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## Unitary Method and Variation
Subject: Compulsory Maths
#### Overview
The unitary method is a technique in elementary algebra for solving a class of problems in variation. It consists of altering one of the variables to a single unit, i.e. 1, and then performing the operation necessary to alter it to the desired value.
#### Unitary Method
The unitary method is a method, in which the value of a quantity is first obtained to find the value of any required quantity.
There are two types of variation while solving the problem of unitary method. They are
• Direct variation
• Inverse variation
### Direct Variation
When the increase or decrease in the value of one quantity causes the increase or decrease in the value of other quantity than they are said to be direct variation. For example, the cost of goods varies directly to the numbers of goods. More good, more cost. Less good, less cost.
### Inverse Variation
When the increase or decrease in the value of one quantity causes the decrease or increase in the value of other quantity then they are said to be inverse variation. For example, time and work. More time consumes to complete a piece of work with less number of men's.
##### Things to remember
• The Unitary Method provides an alternative approach to solving problems.
• Percentages and simple percentage problems would normally precede the unitary method approach.
• Ratio calculations are an aspect of the unitary method.
• It includes every relationship which established among the people.
• There can be more than one community in a society. Community smaller than society.
• It is a network of social relationships which cannot see or touched.
• common interests and common objectives are not necessary for society.
Solution:
12 men can make a furniture in 9 days.
1man can make it in 9 × 12 days = 108 days.
18 men can make the furniture in $\frac{108}{18}$ days = 6 days
Solution:
Since 6 pens cost = Rs 450.
So, cost of 1 pens = Rs $\frac{450}{6}$ = Rs 75
So, cost of 8 pens = Rs 75×8 = Rs 600
Solution:
10 Japanese Yen = Rs 6.60
1 Japanese Yen = Rs $\frac{6.60}{10}$
65,300 Japanese Yen = Rs $\frac{6.60}{10}$× 65300 = Rs 43,098
Solution:
Given information,
From the table,
Cost of 5 Chairs = Rs 150
Cost of 1 Chair = Rs $\frac{150}{5}$ = Rs 30
Cost of 10 Chairs = Rs 30 × 10 = Rs 300
Cost of 6 Chairs= Rs 30 × 6 = Rs 180
Cost of 4 Chairs = Rs 30 × 4 = Rs120
So, Cost of x Chairs = Rs 30 × x = Rs 30x
Solution:
10 Kg Oranges cost Rs 750
1 kg Orange cost Rs $\frac{750}{10}$ = Rs 75
6 kg Oranges cost Rs 75×6 = Rs 450
∴ 6 kg Oranges cost Rs 450
Solution:
18 days is required to complete the work for 10 workers.
1 day is required to complete the same work for = 10×18 = 180 workers.
15 days is required to complete the work = $\frac{180}{15}$ = 12 workers.
∴ 12 workers can do the same work in 15 days.
Solution:
Quantity Distance (D) 7 112 12 X
Here,
$\frac{7}{12}$ = $\frac{112}{x}$
or, 7x = 112× 12
or, x = $\frac{112×12}{7}$
or, x = 192 km
∴ 12 litre can travel 192 km.
Solution:
Days Total Workers 24 20 x 15
Here, $\frac{24}{x}$ = $\frac{15}{20}$
or, 15 x = 480
or, x = $\frac{480}{15}$
or, x = 32
∴ 15 men do the work in 32 days.
Solution:
300 students required = 12 rooms
1 student required = $\frac{12} {300}$ room
375 students required = $\frac{12}{300}$× 375 = 15 rooms
∴ Required rooms = (15 -12) = 3 rooms.
Alternative,
Number of Students Class Room 300 12 375 x=?
$\frac{300}{375}$ = $\frac{12}{x}$
Or, 300×x = 12×375
Or, x = $\frac{12×375}{300}$
0r, x = 15
∴ 15 rooms are required for 375 students.
Solution:
Rs 576 = 4 Dozen(48) pens
Rs 1 = $\frac{48}{576}$pens
Rs 228 = $\frac{48}{576}$× 228 = 19 pens
Alternatively,
Pen's cost Quantity of Pen 576 4×12 = 48 228 x=?
$\frac{576}{228}$ = $\frac{48}{x}$
or, x = $\frac{48×228}{576}$
or, x = 19 pens
∴19 pens cost Rs 228
Solution:
A truck travels 48km per hour and covers the distance in 6 hours
If a truck travels 1 km per hour and covers the distance in 6×48 =288 hours
If a truck travels 36 km per hour and cover the distance = $\frac{288}{36}$ hours = 8 hours
∴ 8 hours is required to cover the distance if the speed of truck decreases by 36 km per hour.
Solution:
Since, cost of 4 Computer chairs = Rs 1,544
So, cost of 1 Computer chairs = Rs $\frac{1544}{4}$ = Rs 386
And cost of 14 Computer Chairs = Rs 386×14 = Rs 5,404
Solution:
In 10 days, a worker earns = Rs 1,850
In 1 day, a worker earns = Rs $\frac{1850}{10}$ = Rs 185
In 3 days, a worker earns = Rs 185× 3 = Rs 555
Hence, he will earn Rs 555 in 3 days.
Solution:
Cost of 93m cloth = Rs 1395
Cost of 1m cloth = Rs $\frac{1395}{93}$ Rs 15
Cost of 105m cloth = Rs 15×105 = Rs 1575
∴ The Cost of 105m cloth is Rs 1575
Solution:
Since, 32 men can reap a field in 15 days.
1 men can reap a field in 15× 32 = 480 days
20 men can reap a field in $\frac{480}{20}$ = 24 days
∴ 20 men can reap a field in 24 days.
Solution:
If, 6 hours is required to finish the work for 4 men
1 hour is required to finish the work for 4×6 = 24 men
4 hours is required to finish the work for $\frac{24}{4}$ = 6 men
∴ 6 men is required to finish the work in 4 hours.
Solution:
5 Euro = Rs 625
1 Euro = Rs $\frac{625}{5}$ = Rs 125
Rs 8,946 = $\frac{8,946}{125}$ = 71.568 Euro
Solution:
Cost of 5 goats and 2 Cows = Rs 1,35,000
Cost of 1 cow = Rs 17,500
Cost of 2 cows = Rs 17,500×2 = Rs 35,000
Cost of 5 goats = Total Amount - 2 Cows Amount
=Rs 1,35,000 - Rs 35,000
= Rs 100,000
∴ The Cost of 1 goat = Rs $\frac{1,00,000}{5}$
= Rs 20,000
∴ The Cost of 1 goat = Rs 20,000
Solution:
Since, Cost of 5 computer mouse = Rs1250.
So, Cost of 1 computer mouse = Rs $\frac{1250}{5}$ = Rs 250
And Cost of 10 computer mouse = Rs 250×10 = Rs 2,500
∴ The cost of 10 computer mouse is Rs 2,500
Solution:
Since 15 days is required to finish the work for 20 workers
1 day is required to finish the work = 20 $\times$15 = 300
10 days is required to finish the work $\frac{20\times15}{10}$ = $\frac{300}{10}$ for 30 workers
Hence, 30 workers should be added to finish the work in 10 days.
Solution:
The price of 3 chairs and 4 tables = Rs 7,540
The price of 1 chair = Rs 220
The price of 3 chairs = Rs 220$\times$3 = Rs 660
Now,
Price of 4 tables = Total amount - Price of 3 chairs
= Rs 7540 - Rs 660
= Rs 6880
Hence, Price of 1 table = $\frac{Rs 6880}{4}$
= Rs 1720 |
Multiplication of two complex numbers is also a complex number. In other words, the product of two complex numbers can be expressed in the standard form A + iB where A and B are real. z1z2 = (pr – qs) + i(ps + qr).
## What is the product of two complex numbers?
Multiplication of two complex numbers is also a complex number. In other words, the product of two complex numbers can be expressed in the standard form A + iB where A and B are real. z1z2 = (pr – qs) + i(ps + qr).
## What is argument in complex number?
The argument of a complex number is defined as the angle inclined from the real axis in the direction of the complex number represented on the complex plane. It is denoted by “θ” or “φ”. It is measured in the standard unit called “radians”.
## Why is it important to learn about complex solutions?
Answer: it is important to learn about complex solution because polynomial equation formed over a complex number can only be solved by a complex number. this is so because, the fundamental theorem of algebra state that every polynomial equation in one variable with complex coefficient has at least one complex solution.
## What is phase complex number?
Abstract. Every nonzero complex number can be expressed in terms of its magnitude and angle. This angle is sometimes called the phase or argument of the complex number. Although formulas for the angle of a complex number are a bit complicated, the angle has some properties that are simple to describe.
## What happens when you square a complex number?
Multiplying complex numbers creates a kind of rotation in the complex plane. Complex numbers that lie on the imaginary (vertical) axis have a different behavior than those that lie on the real axis. Instead of moving along the axis as a result of multiplication or squaring, the numbers rotate away from the axis.
## How do you distribute complex numbers?
Step 1: Distribute (or FOIL) using only the first two complex numbers. Step 2: Simplify the powers of i, specifically remember that i2 = –1. Step 3: Combine like terms, that is, combine real numbers with real numbers and imaginary numbers with imaginary numbers. Step 4: Distribute (or FOIL) to remove the parenthesis.
## What is the importance of complex numbers?
If the formula provides a negative in the square root, complex numbers can be used to simplify the zero. Complex numbers are used in electronics and electromagnetism. A single complex number puts together two real quantities, making the numbers easier to work with.
## Can a complex number be a real number?
No BUT — ALL REAL numbers ARE COMPLEX numbers. It just so happens that many complex numbers have 0 as their imaginary part. When 0 is the imaginary part then the number is a real number, and you might think of a real number as a 1-dimensional number.
## What is the standard form of this complex number?
A complex number is the sum of a real number and an imaginary number. A complex number is expressed in standard form when written a+bi where a is the real part and bi is the imaginary part. For example, 5+2i is a complex number.
## How do you solve triangles?
1. The angles always add to 180°: A + B + C = 180°
2. Law of Sines (the Sine Rule): When there is an angle opposite a side, this equation comes to the rescue.
3. Law of Cosines (the Cosine Rule): This is the hardest to use (and remember) but it is sometimes needed.
## How do you find missing angles?
To determine to measure of the unknown angle, be sure to use the total sum of 180°. If two angles are given, add them together and then subtract from 180°. If two angles are the same and unknown, subtract the known angle from 180° and then divide by 2.
## How do you find the angle on a calculator?
To set your calculator to work in degrees, use the key sequence (SETUP) (Deg). To calculate the sine, cosine or tangent of an angle, press the , or key and then type in the size of the angle. Note that the , and keys automatically open a bracket for you.
## How do you convert complex numbers to standard form?
1. Step 1: Multiply the complex numbers in the same manner as polynomials.
2. Step 2: Simplify the expression.
3. Step 3: Write the final answer in standard form.
4. Step 1: Multiply the complex numbers in the same manner as polynomials.
5. Step 2: Simplify the expression.
6. Step 3: Write the final answer in standard form.
## How do you convert complex numbers to Polar on a calculator?
Complex numbers that are multiplied are displayed in complex format. Press (3-2bU) (5+6bU)p. The argument of the complex number 1+2i, can be found by taking the arctan (y/x) = 63.4349° or by using the Argument command. To convert a complex number into polar form, press 2+5bUiR1(: r∠θ)p.
## How do you subtract complex numbers?
When you subtract complex numbers, you first need to distribute in the minus sign into the second complex number. Then, regroup the terms so like terms are next to each other. Combine those like terms, and you have the answer!
## Are complex numbers associative?
All complex numbers are commutative and associative under addition and multiplication, and multiplication distributes over addition.
## What is Z in complex numbers?
Complex numbers A complex number z is defined as an ordered pair z = (x, y), where x and y are a pair of real numbers.
## How do I multiply complex numbers?
Multiplying a complex number by a real number (x + yi) u = xu + yu i. In other words, you just multiply both parts of the complex number by the real number. For example, 2 times 3 + i is just 6 + 2i. Geometrically, when you double a complex number, just double the distance from the origin, 0.
## How do you find the angle of a complex number?
We find the real and complex components in terms of r and θ where r is the length of the vector and θ is the angle made with the real axis. By using the basic trigonometric ratios : cosθ=ar and sinθ=br . z=a+bi .
## Can the product of two complex numbers be a real number?
So, in general, for the product of two complex numbers to be real, the ratio of the real to imaginary parts of each complex number must be equal up to a minus sign.
## How do you convert polar numbers to complex numbers?
If you want to go from Polar Coordinates to Cartesian Coordinates, that is just: (r*cos(θ), r*sin(θ)) . And since the rectangular form of a Complex Number is a + bi , just replace the letters: a + bi = r*cos(θ) + r*sin(θ)i ← The right-hand side is a Complex Number in Polar Form. |
# Introduction to Vedic Maths for Kids
Vedic maths is not a replacement for traditional mathematics but rather a complementary system. Both traditional maths and Vedic maths can be useful and should be learned together for a well-rounded understanding of mathematics
## Importance of Vedic Maths
Vedic maths is an important subject for kids to learn for several reasons:
• ### Improves problem-solving skills
• It offers efficient and simple methods for solving mathematical problems, which can help kids to develop their problem-solving skills. Kids learn to think critically and creatively, and find solutions to complex problems in an efficient and effective manner.
• ### Enhances mental agility
• Vedic maths involves mental calculation methods that can help kids to perform complex arithmetic operations in their heads. This can enhance their mental agility and improve their mental arithmetic skills.
• ### Develops confidence
• It is a simple and efficient system of mathematics that can help kids to develop confidence in their ability to solve mathematical problems. This can have a positive impact on their self-esteem and their overall attitude towards mathematics.
• ### Increases interest in mathematics
• Vedic maths can make the learning of mathematics more interesting and engaging for kids. By using interactive resources, such as online tutorials and games, kids can develop a love for mathematics and a desire to learn more about the subject.
• ### Provides a foundation for future studies
• Vedic maths provides a strong foundation in mathematics that can be useful for kids in their future studies and careers. The concepts and techniques of Vedic Mathematics can be applied to various fields, including science, engineering, and finance.
## Vedic Maths Tricks for Kids
Here are some Vedic Mathematics tricks for kids:
• Multiplication of 11 – To multiply a number by 11, add the digits of the number and place the result between the original digits. For example, 11 x 12 = 132.
• Square of a number ending in 5 – To find the square of a number that ends in 5, simply append 25 to the number. For example, the square of 25 is 625.
• Division by 9 – To divide a number by 9, simply add the digits of the number and check if the result is a multiple of 9. If it is, then the original number is also a multiple of 9.
• Multiplication of 9 – To find the product of a number and 9, simply add the number to the number formed by its digits in reverse. For example, 9 x 17 = 153.
• Cube of a number – To find the cube of a number, simply find the square of the number and then multiply it by the number again.
These are just a few examples of Vedic Mathematics tricks for kids. By learning these tricks, kids can improve their mental arithmetic skills, develop their problem-solving skills, and have fun with mathematics.
## How to Introduce Vedic Maths to Kids?
Introducing Vedic mathematics to kids in the context of coding can be a fun and engaging experience. As it is a system of mathematics that offers efficient and simple methods for solving math problems, it can be applied to simplify and streamline the problem-solving process.
Here are some steps to help you introduce Vedic maths to kids in coding:
Begin by introducing the concepts of Vedic maths, such as mental calculation methods, addition, subtraction, multiplication, and division. It’s important to make sure that kids have a good understanding of these basic concepts before moving on to more advanced topics.
### 2. Use interactive resources
There are various interactive resources available, such as online tutorials, interactive games, and videos, that can help kids learn Vedic mathematics and apply it to coding. These resources can make the learning experience more interesting and engaging for kids.
### 3. Incorporate coding concepts
Once kids have a good understanding of Vedic maths, you can start incorporating coding concepts, such as algorithms, logic, and pattern recognition. Through this kids can develop a strong foundation in both subjects.
### 4. Encourage experimentation
Encourage kids to experiment and play with Vedic mathematics and coding. This can help them to understand the concepts better and develop their problem-solving skills.
### 5. Use real-world examples
To make Vedic maths more relevant to kids, you can use real-world examples, such as creating a program that performs arithmetic operations or solving mathematical problems. This can help kids to see the practical applications of Vedic maths and coding.
### 6. Keep it fun and interactive
Keep the learning experience fun and interactive for kids. Encourage them to ask questions and participate in group activities and discussions. You can also use games, puzzles, and other interactive resources to keep kids engaged and motivated.
## Conclusion
In a nutshell, Vedic maths is an important subject for kids to learn. It can help them develop their problem-solving skills, mental agility, confidence, interest in mathematics, and provide a foundation for their future studies and careers.
PurpleTutor offers Vedic maths classes conducted by highly qualified and experienced Vedic maths instructors. So, explore online Vedic Classes today at PurpleTutor right away.
## Frequently Asked Questions (FAQs)
1. Is there a free demo class?
A: Yes. At PurpleTutor, we give one free demo class, which can be booked from the booking link. We encourage you to take the class and assess the experience.
2. Can I select my schedule for the classes?
A: Yes. We have flexible days and times. You can select any time and any day that suits your timetable.
3. What are the fees and charges?
A: We suggest you take a complimentary trial class and discuss the requirements with our teachers and counsellors. Once this is done, the requirement can be finalized and subsequently the pricing.
4. Do you follow the school curriculum?
A. Yes, of course we do! We cover the entire grade curriculum in an average of 100 -200 hours that are spread over one academic year. Some children may take less than or more than 100 hours for the same, so the duration is customized as per the individual requirement.
5. What do you require for learning Maths from PurpleTutor?
A: It is necessary to have a laptop or computer with a webcam and a stable internet connection.
6. Do you have assessments during the course?
A. Yes, we assess the student periodically during the progress of classes and give feedback on the student’s performance.
7. What about worksheets & practice problems?
A. Yes, for each grade we have a number of worksheets and practice problems that we make available to the students. Our method teaching ensures conceptual understand while changing context of the problem.
8. Do you provide after-school & homework support?
A. Yes, depending on the requirements, our instructors ensure that the students’ needs are addressed to.
9. Will you prepare my child for exams?
A. Yes, if needed by the student we will take extra sessions to prepare the child for the exams.
10. What certificate will my child get? Will my child get any certificate after completing the course?
A. The student gets a course completion certificate from PurpleTutor after completing the course for the grade he/she has registered for. Our courses and certificates are accredited by STEM.org which is a STEM education research and credentialing organization in USA.
11. What are the courses that PurpleTutor offers?
A: PurpleTutor provides Cutting edge courses to make the student’s future ready. We offer maths courses across the grades from elementary school to high school. Our teachers are vetted through stringent checks to ensure teaching quality levels are very high.
We also have coding courses like – Python, Web Development, Machine Learning and Artificial Intelligence CoursesCyber SecurityRoblox Games & many more on offer.
Please visit our courses section for more information or talk to a counsellor. We encourage you to book a complimentary class with us, enjoy & assess the in-class experience. One can also discuss courses with our teachers in-person too during the class.
## 2 thoughts on “Introduction to Vedic Maths for Kids”
1. Swetha says:
How to book a free trial Vedic mathematics class? |
# An Account of Similar and Congruent Triangles
Triangles that have the same shapes but with different sizes are known as similar triangles. Two triangles are said to be congruent if both of them have the exact same 3 sides and the exact same 3 angles. Let's go through the various attributes of these triangles in more details.
## Similar Triangles
Definition
Two triangles are said to be similar if both of them have identical shapes, but both of them can have different sizes.(Note: Both the triangles can be similar even if one of them is rotated. They are also similar if one is placed as a mirror image of another.) Go through the following picture that depicts 2 similar triangles.
Similar Triangles
Image courtesy - https://www.mathplanet.com/Oldsite/media/43915/similar_triangles.jpg
The triangles in the above picture have identical shapes but they aren't of the same size. They are similar triangles. The formal written notation through which we can represent their similarity is:
?ABC ~ ?DEF.
Properties
• Corresponding angles of similar triangles are congruent (means that the angles should be exactly same). Hence, according to the figure depicted above, ∠ABC = ∠DEF, ∠ACB = ∠DFE, ∠BAC = ∠EDF.
• Corresponding sides of similar triangles should be in identical proportion. According to the above figure,
How to determine similarities of Triangles
Each triangle has six forms of measurement values that are three sides and three angles. It's not necessary for you to know each and every values of the three sides and three angles of each triangle among the two triangles to determine the similarity of those two triangles. A few conditions comprising of 3 values each should be enough to determine that fact. These conditions are:
• AAA (Angle, Angle, Angle): All 3 pairs of their corresponding angles should be same.
• SSS (Side, Side, Side) in the same proportion: All 3 pairs of corresponding sides should be in the exact same proportion.
• SAS (Side, Angle, Side): 2 pairs of sides should be in the same proportion, the included angle should also be equal.
If the two triangles satisfy at least one of the above three conditions, then the two triangles are similar triangles.
## Congruent Triangles
Definition
Two triangles are said to be congruent if all of their corresponding sides as well as the interior angles are proved to be congruent. The size of the two triangles as well as their shapes should be same. (Note: If one triangle is the mirror image of another, the two triangles may be deemed congruent.) The following picture depicts 2 congruent triangles.
Congruent Triangles
The two triangles in the above picture have same corresponding sides as well as same interior angles. The two triangles depicted above are congruent triangles. Mathematically, it can be represented as,
?ABC ≅ ?DEF.
An easier way to understand congruency of triangles is to imagine that the two triangles are made of cardboard. If you place one triangle on top of the other and you find that the two triangles fit perfectly, the two triangles can be deemed congruent.
Properties
• If 2 triangles are deemed congruent, each of the parts (side/angle) of one triangle should be congruent to the respective corresponding part of another triangle.
• Another point to be noted here is that if it's proved that the two triangles are congruent to one another, the angles as well as the sides of any one of the two triangles can be found from the other one. This point can be easily remembered through an acronym CPCTC that stands for “Corresponding Parts of Congruent Triangles are Congruent”.
How to determine Congruency of Triangles
Two triangles can be deemed congruent by the following conditions:
• SSS (Side, Side, Side): The 3 corresponding sides should have equal length.
• SAS (Side, Angle, Side): A pair of sides that are corresponding to each other along with the included angle should be equal.
• ASA (Angle, Side, Angle): A pair of angles that are corresponding to each other along with the inclusive side should be equal.
• AAS (Angle, Angle, Side): A pair of angles that are corresponding to each other along with a non-inclusive side should be equal.
• HL (Hypotenuse and leg of a right-angled triangle): This is possible only in case of two right-angled triangles. Two right-angled triangles are said to be congruent if their hypotenuses and a leg are equal. You can go through an example here.
Congruency and similarity of triangles are an important part of the geometry curriculum. Students need to understand the underlying concepts clearly to solve certain type of problems based on the same topics. Apart from school lessons, there are after-school online mathematics tutorials that can help the students develop a clear concept of these type of topics. That would not only enhance their critical-thinking abilities but will also enhance their performance in mathematics.
### Sudipto Das
Sudipto writes technical and educational content periodically for wizert.com and backs it up with extensive research and relevant examples. He's an avid reader and a tech enthusiast at the same time with a little bit of “Arsenal Football Club” thrown in as well. He's got a B.Tech in Electronics and Instrumentation. |
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# How do you solve k = 8k + 28?
Last updated date: 12th Sep 2024
Total views: 398.4k
Views today: 4.98k
Verified
398.4k+ views
Hint:This is a linear equation in one variable as there is only one variable in an equation. In the given question, the variable is the letter ‘k’, to solve this question we need to get ‘k’ on one side of the “equals” sign, and all the other numbers on the other side. To solve this equation for a given variable ‘k’, we have to undo the mathematical operations such as addition, subtraction, multiplication and division that have been done to the variables.
Complete step by step solution:
We have the given equation;
k = 8k + 28
Subtract k from both the sides of the equation,
k - k = 8k + 28 - k
Simplify, subtract the numbers
0 = 7k + 28
Subtract 28 from both the sides of the equation,
0 – 28 = 7k + 28 - 28
Simplify the numbers, we get
-28 = 7k
Rewriting the equation, we get
7k = -28
Dividing both the sides by 7, we get
k = -4
Therefore, the possible value of k is equals to 4. |
# Kelly Criterion
Home » Kelly Criterion
## Kelly Criterion
Kelly Criterions is based on a theory enunciated in 1956 by the American mathematician John Larry Kelly Jr., who then gave the method its name, in the Bell System Technical Journal. The usefulness of this system is due to its ability to determine the amount to invest in a bet starting from the entire budget available to the bettor.
Here are the four parameters to set in the Kelly Criterion for betting:
• The bankroll available to invest
• A match we want to bet on
• The probability that a prediction will come true (percentage success)
• The odds of the bet we want to bet
Starting from the four variables that you see above, the Kelly criterion will indicate the exact amount that we must invest to minimize risks and minimize any losses.
### Kelly’s Mathematical formula
The mathematical formula is as follows: ((Q x P) – 1) / (Q – 1), where Q stands for the odds of the bet we want to bet and P for the probability that the prediction will come true.
#### How does the Kelly criterion for betting work?
We realize that from just explaining the parameters to use and the mathematical formula it is difficult to understand how Kelly works. So let’s see step by step all the steps to follow to successfully use this system. In this article we have summarized and explained the six main steps of the Kelly system:
1. Establish an initial budget
Based on your financial resources and the amount you want to invest, decide the amount of your bankroll to use in Kelly. Remember that if you want to apply the fractional version of the method you will need to divide your initial budget in half and plug that number into the formula.
2. Choose a game to bet
Consult the schedule of the bookmaker chosen from those proposed on this page and find a match that is right for you. It’s important that you choose an event of a sport and a championship that you know well.
3. Select a tip
Choose a market and an outcome for your bet. In this step, your betting skills are very important for the success of the system.
4. Evaluate the probability of the outcome
With your knowledge of the chosen sport and after having carefully consulted the statistics, establish the probability in percentage terms of the outcome you have chosen.
5. Calculate the stake for this game
Enter the bankroll data, the odds of the chosen bet and the probability of the prediction in the formula and calculate the stake to bet, thus determining the amount to bet.
6. Repeat the operation on the next bet
Once the bet event is finished, remember to update your bankroll with the amount won or lost and repeat the operation with the next bet.
### Kelly Criterion example with a soccer game
An example of using Kelly’s criterion for soccer bets allows you to understand even better how this method is used. Assume a match to be played between Barcelona and Real Madrid with a hypothesis equal to 30% probability of winning the second team.
The bookmaker you rely on quotes a Real Madrid win at 2.5. Your budget is available. €200. Recalling that Kelly’s formula is [(qxp-1)/(q-1)]=f, then: [(2.5×30/100)]-[1/(2.5-1)]x100=8, 33. Thus, the percentage of the bankroll that you have to invest in winning Real Madrid is 8.33% of the budget you have available (€200), which is equivalent to €16.50.
• Match Barcelona and Real Madrid, with probability equal to 30%.
• Real Madrid have odds of 2.50.
• You have a budget of €200.
• You use the formula we gave you: [(2.5×30/100)]-[1/(2.5-1)]x100=8.33
• You have to invest 8.33% of your €200 budget, which is €16.50.
As you can see, the Kelly criterion allows you to better manage your budget, maximize the growth of your money, and prevent it from ending abruptly following hasty, unreasonable and/or incorrect bets. In the long run, it’s always best to use these types of strategies.
### Pros and cons of the Kelly criterion
Given the previous example we can easily understand why the most expert gamblers use this betting system and we can also realize how complicated it is to use it. Like any gaming system, in fact, this too has its advantages and disadvantages.
An advantage of Kelly’s criterion is that it certainly establishes the amount to be bet to play on a given odds. Despite the difficulties that can be encountered at the beginning in using the formula, it later becomes much more mechanical and immediate. Kelly’s criterion also measures your bankroll and that’s definitely a good thing. Always with a view to maximizing profits with a minimum bet, the percentage or decimal value of the amount to bet would prove to be the effectively optimal one to win more while losing significantly much less.
There are, however, some disadvantages to using the Kelly criterion, two in particular. The first is something we’ve already referred to: it’s only really useful when you’re able to accurately calculate the odds of any proposed bet. If you can’t do it reasonably well, then the whole system falls like a house of cards. Wrong amounts will only make you lose your bankroll.
The second of these disadvantages is that the Kelly criterion could be considered too “aggressive”. In fact, betting 10% of your bankroll is a very high sum and, in fact, many expert players prefer to play no more than 5%, some even go down to 2%. Precisely, however, this aggressiveness can be fought by assuming a more cautious relationship with the results.
### Who is the Kelly Criterion suitable for?
Even if there are enthusiastic comments on the Kelly Criterion applied to betting on the net, it is not said that it is always the right method to use. We recommend the use of this system only to very experienced players, who are familiar with both the logic of betting and the dynamics of the sport they want to bet on: the ability to develop intuition for value is acquired only over time.
The biggest obstacle is not in itself the complicated mathematical formula, which can be tackled with the tools we have indicated above. The experience of a bettor is an essential fact here and is necessary in assessing the probability that an outcome will come true. Only by managing to establish this percentage as precisely as possible will the method give the best results. This system could also be useful for those who play betting exchanges, as it can help determine whether it is more convenient to bet or lay on a given event.
Have you tried this method but it’s not for you? Try the Masaniello method. Go to the dedicated article->
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# Power of 10 Positive Exponent Online Quiz
Following quiz provides Multiple Choice Questions (MCQs) related to Power of 10: Positive Exponent. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using Show Answer button. You can use Next Quiz button to check new set of questions in the quiz.
Q 1 - Evaluate 101
A - 1
B - 10
C - 100
D - 101
### Explanation
Step 1:
The exponent here is 1. So writing base 10 one time we get
101 = 10
Step 2:
So 101 = 10
Q 2 - Evaluate 103
A - 30
B - 100
C - 103
D - 1000
### Explanation
Step 1:
The exponent here is 3. So multiplying base 10 three times we get
103 = 10 × 10 × 10 = 1000
Step 2:
So 103 = 1000
Q 3 - Evaluate 102
A - 20
B - 100
C - 102
D - 1000
### Explanation
Step 1:
The exponent here is 2. So multiplying base 10 two times we get
102 = 10 × 10 = 100
Step 2:
So 102 = 100
Q 4 - Evaluate 104
A - 40
B - 1000
C - 10000
D - 100000
### Explanation
Step 1:
The exponent here is 4. So multiplying base 10 four times we get
104 = 10 × 10 × 10 × 10 = 10000
Step 2:
104 = 10000
Q 5 - Evaluate 106
A - 1000000
B - 60
C - 10000
D - 100000
### Explanation
Step 1:
The exponent here is 6. So multiplying base 10 six times we get
106 = 10 × 10 × 10 × 10 × 10 × 10 = 1000000
Step 2:
So 106 = 1000000
### Explanation
Step 1:
The exponent here is 5. So multiplying base 10 five times we get
105 = 10 × 10 × 10 × 10 × 10 = 100000
Step 2:
So 105 = 100000
Q 7 - Evaluate 100
A - 0
B - 1
C - 10
D - 100
### Explanation
Step 1:
The exponent here is 0.
By definition 100 = 1, as any number raised to the power of zero is equal to 1.
Step 2:
So 100 = 1
### Explanation
Step 1:
The exponent here is 7. So multiplying base 10 seven times we get
107 = 10 × 10 × 10 × 10 × 10 × 10 × 10 = 10000000
Step 2:
107 = 10000000
### Explanation
Step 1:
The exponent here is 8. So multiplying base 10 eight times we get
108 = 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 = 100000000
Step 2:
So 108 = 100000000
### Explanation
Step 1:
The exponent here is 9. So multiplying base 10 nine times we get
109 = 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 = 1000000000
Step 2:
So 109 = 1000000000
power_of_10_positive_exponent.htm |
# Question Video: Writing the Decimal Numbers Marked on a Number Line Mathematics • 4th Grade
What decimals are marked on the number line?
03:31
### Video Transcript
What decimals are marked on the number line?
We’re given a number line with four decimals that have been marked. We need to work out, using the information that we’ve been given, what those decimals are. The only numbers that we have to help us have been marked already. And they are 21.67, 21.77, and 21.87. When we look at these numbers closely, we can see that they increase by one-tenth each time. Six-tenths becomes seven-tenths becomes eight-tenths. So, the interval between these two numbers on the number line is worth one-tenth, or 0.1.
And the same is true between 21.77 and 21.87. So, if the distance between the numbers we know already is one-tenth, then what’s the value of each of the small intervals that are marked but not labelled? They’re not worth tenths. Are they worth hundredths? Let’s count in hundredths to see. We’ll start with 21.67, 21.68, 21.69, 21.70. Now if we’re correct and each interval is worth one hundredth, then there should be seven intervals between here and 21.77, one, two, three, four, five, six, seven. We were right. Each interval on the number line is worth one hundredth, or 0.01. We can now use this information to help us.
If this position is 21.77, then the decimal before it will be one hundredth less than 21.77. This decimal is 21.76. Let’s continue to count backwards in hundredths to find the first missing decimal, 21.76, 21.75, 21.74. This must mean that our next missing decimal is 21.73. To find the final two missing decimals, let’s start at 21.77 and count up, 21.78, 21.79, 21.80, 21.81. So, that makes this missing decimal 21.82. 21.83, 21.84, and finally 21.85.
We used the numbers that we can see already and also the number of notches in-between each of them to work out that each interval was worth one hundredth. And we then used this to find the missing decimals. From left to right, they are 21.73, 21.76, 21.82, and finally 21.85. |
# Numerical Problems on Banking of Road
Science > You are Here
#### Example – 1:
• To what angle must a racing track of radius of curvature 600 m be banked so as to be suitable for a maximum speed of 180 km/h?
• Solution:
• Given: maximum speed v = 180 km/hr = 180 x 5/18 = 50 m/s ,Radius of curvature = r = 600 m, g = 9.8 m/s2
• To Find: Angle of banking = θ = ?
Ans: Angle of banking = 23° 18’
#### Example – 2:
• A curve in the road is in the form of an arc of a circle of radius 400 m. At what angle should the surface of the road be laid inclined to the horizontal so that the resultant reaction of the surface acting on a car running at 120 km/h is normal to the surface of the road?
• Solution:
• Given: speed of the car = v = 120 km/hr = 120 x 5/18 = 33.33 m/s, radius of curve = r = 400m, g = 9.8 m/s2
• To Find: Angle of banking = θ = ?
Ans: Angle of banking = 15° 49’
#### Example – 3:
• A vehicle enters a circular bend of radius 200 m at 72 km/h. The road surface at the bend is banked at 10°. Is it safe? At what angle should the road surface be ideally banked for safe driving at this speed ? If the road is 5m wide, what should be the elevation of the outer edge of road surface above the inner edge?
• Solution:
• Given: Speed of vehicle = v = 72 km/hr = 72 x 5/18 = 20 m/s , Angle of banking = θ = 10°, radius of curve = r = 200 m, g = 9.8 m/s2
• To Find: Part – I: To check safety, part – II: nagle of banking = θ = ? for given speed, elevation = h =?
• Part – I:
Now, the velocity of the vehicle (20 m/s) is greater than the safe velocity (18.59 m/s).
Hence it is unsafe to drive at a speed 72 km/hr.
• Part – II:
Ans: It is unsafe to drive at the speed 72 km/hr. The angle of banking for speed 72 km/hr = 11°32’,
Elevation of the outer edge over inner edge = 1m
#### Example – 4:
• Find the minimum radius of an arc of a circle that can be negotiated by a motorcycle riding at 21 m/s if the coefficient of friction between the tyres and the ground is 0.3. What is the angle made with the vertical by the motorcyclist? g = 9.8 m/s2.
• Solution:
• Given: Velocity of motor cycle = v = 21 m/s , Coefficient of friction = μ = 0.3, g = 9.8 m/s2
• To Find: radius of curvature = r = ?, Angle with vertical = θ= ?
Necessary centripetal force is provided by the friction between the road and tyres.
Ans: Radius of the circular path = 150 m, Angle of banking = 16°42’
#### Example – 5:
• What is the angle of banking necessary for a curved road of 50 m radius for safe driving at 54 km/h? If the road is not banked, what is the coefficient of friction necessary between the road surface and tyres for safe driving at this speed?
• Solution:
• Given: velocity of vehicle = v = 54 km/hr = 54 x 5/18 = 15 m/s, radius of curve = r = 50m, g = 9.8 m/s2
• To Find: Angle of banking = θ = ?, coefficient of friction = μ = ?
When the road is not banked, the necessary centripetal force is provided by the friction between the road and tyres.
Ans: The angle of banking = 24°42’, the coefficient of friction = 0.4592
#### Example – 6:
• A train of mass 105 kg rounds a curve of radius 150 m at a speed of 20m/s. Find the horizontal thrust on the outer rail if the track is not banked. At what angle must the track be banked in order that there is no thrust on the rail? g= 9.8 m/s2.
• Solution:
• Given: mass of train = m = 105 kg, velocity of train = v = 20 m/s, radius of curve = r = 400m, g = 9.8 m/s2
• To Find: Horizontal thrust = F = ?, Angle of banking = θ = ?
The horizontal thrust is equal to the centripetal force in magnitude.
Ans: Horizontal thrust = 2.67 x 105 N, the angle of banking = 15° 13’
#### Example – 7:
• The radius of curvature of a metre gauge railway line at a place where the train is moving at 36 km/h is 50m. If there is no side thrust on the rails find the elevation of the outer rail above the inner rail.
• Solution:
• Given: speed of train = v = 36 km/hr = 36 x 5/18 = 10 m/s, radius of curve = r = 50m, g = 9.8 m/s2 , For metre gauge, distance between rail = l = 1 m.
• To Find: elevation = h = ?,
Ans: The elevation of outer rail over inner rail = 0.2 m
#### Example – 8:
• Find the angle of banking of the railway track of radius of curvature 3200 m if there is no side thrust on the rails for a train running at 144 km/h. Find the elevation of the outer rail above the inner one if the distance between the rails is 1.6 m.
• Solution:
• Given: Speed of train = v = 144 km/hr = 144 x 5/18 = 40 m/s, radius of curve = r = 3200m, g = 9.8 m/s2 , distance between rail = l = 1.6 m.
• To find: Angle of banking = θ =? elevation = h = ?,
Ans: The angle of banking = 2°55’, the elevation of outer rail over inner rail = 81 mm
#### Example – 9:
• A motorcyclist at a speed of 5 m/s is describing a circle of radius 25 m. Find his inclination with the vertical. What is the value of the coefficient of friction between the road and tyres?
• Solution:
• Given: Speed of motor cycle = v = 5 m/s, radius of circle = r = 25 m, g = 9.8 m/s2 ,
• To find: Angle of inclination = θ= ?, Coefficient of friction = μ = ?
Necessary centripetal force is provided by the friction between the road and tyres.
Ans: Angle made with the vertical = 5°49’, The coefficient of friction = 0.1020
#### Example – 10:
• A motor van weighing 4400 kg rounds a level curve of radius 200 m on the unbanked road at 60 km/hr. What should be the minimum value of the coefficient of friction to prevent skidding? At what angle the road should be banked for this velocity?
• Solution:
• Given: Mass of vehicle = m = 4400 kg, velocity of vehicle = v = 60 km/hr = 60 x 5/18 = 16.67 m/s , r = 200m, g = 9.8 m/s2
• To Find: Coefficient of friction = μ = ?, Angle of banking = θ = ?,
Necessary centripetal force is provided by the friction between the road and tyres.
Ans: The coefficient of friction = 0.1418, The angle of banking = 8°4’
#### Example – 11:
• A circular road course track has a radius of 500 m and is banked to 10°. If the coefficient of friction between the road and tyre is 0.25. Compute (i) the maximum speed to avoid slipping (ii) optimum speed to avoid wear and tear of the tyres.
• Solution:
• Given: radius of curve = r = 500m, Angle of banking = θ = 10°, coefficient of friction = μ = 0.25, g = 9.8 m/s2 ,
• To find: safe velocity = v = ?, to avoid wear and tear v = ?
Maximum sped to avoid slipping.
Maximum speed to avoid wear and tear
Ans: The maximum velocity to avoid skidding = 46.74 m/s,
The maximum velocity to avoid wear and tear of tyres. = 29.39 m/s.
#### Example – 12:
• A rotor has a diameter of 4.0 m. The rotor is rotated about the central vertical axis. The occupant remains pinned against the wall. When the linear velocity of the drum is 8 m/s. Compute the coefficient of static friction between the wall of the rotor and the clothing of occupant. Also, calculate the angular velocity of the drum. How many revolutions will it make in a minute?
• Solution:
• Given: diameter = d = 4 m, radius = r = 4/2 = 2 m, linear speed = v =.8 m/s , g = 9.8 m/s2 ,
• To find: Coefficient of friction = μ = ?, Angular velocity = ω = ?, rpm = N = ?
As the occupant remains pinned against the wall, his weight is equal to the frictional force.
Frictional force = Weight of a body
Ans: Coefficient of friction = 0.3063, Angular velocity = 4 rad/s, Number of revolutions per minute = 38.21.
Science > You are Here |
# 7.4: Double-Angle, Half-Angle, and Reduction Formulas
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##### Learning Objectives
In this section, you will:
• Use double-angle formulas to find exact values.
• Use double-angle formulas to verify identities.
• Use reduction formulas to simplify an expression.
• Use half-angle formulas to find exact values.
Figure 1 Bicycle ramps for advanced riders have a steeper incline than those designed for novices.
Bicycle ramps made for competition (see Figure 1) must vary in height depending on the skill level of the competitors. For advanced competitors, the angle formed by the ramp and the ground should be θθ such that tanθ=53.tanθ=53. The angle is divided in half for novices. What is the steepness of the ramp for novices? In this section, we will investigate three additional categories of identities that we can use to answer questions such as this one.
### Using Double-Angle Formulas to Find Exact Values
In the previous section, we used addition and subtraction formulas for trigonometric functions. Now, we take another look at those same formulas. The double-angle formulas are a special case of the sum formulas, where α=β.α=β. Deriving the double-angle formula for sine begins with the sum formula,
sin(α+β)=sinαcosβ+cosαsinβsin(α+β)=sinαcosβ+cosαsinβ
If we let α=β=θ,α=β=θ, then we have
sin(θ+θ)=sinθcosθ+cosθsinθ sin(2θ)=2sinθcosθsin(θ+θ)=sinθcosθ+cosθsinθ sin(2θ)=2sinθcosθ
Deriving the double-angle for cosine gives us three options. First, starting from the sum formula, cos(α+β)=cosαcosβ−sinαsinβ,cos(α+β)=cosαcosβ−sinαsinβ, and letting α=β=θ,α=β=θ, we have
cos(θ+θ)=cosθcosθ−sinθsinθ cos(2θ)=cos2θ−sin2θcos(θ+θ)=cosθcosθ−sinθsinθ cos(2θ)=cos2θ−sin2θ
Using the Pythagorean properties, we can expand this double-angle formula for cosine and get two more interpretations. The first one is:
cos(2θ)=cos2θ−sin2θ =(1−sin2θ)−sin2θ =1−2sin2θcos(2θ)=cos2θ−sin2θ =(1−sin2θ)−sin2θ =1−2sin2θ
The second interpretation is:
cos(2θ)=cos2θ−sin2θ =cos2θ−(1−cos2θ) =2cos2θ−1cos(2θ)=cos2θ−sin2θ =cos2θ−(1−cos2θ) =2cos2θ−1
Similarly, to derive the double-angle formula for tangent, replacing α=β=θα=β=θ in the sum formula gives
tan(α+β)=tanα+tanβ1−tanαtanβtan(θ+θ)=tanθ+tanθ1−tanθtanθtan(2θ)=2tanθ1−tan2θtan(α+β)=tanα+tanβ1−tanαtanβtan(θ+θ)=tanθ+tanθ1−tanθtanθtan(2θ)=2tanθ1−tan2θ
##### DOUBLE-ANGLE FORMULAS
The double-angle formulas are summarized as follows:
sin(2θ)=2sinθcosθsin(2θ)=2sinθcosθ
cos(2θ)=cos2θ−sin2θ =1−2sin2θ =2cos2θ−1cos(2θ)=cos2θ−sin2θ =1−2sin2θ =2cos2θ−1
tan(2θ)=2tanθ1−tan2θtan(2θ)=2tanθ1−tan2θ
##### HOW TO
Given the tangent of an angle and the quadrant in which it is located, use the double-angle formulas to find the exact value.
1. Draw a triangle to reflect the given information.
2. Determine the correct double-angle formula.
3. Substitute values into the formula based on the triangle.
4. Simplify.
### EXAMPLE 1
#### Using a Double-Angle Formula to Find the Exact Value Involving Tangent
Given that tanθ=−34tanθ=−34 and θθ is in quadrant II, find the following:
1. ⓐ sin(2θ)sin(2θ)
2. ⓑ cos(2θ)cos(2θ)
3. ⓒ tan(2θ)tan(2θ)
##### TRY IT #1
Given sinα=58,sinα=58, with θθ in quadrant I, find cos(2α).cos(2α).
### EXAMPLE 2
#### Using the Double-Angle Formula for Cosine without Exact Values
Use the double-angle formula for cosine to write cos(6x)cos(6x) in terms of cos(3x).cos(3x).
#### Analysis
This example illustrates that we can use the double-angle formula without having exact values. It emphasizes that the pattern is what we need to remember and that identities are true for all values in the domain of the trigonometric function.
### Using Double-Angle Formulas to Verify Identities
Establishing identities using the double-angle formulas is performed using the same steps we used to derive the sum and difference formulas. Choose the more complicated side of the equation and rewrite it until it matches the other side.
### EXAMPLE 3
#### Using the Double-Angle Formulas to Establish an Identity
Establish the following identity using double-angle formulas:
1+sin(2θ)=(sinθ+cosθ)21+sin(2θ)=(sinθ+cosθ)2
#### Analysis
This process is not complicated, as long as we recall the perfect square formula from algebra:
(a±b)2=a2±2ab+b2(a±b)2=a2±2ab+b2
where a=sinθa=sinθ and b=cosθ.b=cosθ. Part of being successful in mathematics is the ability to recognize patterns. While the terms or symbols may change, the algebra remains consistent.
##### TRY IT #2
Establish the identity: cos4θ−sin4θ=cos(2θ).cos4θ−sin4θ=cos(2θ).
### EXAMPLE 4
#### Verifying a Double-Angle Identity for Tangent
Verify the identity:
tan(2θ)=2cotθ−tanθtan(2θ)=2cotθ−tanθ
#### Analysis
Here is a case where the more complicated side of the initial equation appeared on the right, but we chose to work the left side. However, if we had chosen the left side to rewrite, we would have been working backwards to arrive at the equivalency. For example, suppose that we wanted to show
2tanθ1−tan2θ=2cotθ−tanθ2tanθ1−tan2θ=2cotθ−tanθ
Let’s work on the right side.
2cotθ−tanθ=21tanθ−tanθ(tanθtanθ) =2tanθ1tanθ(tanθ)−tanθ(tanθ) =2tanθ1−tan2θ2cotθ−tanθ=21tanθ−tanθ(tanθtanθ) =2tanθ1tanθ(tanθ)−tanθ(tanθ) =2tanθ1−tan2θ
When using the identities to simplify a trigonometric expression or solve a trigonometric equation, there are usually several paths to a desired result. There is no set rule as to what side should be manipulated. However, we should begin with the guidelines set forth earlier.
##### TRY IT #3
Verify the identity: cos(2θ)cosθ=cos3θ−cosθsin2θ.cos(2θ)cosθ=cos3θ−cosθsin2θ.
### Use Reduction Formulas to Simplify an Expression
The double-angle formulas can be used to derive the reduction formulas, which are formulas we can use to reduce the power of a given expression involving even powers of sine or cosine. They allow us to rewrite the even powers of sine or cosine in terms of the first power of cosine. These formulas are especially important in higher-level math courses, calculus in particular. Also called the power-reducing formulas, three identities are included and are easily derived from the double-angle formulas.
We can use two of the three double-angle formulas for cosine to derive the reduction formulas for sine and cosine. Let’s begin with cos(2θ)=1−2sin2θ.cos(2θ)=1−2sin2θ. Solve for sin2θ:sin2θ:
cos(2θ)=1−2sin2θ2sin2θ=1−cos(2θ) sin2θ=1−cos(2θ)2cos(2θ)=1−2sin2θ2sin2θ=1−cos(2θ) sin2θ=1−cos(2θ)2
Next, we use the formula cos(2θ)=2cos2θ−1.cos(2θ)=2cos2θ−1. Solve for cos2θ:cos2θ:
cos(2θ)=2cos2θ−11+cos(2θ)=2cos2θ1+cos(2θ)2=cos2θ cos(2θ)=2cos2θ−11+cos(2θ)=2cos2θ1+cos(2θ)2=cos2θ
The last reduction formula is derived by writing tangent in terms of sine and cosine:
tan2θ=sin2θcos2θ =1−cos(2θ)21+cos(2θ)2 =(1−cos(2θ)2)(21+cos(2θ)) =1−cos(2θ)1+cos(2θ)Substitute the reduction formulas.tan2θ=sin2θcos2θ =1−cos(2θ)21+cos(2θ)2Substitute the reduction formulas. =(1−cos(2θ)2)(21+cos(2θ)) =1−cos(2θ)1+cos(2θ)
##### REDUCTION FORMULAS
The reduction formulas are summarized as follows:
sin2θ=1−cos(2θ)2sin2θ=1−cos(2θ)2
cos2θ=1+cos(2θ)2cos2θ=1+cos(2θ)2
tan2θ=1−cos(2θ)1+cos(2θ)tan2θ=1−cos(2θ)1+cos(2θ)
### EXAMPLE 5
#### Writing an Equivalent Expression Not Containing Powers Greater Than 1
Write an equivalent expression for cos4xcos4x that does not involve any powers of sine or cosine greater than 1.
#### Analysis
The solution is found by using the reduction formula twice, as noted, and the perfect square formula from algebra.
### EXAMPLE 6
#### Using the Power-Reducing Formulas to Prove an Identity
Use the power-reducing formulas to prove
sin3(2x)=[12sin(2x)][1−cos(4x)]sin3(2x)=[ 12sin(2x) ][ 1−cos(4x) ]
#### Analysis
Note that in this example, we substituted
1−cos(4x)21−cos(4x)2
for sin2(2x).sin2(2x). The formula states
sin2θ=1−cos(2θ)2sin2θ=1−cos(2θ)2
We let θ=2x,θ=2x, so 2θ=4x.2θ=4x.
##### TRY IT #4
Use the power-reducing formulas to prove that 10cos4x=154+5cos(2x)+54cos(4x).10cos4x=154+5cos(2x)+54cos(4x).
### Using Half-Angle Formulas to Find Exact Values
The next set of identities is the set of half-angle formulas, which can be derived from the reduction formulas and we can use when we have an angle that is half the size of a special angle. If we replace θθ with α2,α2, the half-angle formula for sine is found by simplifying the equation and solving for sin(α2).sin(α2). Note that the half-angle formulas are preceded by a ±± sign. This does not mean that both the positive and negative expressions are valid. Rather, it depends on the quadrant in which α2α2 terminates.
The half-angle formula for sine is derived as follows:
sin2θ=1−cos(2θ)2sin2(α2)=1−cos(2⋅α2)2=1−cosα2sin(α2)=±1−cosα2−−−−−−√ sin2θ=1−cos(2θ)2sin2(α2)=1−cos(2⋅α2)2=1−cosα2sin(α2)=±1−cosα2
To derive the half-angle formula for cosine, we have
cos2θ=1+cos(2θ)2cos2(α2)=1+cos(2⋅α2)2 =1+cosα2 cos(α2)=±1+cosα2−−−−−−√ cos2θ=1+cos(2θ)2cos2(α2)=1+cos(2⋅α2)2 =1+cosα2 cos(α2)=±1+cosα2
For the tangent identity, we have
tan2θ=1−cos(2θ)1+cos(2θ)tan2(α2)=1−cos(2⋅α2)1+cos(2⋅α2) =1−cosα1+cosα tan(α2)=±1−cosα1+cosα−−−−−−√ tan2θ=1−cos(2θ)1+cos(2θ)tan2(α2)=1−cos(2⋅α2)1+cos(2⋅α2) =1−cosα1+cosα tan(α2)=±1−cosα1+cosα
##### HALF-ANGLE FORMULAS
The half-angle formulas are as follows:
sin(α2)=±1−cosα2−−−−−−−−√sin(α2)=±1−cosα2
cos(α2)=±1+cosα2−−−−−−−−√cos(α2)=±1+cosα2
tan(α2)=±1−cosα1+cosα−−−−−−√=sinα1+cosα=1−cosαsinαtan(α2)=±1−cosα1+cosα=sinα1+cosα=1−cosαsinα
### EXAMPLE 7
#### Using a Half-Angle Formula to Find the Exact Value of a Sine Function
Find sin(15∘)sin(15∘) using a half-angle formula.
#### Analysis
Notice that we used only the positive root because sin(15o)sin(15o) is positive.
##### HOW TO
Given the tangent of an angle and the quadrant in which the angle lies, find the exact values of trigonometric functions of half of the angle.
1. Draw a triangle to represent the given information.
2. Determine the correct half-angle formula.
3. Substitute values into the formula based on the triangle.
4. Simplify.
### EXAMPLE 8
#### Finding Exact Values Using Half-Angle Identities
Given that tanα=815tanα=815 and αα lies in quadrant III, find the exact value of the following:
1. ⓐ sin(α2)sin(α2)
2. ⓑ cos(α2)cos(α2)
3. ⓒ tan(α2)tan(α2)
##### TRY IT #5
Given that sinα=−45sinα=−45 and αα lies in quadrant IV, find the exact value of cos(α2).cos(α2).
### EXAMPLE 9
#### Finding the Measurement of a Half Angle
Now, we will return to the problem posed at the beginning of the section. A bicycle ramp is constructed for high-level competition with an angle of θθ formed by the ramp and the ground. Another ramp is to be constructed half as steep for novice competition. If tanθ=53tanθ=53 for higher-level competition, what is the measurement of the angle for novice competition?
##### MEDIA
Access these online resources for additional instruction and practice with double-angle, half-angle, and reduction formulas.
### 7.3 Section Exercises
#### Verbal
1.
Explain how to determine the reduction identities from the double-angle identity cos(2x)=cos2x−sin2x.cos(2x)=cos2x−sin2x.
2.
Explain how to determine the double-angle formula for tan(2x)tan(2x) using the double-angle formulas for cos(2x)cos(2x) and sin(2x).sin(2x).
3.
We can determine the half-angle formula for tan(x2)=1−cosx√1+cosx√tan(x2)=1−cosx1+cosx by dividing the formula for sin(x2)sin(x2) by cos(x2).cos(x2). Explain how to determine two formulas for tan(x2)tan(x2) that do not involve any square roots.
4.
For the half-angle formula given in the previous exercise for tan(x2),tan(x2), explain why dividing by 0 is not a concern. (Hint: examine the values of cosxcosx necessary for the denominator to be 0.)
#### Algebraic
For the following exercises, find the exact values of a) sin(2x),sin(2x), b) cos(2x),cos(2x), and c) tan(2x)tan(2x) without solving for x.x.
5.
If sinx=18,sinx=18, and xx is in quadrant I.
6.
If cosx=23,cosx=23, and xx is in quadrant I.
7.
If cosx=−12,cosx=−12, and xx is in quadrant III.
8.
If tanx=−8,tanx=−8, and xx is in quadrant IV.
For the following exercises, find the values of the six trigonometric functions if the conditions provided hold.
9.
cos(2θ)=35cos(2θ)=35 and 90∘≤θ≤180∘90∘≤θ≤180∘
10.
cos(2θ)=12√cos(2θ)=12 and 180∘≤θ≤270∘180∘≤θ≤270∘
For the following exercises, simplify to one trigonometric expression.
11.
2sin(π4)2cos(π4)2sin(π4)2cos(π4)
12.
4sin(π8)cos(π8)4sin(π8)cos(π8)
For the following exercises, find the exact value using half-angle formulas.
13.
sin(π8)sin(π8)
14.
cos(−11π12)cos(−11π12)
15.
sin(11π12)sin(11π12)
16.
cos(7π8)cos(7π8)
17.
tan(5π12)tan(5π12)
18.
tan(−3π12)tan(−3π12)
19.
tan(−3π8)tan(−3π8)
For the following exercises, find the exact values of a) sin(x2),sin(x2), b) cos(x2),cos(x2), and c) tan(x2)tan(x2) without solving for x,x, when 0∘≤x≤360∘0∘≤x≤360∘
20.
If tanx=−43,tanx=−43, and xx is in quadrant IV.
21.
If sinx=−1213,sinx=−1213, and xx is in quadrant III.
22.
If cscx=7,cscx=7, and xx is in quadrant II.
23.
If secx=−4,secx=−4, and xx is in quadrant II.
For the following exercises, use Figure 5 to find the requested half and double angles.
Figure 5
24.
Find sin(2θ),cos(2θ),sin(2θ),cos(2θ), and tan(2θ).tan(2θ).
25.
Find sin(2α),cos(2α),sin(2α),cos(2α), and tan(2α).tan(2α).
26.
Find sin(θ2),cos(θ2),sin(θ2),cos(θ2), and tan(θ2).tan(θ2).
27.
Find sin(α2),cos(α2),sin(α2),cos(α2), and tan(α2).tan(α2).
For the following exercises, simplify each expression. Do not evaluate.
28.
cos2(28∘)−sin2(28∘)cos2(28∘)−sin2(28∘)
29.
2cos2(37∘)−12cos2(37∘)−1
30.
1−2sin2(17∘)1−2sin2(17∘)
31.
cos2(9x)−sin2(9x)cos2(9x)−sin2(9x)
32.
4sin(8x)cos(8x)4sin(8x)cos(8x)
33.
6sin(5x)cos(5x)6sin(5x)cos(5x)
For the following exercises, prove the identity given.
34.
(sint−cost)2=1−sin(2t)(sint−cost)2=1−sin(2t)
35.
sin(2x)=−2sin(−x)cos(−x)sin(2x)=−2sin(−x)cos(−x)
36.
cotx−tanx=2cot(2x)cotx−tanx=2cot(2x)
37.
