known as the par value or nominal value of the bond. The face value determines the interest payments and the principal repayment to the bondholder.
Similarly, in the stock market, the face value of a stock represents the initial value assigned to each share when it is issued. It is also known as the nominal value or par value of the stock. The face value of a stock does not necessarily reflect its market value, which is determined by supply and demand factors.
### 2. Currency and Coins
In the realm of currency and coins, face value plays a crucial role. The face value of a banknote or a coin represents the value assigned to it by the issuing authority. It is the value that the currency or coin can be exchanged for in goods and services.
For example, a \$10 banknote has a face value of \$10, which means it can be exchanged for goods and services worth \$10. The face value of a coin determines its purchasing power and is used as a standard measure of value in economic transactions.
### 3. Tickets and Coupons
Face value is also relevant in the context of tickets and coupons. The face value of a ticket represents the price at which it is sold to the customer. It is the value printed on the ticket and is used as a reference for pricing and revenue calculations.
Similarly, coupons often have a face value that represents the discount or savings offered to the customer. The face value of a coupon determines the amount that can be deducted from the total purchase price.
## Examples of Face Value in Action
Let’s consider a few examples to illustrate the concept of face value:
Suppose we want to add the numbers 345 and 678. By aligning the digits based on their place value, we can add the corresponding digits together:
• Face value of the digit 5 in 345: 5
• Face value of the digit 7 in 678: 7
The sum of the digits is 5 + 7 = 12. Therefore, the face value of the digit 2 in the result is 2.
### Example 2: Bonds
Consider a bond with a face value of \$1,000 and an annual interest rate of 5%. The bondholder will receive interest payments based on the face value of the bond. If the bond pays interest semi-annually, the bondholder will receive \$25 every six months (5% of \$1,000 divided by 2).
### Example 3: Currency
Suppose you have a \$20 bill. The face value of the bill is \$20, which means it can be exchanged for goods and services worth \$20.
## Summary
The face value of a number is the numerical value that a digit holds in a given number. It is determined by the digit’s place value and plays a crucial role in mathematics, finance, currency, and various other fields. Understanding the face value of a number allows us to perform mathematical operations accurately and interpret the significance of digits in different contexts.
By recognizing the face value of numbers, we can navigate the complexities of mathematics and apply this knowledge to real-world scenarios. Whether it’s calculating interest on a bond, determining the purchasing power of currency, or solving mathematical equations, the face value of a number provides valuable insights and applications.
## Q&A
### 1. What is the difference between face value and place value?
Face value refers to the numerical value that a digit holds in a given number, while place value is the value of the digit based on its position within the number. Face value focuses on the individual digit, while place value considers the digit’s position and its contribution to the overall value of the number.
### 2. How is face value used in the stock market?
In the stock market, face value represents the initial value assigned to each share when it is issued. It is used as a reference point for pricing and determining the number of shares outstanding. However, the face value of a stock does not necessarily reflect its market value, which is determined by supply and demand factors.
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# How do you simplify (7-2i)(4-i)?
Dec 8, 2016
The answer is $= 26 - 15 i$
#### Explanation:
This is the multiplication of 2 complex numbers
${z}_{1} = {a}_{1} + i {b}_{1}$
and ${z}_{2} = {a}_{2} + i {b}_{2}$
and ${i}^{2} = - 1$
${z}_{1} \cdot {z}_{2} = \left({a}_{1} + i {b}_{1}\right) \cdot \left({a}_{2} + i {b}_{2}\right)$
$= {a}_{1} {a}_{2} + i {a}_{1} {b}_{2} + i {a}_{2} {b}_{1} + {i}^{2} {b}_{1} {b}_{2}$
$= {a}_{1} {a}_{2} - {b}_{1} {b}_{2} + i \left({a}_{1} {b}_{2} + {a}_{2} {b}_{1}\right)$
Here we have,
$\left(7 - 2 i\right) \left(4 - i\right) = 28 - 7 i - 8 i + 2 {i}^{2}$
$= 28 - 2 - 15 i$
$= 26 - 15 i$ |
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# Solving quadratics by factoring
Learn how to solve quadratic equations like (x-1)(x+3)=0 and how to use factorization to solve other forms of equations.
## What you will learn in this lesson
So far you have solved linear equations, which include constant terms—plain numbers—and terms with the variable raised to the first power, ${x}^{1}=x$.
You may have also solved some quadratic equations, which include the variable raised to the second power, by taking the square root from both sides.
In this lesson, you will learn a new way to solve quadratic equations. Specifically you will learn
• how to solve factored equations like $\left(x-1\right)\left(x+3\right)=0$ and
• how to use factorization methods in order to bring other equations $\left($like ${x}^{2}-3x-10=0\right)$ to a factored form and solve them.
## Solving factored quadratic equations
Suppose we are asked to solve the quadratic equation $\left(x-1\right)\left(x+3\right)=0$.
This is a product of two expressions that is equal to zero. Note that any $x$ value that makes either $\left(x-1\right)$ or $\left(x+3\right)$ zero, will make their product zero.
$\begin{array}{rl}\left(x-1\right)& \left(x+3\right)=0\\ \\ ↙\phantom{\rule{1em}{0ex}}& \phantom{\rule{1em}{0ex}}↘\\ \\ x-1=0\phantom{\rule{1em}{0ex}}& \phantom{\rule{1em}{0ex}}x+3=0\\ \\ x=1\phantom{\rule{1em}{0ex}}& \phantom{\rule{1em}{0ex}}x=-3\end{array}$
Substituting either $x=1$ or $x=-3$ into the equation will result in the true statement $0=0$, so they are both solutions to the equation.
Now solve a few similar equations on your own.
Solve $\left(x+5\right)\left(x+7\right)=0$.
Solve $\left(2x-1\right)\left(4x-3\right)=0$.
### Reflection question
Can the same solution method be applied to the equation $\left(x-1\right)\left(x+3\right)=6$?
### A note about the zero-product property
How do we know there are no more solutions other than the two we find using our method?
The answer is provided by a simple but very useful property, called the zero-product property:
If the product of two quantities is equal to zero, then at least one of the quantities must be equal to zero.
Substituting any $x$ value except for our solutions results in a product of two non-zero numbers, which means the product is certainly not zero. Therefore, we know that our solutions are the only ones possible.
## Solving by factoring
Suppose we want to solve the equation ${x}^{2}-3x-10=0$, then all we have to do is factor ${x}^{2}-3x-10$ and solve like before!
${x}^{2}-3x-10$ can be factored as $\left(x+2\right)\left(x-5\right)$.
The complete solution of the equation would go as follows:
$\begin{array}{rl}{x}^{2}-3x-10& =0\\ \\ \left(x+2\right)\left(x-5\right)& =0& & \text{Factor.}\end{array}$
$\begin{array}{rlr}& ↙& ↘\\ \\ x+2& =0& x-5& =0\\ \\ x& =-2& x& =5\end{array}$
Now it's your turn to solve a few equations on your own. Keep in mind that different equations call for different factorization methods.
### Solve ${x}^{2}+5x=0$ .
Step 1. Factor ${x}^{2}+5x$ as the product of two linear expressions.$\phantom{\rule{1em}{0ex}}$
Step 2. Solve the equation.
### Solve ${x}^{2}-11x+28=0$ .
Step 1. Factor ${x}^{2}-11x+28$ as the product of two linear expressions.$\phantom{\rule{1em}{0ex}}$
Step 2. Solve the equation.
### Solve $4{x}^{2}+4x+1=0$ .
Step 1. Factor $4{x}^{2}+4x+1$ as the product of two linear expressions.$\phantom{\rule{1em}{0ex}}$
Step 2. Solve the equation.
### Solve $3{x}^{2}+11x-4=0$ .
Step 1. Factor $3{x}^{2}+11x-4$ as the product of two linear expressions.$\phantom{\rule{1em}{0ex}}$
Step 2. Solve the equation.
## Arranging the equation before factoring
### One of the sides must be zero.
This is how the solution of the equation ${x}^{2}+2x=40-x$ goes:
$\begin{array}{rlr}& ↙& ↘\\ \\ x+8& =0& x-5& =0\\ \\ x& =-8& x& =5\end{array}$
Before we factored, we manipulated the equation so all the terms were on the same side and the other side was zero. Only then were we able to factor and use our solution method.
### Removing common factors
This is how the solution of the equation $2{x}^{2}-12x+18=0$ goes:
$\begin{array}{rl}2{x}^{2}-12x+18& =0\\ \\ {x}^{2}-6x+9& =0& & \text{Divide by 2.}\\ \\ \left(x-3{\right)}^{2}& =0& & \text{Factor.}\\ \\ & ↓\\ \\ x-3& =0\\ \\ x& =3\end{array}$
All terms originally had a common factor of $2$, so we divided all sides by $2$—the zero side remained zero—which made the factorization easier.
Now solve a few similar equations on your own.
Find the solutions of the equation.
$2{x}^{2}-3x-20={x}^{2}+34$
Choose all answers that apply:
Find the solutions of the equation.
$3{x}^{2}+33x+30=0$
Choose all answers that apply:
Find the solutions of the equation.
$3{x}^{2}-9x-20={x}^{2}+5x+16$
Choose all answers that apply:
## Want to join the conversation?
• In the above equation 3x^2+11x-4 = 0, I understand where we need to find two numbers were a+b need to equal 11 to satisfy the 11x, however, I'm having trouble connecting where -12 came from where it states that we need to find numbers to satisfy (a)(b) = -12. I'm seeing a -4 at the end of the equation. Not sure where -12 came from. Was it from multiplying -4 to the co-effiecient of the 3 in 3x^2?
• In the standard form of quadratic equations, there are three parts to it: ax^2 + bx + c where a is the coefficient of the quadratic term, b is the coefficient of the linear term, and c is the constant. The -4 at the end of the equation is the constant. This hopefully answers your last question. Now, your first question.
So the problem is that you need to find two numbers (a and b) such that the sum of a and b equals 11 and the product equals -12. Correct? Correct indeed. I use a pretty straightforward mental method but I'll introduce my teacher's method of factors first. What you need to do is find all the factors of -12 that are integers. Let's start with 1.
1 * -12
2 * -6
3 * -4
4 * -3
6 * -2
12 * -1
Here we see 6 factor pairs or 12 factors of -12. Let's see which one adds up to 11. It seems like 12 + (-1) = 11. So we'll split 11 to 12 and -1.
My other method is straight out recognising the middle terms. This works well with small numbers. I can clearly see that 12 is close to 11 and all I need is a change of 1. So that leaves out 12 * -1 and -12 * 1. I can see that -12 * 1 makes -11 which is not what I want so I go with 12 * -1.
If you misunderstand something I said, just post a comment.
• Sometimes I don't understand some of the problems. :(
• Just always remember to simplify the expression before u do anything and then make one side equal to zero my subtracting or adding etc. Its easy once u get the hang of it.
• still confused
• Think about what is confusing you.
1) Is it the factoring? If it is, you need to review the lessons and practice problems for factoring. There are multiple factoring techniques. You need to know each of them and when to apply them.
2) If you are ok with the factoring, then what pay close attention to the other topics on the page. They cover the other steps you need to do in quite some detail.
• I don't get 3x^2-9x-20=x^2+5x+16. What do we do in the first place?
• Before trying to factor, you need to put the equation in the standard form: Ax^2+Bx+C=0. To do this, use opposite operations to move each term on the right side to the left side. Then factor.
Hope this helps.
• In the above equation 3x^2+11x-4 = 0, I understand where we need to find two numbers were a+b need to equal 11 to satisfy the 11x, however, I'm having trouble connecting where -12 came from where it states that we need to find numbers to satisfy (a)(b) = -12. I'm seeing a -4 at the end of the equation. Not sure where -12 came from. Was it from multiplying -4 to the co-effiecient of the 3 in 3x^2?
• Yes, the -12 comes from multiplying -4 with the 3 from 3x^2.
• I know this is not related to the above questions, but i could not find where to ask normal day to day questions. If someone could please help me answer this would be great.:
Factorise the expression as far as possible to prime factors:
(x-1)^2-9(y-1)^2 ...
^2(power of 2)
(1 vote)
• Expand the squares, then distribute the 9. Combine like terms then factor. Post comment if you still feel stuck.
• How do you factor it when the leading coefficient is more than 1? For example, something like 3xsquared+ x-2.
• My equation is x^2-2x+1=0 ( find x) How do i factor out the 1? Could i make it 1^2?
• You need 2 factors of 1 that add to -2.
Your choices are to use: 1*1 or (-1)(-1). Which pair adds to -2?
Hope this helps.
• I figure out that when A+B=[some number] and AB=[some number] combines, it could be an linear equation. But I don't know how to solve it, because when I'm solving it, a new quadratic equation comes out! For example:
Problem:
Factoring x²+3·x-10=(x+a)(x+b)
Linear equation:
①a+b=3
②a·b=-10
Solve:
a=3-b
Substitute 3-b into equation ②
(3-b)b=-10
3·b-b²=-10
And I don't know how to solve. |
# General and Principal Values of csc$$^{-1}$$ x
How to find the general and principal values of ccs$$^{-1}$$ x?
Let csc θ = x (| x |≥ 1 i.e., x ≥ 1 or, x ≤ - 1) then θ = csc$$^{-1}$$ x .
Here θ has infinitely many values.
Let – $$\frac{π}{2}$$ ≤ α ≤ $$\frac{π}{2}$$, where α is non-zero(α ≠ 0) positive or negative smallest numerical value of these infinite number of values and satisfies the equation csc θ = x then the angle α is called the principal value of csc$$^{-1}$$ x.
Again, if the principal value of csc$$^{-1}$$ x is α (– $$\frac{π}{2}$$ < α < $$\frac{π}{2}$$) and α ≠ 0 then its general value = nπ + (- 1) n α, where, | x | ≥ 1.
Therefore, tan$$^{-1}$$ x = nπ + α, where, (– $$\frac{π}{2}$$ < α < $$\frac{π}{2}$$), | x | ≥ 1 and (- ∞ < x < ∞).
Examples to find the general and principal values of arc csc x:
1. Find the General and Principal Values of csc $$^{-1}$$ (√2).
Solution:
Let x = csc$$^{-1}$$ (√2)
⇒ csc x = √2
⇒ csc x = csc $$\frac{π}{4}$$
⇒ x = $$\frac{π}{4}$$
⇒ csc$$^{-1}$$ (√2) = $$\frac{π}{4}$$
Therefore, principal value of csc$$^{-1}$$ (√2) is $$\frac{π}{4}$$ and its general value = nπ + (- 1)$$^{n}$$ ∙ $$\frac{π}{4}$$.
2. Find the General and Principal Values of csc $$^{-1}$$ (-√2).
Solution:
Let x = csc$$^{-1}$$ (-√2)
⇒ csc x = -√2
⇒ csc x = csc (-$$\frac{π}{4}$$)
⇒ x = -$$\frac{π}{4}$$
⇒ csc$$^{-1}$$ (-√2) = -$$\frac{π}{4}$$
Therefore, principal value of csc$$^{-1}$$ (-√2) is -$$\frac{π}{4}$$ and its general value = nπ + (- 1)$$^{n}$$ ∙ (-$$\frac{π}{4}$$) = nπ - (- 1)$$^{n}$$ ∙ $$\frac{π}{4}$$.
`
Inverse Trigonometric Functions |
# Question Video: Determining Whether Two Curves Defined Implicitly Are Orthogonal or Not Mathematics
Do the curves 9𝑦⁴ − 8𝑦 = 6𝑥 and −5𝑥² − 3𝑦 = −4𝑥 intersect orthogonally at the origin?
05:18
### Video Transcript
Do the curves nine 𝑦 to the fourth power minus eight 𝑦 equals six 𝑥 and negative five 𝑥 squared minus three 𝑦 equals negative four 𝑥 intersect orthogonally at the origin?
Before we start, we can verify that the curves intersect at the origin by checking both equations evaluated at zero, zero. When 𝑥 equals zero and 𝑦 equals zero, we can see that both equations become zero equals zero, which tells us that both curves pass through the origin. Remember, two curves intersect orthogonally at a point if their tangents are well defined at that point and are orthogonal to each other.
Let us find the slopes of the tangents at the origin for each curve by implicitly differentiating the given equations. For the first curve, we have d by d𝑥 of nine 𝑦 to the fourth power minus eight 𝑦 is equal to d by d𝑥 of six 𝑥. The expression on the left-hand side of the equation is in terms of the variable 𝑦. So we need to apply the chain rule. d by d𝑥 of nine 𝑦 to the fourth power minus eight 𝑦 is equal to d by d𝑦 of nine 𝑦 to the fourth power minus eight 𝑦 multiplied by d𝑦 by d𝑥, which equals 36𝑦 cubed minus eight multiplied by d𝑦 by d𝑥.
On the other hand, the right-hand side is d by d𝑥 of six 𝑥, which is equal to six. Hence, we have 36𝑦 cubed minus eight multiplied by d𝑦 by d𝑥 is equal to six. Dividing through by 36𝑦 cubed minus eight, we have d𝑦 by d𝑥 is equal to six over 36𝑦 cubed minus eight. We know that d𝑦 by d𝑥 evaluated at the origin gives the slope of the tangent.
Substituting the origin zero, zero into the equation gives us the following. Calculating the denominator, we have d𝑦 by d𝑥 is equal to six over negative eight, which is equal to negative three-quarters. Thus, the slope of the tangent of the first curve at the origin is negative three-quarters.
Next, let us find the slope of the tangent of the second curve. Implicitly differentiating the second equation gives us d by d𝑥 of negative five 𝑥 squared minus d by d𝑥 of three 𝑦 is equal to d by d𝑥 of negative four 𝑥. The first term on the left-hand side and the term on the right-hand side of the equation are regular derivatives, since the variable of these expressions matches the variable of differentiation 𝑥.
For the second term on the left-hand side, we need to use the chain rule. We have d by d𝑥 of negative five 𝑥 squared is equal to negative 10𝑥. d by d𝑥 of three 𝑦 is equal to d by d𝑦 of three 𝑦 multiplied by d𝑦 by d𝑥, which is equal to three multiplied by d𝑦 by d𝑥. And d by d𝑥 of negative four 𝑥 is equal to negative four. This gives us negative 10𝑥 minus three multiplied by d𝑦 by d𝑥 is equal to negative four.
Rearranging to make d𝑦 by d𝑥 the subject, we have d𝑦 by d𝑥 is equal to negative four plus 10𝑥 divided by negative three. Evaluating this equation at the origin zero, zero and simplifying, we have the following, which simplifies such that d𝑦 by d𝑥 is equal to four-thirds. Thus, the slope of the tangent to the second curve at the origin is four-thirds.
We now have the slopes of both tangents at the origin. Remember, if the slopes of two lines multiply to give negative one, then they are orthogonal. We obtained the slopes of the tangents for these two curves: negative three-quarters and four-thirds. And we see that negative three-quarters multiplied by four-thirds is equal to negative one.
We can therefore conclude that the two curves intersect orthogonally at the origin. |
# Sum of the Cubes of First n Natural Numbers
We will discuss here how to find the sum of the cubes of first n natural numbers.
Let us assume the required sum = S
Therefore, S = 1$$^{3}$$ + 2$$^{3}$$ + 3$$^{3}$$ + 4$$^{3}$$ + 5$$^{3}$$ + ................... + n$$^{3}$$
Now, we will use the below identity to find the value of S:
n$$^{4}$$ - (n - 1)$$^{4}$$ = 4n$$^{3}$$ - 6n$$^{2}$$ + 4n - 1
Substituting, n = 1, 2, 3, 4, 5, ............., n in the above identity, we get
1$$^{4}$$ - 0$$^{4}$$ = 4 ∙ 1$$^{3}$$ - 6 ∙ 1$$^{2}$$ + 4 ∙ 1 - 1
2$$^{4}$$ - 1$$^{4}$$ = 4 ∙ 2$$^{3}$$ - 6 ∙ 2$$^{2}$$ + 4 ∙ 2 - 1
3$$^{4}$$ - 2$$^{4}$$ = 4 ∙ 3$$^{3}$$ - 6 ∙ 3$$^{2}$$ + 4 ∙ 3 - 1
4$$^{4}$$ - 3$$^{4}$$ = 4 ∙ 4$$^{3}$$ - 6 ∙ 4$$^{2}$$ + 4 ∙ 4 - 1
........ .................... ...............
n$$^{4}$$ - (n - 1)$$^{4}$$ = 4 . n$$^{3}$$ - 6 ∙ n$$^{2}$$ + 4 ∙ n - 1
Adding we get, n$$^{4}$$ - 0$$^{4}$$ = 4(1$$^{3}$$ + 2$$^{3}$$ + 3$$^{3}$$ + 4$$^{3}$$ + ........... + n$$^{3}$$) - 6(1$$^{2}$$ + 2$$^{2}$$ + 3$$^{2}$$ + 4$$^{2}$$ + ........ + n$$^{2}$$) + 4(1 + 2 + 3 + 4 + ........ + n) - (1 + 1 + 1 + 1 + ......... n times)
n$$^{4}$$ = 4S - 6 ∙ $$\frac{n(n + 1)(2n + 1)}{6}$$ + 4 ∙ $$\frac{n(n + 1)}{2}$$ - n
⇒ 4S = n$$^{4}$$ + n(n + 1)(2n + 1) - 2n(n + 1) + n
⇒ 4S = n$$^{4}$$ + n(2n$$^{2}$$ + 3n + 1) – 2n$$^{2}$$ - 2n + n
⇒ 4S = n$$^{4}$$ + 2n$$^{3}$$ + 3n$$^{2}$$ + n - 2n$$^{2}$$ - 2n + n
⇒ 4S = n$$^{4}$$ + 2n$$^{3}$$ + n$$^{2}$$
⇒ 4S = n$$^{2}$$(n$$^{2}$$ + 2n + 1)
⇒ 4S = n$$^{2}$$(n + 1)$$^{2}$$
Therefore, S = $$\frac{n^{2}(n + 1)^{2}}{4}$$ = {$$\frac{n(n + 1)}{2}$$}$$^{2}$$ = (Sum of the first n natural numbers)$$^{2}$$
i.e., 1$$^{3}$$ + 2$$^{3}$$ + 3$$^{3}$$ + 4$$^{3}$$ + 5$$^{3}$$ + ................... + n$$^{3}$$ = {$$\frac{n(n + 1)}{2}$$}$$^{2}$$
Thus, the sum of the cubes of first n natural numbers = {$$\frac{n(n + 1)}{2}$$}$$^{2}$$
Solved examples to find the sum of the cubes of first n natural numbers:
1. Find the sum of the cubes of first 12 natural numbers.
Solution:
Sum of the cubes of first 12 natural numbers
i.e., 1$$^{3}$$ + 2$$^{3}$$ + 3$$^{3}$$ + 4$$^{3}$$ + 5$$^{3}$$ + ................... + 12$$^{3}$$
We know the sum of the cubes of first n natural numbers (S) = {$$\frac{n(n + 1)}{2}$$}$$^{2}$$
Here n = 12
Therefore, the sum of the cubes of first 12 natural numbers = {$$\frac{12(12 + 1)}{2}$$}$$^{2}$$
= {$$\frac{12 × 13}{2}$$}$$^{2}$$
= {6 × 13}$$^{2}$$
= (78)$$^{2}$$
= 6084
2. Find the sum of the cubes of first 25 natural numbers.
Solution:
Sum of the cubes of first 25 natural numbers
i.e., 1$$^{3}$$ + 2$$^{3}$$ + 3$$^{3}$$ + 4$$^{3}$$ + 5$$^{3}$$ + ................... + 25$$^{3}$$
We know the sum of the cubes of first n natural numbers (S) = {$$\frac{n(n + 1)}{2}$$}$$^{2}$$
Here n = 25
Therefore, the sum of the cubes of first 25 natural numbers = {$$\frac{25(25 + 1)}{2}$$}$$^{2}$$
{$$\frac{12 × 26}{2}$$}$$^{2}$$
= {25 × 13}$$^{2}$$
= (325)$$^{2}$$
= 105625
Arithmetic Progression
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Education
# Know About Set Theory Symbols and Types like Equal Set Here
Set Theory is the study of sets and their properties in mathematics. A set is a collection of items. The objects of a given set are its elements. The study of such sets and the relationships that exist between them is known as set theory. Set theory has shown to be a very valuable tool for defining some of mathematics’ most complex and significant structures, and it is thus an important part of the numerical ability/quantitative aptitude syllabus in various competitive tests.
On that note, let’s learn about various symbols used in Set theory as well as an important set type i.e., Equal sets in detail to further your knowledge of sets.
### Common Symbols used in Set Theory
A variety of Set Theory Symbols are used to represent sets. Let’s learn about each one of them in detail.
Symbol Corresponding set N It refers to the collection of all positive integers that are natural numbers. Examples: 1, 15, 192, 723 and so on. Z The entire integer number set is represented by it. From the Greek word “Zahl,” which means “number,” comes this symbol. Z+ stands for positive integers, while Z– is used to denote negative integers. Examples: -13, 0, 1327 etc. Q It represents a set of rational numbers. The sign’s root is the word “Quotient.” Positive and negative rational numbers are represented by Q+ and Q–, respectively, as the quotient of two integers (with a non-zero denominator). Examples: 13/17, -4/5 etc. R In addition to representing other numbers that might be written on a number line, it is used to represent real numbers. R+ and R–, respectively, are used to represent positive and negative real numbers. Examples: 2.56, π, 5√7, etc. C It is utilised to represent a set of complex numbers. Examples: 3+ 4i, i, etc.
Other symbols:
Symbols Symbol Name {} Set U Union ∩ Intersection ⊆ Subset ⊄ Not a subset ⊂ Proper subset ⊃ Proper superset ⊇ Superset ⊅ Not superset Ø Empty set P (C) Power set = Equal Set Ac Complement ∈ ‘An element of’ or ‘belongs to’ ∉ ‘Not an element of’ or ‘does not belong to’
### Equal Sets
All of the elements of Equal sets are equal and they all have the same cardinality. Usually, they are denoted by a capital letter and the braces “. In other words, two or more sets are considered equal if they include the same items and have the same number of elements. Take Set A = {2,4,6,8} and B = {2,4,6,8} as an example. As a result of having identical elements and cardinality, sets A and B are viewed as being equal.
Now, two sets are referred to be unequal sets if all of their elements are different, and they are referred to as equivalent sets if they contain the same amount of components. As an illustration, suppose A = {1, 2, 3, 4, 5}, C = {2, 4, 6, 7, 9}, and D = {2, 5, 6}. The number of elements in Sets A and C is the same, but not all of them are identical. A and C are hence equivalent sets. Now, the elements in sets A and D are not equal, nor do they have the same cardinality. Sets A and D are therefore not equal sets.
### Properties of Equal Sets
Equal sets are now clear to us. We’ll then look at some of its crucial properties to aid in understanding and identification of equal sets,
• The equality of the two sets is unaffected by the elements’ order.
• The total number of elements in equal sets, or cardinality, is equal.
• The set notation is A ⊆ B and B ⊆ A. when two sets are subsets of one another and the two sets are equal. As a result, A = B.
• There must be equality between equal collections of components.
• The power set of equal sets has the same cardinal number.
• Equal and equivalent sets both share the quality of having an equal number of elements.
• The converse is not true: all equal sets are not equivalent sets. |
# Determining the End Behavior of a Function
How do you determine the end behavior of a function? And, what does this mean?
When looking at a graph, the “end behavior” is referring to what is happening all the way to the far left of the graph and all the way to the far right of the graph. Your goal is to analyze the y-value (height) of the function when x is really large and negative, and then again when x is really large and positive. What is the pattern on each end? What is the “end behavior”?
Notationally, we are thinking:
1. As x → -∞, y → ?
2. As x → +∞, y → ?
OK, so let’s try this on a polynomial example:
Q: What is the end behavior of the function y=5x3+7x2-2x-1
A: OK. Let’s look at the left end behavior first:
As x approaches -∞, what is the function (y-value) doing?
Imagine x=-1000000 (some super large and super negative number, like the idea of -∞), we have:
y=5(-1000000)3+7(-1000000)2-2(-1000000)-1
Don’t do the actual math. Just think:
Is this number large or small?
Is it positive or negative?
I can look at the x3 term and see that it dominates this function. x2 and x are small peanuts compared to x3. So, in reaity, in polynomials, I can focus on the term of the largest degree:
y=5(-1000000)3+7(-1000000)2-2(-1000000)-1
y=5(-1000000)3
This number gives y = negative and super large.
As x → -∞, y → -∞
(As x approaches negative infinity, y approaches negative infinity).
Now, let’s look at the right end behavior:
As x approaches +∞, what is the function (y-value) doing?
Imagine x=+1000000 (some super large and super positive number, like the concept of +∞), we have:
y=5(+1000000)3+7(+1000000)2-2(+1000000)-1
And, by the same reasoning, we can focus on the term of largest degree:
y=5(+1000000)3+7(+1000000)2-2(+1000000)-1
y=5(+1000000)3 = super large and super positive
So, as x → +∞, y → +∞
(As x approaches positive infinity, y approaches positive infinity)
Note: in this example, y behavior mimicked x behavior, this isn’t always the case!
# End Behavior
1. Find the end behavior of the following functions:
(a) Consider: y = (x²-2x)/(-x³-5x²+4)
as x goes to infinity, y goes to ?
as x goes to – infinity, y goes ?
(b) Consider y = (2x+1)/(-4x+1)
as x goes to infinity, y goes ?
as x goes to – infinity, y goes ?
# Generating Function
Q: Find the generating function for the sequence:
a0 = 1
an = 2an-1 + 1
Also, find a closed form solution for an.
# Evaluating Functions
Q: Find the following values of the function: f(x) = 3x + 1
a. f(0) b. f(3)
c. f(-2) d.f(-x)
e. -f(x) f. f(x+2)
g. f(2x) h. f(x+h)
# Domain Example
Q: Find the domain of the function:
d(x) = √(x-1)/(x-6)
# Finding the Domain
Q: Find the domain of h(x):
h(x) = 27 / (3-x)
# Functions & points
Q: Evaluate f(3) for each of the following:
a) (1,2) (2,0) (3,1) (4,2)
b) (1,2) (2,3) (3,4) (4,5) |
# What is the probability of getting a head in a fair coin toss?
• Last Updated : 17 Jan, 2022
A branch of mathematics that deals with the happening of a random event is termed probability. It is used in Maths to predict how likely events are to happen. The probability of any event can only be between 0 and 1 and it can also be written in the form of a percentage.
### Probability
The probability of event A is generally written as P(A). Here P represents the possibility and A represents the event. It states how likely an event is about to happen. The probability of an event can exist only between 0 and 1 where 0 indicates that event is not going to happen i.e. Impossibility and 1 indicates that it is going to happen for sure i.e. Certainty.
If the outcome of an event is not sure, take help of the probabilities of certain outcomes, how likely they occur. For a proper understanding of probability, we take an example as tossing a coin, there will be two possible outcomes – heads or tails.
The probability of getting heads is half. It is already known that the probability is half/half or 50% as the event is an equally likely event and is complementary so the possibility of getting heads or tails is 50%.
Formula of Probability
Probability of an event = Favorable outcomes / Total number of outcomes
P(A) = Favorable outcomes / Total number of outcomes
### Some Terms of Probability Theory
• Experiment: An operation or trial done to produce an outcome is called an experiment.
• Sample Space: An experiment together constitutes a sample space for all the possible outcomes. For example, the sample space of tossing a coin is head and tail.
• Favorable Outcome: An event that has produced the required result is called a favorable outcome. For example, If two dice are rolled at the same time then the possible or favorable outcomes of getting the sum of numbers on the two dice as 4 are (1, 3), (2, 2), and (3, 1).
• Trial: A trial means doing a random experiment.
• Random Experiment: A random experiment is an experiment that has a well-defined set of outcomes. For example, when we toss a coin, we would get ahead or tail but the outcome is not confirmed that which one will appear.
• Event: An event is the outcome of a random experiment.
• Equally Likely Events: Equally likely events are rare events that have the same chances or probability of occurring. Here The outcome of one event is independent of the other. For instance, when a coin is tossed, there are equal chances of getting ahead or a tail.
• Exhaustive Events: An exhaustive event is when the set of all outcomes of an experiment is equal to the sample space.
• Mutually Exclusive Events: Events that cannot happen simultaneously are called mutually exclusive events. For example, the climate can be either cold or hot. The same weather again and again can not be experienced.
• Complementary Events: The Possibility of only two outcomes which is an event will occur or not, like a person will eat or not eat the food, buying a bike or not buying a bike, etc. are examples of complementary events.
Some Probability Formulae
• Addition rule: Union of two events, say A and B, then,
P(A or B) = P(A) + P(B) – P(A∩B)
P(A ∪ B) = P(A) + P(B) – P(A∩B)
• Complementary rule: If there are two possible events of an experiment so the probability of one event will be the Complement of another event. For example, if A and B are two possible events, then,
P(B) = 1 – P(A) or P(A’) = 1 – P(A).
P(A) + P(A′) = 1.
• Conditional rule: When the probability of an event is given and the second is required for which first is given, then P(B, given A) = P(A and B), P(A, given B). It can be vice versa,
P(B∣A) = P(A∩B)/P(A)
• Multiplication rule: Intersection of two other events i.e. events A and B need to occur simultaneously. Then,
P(A and B) = P(A) P(B).
P(A∩B) = P(A) P(B∣A)
### What is the probability of getting a head in a fair coin toss?
Solution:
If a fair coin is tossed then the sample space will be {H, T}
Therefore total number of event = 2
Probability of having head = 1
Probability of an event = (number of favorable event) / (total number of event)
P(B) = (occurrence of Event B) / (total number of event).
Probability of getting one head = 1/2.
### Similar Problems
Question1: What are the chances of flipping 5 heads in a row?
Solution:
Probability of an event = (number of favorable event) / (total number of event).
P(B) = (occurrence of Event B) / (total number of event).
Probability of getting one head = 1/2.
Here, tossing a coin is an independent event, its not dependent on how many times it has been tossed.
Probability of getting 2 heads in a row = probability of getting head first time × probability of getting head second time.
Probability of getting 2 head in a row = (1/2) × (1/2).
Therefore, the probability of getting 5 heads in a row = (1/2)5.
Question 2: What are the chances of flipping 20 heads in a row?
Solution:
Probability of an event = (number of favorable event) / (total number of event).
P(B) = (occurrence of Event B) / (total number of event).
Probability of getting one head = 1/2.
Here, tossing a coin is an independent event, its not dependent on how many times it has been tossed.
Probability of getting 3 heads in a row = probability of getting head first time × probability of getting head second time × probability of getting head third time
Probability of getting 3 head in a row = (1/2) × (1/2) × (1/2)
Therefore, the probability of getting 20 heads in a row = (1/2)20
Question 3: What are the chances of flipping 10 tails in a row?
Solution:
Probability of an event = (number of favorable event) / (total number of event).
P(B) = (occurrence of Event B) / (total number of event).
Probability of getting one tail = 1/2.
Here, if Tossing a coin is an independent event, its not dependent on how many times it has been tossed.
Probability of getting 3 tails in a row = probability of getting tail first time × probability of getting tail second time × probability of getting tail third time
Probability of getting 3 tails in a row = (1/2) × (1/2) × (1/2)
Therefore, the probability of getting 10 tails in a row = (1/2)10
Question 4: When a fair coin is tossed what is the probability of getting a tail?
Solution:
If a fair coin is tossed then the sample space will be {H, T}
Therefore total number of event = 2
Probability of having tail = 1
Probability of an event = (number of favorable event) / (total number of event)
P(B) = (occurrence of Event B) / (total number of event).
Probability of getting one tail = 1/2.
Question 5: What are the chances of flipping 20 tails in a row?
Solution:
Probability of an event = (number of favorable event) / (total number of event).
P(B) = (occurrence of Event B) / (total number of event).
Probability of getting one tail = 1/2.
Here, tossing a coin is an independent event, its not dependent on how many times it has been tossed.
Probability of getting 3 tails in a row = probability of getting tail first time × probability of getting tail second time × probability of getting tail third time
Probability of getting 3 tails in a row = (1/2) × (1/2) × (1/2)
Therefore, the probability of getting 20 tails in a row = (1/2)20
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# Perpendicular Bisector Theorem
Any point on a perpendicular bisector of a line segment is equidistant from the endpoints of the line segment.
Based on the characteristics of the diagram, is the perpendicular bisector of Therefore, is equidistant from and
### Proof
Geometric Approach
Suppose is the perpendicular bisector of Then is the midpoint of
Consider a triangle with vertices and and another triangle with vertices and and
Both and have a right angle and congruent legs and Since all right angles are congruent, Furthermore, by the Reflexive Property of Congruence, is congruent to itself.
By the Side-Angle-Side Congruence Theorem, the triangles are congruent. Therefore, since corresponding parts of congruent figures are congruent, their hypotenuses and are also congruent. By the definition of congruent segments, and have the same length. This means that is equidistant from and
Using this reasoning it can be proven that any point on a perpendicular bisector is equidistant from the endpoints of the segment.
### Two-Column Proof
The proof can be summarized in the following two-column table.
Statements Reasons and are right angles Definition of a perpendicular bisector. All right angles are congruent. Reflexive Property of Congruence. SAS Congruence Theorem. Corresponding parts of congruent figures are congruent. Definition of congruent segments.
### Proof
Perpendicular Bisector Theorem
Suppose is the perpendicular bisector of and that is the midpoint of
Two triangles can be created by connecting points and and and
These triangles both have a right angle and one of the legs measures half of They also share one leg,
According to the SAS Congruence Theorem, the triangles are congruent. Thus, their hypotenuses are also congruent.
Therefore, any point on a perpendicular bisector is equidistant from the endpoints of the segment. This can be summarized in a two-column proof.
Statement Reason Given SAS congruence theorem Definition of congruent segments
Note that and are not triangles if is the point of intersection, However, since is the midpoint of it is, by definition, equidistant from and |
# 3 Valentine’s Day Math Crafts
Valentine’s Day in the classroom has a different energy than a typical school day. Meet your students and their energy with a fun and novel math activity that keeps them learning. Try one of these 3 Valentine’s Day math crafts this upcoming holiday!
#### Valentine’s Day Love Bug Place Value Math Craft
To do this activity with your students, you will need to provide your students with bug bodies. Accordion fold strips of paper for arms and legs and your students are ready to create this cute craft! Allow your students to choose a 3-digit number and ask them to show different representations of their number on the bug’s body, arms and legs. If you want to add space for even more representations, you can also add hands and feet to your Love Bug. Representations can include:
• Unit Form
• Word Form
• Standard Form
• Expanded Form
• Place Value Drawing
• Show the Number in Money
• Even or Odd
• Equation
• Etc!
#### Fact Family Valentine’s Day Math Craft
This activity is perfect for a kindergarten or first grade class. Give your students 2 heart templates. On one template, ask your students to create a number bond. On the second template, ask your students to write 4 (or 8!) related number sentences. I HIGHLY recommend allowing your students to use linking cubes in two colors to help to generate the related number sentences. For example, a 4,5,9 number bond can be explored with 4 red cubes and 5 blue cubes. As students put the cubes together and take the cubes apart they can list the related facts.
#### Fraction Chocolate Box Valentine’s Day Activity
Third or Fourth grade students would benefit from this Valentine’s Day activity. Give your students a chocolate box template with a set number of chocolates. Ask your students to “fill” the chocolate box with a small selection of chocolates they design themselves. I like having 8 chocolates and asking the students to fill the box with 3 types of chocolates they design themselves. Eight is a nice number because it plays will with equivalent fractions (1/2 and 1/4) so there is room for students to explore. Next, ask your students to describe their chocolate box in terms of fractions, compare the fraction of different types of chocolates and look for equivalent fractions.
If you are looking for a QUICK way to introduce any of these activities to your students, I have created templates and planning pages. A quick “print and go” will have your students engaged and ready to go!
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# 6+ Addition And Subtraction Word Problems Worksheets
What is a word problem 2nd grade?
What is a word problem for addition?
The key words used in problems involving addition are: sum; total; in all; all together.
What are words for adding and subtracting?
Addition-sum, altogether, all, in all, together, total, total number, add, increase, increased by, more than. Subtraction-minus, greater than, take away, fewer than, less than, subtract, decreased by.
## How do I get my child to understand word problems?
• Solve word problems regularly.
• Teach problem-solving routines.
• Visualize or model the problem.
• Make sure they identify the actual question.
• Remove the numbers.
• Try the CUBES method.
• Show word problems the LOVE.
• ## How do you solve math word problems?
• Identify and list the facts.
• Figure out exactly what the problem is asking for.
• Eliminate excess information.
• Pay attention to units of measurement.
• Draw a diagram.
• Find or develop a formula.
• Consult a reference.
## What are the steps to solving word problems?
• Read the word problem. Make sure you understand all the words and ideas.
• Identify what you are looking for.
• Name what you are looking for.
• Translate into an equation.
• Solve the equation using good algebra techniques.
• Check the answer in the problem.
• Answer the question with a complete sentence.
• ## What is addition keywords?
Specifically, words like sum, total, increase, add, plus are commonly used for addition, as well as words like altogether, combine, more than, all, both, and so on. The same goes for subtraction: decrease, minus, fewer, take away, difference, left, less than, subtract, remain, and so on.
## What are the keywords in math word problems?
Below is a partial list.
• Addition: increased by. more than. combined, together. total of.
• Subtraction: decreased by. minus, less. difference between/of.
• Multiplication: of. times, multiplied by. product of.
• Division: per, a. out of. ratio of, quotient of.
• Equals. is, are, was, were, will be. gives, yields. sold for, cost.
• ## What goes first addition or subtraction?
If needed, remind them that in the order of operations, multiplication and division come before addition and subtraction.
## What is a subtraction word problem?
Subtraction word problems arise in any situations where there is a loss or a decrease of something as a result of deducting a number from another. Sebastian is experiencing a loss of 5 pencils, so the subtraction problem to solve to get the answer is 8 - 5 and 8 - 5 = 3.
## What is the answer to a subtraction problem called?
In the subtraction problem, the bigger number is called minuend and the number subtracted from it is called subtrahend. The answer in subtraction is called difference.
## What are the examples of addition and subtraction?
Subtraction and addition are inverse operations. For example, 6 = 4 + 2 is equivalent to 6 − 4 = 2 and also 6 − 2 = 4.
## What are the 3 types of subtraction?
But there are actually three different interpretations of subtraction:
• Taking away.
• Part-whole.
• Comparison.
## Why do kids struggle with math word problems?
Children often struggle with math word problems because they require an ability to analyze information and extract only the useful elements. Instead of being told directly what operation they need to do, they have to discover it themselves before they can even begin to figure out the solution.
## Why do children find word problems difficult?
Word problems tend to be complicated in part because of their descriptive language. Students often don't understand what exactly they're being asked, especially when the problem includes abstract concepts.
## What age do kids learn math word problems?
Ages 6 to 10 years: Learning math
Understand fractions and word problems by fourth grade.
## What are the 4 steps in solving word problems?
Polya created his famous four-step process for problem solving, which is used all over to aid people in problem solving:
• Step 1: Understand the problem.
• Step 2: Devise a plan (translate).
• Step 3: Carry out the plan (solve).
• Step 4: Look back (check and interpret).
• ## What are the 5 steps in solving word problems?
5 Steps to Word Problem Solving
• Identify the Problem. Begin by determining the scenario the problem wants you to solve.
• Gather Information.
• Create an Equation.
• Solve the Problem.
• ## What is an example of a word problem?
Word problems commonly include mathematical modelling questions, where data and information about a certain system is given and a student is required to develop a model. For example: Jane had \$5.00, then spent \$2.00. How much does she have now?
## How do you introduce a word problem to grade 2?
• Write a Number Sentence.
• Use a Strategy to solve. I explain to students that I can't hop inside their brains to see what they're thinking, and if I don't know what they're thinking, I can't help them become better word problem solvers.
• Computation.
## What is typically the first step in solving a word problem?
Generally, solving a word problem involves four easy steps: Read through the problem and set up a word equation — that is, an equation that contains words as well as numbers. Use math to solve the equation. Answer the question the problem asks.
## What are the 3 steps in solving word problems?
3-Step System
• Plan: Think about what the story is asking you to do.
• Solve: What strategy could you use to find the missing information: addition, subtraction, multiplication, or division?
• ## Which of the following is the first step in solving a word problem?
The first step to solving a math word problem is to read the problem in its entirety to understand what you are being asked to solve. After you read it, you can decide the most relevant aspects of the problem that need to be solved and what aspects are not relevant to solving the problem.
## Can you use the word sum for subtraction?
Because subtraction is based on addition, let's look at some basics of addition first. As illustrated above, the number that is the “sum” in addition becomes the “minuend” in subtraction. The “addends” in addition become the “subtrahend” and the “difference” in subtraction.
## How do you write subtraction in words?
Word problems will often use "is left," "difference," "take away," or "less" to denote subtraction.
## What are the parts of an addition problem called?
There are 3 parts of addition, the addend, the equal sign, and the sum.
• The Equal Sign. The equal sign indicates that the two halves of the equation are equivalent.
• The Sum.
## What is the term for subtraction?
The terms of subtraction are called minuend and subtrahend, the outcome is called the difference. The minuend is the first number, it is the number from which you take something and it must be the larger number. The subtrahend is the number that is subtracted and it must be the smaller number.
## What does how much mean in math word problems?
In mathematics, the term 'how much' usually refers to a quantity of some sort, often a numerical quantity.
## What is the order of math problems?
The order of operations is a rule that tells the correct sequence of steps for evaluating a math expression. We can remember the order using PEMDAS: Parentheses, Exponents, Multiplication and Division (from left to right), Addition and Subtraction (from left to right).
## Does addition always come before subtraction?
Order of operations tells you to perform multiplication and division first, working from left to right, before doing addition and subtraction. (Note that addition is not necessarily performed before subtraction.)
## What are the four rules of maths?
The four rules of mathematics are adding, subtracting, multiplying and dividing.
## Why is the answer to a subtraction problem called difference?
Why is the answer to a subtraction problem called difference? The value being subtracted is called the subtrahend, and the value from which the subtrahend is being subtracted is called the minuend. For example, in the subtraction problem 5 – 3 = 2, 5 is the minuend, 3 is the subtrahend and 2 is the difference.
## Why should I use addition to check the answer to a subtraction problem?
Difference of two numbers is correct when the sum of the subtrahend number and the difference is equal to the minuend. How can you use addition to check the answer of subtraction? Addition (+) and subtraction (-) are opposite operations in math, so you can use one to check the answer of the other one.
## What is the connection between addition and subtraction?
What is the relationship between addition and subtraction? Addition and subtraction are the inverse operations of each other. Put simply, this means that they are the opposite. You can undo an addition through subtraction, and you can undo a subtraction through addition.
## How do you improve addition and subtraction?
• Count From A Number Upwards. Example: 6 + 3.
• Jump Strategy.
• Do The Tens Last.
• Aim for Ten.
• Compensation Method.
• Double when the numbers are the same.
• Double if the numbers are close, then fix.
• ## Images for 6+ Addition And Subtraction Word Problems Worksheets
### Addition subtraction word problem worksheet
Students look for verbal clues when solving word problems. “More” usually (but not always) suggests addition and “less” usually (but not always) suggests subtraction. Look out for problems where these words suggest the opposite of what they usually do.
• Solve word problems regularly.
• Teach problem-solving routines.
• Visualize or model the problem.
• Make sure they identify the actual question.
• Remove the numbers.
• Try the CUBES method.
• Show word problems the LOVE. |
# 7d. Integration by Substitution
## Integration Mini Video Lecture
This video explains how to use substitution to make an integral that looks difficult, into one that is easier.
## Video transcript
[Music...]
Let's start by looking at an example with fractional exponents, just a nice, simple one.
Before I start that, we're going to have quite a lot of this sort of thing going on, where we get some kind of fraction on the bottom of a fraction, and it gets confusing.
So what we should do is think about this as, like, 1, divide, and then 3 over 2.
When I divide by a fraction, what I should do is multiply by the reciprocal, and this just gives me two-thirds (2/3).
Now, here's a simple integration here, involving a fraction. So when I'm integrating this sort of thing I should add 1 to the index and then divide by the new number.
So I'm going to have something like this: I'm going to have on the top v and then
... 3 over 2 (3/2),
... over 3/2 - because
... a half (1/2)
... plus 1 is "1 and a 1/2" (1 1/2),
... and then divide by the new number - plus K.
Now, this thing looks a bit untidy, but I can use what I had up here, I can just type this as 2/3 of v to the 3/2 (2/3 v^(3/2)), and then, plus K.
Now, normally I don't even do it like this. I don't have this step in the middle, I just go straight to this, yep, add 1 to the index here, you get 3/2, turn it upside down and you hit 2/3 times v^(3/2) plus K.
You put this step in the middle if you're not sure what's going on.
Great. Let's move on to the main problem now.
What we want to do is we want to integrate the square root of 5x minus 4 (sqrt(5x-4)).
So the first thing we should do here is to realize that sqrt(5x-4) can be written as 5x − 4 in brackets, to the power 1/2, and then put dx.
Now, with this sort of problem, we should do some substituting. And the normal substitution to do is u equals 5x minus 4 (u = 5x-4).
I've chosen 5x − 4 because that's what's in the brackets here.
The step that you do after that is always (du)/dx, and (du)/dx in this case is 5.
Now, as part of the substitution, I'm going to replace this with u, and this will become u to 1/2 (u^(1/2)), but I've got a dx here, and I can't have an extra u and dx, I've got to do something about it.
So I write the expression I've just done, du = 5dx, and one more step I'm going to have one-fifth [1/5] of du equals dx (1/5 du = dx), and all I've done now is divide both sides of this by 5.
Okay. Now, what I've got to do next is do the substitution steps. So I take this and just paste it in here, and now really put the substitution: 5x − 4 (I've said it's going to become u, and 5x − 4 was to the power 1/2, so I must do it to the power 1/2.
dx is equal to 1/5 du, so I do a du, and where should I put this 1/5? Well, the best place to put it is out in the front, like this. It's just a constant, and so I can just put it out in the front.
So here I've got some expression involving x with a dx. This is correct. I've got an expression with u and du. This is also correct. In this work, don't mix up your x's and du's - it gets very confusing and probably incorrect.
Okay. So I just do that 1/5, the end, and now what I have to do is I have to integrate u^(1/2) du.
But this is just like the problem I did above, isn't it? Let's go and have another look quickly. Yes, look.
My integral of v^(1/2) dv and I did it like this. Well, do the same thing down here, integral of u^(1/2) du is going to be 2/3 what we got before, 2/3 u^(3/2) and then plus K. The 3/2 comes from adding 1 to this, 1 plus 1/2 is 1 1/2 or 3/2, and the 2/3 comes from dividing 1 by 3/2 or multiplying by 2/3.
Then, let's tidy this up, we're going to have fraction out in the front, 2/15, and it's going to be u^(3/2) plus K. Now, we're almost finished, but not quite. What we want to do now is express this back in terms of x. If we don't do that, that's incorrect - we started with x, we should finish with x. Now, it's always a very good idea to have a clear statement of u, because you always have to substitute back again.
I go and look up here, here is my clear statement of what u is, and so I substitute back into here: 5x − 4 and raise it to the power 3/2, because that's what it is, and plus K. And that's the end of the story.
[Music...]
top
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### Calculus Lessons on DVD
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# SCC Education
## Direct Proportion
Introduction:-
1. Direct Proportion: Two quantities are said to be directly proportional, if on the increase (or decrease) of the one, the other increases (or decreases) to the same
Ex. 1. Cost is directly proportional to the number of articles.(More Articles, More Cost)
Ex. 2. Work done is directly proportional to the number of men working on it (More Men, More Work)
2. Indirect Proportion: Two quantities are said to be indirectly proportional,if on the increase of the one, the other decreases to the same extent and vice-versa.
Ex. 1. The time taken by a car in covering a certain distance is inversely proportional to the speed of the car. (More speed, Less is the time taken to cover a distance)
Ex. 2. Time taken to finish a work is inversely proportional to the number of persons working at it. (More persons, Less is the time taken to finish a job)
# Chain Rule Examples
Ex. 1. If 15 toys cost Rs, 234, what do 35 toys cost?
Sol. Let the required cost be Rs. x. Then, More toys, More cost (Direct Proportion)
. 15 : 35 : : 234 : x = (15 x x) = (35 x 234) = x=(35 X 234)/15 =546
Hence, the cost of 35 toys is Rs. 546.
Ex. 2. If 36 men can do a piece of work in 25 hours, in how many hours will 15 men do it ?
Sol. Let the required number of hours be x. Then, Less men, More hours (Indirect Proportion)
15 : 36 : : 25 : x =(15 x x) = (36 x 25) =(36 x 25)/15 = 60
Hence, 15 men can do it in 60 hours.
Ex. 3. If the wages of 6 men for 15 days be Rs.2100, then find the wages of for 12 days.
Sol. Let the required wages be Rs. x.
More men, More wages (Direct Proportion)
Less days, Less wages (Direct Proportion)
Men 6: 9 : :2100:x
Days 15:12
Therefore (6 x 15 x x)=(9 x 12 x 2100) = x=(9 x 12 x 2100)/(6 x 15)=2520
Hence the required wages are Rs. 2520.
Ex. 4. If 20 men can build a wall 66 metres long in 6 days, what length of a similar can be built by 86 men in 8 days?
Sol. Let the required length be x metres
More men, More length built (Direct Proportion)
Less days, Less length built (Direct Proportion)
Men 20: 35
Days 6: 3 : : 56 : x
Therefore (20 x 6 x x)=(35 x 3 x 56)x=(35 x 3 x 56)/120=49
Hence, the required length is 49 m.
Ex. 5. If 15 men, working 9 hours a day, can reap a field in 16 days, in how many days will 18 men reap the field, working 8 hours a day?
Sol. Let the required number of days be x.
More men, Less days (indirect proportion)
Less hours per day, More days (indirect proportion)
Men 18 : 15
Hours per day 8: 9 } : :16 : x
(18 x 8 x x)=(15 x 9 x 16) x=(44 x 15)144 = 15
Hence, required number of days = 15.
Ex. 6. If 9 engines consume 24 metric tonnes of coal, when each is working 8 hours day, bow much coal will be required for 8 engines, each running 13hours a day, it being given that 3 engines of former type consume as much as 4 engines of latter type?
Sol. Let 3 engines of former type consume 1 unit in 1 hour.
Then, 4 engines of latter type consume 1 unit in 1 hour.
Therefore 1 engine of former type consumes(1/3) unit in 1 hour.
1 engine of latter type consumes(1/4) unit in 1 hour.
Let the required consumption of coal be x units.
Less engines, Less coal consumed (direct proportion)
More working hours, More coal consumed (direct proportion)
Less rate of consumption, Less coal consumed(direct prportion)
Number of engines 9: 8
Working hours 8 : 13 } :: 24 : x
Rate of consumption (1/3):(1/4)
[ 9 x 8 x (1/3) x x) = (8 x 13 x (1/4) x 24 ) = 24x = 624 = x = 26.
Hence, the required consumption of coal = 26 metric tonnes.
Ex. 7. A contract is to be completsd in 46 days sad 117 men were said to work 8 hours a day. After 33 days, (4/7) of the work is completed. How many additional men may be employed so that the work may be completed in time, each man now working 9 hours a day?
Sol. Remaining work = (1-(4/7) =(3/7)
Remaining period = (46 – 33) days = 13days
Let the total men working at it be x.
Less work, Less men (Direct Proportion)
Less days, More men (Indirect Proportion)
More Hours per Day, Less men (Indirect Proportion)
Work (4/7): (3/7)
Days 13:33 } : : 117: x
Hrs/day 9 : 8
Therefore (4/7) x 13 x 9 x x =(3/7) x 33 x 8 x 117 or x=(3 x 33 x 8 x 117)/(4 x 13 x 9)=198 Additional men to be employed = (198 – 117) = 81.
Ex. 8. A garrison of 3300 men had provisions for 32 days, when given at the rate of 860 gns per head. At the end of 7 days, a reinforcement arrives and it was for that the provisions wi1l last 17 days more, when given at the rate of 826 gms per head, What is the strength of the reinforcement?
Sol. The problem becomes:
3300 men taking 850 gms per head have provisions for (32 – 7) or 25 days,
How many men taking 825 gms each have provisions for 17 days?
Less ration per head, more men (Indirect Proportion)
Less days, More men (Indirect Proportion)
Ration 825 : 850
Days 17: 25 } : : 3300 : x
(825 x 17 x x) = 850 x 25 x 3300 or x = (850 x 25 x 3300)/(825 x 17)=5000
Strength of reinforcement = (5500 – 3300) = 1700. |
## Precalculus (6th Edition) Blitzer
In order to verify the partial fraction decomposition, rational fractions are to be solved by the least common multiple and their solution is equal to the partial fraction decomposition of this solution. Let us take an example, Consider $\frac{3}{x}+\frac{5}{x+1}$ Normally, we solve this question by taking the least common multiple (L.C.M.). And the common polynomial of the denominator is $x\left( x+1 \right)$. So, \begin{align} & \frac{3}{x}+\frac{5}{x+1}=\frac{3\left( x+1 \right)+5x}{x\left( x+1 \right)} \\ & =\frac{3x+3+5x}{x\left( x+1 \right)} \\ & =\frac{8x+3}{x\left( x+1 \right)} \end{align} Then, operate the partial fraction decomposition of the above polynomial, $\frac{8x+3}{x\left( x+1 \right)}=\frac{A}{x}+\frac{B}{x+1}$ And multiply both sides by $x\left( x+1 \right)$, \begin{align} & 8x+3=A\left( x+1 \right)+Bx \\ & 8x+3=Ax+A+Bx \\ & 8x+3=\left( A+B \right)x+A \end{align} And equate the equation, \begin{align} & A=3 \\ & A+B=8 \end{align} Solving the above equation, $A=3,B=5$ Substitute these values in $\frac{8x+3}{x\left( x+1 \right)}=\frac{A}{x}+\frac{B}{x+1}$, \begin{align} & \frac{8x+3}{x\left( x+1 \right)}=\frac{3}{x}+\frac{5}{x+1} \\ & =\frac{3\left( x+1 \right)+5x}{x\left( x+1 \right)} \\ & =\frac{3x+3+5x}{x\left( x+1 \right)} \\ & =\frac{8x+3}{x\left( x+1 \right)} \end{align} Thus, in both the cases, the solution is the same. So, the result is verified. |
Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :
Application of Integrals
There is a number of methods of calculations among which are functions, differentiation and integration. Application of Integrals is applied in various fields like Mathematics, Science, Engineering etc. For the calculation of areas, we use majorly integrals formulas. So let us give here a brief introduction on integrals based on the Mathematics subject to find areas under simple curves, areas bounded by a curve and a line and area between two curves, and also the application of integrals in the mathematical disciplines along with the solved problem.
Integral Definition
An integral is a function, of which a given function is the derivative. Integration is basically used to find the areas of the two-dimensional region and computing volumes of three-dimensional objects. Therefore, finding the integral of a function with respect to x means finding the area to the X-axis from the curve. The integral is also called as anti-derivative as it is the reverse process of differentiation.
Types of Integrals
There are basically two types of integrals, Definite and Indefinite. Definite Integral is defined as the integral which contains definite limits,i.e., upper limit and lower limit. It is also named as Riemann Integral. It is represented as;
$$\begin{array}{l}\int_{a}^{b}\end{array}$$
f(x)d(x)
Indefinite Integral is defined as the integral whose upper and lower limits are not defined.
$$\begin{array}{l}\int\end{array}$$
f(x)d(x)=F(x)+C, is defined as the indefinite integral, where C is the constant value.
Application of Integrals
There are many applications of integrals, out of which some are mentioned below:
In Maths
• To find the centre of mass(Centroid) of an area having curved sides
• To find the area between two curves
• To find the area under a curve
• The average value of a curve
In Physics
Integrals are used to calculate
• Centre of gravity
• Mass and momentum of inertia of vehicles
• Mass and momentum of satellites
• Mass and momentum of a tower
• The centre of mass
• The velocity of a satellite at the time of placing it in orbit
• The trajectory of a satellite at the time of placing it in orbit
• To calculate Thrust
Video Lesson
Definite Integral Problem
Let us discuss here how the application of integrals can be used to solve certain problems based on scenarios to find the areas of the two-dimensional figure.
Example: Find the area enclosed by the circle x2+y2=r2, where r is the radius of the circle.
Solution: Let us draw a circle in the XY plane with a radius as r.