1+cos(2θ)sin(2θ)tan2θ=tanθ1+cos(2θ)sin(2θ)tan2θ=tanθ
For the following exercises, rewrite the expression with an exponent no higher than 1.
38.
cos2(5x)cos2(5x)
39.
cos2(6x)cos2(6x)
40.
sin4(8x)sin4(8x)
41.
sin4(3x)sin4(3x)
42.
cos2xsin4xcos2xsin4x
43.
cos4xsin2xcos4xsin2x
44.
tan2xsin2xtan2xsin2x
#### Technology
For the following exercises, reduce the equations to powers of one, and then check the answer graphically.
45.
tan4xtan4x
46.
sin2(2x)sin2(2x)
47.
sin2xcos2xsin2xcos2x
48.
tan2xsinxtan2xsinx
49.
tan4xcos2xtan4xcos2x
50.
cos2xsin(2x)cos2xsin(2x)
51.
cos2(2x)sinxcos2(2x)sinx
52.
tan2(x2)sinxtan2(x2)sinx
For the following exercises, algebraically find an equivalent function, only in terms of sinxsinx and/or cosx,cosx, and then check the answer by graphing both equations.
53.
sin(4x)sin(4x)
54.
cos(4x)cos(4x)
#### Extensions
For the following exercises, prove the identities.
55.
sin(2x)=2tanx1+tan2xsin(2x)=2tanx1+tan2x
56.
cos(2α)=1−tan2α1+tan2αcos(2α)=1−tan2α1+tan2α
57.
tan(2x)=2sinxcosx2cos2x−1tan(2x)=2sinxcosx2cos2x−1
58.
(sin2x−1)2=cos(2x)+sin4x(sin2x−1)2=cos(2x)+sin4x
59.
sin(3x)=3sinxcos2x−sin3xsin(3x)=3sinxcos2x−sin3x
60.
cos(3x)=cos3x−3sin2xcosxcos(3x)=cos3x−3sin2xcosx
61.
1+cos(2t)sin(2t)−cost=2cost2sint−11+cos(2t)sin(2t)−cost=2cost2sint−1
62.
sin(16x)=16sinxcosxcos(2x)cos(4x)cos(8x)sin(16x)=16sinxcosxcos(2x)cos(4x)cos(8x)
63.
cos(16x)=(cos2(4x)−sin2(4x)−sin(8x))(cos2(4x)−sin2(4x)+sin(8x))
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# ln
The natural logarithm is a logarithm in which the base is the mathematical constant, e. It is written as ln(x) or loge(x). In certain contexts, log(x) is also used to refer to the natural log. However, log(x) is more commonly used to refer to log10(x). Using ln(x) or loge(x) to refer to the natural log removes this ambiguity.
The natural log is used widely throughout mathematics and physics. Logarithms in general are particularly useful for solving equations that involve variables in the exponent, since a logarithm can be applied to both sides of the equation to "bring the exponent down."
Example
Find x for ex = 12.
ex = 12
ln(ex) = ln(12)
x = ln(12)
This works because ln(x) is the inverse function of the exponential function, ex. In other words:
eln(x) = x, if x > 0
ln(ex) = x
## Natural logarithm function
The natural logarithm can be defined as the area under the curve of the graph y = . For a given real number, a, the natural logarithm is the area between x = 1 and x = a. This can be more clearly defined using calculus:
Below is a graph of the natural logarithm function.
Like the graph of the logarithm of any other base, the natural logarithm function has an x-intercept, or zero/root at x = 1, as shown by the point (1, 0) in green. At x = e), the function has a value of 1. In other words:
ln(e) = 1
ln(1) = 1
As x approaches 0, ln(x) approaches -∞. As x approaches ∞, ln(x) approaches ∞.
## Natural logarithm rules/properties
Natural logarithms share the same basic logarithm rules as logarithms with other bases.
• Product rule: ln(mn) = ln(m) + ln(n), for x > 0 and y > 0
• Quotient rule: ln() = ln(m) - ln(n)
• Power rule: ln(mn) = n·ln(m), for x > 0
Another useful property of logarithms is that they can be expressed in terms of logarithms of other bases multiplied by a constant. For example, to convert between a base 10 logarithm and the natural logarithm, use the following formulas:
Example
Convert the following logarithms between base 10 and base e.
1. ln(2):
2. log10(e): |
# Chapter 6 - Applications of Integration - 6.8 Logarithmic and Exponential - 6.8 Exercises: 12
$$\frac{{18x{{\ln }^2}\left( {3{x^2} + 2} \right)}}{{3{x^2} + 2}}$$
#### Work Step by Step
\eqalign{ & \frac{d}{{dx}}\left( {{{\ln }^3}\left( {3{x^2} + 2} \right)} \right) \cr & {\text{use the chain rule}} \cr & = 3{\ln ^2}\left( {3{x^2} + 2} \right)\frac{d}{{dx}}\ln \left( {3{x^2} + 2} \right) \cr & {\text{use the rule }}\frac{d}{{dx}}\left( {\ln u} \right) = \frac{1}{u}\frac{{du}}{{dx}},{\text{ letting }}u = 3{x^2} + 2 \cr & = 3{\ln ^2}\left( {3{x^2} + 2} \right)\left( {\frac{1}{{3{x^2} + 1}}} \right)\frac{d}{{dx}}\left( {3{x^2} + 2} \right) \cr & = 3{\ln ^2}\left( {3{x^2} + 2} \right)\left( {\frac{1}{{3{x^2} + 1}}} \right)\left( {6x} \right) \cr & {\text{simplify}} \cr & = \frac{{18x{{\ln }^2}\left( {3{x^2} + 2} \right)}}{{3{x^2} + 2}} \cr}
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. |
# 20 Year 6 Ratio Questions And Answers: From Easy To Hard
Year 6 ratio questions are an important aspect of KS2 maths and pupils must have secure place value, division facts and times tables knowledge to understand ratio.
Here we provide a collection of 20 problem solving ratio and proportion questions that cover the concepts pupils may encounter and can be used as practice questions ahead of the Key Stage 2 SATs.
This article also looks at the key terminology children need to confidently answer Year 6 ratio questions.
## How often do ratio questions appear in the KS2 SATs?
Ratio questions appear less frequently in the SATs exams than some of the other national curriculum objectives. However, it is important that children can approach these questions confidently to help prepare them for wider life and deepen their knowledge of ratio in secondary school.
When looking at the SATs questions by topic, in the 2023 papers, ratio and proportion accounted for 7% of all questions across the three maths papers.
### Year 6 ratio questions key terminology
To obtain a secure knowledge of ratio and answer these maths questions in Year 6 securely, children must have a strong understanding of the terminology associated with ratio and proportion.
Key terms Year 6 pupils will need to have an understanding of include:
• Ratio: Explains how much of one thing there is in relation to another thing.
For example, the ratio of boys to girls is 2:3. This means there are 2 boys for every 3 girls.
• Proportion: This explains what the fraction is compared to the whole.
For example, there are 20 pieces of fruit (the whole) and 1 in 4 of those pieces of fruit are bananas (the proportion).
• Equivalent ratios: Ratios can have the same relationship but be written differently.
For example, 1:2 and 2:4 are equivalent ratios.
• Factors: Ths is a number which divides into a number exactly.
For example, 4 is a factor of 12 and 4 divides into 12 exactly 3 times.
• Highest common factor: The highest number that can be divided exactly into 2 or more numbers.
For example, the highest common factor of 12 and 18 is 6. It is the highest number that can be divided into 12 and 18.
• Simplest form: This is the simplest way of showing the ratio. Both both parts of the ratio must be divided by the highest common factor.
For example, 12:18 in its simplest form is 2:3 when both parts are divided by the highest common factor of 12 and 18, which is 6.
• Relative size: How large something is compared to another object.
• Missing value: The quantity or part of the ratio that isn’t initially known or provided. If you only have one part of the ratio, the other part is the missing value.
• Percentage: A ratio expressed out of 100, making it easy to compare things.
For example 75% means 75 out of 100.
• Comparing ratios: Establishing whether one ratio is less than, greater than or equal to another ratio.
For example 1:2 = 2:4 but 1:2 > 1:3
• Comparison of ratio: Comparing the relationship between two or more ratios.
• Scale factor: For 2 similar shapes, the ratio of the corresponding lengths.
• Scaling ratios: Making a ratio bigger or smaller by multiplying or dividing each part by the same number.
• Unequal sharing: Using ratio to share things but not in an equal way.
For example, John gets 2 red sweets for each of Daisy’s 3 green sweets.
200 arithmetic and reasoning questions for Year 6
Download 100 free arithmetic questions and 100 free reasoning questions for Year 6. Includes answers and mark scheme.
### How will these Year 6 ratio word problems help pupils deepen their understanding?
KS2 maths assessments may present ratio questions in different formats:
• Multiple choice
• Short answer questions
• Extended response questions
Exposure to a variety of ratio word problems will help pupils build familiarity with the different types of SATs questions and develop strategies for solving them.
Practising ratio and proportion problems in real-life, relatable contexts helps ensure pupils understand and secure concepts.
Frequent exposure to a range of ratio questions also helps develop pupils’ confidence when faced with them in their Key Stage 2 maths assessments.
This sequence of Year 6 ratio questions becomes progressively harder, enabling pupils to deepen their understanding. Questions may also challenge students who are working at greater depth.
## 20 Year 6 ratio questions and answers
#### Year 6 ratio questions: problems involving the relative sizes of 2 quantities where missing values can be found by using integer multiplication and division facts
This first National Curriculum objective requires pupils to compare the 2 sizes or quantities. One value in the comparison may be missing or unknown.
To complete the ratio, children must determine the missing value.
Year 6 ratio question 1
The ratio of red counters to yellow counters is 3:1. There are 12 counters altogether. How many are there of each colour?
Answer: 9 red counters and 3 yellow counters.
3:1 = 4 parts. 4 × 3 = 12
3 × 3 = 9
1 × 3 = 3
Year 6 ratio question 2
Chloe is making a necklace using purple and silver beads in a ratio of 2:1. If she uses 16 purple beads, how many silver beads will she use?
Answer: 8 silver beads
2 × 8 = 16
1 × 8 = 8
Year 6 ratio question 3
Jared is making vegetable soup, using 200g of tomatoes, 120g of onions and 60g of mushrooms. If Jared uses 600g of mushrooms, what weight of onions will he need to use?
Answer: 360g of onions
200 × 3 = 600g
120 × 3 = 360g
Year 6 ratio question 4
If the ratio of green sweets to red sweets in a bag is 2:3. How many of each colour are in a bag of 30 sweets?
Answer: 12 green sweets and 18 red sweets.
2:3 is a total of 5. For a total of 30, we need to work out 2:3 six times.
2 × 6 = 12
3 × 6 = 18
Year 6 ratio question 5
Students in year 6 have chosen their favourite sport. They chose tennis and football in a ratio of 3:5. If there are 80 children in the year group, how many chose football?
Answer: 50 chose football
3:5 = 8 parts.
8 × 10 = 80
3 × 10 = 30
5 × 10 = 50
#### Year 6 ratio questions: problems involving the calculation of percentages and percentages for comparison
Calculating percentages involves expressing a part of a whole as a percentage.
Children need to be confident applying the concept of percentages to different measures and quantities.
Once pupils have mastered the basic percentage calculations, they can then compare quantities and values and apply their understanding of percentages to interpret pie charts.
Year 6 ratio question 6
Calculate 45% of 740
Answer: 333
10% = 74
40% = 74 × 4 = 296
5% = 37
40% = 296 + 37 = 333
Year 6 ratio question 7
Sam arrives at school at 9am each day and leaves at 3pm. What percentage of the day is he at school?
Answer: 25% of the day.
9 – 3 = 6 hours
\cfrac{6}{24} hours of the day = \cfrac{1}{4} of the day
\cfrac{1}{4} = 25%
Year 6 ratio question 8
In a school, Year 6 has 24 boys and 36 girls. What percentage of the year group are boys?
Answer: 40% are boys
24 + 36 = 60
\cfrac{24}{60} = \cfrac{4}{10} = 40%
Year 6 ratio question 9
Mason wanted to buy a pair of jeans in the sale. The original price was £45 but they were reduced by 15%. How much did he pay?
Answer: £38.35
10% of £45 = £4.50
5% of £45 = £2.25
15% of 45 = 6.75
45 – 6.75 = £38.25
Year 6 ratio question 10
A shop is selling hoodies. One has a full price of £18 but has a 20% discount, the other has a full price of £20 but has a 35% discount. Which hoodie is the cheapest in the sale?
Answer: The second hoodie is cheaper
10% of £18 = £1.80
20% of £18 = £3.60
20% discount = £18 – £3.60 = £14.40
10% of £20 = £2
30% of £20 = £6
5% of £20 = £1
35% of £30 = £7
35% discount = £20 – £7 = £13
Free resource: Year 6 Calculating Ratio Worksheet
#### Year 6 ratio questions: problems involving similar shapes where the scale factor is known or can be found
Shapes may be considered similar if they have the same shape but not the same size. For example, their corresponding angles are equal and their corresponding sides are proportional.
Here, the scale factor refers to a ratio that describes how much one shape has been scaled, either enlarged or reduced, in relation to another similar shape.
Year 6 pupils need to be able to solve problems related to similar shapes when the scale factor is either known or can be determined if the scale factor can be found.
Year 6 ratio question 11
Oliver says that a scale factor of 4 means the new shape is four times bigger than the original shape. Is Oliver correct? Explain your answer
Answer: No. A scale factor of 4 means each side is multiplied by 4. For example, if a square with sides of 1cm was 4 times bigger, the area would go from 1cm² to 4cm². If each side was 4 times bigger, the square would have an area of 16cm²
Year 6 ratio question 12
Draw a rectangle 4cm × 6cm
Now enlarge the rectangle by a scale factor of 3. What are the new measurements of the rectangle?
Answer: A rectangle drawn with a length of 18cm and width of 12cm
Both sides will be 3 x as long
4cm × 3 = 12cm
6cm × 3 = 18cm
Year 6 ratio question 13
What would the dimensions of a rectangle with a length of 6.5cm and a width of 4.5cm be, if it was enlarged by a scale factor of 3?
Answer: length 19.5cm, width 13.5cm.
Year 6 ratio question 14
Draw a right-angled triangle with a base of 7cm and a height of 3cm.
No enlarge the right-angled triangle by a scale factor of 3.5
Answer: A right-angle triangle with a base of 24.5cm and a height of 10.5cm.
Year 6 ratio question 15
Draw a right-angled triangle with a base of 9cm and a height of 5cm.
Draw a new right-angled triangle using a scale factor of \cfrac{1}{2}
Answer: A triangle drawn with a base of 4.5cm and height of 2.5cm
More Questions: Year 6 Using Scale Factors Worksheet
#### Year 6 ratio questions: solve problems involving unequal sharing and grouping using knowledge of fractions and multiples
Pupils must apply their understanding of fractions and multiples to solve problems involving sharing or grouping unequal quantities.
This includes:
• Distributing a quantity among a group of people or containers with different capacities
• Finding the total quantity when given fractional parts of a whole
• Exploring relationships between fractions and multiples when dealing with unequal sharing or grouping situations.
Year 6 ratio question 16
If 600g of tomatoes are needed in a tomato soup recipe for 6 people. How many grams of tomatoes would be needed if there were only 4 people?
Answer: 400g
600 ÷ 6 = 100g
100 × 4 = 400g
Year 6 ratio question 17
A book has 144 pages in total and is split into 6 chapters. How many pages are there in \cfrac{2}{6} of the chapters?
Answer: \cfrac{2}{6} = 48 pages
144 ÷ 6 = 24
2 × 24 = 48
Year 6 ratio question 18
Sam and Amy received £480 between them for Christmas. If Sam received £60 more than Amy, how much did they each receive?
Answer: Sam: £270 Amy: £210
£480 – £60 = £420
£420 ÷ 2 = £210
£210 + £60 = £270
Year 6 ratio question 19
Rosie planted some seeds in the garden. For every 5 seeds she planted, only 3 grew into flowers. If 15 flowers grew, how many seeds did she plant?
Answer: 25 seeds planted
3 × 5 = 15
5 × 5 = 25
Year 6 ratio question 20
A cinema has sold \cfrac{3}{8} of the tickets for the 9pm film. So far they have sold 114 tickets. How many tickets will they have sold if they sell all of them?
Answer: 304 tickets
114 tickets = \cfrac{3}{8}
\cfrac{1}{8} = 114 ÷ 3 = 38
38 × 8 = 304
Free resources for ratio and proportion:
#### FAQs
What is the difference between ratio and proportion in Year 6?
Ratio is how much of one thing there is in relation to something else. For example, for every 2 girls in the class, there are 3 boys. The ratio of girls to boys is 2:3.
Proportion is how much there is of one thing in relation to the whole amount. For example, there are 30 sweets in a bag. 1 in 5 of the sweets are red’.
What are the objectives of ratio and proportion in Year 6?
The four main objectives listed in the National Curriculum are:
• solve problems involving the relative sizes of 2 quantities where missing values can be found by using integer multiplication and division facts
• solve problems involving the calculation of percentages [for example, of measures such as 15% of 360 and the use of percentages for comparison]
• solve problems involving similar shapes where the scale factor is known or can be found
• solve problems involving unequal sharing and grouping using knowledge of fractions and multiples.
How do you calculate Year 6 ratio?
There are 40 sweets in a bag. There are 3 red sweets for every 5 orange sweets. How many red and orange sweets are there?
1. Work out the total number of parts.
For example, 3:5 is a total of 8 parts.
2. Find out how much one part is worth, by dividing by the total number of parts.
For example, if there are 40 sweets, split into a ratio of 3:5 red: orange sweets, 1 part would be 40÷8 = 5
3. Multiply each part of the ratio by the value of one part.
For example, red sweets: orange sweets = 3:5
1 part = 5
3 x 5 = 15 red sweets
5 x 5 = 25 orange sweets:
x
#### FREE Year 6 Survival Pack
A selection of fun activities to use with your Year 6 class after SATs plus our most downloaded free SATs resources to support next year's cohort.
Includes KS2 maths games and investigations and our most popular resource, Fluent in Five! |
For a triangle: The exterior angle d equals the angles a plus b.; The exterior angle d is greater than angle a, or angle b. Define an exterior angle of a triangle and identify all the exterior angles. Also, each interior angle of a triangle is more than zero degrees but less than 180 degrees. Same goes for exterior angles. Types of triangles. Similarly, this property holds true for exterior angles as well. Properties. If the figure is regular all the exterior angles will be equal. The exterior angle theorem is Proposition 1.16 in Euclid's Elements, which states that the measure of an exterior angle of a triangle is greater than either of the measures of the remote interior angles. One of the basic theorems explaining the properties of a triangle is the exterior angle theorem. Prove that it is equal to the sum of the interior angles on the other two arms. ∴ Exterior angle is equal to sum of interior opposite angles Find exterior angle ∠ 1 Here, ∠1 = ∠B + ∠C ( Exterior angle property) ∠1 = 45° + 60° ∠1 = 105° Find exterior angle In ∆ABC, ∠BCD = ∠CAB + ∠ABC (Exterior angle property) ∠BCD = 85° + 25° ∠BCD = 110° Find exterior angle In ∆PQR, Before we begin the discussion, let us have look at what is a triangle. Triangles can be classified in 2 major ways: Classification according to internal angles Exterior Angle Property - Displaying top 8 worksheets found for this concept.. In this article, we will learn about: Triangle exterior angle theorem, exterior angles of a triangle, and, how to find the unknown exterior angle of a triangle. This is a fundamental result in absolute geometry because its proof does not depend upon the parallel postulate.. Example: . ; If the equivalent angle is taken at each vertex, the exterior angles always add to 360° In fact, this is true for any convex polygon, not just triangles. This is called the exterior angle property of a triangle. Property 3: Exterior Angle Theorem An exterior angle of a triangle is equal to the sum of the two opposite interior angles. A polygon is called a plane figure that is bounded by the finite number of line segments for forming a closed figure. Any exterior angle of the triangle is equal to the sum of its interior opposite angles. For more on this see Triangle external angle theorem. Apply this relationship to find unknown angles. In any geometrical figure, whether regular or irregular, and with any number of sides, the sum of the exterior angles will always be 360 degrees. Exterior Angle Theorem. Learn the concepts of Class 7 Maths The Triangle and Its Properties with Videos and Stories. If the equivalent angle is taken at each vertex, the exterior angles always add to 360° In fact, this is true for any convex polygon, not just triangles. Exterior Angle Property. Exterior Angle Theorem. Let us discuss this theorem in detail. Properties. An exterior angle of a triangle is equal to the sum of the opposite interior angles. For more on this see Triangle external angle theorem. The side opposite to the largest angle of a triangle is the largest side. The exterior angle theorem is amongst the most basic theorems of triangles in geometry. Property 2: The sum of an interior angle and its adjacent exterior angle is 180°. An exterior angle of a triangle is equal to the sum of the opposite interior angles. To internal than 180 degrees the sum of the interior angles bounded by the number. Is regular all the exterior angles will be equal basic theorems of triangles in geometry interior! Begin the discussion, let us have look at what is a triangle equal! 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exterior angle property 2021 |
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# Properties of Polygons(pentagon, Hexagon)
Polygons are two dimensional shapes with a number of sides. If all the sides of a polygon, the polygon is said to be regular polygon.
Next to triangles and quadrilaterals, the prominent polygons are pentagons and hexagons. A Bucky ball is a combination of 12 pentagons and 20 hexagons. As the names suggest, a pentagon has five sides and a hexagon has six sides.
The regular pentagons and regular hexagons have some important and special properties and let us take a look at them.
## Properties of Pentagons
Take a look at the following regular pentagon.
O is the center of the pentagon, s is the measure of the side and a is the measure of the apothem.
From the formula for the sum of interior angles, the sum of the interior angles of a pentagon is (1800)(5 – 2) = 5400 and for a regular pentagon, each interior angle is, (5400)/(5) = 1080
The center of the pentagon is the intersection of all the perpendicular bisectors of the sides and it is also the center of the circumscribing circle of the pentagon. Therefore, the line joining the center to a vertex is the radius R of the circle.
The radius R also bisects the interior angle.
The apothem of the pentagon can be worked out as,
a = (s/2)(tan 540 )
The area of the regular pentagon can be worked out from the formula,
A = (1/2)(P)a = (1/2)(5s)( (s/2)(tan 540) = (5/4)(s2)(tan 540)
## Properties of Heagons
Now let us consider the case of a regular hexagon from geometry help
O is the center of the hexagon, s is the measure of the side and a is the measure of the apothem.
From the formula for the sum of interior angles, the sum of the interior angles of a pentagon is (1800)(6 – 2) = 7200 and for a regular pentagon, each interior angle is, (7200)/(6) = 1200
The center of the hexagon is the intersection of all the long diagonals and it is also the center of the circumscribing circle of the pentagon. Therefore, the line joining the center to a vertex is the radius R of the circle.
The radius R bisects the interior angle and the apothem bisects the side.
The apothem of the hexagon can be worked out as,
a = (s/2)(tan 600 ) = ?3(s/2) = (?3/2)(s)
The area of the regular hexagon can be worked out from the formula,
A = (1/2)(P)a = (1/2)(6s)(?3/2)(s) = (3?3/2)(s2)
## Properties of Polygons – Construction of Polygons
The properties of the polygons are used to construct the polygons. The exterior angle properties is one of them. Let us see how it is used to construct a pentagon. The method is exactly the same for a hexagon also.
For a regular pentagon, the exterior angle at any vertex is (1800 – 1080) = 720
Draw a horizontal line l and mark a point A. With A as center draw an arc of radius (equal to the measure of the side of the pentagon) to cut the line at B. Mark the point B.
At point B draw a ray m at an angle of to the left of the line l. With B as center strike an arc of the same radius to cut the ray m at C. Mark the point C.
Repeat the same steps to mark the points D and E.
Finally join E and A.
ABCDE is the required pentagon. |
# Parallel Lines and Proportional Parts
## Presentation on theme: "Parallel Lines and Proportional Parts"— Presentation transcript:
Parallel Lines and Proportional Parts
Lesson 6-4 Parallel Lines and Proportional Parts
Ohio Content Standards:
Ohio Content Standards:
Estimate, compute and solve problems involving real numbers, including ratio, proportion and percent, and explain solutions.
Ohio Content Standards:
Estimate, compute and solve problems involving rational numbers, including ratio, proportion and percent, and judge the reasonableness of solutions.
Ohio Content Standards:
Use proportional reasoning and apply indirect measurement techniques, including right triangle trigonometry and properties of similar triangles, to solve problems involving measurements and rates.
Ohio Content Standards:
Use scale drawings and right triangle trigonometry to solve problems that include unknown distances and angle measures.
Ohio Content Standards:
Apply proportional reasoning to solve problems involving indirect measurements or rates.
Theorem 6.4 Triangle Proportionality Theorem
Theorem 6.4 Triangle Proportionality Theorem
If a line is parallel to one side of a triangle and intersects the other two sides in two distinct points, then it separates these sides into segments of proportional lengths.
Theorem 6.4 Triangle Proportionality Theorem
C B D A E
R 8 V 3 S x U 12 T
Theorem 6.5 Converse of the Triangle Proportionality Theorem
Theorem 6.5 Converse of the Triangle Proportionality Theorem
If a line intersects two sides of a triangle and separates the sides into corresponding segments of proportional lengths, then the line is parallel to the third side.
Theorem 6.5 Converse of the Triangle Proportionality Theorem
D B E A
D G H F E
Midsegment
Midsegment A segment whose endpoints are the midpoints of two sides of the triangle.
Theorem 6.6 Triangle Midsegment Theorem
Theorem 6.6 Triangle Midsegment Theorem
A midsegment of a triangle is parallel to one side of the triangle, and its length is one-half the length of that side.
Theorem 6.6 Triangle Midsegment Theorem
C D B E A
Triangle ABC has vertices A(-2, 2), B(2, 4), and C(4, -4)
Triangle ABC has vertices A(-2, 2), B(2, 4), and C(4, -4). DE is a midsegment of ABC. y B D A O x E C
Triangle ABC has vertices A(-2, 2), B(2, 4), and C(4, -4)
Triangle ABC has vertices A(-2, 2), B(2, 4), and C(4, -4). DE is a midsegment of ABC. y B D Find the coordinates of D and E. A O x E C
Triangle ABC has vertices A(-2, 2), B(2, 4), and C(4, -4)
Triangle ABC has vertices A(-2, 2), B(2, 4), and C(4, -4). DE is a midsegment of ABC. y B D Verify that BC ll DE. A O x E C
Triangle ABC has vertices A(-2, 2), B(2, 4), and C(4, -4)
Triangle ABC has vertices A(-2, 2), B(2, 4), and C(4, -4). DE is a midsegment of ABC. y B D Verify that DE = ½ BC. A O x E C
Corollary 6.1
Corollary 6.1 If three or more parallel lines intersect two transversals, then they cut off the transversals proportionally.
Corollary 6.1 D F E C B A
Corollary 6.2
Corollary 6.2 If three or more parallel lines cut off congruent segments on one transversal, then they cut off congruent segments on every transversal.
Corollary 6.2 D F E C B A
In the figure, Larch, Maple, and Hatch Streets are all parallel
In the figure, Larch, Maple, and Hatch Streets are all parallel. The figure shows the distances in blocks that the streets are apart. Find x. 26 13 16 Larch x Maple Hatch
Find x and y. 2x + 2 3x - 4 5y
Assignment: Pgs. 312-315 14-28 evens, 50-56 evens |
# Areas Of Parallelograms And Triangles Class 9 Notes: Chapter 9
## CBSE Class 9 Maths Areas Of Parallelograms And Triangles Notes:-
Areas of parallelograms and triangles for class 9 notes are given here. Here, we are going to discuss some of the geometrical figures. A detailed explanation of areas of parallelograms and triangles are explained below with many theorems and proofs. Go through the below article and get the complete notes here to learn in an easy way.
### Introduction
The area represents the amount of planar surface being covered by a closed geometric figure.
Areas of closed figures
### Figures on the Common Base and Between the Same Parallels
Two shapes are stated to be on the common base and between the same parallels if:
a) They have a common side.
b) The sides parallel to the common base and vertices opposite the common side lie on the same straight line parallel to the base.
For example Parallelogram ABCD, Rectangle ABEF and Triangles ABP and ABQ
### Area of a parallelogram
Parallelogram
Area of a parallelogram = b×h
Where ‘b′ is the base and ‘h′ is the corresponding altitude(Height).
To know more about Area of a Parallelogram, visit here.
### Area of a triangle
Area of triangle
Area of a triangle = (1/2)×b×h
Where “b” is the base and “h” is the corresponding altitude.
To know more about Area of a Triangle, visit here.
## Theorems
### Parallelograms on the Common Base and Between the Same Parallels
Two parallelograms are said to be on the common/same base and between the same parallels if
a) They have a common side.
b) The sides parallel to the common side lie on the same straight line.
Parallelogram ABCD and ABEF
Theorem: Parallelograms that lie on the common base and between the same parallels are said to have equal in area.
Here, ar(parallelogram ABCD)=ar(parallelogram ABEF)
### Triangles on the Common Base and Between the Same Parallels
Two triangles are said to be on the common base and between the same parallels if
a) They have a common side.
b) The vertices opposite the common side lie on a straight line parallel to the common side.
Triangles ABC and ABD
Theorem: Triangles that lie on the same or the common base and also between the same parallels are said to have an equal area.
Here, ar(ΔABC)=ar(ΔABD)
### Two Triangles Having the Common Base & Equal Areas
If two triangles have equal bases and are equal in area, then their corresponding altitudes are equal.
### A Parallelogram and a Triangle Between the Same parallels
A triangle and a parallelogram are said to be on the same base and between the same parallels if
a) They have a common side.
b) The vertices opposite the common side lie on a straight line parallel to the common side.
A triangle ABC and a parallelogram ABDE
Theorem: If a triangle and a parallelogram are on the common base and between the same parallels, then the area of the triangle is equal to half the area of the parallelogram.
Here,  ar(ΔABC)=(1/2) ar(parallelogarm ABDE) |
## College Algebra (11th Edition)
$x=\{ 1, 10 \}$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $\log x=\sqrt{\log x} ,$ square both sides of the equation. Then use concepts of solving quadratic equations to solve for the values of $x$. Finally, do checking of the solutions with the original equation. $\bf{\text{Solution Details:}}$ Squaring both sides of the equation above results to \begin{array}{l}\require{cancel} \left( \log x \right)^2=\left( \sqrt{\log x} \right)^2 \\\\ \left( \log x \right)^2=\log x .\end{array} Transposing $\log x$ and factoring the $GCF$ of the resulting expression result to \begin{array}{l}\require{cancel} \left( \log x \right)^2-\log x=0 \\\\ \log x(\log x-1)=0 .\end{array} Equating each factor to $0$ (Zero Product Property), the solutions are \begin{array}{l}\require{cancel} \log x=0 \\\\\text{OR}\\\\ \log x-1=0 .\end{array} Since $\log_by=x$ is equivalent to $y=b^x$, the equations above, in exponential form, are equivalent to \begin{array}{l}\require{cancel} \log x=0 \\\\ \log_{10} x=0 \\\\ x=10^0 \\\\ x=1 \\\\\text{OR}\\\\ \log x-1=0 \\\\ \log_{10} x=1 \\\\ x=10^1 \\\\ x=10 .\end{array} Upon checking, both $x=\{ 1, 10 \} ,$ satisfy the original equation. |
# Series questions pdf
#### series concept
Concepts | Practice set 1 | Practice set 2
Non verbal series concept.pdf
Series Quizzes
#### Non – verbal reasoning (Series)
Non-Verbal reasoning appears in Bank exams, Infosys, MAT exams constantly. There are 5 Problem Figures (PF) will be given with 5 Answer Figures (AF). We need to determine the next figure in the series. There are certain rules which make solving these problems easy. So study the rules and solved examples.
Step 1:
For all the series problems the following rules apply. If problem figures A and E are equal our answer is problem figure B. Similarly, the other rules as follows.
1. PF(A) = PF(E) then answer is PF(B)
2. PF(D) = PF(E) then answer is PF(C)
3. PF(A) = PF(C) = PF(E) then answer is PF(B)
4. PF(A) = PF(D), PF(B) = PF(E) then answer is PF(C)
5. PF(D) = inverse of PF(A) and PF(E) = inverse of PF(2) then answer is inverse of PF(C)
Step 2:
In general, the items in the box takes different positions in the subsequent figures. They may rotate certain degrees either clock wise or anti-clockwise. Look at the following diagram. In some problems new items add to the existing figures and some existing figures vanish.
Solved Examples
1.
In this problem If PF(A) = PF(E) then answer is PF(B). In the answer options AF(4) is same as PF(B) so option 4
2.
Here PF(C) and PF(E) are equal. So Answer figure should be PF(B). So correct option is c.
3.
The arrow is changing its positions clock wise 90o, 45o, 135o, 45o, ….next should be 180o. So option 3.
4.
Simple one. A new arrow and a new line are adding alternatively. In PF(E) a new line has added. So in the next figure a new arrow must be added. And total lines should be 6. Option 5
5.
Small hand is moving anticlock wise 90o, 45o, 90o, 45o,… and Big hand is moving clock wise 135o constantly. So in the next figure, small hand must move 90o anti clockwise, and big hand must move 135o. So option 4
6.
Here the symbol is changing positions anti clockwise by 45o and every time a new symbol is adding. The “C”s in the middle are rotating clock wise by 90o. So the next figure must be option 4
7.
This is a simple analogy. There is a relationship between 1 and 2, 3 and 4. the small figures in the first diagram are getting bigger and vice-versa. So Option 3
8.
All the three symbols in the dice are rotating clockwise. So option 3
Alternative method:
We know that if PF(A) = PF(D), PF(B) = PF(E) then answer is PF(C). So option 3
9.
A new symbols is appearing in the middle of the previous figure and the previous figure is getting bigger. So option 4 is the right option. 3 and 5 options are ruled out as the figures in the middle are appeared already.
10.
A dot and line are adding constantly to the figures in left and right sides alternatively. So option 3
11.
There appears to be no pattern on immediate look, but his problem can be solved by simple observation. Have a look at the diagram below..
The positions of two symbols are not changing in 2 consecutive figures. So option 5
12.
the arrow and small line inside the small square are rotating constantly anti clockwise and clockwise respectively by 90o, 45o, 90o, 45o,… and 45o, 90o, 45o, 90o . So next figure would be option 3.
13.
The line is rotating anti clock wise by 90o, 180o, 270o, 360o so next figure should be 90o from figure E and a new symbol must appear. So option 1 is the correct.
14.
Symbols X is rotating clockwise by 45o, 90o, 45o, 90o. So our options will be either 1 or 3 as in the next figure symbol X must move 45o. A new symbols is being added to X each time one at front and next time at back. So option 3 is right one.
15.
Simple. Observe PF(A) and PF(E) are equal. So next figure will be PF(B). So option 5
16.
the symbols are changing constantly in clockwise direction and a new symbol is being added.
The red rounded circle is a place whenever a symbol appear in that position must not appear in the next. And remaining positions are moving clockwise by 90o. A new symbol must come at the place shown by green arrow. So our option will be 1. Option 2 is ruled out as + symbol appeared earlier.
17.
Circle is moving diagonally and triangle is moving clockwise by 90o. So option 1 is correct one.
18.
Here you can easily observe that the lines are rotating 90o clockwise. also in PF(B) and PF(D), half line has added at the right most side and in figures PF(C) and PF(E) a new line has added. So in our answer half line has to be added and lines should rotate 90o. So answer option 2.
19.
Simple one. Figures A and B changed their symbols opposite them. C and D also did So. So option 1
20.
Symbols in A, B are same except Symbols at bottom. A new symbol is coming there. Similarly in C, D. So option 3. Option 2 is ruled out as C appeared earlier. |
Class 11 NCERT Math Solution
TOPICS
Exercise - 3.4
Question-1 :- Find the principal and general solutions of tan x = √3.
Solution :-
``` We know that tan π/3 = √3 and
tan(4π/3) = tan(π + π/3) = tan π/3 = √3
Therefore, principal solutions are π/3 and 4π/3.
Now, tan x = tan π/3
x = nπ + π/3, where n ∈ Z.
```
Question-2 :- Find the principal and general solutions of sec x = 2.
Solution :-
``` We know that sec π/3 = 2 and
sec(5π/3) = sec(2π - π/3) = sec π/3 =2
Therefore, principal solutions are π/3 and 5π/3.
Now, sec x = sec π/3
cos x = cos π/3 [sec x = 1/cos x]
x = 2nπ ± π/3, where n ∈ Z.
```
Question-3 :- Find the principal and general solutions of cot x = -√3.
Solution :-
``` We know that cot π/6 = √3
cot(5π/6) = cot(π - π/6) = -cot π/6 = -√3
cot(11π/6) = cot(2π - π/6) = -cot π/6 = -√3
Therefore, principal solutions are 5π/6 and 11π/6.
Now, cot x = cot(5π/6)
tan x = tan(5π/6) [cot x = 1/tan x]
x = nπ + 5π/6, where n ∈ Z.
```
Question-4 :- Find the principal and general solutions of cosec x = – 2.
Solution :-
``` We know that cosec π/6 = 2
cosec 7π/6 = cosec(π + π/6) = -cosec π/6 = -2
cosec 11π/6 = cosec(2π - π/6) = -cosec π/6 = -2
Therefore, principal solutions are 7π/6 and 11π/6.
Now, cosec x = cosec(7π/6)
sin x = sin(7π/6) [cosec x = 1/sin x]
x = nπ + (-1)ⁿ 7π/6, where n ∈ Z.
```
Question-5 :- Find the general solutions of cos 4x = cos 2x.
Solution :-
``` cos 4x = cos 2x
cos 4x - cos 2x = 0
-2 sin 6x/2 sin 2x/2 = 0 [cos a - cos b = -2 sin(a + b)/2 . sin(a - b)/2]
sin 3x sin x = 0
sin 3x = 0 or sin x = 0
3x = nπ or x = nπ, where n ∈ Z.
x = nπ or x = nπ, where n ∈ Z.
```
Question-6 :- Find the general solutions of cos 3x + cos x – cos 2x = 0.
Solution :-
``` cos 3x + cos x – cos 2x = 0
2 cos 4x/2 cos 2x/2 - cos 2x = 0 [cos a + cos b = 2 cos(a + b)/2 . cos(a - b)/2]
2 cos 2x cos x - cos 2x = 0
cos 2x(2 cos x - 1) = 0
cos 2x = 0 or 2 cos x - 1 = 0
cos 2x = 0 or cos x = 1/2
2x = π/2 or x = π/3
2x = (2n + 1)π/2 or x = 2nπ ± π/3, where n ∈ Z.
x = (2n + 1)π/4 or x = 2nπ ± π/3, where n ∈ Z.
```
Question-7 :- Find the general solutions of sin 2x + cosx = 0.
Solution :-
``` sin 2x + cosx = 0
2 sin x cos x + cos x = 0
cos x(2 sin x + 1) = 0
cos x = 0 or 2 sin x + 1 = 0
cos x = 0 or sin x = -1/2
x = (2n + 1)π/2, where n ∈ Z.
Now,
sin x = -1/2
sin x = sin(π + π/6)
sin x = sin 7π/6
x = 7π/6
x = nπ + (-1)ⁿ 7π/6, where n ∈ Z.
Therefore,x = (2n + 1)π/2 or x = nπ + (-1)ⁿ 7π/6, where n ∈ Z.
```
Question-8 :- Find the general solutions of sec² 2x = 1– tan 2x
Solution :-
``` sec² 2x = 1 – tan 2x
1 + tan² 2x - 1 + tan 2x = 0
tan² 2x + tan 2x = 0
tan 2x(tan 2x + 1) = 0
tan 2x = 0 or tan 2x + 1 = 0
tan 2x = 0
2x = 0
2x = nπ + 0, where n ∈ Z
x = nπ/2, where n ∈ Z
Now,
tan 2x + 1 = 0
tan 2x = -1
tan 2x = -tan π/4
tan 2x = tan(π - π/4)
tan 2x = tan 3π/4
2x = 3π/4
2x = nπ + 3π/4, where n ∈ Z
x = nπ/2 + 3π/8, where n ∈ Z
Therefore, x = nπ/2 or x = nπ/2 + 3π/8, where n ∈ Z
```
Question-9 :- Find the general solutions of sin x + sin 3x + sin 5x = 0
Solution :-
``` sin x + sin 3x + sin 5x = 0
(sin x + sin 5x) + sin 3x = 0
2 sin 6x/2 . cos (-4x)/2 + sin 3x = 0 [sin a + sin b = 2 sin (a + b)/2 . cos (a - b)/2]
2 sin 3x . cos (-2x) + sin 3x = 0
2 sin 3x . cos 2x + sin 3x = 0 [cos(-x) = cos x]
sin 3x(2 cos 2x + 1) = 0
sin 3x = 0 or 2 cos 2x + 1 = 0
sin 3x = 0
3x = 0
3x = nπ + 0, where n ∈ Z
x = nπ/3, where n ∈ Z
Now,
2 cos 2x + 1 = 0
cos 2x = -1/2
cos 2x = -cos π/3
cos 2x = cos(π - π/3)
cos 2x = cos 2π/3
2x = 2π/3
2x = 2nπ ± 2π/3, where n ∈ Z.
x = nπ ± π/3, where n ∈ Z.
Therefore, x = nπ/3 or x = nπ ± π/3, where n ∈ Z.
```
CLASSES |
# Problem of the Week Problem D and Solution A Big Leap
## Problem
Most people think of a year as $$365$$ days, however it is actually slightly more than $$365$$ days. To account for this extra time we use leap years, which are years containing one extra day.
Mara uses the flowchart shown to determine whether or not a given year is a leap year. She has concluded the following:
• $$2018$$ was not a leap year because $$2018$$ is not divisible by $$4$$.
• $$2016$$ was a leap year because $$2016$$ is divisible by $$4$$, but not $$100$$.
• $$2100$$ will not be a leap year because $$2100$$ is divisible by $$4$$ and $$100$$, but not $$400$$.
• $$2000$$ was a leap year because $$2000$$ is divisible by $$4$$, $$100$$, and $$400$$.
If Mara chooses a year greater than $$2000$$ at random, what is the probability that she chooses a leap year?
## Solution
The probability of an event occurring is calculated as the number of favourable outcomes (that is, the number of outcomes where the event occurs) divided by the total number of possible outcomes. This is an issue in our problem because the number of years greater than $$2000$$ is infinite. However, the cycle of leap years repeats every $$400$$ years. For example, since $$2044$$ is a leap year, so is $$2444$$.
Thus, to determine the probability, we need to count the number of leap years in a $$400$$-year cycle. From the flowchart we can determine that leap years are either
• multiples of $$4$$ that are not also multiples of $$100$$, or
• multiples of $$4,~100,$$ and $$400$$.
Note that we can simplify the second case to just multiples of $$400$$, since any multiple of $$400$$ will also be a multiple of $$4$$ and $$100$$.
The number of multiples of $$4$$ in a $$400$$-year cycle is $$\frac{400}{4}=100$$. However, we have included the multiples of $$100$$, so we need to subtract these multiples. There are $$\frac{400}{100}=4$$ multiples of $$100$$ in a $$400$$-year cycle. Thus, there are $$100-4=96$$ multiplies of $$4$$ that are not multiples of $$100$$. We now need to add back the the multiples of $$400$$. There is $$\frac{400}{400}=1$$ multiple of $$400$$ in a $$400$$-year cycle. Thus, there are $$96+1=97$$ numbers that are multiples of $$4$$ and are not multiples of $$100$$, or that are multiples of $$400$$.
Therefore, for every $$400$$-year cycle, $$97$$ of these years will be a leap year.
Therefore, the probability of Mara choosing a leap year is $$\frac{97}{400} = 0.2425$$. |
## Geometry Tutors
Love making paper airplanes, tangram puzzles, and origami? Geometry is the math behind them all. Students begin learning about geometry when they are taught different shapes and angles and how to create new objects from them. They are still interested even when it is time to learn how to determine how much water it will take to fill the bucket or paint they will need to paint the house, but when students get into theorems and postulates, the fun with geometry starts to fade. Terms like hyperbolic, conical, and elliptical send many student straight to panic mode and they struggle in their classes. Getting the help of a geometry tutor can provide the student with an understanding they may miss in the classroom.
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Euclidean/Plane Geometry is the study of flat space. Between every pair of points there is a unique line segment which is the shortest curve between those two points. These line segments can be extended to lines. Lines are infinitely long in both directions and for every pair of points on the line the segment of the line between them is the shortest curve that can be drawn between them. All of these ideas can be described by drawing on a flat piece of paper. From the laws of Euclidean Geometry, we get the famous Pythagorean Theorem.
Non-Euclidean Geometry is any geometry that is different from Euclidean geometry. It is a consistent system of definitions, assumptions, and proofs that describe such objects as points, lines and planes. The two most common non-Euclidean geometries are spherical geometry and hyperbolic geometry. The essential difference between Euclidean geometry and these two non-Euclidean geometries is the nature of parallel lines: In Euclidean geometry, given a point and a line, there is exactly one line through the point that is in the same plane as the given line and never intersects it. In spherical geometry there are no such lines. In hyperbolic geometry there are at least two distinct lines that pass through the point and are parallel to (in the same plane as and do not intersect) the given line.
Riemannian Geometry is the study of curved surfaces and higher dimensional spaces. For example, you might have a cylinder, or a sphere and your goal is to find the shortest curve between any pair of points on such a curved surface, also known as a minimal geodesic. Or you may look at the universe as a three dimensional space and attempt to find the distance between/around several planets.
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## Justin J
### Teaching Style
I am a very enthusiastic tutor and, as I stated in the previous section, believe teaching in such a way that the student gains a true mastery of the given subject. In a sense, I believe in each student understanding a particular concept in their own unique way that is consistent with their unique thinking processes. I strive to do this by relating a given concept analogously to something the student already fundamentally understands. I also create new problems to solve that probe the students’ progress and comprehension. I try to make these problems as realistic as possible to make them interesting to the student.
### Experience Summary
Being a Ph.D. candidate and appointed research assistant, I very much realize the importance of true comprehension of a subject. Not only do I hold two B.S. degrees in Applied Mathematics and Chemical Engineering but I also carry credits to the equivalent of a Non-Thesis Masters in Applied Mathematics and am one course shy of a minor in Chemistry. I have tutored groups of students for Sylvan Learning Centers, conducted one-on-one sessions with Educational Enrichment and independently, and prepared and delivered lectures for advanced undergraduate chemical engineering courses. I am capable and experienced in tutoring all levels of mathematics, chemistry, chemical engineering, physics and materials science.
### Credentials
Type Subject Issued-By Level Year
Degree Chemical Engineering University of Florida Ph.D.--in progress 2010
Degree Chemical Engineering NC State University B.S. with Honors 2005
Degree Applied Mathematics NC State University B.S. with Honors 2005
## Jennifer J
### Teaching Style
I enjoy working with a student one-on-one. We know that every child develops on a different learning curve, which is why the tutoring process is so crucial. Working individually with a child can help them overcome some of the frustration and confusion that comes along with learning in a crowded classroom. I have a great amount of patience and understanding to help the child develop his own learning style and guide him to work through the problems at his own pace to ensure that he truly understands the material. Learning is a step by step process and I know that I can guide each child through the steps they need to take to complete any and all challenges they may face, not just in school, but in life.
### Experience Summary
I began tutoring algebra when I was in high school for my boss' kids in the back of the restaurant where I worked. I then went on to get my BA in mathematics at La Salle University, with a minor in Education. While in college, I continued tutoring my fellow students in math and chemistry for a peer tutoring program set up by the school. In addition I observed/student taught at various schools in the Philadelphia area. I have knowledge of a wide range of mathematics, my specialty being algebra and calculus. Due to my experience, I am also proficient in English and can assist in research paper writing.
### Credentials
Type Subject Issued-By Level Year
Degree Mathematics La Salle University Bachelor's 2007
## Timothy T
### Teaching Style
When I tutor a student, I seek first to understand the student and how he/she thinks. I find it is very important to have good rapport and communication with the student so I understand how he/she views the subject and the difficulties of it. Next I try to make "conceptual bridges" from what they know to what they are having difficulty understanding. This process usually teaches me about seeing the subject from a new point of view. I try to achieve a fine balance between guiding and directing the student’s thoughts on the topic with following the student in their own line of thinking of the subject. The student needs to learn to have confidence in his own thoughts on the subject and in his own ability to master it.
### Experience Summary
Over the past four years, I have tutored high school and middle school students in math, algebra, calculus, chemistry, SAT Math, and general study skills. My preference is to tutor math, algebra, calculus, physics, physical science, chemistry, and programming. Math is the subject for which I have the greatest passion. I also participate in the homeschooling of four of my children (13, 11, 8, 6). I have mentored my 13 yr old son in Algebra I & II, Chemistry, Elementary Math, and Middle-school Physical Science, and taught elementary math to my 11, 8, and 6 year olds. Additionally, I read and review history lessons to my kids. I completed my MS in Electrical Engineering in 2006 from The University of Texas at Arlington and my BS in Electrical Engineering and a BA in Philosophy from Rice University. I have recent experience as a student having completed Cellular Biology II at St. Petersburg College in Fall 2011.
### Credentials
Type Subject Issued-By Level Year
Degree Electrical Engineering Univ. of Texas - Arlington Masters 2006
Certification Design for Six Sigma Honeywell International DFSS - Green Belt 2003
Degree Electrical Engineering Rice University BSEE 1989
Degree Philosophy Rice University BA 1989
## Herbert H
### Teaching Style
The teaching style I use is completely dependent on the particular student being taught. I'm a big fan of the adage 'Give a man a fish, and he will eat for a day. Teach a man to fish, and he will eat for the rest of his life'. My objective is to teach a child not only the material, but how they best learn so that they can teach themselves. I enjoy using applicable references that are relevant to the child so that they will more easily grasp the concepts. I believe a personalized approach works best.
### Experience Summary
I have earned a bachelor's degree in Computer Engineering with a software focus along with a Sales Engineering minor. I was 3 credit hours away from both my Math and Business minors. In the last 5 years, I have tutored students in Basic Math, Algebra 1 and 2, Geometry, Calculus, Visual Basic, Physics, and Differential Equations. I have taught math for the past 4 years at the Professional Academies Magnet at Loften High School in the subjects of Algebra I, Algebra I Honors, Geometry, Geometry Honors, Algebra II, Algebra II Honors, Statistics, Mathematics for College Readiness, and Liberal Arts Math.
### Credentials
Type Subject Issued-By Level Year
Other Mathematics Professional Academies Magnet 9-12 2007-current
Other Tutor Starke Church of God by Faith 9-12 2005-2007
Degree Computer Engineering University of Florida BA 2004
## Vinod V
### Teaching Style
The cornerstone of my teaching philosophy and personal teaching goals is to help students develop their own thinking skills. I believe all students should leave the school armed with the ability to think for them selves, to think critically and to think creatively. Understanding how people learn is one of the significant aspects of teaching. This is linked to their “knowledge” background and maturity. The key to teaching is to relate to the audience by starting from what they know and building upon it. As a teacher I am totally involved with the class, dedicated to my students and 100% prepared to devote time and energy for their intellectual growth. Love for teaching evokes passion and dedication within me. I believe that the enthusiasm of a motivated teacher rubs off on his/her students, who derive the inspiration and encouragement which actuates their desire to learn. A good teacher should have sound fundamentals and command over the concepts. Fundamentals are the foundation intrinsic for mastering the subject; only teachers who are strong in fundamentals will be able to pass it on to their students. I believe that my strong command over the fundamentals will rub off on my students. I believe that the role of a teacher is that of a leader where you have to show the path, motivate, encourage, and lead by example. In short, my success lies in seeing my students succeed.
### Experience Summary
My enthusiasm and love for education can be gauged from the fact that I pursued three Masters degrees in three distinct but related fields. One cannot pursue engineering as a profession without having an affinity for Math and Analysis. Math was a passion for me from my young days and still very much remains so. I have a thorough knowledge and understanding of math. Right from my school days I was involved and loved to teach math. I invariably obtained A+ scores in whatever math test I took in my lifetime. For instance my GRE math score was above 95% of test takers' scores. I have taught Middle school, High school and under-graduate students in Algebra, Geometry, Trigonometry, Quadratic Equations, Applied Probability and Calculus.
### Credentials
Type Subject Issued-By Level Year
Degree City Planning Kansas State University MRCP 2002
Degree Engineering Anna University ME 2000
Degree Civil Engineering Institution of Engineers BE 1994
## Boris B
### Teaching Style
I believe that excellence in teaching comes from the teacher's adaptation to student's specific needs. In particular, for some students, visualization may be the essential component for them to understand a certain concept or an idea in mathematics, and by furnishing examples and various proofs with pictures enables the student to learn the concept, whereas for other students it may be the algebraic equation that allows them to see a certain idea. In the first couple of sessions, I probe for the specific needs of the student and then am able to connect with that student so that he/she feels comfortable with the subject. I am a patient teacher and believe that all students are able to grasp the subject. I teach in a disciplined manner, so that the topic presented is coherent and follows a logical flow. I make sure that the theoretical concepts are internalized in a concrete example for the student. Above all, I carry a positive disposition wherever I go and encourage students to enjoy math.
### Experience Summary
I am a graduate of GaTech, with a Math and Psychology BA degrees. Aside from my two majors I have minors in Philosophy and Cognitive Science. During my years of high school and college, I have tutored students in Mathematics -- be it in calculus or statistics, or the math portion of the SAT. At the moment I work part-time at the Korean after school program, called Daekyo America Inc., as a math instructor for both high school and middle school. I have participated in various Mathematical Competitions, and have won numerous awards, including the Grand Prize Winner in USAMTS (United States Mathematical Talent Search).
### Credentials
Type Subject Issued-By Level Year
Degree Applied Mathematics GaTech Bachelors 2004
Degree Psychology GaTech BA 2004
## Janet J
### Teaching Style
I enjoy teaching; it is my best talent. I understand that math is not a "favorite subject" to a lot of students. Working one on one in a tutoring situation gets past the negative block students have in a math classroom. Several of my private students have worked with me for several years. Their parents said that they felt more confident in the class room and in testing situations because of their work with me. In addition to tutoring in traditional school math subjects, I have also worked with several students on PSAT and SAT preparation. All of these students achieved a higher score on these tests after working with me one on one.
### Experience Summary
I taught 34 years with the Guilford County School System, in Greensboro, NC. I retired in June 2007, and was asked to fill 2 interim positions in the Spring and Fall semesters of 2008. I taught all courses from general math through Honors Pre-calculus. I have been privately tutoring at home for the last 7 or 8 years, to supplement my income.
### Credentials
Type Subject Issued-By Level Year
Degree Secondary Mathematics University of NC-Greensboro BS 1972
## Maria M
### Teaching Style
I love tutoring and consider myself to be effective at it. I approach it with enthusiasm. I apply theory to practical applications in my career and life. My approach is one of ease. I consider mathematics to be easy, as long as you accept what I call "the rules of the game"; i.e., there are certain principles that have to be accepted, and not questioned, then everything else falls in place. I believe that all people have the capability of learning, and I love the opportunity to provide a positive experience to students.
### Experience Summary
I have been tutoring for more than 5 years. My main focus is mathematics, but I also tutor students in Spanish I through IV. I have tutored Spanish, Algebra I, Algebra II, Geometry, Trigonometry, and Pre-Calculus. I have recently helped several students prepare for the SAT math test. I enjoy making a difference in the student’s life.
### Credentials
Type Subject Issued-By Level Year
Other Spanish Native speaker Fluent Current
Certification Project Management Studies Project Management Institute PMP Certification 2006
Degree Civil Engineering California State University at Long Beach MSCE 1987
Other Mathematics El Camino College AA 1981
## Tom & Cynthia H.
Woodbury, MN
We are happy with Parmanand and the help he is providing Paul in Math. Theresa is also doing a good job helping him with English.
## Mercy P.
Miami, FL
Lori was fantastic. I was really impressed by her attitude, patience and ability. My son really enjoyed her lessons and and feels she helped him overcome the blockage that has always made him fear Math. I am planning on requesting her again right bef... |
## The area of a circle
We want to find the area of a circle. It can be calculated as . Let us explain how we arrived at this formula and the derivation of Pi ().
## Derivation of Pi
Consider the unit circle which is a circle with radius . One way of finding its area is to use other geometrical shapes whose area we can already calculate such as a rectangle. Let us decide on a width of the rectangle and place as many as we can inside the circle. The height of the rectangle depends on where it touches the circle. In this example we fit ten rectangles inside the circle:
By calculating the area of those rectangles, we can approximate the area of the circle. The width of the rectangle is decided by us. We only need to calculate its height to calculate the area of it as . With the radius going from the center to one point on the rectangle, we get a right triangle and can use the Pythagorean theorem () to find :
For the first rectangle, we get . Solved for , we get . For the second strip, we get and solved for , we get . The area of the rectangles can then be calculated as:
(1)
The same rectangle is present four times in the circle (once in each quarter of it). By adding all areas of the rectangles and multiplying this by four, we can approximate the area of the circle. In our example we fit five rectangles into the circle. Thus, the width is . In our unit circle, , so .
(2)
Our result of is fairly imprecise. This is because of all the space in the circle that is not covered by rectangles. We can increase the number of rectangles and this space will become smaller. Hence, the more rectangles we fit into the circle (the smaller the ), the more precise our area approximation will be. By observation of our formula above, we can find a pattern so that we can easily calculate the area for an arbitrary number of rectangles:
(3)
Theoretically, if we use infinitely many rectangles (), we can get the exact area of the rectangle.
The area of the unit circle is called . We can approximate with a computer to an arbitrary precision by choosing a very large . With this the derivation of Pi is complete.
To find the area of an arbitrary radius circle in terms of (), we can factor out the :
(4) |
# Solving the Pythagorean Theorem Algebraically Completed
5 teachers like this lesson
Print Lesson
## Objective
SWBAT understand the structure of the Pythagorean Theorem algebraically in order to solve for an unknown side length as an equation.
#### Big Idea
Make sense of structure and move into solving the Pythagorean Theorem Algebraically
## Bellringer
10 minutes
As a review of solving equations, put the following two equations on the board and ask students to solve each equation. Solutions should be rounded to the nearest hundredth.
Solve for x: 45 = 7(x – 3) (5x)2 + 30 = 155
While students are working to solve each equation, move about the room formatively assessing students, providing feedback to struggling students, and choosing correct work for students to share on the board to each question. If possible, choose students to present who wrote or thought about the equations in different ways in order to view a variety of thinking. Have each student present their own work and explain their thinking.
Another great method of assessing every student, is to have students use dry erase markers and marker boards to show their work. You could show the expressions one at a time and put one minute or thirty seconds even on a clock and allow only that much time to simplify then hold up their boards when the time was up. This way, you can quickly see everyone’s thinking without needing to walk about the room so much. The marker boards are faster and you can keep a scratch sheet of paper to list the names of all students who seem to struggle through the bellringer as your formative assessment.
Again, because the activity today is focused around connecting the Pythagorean Theorem to solving equations, this review of solving equations that are both linear and quadratic are important as the lesson opener. You might want to just mention that not all quadratic equations are so simple to solve. Sometimes, a quadratic equation takes on the form 3x2 – 7x + 42 = 18. What is it about this quadratic equation that makes it more difficult to solve for the unknown value?
## Beginning the Activity
20 minutes
Clarifying and Sharing Learning Goals
Always begin by clarifying for the students what it is they will be learning from the activity today.
The learning question for today is: “How do I think about using the Pythagorean Theorem algebraically to solve for missing side lengths?” I want to push as many students as possible into thinking about the Pythagorean Theorem algebraically and connecting the solving for a missing side length to solving equations, even liner equations as in previous unit. However, I always have some students who are just more successful with visual strategies. I do not force my students to write and solve the Pythagorean Theorem algebraically. Many visual learners like to draw the squares on the diagram and still consider area, moving area of squares, and then finding the side lengths of a square to find the missing sides of the right triangle. I allow these students to continue using this strategy, which is why I have allowed so much exterior space around each diagram. It is my hope that most students will begin to transition into solving algebraically as an equation, but I do still allow choice in solving. Later when trying to develop the distance formula from the Pythagorean Theorem, a sound algebraic understanding of the Pythagorean Theorem is very useful. In high school geometry, students will be required to use the distance formula algebraically because it is also directly connected to the equation of a circle as I mentioned in the lesson rational video.
Starting the Activity
Once your expectations and learning goals are clear, you may want to spend about three minutes showing answers to the homework problems in question six. Take questions from the students about problems that were difficult and put a conclusion on finding the missing hypotenuse lengths. If I know I want to review these answers because students were not asked to work in Edmodo the night before, then I give candy to students who voluntarily put answers with work shown on the board as part of the bellringer process. This makes looking at answers to the homework much faster.
Next, allow students about five minutes to work within their cooperative groups to answer questions seven through ten. Prepare students that the diagram in these questions has changed and they must now decide if the change in the diagram also changes the student work. Students must decide which student has worked the problem correctly and then describe in full sentences the exact mistakes of the other two students. It is important for students to write in full sentences and coherent thoughts what the specific mistakes are for the two incorrect students. Again, a goal of this activity is to bring to light all the most common mistakes students make, so that your current students will not repeat these incorrect actions.
As students are working within groups to complete questions seven through ten, move about the room providing feedback that moves learning forward. It is common for some groups to say, “I don’t know who is right. They all look the same.” I usually ask these groups what is the correct answer to the question? Have you worked out the problem your way yet to consider who might be correct?” Usually these groups have not begun to solve the problem for themselves and this suggestion is enough to get them started in a productive direction.
Also, as you move about the room from group to group, select students or groups for presentation during the mini-wrap-up session. Choose student work that is correct and possibly worked or explained in different ways.
After allowing about five minutes or a little longer if needed for groups to working through questions seven through ten bring the whole class back together and hold a mini wrap up for students to share what they have learned. This mini wrap up should be student lead and student focused so students feel that they own their own learning and can discuss and defend it to others. Click below to watch a short video about how I incorporate student ownership of learning in my classroom.
I call this time of students presenting their work a “mini wrap-up” because I do not spend long periods of time closing a lesson at the end of the class period. We use small lesson closers after a small chunk of material has been completed. If you are unsure of what a mini-wrap up looks like, click below to watch a short video on how I use the mini wrap-up strategy.
Throughout the mini-wrap up, either the student groups or you should be scripting important ideas from the presentation onto the whiteboard for preservation. Many students zone out or cannot follow verbal explanations because there are visual learners. Scripting important ideas on the board for all students to use a guide map throughout the class period is very important. For further explanation of scripting, click on the follow link to watch a short video about how and why I believe scripting is vital to student success.
## Group Work
20 minutes
Working in Cooperative Groups to Make Connections Continued
Once you conclude a mini wrap up of student work, allow students about five minutes to complete questions 11, 12, and 13 Prepare students for these questions by saying, “The following two questions are scaffolded questions designed to help you ease into setting up and solving the Pythagorean Theorem algebraically. The blank lines are missing values in each step of solving the Pythagorean Theorem as an equation. Question 13 is designed for you to think about how the first example in question one and second example in question seven are both alike and different in how you think about them and solve them.” Students should work together to solve these equations as you also move about the room providing feedback.
After allowing students time to finish questions 11, 12, and 13 pull three student groups to the board to present their work as a mini wrap up for the class. Make sure either the students or you script the entire solution onto the white board for notes and reference as a map throughout class. Discuss question 13 well so that students understand you cannot always use a and b for the sides you know and simply solve for c in every triangle. This misunderstanding is very common and students do not understand why sometime you subtract and sometimes you add before taking the square root. Make sure students understand the structure algebraically of each equation.
Where to End the Lesson
I try to make it to question 14 which contains practice problems solving for the hypotenuse and sometimes solving for missing legs, with about five to ten minutes left in class so that students can begin to work together to solve for the missing hypotenuse. Homework is to finish all of the practice problems in question 14. I again allowed extra white space around each diagram for students who plan to draw the three squares on each side in order to solve. I encourage the algebraic equation method but do not force it. For my classes with technology at home, I often require students to take pictures of their homework answers to question six and post these images through their Edmodo account to the rest of the class. I then require students to logo into Edmodo and also post at least two comments to another student that night in Edmodo.com. If you are unsure of how to use Edmodo.com or even what it is, click below to watch a short video of how I have used Edmodo.com to link my students this school year.
## Student Work and Standards Addressed
Here are a few images of sample work produced by my students throughout this school year. The samples are of answers to the first five questions only.
Standards Applied in this Lesson
The math standard addressed by this lesson is 8.G.B.7 Apply the Pythagorean Theorem to determine unknown side lengths in right triangles in real-world and mathematical problems in two and three dimensions.
MP3 Construct viable arguments and critique the reasoning of others
Asking students to think algebraically about setting up and solving the Pythagorean Theorem to the nearest hundredth brings in math practice standards 6 and 8.
MP6 Attend to precision. (Click here to watch a short video on MP6)
MP8 Regularity in Math |
# Physics with Calculus/Mechanics/Scalar and Vector Quantities
## Introduction
The study of Mitchel Joens is intimately tied in with the study of mathematics. Sometimes, the direction of a number or quantity is as important as the number itself. Mathematicians in the 19th century developed a convenient way of describing and interacting with quantities with and without direction by dividing them into two types: scalar quantities and vector quantities. Scalar quantities have a magnitude but no direction. Vector quantities have both a magnitude and a direction. For instance, one might describe a plane as flying at 400 miles per hour. However, simply knowing the speed of the airplane is not nearly as useful as knowing the speed and direction of the airplane, so a more accurate description may be a plane flying at 400 miles per hour southeast. we use the concept of scalars and vectors by applying it to the physical quantities so that we can understand their properties and characteristics in a more convenient way.
## Scalar Quantities
Scalar quantities are numbers that have a magnitude but no direction. Scalars are represented by a single letter, such as ${\displaystyle a}$ . Some examples of scalar quantities are mass (five kilograms), temperature (twenty-two degrees Celsius), and numbers without units (such as three).
## Vector Quantities
Vectors are a geometric way of representing quantities that have direction as well as magnitude. An example of a vector is force. If we are to fully describe a force on an object we need to specify not only how much force is applied, but also in which direction. Another example of a vector quantity is velocity -- an object that is traveling at ten meters per second to the east has a different velocity than an object that is traveling ten meters per second to the west. This vector is a special case, however,sometimes people are interested in only the magnitude of the velocity of an object. This quantity, a scalar, is called speed which has magnitude but no given direction.
When vectors are written, they are represented by a single letter in bold type or with an arrow above the letter, such as ${\displaystyle \mathbf {A} }$ or ${\displaystyle {\vec {A}}}$ . Some examples of vectors are displacement (e.g. 120 cm at 30°) and velocity (e.g. 12 meters per second north). The only basic SI unit that is a vector is the meter. All others are scalars. Derived quantities can be vector or scalar, but every vector quantity must involve meters in its definition and unit.
## Unit Vectors
An illustration of common choice of unit vectors in a Cartesian coordinate system
Strictly speaking, vectors exist separately from any coordinate systems. As vectors are geometric objects, we do not need to define a coordinate system in order to talk about vectors—or even to perform most operations on vectors. For example, consider the triple of numbers: number of apples, number of bananas, and number of carrots you have. Say that you calculate the triple in one coordinate system and get (1,2,3). If you rotate your coordinate system, and recalculate, you will have (1,2,3) again. Thus, the triple does not have the most important property of a vector -- that is transform like the coordinate system.
Nevertheless, it is often convenient to introduce a coordinate system. In three dimensions, for many problems the rectangular, or Cartesian coordinate system (named after French mathematician René Descartes) turns out to be convenient, and this coordinate system can be defined in terms of unit vectors.
A unit vector is a vector pointing in a given direction with a magnitude of one. Essentially, it merely indicates direction. In a Cartesian system the three unit vectors are called i, j, and k (or, in handwriting, with a little "hat" on top, as ${\displaystyle {\hat {i}}}$ , ${\displaystyle {\hat {j}}}$ , and ${\displaystyle {\hat {k}}}$ ). Colloquially, you might refer to the directions of the unit vectors as "east", "north", and "up". One could just have easily chosen i as up, j as east, and k as north. In choosing i, j, and k, once i and j are chosen, k must point to a particular direction, so that a common convention called "right-hand rule" holds. Mathematically, this can be compactly expressed as,
${\displaystyle {\hat {k}}={\hat {i}}\times {\hat {j}}}$ ,
but we will expand more on this as we describe "cross products" later on.
Unit vectors are generally chosen to be orthogonal. That is, each unit vector is perpendicular to each of the others. While unit vectors do not need to be orthogonal, working with a coordinate system defined by orthogonal unit vectors will be convenient in most cases. There are two other major coordinate systems used in physics—cylindrical coordinates and spherical coordinates. These will be introduced at a later time as necessary.
## Vector Components
Every vector may be expressed as the sum of its n unit vectors.
${\displaystyle {\vec {A}}=a_{x}~{\hat {i}}+a_{y}~{\hat {j}}+a_{z}~{\hat {k}}}$
The quantities ax, ay, and az are called the vector components of vector A. Sometimes they are represented simply as an ordered triple (e.g. (ax,ay,az)) especially when the choice and ordering of three unit vectors are not ambiguous.
## Vector Algebra
### Negation
Illustration of vector negation and scalar multiplication
Negation can also mean one vector isn't equal to the other so it can be expained like this: ${\displaystyle {\vec {A}}\neq {\vec {B}}}$
${\displaystyle -{\vec {A}}=-(a_{x}~{\hat {i}}+a_{y}~{\hat {j}}+a_{z}~{\hat {k}})=-a_{x}~{\hat {i}}-a_{y}~{\hat {j}}-a_{z}~{\hat {k}}}$
Considering a vector represented graphically by an arrow, the negative of a vector would be represented by a vector of the same length but opposite direction.
### Scalar Multiplication
${\displaystyle k{\vec {A}}=ka_{x}~{\hat {i}}+ka_{y}~{\hat {j}}+ka_{z}~{\hat {k}}}$
Note that vector negation is merely multiplication by a scalar, where that scalar is -1. A scaled vector represented graphically would point in the same direction as the original vector but have its magnitude scaled by a factor of k.
{\displaystyle {\begin{aligned}{\vec {A}}+{\vec {B}}&=(a_{x}~{\hat {i}}+a_{y}~{\hat {j}}+a_{z}~{\hat {k}})+(b_{x}~{\hat {i}}+b_{y}~{\hat {j}}+b_{z}~{\hat {k}})\\&=(a_{x}+b_{x})~{\hat {i}}+(a_{y}+b_{y})~{\hat {j}}+(a_{z}+b_{z})~{\hat {k}}\end{aligned}}}
Two vectors can be added graphically by placing the tail of the second vector (here, B) coincidental with the tip of the first vector (A). The resultant vector A + B is the vector drawn from the tail of A to the tip of B.
Any number of vectors can be added in this fashion. Vector addition is commutative:
${\displaystyle {\vec {A}}+{\vec {B}}+{\vec {C}}={\vec {C}}+{\vec {B}}+{\vec {A}}}$
and associative:
${\displaystyle ({\vec {A}}+{\vec {B}})+{\vec {C}}={\vec {A}}+({\vec {B}}+{\vec {C}})}$
### Dot Product
Calculating bond angles of a symmetrical tetrahedral molecule such as methane using a dot product
When we multiply two vectors, we can either apply a multiplication rule that produces a scalar as the end result, or one that produces a vector as the end result. The first one that produces a scalar is called dot product. In mathematical texts, this is often called inner product, and some older texts will refer to this as scalar product (not to be confused with scalar multiplication); they are all the same. Dot product has all the usual properties of products, such as associativity, commutativity, and the distributive property. Geometrically, dot product is defined as:
${\displaystyle {\vec {A}}\cdot {\vec {B}}=AB\cos(\theta )}$ ,
where ${\displaystyle \theta }$ is the angle between ${\displaystyle {\vec {A}}}$ and ${\displaystyle {\vec {B}}}$ . Note that since ${\displaystyle \cos(0)=1}$ , if ${\displaystyle {\vec {A}}}$ is parallel to ${\displaystyle {\vec {B}}}$ , then ${\displaystyle {\vec {A}}\cdot {\vec {B}}=AB}$ . On the other hand, since ${\displaystyle \cos(90^{\circ })=0}$ if ${\displaystyle {\vec {A}}}$ is perpendicular to ${\displaystyle {\vec {B}}}$ , then ${\displaystyle {\vec {A}}\cdot {\vec {B}}=0}$ . Using this as the guiding rule, we find below relationship:
{\displaystyle {\begin{aligned}{\hat {i}}\cdot {\hat {i}}={\hat {j}}\cdot {\hat {j}}={\hat {k}}\cdot {\hat {k}}=1\\{\hat {i}}\cdot {\hat {j}}={\hat {j}}\cdot {\hat {k}}={\hat {k}}\cdot {\hat {i}}=0\end{aligned}}} .
Using this, we can define dot product in terms of component vectors as follows:
${\displaystyle {\vec {A}}\cdot {\vec {B}}=(A_{x}~{\hat {i}}+A_{y}~{\hat {j}}+A_{z}~{\hat {k}})\cdot (B_{x}~{\hat {i}}+B_{y}~{\hat {j}}+B_{z}~{\hat {k}})=A_{x}B_{x}+A_{y}B_{y}+A_{z}B_{z}}$ .
You are encouraged to expand out the multiplication explicitly, using the distributive property and find which terms cancel to zero and which products become 1.
### Cross Product
The second multiplication rule for product of two vectors yields yet another vector. This multiplication rule is a very special one—in fact, it is a special property of 3-dimensional space that we can define a vector multiplication is this way and still obtain a vector. This rule will not work when limited to 2-D, and in any dimensions greater than 3, an extension of this rule will not result in another vector (cf. dot product can be naturally extended or limited to any dimensions to produce a scalar). This multiplication is called cross product, and in other texts, you may find terms outer product and vector product. The product can be defined with the two rules, first specifying the product vector's direction, and the second specifying its magnitude:
1. ${\displaystyle {\vec {A}}\times {\vec {B}}}$ is perpendicular to ${\displaystyle {\vec {A}}}$ and ${\displaystyle {\vec {B}}}$ (that is, perpendicular to the plane defined by these two vectors). This leaves two possible directions along the line perpendicular to the plane. One of the two directions is called by a "right-hand rule": Hold out index finger, middle finger, and the thumb so that they are all perpendicular to each other. Let the index finger point towards direction of ${\displaystyle {\vec {A}}}$ , and the middle finger towards ${\displaystyle {\vec {B}}}$ . Then the thumb points towards the direction of ${\displaystyle {\vec {A}}\times {\vec {B}}}$ . The ordering is important here (note exchanging A and B makes the thumb point in the opposite direction).
2. ${\displaystyle |{\vec {A}}\times {\vec {B}}|=AB\sin(\theta )}$ , where ${\displaystyle \theta }$ is again the angle between ${\displaystyle {\vec {A}}}$ and ${\displaystyle {\vec {B}}}$ .
Applying this definition to unit vectors again, we find following relationships:
{\displaystyle {\begin{aligned}{\hat {i}}\times {\hat {j}}&=-{\hat {j}}\times {\hat {i}}={\hat {k}}\\{\hat {j}}\times {\hat {k}}&=-{\hat {k}}\times {\hat {j}}={\hat {i}}\\{\hat {k}}\times {\hat {i}}&=-{\hat {i}}\times {\hat {k}}={\hat {j}}\\{\hat {i}}\times {\hat {i}}&={\hat {j}}\times {\hat {j}}={\hat {k}}\times {\hat {k}}=0\end{aligned}}} .
And in terms of components, we have (after a tedious algebra):
${\displaystyle {\vec {A}}\times {\vec {B}}=(A_{y}B_{z}-B_{y}A_{z})~{\hat {i}}+(A_{z}B_{x}-B_{z}A_{x})~{\hat {j}}+(A_{x}B_{y}-B_{x}A_{y})~{\hat {k}}}$ .
It turns out we can write this complicated relationship as a determinant of a 3 x 3 matrix:
${\displaystyle {\vec {A}}\times {\vec {B}}=\left|{\begin{matrix}{\hat {i}}&{\hat {j}}&{\hat {k}}\\A_{x}&A_{y}&A_{z}\\B_{x}&B_{y}&B_{z}\end{matrix}}\right|}$ .
Some properties of cross product, such as ${\displaystyle {\vec {A}}\times {\vec {B}}=-{\vec {B}}\times {\vec {A}}}$ and ${\displaystyle {\vec {A}}\times {\vec {A}}=0}$ can be derived as a property of the determinant of the matrix.
### Useful Properties of Dot Product and Cross Product
Both the dot product and the cross product distribute over vector addition.
${\displaystyle {\vec {A}}\cdot ({\vec {B}}+{\vec {C}})={\vec {A}}\cdot {\vec {B}}+{\vec {A}}\cdot {\vec {C}}}$
${\displaystyle {\vec {A}}\times ({\vec {B}}+{\vec {C}})={\vec {A}}\times {\vec {B}}+{\vec {A}}\times {\vec {C}}}$
${\displaystyle ({\vec {A}}+{\vec {B}})\times {\vec {C}}={\vec {A}}\times {\vec {C}}+{\vec {B}}\times {\vec {C}}}$
The dot product of two vectors is proportional to the cosine of the angle between them, and their cross product is proportional to the sine of the angle between them.
${\displaystyle {\vec {A}}\cdot {\vec {B}}=\|{\vec {A}}\|\|{\vec {B}}\|\cos(\theta )}$
${\displaystyle \|{\vec {A}}\times {\vec {B}}\|=\|{\vec {A}}\|\|{\vec {B}}\|\sin(\theta )}$
As we have seen already, the dot product is associative and commutative.
${\displaystyle {\vec {A}}\cdot ({\vec {B}}\cdot {\vec {C}})=({\vec {A}}\cdot {\vec {B}})\cdot {\vec {C}}}$
${\displaystyle {\vec {A}}\cdot {\vec {B}}={\vec {B}}\cdot {\vec {A}}}$
It is important to remember that the cross product has neither of these properties. Instead of being commutative, it is anticommutative.
${\displaystyle {\vec {A}}\times {\vec {B}}=-{\vec {B}}\times {\vec {A}}}$
The cross product is not even associative. For example, consider ${\displaystyle ({\vec {A}}\times {\vec {A}})\times {\vec {B}}}$ . Since the sine of the angle between ${\displaystyle {\vec {A}}}$ and itself is 0, ${\displaystyle {\vec {A}}\times {\vec {A}}={\vec {0}}}$ , and so ${\displaystyle ({\vec {A}}\times {\vec {A}})\times {\vec {B}}={\vec {0}}}$ . On the other hand, ${\displaystyle {\vec {A}}\times ({\vec {A}}\times {\vec {B}})}$ is not zero, since ${\displaystyle {\vec {A}}}$ and ${\displaystyle {\vec {A}}\times {\vec {B}}}$ are perpendicular. In fact, if ${\displaystyle {\vec {A}}}$ and ${\displaystyle {\vec {B}}}$ were perpendicular, its direction would be opposite to that of ${\displaystyle {\vec {B}}}$ . Check this yourself using the right hand rule.
The component of ${\displaystyle {\vec {A}}}$ parallel to ${\displaystyle {\vec {B}}}$ is given by
${\displaystyle \left({\frac {{\vec {B}}\cdot {\vec {A}}}{{\vec {B}}\cdot {\vec {B}}}}\right){\vec {B}}}$
and the perpendicular component of ${\displaystyle {\vec {A}}}$ is given by
${\displaystyle \left({\frac {{\vec {B}}\times {\vec {A}}}{{\vec {B}}\cdot {\vec {B}}}}\right)\times {\vec {B}}}$ .
This leads to some interesting properties involving combinations of the products, such as
${\displaystyle ({\vec {B}}\cdot {\vec {B}}){\vec {A}}=({\vec {A}}\cdot {\vec {B}}){\vec {B}}+({\vec {B}}\times {\vec {A}})\times {\vec {B}}}$ ,
${\displaystyle {\vec {A}}\times ({\vec {B}}\times {\vec {C}})=({\vec {A}}\cdot {\vec {C}}){\vec {B}}-({\vec {A}}\cdot {\vec {B}}){\vec {C}}}$ , and
${\displaystyle ({\vec {A}}\times {\vec {B}})\cdot ({\vec {C}}\times {\vec {D}})=({\vec {A}}\cdot {\vec {C}})({\vec {B}}\cdot {\vec {D}})-({\vec {A}}\cdot {\vec {D}})({\vec {B}}\cdot {\vec {C}})}$ . |
# Mode in Statistics – Definition, Formula, Examples | How to find Mode?
In this article, we can learn how to calculate the Mode in statistics, its definition, Types of Modes, Advantages, and Example Problems. The purpose of statistics is to make us learn to utilize a restricted sample to make accurate determinations. Statistics deals with the presentation of data, collection of data, and analysis of data or information for a particular purpose.
In statistics to represent the data, we use bar graphs, piecharts, tables, graphs, pictorial representation, and so on. The frequency in statistics tends to represent a set of data by a representative value which would define the entire collection of data. This representative value is known as the measure of central tendency. One such measure of central tendency is the mode of data.
Also, Check Related Articles:
### Mode – Definition
A Mode is defined as the value that appears most frequently in a data set. It is the value that appears the most number of times. A set of data or values may have one mode, more than one mode, or no mode at all. The central tendency includes the mean, or the average of a set, and the median, the middle value in a set.
Consider an example of mode on statistics is,
Example: Find the mode of a given set of data 4, 3, 3, 6, 7.
Solution: Given the data set is 4, 3, 3, 6, 7.
The mode of the data is 3 because it has a higher frequency it means it will be repeated more time. In this example, the mode value will be repeated two times.
The mode’s main advantage is, it can be applied to any type of data set, and the remaining two central tendencies mean and median can not be applied to nominal data and it is also not affected by outliers. The disadvantage of mode is, it cannot be used for more detailed analysis.
### Types of Mode
Based on modes in a data set, Modes are of three types namely:
1. Unimodal
2. Bimodal
3. Trimodal
4. Multimodal
Unimodal: When the given data gas one mode is called Unimodal.
Bimodal: When the given data set has two modes, it is called Bimodal. Consider the example of bimodal is as shown below,
Example: Find the mode of set X = { 1, 2, 3, 3, 4, 4, 4, 5, 6, 6, 6}
Solution: Given the data set is 1, 2, 3, 3, 4, 4, 4, 5, 6, 6, 6
Mode = {4, 6}
The mode of X is 4 and 5 because both 4 and 5 are repeated in the given set those are two modes called bimodal.
Trimodal: When the given data set has three modes, it is called Trimodal. Consider the example of trimodal is as shown below,
Example: Find the mode of the set A = {1, 2, 2, 2, 3, 3, 4, 4, 5, 5, 5, 6, 6, 6}
Solution: Given the data set is 1, 2, 2, 2, 3, 3, 4, 4, 5, 5, 5, 6, 6, 6
Therefore, the Mode is = {2, 5, 6}
The mode of A is 2, 5, 6 because these values repeated, these three modes called Trimodal.
Multimodal: When the given data set has four or more modes is called Multimodal. Consider an example of Multimodal is as shown below,
Example: Find the mean of data set is 2, 3, 4, 5, 5, 5, 7, 6, 9, 9, 9, 12, 12, 12, 10, 1, 1, 1
Solution: Given the data set is 2, 3, 4, 5, 5, 5, 7, 6, 9, 9, 9, 12, 12, 12, 10, 1, 1, 1
Therefore, the Mode is = {5, 9, 12, 1}
The mode of given data is 5, 9, 12, 1 because these values are repeated three times in the data. So, it has more than three modes (or) four modes it is called Multimodal.
### Data Series of Mode
Mode is the value that occurs maximum time in the data set or data series. The series are 3 types:
1. Individual Series – The mode of an individual series is defined as simply finding the value or data that occurs a maximum number of times.
2. Discrete Series – In this mode, the value has higher frequencies.
3. Continuous Series or Frequency distribution Series – In this mode, first we need to find the modal class. Modal class is one of the highest frequency classes.
### Mode Formula for grouped Frequency Distribution
In a grouped frequency distribution, the mode calculation is not possible for the frequency. To determine the mode of data in such cases we can calculate the modal class. Mode lies inside the modal class. The grouped frequency distribution mode formula is given as,
The mode of grouped frequency distribution formula is shown as above. In the formula,
where,
f0 = frequency of the category preceding the modal class
f1Â = frequency of the modal class
f2 = frequency of the category succeeding the modal class
l = lower limit of the modal class
h = size of the class interval
### Ungrouped Frequency Distribution Mode
In Ungrouped data, the observations that occur the most will be the mode of the observation. Observations could also be bimodal or multimodal. With frequency distribution, the observations with the highest frequency will be the modal observation.
### Example Problems on Mode
Example 1:
Find the mode of 3, 5, 7, 9, 2, 1?
Solution:
Given the values are 3, 5, 7, 9, 2, 1
Now we are finding the mode of given data.
Mode means high-frequency value. But in this example, no value in the data set is repeated more than one time.
Hence, the set of a given data has no mode value.
Example 2:
Find the mode of the data set {2, 3, 3, 3, 4, 5, 5, 5, 7, 8, 8, 8}
Solution:Â
Given the data set is 2, 3, 3, 3, 4, 5, 5, 5, 7, 8, 8, 8
Now we find the mode of the given data set.
Mode = {3, 5, 8}
In this, the mode sets are 3, 5, 8, because the values are 3, 5, 8 are repeated. So it has three mode sets. Three mode sets are called Trimodal.
Hence, the mode of given data is 3, 5, 8.
Example 3:
Marks obtained by 30 students of a class are tabulated below. The highest mark is 25. Find the mode?
          Marks Obtained       Number of Students
             0 – 5                1
             5 – 10                4
            10 – 15                7
            15 – 20                14
            20 – 25                 4
Solution:
Given the data in the form of tabular format
The total number of students is 30.
The maximum class frequency is 14 and the class interval corresponding frequency is 15 – 20. So, the modal class is 15 – 20.
The maximum class frequency is 12 and the class interval corresponding to this frequency is 20 – 30. Thus, the modal class is 20 – 30.
The modal class of lower limit (l) = 15
Size of the class interval (h) = 5
Frequency of the modal class (f1) = 14
Frequency of the category preceding the modal class (f0) = 1
Frequency of the category succeeding the modal class (f2)= 7
We know the formula of grouped frequency Mode.
Substituting the given values within the formula we get;
Mean = 15 + ((14 – 1) / (2 x 14 – 1 – 7)) = 15 + ((13) / (6)) = 15 + 2. 166 = 17. 166
Therefore, the mode of a given data set is 17. 166.
Example 4: Find the mode of 5, 5, 5, 6, 7, 15, 15, 15, 28, 49 data set.
Solution: Given the data set is 5, 5, 5, 6, 7, 15, 15, 15, 28, 49
Now, we find the mode of the given data set.
As we know, the data set or values have more than one mode it is the mode. If more than one value occurs which is equal to frequency and number of time compares with other values in the data set.
So, Mode = {5, 15}
Therefore, the mode of the given data set is 5, 15.
### Frequently Asked Questions on Mode
1. Define the difference between Mean, Mode, and Median?
The difference between the mean, mode, and median is, the mean of a data set is adding all numbers in the data set and then dividing by the total number of values in the set. The median is the middle value when a data set is ordered from least to greatest, and the mode is the number that occurs most often in a data set.
2. How do you find the mode or modal?
To find the modal, or mode, the best way is to put the numbers in order. After that count how many of each number. If the number appears most often that is the mode.
3. List the types of Modes?
Modes are of four types namely:
1. Unimodal
2. Bimodal
3. Trimodal
4. Multimodal
4. What happens when you have two mode sets?
If we have two numbers are appear most often, then the data or value has two modes. This is called Bimodal. If more than two modes then the data will be called multimodal.
5. What are the properties of mode in statistics?
In statistics, the mode is the value that repeatedly occurs in a given set of data. It can also say that the value or number in a data set, which has a high frequency or frequently occurs is called mode or modal value. Mode is one of the three measures of central tendency.
6. What are the merits of Mode?
The mode has many merits, some of them are listed below:
1. Mode is easy to calculate and simple to understand.
2. It is not affected by extremely larger values or smaller values.
3. It can be computed in an open-end frequency table.
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# 3. Some Properties of Integrals
Suppose an object moves so that its speed, or more properly velocity, is given by $\ds v(t)=-t^2+5t$, as shown in figure 7.3.1. Let's examine the motion of this object carefully. We know that the velocity is the derivative of position, so position is given by $\ds s(t)=-t^3/3+5t^2/2+C$. Let's suppose that at time $t=0$ the object is at position 0, so $\ds s(t)=-t^3/3+5t^2/2$; this function is also pictured in figure 7.3.1.
Figure 7.3.1. The velocity of an object and its position. You can drag the blue dot and the red dot will follow.
Between $t=0$ and $t=5$ the velocity is positive, so the object moves away from the starting point, until it is a bit past position 20. Then the velocity becomes negative and the object moves back toward its starting point. The position of the object at $t=5$ is exactly $s(5)=125/6$, and at $t=6$ it is $s(6)=18$. The total distance traveled by the object is therefore $125/6 + (125/6 - 18) = 71/3\approx 23.7$.
As we have seen, we can also compute distance traveled with an integral; let's try it. $$\int_0^6 v(t)\,dt = \int_0^6 -t^2+5t\,dt = \left.{-t^3\over 3}+{5\over2}t^2\right|_0^6 = 18.$$ What went wrong? Well, nothing really, except that it's not really true after all that "we can also compute distance traveled with an integral''. Instead, as you might guess from this example, the integral actually computes the {\it net\/} distance traveled, that is, the difference between the starting and ending point.
As we have already seen, $$\int_0^6 v(t)\,dt=\int_0^5 v(t)\,dt+\int_5^6 v(t)\,dt.$$ Computing the two integrals on the right (do it!) gives $125/6$ and $-17/6$, and the sum of these is indeed 18. But what does that negative sign mean? It means precisely what you might think: it means that the object moves backwards. To get the total distance traveled we can add $125/6+17/6=71/3$, the same answer we got before.
Remember that we can also interpret an integral as measuring an area, but now we see that this too is a little more complicated that we have suspected. The area under the curve $v(t)$ from 0 to 5 is given by $$\int_0^5 v(t)\,dt={125\over6},$$ and the "area'' from 5 to 6 is $$\int_5^6 v(t)\,dt=-{17\over 6}.$$ In other words, the area between the $x$-axis and the curve, but under the $x$-axis, "counts as negative area''. So the integral $$\int_0^6 v(t)\,dt=18$$ measures "net area'', the area above the axis minus the (positive) area below the axis.
If we recall that the integral is the limit of a certain kind of sum, this behavior is not surprising. Recall the sort of sum involved: $$\sum_{i=0}^{n-1} v(t_i)\Delta t.$$ In each term $v(t)\Delta t$ the $\Delta t$ is positive, but if $\ds v(t_i)$ is negative then the term is negative. If over an entire interval, like 5 to 6, the function is always negative, then the entire sum is negative. In terms of area, $v(t)\Delta t$ is then a negative height times a positive width, giving a negative rectangle "area''.
So now we see that when evaluating $$\ds\int_5^6 v(t)\,dt=-{17\over 6}$$ by finding an antiderivative, substituting, and subtracting, we get a surprising answer, but one that turns out to make sense.
Let's now try something a bit different: $$\int_6^5 v(t)\,dt=\left.{-t^3\over 3}+{5\over2}t^2\right|_6^5 = {-5^3\over 3}+{5\over2}5^2-{-6^3\over 3}-{5\over2}6^2 ={17\over 6}.$$ Here we simply interchanged the limits 5 and 6, so of course when we substitute and subtract we're subtracting in the opposite order and we end up multiplying the answer by $-1$. This too makes sense in terms of the underlying sum, though it takes a bit more thought. Recall that in the sum $$\sum_{i=0}^{n-1} v(t_i)\Delta t,$$ the $\Delta t$ is the "length'' of each little subinterval, but more precisely we could say that $\ds \Delta t = t_{i+1}-t_i$, the difference between two endpoints of a subinterval. We have until now assumed that we were working left to right, but could as well number the subintervals from right to left, so that $\ds t_0=b$ and $\ds t_n=a$. Then $\ds \Delta t=t_{i+1}-t_i$ is negative and in $$\int_6^5 v(t)\,dt=\sum_{i=0}^{n-1} v(t_i)\Delta t,$$ the values $\ds v(t_i)$ are negative but also $\Delta t$ is negative, so all terms are positive again. On the other hand, in $$\int_5^0 v(t)\,dt=\sum_{i=0}^{n-1} v(t_i)\Delta t,$$ the values $\ds v(t_i)$ are positive but $\Delta t$ is negative,and we get a negative result: $$\int_5^0 v(t)\,dt=\left.{-t^3\over 3}+{5\over2}t^2\right|_5^0 = 0-{-5^3\over 3}-{5\over2}5^2 = -{125\over6}.$$
Finally we note one simple property of integrals: $$\int_a^b f(x)+g(x)\,dx=\int_a^b f(x)\,dx+\int_a^b g(x)\,dx.$$ This is easy to understand once you recall that $(F(x)+G(x))'=F'(x)+G'(x)$. Hence, if $F'(x)=f(x)$ and $G'(x)=g(x)$, then \eqalign{ \int_a^b f(x)+g(x)\,dx&=\left.(F(x)+G(x))\right|_a^b\cr &=F(b)+G(b)-F(a)-G(a)\cr &=F(b)-F(a)+G(b)-G(a)\cr &=\left.F(x)\right|_a^b+\left.G(x)\right|_a^b\cr &=\int_a^b f(x)\,dx+\int_a^b g(x)\,dx.\cr }
In summary, we will frequently use these properties of integrals: $$\displaylines{ \int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx\cr \int_a^b f(x)+g(x)\,dx=\int_a^b f(x)\,dx+\int_a^b g(x)\,dx\cr \int_a^b f(x)\,dx=-\int_b^a f(x)\,dx\cr }$$ and if $a < b$ and $f(x)\le 0$ on $[a,b]$ then $$\int_a^b f(x)\,dx\le 0$$ and in fact $$\int_a^b f(x)\,dx=-\int_a^b |f(x)|\,dx.$$
## Exercises 7.3
Ex 7.3.1An object moves so that its velocity at time $t$ is $v(t)=-9.8t+20$ m/s. Describe the motion of the object between $t=0$ and $t=5$, find the total distance traveled by the object during that time, and find the net distance traveled. (answer)
Ex 7.3.2An object moves so that its velocity at time $t$ is $v(t)=\sin t$. Set up and evaluate a single definite integral to compute the net distance traveled between $t=0$ and $t=2\pi$. (answer)
Ex 7.3.3An object moves so that its velocity at time $t$ is $v(t)=1+2\sin t$ m/s. Find the net distance traveled by the object between $t=0$ and $t=2\pi$, and find the total distance traveled during the same period. (answer)
Ex 7.3.4Consider the function $f(x)=(x+2)(x+1)(x-1)(x-2)$ on $[-2,2]$. Find the total area between the curve and the $x$-axis (measuring all area as positive). (answer)
Ex 7.3.5Consider the function $\ds f(x)=x^2-3x+2$ on $[0,4]$. Find the total area between the curve and the $x$-axis (measuring all area as positive). (answer)
Ex 7.3.6Evaluate the three integrals: $$A=\int_0^3 (-x^2+9)\,dx\qquad B=\int_0^{4} (-x^2+9)\,dx\qquad C=\int_{4}^3 (-x^2+9)\,dx,$$ and verify that $A=B+C$. (answer) |
1. Chapter 6 Class 12 Application of Derivatives
2. Serial order wise
Transcript
Ex 6.2,3 Find the intervals in which the function f given by f (𝑥) = Sin 𝑥 is (a) strictly increasing in 0 , 𝜋2 f (𝑥) = sin 𝑥 f’(𝑥) = cos 𝑥 Since cos 𝑥 > 0 for 𝑥 ∈ 0 , 𝜋2 So, f’ (𝑥) > 0 for 𝑥 ∈ 0 , 𝜋2 Thus, f is strictly increasing in 0 , 𝜋2 Ex 6.2,3 Find the intervals in which the function f given by f (𝑥) = Sin x is (b) strictly decreasing 𝜋2,𝜋 f (𝑥) = sin 𝑥 f’(𝑥) = cos 𝑥 Since cos 𝑥 < 0 for 𝑥 ∈ 𝜋2 , 𝜋 So, f’ (𝑥) < 0 for 𝑥 ∈ 𝜋2 , π Thus, f is strictly decreasing in 𝜋2 π Ex 6.2,3 Find the intervals in which the function f given by f (𝑥) = Sin x is (c)neither increasing nor decreasing in (0 ,π ) Step 1: f (𝑥) = sin 𝑥 f’ (𝑥) = cos 𝑥 Step 2: Putting f’(𝑥) = 0 cos x= 0 𝑥 = 𝜋2 Step 3: Since 𝑥 ∈ (0 , π) we start number line from 0 to π The point 𝑥= 𝜋2 divides the interval (0 , 𝜋) into two disjoint interval i.e. 0 , 𝜋2 , 𝜋2 𝜋 From 1st part , f’(𝑥) is strictly increasing in 0 𝜋2 & from 2nd part, f(𝑥) is strictly decreasing in 𝜋2 𝜋 Hence f(𝑥) is neither increasing nor decreasing in (0 , π) |
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If $x - \dfrac{1}{x} = - \sqrt 3$, then find ${x^3} - \dfrac{1}{{{x^3}}}$A) $6\sqrt 3$B) $2\sqrt 3$C) $3\sqrt 3$D) $- 6\sqrt 3$
Last updated date: 22nd Jun 2024
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Hint:
Here we will take the cube on both sides of the given equation and then we will apply the formula of the cube of difference. We will simplify the equation further and from there, we will get the value of the required expression.
Formula Used:
We will use the formula of cube of difference which is given by ${\left( {a - b} \right)^3} = {a^3} - {b^3} - 3ab\left( {a - b} \right)$.
Complete step by step solution:
It is given that $x - \dfrac{1}{x} = - \sqrt 3$.
Now, taking cube on both sides, we get
$\Rightarrow {\left( {x - \dfrac{1}{x}} \right)^3} = {\left( { - \sqrt 3 } \right)^3}$
Applying the exponent on the term, we get
$\Rightarrow {\left( {x - \dfrac{1}{x}} \right)^3} = - 3\sqrt 3$
Now, using the formula ${\left( {a - b} \right)^3} = {a^3} - {b^3} - 3ab\left( {a - b} \right)$ in the above equation, we get
$\Rightarrow {x^3} - \dfrac{1}{{{x^3}}} - 3x \cdot \dfrac{1}{x} \cdot \left( {x - \dfrac{1}{x}} \right) = - 3\sqrt 3$
Simplifying the above equation, we get
$\Rightarrow {x^3} - \dfrac{1}{{{x^3}}} - 3 \cdot \left( {x - \dfrac{1}{x}} \right) = - 3\sqrt 3$
We know the value $x - \dfrac{1}{x} = - \sqrt 3$. Now, we will substitute this value in the above equation. Therefore, we get
$\Rightarrow {x^3} - \dfrac{1}{{{x^3}}} - 3 \cdot \left( { - \sqrt 3 } \right) = - 3\sqrt 3$
On multiplying the terms, we get
$\Rightarrow {x^3} - \dfrac{1}{{{x^3}}} + 3\sqrt 3 = - 3\sqrt 3$
On subtracting $3\sqrt 3$ on both sides, we get
$\Rightarrow {x^3} - \dfrac{1}{{{x^3}}} + 3\sqrt 3 - 3\sqrt 3 = - 3\sqrt 3 - 3\sqrt 3$
$\Rightarrow {x^3} - \dfrac{1}{{{x^3}}} = - 6\sqrt 3$
Hence, the correct option is option D.
Note:
Here, we have used algebraic identity to solve the given algebraic equation. We generally use the algebraic identities for factorization of polynomials. But we should remember that algebraic expressions and algebraic identities are different. An algebraic expression is defined as an expression which consists of constants and variables. In algebraic expressions, a variable can take any value. Thus, the expression value can change if the variable values are changed. But algebraic identity is defined as an equality which is true for all the values of the variables.
Here, we might make a mistake by writing ${\left( { - \sqrt 3 } \right)^3}$ as $3\sqrt 3$ and forgot the minus sign. When a negative number is multiplied three times then the resulting number will be a negative and not positive. |
# How do you use the integral test to determine whether int dx/lnx converges or diverges from [2,oo)?
Jun 25, 2018
The integral
${\int}_{2}^{\infty} \frac{\mathrm{dx}}{\ln} x$
is divergent.
#### Explanation:
Note that in the interval $x \in \left[2 , \infty\right)$ the function:
$f \left(x\right) = \frac{1}{\ln} x$
is:
1) Infinitesimal as ${\lim}_{x \to \infty} f \left(x\right) = 0$
2) Positive as $f \left(x\right) > 0$ for $x > 1$
3) Decreasing. In fact $f ' \left(x\right) = - \frac{1}{x {\ln}^{2} x} < 0$
4) $f \left(n\right) = \frac{1}{\ln} n$
So, based on the integral test, the convergence of the integral:
${\int}_{2}^{\infty} \frac{\mathrm{dx}}{\ln} x$
is equivalent to the convergence of the series:
${\sum}_{n = 2}^{\infty} \frac{1}{\ln} n$
Now we can easily demonstrate that:
$\ln n < n$
so that:
$\frac{1}{\ln} n > \frac{1}{n}$
and as we know that the harmonic series:
${\sum}_{n = 1}^{\infty} \frac{1}{n}$
is divergent, we can conclude that:
${\sum}_{n = 2}^{\infty} \frac{1}{\ln} n$
is divergent by direct comparison, and hence also:
${\int}_{2}^{\infty} \frac{\mathrm{dx}}{\ln} x$
is divergent. |
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# GRE Math Shortcut: Statistics
Consider the following question:
Set T consists of all multiples of 5 from 30 to 225 inclusive
Column A Column B
Mean of Set T Median of Set T
A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
As you can imagine, we really don’t want to calculate the mean and median of set T. Fortunately, we can apply a nice rule that says:
If the numbers in a set are equally spaced, then the mean and median of that set are equal.
What does it mean to have “equally spaced” numbers?
“Equally spaced” means that, if the numbers in a set are arranged in ascending order, then the difference between any two adjacent numbers will always be the same.
So, in the set {30, 35, 40, . . . , 215, 220, 225}, the difference between any two adjacent numbers will always equal 5. As such, the mean of this set will equal the median of this set (which means the answer to the above question is C). Notice that we can answer this question without making any calculations whatsoever. We can answer it by recognizing that the numbers in the set are equally spaced.
Similarly, the numbers in the set {5, 8, 11, 14, 17, 20, 23} are equally spaced. As such, the mean of this set will equal the median of this set.
Set X: {-27, -20, -13, -6, 1, 8, 15}
Set Y: {-23, -19, -15, -11, -7, -3, 1, 5, 9, 13}
Column A Column B
Mean of set X Mean of set Y
A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Column A: Since the numbers in set X are equally spaced, the mean will equal the median. We can quickly find the median by locating the middlemost element: {-27, -20, -13, -6, 1, 8, 15}. Set X has 7 elements and the middlemost element is -6. So, the median of set X is -6, which means the mean of set X is -6 as well.
Column B: The numbers in set Y are also equally spaced, so the mean will equal the median. Set Y has 10 elements, so there are two middlemost elements: {-23, -19, -15, -11, -7, -3, 1, 5, 9, 13}. As such, the median of set Y will equal the average (mean) of -7 and -3, which is -5. So, if the median of set Y is -5, then the mean of set Y is -5 as well
So, using our rule, we found that Column A = -6, and Column B= -5. As such, the answer to the question is B.
By the way, students who use Magoosh GRE improve their scores by an average of 8 points on the new scale (150 points on the old scale.) Click here to learn more.
### 12 Responses to GRE Math Shortcut: Statistics
1. brit123 August 18, 2014 at 10:04 am #
Hello,
For the first example, I selected choice D since I was unsure if any of the numbers repeated. For questions like this, is it safe to assume none of the numbers occur more than once?
• Chris Lele August 18, 2014 at 1:17 pm #
Hmm…that’s a good question. I’m not a 100% sure on that, and it does seem a bit misleading. I’d be inclined to edit this question so it is ambiguous that each of the multiples appears exactly once. Sorry for any confusion 🙂
• brit123 August 18, 2014 at 1:50 pm #
Thank you so much! 😀
2. Hema June 18, 2014 at 6:30 am #
Great tip 🙂 i used to skip such questions(example 1), due to lack of time. with this tip, it sounds so simple..Thank !!
3. arya January 19, 2014 at 10:31 am #
absolutely, love the tactics!!! can u please post, more of such nice tactics..it will be very much helpful.
• Chris Lele January 23, 2014 at 10:56 am #
Great! Happy you found that helpful :). We’ll try to keep posting such helpful tips.
4. akanksha October 9, 2013 at 12:14 am #
its great shortcut.. reduces whole calculation.. thanks Brent 🙂
5. Inny July 31, 2013 at 6:42 am #
Wow, excellent shortcut! Super helpful 🙂
6. naga June 21, 2012 at 9:41 am #
nice tactic..wish to learn many tactics like this…
7. abhay May 26, 2012 at 12:05 pm #
thanks…nice shortcut particularly the second application.
• Margarette May 29, 2012 at 10:56 am #
You’re welcome, glad we could help! 🙂
Magoosh blog comment policy: To create the best experience for our readers, we will only approve comments that are relevant to the article, general enough to be helpful to other students, concise, and well-written! 😄 Due to the high volume of comments across all of our blogs, we cannot promise that all comments will receive responses from our instructors.
We highly encourage students to help each other out and respond to other students' comments if you can!
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Math
# Difference between Rhombus and parallelogram with examples
Just as the pyramid and the prism have differences, a rhombus, and a parallelogram also. They are very similar geometric figures with some differences between them. The basic difference between rhombus and parallelogram is that Rhombus has all 4 sides are equal while Parallelogram has opposite sides are equal in length.
## Parallelogram
The rhombus is a quadrilateral with 4 sides and vertices. It is a parallelogram with four sides of equal length and diagonals that intersect perpendicularly.
A parallelogram is considered a rhombus if its sides are equal, the diagonals are perpendicular, and also are bisectors of the angles of the parallelogram.
In a rhombus, the line joining opposite vertices is an axis of symmetry.
The point of intersection of the diagonals is the center of the rhombus.
## Parallelogram
It is a parallelogram that is not a rhombus or a rectangle. This parallelogram has two angles and with respect to its four sides, two and two are equal to each other.
For the followers of the Jorge Rey Pastor school, a rhomboid is that quadrilateral that has two pairs of equal consecutive sides.
Rhomboids have the following characteristics:
• Two pairs of opposite angles are equal and parallel to each other.
• Continuous angles are supplementary.
• Its diagonals are not perpendicular.
• It has two obtuse angles and 2 acute angles.
• It has 4 vertices. The sum of all its interior angles in 360º.
## Difference between Rhombus and parallelogram
• A rhombus is a parallelogram with four sides of equal length whose diagonals intersect perpendicularly. The diagonals are bisectors of the angles of the rhombus.
• A rhomboid is a parallelogram where two and two sides are equal, two angles are equal (acute), and two are obtuse and equal. The diagonals are not perpendicular to each other.
• The similarities between both figures are that they are both parallelograms whose sum of their internal angles gives a total of 360º. |
Edit Article
# wikiHow to Calculate Pearson Correlation Coefficient
The Pearson Correlation Coefficient (which used to be called the Pearson Product-Moment Correlation Coefficient) was established by Karl Pearson in the early 1900's. It tells us how strongly things are related to each other, and what direction the relationship is in! The formula is: r = Σ(X-Mx)(Y-My) / (N-1)SxSy
Want to simplify that? Let's say our hypothesis is that as consumption of chocolate increases, so does a person's self-reported happiness on a scale of 1 (unhappy) to 7 (happy). Everyone knows that eating chocolate makes you happier, right? Before we get started, identify your two variables (X and Y). Let's say we had information about how many pieces of chocolate a person eats per day (X) and what their level of happiness was (Y).
## Steps
1. 1
Find the average of chocolate consumption (Mx) by adding up all of the people's scores and dividing by the number of people. Then we would subtract each individual score (X) from the mean. This tells us how far away this person is from the average. You should have a new score for each person.
• In the formula that is: X-Mx
2. 2
Do the same for happiness. We find the average level of happiness (My); then subtract each individual score (Y) from the mean. Again, you'll have a score for each person.
• In the formula that is: Y-My
3. 3
Multiply each person's deviation from the mean for their X score by their deviation from the mean for their Y score. Once again, you'll have a new score for each person.
• In the formula that is: (X-Mx)(Y-My)
4. 4
Add up all of the people's multiplied scores. That's what the funny-shaped "E" means in the formula. "Σ" is the Greek symbol for Sigma, and is used in stats to signify that you should add everything up.
• In the formula that is: Σ(X-Mx)(Y-My)
5. 5
Take the number of people in the sample (N) and subtract by 1.
• In the formula that is: N-1
6. 6
Multiply the standard deviation of chocolate consumption (Sx) by the standard deviation of happiness (Sy).
7. 7
Multiply that number by the number of people in your sample minus one.
• In the formula that is: (N-1)SxSy
8. 8
Take the number you calculated first [Σ(X-Mx)(Y-My)] and divide it by the number you just got [(N-1)SxSy].
9. 9
Interpret your result. r is the symbol used to denote the Pearson Correlation Coefficient).
• A score of .1-.3 indicates a small relationship
• 31-.5 is a moderate relationship
• .51-.7 is a large relationship
• anything above .7 is a very strong (sometimes called "isomorphic") relationship.
• A positive number means they move the same direction (as chocolate consumption goes up, so does a person’s happiness, and if chocolate consumption goes down, so does happiness). It doesn’t mean they both go UP—but it means they move together.
• A negative number means the variables move in opposite directions. That would mean that people were less happy as they ate chocolate, or people ate less chocolate when they were happy.
## Community Q&A
200 characters left
## Tips
• The Pearson Correlation Coefficient is no longer called the Pearson Product-Moment Correlation Coefficient as commonly as it once was. The use of the word "moment" was borrowed from physics, and referred to the distance of a point away from the center point (does that sound familiar? You just calculated it!). The "product moment" is multiplying together the distances away from the center point. So the name makes sense, but thank goodness they simplified it!
## Warnings
• There are two assumptions when calculating a Pearson Correlation Coefficient. These are (a) that there are at least 2 variables and data are at minimum interval level, and (b) the data are normally distributed.
## Sources and Citations
• Andy Field's "Discovering Statistics Using SPSS"
## Article Info
Categories: Mathematics | Probability and Statistics
Thanks to all authors for creating a page that has been read 4,518 times. |
# Trigonometry: The Unit Circle
This post is part of a series of posts on Trigonometry. To see all the posts, click on the tag #TrigonometryTutorials below. This is the post you should read before you read this.
So far we have only defined the trig functions on angles that are strictly less than 90 degrees. If you 'played' with your calculator you will notice that the calculator and calculate sines and cosines and tangents of angles larger than 90 degrees. Why does the calculator output "Math Error" when you try calculating the Tangent of 90? Why is Sine 90 degrees 1?
To answer this we need to Unit Circle. In a cartesian Co-ordinate plane, imagine a circle centred at the origin with radius 1. A line segment is drawn from the Origin to the boundaries of the circle, making an angle $\theta$ with the $x$-axis. Then, you drop a perpendicular down onto the $x$ axis. Given that $theta$ is between 0 degrees and 90 degrees, you basically have a normal right triangle! Let the point where the line segment intersects the circle be $(x,y)$. The Radius of the Circle is the hypotenuse, the opposite of $\theta$ is $y$ and the adjacent is $x$. Since the Hypotenuse is one, Using the Unit circle, we can redefine the trig ratios as:
$\sin \theta = y$ $\cos\theta = x$ $\tan\theta = \frac yx$
Now we have defined the Trig ratios for angles larger than 90, it is the $x$ and $y$ co-ordinate readings of the circle that creates the angle.
The is why the Trig functions are also sometimes called the Circular functions, because they can be defined by the co-ordinate readings on the circle
The Next post in this series is here
Note by Yan Yau Cheng
7 years, 5 months ago
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Lesson Video: Definition of the Derivative Mathematics
In this video, we will learn how to calculate the derivative of a function using the formal definition of the derivative as a limit.
16:24
Video Transcript
Definition of The Derivative
In this video, we will learn how to calculate the derivative of a function using the formal definition of its derivative as a limit. Weβll be covering the definition of a derivative in greater detail and then looking at some examples. Letβs start by exploring the definition of a derivative at a particular point.
We know that the derivative of some function π of π₯ tells us about the rate of change of that function at a particular point. Letβs choose that point to be π₯ nought. And this could be any point. And we know that the rate of change of the function at this point is equal to the slope of the function at this point. Therefore, we can find an estimate for the derivative by approximating the tangent to the curve at this point and then finding the slope of this tangent. Letβs now consider a new graph of a different π of π₯.
If weβre trying to find the derivative of π of π₯ at π₯ nought, then we could estimate it by approximating a tangent at π₯ nought. One way in which we can draw an approximate tangent is to pick another π₯-value which is close to π₯ nought, say π₯ nought plus β where β is a constant. Then, we can draw our approximate tangent as the line passing through π₯ nought, π of π₯ nought and π₯ nought plus β, π of π₯ nought plus β. This is what our approximate tangent would look like. However, this is not a very good approximation.
One way we could improve our approximation for the tangent at this point is to decrease the value of β. Letβs say that we decreased β such that π₯ nought plus β lies at this point on our π₯-axis, then the point π₯ nought plus β, π of π₯ nought plus β, would have shifted to here. Then, our new approximation for the tangent will pass through this point and the point π₯ nought, π of π₯ nought. Therefore, it will look something like this, which is a better approximation for the tangent of π of π₯ at π₯ nought.
We could improve the approximation even more if we decrease the value of β again such that π₯ nought plus β lied at this point on our π₯-axis. This would again shift the point π₯ nought plus β, π of π₯ nought plus β. And we could then draw a new approximation for the tangent at π₯ nought, which we can see is even closer to the actual tangent at π₯ nought. Now, what this method for approximating the tangent at π₯ nought is telling us is that the smaller the value of β, the closer our approximate tangent gets to the real tangent at π₯ nought.
And so, we can say that the limit, as β tends to zero, of our approximate tangent is in fact equivalent to the real tangent of π of π₯ at π₯ is equal to π₯ nought. Now, if we recall what a derivative is, weβll remember that itβs the value of the slope of the function at that particular point. Since the slope of the tangent at any point on our curve represents the slope of the function at that point, we can use the slope of our approximated tangents in order to help us define a derivative.
Letβs recall what the points were which we were using in order to approximate our tangents. We used π₯ nought, π of π₯ nought, and π₯ nought plus β, π of π₯ nought plus β. We know that the slope of any straight line is the change in π¦ over the change in π₯. Therefore, the slope of any of our tangents can be given by π of π₯ nought plus β minus π of π₯ nought over π₯ nought plus β minus π₯ nought. In the denominator, we have π₯ nought plus β minus π₯ nought. So, the two π₯ noughts cancel out, leaving us with the slope of our tangent being equal to π of π₯ nought plus β minus π of π₯ nought all over β.
We will combine the slope of the tangent along with the fact that as π₯ nought gets closer to zero, our approximation of that tangent gets closer to the real tangent at π₯ nought in order to define a derivative. And so, we arrive at the definition of the derivative. We say that the derivative of a function π of π₯, at a point π₯ nought, is defined as the limit as β goes to nought of π of π₯ nought plus β minus π of π₯ nought all over β if the limit exists.