From the graph,
OA=OB=r
A has coordinates(0,r) on the x-axis and B has coordinates(r,0) on y-axis.
Area of Circle=4*Area of region OBAO
Area of Circle= 4*
$$\begin{array}{l}\int_{0}^{r}\end{array}$$
y.dx
Now, from the equation of circle,
x2+y2=r2
y2=r2-x2
y=r2-x2
The region OABO lies in the first quadrant of the x-y plane.
Therefore, y=
$$\begin{array}{l}\pm \sqrt{r^2-x^2}\end{array}$$
Now we can write, Area of circle=4*
$$\begin{array}{l}\int_{0}^{r}\sqrt{r^2-x^2}\end{array}$$
.dx
From the differentiation formula,
$$\begin{array}{l}\sqrt{r^2-x^2}\end{array}$$
.dx=½
$$\begin{array}{l}\sqrt{r^2-x^2}\end{array}$$
+r^2/2
$$\begin{array}{l}sin^{-1}\end{array}$$
x/4+c
Therefore,
Area of circle=
$$\begin{array}{l}4[x/2 \sqrt{r^2-x^2}+r^2/2 sin^{-1} x/r]_{0}^{r}\end{array}$$
$$\begin{array}{l}=4[r/2[x/2\sqrt{r^2-r^2}+r^2/2 sin^{-1} r/r]-0/2 \sqrt{r^2-0} -0^2/2 sin^{-1}0\end{array}$$
=4[0+r2/2sin−1 (1)−0−0]
=4r2/2 pi/2
=pi r2 |
# Boolean Algebra Basics
Boolean algebra basics contains a set of axioms, postulates, and theorems. It also contains a set of elements and operators. The operators are called the binary operators.
A set of elements is a collection of objects that have similar properties. This definition of a set is debatable, but that is not the scope of this article. The general notion of set is that it contains elements of similar nature.
Example of Set:
S = \{2, 4, 6, 8 \}
The notation mean “element x is in set S”.
The * is a binary operator if for any pair it define the rule to find and also, . It is not a binary operator is and .
## Fundamental Postulates of Boolean Algebra
The fundamental postulates for Boolean Algebra are given below. Thought Boolean algebra was introduced by George Boole in 1845, the postulates were formulated by E.V.Huntington in 1904, that every Boolean algebra must satisfy.
We only use two variable Boolean algebra as far as digital design is concerned. The 6 postulates in Boolean algebra basics are listed below.
1. Closure
2. Associative law
3. Commutative law
4. Identity element
5. Inverse
6. Distributive law
Let us discuss each of them, one at a time.
## Closure
A set S is closed with respect to binary operator * if for every pair of an element in set S, the binary operator specifies a rule to find an element in S.
Example,
N is set of natural numbers,
\begin{aligned}
&N = \{1, 2, 3, 4, ...\}
\end{aligned}
Suppose + is the binary operator, then you take any number a and b in N,
\begin{aligned}
&a + b \in N\\
&3 + 2 = 5\\
&2, 3 \in N \\
&5 \in N.\\
\end{aligned}
but when we do subtraction.
\begin{aligned}
&a - b \notin N\\
&2 - 4 = -2\\
&2, 4 \in N \hspace{3px} and \hspace{3px} -2 \notin N\\
\end{aligned}
The subtraction (-) is not a binary operator because it does not satisfy closure property.
## Associativity Law
A binary operator (*) is associative on a set S whenever following is true.
\begin{aligned}&x*(y*z) = (x * y) * z , \\\\
&\forall x, y, z \in S
\end{aligned}
For example, the multiplication is an associative binary operator on a set of natural numbers N.
\begin{aligned}
&= 2 \times ( 4 \times 1)\\
&= 2 \times 4 = 8\\
\end{aligned}
It is equal to the following.
\begin{aligned}
& = (2 \times 4) \times 1\\
& = 8 \times 1 = 8
\end{aligned}
But addition is also an associative binary operator.
\begin{aligned}
&= 2 + ( 4 + 5)\\
&= 2 + 9 = 114
\end{aligned}
Is equal to the following equation.
\begin{aligned}
&= (2 + 4) + 5\\
& = 6 + 5 = 11
\end{aligned}
## Communicative Law
A binary operator * is communicative on a set S, whenever
\begin{aligned}
&x * y = y * x, \\\\
&\forall x, y \in S
\end{aligned}
The addition operator (+) and multiplication (x) are communicative on a set of natural numbers (N).
For example,
\begin{aligned}
&5 + 4 = 4 + 5 = 9\\\\
&and\\\\
&4 \times 2 = 2 \times 4 = 8
\end{aligned}
## Identity Element
A set S has an identity element with respect to a binary operation * on S, if there exists an element such that
\begin{aligned}
&x * e = e * x = x,\\\\
&for \hspace{3px} any \hspace{3px} x \in S
\end{aligned}
The identity element for a set S is always unique to the set. Consider the following example, where + is the binary operation on a set of integers, Z and 0 is the identity element.
\begin{aligned}
Z = 4 + 0 = 0 + 4 = 4\\
Z = 8 + 0 = 0 + 8 = 8
\end{aligned}
## Inverse
A set S with identity element is said to have an inverse with respect to binary operation *, whenever, for all
\begin{aligned}
&x \in S,\\
&\exists y \in S \mid x * y = e\end{aligned}
Let us take the previous example of addition + on a set of integers Z, with identity element 0.
Let x be an element in Z,
\begin{aligned}
&x \in Z\\
&x = 5\\
\end{aligned}
Then, -x also belong to Z, let us call it y.
\begin{aligned}
&y \in Z\\
&y = -5\\
&x + y = 0
\end{aligned}
Therefore, a set of integers have inverse with respect to binary operation – addition.
## Distributive Law
The distributive law is a very important property of algebraic structures. If * and. are two binary operations on set S, then * is said to distributive over. whenever
\begin{aligned}
&x * ( y . z) = (x * y) . (x * z)\\\\
&\forall x,y,z \in S
\end{aligned}
Also, . is distributive over * whenever
\begin{aligned}
&x \cdot (y \cdot z) = (x \cdot y) \cdot (x \cdot z)\\\\
&\forall x, y, z \in S
\end{aligned}
From the above postulate in Boolean algebra basics, it clear that all binary operators cannot hold all six properties. In fact, the modern algebra studies these algebraic structures formally and trying to find interesting properties. The topics of study are groups, sub-groups, monoid, rings, and fields. |
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Enter a matrix and click the Inverse button.
Matrix 5,3,7 2,4,9 3,6,4
# MULTIPLICATIVE INVERSES OF MATRICES
In this section multiplicative identity elements and multiplicative inverses are introduced and used to solve matrix equations. This leads to another method for solving systems of equations.
IDENTITY MATRICES The identity property for real numbers says that a * I = a and I * a = a for any real number a. If there is to be a multiplicative identity matrix I, such that:
AI = A and IA = A,
for any matrix A, then A and I must be square matrices of the same size. Otherwise it would not be possible to find both products. For example, let A be the 2 X 2 matrix and let
represent the 2 X 2 identity matrix. To find I, use the fact that IA = A, or
Multiplying the two matrices on the left side of this equation and setting the elements of the product matrix equal to the corresponding elements of A gives the following system of equations with variables x11, x12, x21, and x22
Notice that this is really two systems of equations in two variables. Use one of the methods of the previous chapter to find the solution of this system: x11 = 1, x12 = x21 = 0, and X22 = 1. From the solution of the system, the 2 X 2 identity matrix is
Check that with this definition of I, both Al = A and IA = A.
Example 1 VERIFYING THE IDENTITY PROPERTY
Let
Verify that MI = M and IM = M
The 2 X 2 identity matrix found above suggests the following generalization:
n x n IDENTITY MATRIX
For any value of n there is an n X n identity matrix having l's down the diagonal and 0's elsewhere. The n x n identity matrix is given by l where:
Here aij = 1 when i = j (the diagonal elements) and aaj = 0 otherwise.
Example 2 - STATING AND VERIFYING THE 3 X 3 IDENTITY MATRIX
Let K =
Given the 3 X 3 identity matrix I and show that KI = K.
The 3 X 3 identity matrix is
By the definition of matrix multiplication,
MULTIPLICATIVE INVERSES For every nonzero real number a, there is a multiplicative inverse l/a such that
Recall that l/a can also be written a^(-1). In the rest of this section, a method is developed for finding a multiplicative inverse for square matrices. The multiplicative inverse of a matrix A is written A^(-1). This matrix must satisfy the statements
The multiplicative inverse of a matrix can be found using the matrix row transformations given in the previous tutorial and repeated here for convenience.
MATRIX ROW TRANSFORMATIONS
The matrix row transformations are:
1. interchanging any two rows of a matrix;
2. multiplying the elements of any row of a matrix by the same nonzero scalar k; and
3. adding a multiple of the elements of one row to the elements of another row.
As an example, let us find the inverse of
Let the unknown inverse matrix be
By the definition of matrix inverse, AA^(-1) = 1, or
By matrix multiplication,
Setting corresponding elements equal gives the system of equations
Since equations (1) and (3) involve only x and z, while equations (2) and (4) involve only y and w, these four equations lead to two systems of equations,
2x + 4z = 1
x-z=0
and
2y + 4w = 0
y-w=1.
Writing the two systems as augmented matrices gives
Each of these systems can be solved by the Gauss-Jordan method. However, since the elements to the left of the vertical bar are identical, the two systems can be combined into the one augmented matrix,
and solved simultaneously as follows. Exchange the two rows to get a 1 in the upper left comer.
Multiply the first row by -2 and add the results to the second row to get
Now, to get a I in the second-row, second-column position, multiply the second row by 1/6.
Finally, add the second row to the first row to get a 0 in the second column above the 1.
The numbers in the first column to the right of the vertical bar give the values of x and z. The second column gives the values of y and w. That is,
so that
To check, multiply A by A^(-1). The result should be I
Verify that A - 1 A = I, also. Finally,
The process for finding the multiplicative inverse A^(-1) n x n matrix A that has an inverse is summarized below.
FINDING AN INVERSE MATRIX
To obtain A^(-1) n x n matrix A for which A^(-1) exists, follow these steps.
1. Form the augmented matrix [A/I], where I is the n x n identity matrix.
2. Perform row transformations on [A|I] to get a matrix of the form [I|B].
3. Matrix B is A^(-1).
4. Verify by showing that BA = AB = I.
CAUTION Only square matrices have inverses, but not every square matrix has an inverse. If an inverse exists, it is unique. That is, any given square matrix has no more than one inverse. Note that the symbol A^(-1) does not mean 1/A; the symbol A^(-1) is just the notation for the inverse of matrix A. |
# The Collatz Conjecture
This is also known as the ‘half or triple plus one’ problem.
Let $$n$$ be any arbitrary positive integer.
If $$n$$ is even, then divide it by two.
Otherwise, if $$n$$ is odd, then multiply it by three and add one.
We have the piecewise function :
$$f(n) = \begin{cases} n/2, & \text{if n is even}\\ 3n + 1, & \text{if n is odd} \end{cases}$$
Now iterate these steps to form a sequence.
The Collatz conjecture states that the sequence will eventually reach the number one for all $$n$$.
Like many open conjectures in arithmetic (number theory) it is easily understood but very difficult to prove.
The first ten sequences :
$$n = 1 : 1, 4, 2, 1.\\ n = 2 : 2, 1.\\ n = 3 : 3, 10, 5, 16, 8, 4, 2, 1.\\ n = 4 : 4, 2, 1.\\ n = 5 : 5, 16, 8, 4, 2, 1.\\ n = 6 : 6, 3, 10, 5, 16, 8, 4, 2, 1.\\ n = 7 : 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1.\\ n = 8 : 8, 4, 2, 1.\\ n = 9 : 9, 28, 14, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1.\\ n = 10 : 10, 5, 16, 8, 4, 2, 1.\\$$
These sequences are often referred to as the hailstone numbers because they rise and fall before eventually dropping down to one.
The sequence for $$n = 27$$ takes $$111$$ steps rising to $$9,232$$ before falling to $$1$$.
All positive integers up to $$2^{64}$$ have been tested and no counter-example has been found, however, this does not constitute a proof.
We can look at the conjecture extended over the complex plane and see the fractal generated by the iterations. The Collatz function over the real numbers is defined as :
$$f(x) = \bigl(1 + (-1)^x\bigr) \frac{x}{4} + \bigl(1 \, – (-1)^x \bigr) \frac{3x +1}{2}$$
Over the complex plane it is defined as :
$$f(z) = \frac{1}{4} \bigl(2 + 7z \, – (2 + 5z)\cos(\pi z) \bigr)$$
since $$(-1)^x = \cos(nx)$$.
The Collatz $$f(z)$$ fractal :
The large black region is around $$z = 0$$. The spikes to the left and right are $$z = -2, -1, 1$$ and $$1$$.
Nathaniel Johnston introduces a modified Collatz function :
$$g(z) = \frac{z}{4} \bigl(1 + \cos(\pi z) \bigr) + \bigl(\frac{3z + 1}{16} \bigr) \bigl(1 \, – \cos(\pi z) \bigr) \bigl(3 \,- \sqrt2\cos \bigl((2z \, -1)\frac{\pi}{4} \bigr) \bigr)$$
His reason for doing so being that $$g(1) = 1$$. That is, $$1$$ is a fixed point of $$g(z)$$.
The Collatz $$g(z)$$ fractal :
Close-up of the $$g(z)$$ fractal at $$z = 1$$ :
$$g(z)$$ at $$z = 4$$ :
$$g(z)$$ at $$z = 8$$ :
$$g(z)$$ at $$z = 16$$ :
$$g(z)$$ at $$z = -8$$ :
© OldTrout $$2018$$ |
# Special Lines in Triangles
### New York State Common Core Math Geometry, Module 1, Lesson 29
Worksheets for Geometry, Module 1, Lesson 29
Student Outcomes
• Students examine the relationships created by special lines in triangles, namely mid-segments.
Special Lines in Triangles
Classwork
Opening Exercise
Construct the midsegment of the triangle below. A midsegment is a line segment that joins the midpoints of two sides of a triangle or trapezoid. For the moment, we will work with a triangle.
a. Use your compass and straightedge to determine the midpoints of π΄B and π΄C as π and π, respectively.
b. Draw midsegment XY.
Compare β AXY and β ABC; compare β AYX and β ACB. Without using a protractor, what would you guess is the relationship between these two pairs of angles? What are the implications of this relationship?
Discussion
Note that though we chose to determine the midsegment of AB and AC, we could have chosen any two sides to work with. Let us now focus on the properties associated with a midsegment.
The midsegment of a triangle is parallel to the third side of the triangle and half the length of the third side of the triangle.
We can prove these properties to be true. Continue to work with the figure from the Opening Exercise.
Given: XY is a midsegment of β³ ABC.
Prove: XY β₯ BC and XY = 1/2 BC
Construct the following: In the Opening Exercise figure, draw β³ YGC according to the following steps. Extend XY to point πΊ so that YG = XY. Draw GC.
(1) What is the relationship between XY and YG? Explain why.
(2) What is the relationship between β AYX and β GYC? Explain why.
(3) What is the relationship between AY and YC? Explain why.
(4) What is the relationship between β³ AXY and β³ CGY? Explain why.
(5) What is the relationship between GC and AX? Explain why.
(6) Since AX = BX, what other conclusion can be drawn? Explain why.
(7) What is the relationship between πβ AXY and πβ YGC? Explain why
(8) Based on (7), what other conclusion can be drawn about AB and GC? Explain why.
(9) What conclusion can be drawn about BXGC based on (7) and (8)? Explain why.
(10) Based on (9), what is the relationship between XG and BC?
(11) Since YG = XY, XG = __ XY. Explain why.
(12) This means BC = __ XY. Explain why.
(13) Or by division, XY = __ BC.
Note that Steps (9) and (13) demonstrate our Prove statement
Example 1
If a quadrilateral is a parallelogram, then its opposite sides and angles are equal in measure. Complete the diagram, and develop an appropriate Given and Prove for this case. Use triangle congruence criteria to demonstrate why opposite sides and angles of a parallelogram are congruent.
Given: _____
Prove: _____
Construction: Label the quadrilateral π΄π΅πΆπ·, and mark opposite sides as parallel. Draw diagonal π΅π·.
Example 2
If a quadrilateral is a parallelogram, then the diagonals bisect each other. Complete the diagram, and develop an appropriate Given and Prove for this case. Use triangle congruence criteria to demonstrate why diagonals of a parallelogram bisect each other. Remember, now that we have proved opposite sides and angles of a parallelogram to be congruent, we are free to use these facts as needed (i.e., π΄π· = πΆπ΅, π΄π΅ = πΆπ·, β π΄ β
β πΆ, β π΅ β
β π·).
Given: _____
Prove: _____
Construction: Label the quadrilateral π΄π΅πΆπ·. Mark opposite sides as parallel. Draw diagonals π΄πΆ and π΅π·.
Now we have established why the properties of parallelograms that we have assumed to be true are in fact true. By extension, these facts hold for any type of parallelogram, including rectangles, squares, and rhombuses. Let us look at one last fact concerning rectangles. We established that the diagonals of general parallelograms bisect each other. Let us now demonstrate that a rectangle has congruent diagonals.
Exercises 1β4
Apply what you know about the properties of midsegments to solve the following exercises.
1. π₯ =
Perimeter of β³ ABC =
2. π₯ =
π¦ =
3. In β³ RST, the midpoints of each side have been marked by points π, π, and π.
• Mark the halves of each side divided by the midpoint with a congruency mark. Remember to distinguish congruency marks for each side.
• Draw midsegments πY, πZ, and πZ. Mark each midsegment with the appropriate congruency mark from the sides of the triangle.
a. What conclusion can you draw about the four triangles within β³ RST? Explain why.
b. State the appropriate correspondences among the four triangles within β³ RST.
c. State a correspondence between β³ RST and any one of the four small triangles
1. Find π₯.
π₯ =
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. |
Home » Class 10 » Time and Distance Formula- Speed Distance...
Time and Distance Formula, Speed Formula Tricks
Time and Distance Formula
Learning the Time and Distance formula is crucial, As Questions based on Time and Distance formula are very common and frequent for school as well as competitive exams. The Time and Distance Formula is frequently utilized in various question types, including motion in a straight line, motion in a circle, boats, streams, races, clocks, etc.
The fundamental ideas of the Time and Distance Formula, are discussed in this article. A few solved examples have also been provided, which will help students the Time and Distance Formula in a better way.
Time and Distance Formula for Competitive Exams
All the important Time and Distance formulas are given here. Using the Time and Distance formulas you can easily solve the Questions based on Speed Time and Distance.
Speed Time and Distance Formula List
1. Speed = Distance/Time
2. Time =Distance/Speed
3. Distance = (Speed × Time)
4. Average Speed= Total Distance / Total Time
5. 1 km/hr = 5/18 m/sec
6. 1 m/sec = 18/5 km/hr
7. If the ratio of the speeds of A and B is a: b, then the ratio of the times taken by them to cover the same distance is 1/a: 1/b =b: a
8. Suppose a man covers a certain distance at x km/hr and an equal distance at y km/hr. Then, the average speed during the whole journey is (2xy /x+y) km/hr.
9. If two people A and B set out from two points P and Q at the same time and cross paths after spending T1 and T2 hours getting to P and Q, respectively, then (A’s speed) / (B’s speed) equals √(T2 / T1).
Time and Distance Formula- Trains
The time and Distance formula can be implemented on the problems on Trains. Although the basic concept of Time and distance is the same, some changes are there due to train length. Let’s give a look at the Time and Distance formula for Trains.
1. If the Speed of the two trains is S1 and S2 respectively and lengths are L1 and L2, then,
While moving in the opposite direction Relative speed = S1+S2 Time taken = [(L1 + L2)/( S1+S2)] While moving in the same direction Relative speed = S1-S2 Time taken = [(L1 + L2)/ (S1- S2)]
2. When two trains of lengths l1 and l2 cross each other at speeds of S1 and S2, respectively, in time t, the equation is given as S1+S2 = (L1+L2)/t.
3. When a train of length L1 passing another train of length l2 passes another train of length L2 at a speed formula is expressed as S1 = (L1+L2)/t
4. when a train of length l1 traveling at a speed of S1 traverses a platform, bridge, or tunnel of length L2 in time t, the equation is stated as S1-S2 = (L1+L2)/t.
5. If the train passes an electric pole than
Length of the Train= Train’s speed × Time Time = Length of the Train/speed Speed = Length of the Train / Time
Speed Time Distance Formula
The formula that relates speed, distance, and time is as follows:
Speed () = Distance () / Time ()
Where:
• is the speed or velocity of the object.
• is the distance traveled.
• is the time taken to cover that distance.
This formula allows you to calculate the speed of an object if you know the distance it has traveled and the time it took to cover that distance. Similarly, you can rearrange the formula to solve for distance or time if you know the speed and one of the other variables.
Just remember to use consistent units for distance and time in order to get accurate results. If the speed is given in, for example, kilometers per hour (km/h), then the distance and time should also be in compatible units such as kilometers and hours, respectively.
Distance Formula Speed Time
The distance formula relates to distance, speed (or velocity), and time. It’s commonly used to calculate the distance traveled by an object when you know its speed and the time it has been in motion. The formula is:
Distance () = Speed () × Time ()
Where:
• is the distance traveled.
• is the speed or velocity of the object.
• is the time the object has been in motion.
This formula assumes constant speed and no changes in direction during the motion. It’s important to ensure that the units of speed and time are consistent when using this formula. If you want to calculate the distance for a segment of motion where the speed changes, you might need to break the motion into segments and calculate the distance for each segment separately.
Remember that speed can be expressed in various units (e.g., meters per second, kilometers per hour, miles per hour), and time can be in seconds, minutes, hours, etc., depending on the context. Always make sure to use the appropriate units to match the problem you are solving.
Time and Distance Aptitude- Speed Time and Distance Conversions
Here. we have given Some Speed, Time & Distance conversions which are very useful while solving numerical based on the Time and Distance formula.
• 1 kilometer= 1000 meters = 0.6214 mile
• 1 mile= 1.609 kilometer
• 1 hour= 60 minutes= 60*60 seconds= 3600 seconds
• 1 mile = 1760 yards
• 1 yard = 3 feet
• 1 mile = 5280 feet
• 1 km / hour = 5 / 18 m / sec
• 1 m / sec = 18 / 5 km / hour = 3.6 km / hour
• 1 km/hr = 5/8 miles/hour
• 1 yard = 3 feet
• 1 mph = (1 x 1760) / (1 x 3600) = 22/45 yards/sec
• 1 mph = (1 x 5280) / (1 x 3600) = 22/15 ft/sec
Time and Distance Formula Tricks
We have learned how to calculate speed and time using the fundamental time and distance formula. Let’s now know some tricks to solve Time and Distance problems quickly.
1. Speed = Distance/Time
2. Time =Distance/Speed
3. Distance = (Speed × Time)
4. Average Speed= Total Distance / Total Time
5. 1 km/hr = 5/18 m/sec
6. 1 m/sec = 18/5 km/hr
7. If the Speed of the two trains is S1 and S2 respectively and lengths are L1 and L2, then, While moving in the opposite direction 1)Relative speed = S1+S2. Time taken = [(L1 + L2)/( S1+S2)]
8. If the Speed of the two trains is S1 and S2 respectively and lengths are L1 and L2, then, While moving in the same direction 1)Relative speed = S1-S2. Time taken = [(L1 + L2)/( S1-S2)]
9. If the train passes an electric pole or a man, Speed = Length of the Train / Time
10. If two trains of lengths L1 and L2 cross each other at speeds of S1 and S2, respectively, in time t, the equation is given as S1+S2 = (L1+L2)/t.
Time and Distance Formula- Solved Problems
Now, Let’s some problems based on Time and Distance formula which will help us to understand the formulas in a good way.
Q.1. If he runs at a speed of 20Km/hr, how much time does Aditiya take to cover a distance of 400 meters?
Solution: Aditya’s Speed = 20 km/hr = [ 20 × (5/18)]m/sec = 50/9 m/sec
Time taken to cover 400m = 400 ÷ 50/9 = 400× 9/50 = 72 sec = 1. 2 min.
Q.2. Rasid travels at a speed of 20 kmph from point A to point B and returns to point A at a speed of 30 kmph. Find his overall journey’s average speed.
Solution: Assume, Distance between A and B to be “d”
Time is taken to travel point A to B = d/ 20h.
Time is taken to travel from point B to A = d/ 30h.
Total Distance traveled by Rasid = 2d km
Average Speed= Total Distance x Total Time
Or. Average Speed= 2d / [(d/20) + (d/30)]
Or. Average Speed= 2d / [(d/20) + (d/30)]
Or. (2d) / [5d/60] = 24 kmph
Q.3.A train 100m long is running at the speed of 30 km/h. Find the time taken by it to pass a man standing near the railway line.
Solution:Speed of the train [ 30 × (5/18)]m/sec = 25/3 m/s.
Distance covered = length of the train = 100m ( because the train crossed a man )
Time taken = [ 100÷(25/3)]=[100× (3/25)]s= 12 sec.(Answer)
Time and Distance Questions Based on Formula
1. Rafiq cycled 2 km in 12 minutes. reaches a distant station. Find out the speed of the bicycle.
2. A motor vehicle at the same speed travels 217 km in 6 hours and 12 minutes. How much time will it need to go 273 km?
3. A motorcycle rider covers a distance of 100 km in 2 hours and 5 minutes and a bi-cyclist takes 6 hours and 40 minutes to cover that distance. Find the ratio of the speed of the motorcycle and bicycle.
4. A freight train travels 49.5 km in 2 hours 45 minutes at a constant speed. reaches a distant station. How much time will the train take to reach the 58.5 km? far station?
5. Sandeepan goes by bicycle to a friend’s house in 45 minutes. But on the way back 32 km per hour because there is no air in the bicycle wheel. It takes him 3 hours and 45 minutes to walk at speed. find the Speed of the bicycle.
6. Your father goes from home to one place on a motorcycle and returns home after finishing work in an hour. It took him a total of 3 hours and 30 minutes. If the speed of the motorcycle is 40 km per hour. How far was the place from the house?
7. A bus leaves Kolkata at 7:30 AM and reaches Digha at 12 PM without stopping anywhere. If the speed of the bus is 45 km per hour, then what is the distance from Calcutta to Digha?
8. Two friends travel 39 km from the same place at the same time, one by bicycle and the other by rickshaw. The first person arrives after 2 hours and 10 minutes. The second person arrives after another 1 hour and 2 minutes. Determine the speed of the cycle and rickshaw.
Time and Distance Formula Problems- Trains
1. A train 1.75 m long travels at 60 km per hour. How much time will the train take to cross a tree?
2. The speed of a train is 48 km per hour, and the train can cross a platform 120 m long in 15 seconds. Find the length of the train.
3. At what time does a train 85 m long cross a bridge 65 m long at a speed of 85 m?
4. A train 150 m long takes 30 seconds to cross a bridge 250 m long; How much time will the train take to cross a platform 130 m long?
5. A train can cross a telegraph post in 4 seconds and a bridge 264 meters long in 20 seconds. Find the length and speed of the train.
6. A train crosses two bridges 210 m and 122 m long in 25 seconds and 17 seconds respectively. Find the length and speed of the train.
Time Speed Distance Aptitude Tricks
When it comes to solving time, speed, and distance problems in aptitude tests, having some tricks and strategies can help you solve them more efficiently. Here are a few tips to consider:
1. Use Standard Units: Make sure to convert all units to a common standard. For example, if speed is given in kilometers per hour (km/h), convert distances to kilometers and times to hours. This helps in avoiding unit conversion errors.
2. Understand the Formulas: Familiarize yourself with the basic formulas: Speed = Distance / Time and Distance = Speed × Time. Being comfortable with these formulas can help you set up the problem correctly.
3. Work with Ratios: Often, these problems involve ratios between speed, time, and distance. If one of these values is constant, the others will vary inversely. Use this concept to your advantage.
4. Use Fractions: Sometimes, fractions can simplify the calculations. For example, if you need to calculate the time taken to cover a certain distance at a given speed, see if any fractions cancel out to simplify the calculation.
5. Use Visualization: Draw diagrams or visualize the situation. Drawing a simple line to represent the distance and using arrows to represent the speeds can help you see the relationships more clearly.
6. Use Approximations: In some cases, you can use approximations to quickly estimate the answer. For example, if the speed is around 100 km/h and the time is around 2 hours, you can quickly estimate that the distance covered would be roughly 200 km.
7. Inverse Proportions: If the speed and time are inversely proportional (one increases while the other decreases), consider using the product of speed and time as a constant to solve problems.
8. Practice with Sample Problems: Practice solving a variety of time, speed, and distance problems to build your problem-solving skills and improve your speed.
9. Plug and Play: If you’re given multiple values for speed, distance, or time and need to find another value, try plugging the given values into the formula to find the unknown value.
10. Elimination Technique: If you’re given answer choices, you can often eliminate unreasonable options based on your understanding of the problem and the given data.
Remember that practice is key. The more you practice these types of problems, the more comfortable and efficient you’ll become at solving them. As you practice, you’ll develop your own preferred strategies and tricks that work best for you.
Time and Distance Formula Questions PDF
Here we have provided the Time and Distance Questions pdf so that students can download and practice that will helps them to remember the Time and Distance formulas. Click on the link provided below.
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FAQs
What is the formula for time speed and distance?
The formula for time speed and distance are
Speed = Distance/Time
Time =Distance/Speed
Distance = (Speed × Time)
What is the difference between time speed and distance?
Different units can be used to express speed, distance, and time. In general, seconds, minutes, and hours can be used to express time (hr). although the measurement of distance is typically done in metres (m), kilometres (km), centimetres (cm), miles (m), feet (ft), etc. m/s and km/hr are frequently used to express speed.
How do you solve for distance?
You can solve the problems based on Distance by the formula, Distance = (Speed × Time)
What is the basic formula for time?
The general time formula is [Time = Distance Speed] for any task. The SI time unit is seconds (s). |
# Multi-Step Equations Unit 3.08 I can solve multi-step equations that use the distributive property.
## Presentation on theme: " Multi-Step Equations Unit 3.08 I can solve multi-step equations that use the distributive property."— Presentation transcript:
Multi-Step Equations Unit 3.08 I can solve multi-step equations that use the distributive property.
Solving Multi-Step Equations: To solve some multi- step equations, you first need to use the distributive property. Vocabulary Example: 4(x + 2) = 4x + 8
Vocabulary Example: Step 1: ________________ 3(y – 5) = 7y + 9 Step 2: ________________ Step 3: ________________ Step 4: ________________ y = ____________ y = –6 – 3y -6 Subt. 3y from both sides Subt. 9 from both sides 3y – 15 = 7y + 9 44 Distribute 3 –9 –24 = 4y –15 = 4y + 9 Divide by 4 on both sides
Let’s Practice! Solve each multi-step equation. 1. Step 1: ________________ –6 + 7h = –2(h – 6) Step 2: ________________ Step 3: ________________ Step 4: ________________ h = ___________ 2 Add 2h to both sides Add 6 to both sides Distribute -2 Divide by 9 on both sides
Let’s Practice! Solve each multi-step equation. 2. Step 1: ________________ –63 = 8b – 3(b + 1) Step 2: ________________ Step 3: ________________ Step 4: ________________ b = ___________ -12 Combine like terms Add 3 to both sides Distribute -3 Divide by 5 on both sides
Let’s Practice! Solve each multi-step equation. 3. Step 1: ________________ 5m + 24 = 7(m + 2) Step 2: ________________ Step 3: ________________ Step 4: ________________ m = ___________ 5 Subt. 5m from both sides Subt. 14 from both sides Distribute 7 Divide by 2 on both sides
Let’s Practice! Solve each multi-step equation. 4. Step 1: ________________ 7(–p + 3) – 14p = 108 Step 2: ________________ Step 3: ________________ Step 4: ________________ p = _____________ -4.143 Combine like terms Subt. 21 from both sides Distribute 7 Divide by -21 on both sides
Let’s Practice! Solve each multi-step equation. 5. Step 1: ________________ 3(2w + 2) + 15w = 174 Step 2: ________________ Step 3: ________________ Step 4: ________________ w = ___________ 8 Combine like terms Subt. 6 from both sides Distribute 3 Divide by 21 on both sides
Homework Time! 3.08 Distribute This! part 1 WS I can solve multi-step equations that use the distributive property.
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Question Video: Completing a Proof About Triangles Mathematics • First Year of Preparatory School
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In the following triangle π΄π΅πΆ, if πβ πΆ = πβ πΆπ΄π· = 43Β° and πβ π΅ = πβ π΅π΄π·, find πβ π΅π΄πΆ.
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In the following triangle π΄π΅πΆ, if the measure of angle πΆ equals the measure of angle πΆπ΄π· equals 43 degrees and the measure of angle π΅ equals the measure of angle π΅π΄π·, find the measure of angle π΅π΄πΆ.
We can start this question by identifying the two pairs of congruent angle measures. We have that the measure of angle πΆ is equal to the measure of angle πΆπ΄π·, and those are both 43 degrees. We also have that the measure of angle π΅ is equal to the measure of angle π΅π΄π·, although we arenβt given an exact measurement for those. We can then identify that the angle that we wish to calculate is that of the measure of angle π΅π΄πΆ, which occurs at the vertex π΄ in the larger triangle π΄π΅πΆ.
A property that we can apply in this question is that the sum of the measures of the interior angles in a triangle is 180 degrees. So then, if we consider the large triangle π΄π΅πΆ, we can say that the measure of angle π΅π΄πΆ plus the measure of angle π΅ plus the measure of angle πΆ is equal to 180 degrees. From the diagram then, we can observe that the measure of angle π΅π΄πΆ actually consists of an angle of 43 degrees and the measure of angle π΅π΄π·.
Then, we are given in the question that the measure of angle π΅ is equal to the measure of angle π΅π΄π·. Adding in the measure of angle πΆ, which is 43 degrees, we can add the left-hand side, and it will be equal to 180 degrees. We can then simplify this by adding the two 43 degrees, which is 86 degrees. And we know that there will be two lots of the measure of angle π΅π΄π·. Subtracting 86 degrees from both sides, we have that two times the measure of angle π΅π΄π· is equal to 94 degrees. Finally, dividing through by two, we have that the measure of angle π΅π΄π· is 47 degrees. Now that we know the measure of this angle, we can calculate the measure of angle π΅π΄πΆ. It will be equal to 43 degrees plus 47 degrees, which gives us a final answer of 90 degrees.
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1. ## algebra
3x^2-5x-12=0
3x^2-5x-12=0
What have you learned in school so far on this?
There is factoring a trinomial in quadratic form into the product of 2 binomials.
3. i think we have learned both ways but i am unsure how to do it both ways
i think we have learned both ways but i am unsure how to do it both ways
Let's do the quadratic method first:
The first step is to get your equation into quadratic form which is;
$\displaystyle ax^2 + bx + c = 0$
You have already done that. Next, we need to find our inputs to the equation. For the equation you entered of $\displaystyle 3x^2-5x-12=0$, we see that $\displaystyle a = 3, b = -5, c = -12$
Try doing that method. Then go here to check your answer and the math:
Enter a,b,c values and press calculate. Memorize the formula and how to solve for the roots.
When you are ready to learn the factoring method or if you have any question on the quadratic method, let me know.
5. Factoring takes practice and at first it may seem difficult but once you get the method down every problem is essentially the same. What I do is first I write down two sets of parentheses:
(____)(____)
Then I look at the x^2 term and decide what two x terms can I multiply together to get the x^2 term. In this case the coefficient on the x^2 term is 3 so I need to have a 3x times an x since 3x*x = 3x^2. Write those in the first position.
(3x___)(x___)
Now I need to multiply two numbers together to get the x^0 term (the term within an x on it) in your case it is -12. Now there are several pairs of numbers I can multiply together to get -12. They are: -6*2, -2*6, -12*1, -1*12, -3*4, -4*3. Not just any of these numbers will work though. We need to choose the pair that when multiplied by the correct x-terms and added together give us the middle term in our quadratic (in your case the -5x).
Lets try a few:
first try, -4*3.
remember FOIL (first outer inner last when multiplying out two binomials)
(3x - 4)(x + 3) = 3x^2 +9x - 4x -12 = 3x^2 +5x -12.
That is pretty close, the problem was that instead of -5x our middle term is +5x. How can we fix this? Instead of -4, and 3 lets try -3 and 4.
(3x + 4)(x-3) = 3x^2 - 9x + 4x -12 = 3x^2 -5x - 12
It worked!
Now we set 3x+4 = 0 and x-3 = 0 and solve for x in each equation.
_____________________
It can get more complicated because maybe we have 6x^2-2x-1. In this case we could choose 3x and 2x or 6x andx for the first values in our binomials since when we multiply them together both come out to 6x^2.
With practice you will be able to see right away which one is more likely to work and you won't usually have to try every possibility. Mathceleb might have some other tricks to factoring faster. Just remember it is like FOILing in reverse.
6. so the 2 x values would be as followed?
3x+4=0
-4 -4
3x=-4 / both sides by 3
x=-1.33333333
and
x-3=0
+3 +3
x=3
7. Exactly correct.
3x^2-5x-12=0
You could also "complete the square:"
$\displaystyle 3x^2 - 5x - 12 = 0$
Isolate the quadratic and linear terms:
$\displaystyle 3x^2 - 5x = 12$
Factor so that you have a coefficient of 1 on the quadratic term:
$\displaystyle 3 \left ( x^2 - \frac{5}{3}x \right ) = 12$
The expression in the parenthesis is now in the form of $\displaystyle x^2 + 2ax$. If we add $\displaystyle a^2$ to this the term becomes a perfect square. And what we add to one side we must also subtract so we aren't changing anything:
$\displaystyle 3 \left ( x^2 - \frac{5}{3}x + \frac{5^2}{6^2} - \frac{5^2}{6^2} \right ) = 12$
Separate out the first three terms:
$\displaystyle 3 \left ( x^2 - \frac{5}{3}x + \frac{5^2}{6^2} \right ) - 3 \cdot \frac{5^2}{6^2} = 12$
Now solve for x:
$\displaystyle 3 \left ( x - \frac{5}{6} \right ) ^2 - \frac{125}{36} = 12$
$\displaystyle 3 \left ( x - \frac{5}{6} \right ) ^2 = 12 + \frac{125}{36}$
$\displaystyle 3 \left ( x - \frac{5}{6} \right ) ^2 = \frac{169}{12}$
$\displaystyle \left ( x - \frac{5}{6} \right ) ^2 = \frac{169}{36}$
$\displaystyle x - \frac{5}{6} = \pm \sqrt{\frac{169}{36}}$
$\displaystyle x = \frac{5}{6} \pm \frac{13}{6}$
So $\displaystyle x = 3$ or $\displaystyle x = -\frac{4}{3}$
-Dan
9. There is another method that only works if the quadratic can be factored. It is called the "ac" method.
$\displaystyle 3x^2 - 5x - 12 = 0$
Multiply the 3 and the -12 (the "a" and "c" coefficients of the quadratic):
$\displaystyle 3 \cdot (-12) = -36$
Now list all the pairs of factors of -36:
1, -36
2, -18
3, -12
4, -9
6, -6
9, -4
12, -3
18, -2
36, -1
Now, if this quadratic factors there will be a pair of these that adds up to be the "b" coefficient, in this case -5. And in fact, there is: (4) + (-9) = -5.
So rewrite the quadratic using the -5 as 4 + -9:
$\displaystyle 3x^2 - 5x - 12 = 0$
$\displaystyle 3x^2 + (4 - 9)x - 12 = 0$
$\displaystyle 3x^2 + 4x - 9x - 12 = 0$
Now you want to "factor by grouping." Factor the first two terms together, and the last two terms together:
$\displaystyle (3x^2 + 4x) + (-9x - 12) = 0$
$\displaystyle x(3x + 4) - 3(3x + 4) = 0$
Now you have a common factor (3x + 4) on both terms, so you can factor this:
$\displaystyle (x - 3)(3x + 4)$
Now solve normally and get
$\displaystyle x = 3$ and $\displaystyle x = -\frac{4}{3}$
-Dan |
# Visual Calculus
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Calculus is a very important section in mathematics which analyzes the behavior of different functions and with the help of the concepts in calculus; we can evaluate the values of the functions in different cases. Visual Calculus is a method of solving complex functions by using a visual approach and we can get the solutions of the functions using simple techniques. In order to find the area covered under the graph, we can easily use the method of integration and find the area covered from one point to another point.
Example 1: Find the area covered by the curve y = x2 and the X-axis(y = 0) between x = 0 and x = 2.
To find the area covered under the curve, we find the integral of the functions.
∫xn dx= x(n+1)/ (n+1)
∫(x2 – 0)dx = x2+1/(2 + 1)= x3/3
First substitute x = 0 and x= 2 in the above answer.
When x= 0, ∫f(x) dx= 03/3= 0
When x=2, ∫f(x)dx= 23/3= 8/3
8/3 - 0= 8/3units2
Hence the area covered between x= 0 and x= 2 is 8/3units2
Example 2: Find the area covered by the curve y = x3 and the X-axis(y = 0) between x = 0 and x = 2.
To find the area covered under the curve, we find the integral of the functions.
∫xn dx= x(n+1)/ (n+1)
∫(x3 – 0)dx = x3+1/(3 + 1)= x4/4
First substitute x = 0 and x= 2 in the above answer.
When x= 0, ∫f(x) dx= 04/4= 0
When x=2, ∫f(x)dx= 24/4= 4
4 - 0= 4units2
Hence the area covered between x= 0 and x= 2 is 4units2 |
1. ## proof by induction
prove by induction that 4^n-1 is divisible by three for all posotive integers of n
any help would be appriciated thanks.
2. Hi
Initialization of induction is easy.
Let's suppose that for a given n $\displaystyle 4^{n}-1$ is divisible by 3.
You need to demonstrate that $\displaystyle 4^{n+1}-1$ is divisible by 3.
Hint : $\displaystyle 4^{n+1}-1 = 4(4^{n}-1)+3$.
3. prove $\displaystyle 4^n - 1$ is divisible by 3
true for $\displaystyle n = 1$
assume true for $\displaystyle n$, show true for $\displaystyle (n+1)$ ...
since $\displaystyle 4^n - 1$ is divisible by 3, then $\displaystyle 4^n - 1 = 3k$ , $\displaystyle k$ a positive integer.
$\displaystyle 4^n - 1 = 3k$
$\displaystyle 4(4^n - 1) = 4(3k)$
$\displaystyle 4^{n+1} - 4 = 12k$
$\displaystyle 4^{n+1} - 1 - 3 = 12k$
$\displaystyle 4^{n+1} - 1 = 12k + 3$
$\displaystyle 4^{n+1} - 1 = 3(4k + 1)$
since $\displaystyle (4k+1)$ is also a positive integer, $\displaystyle 4^{n+1} - 1$ is a multiple of 3, and therefore, is divisible by 3.
4. Hello, hmmmm!
Here's another way . . .
Prove by induction that $\displaystyle 3|(4^n-1)$ for all positive integers $\displaystyle n.$
Verify $\displaystyle S(1)\!:\;\;4^1-1 \:=\:3$ . . . true
Assume $\displaystyle S(k)\!:\;\;4^k-1 \:=\:3a\,\text{ for some integer }a.$
Add $\displaystyle 3\!\cdot\!4^k$ to both sides: .$\displaystyle {\color{blue}3\!\cdot\!4^k} + 4^k - 1 \;=\;3a + {\color{blue}3\!\cdot\!4^k}$
Factor: .$\displaystyle (3 + 1)4^k - 1 \;=\;3\left(a + 4^k\right)$
. . . . . . . . $\displaystyle 4\!\cdot4^k - 1 \;=\;3\left(a + 4^k\right)$
. . . . . . . . $\displaystyle 4^{k+1}-1 \;=\;\underbrace{3\left(a + r^k\right)}_{\text{a multiple of 3}}$
We have proved $\displaystyle S(k+1)$ . . . The inductive proof is complete. |
# Multiplying Polynomials and FFFT!
Most folks learn to multiply increasingly complex quantities gradually over time, starting with constants in elementary school, and eventually continuing on to polynomials in high school. As the quantities become more complex, students master “the distributive property”, “collecting like terms”, and perhaps even procedures made more memorable with acronyms like “FOIL”.
When learning to multiply binomials and polynomials, people often focus more on the process than the reasoning behind it – which can makes things feel complex. Once you understand the reasoning, multiplying polynomials will hopefully become straightforward. And with a small measure of melodrama, I will describe “FFFT!”, a trivial technique with a silly name that can help make multiplying polynomials easy.
Way back in elementary school, perhaps in first grade, you were probably taught to multiply two integers:
$8\cdot 7\\*~\\*=56$
From there, you were probably asked to learn your multiplication tables. It is extremely useful, particularly when studying algebra, to know your multiplication tables by rote. This “reflex knowledge” will help you work faster, rely on a calculator less, and verify that solutions are correct with greater speed and confidence. No matter how old you may be, no matter whether you are still in school or not, I recommend mastering any gaps your multiplication tables.
Having said that, I confess that I have a few gaps in my multiplication tables (memorizing has never been a talent of mine). Despite the gaps, I am confident that I can figure out a value I need quickly. For example, if I do not recall the answer to 8 x 7, but I do recall 8 x 8, I can subtract one eight from the product I recall to get to the one I need:
$8\cdot 7=\\*~\\*8\cdot (8-1)=\\*~\\*8\cdot 8-8\cdot 1=\\*~\\*64-8=\\*~\\*56$
or, if I recall 8 x 6, add one eight to the result:
$8\cdot 7=\\*~\\*8\cdot (6+1)=\\*~\\*8\cdot 6+8\cdot 1=\\*~\\*48+8=\\*~\\*56$
### The Distributive Property
As you can see above, I rely on the distributive property of multiplication over addition or subtraction to fill in my gaps. Understanding why the distributive property of multiplication works as it does helps make polynomial multiplication easy… so let’s explore the distributive property a bit.
What does the expression (3)(5) represent? Five scaled by a factor of three (using the scaling model of multiplication), or three fives added together (using the repeated addition model of multiplication). Either way, the result is 15.
What happens to the above if we substitute an expression that equals five for the number five in the above: (3)(4+1)? Clearly, we are still being asked to evaluate three times five… but five is now expressed as a sum instead of as a single quantity.
If we use the scaling model of multiplication, the above expression asks us to scale what is inside the parentheses, everything inside the parenthesesby a factor of three:
$3\cdot (4+1)=\\*~\\*3\cdot 4+3\cdot 1=\\*~\\*12+3=\\*~\\*15$
If we use the repeated addition model of multiplication, the expression asks us to duplicate “the parentheses”, along with everything inside of them, three times and add the results:
$3\cdot (4+1)=\\*~\\*(4+1)+(4+1)+(4+1)=\\*~\\*5+5+5=\\*~\\*15$
In addition to verifying the above two approaches produce the same result (15), we can show that the scaling and repeated addition models must be algebraically equivalent (when multiplying by a whole number) by using the commutative and associative properties of addition:
$3\cdot (4+1)=\\*~\\*(4+1)+(4+1)+(4+1)=\\*~\\*4+1+4+1+4+1=\\*~\\*4+4+4+1+1+1=\\*~\\*(4+4+4)+(1+1+1)=\\*~\\*3\cdot 4+3\cdot 1=\\*~\\*3\cdot (4+1)$
From here on I will use the scaling model of multiplication because it is a bit easier on the eyes (as you may have noticed above).
Even if that first factor is in parentheses, the process remains the same:
$(3)\cdot (3+2)=\\*~\\*(3)\cdot 3+(3)\cdot 2=\\*~\\*9+6=\\*~\\*15$
or in a slightly more abstract form to help highlight the pattern:
$(~~)\cdot (a+b)=\\*~\\*(~~)\cdot a+(~~)\cdot b$
And now for the final step… what if the number/blank in first set of parentheses above happens to be a sum or a difference, or in other words, has more than one term? The process of distributing does not change. The entire quantity in the first set of parentheses is multiplied by each of the terms in the second set of parentheses.
$(1+2)\cdot (3+4)=\\*~\\*(1+2)\cdot 3+(1+2)\cdot 4$
Note that the above example will require us to use the distributive property two more times before we can get rid of all parentheses:
$(1+2)\cdot (3+4)=\\*~\\*(1+2)\cdot 3+(1+2)\cdot 4=\\*~\\*1\cdot 3+2\cdot 3+1\cdot 4+2\cdot 4=\\*~\\*~\\*3+6+4+8=\\*~\\*21=\\*~\\*3\cdot 7$
Once we have exhausted our opportunities to use the distributive property above, if we look carefully at the result (please review the line before the gap in the work) we see the following pattern:
Every term in the first set of parentheses has been
multiplied by every term in the second set of parentheses.
### Geometric Interpretation of the Distributive Property
The distributive property of multiplication over addition has a geometric interpretation as well. Suppose I seek to multiply two quantities, each of which is a sum:
$(a+b+c)\cdot (d+e+f)$
This corresponds to stretching a line of length (a+b+c) through a distance (d+e+f) in a direction perpendicular to its length – creating a rectangle as shown below. The total area of this rectangle is equal to the product of the two sums:
Yet the total area of the rectangle can also be calculated by adding the areas of the nine smaller rectangles that piece together to create the large one. Each smaller rectangle has an area that equals its length times its width (as shown in its center). So,
$ad+ae+af\\*~\\*+bd+be+bf\\*~\\*+cd+ce+cf\\*~\\*=(a+b+c)\cdot (d+e+f)$
Multiplying the two polynomials using the approach described above produces the exact same result as calculating the area of the large rectangle by adding together all of its component rectangles.
### Where have you seen this before?
Do you remember multiplying multi-digit numbers in elementary school? For example:
$123\cdot 45=5,535$
Doing this by hand involves multiplying every digit in the first number by every digit in the second one… but in a way that also keeps track of the position of each number (ones, tens, hundreds, etc.). You can also do long multiplication algebraically, if you write each digit in a way that also makes its place value explicit:
$123=1\cdot 10^2+2\cdot 10^1+3\cdot 10^0\\*~\\*45=4\cdot 10^1+5\cdot 10^0$
so the product of these two numbers would be:
$(1\cdot 10^2+2\cdot 10^1+3\cdot 10^0)\cdot (4\cdot 10^1+5\cdot 10^0)$
Which provides a wonderful excuse to introduce a quick, easy, and reliable way of multiplying expressions that each have more than one term.
### FFFT!
FFFT, pronounced “fffffffT”, is an acronym for the “Famous Ford Finger Technique.” At this point, a disclaimer is in order: this technique is not famous, and is not something I invented… but the use of a finger is required.
Why does FFFT! work? Because it provides a physical aid (your finger) to help you ensure that every term in one set of parentheses is indeed multiplied by all terms in the other set of parentheses. If someone next to you sneezes loudly as you are in the middle of the problem, you finger helps you recall where you were in the problem. If one particular product took so much concentration that you are not sure quite where it came from once you finish calculating it, your finger helps you quickly find the first factor that led to it. So, without further obfuscation, here is the procedure:
For right-handed writers, when multiplying two polynomials do the following:
1. Place your left index finger under the left-most term in the left set of parentheses.
2. Here’s the hard part: Don’t move your finger until step 4.
3. Multiply the term above your finger by every term in the next set of parentheses, and write each product down with the appropriate sign.
4. Now that the term above your finger has been multiplied by all terms in the second set of parentheses, make a loud tearing sound as you laboriously move your finger to rest under the next term to the right.
5. Repeat steps 3-5 until you have reached the right side of the left set of parentheses
6. Count the number of products you have written down… it had better be equal to the number of terms in the first set of parentheses times the number of terms in the second set of parentheses.
7. Collect like terms to simplify your answer
For left-handed writers, when multiplying two polynomials do the following:
1. Place your right index finger under the right-most term in the right set of parentheses.
2. Here’s the hard part: Don’t move your finger until step 4.
3. Multiply the term above your finger by every term in the left set of parentheses, and write each product down with the appropriate sign.
4. Now that the term above your finger has been multiplied by all terms in the first set of parentheses, make a loud tearing sound as you laboriously move your finger to rest under the next term to the left.
5. Repeat steps 3-5 until you have reached the left side of the right set of parentheses
6. Count the number of products you have written down… it had better be equal to the number of terms in the first set of parentheses times the number of terms in the second set of parentheses.
7. Collect like terms to simplify your answer
### Practice
Now, practice using FFFT! on the following:
1. Multiply a monomial by a binomial (easy):
$(2)\cdot (3x-5)=$
Your initial answer should have had 1 x 2 = 2 terms, and your simplified answer should be equal to: $6x-10$
2. Multiply two binomials (medium):
$(x+1)\cdot (2y-3)=$
Your initial answer should have had 2 x 2 = 4 terms, and your simplified answer should be equal to : $2y+2xy-3x-3$
3. And, for something a bit more complex to test your complete mastery of FFFT!, multiply two trinomials (ugly):
$(x+y+2)\cdot (2x-y+3)=$
Your initial answer should have had 3 x 3 = 9 terms, and your simplified answer should be equal to: $2x^2+7x+xy+y-y^2+6$
Problems like this last one illustrate why it can be very helpful to organize your work. As the number of terms you have to keep track of grows, it is much easier to verify you have the correct total number of terms, and avoid overlooking a term as you collect like terms, if you:
– work neatly
– organized your work in some way (such as like terms each in their own column)
– left enough white space between terms that they don’t run together
– put a check mark over each term after you have “collected” it with like terms
– etc.
4. And finally, try the multiplication problem described above:
$(1\cdot 10^2+2\cdot 10^1+3\cdot 10^0)\cdot (4\cdot 10^1+5\cdot 10^0)=$
Once you have finished multiplying, add the terms together and verify you get the right answer: 5,535.
At this point, you may wish to make up a few of your own polynomial multiplication problems, then solve them, to give yourself a bit more practice and ensure you have both mastered the process and understand why it works.
### Whit Ford
Math Tutor in Yarmouth, Maine
## 4 thoughts on “Multiplying Polynomials and FFFT!”
1. This is exactly how I teach polynomial multiplication, but I never had a name for it. I, though, have students cover up the terms their not multiplying with yet, so I have them cover up the 5 in (x+5)(x+6) say, and distribute the x, then I have them cover up the x and distribute the 5. I’m glad I’ll have a name for it. If you get a chance, I recently through together an activity I hope helps them understand the geometric interpretation of multiplying binomials like this intuitively that I put up on my blog. I would love any advice on improving it. I’m still not sure it’s the best way to introduce the topic, but I’ve tried giving the distributive property argument you have above and the students got pretty lost with it.
1. I have had much more luck explaining the “distributive property” approach to individual students while tutoring. I agree that it does not always work well when attempting to explain it to an entire class. However, if you were to turn it into an activity for student groups – that might work more smoothly… it could look something like:
In your groups, first show each other how to use the distributive property to expand:
(3x)(2x + 5)
Once you have agreed upon how to complete that multiplication problem and are certain that you have arrived at a correct answer, discuss how the distributive property would be applied to a situation like the following, where the contents of the first set of parentheses are not known yet:
( )(2x + 5)
When you have agreed upon how this situation should be handled, use the process you described to multiply the following:
(3x + 1)(2x + 5)
(Hint: you will need to use the distributive property a total of three times before you will have finished expanding this product.) |
SOLUTION 10: $$\displaystyle{ \lim_{x \to 0} \ { e^{-1/x^2} \over x^2 } } = \displaystyle{ \ e^{-1/0^2} \ " \over (0)^2 } = \displaystyle{ \ e^{-\infty} \ " \over 0 } = \displaystyle{ \ { 1 / e^{\infty} } \ " \over 0 } = \displaystyle{ \ { 1 / \infty } \ " \over 0 } = \displaystyle{ \ 0 \ " \over 0 }$$ (Apply Theorem 1 for l'Hopital's Rule. Differentiate the top using the chain rule and the bottom. Recall that $D \{ e^{f(x)} \} = e^{f(x)} \cdot f'(x)$ .) $$= \displaystyle{ \lim_{x \to 0} \ { e^{-1/x^2} \cdot 2/x^3 \over 2x } }$$ $$= \displaystyle{ \lim_{x \to 0} \ { 2e^{-1/x^2} \over x^3} \cdot { 1 \over 2x } }$$ $$= \displaystyle{ \lim_{x \to 0} \ { e^{-1/x^2} \over x^4 } } = \displaystyle{ \ 0 \ " \over 0 }$$ (Applying Theorem 1 for l'Hopital's Rule again leads to $\displaystyle{ \lim_{x \to 0} \ { e^{-1/x^2} \over 2x^6 } }$. Stop and think. This is getting nowhere. Go back to the beginning and algebraically rewrite the problem by flipping" both numerator and denominator.) $$= \displaystyle{ \lim_{x \to 0} \ { e^{-1/x^2} \over x^2 } }$$ $$= \displaystyle{ \lim_{x \to 0} \ { 1/x^2 \over e^{1/x^2} } } = \displaystyle{ { \ 1/0^2 \ " \over e^{1/0^2} } } = \displaystyle{ { \ \infty \ " \over e^{\infty} } } = \displaystyle{ \ \infty \ " \over \infty }$$ (Apply Theorem 2 for l'Hopital's Rule.) $$= \displaystyle{ \lim_{x \to 0} \ { -2/x^3 \over e^{1/x^2} \cdot -2/x^3 } }$$ $$= \displaystyle{ \lim_{x \to 0} \ { { 1 \over e^{1/x^2} } } }$$ $$= \displaystyle{ \lim_{x \to 0} \ { { 1 \over e^{1/(0)^2} } } }$$ $$= \displaystyle{ \lim_{x \to 0} \ { { 1 \over e^{\infty} } } }$$ $$= \displaystyle{ \ 1 \ " \over \infty }$$ $$= 0$$ |
Fraction Problems
• # Fraction Problems
By Harold Reiter, posted February 13, 2017 —
Failure to understand fractions and lack of fluency with their arithmetic often blocks students’ entry to algebra and higher-level mathematics. Perhaps wrestling with some interesting fraction problems can help. Here are three.
1. First, find a fraction between 1/2 and 2/3. Of course, many correct answers exist. I always ask my students to explain how they solved the problem. The most popular answer among teachers and middle school students is 7/12, the midpoint of the segment [1/2, 2/3], or the average of the original fractions. It is the most popular answer among people who have been taught how to solve problems like this. Among adults over 50 years of age, the most popular answer is 5/8, perhaps because they’ve worked with rulers graded in eighths, or possibly they were stock market investors before the big board went to penny increments for stocks.
My favorite two answers are 5/9 and 3/5. I got the answer 5/9 from a precocious six-year-old. He followed his answer with the explanation that 5/9 is greater than 5/10, which is 1/2, and 5/9 is less than 6/9, which is 2/3.
The wonderful answer 3/5 gives the teacher the opportunity to bring up the Water-Sharing model:
Ashley and Betty have a one-liter bottle of water to share; Carl, Dick, and Ethan have two one-liter bottles among the three of them. The girls each get less than each of the boys (1/2 < 2/3). But the five friends meet and agree to share all the water equally. How much does each person get now?
The new fraction 3/5, called the mediant of 1/2 and 2/3, lies between 1/2 and 2/3. More generally, using the symbol ⊕ to represent the mediant, we can write
for positive integers a, b, c, and d with a/b < c/d. Note also that the mediant of two fractions depends on their representations and not their values. For example,
This oddity is the basis for the interesting Simpson’s paradox in probability and statistics. [Recent MT articles that address the mediant include “Mediants Make (Number) Sense of Fraction Foibles” by Eric L. McDowell, August 2016, p. 18; and “Curiosity and Connections” by Kien H. Lim, November 2014, p. 288.]
2. The second problem is actually a sequence of questions. First, ask your students to find a fraction a/b with the two properties
• a and b are different, one-digit numbers.
• a/b is as large as possible but less than 1.
It won’t take long before students name 8/9 as the fraction. To prove the answer is best, recognize that 1 − a/b is as small as it can be.
Next, ask students to find four distinct a, b, c, and d so that a/b + c/d < 1 and the sum is otherwise as large as possible. Someone will realize that we cannot hope to do better than 71/72, so we should try solving a/8 + c/9 < 1, which succumbs to the Euclidean algorithm or trial and error. What about six different digits a, b, c, d, e, and f such that a/b + c/d + e/f is less than 1 but as large as possible? What about an eight-digit problem? (For eight digits, replace the upper bound 1 with a 2.)
3. Our final fraction problem comes from The Art of Problem Solving’s Beast Academy (https://www.beastacademy.com/), Book 5C. Find the next fraction in the sequence below:
One of my fabulous fourth graders looked at the sequence of differences,
1/3 – 1/5 = 2/15
3/7 – 1/3 = 2/21
1/2 – 3/7 = 2/28
5/9 – 1/2 = 2/36
and recognized that that the denominators are all triangular numbers! The next difference should be 2/45, which makes the next in the sequence 2/45 + 5/9 = 3/5. With this additional value, we can perhaps see that the sequence of values is really just
Harold Reiter has taught mathematics for more than fifty-two years. In recent years, he has enjoyed teaching at summer camps, including Epsilon, MathPath, and MathZoom. His favorite current activity is teaching fourth and fifth graders two days each week. |
# How do you solve -5+ ( - 7) ( - 2) - ( - 3)^{3}?
Sep 13, 2016
$- 5 + \left(- 7\right) \left(- 2\right) - {\left(- 3\right)}^{3} = 36$
#### Explanation:
The order of operations in such a case is given by PEMDAS which lists the operations which should be done first. It stands for Parentheses, Exponents, Multiplication, Division, Addition and Subtraction. Hence we will do operations in this order.
$- 5 + \left(- 7\right) \left(- 2\right) - {\left(- 3\right)}^{3}$ (Here $\left(- 7\right) \left(- 2\right)$ means $\left(- 7\right) \times \left(- 2\right)$
= $- 5 + \left(- 7\right) \times \left(- 2\right) - \left(- 3 \times - 3 \times - 3\right)$
= $- 5 + \left(- 7\right) \times \left(- 2\right) - \left(- 27\right)$
= $- 5 + 14 + 27$
= $- 5 + 41$
= $36$ |
What are the extrema and saddle points of f(x,y) = xy(e^(y^2)-e^(x^2))?
Feb 22, 2017
{: ("Critical Point","Conclusion"), ((0,0,0),"saddle") :}
Explanation:
The theory to identify the extrema of $z = f \left(x , y\right)$ is:
1. Solve simultaneously the critical equations
$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y} = 0 \setminus \setminus \setminus$ (ie ${f}_{x} = {f}_{y} = 0$)
2. Evaluate ${f}_{x x} , {f}_{y y} \mathmr{and} {f}_{x y} \left(= {f}_{y x}\right)$ at each of these critical points. Hence evaluate $\Delta = {f}_{x x} {f}_{y y} - {f}_{x y}^{2}$ at each of these points
3. Determine the nature of the extrema;
$\left.\begin{matrix}\Delta > 0 & \text{There is minimum if " f_(x x)<0 \\ \null & "and a maximum if " f_(yy)>0 \\ Delta<0 & "there is a saddle point" \\ Delta=0 & "Further analysis is necessary}\end{matrix}\right.$
So we have:
$f \left(x , y\right) = x y \left({e}^{{y}^{2}} - {e}^{{x}^{2}}\right)$
$\text{ } = x y {e}^{{y}^{2}} - x y {e}^{{x}^{2}}$
Let us find the first partial derivatives:
$\frac{\partial f}{\partial x} = y {e}^{{y}^{2}} + \left\{\left(- x y\right) \left(2 x {e}^{{x}^{2}}\right) + \left(- y\right) \left({e}^{{x}^{2}}\right)\right\}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = y {e}^{{y}^{2}} - 2 {x}^{2} y {e}^{{x}^{2}} - y {e}^{{x}^{2}}$
$\frac{\partial f}{\partial y} = \left\{\left(x y\right) \left(2 y {e}^{{y}^{2}}\right) + \left(x\right) \left({e}^{{y}^{2}}\right)\right\} - x {e}^{{x}^{2}}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = 2 x {y}^{2} {e}^{{y}^{2}} + x {e}^{{y}^{2}} - x {e}^{{x}^{2}}$
So our critical equations are:
$y {e}^{{y}^{2}} - 2 {x}^{2} y {e}^{{x}^{2}} - y {e}^{{x}^{2}} = 0 \implies y \left({e}^{{y}^{2}} - 2 {x}^{2} {e}^{{x}^{2}} - {e}^{{x}^{2}}\right) = 0$
$2 x {y}^{2} {e}^{{y}^{2}} + x {e}^{{y}^{2}} - x {e}^{{x}^{2}} = 0 \implies x \left(2 {y}^{2} {e}^{{y}^{2}} + {e}^{{y}^{2}} - {e}^{{x}^{2}}\right) = 0$
From the these equations we have:
$y = 0$ or ${e}^{{y}^{2}} - {e}^{{x}^{2}} = 2 {x}^{2} {e}^{{x}^{2}}$
$x = 0$ or ${e}^{{y}^{2}} - {e}^{{x}^{2}} = - 2 {y}^{2} {e}^{{y}^{2}}$
And the only simultaneous solution is $x = y = 0$
And so we have one critical point at the origin
So, now let us look at the second partial derivatives so that we can determine the nature of the critical point (I'll just quote these results):
$\setminus \setminus \setminus \frac{{\partial}^{2} f}{\partial {x}^{2}} = - 4 {x}^{3} y {e}^{{x}^{2}} - 6 x y {e}^{{x}^{2}}$
$\setminus \setminus \setminus \frac{{\partial}^{2} f}{\partial {y}^{2}} = 4 x {y}^{3} {e}^{{y}^{2}} + 6 x y {e}^{{y}^{2}}$
$\frac{{\partial}^{2} f}{\partial x \partial y} = {e}^{{y}^{2}} - {e}^{{x}^{2}} - 2 {x}^{2} {e}^{{x}^{2}} + 2 {y}^{2} {e}^{{y}^{2}} \setminus \setminus \setminus \setminus \left(= \frac{{\partial}^{2} f}{\partial y \partial x}\right)$
And we must calculate:
$\Delta = \frac{{\partial}^{2} f}{\partial {x}^{2}} \frac{{\partial}^{2} f}{\partial {y}^{2}} - {\left(\frac{{\partial}^{2} f}{\partial x \partial y}\right)}^{2}$
at each critical point. The second partial derivative values, $\Delta$, and conclusion are as follows:
{: ("Critical Point",(partial^2f) / (partial x^2),(partial^2f) / (partial y^2),(partial^2f) / (partial x partial y),Delta,"Conclusion"), ((0,0,0),0,0,0,= 0,"incluclusive") :}
So after all that work it is rather disappointing to get an inclusive result, but if we examine the behaviour around the critical point we can readily establish that it is a saddle point.
We can see these critical points if we look at a 3D plot: |
# Ratios and Proportions Outline Ratios What is a
• Slides: 23
Ratios and Proportions
Outline: • Ratios! What is a Ratio? How to Use Ratios? How to Simplify? Proportions! What is a proportion? Properties of proportions? How to use proportions? • Mysterious Problems…
What is a Ratio? • A ratio is a comparison of two numbers. • Ratios can be written in three different ways: a to b a: b Because a ratio is a fraction, b can not be zero Ratios are expressed in simplest form
How to Use Ratios? • The ratio of boys and girls in the class is 12 to 11. This means, for every 12 boys you can find 11 girls to match. How could manybe dogs cats do • There justand 12 boys, 11 I have? We don’t know, all we girls. knowcould is if they’d fight, • There be 24 start boys, a 22 girls. eachcould dog has to fight 2 cats. • There be 120 boys, 110 4 cm girls…a huge class 1 cm • The ratio of length and width of this rectangle is 4 to 1. What is the ratio if the rectangle is 8 cm long and 2 cm wide? Still 4 to 1, because for every 4 cm, you can find 1 cm to match . • The ratio of cats and dogs at my home is 2 to 1
How to simplify ratios? • The ratios we saw on last slide were all simplified. How was it done? Ratios can be expressed The ratio of boys and girls in the class is The ratio of the rectangle is in fraction form… This allows us to do math on them. The ratio of cats and dogs in my house is
How to simplify ratios? • Now I tell you I have 12 cats and 6 dogs. Can you simplify the ratio of cats and dogs to 2 to 1? = = Divide both numerator and denominator by their Greatest Common Factor 6.
How to simplify ratios? A person’s arm is 80 cm, he is 2 m tall. Find the ratio of the length of his arm to his total height To compare them, we need to convert both numbers into the same unit …either cm or m. • Let’s try cm first! Once we have the same units, we can simplify them.
How to simplify ratios? • Let’s try m now! Once we have the same units, they simplify to 1. To make both numbers integers, we multiplied both numerator and denominator by 10
How to simplify ratios? • If the numerator and denominator do not have the same units it may be easier to convert to the smaller unit so we don’t have to work with decimals… 3 cm/12 m = 3 cm/1200 cm = 1/400 2 kg/15 g = 2000 g/15 g = 400/3 5 ft/70 in = (5*12)in / 70 in = 60 in/70 in = 6/7 2 g/8 g = 1/4 Of course, if they are already in the same units, we don’t have to worry about converting. Good deal
More examples… = = =
Now, on to proportions! What is a proportion? A proportion is an equation that equates two ratios The ratio of dogs and cats was 3/2 The ratio of dogs and cats now is 6/4=3/2 So we have a proportion :
Properties of a proportion? Cross Product Property 2 x 6=12 3 x 4 = 2 x 6
Properties of a proportion? • Cross Product Property ad = bc means extremes
Properties of a proportion? Let’s make sense of the Cross Product Property… For any numbers a, b, c, d:
Properties of a proportion? • Reciprocal Property If Then Can you see it? If yes, can you think of why it works?
How about an example? Solve for x: 7(6) = 2 x 42 = 2 x 21 = x Cross Product Property
How about another example? Solve for x: 7 x = 2(12) Cross Product Property 7 x = 24 x= Can you solve it using Reciprocal Property? If yes, would it be easier?
Can you solve this one? Solve for x: 7 x = (x-1)3 Cross Product Property 7 x = 3 x – 3 4 x = -3 x= Again, Reciprocal Property?
Now you know enough about properties, let’s solve the Mysterious problems! If your car gets 30 miles/gallon, how many gallons of gas do you need to commute to school everyday? 5 miles to home 5 miles to school Let x be the number gallons we need for a day: x= Gal Can you solve it from here?
5 miles to home 5 miles to school So you use up 1/3 gallon a day. How many gallons would you use for a week? Let t be the number of gallons we need for a week: What property is this? Gal
So you use up 5/3 gallons a week (which is about 1. 67 gallons). Consider if the price of gas is 3. 69 dollars/gal, how much would it cost for a week? Let s be the sum of cost for a week: 3. 69(1. 67) = 1 s s = 6. 16 dollars 5 miles to home 5 miles to school
So what do you think? 5 miles 10 miles You pay about 6 bucks a week just to get to school! What about weekends? If you travel twice as much on weekends, say drive 10 miles to the Mall and 10 miles back, how many gallons do you need now? How much would it cost totally? How much would it cost for a month? Think proportionally!. . . It’s all about proportions! |
A text-based proof (not video) of the quadratic formula.
$x=\dfrac{-\goldD{b}\pm\sqrt{\goldD{b}^2-4\purpleD{a}\redD{c}}}{2\purpleD{a}}$
$\purpleD{a}x^2 + \goldD{b}x + \redD{c} = 0$
If you've never seen this formula proven before, you might like to watch a video proof, but if you're just reviewing or prefer a text-based proof, here it is:
## The proof
We'll start with the general form of the equation and do a whole bunch of algebra to solve for $x$. At the heart of the proof is the technique called $\blueD{\text{completing the square}}$. If you're unfamiliar with this technique, you may want to brush up by watching a video.
### Part 1: Completing the square
\begin{aligned} \purpleD{a}x^2 + \goldD{b}x + \redD{c} &= 0&(1)\\\\ ax^2+bx&=-c&(2)\\\\ x^2+\dfrac{b}{a}x&=-\dfrac{c}{a}&(3)\\\\ \blueD{x^2+\dfrac{b}{a}x+\dfrac{b^2}{4a^2}}&\blueD{=\dfrac{b^2}{4a^2}-\dfrac{c}{a}}&(4)\\\\ \blueD{\left (x+\dfrac{b}{2a}\right )^2}&\blueD{=\dfrac{b^2}{4a^2}-\dfrac{c}{a}}&(5) \end{aligned}
## Part 2: Algebra! Algebra! Algebra!
Remember, our goal is to solve for $x$.
\begin{aligned} \left (x+\dfrac{b}{2a}\right )^2&=\dfrac{b^2}{4a^2}-\dfrac{c}{a}&(5) \\\\ \left (x+\dfrac{b}{2a}\right )^2&=\dfrac{b^2}{4a^2}-\dfrac{4ac}{4a^2} &(6)\\\\ \left (x+\dfrac{b}{2a}\right )^2&=\dfrac{b^2-4ac}{4a^2}&(7)\\\\ x+\dfrac{b}{2a}&=\pm \dfrac{\sqrt{b^2-4ac}}{\sqrt{4a^2}}&(8)\\\\ x+\dfrac{b}{2a}&=\pm \dfrac{\sqrt{b^2-4ac}}{2a}&(9)\\\\ x&=-\dfrac{b}{2a}\pm \dfrac{\sqrt{b^2-4ac}}{2a}&(10)\\\\ x&=\dfrac{-\goldD{b}\pm\sqrt{\goldD{b}^2-4\purpleD{a}\redD{c}}}{2\purpleD{a}}&(11) \end{aligned}
Ecco fatto! |
# MATH 399N Statistics for Decision Making
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MATH 399N Statistics for Decision Making
Week 6 iLab
Name:___AMINA ISSA____________________
Statistical Concepts:
• Data Simulation
• Confidence Intervals
• Normal Probabilities
All answers should be complete sentences.
We need to find the confidence interval for the SLEEP variable. To do this, we need to find the mean and then find the maximum error. Then we can use a calculator to find the interval, (x – E, x + E).
First, find the mean. Under that column, in cell E37, type =AVERAGE(E2:E36). Under that in cell E38, type =STDEV(E2:E36). Now we can find the maximum error of the confidence interval. To find the maximum error, we use the “confidence” formula. In cell E39, type =CONFIDENCE.NORM(0.05,E38,35). The 0.05 is based on the confidence level of 95%, the E38 is the standard deviation, and 35 is the number in our sample. You then need to calculate the confidence interval by using a calculator to subtract the maximum error from the mean (x-E) and add it to the mean (x+E).
1. Give and interpret the 95% confidence interval for the hours of sleep a student gets. (5 points)
Then, you can go down to cell E40 and type =CONFIDENCE.NORM(0.01,E38,35) to find the maximum error for a 99% confidence interval. Again, you would need to use a calculator to subtract this and add this to the mean to find the actual confidence interval.
1. Give and interpret the 99% confidence interval for the hours of sleep a student gets. (5 points)
1. Compare the 95% and 99% confidence intervals for the hours of sleep a student gets. Explain the difference between these intervals and why this difference occurs. (10 points)
1. Find the mean and standard deviation of the DRIVE variable by using =AVERAGE(A2:A36) and =STDEV(A2:A36). Assuming that this variable is normally distributed, what percentage of data would you predict would be less than 40 miles? This would be based on the calculated probability. Use the formula =NORM.DIST(40, mean, stdev,TRUE). Now determine the percentage of data points in the dataset that fall within this range. To find the actual percentage in the dataset, sort the DRIVE variable and count how many of the data points are less than 40 out of the total 35 data points. That is the actual percentage. How does this compare with your prediction? (15 points)
Mean ______________ Standard deviation ____________________ Predicted percentage ______________________________ Actual percentage _____________________________ Comparison ___________________________________________________ ______________________________________________________________
1. What percentage of data would you predict would be between 40 and 70 and what percentage would you predict would be more than 70 miles? Subtract the probabilities found through =NORM.DIST(70, mean, stdev, TRUE) and =NORM.DIST(40, mean, stdev, TRUE) for the “between” probability. To get the probability of over 70, use the same =NORM.DIST(70, mean, stdev, TRUE) and then subtract the result from 1 to get “more than”. Now determine the percentage of data points in the dataset that fall within this range, using same strategy as above for counting data points in the data set. How do each of these compare with your prediction and why is there a difference? (15 points)
Predicted percentage between 40 and 70 ______________________________ Actual percentage _____________________________________________ Predicted percentage more than 70 miles ________________________________ Actual percentage ___________________________________________ Comparison ____________________________________________________ _______________________________________________________________ Why? __________________________________________________________ ________________________________________________________________
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# GRADE 2 | MODULE 3 | TOPIC E | LESSONS 11–15
## KEY CONCEPT OVERVIEW
During the next week, our math class will learn about place value. Using place value disks, or
number disks, we will learn how to count the total value of ones, tens, and hundreds in a given
number. We will also learn to exchange smaller and larger units of equal value (e.g., exchange
10 ones for 1 ten), read and write numbers up to 1,000, and model numbers in standard form,
expanded form, and unit form.
You can expect to see homework that asks your child to do the following:
▪▪ Use place value disks and drawings to model two- and three-digit numbers.
▪▪ Say numbers in standard form (e.g., 349) and unit form (3 hundreds 4 tens 9 ones).
▪▪ Exchange units of equal value, for example, 30 ones for 3 tens, or 4 hundreds for 40 tens.
▪▪ Use the RDW process to solve word problems involving three-digit numbers.
## For more resources, visit » Eureka.support
GRADE 2 | MODULE 3 | TOPIC E | LESSONS 11–15
## HOW YOU CAN HELP AT HOME
▪▪ Practice counting in unit form. Partner A says a number (e.g., 234), and Partner B repeats it in
unit form (2 hundreds 3 tens 4 ones). Take turns with your child being Partner A and Partner B.
▪▪ Play 10 More/10 Less. Partner A says a number (e.g., 30), and Partner B says the number that is
10 less (20). After every few turns, alternate between 10 more and 10 less. You can also play 100
More/100 Less. Take turns with your child being Partner A and Partner B.
▪▪ Play How Many Tens? Partner A says a number (e.g., 23 ones). Partner B tells how many tens
are in the number (2 tens). You can also play How Many Hundreds? or How Many Hundreds
and How Many Tens? Encourage your child to give each answer in both unit form (e.g., 3
hundreds 2 tens 5 ones) and standard form (325). Again, take turns with your child being
Partner A and Partner B.
TERMS
RDW process: A three-step process used in solving word problems. RDW stands for Read, Draw,
Write: Read the problem for understanding; Draw a picture to help make sense of the problem;
Write an equation and a statement of the answer.
MODELS
Place Value Disks or Number Disks: Circles, or disks, that have a value of 1, 10, or 100. (In later
grades, disks may have a larger or smaller value, such as 1,000 or 0.1.)
## For more resources, visit
» Eureka.support © 2016, GREAT MINDS® |
# Side splitter theorem solver
Here, we will be discussing about Side splitter theorem solver. We will give you answers to homework.
## The Best Side splitter theorem solver
Best of all, Side splitter theorem solver is free to use, so there's no reason not to give it a try! The first step in solving the system is to identify its underlying assumptions. For example, an employee might assume that “people will always work harder if they believe their work is important.” Or another employee might assume that “management is fair and treats everyone equally.” These are just two examples of assumptions that can be made about the system. In order for a system to be successful, all of its underlying assumptions must be true. If one assumption is false, the entire system will fail. So it is critical to start with a clear understanding of each assumption before designing a solution. Once the assumptions have been identified, they must be tested and validated. If the assumptions are not true, then the solution will not solve the problem at hand. In this case, it may be necessary to rework the existing system or even start from scratch.
To find the domain and range of a given function, we can use a graph. For example, consider the function f(x) = 2x + 1. We can plot this function on a coordinate plane: As we can see, the function produces valid y-values for all real numbers x. Therefore, the domain of this function is all real numbers. The range of this function is also all real numbers, since the function produces valid y-values for all real numbers x. To find the domain and range of a given function, we simply need to examine its graph and look for any restrictions on the input (domain) or output (range).
There are a few different methods that can be used to solve multi step equations. The most common method is to use the distributive property to simplify the equation and then solve it by using the order of operations. Another method is to use inverse operations to solve the equation.
Word problems are one of the most common types of math questions that students encounter on standardized tests. Understanding how to solve word problems is essential for success on any test, regardless of the subject matter. Word problems typically involve a combination of math skills, including reading comprehension, problem solving, and reasoning. To solve a word problem, start by breaking the problem down into its individual parts. The first step is to read the question carefully. Be sure to understand what the question is asking, as well as what is being asked for in each part of the problem. Then, break down each part into smaller steps by filling in information from the sentence. Once you have identified all of the pieces of information needed to solve the problem, match them up and organize them into logical order to complete the solution.
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# Proper Fractions
Before getting into the proper fractions, let us know what a fraction is? When is the fraction called the proper fraction? Fraction is defined as the part of the whole thing. Mathematically, a fraction is defined as the ratio of two numbers. The general form of a fraction is a/b such that b ≠ 0 and a, b are whole numbers; a is the numerator, and b is the denominator. Fractions are categorized into various types based on the numerical value of numerator and denominator. Also, a fraction is called a proper fraction if the denominator is greater than the numerator. If the numerator is greater than the denominator, then it is an improper fraction.
The figure above shows the fractions which are also proper fractions since the denominator is greater than the numerator.
Let’s learn in detail about What are Proper fractions? How to identify the proper fraction, how to work with proper fractions in different situations along with examples now.
Let’s have a look at the definition of proper fractions.
## Proper Fractions Definition
When the numerator of a fraction is less than the denominator, it is called a Proper fraction. That means, for proper fractions, denominators will be always greater than the numerator.
Fractions are called Proper Fractions when:
Numerator < Denominator
(Or)
Denominator > Numerator
Go through the meaning of a fraction, denominator and numerator along with the example given below. This will help you in understanding the exact meaning of proper fractions.
• If you want to describe a part of a whole, it is called a Fraction.
• Denominator – The number on the bottom of the fraction and it shows the number of equal parts, where the whole is divided into.
• Numerator – The numerator is the number on the top of the fraction, and it shows the number of the parts we are considering.
• Example, in the fraction 5/6, it shows “5 of 6 equal parts.” 5 is the numerator, and 6 is the denominator.
Example of a Proper Fraction:
$$\frac{3}{5}$$ where 3 is a numerator less than 5, which is the denominator.
Few more Proper Fraction examples are :
$$\frac{1}{9} , \frac{2}{7}, \frac{4}{9}$$.
Note: In all the above examples, the number on the top (numerator) is smaller than the number at the bottom(denominator).
So, basically, it is a way to divide or cut any object in smaller parts. Example, if you divide a bar of chocolate into two equal parts, it would be called two halves.
It can be denoted mathematically as
$$\frac{1}{2} +\frac{1}{2}= 1$$
This expression is called a Fraction. You may divide the chocolate bar into more pieces too.
Also, it is important to note that proper fractions are always less than 1. In other words, we can say that all proper fractions are less than 1. That means, when we convert the given proper fraction into decimal we always get the value which is less than 1.
## Types of fractions:
There are 3 types of fractions, and an overview of them is given below:
Types of Fractions Proper Fractions / Common Fractions Improper Fractions Mixed Fractions Property In this, the numerator is smaller than the denominator. In this, the numerator is larger or equal than the denominator. It’s a combination of a whole number and a proper fraction. Example $$\frac{2}{3}$$ $$\frac{2}{1} or \frac{5}{5}$$ $$5\frac{1}{2}$$
Point to know If the numerator and denominator of a fraction are multiplied by the same number, its value doesn’t change.
## Proper Fractions Examples
Let’s Solve a few problems on Proper fractions:
Question Answer Is $$\frac{1}{2}$$ a proper fraction? Yes, It is a proper fraction because the numerator 1 in this fraction is less than the denominator 2. Is $$\frac{4}{2}$$ a proper fraction? No, It is not a proper fraction because the numerator 4 in this fraction is more than the denominator 2. It is an improper fraction. Is $$\frac{9}{8}$$ a proper fraction? No, It is not a proper fraction because the numerator 9 in this fraction is more than the denominator 8. It is an improper fraction. Is $$\frac{6}{5}$$ a proper fraction? No, It is not a proper fraction because the numerator 6 in this fraction is more than the denominator 5. It is an improper fraction. Is $$\frac{5}{9}$$ a proper fraction? Yes, It is a proper fraction because the numerator 5 in this fraction is less than the denominator 9.
### How to Solve Proper Fractions?
As we know numbers can undergo different arithmetic operations such as addition, subtraction, multiplication and division. Similarly, we can perform these operations on proper fractions too. Let us learn how to solve proper fractions involving arithmetic operations.
### Addition and subtraction of Proper Fractions
In this section, you will learn How to add Proper fractions and how to subtract proper fractions with the help of examples.
Example 1: Add 2/4 and 3/4.
Solution:
(2/4) + (3/4)
Here, the denominators are the same.
(2/4) + (3/4)= (2 + 3)/4
= 5/4
Example 2: Subtract 6/10 from 3/4.
Solution:
(3/4) – (6/10)
Here, the denominators are not equal.
So, let us make the denominators equal.
LCM of 4 and 10 = 20
Thus,
(3/4) = (3/4) × (5/5) = 15/20
(6/10) = (6/10) × (2/2) = 12/20
Therefore, (3/4) – (6/10) = (15/20) – (12/20) = (15 – 12)/20 = 3/20
In the same way, we multiply and divide given proper fractions.
Learn how to simplify fractions here.
Note: The addition of two proper fractions may also give the improper fraction as the result and the subtraction of two proper fractions may result in the negative value too.
## Frequently Asked Questions on Proper Fractions
### What is the proper fraction? Give two examples.
A fraction is called a proper fraction if the value of the numerator is less than the denominator. Some of the examples of proper fractions are: 9/17, 2/7, 1/6, 3/14,…
### What will be the simplified value of any proper fraction?
The simplified value of any proper fraction is always less than 1.
### Is 1/8 a proper fraction?
Yes, 1/8 is a proper fraction since the numerator, i.e. 1 is less than the denominator, i.e. 8.
### Is 4/4 a proper fraction?
No, 4/4 is not a proper fraction since the value of 4/4 is 1 and we know that all proper fractions are less than 1.
### Is 5/3 a proper fraction?
No, 5/3 is not a proper fraction but it is an improper fraction. Here, the numerator, i.e. 5 is not less than the denominator, i.e. 3. |
# ∆ABC ∼ ∆PQR such that ar(∆ABC) = 4 ar(∆PQR). If BC = 12 cm, then QR =
Question:
$\triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$ such that $\operatorname{ar}(\triangle \mathrm{ABC})=4 \operatorname{ar}(\triangle \mathrm{PQR})$. If $\mathrm{BC}=12 \mathrm{~cm}$, then $\mathrm{QR}=$
(a) 9 cm
(b) 10 cm
(c) 6 cm
(d) 8 cm
Solution:
Given: In Δ ABC and ΔPQR
$\triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$
$\operatorname{Ar}(\triangle \mathrm{ABC})=4 \mathrm{Ar}(\triangle \mathrm{PQR})$
$\mathrm{BC}=12 \mathrm{~cm}$
To find: Measure of QR
We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.
$\frac{\operatorname{Ar}(\triangle \mathrm{ABC})}{\operatorname{Ar}(\Delta \mathrm{PQR})}=\frac{\mathrm{BC}^{2}}{\mathrm{QR}^{2}}$
$\frac{4 \operatorname{Ar}(\Delta \mathrm{PQR})}{\operatorname{Ar}(\triangle \mathrm{PQR})}=\frac{12^{2}}{\mathrm{QR}^{2}}(\operatorname{Given} \mathrm{Ar}(\triangle \mathrm{ABC})=4 \mathrm{Ar}(\triangle \mathrm{PQR}))$
$\frac{4}{1}=\frac{12^{2}}{Q R^{2}}$
$\frac{2}{1}=\frac{12}{\mathrm{QR}}$
$\mathrm{QR}=6 \mathrm{~cm}$
Hence the correct answer is (c) |
2 dimensional polygons: General Data Protection Regulation(GDPR) Guidelines BYJU’S
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2-Dimensional Figures: Definition, Area & Example
What is a two-dimensional figure in math? Maybe the term 2D sounds familiar and makes you think of a video game, like Tetris, or a movie. This concept is similar to the understanding of 2D that we have in math. Two-dimensional (2D) figures are shapes formed by closed lines in a plane, and they have two dimensions: length and width. As they do not exist in three-dimensional (3D) space, these shapes do not have depth.
Definition of 2-dimensional figures
Two-dimensional figures are the flat plane shapes or figures that have two dimensions (length and width) in the same plane.
For example, if we drew three lines on a 2D plane surface, like a piece of paper, we could obtain a triangle, which is an example of a 2D shape. We just need one plane to show these 2D figures, as they do not have depth. In math, there are as many 2D shapes as you can imagine, as you just have to link one line with another in a plane.
These lines that form the shapes are called the sides of the plane figure. All sides do not have to be connected, as we can distinguish between closed shapes or open shapes, depending on whether they form vertices or not. We will mainly focus on closed shapes, as they are the most common in math.
Examples of 2-dimensional figures, StudySmarter Originals
Examples of 2-dimensional figures
Now, think about the popular game, Tetris, which is played in 2D. All of the shapes we can see in this game are two-dimensional figures that have lengths and widths. In Tetris, there are numerous two-dimensional shapes, but in math, there are four distinguished two-dimensional figures we work often with:
• Triangle
• Square
• Rectangle
• Circle
Let’s consider each of these four two-dimensional shapes in more detail.
Triangle
As a 2D shape, the triangle consists of three sides and three vertices. The summation of all the internal angles in a triangle is equal to 180º. We can distinguish between different types of triangles depending on whether the sides are equal or not. We can also distinguish types of triangles by the angles they form with one another.
For example, the triangles with all sides of the same length are called equilateral triangles, while if they have just two equal sides, they are called isosceles triangles. If none of the sides are the same length, the triangle is called a scalene triangle. On the other hand, an example of a triangle classified by its internal angles is a right triangle, which has an angle of 90º.
Different triangles, StudySmarter Originals
Square
As 2-dimensional figures, squares are formed by four equal sides with four vertices. All of the internal angles formed by the vertices are equal to 90º. We can label a 2D shape as a square only if all four sides are of the same length.
Square, StudySmarter Originals
Rectangle
These shapes are formed by four sides, with each side equal only to its opposite side. Therefore, both of these two pairs have the same length between them. In a rectangle, all internal angles formed by the vertices are equal to 90º, like in the square. If all the sides’ lengths were equal as well, the 2-dimensional figure would be a square.
Illustration of a rectangle, StudySmarter Originals
Circle
In a 2D plane, the circle consists of points that are all equally distanced with respect to one point in the shape’s center. This means that it has no vertices. In other words, we can also understand a circle as a uniquely curved line that is equally distanced from the center at all of its points.
The distance from the circle’s points to its center is called the radius. Also, if we measure from one point of the circle to another, passing through the circle’s center, the distance is called the diameter. The diameter is always twice the length of the radius.
Circle, StudySmarter Originals
There are more two-dimensional shapes in math that we can classify based on aspects like the number of sides and vertices as well as their structure.
Perimeter of a 2-dimensional figure
In math, the perimeter of a 2D figure is the total sum of the length of all of its sides. Therefore, if the sides of the plane figure are expressed in the length unit of meters, for example, the perimeter of the shape is also expressed with meters. We can express the perimeter with the following formula:
P=a1+a2+a3+…+an=∑ani=1n
In the perimeter formula, the terms a1+a2+a3 (and so on) represent the different sides in the two-dimensional figure. In the second part of the equation is a symbol (∑) which indicates that all these side lengths should be summed up.
Perimeter of a triangle
Let’s take a look at the 2D shapes with the lowest number of sides: the triangles. The triangle has three sides; therefore, the perimeter of the triangle is equal to the sum of those three sides. Let’s take a look at an example of the perimeter calculation below.
Isosceles triangle with side lengths, StudySmarter Originals
In the picture above, we have an isosceles triangle in 2D. This type of triangle has two sides of the same length and a third side with a different length. If we compute the perimeter of this 2D figure, we obtain:
P=a+b+c= 3m+3m+1m=7m
Perimeter of squares and rectangles
Even though a triangle, a square, and a rectangle are not the same, we can still calculate their perimeters with the same formula given above. And if we have any other 2D figure, this process of summing up all sides remains the same as well.
For squares and rectangles, we have to sum up four sides to calculate the perimeter. The perimeter of the square is a+a+a+a, where a is the side length of all four sides. The perimeter of a rectangle is a+a+b+b, where a and b are the two different side lengths of the equal opposite pairs. Let’s see some examples.
Eva has a whiteboard that measures 46 cm by 60 cm. what is the perimeter of this board?
Solution: Two different side lengths are given, and we know that a whiteboard has four sides. So, the figure will be a rectangle. The perimeter of this rectangle =46+46+60+60=212cm
Find the perimeter of the given figure.
2D square figure, StudySmarter Originals
Solution: The perimeter of the above square figure is:
Perimeter=a+a+a+a=25+25+25+25=100cm
Perimeter of a circle
Now you may be wondering, «But what about the circle?» Calculating a circle’s perimeter of course cannot be done with side lengths! We defined the circle as a 2D shape formed by points that are all equally distanced from the center. To calculate the perimeter of a circle in 2D (also called the circumference), we use a different formula:
P= 2πr
In this formula, r is equal to the radius of the circle and π is the number pi, which has a fixed value. From this formula, we see that the perimeter of a circle is proportional to its radius. So, if we increase the radius of a circle, we also increase its perimeter.
The diameter of a circle is given as 14 cm. What is the perimeter or circumference of this circle?
Circle with diameter, StudySmarter Originals
Solution: The circle’s diameter was given as d=14cm.To calculate the perimeter, we need to find the radius. And we know that the diameter is twice the length of the radius.
⇒d=2r⇒r=d2=142=7cm
So, the perimeter of a circle is:
P=2πr=2×π×7=44cm
Hence, the perimeter of the circle is 44 cm.
Area of 2-dimensional figures
In math, the area of a two-dimensional figure is the quantity of surface delimited by the perimeter of a figure in a plane. In other words, the area in 2D is the space inside the lines we use to draw a figure. We use square units to describe area, like square meters (m2) or square feet (ft2).
Now, take a look out from your computer at the floor of the room. Imagine the walls as lines of a shape in 2D. The surface of the floor you are observing is its area because it is the space inside the perimeter (in this case, the room’s walls).
Depending on two-dimensional figure and its shape, we have different formulas to compute area.
Area of a triangle
Starting again with the 2D shape with the lowest number of vertices, the area of the triangle is calculated with the following math formula:
A=12bh
Isosceles triangle with base and height, StudySmarter Originals
The area of the triangle depends on the base b of the triangle and its height h, which is the distance from the middle of the base to the opposite vertex. The base of the triangle does not need to be its shortest side: it can be any side. However, we then need to measure the height from the side chosen as the base to the opposite vertex.
A triangle has a base of 13 inches and a height of 6 inches. What is the area of this triangle?
Solution: Here, base b=13inches and height h=6inches. So the area is:
A=12×b×h=12×13×6=13×3=39
So, the area of the given triangle is 39 inches2.
Area of squares and rectangles
The area measurement for the square and the rectangle are the same, but we will describe the area of the rectangle first, as it is more general with this math formula:
A=bh
In this case, b is one side and h is another side with a different value. This area computation works for any 2D figure with four sides that are parallel to each other, called a parallelogram. Therefore, it also works for the square, but as all the sides have the same length in a square, we can also calculate its area as:
A=b2=b×b
Where b is the length of any side.
We have a tablecloth of size 70 inches by 70 inches. What is its area?
Solution: Here, both sides are the same length, so it is a square with length b=70inches. The area of the square tablecloth is:
A=b2=(70)2=4900
The area of the table cloth is 4900 inches2.
Area of a circle
Lastly, we have the area of the circle. As with the perimeter, its area also depends on the radius. The area of a circle can be calculated with the following equation:
A=πr2
Again, the r corresponds to the radius of the circle, and π is the number pi. From the formula, we see that if we make the radius bigger and bigger, the area of the circle also grows (in this case, by the power of two).
For example, you could see how this relationship works in real life in a garden. Imagine you attach a rope to some point and spin it in circles around that point. This motion would describe the shape of a 2D circle. If you moved the spinning rope further away from its center point, increasing the radius of the circle, you would see that the area of the spinning rope is now bigger.
Find the area of a circle with radius r=5.2cm and round it to the nearest tenth.
Solution: The area of the circle is:
A=πr2=3.14×5.22=3.14×5.2×5.2=84.9056≈84.9cm2
Further representations of 2-dimensional figures
We have previously seen some 2D shapes such as the triangle, the square, the rectangle, and the circle. But an infinite number of figures exist that you could describe. In general, we classify two-dimensional figures by their number of sides and vertices as well as their internal angles (formed by the vertices).
If we increased a rectangle’s sides by one, it would have five sides, making it a pentagon. With six sides, it’d be a hexagon, and so on.
Polygons with different numbers of sides, StudySmarter Originals
There are also different types of four-sided two-dimensional figures. Apart from the rectangle and the square, if a 2D shape has at least two equal sides and its angles are not 90º, it is a rhombus, with a shape similar to a diamond.
Rhombus, StudySmarter Originals
There are a lot of different 2D shapes, with regular sides, irregular sides, equal angles, etc. Now you just have to use some imagination and try to search for examples of them!
2 Dimensional Figures — Key takeaways
• In math, two-dimensional figures consist of figures with two dimensions: length and width. They are also called polygons.
• We can classify two-dimensional figures by the number of sides and vertices, the sides’ lengths, and the internal angles they form.
• Some of the most-used shapes in math are the triangle, the square, the rectangle, and the circle.
• The circle consists of points that are all equally distanced with respect to one point in the shape’s center. This means it has no vertices.
• The perimeter is the sum of all side lengths of the shape. For the circle, it is directly proportional to its radius.
• The area of the figure is the 2D surface delimited by its sides. Depending on the figure, we use different math formulas to compute its area.
• There are shapes with five sides called pentagons, six sides called hexagons, and more. Also, there are more examples of shapes with four sides, such as the rhombus.
2D
In geometry, a dimension can be defined as the minimum number of coordinates necessary to specify a point within a mathematical space. Based on this definition, a two-dimensional object is an object in which a point on the object can be specified using 2 coordinates; in other words, the object has 2 separate dimensions that can be measured, as opposed to a 1D object such as a line, where only one dimension can be measured.
A two-dimensional (2D) object is often described as having length and width, but no depth/thickness. 2D objects are also referred to as 2D shapes, 2D figures, plane figures, and more. The study of 2D objects is also referred to as plane geometry, and as the name suggests, plane geometry deals with objects that lie entirely within one plane.
Types of 2D objects
There are many different types of 2D objects. Two broad categories of 2D objects are polygons and curved shapes.
Polygons
Polygons are a type of closed (all the segments form a closed boundary) shape formed by a finite number of straight line segments. A square, hexagon, and pentagon are a few examples of polygons.
All of the above are formed only using straight line segments: a square has 4, a hexagon has 6, and a pentagon has 5. In general an n-gon is made up of n line segments, or sides. Polygons can be further distinguished into many different categories.
Curved shapes
Curved objects are another type of 2D object. They may or may not be closed objects, and not all of their «sides» need to be curved to be considered a curved object. The circle, ellipse, and other objects below are just a few examples.
Regardless of the type of object (curved, polygon, etc.), to be classified as a 2D object, the object must lie entirely within one plane and can only be measured in 2 different dimensions. When a third dimension exists, we are instead dealing with 3D objects and what is referred to as solid geometry.
Coordinate geometry
Coordinate geometry, also referred to as analytic geometry or Cartesian geometry is the study of geometry in the context of a coordinate system. In two dimensions, a Cartesian coordinate system is referred to as a rectangular (or orthogonal) coordinate system. It is comprised of two perpendicular number lines referred to as axes such that both axes share the same unit of length. The position of a point in the coordinate plane is given by an ordered pair of numbers (x1, y1), where x1 indicates horizontal position and y1 indicates vertical position, as shown in the figure below.
Coordinate geometry enables us to define shapes numerically, which in turn allows us to use other mathematical concepts such as algebra to formulate equations that we can use to study shapes and solve problems. For example, the distance formula can be used to determine the lengths of the sides of a shape if the points at the vertices of the shape are known.
Other dimensions
In geometry, we usually consider 0-dimensional, 1-dimensional, 2-dimensional, and 3-dimensional space.
0D
A point is a 0D object; it has no dimensions, only a position in space.
Of course, a point on a page (like the one above) does have some dimension, since for practical purposes, we have to use something to represent 0D.
1D
A one-dimensional object only has 1 dimension. Usually, this dimension is length. A line, ray, or line segment, are all examples of 1D objects. Only one coordinate (3) is necessary to specify the position of point A on the number line below.
3D
A cube is a 3D object; it has length, width, and height. The dotted lines represent the 3 dimensions of the cube, which shows that 3 coordinates are necessary to specify the position of point A within the cube.
Three-dimensional modeling in the modern world / Sudo Null IT News
Today I will tell you about what 3D modeling is, how it happens, where it is used and what it is eaten with. This article is primarily aimed at those who have only vaguely heard what 3D modeling is, or are just trying their hand at it. Therefore, I will explain the maximum «on the fingers.»
I myself am a technical specialist and have been working with 3D models for more than 10 years, I have worked in more than 10 different programs of different classes and purposes, as well as in various industries. In this regard, a certain helicopter view of this industry has accumulated, with which I would like to share with you.
3D modeling has firmly entered our lives, partially or completely restructuring some types of business. Every industry that 3D modeling has brought change to has its own set of standards and unspoken rules. But even within the same industry, the number of software packages can be so numerous that it can be very difficult for a beginner to figure out and navigate where to start. Therefore, for starters, let’s look at what types of 3D modeling are and where they are used.
There are 3 major industries that today can not be imagined without the use of three-dimensional models. They are:
• Entertainment
• Medicine (surgery)
• Industry
The first one we encounter almost every day. These are films, animation and 90% of computer games. All virtual worlds and characters are created using the same principle — polygonal modeling .
These triangles and quadrangles are called polygons.
The more polygons per model area, the more accurate the model. However, this does not mean that if the model contains few polygons (low poly), then this is a bad model, and the person’s hands are not from there. The same cannot be said about the fact that if the model has Over999999 polygons (High poly), then this is cool. It all depends on the destination. If, for example, we are talking about massive multiplayer, then imagine what it will be like for your computer when you need to process 200 characters around if they are all high poly?
Polygon modeling occurs by manipulating polygons in space. Pulling, rotating, moving, etc.
A pioneer in this industry is Autodesk (known to many for its AutoCAD product, but more on that later).
The Autodesk 3Ds Max and Autodesk Maya products have become the de facto industry standard. And as a 15-year-old teenager, I started my acquaintance with 3D models with 3Ds Max.
What do we get as an output by making such a model? We get visual IMAGE . Gamers sometimes say «I fell through textures» in a game. In fact, you are falling through the polygons that these textures are applied to. And the fall into infinity occurs precisely because there is nothing behind the image. Basically, the resulting images are used for RENDER (final rendering of the image), in the game / in the movie / for the desktop image.
Actually, I once tried to “blind” something to make a cool render (then it was much more difficult).
Speaking of modeling. There is such a direction as 3D-sculpting. In essence, the same polygonal modeling, but aimed at creating mostly complex biological organisms. It uses other polygon manipulation tools. The process itself is more like chasing than 3D modeling.
If a polygonal model is made in the form of a closed volume, such as the same sculptures, then thanks to modern 3D printing technology (which can chew through almost any shape), they can be brought to life.
In fact, this is the only way for polygonal 3D models to get into the real world. From the above, we can conclude that polygonal modeling is needed exclusively for creative people (artists, designers, sculptors). But this is not clear. So, for example, another major area of application of 3D models is medicine , namely, surgery. It is possible to grow a prosthesis of a bone instead of a fragmented one. For example, the lower jaw for a turtle.
I have no medical education and I have never modeled anything for medicine, but given the nature of the forms of the model, I am sure that polygonal modeling is used there. Medicine has now stepped very far, and as the following video shows, you can fix almost everything (if you had money).
Of course, using polygonal modeling, it is possible to build all these restoring and reinforcing elements, but it is impossible to control the necessary gaps, sections, take into account the physical properties of the material and manufacturing technology (especially the shoulder joint). For such products, industrial design methods are used.
Correctly they are called: CAD (Computer-Aided Design) or in English CAD (Computer-Aided Design) . This is a fundamentally different type of modeling. That’s what I’ve been doing for 8 years now. And it is about him that I will tell you in the future. How is this method different from polygonal? The fact that there are no polygons. All forms are integral and are built according to the principle profile + direction.
The base type is solid modeling . From the name you can understand that if we cut the body, inside it not will be empty. Solid modeling is available in any CAD system. It is great for designing frames, gears, engines, buildings, airplanes, cars, and anything that comes out of industrial production. But in it (unlike polygonal modeling) it is impossible to make a model of a package with products from a supermarket, a copy of a neighbor’s dog or crumpled things on a chair.
The purpose of this method is to obtain not only a visual image, but also measurable and operational information about the future product.
CAD is an accurate tool and when working with CAD, you must first imagine the topology of the model in your head. This is the algorithm of actions that forms the shape of the model. Here, just by topology, you can distinguish an experienced specialist from a crooked one. Not always conceived topology and shape complexity can be implemented in a solid state, and then an integral part of industrial design comes to our aid — surface modeling .
Topology in surfaces is 10 times more important than in solid modeling. Incorrect topology — the collapse of the model. (I remind you that this article is a review and for beginners, I do not paint the nuances here). Mastering the topology of surfaces at a high level, closes 70% of issues in industrial modeling. But it takes a lot of practice and constant practice. In the end, the surfaces are still closed into a solid model.
Over time comes the understanding of the most convenient method for modeling a particular product. There are a lot of life hacks here, and each specialist has his own.
IMPORTANT: using CAD without specialized education is not productive! I myself have seen many times how creative people, or jacks of all trades, tried to design. Yes, of course they modeled something, but it was all “a spherical horse in a vacuum”.
When modeling in CAD, in addition to topology, it is necessary to have design skills. Know the properties of materials and production technology. Without this, it’s the same as hammering nails with a pillow, or ironing with a vacuum cleaner.
In CAD we get the electronic-geometric model of product .
(I remind you that with polygonal modeling we get a visual image)
From it you can:
• Make drawings
• It can be used to write a program for CNC machines,
• It can be parameterized (this is when changing 1 parameter you can change the model without alteration)
• Strength and other calculations can be carried out.
• It can also be sent for 3D printing (and the quality will be better)
• Make a render.
I think this is enough for you. We sorted out:
• 2 main types of modeling.
• Dismantled the application areas.
• Analyzed the capabilities of each method and its purpose.
• Discussed the basic types of modeling in CAD and some of the nuances.
Hope you enjoyed!
Ministry of Construction and Housing Policy of the UR
April 10, 2017 Trud newspaper
Chemical weapons destroyed irrevocably
Nikolai Ivanov
9015 6 Russia has once again demonstrated to the world that it remains a high-tech power
Russia is completing the destruction of its stockpiles of chemical weapons (CW). The corresponding federal target program is being implemented in an exemplary manner, overcoming all the difficulties that arose on this difficult and dangerous path, which began twenty years ago (in 19On the 97th, our country ratified the international Convention on the Prohibition of the Development, Production, Stockpiling and Use of Chemical Weapons and on Their Destruction). There were many problems: lack of funding, especially in the 1990s, lack of special technologies, trained personnel, industrial capacities, distrust and protests of the population of the regions where chemical weapons are stored, a crisis in industry and the construction industry…
Everything was successfully overcome. Seven high-tech facilities for the destruction of chemical weapons have been created, unique technologies that have no analogues in the world have been developed, which in practice have confirmed high efficiency, reliability and safety, and professional personnel have been trained. The local population has completely changed its attitude towards the chemical disarmament program, not only because there are no more threatening arsenals of weapons of mass destruction in the regions, but also because many social infrastructure facilities have been built there, high-tech enterprises have appeared that should provide people with work for many years to come.
What the staff of the Federal Directorate for the Safe Storage and Destruction of Chemical Weapons (Federal Directorate) is working on today, we are talking with its creator and permanent head Doctor of Technical Sciences, professor, multiple winner of the Russian Federation Government Prize in Science and Technology, holder of many orders, honorary citizen of the regions in which chemical weapons were stored and destroyed, Colonel General Valery Kapashin.
— Valery Petrovich, how many chemical warfare agents are left for the Federal Administration team to destroy and how much time will it take?
— As of April 2017, about 1,000 tons of chemical warfare agents remain in warehouses, all of them are located at one facility in the village of Kizner of the Udmurt Republic. Of the 40,000 tons of chemical agents available in our country, almost 39,000 tons have been safely destroyed by now. Six out of seven facilities completely destroyed their stocks of chemical weapons. At the Kizner facility, 5,744 tons of OM were stored, after putting it into operation, 4,744 tons were destroyed here in three years. We have been ordered not to reduce the rate of destruction, so that the complete completion of the destruction of chemical weapons in the Russian Federation will occur at the end of the current 2017. Mikhail Babich, Chairman of the State Commission for Chemical Disarmament, recently reported this to Russian President Vladimir Putin. With the commissioning of the last two production buildings at the Kizner facility in May and July, there will be no technical, financial, organizational or other unresolved issues to complete the combat mission assigned to us by the Supreme Commander of the country. And we will fulfill it with honor and, I want to emphasize especially, safely!
— This event will undoubtedly cause a wide resonance: Russia has once again demonstrated to the whole world that it remains a high-tech power. Are you interested in the further fate of the facilities created for the destruction of chemical weapons, because these are modern enterprises equipped with the latest technology and technology?
— They will be re-profiled and used to organize new enterprises on their production base. This was originally stipulated in the federal target program «Destruction of stockpiles of chemical weapons in the Russian Federation» (Program). By the way, investors were quickly found for our facilities, having learned about their high potential. In the city of Pochep, Bryansk region, on the basis of the facility for the destruction of chemical weapons, it is planned to expand the production of medical preparations, in the village. Leonidovka, Penza region — production of building and finishing materials, facility capacity in the village. Maradykovsky Kirov region liked «Rosatom». There are applicants for objects in the city of Shchuchye, Kurgan Region, in the city of Kambarka, Udmurt Republic, pos. Mountain Saratov region. However, beforehand, a set of works must be completed everywhere to eliminate the consequences of the activities of storage facilities and facilities for the destruction of chemical weapons. In other words, the territories where chemical warfare agents were stored, the workshops in which they were destroyed, the technological lines through which they were transported, must be brought to a guaranteed safe state. Something will have to be disassembled and subjected to degassing, something will be processed on the spot, and then garbage and the resulting waste will be collected and stored.
— Speaking of waste, many citizens are concerned about their future fate. First, what is this waste? Where and how will they be stored? Are they dangerous for the local population and the environment? What is their total mass? Is it possible, in principle, to use waste from destroyed chemical weapons somewhere?…
— As the organizer and leader of the chemical disarmament process in the Russian Federation from the first kilograms of chemical weapons, I declare with all responsibility: there is no danger to the population and the environment in the places of storage of waste from the destruction of chemical weapons. There is no danger from the chemical composition of the waste: these substances belong to the III and IV hazard classes, like the same fuels and lubricants, household chemicals, mineral fertilizers or ordinary table salt. There is no danger from the fact that they will be taken away by someone or washed away by precipitation, dispelled by winds. All waste is packaged in special waterproofing containers and stored in closed landfills, to which there is no free access. I want to especially emphasize that it is impossible to use these wastes for the preparation of chemical warfare agents, chemical weapons have been irretrievably destroyed.
— Are these landfills currently under protection and will they be protected in the future?
— Yes. And while we continue to destroy the remnants of chemical weapons at the Kizner facility in the Udmurt Republic, process the reaction masses, while we carry out work to eliminate the consequences of the activities of the facilities, in general, as long as the Federal Directorate exists, we will protect the waste landfills. There is no doubt that the state, together with the regional authorities, will find a solution to this issue even after the mission of the Federal Office is completed.
— How long can waste from the destruction of chemical warfare agents be stored?
— No time limit set. Packed in special containers and sheltered from snow, rain, sunlight and winds, these wastes can be stored for decades.
— Are there technologies for their processing and use?
— The question is not quite the right one, the task of the Federal Office is the safe destruction of chemical weapons, which we are scrupulously and responsibly working on. How the waste will be used is not up to us. However, as a specialist, I see no problems here. These wastes can and should be used. And scientists have already proposed several technologies, including, for example, obtaining ultra-pure arsenic from lewisite waste. Over time, of course, other technologies will be offered. So the waste will not interfere with anyone, there is no threat from them either, they can lie down until their time.
— Valery Petrovich, tell us in more detail what is the waste from the destruction of chemical agents?
— Warfare agents are highly toxic lethal chemical compounds. They were destroyed with the help of special degassing reagents, which required 3-4 times more mass than the destroyed agents. During degassing (passage of the neutralization reaction), a non-toxic substance was formed, the so-called reaction mass (RM). However, the reaction masses could still contain minor residues of OM, so PM was subjected to heat treatment in special furnaces. When exposed to high temperatures, the reaction masses completely decomposed with the formation of off-gases and a solid. Gaseous wastes were passed through special absorbent filters, and in the end only solid wastes remained in the form of various salts. In other words, wastes from the destruction of OM are dry salts.
During the destruction of chemical weapons, waste of III and IV hazard classes (low and moderate hazard) accumulated at the facilities, which, I repeat, will still be generated in the future when performing work to eliminate the consequences of the activity and bring the facilities to a safe state. All waste is subject to disposal at special landfills, the construction of which we completed last year. The technology for storing waste from the destruction of chemical weapons completely excludes their contact with the environment.
— And how much waste from the processing of organic matter did you get?
— This is not difficult to calculate. If for guaranteed and effective destruction of toxic substances, 3-4 times more reagents were required by mass, then out of 40 thousand tons of available agents, 3-4 times more waste was obtained. There is also waste of building materials (concrete, brick, mineral wool, as well as aluminum oxide and some others) that are generated during the dismantling of some building structures and waste of ferrous (cases of destroyed chemical munitions) and non-ferrous metals. The latter, after undergoing thermal decontamination, are not sent to the disposal site, but will be used as recyclable materials at metallurgical enterprises.
— What do waste disposal sites look like at chemical weapons storage and destruction facilities?
— These landfills are specialized, they are built according to exclusive, specially developed projects. A separate and protected area of about 10 hectares is located near the industrial zone of the facility. About a dozen specially equipped covered storage facilities, a two-section control and regulation pond for collecting rain and melt water and several control wells, as well as a waste repacking plant have been built there.
The landfill has a metal fence with a security alarm. There are two gates to enter the site.
Ground structures were erected for waste disposal, consisting of compartments isolated from each other with dimensions of 12 by 36 m each and a height (along the bottom of floor slabs) of 3.2 and 5.0 m. The structures are made of prefabricated concrete and monolithic reinforced concrete structures.
In order to comply with the regulatory requirements to ensure a height of at least 2 m from the groundwater level (at its highest rise) to the lower level of the disposed waste, the structures are built on sand bedding with a layer of 2.0-2.5 m. base, which increases their environmental reliability.
In order to increase tightness and prevent waste from entering the ground in each compartment, the floors have a two-layer waterproofing of hydroisol on hot bituminous mastic over a cold bituminous primer. Waterproofing is also installed on the walls to a height of 300 mm and covered with flat asbestos-cement sheets on hot bituminous mastic, thus ensuring the creation of an impervious sump and eliminating the possibility of contamination of ground and surface water.
— All in accordance with modern environmental requirements.
— Waste is currently packed in steel drums with polyethylene liners. When laid down for long-term storage, some of them will be repacked from barrels into flexible containers. Such wastes include: a mixture of sodium salts, a free-flowing mixture of salts, waste from the thermal destruction of RM and wastewater, melted salts, spent aluminum oxide. Certain types of waste are sent for burial in steel barrels with a volume of 200 liters and 216.5 liters. Such wastes include: ash and slag from solid waste incineration, scale from metal roasting, and some others.
Ventilation pipes are installed in the walls of each storage compartment. After filling the compartment with waste, the end openings are tightly sealed with brickwork, and the ventilation pipes are closed with plugs. To ensure control of the state of the air environment inside the compartment filled with waste and covered by a compartment, a device for introducing control devices is embedded in each branch pipe. Through this device, air samples can be taken from the compartment using a portable sampler.
— During the waste storage process, will air quality be monitored not only near the storage facilities, but also in the compartments themselves?
— No chemical reactions can take place there, any specialist will tell you, however, air condition control is provided. This is another guarantee of security. Continuous monitoring will allow monitoring the state of the environment, confirming compliance with environmental safety requirements.
The capacity of the landfills makes it possible to completely bury all the received waste. Their storage will not lead to violation of hygiene standards and deterioration of the sanitary and epidemiological situation.
— The local population should not have concerns for their safety and the safety of the environment, the issue of waste storage is as well thought out as the entire chemical weapons destruction program. I would like to return to the beginning of our conversation so that you can tell us in more detail about the work to eliminate the consequences of the activities of facilities and prepare them for re-profiling. How long will it take, from what sources will funding come, what types of specific work will be done?
— A new targeted federal program «Liquidation of the consequences of the activities of storage facilities and facilities for the destruction of chemical weapons» has been developed, which has passed almost all approvals and is being approved by the Government of the Russian Federation. So the financing of these works will come from the federal budget. By the way, without waiting for the adoption of the new program, we have already launched preparations for the liquidation of the consequences of activities at all the facilities that have completed the destruction of chemical weapons and have begun some liquidation work that does not require capital investments.
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## Combinatorics - Basic Counting Principle for Linear Arrangements
In the previous post on combinatorics, we learned the basic counting principle.
Using that we can solve many simple questions for example:
Question 1: In how many ways can 3 people sit on 3 adjacent seats of the front row of the theatre?
We know that it is a straight forward basic counting principle application. On the leftmost seat, a person can sit in 3 ways (choose any one of the 3 people). On the middle seat, a person can sit in 2 ways (since 1 person has already been seated). There is only 1 person left for the rightmost seat so he can sit there in 1 way. Total number of arrangements = 3*2*1 = 3! = 6
Now, let’s try to add some constraints here. I will start will some very simple constraints and go on to some fairly advanced constraints.
Question 2: In how many ways can 6 people sit on 6 adjacent seats of the front row of the theatre if two of them, A and B, cannot sit together?
Solution: This is a very simple constraint question. First tell me, what if there was no constraint i.e. what is the total number of arrangements in which 6 people can sit in a row? You should know by now that it is 6! (using Basic Counting Principle). Now, rather than counting the number of ways in which they will not sit next to each other, we can count the number of ways in which they will sit next to each other and subtract that from the total number of arrangements. Why? Because it is easier to group them together (think that we have handcuffed them together) and treat them as a single person to get the arrangements where they will sit together.
So let’s deal with a different question first: In how many ways can 6 people sit on 6 consecutive seats of the front row of the theatre if two of them, A and B, must sit together?
There are 5 individuals/groups: AB C D E F
These 5 can be arranged in a line in 5! ways. But the group AB itself can be arranged in 2 ways AB or BA i.e. B could be to the right of A or to the left of A.
Total number of arrangements = 2*5! (Notice the multiplication sign here. We have to arrange the 5 individuals/groups AND A and B.)
This is the number of arrangements in which A and B will sit together.
Therefore, the number of arrangements in which they will not sit together = 6! – 2*5!
Now let’s discuss some trickier variations of this question.
Question 3: 7 people, A, B, C, D, E, F and H, go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. A and F will not sit next to each other in how many different arrangements?
Solution: Here, there is an additional vacant spot since there are only 7 people but 8 seats. You might think that it is a little confusing since you will need to deal with the vacant spot separately. Actually, this be done in a very straight forward way.
7 people including A and F have to be seated such that A and F are not next to each other. So an arrangement where A and F have the vacant spot between them is acceptable. I will just imagine that there is an invisible person called Mr. V. He takes the vacant spot. If A and F have V between them, that arrangement is acceptable to us. Now this question is exactly like the question above. We have 8 people sitting in 8 distinct seats. 8 people (including our imaginary Mr. V) can sit in 8 seats in 8! arrangements.
A and F can sit together in 2*7! arrangements (similar to question no. 2)
Hence, the number of arrangements in which A and F will not sit together = 8! – 2*7!
Question 4: 7 people, A, B, C, D, E, F and H, go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. In how many different arrangements will there be at least one person between A and F?