If we allow π₯ one to be equal to π₯ nought plus β, then we have that β is equal to π₯ one minus π₯ nought. And as β goes to zero, π₯ one minus π₯ nought goes to zero, which means that π₯ one tends to π₯ nought. So, we can rewrite this limit in terms of π₯ one and π₯ nought instead of π₯ nought and β. This equivalent definition is the limit as π₯ one tends to π₯ nought of π of π₯ one minus π of π₯ nought over π₯ one minus π₯ nought. And this is again only if the limit exists.
There are a few ways in which we can denote a derivative. The first is using prime notation. This is written as π prime of π₯. We can say that itβs the derivative of π with respect to π₯. Another way we can denote a derivative is using Leibnizβs notation. Which tells us that if we write π of π₯ as π¦, then the derivative is dπ¦ by dπ₯, which also means the derivative of π¦ with respect to π₯. Both of the definitions and notations here will be very useful for us while studying calculus. And so, itβs important for us to be comfortable using both of the definitions and both sets of notation. Letβs now look at some examples of how we can use the definition to find derivatives.
Find the derivative of π of π₯ which is equal to π₯ squared at the point π₯ is equal to two from first principles.
What this question means when it says from first principles is to use the definition of a derivative. Which tells us that the derivative of π with respect to π₯ at π₯ nought is equal to the limit as π₯ one tends to π₯ nought of π of π₯ one minus π of π₯ nought over π₯ one minus π₯ nought. From the question, we can see that π of π₯ is equal to π₯ squared. And weβve been asked to find the derivative at the point π₯ is equal to two. Therefore, π₯ nought is equal to two.
From this, we can say that π prime of two is equal to the limit as π₯ one tends to two of π₯ one squared minus two squared over π₯ one minus two. We can rewrite two squared as four. Now, we see that in the numerator of our fraction we have π₯ one squared minus four, which is the difference of two squares. And so, it can be factorized to π₯ one minus two multiplied by π₯ one plus two.
Here, we notice we have a factor of π₯ one minus two in both the numerator and denominator. And since π₯ one minus two is not equal to zero, we can cancel these here. And so, we find that π prime of two is equal to the limit as π₯ one tends to two of π₯ one plus two. Using direct substitution, we obtain that this is equal to two plus two. And so, we reach our solution, which is that the derivative of π₯ squared at the point π₯ is equal to two is four.
Now, if we remember back at earlier in this video, when we were defining the derivative, we used the slopes of approximations of tangents to get closer and closer to the derivative of the function. In the next example, weβll see how we can use the definition of a derivative to accurately find the slope of a tangent to a function at a point.
Let π of π₯ be equal to eight π₯ squared minus six π₯ plus nine. Use the definition of the derivative to determine π prime of π₯. What is the slope of the tangent to its graph at one, two?
The first thing we need to do is to use the definition of the derivative to determine π prime of π₯. We can recall the definition for the derivative, which tells us that π prime of π₯ is equal to the limit as β goes to zero of π of π₯ plus β minus π of π₯ all over β. Now, in our case, π of π₯ is equal to eight π₯ squared minus six π₯ plus nine, so we can substitute this into our limit. We obtain that π prime of π₯ is equal to the limit as β tends to zero of eight multiplied by π₯ plus β squared minus six multiplied by π₯ plus β plus nine minus eight π₯ squared minus six π₯ plus nine all over β.
Our next step is to expand the brackets. We obtain this limit here. We notice there are a few things we can cancel out in the numerator of our fraction. We have an eight π₯ squared and a negative eight π₯ squared. We have a negative six π₯ and a six π₯. And we also have a nine and a negative nine. Removing these six terms, we are left with the limit as β tends to zero of 16βπ₯ plus eight β squared minus six β all over β.
And we notice we have a factor of β in both the numerator and denominator. Since β is not equal to zero, we can cancel it out. Giving us that π prime of π₯ is equal to the limit as β tends zero of 16π₯ plus eight β minus six. Applying direct substitution to this limit, we reach the solution to the first part of the question, which is that π prime of π₯ is equal to 16π₯ minus six. We can now move on to the second part of the question, which is to find the slope of the tangent to its graph at one, two.
Now, we know that the derivative of π will tell us the slope at any point on π. Therefore, the slope at one, two will be the value of π prime of π₯ when π₯ is equal to one. So, we find π prime of one. This gives us 16 multiplied by one minus six, which simplifies to give us a slope of 10. In the next example, weβll see how we can use the definition of a derivative in order to find the derivative of a reciprocal function.
Using the definition of the derivative, evaluate d by dπ₯ of one over one plus π₯.
We can recall the definition of the derivative. And that is that if we set π¦ to be equal to π of π₯, then dπ¦ by dπ₯ is equal to the limit as β tends to zero of π of π₯ plus β minus π of π₯ all over β. Now, in our case, π¦ is equal to one over one plus π₯. Therefore, we can say that dπ¦ by dπ₯ is equal to the limit as β tends to zero of one over one plus π₯ plus β minus one over one plus π₯ all over β.
We need to express our fraction as a single fraction over a common denominator. And we can do this by finding a common denominator for the two fractions in the numerator. That common denominator is one plus π₯ multiplied by one plus π₯ plus β, leaving us with this limit. And we can combine the two fractions in the numerator, leaving us with the limit as β tends to zero of negative β over β multiplied by one plus π₯ multiplied by one plus π₯ plus β. Since β is not equal to zero, we can cancel β in the numerator and denominator.
Next, we can apply limit rules. We have that the limit of a quotient is equal to the quotient of the limit. Since negative one is a constant, we are left with this. And we can simply apply direct substitution, giving us that dπ¦ by dπ₯ is equal to negative one over one plus π₯ multiplied by one plus π₯. And so, we reach our solution, which is that d by dπ₯ of one over one plus π₯ is equal to negative one over one plus π₯ squared. In our final example, weβll see a different use for the definition of the derivative of a function.
Evaluate the limit as β tends to zero of π of β plus four minus π of β minus two plus π of negative two minus π of four all over β.
Here, we can see that the limit weβve been asked to evaluate looks very similar to the definition of the derivative. The definition of the derivative tells us that π prime of π₯ is equal to the limit as β tends to zero of π of π₯ plus β minus π of π₯ all over β. Letβs try to rearrange the expression within our limit to see if we can try to isolate the definition of the derivative of π at some point.
The first thing we notice is that in the numerator, we have an π of β plus four and an π of four. So, we can group these two terms together. We also have an π of β minus two and an π of negative two. So, we can also group these two terms together. Now, that weβve done this grouping in the numerator, we can split our fraction into two accordingly, giving us the limit as β tends to zero of π of β plus four minus π of four all over β plus the limit as β tends to zero of π of negative two minus π of β minus two all over β.
We can notice that this first limit is looking very close to the definition of π prime of four. If we were to write out π prime of four, we would see that it is equal to the limit as β tends to zero of π of four plus β minus π of four all over β. The only difference between this limit and the limit we found in the question is that β plus four and four plus β are the other way around. However, since the order of addition does not matter, these two things are in fact equal. And therefore, the limit as β tends to zero of π of β plus four minus π of four over β is, in fact, π prime of four.
Now, letβs look at the second limit. We can see in the definition of the derivative that we subtract π of π₯ from π of π₯ plus β. However, in our limit weβre subtracting π of β minus two from π of negative two. In order to get this in the right order, we need to multiply our fraction by negative one. Weβre able to do this if we stick a minus sign in front of our limit. We obtained the negative limit as β tends to zero of π of β minus two minus π of negative two over β.
Now, this is looking very close to the definition of the derivative of π at π₯ is equal to negative two, since π prime of negative two is equal to the limit as β tends to zero of π of negative two plus β minus π of negative two all over β. And again, we can see that this is identical to our limit except for the fact that β and negative two are the other way around. But we know that these two things are equivalent.
And so, we can say that our second limit is equal to π prime of negative two. And therefore, we have reached our solution. And that is that the limit as β tends to zero of π of β plus four minus π of β minus two plus π of negative two minus π of four all over β is equal to π prime of four minus π prime of negative two. We have now covered the definition of a derivative and a variety of examples including it. Letβs recap the key points of the video.
Key Points
The derivative of a function is defined as the limit as β tends to zero of π of π₯ plus β minus π of π₯ all over β. An alternative but equivalent definition of the derivative is the limit as π₯ one tends to π₯ zero of π of π₯ one minus π of π₯ zero over π₯ one minus π₯ zero. There are two common ways to denote derivatives β prime notation, which is π prime of π₯, and Leibnizβs notation, which is dπ¦ by dπ₯. The derivative defines a function which is equal to the slope of the tangent at each point on the curve. |
# HBSE 8th Class Maths Solutions Chapter 6 Square and Square Roots Ex 6.4
Haryana State Board HBSE 8th Class Maths Solutions Chapter 6 Square and Square Roots Ex 6.4 Textbook Exercise Questions and Answers.
## Haryana Board 8th Class Maths Solutions Chapter 6 Square and Square Roots Exercise 6.4
Question 1.
Find the square root of each of the following numbers by Division method :
(i) 2304
(ii) 4489
(iii) 3481
(iv) 529
(v) 3249
(vi) 1369
(vii) 5776
(viii) 7921
(ix) 576
(x) 1024
(xi) 3136
(xii) 900
Solution:
Question 2.
Find the number of digits in the square root of each of the following numbers (without any calculation).
(i) 64
(ii) 144
(iii) 4489
(iv) 27225
(v) 390625.
Solution:
(i) 64, Here n = 2 (even)
Number of digits in the square root of 64
= $$\frac{n}{2}$$ = $$\frac{2}{2}$$ = 1
(ii) Here, n = 3 (odd)
No. of digits in the square root of 144
= $$\frac{n+1}{2}$$ = $$\frac{3+1}{2}$$ = 2
(iii) 4489, Here n = 4 (even)
∴ No. of digits in the square root of 4489
= $$\frac{n}{2}$$ = $$\frac{4}{2}$$ = 2
(iv) 27225, Here n = 5 (odd)
No. of digits in the square root of 27225 = n+1 5+1
= $$\frac{n+1}{2}$$ = $$\frac{5+1}{2}$$ = 3
(v) 390625, Here n = 6 (even)
∴ The number of digits in the square root of
390625 = $$\frac{n}{2}$$ = $$\frac{6}{2}$$ = 3
Question 3.
Find the square root of the following decimal numbers :
(i) 2.56
(ii) 7.29
(iii) 51.84
(iv) 42.25
(v) 31.36
Solution:
Question 4.
Find the least number which must be subtracted from each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.
(i) 402
(ii) 1989
(iii) 3250
(iv) 825
(v) 4000.
Solution:
(i)
We get remainder 2. It shows that 202 is less than 402 by 2.
Thus, the number to be subtracted so as to make it a perfect a square is 2.
∴ Required perfect square
= 402 – 2 = 400
and $$\sqrt {400}$$ = 20
(ii)
We get remainder 53. It shows that 442 is less than 1989 by 53.
Thus, the number to be subtracted so as to make it a perfect square is 3.
∴ Required perfect square
= 1989 – 53 = 1936
and $$\sqrt {1936}$$ = 44.
(iii) We get the remainder 1.
It shows that 572 is less than 3250 by 1.
Thus, the number to be subtracted so as to make 3250 a perfect square is 1.
∴ Required number
= 3250 – 1 = 3249
and $$\sqrt {3249}$$ = 57
(iv) We get the remainder 41.
It shows that 282 is less than 825 by 41.
Thus, the smallest number that should be subtracted to make it a perfect square is 41.
∴ Required number = 825 – 41 = 784
and $$\sqrt {784}$$ = 28
(v) Since remainder 31.
∴ The smallest number that should be subtracted to make it a perfect square is 31.
and Required number
= 4000 – 31
= 3969
and $$\sqrt {3669}$$ = 63.
Question 5.
Find the least number which must be added to each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.
(i) 525
(ii) 1750
(iii) 252
(iv) 1825
(v) 6412.
Solution:
(i) Since remainderis 41.
∴ 222 < 525
Next perfect square number is 232 = 529
Hence, the number to be added is
232 – 525 = 529 – 525 = 4
∴ The required no. = 525 + 4 = 529
and $$\sqrt {529}$$ = 23
(ii) It is clear by long division method of $$\sqrt {1750}$$ that remainder is 69.
It shows that 412 < 1750
Next perfect square is 422
= 1764
Hence, the number to be added is 422 – 1750
= 1764 – 1750 = 14.
∴ Required number
= 1750 + 14 = 1764
and $$\sqrt {1764}$$ = 42
(iii) ∵ Remainder = 27
∴ 152 < 252 < 162.
Hence, the required number to be added to make it a perfect square
= 162 – 252
= 256 – 252 = 4
and $$\sqrt {256}$$ = 16.
(iv) ∵ Remainder = 61
∴ 422 < 1825 < 432
432 = 1849
Hence, the required number to be added to make it a perfect square
= 432 – 1825
= 1849 – 1825 = 24
∴ $$\sqrt {1849}$$ = 43.
(v) Since remainder = 12.
∴ 802 < 6412 < 812
∴ Required no. to be added to make it a perfect square is
812 – 6412 = 6561 – 6412
= 149
∴ $$\sqrt {6561}$$ = 81.
Question 6.
Find the length of the side of a square whose area is 441 m2.
Solution:
Let the side of square be x.
then, area of a square = x2
∴ x2 = 441
⇒ x = $$\sqrt {441}$$
= 21 m.
Question 7.
In a right triangle ABC, ∠B = 90°.
(а) If AB = 6cm, BC = 8cm, findAC.
(б) If AC = 13 cm, BC = 5 cm, find AB.
Solution:
(i) Using Pythagoras theorem
AC2 = AB2 + BC2
= 62 + 82
= 36 + 64
= 100
⇒ AC = $$\sqrt {100}$$
= 10 cm
(ii) AC = 13 cm
BC = 5 cm
AC2 = AB2 + BC2
(Using Pythagoras theorem)
(13)2 = x2 + (5)2
⇒ 169 = x2 + 25
⇒ 169 – 25 = x2
⇒ 144 = x2
⇒ x = $$\sqrt {144}$$ = 12 cm
Question 8.
A gardener has 1000 plants. He wants to plant these in such a way that the number of rows and the number of columns remain same. Find the minimum number of plants he needs more for this.
Solution:
By prime factorisation, we get
1000 = 2 × $$\underline{2 \times 2}$$ × 5 × $$\underline{5 \times 5}$$
We see that 2 and 5 are still unpaired.
∴ We have to multiply 1000 by 2 × 5 i.e., 10 to make it a perfect square.
Required number of plants
∴ 1000 × 10 = 10000
Thus, he needed 10 more plants.
Question 9.
There are 500 children in a school. For a P.T. drill they have to stand in such a manner that the number of rows is equal to number of columns. How many children would be left out in this arrangement ?
Solution:
By prime factorisation, we get
500 = $$\underline{2 \times 2}$$ × $$\underline{5 \times 5}$$ × 5
Since 5 is not in pair.
∴ 5 children would be left out in this arrangement. |
Teacher resources and professional development across the curriculum
Teacher professional development and classroom resources across the curriculum
Lesson Plans: Introduction Lesson Plan 1: Mathematical Modeling, Circular Movement and Transmission Ratios Lesson Plan 2: Skeeter Populations and Exponential Growth
Lesson Plan 1: Mathematical Modeling, Circular Movement and Transmission Ratios
Supplies:
Teachers will need the following:
• Overhead transparency with a diagram of the fan, the crankshaft, and the alternator pulleys in an engine
Students will need the following:
• Graphing calculator
• Peg board
• Drivers and followers (at a minimum, circles with radii of 1 cm, 2 cm, 4 cm) that can be inserted into the peg board
• Rubber bands
Note: Sarah Wallick obtained the pulleys she used in the video lesson for this workshop from Glencoe McGraw Hill, publisher of the Core Plus curriculum materials.
Steps
Introductory Activity:
1. Display the transparency diagram on the overhead projector. Briefly discuss how the fan, crankshaft, and alternator pulleys in an engine work together.
2. With the class, discuss the following questions:
• How does the speed of the crankshaft affect the speed of the fan? How does it affect the speed of the alternator?
If the idling speed of the crankshaft of a four-cylinder sports car is about 850 rpm, how far, in centimeters, would a point on the edge of the fan pulley travel in one minute? Do you think that a point on the alternator pulley would travel the same distance in one minute? Why or why not?
• Describe another situation in which the speed of one rotating object affects the speed of another object.
Learning Activities:
1. Introduce the setup of a follower and driver, and explain that students will rotate the driver pulley to determine the affect on the follower pulley. Specifically, students will determine the distance traveled by the follower when the driver pulley makes one complete revolution.
Ask students to experiment with different sizes of followers and record their observations in a two row table. They should record the number of driver rotations in the first row and the corresponding number of follower rotations in the second row.
2. Have students experiment with a driver of radius 2 cm and a follower of radius 1 cm. Check to be sure their data is close to the following:
Rotations of driver 1 2 3 4 5 Rotations of follower 2 4 6 8 10
Students should graph the data in a scatterplot using a graphing calculator. Have them describe the patterns in the data, and, as a class, find algebraic models that might fit the data. Discuss various equations or representations they could use; these might include the following:
y = 2x
Next = Now + 2, Start at 0 (the value for 0 rotations of the driver)
3. Have students repeat the experiment with a driver of radius 1 cm and a follower of radius 2 cm. Their data should be close to the following:
Rotations of driver 1 2 3 4 5 Rotations of follower 0.5 1 1.5 2 2.5
Students should look at the patterns in this second set of driver follower data. Again, they should graph the data in a scatterplot using a graphing calculator. Have them describe the patterns in the data, and, as a class, find algebraic models that might fit the data. Discuss various equations or representations they could use; these might include the following:
y = 0.5 x
y = 1/2 x
y = x/2
Next = Now + 0.5, Start at 0 (the value for 0 rotations of the driver)
4. Ask students to work in groups to generate a formula to determine the transmission factor for any ratio of driver-to-follower size. Reconvene the class for a discussion and select students to share and justify their answers
5. Assign student groups a set of problems related to the diagram they investigated in the introductory activity and have them prepare their solutions on poster paper. Choose several students to present solutions to the whole class. As part of this discussion, students should be able to correct, or justify, their intuitive answers to Question #2 from the introductory activity.
Culminating Activity/Assessment:
At the end of class, have students recount what they learned in their math journals. Then ask students to share their thoughts with the whole class. To summarize, create a list of things students have learned on the overhead projector. |
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# High School: Algebra
### Reasoning with Equations and Inequalities HSA-REI.C.8
8. Represent a system of linear equations as a single matrix equation in a vector variable.
Students should understand that a system of equations can be expressed as a matrix of coefficients multiplied by a vector matrix of variables. Sometimes, it's even better to do it that way.
Your students will probably curse you for stressing matrices time and time again, but just reassure them that it's for their own good. In order to work with matrices, your students should already understand what they are and how to perform various functions with them. The same goes for linear equations and systems of linear equations.
Students should know that a matrix equation takes the form AX = B, where A represents the coefficient of our variables, X represents our variables, and B represents the output to our equations.
In order to turn linear equations into matrices, we have to rearrange them all into the same format. Among the preferred formats are ax + by + cz = d. The linear equations y = 2x + 5 and y = 3x – 2 become -2x + y = 5 and -3x + y = -2. Painless, no?
Then we take each term in the equations and split them up into their proper matrix. Taking the equations from above, we'll have the A matrix (the one with all the coefficients) equaling
Since our only variables are x and y (and we put the coefficients into the A matrix in that order), our X matrix becomes:
The final matrix, B, is what's on the other side of our equal signs. That means it turns into:
In matrix form, it looks like this:
This standard only indicates that students should know how to represent systems of linear equations in matrix equation form, but not why doing so is useful (aside from organizational purposes). That is covered in the next standard.
#### Drills
1. What is the correct matrix form of the equations 3x + 15y – 37z = 45, 48x + 19z + 66y = 49, and 23x + 66y + 13 +19z = 66?
Correct Answer:
Answer Explanation:
You first have to be careful and actually put the equations into proper form first because some of them have the variable switched around and the last equation needs 13 subtracted from both sides.
2. What are the corresponding linear equations to the following matrix?
Correct Answer:
16x + 88y = 2; 19x + y + 3z = 19; 21x + 3z = 9
Answer Explanation:
This problem is simply the "art" of taking the numbers in matrix equation form and assigning the proper coefficients to the variables and the output. The only mistake that you may have made is assigning a 0 value in the matrix a variable when there shouldn't be one. But Shmoopers like you are smart, so it's unlikely that you fell for that.
3. Put the equations y = 3x + z, z = 77x – 2y + 34, and x = 55x + 31y into proper matrix equation form.
Correct Answer:
Answer Explanation:
The first order of business is to make sure all the equations are in the same format. That means 0 = 3xy + z, -34 = 77x – 2yz, and 0 = 54x + 31y. Then, all it takes is transferring them over into matrix form. Also, don't get confused by (D) because the variables are in the incorrect order. Variables are always changing things up on you.
4. What is the correct matrix form for the equations x + y = 3 and 2x + 3y = 4?
Correct Answer:
Answer Explanation:
In the first equation, the coefficient for both x and y is one. The rest is simply application of the matrix equation AX = B.
5. If the equations 34z + 24b = 13 and 24z + 34b = 13 were put into matrix form, what would the A matrix look like?
Correct Answer:
Answer Explanation:
All other answers do not put these equations into the correct matrix format. This is a simple a matter of putting the coefficients into the right places.
6. What is the proper matrix form of the equations y = 2x + 3 and y = 3x + 5?
Correct Answer:
Answer Explanation:
If we organize the equations into the proper format, we will have y – 2x = 3 and y – 3x = 5. If we put this into matrix form we will see that (A) is the only correct answer, since (B) switches the variables, (D) switches the entire product matrix, and (C) has the incorrect numbers and order of numbers.
7. The correct form of a matrix equation is:
Correct Answer:
AX = B
Answer Explanation:
The only answer that makes sense is (C), since (A) is the slope-intercept form of a linear equation, (B) is the equation for a circle, and (D) is the standard form of a quadratic equation.
8. A matrix equation is used for:
Correct Answer:
Both (B) and (C)
Answer Explanation:
We use it to solve for the variable in linear equations. So what if we don't know exactly how yet? We can translate linear equations to matrices and back. Isn't that enough?
9. In the matrix equation AX = B, X represents the:
Correct Answer:
Variable
Answer Explanation:
In our matrix equation, matrix A represents the coefficients in the linear equations and matrix B represents the output.
10. What would the B matrix for the equations x + 2y = 3 and x + 2y = 4 be?
Correct Answer:
Answer Explanation:
The B matrix in our matrix equation is the solution of the two linear equations. Since 3 and 4 are our values for the two equations, we list them out in matrix B. |
# The Nature of Force.
## Presentation on theme: "The Nature of Force."— Presentation transcript:
The Nature of Force
“An object in motion will remain in motion and an object at rest will remain at rest unless acted on by an unbalanced force.” Newton’s First Law Read Newton’s First Law above and then look at the picture. Write a paragraph predicting the motion of the soccer ball. Will the ball continue to move upward forever? Will it continue to move forward forever? If not, what do you think causes the motion of the ball to change after the player kicks it?
Forces acting on an object can change the object’s motion!
Change direction Slow down Speed up
What is a FORCE? A force is a push or pull
The strength of a force is measured in Newtons (N) Every force has two components: How strong the force is (in N) The direction in which the force acts
FORCE Forces acting on an object can be drawn using arrows. The length of the arrow represents the strength of the force, and the direction of the arrow shows the direction of the force.
There is usually more than one force acting on an object.
NET FORCE There is usually more than one force acting on an object. When two forces act in the same direction, they add together. When forces act in opposite directions, they are combined by subtracting the smaller force from the large force. The direction will be in the direction of the larger force. The combination of all the forces acting on an object is called the net force. Which of these three arrows represents the correct amount of net force of the above two arrows?
NET FORCE If equal forces act on an object in opposite directions, the net force will be zero. If the net force on an object is zero, the forces are said to be balanced. Balanced forces will NOT cause a change in an object’s motion.
REMEMBER NEWTON’S FIRST LAW:
NET FORCE If there is a non-zero net force acting on an object, the forces are said to be unbalanced. Unbalanced forces WILL cause a change in an object’s motion (speed up, slow down, change direction). REMEMBER NEWTON’S FIRST LAW: “An object in motion will remain in motion and an object at rest will remain at rest unless acted on by an unbalanced force.”
NET FORCE Unbalanced forces will cause an object to speed up, slow down, or change direction. In other words, an unbalanced force acting on an object will cause the object to ACCELERATE!
NET FORCE Net Force The change in motion of an object is determined by the net force acting on the object. What is the net force for each situation?
NET FORCE
GRAVITY Gravity is a force that pulls all objects toward each other.
Gravity acts between any two objects in the universe. Gravity is affected by two factors: The mass of the object (the more mass an object has, the greater its gravitational force) The distance between two objects (the greater the distance between the objects, the weaker the gravitational force between them)
GRAVITY Gravitational Attraction
Gravitational attraction depends on two factors: mass and distance. Compare the gravitational force between different planets and their sun.
GRAVITY All objects, regardless of their masses, fall toward Earth with the same acceleration. The force of gravity causes all falling objects to speed up as they fall. The Earth’s gravity causes all objects to be pulled toward the center of the Earth.
The force of gravity is what causes objects to have weight
The Earth’s gravity pulls objects toward the Earth. The more mass the object has, the greater the pull of gravity. Therefore, objects with more mass will “weigh” more. Also, the same object will have different “weights” on planets that have different masses. For example, the mass of the moon is less than the mass of the Earth, so the moon’s gravitational force is less. Therefore you would weigh less on the moon than on Earth even though your mass didn’t change.
GRAVITY On which planet would you weigh the most? On which planet would objects fall with the slowest speed?
FRICTION Friction is the force that two surfaces exert on each other when they rub against each other. Friction acts in the direction opposite an object’s motion. The force of friction causes an object to slow down and/or stop. Friction is affected by two factors: the types of surfaces involved how hard the surfaces are pushed together
There are 4 types of Friction:
Find examples of sliding, static, fluid and rolling frictions.
GRAVITY AND FRICTION TERMINAL VELOCITY
The force of gravity causes a falling object to speed up (accelerate) as it falls. However, as the object speeds up, the force of friction (air resistance) also becomes greater. When the force of friction acting upward equals the force of gravity pulling down, the object will stop accelerating and will then fall at a constant speed. The speed at which this occurs is called terminal velocity.
Draw a force diagram showing all of the forces that are acting on each of the objects below.
the rocket as it moves upward the ball as it is hit by the bat the skydiver as it falls down
Write a paragraph describing the motion of the soccer ball
Write a paragraph describing the motion of the soccer ball. Begin with the moment the ball was kicked by the player until the moment the ball stopped rolling on the ground. Include all of the following terms: gravity, friction, unbalanced force, accelerate, at rest, change direction. Draw a force diagram of the ball traveling through the air.
Motion of a Soccer Ball The ball begins at rest. When the player kicks the ball, he exerts an unbalanced force on the ball that causes the ball to accelerate (speed up). As the ball travels through the air, the unbalanced force of air resistance (fluid friction) acts on the ball which causes it to slow down. At the same time, gravity acts on the ball, causing it to begin to move down toward the ground (change direction). As the ball begins to move down toward the ground, it speeds up. So overall, the combined forces of gravity and friction cause the ball to slow down, change direction, and then speed up. Once the ball hits the ground, it rolls over the grass. The force of rolling friction acts on the ball in the direction opposite the ball’s motion, which causes the ball to slow down. Eventually the ball is at rest again. |
# Step-by-Step Selection Sort Algorithm
## Selection Sort: Step-by-Step Selection Sort Algorithm
Sorting algorithms are an essential part of programming, allowing us to organize data in a specific order. One such algorithm is the selection sort algorithm. In this tutorial, we will take an in-depth look at how selection sort works and guide you through the step-by-step process of implementing it in your code.
### What is Selection Sort?
Selection sort is a simple and intuitive sorting algorithm that comprehensively iterates through a list of data elements and gradually builds a sorted output. This algorithm works by repeatedly finding the minimum or maximum element and swapping it with the current element in the iteration. The process continues until the entire list is sorted.
Selection sort has an average and worst-case time complexity of O(n^2), making it less efficient than other sorting algorithms for large datasets. However, it is a great algorithm to understand due to its simplicity and easy implementation.
### Step-by-Step Selection Sort Algorithm Explanation
Let's dive into the step-by-step process of the selection sort algorithm using an example to illustrate each step along the way. Consider an unsorted list of integers:
``````[7, 3, 1, 9, 5]
``````
Step 1: Find the minimum element in the unsorted part of the list.
To begin, we assume the first element of the list is the minimum. We then compare it with the remaining elements and update our assumption if we find a smaller element. In our example, after iterating over the list, we identify 1 as the minimum.
Step 2: Swap the minimum element with the current element.
Once we have found the minimum element, we swap it with the current element. In our case, we swap 7 with 1, resulting in:
``````[1, 3, 7, 9, 5]
``````
Now, the first element is in its correct position, and the remaining unsorted part of the list is reduced.
Step 3: Repeat steps 1 and 2 for the remaining unsorted part.
We continue the process by focusing on the remaining unsorted part of the list, excluding the elements already sorted. With each iteration, the sorted part grows, and the unsorted part shrinks.
In our example, the next minimum element is 3. We swap it with the current element, resulting in:
``````[1, 3, 7, 9, 5]
``````
By repeating the process, we identify 5 as the next minimum, resulting in:
``````[1, 3, 5, 9, 7]
``````
Finally, we find 7 as the last minimum element and swap it with the current element, yielding the sorted list:
``````[1, 3, 5, 7, 9]
``````
### Implementing Selection Sort in Code
Now that we understand the step-by-step process of the selection sort algorithm, let's take a look at an implementation in Python:
``````def selection_sort(arr):
n = len(arr)
for i in range(n-1):
min_index = i
for j in range(i+1, n):
if arr[j] < arr[min_index]:
min_index = j
arr[i], arr[min_index] = arr[min_index], arr[i]
return arr
``````
In this code snippet, we define a function `selection_sort` that takes an array `arr` as input. We iterate through the array, finding the minimum element and swapping it with the current element. The sorted array is then returned.
### Conclusion
Selection sort is a relatively simple sorting algorithm that is easy to understand and implement. Although it might not be the most efficient algorithm for large datasets, it serves as an excellent learning tool to grasp the fundamentals of sorting algorithms.
By following the step-by-step explanation provided in this tutorial, you should now have a clear understanding of how selection sort works. Remember to practice implementing the algorithm in your programming language of choice to reinforce your knowledge.
Now that you have learned about selection sort, you can explore other sorting algorithms such as bubble sort, insertion sort, or even more advanced ones like merge sort or quicksort.
Happy coding!
Please note that this blog post is written in Markdown format and can be converted to HTML using any Markdown to HTML converter tool. |
# Find the sum of the series whose nth term is:
Question:
Find the sum of the series whose nth term is:
(i) 2n2 − 3n + 5
(ii) 2n3 + 3n2 − 1
(iii) n3 − 3n
(iv) n (n + 1) (n + 4)
(v) (2n − 1)2
Solution:
Let $T_{n}$ be the $n$th term of the given series.
Thus, we have:
(i) $T_{n}=2 n^{2}-3 n+5$
Let $S_{n}$ be the sum of $n$ terms of the given series.
Now,
$S_{n}=\sum_{k=1}^{n} T_{k}$
$\Rightarrow S_{n}=\sum_{k=1}^{n}\left(2 k^{2}-3 k+5\right)$
$\Rightarrow S_{n}=2 \sum_{k=1}^{n} k^{2}-3 \sum_{k=1}^{n} k+\sum_{k=1}^{n} 5$
$\Rightarrow S_{n}=\frac{2 n(n+1)(2 n+1)}{6}-\frac{3 n(n+1)}{2}+5 n$
$\Rightarrow S_{n}=\frac{2 n(n+1)(2 n+1)-9 n(n+1)+30 n}{6}$
$\Rightarrow S_{n}=\frac{\left(2 n^{2}+2 n\right)(2 n+1)-9 n^{2}-9 n+30 n}{6}$
$\Rightarrow S_{n}=\frac{4 n^{3}+4 n^{2}+2 n^{2}+2 n-9 n^{2}-9 n+30 n}{6}$
$\Rightarrow S_{n}=\frac{4 n^{3}-3 n^{2}+23 n}{6}$
$\Rightarrow S_{n}=\frac{n\left(4 n^{2}-3 n+23\right)}{6}$
(ii) $T_{n}=2 n^{3}+3 n^{2}-1$
Let $S_{n}$ be the sum of $n$ terms of the given series.
Now,
$S_{n}=\sum_{k=1}^{n} T_{k}$
$\Rightarrow S_{n}=\sum_{k=1}^{n}\left(2 k^{3}+3 k^{2}-1\right)$
$\Rightarrow S_{n}=2 \sum_{k=1}^{n} k^{3}+3 \sum_{k=1}^{n} k^{2}-\sum_{k=1}^{n} 1$
$\Rightarrow S_{n}=\frac{2 n^{2}(n+1)^{2}}{4}+\frac{3 n(n+1)(2 n+1)}{6}-n$
$\Rightarrow S_{n}=\frac{n^{2}(n+1)^{2}}{2}+\frac{n(n+1)(2 n+1)}{2}-n$
$\Rightarrow S_{n}=\frac{n^{2}(n+1)^{2}+n(n+1)(2 n+1)-2 n}{2}$
$\Rightarrow S_{n}=\frac{n^{2}\left(n^{2}+1+2 n\right)+\left(n^{2}+n\right)(2 n+1)-2 n}{2}$
$\Rightarrow S_{n}=\frac{\left(n^{4}+n^{2}+2 n^{3}\right)+\left(2 n^{3}+n^{2}+2 n^{2}+n\right)-2 n}{2}$
$\Rightarrow S_{n}=\frac{n^{4}+4 n^{2}+4 n^{3}-n}{2}$
$\Rightarrow S_{n}=\frac{n\left(n^{3}+4 n+4 n^{2}-1\right)}{2}$
(iii) $T_{n}=n^{3}-3^{n}$
Let $S_{n}$ be the sum of $n$ terms of the given series.
Now,
$S_{n}=\sum_{k=1}^{n} T_{k}$
$\Rightarrow S_{n}=\sum_{k=1}^{n}\left(k^{3}-3^{k}\right)$
$\Rightarrow S_{n}=\sum_{k=1}^{n} k^{3}-\sum_{k=1}^{n} 3^{k}$
$\Rightarrow S_{n}=\frac{n^{2}(n+1)^{2}}{4}-\left(3+3^{2}+3^{3}+3^{4}+\ldots+3^{n}\right)$
$\Rightarrow S_{n}=\frac{n^{2}(n+1)^{2}}{4}-\left[\frac{3\left(3^{n}-1\right)}{3-1}\right]$
$\Rightarrow S_{n}=\frac{n^{2}(n+1)^{2}}{4}-\frac{3}{2}\left(3^{n}-1\right)$
(iv) $T_{n}=n(n+1)(n+4)=\left(n^{2}+n\right)(n+4)=n^{3}+5 n^{2}+4 n$
Let $S_{n}$ be the sum of $n$ terms of the given series.
Now,
$S_{n}=\sum_{k=1}^{n} T_{k}$
$\Rightarrow S_{n}=\sum_{k=1}^{n}\left(k^{3}+5 k^{2}+4 k\right)$
$\Rightarrow S_{n}=\sum_{k=1}^{n} k^{3}+5 \sum_{k=1}^{n} k^{2}+4 \sum_{k=1}^{n} k$
$\Rightarrow S_{n}=\frac{n^{2}(n+1)^{2}}{4}+\frac{5 n(n+1)(2 n+1)}{6}+\frac{4 n(n+1)}{2}$
$\Rightarrow S_{n}=\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}+\frac{5(2 n+1)}{3}+4\right]$
$\Rightarrow S_{n}=\frac{n(n+1)}{12}[3 n(n+1)+10(2 n+1)+24]$
$\Rightarrow S_{n}=\frac{n(n+1)}{12}\left(3 n^{2}+23 n+34\right)$
(v) $T_{n}=(2 n-1)^{2}$
Let $S_{n}$ be the sum of $n$ terms of the given series.
Now,
$S_{n}=\sum_{k=1}^{n} T_{k}$
$\Rightarrow S_{n}=\sum_{k=1}^{n}(2 k-1)^{2}$
$\Rightarrow S_{n}=\sum_{k=1}^{n}\left(4 k^{2}+1-4 k\right)$
$\Rightarrow S_{n}=4 \sum_{k=1}^{n} k^{2}+\sum_{k=1}^{n} 1-4 \sum_{k=1}^{n} k$
$\Rightarrow S_{n}=\frac{4 n(n+1)(2 n+1)}{6}+n-\frac{4 n(n+1)}{2}$
$\Rightarrow S_{n}=\frac{n(n+1)}{2}\left[\frac{4(2 n+1)}{3}-4\right]+n$
$\Rightarrow S_{n}=\frac{n(n+1)}{2}\left(\frac{8 n+4-12}{3}\right)+n$
$\Rightarrow S_{n}=\frac{n(n+1)}{2}\left(\frac{8 n-8}{3}\right)+n$
$\Rightarrow S_{n}=4 n(n+1)\left(\frac{n-1}{3}\right)+n$
$\Rightarrow S_{n}=\frac{n(4 n+4)(n-1)+3 n}{3}$
$\Rightarrow S_{n}=\frac{n}{3}\left(4 n^{2}+4 n-4 n-4+3\right)$
$\Rightarrow S_{n}=\frac{n}{3}\left(4 n^{2}-1\right)$
$\Rightarrow S_{n}=\frac{n}{3}(2 n-1)(2 n+1)$ |
Both sets (top and bottom) are supplementary but only the top ones are linear pairs because these ones are also adjacent. two angles with one common arm. Because, we know that the measure of a straight angle is 180 degrees, so a linear pair of angles must also add up to 180 degrees. Because in a linear pair the sum of the 2 angles is 180 degrees and both the angles are of measure 90 degrees. 6. This concept will introduce students to linear pairs of angles. Supplementary Angles. If two angl… chanybonilla chanybonilla 10/21/2019 Mathematics Middle School Which statement is the converse of the following? But they are not a linear pair because they are not connected, and they do not share a common side. Adjacent Angles - Adjacent angles are angles that share a leg. Find the measure of each angle in the diagram. Also, in doing proofs, when do I know to use: Definition of supplementary angles. In this picture there is a flatscreen Toshiba television. Consider the following figure in which a ray $$\overrightarrow{OP}$$ stand on the line segment $$\overline{AB}$$ as shown: The angles ∠POB and ∠POA are formed at O. These are examples of adjacent angles. So an angle that forms a linear pair will be an angle that is adjacent, where the two outer rays combined will form a line. Equate the sum of these measures with 90° or 180° and solve for the value of x. If the sum of two angles is 1800 then they are called supplementary angles. Name a pair of complementary angles. PQR and ABC are supplementary, because 1000 + 800 = 1800 RQ P A B C 1000 800 PQR + ABC Supplementary Angles 6. (i) Obtuse vertically opposite angles (ii) Adjacent complementary angles (iii) Equal supplementary angles (iv) Unequal supplementary angles (v) Adjacent angles that do not form a linear pair 5.3 PAIRS OF LINES 5.3.1 Intersecting Lines LINES AND ANGLES In the Fig 5.22, p is a transversal to the lines l and m. * another name for a supplementary angle is "linear pair" * unlike a complentary angle, it takes 2 right angles to . Obtuse vertically opposite angles. The angles in a linear pair are supplementary. Therefore, they are supplementary. Corresponding angles are pairs of angles that lie on the same side of the transversal in matching corners. All linear pairs are adjacent angles but all adjacent angles are not linear pairs. The adjacent angles are the angles that have a common vertex. Also, there is a common arm that represents both the angles of the linear pair. Every linear pair is a supplementary angle but every supplementary angle is not a linear pair. If two angles form a linear pair, then they are supplementary. A supplementary angle can be either adjacent or non-adjacent.A linear pair must be adjacent and is never non-adjacent.NOTE: They both add up to 180°. Find the measure of each angle. They are supplementary because each angle is 90 degrees so they add up to 180 degrees. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Also, know about linear pair, vertically opposite angles and more. Start studying Angle Pairs (Vertical, linear pair, adjacent, complementary & supplementary). Such angles are also known as supplementary angles. they share a same vertice and a same arm. 45º 15º 3. 4. 3 = 55(5 = 88(m4 = m9. Supplementary Angles. And two intersection lines always create four linear pairs of angles (draw a picture). But two angles may be separated and yet add 180°, then they are supplementary but are not linear angle pairs. Linear angle pairs are adjacent angles (they are together) and they form a straight angle, which is 180°. No. And in a supplementary angle the sum of the 2 angles is 180 degrees but the measures of the both angles can be different. Step-by-step explanation: please follow me only 3 followers are remaining for 500 followers. Adjacent angles are “side by side” and share a common ray. When the sum of the measure of two angles is 180 0, then the pair of angles is said to be supplementary angles.Here the supplementary meaning is one angle is supplemented to another angle to make a sum of 180 0.. From the figure, we can say that ∠1 + ∠2 = 180 0 Image Will be uploaded soon Name a different pair of supplementary angles. Name a pair of supplementary angles. 7. Click Create Assignment to assign this modality to your LMS. One of the supplementary angles is said to be the supplement of the other. Since the non-adjacent sides of a linear pair form a line, a linear pair of angles is always supplementary. 55º … Linear Pair Definition. Our next theorem relates these two definitions. Pair of adjacent angles whose measures add up to form a straight angle is known as a linear pair. A. Non-overlapping supplementary angles sharing a vertex and one side are a linear pair. So linear pair with angle DGF, so that's this angle right over here. In fact, a linear pair forms supplementary angles. Supplementary Angles Definition. A 600 1200 PC D 600 + 1200 = 1800 APC + APD Linear Pair Of Angles 7. Positive: 28.571428571429 % The angles are said to be linear if they are adjacent to each other after the intersection of the two lines. Adjacent complementary angles. 2. One of the angles in the pair is an exterior angle and one is an interior angle. A linear pair is precisely what its name indicates. 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how are supplementary angles and linear pairs different 2021 |
# Age Calculation Equations
When you do the age calculation, you can see what age group most people belong to. This statistic can be used in a variety of situations.
## Age Equations
The section on quantitative aptitude includes Age Equation Problems. The questions in age equation problems are structured so that they result in equations. These equations will contain solutions that indicate the age of the people in the question, and they will be either linear or nonlinear.
Comparing two people’s ages at different times, such as now, in the past, or the future is a common theme in age word problems.
### Points to Note
• After reading the age issues, assume the unknown age is a variable, such as ‘X.’
• Make mathematical equations out of the statements in the question.
• Calculate the variable by solving the equations, and the resulting value must meet the problem’s requirements.
## Age calculations: How to Solve Them
Word problems can be perplexing since they include a lot of information, and we’re not always sure what to do with it. The procedure is as follows:
1. Make a variable out of what we don’t know.
2. Create an equation using the data you’ve been given.
3. Find the unknown variable and solve for it.
4 Substitute our solution back into the equation to determine if the left and right sides of the equation are equal.
## Important age calculator formulas:
1. When the current age is x, n times the age = nx.
2. When the current age is x, age n years later = x + n.
3. When the current age is x, age n years ago = x – n.
4. The ages in a ratio a: b will be ax and bx.
5. when the current age is x,1/n of the age is x/n
## What Is The Best Way for an Exact Age calculator?
If only one individual is involved, the problem is analogous to an Integer Problem. To figure out the relationship between the numbers, carefully read the problem. This is demonstrated in the single-person instances.
If 2 or more people are there in the age dilemma, employing a table is an excellent option. A table will assist you in organizing the data and writing the equations.
Example
Divya is two years older than Ronak, and their combined ages are 18. What are Ronak and Divya’s ages?
The first step in answering a word problem involving age is identifying the unknown and trying to express it as a variable, an alphabetical letter representing the unknown information. We don’t know Ronak’s or Divya’s ages in this problem. Because Ronak’s age is expressed as a percentage of Divya’s, our variable will be based on Divya’s age. In another way, let d be Divya’s age. Ronak’s age must be (d + 2) if he is two years older than Divya.
We can create our equation now that we have our variables. Because Ronak is two years older than Divya and their ages add up to 18, the calculation is:
d + d + 2 = 18
The next step is to solve the unknown variable, d, in the equation.
d + d + 2 = 18
2d + 2 = 18
2d = 16
d = 8
Therefore, Divya is 8 years old. We must substitute into Ronak’s equation to determine his age. Recall that Ronak’s age = d + 2. If d = 8, then Ronak’s age = 8 + 2 = 10.
The next is to verify that the left side of the equation equals the right side of the equation by putting our solution into the original equation.
d + d + 2 = 18
If d = 8, then:
8+8+2= 18
18=18
Therefore, the left and right sides are equal, Divya must be 8 years old, and Ronak must be 10 years old.
## Short tricks
1. The total age of the brother and sister is 46. Seven years ago, the sum of their ages was four times the brother’s age. What is the recent age of the brother and sister?
Solution:
(B-7) (S-7) = 4(B-7)
S = 7 = 4
S = 11 years (present)
B=35 years (present)
1. When two of the eight guys, aged 21 and 23, are replaced by two new men, the group’s average age rises by two years. What is the average of two new men?
Solution:
The numerator must be modified by 16 to modify the average age by 2 years. 2 years increase = 8 2 = 16
Two men = 21 + 23 = 44
= 16 +44 = 60
Average 60/2 = 30 years
## Question for practice
1. Amy’s mother, Aashi, is 23 years her senior. Aashi will be twice Amy’s age in six years.?
2. Rahul is 4 years older than Mukesh. Their combined ages were 48 five years ago?
3. Shreya is four times Martha’s age. Their combined ages were 50 five years ago?
4. A vase is 22 years old, while a pitcher is 30 years old. How many years ago, the pitcher was twice as old as the vase?
5. Pankaj is two years younger than Rahul. Seven years ago, their ages added up to 13. What are their current ages?
## Conclusion
Age calculations are a useful tool for expressing conditions or relationships between two or more variables. There could be one, two, or three unknowns in an equation. The essential rule is that these equations are solvable if the number of unknowns equals the number of conditions; otherwise, they are not. |
# How do you find two consecutive odd integers whose sum is 36?
Apr 6, 2018
$17$ and $19$.
#### Explanation:
Let $n$ denote the first odd integer.
If the two integers are consecutive and odd, then the second number is $n + 2$.
Hence, $n + n + 2 = 36$
$\setminus \implies 2 n = 34$
$\setminus \implies n = 17$.
Therefore, the numbers are $17$ and $19$.
Apr 6, 2018
$\therefore 17 + 19 = 36$
#### Explanation:
If the odd numbers are consecutive, they must be on either side of half of 36.
$18 + 18 = 36$
$\therefore 17 + 19 = 36$ |
# Pascal’s triangle – Definition, Patterns, and Applications
Pascal’s Triangle is a number pattern that is known for its shape – yes, a triangle! This interesting pattern and property is named after Blaise Pascal and has been a famous triangle in mathematics due to its extensive application in algebra, number theory, and statistics.
Pascal’s Triangle is a number pattern that returns the values or coefficients used in binomial expansions. The numbers in the next layer will depend on the sum of two terms positioned above them in the previous layer.
You may already have encountered this unique triangle in your earlier math classes, so in this article, we’ll dive deeper into understanding how we can apply Pascal’s Triangle in algebra and statistics.
For now, let’s refresh our knowledge and take a look at the important properties of a Pascal’s Triangle.
What is Pascal’s Triangle?
The Pascal’s Triangle is a number pattern that shows a significant number of patterns used to expand binomial expressions. Before we dive right into its application in statistics, let’s observe the image shown below.
The best way to remember the definition of Pascal’s Triangle and its properties is by constructing the Pascal’s Triangle itself, as shown above.
As can be seen, the values in the middle of the third row can be determined by adding the two consecutive terms in the second row. Similarly, the center of the fourth row’s values are the results of the consecutive terms from the third row are added.
Here’s an extended version of Pascal’s Triangle shown, and it shows how the sixth row can be formed by:
• Using $1$ at the ends of the row.
• Adding two consecutive terms to find the term found below them in the next row.
The arrows guide the two numbers that were added to find the next row’s term.
Pascal’s Triangle definition and hidden patterns
Generalizing this observation, Pascal’s Triangle is simply a group of numbers that are arranged where each row of values represents the coefficients of a binomial expansion, $(a+ b)^n$.
The rows’ values can be determined by adding two consecutive numbers above each value, as shown in the earlier section.
Why don’t we go ahead and arrange Pascal’s Triangle’s values and justify them to the left? This will help us observe the most interesting patterns there are within these numbers.
\begin{aligned}&1\\&1\phantom{xxxx}1\\&1\phantom{xxxx}2\phantom{xxxx}1\\&1\phantom{xxxx}3\phantom{xxxx}3\phantom{xxxx}1\\&1\phantom{xxxx}4\phantom{xxxx}6\phantom{xxxx}4\phantom{xxxx}1\\&.\\&.\\&.\end{aligned}
We can begin by adding all the terms in each row and observe the result. What can you observe?
\begin{aligned}&1\phantom{x} = \phantom{x} 1\\&1\phantom{x}+\phantom{x}1\phantom{x} = \phantom{x}2\\&1\phantom{x}+\phantom{x}2\phantom{x}+\phantom{x}1\phantom{x} = \phantom{x} 4\\&1\phantom{x}+\phantom{x} 3\phantom{x}+\phantom{x}3\phantom{x}+\phantom{x} 1\phantom{x}=\phantom{x} 8\\&1\phantom{x}+\phantom{x} 4\phantom{x}+\phantom{x} 6\phantom{x}+\phantom{x} 4\phantom{x}+\phantom{x}1\phantom{x}=\phantom{x}16\\&.\\&.\\&.\end{aligned}
We can see that each row’s sum is actually equal to the power of $2$ that depends on the number of rows.
Row Number Sum $1$ $2^0 = 1$ $2$ $2^1 = 2$ $3$ $2^2 = 4$ $4$ $2^3 = 8$ $5$ $2^4 = 16$
From this, we can see that the sum of the given Pascal’s Triangle’s row is in fact equal to $2^{n -1}$, where $n$ is the row number.
Here’s another interesting pattern about Pascal’s Triangle: when we treat them as digits and combine one row’s values into one number, we will end up with a power of $11$.
\begin{aligned}1&= 11^0\\11&=11^1\\121&=11^2\\1331&=11^3\\14641 &= 11^4\end{aligned}
Does this still apply for rows with two-digit values? Yes, what can be done is to carry over the tens place to the previous value.
Amazing isn’t it? Here’s a third one for now:
\begin{aligned}&1 \\&1 \phantom{xxxx}\color{blue} 1\\&{\color{blue}1 }\phantom{xxxx}{\color{red} 2}\phantom{xxxx}{\color{green} 1}\\&{\color{red} 1}\phantom{xxxx}{\color{green} 3}\phantom{xxxx}{\color{blue} 3}\phantom{xxxx} 1\\&{\color{green}1}\phantom{xxxx}{\color{blue} 3}\phantom{xxxx} 6\phantom{xxxx} 4\phantom{xxxx}1\\&{\color{blue}1}\phantom{xxxx} 5\phantom{xxxx} 10\phantom{xxxx} 10\phantom{xxxx} 5\phantom{xxxx}1\\&.\\&.\\&.\end{aligned}
The first few sums of the terms along the diagonal of Pascal’s Triangle are $1, 1, 2, 3, 5, 8,…$. These are the first six terms of the Fibonacci sequence. Amazing how different math concepts can be observed in this simple number pattern?
One important pattern from Pascal’s Triangle has been extensively studied and also has important applications in binomial expansion and combinatorial numbers.
How to use Pascal’s Triangle?
Here’s an interesting way to view the Pascal’s Triangle in terms of combinatorial numbers. Before we proceed, recall that $\binom{n}{r} = \dfrac{n!}{r! N!}$. If you need a quick refresher on what these symbols represent, check out this article as well.
\begin{aligned} \binom{0}{0}\phantom{xxxxxxxxxxxx}\\\binom{1}{0} \phantom{xxxx}\binom{1}{1}\phantom{xxxxxxxx}\\\phantom{xx}\binom{2}{0} \phantom{xxx}\binom{2}{1}\phantom{xxx}\binom{2}{2}\phantom{xzxxx}\\\binom{3}{0} \phantom{xxx}\binom{3}{1}\phantom{xxx}\binom{3}{2}\phantom{xx}\binom{3}{3}\phantom{xxx}\end{aligned}
\begin{aligned}\text{First Row:}\phantom{xxxxxxxxxxxxx} 1\phantom{xxxxxxxxxx}\\\text{Second Row:}\phantom{xxxxxxxxxxx} 1\phantom{xxxx} 1\phantom{xxxxxxx}\\\text{Third Row:}\phantom{xxxxxxxxx} 1\phantom{xxxx} 2\phantom{xxxx} 1\phantom{xxxx}\\\text{Fourth Row:}\phantom{xxxxxxx} 1\phantom{xxxx} 3\phantom{xxx}{\color{blue} 3}\phantom{xxx} 1\phantom{xxx}\\\text{Fifth Row:}\phantom{xxx}1\phantom{xxxx} 4\phantom{xxxx} 6\phantom{xxxx} 4\phantom{xxxx}1\\.\phantom{xxxxxxxxxx}\\\phantom{xxxxxxxxxx}.\phantom{xxxxxxxxxx}\\\phantom{xxxxxxxxxx}.\phantom{xxxxxxxxxx}\end{aligned}
This means that if we want to find the value of $\binom{\color{blue}3}{\color{red}2}$, we simply inspect the value that is located at the $\color{blue} 3 + 1 = 4$th row and $\color{red} 2 + 1 =3$rd value to the right. This means that $\binom{3}{2} = 2$.
How to apply Pascal’s Triangle formula?
Let’s use what we’ve learned about Pascal’s Triangle and apply these in expanding binomial theorems and combinatorics. Recall that $\binom{n}{r}$ simply means that we choose $n$ from $r$.
Applying the Pascal’s Triangle to Find $\boldsymbol{\binom{n}{r}}$
From what we’ve observed, we can find the value of $\binom{n}{r}$ using the values found in the Pascal’s Triangle:
• Complete the Pascal’s Triangle up to $(n + 1)$ rows.
• Focus on the $(n +1)$th row of the triangle.
• Find the value located at the $(r + 1)$th position of the row we’re focusing on.
Once we locate that particular value, we now have $\binom{n}{r}$. Yes, it’s that easy!
This means that $\binom{4}{2}$ is equivalent to the value of the Pascal’s Triangle located at the $5$th row and the $3$rd term from the right.
\begin{aligned}\text{First Row:}\phantom{xxxxxxxxxxxxx} 1\phantom{xxxxxxxxxx}\\\text{Second Row:}\phantom{xxxxxxxxxxx} 1\phantom{xxxx} 1\phantom{xxxxxxx}\\\text{Third Row:}\phantom{xxxxxxxxx} 1\phantom{xxxx} 2\phantom{xxxx}1\phantom{xxxx}\\\text{Fourth Row:}\phantom{xxxxxxx}1 \phantom{xxxx} 3\phantom{xxx} 3\phantom{xxx}1\phantom{xxx}\\\text{Fifth Row:}\phantom{xxx}1\phantom{xxxx}4\phantom{xxxx}{\color{blue}6}\phantom{xxxx}4\phantom{xxxx}1\\.\phantom{xxxxxxxxxx}\\\phantom{xxxxxxxxxx}.\phantom{xxxxxxxxxx}\\\phantom{xxxxxxxxxx}.\phantom{xxxxxxxxxx}\end{aligned}
This means that $\binom{4}{2}$ is equal to $6$.
Applying the Pascal’s Triangle in Binomial Expansion
We can extend this formula when expanding binomials to the $n$th power. Let’s do a quick recap of the binomial expansion’s formula:
\begin{aligned}(a+ b)^n &= \sum_{k=0}^{n} \binom{n}{k}a^{n-k}b^k\\&=a^n + \binom{n}{1}a^{n-1}b+\binom{n}{2}a^{n-2}b^2 + … +\binom{n}{n -1}a^{1}b^{n- 1} + b^n \end{aligned}
Remember that we can find the values of $\binom{n}{k}$ using the Pascal’s Triangle. In fact, the entire coefficients of the expansion can be determined by using the coefficients of the $(n + 1)$ row of the Pascal’s Triangle.
For example, if we want to find the coefficients of $(a + b)^3$ once it’s expanded, we inspect the values found on the $3 + 1=4$th or fourth row of the given Pascal’s triangle.
\begin{aligned}\text{First Row:}\phantom{xxxxxxxxxxxx}1 \phantom{xxxxxxxxxx}\\\text{Second Row:}\phantom{xxxxxxxxxx} 1\phantom{xxxx} 1\phantom{xxxxxxx}\\\text{Third Row:}\phantom{xxxxxxxx} 1\phantom{xxxx}2 \phantom{xxxx} 1\phantom{xxxx}\\\color{green}\text{Fourth Row:}\phantom{xxxxxxx}1\phantom{xxxx}3\phantom{xxx} 3\phantom{xxx} 1\phantom{xx}\\\text{Fifth Row:}\phantom{xxx}1\phantom{xxxx} 4\phantom{xxxx} 6\phantom{xxxx} 4\phantom{xxx}1\\.\phantom{xxxxxxxxxx}\\\phantom{xxxxxxxxxx}.\phantom{xxxxxxxxxx}\\\phantom{xxxxxxxxxx}.\phantom{xxxxxxxxxx}\end{aligned}
From this, we can see that the coefficients of the expansion are $1, 3, 3, 1$ in a particular order. These are the coefficients of the expansion, so we can now complete the expansion.
\begin{aligned}(a + b)^3 &= \underline{\phantom{xxx}}a^3 + \underline{\phantom{xxx}}a^2b + \underline{\phantom{xxx}}ab^2+\underline{\phantom{xxx}}b^3\end{aligned}
This means that we have expanded $(a + b)^3$ using Pascal’s Triangle’s values, as shown below.
\begin{aligned}(a + b)^3 &= {\color{green}1}a^3 + {\color{green}3}a^2b + {\color{green}3}ab^2+{\color{green}1}b^3\\&=a^3 + 3a^2b + 3ab^2 + b^3\end{aligned}
Here’s a quick recap of when we want to expand $(a + b)^n$ and use Pascal’s Triangle values.
Step 1: Write down the first $(n +1)th row of the Pascal’s Triangle. Step 2:The terms of the combination will have$(a + b)^n = \underline{\phantom{xxx}}a^n + \underline{\phantom{xxx}}a^{n-1}b^1 + … + \underline{\phantom{xxx}}ab^{n -1}+\underline{\phantom{xxx}}b^n$. Step 3:Use the values of the$(n +1)$th row to fill in the coefficients of the partial expansion from the previous step. That’s it! Don’t worry; we’ll review what we’ve learned so far and apply them to answer the examples shown below. Example 1 Write down the first seven rows of the known Pascal’s Triangle to answer the following expression. a. What is the sum of the values found in the seventh row? b. What is the value of$11^6? Solution Let’s write down the first seven rows of the Pascal’s Triangle first and highlight the seventh row: \begin{aligned}\text{First Row:}\phantom{xxxxxxxxxxxxxxxxxxxxxx}1\phantom{xxxxxxxxxxxxxx}\\\text{Second Row:} \phantom{xxxxxxxxxxxxxxxxxxx}1\phantom{xxxxx} 1\phantom{xxxxxxxxxxx}\\\text{Third Row:}\phantom{xxxxxxxxxxxxxxxxx} 1\phantom{xxxx} 2\phantom{xxxx} 1\phantom{xxxxxxxxx}\\\text{Fourth Row:}\phantom{xxxxxxxxxxxxxx} 1\phantom{xxxx} 3\phantom{xxxx}3\phantom{xxxx}1\phantom{xxxxxxx}\\\text{Fifth Row:}\phantom{xxxxxxxxsxx} 1\phantom{xxxx} 4\phantom{xxxx} 6 \phantom{xxxx} 4\phantom{xxxx}1 \phantom{xxxxx}\\\text{Sixth Row:}\phantom{xxxxxxx}1\phantom{xxxx} 5\phantom{xxxx}10\phantom{xxxx} 10\phantom{xxxx}5\phantom{xxxx}1\phantom{xx}\\\color{green}\text{Seventh Row:}\phantom{xxx} 1\phantom{xxxx} 6\phantom{xxxx}15\phantom{xxxx} 20\phantom{xxxx} 15\phantom{xxxx} 6 \phantom{xxxx}1\end{aligned} As we have discussed in the earlier section, the sum of the values found at then$th row is equivalent to$2^{n – 1}$. Since we want to find the sum of the seventh row’s values, we find the value of$2^{7 – 1}$equal to$2^6 = 64. We can also confirm this by actually adding the terms from the seventh row: \begin{aligned}&{\color{green} \text{Seventh Row}}: 1 \phantom{xx} 6\phantom{xx}15\phantom{xx} 20\phantom{xx}15\phantom{xx} 6 \phantom{xx}1\\&\phantom{xxxxxxxxx}=1 \phantom{x} +\phantom{x} 6\phantom{x} +\phantom{x} 15\phantom{xx} 20\phantom{x} +\phantom{x}15\phantom{x} +\phantom{x} 6\phantom{x} +\phantom{x}1\\&\phantom{xxxxxxxxx}=64\end{aligned} a. Hence, the sum values of found at the seventh row is equal to64$. To find the value of$11^6$, we inspect the values found at the$(6 + 1)= 7$th row of the Pascal’s Triangle. We can see that the values of the seventh row are:$1 \phantom{x} 6\phantom{x}15\phantom{x} 20\phantom{x}15\phantom{x} 6 \phantom{x}1$. We can use these to find the value of$11^6: use these values and carry the tens place over for instances where the values have two digits. \begin{aligned}&\phantom{xx}1 \phantom{x} 6\phantom{x}15\phantom{x} 20\phantom{x}15\phantom{x} 6 \phantom{x}1\\&=1 \phantom{x} 6^{\color{green} + 1}\phantom{x}5^{\color{green} + 2}\phantom{x} 0^{\color{green} + 1}\phantom{x}5\phantom{x} 6 \phantom{x}1\\&=1 \phantom{x}7 \phantom{x}7 \phantom{x}1 \phantom{x}5 \phantom{x}6 \phantom{x}1\phantom{xxx}\end{aligned} b. This means that11^6$is equal to$1, 771, 561$. Example 2 Determine the values of the following using a Pascal’s Triangle. a.$\binom{5}{4}$b.$\binom{4}{1}$c.$\binom{6}{2}$Solution When we have a Pascal’s Triangle, we can actually find the value of$\binom{n}{r}$by finding the value located at the$(n + 1)$th term and$(r + 1)th unit/s to the right. Let’s use the same Pascal Triangle for all three, so it helps to write down the first seven rows for this item. \begin{aligned}\text{First Row:}\phantom{xxxxxxxxxxxxxxxxxxxxxx}1\phantom{xxxxxxxxxxxxxx}\\\text{Second Row:} \phantom{xxxxxxxxxxxxxxxxxxx}1\phantom{xxxxx} 1\phantom{xxxxxxxxxxx}\\\text{Third Row:}\phantom{xxxxxxxxxxxxxxxxx} 1\phantom{xxxx} 2\phantom{xxxx} 1\phantom{xxxxxxxxx}\\\text{Fourth Row:}\phantom{xxxxxxxxxxxxxx} 1\phantom{xxxx} 3\phantom{xxxx}3\phantom{xxxx}1\phantom{xxxxxxx}\\\text{Fifth Row:}\phantom{xxxxxxxxsxx} 1\phantom{xxxx} 4\phantom{xxxx} 6 \phantom{xxxx} 4\phantom{xxxx}1 \phantom{xxxxx}\\\text{Sixth Row:}\phantom{xxxxxxx}1\phantom{xxxx} 5\phantom{xxxx}10\phantom{xxxx} 10\phantom{xxxx}5\phantom{xxxx}1\phantom{xx}\\\text{Seventh Row:}\phantom{xxx} 1\phantom{xxxx} 6\phantom{xxxx}15\phantom{xxxx} 20\phantom{xxxx} 15\phantom{xxxx} 6 \phantom{xxxx}1\end{aligned} For the first item, we simply find the value located at the6+1 = 6$th or sixth row and$4 + 1 = 5units to the right. \begin{aligned}\text{First Row:}\phantom{xxxxxxxxxxxxxxxxxxxxxx}1\phantom{xxxxxxxxxxxxxx}\\\text{Second Row:} \phantom{xxxxxxxxxxxxxxxxxxx}1\phantom{xxxxx} 1\phantom{xxxxxxxxxxx}\\\text{Third Row:}\phantom{xxxxxxxxxxxxxxxxx} 1\phantom{xxxx} 2\phantom{xxxx} 1\phantom{xxxxxxxxx}\\\text{Fourth Row:}\phantom{xxxxxxxxxxxxxx} 1\phantom{xxxx} 3\phantom{xxxx}3\phantom{xxxx}1\phantom{xxxxxxx}\\\text{Fifth Row:}\phantom{xxxxxxxxsxx} 1\phantom{xxxx} 4\phantom{xxxx} 6 \phantom{xxxx} 4\phantom{xxxx}1 \phantom{xxxxx}\\\text{Sixth Row:}\phantom{xxxxxxx}1\phantom{xxxx} 5\phantom{xxxx}10\phantom{xxxx} 10\phantom{xxxx}{\color{red}5}\phantom{xxxx}1\phantom{xx}\\\text{Seventh Row:}\phantom{xxx} 1\phantom{xxxx} 6\phantom{xxxx}15\phantom{xxxx} 20\phantom{xxxx} 15\phantom{xxxx} 6 \phantom{xxxx}1\end{aligned} We’ll apply a similar process for b and c – note that for Pascal’s Triangle shown below, the value of\binom{4}{1}$is marked in green, and$\binom{6}{2}$is marked in blue. • For$\color{green}\binom{4}{1}$, find the second value in the fifth row. • Similarly, to find$\color{blue}\binom{6}{2}, find the third value of the seventh row. \begin{aligned}\text{First Row:}\phantom{xxxxxxxxxxxxxxxxxxxxxx}1\phantom{xxxxxxxxxxxxxx}\\\text{Second Row:} \phantom{xxxxxxxxxxxxxxxxxxx}1\phantom{xxxxx} 1\phantom{xxxxxxxxxxx}\\\text{Third Row:}\phantom{xxxxxxxxxxxxxxxxx} 1\phantom{xxxx} 2\phantom{xxxx} 1\phantom{xxxxxxxxx}\\\text{Fourth Row:}\phantom{xxxxxxxxxxxxxx} 1\phantom{xxxx} 3\phantom{xxxx}3\phantom{xxxx}1\phantom{xxxxxxx}\\\text{Fifth Row:}\phantom{xxxxxxxxsxx} 1\phantom{xxxx} {\color{green} 4}\phantom{xxxx} 6 \phantom{xxxx} 4\phantom{xxxx}1 \phantom{xxxxx}\\\text{Sixth Row:}\phantom{xxxxxxx}1\phantom{xxxx} 5\phantom{xxxx}10\phantom{xxxx} 10\phantom{xxxx}5 \phantom{xxxx}1\phantom{xx}\\\text{Seventh Row:}\phantom{xxx} 1\phantom{xxxx} 6\phantom{xxxx}{\color{blue} 15}\phantom{xxxx} 20\phantom{xxxx} 15\phantom{xxxx} 6 \phantom{xxxx}1\end{aligned} Hence, we have the following values: a.\binom{5}{4} = 6$b.$\binom{4}{1} = 4$c.$\binom{6}{2} = 15$Example 3 Expand the following expressions using the known Pascal’s Triangle. a.$(a – b)^3$b.$(4x + 3y)^4$Solution Since we’re working with expanding binomials to powers of$3$and$4$, we can begin by writing the first five rows of the Pascal’s Triangle. Highlight the$(3 + 1) =4$th and$(4 + 1) = 5\$th rows.
\begin{aligned}\text{First Row:}\phantom{xxxxxxxxxxxxxxxxxxxxxx}1\phantom{xxxxxxxxxxxxxx}\\\text{Second Row:} \phantom{xxxxxxxxxxxxxxxxxxx}1\phantom{xxxxx} 1\phantom{xxxxxxxxxxx}\\\text{Third Row:}\phantom{xxxxxxxxxxxxxxxxx} 1\phantom{xxxx} 2\phantom{xxxx} 1\phantom{xxxxxxxxx}\\\color{blue}\text{Fourth Row:}\phantom{xxxxxxxxxxxxxx} 1\phantom{xxxx} 3\phantom{xxxx}3\phantom{xxxx}1\phantom{xxxxxxx}\\\color{green}\text{Fifth Row:}\phantom{xxxxxxxxsxx} 1\phantom{xxxx} 4\phantom{xxxx} 6 \phantom{xxxx} 4\phantom{xxxx}1 \phantom{xxxxx}\end{ |
# Pairs of Angles
Pairs of angles are discussed here in this lesson.
1. Complementary Angles:
Two angles whose sum is 90° (that is, one right angle) are called complementary angles and one is called the complement of the other.
Here, ∠AOB = 40° and ∠BOC = 50°
Therefore, ∠AOB + ∠BOC = 90°
Here, ∠AOB and ∠BOC are called complementary angles.
∠AOB is complement of ∠BOC and ∠BOC is complement of ∠AOB.
For Example:
(i) Angles of measure 60° and 30° are complementary angles because 60° + 30° = 90°
Thus, the complementary angle of 60° is the angle measure 30°. The complementary angle angle of 30° is the angle of measure 60°.
(ii) Complement of 30° is → 90° - 30° = 60°
(iii) Complement of 45° is → 90° - 45° = 45°
(iv) Complement of 55° is → 90° - 55° = 35°
(v) Complement of 75° is → 90° - 75° = 15°
Working rule: To find the complementary angle of a given angle subtract the measure of an angle from 90°.
So, the complementary angle = 90° - the given angle.
2. Supplementary Angles:
Two angles whose sum is 180° (that is, one straight angle) are called supplementary angles and one is called the supplement of the other.
Here, ∠PQR = 50° and ∠RQS = 130°
∠PQR + ∠RQS = 180° Hence, ∠PQR and ∠RQS are called supplementary angles and ∠PQR is
supplement of ∠RQS and ∠RQS is supplement of ∠PQR.
For Example:
(i) Angles of measure 100° and 80° are supplementary angles because 100° + 80° = 180°.
Thus the supplementary angle of 80° is the angle of measure 100°.
(ii) Supplement of - 55° is 180° - 55° = 125°
(iii) Supplement of 95° is 180° - 95° = 85°
(iv) Supplement of 135° is 180° - 135° = 45°
(v) Supplement of 150° is 180° - 150° = 30°
Working rule: To find the supplementary angle of a given angle, subtract the measure of angle from 180°.
So, the supplementary angle = 180° - the given angle.
Two non – overlapping angles are said to be adjacent angles if they have:
(a) a common vertex
(b) a common arm
(c) other two arms lying on opposite side of this common arm, so that their interiors do not overlap.
In the above given figure, ∠AOB and ∠BOC are non – overlapping, have OB as the common arm and O as the common vertex. The other arms OC and OA of the angles ∠BOC and ∠AOB are an opposite sides, of the common arm OB.
Hence, the arm ∠AOB and ∠BOC form a pair of adjacent angles.
4. Vertically Opposite Angles:
Two angles formed by two intersecting lines having no common arm are called vertically opposite angles.
In the above given figure, two lines $$\overleftrightarrow{AB}$$ and $$\overleftrightarrow{CD}$$ intersect each other at a point O.
They form four angles ∠AOC, ∠COB, ∠BOD and ∠AOD in which ∠AOC and ∠BOD are vertically opposite angles. ∠COB and ∠AOD are vertically opposite angle.
∠AOC and ∠COB, ∠COB and ∠BOD, ∠BOD and ∠DOA, ∠DOA and ∠AOC are pairs of adjacent angles.
Similarly we can say that, ∠1 and ∠2 form a pair of vertically opposite angles while ∠3 and ∠4 form another pair of vertically opposite angles.
When two lines intersect, then vertically opposite angles are always equal.
∠1 = ∠2
∠3 = ∠4
5. Linear Pair:
Two adjacent angles are said to form a linear pair if their sum is 180°.
These are the pairs of angles in geometry.
Interior and Exterior of an Angle.
Measuring an Angle by a Protractor.
Types of Angles.
Pairs of Angles.
Bisecting an angle.
Construction of Angles by using Compass.
Worksheet on Angles.
Geometry Practice Test on angles. |
# PRACTICAL PROBLEMS USING PYTHAGOREAN THEOREM
Practical Problems Using Pythagorean Theorem
Here we are going to see some practice problems on pythagorean theorem.
## Practical Problems Using Pythagorean Theorem
Question 1 :
5 m long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4 m high. If the foot of the ladder is moved 1.6 m towards the wall, then find the distance by which the top of the ladder would slide upwards on the wall.
Solution :
In triangle ABC,
AC2 = AB2 + BC2
52 = AB2 + 42
AB2 = 25 - 16
AB2 = 9
AB = 3
DB = 3 - 1.6
DB = 1.4
In triangle DBE,
ED2 = EB2 + DB
52 = EB2 + (1.4)
EB2 = 25 - 1.96
EB = √23.04
EB = 4.8
EC = EB - BC
= 4.8 - 4
EC = 0.8 m
Hence the required distance is 0.8 m.
Question 2 :
The perpendicular PS on the base QR of a tiranlge PQR intersects QR at S, such that QS = 3 SR. Prove that 2PQ2 = 2PR2 + QR2
Solution :
QR = QS + RS
QR = 3RS + RS
QR = 4RS
RS = QR/4
By applying pythagorean theorem in a triangle PQS, we get
PQ2 = PS2 + QS2
PS= PQ2 - QS -----(1)
By applying pythagorean theorem in a triangle PRS, we get
PR2 = PS2 + SR2
PS= PR- SR2 -----(2)
PQ2 - QS = PR- SR2
PQ2 - QS = PR- SR2
PQ2 = PR- SR2 + QS
PQ2 = PR- SR2 + (QR - RS)
PQ2 = PR- SR2 + QR2 + RS2 - 2QR x RS
PQ2 = PR+ QR2 - 2QR x RS
PQ2 = PR+ QR2 - 2 QR x (QR/4)
PQ2 = PR+ QR2 - (QR2/2)
PQ2 = (2PR+ 2QR2 - QR2)/2
2PQ2 = 2PR+ QR2
Question 3 :
In the adjacent figure, ABC is a right angled triangle with right angle at B and points D, E trisect BC. Prove that 8AE2 = 3AC2 + 5AD2
Solution :
AE² = AB² + (2x)²
AE² = AB² + 4x²
In Δ ABC,
AC² = AB² + BC²
AC² = AB + (3x)²
AC² = AB² + 9x²
Now,
3AC² + 5AD² = 3(AB² + 9x²) + 5(AB² + x²)
= 8AB² + 32x²
= 8(AB² + 4x²)
= 8AE²
Hence proved.
After having gone through the stuff given above, we hope that the students would have understood, "Pythagorean Theorem Word Problems for Grade 10".
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# 2008 UNCO Math Contest II Problems/Problem 9
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
Let $C_n = 1+10 +10^2 + \cdots + 10^{n-1}.$
(a) Prove that $9C_n = 10^n -1.$
(b) Prove that $(3C_3+ 2)^2 =112225.$
(c) Prove that each term in the following sequence is a perfect square: $$25, 1225, 112225, 11122225, 1111222225,\ldots$$
## Solution
(a) We know that $C_n$ is a geometric series, so we can define it explicitly as follows
$C_n=\frac{10^n-1}{9}$
multiplying both sides by 9 yields our answer.
(b) We have
$(3*111+2)^2=335^2$,
yielding $112,225$.
(c) We say that the nth member of the sequence equals $(3*C_n+2)^2$. Expanding yields
$(3*\frac{10^n-1}{9}+2)^2$,
$=(\frac{10^n+5}{3})^2$,
$=\frac{10^{2n}}{9}+\frac{10^{n+1}}{9}+\frac{25}{9}$.
Dividing each term separately, we know that the first term will add $2n$ $1$s and $\frac{1}{9}$, the second term will add $n+1$ $1$s and $\frac{1}{9}$, and the third will add $\frac{25}{9}$, giving
$\overbrace{11...11}^{2n}+\overbrace{11...11}^{n+1}+\frac{27}{9}$,
$\overbrace{11...11}^{n-1}\overbrace{22...22}^{n}5$,
which is exactly what we wanted.
(a) $\frac{10^n-1}{9}$ (b) $112,225$ (c) $11,122,225$
## Solution 2
(a) $9=10-1$. Multiply these two binomials and we have reach our answer (remember the formula -- it's like Difference of Cubes)
(b)$C_3=111$. The original expression is equal to $\boxed{112225}$. (Just brute force this out).
(c) Now notice that each term in the sequence is $(3C_n)^2$. As seen in part (a), we see that $C_n=\frac{10^n-1}{9}$. Follow Solution 1 above.
~hastapasta |
Integrating a Function as a Power Series
Integrating a Function as a Power Series
Why there is no need to decrease n by 1 in the case of integration, while it increases 1 for n in the case of differentiation in this theorem?
Does using integration for representing power series requires n-1 as the starting point?
• $\begingroup$ Can you clarify your question? When integrating or differentiating a power series, you do so term by term following the usual rules. The result is the power series expansion for $\int f$ or $f'(x)$, respectively. $\endgroup$– gabeNov 22, 2016 at 5:05
• $\begingroup$ Look at the derivative and integral of $x^n$. That might help. Although you are a CS student, I would suggest learn calculus nicely…it helps a lot. $\endgroup$ Nov 22, 2016 at 5:10
$$f'(x) = \sum_{n=0}^{\infty} \frac{d}{dx} \left[ c_n (x-a)^n \right]$$
For $n=0$, we have $c_n (x-a)^n = c_0 (x-a)^0 = x_0 \cdot 1 =c_0$. Thus $\frac{d}{dx} c_0 =0$ because the derivative of a constant is always zero. Therefore, we can rewrite the above as: $$f'(x) = \sum_{n=0}^{\infty} \frac{d}{dx} \left[ c_n (x-a)^n \right] =\frac{d}{dx}\left[c_0 \right]+ \sum_{n=1}^{\infty} \frac{d}{dx}\left[c_n (x-a)^n \right] = 0+ \sum_{n=1}^{\infty} n\cdot c_n(x-a)^{n-1}.$$
In other words, the book is being lazy to some extent and just ignoring the $0$th term in the power series corresponding to the derivative because it is equal to $0$. There isn’t any “need” to “increase the dummy index by $1$” for differentiation. |
How do you solve and write the following in interval notation:−12 <3 − 3x ≤ 15?
Jun 29, 2017
See a solution process below:
Explanation:
First, subtract $\textcolor{red}{3}$ from each segment of the system of inequalities to isolate the $x$ term while keeping the system balanced:
$- \textcolor{red}{3} - 12 < - \textcolor{red}{3} + 3 - 3 x \le - \textcolor{red}{3} + 15$
$- 15 < 0 - 3 x \le 12$
$- 15 < - 3 x \le 12$
Now, divide each segment by $\textcolor{b l u e}{- 3}$ to solve for $x$ while keeping the system balanced. However, because we are multiplying or dividing inequalities by a negative number we must reverse the inequality operators:
$\frac{- 15}{\textcolor{b l u e}{- 3}} \textcolor{red}{>} \frac{- 3 x}{\textcolor{b l u e}{- 3}} \textcolor{red}{\ge} \frac{12}{\textcolor{b l u e}{- 3}}$
$5 \textcolor{red}{>} \frac{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{- 3}}} x}{\cancel{\textcolor{b l u e}{- 3}}} \textcolor{red}{\ge} - 4$
$5 \textcolor{red}{>} x \textcolor{red}{\ge} - 4$
Or
$x < 5$ and $x \ge - 4$
Or, in interval notation:
$\left[- 4 , 5\right)$ |
# Solve the inequality x-5#3x+10+
## x-5\leq +3x+10
Go!
1
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8
9
0
x
y
(◻)
◻/◻
2
e
π
ln
log
lim
d/dx
d/dx
>
<
>=
<=
sin
cos
tan
cot
sec
csc
asin
acos
atan
acot
asec
acsc
sinh
cosh
tanh
coth
sech
csch
asinh
acosh
atanh
acoth
asech
acsch
$x\geq \frac{1}{2}-\frac{15}{2}$
## Step by step solution
Problem
$x-5\leq +3x+10$
1
Grouping terms
$-3x-5+x\leq 10+$
2
Adding $-3x$ and $x$
$-2x-5\leq 10+$
3
Moving the term $-5$ to the other side of the inequation with opposite sign
$-2x\leq 5+10+$
4
Add the values $10$ and $5$
$-2x\leq 15$
5
Multiply both sides of the inequality by $-1$, reversing the sign
$2x\geq \left(15\right)\left(-1\right)$
6
Multiply $\left(15+\right)$ by $-1$
$2x\geq -15-1$
7
Divide both sides of the inequation by $2$
$x\geq \frac{-15-1}{2}$
8
Split the fraction $\frac{-15+-1}{2}$ in two terms with same denominator
$x\geq \frac{-1}{2}+\frac{-15}{2}$
9
Divide $-15$ by $2$
$x\geq \frac{-1}{2}-\frac{15}{2}$
10
Apply the formula: $\frac{b\cdot a}{c}$$=b\frac{a}{c}$, where $a=-1$, $b=$ and $c=2$
$x\geq -\frac{15}{2}-1\cdot \left(-\frac{1}{2}\right)$
11
Multiply $-\frac{1}{2}$ times $-1$
$x\geq \frac{1}{2}-\frac{15}{2}$
$x\geq \frac{1}{2}-\frac{15}{2}$
Polynomials
0.23 seconds
74 |
Home Fashion & Lifestyle How to solve a Rubik’s cube in 1 minute? A Beginner’s Guide
# How to solve a Rubik’s cube in 1 minute? A Beginner’s Guide
0
Everyone in the world knows what a Rubik’s cube is, and it’s a dream for everyone to solve it once. It has more than 43 quintillion possible permutations which makes it impossible to solve with some random moves.
Yet it is not that difficult to solve if you know the proper method.
There are many different types of method out there, which enables people to solve it within even 5 seconds. It’s what no one believes until they see it with their own eyes.
So, we will be discussing the easiest method for a beginner to solve the Rubik’s cube. This method uses very less number of algorithms, which you can remember very easily.
If you want to learn an Advanced guide, then go to my other page: How to solve a Rubik’s cube in 30 seconds? Advanced guide. Or “How to solve a Rubik’s cube in 10 seconds? The World Record Guide”
## Concept used to solve Rubik’s Cube
Now you have to believe in the fact that the cube cannot be solved by just making any random moves. If you want to solve it, you need to use a proper method.
So now let’s just understand how the methods work.
The cube is solved layer by layer. You can start with any colour and then end with the colour opposite to it. We’ll take white colour as our first layer.
ALSO READ : “Tricks of Rubik’s Cube (Patterns) – Impressing your friends”
## Standard Notationsof rubik’s cube
Now that you have understood the concept of solving the Rubik’s cube, you need to know the notations.
Notations are basically the commands which lets you know which move to make. Algorithms are formed by the sequence of notations. There are total 6 sides in the cube, which we can rotate freely. Therefore, we get 6 notations from this.
From the picture above you can see the notations along with their respective move. For example, if you need to perform the notation R, then you need to rotate the right-side vertical layer by 90° clockwise, and if you need to perform Ri, then you need to rotate the right-side vertical layer by 90° anti-clockwise.
In the notations “i” or “ ” (apostrophe symbol) denotes anti-clockwise rotation and is called “inverse”. When the notation is written like R2, then you need to perform notation R twice, means rotate right layer by 180°.
Similarly, all the notations work, you just need to remember the name of the side. From the image, you can notice that only upper layer (U) and down layer (D) are horizontal, and the others are vertical layer.
So, when you get a whole algorithm written like: R Ui Ri U
You need to perform all the notations in the sequence it is written, first R then Ui, then Ri, and then lastly U.
Now we know all the basics of the Rubik’s cube, let us start with the solution.
## Making the Cross
We need to select one colour for the first layer, and here we will be taking white as the first layer. Making a cross is not easy for everyone, because I can’t give you any fixed algorithm which will automatically make the cross, you need to hit and try.
Let’s start:
• First you need to hold the cube with white center facing up.
• Now find any one edge piece with white color, the other color may be green, orange, blue, or red. Here let us take the edge piece with white and blue color. Then bring that piece to the side between the white centerpiece and blue centerpiece. You will be getting either of the two cases.
• If you get the first case, then you are good to go with other edge pieces.
• If you get the second case, then you need to perform an algorithm. Face the cube, with white centerpiece upwards, and the desired edge piece to the right side, then perform the algorithm: Ri U Fi Ui.
• Do the same step with other 3 edge pieces. Remember matching the colour of the edge pieces adjacently with the centrepieces of white and other four colours.
• After making the cross you can see something like this. If you get this, then you are good to go.
## Placing the corners
In this step, we need to push the corner white pieces to their respective locations, which will ultimately complete the first white layer.
• Hold the cube with a white centerpiece facing up.
• Now you need to find the corner pieces with white color. These corner pieces will have white and other 2 colors.
• Here let us take the corner piece with white, blue, and red color, now we need to bring it to the bottom layer, just below the original position of that corner piece, just like in the image.
• If the corner piece is in the upper layer, then you need to perform this algorithm to bring it down to the bottom layer, hold the cube with white facing up and the corner to the right, and perform: Ri D R
• Now that we have the corner piece in the bottom layer, perform the algorithm: Ri D R Di
Repeat this algorithm until the corner piece is aligned correctly.
• Repeat these steps with all the white corner pieces. after placing all the corners, you can see the cube will look like this. Now you’re good to go to another Step.
Now we are done with the First white layer, and I know for the beginners it is quite confusing, but you’ll better with practice.
Note: After the second layer is complete now remember you have to hold the cube with yellow centerpiece facing upwards until the cube is solved.
## The Second Layer
Now in the second layer, we need to fix just 4 edge pieces. These pieces have 2 adjacent colours of the four colours green, orange, blue and red.
Hold the cube with yellow centrepiece facing upwards, and find the non-yellow edge pieces.
• Here let us take the edge piece with blue and red colour, now you need to see which colour is facing front, and move the piece accordingly, by rotating the upper layer, to its respective side making a vertical line. Like in the picture the edge piece with blue colour facing front is aligned with the blue centrepiece to make a vertical line.
• Now the upward facing colour will decide whether the piece will move to left or right. Like in the picture red colour is facing up, and in the cube the red colour is to the right of the blue, so we must move the piece to the right middle layer between the blue and the red. It can be done by holding the cube with the yellow centrepiece facing up and the desired edge piece facing front, and perform the algorithm: U R Ui Ri Ui Fi U F
• Now in the other case with edge piece moving to the left like in the picture, you need to perform another set of algorithms: Ui Li U L U F Ui Fi
• In case the edge piece is already in the middle layer but in wrong position, then you need to perform either of the two algorithms, to replace it with a yellow edge piece, which will ultimately bring the edge piece to the upper layer. Now you can place the edge piece in its desired location using the algorithm.
• Perform the same procedure with the other 3 edge pieces and this will ultimately complete the middle layer and look like this.
## Orienting Last Layer
So, here comes the last layer, the yellow layer. Now in this step we need to orient all the yellow pieces to face up.
### The Yellow Cross
First, we will make a yellow cross. Don’t worry you need not match the edge with the adjacent side colors, we will do that in the Permutating step.
Now just look at the yellow center and the edge pieces, ignore the corner pieces for a second, you will get one of the four cases shown in the picture (you need to rotate the upper layer to match with the Cases exactly as shown).
Case 1. Proceed to the “Orienting the Yellow Corners” step.
Case 2. Perform the algorithm: F U R Ui Ri Fi. Now match the position with Case 3 and Case 4 and proceed accordingly.
Case 3. Perform the algorithm: F U R Ui Ri Fi.
Case 4. Perform the algorithm: F R U Ri Ui Fi.
### Orienting the Yellow Corners
Now that we have the Yellow cross, we need to orient the corners to complete the layer orientation. After making the cross, you will get many cases like
Case 1. None of the corners has a yellow color facing up. Then rotate the upper layer to get the left facing yellow corner piece in the left, as shown in the upper picture, and perform the algorithm given below.
Case 2. One Corner has the yellow colour facing up. Then hold the cube as shown in the middle picture and perform the algorithm given below.
Case 3. Two corners have the yellow color facing up. Then rotate the upper layer to get the front-facing yellow corner piece in the left, as shown in the bottom picture, and perform the algorithm given below.
R U Ri U R U U Ri
After performing the algorithm once, match the positions with the cases above and perform the algorithm accordingly. Repeat it until all the yellow corners are oriented. The cube will look like this. Now proceed to the permutation step.
Note: There are no such cases of three corner pieces with yellow color facing up, if such case arises with you, then it’s obvious that any one of the 8 corner pieces is turned. To fix this you need to turn that corner piece to face the yellow color up.
## Permutating the Yellows
Now that we have all the yellows facing up, we need to position the yellow pieces in their correct locations. For this we have to first position the corners and then the edges.
### Corners
First we need to observe the correctly positioned two corner pieces, it may be either adjacent like A and B or diagonally placed lke A and D, or B and C. You need to rotate the upper layer to check the correctly positioned pieces.
Case 1. If the two adjacent corners are placed corectly, then move them to the back then perform the algorithm
Ri F (Ri B B R) Fi (Ri B B R) R Ui
Case 2. If the two diagonal corners are placed correctly then perform the above algorithm once then look for two adjacent corners fixed correctly, place them back and perform the same algorithm again. It will fix all the corners in their correct position.
After successful corner permutation, cube will look like this.
### Edge pieces
Now comes the last part, fixing the edge pieces.
Here you will see either one of the edge piece is placed correctly or none of them is placed correctly, and if lucky enough all the edge pieces may be placed correctly, which means the cube has been solved.
Case 1. One edge piece is correctly placed. Then perform the algorithm given below accordingly, while holding the cube with the correct piece at the back side of the cube.
Case 2. None of the edge pieces is placed correctly. In this case, perform any of the algorithm given below then follow Case 1.
Algorithm to rotate EGF clockwise:
F F U L Ri F F Li R U F F
Algorithm to rotate EFG anti-clockwise:
F F Ui L Ri F F Li R Ui F F
Congratulations, finally we have solved the cube using just 11 algorithms.
Learning these 11 algorithms is not too hard, you need not learn the notations in your mind, you just need to practice and practice, your muscle memory will automatically memorise the steps in sequence.
If you want to learn all the algorithms of the Rubik’s cube then go to AlgDb
The beginners guide will allow you to solve Rubik’s cube in minimum 1 minute. To learn solving it within 30 seconds, head towards my advanced guide How to solve a Rubik’s cube under 30 seconds? Advanced guide |
# Quadratic Formula Completing And Not Completing the Square
### Introduction
The importance of the investigations connected with the expression $ax^{2} + bx + c,$ can hardly be overrated, at least to those students who pursue mathematics to any extent. In the higher branches, great familiarity with these results is indispensible. A. De Morgan (1806-1871) On the Study and Difficulties of Mathematics Dover, 2005, p. 149
By the Fundamental Theorem of Algebra, every polynomial $P_{n}(x)$ of degree $n$ with complex coefficients has $n$ (perhaps repeated) complex roots. Here we are concerned with the case of $n = 2$ and the polynomials of second degree. Such polynomials are in the form
$P(x)=ax^{2} + bx + c,$
where $a$ is assumed not to be zero: $a\ne 0.$ A quadratic polynomial is assured to have two roots, say, $x_{1}$ and $x_{2},$ such that it admits a factorization
$P(x)=ax^{2} + bx + c =a(x- x_{1})(x- x_{2}).$
The two roots can be found by means of what is known as a Quadratic Formula:
Below we offer several derivations of that formula.
### Example
Solve $2x^{2} - 7x + 3 = 0,$ with $a = 2,$ $b = -7,$ $c = 3.$ The quadratic formula gives:
\begin{align} x &= (7 \pm \sqrt{(-7)^{2} - 4\cdot 2\cdot 3}) / 2\cdot 2\\ &= (7 \pm \sqrt{49 - 24}) / 4\\ &= (7 \pm \sqrt{25}) / 4\\ &= (7 \pm 5) / 4\\ \end{align}
producing two roots, say, $x_{1} = (7 + 5) / 4 = 3$ and $x_{2} = (7 - 5) / 4 = 1/2.\;$ It can be also verified that
$2x^{2} - 7x + 3 = 2(x - 3)(x - 1/2).$
### First Derivation
The formula is often derived by the process of completing a square, based on a well known (and easily verifiable) algebraic identity
$(u+ v)^{2} =u^{2} + 2uv+ v^{2}.$
So, for example, if we can identify an expression as $u^{2} + 2uv,$ then to complete the squarewe need to add to it $v^{2}$ which converts the sum of two terms $u^{2} + 2uv$ into a single square $(u + v)^{2}.$
We start with the quadratic equation $ax^{2} + bx + c = 0$ and move the free termto the right:
$ax^{2} + bx = -c.$
The left hand side can be seen to be an incomplete square in the form $u^{2} + 2uv$. Indeed,
\begin{align} ax^{2} + bx &= (\sqrt{a} x)^{2} + 2(\sqrt{a} x)(b / 2\sqrt{a}) \\ &= (\sqrt{a} x)^{2} + 2(\sqrt{a} x)(b / 2\sqrt{a}) + (b / 2\sqrt{a})^{2} - (b / 2\sqrt{a})^{2}. \end{align}
(Here $u= \sqrt{a} x$ and $v=b/ 2\sqrt{a}$). We can continue to transform the quadratic equation step by step:
1. $ax^{2} + bx + c = 0,$
2. $ax^{2} + bx = -c,$
3. $(\sqrt{a} x)^{2} + 2(\sqrt{a} x)(b / 2\sqrt{a}) + (b / 2\sqrt{a})^{2} = - c + (b / 2\sqrt{a})^{2},$
4. $(\sqrt{a} x + b / 2\sqrt{a})^{2} = - c + (b / 2\sqrt{a})^{2},$
5. $(2ax + b)^{2} = - 4ac+ b^{2},$
6. $2ax + b = \pm \sqrt{b^{2} - 4ac},$
7. $2ax = \pm \sqrt{b^{2} - 4ac} - b,$
8. $x = (-b \pm \sqrt{b^{2} - 4ac}) / 2a.$
Note that the derivation may be simplified somewhat typographically by first converting the equation to $\displaystyle x^2+\frac{b}{a}x+\frac{c}{a}=0.$
### Second Derivation
There is a much simpler way to derive the quadratic formula. Multiply the equation $ax^{2} + bx + c = 0$ by $4a$ to obtain $4a^{2}x^{2} + 4abx + 4ac= 0.$ But then
\begin{align} 4a^{2}x^{2} + 4abx + 4ac &= (2ax)^{2} + 2\cdot (2ax)(b) + 4ac\\ &= (2ax + b)^{2} - b^{2} + 4ac.\\ \end{align}
In other words, $(2ax + b)^{2} =b^{2} - 4ac.$ Taking the square root,
$2ax + b = \pm \sqrt{b^{2} - 4ac}.$
Rearranging the terms and dividing by $2$ adirectly leads to the quadratic formula.
### Third Derivation
I am grateful to Michael Brozinsky for bringing this derivation to my attention.
The factorization $a(x-x_{1})(x-x_{2})$ of a quadratic polynomial $ax^2+bx+c,$ with two roots $x_1$ and $x_2,$ leads Viète's formulas: $a(x_1+x_2)=-b$ and $a(x_1\cdot x_2)=c.$ A change of variables, say, $\displaystyle z=x+\frac{b}{2a},$ yields two numbers $\displaystyle z_{1}=x_{1}+\frac{b}{2a}$ and $\displaystyle z_{2}=x_{2}+\frac{b}{2a},$ that add up to $0:$ $z_1+z_2=0,$ implying they are the roots of a quadratic equation $z^2 + z_1\cdot z_2=0.$ So we have
\begin{align}\displaystyle 0 &= z^2 + z_1\cdot z_2 \\ &= z^2 + (x_{1}+\frac{b}{2a})(x_{2}+\frac{b}{2a}) \\ &= z^2 + x_{1}x_{2} + (x_{1}+x_{2})\frac{b}{2a} + \frac{b^2}{4a^2} \\ &= z^2 + \frac{c}{a} - \frac{b}{a}\frac{b}{2a} + \frac{b^2}{4a^2} \\ &= z^2 + \frac{4ac - 2b^2 + b^2}{4a^2} \\ &= z^2 + \frac{4ac - b^2}{4a^2}. \end{align}
In other words, $\displaystyle z^2 = \frac{b^2 - 4ac}{4a^2}$ and, as a consequence, $\displaystyle z_{1,2} = \frac{\pm\sqrt{b^2 - 4ac}}{2a}$ and, finally, with the back substitution,
$\displaystyle x_{1,2} = \frac{\pm\sqrt{b^2 - 4ac}}{2a}-\frac{b}{2a}=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}.$
### Fourth Derivation
Let it be required to solve the equation $ax^2+bx+c=0.\;$ We proceed in several steps:
\begin{align} &ax^2+c=-bx,\\ &(ax^2+c)^2=(-bx)^2,\\ &a^2x^4+2acx^2+c^2=b^2x^2,\\ &(a^2x^4+2acx^2+c^2)-4acx^2=b^2x^2)-4acx^2,\\ &a^2x^4-2acx^2+c^2=(b^2-4ac)x^2,\\ &(ax^2-c)^2=(b^2-4ac)x^2,\\ &ax^2-c=\pm\sqrt{b^2-4ac}x. \end{align}
Adding to this $ax^2+c=-bx\;$ we get $2a^2x^2=(-b\pm\sqrt{b^2-4ac})x\;$ such that $\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$
References
1. H. Heaton, A Method of Solving Quadratic Equations, The American Mathematical Monthly, Vol. 3, No. 10 (Oct., 1896), pp. 236-237
### Fifth Derivation
The strating point is the equation $x^2+x=a$ where the left-hand side is the product of two factors that differe by $1.\;$ The two factors $\displaystyle\left(a+\frac{1}{2}\right)\left(a-\frac{1}{2}\right)=a^2-\frac{1}{4}\;$ give such an example and can be amended to satisfy the entire requirement:
$\displaystyle\left(\sqrt{a+\frac{1}{4}}+\frac{1}{2}\right)\left(\sqrt{a+\frac{1}{4}}-\frac{1}{2}\right)=a.$
Here the smaller of the two factors, viz., $\displaystyle x=\sqrt{a+\frac{1}{4}}-\frac{1}{2}\;$ is the root of $x^2+x=a.$
The more general equation - $x^2+bx=a\;$ - is solved similarly:
\displaystyle\begin{align} &x(x+b)=a,\\ &\sqrt{a}(\sqrt{a}+b)\ne a,\\ &\left(\sqrt{a}+\frac{b}{2}\right)\left(\sqrt{a}-\frac{b}{2}\right)=a-\frac{b^2}{4},\\ &\left(\sqrt{a+\frac{b^2}{4}}+\frac{b}{2}\right)\left(\sqrt{a+\frac{b^2}{4}}-\frac{b}{2}\right)=a,\\ &x=\sqrt{a+\frac{b^2}{4}}-\frac{b}{2}. \end{align}
Getting the larger root is left to the reader.
References
1. J. W. Cirul, A Method of Solving Quadratic Equations, The American Mathematical Monthly, Vol. 44, No. 7 (Aug. - Sep., 1937), pp. 462-463
### Sixth Derivation
This derivation has been suggested by Uzair Baig. It is simple but not elementary in the sense that it assumes knowledge of derivatives.
Let $f(x)=a x^2 + b x +c.\;$ Then $f'(x^*) =0\;$ where $\displaystyle x^*= -\frac{b}{2a}.\;$ Define $g(x)=f(x+x^*),\;$ i.e., $\displaystyle g(x)= ax^2 + \left(c- \frac{b^2}{4 a}\right).\;$ Now, $g(x)=0\;$ is easily solved, giving two roots, $\displaystyle x_{1,2}= \pm\frac{\sqrt{b^2-4ac}}{2a}.\;$ This tells us that $f(x)=0\;$ for $\displaystyle x_{1,2}= \pm\frac{\sqrt{b^2-4ac}}{2a}-\frac{b}{2a}\;$ which is the quadratic formula.
### Discriminant
Obviously, the expression $D=b^{2} - 4ac$ that appears in the quadratic formula under the square root plays an important role in solving quadratic equations. Because of its importance it was given a name. $D=b^{2} - 4ac$ is called the discriminant of the quadratic equation $ax^{2} + bx + c = 0.$ As the quadratic formula shows, there are three possible cases:
1. $D\gt 0.$ In this case, the equation has two distinct real roots.
2. $D= 0.$ In this case, the two roots coalesce into one, which is called a double root implying that the are still two roots albeit equal.
3. $D\lt 0.$ In this case, the equation has to complex roots which, for real a, b, c, are conjugate.
### An Aside
For quadratic polynomials with real coefficients, the roots come in pairs as manifest by the "$\pm$" symbol. For the polynomials with real coefficients but complex roots, the roots in such pairs are conjugate. (An interactive illustration of the behavior of the roots as the function of the coefficients is available elsewhere.) We now know that the same holds for all polynomials with real coefficients of degree higher than $1.$ This means that any polynomial with real coefficients admits a decomposition into a product of linear and quadratic terms. However, this fact was not always known of course. In the 17th century, this was still a conjecture. The Fundamental Theorem of Algebra is the product of the 19th century. According to [W. Dunham, p. 109],
No less an authority than Leibniz doubted that every real polynomial can be factored into real linear and/or real quadratic pieces. Worse, Nicolaus Bernoulli (1687-1759) claimed to have found a counter example - namely, $x^{4} - 4x^{3} + 2x^{2} + 4x + 4$ - that could not be so factored. If he were correct, the game was over: the fundamental "theorem" of algebra would have been automatically disproved.
Euler, rising to the defence of this conjecture, showed that Bernoulli was wrong. In a 1742 letter to Christian Goldbach, he factored the supposedly unfactorable, splitting the quartic into the product of quadratics
$x^{2} - (2 + \sqrt{4 + 2\sqrt{7}})x + (1 + \sqrt{4 + 2\sqrt{7}} + \sqrt{7})$
and
$x^{2} - (2 - \sqrt{4 + 2\sqrt{7}})x + (1 - \sqrt{4 + 2\sqrt{7}} + \sqrt{7}).$
This factorization appears to lie somewhere between miraculous and preposterous. It looks ever so much like a misprint - but it is perfectly correct.
Every one is welcome to verity the correctness of Euler's feat.
References
1. W. Dunham, Euler: The Master of Us All, MAA, 1999
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Subtracting Negative Numbers
```
Date: 05/11/98 at 01:22:06
From: William
Subject: Subtracting Negative And Positive Numbers, like 3-(-4)=?
How do you subtract a negative number from a positive number or a
positive number from a negative number? I got stuck when I saw it. I
can do this kind of problem:
3 - (-4) = 3 + 4 = 7
That one was easy. The opposite makes me stuck.
```
```
Date: 05/18/98 at 11:51:02
From: Doctor Jeremiah
Subject: Re: Subtracting Negative And Positive Numbers, like 3-(-4)=?
Hi William:
The easy way to remember how to do this is with a number line. You
don't have to have a real one, you just have to imagine one in your
mind. A number line looks like this:
+---+---+---+---+---+---+---+---|---+---+---+---+---+---+---+---+---+
-8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9
When you need to answer a problem, go through these steps.
Step 1:
You already know how to change -(-4) into + 4. That is step one.
Step 2:
Now you have something that looks like:
3 + 4, 3 - 4, -3 + 4 or -3 - 4
Here is where you use the number line. Take the first number and
figure out where it goes on the number line.
Step 3:
Then move from that spot however far the second number says to go. The
direction you move is left if the second number is being subtracted
and right if the second number is being added.
Example 1: -3 - (-4)
Step 1: Change the -(-4) into a + 4:
-3 - (-4) = -3 + 4
Step 2: Start at -3
O
+---+---+---+---+---+---+---+---+---|---+---+---+---+---+---+---+---+
-8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9
Step 3: We are adding four because the expression is -3 + 4 so we
want to move to the right 4 numbers:
O--- --- --- -->O
+---+---+---+---+---+---+---+---+---|---+---+---+---+---+---+---+---+
-8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9
Example 2: -3 - 4
Step 1: There is no double negative so we don't do anything.
Step 2: Start at -3
O
+---+---+---+---+---+---+---+---+---|---+---+---+---+---+---+---+---+
-8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9
Step 3: We are subtracting four because the expression is -3 - 4 so
we want to move to the left 4 numbers:
O<-- --- --- ---O
+---+---+---+---+---+---+---+---+---|---+---+---+---+---+---+---+---+
-8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9
See how that works? I usually just imagine a number line in my head.
It works great for small numbers. But what about large numbers?
When you do 34 - 86, you can't draw a number line big enough. You have
to think: what if I start at 34 and subtract 86? I need to move left.
If I move left 86 I will cross zero because 86 is bigger than 34. How
far past zero will I go? Well, the first 34 numbers out of 86 are on
the right side of the zero and the rest (86 - 34) are on the left side
of the zero. So I am on the left side of zero by 86 - 34. Being on the
left side in the end means that the answer is negative. So the answer
to 34 - 86 is 52 on the left side of zero or -52.
When you do -34 - (-86), again, you can't draw a number line big
enough, so change -(-86) to + 86. Then you have to think: what if I
start at -34 and add 86? I start on the left side of the zero and I
need to move right. If I move right 86 I will cross zero because 86 is
bigger than 34. How far past zero will I go? Well, the first 34
numbers out of 86 are on the left side of the zero and the rest
(86 - 34) are on the right size of the zero, so I am on the right side
of zero by 86 - 34. Being on the right side in the end means that the
answer is positive. So the answer to -34 - (-86) is 52 on the right
side of zero or 52.
Does that help explain it? If you need more help please write me back
and I will give you more details.
-Doctor Jeremiah, The Math Forum
Check out our web site! http://mathforum.org/dr.math/
```
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Elementary Subtraction
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Question
# A right triangle ABC right angled at A with lengths of perpendicular sides 4 cm and 3 cm is constructed. Keeping the third side as the base, another triangle BCD is constructed such that ∠DBC = 40∘ and ∠BCD = 30∘. Then the perimeter of triangle BCD is
A
8.4 cm
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B
7.7 cm
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C
11.1 cm
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D
6.1 cm
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Solution
## The correct option is A 11.1 cm Step 1: Draw a pair of perpendicular lines and name the point where they meet as A. Step 2: Mark B on one of the lines, 3 cm from A and mark C on the other line, 4 cm from A. Step 3: Join the points B and C to get the required right angles triangle. From the above construction, we see that the length of side BC is 5 cm. Keeping side BC as the base, triangle BCD is constructed, as follows: Step 4: Measure the ∠BCD with your protractor. Make a mark at 30 degrees and draw a line from C passing through the mark. Step 5: Measure the ∠CBD with your protractor. Make a mark at 40 degrees and draw a line from B passing through the mark. Step 6: Name the point of intersection of these two lines as D and △BCD is complete. Now, from the above construction, we see that the length of the side CD is 3.4 cm and the length of the side BD is 2.7 cm. Now we shall find the perimeter of triangle BCD. Perimeter of triangle BCD = BC + CD + BD = 5 cm + 3.4 cm + 2.7 cm = 11.1 cm
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Q:
# An acute triangle has side lengths 21 cm, x cm, and 2x cm. If 21 is one of the shorter sides of the triangle, what is the greatest possible length of the longest side, rounded to the nearest tenth?18.8 cm24.2 cm42.0 cm72.7 cm
Accepted Solution
A:
Answer:B. 24.2 cmStep-by-step explanation:We are given that,Lengths of the sides of a triangle are 21 cm, x cm and 2x cm.Using Triangle Inequality Theorem, which states that,'The sum of measure of any two sides must be greater than the measure of the third side'.Since, 21 cm is the shorter side. We have,A) $$21+x>2x$$ i.e. $$21>x$$ or $$x<21$$.B) $$x+2x>21$$ i.e. $$3x>21$$ or $$x>7$$So, $$7<x<21$$That is, $$14<2x<42$$Thus, the length of the larger side has measure between 14 cm and 42 cm.Hence, the possible length of the longest side is 24.2 cm. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
5.9: Possible Triangles with Side-Side-Angle
Difficulty Level: At Grade Created by: CK-12
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Your team has just won the flag in a flag football tournament at your school. As a reward, you get to take home the flag and keep it until the next game, when the other team will try to win it back. The flag looks like this:
It makes an isosceles triangle. You start to wonder how many different possible triangles there are for different lengths of sides. For example, if you make an oblique triangle that has a given angle greater than ninety degrees, how many ways are there to do this? Can you determine how many different possible triangles there are if the triangle is an isosceles triangle?
SSA Triangles
In Geometry, you learned that two sides and a non-included angle do not necessarily define a unique triangle.
Consider the following cases given a,b\begin{align*}a, b\end{align*}, and A\begin{align*}\angle{A}\end{align*}:
Case 1: No triangle exists (a<b)\begin{align*}(a < b)\end{align*}
In this case a<b\begin{align*}a < b\end{align*} and side a\begin{align*}a\end{align*} is too short to reach the base of the triangle. Since no triangle exists, there is no solution.
Case 2: One triangle exists (a<b)\begin{align*}(a < b)\end{align*}
In this case, a<b\begin{align*}a < b\end{align*} and side a\begin{align*}a\end{align*} is perpendicular to the base of the triangle. Since this situation yields exactly one triangle, there is exactly one solution.
Case 3: Two triangles exist (a<b)\begin{align*}(a < b)\end{align*}
In this case, a<b\begin{align*}a < b\end{align*} and side a\begin{align*}a\end{align*} meets the base at exactly two points. Since two triangles exist, there are two solutions.
Case 4: One triangle exists (a=b)\begin{align*}(a = b)\end{align*}
In this case a=b\begin{align*}a = b\end{align*} and side a\begin{align*}a\end{align*} meets the base at exactly one point. Since there is exactly one triangle, there is one solution.
Case 5: One triangle exists (a>b)\begin{align*}(a > b)\end{align*}
In this case, a>b\begin{align*}a > b\end{align*} and side a\begin{align*}a\end{align*} meets the base at exactly one point. Since there is exactly one triangle, there is one solution.
Case 3 is referred to as the Ambiguous Case because there are two possible triangles and two possible solutions. One way to check to see how many possible solutions (if any) a triangle will have is to compare sides a\begin{align*}a\end{align*} and b\begin{align*}b\end{align*}. If you are faced with the first situation, where a<b\begin{align*}a < b\end{align*}, we can still tell how many solutions there will be by using a\begin{align*}a\end{align*} and bsinA\begin{align*}b \sin A\end{align*}.
If: Then:
a. a<b\begin{align*}a < b\end{align*} No solution, one solution, two solutions
i. a<bsinA\begin{align*}a < b \sin A\end{align*} No solution
ii. a=bsinA\begin{align*}a = b \sin A\end{align*} One solution
iii. a>bsinA\begin{align*}a > b \sin A\end{align*} Two solutions
b. a=b\begin{align*}a = b\end{align*} One solution
c. a>b\begin{align*}a > b\end{align*} One solution
Identifying Triangles
For the following problems, determine if the sides and angles given determine no, one or two triangles.
1. The set contains an angle, its opposite side and the side between them.
a=5,b=8,A=62.19\begin{align*}a = 5, b = 8, A = 62.19^\circ\end{align*}
5<8,8sin62.19=7.076\begin{align*}5 < 8, 8 \sin 62.19^\circ = 7.076\end{align*}. So 5<7.076\begin{align*}5 < 7.076\end{align*}, which means there is no solution.
2. The set contains an angle, its opposite side and the side between them.
c=14,b=10,B=15.45\begin{align*}c = 14, b = 10, B = 15.45^\circ\end{align*}
Even though a,b\begin{align*}a, b\end{align*} and A\begin{align*}\angle{A}\end{align*} is not used in this example, follow the same pattern from the table by multiplying the non-opposite side (of the angle) by the angle.
10<14,14sin15.45=3.73\begin{align*}10 < 14, 14 \sin 15.45^\circ = 3.73\end{align*}. So 10>3.73\begin{align*}10 > 3.73\end{align*}, which means there are two solutions.
3. The set contains an angle, its opposite side and the side between them.
d=16,g=11,D=44.94\begin{align*}d = 16, g = 11, D = 44.94^\circ\end{align*}
Even though a,b\begin{align*}a, b\end{align*} and A\begin{align*}\angle{A}\end{align*} is not used in this example, follow the same pattern from the table by multiplying the non-opposite side (of the angle) by the angle.
16>11\begin{align*}16 > 11\end{align*}, there is one solution.
Examples
Example 1
Earlier, you were given a problem about a triangle.
As you now know, when two sides of a triangle with an included side are known, and the lengths of the two sides are equal, there is one possible solution. Since an isosceles triangle meets these criteria, there is only one possible solution.
Example 2
Determine how many solutions there would be for a triangle based on the given information and by calculating bsinA\begin{align*}b \sin A\end{align*} and comparing it with a\begin{align*}a\end{align*}. Sketch an approximate diagram for each problem in the box labeled “diagram.”
A=32.5,a=26,b=37\begin{align*}A = 32.5^\circ, a = 26, b = 37\end{align*}
A=32.5,a=26,b=37\begin{align*}A = 32.5^\circ, a = 26, b = 37\end{align*} 26>19.9\begin{align*}26 > 19.9\end{align*} 2 solutions
Example 3
Determine how many solutions there would be for a triangle based on the given information and by calculating bsinA\begin{align*}b \sin A\end{align*} and comparing it with a\begin{align*}a\end{align*}. Sketch an approximate diagram for each problem in the box labeled “diagram.”
A=42.3,a=16,b=26\begin{align*}A = 42.3^\circ, a = 16, b = 26\end{align*}
A=42.3,a=16,b=26\begin{align*}A = 42.3^\circ, a = 16, b = 26\end{align*} 16<17.5\begin{align*}16 < 17.5\end{align*} 0 solutions
Example 4
Determine how many solutions there would be for a triangle based on the given information and by calculating \begin{align*}b \sin A\end{align*} and comparing it with \begin{align*}a\end{align*}. Sketch an approximate diagram for each problem in the box labeled “diagram.”
\begin{align*}A = 47.8^\circ, a = 13.48,b = 18.2\end{align*}
\begin{align*}A = 47.8^\circ, a = 13.48,b = 18.2\end{align*} \begin{align*}13.48 = 13.48\end{align*} 1 solution
Review
Determine if the sides and angle given determine no, one or two triangles. The set contains an angle, its opposite side and another side of the triangle.
1. \begin{align*}a = 6, b = 6, A = 45^\circ\end{align*}
2. \begin{align*}a = 4, b = 7, A = 115^\circ\end{align*}
3. \begin{align*}a = 5, b = 2, A = 68^\circ\end{align*}
4. \begin{align*}a = 7, b = 6, A = 34^\circ\end{align*}
5. \begin{align*}a = 5, b = 3, A = 89^\circ\end{align*}
6. \begin{align*}a = 4, b = 4, A = 123^\circ\end{align*}
7. \begin{align*}a = 6, b = 8, A = 57^\circ\end{align*}
8. \begin{align*}a = 4, b = 9, A = 24^\circ\end{align*}
9. \begin{align*}a = 12, b = 11, A = 42^\circ\end{align*}
10. \begin{align*}a = 15, b = 17, A = 96^\circ\end{align*}
11. \begin{align*}a = 9, b = 10, A = 22^\circ\end{align*}
12. In \begin{align*}\triangle ABC\end{align*}, a=4, b=5, and \begin{align*}m\angle A=32^\circ\end{align*}. Find the possible value(s) of c.