Solution: This variant wants you to put at least one person between A and F. This means that all those cases where A and F are together are not acceptable and all those cases where A and F have Mr. V (the vacant spot) between them are also not acceptable.
8 people (including our imaginary Mr. V) can sit in 8 seats in 8! ways.
A and F can sit together in 2*7! arrangements (similar to question no. 2). Number of arrangements in which A and F have Mr. V between them = 2*6!
How? Now we group AVF together and consider this group one person. So there are 6 distinct individuals/groups which can be arranged in 6! ways. But we have 2 arrangements in this group: AVF and FVA. So total number of arrangements here = 2*6!
These cases are not acceptable.
Hence, the number of arrangements in which A and F will have a person between them = 8! – 2*7! – 2*6! = 8! – 16*6!
Compare question no. 3 with question no. 4: one where you don’t want them to be together, the other where you don’t want them to be together and you don’t want the vacant spot between them.
Obviously, in the second case, the number of cases you do not want are higher. So you subtract a higher number out of the total number of cases.
Let me leave you with a question now. Make sure you answer exactly what is asked.
Question 5: 6 people go to a movie and sit next to each other in 6 adjacent seats in the front row of the theatre. If A cannot sit to the right of F, how many different arrangements are possible? |
### Home > CC3MN > Chapter 4 > Lesson 4.1.7 > Problem4-70
4-70.
Josie and Jules are building a model car. They find that the real car is $54$ inches tall and $180$ inches long. They decide to make their model $3$ inches tall, but now they are having a disagreement. Josie thinks that their model should be $10$ inches long and Jules thinks it should be $129$ inches long. Help them settle their argument by deciding if either of them is correct. Explain how you know exactly how long their model should be.
Set up a proportion
$\frac{\text{car height}}{\text{car length}}=\frac{\text{ model height}}{\text{model length}}$
Substitute in the given information.
$\frac{54}{180}=\frac{3}{x}$
Multiply both sides by $x$ to get it in the numerator.
$x\left(\frac{54}{180}\right)=\left(\frac{3}{x}\right)x$
$\frac{54x}{180}=3$
Multiply both sides by $180$ to remove the fractions.
$180\left(\frac{54x}{180}\right)=3(180)$
Solve for $x$.
$54x=540$
Josie is correct.
$x=10$ inches |
### Assignments:
Unfinished Assignment Study Questions for Lesson 24
### Lesson Objectives:
- Find the inverse of a one-to-one function.
- Simplify the composition of a function and its inverse.
Given a relation defined by a set of ordered pairs, the inverse relation is obtained by interchanging the first and second coordinates of each ordered pair.
If the relation is defined by an equation, then the inverse relatino is obtained by interchanging the values.
Find the inverse of the relation.
{(6, 7), (-3,7), (2, -5), (7, -9)}.
We have to interchange the first and second coordinates of ach ordered pair in order to produce the inverse relation.
So the inverse relation is the set of ordered pairs, (7,6), (7, -3), (-5, 2), and (-9, 7).
Find an equation of the inverse relation.
x = y^2-3y. Now we need to interchange the variables x and y in order to produce the inverse relation. So we have y = x^2-3x. Now let's graph both of these equations.
We'll label the first equation 1, and the second equation 2. So this is the graph of our first equation, and this is the graph of our second equation. This is the line, y = x. The graphs of a relation and its inverse are always reflections of each other over the line y = x.
One-to-One Functions
A function is one-to-one if the outputs are the same when the inputs are the same, and the outputs are different, when the inputs are different.
One-to-One Functions and Inverses
The inverse function is called "f inverse", with an f and a -1 in the corner, where -1 is not a power. So "f inverse" is not equivalent to 1/f.
If f is a one-to-one function, then its inverse is also a function. And the domain of f is the range of its inverse. The range of f is the domain of its inverse.
Now prove that f is one-to-one, using the definition of a one-to-one function.
f(x) = (1/4)x-6. We'll need to show that f(a) = f(b), when a = b. So, f(a) = 1/4(a-6), and f(b) = 1/4(b-6). So if we add 6 to both sides, that cancels, and we're left with 1/4a = 1/4b. And then multiply both sides by 4 and we're left with a = b.
So if f(a) = f(b), then a = b. This proves that f is one-to-one.
Prove that g is not one-to-one, using the definition of a one-to-one function.
We're given g(x) = 5-x^2. Now we need to show that when g(a) = g(b), a != b. Because of this x^2 term here, we know that if we let a = -2 and b = +2; those are two terms which don't equal each other, or which g(a) and g(b) will be equal to each other. 5-(-2)^2 is the same as 5-(2)^2. We get 1 = 1.
So since g(a) = g(b), when a != b, then g(x) is not one-to-one.
The Horizontal Line Test.
If a horizontal line can cross the graph of a function more than once, then the function is not one-to-one.
And remember that if a function is not one-to-one, then its inverse is not a function.
Using the horizontal line test, determine whether each function is one-to-one.
So for our first graph, we're given f(x) = (7.2)^x. Now if we draw a horizontal line, it strikes only once. So, yes, this is a one-to-one function.
Now for our second graph, we're given f(x) = 5-x^2. And if we draw a horizontal line here, it strikes the graph more than once. So, no, this is not a one-to-one function.
Since f(x) = (7.2)^x is one-to-one, we know that its inverse is a function.
For the function, determine if it is one-to-one, and if it is, find a formula for the inverse.
f(x) = 3x-1. This is of the form, y = mx+b. It's a straight line, so we know that the y-intercept occurs at (0, b) or (0, -1). And the slope, m, is equal to 3/1.
So now if we draw the graph, we have our y-intercept at (0, -1), and our slope is 3 units up, 1 unit to the right. So now if we use our horizontal line test, we can see that a horizontal line, drawn anywhere on this graph, can only strike the graph once. So it is a one-to-one function.
Now let's find the formula for the inverse function. We'll start by replace f(x) with y. So we have y = 3x-1. And then interchange the variables, so we have x = 3y-1, and solve for y. So we have x+1 = 3y, when we add 1 to both sides. And then when we divide both sides by 3, we get (x+1)/3 = y. So we have f^(-1)(x) = (x+1)/3.
Given the graph of the one-to-one function, f, sketch the graph of f inverse.
So f contains the points: (-2, -7), (-1, 0), (0, 1), (1, 2), and (2, 9). So f inverse will contain the points: (-7, -2), (0, -1), (1, 0), (2, 1), and (9, 2).
This is the graph of f inverse. It's the reflection of the graph of f, across the line y = x.
Inverse Function and Composition
If f is a one-to-one function, then the composition of f inverse and f is equal to x, for every x in the domain of f. And the composition of f and f inverse is equal to x, for every x in the domain of f inverse.
For the function, f, use composition of functions to show that f inverse is as follows:
If f(x) = (2/3)x+1 then f^(-1)(x) = (3x-3)/2. So first we need to find the composition of f^(-1) and f, or input f into f^(-1). So we have (3(2/3x+1)-3)/2. We can simplify this as (2x+3-3)/2, and 3-3 cancels, and we're left with 2x/2. And then the 2's cancel, so all of this is equal to x.
And now we need to find the composition of f and f^(-1). This is the same as inputting f^(-1) into f. So we have f((3x-3)/2), which is equal to 2/3((3x-3)/2)+1. 2's cancel, the 3's cancel, and we're left with x-1+1, which equals x.
So since the composition of f^(-1) and f, and the composition of f and f^(-1) both equal x, that means that f^(-1) is as given. |
A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is opened at the top. Ignoring the thickness of the plastic sheet, determine: (i) The area of the sheet required for making the box.(ii) The cost of sheet for it, if a sheet measuring 1 m2 costsर 20. from Mathematics Surface Areas and Volumes Class 9 CBSE
The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of ` 10 per m2 is ` 15000, find the height of the hall. [Hint : Area of the four walls = Lateral surface area.]
Let the length, breadth and height of the rectangular hall be l m, b m and h m respectively.
Perimeter = 250 m
⇒ 2(l + b) = 250
⇒ l + b = 125 ...(1)
Area of the four walls
Hence, the height of the hall is 6 m.
1042 Views
A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cm wide and 8 cm high.
(i) Which box has the greater lateral surface area and by how much?
Each edge of the cubical box (a) = 10 cm
∴ Lateral surface area of the cubical box = 4a2 = 4(10)2 = 400 cm2.
For cuboidal box
l = 12.5 cm b = 10 cm h = 8 cm
Lateral surface area of the cuboidal box = 2(l + b) h
= 2(12.5+ 10)(8) = 360 cm2.
∴ Cubical box has the greater lateral surface area than the cuboidal box by (400 – 360) cm2, i.e., 40 cm2.
873 Views
A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is opened at the top. Ignoring the thickness of the plastic sheet, determine:(i) The area of the sheet required for making the box.(ii) The cost of sheet for it, if a sheet measuring 1 m2 costsर 20.
(i) l = 1.5 m
b= 1.25 m
h = 65 cm = 0.65 m.
∴ The area of the sheet required for making the box = lb + 2(bh + hl)
= (1.5)(1.25) + 2 {(1.25)(0.65) + (0.65)(1.5)} = 1.875 + 2{0.8125 + 0.975}
= 1.875 + 2(1.7875) = 1.875 + 3.575 = 5.45 m2.
(ii) The cost of sheet for it = र 5.45 x 20 = र 109.
2110 Views
The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of र 7.50 per m2.
l = 5 m
b = 4 m
h = 3 m
Area of the walls of the room = 2(l + b) h
= 2(5 + 4) 3 = 54 m2
Area of the ceiling = lb
= (5) (4) = 20 m2
∴ Total area of the walls of the room and the ceiling = 54 m2 + 20 m2 = 74 m2
∴ Cost of white washing the walls of the room and the ceiling = 74 x 7.50 = र 555.
1125 Views
The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of dimensions 22.5 cm × 10 cm × 7.5 cm can be painted out of this container?
For a brick
l = 22.5 cm b = 10 cm h = 7.5 cm
∴ Total surface area of a brick = 2 (lb + bh + hl)
= 2(22.5 x 10 + 10 x 7.5 + 7.5 x 22.5)
= 2(225 + 75 + 168.75)
= 2(468.75) = 937.5 cm2 = .09375 m2
∴ Number of bricks that can be painted out
1324 Views |
# Search by Topic
#### Resources tagged with Odd and even numbers similar to Even and Odd:
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### There are 61 results
Broad Topics > Numbers and the Number System > Odd and even numbers
### Even and Odd
##### Stage: 1 Challenge Level:
This activity is best done with a whole class or in a large group. Can you match the cards? What happens when you add pairs of the numbers together?
### The Set of Numbers
##### Stage: 1 Challenge Level:
Can you place the numbers from 1 to 10 in the grid?
### Multiplication Series: Number Arrays
##### Stage: 1 and 2
This article for teachers describes how number arrays can be a useful reprentation for many number concepts.
### Ring a Ring of Numbers
##### Stage: 1 Challenge Level:
Choose four of the numbers from 1 to 9 to put in the squares so that the differences between joined squares are odd.
### Becky's Number Plumber
##### Stage: 2 Challenge Level:
Becky created a number plumber which multiplies by 5 and subtracts 4. What do you notice about the numbers that it produces? Can you explain your findings?
### Odd Times Even
##### Stage: 1 Challenge Level:
This problem looks at how one example of your choice can show something about the general structure of multiplication.
### Numbers as Shapes
##### Stage: 1 Challenge Level:
Use cubes to continue making the numbers from 7 to 20. Are they sticks, rectangles or squares?
### Two Numbers Under the Microscope
##### Stage: 1 Challenge Level:
This investigates one particular property of number by looking closely at an example of adding two odd numbers together.
### Number Round Up
##### Stage: 1 Challenge Level:
Arrange the numbers 1 to 6 in each set of circles below. The sum of each side of the triangle should equal the number in its centre.
### How Odd
##### Stage: 1 Challenge Level:
This problem challenges you to find out how many odd numbers there are between pairs of numbers. Can you find a pair of numbers that has four odds between them?
### Sets of Numbers
##### Stage: 2 Challenge Level:
How many different sets of numbers with at least four members can you find in the numbers in this box?
### Crossings
##### Stage: 2 Challenge Level:
In this problem we are looking at sets of parallel sticks that cross each other. What is the least number of crossings you can make? And the greatest?
### Domino Pick
##### Stage: 1 Challenge Level:
Are these domino games fair? Can you explain why or why not?
### Number Tracks
##### Stage: 2 Challenge Level:
Ben’s class were cutting up number tracks. First they cut them into twos and added up the numbers on each piece. What patterns could they see?
### What Number?
##### Stage: 1 Short Challenge Level:
I am less than 25. My ones digit is twice my tens digit. My digits add up to an even number.
### Largest Even
##### Stage: 1 Challenge Level:
How would you create the largest possible two-digit even number from the digit I've given you and one of your choice?
### Grouping Goodies
##### Stage: 1 Challenge Level:
Pat counts her sweets in different groups and both times she has some left over. How many sweets could she have had?
### Lots of Biscuits!
##### Stage: 1 Challenge Level:
Help share out the biscuits the children have made.
### Three Spinners
##### Stage: 2 Challenge Level:
These red, yellow and blue spinners were each spun 45 times in total. Can you work out which numbers are on each spinner?
### More Numbers in the Ring
##### Stage: 1 Challenge Level:
If there are 3 squares in the ring, can you place three different numbers in them so that their differences are odd? Try with different numbers of squares around the ring. What do you notice?
### What Do You Need?
##### Stage: 2 Challenge Level:
Four of these clues are needed to find the chosen number on this grid and four are true but do nothing to help in finding the number. Can you sort out the clues and find the number?
### Curious Number
##### Stage: 2 Challenge Level:
Can you order the digits from 1-3 to make a number which is divisible by 3 so when the last digit is removed it becomes a 2-figure number divisible by 2, and so on?
### The Thousands Game
##### Stage: 2 Challenge Level:
Each child in Class 3 took four numbers out of the bag. Who had made the highest even number?
### A Mixed-up Clock
##### Stage: 2 Challenge Level:
There is a clock-face where the numbers have become all mixed up. Can you find out where all the numbers have got to from these ten statements?
### Square Subtraction
##### Stage: 2 Challenge Level:
Look at what happens when you take a number, square it and subtract your answer. What kind of number do you get? Can you prove it?
### Odds and Threes
##### Stage: 2 Challenge Level:
A game for 2 people using a pack of cards Turn over 2 cards and try to make an odd number or a multiple of 3.
### Down to Nothing
##### Stage: 2 Challenge Level:
A game for 2 or more people. Starting with 100, subratct a number from 1 to 9 from the total. You score for making an odd number, a number ending in 0 or a multiple of 6.
### Always, Sometimes or Never?
##### Stage: 1 and 2 Challenge Level:
Are these statements relating to odd and even numbers always true, sometimes true or never true?
### Carroll Diagrams
##### Stage: 1 Challenge Level:
Use the interactivities to fill in these Carroll diagrams. How do you know where to place the numbers?
### One of Thirty-six
##### Stage: 1 Challenge Level:
Can you find the chosen number from the grid using the clues?
### Sets of Four Numbers
##### Stage: 2 Challenge Level:
There are ten children in Becky's group. Can you find a set of numbers for each of them? Are there any other sets?
### Magic Vs
##### Stage: 2 Challenge Level:
Can you put the numbers 1-5 in the V shape so that both 'arms' have the same total?
### Number Detective
##### Stage: 2 Challenge Level:
Follow the clues to find the mystery number.
### Pairs of Legs
##### Stage: 1 Challenge Level:
How many legs do each of these creatures have? How many pairs is that?
### Take Three Numbers
##### Stage: 2 Challenge Level:
What happens when you add three numbers together? Will your answer be odd or even? How do you know?
### Number Differences
##### Stage: 2 Challenge Level:
Place the numbers from 1 to 9 in the squares below so that the difference between joined squares is odd. How many different ways can you do this?
### Break it Up!
##### Stage: 1 and 2 Challenge Level:
In how many different ways can you break up a stick of 7 interlocking cubes? Now try with a stick of 8 cubes and a stick of 6 cubes.
### Sorting Numbers
##### Stage: 1 Challenge Level:
Use the interactivity to sort these numbers into sets. Can you give each set a name?
### Seven Flipped
##### Stage: 2 Challenge Level:
Investigate the smallest number of moves it takes to turn these mats upside-down if you can only turn exactly three at a time.
### Part the Piles
##### Stage: 2 Challenge Level:
Try to stop your opponent from being able to split the piles of counters into unequal numbers. Can you find a strategy?
### More Carroll Diagrams
##### Stage: 2 Challenge Level:
How have the numbers been placed in this Carroll diagram? Which labels would you put on each row and column?
### Odd Tic Tac
##### Stage: 1 Challenge Level:
An odd version of tic tac toe
### Domino Sorting
##### Stage: 1 Challenge Level:
Try grouping the dominoes in the ways described. Are there any left over each time? Can you explain why?
### Play to 37
##### Stage: 2 Challenge Level:
In this game for two players, the idea is to take it in turns to choose 1, 3, 5 or 7. The winner is the first to make the total 37.
### What Could it Be?
##### Stage: 1 Challenge Level:
In this calculation, the box represents a missing digit. What could the digit be? What would the solution be in each case?
### Red Even
##### Stage: 2 Challenge Level:
You have 4 red and 5 blue counters. How many ways can they be placed on a 3 by 3 grid so that all the rows columns and diagonals have an even number of red counters?
### Always, Sometimes or Never? Number
##### Stage: 2 Challenge Level:
Are these statements always true, sometimes true or never true?
### Odd Squares
##### Stage: 2 Challenge Level:
Think of a number, square it and subtract your starting number. Is the number you’re left with odd or even? How do the images help to explain this?
### Diagonal Trace
##### Stage: 2 Challenge Level:
You can trace over all of the diagonals of a pentagon without lifting your pencil and without going over any more than once. Can the same thing be done with a hexagon or with a heptagon?
### I Like ...
##### Stage: 1 Challenge Level:
Mr Gilderdale is playing a game with his class. What rule might he have chosen? How would you test your idea? |
# Elimination method
System of linear equations can also be solved using the elimination method. We will show how to solve them with some carefully chosen examples. Before you learn this lesson, make sure you understand how to solve linear equations.
## Here are the steps to follow when solving with the elimination method.
Step 1
Try to eliminate a variable as you add the left sides and the right sides of the two equations.
Step 2
Set the sum resulting from adding the left sides equal to the sum resulting from adding the right sides.
Step 3
Solve for the variable that was not cancelled or eliminated.
Step 4
Use the answer found in step 3 to solve for the other variable by substituting this value in one of the two equations.
First, read the example in the figure carefully and then keep reading to see more detailed explanations on how to solve using the elimination method
## Examples showing how to solve a system of linear equations by elimination using the 4 steps outlined above.
Example #1: Solve the following system using the elimination method.
x + y = 20
x − y = 10
Step 1
Examine the two equations carefully. After a close examination, notice that you can eliminate or cancel the variable y by adding the left sides (x + y and x − y).
However, since you are adding the left sides, you have to add the right sides (20 and 10) of the two equations also to keep the equation balanced.
For the system above, it turns out that it is easy to eliminate y while adding the left sides since x + y + x − y = x + x + y − y = x + x + 0 = 2x.
Step 2
The sum for the left sides is 2x and the sum for the right sides is 20 + 10 = 30
Setting them equal, we get 2x = 30
Step 3
Solve 2x = 30 for x by dividing both sides of this equation by 2.
2x = 30
(2/2)x = 30/2
x = 15
Step 4
You can substitute 15 for x in either x + y = 20 or x − y = 10 to get y.
Choosing the first one, we get 15 + y = 20
Minus 15 from both sides to get y = 5
Now check yourself that the answer is still the same if you had chosen to substitute 15 for x in x − y = 10
Example #2: Solve the following system using the elimination method.
3x + y = 10
-4x − 2y = 2
Step 1
First, notice that nothing can be eliminated when adding the left sides since
3x + y + -4x − 2y = -1x + -1y
However, in 3x + y = 10, if I can turn y into 2y, I could eliminate the variable y by adding 2y to -2y in -4x − 2y = 2
Therefore, turn y into 2y by multiplying the whole equation by 2
2 ×( 3x + y = 10) gives the new equation 6x + 2y = 20
Adding the left side of 6x + 2y = 20 to the left side of -4x − 2y = 2 gives what you see below:
6x + 2y + -4x − 2y = 6x + -4x + 2y − 2y = 2x + 0 = 2x
Adding the right gives us 20 + 2 = 22
Step 2
The sum for the left sides is 2x and the sum for the right sides is 22
Setting them equal, we get 2x = 22
Step 3
2x = 22
Solve for x by dividing both sides of this equation by 2
2x = 22
(2/2)x = 22/2
x = 11
Step 4
You can substitute 11 for x in either 3x + y = 10 or -4x − 2y = 2 to get y
Choosing the first one, we get 3 × 11 + y = 10
33 + y = 10
Minus 33 from both sides to get y = -23
You should have noticed that the reason we call this method the elimination method is because the first thing you do is eliminate a variable.
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# 2.1 Vectors in the plane
Page 1 / 29
• Describe a plane vector, using correct notation.
• Perform basic vector operations (scalar multiplication, addition, subtraction).
• Express a vector in component form.
• Explain the formula for the magnitude of a vector.
• Express a vector in terms of unit vectors.
• Give two examples of vector quantities.
When describing the movement of an airplane in flight, it is important to communicate two pieces of information: the direction in which the plane is traveling and the plane’s speed. When measuring a force, such as the thrust of the plane’s engines, it is important to describe not only the strength of that force, but also the direction in which it is applied. Some quantities, such as or force, are defined in terms of both size (also called magnitude ) and direction. A quantity that has magnitude and direction is called a vector . In this text, we denote vectors by boldface letters, such as v .
## Definition
A vector is a quantity that has both magnitude and direction.
## Vector representation
A vector in a plane is represented by a directed line segment (an arrow). The endpoints of the segment are called the initial point and the terminal point of the vector. An arrow from the initial point to the terminal point indicates the direction of the vector. The length of the line segment represents its magnitude . We use the notation $‖\text{v}‖$ to denote the magnitude of the vector $\text{v}.$ A vector with an initial point and terminal point that are the same is called the zero vector , denoted $0.$ The zero vector is the only vector without a direction, and by convention can be considered to have any direction convenient to the problem at hand.
Vectors with the same magnitude and direction are called equivalent vectors. We treat equivalent vectors as equal, even if they have different initial points. Thus, if $\text{v}$ and $\text{w}$ are equivalent, we write
$\text{v}=\text{w}.$
## Definition
Vectors are said to be equivalent vectors if they have the same magnitude and direction.
The arrows in [link] (b) are equivalent. Each arrow has the same length and direction. A closely related concept is the idea of parallel vectors. Two vectors are said to be parallel if they have the same or opposite directions. We explore this idea in more detail later in the chapter. A vector is defined by its magnitude and direction, regardless of where its initial point is located.
The use of boldface, lowercase letters to name vectors is a common representation in print, but there are alternative notations. When writing the name of a vector by hand, for example, it is easier to sketch an arrow over the variable than to simulate boldface type: $\stackrel{\to }{v}.$ When a vector has initial point $P$ and terminal point $Q,$ the notation $\stackrel{\to }{PQ}$ is useful because it indicates the direction and location of the vector.
## Sketching vectors
Sketch a vector in the plane from initial point $P\left(1,1\right)$ to terminal point $Q\left(8,5\right).$
See [link] . Because the vector goes from point $P$ to point $Q,$ we name it $\stackrel{\to }{PQ}.$
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
Abigail
for teaching engĺish at school how nano technology help us
Anassong
How can I make nanorobot?
Lily
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
how can I make nanorobot?
Lily
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
Got questions? Join the online conversation and get instant answers! |
5th Class Mathematics Operations on Numbers Subtraction
Subtraction
Category : 5th Class
Subtraction
Subtraction: Under the operation of subtraction, difference between two numbersis found.
Minuend: The number from which other number is subtracted is called minuend.
Subtrahend: The number which is subtracted from a number is called subtrahend.
Difference: The result we get after subtraction is called difference.
Subtract 1254 from 7889
Solution:
Thousands Hundreds Tens Ones 7 8 8 9 Minuend -1 2 5 4 Subtrahend 6 6 3 5 Difference
How to Perform Subtraction
Just like addition digits of both minuend and subtrahend are arranged in the place value chart. Digits of the subtrahend comes below to the digits of minuend. Now subtract ones from ones, tens from tens and so on.
Find the difference between 4568 and 2795
Solution:
Thousands Hundreds Tens Ones 4 5 6 8 -2 7 9 5 1 7 7 3
Word Problem Based on Subtraction
A person earns Rs. 36000 per month. He spends Rs. 1500 on education, Rs. 3600 on house rent, and Rs. 7500 on fooding and he saves the rest of money. Find his saving in a month.
(a) Rs. 23400
(b) Rs. 90000
(c) Rs. 23600
(d) Rs. 12600
(e) None of these
Explanation
His total income in a month = Rs. 36000
His total expenditure in a month = Rs. 1500 + Rs. 3600 + Rs. 7500 = Rs. 12600
Therefore, his saving in a month = Rs. 36000 - Rs. 12600 = Rs. 23400
A book contains 1256 pages. 256 pages are blue, 186 pages are green and rest of the pages are white. Find the difference between the numbers of white pages and blue pages:
(a) 814
(b) 1000
(c) 628
(d) 558
(e) None of these
Explanation
Total number of pages in the book = 1256
Total number of blue and green pages = 256 + 186 = 442
Therefore, total number of white pages = 1256 - 442 = 814
Difference between the numbers of white and blue pages = 814 - 256
= 558
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# 10th Maths Chapter 8 Statistics and Probability Exercise 8.3
10th Standard Maths Chapter 8 Statistics and Probability Exercise 8.3 Guide. TN SSLC Samacheer Kalvi Guide Chapter 8 Exercise 8.3 Book Back Answers & Solutions. 10th All Subject Guide – Click Here. Class 1 to 12 All Subject Book Back Answers – Click Here.
## 10th Maths Chapter 8 Statistics and Probability Exercise 8.3
Solution:
### 2. Write the sample space for selecting two balls from a bag containing 6 balls numbered 1 to 6 (using a tree diagram).
Solution:
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
### 3.If A is an event of a random experiment such that P(A) : P($$\overline{\mathbf{A}}$$) =17 : 15 and n(S) = 640 then find
(i) P($$\overline{\mathbf{A}}$$)
(ii) n(A).
Solution:
P(A): P($$\overline{\mathbf{A}}$$) = 17 : 15
### 4. A coin is tossed thrice. What is the probability of getting two consecutive tails?
Solution:
Outcomes {O}: {(HHH), (THH), (HTH), (HHT), (HTT), (THT), (TTH), (TTT)}
Two consecutive tails {F} : {(HTT), (TTH), (TTT)}
n{F} = 3
n{O} = 8
### 5. At a fete, cards bearing numbers 1 to 1000, one number on one card are put in a box. Each player selects one card at random and that card is not replaced. If the selected card has a perfect square number greater than 500, the player wins a prize. What is the probability that
• (i) the first player wins a prize
• (ii) the second player wins a prize if the first has won?
Sample space = {1, 2, 3,… ,1000}
n(S) = 1000
(i) Let A be the event of setting square number greater than 500
A = {529, 576, 625, 676, 729, 784, 841, 900, 961}
n(A) = 9
P(A) = $$\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{9}{1000}$$
The probability that the first player wins prize = $$\frac{9}{1000}$$
(ii) If the first player wins, the number is excluded for the second player.
n(A) = 8 and n(S) = 999
P(A) = $$\frac{n(A)}{n(S)}=\frac{8}{999}$$
Probability the second player wins a prize = $$\frac{8}{999}$$
### 6. A bag contains 12 blue balls and x red balls. If one ball is drawn at random
• (i) what is the probability that it will be a red ball?
• (ii) If 8 more red balls are put in the bag, and if the probability of drawing a red ball will be twice that of the probability in (i), then find x.
Solution:
12 ➝ blue balls
x ➝ red balls
(i) P (red ball) = $$\frac{x}{x+12}$$
(ii) 8 red balls are added to the bag.
∴ 12 ➝ blue balls
x + 8 ➝ red balls
Given that P(ii) = 2 × P(i)
⇒ (x + 8)(x + 12) = 2x(x + 20)
⇒ (x2 + 20x + 96) = 2×2 + 40x
⇒ x2 + 20x – 96 = 0
⇒ x2 + 24x – 4x – 96 = 0
⇒ x(x + 24) – 4(x + 24) = 0
⇒ (x – 4)(x + 24) = 0
∴ x = 4 (or) x = -24
x cannot be negative ⇒ x = 4
Substituting x = 4 in (i),
### 7. Two unbiased dice are rolled once. Find the probability of getting
• (i) a doublet (equal numbers on both dice)
• (ii) the product as a prime number
• (iii) the sum as a prime number
• (iv) the sum as 1
Solution:
Doublet = {(1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6,6)}
Total number of outcomes = 6 × 6
n(S) = 36
Number of favourable outcomes = 6
(ii) Number of favourable outcomes = 6
as favourable outcomes = (1, 2), (2, 1), (1, 3), (3, 1),(1, 5),and (5, 1)
(iii) Sum as prime numbers = {(1, 1), (1, 2), (2, 3), (1, 4), (1, 6), (4, 3), (5, 6)}
Number of favourable outcomes = 7
⇒ Probability = $$\frac{7}{36}$$
(iv) With two dice, minimum sum possible = 2
∴ Prob (sum as 1) = 0 [Impossible event]
### 8. Three fair coins are tossed together. Find the probability of getting
• (ii) atleast one tail
• (iv) atmost two tails
Three fair coins are tossed together
Sample spade = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
n(S) = 8
(i) Let A be the event of getting all heads
A = {HHH}
n(A) = 1
$$P(A)=\frac{n(A)}{n(S)}=\frac{1}{8}$$
(ii) Let B be the event of getting atleast one tail.
B = {HHT, HTH, HTT, THH, THT, TTH, TTT}
n(B) = 7
$$P(B)=\frac{n(B)}{n(S)}=\frac{7}{8}$$
(iii) Let C be the event of getting atmost one head
C = {HTT, THT, TTH, TTT}
n(C) = 4
$$P(C)=\frac{n(C)}{n(S)}=\frac{4}{8}=\frac{1}{2}$$
(iv) Let D be the event of getting atmost two tails.
D = {HTT, TTT, TTH, THT, THH, HHT, HTH}
n(D) = 7
$$P(D)=\frac{n(D)}{n(S)}=\frac{7}{8}$$
### 9.Two dice are numbered 1,2,3,4,5,6 and 1,1,2,2,3,3 respectively. They are rolled and the sum of the numbers on them is noted. Find the probability of getting each sum from 2 to 9 separately.
Solution:
Dice 1
S = {1,2, 3, 4, 5, 6}
Dice 2
S = {1,1,2, 2, 3, 3}
Total possible outcomes when they are rolled
n(S) = 36
Event of sum (2) = A = {(1,1), (1,1)},
n(A) = 2,P(A) = $$\frac{2}{36}$$
Event of sum 3 is B = {(1, 2), (1, 2), (2, 1), (2, 1)}
Event of sum 4 is C= {(1, 3), (1, 3), (2, 2), (2, 2), (3, 1) (3, 1)}
n(C) = 6
Event of getting the sum 5 is
D = {(2, 3), (2, 3), (3, 2), (3, 2), (4, 1), (4, 1)}
n(D) = 6, P(D) = $$\frac{6}{36}$$ .
Event of getting the sum 6 is
E = {(3, 3), (3, 3), (4, 2), (4, 2), (5, 1), (5, 1)}
n(E) = 6, P(E) = $$\frac{6}{36}$$
Event of getting the sum 7 is
F = {(4, 3), (4, 3), (5, 2), (5, 2), (6, 1), (6, 1)}
n(F) = 6
P(F) = $$\frac{6}{36}$$
Event of getting the sum 8 is
G = {(5, 3), (5, 3), (6, 2), (6, 2)}
Event of getting the sum 9 is
H = {(6, 3), (6, 3), n(H) = 2
### 9. A bag contains 5 red balls, 6 white balls, 7 green balls, and 8 black balls. One ball is drawn at random from the bag. Find the probability that the ball is drawn
• (i) white
• (ii) black or red
• (iii) not white
• (iv) neither white nor black
Solution:
5 red 6 white 7 green 8 black total no. of balls = 5 + 6 + 7 +8= 26
### 10. In a box there are 20 non-defective and some defective bulbs. If the probability that a bulb selected at random from the box found to be defective is 3/8 then, find the number of defective bulbs.
Solution:
Let the number of defective bulbs be ‘x’
Total number of bulbs = x + 20
⇒ 8x = 3x + 60
⇒ 5x = 60
⇒ x = 12
∴ No.of defective bulbs are = 12.
### 11. The king and queen of diamonds, queen and jack of hearts, jack and king of spades are removed from a deck of 52 playing cards and then well shuffled. Now one card is drawn at random from the remaining cards. Determine the probability that the card is
• (i) a clavor
• (ii) a queen of red card
• (iii) a king of black card
Solution:
(i.e) remaining number of cards = 52 – 6 = 46 13
(i) P(a clavor) = $$\frac{13}{46}$$
(ii) P(queen of red card) = 0 as both Queen of diamond and heart have been removed.
(iii) only K of clavor is in the deck
⇒ P(king of black card) = $$\frac{1}{46}$$
Solution:
### 13. Two customers Priya and Amuthan are visiting a particular shop in the same week (Monday to Saturday). Each is equally likely to visit the shop on any one day as on another day. What is the probability that both will visit the shop on
• (i) the same day
• (ii) different days
• (iii) consecutive days?
Solution:
### 14. In a game, the entry fee is ₹ 150. The game consists of tossing a coin 3 times. Dhana bought a ticket for entry. If one or two heads show, she gets her entry fee back. If she throws 3 heads, she receives double the entry fees. Otherwise, she will lose. Find the probability that she
(i) gets a double entry fee
(ii) just gets her entry fee
(iii) loses the entry fee.
Solution: |
# Ch.17 :- Arithmetical Reasoning : Introduction & Formula
Example 1
Ex. 1. The number of boys in a class are three times the number of girls. Which one of the following number can’t represent the total number of children in the class?
(a) 48
(b) 44
(c) 42
(d) 40
Sol. Let the number of girls be x, then from the question it is clear that number of bays are 3x.
Therefore, total no. of students
= Number of boys + Number of girls
= 3x + x
= 4x
Now, the total number of children in the class must be a multiple of 4. Out of the four options given (c) does not qualify this condition. Therefore, 42 does not represent the total number of children in the class. Hence, the correct answer is (c).
Example 2
Ex. 2. The sum of the ages of a son and father is 56 years. After four years, the age of the father will be three times that of the son. Their ages respectively are:
(a) 12 years, 44 years
(b) 16 years, 42 years
(c) 16 years, 48 years
(d) 18 years, 36 years
Sol. Let the age of the father be x, then the age of the son would be (56 – x). After four years, the age of father would be (x + 4) and that of son would be (56 -x + 4) years. Now, from the information given in the question,we have
(x + 4) = 3 (56 -x + 4)
=> x + 4 = 168 – 3x + 12
=> 4x = 168 + 12 – 4 = 176
=> x = 44 years
Therefore, the age of father and son is 44 years and 12 years, respectively. Hence, the correct answer is (a).
Example 3
Ex. 3. In 10 years, A will be twice as old as B was 10 years ago. If at present A is 9 years older than B, the present age of B is:
(a) 19 years
(b) 29 years
(c) 39 years
(d) 49 years
Sol. Let the present age of B be x years. Then, the present age of A would be (x + 9) years. After 10 years, the age of A would be
(x + 9 + 10)
= (x + 19) years
and before ten years, the age of B was (x – 10) years.
Now, from the information given in the question,
(x + 19) = 2 (x – 10)
or x + 19 = 2 x – 20
or x = 19 + 20 = 39 years
Therefore, the present age of B is 39 years.Hence, the correct answer is (c).
Example 4
Ex. 4. The 1st bunch of bananas has 1/4 excess to as many as bananas in 2nd bunch. If the 2nd bunch has 3 bananas less than the 1st bunch, then the number of bananas in 1st bunch is :
(a) 9
(b) 10
(c) 12
(d) 15
Sol. Let the number of bananas in 2nd bunch be x.
Then, the number of bananas in 1st bunch = x + x4 = 5x4
Therefore,
5x4 – x = 3 => 5x – 4x = 12 => x = 12.
Then, the number of bananas in 1st bunch
=54 x 12 = 15
Hence, the correct answer is (d).
Example 5
Ex. 5. In a town. 65% people watch the news on television, 40% read a newspaper and 25% read a newspaper and watch the news on television also. What percentage of the people neither watch the news on television nor read a newspaper?
(a) 5%
(b) 10%
(c) 15%
Sol. Let the total number of people Let be 100.
Let circle A represents people
who watched television and B represents people who read newspaper.
Then, x + y = 65, Y + z = 40,
We get, x = 40, y = 25,
Then, the number of people who television nor read newspaper
= 100 – (x + y + z)
= 100- (40 + 25 + 15)
= 100-80 = 20
Therefore, the required percentage is 20%. Hence, the correct answer is (d).
Example 6
Ex. 6. In a group of 15 people, 7 read French, 8 read English while 3 of them read none of these two. How many of them read French and English both?
(a) 0
(b) 3
(c) 4
(d) 5
Sol. Let circles x and y represent people who read French and English, respectively.
Area A shows the people who read French only.
Area C shows the people who read English only.
Area B shows the people who read French and English both.
Now,
(A + B + C) + 3 = 15 or A + B + C = 12 … (i)
A + B = 7, B + C = 8
Adding these two, we get A + 2B + C = 15 … (ii)
Subtracting (i) from (ii), we get
B = 15 – 12 = 3
Therefore, number of people who read French and English both is 3.
Hence, the correct answer is (b).
Example 7
Ex. 7. In a chess tournament each of six players will play with all other players exactly once. How many matches will be played during the tournament?
(a) 12
(b) 15
(c) 30
(d) 36
Sol. The situation in which the matches will be played would be as follows:
(i) 1st player will play matches with other 5 players.
(ii) 2nd player will play matches with 4 players other than the 1st player.
(iii) 3rd player will play matches with 3 players other than 1st and 2nd players.
(iv) 4th player will play matches with 2 players other than 1st, 2nd and 3rd players.
(v) 5th player will play match with 6th player only.
Therefore, the number of matches played during the tournament is
= 5 + 4 + 3 + 2 + 1 = 15.
Example 8
Ex.8. Consider the diagram given below:
Five hundred candidates appeared in the examination conducted for the tests in English, Hindi and Mathematics. The diagram gives the number of candidates who failed in different tests. What is the percentage of candidates who failed in at least two subjects ?
(a) 0.078%
(b) 1.0%
(c) 6.8%
(d) 7.8%
Sol. From the diagram, it is clear that number of candidates who failed in at least two subjects
= number of candidates who failed in two or more subjects.
= (10 + 12 + 12 + 5)
= 39.
Therefore, the required percentage
= (39500 x 100)%
= 7.8%
Therefore, the option (d) is the correct answer.
Example 9
Ex. 9. In an examination, 42% students failed in Hindi and 52% failed in English. If 17% failed in both the subjects, the percentage of those who passed in both the subjects, is :
(a) 23%
(b) 27%
(c) 34%
(d) 40%
Sol. Let the total number of students who appeared for the examination be 100. Circles X and Y represent the students who failed in Hindi and English, respectively .
Now, number of students who failed in Hindi only
= (42-17)
= 25%
Number of students who failed in English only
(52 -17) = 35%
Total number of students failed
= students failed in Hindi only + students failed in English only
= 25 + 35 = 60%.
Number of students passed = 100 – 60 = 40%
Hence, option (d) is the correct answer.
Example 10
Ex.10. A shepherd had 17 sheep. All but nine died. How many was he left with ?
(a) Nil
(b) 8
(c) 9
(d) 17
Sol. “All but nine died” means “All except nine died”. It means that nine sheep remained alive and other died. Hence, option (c) is the correct answer. |
Question
# If the point P on the parabola ${y^2} = 4ax$ for which |PR - PQ|Â is maximum, where $R = (- a,0)$, $Q = (0,a)$ is represented as $( {{k_1}a,\;{k_2}a})$, then find the value of $( {{k_1} + {k_2}})$
Hint: Use triangle inequality to find the general points of parabola in parametric form to find the sum ofÂ
$K_1$Â and $K_2$.
Â
P point is given as $(K_1a,K_2a)$,Q point is given as (0,a) and R point is given as (−a,0)
When you join these points it makes a triangle, So according to triangle inequality the modulus of difference of two sides is always less than or equal to the third side.
$\Rightarrow |PR−PQ|⩽|QR|$
Maximum value ofÂ
|PR−PQ|=QR because everything is less than or equal to QR.
$\Rightarrow |PR−PQ|=QR$, when P, Q, R are collinear
So, the equation of line joining R and Q is
$\Rightarrow y−0= \dfrac {a−0}{0−(−a)}(x−(−a))$
$\Rightarrow y=x+a$
Equation of parabola is $y^2=4ax$
In parametric form the coordinates of parabola is $(at^2,2at)$
Where t is a parameter and these parametric coordinates represent every point on parabola.
$\Rightarrow$ p point is $(at^2,2at)$
Now P, Q, R is in straight line, then this p point satisfy the equation of straight line
$\Rightarrow 2at−at^2=a$
$\Rightarrow t^2−2t+1=0$
$\Rightarrow (t−1)^2=0$
$\Rightarrow (t−1)=0$
$\Rightarrow t=1$
So, put the value of t in P points
$\Rightarrow$ P(a,2a), So compare this points with given points $(K_1a,K_2a)$
So, on comparing $K_1=1,K_2=2$
$\Rightarrow$ The value of $(K_1+K_2)=1+2=3$.
So, this is the required answer.
Â
NOTE: - In this type of problem always apply inequality identity, after that find out the general points of parabola in parametric form and satisfy these points in the equation which pass from this point. |
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# How do you solve and graph x – 3 > 7?
Last updated date: 13th Jun 2024
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Hint: Rearrange the terms by taking the constant terms to the R.H.S. and leaving the terms containing the variable x to the L.H.S. Now, simplify the R.H.S. by simple addition or subtraction, whichever needed to get a particular numerical value. Leave the inequality sign as it is. Draw a line x = 10 and consider the suitable part of the graph according to the simplified inequality obtained.
Complete step-by-step solution:
Here, we have been provided with the inequality x – 3 > 7 and we are asked to solve it and draw the graph.
Now, rearranging the given inequality by taking the constant terms to the R.H.S. and leaving the terms containing the variable x in the L.H.S., we get,
\begin{align} & \Rightarrow x-3>7 \\ & \Rightarrow x>7+3 \\ \end{align}
$\Rightarrow x>10$ - (1)
Here, as you can see, the direction of the inequality sign does not change. This is because the direction of inequality sign only changes when we multiply or divide both the sides with a negative number or take reciprocal on both the sides.
Now, let us draw the graph of inequality obtained in (1). To do this first we have to remove the inequality sign and replace it with ‘=’ sign and draw the required line, so we have,
$\Rightarrow x=10$
Drawing the line x = 10, we get,
Now, considering equation (1), i.e., x > 10, here we can clearly see that we have to select that part of the graph in which x will be greater than 10. So, in the above graph we have to select the right side of the graph x = 10. Therefore, we have,
So, the above graph represents the graphical solution of our inequality.
Note: One may note that there are not many differences in solving and graphing an inequality and an equality. We need the help of equality while drawing the graph. One thing you can note is we do not have to consider the value x = 10 in our graph, so it will be better to use the dashed line instead of a solid line, however we have used a solid line. Always remember the rules of reversing the direction of inequality. The direction is only reversed when we take reciprocal or divide and multiply both the sides with a negative number. |
# Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 3
Sets
Mathematical structures like numbers are described as sets. A set is a collection of distinct objects which are called the elements of the set. By distinct we mean that it is clear which objects are considered to be equal and which are considered to be different. The containment of an element ${\displaystyle {}x}$ to a set ${\displaystyle {}M}$ is expressed by
${\displaystyle {}x\in M\,,}$
the noncontainment by
${\displaystyle {}x\notin M\,.}$
For every element, exactly one of these possibilities holds. For example, we have ${\displaystyle {}{\frac {3}{7}}\notin \mathbb {N} }$ and ${\displaystyle {}{\frac {3}{7}}\in \mathbb {Q} }$. An important principle for sets is the principle of extensionality, i.e. a set is determined by the elements it contains, beyond that it bears no further information. In particular, two sets coincide if they contain the same elements.
The set which does not contain any element is called the empty set and is denoted by
${\displaystyle \emptyset .}$
A set ${\displaystyle {}N}$ is called a subset of a set ${\displaystyle {}M}$ if every element from ${\displaystyle {}N}$ does also belong to ${\displaystyle {}M}$. For this relation we write ${\displaystyle {}N\subseteq M}$ (some people write ${\displaystyle {}N\subset M}$ for this). One also says that the inclusion ${\displaystyle {}N\subseteq M}$ holds. For the number sets, the inclusions
${\displaystyle {}\mathbb {N} \subseteq \mathbb {Z} \subseteq \mathbb {Q} \subseteq \mathbb {R} \,}$
hold. The subset relation ${\displaystyle {}N\subseteq M}$ is a statement using for all, as it makes a claim about all elements from ${\displaystyle {}N}$. If we want to show ${\displaystyle {}N\subseteq M}$, then we have to show for an arbitrary element ${\displaystyle {}x\in N}$ that also the containment ${\displaystyle {}x\in M}$ holds. In order to show this, we are only allowed to use the property ${\displaystyle {}x\in N}$. For us, sets will be either number sets or sets constructed from such number sets. A set is called finite if its elements may be counted by the natural numbers ${\displaystyle {}1,2,3,\ldots ,n}$ for a certain ${\displaystyle {}n\in \mathbb {N} }$. In this case, the number ${\displaystyle {}n}$ is called the number (or the cardinality) of the set.
Possible descriptions for sets
There are several ways to describe a set. The easiest one is to just list the elements of the set, here the order of the listing is not important. For finite sets this is possible, however, for infinite sets one has to describe a "construction rule“ for the elements.
The most important set given by an infinite listing is the set of natural numbers
${\displaystyle {}\mathbb {N} =\{0,1,2,3,\ldots \}\,.}$
Here a certain set of numbers is described by a list of starting elements in the hope that this makes it clear which numbers belong to the set. An important point is that ${\displaystyle {}\mathbb {N} }$ is not a set of certain digits, but the set of values represented by these digits or sequences of digits. For a natural number there are many possibilities to represent it, the decimal expansion is just one of them.
We discuss now the description of sets by properties. Let ${\displaystyle {}M}$ denote a given set. In ${\displaystyle {}M}$ there are certain elements which fulfil a certain property ${\displaystyle {}E}$ (a predicate) or not. Hence, for the property ${\displaystyle {}E}$ we have within ${\displaystyle {}M}$ the subset consisting of all the elements from ${\displaystyle {}M}$ which fulfil this property. We write for this subset given by ${\displaystyle {}E}$
${\displaystyle {}{\left\{x\in M\mid E(x)\right\}}={\left\{x\in M\mid x{\text{ fulfils property }}E\right\}}\,.}$
This only works for such properties for which the statement ${\displaystyle {}E(x)}$ is well-defined for every ${\displaystyle {}x\in M}$. If one introduces such a subset then one gives a name to it which often reflects the name of the property, like
${\displaystyle {}E={\left\{x\in \mathbb {N} \mid x{\text{ is even}}\right\}}\,,}$
${\displaystyle {}O={\left\{x\in \mathbb {N} \mid x{\text{ is odd}}\right\}}\,,}$
${\displaystyle {}S={\left\{x\in \mathbb {N} \mid x{\text{ is a square number}}\right\}}\,}$
${\displaystyle {}{\mathbb {P} }={\left\{x\in \mathbb {N} \mid x{\text{ is a prime number}}\right\}}\,.}$
For the sets occurring in mathematics, a multitude of mathematical properties is relevant and therefore there is a multitude of relevant subsets. But also in the sets of everyday life like the set ${\displaystyle {}C}$ of the students in a course, there are many important properties which determine certain subsets, like
${\displaystyle {}O={\left\{x\in C\mid x{\text{ lives in Osnabr}}{\ddot {\rm {u}}}{\text{ck}}\right\}}\,,}$
${\displaystyle {}P={\left\{x\in C\mid x{\text{ studies physics}}\right\}}\,,}$
${\displaystyle {}D={\left\{x\in C\mid x{\text{ has birthday in December}}\right\}}\,.}$
The set ${\displaystyle {}C}$ itself is also given by a property, since
${\displaystyle {}C={\left\{x\mid x{\text{ is a student in the course}}\right\}}\,.}$
Set operations
Similar to the connection of statements to get new statements, there are operations to make new sets from old ones. The most important operations are the following.[1]
1. Union
${\displaystyle {}A\cup B:={\left\{x\mid x\in A{\text{ or }}x\in B\right\}}\,.}$
2. Intersection
${\displaystyle {}A\cap B:={\left\{x\mid x\in A{\text{ and }}x\in B\right\}}\,.}$
3. Difference set
${\displaystyle {}A\setminus B:={\left\{x\mid x\in A{\text{ and }}x\notin B\right\}}\,.}$
For these operations to make sense, the sets need to be subsets of a common basic set. This ensures that we are talking about the same elements. Quite often this basic set is not mentioned explicitly and has to be understood from the context. A special case of the difference set is the complement of a subset ${\displaystyle {}A\subseteq G}$ in a given base set ${\displaystyle {}G}$, also denoted as
${\displaystyle {}\complement A:=G\setminus A={\left\{x\in G\mid x\not \in A\right\}}\,.}$
If two sets have an empty intersection, meaning ${\displaystyle {}A\cap B=\emptyset }$, we also say that they are disjoint.
Product set
We want to describe within set theory the arithmetic operations on the number sets mentioned above like addition and multiplication. For the addition (say on ${\displaystyle {}\mathbb {N} }$), two natural numbers ${\displaystyle {}a}$ and ${\displaystyle {}b}$ are added to yield another natural number, namely ${\displaystyle {}a+b}$. The set of pairs constitute the product set and the adding is interpreted as a mapping on the product set.
We define.[2]
## Definition
Suppose that two sets ${\displaystyle {}L}$ and ${\displaystyle {}M}$ are given. Then the set
${\displaystyle {}L\times M={\left\{(x,y)\mid x\in L,\,y\in M\right\}}\,}$
is called the product set of the sets.
The elements of a product set are called pairs and denoted by ${\displaystyle {}(x,y)}$. Here the ordering is essential. The product set consists of all pair combinations, where in the first component there is an element of the first set and in the second component there is an element of the second set. Two pairs are equal if and only if they are equal in both components.
If one of the sets is empty, then so is the product set. If both sets are finite, say the first with ${\displaystyle {}n}$ elements and the second with ${\displaystyle {}k}$ elements, then the product set has ${\displaystyle {}n\cdot k}$ elements. It is also possible to form the product set of more than two sets.
## Example
Let ${\displaystyle {}F}$ be the set of all first names, and ${\displaystyle {}L}$ be the set of all last names. Then
${\displaystyle F\times L}$
is the set of all names. The elements of this set are in pair notation ${\displaystyle {}({\text{Heinz}},{\text{Miller}})}$, ${\displaystyle {}({\text{Petra}},{\text{Miller}})}$ and ${\displaystyle {}({\text{Lucy}},{\text{Sonnenschein}})}$. From a name, one can deduce easily the first name and the last name by looking at the first or the second component. Even if all first names and all last names do really occur, not every combination of a first name and a last name does occur. For the product set, all possible combinations are allowed.
For a product set it is also possible that both sets are equal. Then one has to be careful not to confuse the components.
## Example
A chess board (meaning the set of squares of a chess board where a chess piece may stand) is the product set ${\displaystyle {}\{a,b,c,d,e,f,g,h\}\times \{1,2,3,4,5,6,7,8\}}$. Every square is a pair, e.g. ${\displaystyle {}(a,1),(d,4),(c,7)}$. Because the two component sets are different, one may write instead of pair notation simply ${\displaystyle {}a1,d4,c7}$. This notation is the starting point to describe chess positions, and complete chess games.
The product set ${\displaystyle {}\mathbb {R} \times \mathbb {R} }$ is thought of as a plane, one denotes it also by ${\displaystyle {}\mathbb {R} ^{2}}$. The product set ${\displaystyle {}\mathbb {Z} \times \mathbb {Z} }$ is a set of lattice points.
## Example
Let ${\displaystyle {}S}$ denote a circle (the circumference), and let ${\displaystyle {}I}$ be a line segment. The circle is a subset of a plane ${\displaystyle {}E}$, and the line segment is a subset of a line ${\displaystyle {}G}$, so that for the product sets, we have the relation
${\displaystyle {}S\times I\subseteq E\times G\,.}$
The product set ${\displaystyle {}E\times G}$ is the three-dimensional space, and the product set ${\displaystyle {}S\times I}$ is the surface of a cylinder.
Mappings
Suppose that a particle is moving in space. This process is described by assigning to every point in time ${\displaystyle {}t\in \mathbb {R} }$ the point in space ${\displaystyle {}z(t)\in \mathbb {R} ^{3}}$ where the particle is at that time. The running of a computer program that uses altogether ${\displaystyle {}s}$ memory units can be described by saying for every computing step (which means the execution of a program line) what the content of the memory units is. So to the number ${\displaystyle {}n}$ counting the computing steps, we assign the tuple
${\displaystyle {}(b_{1}(n),\ldots ,b_{s}(n))\in \mathbb {N} ^{s}\,.}$
In an election, every voter has to decide for exactly one party ( or for not voting). The temperature profile on the earth is described by assigning to every point in time and every point on the surface its temperature. Such and many other situations are captured by the concept of a mapping.
## Definition
Let ${\displaystyle {}L}$ and ${\displaystyle {}M}$ denote sets. A mapping ${\displaystyle {}F}$ from ${\displaystyle {}L}$ to ${\displaystyle {}M}$ is given by assigning, to every element of the set ${\displaystyle {}L}$, exactly one element of the set ${\displaystyle {}M}$. The unique element which is assigned to ${\displaystyle {}x\in L}$, is denoted by ${\displaystyle {}F(x)}$. For the mapping as a whole, we write
${\displaystyle F\colon L\longrightarrow M,x\longmapsto F(x).}$
If a mapping ${\displaystyle {}F\colon L\rightarrow M}$ is given, then ${\displaystyle {}L}$ is called the domain (or domain of definition) of the map and ${\displaystyle {}M}$ is called the codomain (or target range) of the map. For an element ${\displaystyle {}x\in L}$, the element
${\displaystyle {}F(x)\in M\,}$
is called the value of ${\displaystyle {}F}$ at the place (or argument) ${\displaystyle {}x}$.
Two mappings ${\displaystyle {}F\colon L_{1}\rightarrow M_{1}}$ and ${\displaystyle {}G\colon L_{2}\rightarrow M_{2}}$ are equal if and only if their domains coincide, their codomains coincide and if for all ${\displaystyle {}x\in L_{1}=L_{2}}$ the equality ${\displaystyle {}F(x)=G(x)}$ in ${\displaystyle {}M_{1}=M_{2}}$ holds. So the equality of mappings is reduced to the equalities of elements in a set. Mappings are also called functions. However, we will usually reserve the term function for mappings where the codomain is a number set like the real numbers ${\displaystyle {}\mathbb {R} }$.
For every set ${\displaystyle {}L}$, the mapping
${\displaystyle L\longrightarrow L,x\longmapsto x,}$
which sends every element to itself, is called the identity (on ${\displaystyle {}L}$). We denote it by ${\displaystyle {}\operatorname {Id} _{L}}$. For another set ${\displaystyle {}M}$ and a fixed element ${\displaystyle {}c\in M}$, the mapping
${\displaystyle L\longrightarrow M,x\longmapsto c,}$
which sends every element ${\displaystyle {}x\in L}$ to the constant value ${\displaystyle {}c}$, is called the constant mapping (with value ${\displaystyle {}c}$). It is usually again denoted by ${\displaystyle {}c}$.[3]
There are several ways to describe a mapping, like value table, bar chart, pie chart, arrow diagram, the graph of the mapping. In mathematics, a mapping is most often given by a mapping rule, which allows computing the values of the mapping for every argument. Such rules are e.g. (from ${\displaystyle {}\mathbb {R} }$ to ${\displaystyle {}\mathbb {R} }$) ${\displaystyle {}x\mapsto x^{2}}$, ${\displaystyle {}x\mapsto x^{3}-e^{x}+\sin x}$, etc. In the sciences and in sociology also empirical functions are important which describe real movements or developments. But also for such functions, one wants to know whether they can be described (approximated) in mathematical manner.
${\displaystyle {}x}$ ${\displaystyle {}1}$ ${\displaystyle {}2}$ ${\displaystyle {}3}$ ${\displaystyle {}4}$ ${\displaystyle {}5}$ ${\displaystyle {}6}$ ${\displaystyle {}\pi (x)}$ ${\displaystyle {}2}$ ${\displaystyle {}4}$ ${\displaystyle {}6}$ ${\displaystyle {}5}$ ${\displaystyle {}3}$ ${\displaystyle {}1}$
${\displaystyle {}\cdot }$ ${\displaystyle {}0}$ ${\displaystyle {}1}$ ${\displaystyle {}2}$ ${\displaystyle {}3}$ ${\displaystyle {}4}$ ${\displaystyle {}5}$ ${\displaystyle {}6}$
${\displaystyle {}0}$ ${\displaystyle {}0}$ ${\displaystyle {}0}$ ${\displaystyle {}0}$ ${\displaystyle {}0}$ ${\displaystyle {}0}$ ${\displaystyle {}0}$ ${\displaystyle {}0}$
${\displaystyle {}1}$ ${\displaystyle {}0}$ ${\displaystyle {}1}$ ${\displaystyle {}2}$ ${\displaystyle {}3}$ ${\displaystyle {}4}$ ${\displaystyle {}5}$ ${\displaystyle {}6}$
${\displaystyle {}2}$ ${\displaystyle {}0}$ ${\displaystyle {}2}$ ${\displaystyle {}4}$ ${\displaystyle {}6}$ ${\displaystyle {}1}$ ${\displaystyle {}3}$ ${\displaystyle {}5}$
${\displaystyle {}3}$ ${\displaystyle {}0}$ ${\displaystyle {}3}$ ${\displaystyle {}6}$ ${\displaystyle {}2}$ ${\displaystyle {}5}$ ${\displaystyle {}1}$ ${\displaystyle {}4}$
${\displaystyle {}4}$ ${\displaystyle {}0}$ ${\displaystyle {}4}$ ${\displaystyle {}1}$ ${\displaystyle {}5}$ ${\displaystyle {}2}$ ${\displaystyle {}6}$ ${\displaystyle {}3}$
${\displaystyle {}5}$ ${\displaystyle {}0}$ ${\displaystyle {}5}$ ${\displaystyle {}3}$ ${\displaystyle {}1}$ ${\displaystyle {}6}$ ${\displaystyle {}4}$ ${\displaystyle {}2}$
${\displaystyle {}6}$ ${\displaystyle {}0}$ ${\displaystyle {}6}$ ${\displaystyle {}5}$ ${\displaystyle {}4}$ ${\displaystyle {}3}$ ${\displaystyle {}2}$ ${\displaystyle {}1}$
Injective and surjective mappings
## Definition
Let ${\displaystyle {}L}$ and ${\displaystyle {}M}$ denote sets, and let
${\displaystyle F\colon L\longrightarrow M,x\longmapsto F(x),}$
be a mapping. Then ${\displaystyle {}F}$ is called injective, if for two different elements ${\displaystyle {}x,x'\in L}$, also ${\displaystyle {}F(x)}$ and ${\displaystyle {}F(x')}$
are different.
If we want to show that a certain mapping is injective then we may show the following: For any two elements ${\displaystyle {}x}$ and ${\displaystyle {}x'}$ fulfilling the condition ${\displaystyle {}F(x)=F(x')}$, we can deduce that ${\displaystyle {}x=x'}$. This is often easier to show than the statement that ${\displaystyle {}x\neq x'}$ implies ${\displaystyle {}F(x)\neq F(x')}$.