13. In \begin{align*}\triangle DEF\end{align*}, d=7, e=5, and \begin{align*}m\angle D=67^\circ\end{align*}. Find the possible value(s) of f.
14. In \begin{align*}\triangle KQD\end{align*}, \begin{align*}m\angle K=20^\circ\end{align*}, k=24, and d=31. Find \begin{align*}m\angle D\end{align*}.
15. In \begin{align*}\triangle MRS\end{align*}, \begin{align*}m\angle M=70^\circ\end{align*}, m=44, and r=25. Find \begin{align*}m\angle R\end{align*}.
To see the Review answers, open this PDF file and look for section 5.9.
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Vocabulary Language: English
TermDefinition
Side Side Angle Triangle A side side angle triangle is a triangle where the length of two sides and one of the angles that is not between the two sides are known quantities.
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# 1661 A Little About Palindromes
### Today’s Puzzle:
A number is a palindrome if it reads the same backward as it does forward. For example, 1661 is a 4-digit palindrome. Today’s puzzle asks you to explore number palindromes and their factors.
Half of the time when a 4-digit palindrome is divided by 11, we get a 3-digit palindrome. Why does that happen?
Are there 3-digit palindromes that were NOT included in the Divided by Eleven column in the table above?
All 2-digit palindromes are divisible by 11. They are 11, 22, 33, 44, 55, 66, 77, 88, 99.
There are only eight 3-digit palindromes that are divisible by 11. They are 121, 242, 363, 484, 616, 737, 858, 979. Some 3-digit palindromes are prime numbers. Others are divisible by 101 or 111. Still, there are plenty of 3-digit palindromes that are composite numbers but not divisible by 11, 101, or 111.
Most 5-digit palindromes are NOT divisible by 11. I was able to construct one that is, 76967, because the red digits minus the blue digits are 2312 = 11, a number divisible by 11.
Here’s another: 81818. It works because 242 = 22, a number divisible by 11.
What 5-digit palindrome can you construct that is divisible by 11?
Will an N-digit palindrome be divisible by 11? What difference does it make if N is an even number or if N is an odd number?
### Factors of 1661:
1661 is divisible by 11 because the red digits minus the blue digits equal 0, a number divisible by 11.
• 1661 is a composite number.
• Prime factorization: 1661 = 11 × 151.
• 1661 has no exponents greater than 1 in its prime factorization, so √1661 cannot be simplified.
• The exponents in the prime factorization are 1 and 1. Adding one to each exponent and multiplying we get (1 + 1)(1 + 1) = 2 × 2 = 4. Therefore 1661 has exactly 4 factors.
• The factors of 1661 are outlined with their factor pair partners in the graphic below.
### More about the Number 1661:
1661 is also a palindrome in base 18:
1661₁₀ = 525₁₈ because
5(18²) + 2(18¹) + 5(18º) =
5(324) + 2(18) + 5(1) =
1620 + 36 + 5 =
1661.
# 1462 Has a Palindromic Factor Pair
There are only forty-five numbers between 100 and 10,000 that have palindromic factor pairs. That’s really not very many.
1462 is one of those numbers. That is pretty amazing, all by itself, but 1462 is also only four numbers more than another number on the list. That’s even more amazing!
Here are some more facts about the number 1462:
• 1462 is a composite number.
• Prime factorization: 1462 = 2 × 17 × 43
• 1462 has no exponents greater than 1 in its prime factorization, so √1462 cannot be simplified.
• The exponents in the prime factorization are 1, 1, and 1. Adding one to each exponent and multiplying we get (1 + 1)(1 + 1)(1 + 1) = 2 × 2 × 2 = 8. Therefore 1462 has exactly 8 factors.
• The factors of 1462 are outlined with their factor pair partners in the graphic below.
1462 is the hypotenuse of one Pythagorean triple:
688-1290-1462 which is (8-15-17) times 86
# 898 and Level 2
27² + 13² = 898. That means that 898 is the hypotenuse of a Pythagorean triple:
• 560-702-898, which is 2 times (280-351-449).
898 reads the same way frontwards and backwards so it is a palindrome in base 10.
It is also palindrome 747 in BASE 11 because 7(11²) + 4(11) + 7(1) = 898
AND it is palindrome 1G1 in BASE 23 (G is 16 in base 10) because 1(23²) + 16(23) +1(1) =898.
Print the puzzles or type the solution on this excel file: 10-factors-897-904
• 898 is a composite number.
• Prime factorization: 898 = 2 × 449
• The exponents in the prime factorization are 1 and 1. Adding one to each and multiplying we get (1 + 1)(1 + 1) = 2 × 2 = 4. Therefore 898 has exactly 4 factors.
• Factors of 898: 1, 2, 449, 898
• Factor pairs: 898 = 1 × 898 or 2 × 449
• 898 has no square factors that allow its square root to be simplified. √898 ≈ 29.9666
# Palindrome 868 Has a Fun Square Root
Obviously, 868 is a palindrome in base 10. It has some interesting representations in some other bases, too:
• 868 is 4004 in BASE 6, because 4(6³) + 0(6²) + 0(6¹) + 4(6º) = 868
• SS in BASE 30 (S is 28 base 10), because 28(30) + 28(1) = 28(31) = 868
• S0 in BASE 31, because 28(31) = 868
From OEIS.org I learned that all the digits from 1 to 9 make up the first nine decimal places of √868, so I decided to make a gif showing that fun fact:
make science GIFs like this at MakeaGif
• 868 is a composite number.
• Prime factorization: 868 = 2 × 2 × 7 × 31, which can be written 868 = 2² × 7 × 31
• The exponents in the prime factorization are 2, 1, and 1. Adding one to each and multiplying we get (2 + 1)(1 + 1)(1 + 1) = 3 × 2 × 2 = 12. Therefore 868 has exactly 12 factors.
• Factors of 868: 1, 2, 4, 7, 14, 28, 31, 62, 124, 217, 434, 868
• Factor pairs: 868 = 1 × 868, 2 × 434, 4 × 217, 7 × 124, 14 × 62, or 28 × 31,
• Taking the factor pair with the largest square number factor, we get √868 = (√4)(√217) = 2√217 ≈ 29.461839725
# 858 and Level 4
There are sixteen numbers less than 1000 that have four different prime factors. 858 is one of them, and it is the ONLY one that is also a palindrome. Thank you, OEIS.org for alerting us to that fact. No smaller palindrome has four different prime factors!
The sixteen products on that chart each have exactly sixteen factors!
Here’s a Find the Factors 1-10 puzzle for you to solve:
Print the puzzles or type the solution on this excel file: 10-factors-853-863
Here’s a little more about the number 858:
858 is the hypotenuse of a Pythagorean triple: 330-792-858
• 858 is a composite number.
• Prime factorization: 858 = 2 × 3 × 11 × 13
• The exponents in the prime factorization are 1, 1, 1, and 1. Adding one to each and multiplying we get (1 + 1)(1 + 1)(1 + 1)(1 + 1) = 2 × 2 × 2 × 2 = 16. Therefore 858 has exactly 16 factors.
• Factors of 858: 1, 2, 3, 6, 11, 13, 22, 26, 33, 39, 66, 78, 143, 286, 429, 858
• Factor pairs: 858 = 1 × 858, 2 × 429, 3 × 286, 6 × 143, 11 × 78, 13 × 66, 22 × 39, or 26 × 33
• 858 has no square factors that allow its square root to be simplified. √858 ≈ 29.291637
# 797 and Mathematical Ways to Love
Mathwithbaddrawings.com has some thoughtful and entertaining Ways to Tell a Mathematician that you love them.
Artful Maths wrote a post that includes beautiful mathematical origami valentines and a “string art” cardioid that is made with a pencil instead of string.
This puzzle could be another mathematical way to show some love:
Print the puzzles or type the solution on this excel file: 12-factors-795-799
Here are a few facts about the number 797:
797 is a palindrome in three bases:
• 797 BASE 10 because 7(100) + 9(10) + 7(1) = 797
• 565 BASE 12 because 5(144) + 6(12) + 5(1) = 797
• 494 BASE 13 because 4(169) + 9(13) + 4(1) = 797
But there’s one more palindromic fact about the number 797: It is the sum of two square numbers that are also palindromes!
• 797 = 121 + 676. Note that 11² = 121 and 26² = 676.
Since it is the sum of two squares, 797 will also be the hypotenuse a Pythagorean triple:
• 555 – 572 – 797 calculated from 26² – 11², 2(26)(11), 26² + 11².
797 is the sum of three squares seven different ways:
• 28² + 3² + 2² = 797
• 27² + 8² + 2² = 797
• 24² + 14² + 5² = 797
• 24² + 11² + 10² = 797
• 22² + 13² + 12² = 797
• 21² + 16² + 10² = 797
• 20² + 19² + 6² = 797
797 is also the sum of the 15 prime numbers from 23 to 83:
• 23 + 29 + 31 + 37+ 41 + 43 + 47 + 53 + 59 + 61 + 67 + 71 + 73 + 79 + 83 = 797
Here is the most basic information about the number 797:
• 797 is a prime number.
• Prime factorization: 797 is prime.
• The exponent of prime number 797 is 1. Adding 1 to that exponent we get (1 + 1) = 2. Therefore 797 has exactly 2 factors.
• Factors of 797: 1, 797
• Factor pairs: 797 = 1 x 797
• 797 has no square factors that allow its square root to be simplified. √797 ≈ 28.231188.
How do we know that 797 is a prime number? If 797 were not a prime number, then it would be divisible by at least one prime number less than or equal to √797 ≈ 28.2. Since 797 cannot be divided evenly by 2, 3, 5, 7, 11, 13, 17, 19, or 23, we know that 797 is a prime number.
Here’s another way we know that 797 is a prime number: Since its last two digits divided by 4 leave a remainder of 1, and 26² + 11² = 797 with 26 and 11 having no common prime factors, 797 will be prime unless it is divisible by a prime number Pythagorean triple hypotenuse less than or equal to √797 ≈ 28.2. Since 797 is not divisible by 5, 13, or 17, we know that 797 is a prime number.
# 727 Enjoy Some Sparkling Cider!
Have a happy New Year’s Eve! My husband and I will be enjoying some Sparkling Apple-Pear. I invite you to have some, too.
Perhaps the goblet in the puzzle can start off your festivities.
Print the puzzles or type the solution on this excel file: 12 Factors 2015-12-28
—————————————————————————————————
• 727 is a prime number.
• Prime factorization: 727 is prime.
• The exponent of prime number 727 is 1. Adding 1 to that exponent we get (1 + 1) = 2. Therefore 727 has exactly 2 factors.
• Factors of 727: 1, 727
• Factor pairs: 727 = 1 x 727
• 727 has no square factors that allow its square root to be simplified. √727 ≈ 26.9629375.
727 is a palindrome in base 10 and one other base:
• 727 BASE 10; note that 7(100) + 2(10) + 7(1) = 727.
• 1B1 BASE 22 (B = 11 base 10); note that 1(22²) + 11(22) + 1(1) = 727.
OEIS.org informs us that 727² = 528529, a number whose digits can be split in half to make two consecutive numbers.
Since 727 is a prime number, there is only one way it can be expressed as the sum of consecutive numbers: 363 + 364 = 727.
—————————————————————————————————
# When is 690 a Palindrome? In Base 16 and Base 29.
• 690 is a composite number.
• Prime factorization: 690 = 2 x 3 x 5 x 23
• The exponents in the prime factorization are 1, 1, 1, and 1. Adding one to each and multiplying we get (1 + 1)(1 + 1)(1 + 1)(1 + 1) = 2 x 2 x 2 x 2 = 16. Therefore 690 has exactly 16 factors.
• Factors of 690: 1, 2, 3, 5, 6, 10, 15, 23, 30, 46, 69, 115, 138, 230, 345, 690
• Factor pairs: 690 = 1 x 690, 2 x 345, 3 x 230, 5 x 138, 6 x 115, 10 x 69, 15 x 46, or 23 x 30
• 690 has no square factors that allow its square root to be simplified. √690 ≈ 26.267851
Here is today’s factoring puzzle:
Print the puzzles or type the solution on this excel file: 10 Factors 2015-11-23
———————————————————————————
Here is a little more about the number 690:
690 is the sum of the six prime numbers from 103 to 131. Do you know what all of those prime numbers are?
690 is also the hypotenuse of Pythagorean triple 414-552-690. What is the greatest common factor of those three numbers?
In BASE 10 we use the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. For bases smaller than 10 we use only the digits less than that base. For bases greater than 10, we might use A to represent 10, B to represent 11, and so forth all the way to Z representing 35 in BASE 36.
690 is a palindrome in two bases that require us to use letters of the alphabet to represent it:
• 2B2 in BASE 16; note that 2(256) + 11(16) + 2(1) = 690
• NN in BASE 29; note that 23(29) + 23(1) = 690. (N is the 14th letter of the alphabet and 14 + 9 = 23)
NN looks like it is divisible by 11, but remember that 11 base 29 is the same as 30 in base 10.
———————————————————————————
# 686 and Level 5
686 is divisible by 2, not by 3, and not by 5. Is it divisible by 7? Let’s try a divisibility rule for 7’s: 68 – 2(6) = 56, a multiple of 7. Yes, 686 is divisible by 7.
686 is a palindrome in two bases:
• 686 in base 10; note that 6(100) + 8(10) + 6(1) = 681
• 222 in base 18; note that 2(324) + 2(18) + 2(1) = 681
Print the puzzles or type the solution on this excel file: 12 Factors 2015-11-16
—————————————————————————————————
• 686 is a composite number.
• Prime factorization: 686 = 2 x 7 x 7 x 7, which can be written 686 = 2 x (7^3)
• The exponents in the prime factorization are 1 and 3. Adding one to each and multiplying we get (1 + 1)(3 + 1) = 2 x 4 = 8. Therefore 686 has exactly 8 factors.
• Factors of 686: 1, 2, 7, 14, 49, 98, 343, 686
• Factor pairs: 686 = 1 x 686, 2 x 343, 7 x 98, or 14 x 49
• Taking the factor pair with the largest square number factor, we get √686 = (√49)(√14) = 7√14 ≈ 26.1916017.
Here is a little cake to simplify √686:
Just take the square root of everything on the outside of the cake and multiply them together to get √686 = (√2)(√7)(√49) = 7√14
—————————————————————————————————
# 626 Semordnilaps, Palindromes and Level 4
626 is the hypotenuse of the Pythagorean triple 50-624-626. What is the greatest common factor of those three numbers?
Today my son posted on facebook, “A word that when spelled backwards spells a different word is called a Semordnilap.” That’s a word I hadn’t heard before, and it also applies to phrases and sentences that form different phrases or sentences when read backwards.
“Semordnilap” is a semordnilap for the word “palindromes” which are words, phrases, sentences, and numbers that read the same forward and backward.
Palindrome sentences are sometimes made with words that are semordnilaps: was, saw, live, evil, on, no, desserts, stressed, stop, pots, tops, spot, diaper, repaid.
626 is a number that is a palindrome in several different bases:
• 10001 in base 5; note that (5^4) + 1 = 626.
• 626 in base 10
• 272 in base 16; note that 2(16^2) + 7(16) + 2 = 626
• 1DI in base 19, if “1” and “I” look the same, and too often they do. Note that 1(19^2) + 13(19) + 18 = 626
• 101 in base 25; note that (25^2) + 1 = 626
• 11 in base 625; note that 626 + 1 = 626
I guess we could say that in all other bases 626 is a semordnilap.
All of 626’s factors are palindromes, too.
Print the puzzles or type the solution on this excel file: 12 Factors 2015-09-21
—————————————————————————————————
• 626 is a composite number.
• Prime factorization: 626 = 2 x 313
• The exponents in the prime factorization are 1 and 1. Adding one to each and multiplying we get (1 + 1)(1 + 1) = 2 x 2 = 4. Therefore 626 has exactly 4 factors.
• Factors of 626: 1, 2, 313, 626
• Factor pairs: 626 = 1 x 626 or 2 x 313
• 626 has no square factors that allow its square root to be simplified. √626 ≈ 25.019992.
————————————————————————————————— |
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# Comparing Equations of Parallel and Perpendicular Lines
## Use slope to identify parallel and perpendicular lines
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Practice Comparing Equations of Parallel and Perpendicular Lines
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Comparing Equations of Parallel and Perpendicular Lines
What if you had two lines? One line passes through the points (1, -2) and (3, 5). The other passes through the points (0, 2) and (7, 0). How could you determine if the two lines are parallel or perpendicular? After completing this Concept, you'll be able to make such a determination.
### Guidance
In this section you will learn how parallel lines and perpendicular lines are related to each other on the coordinate plane. Let’s start by looking at a graph of two parallel lines.
We can clearly see that the two lines have different intercepts: 6 and –4.
How about the slopes of the lines? The slope of line is , and the slope of line is . The slopes are the same.
Is that significant? Yes. By definition, parallel lines never meet. That means that when one of them slopes up by a certain amount, the other one has to slope up by the same amount so the lines will stay the same distance apart. If you look at the graph above, you can see that for any value you pick, the values of lines and are the same vertical distance apart—which means that both lines go up by the same vertical distance every time they go across by the same horizontal distance. In order to stay parallel, their slopes must stay the same.
All parallel lines have the same slopes and different intercepts.
Now let’s look at a graph of two perpendicular lines.
We can’t really say anything about the intercepts. In this example, the intercepts are different, but if we moved the lines four units to the right, they would both intercept the axis at (0, -2). So perpendicular lines can have the same or different intercepts.
What about the relationship between the slopes of the two lines?
To find the slope of line , we pick two points on the line and draw the blue (upper) right triangle. The legs of the triangle represent the rise and the run. We can see that the slope is , or 2.
To find the slope of line , we pick two points on the line and draw the red (lower) right triangle. Notice that the two triangles are identical, only rotated by . Where line goes 8 units up and 4 units right, line goes 8 units right and 4 units down. Its slope is , or .
This is always true for perpendicular lines; where one line goes units up and units right, the other line will go units right and units down, so the slope of one line will be and the slope of the other line will be .
The slopes of perpendicular lines are always negative reciprocals of each other.
The Java applet at http://members.shaw.ca/ron.blond/perp.APPLET/index.html lets you drag around a pair of perpendicular lines to see how their slopes change. Click “Show Grid” to see the and axes, and click “Show Constructors” to see the triangles that are being used to calculate the slopes of the lines (you can then drag the circle to make it bigger or smaller, and click on a triangle to see the slope calculations in detail.)
Determine When Lines are Parallel or Perpendicular
You can find whether lines are parallel or perpendicular by comparing the slopes of the lines. If you are given points on the lines, you can find their slopes using the formula. If you are given the equations of the lines, re-write each equation in a form that makes it easy to read the slope, such as the slope-intercept form.
#### Example A
Determine whether the lines are parallel or perpendicular or neither. One line passes through the points (2, 11) and (-1, 2); another line passes through the points (0, -4) and (-2, -10).
Solution
Find the slope of each line and compare them.
The slopes are equal, so the lines are parallel.
#### Example B
Determine whether the lines are parallel or perpendicular or neither. One line passes through the points (-2, -7) and (1, 5); another line passes through the points (4, 1) and (-8, 4).
Solution:
The slopes are negative reciprocals of each other, so the lines are perpendicular.
#### Example C
Determine whether the lines are parallel or perpendicular or neither. One line passes through the points (3, 1) and (-2, -2); another line passes through the points (5, 5) and (4, -6).
Solution:
The slopes are not the same or negative reciprocals of each other, so the lines are neither parallel nor perpendicular.
Watch this video for help with the Examples above.
### Vocabulary
• All parallel lines have the same slopes and different intercepts.
• The slopes of perpendicular lines are always negative reciprocals of each other.
### Guided Practice
Determine whether the lines are parallel or perpendicular or neither:
a) and
b) and
c) and
Solution
Write each equation in slope-intercept form:
a) line 1:
line 2:
The slopes are negative reciprocals of each other, so the lines are perpendicular.
b) line 1:
line 2:
The slopes are not the same or negative reciprocals of each other, so the lines are neither parallel nor perpendicular.
c) line 1:
line 2:
The slopes are the same, so the lines are parallel.
### Practice
For 1-10, determine whether the lines are parallel, perpendicular or neither.
1. One line passes through the points (-1, 4) and (2, 6); another line passes through the points (2, -3) and (8, 1).
2. One line passes through the points (4, -3) and (-8, 0); another line passes through the points (-1, -1) and (-2, 6).
3. One line passes through the points (-3, 14) and (1, -2); another line passes through the points (0, -3) and (-2, 5).
4. One line passes through the points (3, 3) and (-6, -3); another line passes through the points (2, -8) and (-6, 4).
5. One line passes through the points (2, 8) and (6, 0); another line has the equation .
6. One line passes through the points (-5, 3) and (2, -1); another line has the equation .
7. Both lines pass through the point (2, 8); one line also passes through (3, 5), and the other line has slope 3.
8. Line 1: Line 2:
9. Line 1: Line 2:
10. Line 1: Line 2:
11. Lines , and all pass through the point (3, 6). Line also passes through (7, 12); line passes through (8, 4); line passes through (-1, -3); line passes through (1, 1); and line passes through (6, 12).
1. Are any of these lines perpendicular? If so, which ones? If not, why not?
2. Are any of these lines parallel? If so, which ones? If not, why not?
### Vocabulary Language: English
Parallel
Parallel
Two or more lines are parallel when they lie in the same plane and never intersect. These lines will always have the same slope.
Perpendicular
Perpendicular
Perpendicular lines are lines that intersect at a $90^{\circ}$ angle. The product of the slopes of two perpendicular lines is -1.
standard form
standard form
The standard form of a quadratic function is $f(x)=ax^{2}+bx+c$. |
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Quick Homework Help
Area of Regular Polygons
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If radii are drawn from the center of a regular polygon to the vertices, congruent isosceles triangles are formed. Using the apothem as the height and the polygon side as the base, the area of each triangle can be calculated and summed. Therefore, the area regular polygons is equal to the number of triangles formed by the radii times their height: (side length)(apothem length)(number of sides)/2.
If we want to calculate the area of any regular polygon, the first thing we need to do is to write it inscribed in a circle, and by doing that we've created the ability to draw an apothem. Now an apothem is a perpendicular segment from the center of the circumscribed circle to one of the sides of your polygon. Now the reason why this is important is because if we want to calculate this hexagon, so again this will apply to any regular polygon. What we're going to do is we're going to think, "well Mr. McCall, I have no idea how to calculate that area. But what I do know how to calculate is the area of a triangle." So if I divide this hexagon into congruent triangles, and they'll all be congruent because it's regular, then all I have to do is add up the area of my 6 triangles. So you can say the area of the hexagon, is equal to 6 times the area of one of those triangles. But how do we calculate the area of one of those triangles?
Well, if you recall in order to calculate the area of a triangle, you need two things. You need the base and its corresponding height. So that's why we need the apothem because the apothem is that corresponding height. So specifically for a polygon, we're going to call this a and we're going to call that base x where, if we go back here s is our side length. So notice that the sides will all be the same for a regular polygon. So the area of this triangle is going to be the apothem which is the height. So I'm going to say 6 times the apothem times the base which is s divide by 2. So the area of a hexagon is equal to 6 times the area of one of your triangles. But that's not very useful because this is only going to apply to a hexagon. So let's make this area formula for any type of polygon. So instead of writing 6, I'm going to write n for the number of sides. So I can write this as apothem times side length times number of sides all divided by 2.
Now if we look at this, I see that I can simplify this a little bit more. Getting back to our regular polygon. The perimeter, capital p of this polygon is going to be 1, 2, 3, 4, 5, 6s. If I had a pentagon, that would be 5s. So what I'm going to say that the perimeter of any regular polygon is n times s. So I can substitute in for n times s, capital p which is going to stand for perimeter. So there's going to be two ways to write your area formula and I'm going to erase this to make it a little bit clear here.
So we're going to say the area of any regular polygon is equal to apothem times side length times the number of sides divided by 2, or if we substitute in for n times s, it's going to be the apothem times the perimeter divide by 2. So you have 2 different formulas, both of which will calculate the area of any regular polygon. |
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# 7.10: XYZ
Difficulty Level: At Grade Created by: CK-12
XYZ - Solve Systems of Equations
Teacher Notes
Students are presented with three equations with three unknowns. All equations have more than one variable. In all cases, all variables in one equation are in a second equation. The second equation contains at least one other variable. Students learn to replace the set of variables with its value in order to find the value of the other variable in the second equation. Once the value of one variable is known, its value can be used to figure out the values of the other variables. Encourage students to check solutions by replacing variables with their values.
Solutions
1. x=4, y=7, z=6\begin{align*}x = 4, \ y = 7, \ z = 6\end{align*}
2. x=6, y=8, z=9\begin{align*}x = 6, \ y = 8, \ z = 9\end{align*}
3. x=3, y=6, z=5\begin{align*}x = 3, \ y = 6, \ z = 5\end{align*}
4. x=5, y=4, z=7\begin{align*}x = 5, \ y = 4, \ z = 7\end{align*}
5. x=9, y=5, z=8\begin{align*}x = 9, \ y = 5, \ z = 8\end{align*}
6. x=8, y=10, z=7\begin{align*}x = 8, \ y = 10, \ z = 7\end{align*}
7. x=12, y=9, z=11\begin{align*}x = 12, \ y = 9, \ z = 11\end{align*}
8. x=7, y=5, z=9\begin{align*}x = 7, \ y = 5, \ z = 9\end{align*}
Eric wrote these equations to represent pictures of scales with blocks. Figure out the value of each unknown.
A:x+y=12x=B:x+x+y=19 y=C:x+z=10z=\begin{align*}A: x + y = 12 \qquad \qquad \qquad x = \underline{\;\;\;\;\;\;\;\;\;\;}\\ B: x + x + y = 19 \ \qquad \qquad y = \underline{\;\;\;\;\;\;\;\;\;\;}\\ C: x + z = 10 \qquad \qquad \qquad z = \underline{\;\;\;\;\;\;\;\;\;\;}\end{align*}
\begin{align*}& \mathbf{Describe} && \text{There are three equations. There are 3 unknowns.}\\ &&& \text{All equations have more than one unknown.}\\ & \mathbf{My \ Job} && \text{Figure out the values of} \ x, y, \ \text{and} \ z.\\ & \mathbf{Plan} && \text{All equations have more than one unknown.}\\ &&& \text{Equations} \ A \ \text{and} \ B \ \text{are related. In Equation} \ A, x + y = 12\\ &&& \text{That same} \ x + y \ \text{can be replaced by its value} \ 12 \ \text{in Equation} \ B.\\ &&& \text{Replace} \ x + y \ \text{with} \ 12. \ \text{Figure out the value of the extra} \ x.\\ &&& \text{Replace all} \ x's \ \text{with that value in all equations and continue to solve for the}\\ &&& \text{other unknowns.}\\ & \mathbf{Solve} && A: x + y = 12.\\ &&& B: \text{Replace} \ x + y \ \text{with} \ 12. \ x + 12 = 19.\\ &&& \text{So,} \ x = 19 - 12, \ \text{or} \ 7.\\ &&& \text{Replace all} \ x's \ \text{with} \ 7.\\ &&& C: 7 + z = 10, \ \text{so} \ z = 10 - 7, \ \text{or} \ 3.\\ &&& A: 7 + y = 12, \ \text{so} \ y = 12 - 7, \ \text{or} \ 5.\\ &&& \text{So} \ x = 7, \ y = 5, \ z = 3\\ & \mathbf{Check} && \text{Replace each variable with its value.}\\ &&& A: 7 + 5 = 12; \ B: 7 + 7 + 5 = 19; \ C: 7 + 3 = 10.\end{align*}
Eric wrote these equations to represent pictures of scales with blocks. Figure out the value of each unknown.
\begin{align*}1. \quad y + z + z = 19 \qquad \qquad \qquad x = \underline{\;\;\;\;\;\;\;\;\;\;}\!\\ {\;} \quad \ y + z = 13 \qquad \qquad \qquad \qquad y = \underline{\;\;\;\;\;\;\;\;\;\;}\!\\ {\;} \quad \ x + y + y = 18 \qquad \qquad \qquad z = \underline{\;\;\;\;\;\;\;\;\;\;}\end{align*}
\begin{align*}2. \quad x + x + y = 20 \qquad \qquad \qquad x = \underline{\;\;\;\;\;\;\;\;\;\;}\!\\ {\;} \quad \ y + z + z = 26 \qquad \qquad \qquad \ y = \underline{\;\;\;\;\;\;\;\;\;\;}\!\\ {\;} \quad \ y + z = 17 \qquad \qquad \qquad \qquad z = \underline{\;\;\;\;\;\;\;\;\;\;}\end{align*}
\begin{align*}3. \quad x + z = 8 \qquad \qquad \qquad \qquad x = \underline{\;\;\;\;\;\;\;\;\;\;}\!\\ {\;} \quad \ x + z + z = 13 \qquad \qquad \qquad y = \underline{\;\;\;\;\;\;\;\;\;\;}\!\\ {\;} \quad \ x + y + z = 14 \qquad \qquad \qquad z = \underline{\;\;\;\;\;\;\;\;\;\;}\end{align*}
\begin{align*}4. \quad x + z + z = 19 \qquad \qquad \qquad x = \underline{\;\;\;\;\;\;\;\;\;\;}\!\\ {\;} \quad \ y + y + z = 15 \qquad \qquad \qquad y = \underline{\;\;\;\;\;\;\;\;\;\;}\!\\ {\;} \quad \ x + z = 12 \qquad \qquad \qquad \quad \ \ z = \underline{\;\;\;\;\;\;\;\;\;\;}\end{align*}
Eric wrote these equations to represent pictures of scales with blocks. Figure out the value of each unknown.
\begin{align*}5. \quad x + y + z = 22 \qquad \qquad \qquad x = \underline{\;\;\;\;\;\;\;\;\;\;}\!\\ {\;} \quad \ x + z = 17 \qquad \qquad \qquad \quad \ \ y = \underline{\;\;\;\;\;\;\;\;\;\;}\!\\ {\;} \quad \ y + z + z = 21 \qquad \qquad \qquad z = \underline{\;\;\;\;\;\;\;\;\;\;}\end{align*}
\begin{align*}6. \quad x + y + z = 25 \qquad \qquad \qquad x = \underline{\;\;\;\;\;\;\;\;\;\;}\!\\ {\;} \quad \ y + z = 17 \qquad \qquad \qquad \quad \ \ y = \underline{\;\;\;\;\;\;\;\;\;\;}\!\\ {\;} \quad \ x + x + z = 23 \qquad \qquad \qquad z = \underline{\;\;\;\;\;\;\;\;\;\;}\end{align*}
\begin{align*}7. \quad x + x + y = 33 \qquad \qquad \qquad x = \underline{\;\;\;\;\;\;\;\;\;\;}\!\\ {\;} \quad \ x + z + z = 34 \qquad \qquad \qquad \ y = \underline{\;\;\;\;\;\;\;\;\;\;}\!\\ {\;} \quad \ x + y = 21 \qquad \qquad \qquad \quad \ \ z = \underline{\;\;\;\;\;\;\;\;\;\;}\end{align*}
\begin{align*}8. \quad y + y + z + z = 28 \qquad \qquad x = \underline{\;\;\;\;\;\;\;\;\;\;}\!\\ {\;} \quad \ x + x + y = 19 \qquad \qquad \quad \ \ y = \underline{\;\;\;\;\;\;\;\;\;\;}\!\\ {\;} \quad \ x + x + y + y = 24 \qquad \qquad z = \underline{\;\;\;\;\;\;\;\;\;\;}\end{align*}
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How to solve this problem
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This can help the student to understand the problem and How to solve this problem. Solve domain and range are two important words in the programming world. Solving domain and range means solving a problem with a set of variables including the value of each variable and ranges that connect each variable. For example, if you want to know how many students live in Los Angeles, you can use a program like Google Maps to find all the places where Los Angeles is mentioned on a map. You can also use this technique to solve problems with an array of numbers, such as computing the distance between two points. If you want to solve a problem with more than one set of variables, you can use a loop. A loop looks at one set of variables at a time until it runs out of values or reaches a stopping condition such as an end-of-file or an empty list. If you are not sure how to solve your problem with all your variables, check out some examples online or ask your teacher for help.
A trigonometric function is a mathematical function that relates two angles. Trig functions are used in trigonometry, which is the study of triangles. There are many trig functions, including sine and cosine. A trigonometric function is represented by an angle (theta) and a side (the length of the hypotenuse). The angle is measured from left to right, so if you have an angle of 60 degrees, the hypotenuse would be 4 times as long as the other side. Another way to look at it is based on the 90-degree difference between adjacent angles: angles adjacent to a 90 degree angle are 180 degrees apart; angles adjacent to a 45 degree angle are 135 degrees apart; and angles adjacent to a 0 degree angle are 90 degrees apart. The first derivative of a trig function is called its "derivative." The derivative of sin(x) = x - x^2 The second derivative of a trig function is called its "second derivative." The second derivative of sine(x) = 2x You can find these values by taking the derivative with respect to x, then plugging in your initial value for x. If you know how to do these derivatives, you can use them to solve equations. For example, if y = sin(x), then dy/dx = 2sin(x)/(
Long division is the process of dividing a large number by a smaller number. Long division can be done with paper and pencil, or it can be done online using a calculator. If you need to divide a number by a whole-number factor, such as 7, you will multiply that number by the divisor (e.g., 7 x 5 = 35). Then, you will divide the larger number by the result of the multiplication (e.g., 35 ÷ 5 = 12). Finally, you will add the two numbers that were divided (e.g., 12 + 35 = 49). If you need to divide a number by a fractional factor, such as 1/3, you will divide the larger number by the result of the multiplication (e.g., 35 ÷ 3 = 12) and then multiply the resulting fraction by the divisor (e.g., 12 x 1/3 = 4). Then, you will divide the larger number by the result of the multiplication (e.g., 12 ÷ 1/3 = 4) and add this answer to your original one (e.g., 4 + 4 = 8). IMPORTANT: If you are trying to solve long division using pencil and paper or on an online calculator, it is important to follow these steps in order: first, multiply; then divide; then subtract; then check
There are a number of different ways to solve a tangent problem. The most straightforward method is to let a computer solve the problem for you. However, it may not be the best approach if you are in a hurry or don't have access to a computer. A better option is to solve the problem by hand. The main advantage to this approach is that you can try different strategies and take breaks while you are solving the problem. You also get to practice using your skills in another area. Another advantage is that it can be easier to spot when you are off track with your solution. This is because you will notice more errors as soon as you start making mistakes. Another option is to use a tangent calculator (a software program that solves for tangents). These can be helpful when trying to learn new techniques, but they may not be accurate enough to use in an actual application.
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# Famous Theorems of Mathematics/Pythagoras theorem
The Pythagoras Theorem or the Pythagorean theorem, named after the Greek mathematician Pythagoras states that:
In any right triangle, the area of the square whose side is the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares whose sides are the two legs (the two sides that meet at a right angle).
This is usually summarized as follows:
The square of the hypotenuse of a right triangle is equal to the sum of the squares on the other two sides.
If we let c be the length of the hypotenuse and a and b be the lengths of the other two sides, the theorem can be expressed as the equation:
$a^2 + b^2 = c^2\,$
or, solved for c:
$c = \sqrt{a^2 + b^2}. \,$
If c is already given, and the length of one of the legs must be found, the following equations can be used (The following equations are simply the converse of the original equation):
$c^2 - a^2 = b^2\,$
or
$c^2 - b^2 = a^2.\,$
This equation provides a simple relation among the three sides of a right triangle so that if the lengths of any two sides are known, the length of the third side can be found. A generalization of this theorem is the law of cosines, which allows the computation of the length of the third side of any triangle, given the lengths of two sides and the size of the angle between them. If the angle between the sides is a right angle it reduces to the Pythagorean theorem.
## History
The history of the theorem can be divided into four parts: knowledge of Pythagorean triples, knowledge of the relationship between the sides of a right triangle, knowledge of the relationship between adjacent angles, and proofs of the theorem.
Megalithic monuments from circa 2500 BC in Egypt, and in Northern Europe, incorporate right triangles with integer sides. Bartel Leendert van der Waerden conjectures that these Pythagorean triples were discovered algebraically.
Written between 2000 and 1786 BC, the Middle Kingdom Egyptian papyrus Berlin 6619 includes a problem whose solution is a Pythagorean triple.
During the reign of Hammurabi the Great, the Mesopotamian tablet Plimpton 322, written between 1790 and 1750 BC, contains many entries closely related to Pythagorean triples.
The Baudhayana Sulba Sutra, the dates of which are given variously as between the 8th century BC and the 2nd century BC, in India, contains a list of Pythagorean triples discovered algebraically, a statement of the Pythagorean theorem, and a geometrical proof of the Pythagorean theorem for an isosceles right triangle.
The Apastamba Sulba Sutra (circa 600 BC) contains a numerical proof of the general Pythagorean theorem, using an area computation. Van der Waerden believes that "it was certainly based on earlier traditions". According to Albert Bŭrk, this is the original proof of the theorem; he further theorizes that Pythagoras visited Arakonam, India, and copied it.
Pythagoras, whose dates are commonly given as 569–475 BC, used algebraic methods to construct Pythagorean triples, according to Proklos's commentary on Euclid. Proklos, however, wrote between 410 and 485 AD. According to Sir Thomas L. Heath, there is no attribution of the theorem to Pythagoras for five centuries after Pythagoras lived. However, when authors such as Plutarch and Cicero attributed the theorem to Pythagoras, they did so in a way which suggests that the attribution was widely known and undoubted.
Around 400 BC, according to Proklos, Plato gave a method for finding Pythagorean triples that combined algebra and geometry. Circa 300 BC, in Euclid's Elements, the oldest extant axiomatic proof of the theorem is presented.
Written sometime between 500 BC and 100 AD, the Chinese text Chou Pei Suan Ching (周髀算经), (The Arithmetical Classic of the Gnomon and the Circular Paths of Heaven) gives a statement of the Pythagorean theorem — in China it is called the "Gougu Theorem" (勾股定理) — for the (3, 4, 5) triangle. A visual proof is recorded in a Ming dynasty text though it is unclear as to when it was originally provided. During the Han Dynasty, from 202 BC to 220 AD, Pythagorean triples appear in The Nine Chapters on the Mathematical Art, together with a mention of right triangles.
The first recorded use is in China, known as the "Gougu theorem" (勾股定理) and in India known as the Bhaskara Theorem.
There is much debate on whether the Pythagorean theorem was discovered once or many times. Boyer (1991) thinks the elements found in the Shulba Sutras may be of Mesopotamian derivation.
## Proofs
This is a theorem that may have more known proofs than any other; the book Pythagorean Proposition, by Elisha Scott Loomis, contains 367 proofs.
### Proof using similar triangles
Proof using similar triangles.
Like most of the proofs of the Pythagorean theorem, this one is based on the proportionality of the sides of two similar triangles.
Let ABC represent a right triangle, with the right angle located at C, as shown on the figure. We draw the altitude from point C, and call H its intersection with the side AB. The new triangle ACH is similar to our triangle ABC, because they both have a right angle (by definition of the altitude), and they share the angle at A, meaning that the third angle will be the same in both triangles as well. By a similar reasoning, the triangle CBH is also similar to ABC. The similarities lead to the two ratios..: As
$BC=a, AC=b, \text{ and } AB=c, \!$
so
$\frac{a}{c}=\frac{HB}{a} \mbox{ and } \frac{b}{c}=\frac{AH}{b}.\,$
These can be written as
$a^2=c\times HB \mbox{ and }b^2=c\times AH. \,$
Summing these two equalities, we obtain
$a^2+b^2=c\times HB+c\times AH=c\times(HB+AH)=c^2 .\,\!$
In other words, the Pythagorean theorem:
$a^2+b^2=c^2.\,\!$
### Euclid's proof
Proof in Euclid's Elements
In Euclid's Elements, Proposition 47 of Book 1, the Pythagorean theorem is proved by an argument along the following lines. Let A, B, C be the vertices of a right triangle, with a right angle at A. Drop a perpendicular from A to the side opposite the hypotenuse in the square on the hypotenuse. That line divides the square on the hypotenuse into two rectangles, each having the same area as one of the two squares on the legs.
For the formal proof, we require four elementary lemmata:
1. If two triangles have two sides of the one equal to two sides of the other, each to each, and the angles included by those sides equal, then the triangles are congruent. (Side - Angle - Side Theorem)
2. The area of a triangle is half the area of any parallelogram on the same base and having the same altitude.
3. The area of any square is equal to the product of two of its sides.
4. The area of any rectangle is equal to the product of two adjacent sides (follows from Lemma 3).
The intuitive idea behind this proof, which can make it easier to follow, is that the top squares are morphed into parallelograms with the same size, then turned and morphed into the left and right rectangles in the lower square, again at constant area.
The proof is as follows:
1. Let ACB be a right-angled triangle with right angle CAB.
2. On each of the sides BC, AB, and CA, squares are drawn, CBDE, BAGF, and ACIH, in that order.
3. From A, draw a line parallel to BD and CE. It will perpendicularly intersect BC and DE at K and L, respectively.
4. Join CF and AD, to form the triangles BCF and BDA.
Illustration including the new lines
1. Angles CAB and BAG are both right angles; therefore C, A, and G are collinear. Similarly for B, A, and H.
2. Angles CBD and FBA are both right angles; therefore angle ABD equals angle FBC, since both are the sum of a right angle and angle ABC.
3. Since AB and BD are equal to FB and BC, respectively, triangle ABD must be equal to triangle FBC.
4. Since A is collinear with K and L, rectangle BDLK must be twice in area to triangle ABD.
5. Since C is collinear with A and G, square BAGF must be twice in area to triangle FBC.
6. Therefore rectangle BDLK must have the same area as square BAGF = AB2.
7. Similarly, it can be shown that rectangle CKLE must have the same area as square ACIH = AC2.
8. Adding these two results, AB2 + AC2 = BD × BK + KL × KC
9. Since BD = KL, BD* BK + KL × KC = BD(BK + KC) = BD × BC
10. Therefore AB2 + AC2 = BC2, since CBDE is a square.
This proof appears in Euclid's Elements as that of Proposition 1.47.
### Garfield's proof
James A. Garfield (later President of the United States) is credited with a novel algebraic proof[1] using a trapezoid containing two examples of the triangle, the figure comprising one half of the figure using four triangles enclosing a square shown below.
Proof using area subtraction.
### Similarity proof
From the same diagram as that in Euclid's proof above, we can see three similar figures, each being "a square with a triangle on top". Since the large triangle is made of the two smaller triangles, its area is the sum of areas of the two smaller ones. By similarity, the three squares are in the same proportions relative to each other as the three triangles, and so likewise the area of the large square is the sum of the areas of the two smaller squares.
### Proof by rearrangement
A proof by rearrangement is given by the illustration and the animation. In the illustration, the area of each large square is (a + b)2. In both, the area of four identical triangles is removed. The remaining areas, a2 + b2 and c2, are equal. Q.E.D.
Animation showing another proof by rearrangement.
Proof using rearrangement.
A square created by aligning four right angle triangles and a large square.
This proof is indeed very simple, but it is not elementary, in the sense that it does not depend solely upon the most basic axioms and theorems of Euclidean geometry. In particular, while it is quite easy to give a formula for area of triangles and squares, it is not as easy to prove that the area of a square is the sum of areas of its pieces. In fact, proving the necessary properties is harder than proving the Pythagorean theorem itself and Banach-Tarski paradox. Actually, this difficulty affects all simple Euclidean proofs involving area; for instance, deriving the area of a right triangle involves the assumption that it is half the area of a rectangle with the same height and base. For this reason, axiomatic introductions to geometry usually employ another proof based on the similarity of triangles (see above).
A third graphic illustration of the Pythagorean theorem (in yellow and blue to the right) fits parts of the sides' squares into the hypotenuse's square. A related proof would show that the repositioned parts are identical with the originals and, since the sum of equals are equal, that the corresponding areas are equal. To show that a square is the result one must show that the length of the new sides equals c. Note that for this proof to work, one must provide a way to handle cutting the small square in more and more slices as the corresponding side gets smaller and smaller.[1]
### Algebraic proof
An algebraic variant of this proof is provided by the following reasoning. Looking at the illustration which is a large square with identical right triangles in its corners, the area of each of these four triangles is given by an angle corresponding with the side of length C.
$\frac{1}{2} AB.$
The A-side angle and B-side angle of each of these triangles are complementary angles, so each of the angles of the blue area in the middle is a right angle, making this area a square with side length C. The area of this square is C2. Thus the area of everything together is given by:
$4\left(\frac{1}{2}AB\right)+C^2.$
However, as the large square has sides of length A + B, we can also calculate its area as (A + B)2, which expands to A2 + 2AB + B2.
$A^2+2AB+B^2=4\left(\frac{1}{2}AB\right)+C^2.\,\!$
(Distribution of the 4) $A^2+2AB+B^2=2AB+C^2\,\!$
(Subtraction of 2AB) $A^2+B^2=C^2\,\!$
### Proof by differential equations
One can arrive at the Pythagorean theorem by studying how changes in a side produce a change in the hypotenuse in the following diagram and employing a little calculus.
Proof using differential equations.
As a result of a change in side a,
$\frac {da}{dc} = \frac {c}{a}$
by similar triangles and for differential changes. So
$c\, dc = a\,da$
upon separation of variables.
which results from adding a second term for changes in side b.
Integrating gives
$c^2 = a^2 + \mathrm{constant}.\ \,\!$
When a = 0 then c = b, so the "constant" is b2. So
$c^2 = a^2 + b^2.\,$
As can be seen, the squares are due to the particular proportion between the changes and the sides while the sum is a result of the independent contributions of the changes in the sides which is not evident from the geometric proofs. From the proportion given it can be shown that the changes in the sides are inversely proportional to the sides. The differential equation suggests that the theorem is due to relative changes and its derivation is nearly equivalent to computing a line integral.
These quantities da and dc are respectively infinitely small changes in a and c. But we use instead real numbers Δa and Δc, then the limit of their ratio as their sizes approach zero is da/dc, the derivative, and also approaches c/a, the ratio of lengths of sides of triangles, and the differential equation results.
## Converse
The converse of the theorem is also true:
For any three positive numbers a, b, and c such that a2 + b2 = c2, there exists a triangle with sides a, b and c, and every such triangle has a right angle between the sides of lengths a and b.
This converse also appears in Euclid's Elements. It can be proven using the law of cosines, or by the following proof:
Let ABC be a triangle with side lengths a, b, and c, with a2 + b2 = c2. We need to prove that the angle between the a and b sides is a right angle. We construct another triangle with a right angle between sides of lengths a and b. By the Pythagorean theorem, it follows that the hypotenuse of this triangle also has length c. Since both triangles have the same side lengths a, b and c, they are congruent, and so they must have the same angles. Therefore, the angle between the side of lengths a and b in our original triangle is a right angle.
A corollary of the Pythagorean theorem's converse is a simple means of determining whether a triangle is right, obtuse, or acute, as follows. Where c is chosen to be the longest of the three sides:
• If a2 + b2 = c2, then the triangle is right.
• If a2 + b2 > c2, then the triangle is acute.
• If a2 + b2 < c2, then the triangle is obtuse.
## Consequences and uses of the theorem
### Pythagorean triples
A Pythagorean triple has 3 positive numbers a, b, and c, such that $a^2 + b^2 = c^2$. In other words, a Pythagorean triple represents the lengths of the sides of a right triangle where all three sides have integer lengths. Evidence from megalithic monuments on the Northern Europe shows that such triples were known before the discovery of writing. Such a triple is commonly written (abc). Some well-known examples are (3, 4, 5) and (5, 12, 13).
### List of primitive Pythagorean triples up to 100
(3, 4, 5), (5, 12, 13), (7, 24, 25), (8, 15, 17), (9, 40, 41), (11, 60, 61), (12, 35, 37), (13, 84, 85), (16, 63, 65), (20, 21, 29), (28, 45, 53), (33, 56, 65), (36, 77, 85), (39, 80, 89), (48, 55, 73), (65, 72, 97)
### The existence of irrational numbers
One of the consequences of the Pythagorean theorem is that irrational numbers, such as the square root of 2, can be constructed. A right triangle with legs both equal to one unit has hypotenuse length square root of 2. The Pythagoreans proved that the square root of 2 is irrational, and this proof has come down to us even though it flew in the face of their cherished belief that everything was rational. According to the legend, Hippasus, who first proved the irrationality of the square root of two, was drowned at sea as a consequence.
### Distance in Cartesian coordinates
The distance formula in Cartesian coordinates is derived from the Pythagorean theorem. If (x0, y0) and (x1, y1) are points in the plane, then the distance between them, also called the Euclidean distance, is given by
$\sqrt{(x_1-x_0)^2 + (y_1-y_0)^2}.$
More generally, in Euclidean n-space, the Euclidean distance between two points, $\scriptstyle A\,=\,(a_1,a_2,\dots,a_n)$ and $\scriptstyle B\,=\,(b_1,b_2,\dots,b_n)$, is defined, using the Pythagorean theorem, as:
$\sqrt{(a_1-b_1)^2 + (a_2-b_2)^2 + \cdots + (a_n-b_n)^2} = \sqrt{\sum_{i=1}^n (a_i-b_i)^2}.$
## References
1. Pythagorean Theorem: Subtle Dangers of Visual Proof by Alexander Bogomolny, retrieved 19 December 2006. |
# How to Find Area A Comprehensive Guide to Calculating Area in Various Shapes
Welcome to markantony.org, your ultimate destination for how-to guides, unlocking knowledge, and mastering skills. In this article, we will delve into the fascinating world of finding area, a fundamental concept in mathematics and geometry. Whether you’re a student, a DIY enthusiast, or simply curious about the subject, this comprehensive guide will equip you with the tools and techniques to calculate area in various shapes. Let’s dive in and unravel the mysteries of area calculations!
## The Basics of Area: Understanding the Concept
Before we embark on our journey to explore the intricacies of finding area in different shapes, let’s start by understanding the basic concept of area. Area refers to the amount of space occupied by a two-dimensional shape or surface. It is typically measured in square units, such as square inches, square feet, or square meters.
Area plays a crucial role in numerous fields, including architecture, engineering, landscaping, and even everyday activities like measuring the floor space of a room. Having a solid understanding of how to find area empowers you to make informed decisions and solve practical problems in a wide range of situations.
## Calculating Area in Rectangles and Squares
Rectangles and squares are among the simplest shapes when it comes to finding area. Let’s explore how to calculate their areas:
### 1. Area of a Rectangle:
To find the area of a rectangle, you need to multiply its length by its width. The formula for calculating the area of a rectangle is:
Area of a Rectangle : Length × Width
For example, suppose you have a rectangle with a length of 6 inches and a width of 4 inches. Using the formula, you would calculate the area as follows:
Area = 6 inches × 4 inches = 24 square inches
The area of the given rectangle is 24 square inches. Remember to include the appropriate unit when expressing the area.
### 2. Area of a Square:
A square is a special case of a rectangle where all sides are equal in length. Calculating the area of a square is straightforward:
Area of a Square : Side Length × Side Length
Suppose you have a square with a side length of 5 meters. Applying the formula, the area would be:
Area = 5 meters × 5 meters = 25 square meters
The area of the given square is 25 square meters. Remember to include the appropriate unit when expressing the area.
## Exploring Area in Circles and Triangles
As we venture further into the realm of finding area, we encounter more complex shapes like circles and triangles. Let’s explore how to calculate the area in these intriguing geometrical forms:
### 3. Area of a Circle:
The area of a circle can be determined using the radius or diameter. The formula for calculating the area of a circle is:
</ tr>
Area of a Circle : π × (Radius)^2
Here, π (pi) represents a mathematical constant approximately equal to 3.14159. Suppose you have a circle with a radius of 8 centimeters. Applying the formula, the area would be:
Area = 3.14159 × (8 centimeters)^2 ≈ 201.06 square centimeters
The area of the given circle is approximately 201.06 square centimeters. Remember to include the appropriate unit when expressing the area.
### 4. Area of a Triangle:
Calculating the area of a triangle involves using its base and height. The formula for finding the area of a triangle is:
Area of a Triangle : (Base × Height) / 2
For example, consider a triangle with a base of 6 meters and a height of 10 meters. Applying the formula, the area would be:
Area = (6 meters × 10 meters) / 2 = 30 square meters
The area of the given triangle is 30 square meters. Remember to include the appropriate unit when expressing the area.
## Advanced Techniques for Finding Area in Irregular Shapes
While the aforementioned shapes have relatively straightforward formulas for finding area, irregular shapes often require advanced techniques. Let’s explore some approaches to calculate area in irregular shapes:
### 5. Area of a Trapezoid:
A trapezoid is a quadrilateral with one pair of parallel sides. To find the area of a trapezoid, you can use the formula:
Area of a Trapezoid : ((Base1 + Base2) × Height) / 2
Suppose you have a trapezoid with Base 1 measuring 5 meters, Base 2 measuring 8 meters, and a height of 6 meters. Applying the formula, the area would be:
Area = ((5 meters + 8 meters) × 6 meters) / 2 = 39 square meters
The area of the given trapezoid is 39 square meters. Remember to include the appropriate unit when expressing the area.
### 6. Area of an Irregular Polygon:
Irregular polygons are shapes that do not conform to standard geometric forms. Calculating the area of an irregular polygon can be challenging, but there are methods to approximate it. One approach involves dividing the irregular polygon into smaller, regular shapes, finding their respective areas, and summing them up.
For instance, suppose you have an irregular polygon composed of a rectangle and a triangle. The rectangle has a length of 5 meters and a width of 3 meters, while the triangle has a base of 4 meters and a height of 2 meters. Calculating the area would involve finding the area of the rectangle and the triangle separately, then summing them up:
Area of Rectangle = 5 meters × 3 meters = 15 square meters
Area of Triangle = (4 meters × 2 meters) / 2 = 4 square meters
Total Area = 15 square meters + 4 square meters = 19 square meters
The area of the given irregular polygon is
19 square meters. Remember to include the appropriate unit when expressing the area.
1. Q: Why is finding area important in real life?
A: Calculating area is essential in various real-life scenarios, such as determining the amount of paint needed to cover a wall, estimating the material required for flooring, or calculating the size of a garden for landscaping purposes. It helps us make informed decisions and solve practical problems in different fields.
2. Q: Are there any shortcuts or tricks for finding area?
A: While there are no universal shortcuts for finding area in all shapes, some regular shapes have specific formulas that make calculations easier. Additionally, breaking down irregular shapes into smaller, regular components can simplify the process.
3. Q: Can you find the area of a three-dimensional object?
A: No, the concept of area applies only to two-dimensional shapes or surfaces. To find the volume of a three-dimensional object, different formulas and techniques are used.
4. Q: How accurate should area calculations be?
A: The level of accuracy required depends on the specific application. In some cases, an approximation may suffice, while in others, precise calculations are necessary. Consider the context and requirements of the problem at hand.
5. Q: Can I use a calculator to find area?
A: Yes, calculators can be helpful, particularly when dealing with complex formulas or large numbers. However, it’s important to understand the underlying concepts and formulas to ensure accurate results. |
# Mobile math apps
Apps can be a great way to help learners with their math. Let's try the best Mobile math apps. Math can be a challenging subject for many students.
## The Best Mobile math apps
Mobile math apps is a mathematical instrument that assists to solve math equations. A math tutor can be an invaluable resource for this. By definition, a word problem is a mathematical problem that involves words rather than numbers or symbols. You might see words like "if it rains tomorrow, how many inches of rain will there be?" Word problems usually involve numbers or quantities, but they also include words that represent concepts such as length, time, area and volume. However, they often look different from standard mathematical problems because they rely more on language than mathematics. For example, you might be given the word "lose" and asked how many pounds of weight you would have to lose to reach a certain weight goal.
To solve a trinomial, first find the coefficients of all of the terms in the expression. In this example, we have ("3x + 2"). Now you can start solving for each variable one at a time using algebraic equations. For example, if you know that x = 0, y = 9 and z = -2 then you can solve for y with an equation like "y = (0)(9)/(-2)" After you've figured out all of the variables, use addition or subtraction to combine them into one final answer.
The matrix 3x3 is a common problem in mathematics. In this case, we have a 3-by-3 square of numbers. We want to find the values of A, B and C that solve the equation AB=C. The solution is: A=C/2 B=C/4 C=C/8 When we multiply B by C (or C by -1), we get A. When we divide A by B, we get C. And when we divide C by -1, we get -B. This is a fairly simple way to solve the matrix 3x3. It's also useful to remember if you have any nonlinear equations with matrices, like x^2 + y^2 = 4x+2y. In these cases, you can usually find a solution by finding the roots of the nonlinear equation and plugging it into the matrix equation.
Linear differential equation solvers are used to find the solution to a linear differential equation. They are useful in applications where the system has a known set of known values that can be used to solve for the unknown output value. The input values may be the product of one or more other variables, but the output value is only dependent on these values. There are two types of linear differential equation solvers: iterative methods and recursive methods. Iterative methods solve an equation by repeatedly solving small subsets of the problem and using these solutions to compute new intermediate solutions. These methods require an initial guess of the solution and may require several iterations to converge on a solution. Recursive methods solve an equation by recursively evaluating specific portions of it. As each portion is evaluated, it is passed back as part of the next evaluation step, which allows this method to converge more quickly than iterative methods. Both types of linear differential equations solvers can be used to solve many different types of problems, including those with multiple unknowns (like nonlinear differential equations) or those involving non-linearities (like polynomial differential equations).
This app helpful for the students that have confusion so he or she can use the app for me it's very much good and helpful for my children I liked this app. Extremely helpful for majority of math topics, solutions/explanations could maybe be a bit more in depth.
Zayda Wright
This app helps me so much! Not only does it give an answer to math problems, especially algebraic ones, but it also gives an explanation of the steps, along with explaining why it's done! The only thing I'd ask be added is something that can show equivalent responses, more specifically in different forms. Thank you!
Zelda Jackson
Basic algebra math problems Trigonometry proofs solver Online free math tutor chat Differential equations solver with steps Algebra homework solver Conditional proof solver |
# Equivalant Fractions with Unlike Denominators
1025 words 5 pages
In elementary math there are several concepts about fractions. One concept students in fourth grade will need to master is learning how to tell if fractions are equivalent with unlike denominators. There are a few prerequisite skills that are necessary in order for the students to understand this concept. The first thing students need to know is what fractions are. Fractions are a way of counting parts of a whole. Secondly, the students need to know how to identify parts of a fraction. The top number in a fraction is the numerator. The numerator is the number of parts in a whole (Eather). The bottom number in a fraction is the denominator. The denominator is the number of parts the whole is divided into (Eather). Lastly, the student will …show more content…
Show the students a couple examples, if i have 3/10 an equivalent fraction is 6/?.. If we multiplied 3 by (what) to get 6, which is 2. We need to also multiply 10 by 2 to get 20. You can double check yourself by divining 6/20 by two to get 3/10. Another way to double check your work is by cross multiplying. If we multiply the numerator of the first problem (3) by the denominator of the second problem (20) and the denominator of the first problem (10) by the numerator of the second problem (6), they should equal the same. 3 times 20 is 60 and 10 times 6 is 60, so we have found equivalent fractions. You could also show the students a way of proving that two fractions aren't equivalent. If you were to take 3/4 and 2/12.. are these equivalent fractions? If we cross multiply 3 times 12 and 4 times 2, do we get the same answer? 3 times 12 is 36 and four times two is 8, these fractions are not equivalent. Once the lesson has been taught the teacher can give the students a worksheet with ten fractions, the students are solve the problems per the written instruction.
1. Find an equivalent fraction for 4/12.
2. Find and equivalent fraction for 8/16.
3. Find an equivalent fraction for 3/5.
4. Find the missing number 2/3 = 6/ __ ?
5. Find the missing number 4/12 = 12/ ___ ?
6. Find the missing number 2/9 = ___/ 36?
7. Are these fractions equivalent 2/5 and 10/25?
8. Are these fractions equivalent 2/11 and |
```Algebra 1 Functions Test STUDY GUIDE
Name: _______________________________________ Date: __________________ Block: ___________
Algebra 1 Function Introduction STUDY GUIDE
SOLs: A.7, A.4
Know how to…
Plot points on a Cartesian coordinate plane.
Find the (x, y) coordinates of points on the Cartesian plane.
Identify the quadrant in which a point is located.
Given a relation in different formats (such as a set of (x, y) pairs, mapping diagram, or
graph), determine whether the relation is a function. Be able to explain why you made
your choice. Remember: a relation is a function if every input has one and only one
(OAOO) output.
Given a relation with a discrete domain, be able to find the range of the relation.
Given a graph of a relation, be able to describe the domain and range in set builder
notation.
Be able to graph linear equations using the table of values method and the intercept
method. Be prepared to write the x- and y-intercepts in (x, y) pair form.
Be able to evaluate functions that use f(x) function notation. Remember: x is the input
to the function, and f(x) is the output.
Be able to identify the zero of a linear function, which is the same as its x-intercept.
Find it by setting the function’s output to 0, and solving for x (find x where f(x) = 0).
Be able to put a linear equation into function form (solve for y).
Be able to solve formulas for a given variable.
Be able to use formulas created by solving for a variable in word problems.
Study Questions
1) Use the graph at right to answer the questions…
a) Plot and label the points P(-2, -3), Q(1, 0), R(0, 3),
and S(4, -5) in the coordinate plane at right.
Identify the quadrants of the points.
b) Identify the coordinates of points A, B, and C.
Identify the quadrants they lie in.
Algebra 1 Functions Test STUDY GUIDE Page 2
2) Determine whether the relations below are functions or not. Explain how you made
a)
{(-4, 2), (0, -4), (4, 2), (1, -4)}
b)
c)
d)
Function? ___________
Explain:
Function? ___________
Explain:
Function? ________
Explain:
Function? ________
Explain:
3) Given the functions below, find the range given their domain.
a) y = -3x + 2 D= {-3, -2, 0, 1, 4}
b) y =
1
x - 3 D= {-4, -2, 0, 1, 6}
2
4) Given the graphs below, determine whether the relations are functions. Then find the
domain and range for each in set builder notation.
a)
b)
c)
d)
Function? ___________
Domain _____________
Range ______________
Function? __________
Domain ____________
Range ______________
Function? __________ Function? __________
Domain ____________ Domain ____________
Range______________ Range______________
Algebra 1 Functions Test STUDY GUIDE Page 3
5) Graph the function with the given domain using the table of values method. Then
identify the range of the function in set builder notation. The domain is the set of all
real numbers, D = {x | x}.
a) y = 4x + 3
b) y = -
1
x -1
2
6) Find the x- and y-intercepts of the graphs of the equations as ordered pairs. Then graph
the functions using the intercept method.
a) 3x – 2y = 12
x-intercept
b) 4x – y = -8
y-intercept
x-intercept
3
x+3
4
x-intercept
y-intercept
c) y = y-intercept
7) Evaluate the functions below.
a) f(x) = -5x + 4
Find: f(1), f(-3), f(0)
3
x 7
4
Find: g(0), g(4), g(-8)
b) g(x) =
c) h(x) = 3x + 9
1
Find: h , h(0), h(9)
3
Find the zero of h (the x-intercept)
Algebra 1 Functions Test STUDY GUIDE Page 4
8) Given f(x) = 4x – 1, complete the table below for the function.
x
-1
f(x)
0
3
0
7
9) Find the zero of each function (the x-intercept) below.
a) f(x) = 4x + 16
b) h(x) =
2
x 10
3
10) Put the linear equations below into function form (solve for y).
a) 15x + 10 = 5y
b) -3x + 4y = 24
c)
3x 6y
2
12
d) 10x – 2y + 20 = 0
11) Solve each equation for the given variable.
a) Solve for t:
v = r + at
b) Solve for x:
5xy n
6
2
c) Solve for a:
4a + b = 3a
d) Solve for n:
n
s (a t)
2
e) Solve for c:
3x y
4
c
f) Solve for C:
9
F = C+32
5
1
h(b1 b 2 ) , where b1 and b2
2
are the lengths of the bases of the trapezoid.
12) The area of a trapezoid is given by A =
a) Solve the formula for h, the height of the trapezoid.
b) Use the formula you created in part a above to find the height of
a trapezoid if its area is 40 square inches, and its bases are 10
and 6 inches in length.
Algebra 1 Functions Test STUDY GUIDE Page 5
2) a) Yes, every input has OAOO output.
1) a) P III, Q no
b) A(-3, -4) III, B(0, -4)
II
b) No, input 5 has more than one output.
c) Yes, every input has OAOO output.
d) No, there exist inputs with more than one
output.
3) a) R ={11, 8, 2, -1, -10}
5
b) R = {-5, -4, -3, , 0}
2
4) a) No; D={x | x ≥ 0}; R = {y x | y}
5) (Tables may vary depending on inputs
5b)
b) Yes; D = {x | x}; R = {y | y}
c) No; D = {x | -5 ≤ x < 0}; R = {y | -1 < y < 4}
d) Yes; D = {x | -4 ≤ x ≤ 4}; R = {y | -3 ≤ y ≤ 4}
chosen)
a)
6) a) x-int: (4, 0); y-int: (0, -6)
6b) x-int: (-2, 0); y-int: (0, 8)
6c)
7) a) f(1)=-1, f(-3)=19, f(0)=4
x-int: (4, 0); y-int: (0, 3)
b) g(0)=-7, g(4)=-4, f(-8)=-13
1
c) h =8, h(0)=9, h(9)=36
3
zero of h occurs at x=-3
8)
9) a) x=-4 b) x=15
10) a) y = 3x + 2 b) y
3
1
x 6 c) y x 4
4
2
d) y = 5x + 10
12) a) h
2A
b) 5 inches
b1 b 2
11) a) t
d) n
vr
-12 - n
b) x
c) a = -b
a
5y
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# Lottery Dream Numbers for the Word “Death”
It’s not fun to dream about anything related to death. However, the experts claim dreaming that you or someone else is dead isn’t necessarily bad. How about using that fact for playing the lottery?
This guide will discover which lottery numbers to use when you dream about death. Check out how to use that to your advantage and boost your odds for the next draw!
## The Best Numbers for the Word “Death”
Did you dream about death, and that’s the word you want to convert to the desired lottery numbers? According to the basic lottery science, you use numbers from 0-9 when deciding the selection for your ticket.
If we convert the word “death” to digits, it would be “45108.” You can use this combination of digits when deciding on the numbers to put on your ticket.
The formula is simple – for the digit “4,” choose a number whose sum of single digits is “4.” Those could be 4, 13, 22, 31, 59, 68, 77, 86, etc. As you can see, the lottery math is applied here. For example, 68 gives us 6+8=14. The lottery math eliminates the first digit and only considers the second.
You use this same principle with other digits provided. That will give you many options when choosing the death dream number for your tickets.
Copy the Best Numbers for the Word “Death” and Play Now!
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### Pick 3 Strategy
The Pick 3 Strategy relies on choosing three numbers for your ticket. You can combine any of those from the digitized formula for the word “death.” So, the formula is “45108.”
Here are some suggestions for the Pick 3 method:
• 451 – 22, 14, 29
• 410 – 13, 38, 28
• 408 – 31, 19, 26
• 510 – 23, 1, 37
• 508 – 5, 19, 35
• 450 – 4, 32, 46
• 458 – 22, 41, 26
• 518 – 14, 29, 17
• 108 – 1, 37, 26
These are only some examples, so don’t hesitate to mix and match based on the number your find attractive. You can combine these numbers with hot, cold, or overdue balls.
### Pick 4 Strategy
Our next example is the Pick 4 Strategy. This approach relies on choosing digits. Since the word death has five – 45108, we’ll have to make a selection from any four of the five digits.
Here are the potential 4-digit lottery number for death to use in the lottery:
• 4510 – 22, 14, 29, 37
• 4518 – 13, 32, 38, 17
• 4508 – 4, 23, 46, 35
• 4108 – 31, 29, 19, 26
• 5108 – 5, 56, 28, 17
As long as you get the desired digits, you can mix and match any way you see fit. Also, don’t forget to use lottery software if you need other numbers for your ticket.
### Pick 5 Strategy
This strategy seems to be the best-fitting out of all the “pick” options. That’s because five letters in the word death generate five digits – 45108.
Here is how your grid could look with the Pick 5 Strategy:
4, 41, 29, 46, 35 4, 32, 1, 19, 35 4, 23, 1, 19, 35 4, 23, 1, 19, 8 13, 32, 38, 19, 26 13, 14, 47, 46, 26 13, 41, 47, 28, 26 13, 41, 29, 19, 35 22, 14, 1, 28, 37 22, 23, 29, 19, 37 22, 32, 29, 19, 37 22, 41, 38, 46, 8 31, 23, 47, 19, 8 31, 41, 38, 28, 8 31, 41, 47, 28, 26 31, 14, 29, 28, 37
You can work out other potential combinations. If you want to shorten the selection, use these tips:
• Pick only combinations with the appropriate median value (Highest potential lotto number/2=Sum of all numbers chosen / 5).
• Use only combinations that have balanced odd and even numbers.
• Use only those options that have balanced high and low numbers.
### Powerball
The trick with US Powerball is that it uses a two-drum concept. The exact formula is 5/69+1/26. Dream numbers for the word death fit the basic drum well. As for the PB number, use any digit from the formula and come up with a suitable number from 1 to 26.
So, our starting position is 45108. Here are some Powerball tickets to try:
• 4, 50, 65, 37, 26 + PB 13.
• 13, 32, 38, 55, 62 + PB 23.
• 22, 41, 56, 64, 8 + PB 19.
### MegaMillions
MegaMillions is similar to Powerball. It has generous jackpots and comes with a two-drum concept. You use the formula “45108” for the lotto’s concept of 5/75 + 1/15.
Here are some suggestions:
• 31, 69, 47, 19, 8 + MB 14.
• 4, 23, 38, 73, 62 + MB 10.
• 40, 32, 74, 28, 17 + MB 8.
## What Does It Mean If You Dream Death?
There are multiple ways to dream “death.” The explanation depends on the dream’s specifics. Check out the list below and find a suitable one:
• You died in a dream. Your life is changing. Perhaps it’s time to say goodbye to an old home, job, or friend. A violent death could indicate a change you can’t wait to happen.
• Someone you knew died in your dream. That indicates you worry about that person more than usual. Perhaps you worry about losing what that person represents in your life.
• Dreaming deceased friends and loved ones. This is common among patients who deal with a terminal disease. They also happen shortly after losing a friend and present a way of coping.
• Seeing a dead body. If you see a dead body, you might be expecting to leave something behind you. That could include a home or job. An autopsy indicates you try to figure out why that happened.
• Pets and plants are dying in your dream. You care about these, so dreaming about them dying indicates you neglected an area of your life.
• You killed someone in your dream. If this happens, you are eager to make some changes. That could be switching jobs, adjusting habits, etc.
## Can You Pick Lottery Numbers Based on Your Dream?
Many people believe that it’s possible to choose lottery numbers based on your dreams. Do you believe that dreams send us messages? If yes, who’s to say they aren’t helping us to win the lottery? Our task is to ensure that we analyze the dreams properly and figure out the right lottery numbers. If you do that, you can boost your winning odds easily.
## Do You Use the Same Numbers for All Lotteries?
You have two approaches here. The first one is to check lottery dream numbers and see which number is connected to what you dreamt. That way, you can use that lucky number(s) for any lottery out there.
Another approach is to use lottery math. It analyzes the keyword from your dream and converts it to digits. Those digits allow you to pick a wide range of lottery numbers. As we mentioned, the digit “4” allows you to play 4, 13, 22, 31, and 40. That gives you more options, so you don’t have to stick to the same ones for all lotteries.
## Does Picking the Numbers Based on This Dream Guarantee a Win?
We always emphasize that nothing can guarantee a win when playing the lottery. You can win by purchasing all combinations, but the odds are that won’t be profitable. If you play a limited number of tickets, you’ll need the luck to win. While strategies and dream numbers might boost your odds, it’s impossible to win without being lucky.
## Final Thoughts
The moment you wake up after dreaming about death can be terrifying. Fortunately, the experts suggest those dreams don’t have to mean anything negative. We hope this guide helped you to utilize your dreams when playing your lottery tickets for today. Don’t hesitate to experiment with different strategies and pick your favorite!
## FAQs
What role do death numbers play in the lottery?
Playing death numbers in the lottery may help you to get lucky in the lottery. However, they do not offer any specific guarantee of winning the lottery. While the lottery number for death may help you to make a calculated lottery decision, you will also need a good amount of luck for success in the lottery.
What is the dead man number in the lottery?
The dead man number is the number to play when someone dies. It is the second number in the fate number, after the date of birth. Suppose someone died on the 15th of October 2022. We will add the digits using lottery mathematics: 1+5 + 1+0 + 2+0+2+2 = 13. The dead number in lottery will be 13.
What is the lottery number for death?
The lottery number for death is “45108”. We get this by using their position on the alphabet and combining it with lottery science. You can employ the lottery number for death to create your Pick 3, Pick 4, and Pick 5 lotteries.
What is the lottery number for talking to the dead?
The lottery number for talking to the dead in your dream is 729. If you only win them without talking to them, the lottery number is 769.
What does seeing the dead play for in the lottery?
Seeing the dead in your dream has its own interpretation in the lottery. It could imply that you could get lucky if you play the dead man numbers in your next lottery game. Simply follow the steps in the article for your next Pick 3, Pick 4, or Pick 5 lottery draw to get the lottery numbers for death.
Get 20% OFF on your Powerball OR Mega Millions Tickets!
\$102 Million
Monday, Jul 22, 2024
\$279 Million
Tuesday, Jul 23, 2024 |
Class 9 Maths
# Number System
## Exercise 1.3 Part 3
Question 5: What can the maximum number of digits be in the repeating block of digits in the decimal expansion of 1/17 ? Perform the division to check your answer.
= 0.0588235294117647
Thus, maximum number of digits in the repeating block is 17.
Question 6: Look at several examples of rational numbers in the form p/q (q ≠ 0), where p and q are integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy.
Answer: For having terminating decimal expansions, the denominator should have either 2 or 5 or both as factor. So, q must have either 2 or 5 or both.
Examples:
1/2=0.5 terminating
1/3 = 0.3 non-terminating repeating
1/4=0.25 terminating
1/5=0.2 terminating
1/6 = 0.16 non-terminating repeating
1/7 = 0.142857 non-terminating repeating
1/8=0.125 terminating
1/9 = 0.1 non-terminating repeating
Question 7: Write three numbers whose decimal expansions are non-terminating non-recurring.
Answer: Non-terminating non-recurring numbers are known as irrational numbers. Irrational numbers cannot be expressed in the form of p/q where q≠0.
Following are the possible numbers:
0.72012001200012000001………
0.73013001300013000001…………
0.7501500150001500001………..
Question 8: Find three different irrational numbers between the rational numbers 5/7 and 9/11.
Answer: 5/7 = 0.714285714285…….. and 9/11 = 0.8181818………
Possible irrational numbers between them can be as follows:
0.72012001200012000001………
0.73013001300013000001…………
0.7501500150001500001………..
Note: Non-terminating non-recurring numbers are known as irrational numbers. Irrational numbers cannot be expressed in the form of p/q where q≠0. Numbers given above cannot be expressed in the form of p/q and hence are irrational. |
# Solving linear equations: the addition property (Page 2/2)
Page 2 / 2
From this, we can suggest the addition/subtraction property of equality .
Given any equation,
1. We can obtain an equivalent equation by adding the same number to both sides of the equation.
2. We can obtain an equivalent equation by subtracting the same number from both sides of the equation.
## The idea behind equation solving
The idea behind equation solving is to isolate the variable on one side of the equation. Signs of operation (+, -, ⋅,÷) are used to associate two numbers. For example, in the expression $5+3$ , the numbers 5 and 3 are associated by addition. An association can be undone by performing the opposite operation. The addition/subtraction property of equality can be used to undo an association that is made by addition or subtraction.
Subtraction is used to undo an addition.
Addition is used to undo a subtraction.
The procedure is illustrated in the problems of [link] .
## Sample set b
Use the addition/subtraction property of equality to solve each equation.
$x+4=6$ .
4 is associated with $x$ by addition. Undo the association by subtracting 4 from both sides.
$\begin{array}{c}x+4-4=6-4\hfill \\ x+0=2\hfill \\ x=2\hfill \end{array}$
Check: When $x=2$ , $x+4$ becomes
The solution to $x+4=6$ is $x=2$ .
$m-8=5$ . 8 is associated with $m$ by subtraction. Undo the association by adding 8 to both sides.
$\begin{array}{c}m-8+8=5+8\hfill \\ m+0=\text{13}\hfill \\ m=\text{13}\hfill \end{array}$
Check: When $m=\text{13}$ ,
becomes
a true statement.
The solution to $m-8=5$ is $m=\text{13}$ .
$-3-5=y-2+8$ . Before we use the addition/subtraction property, we should simplify as much as possible.
$-3-5=y-2+8$
$-8=y+6$
6 is associated with $y$ by addition. Undo the association by subtracting 6 from both sides.
$\begin{array}{c}-8-6=y+6-6\hfill \\ -\text{14}=y+0\hfill \\ -\text{14}=y\hfill \end{array}$
This is equivalent to $y=-\text{14}$ .
Check: When $y=-\text{14}$ ,
$-3-5=y-2+8$
becomes
,
a true statement.
The solution to $-3-5=y-2+8$ is $y=-\text{14}$ .
$-5a+1+6a=-2$ . Begin by simplifying the left side of the equation.
$\underset{-5+6=1}{\underbrace{-5a+1+6a}}=-2$
$a+1=-2$ 1 is associated with $a$ by addition. Undo the association by subtracting 1 from both sides.
$\begin{array}{c}a+1-1=-2-1\hfill \\ a+0=-3\hfill \\ a=-3\hfill \end{array}$
Check: When $a=-3$ ,
$-5a+1+6a=-2$
becomes
,
a true statement.
The solution to $-5a+1+6a=-2$ is $a=-3$ .
$7k-4=6k+1$ . In this equation, the variable appears on both sides. We need to isolate it on one side. Although we can choose either side, it will be more convenient to choose the side with the larger coefficient. Since 8 is greater than 6, we’ll isolate $k$ on the left side.
$7k-4=6k+1$ Since $6k$ represents $+6k$ , subtract $6k$ from each side.
$\underset{7-6=1}{\underbrace{7k-4-6k}}=\underset{6-6=0}{\underbrace{6k+1-6k}}$
$k-4=1$ 4 is associated with $k$ by subtraction. Undo the association by adding 4 to both sides.
$\begin{array}{c}k-4+4=1+4\hfill \\ k=5\hfill \end{array}$
Check: When $k=5$ ,
$7k-4=6k+1$
becomes
a true statement.
The solution to $7k-4=6k+1$ is $k=5$ .
$-8+x=5$ . -8 is associated with $x$ by addition. Undo the by subtracting -8 from both sides. Subtracting -8 we get $-\left(-8\right)\text{=+}8$ . We actually add 8 to both sides.
$-8+x+8=5+8$
$x=\text{13}$
Check: When $x=\text{13}$
$-8+x=5$
becomes
,
a true statement.
The solution to $-8+x=5$ is $x=\text{13}$ .
## Practice set b
$y+9=4$
$y=-5$
$a-4=\text{11}$
$a=\text{15}$
$-1+7=x+3$
$x=3$
$8m+4-7m=\left(-2\right)\left(-3\right)$
$m=2$
$\text{12}k-4=9k-6+2k$
$k=-2$
$-3+a=-4$
$a=-1$
## Exercises
For the following 10 problems, verify that each given value is a solution to the given equation.
$x-\text{11}=5$ , $x=\text{16}$
Substitute $x=4$ into the equation $4x-\text{11}=5$ .
$\begin{array}{}\text{16}-\text{11}=5\\ 5=5\end{array}$
$x=4$ is a solution.
$y-4=-6$ , $y=-2$
$2m-1=1$ , $m=1$
Substitute $m=1$ into the equation $2m-1=1$ .
$m=1$ is a solution.
$5y+6=-\text{14}$ , $y=-4$
$3x+2-7x=-5x-6$ , $x=-8$
Substitute $x=-8$ into the equation $3x+2-7=-5x-6$ .
$x=-8$ is a solution.
$-6a+3+3a=4a+7-3a$ , $a=-1$
$-8+x=-8$ , $x=0$
Substitute $x=0$ into the equation $-8+x=-8$ .
$x=0$ is a solution.
$8b+6=6-5b$ , $b=0$
$4x-5=6x-\text{20}$ , $x=\frac{\text{15}}{2}$
Substitute $x=\frac{\text{15}}{2}$ into the equation $4x-5=6x-\text{20}$ .
$x=\frac{\text{15}}{2}$ is a solution.
$-3y+7=2y-\text{15}$ , $y=\frac{\text{22}}{5}$
Solve each equation. Be sure to check each result.
$y-6=5$
$y=\text{11}$
$m+8=4$
$k-1=4$
$k=5$
$h-9=1$
$a+5=-4$
$a=-9$
$b-7=-1$
$x+4-9=6$
$x=\text{11}$
$y-8+\text{10}=2$
$z+6=6$
$z=0$
$w-4=-4$
$x+7-9=6$
$x=8$
$y-2+5=4$
$m+3-8=-6+2$
$m=1$
$z+\text{10}-8=-8+\text{10}$
$2+9=k-8$
$k=\text{19}$
$-5+3=h-4$
$3m-4=2m+6$
$m=\text{10}$
$5a+6=4a-8$
$8b+6+2b=3b-7+6b-8$
$b=-\text{21}$
$\text{12}h-1-3-5h=2h+5h+3\left(-4\right)$
$-4a+5-2a=-3a-\text{11}-2a$
$a=\text{16}$
$-9n-2-6+5n=3n-\left(2\right)\left(-5\right)-6n$
## Calculator exercises
$y-2\text{.}\text{161}=5\text{.}\text{063}$
$y=7\text{.}\text{224}$
$a-\text{44}\text{.}\text{0014}=-\text{21}\text{.}\text{1625}$
$-0\text{.}\text{362}-0\text{.}\text{416}=5\text{.}\text{63}m-4\text{.}\text{63}m$
$m=-0\text{.}\text{778}$
$8\text{.}\text{078}-9\text{.}\text{112}=2\text{.}\text{106}y-1\text{.}\text{106}y$
$4\text{.}\text{23}k+3\text{.}\text{18}=3\text{.}\text{23}k-5\text{.}\text{83}$
$k=-9\text{.}\text{01}$
$6\text{.}\text{1185}x-4\text{.}\text{0031}=5\text{.}\text{1185}x-0\text{.}\text{0058}$
$\text{21}\text{.}\text{63}y+\text{12}\text{.}\text{40}-5\text{.}\text{09}y=6\text{.}\text{11}y-\text{15}\text{.}\text{66}+9\text{.}\text{43}y$
$y=-\text{28}\text{.}\text{06}$
$0\text{.}\text{029}a-0\text{.}\text{013}-0\text{.}\text{034}-0\text{.}\text{057}=-0\text{.}\text{038}+0\text{.}\text{56}+1\text{.}\text{01}a$
## Exercises for review
( [link] ) Is $\frac{7\text{calculators}}{\text{12}\text{students}}$ an example of a ratio or a rate?
rate
( [link] ) Convert $\frac{3}{8}\text{}$ % to a decimal.
( [link] ) 0.4% of what number is 0.014?
3.5
( [link] ) Use the clustering method to estimate the sum: $\text{89}+\text{93}+\text{206}+\text{198}+\text{91}$
( [link] ) Combine like terms: $4x+8y+\text{12}y+9x-2y$ .
$\text{13}x+\text{18}y$
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
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
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers! |
# Nines of nines
In the operations business we like to talk about nines of things, especially regarding service levels.
If
• “one nine of availability” = available 0.9 of the time,
• “two nines of availability” = available 0.99 of the time,
• and so on…
then generally,
• “$$n$$ nines of availability” = available $$(1 - 10^{-n})$$ of the time,
right?
This works for any whole number n: e.g. 5 nines is \begin{align}1 - 10^{-5} &= 1 - 0.00001 \\ &= 0.99999.\end{align}
There’s a problem with this simple generalisation, and that is, when people say “three and a half nines” the number they actually mean doesn’t fit the pattern. “Three and a half nines” means 0.9995, but
• $$1 - 10^{-3.5} \approx 0.9996838$$, and going the other way,
• $$0.9995 \approx 1 - 10^{-3.30103}$$.
We could resolve this difficulty by saying “3.3ish nines” when we mean 0.9995, or by meaning ~0.9996838 when we say “three and a half nines.” But there’s at least one function that fits the half-nines points as well!
Let’s start with the function above: $$f(n) = 1 - 10^{-n}.$$ For every odd integer, it just has to be lower by a small, correspondingly decreasing amount. We can do this by increasing the exponent of 10 by \begin{align}k &= 0.5 + \log_{10}(0.5) \\ &\approx 0.19897.\end{align}
One function for introducing a perturbation for halfodd integers is $$p(n) = \sin^2(\pi n).$$ When n is a whole integer, $$p(n) = 0$$, and when $$n$$ is half an odd integer, $$p(n) = 1$$. Multiply this function by some constant and you’re in business.
Thus, define a new function $$g(n)$$ for all $$n$$:
$$g(n) := 1 - 10^{-n + k p(n)}$$
i.e.
$$g(n) = 1 - 10^{-n + (0.5 + \log_{10}(0.5))\sin^2(\pi n)}$$
which, when plotted, looks like this:
a negative exponential curve with a negative exponential wiggle. And it has the desired property that at every integer and half-integer it has a value with the traditional number of nines and trailing five (or not). |
# Zero exponent/Neg. exponent Absent Thurs/Fri 10/31-11/1
## Presentation on theme: "Zero exponent/Neg. exponent Absent Thurs/Fri 10/31-11/1"— Presentation transcript:
Zero exponent/Neg. exponent Absent Thurs/Fri 10/31-11/1
Pattern example Zero/Neg. exponent Property
2 21 2 22 4 23 8 What Pattern do we see when we are moving down? As the exponent increases by 1 number than the answer doubles itself. What pattern do we see when we are moving up? As the exponent decreases by 1 number than the answer is reduced by half. What does the 2-1 turn into? The Negative exponent turns into a fraction. Neg. exponents always turn into fractions.
Example #1 50 1 Solution Simplify: (3 + 2)0 What can we do first?
Use GEMA and add what’s inside the groupings. What happens when there is a exponent of 0? Anytime there is an exponent of 0 then the base (factor) becomes 1 1
Example 2 1 Simplify: -64 What is the base (factor)?
The factor is (-4) What is the exponent? The exponent is -3 How do we re-write the problem? We have to write the problem as a fraction because of the neg. exponent. It makes it Positive. What do we do next? Write out the factor in expanded form with ( ). 3 neg signs. Odd # What is the last step? To multiply the factors which are part of the denominator. Simplify: (-4)-3 1 (-4)3 (-4)(-4)(-4) -64 Solution 1 -64
Example 3 -2 = -1 4 2 Simplify: (4)-1 – (3 + 2)0 + (2)-2 1 + -4 + 1
What are the first step? Use GEMA and add what’s inside the groupings. How do we re-write the expression when we have zero’s & neg. exponents? We write all zero exponents as 1 and all neg. exponents as a fraction. What is the next step? We multiply the exponents. What do we have to do first before adding the fractions? We have to get all the denominators to be the same. What's are last step? We can now add all the numerators using the integer rules. Simplify: (4)-1 – (3 + 2)0 + (2)-2 (4)-1 - (5)0 + (2)-2 (4) (2)2 Solution = -1
Example 4 Simplify: 3(1 – 4)-2 + (9)-1 + 120 What are the first step?
3(-3) (9) (-3)2 (9) 13 9 What are the first step? Use GEMA and add what’s inside the groupings. How do we re-write the expression when we have zero’s & neg. exponents? We write all zero exponents as 1 and all neg. exponents as a fraction. What is the next step? We multiply the exponents. There are 2 neg. signs with the (-3) even #. What do we have to do first before adding the fractions? We have to get all the denominators to be the same. What is the last step? We can now add all the numerators using the integer rules. Reduce the answer. 13 9 |
Decameter To Meters Calculator
For a Mathematics student, it is common to get questions related to converting between various units. One of the most common tasks is to convert units related to length. The conversion of multiple units to meters is the main task that many students perform. For some students, it can be a complicated process when they don’t know the relationship between the concerned units.
Among other converting units, we are here to discuss Decameter to Meters conversion. If you think it will be a difficult process, you can use the Decameter to Meters Calculator. This handy tool has been designed to convert a measurement from Decameters to Meters quickly. You can use this Maths Calculator using simple steps that we will discuss later.
What is Decameter?
A Decameter is a particular unit used to measure the length of standard objects. This measure of length is neither used for small objects nor for extra-large ones. It is based on the SI unit that we call meter and discuss in the following section.
Based on the meters, they can be converted to meters within a few minutes. You can do this manually or follow the steps to use Decameter to Meters Calculator. To show a measurement in this unit, “dam” is used.
What is a Meter?
It is the SI unit of length that is used all over the globe. A meter is considered the unit used for small objects like clothes, box sizes, and other similar objects. For showing measurement in meters, “m” is used just beside the number.
For example, if we want to say 5 meters, we will write it as “5 m”. This SI unit has many extended units like centimeters, kilometers, and decimeters. These units are based on the meter and given names as the fraction of this base unit.
How to calculate Decameter to Meter?
To understand the method to calculate Decameter to meters, you need to know the relationship between these units . A decimeter is a bigger unit of meter and includes 10 meters. It means that 10 meters are combined to make a decameter.
1 Decameter = 10 meters
Similarly, we can also say that,
1 meter = 1/10 Decameters
To make it understandable, we have also shared a solved example of this conversion here.
Example 1:
Convert 12 Decameters to Meters.
Solution:
We know that,
1 Dam = 10 m
So,
12 Dam = 12 x 10 m
= 120 m
Example 2:
Convert 900 meters to Decameters.
Solution:
As we know,
1 meter = 1/10 Decameters
So,
900 m = 900/10 Dam
= 90 Dam
DM to M Table
For the ease of our learners, we have also designed a conversion table for Decameters to Meters.
Decameters Meters
1 10
2 20
5 50
10 100
20 200
50 500
100 1000
500 5000
1000 10000
How to use the Decameter to Meters Calculator?
Sometimes, you may get decimals to convert from Decameters to Meters. It will be a difficult process that you can make easier using this tool by Calculator’s Bag. Here are the steps to follow for this purpose.
• Insert the measurement given in Decameters
• This tool will automatically convert the measurement in meters
FAQ | Decameter to Meters
How many meters are in a decameter?
10 meters will be equal to 1 decameter.
Is a meter bigger than a decameter?
No, the meter is smaller as well as the base unit for Decameter.
What is the relation between dm and m?
Here is the relation between dm and m,
1 dam = 10 m
What is the difference between a decameter and a meter?
A decimeter is a bigger unit of length and has been derived from the meter. But a meter is a base unit of length that is used for measuring small lengths.
How many decameters are in a meter?
0.1 decameter will be equal to 1 meter. |
## Properties of Polygons: Overview
A polygon is a simple closed plane figure made up of three or more line segments. Examples of polygons can be found everywhere. Doors, chalkboards, and walls are usually rectangular in shape. Triangles can be found in buildings, bridges, and other structures. Squares can be found in tiles on a floor or a ceiling. Hexagons and other polygons can be found in signs, in windows, and in nature. Understanding the properties of polygons can help students make sense of the world around them.
It is important that students understand the geometric terminology associated with polygons. A point is a location in space. A line is a straight, continuous, unending set of points. Since a point has no dimensions and a line has only one dimension, it is impossible to see either of them. However, we can see representations of these ideas. For example, the corner of a room where two walls and the ceiling come together or the end of a sharp pencil could represent a point. A line segment is a part of a line with two endpoints. The part of a room where two walls meet and the edge of a ruler are representations of line segments. A plane is a collection of points that forms a flat, continuous, and unending surface (imagine a wall or the surface of a book cover continuing forever in all directions). Space is the collection of all points.
A ray is a half-line−a straight, continuous, and unending set of points with one endpoint. If two rays have a common endpoint they form an angle, and the common endpoint is called the vertex of the angle. When two lines intersect, the angles opposite one another are called vertical angles. In the diagram below, a and c are vertical angles, and b and d are vertical angles. Vertical angles are congruent, that is, they have the same measure.
Adjacent angles are angles that share a common vertex and a common ray and whose interiors do not intersect. The angles next to each other in the diagram above-namely, a and b, b and c, c and d, and d and a−are adjacent angles.
Acute angles have measures less than 90 degrees. Obtuse angles have measures greater than 90 degrees but less than 180 degrees. In the diagram above, a and c are acute angles, and b and d are obtuse angles. Angles whose measures add to 180 degrees are supplementary angles. In the diagram above, a and b are supplementary angles.
An angle with measure 90 degrees, such as RSU below, is a right angle. Two angles whose measures add to 90 degrees, such as m and n below, are complementary angles.
A central angle is an angle whose vertex is the center of the circle and whose sides intersect the circle. In the diagram below, AOC is a central angle.
A triangle is a polygon with three sides. The triangle does not include the region enclosed by its sides; that region is called the interior of the triangle. One way to classify triangles is by the lengths of their sides. A scalene triangle, such as triangle RST below, has three sides with different lengths. A triangle with at least two congruent sides is an isosceles triangle. In the diagram below, triangles ABC and XYZ are both isosceles triangles. Triangle ABC is also called an equilateral triangle, since all three of its sides are congruent. In fact, every equilateral triangle is an isosceles triangle.
Triangles can also be classified according to their angle measures. A right triangle has one right angle. An acute triangle has three acute angles. An obtuse triangle has one obtuse angle. The sum of the angle measures for any triangle is 180 degrees. If we combine these two ways to classify triangles, we come up with seven different types of triangles: acute scalene, right scalene, obtuse scalene, acute isosceles, right isosceles, obtuse isosceles, and acute equilateral. There are no right equilateral or obtuse equilateral triangles.
A quadrilateral is a polygon with four sides. There are many types of quadrilaterals, and many of them share some common properties. A quadrilateral with exactly one pair of parallel sides is called a trapezoid. In the diagram below, ABCD is a trapezoid. If a quadrilateral has two pairs of parallel sides, then it is called a parallelogram. In a parallelogram the opposite sides and opposite angles are congruent. All of the figures below except ABCD are parallelograms. A rectangle is a parallelogram with four right angles. In the diagram below, figures IJKL and QRST are both rectangles. A parallelogram with four congruent sides is called a rhombus. Figures MNOP and QRST below are rhombuses. A square is a rectangle that is also a rhombus. In other words, a square is a parallelogram with four right angles and four congruent sides. Figure QRST is a square.
Two polygons are congruent if they are the same size and shape. If two figures are congruent, then their corresponding parts are also congruent. For example, quadrilateral ABCD is congruent to quadrilateral WXYZ, so A is congruent to W, D is congruent to Z, and so on. Also, side AD is congruent to side WZ, side DC is congruent to side ZY, and so on.
Polygons that have the same shape, but that are not necessarily the same size, are said to be similar. For example, all equilateral triangles are similar to one another and all squares are similar to one another. If two figures are similar, the corresponding angles are congruent, and the corresponding sides are proportional. All congruent figures are similar, but not all similar figures are congruent. In the figure below, quadrilateral ABCD is similar to quadrilateral MNOP, since the corresponding angles are congruent and the corresponding sides are proportional, with a ratio of 3 to 2.
A geometric transformation changes the position of a figure. A translation slides a figure along a straight line in one direction. Translations are sometimes referred to as slides. In the diagram below, PQRS was translated to HIJK. The figure resulting from a translation is congruent to the original figure.
A rotation rotates, or turns, a figure about a given point. Rotations are sometimes referred to as turns. A figure and its image after a rotation are congruent to one another. In the diagram below, PQRS has been rotated 90 degrees counterclockwise about point P to figure PQ'R'S'.
A reflection is a transformation that reflects a figure across a line. The figures resulting from a reflection, or flip, are congruent to the original figures. In the figure below, ABCD has been reflected about line MN to figure A'B'C'D'. |
# Common Core: 8th Grade Math : Describe Qualitatively the Functional Relationship Between Two Quantities by Analyzing a Graph: CCSS.Math.Content.8.F.B.5
## Example Questions
### Example Question #1 : Describe Qualitatively The Functional Relationship Between Two Quantities By Analyzing A Graph: Ccss.Math.Content.8.F.B.5
The graph provided displays the speed of a race car during a race.
Based on the graph, what is likely occurring between the and the period of the race?
The race car driver is increasing his speed
The race car driver is waiting to begin the race
The race car driver s decreasing his speed
The race car driver is driving at a consistent speed
The race car driver is driving at a consistent speed
Explanation:
The time is displayed along the x-axis of the provided graph. Between the and the period of the race, the driver is going at each time point. We can see from the straight line between those two points that the race car driver's speed doesn't change; thus, the race car driver is driving at a consistent speed.
### Example Question #2 : Describe Qualitatively The Functional Relationship Between Two Quantities By Analyzing A Graph: Ccss.Math.Content.8.F.B.5
The graph provided displays the speed of a race car during a race.
Based on the graph, what is likely occurring during the last of the race?
The race car driver is driving at a consistent speed
The race car driver is waiting to begin the race
The race car driver is decreasing his speed
The race car driver is increasing his speed
The race car driver is decreasing his speed
Explanation:
The time is displayed along the x-axis of the provided graph. During the last of the race, the driver goes from , displayed on the y-axis of the graph, to . We can see the line between those two points is decreasing in a negative direction, or getting slower; thus, the race car driver is decreasing his speed.
### Example Question #3 : Describe Qualitatively The Functional Relationship Between Two Quantities By Analyzing A Graph: Ccss.Math.Content.8.F.B.5
The graph provided displays the speed of a race car during a race.
Based on the graph, what is likely occurring in the first of the race?
The race car driver is decreasing his speed
The race car driver is increasing his speed
The race car driver is waiting to begin the race
The race car driver is driving at a consistent speed
The race car driver is increasing his speed
Explanation:
The time is displayed along the x-axis of the provided graph. During the first of the race, the driver goes from , displayed on the y-axis of the graph, to . We can see the line between those two points is increasing in a positive direction, or getting faster; thus, the race car driver is increasing his speed.
### Example Question #4 : Describe Qualitatively The Functional Relationship Between Two Quantities By Analyzing A Graph: Ccss.Math.Content.8.F.B.5
The graph provided displays the speed of a race car during a race.
Based on the graph, what is likely occurring when the time is at
The race car driver is increasing his speed
The race car driver is driving at a consistent speed
The race car driver is decreasing his speed
The race car driver is waiting to begin the race
The race car driver is waiting to begin the race
Explanation:
When the time is at , displayed along the x-axis, we see that the speed, displayed along the y-axis is also . If a car has a speed of , the car is not moving. And if the time is at , the race has not started; thus, the race car driver is waiting to begin the race.
### Example Question #1 : Describe Qualitatively The Functional Relationship Between Two Quantities By Analyzing A Graph: Ccss.Math.Content.8.F.B.5
The graph provided displays the speed of a race car during a race.
Based on the graph, what is likely occurring in the first of the race?
The race car driver is increasing his speed
The race car driver is decreasing his speed
The race car driver is waiting to begin the race
The race car driver is driving at a consistent speed
The race car driver is increasing his speed
Explanation:
The time is displayed along the x-axis of the provided graph. During the first of the race, the driver goes from , displayed on the y-axis of the graph, to . We can see the line between those two points is increasing in a positive direction, or getting faster; thus, the race car driver is increasing his speed.
### Example Question #6 : Describe Qualitatively The Functional Relationship Between Two Quantities By Analyzing A Graph: Ccss.Math.Content.8.F.B.5
The graph provided displays the speed of a race car during a race.
Based on the graph, what is likely occurring between the and the period of the race?
The race car driver is driving at a consistent speed
The race car driver is waiting to begin the race
The race car driver is increasing his speed
The race car driver is decreasing his speed
The race car driver is driving at a consistent speed
Explanation:
The time is displayed along the x-axis of the provided graph. Between the and the period of the race, the driver is going at each time point. We can see from the straight line between those two points that the race car driver's speed doesn't change; thus, the race car driver is driving at a consistent speed.
### Example Question #7 : Describe Qualitatively The Functional Relationship Between Two Quantities By Analyzing A Graph: Ccss.Math.Content.8.F.B.5
The graph provided displays the speed of a race car during a race.
Based on the graph, what is likely occurring in the last of the race?
The race car driver is driving at a consistent speed
The race car driver is decreasing his speed
The race car driver is increasing his speed
The race car driver is waiting to begin the race
The race car driver is decreasing his speed
Explanation:
The time is displayed along the x-axis of the provided graph. During the last of the race, the driver goes from , displayed on the y-axis of the graph, to . We can see the line between those two points is decreasing in a negative direction, or getting slower; thus, the race car driver is decreasing his speed.
### Example Question #1 : Describe Qualitatively The Functional Relationship Between Two Quantities By Analyzing A Graph: Ccss.Math.Content.8.F.B.5
The graph provided displays the speed of a race car during a race.
Based on the graph, what is likely occurring when the time is at
The race car driver is driving at a consistent speed
The race car driver is waiting to begin the race
The race car driver is decreasing his speed
The race car driver is increasing his speed
The race car driver is waiting to begin the race
Explanation:
When the time is at , displayed along the x-axis, we see that the speed, displayed along the y-axis is also . If a car has a speed of , the car is not moving. And if the time is at , the race has not started; thus, the race car driver is waiting to begin the race.
### Example Question #1 : Describe Qualitatively The Functional Relationship Between Two Quantities By Analyzing A Graph: Ccss.Math.Content.8.F.B.5
The graph provided displays the speed of a race car during a race.
Based on the graph, what is likely occurring in the first of the race?
The race car driver is decreasing his speed
The race car driver is increasing his speed
The race car driver is driving at a consistent speed
The race car driver is waiting to begin the race
The race car driver is increasing his speed
Explanation:
The time is displayed along the x-axis of the provided graph. During the first of the race, the driver goes from , displayed on the y-axis of the graph, to . We can see the line between those two points is increasing in a positive direction, or getting faster; thus, the race car driver is increasing his speed.
### Example Question #313 : Grade 8
The graph provided displays the speed of a race car during a race.
Based on the graph, what is likely occurred between the and part of the race?
The race car driver increased his speed
The race car driver stopped
The race car driver slowed down
The race car driver finished the race |
# Prime spiral (Ulam spiral)
Ulam spiral
The Ulam spiral, or prime spiral, is a plot in which prime numbers are marked among positive integers that are arranged in a counterclockwise spiral. The prime numbers show a pattern of diagonal lines.
# Basic Description
The prime spiral was discovered by Stanislaw Ulam (1909-1984) in 1963 while he was doodling on a piece of paper during a science meeting. Starting with 1 in the middle, he wrote positive numbers in a grid as he spiraled out from the center, as shown in Image 1. He then circled the prime numbers, and the prime numbers showed patterns of diagonal lines as shown by the grid in Image 2. The grid in Image 2 is a close-up view of the center of the main image such that the green line segments and red boxes in the center of the main image line up with those in Image 2.
Image 1
Image 2
A larger Ulam spiral with 160,000 integers and 14,683 primes is shown in the main image. Black dots indicate prime numbers. In addition to diagonal line segments formed by the black dots, we can see white vertical and horizontal line segments that cross the center of the spiral and do not contain any black dots, or prime numbers. There are also white diagonal line segments that do not contain any prime numbers. Ulam spiral implies that there is some order in the distribution of prime numbers.
# A More Mathematical Explanation
From time immemorial, humans have tried to discover patterns among prime numbers. Currently, there is [...]
From time immemorial, humans have tried to discover patterns among prime numbers. Currently, there is no known simple formula that yields all the primes. The diagonal patterns in the Ulam spiral gives some hint for formulas of primes numbers.
Definition of Half-lines
Many half-lines in the Ulam spiral can be described using quadratic polynomials. A half-line is a line which starts at a point and continues infinitely in one direction. In this page, we only consider half-lines that are horizontal or vertical, or have a slope of -1 or +1. The half-lines that can be described using quadratic polynomials are the ones in which each entry of the diagonal is positioned on a different ring of the spiral.
The ring of a spiral can be considered as the outermost layer of a concentric square or a rectangle centered around the center of the spiral, 1, as defined by the blue line in the grid. Although there is no exact place where we can determine the beginning and end of a ring, we will assume that entries that are positioned on squares or rectangles of the same sizes to be on the same ring. For instance, 24, 25, 26, 27, 28, which are shown in red boxes in Image 3, are on the same ring, whereas 48, 49, 50, 51, 52, as shown in blue boxes, are on a different ring because they are positioned on a bigger square.
Image 3: Diagonal half-lines
The green lines in the Image 3 qualify as diagonal half-lines whereas the red lines do not because the red lines cross the corner of the grid in a way that two entries of the red diagonal line are positioned on the same ring.
Thus, diagonal line segments that are composed of prime numbers can also be expressed as outputs of quadratic polynomials. To learn more about the relation between the diagonal lines and quadratic polynomials, click below.
First, we need to know the relation between quadratic polynomials and the difference table of sequences. Let's choose the diagonal $5, 19, 41, 71, 109, 155$, the diagonal indicated with green dotted line in Image 3, as our original sequence and create a difference table.
As we can see from the table, the 2nd differences are constant, and this implies that the original sequence can be described by a second degree polynomial. For more information about difference tables and about finding specific polynomials for sequences, go to difference tables.
Image 4: Constant 2nd differences
Now, we will see that the second differences are always constant for any segment of a diagonal half-line. Let's choose the same diagonal with entries $5, 19, 41, 71, 109, 155$. (For a better understanding of the following paragraphs, the reader should wait until the animation shows three different blue rings and then watch the rest of the animation.)
In Image 4, the number of the innermost light blue boxes, which is 14, indicate the distance from $5$ to $19$, or the difference between these two numbers. The number of darker blue boxes in the middle indicate the difference between the numbers $19$ and $41$. The darkest blue boxes on the outside indicate the difference between the entries $41$ and $71$. The number of all these blue boxes correspond to the first differences in the difference table.
In Image 4, the light blue boxes are divided into pieces, and each piece is moved to on top of the darker blue boxes. We can see that there are $8$ more darker blue boxes than the innermost light blue boxes. Similarly, the blue boxes in the middle are divided into pieces and are moved to on top of the darkest blue boxes.
Indeed, as Image 4 illustrates, regardless of the exact number of the blue boxes, there are $8$ more boxes in any given ring than there are in the ring that is one layer inwards from it. This means that the the second differences, or the differences between the first differences is always going to be constant at $8$. Thus, the sequence for any diagonal half-line can be described using a quadratic polynomial.
Examples of quadratic polynomials for half-lines
For instance, prime numbers $5, 19, 41, 71, 109$, which are aligned in the same diagonal, can be described through the output of the polynomial $4x^2+10x+5$ for $x=0, 1, 2, 3, 4$. Similarly, numbers $1, 3, 14, 31, 57, 91 \dots$, which are also aligned in a green diagonal in Image 5 starting from the center and continuing to the upper right corner, can be expressed by:
Eq. (1) $4x^2-2x+1$
for $x=0, 1, 2, 3, \dots$. (We will refer back to this polynomial in a later section) In fact, the part of the green diagonal that starts from the center and continues to the bottom left corner can also be described through Eq. (1) for $x=0, -1, -2, -3, \dots$.
Image 5
In fact, even horizontal and vertical line segment in the grid can be described by quadratic polynomials, as long as the lines satisfy the condition that no two entries are positioned on the same ring. For instance, the blue horizontal line segment in Image 5, $10, 27, 52, 85$ can be described by a quadratic polynomial. However, the sequence $9, 10, 27, 85$ on the same horizontal line cannot be described by a polynomial because the entries 9 and 10 are on the same ring.
Moreover, it is not hard to show that the green diagonal line from Image 5 that goes through the center and has a slope of +1 is the only line on which the Ulam numbers in both directions can be described by the same polynomial. Even the red diagonal line that goes through the center and has a slope of -1 cannot be described by one polynomial.
Half of the red diagonal, the segment going up from the center, $5, 17, 37, 67, \dots$ can be described by $x^2+1$ for inputs that are even numbers, while the diagonal going down from the center, $1, 9, 25, 49, \dots$ is a sequence of perfect squares of odd numbers. Thus we cannot find one polynomial that generates both all entries of the red diagonal.
## Euler's Prime Generator
Many have come up with polynomials in one variable that generate prime numbers, although none of these polynomials [...]
Many have come up with polynomials in one variable that generate prime numbers, although none of these polynomials can generate all the prime numbers. One of the most famous polynomials is the one discovered by Leonhard Euler(1707-1783), which is :
$x^2-x+41$
Euler's polynomial generates distinct prime numbers for each integer $x$ from $x=1$ to $x=40$.
Image 6
As we can see in Image 6, we can start an Ulam's spiral with 41 at the center of the grid and get a long, continuous diagonal with 40 prime numbers. One interesting fact is that the 40 numbers in the diagonal line segment are the first 40 prime numbers that are generated through Euler's polynomial. Moreover, the prime numbers are not aligned in order of increasing values. In fact, with 41 in the center, other prime numbers alternate in position between the upper right and lower left part of the diagonal.
We will show why an Ulam's spiral that starts with 41 generates entries aligned in a diagonal that can also be descried as outputs of Euler's polynomial.
First, we found out in the previous section that Eq. (1) is the polynomial for the diagonal that goes through the center 1 and has a slope of +1 in the Ulam's spiral. That is,
$4x^2-2x+1$
describes the upper half of the diagonal as we plug in $x=0, 1, 2, 3, \dots$ and the lower half of the diagonal as we plug in $x=0, -1, -2, -3, \dots$. Thus, $x$'s are positioned in the diagonal as if the diagonal was a number line that starts with 0 at the center, with positive integers on the upper right and negative integers on the lower left direction, as shown in Image 8.
Image 8
Then, if we start the Ulam's spiral at 41 and thus make 41 the center, each entry on the grid, including the entries on the diagonal segment, will increase by 40. Then, the diagonal can be described by the polynomial :
Eq. (2) $4x^2-2x+41$.
In fact, we will see that this polynomial and Euler's polynomial are describing the same entries along the diagonal once we make some changes in numbering the position of $x$'s. Euler's polynomial can be found by reassigning the position of the entries of the diagonals so that $x'=1$ is at the center, $x'=2, 4, 6, 8, \dots$ are the positions up the diagonal line, $x'=1, 3, 5, 7, 9, \dots$ are positions of entries down the diagonal line from the center, as shown in Image 9. Thus, starting with $x'=1$ at the center, we alternate between the right and left of the center as we number the positions $x'=1, 2, 3, 4, 5, \dots$.
Image 9
Then, for the upper half of the diagonal in Image 9, we can use $x'=2, 4, 6, 8, \dots$ instead of $x=1, 2, 3, 4, \dots$ from Image 8. Then,
$x=\frac{1}{2} x'$
If we plug this value in to $4x^2-2x+41$ of Eq. (2), we get :
$4(\frac{1}{2}x')^2-2(\frac{1}{2}x')+41$
$={x'}^2-x'+41$,
which has the same form as Euler's polynomial.
The same method works for the bottom half as well. We use $x'=1, 3, 5, 7, \dots$ in Image 9 while we used $x=0, -1, -2, -3, \dots$ in Image 8. Then,
$x=\frac{-x'+1}{2}$.
By substituting the $x$'s in the polynomial $4x^2-2x+41$ the same way we did above, we get:
${x'}^2-x'+41$
Thus, we can see that Eq. (2), in fact, have the same output as Euler's polynomial after we reassign the position of $x$'s. We have shown that Eq. (1), which describes the diagonal that goes through the center and has a slope of +1, can be transformed to Euler's polynomial that describes the same diagonal that has 41 as its center.
## Sacks Spiral
Sacks spiral is a variation of the Ulam spiral that was devised by Robert Sacks in 1994. Sacks spiral places 0 in th [...]
Sacks spiral is a variation of the Ulam spiral that was devised by Robert Sacks in 1994. Sacks spiral places $0$ in the center and places nonnegative numbers on an Archimedean spiral, whereas the Ulam spiral places $1$ in the center and places other numbers on a square grid. Moreover, Sacks spiral makes one full counterclockwise rotation for each square number $4, 9, 16, 25, 36, 46, \dots$, as shown in the image below. The darker dots indicate the prime numbers.
We can also see that numbers that have blue check marks and are aligned in the left side of the spiral are pronic numbers. For example, $2=1(1+1), 6=2(2+1), 12=3(3+1),\dots$.
Moreover, these numbers are aligned in positions that are a little less than half of one full rotation from one perfect square to the next perfect square, for instance, from 4 to 9, or 9 to 16. Let the $n$th perfect square be $n^2$. Then, going from the $n$th perfect square to the $(n+1)$st perfect square, the difference between the two numbers will be $2n+1$. We can show this by calculating the difference between two consecutive perfect squares,
$(n+1)^2-n^2=n^2+2n+1-n^2=2n+1$
Because the pronic numbers are positioned a little less than half of one full rotation from the perfect squares, their position is a little less than $n+1/2$ from the $n$th perfect square as we go around the spiral. Indeed, any pronic number $n(n+1)=n^2+n$, and the pronic number with the form $n(n+1)$ is aligned at the $(n^2+n)$th position from the origin. From this, we can see that pronic number that has the form $n(n+1)$ appears as the $n$th number along the spiral from the $n$th perfect square.
An interesting pattern can be discovered when we start the spiral at 41. As the image below shows, the red dots are the first 40 prime numbers generated by Euler's polynomial, and they are aligned in the center and positions where pronic numbers used to be in the original Sacks spiral.
## Other Numbers and Patterns
### Triangular Number
A number ${n}$ is a triangular number if UNIQ7e9e8d976dcd50bd-math-00000 [...]
A number $n$ is a triangular number if $n$ number of dots can be arranged into an equilateral triangle evenly filled with the dots. As shown in theimage below, the sequence of triangular numbers continue as $1, 3, 6, 10, 15, 21, \dots$.
The $n^{\rm th}$ triangular number, $T_n$, is given by the formula : $T_n=\frac{n(n+1)}{2}$
When we mark the triangular numbers in a Ulam spiral, a set of spirals are formed as shown in the image below.
### Prime numbers in lines
The Ulam spiral inspired the author of this page to create another table and find a pattern among prime numbers. Firs [...]
Image 10
The Ulam spiral inspired the author of this page to create another table and find a pattern among prime numbers. First, we create a table that has 30 columns and write all the natural numbers starting from 1 as we go from left to right. Thus, each row will start with a multiple of 30 added by 1, such as 1, 31, 61, 91, 121, ... . When we mark the prime numbers in this table, we get Image 10.
We can see that prime numbers appear only on certain columns that had 1, 7, 11, 13, 17, 19, 23, 29 on their first row. This image shows that all prime numbers have the form:
$30n+(1, 7, 11, 13, 17, 19, 23, 29)$
Note that not all numbers generated by this form are prime numbers. Another point to notice in the picture is that the nonprime numbers that appear on prime-concentrated columns are all multiples of prime numbers larger than or equal to 7. For instance, $49=7\times 7, \quad 77=7\times 11, \quad 91=7\times 13, \quad 119=7\times 17, \quad 121=11\times 11, \quad 133=7\times 19 \dots$. In fact any combination of two prime numbers larger than or equal to 7 all appear on these prime-concentrated columns.
Image 11 shows the process of eliminating multiples of prime numbers from the table. First, we eliminate multiples of 2, and then we eliminate multiples of 3. Since 4 is a multiple of 2, any multiple of 4 was already eliminated. Thus, we next eliminate multiples of 5. We continue this process until we eliminate multiples of 19. (We do not eliminate numbers bigger than 19 because we only have numbers up to 390 in this table.) Note that when we eliminate multiples of 2, 3 or 5, the entire column that comes below any multiple of 2, 3, 5 on the first row get eliminated as well.