## Definition
Let ${\displaystyle {}L}$ and ${\displaystyle {}M}$ denote sets, and let
${\displaystyle F\colon L\longrightarrow M,x\longmapsto F(x),}$
be a mapping. Then ${\displaystyle {}F}$ is called surjective, if for every ${\displaystyle {}y\in M}$, there exists at least one element ${\displaystyle {}x\in L}$, such that
${\displaystyle {}F(x)=y\,.}$
## Example
We consider a football game as the mapping which assigns, to every goal of team ${\displaystyle {}A}$, the corresponding goal scorer. Suppose that there are no own goals and no changes. The goals of ${\displaystyle {}A}$ are numbered by ${\displaystyle {}1,2,\ldots ,n}$. Then we have a mapping
${\displaystyle \psi \colon \{1,\ldots ,n\}\longrightarrow A=\{{\text{player }}1,\,{\text{player }}2,\ldots ,{\text{player }}11\},}$
given by
${\displaystyle {}\psi (i)={\text{ the player who has scored the }}i{\text{-th goal}}\,.}$
The injectivity of ${\displaystyle {}\psi }$ means that every player has scored at most one goal, the surjectivity means that every player has scored at least one goal.
## Example
Let ${\displaystyle {}H}$ denote the set of all (living or dead) people. We study the mapping
${\displaystyle \varphi \colon H\longrightarrow H,}$
which assigns to every person his or her (biological) mother. This is well-defined, as every person has a uniquely determined mother. This mapping is not injective, since there exists different people (brothers and sisters) with the same mother. It is also not surjective, since not every person is a mother of somebody.
## Example
The mapping
${\displaystyle \mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto x^{2},}$
is neither injective nor surjective. It is not injective, because the different numbers ${\displaystyle {}2}$ and ${\displaystyle {}-2}$ are both sent to ${\displaystyle {}4}$. It is not surjective, because only nonnegative elements are in the image (a negative number does not have a real square root). The mapping
${\displaystyle \mathbb {R} _{\geq 0}\longrightarrow \mathbb {R} ,x\longmapsto x^{2},}$
is injective, but not surjective. The injectivity can be seen as follows: If ${\displaystyle {}x\neq y}$, then one number is larger, say
${\displaystyle {}x>y\geq 0\,.}$
But then also ${\displaystyle {}x^{2}>y^{2}}$, and in particular ${\displaystyle {}x^{2}\neq y^{2}}$. The mapping
${\displaystyle \mathbb {R} \longrightarrow \mathbb {R} _{\geq 0},x\longmapsto x^{2},}$
is not injective, but surjective, since every nonnegative real number has a square root. The mapping
${\displaystyle \mathbb {R} _{\geq 0}\longrightarrow \mathbb {R} _{\geq 0},x\longmapsto x^{2},}$
is injective and surjective.
## Definition
Let ${\displaystyle {}M}$ and ${\displaystyle {}L}$ denote sets and suppose that
${\displaystyle F\colon M\longrightarrow L,x\longmapsto F(x),}$
is a mapping. Then ${\displaystyle {}F}$ is called bijective if ${\displaystyle {}F}$ is injective as well as
surjective.
## Remark
The question, whether a mapping ${\displaystyle {}F\colon L\rightarrow M}$ has the properties of being injective or surjective, can be understood looking at the equation
${\displaystyle {}F(x)=y\,}$
(in the two variables ${\displaystyle {}x}$ and ${\displaystyle {}y}$). The surjectivity means that for every ${\displaystyle {}y\in M}$ there exists at least one solution
${\displaystyle {}x\in L\,}$
for this equation, the injectivity means that for every ${\displaystyle {}y\in M}$ there exist at most one solution ${\displaystyle {}x\in L}$ for this equation, and the bijectivity means that for every ${\displaystyle {}y\in M}$ there exists exactly one solution ${\displaystyle {}x\in L}$ for this equation. Hence surjectivity means the existence of solutions, injectivity means the uniqueness of solutions. Both questions are everywhere in mathematics and they also can be interpreted as surjectivity or injectivity of suitable mappings.
## Definition
Let ${\displaystyle {}F\colon L\rightarrow M}$ denote a bijective mapping. Then the mapping
${\displaystyle G\colon M\longrightarrow L}$
which sends every element ${\displaystyle {}y\in M}$ to the uniquely determined element ${\displaystyle {}x\in L}$ with ${\displaystyle {}F(x)=y}$,
is called the inverse mapping of ${\displaystyle {}F}$.
## Definition
Let ${\displaystyle {}L,\,M}$ and ${\displaystyle {}N}$ denote sets, let
${\displaystyle F\colon L\longrightarrow M,x\longmapsto F(x),}$
and
${\displaystyle G\colon M\longrightarrow N,y\longmapsto G(y),}$
be mappings. Then the mapping
${\displaystyle G\circ F\colon L\longrightarrow N,x\longmapsto G(F(x)),}$
is called the composition of the mappings
${\displaystyle {}F}$ and ${\displaystyle {}G}$.
So we have
${\displaystyle {}(G\circ F)(x):=G(F(x))\,}$
where the left hand side is defined by the right hand side. If both mappings are given by functional expressions, then the composition is realized by plugging in the first term into the variable of the second term (and to simplify the expression if possible).
## Lemma
Let ${\displaystyle {}L,M,N}$ and ${\displaystyle {}P}$ be sets and let
${\displaystyle F\colon L\longrightarrow M,x\longmapsto F(x),}$
${\displaystyle G\colon M\longrightarrow N,y\longmapsto G(y),}$
and
${\displaystyle H\colon N\longrightarrow P,z\longmapsto H(z),}$
be mappings. Then
${\displaystyle {}H\circ (G\circ F)=(H\circ G)\circ F\,}$
holds.
### Proof
${\displaystyle \Box }$
Footnotes
1. It is easy to memorize the symbols: the ${\displaystyle {}\cup }$ for union looks like u. The intersection is written as ${\displaystyle {}\cap }$. The corresponding logical operations or, and have the analog form ${\displaystyle {}\vee }$ and ${\displaystyle {}\wedge }$ respectively.
2. In mathematics, definitions are usually presented as such and get a number so that it is easy to refer to them. The definition contains the description of a situation where only concepts are used which have been defined before. In this situation, a new concept together with a name for it is introduced. This name is printed in a certain font, typically in italic. The new concept can be formulated without the new name, the new name is an abbreviation for the new concept. Quite often, the concepts depend on parameters, like the product set depends on its component sets. The names are often chosen arbitrarily, the meaning of the word within the mathematical context can be understood only via the explicit definition and not via its meaning in everyday life.
3. Hilbert has said that the art of denotation in mathematics is to use the same symbol for different things.
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# Problems on LCM and HCF Part 1
## Problems on LCM and HCF - 1
1. Factors and Multiples:
2. If number a divided another number b exactly, we say that a is a factor of b.
In this case, b is called a multiple of a.
3. Highest Common Factor (H.C.F.) or Greatest Common Measure (G.C.M.) or Greatest Common Divisor (G.C.D.):
The H.C.F. of two or more than two numbers is the greatest number that divides each of them exactly.
There are two methods of finding the H.C.F. of a given set of numbers:
1. Factorization Method: Express the each one of the given numbers as the product of prime factors. The product of least powers of common prime factors gives H.C.F.
2. Division Method: Suppose we have to find the H.C.F. of two given numbers, divide the larger by the smaller one. Now, divide the divisor by the remainder. Repeat the process of dividing the preceding number by the remainder last obtained till zero is obtained as remainder. The last divisor is required H.C.F.
Finding the H.C.F. of more than two numbers: Suppose we have to find the H.C.F. of three numbers, then, H.C.F. of [(H.C.F. of any two) and (the third number)] gives the H.C.F. of three given number.
Similarly, the H.C.F. of more than three numbers may be obtained.
4. Least Common Multiple (L.C.M.):
The least number which is exactly divisible by each one of the given numbers is called their L.C.M.
There are two methods of finding the L.C.M. of a given set of numbers:
1. Factorization Method: Resolve each one of the given numbers into a product of prime factors. Then, L.C.M. is the product of highest powers of all the factors.
2. Division Method (short-cut): Arrange the given numbers in a rwo in any order. Divide by a number which divided exactly at least two of the given numbers and carry forward the numbers which are not divisible. Repeat the above process till no two of the numbers are divisible by the same number except 1. The product of the divisors and the undivided numbers is the required L.C.M. of the given numbers.
5. Product of two numbers = Product of their H.C.F. and L.C.M.
6. Co-primes: Two numbers are said to be co-primes if their H.C.F. is 1.
7. H.C.F. and L.C.M. of Fractions:
1. H.C.F. = H.C.F. of Numerators L.C.M. of Denominators
2. L.C.M. = L.C.M. of Numerators H.C.F. of Denominators
8. H.C.F. and L.C.M. of Decimal Fractions:
In a given numbers, make the same number of decimal places by annexing zeros in some numbers, if necessary. Considering these numbers without decimal point, find H.C.F. or L.C.M. as the case may be. Now, in the result, mark off as many decimal places as are there in each of the given numbers.
9. Comparison of Fractions:
Find the L.C.M. of the denominators of the given fractions. Convert each of the fractions into an equivalent fraction with L.C.M as the denominator, by multiplying both the numerator and denominator by the same number. The resultant fraction with the greatest numerator is the greatest.
1.
Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.
A. 4 B. 7 C. 9 D. 13
Explanation:
Required number = H.C.F. of (91 - 43), (183 - 91) and (183 - 43)
= H.C.F. of 48, 92 and 140 = 4.
2.
The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is:
A. 276 B. 299 C. 322 D. 345
Explanation:
Clearly, the numbers are (23 x 13) and (23 x 14).
Larger number = (23 x 14) = 322.
3.
Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ?
A. 4 B. 10 C. 15 D. 16
Explanation:
L.C.M. of 2, 4, 6, 8, 10, 12 is 120.
So, the bells will toll together after every 120 seconds(2 minutes).
In 30 minutes, they will toll together 30 + 1 = 16 times. 2
4.
Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:
A. 4 B. 5 C. 6 D. 8
Explanation:
N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)
= H.C.F. of 3360, 2240 and 5600 = 1120.
Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4
5.
The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:
A. 9000 B. 9400 C. 9600 D. 9800
Explanation:
Greatest number of 4-digits is 9999.
L.C.M. of 15, 25, 40 and 75 is 600.
On dividing 9999 by 600, the remainder is 399.
Required number (9999 - 399) = 9600.
6.
The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:
A. 101 B. 107 C. 111 D. 185
Explanation:
Let the numbers be 37a and 37b.
Then, 37a x 37b = 4107
ab = 3.
Now, co-primes with product 3 are (1, 3).
So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111).
Greater number = 111.
7.
Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is:
A. 40 B. 80 C. 120 D. 200
Explanation:
Let the numbers be 3x, 4x and 5x.
Then, their L.C.M. = 60x.
So, 60x = 2400 or x = 40.
The numbers are (3 x 40), (4 x 40) and (5 x 40).
Hence, required H.C.F. = 40.
8.
The G.C.D. of 1.08, 0.36 and 0.9 is:
A. 0.03 B. 0.9 C. 0.18 D. 0.108
Explanation:
Given numbers are 1.08, 0.36 and 0.90. H.C.F. of 108, 36 and 90 is 18,
H.C.F. of given numbers = 0.18.
9.
The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:
A. 1 B. 2 C. 3 D. 4
Explanation:
Let the numbers 13a and 13b.
Then, 13a x 13b = 2028
ab = 12.
Now, the co-primes with product 12 are (1, 12) and (3, 4).
[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]
So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).
Clearly, there are 2 such pairs.
10.
The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:
A. 74 B. 94 C. 184 D. 364
Explanation:
L.C.M. of 6, 9, 15 and 18 is 90.
Let required number be 90k + 4, which is multiple of 7.
Least value of k for which (90k + 4) is divisible by 7 is k = 4.
Required number = (90 x 4) + 4 = 364.
11.
Find the lowest common multiple of 24, 36 and 40.
A. 120 B. 240 C. 360 D. 480
Explanation:
``` 2 | 24 - 36 - 40
--------------------
2 | 12 - 18 - 20
--------------------
2 | 6 - 9 - 10
-------------------
3 | 3 - 9 - 5
-------------------
| 1 - 3 - 5
L.C.M. = 2 x 2 x 2 x 3 x 3 x 5 = 360.
```
12.
The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:
A. 3 B. 13 C. 23 D. 33
Explanation:
L.C.M. of 5, 6, 4 and 3 = 60.
On dividing 2497 by 60, the remainder is 37.
Number to be added = (60 - 37) = 23.
13.
Reduce 128352 to its lowest terms. 238368
A.
3 4
B.
5 13
C.
7 13
D.
9 13
Explanation:
``` 128352) 238368 ( 1
128352
---------------
110016 ) 128352 ( 1
110016
------------------
18336 ) 110016 ( 6
110016
-------
x
-------
So, H.C.F. of 128352 and 238368 = 18336.
128352 128352 ÷ 18336 7
Therefore, ------ = -------------- = --
238368 238368 ÷ 18336 13
```
14.
The least number which when divided by 5, 6 , 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is:
A. 1677 B. 1683 C. 2523 D. 3363
Explanation:
L.C.M. of 5, 6, 7, 8 = 840.
Required number is of the form 840k + 3
Least value of k for which (840k + 3) is divisible by 9 is k = 2.
Required number = (840 x 2 + 3) = 1683.
15.
A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and c in 198 seconds, all starting at the same point. After what time will they again at the starting point ?
A. 26 minutes and 18 seconds B. 42 minutes and 36 seconds C. 45 minutes D. 46 minutes and 12 seconds
Explanation:
L.C.M. of 252, 308 and 198 = 2772.
So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.
16.
The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one of the numbers is 275, then the other is:
A. 279 B. 283 C. 308 D. 318
Explanation:
Other number = 11 x 7700 = 308. 275
17.
What will be the least number which when doubled will be exactly divisible by 12, 18, 21 and 30 ?
A. 196 B. 630 C. 1260 D. 2520
Explanation:
``` L.C.M. of 12, 18, 21 30 2 | 12 - 18 - 21 - 30
----------------------------
= 2 x 3 x 2 x 3 x 7 x 5 = 1260. 3 | 6 - 9 - 21 - 15
----------------------------
Required number = (1260 ÷ 2) | 2 - 3 - 7 - 5
= 630.
```
18.
The ratio of two numbers is 3 : 4 and their H.C.F. is 4. Their L.C.M. is:
A. 12 B. 16 C. 24 D. 48
Explanation:
Let the numbers be 3x and 4x. Then, their H.C.F. = x. So, x = 4.
So, the numbers 12 and 16.
L.C.M. of 12 and 16 = 48.
19.
The smallest number which when diminished by 7, is divisible 12, 16, 18, 21 and 28 is:
A. 1008 B. 1015 C. 1022 D. 1032
Explanation:
Required number = (L.C.M. of 12,16, 18, 21, 28) + 7
= 1008 + 7
= 1015
20.
252 can be expressed as a product of primes as:
A. 2 x 2 x 3 x 3 x 7 B. 2 x 2 x 2 x 3 x 7 C. 3 x 3 x 3 x 3 x 7 D. 2 x 3 x 3 x 3 x 7 |
Home » 6th Roots » 6th Root of 64
# 6th Root of 64
The 6th root of 64 is the number, which multiplied by itself 6 times, is 64. In other words, this number to the power of 6 equals 64.
Besides the real values of
along with an explanation, on this page you can also find what the elements of the 6th root of 64 are called.
In addition to the terminology, we have a calculator you don’t want to miss:
## Calculator
±
Reset
⁶√64 = ±2
If you have been looking for the 6th root of sixty-four, then you are right here, too.
## Sixth Root of 64
In this section we provide you with important additional information about the topic of this post:
The term can be written as ⁶√64 or 64^1/6.
As the index 6 is even and 64 is greater than 0, 64 has two real 4th roots:
⁶√64, which is positive and called principal 6th root of 64, and -⁶√64 which is negative.
Together, they are denominated as
Although the principal sixth root of sixty-four is only one of the two 6th roots, the term “6th root of 64” usually refers to the positive number, that is the principal square root.
If you want to know how to find the value of this root, then read our article 6th Root located in the header menu.
There, we also discuss the properties for index n = 6 by means of examples: multiplication, division, exponentiation etc.
Next, we have a look at the inverse function.
### Inverse of 6th Root of 64
Extracting the 6th root is the inverse operation of ^6:
In the following paragraph, we are going to name the elements of this √.
## What is the 6th Root of 64?
You already have the answer to that question, and you also know about the inverse operation of 64 6th root.
Keep reading to learn what the parts are called.
• ⁶√64 is the 6th root of 64 symbol
• 6 is the index
• 6th root = ±2
6th root of 64 = ±2
As a sidenote: All values on this page have been rounded to ten decimal places.
Now you really know all about ⁶√64, including its values, parts and the inverse.
If you need to extract the 6th root of any other real number use our calculator above.
Simply insert the number of which you want to find the 6th root (e.g. 64); the calculation is done automatically.
If you like our information about ⁶√64, then a similar 6th root you may be interested in is, for example: 6th root of 66.
In the following table you can find the n-th root of 64 for n = 2,3,4,5,6,7,8,9,10.
## Table
The aim of this table is to provide you with an overview of the nth roots of 64.
264Square Root of 64²√64±8
364Cube Root of 64³√644
464Forth Root of 64⁴√64±2.8284271247
564Fifth Root of 64⁵√642.29739671
664Sixth Root of 64⁶√64±2
764Seventh Root of 64⁷√641.8114473285
864Eight Root of 64⁸√64±1.6817928305
964Nineth Root of 64⁹√641.587401052
1064Tenth Root of 64¹⁰√64±1.5157165665
A few lines down from here we review the FAQs.
## Sixth Root of Sixty-Four
If you have been searching for what’s the sixth root of sixty-four or 6-th root of 64, then you are reading the right post as well.
The same is true if you typed 6 root of 64 or 64 6 root in the search engine of your preference, just to name a few similar terms.
Right below you can find the frequently asked questions in the context.
## FAQs About the 6th Root of 64
Click on the question which is of interest to you to see the collapsible content answer.
### How Many Real Sixth Roots Does 64 Have?
64 has two real sixth roots, because the radicand 64 is positive and the index 6 is even.
### What to the 6th Power Equals 64?
The 6th root of 64 to the power of 6 equals 64.
### How Do You Find the Sixth Root of 64?
Start with an initial guess such that 6 times that value equals approximately 64, then keep improving the guess until you have the required precision.
### What is 64 to the 6th Root?
64 to the 6th root = 64^1/64 = ±2.
### What Number is the 6th Root of 64?
The 6th root of 64 = ±2.
### How Do You Solve the 6-th Root of 64?
To compute the 6-th root of 64 use a calculator, and/or employ the Newton–Raphson method.
### What is the Value of 6 Root 64?
The value of 6 root 64 is ±2.
If something remains unclear do not hesitate getting in touch with us.
We are constantly trying to improve our site, and truly appreciate your feedback.
Ahead is the wrap-up of our information.
## Summary
To sum up, the sixth roots of 64 are ±2.
Finding the 6th root of the number 64 is the inverse operation of rising the ⁶√64 to the power of 6. That is to say, (±2)6 = 64.
Further information related to ⁶√64 can be found on our page 6th Root.
Note that you can also locate roots like ⁶√64 by means of the search form in the menu and in the sidebar of this site.
If our article about the sixth √ 64 has been useful to you , then press some of the share buttons located at the bottom of this page.
If you have any questions about the 6-th root of 64, fill in the comment form below.
Websites which are related to this one can be found in the “recommended sites” section in the sidebar.
Last, but not least, don’t forget to install our absolutely free PWA app (see menu or sidebar), or to bookmark us.
And come back soon! |
# Into Math Grade 2 Module 2 Lesson 2 Answer Key Write Equations to Represent Even Numbers
We included HMH Into Math Grade 2 Answer Key PDF Module 2 Lesson 2 Write Equations to Represent Even Numbers to make students experts in learning maths.
## HMH Into Math Grade 2 Module 2 Lesson 2 Answer Key Write Equations to Represent Even Numbers
I Can write an equation to model an even number as the sum of two equal addends.
How can you show Sofia’s toy cars in two equal groups?
Explanation:
4 + 4 = 8
There are 8 toy cars with sofia.
Have children work in groups. Read the following: Sofia has an even number of toy cars. How can you show her toy cars in two equal groups? Complete the equation to show the equal groups and the number of toy cars Sofia has.
Build Understanding
Von finds 14 shells at the beach. He sorts the shells into two equal groups. How many shells are in each group?
A. How can you make a concrete model to show Von’ s shells as two equal groups?
B. An even number of objects can be shown as two equal groups. Is 14 an even number?
Circle Yes or No.
C. How can you write an addition equation to show 14 osa sum of two equal addends?
___ + ___ = ____
D. How many šhells are in each group?
____ shells
Explanation:
7 + 7 = 14
There are 7 shells in each group.
Turn and Talk How did you show Von’s shells as two equal groups? Explain.
Step It Out
1. Mr. Lee plants 12 seeds in a garden. He sorts the seeds into two equal groups. How many seeds are in each group?
A. Show 12 as two equal groups.
B. Write an addition equation to show 12 as a sum of two equal addends.
___ = ___ + ____
C. Solve.
There are ____ seeds in each group.
Explanation:
6 + 6 = 12
There are 6 seeds in each group.
Check Understanding Math Board
Question 1.
Monica and Billy have 8 puzzles in all. They each have the same number of puzzles. How many puzzles do Monica and Billy each have? Write an addition equation to show the equal groups. Solve.
___ = ____ + ___ ____ puzzles
4 + 4 = 8 puzzles
Explanation:
Monica and Billy have 8 puzzles in all
They each have the same number of puzzles
8 = 4 + 4.
Question 2.
There are three different kinds of bluebirds in the United States. But all bluebird eggs are light blue in color. Noah and Elisa see 10 bluebirds in all. They each see the same number of bluebirds. How many bluebirds do Noah and Elisa each see? Write an addition equation to show the equal groups. Solve.
____ = ___ + ____ ____ bluebirds
10 = 5 + 5
Explanation:
There are three different kinds of bluebirds in the United States. But all bluebird eggs are light blue in color.
Noah and Elisa see 10 bluebirds in all. They each see the same number of bluebirds
10 = 5 + 5.
Model with Mathematics Write an addition equation to show the number as the sum of two equal addends.
Question 3.
6
___ = ____ + ____
6 = 3 + 3
Explanation:
Sum of 3 and 3 is 6.
Question 4.
18
____ = ___ + ____
18 = 9 + 9
Explanation:
The sum of 9 and 9 is 18.
Question 5.
Open Ended Can you show the number 5 as a sum of two equal addends? Explain. |
## Intermediate Algebra (12th Edition)
$(2x-3)$
$\bf{\text{Solution Outline:}}$ To factor the given expression, $4x^2-4x-3 ,$ find two numbers whose product is $ac$ and whose sum is $b$ in the quadratic expression $ax^2+bx+c.$ Use these $2$ numbers to decompose the middle term of the given quadratic expression and then use factoring by grouping. $\bf{\text{Solution Details:}}$ In the given expression the value of $ac$ is $4(-3)=-12$ and the value of $b$ is $-4 .$ The possible pairs of integers whose product is $ac$ are \begin{array}{l}\require{cancel} \{1,-12\}, \{2,-6\}, \{3,-4\}, \{-1,12\}, \{-2,6\}, \{-3,4\} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ 2,-6 \}.$ Using these $2$ numbers to decompose the middle term of the given expression results to \begin{array}{l}\require{cancel} 4x^2+2x-6x-3 .\end{array} Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (4x^2+2x)-(6x+3) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 2x(2x+1)-3(2x+1) .\end{array} Factoring the $GCF= (2x+1)$ of the entire expression above results to \begin{array}{l}\require{cancel} (2x+1)(2x-3) .\end{array} The missing factor of the given expression is $(2x-3) .$ |
# Question Video: Simplifying Algebraic Expressions Using Laws of Exponents Mathematics
Simplify 𝑥 × 𝑥 × 𝑥 × 𝑥 × 𝑥 × 𝑥 × 𝑥.
02:17
### Video Transcript
Simplify 𝑥 times 𝑥 times 𝑥 times 𝑥 times 𝑥 times 𝑥 times 𝑥.
In this question, we are asked to simplify a product involving seven factors of 𝑥. There are a few different ways of simplifying this expression. And we will go through two of these.
The first way that we can simplify this expression is to recall how we define raising a number to a positive integer exponent. We can recall that if 𝑛 is a positive integer, then 𝑥 raised to the power of 𝑛 is the same as multiplying 𝑛 lots of 𝑥. In our expression, we can see that we have the product of seven lots of 𝑥. Therefore, this product must be equal to 𝑥 raised to the seventh power.
This is not the only way that we can simplify this product. We can note that 𝑥 is a monomial. So this expression is the product of monomials. We can then recall that we can simplify the product of monomials by using the product rule for exponents, which tells us that if 𝑚 and 𝑛 are nonnegative integers, then 𝑥 raised to the power of 𝑚 times 𝑥 raised to the power of 𝑛 is equal to 𝑥 raised to the power of 𝑚 plus 𝑛.
This result holds true for any number of factors in the product. And we can also recall that raising a number to the first power leaves it unchanged. So 𝑥 is equal to 𝑥 raised to the first power.
Therefore, we can apply the product rule for exponents to simplify this product. We obtain 𝑥 raised to the power of one plus one plus one plus one plus one plus one plus one. We can then evaluate the sum in the exponent to once again obtain an answer of 𝑥 raised to the seventh power. |
# CHSE Odisha Class 12 Math Notes Chapter 7 Continuity and Differentiability
Odisha State Board CHSE Odisha Class 12 Math Notes Chapter 7 Continuity and Differentiability will enable students to study smartly.
## CHSE Odisha 12th Class Math Notes Chapter 7 Continuity and Differentiability
Definition:
A function f is said to be continuous at a point a Df if
(i) f(x) has definite value f(a) at x = a,
(ii) limx->a f(x) exists,
(iii) limx->a f(x) = f(a).
If one or more of the above conditions fail, the function f is said to be discontinuous at x = a. The above definition of continuity of a function at a point can also be formulated as follows:
A function f is said to be continuous at x = a if
(i) holds and for a given ∈ > 0, there exists a δ > 0 depending on ∈ such that
|x – a| < 8 ⇒ |f(x) – f(a)| < ∈.
A function f is continuous on an interval if it is continuous at every point of the interval.
If the interval is a closed interval [a, b] the function f is continuous on [a, b] if it is continuous on (a, b),
limx->a+ f(x) = f(a) and limx->b- f(x) = f(b).
Differentiation of a function:
(a) Differential coefficient (or derivative) of a function y = f(x) with respect to x is
Fundamental theorems:
Then to get $$\frac{d y}{d x}$$ it is convenient to take log of both sides before differentiation.
Derivative of some functions:
Higher order derivative:
Leibnitz Theorem:
If ‘u’ and ‘v’ are differentiable functions having ‘n’th derivative then
$$\frac{d^n}{d x^n}$$(u.v) = C0unv + C1un-1v1 + C2un-2v2 + ….. + Cnuvn
Partial derivatives and Homogeneous functions:
(a) If z = f(x, y) is any function of two variables then the partial derivative of z w.r.t. x and y are given below.
(b) Homogeneous function:
z = f(x > y) is a homogeneous function of degree ‘n’ if f(tx, ty) = tn f(x, y).
Euler’s Theorem:
If z = f(x, y) is a homogeneous function of degree ‘n’ then $$x \frac{\partial f}{\partial x}+y \frac{\partial f}{\partial y}$$ = nf(x, y). |
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### Class Presentation Math 1
1. 1. Seventh Grade Math<br />Math 2<br />1<br />
2. 2. By: Claudia<br />Subtracting Integers <br />Subtracting integers:<br />2 – (-8) <br /> Rule<br /> When we subtract we ADD THE OPPISITE! (never change the first number!)<br />2 + 8 is the same as<br /> 2 – (-8)…..<br />So, the answer is the same, 10 <br />-8 + 2 = -6 <br />The answer is positive because you always use the sign of the number with the highest absolute value. -8 is farther away from zero then 2 is. So -8 has the highest absolute value. <br />Vocab:<br />Integers – the set of whole numbers and their opposites. <br />Absolute Value – the distance the number is from zero on the number line. <br />2<br />
3. 3. How to Multiply Integers<br />When multiplying integers with the same sign the product is always positive.<br />When multiplying integers with different signs the answer will always be negative.<br />If any of the integers is zero the result is always zero.<br />3<br />
4. 4. Ex: 5+3=8 Add the numbers as if they were positive, then add the sign of the numbers.<br />Adding Integers having the same sign= Ex: -5+(-3)= -8<br />2. Adding two integers having signs:<br />Ex: -5+3 Take the difference of the numbers as if they were positive, then add the sign of the number having absolute value.<br />Ex: -5+3=-2<br />Subtracting Integers<br />Adding Integers<br /><ul><li>Ex: -5-(-3)</li></ul> =-5+3 When we subtract, we ADD THE OPPOSITE<br />Rule: 2-1 then turns into 2+(-1) It helps us to get the right answer, too, and less confusing.<br />4<br />
5. 5. Multiplying and Dividing Integers<br />Dividing Integers<br /> <br />If a pair of integers has the same sign, then the answer will have a positive sign. You must calculate the absolute value of each integer and then divide the first integer by the second integer.<br /> <br />Example: -10 / -2 = ?<br /> <br /> Step 1: |-10| / |-2| = 10 / 2 <br /> Step 2: 10 / 2 = 5 <br /> Step 3: Since integers have same sign, answer is <br /> positive: +5<br /> <br /> <br />If a pair of integers have different signs, then the answer will be negative. You must calculate the absolute value of each integer and then divide the first integer by the second integer.<br /> <br />Example: -10 / +2 = ?<br /> <br /> Step 1: |-10| / |+2| = 10 / 2 <br /> Step 2: 10 / 2 = 5<br />Step 3: Since integers have different signs, answer is negative: -5 <br />Multiplying Integers<br /> <br /> <br />When multiplying two integers having the same sign, the product is always positive<br /> <br /> Example 1: -2 · (-5) = 10<br /> Example 2: 2 · 5 = 10<br /> <br />When having two integers with different signs, the product is always negative<br /> <br /> Example 1: -2 · 5 = (-2)+(-2)+(-2)+(-2)+(-2) = -10<br /> Example 2: 2 · (-5) = (-5)+(-5) = -10<br /> <br />When multiplying more than two integers<br /> <br /> Example 1: (-1) · (-2) · (-3) = ?<br />Step 1: group the first two numbers and use rules I and II above to calculate the intermediate step<br /> (-1) · (-2) = +2 (used rule I)<br />Step 2: use result from intermediate step 1 and multiply by the third number.<br />2 · (-3) = -6 (used rule II)<br /> <br />5<br />
6. 6. AddingandSubtractingIntegers<br />When adding Integers with the same sign add them as if<br /> they were positive then add the sign. <br />Example:<br />6 + 3 = 9 -6 + (-3) = -9<br />When adding integers with different signs, subtract them as if they were<br /> positive and add the sign <br />of the number with greatest absolute value. <br />Example:<br />-6 + 3 = -3 <br />When subtracting any integer you add the opposite.<br />Example:<br />-6 – 3 = -9 Change to<br />-6 + (-3) = -9<br />Hannah<br />6<br />
7. 7. Subtracting Integers<br />Ex<br />126-(-176)<br /> 126+176=302 or<br />126-176<br />126+(-176)=(-50) or<br />-126-(-176)<br />-126+176=50<br />(note)<br />when you add integers remember that when you add integers with the same sign the answer is going to be the same as the sign, but if the absolute value of the negative number ishigher than the positive thanthenumbers going to be anegative.<br />Convert the problem to addition. Ex. 12-(-36) to 12+36. remember to change the last number of the sequence from negative to positive or positive negative.<br />Add or subtract the problem like a regular math problem. Ex. 12+36=48.<br />7<br />
8. 8. Solving Equations<br />When the number in the equation is positive you add the opposite to the number. Then you add the opposite to the answer. That way, the variable is alone on the left side of the equation in this example, and the difference of the answer and the opposite number is on the other.<br />When the number in the equation is negative then you convert the number to a positive. Then you change the operation to its opposite. After that, you add the opposite to the number. Then you add the opposite to the answer. That way, the variable is alone on the one side of the equal sign, and the difference of the answer and the opposite number is on the other.<br />Example<br />X+13=26<br />X+13+(-13)=26+(-13)<br />26+(-13)=13<br />13+(-13)=0<br />X=13<br />13+13=26<br />Example<br />X-(-13)=13<br />X+13+(-13)=26+(-13)<br />26+(-13)=13<br />13+(-13)=0<br />X=13<br />13-(-13)=26<br />8<br />
9. 9. Distributive Property For Algebra<br /> Take both numbers in the parentheses and multiply them separately to the number outside of the parentheses, still using the sign in between both numbers in parentheses.<br />Ex. 1 : 5(Y+9) turns into 5y+59 = 5Y+45<br />Ex. 2 : 5(-Y+9) turns into 5(-Y)+59 = 5(- Y)+45<br />Ex. 3: -5(Y-9) turnsinto -5Y-(-59) = -5Y - (-45)<br />9<br />
10. 10. How to Solve Equations<br />Created By:<br />Jonah<br />Step 1<br />A legal move (you have to do the same thing to both sides) is very simple.<br />Step 2<br />Step 3<br />What you are trying to do here is; you want to get the variable alone. All you have to do is add the opposite to the constant <br />Once the constant is gone, you add the same number you added to the sum, then whatever you get from that equation, is what the variable equals<br />Example:<br />X + 5 = 12<br />X + 5 + (-5) = 12 + (-5)<br />Example:<br />X + 5 + (-5) = 12 + (-5)<br /> X = 7<br />10<br />
11. 11. Subtracting Integers<br />By: Cameron<br /><ul><li>When subtracting integers you “add the opposite”.
12. 12. Example: 12-8=4</li></ul> 12+(-8)=4<br /><ul><li> Rule</li></ul>When Subtracting Integers you add the opposite.<br />Example: 10-(-4)=14 10+4=14<br /> Do you want to know how this works~ click to find out.<br />11<br />
13. 13. Just draw a number line if it helps you more.<br />Also when you have a subtraction sign next to a parenthesis.<br />You change the sign to addition and the negative number to<br />a positive. <br />Example2: -10-(4)=6<br />-10+-4=14<br />Example: -10-(-4)=14<br />10+4=14<br />12<br />
14. 14. HOW TO COMBINE LIKE TERMSA.K.A. SIMPLIFYING ALGEBRAIC EXPRESSIONS <br />Congrats you can now simplify algebraic expressions!!!!!!!!<br /><ul><li>Terms --- The algebraic expression separated between each plus or minus sign</li></ul>Ex. 3x, y, 2x, 7<br /><ul><li> Like terms --- Terms that conduct the same variables</li></ul>Ex. 3x & 2x<br /><ul><li> Coefficients --- The numbers that are involved with a variable </li></ul>Ex. 3, 2, 1<br /><ul><li> Constants --- Terms without an variable </li></ul>Ex. 7<br />Step 1<br />Step 3<br />Hint<br />The only like terms are 3x & 2x<br /><ul><li>You begin with an Algebraic Expression to simplify
15. 15. Begin simplifying </li></ul>3x+y+2x+7=?<br />Step 2<br />3x+2x=5x<br />Final answer<br />5x+y+7<br />The Surprise Expression<br />and find…………………..<br /><ul><li>Before we simplify, find the terms, like terms, coefficients, and constants.</li></ul>13<br />
16. 16. Solving an algebraic equation!!!!<br /> You solve an algebraic equation by doing different sets of legal moves. You do a legal move by adding or subtracting and in some cases multiplication and dividing what you do to one side to the other until you cant do anymore moves. <br />example: 3+4+-4=Y+4+-4+3<br />By Lennon Dresnin<br />
17. 17. (-2)2 ≠ -22<br />Exponents<br />Exponents<br />This is where your journey into Exponents begins<br />By Loghan<br />In (-2)2 You can tell (-2) is the base because it is in parenthesis, In -22 there are no parenthesis. Because of that 2 is the base not -2. Another way of doing the problem would be 0-22. In 0-22 you would start off by doing 2 to the second power, which is 2 times 2. The answer would be 4. But after you do that the equation would be 0-4. 0-4=(-4). The answer is different than (-2)2. <br />In (-2)2 you can tell the base is (-2) because there are parenthesis. There is no chance of it being -2 or minus two. Now all we need to find out is what negative 2 multiplied by itself is. <br />When you multiply a negative by a negative what do you get? A positive! So when you multiply -1 by -1 you get 1! Positive 1. So when you multiply -2 by -2<br />[In other words (-2)2] <br />you get… <br />By Loghan<br />I hope you enjoyed your journey into Exponents<br /> 22=(-2)2<br />22=-22<br />So this will sum everything up:<br />(-2)2 = 4<br />Does it work?<br />It Works<br />OH NO IT DOESN’T WORK<br />-22 = -4<br /> |
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## Algebra (all content)
### Unit 2: Lesson 8
Linear equations with parentheses
# Multi-step equations review
To solve an equation we find the value of the variable that makes the equation true. For more complicated, fancier equations, this process can take several steps.
When solving an equation, our goal is to find the value of the variable that makes the equation true.
### Example 1: Two-step equation
Solve for x.
3, x, plus, 7, equals, 13
We need to manipulate the equation to get x by itself.
\begin{aligned} 3x+7&=13 \\\\ 3x+7\redD{-7}&=13\redD{-7} \\\\ 3x&=6 \\\\ \dfrac{3x}{\redD{3}}&=\dfrac{6}{\redD{3}} \\\\ x&=2 \end{aligned}
We call this a two-step equation because it took two steps to solve. The first step was to subtract 7 from both sides, and the second step was to divide both sides by 3. Want an explanation of why we do the same thing to both sides of the equation? Check out this video.
We check the solution by plugging start color #e84d39, 2, end color #e84d39 back into the original equation:
\begin{aligned} 3x+7&=13 \\\\ 3\cdot \redD 2 + 7 &\stackrel?= 13 \\\\ 6+7 &\stackrel?= 13 \\\\ 13 &= 13 ~~~~~~~\text{Yes!} \end{aligned}
### Example 2: Variables on both sides
Solve for a.
5, plus, 14, a, equals, 9, a, minus, 5
We need to manipulate the equation to get a by itself.
\begin{aligned} 5 + 14a &= 9a - 5 \\\\ 5 + 14a \blueD{- 9a} &= 9a - 5 \blueD{- 9a} \\\\ 5 + 5a &= -5 \\\\ 5 + 5a \blueD{-5} &= -5 \blueD{- 5}\\\\ 5a &= -10\\\\ \dfrac{5a}{\blueD5} &= \dfrac{-10}{\blueD5} \\\\ a &= \blueD{-2} \end{aligned}
The answer:
a, equals, start color #11accd, minus, 2, end color #11accd
Check our work:
\begin{aligned} 5 + 14a &= 9a - 5 \\\\ 5 + 14(\blueD{-2}) &\stackrel?= 9(\blueD{-2}) - 5 \\\\ 5 + (-28) &\stackrel?= -18 - 5 \\\\ -23 &= -23 ~~~~~~~\text{Yes!} \end{aligned}
Want to learn more about solving equations with variables on both sides? Check out this video.
### Example 3: Distributive property
Solve for e.
7, left parenthesis, 2, e, minus, 1, right parenthesis, minus, 11, equals, 6, plus, 6, e
We need to manipulate the equation to get e by itself.
\begin{aligned} 7(2e-1)-11 &= 6+6e \\\\ 14e-7 -11&= 6+6e\\\\ 14e-18 &= 6+6e\\\\ 14e-18\purpleD{-6e} &= 6+6e\purpleD{-6e} \\\\ 8e-18&=6\\\\ 8e-18\purpleD{+18} &=6 \purpleD{+18} \\\\ 8e &=24\\\\ \dfrac{8e}{\purpleD{8}}&= \dfrac{24}{\purpleD{8}}\\\\ e &= \purpleD{3} \end{aligned}
The answer:
e, equals, start color #7854ab, 3, end color #7854ab
Check our work:
\begin{aligned} 7(2e-1)-11 &= 6+6e \\\\ 7(2(\purpleD{3})-1) -11&\stackrel?= 6+6(\purpleD{3}) \\\\ 7(6-1)-11 &\stackrel?= 6+18 \\\\ 7(5)-11&\stackrel?=24 \\\\ 35-11&\stackrel?=24 \\\\ 24 &=24 ~~~~~~~\text{Yes!} \end{aligned}
Want to learn more about solving equations with the distributive property? Check out this video.
## Practice
Problem 1
Solve for b.
4, b, plus, 5, equals, 1, plus, 5, b
b, equals
Want more practice? Check out these exercises:
## Want to join the conversation?
• how did you get 5/3b?
(47 votes)
• So as we can see the question starts off as:
2/3b + 5 = 20-b
first, add +b to both sides of the equation
(addition property of equality). Next:
We got 5/3 by adding fractions:
2/3+1/1.
Step 1: Multiply the denominators (x/3)
Step 2: Cross multiply the numerators and denominators (2x1 and 3x1)
Step 3: Add the two products together (2x1=2, 3x1=3 therefore, add 2+3). WITHOUT touching the denominator!
Step 4: 5/3b + 5 = 20. Subtract 5 from both sides of the equation to cancel out 5.
Step 5. divide 5/3 to 15. Keep change Flip
Keep the fraction change the division sign to multiplication and flip the second fraction (example 2/3 to 3/2). So, 5/3 to 3/5 and multiply both sides of the equation, lastly, your answer is 4.
(11 votes)
• What do I do if the variable is equal to 0? How do I check my answer?
(22 votes)
• To check your answer, just plug 0 in for the variable.
For example, let's say you solved this equation:
9x - 1 = 5x - 1
9x - 5x = 1 - 1
4x = 0
x = 0
It looks like the variable is equal to 0.
Now, double check your answer with the original equation by replacing all the x's with 0's:
9x - 1 = 5x - 1
9(0) - 1 = 5(0) - 1
0 - 1 = 0 - 1
-1 = -1
The equation is true, so 0 is valid.
Hope this helps!
(36 votes)
• These comments be getting answers years later. Like bro they already progressed into High School or College, they most likely know after then.
(22 votes)
• how do you solve an equation like 0.5(5-7x) = 8-(4x+6)
(2 votes)
• 1) Use distributive property to remove the parentheses. Distribute the 0.5 and the "-"
2) On the right side, you will have 2 terms that are like terms. Combine them.
3) Move all the "x" terms to the same side of the equation
4) Move the constants to the opposite side from the x's
5) Then divide by the coefficient of "x" (the number in front of x)
See if you can solve the equation. Comment back if you get stuck with what you have done so far.
(11 votes)
• This type of math is a little confusing because sometimes when u have a -3x you divide or subtract at least i think so i don't know why but my teacher said that and can some one explain it please?
(9 votes)
• You divide. you can think of -3x as -3 times x, so using the opposite of multiplication(division), you can get rid of it.
(2 votes)
• If I have one cookie and my friend eats my cookie is he really my friend?
(5 votes)
• It depends on a lot of circumstances; have you already eaten the rest of the bag and are offering him the last one?
If you really wanted the cookie, did you give it away, or did they just take it?
Why is there only one cookie?
I do not feel we have enough information to answer this question.
(5 votes)
• Why is peter so silly
(3 votes)
• Because he has a willy
(5 votes)
• This is still a little confusing for me
(3 votes)
• When all the numbers are negative the answer is negative always or the most times?
(2 votes)
• It depends on what you are doing.
If you are multiplying/dividing two negatives (or any even number of negatives), then the answer is positive. So if you have -2 x = -6, dividing by -2 gives x = -6/-2=3.
If you are adding two negatives, the answer would always be a negative.
(2 votes)
• Why do you multiply by 3/5 on both sides and not jut divide 15 by 5/3 in problem 2 ?
(2 votes)
• If you know how to divide fractions, they end up the same thing. You want to divide 15/(5/3), to divide fractions, you reciprocate the denominator and multiply, so you end up with 15*3/5. So if you have 5/3 x, it is easier to multiply by the reciprocal so that you do not have to go through the extra step of dividing by a fraction.
(1 vote) |
# What is a Vector?
••• jakkaje808/iStock/GettyImages
Print
A vector lets you describe quantities in terms of an amount (called the magnitude) and a direction, making them a handy mathematical tool. Treating quantities as vectors opens up many powerful ways of calculating and analyzing forces, motion and other phenomena where direction plays a role. Vectors are indispensable not only in math itself, but also in hard sciences such as physics, and disciplines such as engineering. Though the math can be complex, the basic ideas behind vectors are not hard to grasp.
#### TL;DR (Too Long; Didn't Read)
A vector is a quantify that has both an amount and a direction. Force and velocity are two examples of vector quantities.
## Scalars and Vectors
Mathematicians call simple quantities scalars; these include properties such as temperature, weight and height, where a single number tells you everything you need. A vector also has an amount, but adds a direction; for example, a plane flies north at a speed of 645 kilometers per hour (400 miles per hour). The amount is the speed, 645 kph, and the direction is north. Both of these pieces of information form the plane’s velocity vector. Similarly, to open a door, you push on it with a force of 50 newtons (11 pounds). Fifty newtons is the magnitude; the direction is “away from the front of your body.” This forms the vector for the pushing force on the door.
## Drawing Vectors
It helps to visualize vectors by drawing them as arrows. The arrow points in the direction of the vector, and has a length that represents the vector’s magnitude. You can combine several vectors in the drawing, each with its own direction and length. In addition, you can choose between Cartesian (x and y) or polar coordinates (magnitude and angle). If your drawing skills are up to it, you can also sketch vectors in three dimensions using perspective and depth.
## Math with Vectors
Just as you can do math with scalar quantities, you can add and subtract vectors as well as perform other operations on them. One approach to adding vectors is to simply add up their x and y coordinates. For example, if you have two vector arrows, one of which has its tail at the origin, (0, 0), and head at (5, 5), and the other which also has its tail at the origin and has its head at (3, 0). Adding the x coordinates gives you 8, and adding the y locations gives 5, so the resulting vector is (8, 5).
Other operations with vectors include the dot product and the cross product; these are functions done in linear algebra that take two vectors and produce a result. The dot product yields a scalar that combines the lengths of the two original vectors. It applies to problems such as finding the energy needed to push a heavy object up a ramp. The cross product yields a third vector that points 90 degrees from either of the first two; it has applications in the forces of electricity and magnetism.
## Physics, Engineering and Other Fields
It should come as no great surprise that you encounter vectors a great deal in physics and engineering. Vectors come in handy for solving problems involving quantities such as force, velocity and acceleration. Wind vectors help weather forecasters chart the progress of storms. These disciplines also make use of “vector fields,” or large groups of vectors spread out the represent phenomena such as the field lines around a magnet or the complex water currents in an ocean.
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Home / Calculus III / 3-Dimensional Space / Tangent, Normal and Binormal Vectors
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### Section 12.8 : Tangent, Normal and Binormal Vectors
1. Find the unit tangent vector for the vector function : $$\vec r\left( t \right) = \left\langle {{t^2} + 1,3 - t,{t^3}} \right\rangle$$
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Start Solution
From the notes in this section we know that to get the unit tangent vector all we need is the derivative of the vector function and its magnitude. Here are those quantities.
$\vec r'\left( t \right) = \left\langle {2t, - 1,3{t^2}} \right\rangle$ $\left\| {\vec r'\left( t \right)} \right\| = \sqrt {{{\left( {2t} \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( {3{t^2}} \right)}^2}} = \sqrt {1 + 4{t^2} + 9{t^4}}$ Show Step 2
The unit tangent vector for this vector function is then,
$\vec T\left( t \right) = \frac{1}{{\sqrt {1 + 4{t^2} + 9{t^4}} }}\left\langle {2t, - 1,3{t^2}} \right\rangle = \require{bbox} \bbox[2pt,border:1px solid black]{{\left\langle {\frac{{2t}}{{\sqrt {1 + 4{t^2} + 9{t^4}} }}, - \frac{1}{{\sqrt {1 + 4{t^2} + 9{t^4}} }},\frac{{3{t^2}}}{{\sqrt {1 + 4{t^2} + 9{t^4}} }}} \right\rangle }}$ |
# How do you express 8,400,000 in scientific notation?
Mar 9, 2017
$8.4 \times {10}^{6}$
#### Explanation:
Expressing a number in scientific notation means you will adjust the given number to result a more compact form of the number that may include rounding it off to match the precision of other numbers given.
Scientific notation is a compact way of writing numbers to reduce complicated computations using very large or very small numbers and to reduce errors.
In our example: $8 , 400 , 000 = 8.4 \times {10}^{6}$ in scientific notation.
To get this result we divided the given larger number repeatedly by 10, dropping a zero each time. While we did that, we added another $1$ to the exponent of the $10$ multiplier that is at the end of every number in scientific notation. It looks like this:
$8 , 400 , 000 = 8400000 \times {10}^{0}$ because we know ${10}^{0} = 1$
$8 , 400 , 000 = 840000 \times {10}^{1}$ drop one $0$ and exponent is $1$.
$8 , 400 , 000 = 84000 \times {10}^{2}$ drop two $0$ and exponent is $2$.
$8 , 400 , 000 = 8400 \times {10}^{3}$ drop three $0$ and exponent is $3$.
$8 , 400 , 000 = 840 \times {10}^{4}$ drop four $0$ and exponent is $4$.
$8 , 400 , 000 = 84 \times {10}^{5}$ drop five $0$ and exponent is $5$.
Having run out of zeros, we could stop here and not be numerically incorrect, but there is a definition of scientific notation called "normalized" that requires the first digit to reside between $0$ and $9$.
So to become normalized, we must divide our number again by $10$ so the first digit becomes $8$.
Remember to add to the exponent.
Then: $8 , 400 , 000 = 8.4 \times {10}^{6}$
If you understand that method the easy way follows:
Place a dot (decimal point) at the right hand side of the original number of $8400000$. Notice we removed the commas.
Now move the decimal point to the left counting each time you pass a digit $8.4 .0 .0 .0 .0 .0 .$
If you count each $m o v e$ correctly you should reach $6$ when you land between $8$ and $4$.
Then: $8 , 400 , 000 = 8.4 \times {10}^{6}$ |
This is the probability that the estimate procedure will generate an interval
# This is the probability that the estimate procedure
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This is the probability that the estimate procedure will generate an interval that does not contain The confidence coefficient (1- α) is the probability that it will contain (.95 for example) Confidence level = 100(1- α) % (aka 95%) 11
Confidence In order to figure out z-scores to use for the confidence intervals, we use the z-score at α/2 We take a confidence we want, find z by doing 1-α/2 12
Example For a 95% interval, α/2 = .05/2 = 0.025 so I look up 1-.025 = . 975 on the table to see what’s to the left of that. I get z= 1.96 For a 90% interval, α/2 = .1/2 = .05 so I look up 1-.05 = .95 on the table and see z = 1.645 What about 99%? 13
For all problems Figure out the z-score from the confidence level and use it in the formula: - z/ <= <= + z/ Margin of error = z/ For a given confidence level, the larger the population standard deviation, the wider the confidence level. 14
Example Let the sd of a population be 0.05 in the last example instead of 0.03 Compute the 95% confidence interval based on the sample information. Our mean was 1.02 and sample size was 25 - z/ <= <= + z/ 15
Example Let the sd of a population be 0.05 in the last example instead of 0.03 Compute the 95% confidence interval based on the sample information. Our mean was 1.02 and sample size was 25 - z/ <= <= + z/ 1.02 +/- 1.96(.05/ ) 1.02 +/- 0.02 = 1 to 1.04 The width is 04 here (.02 X 2) 16
Other Rules For a given confidence level and population standard deviation, the smaller sample n, the wider the confidence interval. For a given sample size n and population standard deviation, the greater the confidence level, the wider the confidence interval. 17
Example Compute the 99% confidence interval instead from the last example. sd =0.03, mean= 1.02, n=25 - z/ <= <= + z/ 18
Example Compute the 99% confidence interval instead from the last example. sd =0.03, mean= 1.02, n=25 - z/ <= <= + z/ 1.02 +/- 2.576 (/ = 1.02 +/- .015 = 1.005 to 1.035 19
Confidence interval for pop mean when sigma is unknown In order to use a confidence interval for a population mean, Xbar needs to be normally distributed. Another standardized statistic T is used which uses S as a replacement for sigma when the population standard deviation is not known. T = This is called a Student’s distribution, or a t distribution 20
T-distribution If a random sample of size n is taken from a normal distribution with a finite variance, then the statistic T = follows the T distribution with (n-1) degrees of freedom. The T distribution is actually a family of distributions which are similar to x in that they are bell shaped and symmetric around 0. However, T distributions all have slightly broader tails. The degrees of freedom define the broadness of the tails of the distribution. The fewer the “df” the broader the tails. They are also the number of independent pieces of information that go into the calculation of a given statistic and can be “freely chosen.” 21
T table Just like the normal table, there is a T-table on Table 2.
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## What is an example of point slope form?
Examples of Applying the Concept of Point-Slope Form of a Line. Example 1: Write the point-slope form of the line with a slope of 3 which passes through the point (2,5). The slope is given as m = 3 m = 3 m=3, and the point (2,5) has coordinates of x 1 = 2 {x_1} = 2 x1=2 and y 1 = 5 {y_1} = 5 y1=5.
## How do you find slope with points?
How to Use the Formula and Calculate Slope. The slope of a line characterizes the direction of a line. To find the slope, you divide the difference of the y-coordinates of 2 points on a line by the difference of the x-coordinates of those same 2 points .
## What are the 3 slope formulas?
There are three major forms of linear equations: point-slope form, standard form, and slope-intercept form.
## Which is the equation of a line that has a slope of 4 and passes through Point 1 6?
Which is the equation of a line that has a slope of 4 and passes through point (1, 6)? a.y = 4x – 2.
## What is standard form slope?
Standard form is another way to write slope-intercept form (as opposed to y=mx+b). It is written as Ax+By=C. You can also change slope-intercept form to standard form like this: Y=-3/2x+3. A, B, C are integers (positive or negative whole numbers) No fractions nor decimals in standard form.
## How do you find the slope intercept form of an equation?
To write an equation in slope-intercept form, given a graph of that equation, pick two points on the line and use them to find the slope. This is the value of m in the equation. Next, find the coordinates of the y-intercept–this should be of the form (0, b). The y- coordinate is the value of b in the equation.
## How do you find slope given two points?
There are three steps in calculating the slope of a straight line when you are not given its equation.Step One: Identify two points on the line.Step Two: Select one to be (x1, y1) and the other to be (x2, y2).Step Three: Use the slope equation to calculate slope.
You might be interested: Equation for time
## How do you solve for slope of a line?
Using the Slope EquationPick two points on the line and determine their coordinates.Determine the difference in y-coordinates of these two points (rise).Determine the difference in x-coordinates for these two points (run).Divide the difference in y-coordinates by the difference in x-coordinates (rise/run or slope).
## What is the Y intercept formula?
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.
## What is the easiest way to find slope?
To find the slope of two given points, you can use the point-slope formula of (y2 – y1) / (x2 – x1). With the points plugged in, the formula looks like (3 – 2) / (4 – 1). Simplify the formula to get a slope of ⅓.
### Releated
#### First order equation
What is first order difference equation? Definition A first-order difference equation is an equation. xt = f(t, xt−1), where f is a function of two variables. How do you solve first order equations? Here is a step-by-step method for solving them:Substitute y = uv, and. Factor the parts involving v.Put the v term equal to […]
#### Energy equation physics
What is the formula for energy? The formula that links energy and power is: Energy = Power x Time. The unit of energy is the joule, the unit of power is the watt, and the unit of time is the second. How do you solve for energy in physics? In classical mechanics, kinetic energy (KE) […] |
# Solving one-step equations and inequalities
In solving one-step equations and inequalities, the student should know all the rules of inequality. For this you can refer the previous page of ask-math or visit the link given below.
Rules for solving linear equations inequalities
## Steps involved while solving one-step equations and inequalities
1) Isolate the given variable. For this use the opposite operation rule.
(iii)Multiplication ------------> Division
(iv)Division ------------------> Multiplication.
2) If you are dividing or multiply both the sides by negative number then flip the sign.
(i)greater than (>) becomes less than (<)
(ii)less than (<) becomes greater than (>)
(iii)greater than and equal to ($\geq$) becomes less than and equal to ($\leq$)
(iv)less than and equal to ($\leq$) becomes greater than and equal to ($\geq$)
Examples on solving one-step equations and inequalities
Example 1 : x + 8 > 11
Solution :
x + 8 > 11
Since there is positive 8 so as to isolate x, we will add negative(-8) on both sides
x + 8 - 8 > 11 - 8
∴ x > 3 ( + 8 - 8 = 0)
Example 2 : x - 11 < 5
Solution :
x - 11 < 5
Since there is negative 11 so as to isolate x, we will add positive 11 (+11) on both sides
x - 11 + 11 < 5 + 11
∴ x < 16 ----- ( - 11 + 11 = 0)
Example 3 : x + 72 $\geq$ 65
Solution :
x + 72 $\geq$ 65
Since there is positive 72 so as to isolate x, we will add negative 72 (-72) on both sides
x + 72 - 72 $\geq$ 65 - 72
∴ x $\geq$ -7 ----- ( 72 - 72 = 0)
Example 4 : 5x $\leq$ 65
Solution :
5x $\leq$ 65
Since there is multiplication between 5 and x so to isolate x, we will divide both side by 5
$\frac{5x}{5}$ $\leq$ $\frac{65}{5}$
∴ x $\leq$ 13 ----- ( 5 ÷ 5 = 1)
Example 5 : $\frac{x}{8}$ $\geq$ 6
Solution :
$\frac{x}{8}$ $\geq$ 6
Since there is division between x and 8 so to isolate x, we will multiply both side by 8
$\frac{x}{8}\times 8$ $\geq$ 8 $\times$ 6
∴ x $\geq$ 48 ----- ( 8 ÷ 8 = 1)
Example 6 : -12x $\leq$ 48
Solution :
-12x $\leq$ 48
Since there is multiplication between negative 12 (-12) and x so to isolate x, we will divide both side by (-12)
$\frac{-12x}{-12}$ $\leq$ $\frac{48}{-12}$
According to inequality rule, we have to flip the inequality sign.
∴ x $\geq$ -4 ----( -12 ÷ -12 =+1 and 48 ÷ (-12)= -4 )
Example 7 : $\frac{x}{-5}$ $\geq$ - 11
Solution :
$\frac{x}{-5}$ $\geq$ - 11
Since there is division between x and (-5) so to isolate x, we will multiply both side by (-5)
$\frac{x}{-5}\times (-5)$ $\geq$ -11 $\times$ (-5)
According to inequality rule, we have to flip the inequality sign.
∴ x $\leq$ 55 ----- ( (-5) ÷ (-5) = 1 and (-11) X (-5) = +55 |
# Calculus problem solver with steps
This Calculus problem solver with steps helps to fast and easily solve any math problems. Our website can help me with math work.
## The Best Calculus problem solver with steps
Apps can be a great way to help students with their algebra. Let's try the best Calculus problem solver with steps. Composite functions can be used to model real-world situations. For example, if f(t) represents the temperature in degrees Celsius at time t, and g(t) represents the number of hours since midnight, then the composite function (fog)(t), which represents the temperature at a certain hour of the day, can be used to predict how the temperature will change over the course of 24 hours. To solve a composite function, it is important to understand the individual functions that make up the composite function and how they interact with each other. Once this is understood, solving a composite function is simply a matter of plugging in the appropriate values and performing the necessary calculations.
Algebra is the branch of mathematics that deals with the solution of equations. In an equation, the unknown quantity is represented by a letter, usually x. The object of algebra is to find the value of x that will make the equation true. For example, in the equation 2x + 3 = 7, the value of x that makes the equation true is 2. To solve an equation, one must first understand what each term in the equation represents. In the equation 2x + 3 = 7, the term 2x represents twice the value of x; in other words, it represents two times whatever number is assigned to x. The term 3 represents three units, nothing more and nothing less. The equal sign (=) means that what follows on the left-hand side of the sign is equal to what follows on the right-hand side. Therefore, in this equation, 2x + 3 is equal to 7. To solve for x, one must determine what value of x will make 2x + 3 equal to 7. In this case, the answer is 2; therefore, x = 2. |
Introduction to Puncture Points (holes)
# INTRODUCTION TO PUNCTURE POINTS (HOLES)
• PRACTICE (online exercises and printable worksheets)
Rational functions can exhibit puncture points (also called ‘holes’).
As you'll learn in Calculus, puncture points are an example of a ‘removable discontinuity’.
Here's the idea.
The function $\displaystyle R(x) := \frac{x^3-2x^2}{x-2}\,$ certainly looks like a typical rational function.
Upon closer inspection, though, we see that there's an extra factor of $\,1\,$ in the formula: $$\frac{x^3 - 2x^2}{x - 2} \quad = \quad \frac{x^2(x-2)}{x-2} \quad = \quad x^2\cdot\frac{x-2}{x-2}$$ Therefore, $\,\displaystyle R(x) = \frac{x^3-2x^2}{x-2}\,$ has exactly the same outputs as the much simpler function, $\,P(x) := x^2\,$,
except that the function $\,R\,$ it isn't defined when $\,x = 2\,$.
The graphs of both $\,P\,$ and $\,R\,$ are shown below—the puncture point (hole) in $\,R\,$ is caused by that extra factor of $\,1\,$.
$P(x) = x^2$ $\displaystyle R(x) = \frac{x^3 - 2x^2}{x-2} = x^2\cdot\frac{x-2}{x-2}$
Puncture points are studied in more detail in a future section.
Master the ideas from this section |
# Section 3.4 – Concavity and the Second Derivative Test
## Presentation on theme: "Section 3.4 – Concavity and the Second Derivative Test"— Presentation transcript:
Section 3.4 – Concavity and the Second Derivative Test
The Second Derivative and the Function
The first derivative tells us where a function is increasing or decreasing. But how can we tell the manner in which a function is increasing or decreasing? For example, if f '(x) = 3x2 +3 then f(x) is always increasing because f '(x) is always positive. But which graph below represents f(x)?
Concavity CONCAVE DOWN CONCAVE UP
If the graph of a function f lies above all of its tangents on an interval I, then it is said to be concave up (cupped upward) on I. If the graph of a function f lies below all of its tangents on an interval I, then it is said to be concave down (cupped downward) on I. CONCAVE DOWN CONCAVE UP
Test for Concavity Slopes are decreasing. CONCAVE UP
If f ''(x) > 0 for all x in I, then the graph of f is concave upward on I. If f ''(x) < 0 for all x in I, then the graph of f is concave downward on I. Slopes are decreasing. CONCAVE UP Slopes are increasing. CONCAVE DOWN
Find where the first derivative is increasing and decreasing.
Procedure for Finding Intervals on which a Function is Concave Up or Concave Down If f is a continuous function on an open interval (a,b). To find the open intervals on which f is concave up or concave down: Find the critical numbers of f ' and values of x that make f '' undefined in (a,b). These numbers divide the x-axis into intervals. Test the sign (+ or –) of the second derivative inside each of these intervals. If f '' (x) > 0 in an interval, then f is concave up in that same interval. If f ''(x) < 0 in an interval, then f is concave down in that same interval. Find where the first derivative is increasing and decreasing.
Example 1 f is concave down when the derivative is decreasing.
Use the graph of f '(x) below to determine when f is concave up and concave down. f is concave down when the derivative is decreasing. f ' (x) x f is concave up when the derivative is increasing. A critical point for the first derivative (f''=0) Concave Up: (1, ∞) Concave Down: (-∞,1)
Example 2 Find where the graph is concave up and where it is concave down. Domain of f: All Reals Find the critical numbers of f ' Find the first derivative. Find where the 2nd derivative is 0 or undefined Find the second derivative. Find the sign of the second derivative on each interval. Answer the question The function is concave down on (-∞,0) because f ''<0 and is concave up on (0,∞) because f ''>0
The Change in Concavity
If a graph changes from concave upward to concave downward (or vice versa), then there must be a point where the change of concavity occurs. This point is referred to as an inflection point. CONCAVE DOWN CONCAVE UP
Inflection Points A point P on a curve is called an inflection point if the graph is concave up on one side of P and concave down on the other side. In calculus terms, (if f is continuous on an interval that contains c) c is an inflection point if f '' changes from positive to negative or vice versa at c. Thus, f '' (c) must equal 0 or be undefined.
Example Determine the points of inflection of . Domain of f:
Find the critical numbers of f ' All Reals Find the derivative. Find where the 2nd derivative is 0 or undefined Find the second derivative. The 2nd derivative is undefined at t=6. NOTE: 6 is not an inflection point since the 2nd derivative does not change sign. Find the sign of the second derivative on each interval. Find the value of the function: 6 7.2 Answer the question
Example: Answer The point of Inflection is (7.2,8.131) because the second derivative changes from negative to positive values at this point.
White Board Challenge Find where the inflection point(s) are for f if f ''(x) = 4cos2x – 2 on [0,π]. Justify.
How the concavity is connected to Relative Minimum and Maximum
When a critical point is a relative maximum, what are the characteristics of the function? When a critical point is a relative minimum, what are the characteristics of the function? The function is concave downward at B. The function is concave upward at C. D f(x) B A C x
The Second Derivative Test
Let f be a function such that f '(c) = 0 (a critical number of a continuous function f(x) ) and the second derivative exists on an interval containing c. (a) If f ''(c) < 0, there is a relative maximum at x = c. f(x) Relative Maximum f '(c) = 0 f ''(c) < 0 c x
The Second Derivative Test
Let f be a function such that f '(c) = 0 (a critical number of a continuous function f(x) ) and the second derivative exists on an interval containing c. (b) If f ''(c) > 0, there is a relative minimum at x = c. f(x) f '(c) = 0 f ''(c) > 0 Relative Minimum c x
The Second Derivative Test
Let f be a function such that f '(c) = 0 (a critical number of a continuous function f(x) ) and the second derivative exists on an interval containing c. (c) If f ''(c) = 0, then the second-derivative test fails (either a maximum, or a minimum, or neither may occur). f(x) Here is a common example of neither a minimum or maximum f '(c) = 0 When the Second-Derivative Test fails, the critical point can often be classified using the First-Derivative Test. f ''(c) = 0 c x
Example 1 Domain of f: Find the relative extrema of . All Reals
Find the critical numbers Apply the 2nd Derivative Test Find the first derivative. Find where the derivative is 0 or undefined FAIL MIN MAX Apply the 1st Derivative Test where the 2nd Derivative Test Fails. Find the value of the function: -1 1 Answer the question No sign change means x=0 is not a relative extrema.
Example 1: Answer The function has a relative maximum of 4 at x = -1 because the second derivative is negative at this point. The function has a relative minimum 0 at x = 1 because the second derivative is positive at this point.
Example 2 The Second Derivative Test is less popular than the First Derivative Test for two reasons: the Second Derivative Test does not always work and a question often requires a student to find where a function is increasing/decreasing before asking for the relative min/max (Thus, the sign chart for the First Derivative Test is done already). Yet, the AP Test will have questions, like the one below, specifically about the Second Derivative Test: Example: If the function f has a horizontal tangent line at x = 4 and f ''(4) = 3, what is true about f(4)? |
Lawrence Technological University
College of Arts and Science
Department of Mathematics and Computer Sciences
### Mathematical Experimenting with LEGO Mindstorms
This is a series of exercises you might use to apply the LEGO Mindstorms kit to a mathematics class or an after-school program. These handouts are intended both for teachers and students.
As with all the series in these handout pages, I will add to it as time permits and your suggestions and submissions are most welcome.
If you are new to using this kit as an educational platform, you might want to also look at the handout on short sessions with LEGO Mindstorms for day camps.
• To explore the addition and subtraction of the small natural numbers between -7 and 7, build a wheeled robot with a separate motor for each of two driving wheels. When the right side motor is turning forward faster than the left side motor, the robot will drive in a circle counterclockwise. The bigger the difference in speeds, the smaller the circle. If you use NQC, it will help to do something like
```#define Speed_1 0
#define Speed_2 1
#define Speed_3 2
#define Speed_4 3
#define Speed_5 4
#define Speed_6 5
#define Speed_7 6
#define Speed_8 7
```
Then speed 0 would be with that motor off. The negative speeds are the same as the positive speeds with that motor's direction reversed. Consider several situations:
```Right 7 + Left 3 = ?
Right 7 + Left 5 = ?
Right 7 + Left 0 = ?
Right 3 + Left 0 = ?
Right 3 + Left -3 = ?
Right 3 + Left -7 = ?
```
• After exploring the small natural numbers as above, it would be interesting to consider differences in the behavior of the robot between what is observed and what is predicted with geometry and arithmetic. Try situations like:
```Right 7 + Left 3 = ?
Right 4 + Left 0 = ?
and
Right 6 + Left 0 = ?
Right 3 + Left -3 = ?
```
Friction and physical differences between pairs of motors are easy answers. Calculating the diameter of a circle from the circumference, time × (speedright − speedleft), is also easy when both driving wheels are the same distance from the center of the circle. Which of these cases should work with this calculation?
• Introduce the binary number system. Mount two touch sensors side by side and connect them to the LEGO RCX Brick with long wires. Use this "code reader" to read the code in two sets of LEGO beams that are either 1 or 2 beams thick. The possible codes are then
```0 0
0 1
1 0
1 1
```
Under what conditions are 0 1 and 1 0 different? Use the Awfully Small Code for Information Interchange to show the correct symbol on the Brick display.
CodeSymbol
0 0A
0 1b
1 0c
1 1d
• Reuse the wheeled robot with 2 drive wheels. Add a felt tip marker taped to the robot and some paper to draw on. Then, starting with a triangle, consider consider polygons as sequences of straight lines and turns. Does the sum of the enclosed angles or the sum of the turns equal 360 degrees? Consider a non Euclidean definition of a circle as a limit of a polygon where the number of sides increases and the length of the sides decreases. The pentagon drawn by this fragment of Logo has sharp corners.
```? to polygon :side_length :sides
> repeat :sides [forward :side_length right 360 / :sides ]
> end
polygon defined
? polygon 20 5
```
What modifications are needed, if any, to avoid rounded corners when drawing with your robot?
• Three exercises in event counting with sensors
1. Press a touch sensor with a finger repeatedly and rapidly. Count the presses with the RCX. Discuss "contact bounce" and any differences between the manual and computer count.
2. With a passive light sensor pointed at a light bulb, count flashes with the RCX. Discuss thresholds and sensors.
3. Attach a light sensor pointed down to the wheeled robot. Then count the number of black lines the robot crosses. Discuss edge detection and hysteresis.
Revised August 28, 2005 |
## Step by step solution for calculating 17 is 25 percent that what number
We already have our very first value 17 and also the 2nd value 25. Let"s i think the unknown value is Y i beg your pardon answer we will discover out.
You are watching: 17 is what percent of 25
As we have all the compelled values us need, now we deserve to put castle in a basic mathematical formula as below:
STEP 117 = 25% × Y
STEP 217 = 25/100× Y
Multiplying both sides by 100 and dividing both political parties of the equation by 25 we will certainly arrive at:
STEP 3Y = 17 × 100/25
STEP 4Y = 17 × 100 ÷ 25
STEP 5Y = 68
Finally, us have found the value of Y which is 68 and that is ours answer.
You can conveniently calculate 17 is 25 percent of what number by using any type of regular calculator, simply get in 17 × 100 ÷ 25 and also you will acquire your answer i m sorry is 68
Here is a portion Calculator come solve comparable calculations such together 17 is 25 percent of what number. You can solve this kind of calculation with your values by beginning them right into the calculator"s fields, and also click "Calculate" to gain the an outcome and explanation.
is
percent of what number
Calculate
## Sample questions, answers, and how to
Question: your friend has actually a bag that marbles, and he speak you that 25 percent of the marbles room red. If there are 17 red marbles. How countless marbles go he have altogether?
How To: In this problem, we recognize that the Percent is 25, and we are also told that the component of the marbles is red, so we understand that the component is 17.
So, that way that it must be the total that"s missing. Below is the way to figure out what the total is:
Part/Total = Percent/100
By making use of a straightforward algebra we can re-arrange ours Percent equation prefer this:
Part × 100/Percent = Total
If we take the "Part" and also multiply that by 100, and then we divide that by the "Percent", us will acquire the "Total".
Let"s try it out on our problem around the marbles, that"s very an easy and it"s just two steps! We know that the "Part" (red marbles) is 17.
So step one is to just multiply that component by 100.
17 × 100 = 1700
In step two, we take that 1700 and also divide it by the "Percent", which we are told is 25.
So, 1700 separated by 25 = 68
And that means that the total number of marbles is 68.
Question: A high institution marching band has actually 17 flute players, If 25 percent the the band members beat the flute, climate how countless members space in the band?
Answer: There space 68 members in the band.
How To: The smaller sized "Part" in this trouble is 17 because there room 17 flute players and also we space told the they comprise 25 percent of the band, for this reason the "Percent" is 25.
Again, it"s the "Total" that"s missing here, and to discover it, we just need to monitor our 2 action procedure together the vault problem.
For step one, we multiply the "Part" by 100.
17 × 100 = 1700
For action two, we division that 1700 by the "Percent", i m sorry is 25.
See more: 3 Geometric Solids With Circular Cross Sections, Pi Day Scavenger Hunt
1700 split by 25 equates to 68
That method that the total number of band members is 68.
## Another step by step method
Step 1: Let"s i think the unknown value is Y
Step 2: very first writing it as: 100% / Y = 25% / 17
Step 3: autumn the portion marks to simplify your calculations: 100 / Y = 25 / 17
Step 4: main point both sides by Y to move Y top top the appropriate side the the equation: 100 = ( 25 / 17 ) Y
Step 5: simple the best side, we get: 100 = 25 Y
Step 6: splitting both sides of the equation through 25, we will certainly arrive at 68 = Y
This leaves us with our last answer: 17 is 25 percent of 68
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# How is geometry related to trigonometry?
## Trigonometric functions
### Meaning of the trigonometric functions
The three sides of a right triangle are given:
• Edge of the angle \ (\ alpha \): 12 cm
• Opposite cathetus of the angle \ (\ alpha \): 5 cm
• Hypotenuse: 13 cm
The sine, i.e. the ratio of the opposite side to the hypotenuse, can be easily calculated:
\ [\ sin \ alpha = \ frac {\ text {Opposite cathet}} {\ text {Hypotenuse}} = \ frac {5 \ text {cm}} {13 \ text {cm}} \ approx 0.385 \]
Now we know that the sine of the angle \ (\ alpha \) of this triangle is (approximately) 0.385 ... but what does that mean? What have we just calculated?
Let us consider a second example. Then it becomes immediately clear what it is all about.
The three sides of a right triangle are given:
• Edge of the angle \ (\ alpha \): 24 cm
• Opposite cathetus of the angle \ (\ alpha \): 10 cm
• Hypotenuse: 26 cm
In case you don't immediately notice, the sides of this triangle are twice as long as the sides of the first triangle. If you were to draw the two triangles, you might find that although they are different sizes, the three angles match.
We calculate the sine again, i.e. the ratio of the opposite side to the hypotenuse:
\ [\ sin \ alpha = \ frac {\ text {Opposite cathet}} {\ text {Hypotenuse}} = \ frac {10 \ text {cm}} {26 \ text {cm}} \ approx 0.385 \]
Although the two triangles under consideration are of different sizes, the sine of the angle \ (\ alpha \) has the same value!
We know that the following applies: \ (\ sin \ alpha \ approx 0.385 \).
If we solve the equation for \ (\ alpha \), we know how big the angle is:
\ (\ alpha = \ sin ^ {- 1} (0.385) \ approx 22.64 ° \)
The angle functions can be used to calculate the angles of a triangle without having to measure a single angle.
Notes on calculating with the calculator
• Your calculator must be set to DEG (Degree).
• The side lengths of the triangle (in our example: opposite cathetus and hypotenuse) must have the same unit - e.g. cm (centimeters) or m (meters).
• To calculate sine (angle \ (\ alpha \) is given), you have to enter the angle in degrees - e.g. 30 ° or 45 °.
• To calculate the angle \ (\ alpha \) (sine is given), you have to use the inverse function of the sine \ (\ sin ^ {- 1} \). There is a corresponding button on your calculator for this.
In the next chapter we will deal with the unit circle. This helps to graphically illustrate the trigonometric functions. We will also see that trigonometric functions are defined for any (positive and negative) angle. So far we have only defined the trigonometric functions using right-angled triangles, which is why our consideration has been limited to angles between 0 ° and 90 °.
### More about trigonometry
You can find more information about trigonometry in the following chapters.
Basics Trigonometric functions Unit circle Trigonometric functions Sine \ (\ sin \ alpha = \ frac {\ text {Opposite cathet}} {\ text {Hypotenuse}} \) Cosine \ (\ cos \ alpha = \ frac {\ text {adjacent}} {\ text {hypotenuse}} \) tangent \ (\ tan \ alpha = \ frac {\ text {opposite side}} {\ text {adjacent side}} \) Reciprocal values Cosekans \ (\ csc \ alpha = \ frac {\ text {Hypotenuse}} {\ text {Opposite cathet}} = \ frac {1} {\ sin \ alpha} \) Secans \ (\ sec \ alpha = \ frac {\ text {hypotenuse}} {\ text {adjacent}} \ phantom {1 \:} = \ frac {1} {\ cos \ alpha} \) Cotangent \ (\ cot \ alpha = \ frac {\ text {adjacent side}} {\ text {opposite side}} = \ frac {1} {\ tan \ alpha} \) |
# Math 6
### Mathematics homework help
Lesson 3.2
Introduction
Course Objectives
This lesson will address the following course outcomes:
· 1. Demonstrate operation sense by communicating in words and symbols the effects of operations on numbers. Apply the correct order of operations in evaluating expressions and formulas.
· 13. Evaluate formulas with multiple variables in a variety of contexts, including science, statistics, geometry, and financial math. Solve simple formulas for a specified variable.
· 15. Solve linear equations in one variable, including problems involving the distributive property and fractions.
Specific Objectives
Students will understand that
· the behavior of a formula can be explored using a table and graph.
Students will be able to
· simplify a formula given values for some parameters.
· solve for a variable in a linear equation, and evaluate an equation.
Problem Situation: Calculating Blood Alcohol Content
Blood alcohol content (BAC) is a measurement of how much alcohol is in someone’s blood. It is usually measured as a percentage. So, a BAC of 0.3% is three-tenths of 1%. That is, there are 3 grams of alcohol for every 1,000 grams of blood. A BAC of 0.05% impairs reasoning and the ability to concentrate. A BAC of 0.30% can lead to a blackout, shortness of breath, and loss of bladder control. In most states, the legal limit for driving is a BAC of 0.08%.
#1 Points possible: 12. Total attempts: 5
Think about the variables that might influence BAC, and how they would influence it. Use your intuition to answer these questions:
If you drink more drinks, your BAC would
If the time since your first drink was longer, your BAC would
If you weighed more, your BAC would
Estimating BAC
BAC is usually determined by a breathalyzer, urinalysis, or blood test. However, Swedish physician, E.M.P. Widmark developed the following equation for estimating an individual’s BAC. This formula is widely used by forensic scientists:
B=−0.015t+2.84NW⋅gB=-0.015t+2.84NW⋅g
Where
B = percentage of BAC
N = number of “standard drinks” (A standard drink is one 12-ounce beer, one 5-ounce glass of wine, or one 1.5-ounce shot of liquor.) N should be at least 1.
W = weight in pounds
= gender constant, 0.68 for men and 0.55 for women
t = number of hours since the first drink
The variables BNWg, and t change depending on the person and situation. The units of B are a percentage. So, 0.08 is not 8%, but eight-hundredths of a percent.
This equation has been simplified from the one found in the reference. The numbers 0.015 and 2.84 are constants based on the average person. It’s also worth noting many high-end beers have twice as much alcohol as the “standard beer,” which assumes 4% alcohol.
#2 Points possible: 5. Total attempts: 5
Consider the case of a male student who has three beers and weighs 120 pounds. For this case, these values can replace the appropriate variables in the formula. What variables are still unknown in the equation? (select all that are still unknown)
· B
· N
· W
· g
· t
#3 Points possible: 5. Total attempts: 5
Consider the case of a male student who has three beers and weighs 120 pounds. Simplify the Widmark equation as much as possible for this case, rounding any constants to 3 decimal places. Hint
#4 Points possible: 12. Total attempts: 5
Using your simplified equation, find the estimated BAC for this student one, three, and five hours after his first drink. Look for patterns in the data.
Hours BAC 1 % 3 % 5 %
#5 Points possible: 6. Total attempts: 5
Complete the following contextual sentence:
Every hour, this student’s BAC by %.
#6 Points possible: 5. Total attempts: 5
Create a graph of the BAC over time for this student. Your graph doesn’t have to be perfect.
Clear All Draw:
#7 Points possible: 10. Total attempts: 5
Using your equation, what would this student’s BAC be after 8 hours?
%
Is this reasonable?
#8 Points possible: 10. Total attempts: 5
How long will it take for this student’s BAC to be 0.08%, the legal limit? Give your answer to 1 decimal place.
hours
How long will it take for the alcohol to be completely metabolized resulting in a BAC of 0.0? Give your answer to 1 decimal place.
hours
Hint: Try this problem on your own. If you’re having trouble after two tries, we’ll give some hints
Another Case
A female student, weighing 110 pounds, plans on going home in two hours. Using the Widmark formula, the simplified equation for this case is
B=−0.03+2.84N60.5B=-0.03+2.84N60.5
#9 Points possible: 8. Total attempts: 5
Compare her BAC for one glass of wine versus three glasses of wine at the time she will leave. Give BAC to 3 decimal places.
Number of Drinks BAC 1 % 3 %
#10 Points possible: 5. Total attempts: 5
In this scenario, determine how many drinks she can have so that her BAC remains less than 0.08%. Since she could finish only part of a drink, give your answer to one decimal place.
drinks
#11 Points possible: 5. Total attempts: 5
Create a graph of the BAC given number of drinks for this student. Your graph doesn’t have to be perfect.
Clear All Draw:
HW 3.2
#1 Points possible: 5. Total attempts: 5
Which of the following was one of the main mathematical ideas of the lesson?
· Blood Alcohol Content (BAC) is affected by many different variables.
· A formula can be explored by evaluating it and solving for specific values.
· When solving an equation, an operation that changes the value of one side must also be done to the other side of the equation.
· Sometimes letters can represent a constant.
#2 Points possible: 20. Total attempts: 5
Find the solution to each of the following:
a. 5+3x=145+3x=14 x =
b. 6x−5=106x-5=10 x =
c. 1=7+2×1=7+2x x =
d. x4+3=8×4+3=8 x =
#3 Points possible: 15. Total attempts: 5
Find the solution to each of the following:
a. 2.5x+1=8.52.5x+1=8.5 x =
b. 3.6+4.2x=9.93.6+4.2x=9.9 x =
c. 14.92=5+1.6×14.92=5+1.6x x =
#4 Points possible: 15. Total attempts: 5
Find the solution to each of the following, giving the answer in reduced form as proper or improper fractions (no mixed numbers).
a. 13x+38=5813x+38=58 x =
b. 12x+14=1612x+14=16 x =
c. 2x+23=82x+23=8 x =
#5 Points possible: 8. Total attempts: 5
Recall that Blood Alcohol Content (BAC) is a measurement of how much alcohol is in someone’s blood as a percentage. However, police and the public typically omit the language for % when quoting the BAC and simply say, “BAC is 0.04.”
Write an interpretation of what each of the following BAC values means in terms of how much alcohol is in the bloodstream in the form of the amount of alcohol per 1,000 grams of blood. You may want to refer back to the example in the lesson.
a. BAC = 0.1 gram of alcohol for every 1,000 grams of blood.
b. BAC = 0.02 gram of alcohol for every 1,000 grams of blood.
#6 Points possible: 1. Total attempts: 5
Use information from the website http://en.wikipedia.org/wiki/Blood_alcohol_content to list effects on an individual having a BAC as given. Give at least one effect for each.
a. BAC = 0.1
b. BAC = 0.5
c. BAC = 0.05
Use the Widmark Equation, B=−0.015t+2.84NW⋅gB=-0.015t+2.84NW⋅g to solve the next two questions. Recall that g = 0.68 for men and g = 0.55 for women.
#7 Points possible: 10. Total attempts: 5
A male student had five glasses of wine at a party. He weighs 155 pounds. Using the Widmark equation,
a) Write the equation you would solve to figure out how long it will take before his BAC is 0.08. Is it okay to round any constants to 3 decimal places.
b) Solve that equation. Round to 2 decimal places.
hours
#8 Points possible: 5. Total attempts: 5
This question will apply the Widmark Equation to you.
Answer the next two questions based on the Widmark Equation using the weight and gender you specified.
If you drink alcohol over a period of 5 hours, how many whole drinks would you be able to consume and still ensure that your BAC is less than the legal limit in Washington? drinks
#9 Points possible: 10. Total attempts: 5
Indicate if each of the following is an expression or an equation:
· 4x+34x+3
· x2−16x−4×2-16x-4
· (x−4)(x+3)=6(x-4)(x+3)=6
· (x−4)(x−3)(x-4)(x-3)
· 4y=34y=3
#10 Points possible: 9. Total attempts: 5
The percentage of Americans who are retired has been increasing over the last decade. This is causing some concern because health care, social security, and other costs will be the responsibility of a smaller group of people. That is, as the percentage of retired people increases, the percentage of working age people decreases. The following model predicts the percentage of retired people based on demographic data1:
R=t873.36−2.15R=t873.36-2.15
where R is the percentage (as a decimal) of Americans who are retired in the year (for example, the year 1995 would be t = 1995). Use this model to complete the table below by solving for the year.
Year % of Retired People 10% 15% 20%
#11 Points possible: 5. Total attempts: 5
Andy’s house is on a large lot. He got 100 yards of chain-link fence on sale. He wants to use all of the material to completely enclose a rectangular area in his backyard. He wants to make the fenced area 60 feet wide and as long as possible. What is the longest length possible for the sides?
ft |
# Whats \frac{d}{dx}\ \frac{d}{dy}sin(4x^(\frac{5}{2})z^2)^*cos(in(y^2))? explain the notation
## Question:
Whats {eq}\frac{d}{dx}\ \frac{d}{dy}sin(4x^(\frac{5}{2})z^2)^*cos(ln(y^2))? {/eq} explain the notation
## Partial Derivative:
The derivative with respect to one variable of the multivariable function is known as the partial derivative. Also, we can use the chain rule to find the derivative of the multivariable function.
In the problem, we have the find the partial derivative, as we are given the d-symbol.
So we have the expression:
{eq}\frac{d}{dx}\ \frac{d}{dy}sin(4x^(\frac{5}{2})z^2)^*cos(ln(y^2)) {/eq}
The notation:
{eq}\frac{d}{dx}\ \frac{d}{dy}sin(4x^(\frac{5}{2})z^2)^*cos(ln(y^2)) {/eq}
means that we will first find the partial derivative of the expression:
{eq}sin(4x^(\frac{5}{2})z^2)^*cos(ln(y^2)) {/eq} with respect to y and then the result is differentiated partially with respect to x.
using the expression: {eq}\frac{d}{dx} {/eq}
Now, in order to solve the expression:
{eq}\frac{d}{dx}\left(\frac{d}{dy}\sin \left(4x^{\frac{5}{2}}z^2\right)\cdot \cos \left(\ln \left(y^2\right)\right)\right)\\ {/eq}
So we solve
{eq}\frac{d}{dy}\sin \left(4x^{\frac{5}{2}}z^2\right)\cdot \cos \left(\ln \left(y^2\right)\right)\\ =\sin \left(4x^{\frac{5}{2}}z^2\right)\left(-\sin \left(\ln \left(y^2\right)\right)\right)\frac{2}{y}~~~~~~~~~~~~~~~~~~~~~~~~~~\left [ \because \frac{d}{du}\left(\cos \left(u\right)\right)=-\sin \left(u\right)~~and~~~\frac{d}{dy}\left(\ln \left(y^2\right)\right)=\frac{2}{y} \right ]\\ =-\frac{2\sin \left(4z^2x^{\frac{5}{2}}\right)\sin \left(\ln \left(y^2\right)\right)}{y}\\ {/eq}
Now we have to find the partial derivative wrt x, as follows:
{eq}\frac{d}{dx}\left(-\frac{2\sin \left(4z^2x^{\frac{5}{2}}\right)\sin \left(\ln \left(y^2\right)\right)}{y}\right)\\ =-\frac{2\sin \left(\ln \left(y^2\right)\right)}{y}\cos \left(4z^2x^{\frac{5}{2}}\right)\cdot \:10z^2x^{\frac{3}{2}}~~~~~~~~~~~~~~~~~~~~~~~\left [ \because \frac{d}{du}\left(\sin \left(u\right)\right)=\cos \left(u\right)~~~and~~~~~\frac{d}{dx}\left(x^a\right)=a\cdot x^{a-1} \right ]\\ =-\frac{20z^2x^{\frac{3}{2}}\cos \left(4z^2x^{\frac{5}{2}}\right)\sin \left(\ln \left(y^2\right)\right)}{y} {/eq} |
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# 1.4 Linear equations and the integrating factor
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### Introduction
One of the most important types of equations we will learn how to solve are the so-called linear equations. In fact, the majority of the course is about linear equations. In this lecture we focus on the first order linear equation. A first order equation is linear if we can put it into the form:
$y' + p(x)y = f(x).$
Here the word “linear” means linear in $$y$$ and $$y'$$; no higher powers nor functions of $$y$$ or $$y'$$ appear. The dependence on $$x$$ can be more complicated.
Solutions of linear equations have nice properties. For example, the solution exists wherever $$p(x)$$ and $$f(x)$$ are defined, and has the same regularity (read: it is just as nice). But most importantly for us right now, there is a method for solving linear first order equations. The trick is to rewrite the left hand side of (1.4.1) as a derivative of a product of $$y$$ with another function. To this end we find a function $$r(x)$$ such that
$r(x)y' + r(x)p(x)y = \frac{d}{dx}\left[r(x)y\right]$
This is the left hand side of (1.4.1) multiplied by $$r(x)$$. So if we multiply (1.4.1) by $$r(x)$$, we obtain
$\frac{d}{dx}\left[r(x)y\right] = r(x)f(x)$
Now we integrate both sides. The right hand side does not depend on $$y$$ and the left hand side is written as a derivative of a function. Afterwards, we solve for$$y$$. The function $$r(x)$$ is called the integrating factor and the method is called the integrating factor method.
We are looking for a function $$r(x)$$, such that if we differentiate it, we get the same function back multiplied by $$p(x)$$. That seems like a job for the exponential function! Let
$r(x) = e^{\int p(x)dx}$
We compute:
$y' + p(x)y = f(x),$
$e^{\int p(x)dx}y' + e^{\int p(x)dx}p(x)y = e^{\int p(x)dx}f(x),$
$\frac{d}{dx}\left[e^{\int p(x)dx}y\right] = e^{\int p(x)dx}f(x),$
$e^{\int p(x)dx}y = \int e^{\int p(x)dx}f(x)dx + C,$
$y = e^{-\int p(x)dx}\left( \int e^{\int p(x)dx}f(x)dx + C \right).$
Of course, to get a closed form formula for $$y$$, we need to be able to find a closed form formula for the integrals appearing above.
Example $$\PageIndex{1}$$:
Solve
$y' + 2xy = e^{x-x^2}, \, \, \, \, y(0) = -1$
First note that$$p(x) = 2x$$ and $$f(x) = e^{x-x^2}$$. The integrating factor is $$r(x) = e^{\int p(x)dx} = e^{x^2}$$. We multiply both sides of the equation by $$r(x)$$ to get
$e^{x^2}y' + 2xe^{x^2}y = e^{x-x^2}e^{x^2},$
$\frac{d}{dx}\left [ e^{x^2}y\right] = e^{x}.$
We integrate
$e^{x^2}y = e^x + C,$
$y = e^{x-x^2} + Ce^{-x^2}.$
Next, we solve for the initial condition $$-1 = y(0) = 1 + C$$, so $$C = -2$$. The solution is
$y = e^{x-x^2} - 2e^{-x^2}.$
Note that we do not care which antiderivative we take when computing $$e^{\int p(x) dx}$$. You can always add a constant of integration, but those constants will not matter in the end.
Exercise $$\PageIndex{1}$$:
Try it! Add a constant of integration to the integral in the integrating factor and show that the solution you get in the end is the same as what we got above. An advice: Do not try to remember the formula itself, that is way too hard. It is easier to remember the process and repeat it.
Since we cannot always evaluate the integrals in closed form, it is useful to know how to write the solution in definite integral form. A definite integral is something that you can plug into a computer or a calculator. Suppose we are given
$y' + p(x)y = f(x), \, \, \, \, \, y(x_0) = y_0$.
Look at the solution and write the integrals as definite integrals.
$y(x) = e^{\int-_{x_0}^x p(s)ds} \left(\int_{x_0}^x e^{\int_{x_0}^t p(s)ds} f(t) dt + y_0 \right)$
You should be careful to properly use dummy variables here. If you now plug such a formula into a computer or a calculator, it will be happy to give you numerical answers.
Exercise $$\PageIndex{2}$$:
Check that $$y(x_0) = y_0$$ in formula (1.4.17).
Exercise $$\PageIndex{3}$$:
Write the solution of the following problem as a definite integral, but try to simplify as far as you can. You will not be able to find the solution in closed form.
$y' + y = e^{x^2-x}, \, \, \, \, \, y(0) = 10$
Remark 1.4.1: Before we move on, we should note some interesting properties of linear equations. First, for the linear initial value problem $$y' + p(x)y = f(x), y(x_0) = y_0$$, there is always an explicit formula (1.4.17) for the solution. Second, it follows from the formula (1.4.17) that if $$p(x)$$ and $$f(x)$$ are continuous on some interval $$(a, b)$$, then the solution $$y(x)$$ exists and is differentiable on $$(a, b)$$. Compare with the simple nonlinear example we have seen previously, $$y' = y^2$$, and compare to Theorem 1.2.1.
Example $$\PageIndex{2}$$:
Let us discuss a common simple application of linear equations. This type of problem is used often in real life. For example, linear equations are used in figuring out the concentration of chemicals in bodies of water (rivers and lakes).
A 100 liter tank contains 10 kilograms of salt dissolved in 60 liters of water. Solution of water and salt (brine) with concentration of 0.1 kilograms per liter is flowing in at the rate of 5 liters a minute. The solution in the tank is well stirred and flows out at a rate of 3 liters a minute. How much salt is in the tank when the tank is full?
Let us come up with the equation. Let $$x$$ denote the kilograms of salt in the tank, let $$t$$ denote the time in minutes. For a small change $$\Delta t$$ in time, the change in $$x$$ (denoted $$\Delta x$$) is approximately
$\Delta x \approx \text{(rate in x concentration in)}\Delta t - \text{(rate out x concentration out)}\Delta t.$
Dividing through by $$\Delta t$$ and taking the limit $$\Delta t \rightarrow 0$$ we see that
$\frac{dx}{dt} = \text{(rate in x concentration in)} - \text{(rate out x concentration out)}$
In our example, we have
$\text{rate in} = 5,$
$\text{concentration in} = 0.1,$
$\text{rate out} = 3,$
$\text{concentration out} = \frac{x}{\text{volume}} = \frac{x}{60 + (5 - 3)t} .$
Our equation is, therefore,
$\frac{dx}{dt} = \text{(5 x 0.1)} - \left ( 3 \frac{x}{60 + 2t} \right)$
Or in the form (1.4.1)
$\frac{dx}{dt} + \frac{3}{60 + 2t}x = 0.5$
Let us solve. The integrating factor is
$r(t) = \text{exp} \left ( \int \frac{3}{60 + 2t} dt \right ) = \text{exp} \left ( \frac{3}{2} \ln (60 + 2t) \right ) = {(60 + 2t)}^{3/2}$
We multiply both sides of the equation to get
$(60 + 2t)^{3/2} \frac{dx}{dt} + (60 + 2t)^{3/2} \frac{3}{60 + 2t}x = 0.5(60 + 2t)^{3/2},$
$\frac{d}{dt} \left [ (60 +2t)^{3/2}x \right ] = 0.5(60 + 2t)^{3/2},$
$(60 + 2t)^{3/2} x= \int 0.5(60 + 2t)^{3/2}dt + C,$
$x = (60 + 2t)^{-3/2} \int \frac {(60 + 2t)^{3/2}}{2}dt + C(60 + 2t)^{-3/2}$,
$x = (60 + 2t)^{-3/2} \frac{1}{10} (60 + 2t)^{5/2} + C(60 + 2t)^{-3/2},$
$x = \frac {(60 + 2t)}{10} + C(60 + 2t)^{-3/2}.$
We need to find $$C$$. We know that at $$t = 0$$, $$x = 10$$. So
$10 = x(0) = \frac {60}{10} + C{(60)}^{-3/2} = 6 + C{(60)}^{-3/2}$
or
$C = 4({60}^{3/2}) \approx 1859.03.$
We are interested in $$x$$ when the tank is full. So we note that the tank is full when $$60 + 2t = 100$$, or when $$t = 20$$. So
$x(20) = \frac{60 + 40}{10} + C{(60 + 40)}^{-3/2} \approx 10 + 1859.03{(100)}^{-3/2} \approx 11.86.$
The concentration at the end is approximately 0.1186 kg/liter and we started with 1/6 or 0.167 kg/liter. |
# Fixed-Point Representation
Positive integers, including zero, can be represented as unsigned numbers. However, to represent negative integers, we need a notation for negative values. In ordinary arithmetic, a negative number is indicated by a minus sign and a positive number by a plus sign. Because of hardware limitations, computers must represent everything with 1's and D's, including the sign of a number. As a consequence, it is customary to represent the sign with a bit placed in the leftmost position of the number. The convention is to make the sign bit equal to 0 for positive and to 1 for negative.
binary point :
In addition to the sign, a number may have a binary (or decimal) point. The position of the binary point is needed to represent fractions, integers, or mixed integer-fraction numbers. The representation of the binary point in a register is complicated by the fact that it is characterized by a position in the register. There are two ways of specifying the position of the binary point in a register: by giving it a fixed position or by employing a floating-point representation. The fixed-point method assumes that the binary point is always fixed in one position. The two positions most widely used are (1) a binary point in the extreme left of the register to make the stored number a fraction, and (2) a binary point in the extreme right of the register to make the stored number an integer. In either case, the binary point is not actually present, but its presence is assumed from the fact that the number stored in the register is treated as a fraction or as an integer. The floating-point representation uses a second register to store a number that designates the position of the decimal point in the first register. Floating-point r.epresentation is discussed further in the next section
## Frequently Asked Questions
+
Ans: Since we are dealing with unsigned numbers, there is really no way to get an unsigned result for the second example. view more..
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Ans: The direct method of subtraction taught in elementary schools uses the borrow concept. In this method we borrow a 1 from a higher significant position when the minuend digit is smaller than the corresponding subtrahend digit. view more..
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Ans: The r's complement of an n-digit number N in base r is defined as r' - N for N * D and D for N = D. Comparing with the (r - I)'s complement, we note that the r's complement is obtained by adding I to the (r - I)'s complement since r' - N = [(r' - I) - N] + I. view more..
+
Ans: Positive integers, including zero, can be represented as unsigned numbers. However, to represent negative integers, we need a notation for negative values. In ordinary arithmetic, a negative number is indicated by a minus sign and a positive number by a plus sign. view more..
+
Ans: When an integer binary number is positive, the sign is represented by 0 and the magnitude by a positive binary number. When the number is negative, the sign is represented by 1 but the rest of the number may be represented in one of three possible ways: view more..
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Ans: The addition of two numbers in the signed-magnitude system follows the rules of ordinary arithmetic. If the signs are the same, we add the two magnitudes and give the sum the common sign. If the signs are different, we subtract the smaller magnitude from the larger and give the result the sign of the larger magnitude. view more..
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Ans: Subtraction of two signed binary numbers when negative numbers are in 2' s complement form is very simple and can be stated as follows: Take the 2's complement of the subtrahend (including the sign bit) and add it to the minuend (including the sign bit). A carry out of the sign bit position is discarded. view more..
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Ans: When two numbers of n digits each are added and the sum occupies n + 1 digits, we say that an overflow occurred. When the addition is performed with paper and pencil, an overflow is not a problem since there is no limit to the width of the page to write down the sum. view more..
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Ans: An overflow condition can be detected by observing the carry into the sign bit position and the carry out of the sign bit position. If these two carries are not equal, an overflow condition is produced. view more..
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Ans: The representation of decimal numbers in registers is a function of the binary code used to represent a decimal digit. A 4-bit decimal code requires four flip-flops for each decimal digit. view more..
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Ans: The floating-point representation of a number has two parts. The first part represents a signed, fixed-point number called the mantissa. The second part designates the position of the decimal (or binary) point and is called the exponent. The fixed-point mantissa may be a fraction or an integer. For exam ple, the decimal number +6132.789 is represented in floating-point with a fraction and an exponent as follows: view more..
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Ans: A floating-point number is said to be normalized if the most significant digit of the mantissa is nonzero. For example, the decimal number 350 is normalized but 00035 is not. Regardless of where the position of the radix point is assumed to be in the mantissa, the number is normalized only if its leftmost digit is nonzero. view more..
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Ans: In previous sections we introduced the most common types of binary-coded data found in digital computers. Other binary codes for decimal numbers and alphanumeric characters are sometimes used. Digital computers also employ other binary codes for special applications. A few additional binary codes encountered in digital computers are presented in this section. view more..
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Ans: Binary codes for decimal digits require a minimum of four bits. Numerous different codes can be formulated by arranging four or more bits in 10 distinct possible combinations. A few possibilities are shown in Table 3-6. view more..
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Ans: The ASCII code (Table 3-4) is the standard code commonly used for the transmission of binary information. Each character is represented by a 7-bit code and usually an eighth bit is inserted for parity (see Sec. 3-6). The code consists of 128 characters. Ninety-five characters represent graphic symbols that include upper- and lowercase letters, numerals zero to nine, punctuation marks, and special symbols view more..
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Ans: Binary information transmitted through some form of communication medium is subject to external noise that could change bits from 1 to 0, and vice versa. An error detection code is a binary code that detects digital errors during transmission. The detected errors cannot be corrected but their presence is indicated. The usual procedure is to observe the frequency of errors. If errors occur infrequently at random, the particular erroneous information is transmitted again. If the error occurs too often, the system is checked for malfunction view more..
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Ans: Parity generator and checker networl<s are logic circuits constructed with exclusive-OR functions. This is because, as mentioned in Sec. 1·2, the exclusive-OR function of three or more varia.bles is by definition an odd function. An odd function is a logic function whose value is binary 1 if, and only if, an odd function number of variables are equal to 1. According to this definition, the P( even) is the exclusive-OR of x, y, and l because it is equal to 1 when either one or all three of the variables are equal to I (Table 3-7). The P(odd) function is the complement of the P(even) function. view more..
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Ans: A digital system Is an interconnection of digital hardware module. that at'ClOinpl.lsh a specific Wormation-proceaslna taslc. Digital systems vary in size and complexi.ty interacting digital &om a few integrated circuits to a complex of interconnected and computers. Digital system design invariably UBeS a modular approach. The modules are constructed &om such digital components as ules registet&, are in decoders, terconnected arithmetic with common elements data and control paths , and control logic. The to fonn various moda digital computer system. view more..
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# In the Given Figure, a is the Centre of the Circle. ∠ Abc = 45° and Ac = 7 √ 2 Cm. Find the Area of Segment Bxc. - Geometry
Sum
In the given figure, A is the centre of the circle. $\angle$ABC = 45° and AC = 7 $\sqrt{2}$cm. Find the area of segment BXC.
#### Solution
In ∆ABC,
AB = AC (Radii of the circle)
∴ ∠ACB = ∠ABC = 45º (Equal sides have equal angles opposite to them)
Using angle sum property, we have
∠ACB + ∠ABC + ∠BAC = 180º
∴ 45º + 45º + ∠BAC = 180º
⇒ ∠BAC = 180º − 90º = 90º
Here,
Radius of the circle, r = $7\sqrt{2}$cm Measure of arc BXC, θ = 90º
∴ Area of segment BXC
$= r^2 \left( \frac{\pi\theta}{360° } - \frac{\sin\theta}{2} \right)$
$= \left( 7\sqrt{2} \right)^2 \left( \frac{22}{7} \times \frac{90°}{360°} - \frac{\sin90°}{2} \right)$
$= 98 \times \frac{22}{7} \times \frac{1}{4} - 98 \times \frac{1}{2}$
$= 77 - 49$
$= 28 {cm}^2$
Thus, the area of the segment BXC is 28 cm2.
Concept: Areas of Sector and Segment of a Circle
Is there an error in this question or solution?
#### APPEARS IN
Balbharati Mathematics 2 Geometry 10th Standard SSC Maharashtra State Board
Chapter 7 Mensuration
Practice set 7.4 | Q 1 | Page 159 |
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# 1.4: Evaluate Numerical and Variable Expressions Using the Order of Operations
Difficulty Level: Basic Created by: CK-12
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The students at Washington Middle School experienced changes with their new school year. Their old gym teacher, Mr. Woullard had retired and there was a new gym teacher who greeted them on the first day of eighth grade.
Mr. Osgrove was young and lively with lots of energy, but he also had some new ideas about how gym class ought to be run.
“We’ll combine two periods of students together,” he explained. “That will give us so many more combinations of students when it comes to teams.”
Jesse looked around. He counted the number of boys and girls in his class. There were eleven boys and fourteen girls in the class. The other class had thirteen boys and fourteen girls in it.
“We can add the boys together and form four teams, and we can add the girls together and form four teams.”
Jesse left gym class with his head full of numbers. If they were to combine all of the boys from the two classes and all of the girls from the two classes, then that would be a lot of students. Jesse began figuring out the different combinations of teams. Jesse knows that he will need to use the order of operations.
Did you know that? Pay attention and you will be able to help Jesse at the end of the Concept.
### Guidance
In mathematics, you will often hear the word “evaluate.” Before we begin, it is important for you to understand what the word “evaluate” means. When we evaluate a mathematical sentence, we figure out the value of the number sentence. If the mathematical sentence has numbers in it, then we figure out the value of the number sentence. Often times we think of evaluating as solving, and it can be that, but more specifically, evaluating is figuring out the value of a sentence.
What do we evaluate?
In mathematics, we can evaluate different types of number sentences. Sometimes we will be working with equations and other times we will be working with expressions.
That is a great question.
An equation is a number sentence with an equals sign. Therefore, the quantity on one side of the equals sign is equal to the quantity on the other side of the equals sign.
An expression is a group of numbers, symbols and variables that represents a quantity; there is not an equal sign. We evaluate an expression to figure out the value of the mathematical statement itself, we are not trying to make one side equal another, as with an equation.
Let’s start by focusing on evaluating expressions.
Two eighth grade math students evaluated the expression: 2+3×4÷2\begin{align*}2 + 3 \times 4 \div 2\end{align*}. Both students approached the expression differently. Macy’s answer was ten. Cole’s answer was eight.
What happened here? How could one student arrive at one answer and another student come up with an entirely different answer? The key is in the order that each student performed each operation.
This is where the order of operations comes in. The order of operations is a rule that tells us which operation we need to perform in what order to achieve the accurate answer. The order of operations applies whenever you have two or more operations in a single expression.
Here is the order of operations.
Order of Operations
P parentheses or grouping symbols
E exponents
MD multiplication and division in order from left to right
AS addition and subtraction in order from left to right
Take a few minutes to write down the order of operations in your notebook.
Let’s look at how Macy and Cole arrived at their answers.
2+3×4÷2\begin{align*}2 + 3 \times 4 \div 2\end{align*}
Cole worked on evaluating this expression using the order of operations. He multiplied 3×4=12\begin{align*}3 \times 4 = 12\end{align*}. Then he divided by 2 which is 6 and finally he added 2 for a final answer of 8. This is correct. It may seem out of order to work this way, but remember you are working according to the order of operations.
What did Macy do? Macy worked on evaluating the expression by working in order from left to right. She simply did not follow the order of operations. In this case, she evaluated the value of the expression as 10. This is incorrect. Next time, Macy needs to follow the order of operations.
Working in this way is called evaluating a numerical expression. It is a numerical expression because it is made up only of numbers and operations.
Here is another one.
Evaluate 3+92÷3+8\begin{align*}3 + 9 \cdot 2 \div 3 + 8\end{align*}
To begin with, we first need to remind ourselves of the order of operations. Notice that there is a dot in between the nine and the two. This is another way to show multiplication. As you get into higher levels of math, you will see that multiplication is often shown in other ways besides using an x\begin{align*}x\end{align*}. Now back to evaluating.
Following the order of operations, we would first multiply and then divide.
92=18\begin{align*}9 \cdot 2 = 18\end{align*}
18÷3=6\begin{align*}18 \div 3 = 6\end{align*}
Next we perform addition and subtraction in order from left to right.
6+3=9\begin{align*}6 + 3 = 9\end{align*}
9+8=17\begin{align*}9 + 8 = 17\end{align*}
You will also find another type of expression. These expressions can have letters in them. These letters are called variables and variables represent an unknown quantity. When you see an expression with a variable in it, we call it a variable expression.
We can evaluate variable expressions using the order of operations too. The key thing with a variable expression is that you will have to substitute a given value into the expression for the unknown variable and then evaluate the expression.
Take a look.
Evaluate the expression 60÷22a+164\begin{align*}60 \div 2 \cdot 2a + 16 - 4\end{align*} when a=5\begin{align*}a = 5\end{align*}.
First, notice that the expression has the letter a\begin{align*}a\end{align*} in it. This is our variable. Also, you can see that you have been given a value for a\begin{align*}a\end{align*}. Our first step is to substitute the value of a\begin{align*}a\end{align*} into the expression.
60÷22(5)+164\begin{align*}60 \div 2 \cdot 2(5) + 16 - 4\end{align*}
Another good question-using the parentheses is another way to show multiplication. So now you have three ways to show multiplication. You can use an x\begin{align*}x\end{align*}, a dot or a set of parentheses around a number. This means that we are multiplying the 2 times the 5.
Now we can use our order of operations. We have division, multiplication and multiplication in this expression right away. We complete multiplication and division in order from left to right.
Another good question! In this case, the set of parentheses is not grouping two numbers and an operation. When talking about parentheses in the order of operations, we need to have an operation inside it. The set of parentheses in this problem is being used to show multiplication. There isn’t multiplication inside the parentheses.
Now back to evaluating the expression by performing multiplication and division in order from left to right.
60÷22(5)+16460÷2=30302(5)=300\begin{align*}& 60 \div 2 \cdot 2(5) + 16 - 4\\ & 60 \div 2 = 30\\ & 30 \cdot 2(5) = 300\end{align*}
Next, we work with addition and subtraction in order from left to right.
300+1643164312\begin{align*}& 300 + 16 - 4\\ & 316 - 4\\ & 312\end{align*}
Now let's add in the grouping symbols. The “grouping symbols” that we are going to be working with are brackets [ ] and parentheses ( ). According to the order of operations, we perform all operations inside of grouping symbols BEFORE performing any other operation in the list.
Evaluate the expression 7+4(15÷5)6\begin{align*}7 + 4(15 \div 5) - 6\end{align*}.
First, notice that we have a set of parentheses in this numerical expression. Remember that it is called a numerical expression because this expression does not have any variable in it.
We perform any and all operations in parentheses first.
15÷5=3\begin{align*}15 \div 5 = 3\end{align*}
Now let’s rewrite the expression.
7+4(3)6\begin{align*}7 + 4(3) - 6\end{align*}
Our next step is to continue with the order of operations. We have multiplication in this expression. We do that next.
7+126\begin{align*}7 + 12 - 6\end{align*}
Now we can perform addition and subtraction in order from left to right.
196=13\begin{align*}19 - 6 = 13\end{align*}
Brackets can be used to group more than one operation. When you see a set of brackets, remember that brackets are a way of grouping numbers and operations. There is a spotlight on brackets too.
Evaluate the expression 6+[5+(46)]17\begin{align*}6 + [5 + (4 \cdot 6)] - 17\end{align*}.
Now we have a set of parentheses within a set of brackets. To work through this, we are going to perform the operation in the parentheses inside the brackets first, and then we will perform the other operation in the brackets.
6+[5+24]176+2917\begin{align*}& 6 + [5 + 24] - 17\\ & 6 + 29 - 17\end{align*}
Next we perform addition and subtraction in order from left to right.
351718\begin{align*}& 35 - 17\\ & 18\end{align*}
#### Example A
Evaluate the expression 6y+3(24)\begin{align*}6y + 3 - (2 \cdot 4)\end{align*} when y=15\begin{align*}y = 15\end{align*}.
Solution: 85\begin{align*}85\end{align*}
#### Example B
7+[4+(32)]5\begin{align*}7 + [4 + (3 \cdot 2)] - 5\end{align*}
Solution: 12\begin{align*}12\end{align*}
#### Example C
9+[6(23)]+15\begin{align*}9 + [6 - (2 \cdot 3)] + 15\end{align*}
Solution: 24\begin{align*}24\end{align*}
Now back to the dilemma from the beginning of the Concept. First, let’s look at the information that we have been given in the problem.
Class one has 11 boys and 14 girls.
Class two has 13 boys and 14 girls.
The boys from the two classes will be added together, and the girls from the two classes will be added together.
11+13\begin{align*}11 + 13\end{align*}
14+14\begin{align*}14 + 14\end{align*}
We can use parentheses to show that the boys will be added and the girls will be added. Both groups will be divided by four.
(11+13)÷4\begin{align*}(11 + 13) \div 4\end{align*}
Or 11+134\begin{align*}\frac{11+13}{4}\end{align*}
This is an expression for the boys.
(14+14)÷4\begin{align*}(14 + 14) \div 4\end{align*}
Or 14+144\begin{align*}\frac{14+14}{4}\end{align*}
This is an expression for the girls.
Now we can solve for the number on each team.
Boys=(11+13)÷4=24÷4=6\begin{align*}\text{Boys} = (11 + 13) \div 4 = 24 \div 4 = 6\end{align*} boys on each team
Girls=(14+14)÷4=28÷4=7\begin{align*}\text{Girls} = (14 + 14) \div 4 = 28 \div 4 = 7\end{align*} girls on each team
Now our work is complete.
### Vocabulary
Evaluate
to figure out the value of a numerical or variable expression.
Equation
a mathematical statement with an equals sign where one side of the equation has the same value as the other side.
Expression
a group of numbers, symbols and variables that represents a quantity.
Numerical Expression
a group of numbers and operations.
Variable Expression
a group of numbers, operations and at least one variable.
Variable
a letter used to represent an unknown quantity.
### Guided Practice
Here is one for you to try on your own.
Evaluate the expression 142÷7+3b4\begin{align*}14 \cdot 2 \div 7 + 3b - 4\end{align*} when b=12\begin{align*}b = 12\end{align*}.
Solution
First, we substitute the value of b\begin{align*}b\end{align*} into our variable expression.
142÷7+3(12)4\begin{align*}14 \cdot 2 \div 7 + 3(12) - 4\end{align*}
Now we follow the order of operations by performing multiplication and division in order from left to right.
142=28\begin{align*}14 \cdot 2 = 28\end{align*}
28÷7=4\begin{align*}28 \div 7 = 4\end{align*}
Let’s rewrite what we have so far so we don’t get confused.
4+3(12)4\begin{align*}4 + 3(12) - 4\end{align*}
OH! There’s more multiplication to do!
3(12)=36\begin{align*}3(12) = 36\end{align*}
Now our expression is:
4+364\begin{align*}4 + 36 - 4\end{align*}
Our last step is to perform addition and subtraction in order from left to right.
4+36=404=36\begin{align*}4 + 36 = 40 - 4 = 36\end{align*}
### Practice
Directions: Evaluate each numerical expression using the order of operations.
1. 4+523\begin{align*}4 + 5 \cdot 2 - 3\end{align*}
2. 6+63÷27\begin{align*}6 + 6 \cdot 3 \div 2 - 7\end{align*}
3. 5+58÷2+6\begin{align*}5 + 5 \cdot 8 \div 2 + 6\end{align*}
4. 1332+82\begin{align*}13 - 3 \cdot 2 + 8 - 2\end{align*}
5. 1753+8÷2\begin{align*}17 - 5 \cdot 3 + 8 \div 2\end{align*}
6. 9+42+71\begin{align*}9 + 4 \cdot 2 + 7 - 1\end{align*}
7. 8+56+243\begin{align*}8 + 5 \cdot 6 + 2 \cdot 4 - 3\end{align*}
8. 19+2432+10\begin{align*}19 + 2 \cdot 4 - 3 \cdot 2 + 10\end{align*}
9. 12+44÷83\begin{align*}12 + 4 \cdot 4 \div 8 - 3\end{align*}
10. 122+16÷212\begin{align*}12 \cdot 2 + 16 \div 2 - 12\end{align*}
Directions: Evaluate each variable expression. Remember to use the order of operations when necessary.
1. 4y+62\begin{align*}4y+6-2\end{align*}, when y=6\begin{align*}y=6\end{align*}
2. 9+3x5+2\begin{align*}9+3x-5+2\end{align*}, when x=8\begin{align*}x=8\end{align*}
3. 6y+2y5\begin{align*}6y+2y-5\end{align*}, when y=3\begin{align*}y=3\end{align*}
4. 8+3y52\begin{align*}8+3y-5 \cdot 2\end{align*}, when y=4\begin{align*}y=4\end{align*}
5. 7x23÷3+12\begin{align*}7x-2 \cdot 3 \div 3+12\end{align*}, when x=5\begin{align*}x=5\end{align*}
6. 3+432y+5\begin{align*}3+4 \cdot 3 - 2y+5\end{align*}, when y=7\begin{align*}y=7\end{align*}
7. 6a+3(2)+54\begin{align*}6a+3(2)+5 - 4\end{align*}, when a=9\begin{align*}a=9\end{align*}
8. 10+35+29b\begin{align*}10+3 \cdot 5+2-9b\end{align*}, when b=2\begin{align*}b=2\end{align*}
9. 14÷2+3a+7a\begin{align*}14 \div 2+3a+7a\end{align*}, when a=2\begin{align*}a=2\end{align*}
10. 5+6y2y+114\begin{align*}5+6y-2y+11-4\end{align*}, when y=3\begin{align*}y=3\end{align*}
Directions: Evaluate each expression using the order of operations. Remember to pay attention to the grouping symbols.
1. 3+(4+5)6(2)\begin{align*}3 + (4 + 5) - 6(2)\end{align*}
2. 4+(6÷3)+2(7)4\begin{align*}4 + (6 \div 3) + 2(7) - 4\end{align*}
3. 3+2(4+2)5(2)\begin{align*}3 + 2(4 + 2) - 5(2)\end{align*}
4. 7+(3+2)5+8(3)\begin{align*}7 + (3 + 2) - 5 + 8(3)\end{align*}
5. 4(2)+(3+9)4\begin{align*}4(2) + (3 + 9) - 4\end{align*}
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
Brackets
Brackets [ ], are symbols that are used to group numbers in mathematics. Brackets are the 'second level' of grouping symbols, used to enclose items already in parentheses.
Equation
An equation is a mathematical sentence that describes two equal quantities. Equations contain equals signs.
Evaluate
To evaluate an expression or equation means to perform the included operations, commonly in order to find a specific value.
Expression
An expression is a mathematical phrase containing variables, operations and/or numbers. Expressions do not include comparative operators such as equal signs or inequality symbols.
Grouping Symbols
Grouping symbols are parentheses or brackets used to group numbers and operations.
nested parentheses
Nested parentheses describe groups of terms inside of other groups. By convention, nested parentheses may be identified with other grouping symbols, such as the braces "{}" and brackets "[]" in the expression $\{ 3 + [ 2 - ( 5 + 4 ) ] \}$. Always evaluate parentheses from the innermost set outward.
Numerical expression
A numerical expression is a group of numbers and operations used to represent a quantity.
Parentheses
Parentheses "(" and ")" are used in algebraic expressions as grouping symbols.
PEMDAS
PEMDAS (Please Excuse My Daring Aunt Sally) is a mnemonic device used to help remember the order of operations: Parentheses, Exponents, Multiplication/Division, Addition/Subtraction.
Real Number
A real number is a number that can be plotted on a number line. Real numbers include all rational and irrational numbers.
Variable
A variable is a symbol used to represent an unknown or changing quantity. The most common variables are a, b, x, y, m, and n.
Variable Expression
A variable expression is a mathematical phrase that contains at least one variable or unknown quantity.
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# Factors of 599: Prime Factorization, Methods, and Examples
The factors that can be divided by 599 in all possible ways are known as 599 factors. The integer whose multiple is 599 is another way to describe the599 factor.
It can also be categorized as a Prime, which denotes the presence of only two Integers.
### Factors of 599
Here are the factors of number 599.
Factors of 599: 1 and 599
### Negative Factors of 599
The negative factors of 599 are similar to their positive aspects, just with a negative sign.
Negative Factors of 599: -1 and -599
### Prime Factorization of 599
The prime factorization of 599 is the way of expressing its prime factors in the product form.
Prime Factorization: 1 x 599
In this article, we will learn about the factors of 599 and how to find them using various techniques such as upside-down division, prime factorization, and factor tree.
## What Are the Factors of 599?
The factors of 599 are 1 and 599. These numbers are the factors as they do not leave any remainder when divided by 599.
The factors of 599 are classified as prime numbers and composite numbers. The prime factors of the number 599 can be determined using the prime factorization technique.
## How To Find the Factors of 599?
You can find the factors of 599 by using the rules of divisibility. The divisibility rule states that any number, when divided by any other natural number, is said to be divisible by the number if the quotient is the whole number and the resulting remainder is zero.
To find the factors of 599, create a list containing the numbers that are exactly divisible by 599 with zero remainders. One important thing to note is that 1 and 599 are the 599’s factors as every natural number has 1 and the number itself as its factor.
1 is also called the universal factor of every number. The factors of 599 are determined as follows:
$\dfrac{ 599}{1} = 599$
$\dfrac{ 599}{ 599} = 1$
Therefore, 1 and 599 are the factors of 599.
### Total Number of Factors of 599
For 599, there are 2 positive factors and 2 negative ones. So in total, there are 4 factors of 599.
To find the total number of factors of the given number, follow the procedure mentioned below:
1. Find the factorization/prime factorization of the given number.
2. Demonstrate the prime factorization of the number in the form of exponent form.
3. Add 1 to each of the exponents of the prime factor.
4. Now, multiply the resulting exponents together. This obtained product is equivalent to the total number of factors of the given number.
By following this procedure, the total number of factors of 599 is given as:
Factorization of 599 is 1 x 599.
The exponent of 1 and 599 is 1.
Adding 1 to each and multiplying them together results in 4.
Therefore, the total number of factors of 599 is 4. 2 is positive, and 2 factors are negative.
### Important Notes
Here are some essential points that must be considered while finding the factors of any given number:
• The factor of any given number must be a whole number.
• The factors of the number cannot be in the form of decimals or fractions.
• Factors can be positive as well as negative.
• Negative factors are the additive inverse of the positive factors of a given number.
• The factor of a number cannot be greater than that number.
• Every even number has 2 as its prime factor, the smallest prime factor.
## Factors of 599 by Prime Factorization
The number 599 is a prime number. Prime factorization is a valuable technique for finding the number’s prime factors and expressing the number as the product of its prime factors.
Before finding the factors of 599 using prime factorization, let us find out what prime factors are. Prime factors are the factors of any given number that are only divisible by 1 and themselves.
To start the prime factorization of 599, start dividing by its most minor prime factor. First, determine that the given number is either even or odd. If it is an even number, then 2 will be the smallest prime factor.
Continue splitting the quotient obtained until 1 is received as the quotient. The prime factorization of 599 can be expressed as:
599 = 1 x 599
## Factors of 599 in Pairs
The factor pairs are the duplet of numbers that, when multiplied together, result in the factorized number. Factor pairs can be more than one depending on the total number of factors given.
For 599, the factor pairs can be found as:
1 x 599 = 599
The possible factor pairs of 599 are given as (1, 599) ).
All these numbers in pairs, when multiplied, give 599 as the product.
The negative factor pairs of 599 are given as:
-1 x – 599 = 599
It is important to note that in negative factor pairs, the minus sign has been multiplied by the minus sign, due to which the resulting product is the original positive number. Therefore,-1 and -599 are called negative factors of 599.
The list of all the factors of 599, including positive as well as negative numbers, is given below.
Factor list of 599: 1,-1, 599, and – 599
## Factors of 599 Solved Examples
To better understand the concept of factors, let’s solve some examples.
### Example 1
How many factors of 599 are there?
### Solution
The total number of Factors of 599 is 4.
Factors of 599 are 1 and 599.
### Example 2
Find the factors of 599 using prime factorization.
### Solution
The prime factorization of 599 is given as:
599 $\div$ 1 = 599
599 $\div$ 599 = 1
So the prime factorization of 599 can be written as:
1 x 599 = 599 |
2016-06-18T21:20:58+08:00
Definition Of Union Of Sets
Combining all the elements of any two sets is called the Union of those sets
Union of two sets A and B is obtained by combining all the members of the sets and is represented as A ∪ B
More About Union of SetsIn the union of sets, element is written only once even if they exist in both the sets.Union of two sets is commutative i.e. if A and B are two sets, then A ∪ B = B ∪ AUnion of sets is also associative. If A, B, and C are three sets, then A ∪ (B ∪ C) = (A ∪ B) ∪ C
Examples of Union of SetsIf A = {1, 2, 3, 4, 5} and B = {2, 4, 6}, then the union of these sets is A ∪ B = {1, 2, 3, 4, 5, 6}
Basic operations
There are several fundamental operations for constructing new sets from given sets.
UnionsThe union of A and B, denotedA ∪ BMain article: Union (set theory)
Two sets can be "added" together. The union of A and B, denoted by A ∪ B, is the set of all things that are members of either A or B.
Examples:
{1, 2} ∪ {1, 2} = {1, 2}.{1, 2} ∪ {2, 3} = {1, 2, 3}.{1, 2, 3} ∪ {3, 4, 5} = {1, 2, 3, 4, 5}
Some basic properties of unions:
A ∪ B = B ∪ A.A ∪ (B ∪ C) = (A ∪ B) ∪ C.A ⊆ (A ∪ B).A ∪ A = A.A ∪ ∅ = A.A ⊆ B if and only if A ∪ B = B.IntersectionsMain article: Intersection (set theory)
A new set can also be constructed by determining which members two sets have "in common". The intersection of A and B, denoted by A ∩ B, is the set of all things that are members of both A and B. If A ∩ B = ∅, then A and B are said to be disjoint.
The intersection of A and B, denoted A ∩ B.
Examples:
{1, 2} ∩ {1, 2} = {1, 2}.{1, 2} ∩ {2, 3} = {2}.
Some basic properties of intersections:
A ∩ B = B ∩ A.A ∩ (B ∩ C) = (A ∩ B) ∩ C.A ∩ B ⊆ A.A ∩ A = A.A ∩ ∅ = ∅.A ⊆ B if and only if A ∩ B = A.ComplementsThe relative complement
of B in AThe complement of A in UThe symmetric difference of A andBMain article: Complement (set theory)
Two sets can also be "subtracted". The relative complement of B in A (also called the set-theoretic difference of A and B), denoted by A \ B (or A − B), is the set of all elements that are members of A but not members of B. Note that it is valid to "subtract" members of a set that are not in the set, such as removing the element green from the set {1, 2, 3}; doing so has no effect.
In certain settings all sets under discussion are considered to be subsets of a given universal set U. In such cases, U \ A is called the absolute complement or simply complement of A, and is denoted by A′.
Examples:
{1, 2} \ {1, 2} = ∅.{1, 2, 3, 4} \ {1, 3} = {2, 4}.If U is the set of integers, E is the set of even integers, and O is the set of odd integers, then U \ E = E′ = O.
Some basic properties of complements:
A \ B ≠ B \ A for A ≠ B.A ∪ A′ = U.A ∩ A′ = ∅.(A′)′ = A.A \ A = ∅.U′ = ∅ and ∅′ = U.A \ B = A ∩ B′.
An extension of the complement is the symmetric difference, defined for sets A, B as
{\displaystyle A\,\Delta \,B=(A\setminus B)\cup (B\setminus A).}
For example, the symmetric difference of {7,8,9,10} and {9,10,11,12} is the set {7,8,11,12}.
Cartesian productMain article: Cartesian product
A new set can be constructed by associating every element of one set with every element of another set. The Cartesian product of two sets A and B, denoted by A × B is the set of all ordered pairs (a, b) such that a is a member of A and b is a member of B.
Examples:
{1, 2} × {red, white} = {(1, red), (1, white), (2, red), (2, white)}.{1, 2} × {red, white, green} = {(1, red), (1, white), (1, green), (2, red), (2, white), (2, green) }.{1, 2} × {1, 2} = {(1, 1), (1, 2), (2, 1), (2, 2)}.
Some basic properties of cartesian products:
A × ∅ = ∅.A × (B ∪ C) = (A × B) ∪ (A × C).(A ∪ B) × C = (A × C) ∪ (B × C).
Let A and B be finite sets. Then
| A × B | = | B × A | = | A | × | B |.
For example,
{a,b,c}×{d,e,f}={(a,d),(a,e),(a,f),(b,d),(b,e),(b,f),(c,d),(c,e),(c,f)}.
2016-06-25T16:19:22+08:00
Is a group of objects,persons or anything..... |
# The Central Limit Theorem. 1. The random variable x has a distribution (which may or may not be normal) with mean and standard deviation. 2. Simple random.
## Presentation on theme: "The Central Limit Theorem. 1. The random variable x has a distribution (which may or may not be normal) with mean and standard deviation. 2. Simple random."— Presentation transcript:
The Central Limit Theorem
1. The random variable x has a distribution (which may or may not be normal) with mean and standard deviation. 2. Simple random samples all of the same size n are selected from the population.
1. The distribution of sample means will, as the sample size increases, approach a normal distribution. 2. The mean of all sample means is the population mean. 3. The standard deviation of all sample means is.
1. For a population with any distribution, if n > 30, then the sample means have an approximately normal distribution. 2. If n ≤ 30 and the original population has a normal distribution, then the sample means have an approximately normal distribution. 3. If n ≤ 30 and the original distribution does not have a normal distribution then the methods of this section do not apply.
The Central Limit Theorem: In Pictures As the sample size increases, the distribution of sample means approaches a normal distribution.
The Central Limit Theorem: In Pictures As the sample size increases, the distribution of sample means approaches a normal distribution.
Important Points to Note As the sample size increases, the distribution of sample means tends to approach a normal distribution. The mean of the sample means is the same as the mean of the original population.
Important Points to Note As the sample size increases, the width of the graph becomes narrower, showing that the standard deviation of the sample mean becomes smaller.
Think of what happens when we have a normal distribution. What can we do? Table A2 Calculate any area, therefore any probability. The central limit theorem shows us that most samples can fit a normal distribution. So our calculation power (and therefore understanding) is endless!!!
INDIVIDUAL VALUESAMPLE OF VALUES When working with an individual value from a normally distributed population, use the methods of Section 6.3. When working with a mean for some sample be sure to use the value of for the standard deviation of the sample mean.
Previously, we talked about the Baltimore water taxi that sank because of old weight limit standards. The water taxi assumed that the average person weight was 140lbs. Given that today the weights of men are normally distributed with a mean of 172 lbs and a standard deviation of 29 lbs. Find the probability that 20 randomly selected men will have a mean weight that is greater than 175lbs (so that their total weight exceeds the current safe capacity of 3500 lbs).
Previously, we talked about the Baltimore water taxi that sank because of old weight limit standards. The water taxi assumed that the average person weight was 140lbs. Given that today the weights of men are normally distributed with a mean of 172 lbs and a standard deviation of 29 lbs. Find the probability that if an individual man is random selected, his weight will be greater than 175lbs.
What does this mean for our water taxi and their current weight limits?
The central limit theorem works if the sample size is greater than 30, or if the original population is normally distributed.
Previously, we talked about the Baltimore water taxi that sank because of old weight limit standards. The water taxi assumed that the average person weight was 140lbs. Given that today the weights of men are normally distributed with a mean of 172 lbs and a standard deviation of 29 lbs. Find the probability that 20 randomly selected men will have a mean weight that is greater than 175lbs (so that their total weight exceeds the current safe capacity of 3500 lbs).
Pg. 295-296 #2, 5-7
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INEQUALITY WORD PROBLEMS
Problem 1 :
Sum of a number and 5 is less than -12. Find the number.
Solution :
Let x be the number.
Step 1 :
Write the inequality.
x + 5 < -12
Step 2 :
Solve the inequality using Subtraction Property of Inequality.
Subtract 5 on from both sides.
(x + 5) - 5 < -12 - 5
x < -17
So, the number is any value less than -17.
Problem 2 :
David has scored 110 points in the first level of a game. To play the third level, he needs more than 250 points. To play third level, how many points should he score in the second level ?
Solution :
Let x be points scored in the second level.
Step 1 :
Points scored scored in the second level = x
Total points in the first two levels = x + 110
Step 2 :
Write the inequality.
To play third level, the total points in the first two levels should be more than 250. So, we have
x + 110 > 250
Subtract 110 on from both sides.
(x + 110) - 110 > 250 - 110
x > 140
So, he has to score more than 140 points in the second level.
Problem 3 :
An employer recruits experienced (x) and fresh workmen (y) for his firm under the condition that he cannot employ more then 9 people. If 5 freshmen are recruited, how many experienced men have to be recruited ?
Solution :
Step 1 :
Write the inequality.
x + y ≤ 9
Step 2 :
Substitute 5 for y.
x + 5 ≤ 9
Subtract 5 from both sides.
(x + 5) - 5 ≤ 9 - 5
x ≤ 4
To meet the given condition, no. of freshmen to be recruited can be less than or equal to 4.
Problem 4 :
On the average, experienced person does 5 units of work while a fresh one (y) does 3 units of work daily. But the employer has to maintain an output of at least 30 units of work per day. How can this situation be expressed ?
Solution :
Let x and y be the number of experienced person and fresh workmen respectively.
Step 1 :
From the given information, we have
Total number of units of work done by experienced person per day is
= 5x
Total number of units of work done by fresh one per day is
= 3y
Step 2 :
Total number of units of work done by both experienced person and fresh one per day is
= 5x + 3y
As per the question, total number of units of work per day should be at least 30 units.
That is, total number of units of work (5x+3y) should be equal to 30 or more than 30.
So, we have 5x + 3y ≥ 30.
Kindly mail your feedback to v4formath@gmail.com
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How do you solve the following system?: -2x+y=1, -x+y=-3
Dec 23, 2015
The solution for the system of equations is
color(blue)(x=-4,y=-7
Explanation:
$- 2 x + \textcolor{b l u e}{y} = 1$......equation $\left(1\right)$
$- x + \textcolor{b l u e}{y} = - 3$....equation $\left(2\right)$
Solving by elimination
Subtracting equation $2$ from $1$
$- 2 x + \cancel{\textcolor{b l u e}{y}} = 1$
$+ x - \cancel{\textcolor{b l u e}{y}} = + 3$
$- x = 4$
color(blue)(x=-4
Finding $y$ by substituting $x$ in equation $1$:
$- 2 x + y = 1$
$- 2 \times \left(- 4\right) + y = 1$
$8 + y = 1$
$y = 1 - 8$
color(blue)(y=-7 |
## The Central Limit Theorem for Sums
### Learning Outcomes
• Apply and interpret the central limit theorem for sums.
Suppose X is a random variable with a distribution that may be known or unknown (it can be any distribution) and suppose:
1. μX = the mean of Χ
2. σΧ = the standard deviation of X
If you draw random samples of size n, then as n increases, the random variable $\sum X$ consisting of sums tends to be normally distributed and
$\displaystyle {\sum{X}{\sim}{N}(n \cdot \mu_X ,\sqrt{n}\sigma_X)}$.
The central limit theorem for sums says that if you keep drawing larger and larger samples and taking their sums, the sums form their own normal distribution (the sampling distribution), which approaches a normal distribution as the sample size increases. The normal distribution has a mean equal to the original mean multiplied by the sample size and a standard deviation equal to the original standard deviation multiplied by the square root of the sample size.
The random variable ΣX has the following z-score associated with it:
1. $\sum x$ is one sum.
2. ${z}=\frac{{\sum{x}-{({n})}{({\mu}_{{X}})}}}{{{(\sqrt{{n}})}{({\mu}_{{X}})}}}$
1. ${({n})}{({\mu}_{{X}})}=\text{the mean of }\sum{X}$
2. ${\left(\sqrt{{n}}\right)}{\left({\sigma}_{{X}}\right)} =\text{the standard deviation of }\sum{X}$
To find probabilities for sums on the calculator, follow these steps.
2nd
DISTR
2:normalcdf
normalcdf(lower value of the area, upper value of the area, (n)(mean), ($\displaystyle\sqrt{{n}}$)(standard deviation))
where: mean is the mean of the original distribution standard deviation is the standard deviation of the original distribution sample size = n
### Example
An unknown distribution has a mean of 90 and a standard deviation of 15. A sample of size 80 is drawn randomly from the population.
1. Find the probability that the sum of the 80 values (or the total of the 80 values) is more than 7,500.
2. Find the sum that is 1.5 standard deviations above the mean of the sums.
Solution:
Let X = one value from the original unknown population. The probability question asks you to find a probability for the sum (or total of) 80 values.
$\displaystyle\sum{X}$ = the sum or total of 80 values. Since $\displaystyle{\mu}_{x}=90,{\sigma}_{x}=15$, and n = 80, $\sum{X}$~ N((80)(90), ($\displaystyle\sqrt{{80}}$)(15))
• mean of the sums = (n)(μX) = (80)(90) = 7,200
• standard deviation of the sums =$\displaystyle{(\sqrt{{n}})}{({\sigma}_{{X}})}={(\sqrt{{80}})}{({15})}$
• sum of 80 values = Σx = 7,500
1. Find Px > 7,500)
Px > 7,500) = 0.0127
normalcdf(lower value, upper value, mean of sums, stdev of sums) The parameter list is abbreviated(lower, upper, (n)(μX, $\displaystyle{(\sqrt{{n}})}$(σX))
normalcdf (7500, 1E99, (80)(90), $\displaystyle{(\sqrt{{80}})}$(15)) = 0.0127
Remember that 1E99 = 1099.
Press the EE key for E.
2. Find Σx where z = 1.5.
$\displaystyle{\sum{x}}={(n)}{({\mu}_{X})}+{(z)}{(\sqrt{n})}{({\sigma}_{X}})=(80)(90)+(1.5)(\sqrt{80})(15)=7401.2$
### Try it
An unknown distribution has a mean of 45 and a standard deviation of eight. A sample size of 50 is drawn randomly from the population. Find the probability that the sum of the 50 values is more than 2,400.
Solution:
0.0040
To find percentiles for sums on the calculator, follow these steps.
2nd DIStR
3:invNormk = invNorm (area to the left of k, (n)(mean), (standard deviation))
where: k is the kth percentilemean is the mean of the original distribution standard deviation is the standard deviation of the original distribution sample size = n
### Example
In a recent study reported Oct. 29, 2012 on the Flurry Blog, the mean age of tablet users is 34 years. Suppose the standard deviation is 15 years. The sample of size is 50.
1. What are the mean and standard deviation for the sum of the ages of tablet users? What is the distribution?
2. Find the probability that the sum of the ages is between 1,500 and 1,800 years.
3. Find the 80th percentile for the sum of the 50 ages.
Solution:
1. $\displaystyle{\mu}_{\sum{X}}={n}{\mu}_{X}={50(34)}={1,700}$ and $\displaystyle{\sigma}_{\sum{X}}={\sigma}_{X}=({\sqrt{50}})({15})={106.01}$, The distribution is normal for sums by the central limit theorem.
2. P(1500< $\displaystyle{\sum{x}}$ = normalcdf(1500, 1800, (50)(30), ($\displaystyle{\sqrt{50}}$)(15)) = 0.7974
3. Let k = the 80th percentile. k = invNorm(0.80, 50, 34, $\displaystyle{\sqrt{50}}$, (15))=1789.3
### Try it
In a recent study reported Oct.29, 2012 on the Flurry Blog, the mean age of tablet users is 35 years. Suppose the standard deviation is ten years. The sample size is 39.
1. What are the mean and standard deviation for the sum of the ages of tablet users? What is the distribution?
2. Find the probability that the sum of the ages is between 1,400 and 1,500 years.
3. Find the 90th percentile for the sum of the 39 ages.
Solution:
1. $\displaystyle{\mu}_{\sum{ X }} = {n}{\mu}_{X } = {1365}$ and ${\sigma}_{\sum{X }} = {\sigma}_{X }=({\sqrt{n{\sigma}_{x }}})({ 15 })={62.4}$. The distribution is normal for sums by the central limit theorem.
2. P(1400<$\displaystyle\sum{x}$<1500) = normalcdf(1400, 1500, (39)(35), 10)= 0.2723
3. Let k = the 90th percentile. invNorm(0.90,(39)(35), (10)) = 1445.0
### Example
The mean number of minutes for app engagement by a tablet user is 8.2 minutes. Suppose the standard deviation is one minute. Take a sample of size 70.
1. What are the mean and standard deviation for the sums?
2. Find the 95th percentile for the sum of the sample. Interpret this value in a complete sentence.
3. Find the probability that the sum of the sample is at least ten hours.
Solution:
1. $\displaystyle{\mu}_{\sum{X}}={n}{\mu}_{X}={70(8.2)} = 574\text{ minutes }$ and $\displaystyle{\sigma}_{\sum{X}}={\sigma}_{X}=({\sqrt{n{\sigma}_{x}}})({\sqrt{70}}){(1)}={\sqrt{70}(1)}=8.37$ minutes.
2. Let k = the 95th percentile., k = invNorm (0.95,(70)(8.2),($\displaystyle\sqrt{70}$)(1)) = 587.76 minutes. Ninety five percent of the app engagement times are at most 587.76 minutes.
3. Ten hours = 600 minutes, P($\displaystyle\sum{x}\geq{600}$) = normalcdf(600,E99,(70)(8.2)$\displaystyle\sqrt{70}$)(1)) =0.0009
### Example
The mean number of minutes for app engagement by a table use is 8.2 minutes. Suppose the standard deviation is one minute. Take a sample size of 70.
1. What is the probability that the sum of the sample is between seven hours and ten hours? What does this mean in context of the problem?
2. Find the 84th and 16th percentiles for the sum of the sample. Interpret these values in context.
Solution:
1. 7 hours = 420 minutes
10 hours = 600 minutes
normalcdf P(420$\displaystyle\leq{\sum{x}}\leq{600}$)=normalcdf(420, 600, (70)(8.2), $\displaystyle\sqrt{70}(1))$= 0.9991. This means that for this sample sums there is a 99.9% chance that the sums of usage minutes will be between 420 minutes and 600 minutes.
2. invNorm (0.84,(70)(8.2),($\displaystyle\sqrt{70}$)(1))=582.32, invNorm (0.16,(70)(8.2),($\displaystyle\sqrt{70}$)(1))=565.68.
3. Since 84% of the app engagement times are at most 582.32 minutes and 16% of the app engagement times are at most 565.68 minutes, we may state that 68% of the app engagement times are between 565.68 minutes and 582.32 minutes.
## References
Farago, Peter. “The Truth About Cats and Dogs: Smartphone vs Tablet Usage Differences.” The Flurry Blog, 2013. Posted October 29, 2012. Available online at http://blog.flurry.com (accessed May 17, 2013).
# Concept Review
The central limit theorem tells us that for a population with any distribution, the distribution of the sums for the sample means approaches a normal distribution as the sample size increases. In other words, if the sample size is large enough, the distribution of the sums can be approximated by a normal distribution even if the original population is not normally distributed. Additionally, if the original population has a mean of μX and a standard deviation of σx, the mean of the sums is x and the standard deviation is $\displaystyle(\sqrt{n})({\sigma}_{x})$ where n is the sample size.
# Formula Review
The Central Limit Theorem for Sums: $\displaystyle\sum{X}{\sim}{N}[{{({n})}{({\mu}_{{X}})},{(\sqrt{{n}})}{({\mu}_{{X}})}}]$
Mean for Sums: $\displaystyle(\sum{X}):({n})({\mu}_{x})$The Central Limit Theorem for Sums z-score and standard deviation for sums: z for the sample mean: $\displaystyle{z}=\frac{{\sum{x}-{({n})}{({\mu}_{{X}})}}}{{{(\sqrt{{n}})}{({\sigma}_{{X}})}}}$
Standard deviation for Sums: $\displaystyle(\sum{X}):({\sqrt{n}})({\sigma}_{x})$ |
# One-step inequality word problem
CCSS Math: 7.EE.B.4b
## Video transcript
A contractor is purchasing some stone tiles for a new patio. Each tile costs $3, and he wants to spend less than$1,000. And it's less than $1,000, not less than or equal to$1,000. The size of each tile is one square foot. Write an inequality that represents the number of tiles he can purchase with a $1,000 limit. And then figure out how large the stone patio can be. So let x be equal to the number of tiles purchased. And so the cost of purchasing x tiles, they're going to be$3 each, so it's going to be 3x. So 3x is going to be the total cost of purchasing the tiles. And he wants to spend less than $1,000. 3x is how much he spends if he buys x tiles. It has to be less than$1,000, we say it right there. If it was less than or equal to, we'd have a little equal sign right there. So if we want to solve for x, how many tiles can he buy? We can divide both sides of this inequality by 3. And because we're dividing or multiplying-- you could imagine we're multiplying by 1/3 or dividing by 3 -- because this is a positive number, we do not have to swap the inequality sign. So we are left with x is less than 1,000 over three, which is 333 and 1/3. So he has to buy less than 333 and 1/3 tiles, that's how many tiles, and each tile is one square foot. So if he can buy less than 333 and 1/3 tiles, then the patio also has to be less than 333 and 1/3 square feet. Feet squared, we could say square feet. And we're done. |
# The Inverse of a 2 x 2 matrix
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0/2
##### Intros
###### Lessons
1. The Inverse of a 2 x 2 matrix overview:
2. Are the two matrices inverses?
3. Finding the inverse of a 2 x 2 matrix
0/7
##### Examples
###### Lessons
1. Checking if the two matrices are inverses
Check that and are inverses.
1. Check that and are inverses.
1. Check that and are inverses.
1. Finding the inverse of a matrix
Find the inverse of the matrix
1. Find the inverse of the matrix
1. Find the inverse of the matrix
1. Find the inverse of the matrix
###### Topic Notes
In this lesson, we will learn how to find the inverse of a 2 x 2 matrix. You will learn that if two matrices are inverses of each other, then the product of the two matrices will result in an identity matrix. Next, you will learn how to find the inverse by using the formula below. You may find that the formula is hard to memorize. There is another way to find a 2 x 2 matrix without memorizing the formula, but it would require matrix row operations. You will see this method in the section "the inverse of 3 x 3 matrices with matrix row operations". Lastly, note that the inverse of a 2 x 2 identity matrix is just the identity matrix itself.
## The inverse of a 2x2 matrix
In our past lesson we learnt that for an invertible matrix there is always another matrix which multiplied to the first, will produce the identity matrix of the same dimensions as them.
In other words, an invertible matrix is that which has an inverse matrix related to it, and if both of them (the matrix and its inverse) are multiplied together (no matter in which order), the result will be an identity matrix of the same order. In general, the condition of invertibility for a nxn matrix $A$ is:
$A \cdot A^{-1}=A^{-1}A \cdot A =I_{n}$
Therefore, if we define a 2x2 matrix $X,$ the condition for the inverse 2x2 matrix is written as:
$X \cdot X^{-1} =X^{-1} \cdot X=I_{2}$
But how do you find the inverse of a 2x2 matrix? We go through the whole process to find the inverse of a 2x2 matrix in our first section for this lesson, and then we follow that concept by making the calculations which prove it is right in the second section. So get ready and lets have some math fun!
#### How to find the inverse of a 2x2 matrix
In order to know what is the inverse of a 2x2 matrix we must start by defining a second order matrix, such as matrix $X$ shown below:
From our lesson on the the determinant of a 2x2 matrix we learnt that the determinant of $X$ is mathematically defined as:
$det (X)=ad-bc$
Now we need to find its matrix inverse $X^{-1}$. When thinking on the expression $X^{-1}$ the first idea that comes to mind is a division since an exponent of minus one in general algebra denotes a division of one by the number that has the exponent. The problem is that the operation of division using matrices does not exist given that a matrix is not a particular value, but a collection or array of multiple values, which geometrically speaking, do not even represent values in the same dimensional plane (depending on the dimensions of the matrix itself), so division by such a range of different variable characteristics cannot be defined and so we say that matrix division is undefined.
This is where the concept of inversion comes to play an important role, and so, although not division per se, a matrix inversion comes to represent a related operation which will allow you to cancel out matrices when solving systems of equations or even simple matrix multiplications.
The formula for the inverse of a 2x2 matrix $X$ is defined as:
Notice that the first factor in the right hand side composed by a division of one by a subtraction of the multiplication of the matrix elements, is equal to have a factor of one divided by the determinant of the matrix. In later lessons you will see how this particular factor occurs in general in the formula for the inverse matrix of any size of matrices.
For now, we continue to focus on the inverse of a 2x2 matrix and the next question arises: Why is an inverse matrix important, and how can it be used?
Imagine you have a matrix multiplication defined as $A \cdot B = C$, where all $A$, $B$, and $C$ are square matrices of the same order (same dimensions) and $A$ and $C$ are both known. Then you are asked to find out what the $B$ matrix is.
Intuitively you will think about dividing out matrix $A$ from both sides of the equation in order to cancel it out from the left hand side and solve for $B$; this is the method that we would use if they were regular variables after all, the problem is, these are matrices and division of matrices does not exist! What do we do? We obtain the inverse of matrix $A$ and multiply it to both sides of the equation:
$A \cdot B = C$
$A^{-1} \cdot A \cdot B=A^{-1} \cdot C$
$I_{n} \cdot B=A^{-1} \cdot C$
$B=A^{-1} \cdot C$
This particular solution allows us to observe how the inversion of matrices is the equivalent to divide one by a matrix, and thus, how it can be used to cancel out matrices in equations which require a division. We have been able to solve for matrix $B$ in equation 6 thanks to use our knowledge from equation 1: the multiplication of a matrix and its inverse, no matter in which order the factors are arranged, produces an identity matrix $I_{n}$ of the same dimensions as the original matrices. Then, applying what we learnt in our lesson about the identity matrix, we know that any matrix multiplied by an identity matrix gives a result the non-identity matrix itself. And so, we can conclude that $B$ is equal to the inverse of $A$ time $C$.
Having learnt the usage and how to get the inverse of a 2x2 matrix, let us go next into a section dedicated to prove that equations 2 and 5 are correct, if other words, let us calculate the inverse of 2x2 matrix proof with an example given matrix so you can observe the formula for inverse of 2x2 matrix in action. Later, in our last section, we will work through a series of exercises in order for you to practice.
#### Inverse of a 2x2 matrix proof
On this section we will prove how a 2x2 matrix and its inverse meet the condition defined in equation 2. For that we define matrix $A$ as shown below:
The first part of our proof is to verify this matrix is in fact an invertible matrix, for that, we obtain its determinant:
Since our matrix $A$ has a determinant which is not equal to zero, we can determine $A$ is an invertible matrix and so, we can finally calculate its inverse. For that, we use equation 5 and obtain:
Hence, now we can finally prove the expression found in equation 2 by multiplying matrix $A$ with is inverse.
And so, as you can see, the matrix multiplication between matrix $A$ and its inverse produces the identity matrix with the same dimensions as them. Therefore we have proved that the expressions providing the conditions for invertibility of a matrix (shown in both equation 1 and equation 2), hold true.
Remember the expression found in equation 1 provides de invertibility condition in general, meaning, it applies to square matrices of any order (dimensions). In later lessons we will look at how to compute the inverse of 3x3 matrices with matrix row operations. Notice we have not used row operations during the calculation of a 2x2 inverse, so, why trying it with a different method when the matrix gets a little bigger?
The reason is that we could use a process to find the inverse of a larger matrix based on the same principle as the one we are using today for a 2x2 matrix (which would also contain a factor of one over the determinant of the matrix), BUT, as the original matrix gets bigger, such process becomes too large and time consuming to be practical. So, we will be looking into a method using row operations because we believe is a much more practical approach.
For now, we will continue to focus on the inverse of 2x2 matrices only, so let us continue.
#### Example 1
Using the matrices $X$ and $Y$ provided below:
Check if the two matrices are inverses.
We attack this problem by remembering that two matrices are inverses of each other they will produce an identity matrix of their same dimensions when being multiplied, just as described in equation 2 for the case of 2x2 matrices. Therefore, let us use that expression and multiply matrices $X$ and $Y$ in order to see what they produce:
As you can see, the product of matrices $X$ and $Y$ happens to be the identity matrix of second order, therefore, these two matrices are inverses of each other.
#### Example 2
Just as done in problem example 1, use the two matrices ($A$ and $B$) defined below and show if they are inverses of each other.
So, we multiply $A$ and $B$:
Since the product of $A$ and $B$ is an identity matrix, then that means $A$ and $B$ are two matrices inverses of each other.
#### Example 3
Once more, we use the matrices $A$ and $B$ provided below:
And check if the two matrices are inverses by multiplying them:
And once more, we have proved that $A$ and $B$ are inverses of each other due the identity matrix being the result of their multiplication.
Now, before we continue with some other exercises, it is important to note something: The order in which you multiply matrices which are inverses of each other does not make a change on their product, their multiplication will still produce the identity matrix.Therefore, if a square matrix is invertible, the matrix multiplication of such matrix and its inverse is commutative (no matter the order in which they are multiplied, they always produce the same result: the identity). You can check this on your own and we recommend that you to do it for practice.
***
For the next four example exercises we will be using the general formula for inverse of 2x2 matrix as shown in equation 5, in order to obtain the result in each case.
#### Example 4
Find the inverse of 2x2 matrix $X$ defined below:
For this, as mentioned before, we use equation 5 (inverse of 2x2 matrix formula) assuming the matrix X follows the element notation from equation 3. Therefore, the computation of the 2x2 inverse matrix goes as follows:
#### Example 5
Having matrix A as defined below:
Calculate its 2x2 matrix inverse:
#### Example 6
Calculate the matrix inverse 2x2 of F, which is defined below:
Using equation 5 we obtain:
#### Example 7
If $C$ is defined as the identity matrix of second order (just as shown below). What is the inverse of matrix 2x2 on this case?
We truly do not need to make any calculations to answer this question since during our lesson on the 2x2 invertible matrix we learnt that the identity matrix is an involutory matrix (or just an involution for simplicity). Remember that an involutory matrix is that which multiplied by itself (squaring the matrix) produces the identity matrix, and so, an involutory matrix is that which is its own inverse. Therefore, we can quickly conclude the inverse of $C$ for this case will be $C$ itself.
We have even proved this before by multiplying the 2x2 identity matrix by itself generating the identity matrix once more, such as shown here:
For the purpose of clarity, let us not work the inverse using the formula for the inverse of a matrix 2x2 shown in equation 5 so we can see the same result will be obtained:
So, as predicted before, the inverse of matrix 2x2 named $C$ is $C$ itself due to involution.
***
Now that we have learnt what an invertible matrix is and how to obtain it, is time to find a use for it and that is what our next lesson is all about, so get ready!
Before we go, let us recommend you an extra resource containing information and extra example exercises about inverse matrices, besides it, you could also visit this handout on the inverse of a matrix and Cramers rule.
So, this is it for today, see you on our next lesson!
Let the matrices $X$ and $Y$ be inverses. Then that means the following is true:
$XY=I$
where $I$ is the identity matrix. To be more precise, we can say that since $X$ and $Y$ are inverses, then $Y$ is the same as $X^{-1}$, and so we can say that
$XX^{-1}=I$
Let $X$be a matrix and you want to find the inverse (denote as $X^{-1}$ ). Then we use the following formula: |
# How do you simplify (7i)/(8+i)?
Sep 2, 2016
$\frac{7 i}{8 + i} = \frac{7}{65} + \frac{56}{65} i$
#### Explanation:
To simplify $\frac{7 i}{8 + i}$, we should multiply numerator and denominator by the complex conjugate of the denominator.
As complex conjugate of the denominator $\left(8 + i\right)$ is $\left(8 - i\right)$, we have
(7i)/(8+i)=(7i×(8-i))/((8+i)(8-i))
= $\frac{56 i - 7 {i}^{2}}{64 + 8 i - 8 i - {i}^{2}}$
= (56i-7×(-1))/(64+1)
= $\frac{7 + 56 i}{65}$
= $\frac{7}{65} + \frac{56}{65} i$
Sep 4, 2016
$\frac{7}{65} + \frac{56}{65} i$
#### Explanation:
We have: $\frac{7 i}{8 + i}$
Let's multiply both the numerator and the denominator by the complex conjugate of the denominator:
$= \frac{7 i}{8 + i} \cdot \frac{8 - i}{8 - i}$
$= \frac{\left(7 i\right) \left(8\right) + \left(7 i\right) \left(- i\right)}{{\left(8\right)}^{2} - {\left(i\right)}^{2}}$
$= \frac{56 i - 7 {i}^{2}}{64 - {i}^{2}}$
Let's apply the fact that ${i}^{2} = - 1$:
$= \frac{56 i - \left(7 \cdot \left(- 1\right)\right)}{64 - \left(- 1\right)}$
$= \frac{56 i + 7}{65}$
$= \frac{7}{65} + \frac{56}{65} i$ |
# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2004 | Oct-Nov | (P2-9709/02) | Q#7
Question
The diagram shows the curve y = 2ex + 3e-2x.
The curve cuts the y-axis at A.
i. Write down the coordinates of A.
ii. Find the equation of the tangent to the curve at A, and state the coordinates of the point where this tangent meets the x-axis.
iii. Calculate the area of the region bounded by the curve and by the lines x = 0, y = 0 and x = 1, giving your answer correct to 2 significant figures.
Solution
i.
We are given that curve with equation cuts the y-axis at point A and we are required to find the coordinates of point A.
It is evident that point A is the y-intercept of the curve.
The point at which curve (or line) intercepts y-axis, the value of . So we can find the value of coordinate by substituting in the equation of the curve (or line).
Therefore;
Hence, coordinates of point A are (0,5).
ii.
We are required to find the equation of the tangent to the curve at point A and find coordinates of its x-intercept.
To find the equation of the line either we need coordinates of the two points on the line (Two-Point form of Equation of Line) or coordinates of one point on the line and slope of the line (Point-Slope form of Equation of Line).
We have found coordinates of point A on the tangent (and curve) as (0,5).
Therefore, we only need slope of the tangent to write its equation.
The slope of a curve at a particular point is equal to the slope of the tangent to the curve at the same point;
Therefore, if we find gradient (slope) of the curve at point A where tangent intersects the curve, then we can find slope of the tangent.
Hence, we need gradient of the curve at point (0,5).
Gradient (slope) of the curve at the particular point is the derivative of equation of the curve at that particular point.
Gradient (slope) of the curve at a particular point can be found by substituting x- coordinates of that point in the expression for gradient of the curve;
We are given equation of the curve as;
Therefore;
Rule for differentiation of is:
Therefore;
Rule for differentiation natural exponential function , or ;
Rule for differentiation of is:
We are looking for gradient of the curve at point A(0,5).
Therefore;
Therefore slope of the curve at point A(0,5) is;
Hence, slope of the tangent to the curve at this point is;
Now we can write equation of the tangent as follows.
Point-Slope form of the equation of the line is;
Next we are required to find the coordinates of x-intercept of this tangent to the curve at point A.
The point at which curve (or line) intercepts x-axis, the value of . So we can find the value of coordinate by substituting in the equation of the curve (or line).
Therefore coordinates of x-intercept of the tangent to the curve at point A(0,5) are .
iii.
To find the area of region under the curve , we need to integrate the curve from point to along x-axis.
It is evident from the given data, area of the region is bounded by the curve and by the lines x = 0, y = 0 and x = 1, that area is under the curve along x-axis from x=0 to x=1.
Therefore;
Rule for integration of is:
Rule for integration of , or ;
Therefore;
The answer correct to 2 significant figures; |
After you’ve solved an equation, you can check the answer by substituting the answer you got for $x$ into the equation. Then, you check to see if you get the same result on both the left-hand and the right-hand side.
Example 1
Check if $x=3$ is a solution to $\frac{x}{3}+1=4-\frac{2x}{3}$
Substituting $x=3$ on the left-hand side gives you
$\frac{x}{3}+1=\frac{3}{3}+1=1+1=2$
Substituting $x=3$ on the right-hand side gives you
$\begin{array}{llll}\hfill 4-\frac{2x}{3}& =4-\frac{2\cdot 3}{3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =4-\frac{6}{3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =4-2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
$4-\frac{2x}{3}=4-\frac{2\cdot 3}{3}=4-\frac{6}{3}=4-2=2$
This means that both the left-hand side and right-hand side are equal to 2 when you substitute 3 for $x$. You’ve now checked the answer, and since both sides are equal, it means that $x=3$ is a correct solution.
Note! It is possible for there to be more than one solution for $x$ in many equations, especially those where $x$ is raised to a power. |
# What's the solution of the functional equation
I need help with this:
“Find all functions $f$, $g : \mathbb{Z} \rightarrow \mathbb{Z}$, with $g$ injective and such that:
$$f(g(x)+y) = g(f(x)+y), \mbox{ for all } x, y \in \mathbb{Z}.$$
#### Solutions Collecting From Web of "What's the solution of the functional equation"
Suppose $f,g:\Bbb Z\to\Bbb Z$ with $g$ injective and $$f(g(x)+y)=g(f(x)+y)\tag{#}$$ for all $x,y\in\Bbb Z$.
Note that $$f(0)=f(g(x)-g(x))=g(f(x)-g(x))\tag{1}$$ for all $x$, so since $g$ is injective, then $f(x)-g(x)$ is constant. In particular, then $$f(x)=g(x)+c\tag{2}$$ for some $c\in\Bbb Z$. Note that it then follows that $f$ is injective.
It follows from $(1)$ and $(2)$ that $$g(c)=g(f(x)-g(x))=f(0)=g(0)+c.$$ Suppose $g(x+c)=g(x)+c$ for some $x\in\Bbb Z$. Then $$g(x+1+c)=g(f(0)-g(0)+x+1)=f(g(0)-g(0)+x+1)=f(x+1)=g(x+1)+c,$$ and similarly, $$g(x-1+c)=g(x-1)+c.$$ By $2$-directional induction on $x$, we have $$g(x+c)=g(x)+c\tag{3}$$ for all $x\in\Bbb Z$. Using $(3)$, another quick $2$-directional induction on $m$ will show that $$g(y+mc)=g(y)+mc\tag{4}$$ for all $m,y\in\Bbb Z.$
Now, if $c=0$, then we must have $f=g$ by $(2)$. Suppose $c\neq 0$. Given any $x\in\Bbb Z$, there is a unique $0\leq y<|c|$ such that $x=y+mc$ for some $m\in\Bbb Z$ (why?)–denote this $y$ by $r(x)$–so by $(4)$, we have $$g(x)=g(r(x)+mc)=g(r(x))+mc=g(r(x))+x-r(x),$$ so $$g(x)-x=g(r(x))-r(x).$$ Defining $h(x):=g(x)-x$, we have $$h(x)=h(r(x))\tag{5}$$ for all $x\in\Bbb Z$. Observing that $r(x+mc)=r(x)$ for all $x,m\in\Bbb Z$ (why?), it follows that $h$ is a periodic function $\Bbb Z\to \Bbb Z$, with $c$ a period of $h$.
The above gives us necessary conditions for $f,g$ to satisfy. We will see that they are sufficient, as well. Specifically, we prove the following:
Claim: Suppose $f,g:\Bbb Z\to\Bbb Z$ with $g$ injective. Then $f,g$ satisfy $(\#)$ if and only if one of the following holds:
(a) $f=g$, or
(b) there is some non-$0$ $c\in\Bbb Z$ and some periodic function $h:\Bbb Z\to\Bbb Z$ such that $h$ has $c$ as a period, $g(x)=h(x)+x$, and $f(x)=g(x)+c$ for all $x\in\Bbb Z$.
Proof: We saw above that if $f,g$ satisfy $(\#)$, then (a) or (b) holds. If (a) holds, then it’s easily seen that $f,g$ satisfy $(\#)$, so suppose that (b) holds. Then \begin{align}f(g(x)+y) &= g(g(x)+y)+c\\ &= h(g(x)+y)+g(x)+y+c\\ &= h(g(x)+y+c)+g(x)+y+c\\ &= g(g(x)+y+c)\\ &= g(g(x)+c+y)\\ &= g(f(x)+y),\end{align} as desired. $\Box$
How about $f(x)=g(x)=x?$…… |
# Cross Multiplication Method
In mathematics, solving linear equations is one of the important topics. We can say concept of linear equations is the base of advance algebra. Many students are scared of math and this anxiety should be the first thing that needs to be fixed as soon as possible. Many times students are afraid of asking any question related to it because they do not want others to think that they did not understand the concept. So we have designed cross multiplication method concept in such a way so that students can learn at their own pace and clear their doubts. Go through the below content and at the end you will be confident about the concept and able to solve problems at our own. Also we provide NCERT Solutions on cross multiplication method problems which are prepared by experts. We provide accurate and easy solutions for all questions covered in NCERT textbooks. Answers have been structured in a logical and easy language for quick revisions during examination or tests.
In this section, we will discuss about simultaneous linear equations by using cross-multiplication method.
## What is Cross Multiplication Method
A method can be used to determine the value of a variable from any equation. Usually in elementary algebra and elementary arithmetic, rational expressions and equations involving fractions are solved by using cross-multiplication method.
## Cross Multiplication Method Pair of Linear Equations
General form of a linear equation in two unknown quantities: ax + by + c = 0, (a, b ≠0)
Assume two linear equation for x and y variables be
$a_1$ x + $b_1$y + $c_1$ = 0 ...(1)
$a_2$ x + $b_2$ y + $c_2$ = 0 ...(2)
The coefficients of x are: $a_1$ and $a_2$
The coefficients of y are: $b_1$ and $b_2$
The constant terms are: $c_1$ and $c_2$
Use method of elimination to solve both the equations:
Equation (1) is multiplied with $b_2$
$b_2$($a_1$ x + $b_1$y + $c_1$ = 0)
$a_1$$b_2 x + b_1$$b_2$y + $c_1$$b_2 = 0 ...(3) Equation (2) is multiplied with b_1 b_1(a_2 x + b_2 y + c_2 = 0) a_2$$b_1$ x + $b_2$$b_1 y + c_2$$b_1$ = 0 ...(4)
Subtract equation 4 from equation 3, we have
($a_1$$b_2 - a_2$$b_1$) x + ($b_1$$b_2 - b_2$$b_1$)y + ($c_1$$b_2 - c_2$$b_1$) = 0
This implies
x = $\frac{ b_1\ c_2 -\ b_2\ c_1}{ b_2\ a_1 -\ a_2\ b_1}$; where ($a_1$$b_2 - a_2$$b_1$) $\neq$ 0
To obtain the value of y, substitute the value of x in equation (1),
y = $\frac{ c_1\ a_2 -\ c_2\ a_1}{a_1\ b_2\ -\ a_2\ b_1}$; where ($a_1$$b_2 - a_2$$b_1$) $\neq$ 0
From the value of x and y we can obtain the result as:
## Examples
Example: Solve below system of linear equations using cross multiplication method
3x + y = 10 and x + 2y = 5
Solution:
3x + y - 10 = 0 ...equation(1)
x + 2y - 5 = 0 ...equation (2)
Here,
a$_1$ = 3, b$_1$ = 1, c$_1$ = -10
a$_2$ = 1, b$_2$ = 2, c$_2$ = -5
Use above derived formula, to find the value of x and y.
x = $\frac{ b_1 c_2- b_2 c_1}{a_1 b_2 - a_2 b_1}$
= $\frac{-5 \times 1\ - ( -10)\ \times 2}{3\ \times 2\ -\ 1\ \times 1}$
$\frac{(-5) - (-20)}{6 -1)}$
= $\frac{15}{5}$
= 3
Value of x is 3
Now,
y = $\frac{ c_1\ a_2 -\ c_2\ a_1}{a_1\ b_2\ -\ a_2\ b_1}$
= $\frac{(-10)(1)-(-5)(3)}{(3)(2) - (1)(1)}$
= $\frac{(-10) - (-15)}{6 - 1}$
= $\frac{5}{5}$
= 1
Therefore, solution is : x = 3 and y = 1
## Practice Problems
Solve below linear equations using cross multiplication method:
Problem 1: Solve -x + y = 10 and 3x - 5y = 1
Problem 2: Find the value of x and y: 3x - 1 = 5 and 7x = y - 10
Problem 3: Solve linear equations: 1/2x - 4y - 7 = 0 and x - y = 1 |
# GRE Math : How to find the angle for a percentage of a circle
## Example Questions
### Example Question #31 : Plane Geometry
An ant begins at the center of a pie with a 12" radius. Walking out to the edge of pie, it then proceeds along the outer edge for a certain distance. At a certain point, it turns back toward the center of the pie and returns to the center point. Its whole trek was 55.3 inches. What is the approximate size of the angle through which it traveled?
149.52°
128.21°
74.76°
91.44°
81.53°
149.52°
Explanation:
To solve this, we must ascertain the following:
1) The arc length through which the ant traveled.
2) The percentage of the total circumference in light of that arc length.
3) The percentage of 360° proportionate to that arc percentage.
To begin, let's note that the ant travelled 12 + 12 + x inches, where x is the outer arc distance. (It traveled the radius twice, remember); therefore, we know that 24 + x = 55.3 or x = 31.3.
Now, the total circumference of the circle is 2πr or 24π. The arc is 31.3/24π percent of the total circumference; therefore, the percentage of the angle is 360 * 31.3/24π. Since the answers are approximations, use 3.14 for π. This would be 149.52°.
### Example Question #11 : Sectors
A study was conducted to determine the effectiveness of a vaccine for the common cold (Rhinovirus sp.). 1000 patients were studied. Of those, 500 received the vaccine and 500 did not. The patients were then exposed to the Rhinovirus and the results were tabulated.
Table 1 shows the number of vaccinated and unvaccinated patients in each age group who caught the cold.
Suppose the scientists wish to create a pie chart reflecting a patient's odds of catching the virus depending on vaccination status and age group.
All 1000 patients are included in this pie chart.
What would be the angle of the arc for the portion of the chart representing vaccinated patients of all age groups who caught the virus?
10°
18°
Insufficient information to answer this question
36°
60°
18°
Explanation:
First, we must determine what proportion of the 1000 patients were vaccinated and caught the virus. The total number of patients who were vaccinated and caught the virus is 50.
18 + 4 + 5 + 4 + 19 = 50
The proportion of the patients is represented by dividing this group by the total number of participants in the study.
50/1000 = 0.05
Next, we need to figure out how that proportion translates into a proportion of a pie chart. There are 360° in a pie chart. Multiply 360° by our proportion to reach the solution.
360° * 0.05 = 18°
The angle of the arc representing vaccinated patients who caught the virus is 18°.
### Example Question #1 : How To Find The Angle For A Percentage Of A Circle
A group of students ate an -inch pizza that was cut into equal slices. What was the angle measure needed to cut this pizza into these equal slices? |
Mathematics Part I Solutions Solutions for Class 9 Maths Chapter 4 Ratio And Proportion are provided here with simple step-by-step explanations. These solutions for Ratio And Proportion are extremely popular among Class 9 students for Maths Ratio And Proportion Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part I Solutions Book of Class 9 Maths Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part I Solutions Solutions. All Mathematics Part I Solutions Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.
#### Question 1:
From the following pairs of numbers, find the reduced form of ratio of first number to second number.
(i) 72, 60 (ii) 38, 57 (iii) 52, 78
#### Answer:
(i)
72 : 60 = $\frac{72}{60}=\frac{72÷12}{60÷12}=\frac{6}{5}$ = 6 : 5 (HCF of 72 and 60 = 12)
Thus, the reduced form of 72 : 60 is 6 : 5.
(ii)
38 : 57 = $\frac{38}{57}=\frac{38÷19}{57÷19}=\frac{2}{3}$ = 2 : 3 (HCF of 38 and 57 = 19)
Thus, the reduced form of 38 : 57 is 2 : 3.
(iii)
52 : 78 = $\frac{52}{78}=\frac{52÷26}{78÷26}=\frac{2}{3}$ = 2 : 3 (HCF of 52 and 78 = 26)
Thus, the reduced form of 52 : 78 is 2 : 3.
#### Question 2:
Find the reduced form of the ratio of the first quantity to second quantity.
(i) 700 rs, 308 rs (ii) 14 rs, 12 rs. 40 paise. (iii) 5 litre, 2500 ml (iv) 3 years 4 months, 5years 8 months (v) 3.8 kg, 1900 gm
(vi) 7 minutes 20 seconds, 5 minutes 6 seconds.
#### Answer:
(i)
Rs 700 : Rs 308 = $\frac{700}{308}=\frac{700÷28}{308÷28}=\frac{25}{11}$ = 25 : 11 (HCF of 700 and 308 = 28)
(ii)
14 rupees = 1400 p
12 rupees 40 paise = 1200 + 40 = 1240 p
∴ 14 rupees : 12 rupees 40 paise = 1400 p : 1240 p = $\frac{1400}{1240}=\frac{1400÷40}{1240÷40}=\frac{35}{31}$ = 35 : 31 (HCF of 1400 and 1240 = 40)
(iii)
5 litre = 5000 mL (1 L = 1000 mL)
∴ 5 litre : 2500 mL = 5000 mL : 2500 mL = $\frac{5000}{2500}=\frac{5000÷2500}{2500÷2500}=\frac{2}{1}$ = 2 : 1 (HCF of 5000 and 2500 = 2500)
(iv)
3 years 4 months = 3 years + 4 months = 3 × 12 + 4 = 36 + 4 = 40 months (1 year = 12 months)
5 years 8 months = 5 years + 8 months = 5 × 12 + 8 = 60 + 8 = 68 months
∴ 3 years 4 months : 5 years 8 months = 40 months : 68 months = $\frac{40}{68}=\frac{40÷4}{68÷4}=\frac{10}{17}$ = 10 : 17 (HCF of 40 and 68 = 4)
(v)
3.8 kg = 3.8 × 1000 = 3800 g (1 kg = 1000 g)
∴ 3.8 kg : 1900 g = 3800 g : 1900 g = $\frac{3800}{1900}=\frac{3800÷1900}{1900÷1900}=\frac{2}{1}$ = 2 : 1 (HCF of 3800 and 1900 = 1900)
(iv)
7 minutes 20 seconds = 7 minutes + 20 seconds = 7 × 60 + 20 = 420 + 20 = 440 seconds (1 minute = 60 seconds)
5 minutes 6 seconds = 5 minutes + 6 seconds = 5 × 60 + 6 = 300 + 6 = 306 seconds
∴ 7 minutes 20 seconds : 5 minutes 6 seconds = 440 seconds : 306 seconds = $\frac{440}{306}=\frac{440÷2}{306÷2}=\frac{220}{153}$ = 220 : 153 (HCF of 220 and 153 = 2)
#### Question 3:
Express the following percentages as ratios in the reduced form.
(i) 75 : 100
(ii) 44 : 100
(iii) 6.25%
(iv) 52 : 100
(v) 0.64%
#### Answer:
(i)
75 : 100 = $\frac{75}{100}=\frac{75÷25}{100÷25}=\frac{3}{4}$ = 3 : 4 (HCF of 75 and 100 = 25)
(ii)
44 : 100 = $\frac{44}{100}=\frac{44÷4}{100÷4}=\frac{11}{25}$ = 11 : 25 (HCF of 44 and 100 = 4)
(iii)
6.25% = $\frac{6.25}{100}=\frac{625}{10000}=\frac{625÷625}{10000÷625}=\frac{1}{16}$ = 1 : 16 (HCF of 625 and 10000 = 625)
(iv)
52 : 100 = $\frac{52}{100}=\frac{52÷4}{100÷4}=\frac{13}{25}$ = 13 : 25 (HCF of 52 and 100 = 4)
(v)
0.64% = $\frac{0.64}{100}=\frac{64}{10000}=\frac{64÷16}{10000÷16}=\frac{4}{625}$ = 4 : 625 (HCF of 64 and 10000 = 16)
#### Question 4:
Three persons can build a small house in 8 days. To build the same house in 6 days, how many persons are required?
#### Answer:
Let the number of persons required to build the same house in 6 days be x.
The number of persons and the number of days required to build the house are in inverse variation. So, the product of number of persons and the number of days required to build the house is constant.
∴ x × 6 = 3 × 8
⇒ x$\frac{3×8}{6}=\frac{24}{6}$ = 4
Thus, 4 persons are required to build the same house in 6 days.
#### Question 5:
Convert the following ratios into percentage.
(i) 15 : 25 (ii) 47 : 50 ​ (iii) $\frac{7}{10}$ (iv) $\frac{546}{600}$ (v) $\frac{7}{16}$
#### Answer:
(i)
15 : 25 = $\frac{15}{25}=\frac{15}{25}×100%$ = 15 × 4% = 60%
(ii)
47 : 50 = $\frac{47}{50}=\frac{47}{50}×100%$ = 47 × 2% = 94%
(iii)
$\frac{7}{10}=\frac{7}{10}×100%$ = 7 × 10%= 70%
(iv)
$\frac{546}{600}=\frac{546}{600}×100%=\frac{546}{6}%$ = 91%
(v)
$\frac{7}{16}=\frac{7}{16}×100%=\frac{700}{16}%$ = 43.75%
#### Question 6:
The ratio of ages of Abha and her mother is 2 : 5. At the time of Abha's birth her mothers age was 27 year. Find the present ages of Abha and her mother.
#### Answer:
The ratio of the present ages of Abha and her mother is 2 : 5.
Let the present age of Abha and her mother be 2x years and 5years, respectively.
∴ Abha's mother age at the time of Abha's birth = 5x − 2x = 3x years
It is given that at the time of Abha's birth her mothers age was 27 year.
∴ 3x = 27
⇒ x = 9
∴ Present age of Abha = 2x = 2 × 9 = 18 years
Present age of Abha's mother = 5x = 5 × 9 = 45 years
Thus, the present age of Abha and her mother is 18 years and 45 years, respectively.
#### Question 7:
Present ages of Vatsala and Sara are 14 years and 10 years respectively. After how many years the ratio of their ages will become 5 : 4?
#### Answer:
Let after x years, the ratio of their ages will become 5 : 4.
Age of Vatsala after x years = (14 + x) years
Age of Sara after x years = (10 + x) years
$\therefore \frac{14+x}{10+x}=\frac{5}{4}\phantom{\rule{0ex}{0ex}}⇒56+4x=50+5x\phantom{\rule{0ex}{0ex}}⇒5x-4x=56-50\phantom{\rule{0ex}{0ex}}⇒x=6$
Thus, the ratio of their ages will become 5 : 4 after 6 years.
#### Question 8:
The ratio of present ages of Rehana and her mother is 2 : 7. After 2 years, the ratio of their ages will be 1 : 3. What is Rehana's present age ?
#### Answer:
Let the present ages of Rehana and her mother be 2x years and 7x years, respectively.
After 2 years,
Age of Rehana = (2x + 2) years
Age of Rehana's mother = (7x + 2) years
It is given that after 2 years, the ratio of their ages will be 1 : 3.
$\therefore \frac{2x+2}{7x+2}=\frac{1}{3}\phantom{\rule{0ex}{0ex}}⇒6x+6=7x+2\phantom{\rule{0ex}{0ex}}⇒7x-6x=6-2\phantom{\rule{0ex}{0ex}}⇒x=4$
∴ Rehana's present age = 2x = 2 × 4 = 8 years
Thus, the present age of Rehana is 8 years.
#### Question 1:
Using the property , fill in the blanks substituting proper numbers in the following.
(i) (ii)
#### Answer:
(i)
28 = 7 × 4, 35 = 5 × 4, 3.5 = 7 × 0.5
Now,
(ii)
4.5 = 9 × 0.5, 42 = 14 × 3, 3.5 = 14 × 0.25
Now,
#### Question 2:
Find the following ratios.
(i) The ratio of radius to circumference of the circle.
(ii) The ratio of circumference of circle with radius r to its area.
(iii) The ratio of diagonal of a square to its side, if the length of side is 7 cm.
(iv) The lengths of sides of a rectangle are 5 cm and 3.5 cm. Find the ratio of its perimeter to area.
#### Answer:
(i)
Let the radius of the circle be r units.
∴ Circumference of the circle = 2$\mathrm{\pi }$r units
Radius of the circle : Circumference of the circle = r : 2$\mathrm{\pi }$r = $\frac{r}{2\mathrm{\pi }r}=\frac{1}{2\mathrm{\pi }}$ = 1 : 2$\mathrm{\pi }$
Thus, the ratio of radius to cirumference of the circle is 1 : 2$\mathrm{\pi }$.
(ii)
Radius of the circle = r units
∴ Circumference of the circle = 2$\mathrm{\pi }$r units
Area of the circle = $\mathrm{\pi }$r2 square units
Circumference of the circle : Area of the circle = 2$\mathrm{\pi }$$\mathrm{\pi }$r$\frac{2\mathrm{\pi }r}{\mathrm{\pi }{r}^{2}}=\frac{2}{r}$ = 2 : r
Thus, the ratio of circumference of circle with radius r to its area is 2 : r.
(iii)
Side of the square = 7 cm
∴ Length of diagonal of the square = $\sqrt{2}$ × Side of the square = $7\sqrt{2}$ cm
Length of diagonal of the square : Side of the square = $7\sqrt{2}$ cm : 7 cm = $\frac{7\sqrt{2}}{7}=\frac{\sqrt{2}}{1}$ = $\sqrt{2}$ : 1
Thus, the ratio of diagonal of the square to its side is $\sqrt{2}$ : 1.
(iv)
Length of the rectangle, l = 5 cm
Breadth of the rectangle, b = 3.5 cm
∴ Perimeter of the rectangle = 2(lb) = 2 × (5 + 3.5) = 2 × 8.5 = 17 cm
Area of the rectangle = l × b = 5 × 3.5 = 17.5 cm2
Perimeter of the rectangle : Area of the rectangle = 17 : 17.5 = $\frac{17}{17.5}=\frac{170}{175}=\frac{170÷5}{175÷5}=\frac{34}{35}$ = 34 : 35
Thus, the ratio of perimeter to area of the rectangle is 34 : 35.
#### Question 3:
Compare the following pairs of ratios.
(i) (ii) (iii) (iv) (v)
#### Answer:
(i)
$\sqrt{5}×\sqrt{7}=\sqrt{5×7}=\sqrt{35}\phantom{\rule{0ex}{0ex}}3×3=9=\sqrt{81}$
Now, $\sqrt{35}<\sqrt{81}$
$\therefore \frac{\sqrt{5}}{3}<\frac{3}{\sqrt{7}}$
(ii)
$3\sqrt{5}×\sqrt{125}=3\sqrt{5}×\sqrt{25×5}=3\sqrt{5}×5\sqrt{5}=75\phantom{\rule{0ex}{0ex}}5\sqrt{7}×\sqrt{63}=5\sqrt{7}×\sqrt{9×7}=5\sqrt{7}×3\sqrt{7}=105$
Now, 75 < 105
$\therefore \frac{3\sqrt{5}}{5\sqrt{7}}<\frac{\sqrt{63}}{\sqrt{125}}$
(iii)
$5×121=605\phantom{\rule{0ex}{0ex}}18×17=306$
Now, 605 > 306
$\therefore \frac{5}{18}>\frac{17}{121}$
(iv)
$\sqrt{80}×\sqrt{27}=\sqrt{16×5}×\sqrt{9×3}=4\sqrt{5}×3\sqrt{3}=12\sqrt{15}\phantom{\rule{0ex}{0ex}}\sqrt{45}×\sqrt{48}=\sqrt{9×5}×\sqrt{16×3}=3\sqrt{5}×4\sqrt{3}=12\sqrt{15}$
Now, $12\sqrt{15}=12\sqrt{15}$
$\therefore \frac{\sqrt{80}}{\sqrt{48}}=\frac{\sqrt{45}}{\sqrt{27}}$
(v)
$9.2×7.1=65.32\phantom{\rule{0ex}{0ex}}5.1×3.4=17.34$
Now, 65.32 > 17.34
$\therefore \frac{9.2}{5.1}>\frac{3.4}{7.1}$
#### Question 4:
(i) $\square$ABCD is a parallelogram. The ratio of $\angle$A and $\angle$B of this parallelogram is 5 : 4. Find the measure of $\angle$B.
(ii) The ratio of present ages of Albert and Salim is 5 : 9. Five years hence ratio of their ages will be 3 : 5. Find their present ages.
(iii) The ratio of length and breadth of a rectangle is 3 : 1, and its perimeter is 36 cm. Find the length and breadth of the rectangle.
(iv) The ratio of two numbers is 31 : 23 and their sum is 216. Find these numbers.
(v) If the product of two numbers is 360 and their ratio is 10 : 9, then find the numbers.
#### Answer:
(i)
Quadrilateral ABCD is a parallelogram.
Let the measure of $\angle$A and $\angle$B be 5x and 4x, respectively.
Now,
$\angle$A + $\angle$B = 180º (Adjacent angles of a parallelogram are supplementary)
∴ 5x + 4x = 180º
⇒ 9x = 180º
⇒ x = 20º
∴ Measure of $\angle$B = 4x = 4 × 20º = 80º
Thus, the measure of $\angle$B is 80º.
(ii)
Let the present ages of Albert and Salim be 5x years and 9x years, respectively.
5 years hence,
Age of Albert = (5x + 5) years
Age of Salim = (9x + 5) years
It is given that five year hence, the ratio of their ages will be 3 : 5.
$\therefore \frac{5x+5}{9x+5}=\frac{3}{5}\phantom{\rule{0ex}{0ex}}⇒25x+25=27x+15\phantom{\rule{0ex}{0ex}}⇒27x-25x=25-15\phantom{\rule{0ex}{0ex}}⇒2x=10\phantom{\rule{0ex}{0ex}}⇒x=5$
∴ Present age of Albert = 5x = 5 × 5 = 25 years
Present age of Salim = 9x = 9 × 5 = 45 years
Thus, the present age of Albert is 25 years and the present age of Salim is 45 years.
(iii)
Let the length and breadth of the rectangle be 3x cm and cm, respectively.
Perimeter of the rectangle = 36 cm
∴ 2(Length + Breadth) = 36 cm
⇒ 2(3xx) = 36
⇒ 2 × 4x = 36
⇒ 8x = 36
⇒ x = 4.5
∴ Length of the rectangle = 3x = 3 × 4.5 = 13.5 cm
Breadth of the rectangle = x = 4.5 cm
Thus, the length and breadth of the rectangle is 13.5 cm and 4.5 cm, respectively.
(iv)
Let the two numbers be 31x and 23x.
Sum of the two numbers = 216
∴ 31x + 23x = 216
⇒ 54x = 216
⇒ x = $\frac{216}{54}$ = 4
∴ One number = 31x = 31 × 4 = 124
Other number = 23x = 23 × 4 = 92
Thus, the two numbers are 92 and 124.
(v)
Let the two numbers be 10x and 9x.
Product of the two numbers = 360
∴ 10x × 9x = 360
⇒ 90x2 = 360
⇒ x2 = 4
⇒ x = 2
∴ One number = 10x = 10 × 2 = 20
Other number = 9x = 9 × 2 = 18
Thus, the two numbers are 18 and 20.
#### Question 5:
If a : b = 3 : 1 and b : c = 5 : 1 then find the value of
(i) ${\left(\frac{{a}^{3}}{15{b}^{2}c}\right)}^{3}$ (ii) $\frac{{a}^{2}}{7bc}$
#### Answer:
ab = 3 : 1
.....(1)
b : c = 5 : 1
.....(2)
From (1) and (2), we have
a = 3 × 5c = 15c
(i)
${\left(\frac{{a}^{3}}{15{b}^{2}c}\right)}^{3}\phantom{\rule{0ex}{0ex}}={\left[\frac{{\left(15c\right)}^{3}}{15×{\left(5c\right)}^{2}×c}\right]}^{3}\phantom{\rule{0ex}{0ex}}={\left(\frac{15×15×15{c}^{3}}{15×25{c}^{2}×c}\right)}^{3}\phantom{\rule{0ex}{0ex}}={\left(9\right)}^{3}\phantom{\rule{0ex}{0ex}}=729$
(ii)
$\frac{{a}^{2}}{7bc}\phantom{\rule{0ex}{0ex}}=\frac{{\left(15c\right)}^{2}}{7×5c×c}\phantom{\rule{0ex}{0ex}}=\frac{225}{35}\phantom{\rule{0ex}{0ex}}=\frac{45}{7}$
#### Question 6:
If then find the ratio $\frac{a}{b}$.
#### Answer:
$⇒\frac{a}{b}=0.4×0.04\phantom{\rule{0ex}{0ex}}⇒\frac{a}{b}=\frac{4}{10}×\frac{4}{100}=\frac{16}{1000}\phantom{\rule{0ex}{0ex}}⇒\frac{a}{b}=\frac{2}{125}$
∴ ab = 2 : 125
#### Question 7:
( x + 3) : ( x + 11) = ( x $-$ 2) : ( x + 1) then find the value of x.
#### Answer:
$\left(x+3\right):\left(x+11\right)=\left(x-2\right):\left(x+1\right)\phantom{\rule{0ex}{0ex}}⇒\frac{x+3}{x+11}=\frac{x-2}{x+1}\phantom{\rule{0ex}{0ex}}⇒\left(x+3\right)×\left(x+1\right)=\left(x-2\right)×\left(x+11\right)\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+x+3x+3={x}^{2}+11x-2x-22$
$⇒4x+3=9x-22\phantom{\rule{0ex}{0ex}}⇒9x-4x=22+3\phantom{\rule{0ex}{0ex}}⇒5x=25\phantom{\rule{0ex}{0ex}}⇒x=5$
Thus, the value of x is 5.
#### Question 1:
If then find the values of the following ratios.
(i) (ii) (iii) (iv)
#### Answer:
(i)
$\frac{5a+3b}{5a-3b}\phantom{\rule{0ex}{0ex}}=\frac{5×7k+3×3k}{5×7k-3×3k}\phantom{\rule{0ex}{0ex}}=\frac{35k+9k}{35k-9k}\phantom{\rule{0ex}{0ex}}=\frac{44k}{26k}\phantom{\rule{0ex}{0ex}}=\frac{22}{13}$
(ii)
$\frac{2{a}^{2}+3{b}^{2}}{2{a}^{2}-3{b}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{2×{\left(7k\right)}^{2}+3×{\left(3k\right)}^{2}}{2×{\left(7k\right)}^{2}-3×{\left(3k\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{98{k}^{2}+27{k}^{2}}{98{k}^{2}-27{k}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{125{k}^{2}}{71{k}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{125}{71}$
(iii)
$\frac{{a}^{3}-{b}^{3}}{{b}^{3}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(7k\right)}^{3}-{\left(3k\right)}^{3}}{{\left(3k\right)}^{3}}\phantom{\rule{0ex}{0ex}}=\frac{343{k}^{3}-27{k}^{3}}{27{k}^{3}}\phantom{\rule{0ex}{0ex}}=\frac{316{k}^{3}}{27{k}^{3}}\phantom{\rule{0ex}{0ex}}=\frac{316}{27}$
(iv)
$\frac{7a+9b}{7a-9b}\phantom{\rule{0ex}{0ex}}=\frac{7×7k+9×3k}{7×7k-9×3k}\phantom{\rule{0ex}{0ex}}=\frac{49k+27k}{49k-27k}\phantom{\rule{0ex}{0ex}}=\frac{76k}{22k}\phantom{\rule{0ex}{0ex}}=\frac{38}{11}$
#### Question 2:
If then find the values of following ratios.
(i) $\frac{a}{b}$ (ii) (iii) (iv)
#### Answer:
(i)
$\frac{15{a}^{2}+4{b}^{2}}{15{a}^{2}-4{b}^{2}}=\frac{47}{7}$
Applying componendo and dividendo, we get
$\frac{\left(15{a}^{2}+4{b}^{2}\right)+\left(15{a}^{2}-4{b}^{2}\right)}{\left(15{a}^{2}+4{b}^{2}\right)-\left(15{a}^{2}-4{b}^{2}\right)}=\frac{47+7}{47-7}\phantom{\rule{0ex}{0ex}}⇒\frac{15{a}^{2}+4{b}^{2}+15{a}^{2}-4{b}^{2}}{15{a}^{2}+4{b}^{2}-15{a}^{2}+4{b}^{2}}=\frac{54}{40}\phantom{\rule{0ex}{0ex}}⇒\frac{30{a}^{2}}{8{b}^{2}}=\frac{27}{20}\phantom{\rule{0ex}{0ex}}⇒\frac{{a}^{2}}{{b}^{2}}=\frac{27×8}{20×30}=\frac{9}{25}\phantom{\rule{0ex}{0ex}}⇒\frac{a}{b}=\sqrt{\frac{9}{25}}=\frac{3}{5}$
(ii)
$=\frac{\frac{21-15}{5}}{\frac{21+15}{5}}\phantom{\rule{0ex}{0ex}}=\frac{6}{36}\phantom{\rule{0ex}{0ex}}=\frac{1}{6}$
(iii)
$=\frac{7}{43}$
(iv)
$=\frac{71}{179}$
#### Question 3:
If then find the value of the ratio
#### Answer:
$\frac{3a+7b}{3a-7b}=\frac{4}{3}$
Applying componendo and dividendo, we get
Now,
#### Question 4:
Solve the following equations.
(i)
(ii)
(iii)
(iv)
(v)
(vi)
#### Answer:
(i)
$\frac{{x}^{2}+12x-20}{3x-5}=\frac{{x}^{2}+8x+12}{2x+3}$
Multiplying both sides by $\frac{1}{4}$, we get
$\frac{{x}^{2}+12x-20}{12x-20}=\frac{{x}^{2}+8x+12}{8x+12}$
Using dividendo, we get
$\frac{\left({x}^{2}+12x-20\right)-\left(12x-20\right)}{12x-20}=\frac{\left({x}^{2}+8x+12\right)-\left(8x+12\right)}{8x+12}\phantom{\rule{0ex}{0ex}}⇒\frac{{x}^{2}+12x-20-12x+20}{12x-20}=\frac{{x}^{2}+8x+12-8x-12}{8x+12}\phantom{\rule{0ex}{0ex}}⇒\frac{{x}^{2}}{12x-20}=\frac{{x}^{2}}{8x+12}$
This equation is true for x = 0. Therefore, x = 0 is a solution of the given equation.
If x ≠ 0, then x2 ≠ 0.
Dividing both sides by x2, we get
$\frac{1}{12x-20}=\frac{1}{8x+12}\phantom{\rule{0ex}{0ex}}⇒12x-20=8x+12\phantom{\rule{0ex}{0ex}}⇒12x-8x=20+12\phantom{\rule{0ex}{0ex}}⇒4x=32\phantom{\rule{0ex}{0ex}}⇒x=8$
Thus, the solutions of the given equation are x = 0 and x = 8.
(ii)
Multiplying both sides by $\frac{1}{5x}$, we get
$\frac{10{x}^{2}+15x+63}{10{x}^{2}+15x}=\frac{5{x}^{2}-25x+12}{5{x}^{2}-25x}$
Using dividendo, we get
$\frac{\left(10{x}^{2}+15x+63\right)-\left(10{x}^{2}+15x\right)}{10{x}^{2}+15x}=\frac{\left(5{x}^{2}-25x+12\right)-\left(5{x}^{2}-25x\right)}{5{x}^{2}-25x}\phantom{\rule{0ex}{0ex}}⇒\frac{63}{10{x}^{2}+15x}=\frac{12}{5{x}^{2}-25x}\phantom{\rule{0ex}{0ex}}⇒\frac{63}{5x\left(2x+3\right)}=\frac{12}{5x\left(x-5\right)}\phantom{\rule{0ex}{0ex}}⇒\frac{63}{2x+3}=\frac{12}{x-5}$
$⇒63x-315=24x+36\phantom{\rule{0ex}{0ex}}⇒63x-24x=315+36\phantom{\rule{0ex}{0ex}}⇒39x=351\phantom{\rule{0ex}{0ex}}⇒x=\frac{351}{39}=9$
Thus, the solution of the give equation is x = 9.
(iii)
Applying componendo and dividendo, we get
$⇒6x+3=10x-5\phantom{\rule{0ex}{0ex}}⇒10x-6x=5+3\phantom{\rule{0ex}{0ex}}⇒4x=8\phantom{\rule{0ex}{0ex}}⇒x=2$
Thus, the solution of the given equation is x = 2.
(iv)
$\frac{\sqrt{4x+1}+\sqrt{x+3}}{\sqrt{4x+1}-\sqrt{x+3}}=\frac{4}{1}$
Applying componendo and dividendo, we get
$\frac{\left(\sqrt{4x+1}+\sqrt{x+3}\right)+\left(\sqrt{4x+1}-\sqrt{x+3}\right)}{\left(\sqrt{4x+1}+\sqrt{x+3}\right)-\left(\sqrt{4x+1}-\sqrt{x+3}\right)}=\frac{4+1}{4-1}\phantom{\rule{0ex}{0ex}}⇒\frac{2\sqrt{4x+1}}{2\sqrt{x+3}}=\frac{5}{3}\phantom{\rule{0ex}{0ex}}⇒\frac{\sqrt{4x+1}}{\sqrt{x+3}}=\frac{5}{3}$
Squaring on both sides, we get
$\frac{4x+1}{x+3}={\left(\frac{5}{3}\right)}^{2}=\frac{25}{9}\phantom{\rule{0ex}{0ex}}⇒36x+9=25x+75\phantom{\rule{0ex}{0ex}}⇒36x-25x=75-9\phantom{\rule{0ex}{0ex}}⇒11x=66\phantom{\rule{0ex}{0ex}}⇒x=6$
Thus, the solution of the given equation is x = 6.
(v)
Applying dividendo, we get
$\frac{{\left(4x+1\right)}^{2}+{\left(2x+3\right)}^{2}-{\left(2x+3\right)}^{2}}{{\left(2x+3\right)}^{2}}=\frac{61-36}{36}\phantom{\rule{0ex}{0ex}}⇒\frac{{\left(4x+1\right)}^{2}}{{\left(2x+3\right)}^{2}}=\frac{25}{36}$
Taking square root on both sides, we get
$\frac{4x+1}{2x+3}=\sqrt{\frac{25}{36}}=\frac{5}{6}\phantom{\rule{0ex}{0ex}}⇒24x+6=10x+15\phantom{\rule{0ex}{0ex}}⇒24x-10x=15-6\phantom{\rule{0ex}{0ex}}⇒14x=9\phantom{\rule{0ex}{0ex}}⇒x=\frac{9}{14}$
Thus, the solution of the given equation is $x=\frac{9}{14}$.
(vi)
$\frac{{\left(3x-4\right)}^{3}-{\left(x+1\right)}^{3}}{{\left(3x-4\right)}^{3}+{\left(x+1\right)}^{3}}=\frac{61}{189}$
Applying componendo and dividendo, we get
$\frac{\left[{\left(3x-4\right)}^{3}-{\left(x+1\right)}^{3}\right]+\left[{\left(3x-4\right)}^{3}+{\left(x+1\right)}^{3}\right]}{\left[{\left(3x-4\right)}^{3}-{\left(x+1\right)}^{3}\right]-\left[{\left(3x-4\right)}^{3}+{\left(x+1\right)}^{3}\right]}=\frac{61+189}{61-189}\phantom{\rule{0ex}{0ex}}⇒\frac{2{\left(3x-4\right)}^{3}}{-2{\left(x+1\right)}^{3}}=\frac{250}{-128}\phantom{\rule{0ex}{0ex}}⇒\frac{{\left(3x-4\right)}^{3}}{{\left(x+1\right)}^{3}}=\frac{125}{64}$
Taking cube root on both sides, we get
$\frac{3x-4}{x+1}=\sqrt[3]{\frac{125}{64}}\phantom{\rule{0ex}{0ex}}⇒\frac{3x-4}{x+1}=\sqrt[3]{{\left(\frac{5}{4}\right)}^{3}}=\frac{5}{4}\phantom{\rule{0ex}{0ex}}⇒12x-16=5x+5\phantom{\rule{0ex}{0ex}}⇒12x-5x=16+5\phantom{\rule{0ex}{0ex}}⇒7x=21\phantom{\rule{0ex}{0ex}}⇒x=3$
Thus, the solution of the given equation is x = 3.
#### Question 1:
Fill in the blanks of the following
(i)
(ii)
#### Answer:
(i)
Also,
$\therefore \frac{x}{7}=\frac{y}{3}=\frac{3x+5y}{\overline{)\mathbf{36}}}=\frac{7x-9y}{\overline{)\mathbf{22}}}$
(ii)
Also,
$\therefore \frac{a}{3}=\frac{b}{4}=\frac{c}{7}=\frac{a-2b+3c}{\overline{)\mathbf{16}}}=\frac{\overline{)\mathbf{2}\mathbf{a}\mathbf{-}\mathbf{2}\mathbf{b}\mathbf{+}\mathbf{2}\mathbf{c}}}{6-8+14}$
#### Question 2:
5 then find the values of the following expressions.
(i)
(ii)
#### Answer:
$5m-n=3m+4n\phantom{\rule{0ex}{0ex}}⇒5m-3m=4n+n\phantom{\rule{0ex}{0ex}}⇒2m=5n\phantom{\rule{0ex}{0ex}}⇒\frac{m}{n}=\frac{5}{2}$
(i)
​
(ii)
$\frac{3m+4n}{3m-4n}\phantom{\rule{0ex}{0ex}}=\frac{\frac{3m}{n}+\frac{4n}{n}}{\frac{3m}{n}-\frac{4n}{n}}\phantom{\rule{0ex}{0ex}}=\frac{3×\frac{5}{2}+4}{3×\frac{5}{2}-4}\phantom{\rule{0ex}{0ex}}=\frac{15+8}{15-8}\phantom{\rule{0ex}{0ex}}=\frac{23}{7}$
#### Question 3:
(i) If a(y+z) = b(z+x) = c(x+y) and out of a, b, c no two of them are equal then show that,
(ii) If and then show that the value of each ratio is equal to 1.
(iii) If and then show that .
(iv) If then show that .
(v) If then show that every ratio = $\frac{x}{y}$.
#### Answer:
(i)
Let $a\left(y+z\right)=b\left(z+x\right)=c\left(x+y\right)=k$
Similarly,
From (1), (2) and (3), we have
(ii)
$\frac{x}{3x-y-z}=\frac{y}{3y-z-x}=\frac{z}{3z-x-y}=\frac{x+y+z}{\left(3x-y-z\right)+\left(3y-z-x\right)+\left(3z-x-y\right)}$ (Theorem of equal ratios)
$⇒\frac{x}{3x-y-z}=\frac{y}{3y-z-x}=\frac{z}{3z-x-y}=\frac{x+y+z}{x+y+z}$
$⇒\frac{x}{3x-y-z}=\frac{y}{3y-z-x}=\frac{z}{3z-x-y}=1$
(iii)
$\frac{ax+by}{x+y}=\frac{bx+az}{x+z}=\frac{ay+bz}{y+z}=\frac{\left(ax+by\right)+\left(bx+az\right)+\left(ay+bz\right)}{\left(x+y\right)+\left(x+z\right)+\left(y+z\right)}$ (Theorem of equal ratios)
$⇒\frac{ax+by}{x+y}=\frac{bx+az}{x+z}=\frac{ay+bz}{y+z}=\frac{\left(a+b\right)x+\left(a+b\right)y+\left(a+b\right)z}{2x+2y+2z}\phantom{\rule{0ex}{0ex}}⇒\frac{ax+by}{x+y}=\frac{bx+az}{x+z}=\frac{ay+bz}{y+z}=\frac{\left(a+b\right)\left(x+y+z\right)}{2\left(x+y+z\right)}\phantom{\rule{0ex}{0ex}}⇒\frac{ax+by}{x+y}=\frac{bx+az}{x+z}=\frac{ay+bz}{y+z}=\frac{a+b}{2}$
(iv)
(By invertendo)
Now,
Also,
From (1), (2) and (3), we have
(v)
Solve
(i)
(ii)
#### Answer:
(i)
$\frac{16{x}^{2}-20x+9}{8{x}^{2}+12x+21}=\frac{4x-5}{2x+3}$
If x = 0, then $\frac{16×0-20×0+9}{8×0+12×0+21}=\frac{4×0-5}{2×0+3}$ ⇒ $\frac{9}{21}=\frac{-5}{3}$, which is not true.
So, x = 0 is not a solution of the given equation.
Now,
$\therefore \frac{4x-5}{2x+3}=\frac{9}{21}\phantom{\rule{0ex}{0ex}}⇒\frac{4x-5}{2x+3}=\frac{3}{7}\phantom{\rule{0ex}{0ex}}⇒28x-35=6x+9\phantom{\rule{0ex}{0ex}}⇒28x-6x=35+9\phantom{\rule{0ex}{0ex}}⇒22x=44\phantom{\rule{0ex}{0ex}}⇒x=2$
Thus, the solution of the given equation is x = 2.
(ii)
$\frac{5{y}^{2}+40y-12}{5y+10{y}^{2}-4}=\frac{y+8}{1+2y}$
If y = 0, then $\frac{5×0+40×0-12}{5×0+10×0-4}=\frac{0+8}{1+2×0}$ ⇒ $\frac{-12}{-4}=\frac{8}{1}$, which is not true.
So, y = 0 is not a solution of the given equation.
Now,
$\therefore \frac{y+8}{1+2y}=\frac{-12}{-4}=3\phantom{\rule{0ex}{0ex}}⇒y+8=3+6y\phantom{\rule{0ex}{0ex}}⇒6y-y=8-3\phantom{\rule{0ex}{0ex}}⇒5y=5\phantom{\rule{0ex}{0ex}}⇒y=1$
Thus, the solution of the given equation is y = 1.
#### Question 1:
Which number should be subtracted from 12, 16 and 21 so that resultant numbers are in continued proportion?
#### Answer:
Le the number x be subtracted from 12, 16 and 21 so that resultant numbers are in continued proportion.
∴ (12 − x), (16 − x), (21 − x) are in continued proportion.
$⇒\frac{12-x}{16-x}=\frac{16-x}{21-x}\phantom{\rule{0ex}{0ex}}⇒{\left(16-x\right)}^{2}=\left(12-x\right)×\left(21-x\right)\phantom{\rule{0ex}{0ex}}⇒256+{x}^{2}-32x=252-12x-21x+{x}^{2}\phantom{\rule{0ex}{0ex}}⇒256-32x=252-33x\phantom{\rule{0ex}{0ex}}⇒33x-32x=252-256\phantom{\rule{0ex}{0ex}}⇒x=-4$
Thus, the number −4 must be subtracted from 12, 16 and 21 so that resultant numbers are in continued proportion.
#### Question 2:
If (28$-x$) is the mean proportional of (23$-x$) and (19$-x$) then find the vaue of x.
#### Answer:
It is given that (28 − x) is the mean proportional of (23 − x) and (19 − x).
$\therefore {\left(28-x\right)}^{2}=\left(23-x\right)×\left(19-x\right)\phantom{\rule{0ex}{0ex}}⇒784+{x}^{2}-56x=437-23x-19x+{x}^{2}\phantom{\rule{0ex}{0ex}}⇒784-56x=437-42x\phantom{\rule{0ex}{0ex}}⇒56x-42x=784-437\phantom{\rule{0ex}{0ex}}⇒14x=347\phantom{\rule{0ex}{0ex}}⇒x=\frac{347}{14}$
Thus, the value of x is $\frac{347}{14}$.
#### Question 3:
Three numbers are in continued proportion, whose mean proportional is 12 and the sum of the remaining two numbers is 26, then find these numbers.
#### Answer:
Let the remaining two numbers be and y.
So, the numbers x, 12, y are in continued proportion.
​∴ xy = (12)2 = 144 .....(1)
Also,
xy = 26 .....(2)
Solving (1) and (2), we get
$x+\frac{144}{x}=26\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+144=26x\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-26x+144=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-18x-8x+144=0\phantom{\rule{0ex}{0ex}}⇒x\left(x-18\right)-8\left(x-18\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-8\right)\left(x-18\right)=0$ ​​
⇒ x − 8 = 0 or x − 18 = 0
⇒ x = 8 or x = 18
When x = 8, y = 26 − 8 = 18 [Using (2)]
When x = 18, y = 26 − 18 = 8
Thus, the numbers are 8, 12, 18 or 18, 12, 8.
#### Question 4:
If (a + b + c) (a $-$ b + c) = a2 + b2 + c2 show that a, b, c are in continued proportion.
#### Answer:
$\left(a+b+c\right)\left(a-b+c\right)={a}^{2}+{b}^{2}+{c}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(a+c\right)}^{2}-{b}^{2}={a}^{2}+{b}^{2}+{c}^{2}\phantom{\rule{0ex}{0ex}}⇒{a}^{2}+{c}^{2}+2ac-{b}^{2}={a}^{2}+{b}^{2}+{c}^{2}\phantom{\rule{0ex}{0ex}}⇒2{b}^{2}=2ac\phantom{\rule{0ex}{0ex}}⇒{b}^{2}=ac$
Therefore, abc are in continued proportion.
#### Question 5:
If and $a,b,c$ > 0 then show that,
(i) (a + b + c) ($-$ c) = ab $-$ c2
(ii) (a2 + b2) (b2 + c2 ) = (ab + bc)2
(iii)
(i)
(ii)
(iii)
#### Question 6:
Find mean proportional of
#### Answer:
If b is the mean proportional of a and c, then ${b}^{2}=ac$ or $b=\sqrt{ac}$.
Mean proportional of $\frac{x+y}{x-y}$ and $\frac{{x}^{2}-{y}^{2}}{{x}^{2}{y}^{2}}$
$=\sqrt{\frac{x+y}{x-y}×\frac{{x}^{2}-{y}^{2}}{{x}^{2}{y}^{2}}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{x+y}{x-y}×\frac{\left(x-y\right)\left(x+y\right)}{{x}^{2}{y}^{2}}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{{\left(x+y\right)}^{2}}{{\left(xy\right)}^{2}}}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(\frac{x+y}{xy}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{x+y}{xy}$
#### Question 1:
Select the appropriate alternative answer for the following questions.
(i) If 6 : 5 = y : 20 then what will be the value of y ?
(A) 15 (B) 24 (C) 18 (D) 22.5
(ii) What is the ratio of 1 mm to 1 cm ?
(A) 1 : 100 (B) 10 : 1 (C) 1 : 10 (D) 100 : 1
(iii * ) The ages of Jatin, Nitin and Mohasin are 16, 24 and 36 years respectively. What is the ratio of Nitin’s age to Mohasin’s age ?
(A) 3 : 2 (B) 2 : 3 (C) 4 : 3 (D) 3 : 4
(iv) 24 Bananas were distributed between Shubham and Anil in the ratio 3 : 5, then how many bananas did Shubham get ?
(A) 8 (B) 15 (C) 12 (D) 9
(v) What is the mean proportional of 4 and 25 ?
(A) 6 (B) 8 (C) 10 (D) 12
#### Answer:
(i)
6 : 5 = y : 20
$\therefore \frac{6}{5}=\frac{y}{20}\phantom{\rule{0ex}{0ex}}⇒y=\frac{6×20}{5}=24$
Thus, the value of y is 24.
Hence, the correct answer is option (B).
(ii)
Ratio of 1 mm to 1 cm
= 1 mm : 1 cm
= 1 mm : 10 mm (1 cm = 10 mm)
= 1 : 10
Hence, the correct answer is option (C).
(iii)
Age of Nitin = 24 years
Age of Mohasin = 36 years
∴ Ratio of Nitin's age to Mohansin's age
= Age of Nitin : Age of Mohasin
= 24 years : 36 years
= 2 : 3
Thus, the ratio of Nitin's age to Mohasin's age is 2 : 3.
Hence, the correct answer is option (B).
(iv)
The number of bananas received by Shubham and Anil is the ratio 3 : 5.
Let the number of bananas received by Shubham and Anil be 3x and 5x, respectively.
∴ 3x + 5x = 24
⇒ 8x = 24
⇒ x = 3
∴ Number of bananas received by Shubham = 3x = 3 × 3 = 9
Thus, Shubham gets 9 bananas.
Hence, the correct answer is option (D).
(v)
Mean proportional of 4 and 25
$=\sqrt{4×25}\phantom{\rule{0ex}{0ex}}=\sqrt{100}\phantom{\rule{0ex}{0ex}}=10$
Thus, the mean proportional of 4 and 25 is 10.
Hence, the correct answer is option (C).
#### Question 2:
For the following numbers write the ratio of first number to second number in the reduced form.
(i) 21, 48 (ii) 36, 90 (iii) 65, 117 (iv) 138, 161 (v) 114, 133
#### Answer:
(i)
21 : 48 = $\frac{21}{48}=\frac{21÷3}{48÷3}=\frac{7}{16}$ = 7 : 16 (HCF of 21 and 48 = 3)
Thus, the reduced form of 21 : 48 is 7 : 16.
(ii)
36 : 90 = $\frac{36}{90}=\frac{36÷18}{90÷18}=\frac{2}{5}$ = 2 : 5 (HCF of 36 and 90 = 18)
Thus, the reduced form of 36 : 90 is 2 : 5.
(iii)
65 : 117 = $\frac{65}{117}=\frac{65÷13}{113÷13}=\frac{5}{9}$ = 5 : 9 (HCF of 65 and 113 = 13)
Thus, the reduced form of 65 : 117 is 5 : 9.
(iv)
138 : 161 = $\frac{138}{161}=\frac{138÷23}{161÷23}=\frac{6}{7}$ = 6 : 7 (HCF of 138 and 161 = 23)
Thus, the reduced form of 138 : 161 is 6 : 7.
(v)
114 : 133 = $\frac{114}{133}=\frac{114÷19}{133÷19}=\frac{6}{7}$ = 6 : 7 (HCF of 114 and 133 = 19)
Thus, the reduced form of 114 : 133 is 6 : 7.
#### Question 3:
Write the following ratios in the reduced form.
(i) Radius to the diameter of a circle.
(ii) The ratio of diagonal to the length of a rectangle, having length 4 cm and breadth 3 cm.
(iii) The ratio of perimeter to area of a square, having side 4 cm.
#### Answer:
(i)
Let the radius of the circle be r units.
Diameter of the circle = 2 × Radius of the circle = 2r units
∴ Ratio of radius to diameter of a circle = Radius of the circle : Diameter of the circle = r : 2r = 1 : 2
(ii)
Length of the rectangle, l = 4 cm
Breadth of the rectangle, b = 3 cm
Now,
Diagonal of the rectangle = $\sqrt{{l}^{2}+{b}^{2}}=\sqrt{{3}^{2}+{4}^{2}}=\sqrt{9+16}=\sqrt{25}$ = 5 cm
∴ Ratio of diagonal to the length of the rectangle = Diagonal of the rectangle : Length of the rectangle = 5 cm : 4 cm = 5 : 4
(iii)
Side of the squre = 4 cm
Perimeter of the square = 4 × Side of the squre = 4 × 4 cm = 16 cm
Area of the squre = (Side of the square)2 = (4 cm)2 = 16 cm2
∴ Ratio of perimeter to area of the square = Perimeter of the square : Area of the squre = 16 : 16 = 1 : 1
#### Question 4:
Check whether the following numbers are in continued proportion.
(i) 2, 4, 8 (ii) 1, 2, 3 (iii) 9, 12, 16 (iv) 3, 5, 8
#### Answer:
The numbers abc are in continued proportion if $\frac{a}{b}=\frac{b}{c}$.
(i)
$\frac{2}{4}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\frac{4}{8}=\frac{1}{2}$
Since $\frac{2}{4}=\frac{4}{8}$, so the numbers 2, 4, 8 are in continued proportion.
(ii)
Since $\frac{1}{2}\ne \frac{2}{3}$, so the numbers 1, 2, 3 are not in continued proportion.
(iii)
$\frac{9}{12}=\frac{3}{4}\phantom{\rule{0ex}{0ex}}\frac{12}{16}=\frac{3}{4}$
Since $\frac{9}{12}=\frac{12}{16}$, so the numbers 9, 12, 16 are in continued proportion.
(iv)
Since $\frac{3}{5}\ne \frac{5}{8}$, so the numbers 3, 5, 8 are not in continued proportion.
#### Question 5:
a, b, c are in continued proportion. If a = 3 and c = 27 then find b.
#### Answer:
It is given that 3, b, 27 are in continued proportion.
$\therefore \frac{3}{b}=\frac{b}{27}\phantom{\rule{0ex}{0ex}}⇒{b}^{2}=3×27\phantom{\rule{0ex}{0ex}}⇒{b}^{2}=81\phantom{\rule{0ex}{0ex}}⇒b=\sqrt{81}=9$
Thus, the value of b is 9.
#### Question 6:
Convert the following ratios into percentages..
(i) 37 : 500 (ii) $\frac{5}{8}$ (iii) $\frac{22}{30}$ (iv) $\frac{5}{16}$ ( v) $\frac{144}{1200}$
#### Answer:
(i)
37 : 500 = $\frac{37}{500}×100%=\frac{37}{5}%=7.4%$
(ii)
$\frac{5}{8}=\frac{5}{8}×100%=\frac{500}{8}%=62.5%$
(iii)
$\frac{22}{30}=\frac{22}{30}×100%=\frac{220}{3}%=73.33%$
(iv)
$\frac{5}{16}=\frac{5}{16}×100%=\frac{500}{16}%=31.25%$
(v)
​$\frac{144}{1200}=\frac{144}{1200}×100%=\frac{144}{12}%=12%$
#### Question 7:
Write the ratio of first quantity to second quantity in the reduced form.
(i) 1024 MB, 1.2 GB [(1024 MB = 1 GB)] (ii) 17 Rupees, 25 Rupees 60 paise (iii) 5 dozen, 120 units
(iv) 4 sq.m, 800 sq.cm (v) 1.5 kg, 2500 gm
#### Answer:
(i)
Ratio of 1024 MB to 1.2 GB
= 1024 MB : 1.2 × 1024 MB
$=\frac{1024}{1.2×1024}\phantom{\rule{0ex}{0ex}}=\frac{1}{1.2}\phantom{\rule{0ex}{0ex}}=\frac{10}{12}\phantom{\rule{0ex}{0ex}}=\frac{5}{6}$
= 5 : 6
(ii)
17 Rupees = 1700 paise (Re 1 = 100 paise)
25 Rupees 60 paise = 25 Rupees + 60 paise = 2500 paise + 60 paise = 2560 paise
∴ Ratio of 17 Rupees to 25 Rupees 60 paise
= 1700 paise : 2560 paise
= 85 : 128
(iii)
Ratio of 5 dozen to 120 units
= 5 × 12 units : 120 units
= 60 units : 120 units
$=\frac{60}{120}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}$
= 1 : 2
(iv)
1 m2 = 10000 cm2
4 m2 = 4 × 10000 cm2 = 40000 cm2
∴ Ratio of 4 m2 to 800 cm2
= 40000 cm: 800 cm2
$=\frac{40000}{800}\phantom{\rule{0ex}{0ex}}=50$
= 50 : 1
(v)
1.5 kg = 1.5 × 1000 g = 1500 g
∴ Ratio of 1.5 kg to 2500 g
= 1500 g : 2500 g
$=\frac{1500}{2500}\phantom{\rule{0ex}{0ex}}=\frac{3}{5}$
= 3 : 5
#### Question 8:
If then find the values of the following expressions.
(i) (ii) (iii) (iv)
#### Answer:
(i)
$\frac{a}{b}=\frac{2}{3}\phantom{\rule{0ex}{0ex}}⇒\frac{4a}{3b}=\frac{4×2}{3×3}=\frac{8}{9}$
Applying componendo, we get
$\frac{4a+3b}{3b}=\frac{8+9}{9}\phantom{\rule{0ex}{0ex}}⇒\frac{4a+3b}{3b}=\frac{17}{9}$
(ii)
Applying componendo and dividendo, we get
$\frac{5{a}^{2}+2{b}^{2}}{5{a}^{2}-2{b}^{2}}=\frac{20+18}{20-18}\phantom{\rule{0ex}{0ex}}⇒\frac{5{a}^{2}+2{b}^{2}}{5{a}^{2}-2{b}^{2}}=\frac{38}{2}\phantom{\rule{0ex}{0ex}}⇒\frac{5{a}^{2}+2{b}^{2}}{5{a}^{2}-2{b}^{2}}=19$
(iii)
Applying componendo, we get
$\frac{{a}^{3}+{b}^{3}}{{b}^{3}}=\frac{8+27}{27}\phantom{\rule{0ex}{0ex}}⇒\frac{{a}^{3}+{b}^{3}}{{b}^{3}}=\frac{35}{27}$
(iv)
$\frac{a}{b}=\frac{2}{3}$
Applying invertendo, we get
Applying componendo and dividendo, we get
$\frac{7b+4a}{7b-4a}=\frac{21+8}{21-8}\phantom{\rule{0ex}{0ex}}⇒\frac{7b+4a}{7b-4a}=\frac{29}{13}$
Applying invertendo, we get
$\frac{7b-4a}{7b+4a}=\frac{13}{29}$
#### Question 9:
If a, b, c, d are in proportion , then prove that
(i)
(ii)
(iii)
#### Answer:
It is given that abcd are in proportion.
(i)
From (1) and (2), we get
$\frac{11{a}^{2}+9ac}{11{b}^{2}+9bd}=\frac{{a}^{2}+3ac}{{b}^{2}+3bd}$
(ii)
From (1) and (2), we get
$\sqrt{\frac{{a}^{2}+5{c}^{2}}{{b}^{2}+5{d}^{2}}}=\frac{a}{b}$
(iii)
From (1) and (2), we get
$\frac{{a}^{2}+ab+{b}^{2}}{{a}^{2}-ab+{b}^{2}}=\frac{{c}^{2}+cd+{d}^{2}}{{c}^{2}-cd+{d}^{2}}$
#### Question 10:
If a, b, c are in continued proportion , then prove that
(i)
(ii)
#### Answer:
abc are in continued proportion.
(i)
From (1) and (2), we get
$\frac{a}{a+2b}=\frac{a-2b}{a-4c}$
(ii)
From (1) and (2), we get
$\frac{b}{b+c}=\frac{a-b}{a-c}$
Solve :
#### Answer:
If x = 0, then $\frac{12×0+18×0+42}{18×0+12×0+58}=\frac{2×0+3}{3×0+2}⇒\frac{42}{58}=\frac{3}{2}$, which is not true.
So, x = 0 is not a solution of the given equation.
Now,
$\therefore \frac{2x+3}{3x+2}=\frac{42}{58}\phantom{\rule{0ex}{0ex}}⇒\frac{2x+3}{3x+2}=\frac{21}{29}\phantom{\rule{0ex}{0ex}}⇒58x+87=63x+42\phantom{\rule{0ex}{0ex}}⇒63x-58x=87-42\phantom{\rule{0ex}{0ex}}⇒5x=45\phantom{\rule{0ex}{0ex}}⇒x=9$
Thus, the solution of the given equation is x = 9.
#### Question 12:
If then prove that every ratio = $\frac{x}{y}$.
#### Answer:
$⇒\frac{2x-3y}{3z+y}=\frac{3\left(z-y\right)}{3\left(z-x\right)}=\frac{x+3z}{2y-3x}=\frac{2x-3y-3z+3y+x+3z}{3z+y-3z+3x+2y-3x}\phantom{\rule{0ex}{0ex}}⇒\frac{2x-3y}{3z+y}=\frac{3\left(z-y\right)}{3\left(z-x\right)}=\frac{x+3z}{2y-3x}=\frac{3x}{3y}\phantom{\rule{0ex}{0ex}}\therefore \frac{2x-3y}{3z+y}=\frac{z-y}{z-x}=\frac{x+3z}{2y-3x}=\frac{x}{y}$
#### Question 13:
(13 * ) If then prove that .
#### Answer:
Again applying theorem of equal ratios, we get
$⇒\frac{-ax}{-{a}^{2}}=\frac{-by}{-{b}^{2}}=\frac{-cz}{-{c}^{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$
View NCERT Solutions for all chapters of Class 9 |
# Frequent question: How is 10 percent discount calculated?
Contents
## How do you calculate 10% of a percent?
To calculate 10 percent of a number, simply divide it by 10 or move the decimal point one place to the left. For example, 10 percent of 230 is 230 divided by 10, or 23. 5 percent is one half of 10 percent. To calculate 5 percent of a number, simply divide 10 percent of the number by 2.
## What does a 10% discount mean?
Let says the original price =\$X. \$X= 100% cause original price. When something is discounted by 10% this means the full price is no longer 100% but rather now becomes 90% . It should be interpreted that 10 % off of something means 90% of the original price.
## How do I calculate a discount percentage?
How do I calculate discount in percentages?
1. Subtract the final price from the original price.
2. Divide this number by the original price.
3. Finally, multiply the result by 100.
4. You’ve obtained a discount in percentages. How awesome!
## What is a 10% discount off of \$60?
Percent Off Table For 60.00
1 percent off 60.00 is 59.40 The difference is 0.60
9 percent off 60.00 is 54.60 The difference is 5.40
10 percent off 60.00 is 54.00 The difference is 6.00
11 percent off 60.00 is 53.40 The difference is 6.60
12 percent off 60.00 is 52.80 The difference is 7.20
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## What is \$20 with 10% off?
Sale Price = \$18 (answer). This means the cost of the item to you is \$18. You will pay \$18 for a item with original price of \$20 when discounted 10%. In this example, if you buy an item at \$20 with 10% discount, you will pay 20 – 2 = 18 dollars.
## What is the 10% discount of 500?
You will pay \$450 for a item with original price of \$500 when discounted 10%. In this example, if you buy an item at \$500 with 10% discount, you will pay 500 – 50 = 450 dollars.
## How do you take 10% off?
One of the easiest ways to determine a 10 percent discount is to divide the total sale price by 10 and then subtract that from the price. You can calculate this discount in your head. For a 20 percent discount, divide by ten and multiply the result by two.
## What is 20% off?
A 20 percent discount is 0.20 in decimal format. Secondly, multiply the decimal discount by the price of the item to determine the savings in dollars. For example, if the original price of the item equals \$24, you would multiply 0.2 by \$24 to get \$4.80.
## What is 20% off a \$15 shirt?
Percent Off Table For 15.00
1 percent off 15.00 is 14.85 The difference is 0.15
18 percent off 15.00 is 12.30 The difference is 2.70
19 percent off 15.00 is 12.15 The difference is 2.85
20 percent off 15.00 is 12.00 The difference is 3.00
21 percent off 15.00 is 11.85 The difference is 3.15
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## What is the formula of discount %?
The formula to calculate the discount rate is: Discount % = (Discount/List Price) × 100.
## How do you take 20% off a price?
How do I take 20 % off a price?
1. Take the original price.
2. Divide the original price by 5.
3. Alternatively, divide the original price by 100 and multiply it by 20.
4. Subtract this new number from the original one.
5. The number you calculated is the discounted value. |
## Circles, Arcs, and Sectors
### Transcript
Arcs and Sectors. In the previous lesson, we said that arc measure was one way to talk about the size of an arc, so the number of degrees that an arc has. That's one way we can say how big an arc is. As long as we're looking at two arcs in the same circle, arc measures perfectly adequate to compare the size of two different arcs.
If we start comparing arcs in different size circles, it becomes clear that arc measure alone is not sufficient. The two arcs here, the arc BC and the arc CD, those are both 30 degree arcs, because they share a central angle of 30 degrees. So, they have the same measure, but clearly they're not identical.
They're not the same in every way, they have different lengths. If these were actual pads outside, it would take us longer to walk from C to D, than it would take us to walk from A to B. Those paths have two different lengths. So the arc measure is more about the curvature of the arc, but we also have to consider the length.
So then it becomes a question, how do find the length of an arc? Well, first of all, this length is called an arc length, and it is found by setting up a proportion, a part-to-whole proportion. So, I'm very much going to emphasize here, don't simply memorize a formula. I want you to understand the logic of it. We're really asking the question how much of the circle do we have, and on the basis of that question we're setting up this proportion.
So the arclength is part of the circumference, and so you compare the arc to the circumference. That's the part to the whole on one side of the equation. Similarly, the arc measure, or the measure of the central angle, those two are interchangeable, they're the same, is the part of 360 degrees all the way around the circle.
So the proposed we set up is arclength/2r = angle/360. For example, in this circle, we have an angle of 120 degrees, well 120 degree is 1/3 of 360. So we have 1/3 of the whole circle here, and because that angle is 1/3 of 360, the arc has to be 1/3 of the circumference.
Well, the circumference will get figured out easily enough, that's 2r, which is 48, because the radius is 24, and then 1/3 of that is 16, so we have an arc length of 16. Here's a very easy practice problem. Pause the video and then we'll talk about this. Okay if the length of the arc is 12, then find the area of the circle.
Well, we're given the angle, so the first thing that we're gonna do is figure how much of the circle do we have, and we're gonna put that angle over 360, and 72/360, if you think about it 72 is two times 36, so we have two times 36 over 10 times 36. 2/10, which is 1/5. That's a handy fraction to know, to know that 72 degrees is 1/5 of 360.
So I have 1/5 of the circle, and so that that means that that arc is 1/5 of the circumference. So the circumference must be 5 times that, or 60, and that 60 has to equal 2r. Well here's one of our big circle strategies. Use the information you're given to find the radius. Once you find the radius, you can find everything else you need.
So we divide. We get the radius equals 30. Then, of course, the area equals r2 = 900. That is the area of the circle. Just as we can divide the circumference into fractions, so we can divide the whole area of the circle.
A slice of the circle like this is called a circular sector. Some people like to think of it as a slice of pie or a slice of pizza. To find the area of a circle, we set up another part-to-whole proportion, in which the area of the sector is part of the area of the whole circle, r2. So again, don't simply memorize a formula, think about the logic of this. How much of the circle do we have?
We set up the ratio for the angles, the ratio of the areas, and we set them equal. So, area of a circle/r2 = angle/360. For example, if we're given a 60 degree angle, well, 60 degrees, 60/360 that's obviously 1/6, we're dealing with 1/6 of the circle. Well, if we want that area, the whole area, the area of the whole circle, of course is r2, we could write that as (18)(18).
Now, I'm not gonna make the mistake of multiplying that out, I'm just gonna leave that as (18)(18), because I'm gonna take 1/6 of that, and when I take 1/6 of it I divide one of those 18's, I divide it down 18/3 = 3, so then I just have to really multiply (3)(18)=54, and that is the area of the sector. Here's a practice problem.
Pause the video and then we'll talk about this. So we're given a warning, the diagram is not drawn to scale. We don't actually know what that means at the beginning, okay? The circle has a radius of 8, all right, so we could figure out the area of the circle, and the sector has an area of 16. Well, that's interesting.
From the radius, we could figure out the area of the whole circle, 64. Now let's figure out how much of the circle we have, 16/64 = 1/4. So that means that the angle, JOK, that's a 90 degree angle. Okay, so that's where the diagram was fooling us. If that angle were drawn to scale, that would actually be a right angle JOK. Okay, but now we know that's a right angle, and so now, we want the length of arc.
Now, this is tricky, JLK. So that's not the little arc that runs along the sector. That's the big wrap around arc that starts at J, goes counterclockwise around through L, and arrives at K. So that would be three-quarters of the circle, three-quarters of the circumference.
Well, the circumference we can figure out easily enough is 16. Three quarters of that circumference would be 12, and that is the length of that gigantic arc. In summary, we find arclength and an area of a sector by setting up part-to-whole proportions. So we'd say arclength over the circumference, 2r = angle/360.
Again, we're just figuring out how much of the circle do we have on each side of the equation, or we set up a similar ratio, area of the sector/r2 = angle/360. We're figuring out how much of the circle do we have. |
# How to Find the Range of a Quadratic Function – Simple Steps to Understand its Scope
To find the range of a quadratic function, you should first understand the basic form of a quadratic function, which is $y = ax^2 + bx + c$, where $a$, $b$, and $c$ are constants, and $a \neq 0$.
This equation represents a parabola when graphed on the coordinate plane, and its range is the set of possible values that $y$ can take.
The domain, on the other hand, generally includes all real numbers since a parabola extends infinitely in either direction along the x-axis.
The range of a quadratic function is dependent on the direction of the parabola. If the parabola opens upwards (meaning $a > 0$), the range will be all real numbers greater than or equal to the minimum value of $y$, which is the y-coordinate of the vertex of the parabola.
Conversely, if the parabola opens downwards ($a < 0$), the range consists of all real numbers less than or equal to the maximum value of $y$. Determining the vertex and knowing whether the parabola opens up or down allows me to establish the range of the function.
As we walk through the process, remember that exploring the behavior of quadratic functions gives us insights into many physical phenomena described by parabolic curves, such as the path of a ball thrown into the air.
Finding the range helps us predict and understand the limitations of these phenomena. Stay tuned to unlock the secrets of these captivating curves.
## Determining the Range of a Quadratic Function
When I’m trying to determine the range of a quadratic function, I consider the shape and properties of its graph, which is a parabola. A quadratic function has the general form $f(x) = ax^2 + bx + c$, where $a$, $b$, and $c$ are constants, and the graph is a u-shaped curve (parabola).
First, I identify the vertex of the parabola, which is the point where the parabola has either its maximum or minimum value.
The vertex has the coordinates $\left(h, k\right)$ and lies on the axis of symmetry of the parabola. The equation for the axis of symmetry is $x = -\frac{b}{2a}$.
The direction in which the parabola opens depends on the leading coefficient $a$. If $a > 0$, the parabola opens upward, and the vertex represents the minimum value of the quadratic function.
If $a < 0$, it opens downward, making the vertex the maximum value.
To state the range in interval notation:
• If the parabola opens upward: $\text{Range} = [k, \infty)$
• If the parabola opens downward: $\text{Range} = (-\infty, k]$
Here’s a table to visualize this concept:
Leading Coefficient $a$Parabola OpensVertex RepresentsInterval Notation
$a > 0$UpwardMinimum Value$[k, \infty)$
$a < 0$DownwardMaximum Value$(-\infty, k]$
To find the vertex, we can use the vertex form of a quadratic function, $f(x) = a(x – h)^2 + k$. Alternatively, we may need to use methods like completing the square or applying the quadratic formula to find the vertex from the standard form.
Remember, the range describes all possible output values of $y$ based on the domain we are interested in. It’s also the set of outputs of the quadratic function where $y$ is a non-negative number.
Using a graphing calculator can also aid in visualizing the vertex and the parabola‘s symmetric properties to confirm the range.
For quadratic functions of degree two, the range is always an interval, with the vertex indicating the extreme point—either the minimum or maximum.
Knowing whether the parabola opens up or opens down helps determine the range appropriately.
## Conclusion
In this discussion, I’ve outlined the process to determine the range of a quadratic function.
Remember that the key step is finding whether the parabola opens upward or downward, which you can deduce from the sign of the coefficient a in the quadratic function $f(x) = ax^2 + bx + c$. If a > 0, the parabola opens upward, indicating that the range is from the y-coordinate of the vertex to infinity.
Conversely, if a < 0, the function opens downward, and the range will be from negative infinity to the y-coordinate of the vertex.
Determining the vertex (( h, k )) is crucial because it represents either the minimum or maximum value of ( f(x) ), depending on the orientation of the parabola. The vertex formula $h = -\frac{b}{2a}$ and $k = f(h)$ is essential in finding these coordinates.
I trust you now feel confident in identifying the range of any quadratic function by applying the appropriate method and using the standard or general form of the quadratic equation.
Be mindful of the value you obtain for k since it is the starting or ending point of your range.
Acquiring the range of a function is not just a mechanical process but a foundational skill in algebra that can elevate your understanding of how functions behave. And remember, practice makes perfect – so don’t hesitate to apply these techniques to as many problems as you can. |
What is the vertex of y=-3x^2-4x+2 ?
Dec 4, 2015
$\left(- \frac{2}{3} , \frac{10}{3}\right)$
Explanation:
The vertex of a quadratic equation can be found through the vertex formula:
$\left(- \frac{b}{2 a} , f \left(- \frac{b}{2 a}\right)\right)$
The letters represent the coefficients in the standard form of a quadratic equation $a {x}^{2} + b x + c$.
Here:
$a = - 3$
$b = - 4$
Find the $x$-coordinate of the vertex.
$- \frac{b}{2 a} = - \frac{- 4}{2 \left(- 3\right)} = - \frac{2}{3}$
The $y$-coordinate is found by plugging $- \frac{2}{3}$ into the original equation.
$- 3 {\left(- \frac{2}{3}\right)}^{2} - 4 \left(- \frac{2}{3}\right) + 2 = - 3 \left(\frac{4}{9}\right) + \frac{8}{3} + 2$
$= - \frac{4}{3} + \frac{8}{3} + \frac{6}{3} = \frac{10}{3}$
Thus, the vertex is located at the point $\left(- \frac{2}{3} , \frac{10}{3}\right)$.
This can also be found through putting the quadratic into vertex form $y = a {\left(x - h\right)}^{2} + k$ by completing the square.
y=-3(x^2+4/3x+?)+2
$y = - 3 \left({x}^{2} + \frac{4}{3} x + \textcolor{b l u e}{\frac{4}{9}}\right) + 2 + \textcolor{b l u e}{\frac{4}{3}}$
$y = - 3 {\left(x + \frac{2}{3}\right)}^{2} + \frac{10}{3}$
Again, the vertex is located at the point $\left(- \frac{2}{3} , \frac{10}{3}\right)$. |
# Math apps that show work
Here, we will be discussing about Math apps that show work. Our website can solving math problem.
## The Best Math apps that show work
There are a lot of Math apps that show work that are available online. One of the best things you can do is to practice. This means that you should try to answer math questions every day. The more practice you get, the better you will become at math. You can also find other ways to practice math, such as by playing games on your phone or tablet. Another thing that you can do is to use a calculator whenever possible. It may seem like math doesn’t need a calculator, but in reality, it does! Not all problems require exact numbers, but they still need to be exact enough so that they can be solved with a calculator.
As the name suggests, a square calculator is used to calculate the area of a square. A square calculator is made up of four basic parts – a base, a top, a pair of sides, and an angle. The area of any four-sided figure can be calculated by using these four components in the correct order. For example, if you want to calculate the area of a square with side lengths $x$, $y$, $z$, and an angle $heta$ (in degrees), then you simply add together the values of $x$, $y$, $z$, and $heta$ in this order: egin{align*}frac{x}{y} + frac{z}{ heta} end{align*}. The above formula can also be expressed as follows: egin{align*}frac{1}{2} x + frac{y}{2} y + frac{z}{4} z = frac{ heta}{4}\end{align*} To find the area of a cube with length $L$ and width $W$, first multiply $L$ by itself twice (to get $L^2$). Next, multiply each side by $W$. Lastly, divide the result by 2 to find the area. For example: egin{align*}left(L
One of the best ways to learn mathematics is to start early, and there are plenty of apps out there that can help you do just that. Whether you're a student or teacher, there's an app for you. If you're looking for free apps, be sure to check out the App Store and Google Play. But if you're looking for paid apps, be sure to look online first. There are plenty of options available, from games that help you practice math to apps that teach basic functions. There are also tools out there for teachers to use with their students. These include a wide range of apps that provide everything from lesson plans to quizzes to a virtual classroom. If your school doesn't have an app yet, it's definitely time to start thinking about one!
Solving exponential equations can be a challenging task for students. However, it is important for students to understand how to solve exponential equations because they will encounter them in many different settings throughout their life. Exponential equations are used in areas such as chemistry and physics when dealing with things like growth and decay. They are also used in topics like biology and economics when discussing topics like population growth. When solving exponential equations, it is important to first determine what type of equation you are dealing with. There are three main types of exponential equations: linear, logarithmic, and power. Each of these equations has a different way of solving them, so it is important to take note of this before beginning the process. Once you have determined the type of equation you are dealing with, you can then begin by breaking down the problem into smaller pieces so that you can work on each piece individually. Once you have solved each piece of the problem individually, you can then combine all the pieces together to form a final solution for the entire problem.
the app is so amazing. Every time I have a problem with a math question the app helps me it's very cool download this app it's very helpful I have this because apps can solve every problem. Still a great one if you're struggling in math. I 100% recommend.
Xavia Simmons
I have struggled in math until a friend recommended the app to me and now, I make straight 100s on tests. the app explains how to solve each problem in detail so that you know how to work problem just like it in the future. I have also discovered that if I take a picture of a photo in my math book it recognizes that it was from my man book and shows me the answers to the problems for the entire page. I really enjoy this app it's extremely enjoyable and reliable.
Imogen Watson
How to solve an equation by factoring Calculator app picture Answer my math homework App that solves calculus problems Algebra math problems with answers Input output table solver online |
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