Image 11
We can see from Image 11 that eliminating multiples of 2, 3, 5 leaves us with columns where prime numbers appear. All of these multiples of 2, 3, and 5, or all the numbers that are not in the prime-concentrated column, have the form
$30n + 2^{a} 3^{b} 5^{c}$
or
Eq. (3) $(2\times 3 \times 5)n+2^{a} 3^{b} 5^{c}$,
where $a \geq 0, b \geq 0, c \geq 0$.
Thus, for a given number with the form of Eq. (3) we are able to factor out 2 or 3 or 5, or any combination of the three, which means that this number is divisible by 2, 3, or 5.
However, we know that prime numbers are not multiples of 2 or 3 or 5. Then, these prime numbers will not have the same form as Eq. (3), or else the prime numbers will be divisible by 2 or 3 or 5. Indeed, we can see that the prime numbers are positioned at :
$(2 \times 3 \times 5)+ (1, 7, 11, 13, 17, 19, 23, 29)$
Thus, prime numbers have to be positioned on the blue columns that do not have the same form as Eq. (3).
Next, we look at nonprime numbers that are products of multiplication of two prime numbers or a perfect square of a prime number. These products will be divisible by the prime numbers they are composed of, but they will not be divisible by 2 or 3 or 5.
For instance, 77, which is a nonprime number that appears on the column of prime numbers 117, 47, 107, 137, ..., is a product of two prime numbers, 7 and 11. Because each of these prime numbers do not have any divisors other than 1 and themselves, 77 is not divisible by 2 or 3 or 5. Thus, 77 and all other products of prime numbers also cannot have the same form as Eq. (3), and they have to appear on the blue columns that are concentrated with prime numbers.
# Why It's Interesting
Image 12
Not much is discovered about the Ulam spiral. For instance, the reason for the diagonal alignment of prime numbers or the vertical and horizontal arrangement of non-prime numbers is not clear yet. Indeed, Ulam spiral is not heavily studied by mathematicians. However, Ulam spiral's importance lies on the fact that it shows a clear pattern among prime numbers.
Some might suspect that we are seeing diagonal lines in the Ulam spiral because the human eye seeks patterns and groups even among random cluster of dots. However, we can compare Image 12 and Image 13 and see that the prime numbers actually have a distinct pattern of diagonal lines that random numbers do not have. Image 12 is a Ulam spiral where the black dots denote for the prime numbers, and Image 13 is a Ulam spiral of random numbers.
Image 13
People are interested in the pattern among prime numbers because the pattern might give enough information for us to discover a new polynomial that will generate more prime numbers than previously-discovered polynomials. The discovery of formula for prime numbers can lead us to have better understanding of other mysterious conjectures and theories involving prime numbers, such as twin prime conjecture or Goldbach's conjecture. For more information about the twin prime conjecture or Goldbach's conjecture, go to Wolfram Math World :Twin Prime Conjecture or Wolfram Math World :Goldbach Conjecture. |
Time and work chapter is most important topic for railway and ssc all kind of competitive exams.
Here we are providing the some short cut techniques from this chapter.
Under this chapter, we will study the following:
A. How to calculate the required number of persons to complete a particular work in a stipulated time period?
B. How to calculate the required time to complete a particular work by certain number of persons?
To find ‘A’ and ‘B’, the 1st and foremost task is to keep following basic point mind:
1. While solving problems, the work done is always supposed to be equal to 1.
2. If a person can do a piece of work in ‘n’ days, then that person’s 1 day work = 1/n.
3. If a person 1 day’s work = 1/n, then the person will complete the work in ‘n’ days.
4. A person works equally every day.
Important Relations:
1. Work and persons directly proportional (more work, more men and conversely more men, more work).
2. Time and person inversely proportional (more men, less time and conversely more time, less men).
3. Work and time directly proportional (more work, more time and conversely more time, more work).
Some important techniques about Time and Work:
TECHNIQUE 1:
If ‘M1’ persons can do ‘W1’ work in ‘D1’ days working T1 h in a day and ‘M2’ persons can do ‘W2’ work in ‘D2’ days working T2 h in a day, then we have a very basic and all in one relationship as
M1D1T1W2 = M2D2T2W1
Example: 16 men can do a piece of a work in 10 days. How many men are needed to complete the work in 20 days?
Solution: Here, M1=16, D1= 10, W1 = W2 = 1, D2 = 20, M2 =?
According to the formula, M1D1W2 = M2D2W1
16X10X1 = M2X20X1
M2 = 16X10/20 = 160/20
M2 = 8 men
Alternate method:
To do a work in 10 days, 16 men are needed.
To do the work in 1 day, 16X10 men are needed.
Here, to do the work in 20 days, = 1610/20 = 160/20
= 8 men are needed.
TECHNIQUE 2:
If A can do a piece of work in ‘x’ days and B can do the same work in ‘y’ days then
(A+B)’s 1 day’s work = 1/x+1/y
Inverse of (A+B)’s 1 day’s work = time taken by (A+B) to complete the work
Note: this formula is also applicable for 3 or more persons
Example: A can do a piece of work in 4 days, B can do the same work in 8 days and C can do the same work in 12 days, then working together, how many days will they take to complete the work?
Answer: A’s 1 day’s work = 1/4
B’s 1 day’s work=1/8
C’s 1 day’s work=1/12
Therefore (A+B+C)’s 1 day’s work =1/4+1/8+1/12= 6+3+2/24
= 11/24
Therefore (A+B+C) complete the whole work in 24/11 days
TECHNIQUE 3:
If A and B can do a piece of work in X days, B and C can do the same work in y days and A and C can do it in z days, then working together, A, B, and C can do that work in
2xyz/ xy+ yz+zx days.
Example: A and B can do a piece of work in 3 days. B and C can do the same work in 9 days while C and A can do it in 12 days. Find the time in which A, B, and C can finish the work, working together.
= 2x3x9x12/3x9+9x12+3x12
= 2x3x9x12/27+108+36
= 2x3x9x12/171
= 6x12/19
= 72/19 days.
TECHNIQUE 4:
If ‘x’ takes ‘a’ days more to complete a work than the time taken by (x+ y) to do same work and ‘y’ takes ‘b’ days more than the time taken by (x+y) to do the same work, then (x+y) do the work in
Example: when A alone does a piece of work, he takes 16 days more then the time taken by (A+B) to complete that work while B alone takes 9 days more than the time taken by (A+B) to finish the work. What time, A and B together will take to finish this work?
Answer: according to the formula, where A= 16 B= 9
Required answer: square root of 16x9 = 4x3 = 12 days.
Technique 5:
A and B, each alone can do a piece of work in ‘a’ and ‘b’ days, respectively. Both begin together and if
i. A leaves the work ‘X’ days before its completion, then total time taken for completion of work will be given as
T = (a+x) b/ (a+b)
ii. B leaves the work ‘x’ days before its completion, then total time taken for completion of work will be given as
T = (b+x)a/(a+b)
Example: A can do a piece of work in 10 days while B can do it in 15 days. They begin together but 5 days before completion of the work, B leaves off. Find the total number ofdays for the work to be completed.
Answer: here, a=10 days, b= 15 days, x= 5, t=?
According to the formula,
Required time T = (b+x)a/(a+b)
= (15+5)10/10+15
= 4X2 = 8 days
Technique 6:
A and B do a piece of work in ‘a’ and ‘b’ days, respectively. Both begin together but after some days, A leaves off and the remaining work is completed by B in ‘x’ days. Then, the time after which A left, is given by
Example: A and B do a piece of work in 40 days and 50 days, respectively. Both begin together but after certain time, A leaves off. In this case B finishes the remaining work in 20 days.After how many days did a leave?
Answer: given that a= 40 days, b=50 days, X=20, t =?
Required time = (b-x) a/ a+b
= (50-20) 40 / 50+40
= 30X40 / 90
= 40 / 3 days
Short cut Technique For Time And Work :: Railway and SSC CHSL And All Competitive exams Reviewed by Mani Babu on 16:40:00 Rating: 5 |
How to find the measure of an interior angleTo find the measurement of the interior angles in a square, you divide the sum of the angles (360°) by the number of sides (4). So, 360° / 4 = 90°. All the angles in a square are equal to 90 ...Common Core Connection for 4th Grade. Recognize that angles are formed when two rays share a common endpoint. Understand concepts of angle measurement. Measure angles in whole-number degrees using a protractor. EXAMPLE 1. Find the measure of the missing angle of the heptagon. Solution: We have the measures of the angles 110°, 150°, 120°, 125°, 155°, and 120°. Therefore, we add these angles: 110°+150°+120°+125°+155°+120° = 780°. Now, we find the missing angle by subtracting the obtained value from 900°: 900°-780° = 120°.Set up the formula for finding the sum of the interior angles. The formula is = (), where is the sum of the interior angles of the polygon, and equals the number of sides in the polygon.. The value 180 comes from how many degrees are in a triangle. The other part of the formula, is a way to determine how many triangles the polygon can be divided into. . So, essentially the formula is ...To find the sum of its interior angles, substitute n = 5 into the formula 180(n - 2) and get 180(5 - 2) = 180(3) = 540° Since the pentagon is a regular pentagon, the measure of each interior angle will be the same. To find the size of each angle, divide the sum, 540º, by the number of angles in the pentagon.So we'll mark the other base angle 72° also. Now we can find the angle at the top point of the star by adding the two equal base angles and subtracting from 180°. 72° + 72° = 144° 180° - 144° = 36° So each point of the star is 36°. You wanted the sum of the points interior angles of the points.Angles are congruent when they are the same size (in degrees or radians). 50° + 50° = 100°. Congruent angles are angles that have the same measure. These angles are congruent.Find the sum of the interior angles of a 35-gon. S = 0 5. If the sum of the interior angles of a polygon Name: Date: Block: 2. What is the formula to find the measure of each interior angle of a regular polygon? 4. Six angles of a heptagon measure 1070 1390, 1310 1100 1450, and 1280. What is theAngles are congruent when they are the same size (in degrees or radians). 50° + 50° = 100°. Congruent angles are angles that have the same measure. These angles are congruent.An angle formed between two adjacent sides at any of the vertices is called an interior angle. An exterior angle is an angle formed outside the polygon's enclosure by one of its sides and the extension of its adjacent side. The sum of the exterior angles of a polygon is 360 degrees. Let us learn in detail the concept of exterior angles of ...The measure of this angle right over here is 60 degrees. And the measure of this angle right over here is x. So let's try to figure out what all of these angles are. So to do that, we have to figure out what x is. And there's a big clue here, because the 60-degree angle plus the x angle, they're adjacent. And if you add these two angles ...In addition, students should find the sum of the measures of central, interior and exterior angles. Then they should form conjectures about the central, interior and exterior angles of regular n polygons. Students will form into groups of no more than 3. Students will work on the investigation (3 parts).Find the sum of the measures of the interior angles using the Polygon Interior Angle Sum Theorem. 1080° (Q1) A corollary to the Triangle Sum Theorem states that the measure of each angle of an equilateral triangle is _____.concentration check pathfinderscottish terrier whiteWhen dealing with polygons in geometry, it sometimes helps to know the total measure of all interior angles. We can also find the exact measure of each angle in a regular polygon.So, how many degrees are in a polygon? If we sum the interior angles of a polygon with N sides (an N-gon), there are 180(N-2) degrees. If the polygon is regular, then ...Find the measures of the angles in each problem below. 1. Find the sum of the interior angles in a 7-gon. 2. Find the sum of the interior angles in an 8-gon. 3. Find the size of each of the interior angles in a regular 12-gon. 4. Find the size of each of the interior angles in a regular 15-gon. 5. Find the size of each of the exteriorStep 1 Add up the 3 angles that are given and simplify the expression. How do you find the measure of the smallest in an acute triangle whose side lengths are 4m, 7m, and 8m? ThenThis question is testing your knowledge of the external angle theorem. $$\displaystyle \angle XZW$$ is an external angle of triangle $$\displaystyle \Delta XZY$$. The theorem: The measure of an external angle is equal to the sum of the measures of the opposite two interior angles. Applying that theorem we get $$\displaystyle 18x+5=(48)+(8x-3)$$.Find the measure of the angle indicated in bold. 5) x x ° 6) x x ° Identify the angle pair as either corresponding angles, alternate interior angles, same side interior angles. Find the measure of each angle indicated. 7) ? ° ° 8) ? ° -1-In planar geometry, an angle is the figure formed by two rays, called the sides of the angle, sharing a common endpoint, called the vertex of the angle. Angle is also used to designate the measure of an angle or of a rotation. In the case of a geometric angle, the arc is centered at the vertex and delimited by the sides.1. Find the measure of each interior angle, each exterior angle, and the sum of the interior angles of a regular triacontagon (30-gon). 2. Each exterior angle of a regular polygon measures 10°. Find the number of sides, the measure of each interior angle, and the sum of the measures of the interior angles. B.How do you find the interior angle measure of a triangle? The sum of these angle measures equals 180 degrees. You can use this fact in order to find missing angles of a triangle. Let's look at two examples from the video. Example 1. Find the missing angle in following triangle. Step 1. 70 + 32 + x =180. Step 2. 102 + x = 180.Find the unknown angles in the figures below. . Solution The sum of all 3 interior angles of a triangle is equal to 180°. Hence 92 + 27 + x = 180 Solve for x x = 180 - (92 + 27) = 61° ; Solution The sum of all 3 interior angles of the right triangle is equal to 180°. Hence y + 34 + 90 = 180 Solve for y y = 180 - (90 + 34) = 56° ; Solution Angle y and angle of measure 56° are supplementary.Common Core Connection for 4th Grade. Recognize that angles are formed when two rays share a common endpoint. Understand concepts of angle measurement. Measure angles in whole-number degrees using a protractor. To find out the sum of the interior angles of a triangle, we draw a parallel line to one of the sides and, notice that the other two sides are transversal, determine alternate interior angles....busted mugshots nc3/4/10 4:15 AM. WonderHowTo. This tutorial the shows how to find out the measure of an exterior angle of a regular polygon. He shows the formula to find it which is 360/n, where n is the number of sides of the regular polygon. He goes on further to explain the formula by taking an 18-sided regular polygon as example and computes its exterior ...To find the measure of each interior angle of a regular polygon, use the formula 360: where n is the number of sides. n: Apply the formula to solve for n. 360 = 36: n: 360 = 36n: 360 = n: 36: n = 10 # Problem Correct Answer Your Answer; 2: An interior angle of a regular polygon measures 60°. How many sides doesFind the measures of unknown angles for a polygon using our new formulas and properties. Video - Lesson & Examples 46 min 00:12:01 - Find the sum of the interior angles and the measure of each interior and exterior angle for a regular polygon (Examples #1-5) 00:23:37 - Find the number of sides of a regular polygon, given an exterior angle ...Finding the Interior Angle Subtract the sum of the two angles from 180° to find the measure of the indicated interior angle in each triangle. Once you've solved for x, plug that This videos provides an example of how to determine the measure of a missing interior angle of a triangle.Site: http://mathispower4u.com3/4/10 4:15 AM. WonderHowTo. This tutorial the shows how to find out the measure of an exterior angle of a regular polygon. He shows the formula to find it which is 360/n, where n is the number of sides of the regular polygon. He goes on further to explain the formula by taking an 18-sided regular polygon as example and computes its exterior ...Explanation. Transcript. If one side of a triangle is extended beyond the vertex, an exterior angle is formed. This exterior angle is supplementary with its adjacent, linear angle. Since the angle sum in a triangle is also 180 degrees, the exterior angle must have a measure equal to the sum of the remaining angles, called the remote interior ...The exterior angle of a triangle is always equal to the sum of the interior opposite angles. The sum of the measure of the three interior angles of a quadrilateral is always 360 o. Sum of any two adjacent angles in a quadrilateral is equal to 180 o. The interior angle of a square or a rectangle at each vertex is 90°.In the quadrilateral above, one of the angles marked in red color is right angle. By Internal Angles of a Quadrilateral Theorem, "The sum of the measures of the interior angles of a quadrilateral is 360°" So, we have. 60 ° + 150° + 3x° + 90° = 360° 60 + 150 + 3x + 90 = 360. Simplify. 3x + 300 = 360Therefore, we can deduce that x + y + z + (measure of angle AED) = 360. Line Segment Angle Calculator This calculator converts between polar and rectangular coordinates. For eachMar 07, 2020 · To Find the measure of each Interior Angle of a regular convex polygon. Ex. 8. Find the measure of each angle in a regular convex octagon. Ex. 9. The measure of each interior angle of a regular polygon is 165º. How many sides does the polygon have? Exterior Angles The Exterior Angle of any polygon forms a linear pair with an Interior angle of ... Understanding Quadrilaterals - Measures of the Exterior Angles of a Polygon. Polygon is a closed, connected shape made of straight lines. It may be a flat or a plane figure spanned across two-dimensions. A polygon is an enclosed figure that can have more than 3 sides. The lines forming the polygon are known as the edges or sides and the ...The measure of angles is made with an extension of a side and its corresponding adjacent sides are the exterior angle. measure of exterior angles Overview of Measure of Exterior Angles: If you can measure the interior and the exterior angle of straight line is 180 degree. The regular polygons the exterior angles are the congruent, and like ...oz. to lbtoyota dealer allentown palyrics to let it goFind the measure of one interior angle in each regular polygon. Round your answer to the nearest tenth if necessary. 22) 140° 23) 90° 24) 128.6° 25) 147.3° 26) 108° 27) 144° Find the measure of one exterior angle in each regular polygon. Round your answer to the nearest tenth if necessary. 28) 32.7° 29) 51.4° 30) 72° 31) 90°Find the sum of the interior angles of a 35-gon. S = 0 5. If the sum of the interior angles of a polygon Name: Date: Block: 2. What is the formula to find the measure of each interior angle of a regular polygon? 4. Six angles of a heptagon measure 1070 1390, 1310 1100 1450, and 1280. What is theWe know that angle 2 and 4 are congruent because they are alternate interior angles. Angles 3, 4, cnd 5 must add uo to 180, 30 angle 5 = 39 0 Use what you know about the sum of angles in a triangle to set up and solve an equation to find the measure of each missing angle. 2. a. 13K -3 4. 3x + 680 Equation: Equation: Equation: 4x —- 20 1What are angles in a quadrilateral? Angles in a quadrilateral are the four angles that occur at each vertex within a four-sided shape; these angles are called interior angles of a quadrilateral.The sum of the interior angles of any quadrilateral is 360° .. We can prove this using the angle sum of a triangle.The sum of the interior angles of any hexagon is always equal to 720°. We can find the measure of each interior angle of a regular hexagon by dividing 720° by 6. Therefore, we have: 720°÷6 = 120°. Each interior angle in a regular hexagon measures 120°. In the following diagram, we have a regular hexagon, which has sides of the same length ...Jun 21, 2019 · Correct answers: 3 question: Find the measure of each interior angle of the regular polygon. In planar geometry, an angle is the figure formed by two rays, called the sides of the angle, sharing a common endpoint, called the vertex of the angle. Angle is also used to designate the measure of an angle or of a rotation. In the case of a geometric angle, the arc is centered at the vertex and delimited by the sides.The measure of each exterior angle of a regular polygon is 15. Ex. 4 Find the measure of ONE exterior angle of a regular 20-gon. 18° 16. Ex. 5 Find the measure of ONE exterior angle of a regular heptagon. 51.4 ° 17. Ex. 6 The sum of the measures of five interior angles of a hexagon is 625.The sum of the exterior angles is always , and that is true for all polygons. The sum of interior angles differs by the number of sides polygons have. Triangles - There are 3 sides and a sum of 180 degrees. Quadrilaterals - There are 4 sides and a sum of 360 degrees. Pentagons - There are 5 sides and a sum of 540 degrees.Find the measure of 1 interior angle of a regular 18-gon. b. Find the sum of the angles of a 50-gon. c. One exterior angle of a regular polygon is 22.5°. Find the number of sides d. Find the measure of angle B . 6.1 (6) Use what you have learned and the information on the "Polygon Angles Pattern & Formula Page" to help you writebmw coupesFinding Angle Measures in Regular Polygons The trampoline shown is shaped like a regular dodecagon. a. Find the measure of each interior angle. b. Find the measure of each exterior angle. SOLUTION a. Use the Polygon Interior Angles Theorem to fi nd the sum of the measures of the interior angles. (n − 2) ⋅ 180° = (12 − 2) ⋅ 180° = 1800°The sum of the exterior angles of any convex polygon is 360°. Example 3: Find the measure of each interior angle of a regular hexagon (Figure 3). Figure 3 An interior angle of a regular hexagon. Method 1: Because the polygon is regular, all interior angles are equal, so you only need to find the interior angle sum and divide by the number of ...Now, creating a Find The Sum Of The Measures Of The Interior Angles Of Each Convex Polygon requires not more than 5 minutes. Our state web-based blanks and crystal-clear guidelines eliminate human-prone faults. Adhere to our easy steps to have your Find The Sum Of The Measures Of The Interior Angles Of Each Convex Polygon ready rapidly:Find the measure of each interior angle of a regular polygon having (i) 10 sides (ii) 15 sides. asked Jul 1, 2021 in Polygons by Anaswara (31.5k points) polygons; class-8; 0 votes. 1 answer. Find the measure of each interior angle of a regular polygon having (i) 10 sides (ii) 15 sides.Find the measure of each interior angle of the regular polygon. Answers: 3 Get Other questions on the subject: Mathematics. Mathematics, 21.06.2019 14:00, crispingolfer7082. What is the equation of the following graph in vertex form? parabolic function going down from the left through the point zero comma twelve and through the point two comma ...a. Determine the measure of each interior angle of a regular octagon. Explain how you found your answer. b. What about the interior angles of other regular polygons? Find the interior angles of a regular nonagon and a regular 100-gon. c. Will the process you used for part (a) work for any regular polygon? Write an expression that will calculate theProtractors. The types of tools to measure angles fall into one of three categories: protractors, squares and compasses. There are many variations of these basic tools. A protractor is one of the most common tools to measure angles. When you know how to use a protractor, you can measure both small and large angles.angular 4 datagridairsoft picschicago pizza with a twistmakita batteryExample 1: Find the interior angle at vertex B in the following figure. Solution: The number of sides of the given polygon is n =6, so it's a hexagon (Hexagon has 6 sides). Thus, the sum of the interior angles of this polygon is 180(n-2)º. ... The sum of measures of interior angles of a 27-gon is 180(27-2)°. It is equal to 180 × 25, which is ...The sum of the interior angle measures of a polygon is 4140°. How many sides does the polygon have? 0 . 7161 . 1 . The sum of the interior angle measures of a polygon is 4140°. How many sides does the polygon have?An interior (inside) angle is an angle considered inside a circle when the vertex is somewhere inside the circle, but not on the center. All angles inside a circle are formed by two intersecting chords. Construction Investigation: Find the Measure of an Angle inside a Circle Tools Needed: pencil, straightedge, protractor, colored pencilsThe sum of the measures of the 5 exterior angles and the 5 interior angles is 5 × 180 o = 900 o. Subtract the sum of the measures of the 5 exterior angles from 900 o to find the sum of the measures of the 5 interior angles. You know that the sum of the measure of two of the interior angles is 240 o.A remote interior angle is an interior angle that is not adjacent to the exterior angle. ∠1, ∠2, and ∠3 are interior angles. ∠4 is an exterior angle. ∠1 and ∠2 are remote interior angles to ∠4. There is a special relationship between the measure of an exterior angle and the measures of its remote interior angles.Finding Angle Measures in Regular Polygons The trampoline shown is shaped like a regular dodecagon. a. Find the measure of each interior angle. b. Find the measure of each exterior angle. SOLUTION a. Use the Polygon Interior Angles Theorem to fi nd the sum of the measures of the interior angles. (n − 2) ⋅ 180° = (12 − 2) ⋅ 180° = 1800°A regular polygon is a flat shape whose sides are all equal and whose angles are all equal. The formula for finding the sum of the measure of the interior angles is (n - 2) * 180. To find the measure of one interior angle, we take that formula and divide by the number of sides n: (n - 2) * 180 / n. Jul 22, 2018 · The best way to measure an angle is to use a protractor. To do this, you’ll start by lining up one ray along the 0-degree line on the protractor. Then, line up the vertex with the midpoint of the protractor. Follow the second ray to determine the angle’s measurement to the nearest degree. 900 seconds. Report an issue. Q. Find angle x. (Remember use your formula (sides -2) x 180 to figure out total interior angle degrees in shape.) answer choices. 247 o. 67 o. 157 o.Before learning the formulas for finding angles, let us see the situations where we may need to use these formulas. There are different formulas for finding angles depending on the available data. Let us learn the formulas of finding angles case by case here. To find the missing angle in a polygon, we use the sum of interior angles formula. To ...The sum of exterior angles in a polygon is always equal to 360 degrees. Therefore, for all equiangular polygons, the measure of one exterior angle is equal to 360 divided by the number of sides in the polygon. exterior angle sum of angles equiangular polygon. Next to your angle is formed by a side. and an extension of an adjacent.As we know, in a polygon, the sum of an exterior angle and its corresponding interior angle is equal to 180° (since they form a linear pair). So, measure of an exterior angle = 180° - 180° × (n-2)/ n. = [180n - 180n + 360]/ n. = 360/n. Hence, the sum of all the exterior angles of the polygon is n × 360/n = 360°.900 seconds. Report an issue. Q. Find angle x. (Remember use your formula (sides -2) x 180 to figure out total interior angle degrees in shape.) answer choices. 247 o. 67 o. 157 o.The sum of the angles of a triangle always add up to 180 degrees. So when you subtract the given angles of a triangle from 180, you'll find the unknown angles. See the triangle below. The known angles are 77 and 35. To find the unknown angle, we must add these up and subtract them from 180: 77 + 35 = 112. 180 - 112 = 68.Figure out the number of sides, measure of each exterior angle, and the measure of the interior angle of any polygon. Simply enter one of the three pieces of information! The sum of the measures of the angles of a convex polygon with n sides is (n - 2)180. Sides on the polygon: ...In the parallelogram shown above, angle is degrees. Find the measure of angle Possible Answers: Correct answer: Explanation: A parallelogram must have equivalent opposite interior angles. Additionally, the sum of all four interior angles must equal degrees. And, the adjacent interior angles must be supplementary angles (sum of degrees). Since,crochet flower blanketThe sum of interior angles = 180 (n-2)º The interior angles of a polygon always lie inside the polygon and the formula to calculate it can be obtained in three ways. Formula 1: For "n" is the number of sides of a polygon, formula is as, Interior angles of a Regular Polygon = [180° (n) - 360°] / nThe sum of the measures of the 5 exterior angles and the 5 interior angles is 5 × 180 o = 900 o. Subtract the sum of the measures of the 5 exterior angles from 900 o to find the sum of the measures of the 5 interior angles. You know that the sum of the measure of two of the interior angles is 240 o.Find the measure of each interior angle of a regular polygon having (i) 10 sides (ii) 15 sides. asked Jul 1, 2021 in Polygons by Anaswara (31.5k points) polygons; class-8; 0 votes. 1 answer. Find the measure of each interior angle of a regular polygon having (i) 10 sides (ii) 15 sides.👉 Learn how to determine the measure of the interior angles of a regular polygon. A polygon is a plane shape bounded by a finite chain of straight lines. A ...A remote interior angle is an interior angle that is not adjacent to the exterior angle. ∠1, ∠2, and ∠3 are interior angles. ∠4 is an exterior angle. ∠1 and ∠2 are remote interior angles to ∠4. There is a special relationship between the measure of an exterior angle and the measures of its remote interior angles.To find the measurement of the interior angles in a square, you divide the sum of the angles (360°) by the number of sides (4). So, 360° / 4 = 90°. All the angles in a square are equal to 90 ...Finding the Interior Angle Subtract the sum of the two angles from 180° to find the measure of the indicated interior angle in each triangle. Once you've solved for x, plug thatIn the parallelogram shown above, angle is degrees. Find the measure of angle Possible Answers: Correct answer: Explanation: A parallelogram must have equivalent opposite interior angles. Additionally, the sum of all four interior angles must equal degrees. And, the adjacent interior angles must be supplementary angles (sum of degrees). Since,The measure of the central angle is the same as the arc of the circle intercepted by this angle. Important fact: The measure of a central angle is the same as the measure of the intercepted arc. Definition: The diagram below shows an additional angle within the circle O.The angle has a vertex F on the circle. This is called an interior angle.a^2+b^2=c^2 is the Pythagorean Theorem . This formula is used to find the area of right triangles. a represents the shortest side of the triangle, b represents the middle side of the triangle, and c represents the longest side of the triangle. You can use this formula to find a missing side of a triangle as long as you have the other two.May 13, 2022 · The length of the adjacent of a right triangle with an angle of 60° and a hypotenuse of 5 cm is 2.5 cm. Next, use this relationship to calculate the measure of . The interior angles of a triangle always add up to 180° while the exterior angles of a triangle are equal to the sum of the two interior angles that are not adjacent to it. View PDF. A triangle has exactly 3 sides and the sum of interior angles sum up to 180°. all right angles are equal in measure). Acute angles along with right angles are probably the easiest type of angles to learn first because they are most commonly seen in day-to-day life. Find the remaining side and angles.glowing man sitting at tableHexagon Interior Angles. Here are a number of highest rated Hexagon Interior Angles pictures on internet. We identified it from well-behaved source. Its submitted by dealing out in the best field. We acknowledge this nice of Hexagon Interior Angles graphic could possibly be the most trending subject with we allowance it in google plus or facebook.In addition, students should find the sum of the measures of central, interior and exterior angles. Then they should form conjectures about the central, interior and exterior angles of regular n polygons. Students will form into groups of no more than 3. Students will work on the investigation (3 parts).Therefore, we can deduce that x + y + z + (measure of angle AED) = 360. Line Segment Angle Calculator This calculator converts between polar and rectangular coordinates. For eachThe exterior angle of a triangle is always equal to the sum of the interior opposite angles. The sum of the measure of the three interior angles of a quadrilateral is always 360 o. Sum of any two adjacent angles in a quadrilateral is equal to 180 o. The interior angle of a square or a rectangle at each vertex is 90°.Finding the Interior Angle Subtract the sum of the two angles from 180° to find the measure of the indicated interior angle in each triangle. Once you've solved for x, plug that We can calculate the angle between two sides of a right triangle using the length of the sides and the sine, cosine or tangent. To do this, we need the inverse functions arcsine, arccosine and arctangent. If you only know the length of two sides, or one angle and one side, this is enough to determine everything of the triangle. ...In addition, students should find the sum of the measures of central, interior and exterior angles. Then they should form conjectures about the central, interior and exterior angles of regular n polygons. Students will form into groups of no more than 3. Students will work on the investigation (3 parts).chanel west coast momThe Sum of all the interior angles of a polygon is equal to the product of a straight angle and two less than the number of sides of the polygon. In a regular polygon, all the interior angles measure the same and hence can be obtained by dividing the sum of the interior angles by the number of sides. Sum of interior angles = (p - 2) 1800 3.The sum of interior angles formula {eq}S_n~=~180 (n~-~2) {/eq} can be used to find the sum of the angles (measured in degrees) of a polygon if the number of edges is given. If the sum itself is...Find the measure of one interior angle in each polygon. Round your answer to the nearest tenth if necessary. 1) 108 ° 2) 135 ° 3) 147.3 ° 4) 120 ° 5) 140 ° 6) 150 ° 7) regular 24-gon 165 ° 8) regular quadrilateral 90 ° 9) regular 23-gon 164.3 ° 10) regular 16-gon 157.5 ° Find the measure of one exterior angle in each polygon.The sum of the exterior angles is always , and that is true for all polygons. The sum of interior angles differs by the number of sides polygons have. Triangles - There are 3 sides and a sum of 180 degrees. Quadrilaterals - There are 4 sides and a sum of 360 degrees. Pentagons - There are 5 sides and a sum of 540 degrees.A regular polygon is a flat shape whose sides are all equal and whose angles are all equal. The formula for finding the sum of the measure of the interior angles is (n - 2) * 180. To find the measure of one interior angle, we take that formula and divide by the number of sides n: (n - 2) * 180 / n.This calculator is designed to give the angles of any regular polygon. Enter the number of sides on the polygon. Then click on Calculate. All angle measurements are in degrees. The information calculated is the name of the polygon, the total of the interior angles, the measure of each interior angle, and the measure of each exterior angle. The ...Finding Angle Measures in Regular Polygons The trampoline shown is shaped like a regular dodecagon. a. Find the measure of each interior angle. b. Find the measure of each exterior angle. SOLUTION a. Use the Polygon Interior Angles Theorem to fi nd the sum of the measures of the interior angles. (n − 2) ⋅ 180° = (12 − 2) ⋅ 180° = 1800°2) Find the measure of an interior and an exterior angle of a regular 46-gon. 3) The measure of an exterior angle of a regular polygon is 2x, and the measure of an interior angle is 4x. a) Use the relationship between interior and exterior angles to find x. b) Find the measure of one interior and exterior angle.An Interior Angle is an angle inside a shape. Example: ... Pentagon. A pentagon has 5 sides, and can be made from three triangles, so you know what ..... its interior angles add up to 3 × 180° = 540° And when it is regular (all angles the same), then each angle is 540° / 5 = 108° (Exercise: make sure each triangle here adds up to 180°, and check that the pentagon's interior angles add up ...The exterior angle of a triangle is always equal to the sum of the interior opposite angles. The sum of the measure of the three interior angles of a quadrilateral is always 360 o. Sum of any two adjacent angles in a quadrilateral is equal to 180 o. The interior angle of a square or a rectangle at each vertex is 90°.A regular polygon is a flat shape whose sides are all equal and whose angles are all equal. The formula for finding the sum of the measure of the interior angles is (n - 2) * 180. To find the measure of one interior angle, we take that formula and divide by the number of sides n: (n - 2) * 180 / n. See the answer. See the answer See the answer done loading. Using optimization, find the measures of the interior angles that maximize the area of an isosceles trapezoid. where the length of the non-parallel sides are each 4 inches and the length the shorter of. the two bases is 6 inches.What is the measure of each base angle of an isosceles triangle if its vertex angle measures 40 degrees and its. 2 congruent sides measure 25 units? a. 70° b. 140° c. 50° d. 40° ____ 16.gm stocktwitsIf you cut a pizza into four big slices and then cut each of those slices in half, you get eight pieces, each of which makes a 45° angle. If you cut the original pizza into 12 slices, each slice makes a 30° angle. So 1/12 of a pizza is 30°, 1/8 is 45°, 1/4 is 90°, and so on. The bigger the fraction of the pizza, the bigger the angle.Angles in the Polygons Worksheets. This assemblage of printable angles in polygons worksheets for grade 6 through high school encompasses a multitude of exercises to find the sum of interior angles of both regular and irregular polygons, find the measure of each interior and exterior angle, simplify algebraic expressions to find the angle ...Let m be the measure of the angle BCD and n be the measure of the angle CDA. Since BC and DA are parallel and CE is a transversal the measure of the angle ADE is also m. Thus. m + n = 180 degrees. Now, what else do you know? the measure of one interior angle of a parallelogram is 42 degrees more than twice the measure of another angleWhat are angles in a quadrilateral? Angles in a quadrilateral are the four angles that occur at each vertex within a four-sided shape; these angles are called interior angles of a quadrilateral.The sum of the interior angles of any quadrilateral is 360° .. We can prove this using the angle sum of a triangle.900 seconds. Report an issue. Q. Find angle x. (Remember use your formula (sides -2) x 180 to figure out total interior angle degrees in shape.) answer choices. 247 o. 67 o. 157 o.Your Answer. 1. What is the sum of the interior angles. of a quadrilateral? °. Solution. To find the sum of the interior angles of a polygon, use the formula. 180 ( n - 2) where n is the number of sides. Sum of the interior angles of a quadrilateral = 180 (4 - 2) = 360 degrees. The sum of the exterior angles of any convex polygon is 360°. Example 3: Find the measure of each interior angle of a regular hexagon (Figure 3). Figure 3 An interior angle of a regular hexagon. Method 1: Because the polygon is regular, all interior angles are equal, so you only need to find the interior angle sum and divide by the number of ...Angles are congruent when they are the same size (in degrees or radians). 50° + 50° = 100°. Congruent angles are angles that have the same measure. These angles are congruent.In order to find the measure of a single interior angle of a regular polygon (a polygon with sides of equal length and angles of equal measure) with n sides, we calculate the sum interior anglesor $$(\red n-2) \cdot 180$$ and then divide that sum by the number of sides or $$\red n$$. The FormulaMath. Geometry. Geometry questions and answers. 2. Find the measure of each interior angle. Round to the nearest hundredth. (3x - 42) (2x - 5)° Find the missing variable for the above quadrilateral * O x=60 x=73.05 Ox=63.31 (3x - 42) (2x - 5)° Find the missing variable for the above quadrilateral * X=60 X=73.05 X=63 31 x=59.23 Other. Question ...The sum of the measures of the interior angles of a convex polygon with n sides is (n - 2) 180°. Example 1 Determine the unknown angle measures. For the nonagon shown, find the unknown angle measure x°. First, use the Polygon Angle Sum Theorem to find the sum of the interior angles: n = 9 (-2)180n ° = (9 -2)180°= (7)180°= 1260°Angles >. Polygons. Try our free pdf angles in polygons worksheets to help children solve a variety of tasks, such as finding a missing angle in a polygon, finding the measures of two or more indicated angles in polygons, finding the sum of interior angles of a polygon, and more. Young students will encounter a mix of concave and convex shapes ...Before learning the formulas for finding angles, let us see the situations where we may need to use these formulas. There are different formulas for finding angles depending on the available data. Let us learn the formulas of finding angles case by case here. To find the missing angle in a polygon, we use the sum of interior angles formula. To ...Finding the Interior Angle Subtract the sum of the two angles from 180° to find the measure of the indicated interior angle in each triangle. Once you've solved for x, plug thatscout autobonefish grill orlandooklahoma deq -fc https://pflegeradar-deutschland.de/msqoya.html |
# How do you find the x and y intercept of 5x-y=35?
Jan 18, 2017
See the entire process below:
#### Explanation:
To find the x-intercept substitute $\textcolor{red}{0}$ for $\textcolor{red}{y}$ and solve for $x$:
$5 x - \textcolor{red}{y} = 35$ becomes:
$5 x - \textcolor{red}{0} = 35$
$5 x = 35$
$\frac{5 x}{\textcolor{red}{5}} = \frac{35}{\textcolor{red}{5}}$
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{5}}} x}{\cancel{\textcolor{red}{5}}} = 7$
$x = 7$
To find the y-intercept substitute $\textcolor{red}{0}$ for $\textcolor{red}{x}$ and solve for $y$:
$5 \textcolor{red}{x} - y = 35$ becomes:
$\left(5 \times \textcolor{red}{0}\right) - y = 35$
$0 - y = 35$
$- y = 35$
$\textcolor{red}{- 1} \times - y = \textcolor{red}{- 1} \times 35$
$y = - 35$
The x-intercept is $7$ or (7, 0)
The y-intercept is $- 35$ or (0, -35) |
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# The Quadratic Formula and the Discriminant
So far, we have used techniques such as graphing, factoring, and applying the Square Root Property to find exact solutions to certain quadratic equations. We have also learnt how to solve quadratic equations by completing the square.
While some of these methods seem to be the best option to solve any type of quadratic equation, it may turn out to be rather difficult if fractions or decimals are involved in the given quadratic equation. However, fear not! There happens to be a solution to solving any form of quadratic equation expressed as the definition above. This is known as the Quadratic Formula.
The Quadratic Formula is an important tool used to determine the solutions for any given quadratic equation. We can apply this concept when solving quadratic equations that cannot be factorized through standard factoring methods.
Note that we can indeed use the Quadratic Formula to find solutions for any form of quadratic equations, even ones that can be factorized.
Before we dive into this topic, let us first recall the standard form of a quadratic equation.
The standard form of a quadratic equation is where
With that in mind, let us now introduce the Quadratic Formula.
For a quadratic equation of the form where the solutions are given by the Quadratic Formula,
.
Notice that the Quadratic Formula has the '±' sign. This means that the formula produces two solutions, namely
.
Given that the Quadratic Formula tells us the roots of a given quadratic equation, we can easily locate these points and plot the graph more accurately.
### Derivation of the Quadratic Formula
The Quadratic Formula is derived via completing the square. This section explains its derivation step-by-step as below.
Given the general form of a quadratic equation :
Step 1: Divide the expression by a
Step 2: Subtract from each side
Step 3: Complete the square
Step 4: Factor the left-hand side and simplify the right-hand side
Step 5: Square root each side
Don't forget the '±' sign!
Step 6: Subtract from each side
Step 7: Simplify the expression
Note: this method of completing the square is explained in detail in the topic Completing the Squares. This discussion contains clearly worked examples that show how this derivation is applied to a given quadratic equation. Check it out if you'd like to explore this in greater depth!
## The Discriminant
In the following sections, we shall look at the properties of roots for given quadratic equations. We will be introduced to a new concept called the discriminant. The discriminant plays a crucial role in understanding the nature of the roots of a quadratic equation.
Before we look into the idea of a discriminant, we need to familiarise ourselves with several important terms to aid our understanding throughout this discussion. Let us begin by defining a rational and irrational root.
A rational root is a solution that can be expressed as a quotient of two integers.
They are represented in the form where p and q are integers where p is the constant of the polynomial and q is the leading coefficient.
An irrational root is a solution that cannot be expressed as a quotient of two integers. They are often represented by infinitely non-repeating decimals or surds.
Next, we shall define what it means to be a perfect square. This concept is crucial when we start using the Quadratic Formula as it determines whether the roots of our given quadratic equation are rational or irrational, as we shall soon see!
A perfect square is an integer that is the square of an integer, that is to say, the product of some integer with itself. This takes the form where p is an integer. Essentially, .
Examples include 9 (32), 16 (42), 25 (52), etc.
Now that we have our key definitions sorted, let us now move on to the concept of a discriminant and its relation to the properties of roots.
### The Discriminant and the Properties of Roots
To find the number of roots in a given quadratic equation, we shall make use of the discriminant. We can also determine the type of roots the expression holds.
The discriminant of a quadratic polynomial is used to find the number and type of solutions a quadratic equation has. It is described by the formula
Notice that this is the component inside the square root in the Quadratic Formula.
The condition of a discriminant has three cases.
#### Case 1: D > 0
When the determinant is more than zero, or in other words, b2 – 4ac > 0, we obtain two real distinct roots. This can be further categorized as the following.
1. If b2 – 4ac is a perfect square then we have two real rational roots;
2. If b2 – 4ac is not a perfect square then we have two real irrational roots.
The graph for this case is shown below.
Discriminant case when D > 0, StudySmarter Originals
#### Case 2: D = 0
When the determinant is equal to zero, or in other words, b2 – 4ac = 0, we obtain one real root. This is also known as a repeated root. The graph for this case is shown below.
Discriminant case when D = 0, StudySmarter Originals
#### Case 3: D < 0
When the determinant is less than zero, or in other words, b2 – 4ac < 0, we obtain two complex conjugate roots. This means that our solution is of the form a + bi where a is the real part and b is the imaginary part. The graph for this case is shown below.
Discriminant case when D < 0, StudySmarter Originals
Recall that the imaginary unit is
## Using the Quadratic Formula and Discriminant to Find Roots
In this section, we shall look at some worked examples that demonstrate the application of the Quadratic Formula and the discriminant to look for solutions to a given quadratic equation.
### Two Real Rational Roots
Calculate the discriminant and identify the number and type of roots this expression holds. Then, use the Quadratic Formula to evaluate its solutions.
Solution
Step 1: Identify a, b and c
Step 2: Calculate the discriminant
As D > 0, there are two real distinct roots.
Step 3: Find the solutions
Using the Quadratic Formula we obtain
Note that the component inside the square root is D, or in other words
Here, is a perfect square so we obtain a pair of rational roots
Thus, the solutions are .
The graph for this quadratic equation is plotted below. The green dots represent the solutions to the expression.
Example 1, StudySmarter Originals
### Two Real Irrational Roots
Calculate the discriminant and identify the number and type of roots this expression holds. Then, use the Quadratic Formula to evaluate their solutions.
Solution
Step 1: Identify a, b and c
Step 2: Calculate the discriminant
As D > 0, there are two real distinct roots.
Step 3: Find the solutions
Using the Quadratic Formula we obtain
Here, is not a perfect square so we obtain a pair of irrational roots
Thus, the solutions are .
The graph for this quadratic equation is plotted below. The green dots represent the solutions to the expression.
Example 2, StudySmarter Originals
Note that you can keep the roots in the exact form and that the decimal places are an approximate answer.
### One Real Repeated Root
Calculate the discriminant and identify the number and type of roots this expression holds. Then, use the Quadratic Formula to evaluate their solutions.
Solution
Step 1: Identify a, b and c
Step 2: Calculate the discriminant
As D = 0, there is one real distinct root.
Step 3: Find the solutions
Using the Quadratic Formula we obtain
Noting that
Thus, the solution is .
The graph for this quadratic equation is plotted below. The green dots represent the solutions of the expression.
Example 3, StudySmarter Originals
### Two Complex Roots
Calculate the discriminant and identify the number and type of roots this expression holds. Then, use the Quadratic Formula to evaluate their solutions.
Solution
Step 1: Identify a, b and c
Step 2: Calculate the discriminant
As D < 0, there are two complex conjugate roots.
Step 3: Find the solutions
Using the Quadratic Formula we obtain
Noting that
Simplifying this, we obtain
Thus, the solutions are .
The graph for this quadratic equation is plotted below. The green dots represent the solutions to the expression.
Example 4, StudySmarter Originals
Notice that there are no solutions labelled on this graph. This is because the solutions are imaginary and cannot be graphed in the standard Cartesian plane. The Cartesian plane is represented by real numbers, not imaginary numbers! In this case, we can essentially 'assume' the shape of the graph based on the coefficient of the x2 term and that the y-intercept given by the initial quadratic equation.
## Discriminant of a Cubic Equation
In this section, we shall look at the discriminant of a cubic equation and identify the types of roots the expression has, given the value of its discriminant.
For a cubic equation of the (general) form
,
where a 0, the discriminant is described by the formula
.
The formula for evaluating the discriminant of cubic equations can be quite lengthy. Questions, where this formula may be applied, are often rare in this syllabus. However, it may be helpful to know how it is done for clarity.
Just like the quadratic case, the discriminant for cubic equations has three conditions.
### Case 1: D > 0
When the discriminant is more than zero, we obtain three (distinct) real roots.
Say we have the cubic equation .
Here, the discriminant is .
Hence, we have three distinct real roots. Factorizing this expression yields
Thus, the roots are .
The graph is shown below.
Example 5, StudySmarter Originals
### Case 2: D = 0
Case 2(a): If the discriminant is equal to zero and b2 = 3ac, we obtain three repeated real roots (distinct triple root).
Say we have the cubic equation .
Here, the discriminant is .
Further, .
Hence, we have three repeated real roots. Factorizing this expression yields
Thus, the roots are .
The graph is shown below.
Example 6, StudySmarter Originals
Case 2(b): If the discriminant is equal to zero and b2 3ac, we obtain two repeated real roots (distinct double root) and one real (distinct) root.
Say we have the cubic equation .
Here, the discriminant is .
Further, .
Hence, we have two repeated real roots and one real root. Factorizing this expression yields
Thus, the roots are .
The graph is shown below.
Example 7, StudySmarter Originals
### Case 3: D < 0
When the discriminant is less than zero, we obtain one (distinct) real root and a pair of complex conjugate roots.
Say we have the cubic equation .
Here, the discriminant is.
Hence, we have one real root and two complex conjugate roots. Factorizing this expression yields
Thus, the roots are .
The graph is shown below.
Example 8, StudySmarter Originals
## The Quadratic Formula and the Discriminant - Key takeaways
• The Quadratic Formula is used to determine the solutions of a given quadratic equation.
• For a quadratic equation of the form, the Quadratic Formula is
• The Discriminant is used to find the number and type of solutions a quadratic equation has. It is given by the formula D = b2 - 4ac.
• The conditions for the discriminant are summarized in the following table.
Value of Discriminant Type and Number of Roots Graph D > 0, D is a perfect square 2 Real Rational Roots Graph when D > 0, Aishah Amri - StudySmarter Originals D > 0, D is not a perfect square 2 Real Irrational Roots D = 0 1 Real Repeated Root Graph when D = 0, Aishah Amri, StudySmarter Originals D < 0 2 Complex Conjugate Roots Graph when D = 0, Aishah Amri, StudySmarter Originals
A positive discriminant means that the quadratic equation has two real roots while a negative discriminant means that the quadratic equation has two complex roots.
The discriminant is a quantity used to find the number and type of solutions a quadratic equation has.
The discriminant is described by the formula D = b2 – 4ac.
The discriminant is described by the formula D = b2 – 4ac and the quadratic formula is used to determine the solutions of a given quadratic equation.
Formula for discriminant of cubic equation is D = b2c2 + 18abcd − 4b3d − 4ac3 − 27a2d2.
## Final The Quadratic Formula and the Discriminant Quiz
Question
What is the Quadratic Formula used for?
To find solutions of a given quadratic equation
Show question
Question
How many solutions does the Quadratic Formula produce? What is the sign in the Quadratic Formula that gives that particular number of solutions?
Two solutions. The '±' sign indicates that there are two solutions when we apply the Quadratic Formula.
Show question
Question
How can the Quadratic Formula help us plot the graph of a given quadratic equation?
Since the Quadratic Formula determines the roots of a quadratic equation, we can locate them and plot the graph more accurately
Show question
Question
When can we use the Quadratic Formula to solve a given quadratic equation?
We can use this for quadratic equations that cannot be factored (however, we can indeed use it to solve any quadratic equation)
Show question
Question
What is another term used to describe a graph with one real root?
Repeated real root
Show question
Question
What are the components of a complex solution?
A complex solution takes the form a+bi where a is the real part and b is the imaginary part
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005 Sample Final A, Question 11
Question Solve the following equation in the interval ${\displaystyle [0,2\pi )}$
${\displaystyle \sin ^{2}(\theta )-\cos ^{2}(\theta )=1+\cos(\theta )}$
Foundations:
1) Which trigonometric identities are useful in this problem?
1) ${\displaystyle \sin ^{2}(\theta )=1-\cos ^{2}(\theta )}$ and
Step 1:
We need to get rid of the ${\displaystyle \sin ^{2}(\theta )}$ term. Since ${\displaystyle \sin ^{2}(\theta )=1-\cos ^{2}(\theta )}$, the equation becomes
${\displaystyle (1-\cos ^{2}(\theta ))-\cos ^{2}(\theta )=1+\cos(\theta )}$.
Step 2:
If we simplify and move all the terms to the right hand side, we have ${\displaystyle 0=2\cos ^{2}(\theta )+\cos(\theta )}$.
Step 3:
Now, factoring, we have ${\displaystyle 0=\cos(\theta )(2\cos(\theta )+1)}$. Thus, either ${\displaystyle \cos(\theta )=0}$ or ${\displaystyle 2\cos(\theta )+1=0}$.
Step 4:
The solutions to ${\displaystyle \cos(\theta )=0}$ in ${\displaystyle [0,2\pi )}$ are ${\displaystyle \theta ={\frac {\pi }{2}}}$ or ${\displaystyle \theta ={\frac {3\pi }{2}}}$.
Step 5:
The solutions to ${\displaystyle 2\cos(\theta )+1=0}$ are angles that satisfy ${\displaystyle \cos(\theta )={\frac {-1}{2}}}$. In ${\displaystyle [0,2\pi )}$, the
solutions are ${\displaystyle \theta ={\frac {2\pi }{3}}}$ or ${\displaystyle \theta ={\frac {4\pi }{3}}}$.
The solutions are ${\displaystyle {\frac {\pi }{2}},{\frac {3\pi }{2}},{\frac {2\pi }{3}},{\frac {4\pi }{3}}}$. |
# 2004 AMC 12A Problems/Problem 17
The following problem is from both the 2004 AMC 12A #17 and 2004 AMC 10A #24, so both problems redirect to this page.
## Problem
Let $f$ be a function with the following properties:
(i) $f(1) = 1$, and
(ii) $f(2n) = n \cdot f(n)$ for any positive integer $n$.
What is the value of $f(2^{100})$?
$\textbf {(A)}\ 1 \qquad \textbf {(B)}\ 2^{99} \qquad \textbf {(C)}\ 2^{100} \qquad \textbf {(D)}\ 2^{4950} \qquad \textbf {(E)}\ 2^{9999}$
## Solution 1 (Forward)
From (ii), note that \begin{alignat*}{8} f(2) &= 1\cdot f(1) &&= 1, \\ f\left(2^2\right) &= 2\cdot f(2) &&= 2, \\ f\left(2^3\right) &= 2^2\cdot f\left(2^2\right) &&= 2^{2+1}, \\ f\left(2^4\right) &= 2^3\cdot f\left(2^3\right) &&= 2^{3+2+1}, \end{alignat*} and so on.
In general, we have $$f\left(2^n\right)=2^{(n-1)+(n-2)+(n-3)+\cdots+3+2+1}$$ for any positive integer $n.$
Therefore, the answer is \begin{align*} f\left(2^{100}\right)&=2^{99+98+97+\cdots+3+2+1} \\ &=2^{99\cdot100/2} \\ &= \boxed{\textbf {(D)}\ 2^{4950}}. \end{align*} ~MRENTHUSIASM
## Solution 2 (Backward)
Applying (ii) repeatedly, we have \begin{align*} f\left(2^{100}\right) &= 2^{99} \cdot f\left(2^{99}\right) \\ &= 2^{99} \cdot 2^{98} \cdot f\left(2^{98}\right) \\ &= 2^{99} \cdot 2^{98} \cdot 2^{97} \cdot f\left(2^{97}\right) \\ &= \cdots \\ &= 2^{99} \cdot 2^{98} \cdot 2^{97} \cdots 2^{3} \cdot 2^{2} \cdot 2^{1} \cdot 1 \cdot f(1) \\ &= 2^{99} \cdot 2^{98} \cdot 2^{97} \cdots 2^{3} \cdot 2^{2} \cdot 2^{1} \cdot 1 \cdot 1 \\ &=2^{99 + 98 + 97 + \cdots + 3 + 2 + 1} \\ &=2^{99\cdot100/2} \\ &= \boxed{\textbf {(D)}\ 2^{4950}}. \end{align*} ~Azjps (Fundamental Logic)
~MRENTHUSIASM (Reconstruction) |
# Question Video: Using Skip Counting to Find Multiples of Five Mathematics • 3rd Grade
Consider the diagram. We notice that when we multiply by 5, each row is 5 more than the previous one. Find the result of the following. 4 × 5 = ＿. Find the result of the following. 5 × 5 = ＿.
01:50
### Video Transcript
Consider the following diagram. One times five equals five. Two times five equals 10. Three times five equals 15. We notice that when we multiply by five, each row is five more than the previous one. Find the result of the following. Four times five equals what. Find the result of the following. Five times five equals what.
In this question, we’re being asked to find four times five and five times five. And we’re told to use the diagram or the model to help. When we multiply by five, each row is five more than the previous one. One times five is five, and two times five is 10. We know that 10 is five more than five. 15 is five more than 10.
So to find four times five, we need to find the number which is five more than 15: five, 10, 15, 20. Four times five equals 20. And we know that five times five is five more than 20. Five times five is 25. One times five is five. Two times five is 10. Three times five is 15. Four times five is 20. And five times five is 25. To find the missing products, we counted in fives. |
# Understand orthogonal circles using examples here!
The geometric field is vast and there are different examples and shapes that you read about daily. The Circle, rectangle, triangle and many more are there. However, hardly many of you have read about Orthogonal Circles. These are some new parts that many students are unaware of. How many of you have gone through Venn diagrams? In your higher mathematics classes, you have read or might be reading about the ‘SETS’ module, in that all Venn diagrams symbolize the orthogonal circle.
These are also types of circles that are intersecting to each other and forming a 90 degree angle at the curve of a circle. to give you more details on it, we have brought this post and here is some amazing information related to the same.
## Circle: Explained in Details
How many have you saw the round shape figure? Whether it is a dish, tyre or a bangle, all are in the circular shape. Thus, if you have closely observed them, you could have noticed that there is an enclosed figure with the round curve boundary. It means there is no angle forming among curves. This figure we call it in simpler terms as the circle. In mathematical geometry, there are many modules based on the circle module and you will have to go through its equational form and other theoretical concepts.
Only talking about the circle, there are several other concepts involved in it. For that purpose, you have to go through your textbooks and other study materials that involve vital concepts. In the circle, there are different elements like the radius (the line segment from the central point of the circle to the circumference.
## What Are Called Orthogonal Circles?
Before jumping to the information of orthogonal circles, it is recommended to understand the meaning of orthogonal first. It means that two figures are involved with right angles which is 90 degrees. You can imagine that shapes like square, triangle, rectangle and others possess the feature of 90 degree angle. But what about the circle? Will it be possible to intersect two circles at 90 degrees?
Yes, it is possible if you have accurate measures of the circle’s radius and it could give you orthogonal circles. For that purpose, you need to draw two circles with two different radii. And ensure that one circle’s tangent is touching the other circle’s circumference. In this manner, you get your orthogonal circles ready.
## How to draw orthogonal circles?
Taking the fix point ‘A’ draws one circle, you have the freedom to choose your radius length. Now, on the exterior of the circle, choose any random point ‘B’ and draw one circle ensuring both cut each other. Draw one imaginary tangent ‘T’ from the center of the circle which you have drawn through the point ‘B’ and this tangent should touch the circle of point ‘A’ circumference. Thus, you can see that both of them are forming orthogonal circles and there you get the actual result which you were looking for.
Always remember this geometric construction of orthogonal circles. It will help you in solving different questions and difficult problems related to the orthogonal circle.
We recommend you to practice daily and keep your geometrical skills up. It will help you in overcoming the phobia of geometry and you can score maximum numbers in your exam.
Tags: |
Q:
# Consider the two triangles.Triangles L M N and X Y Z are shown. Side L M is blank, side M N is 3, and side N L is 2. Side X Y is 12, side Y Z is 9, and side Z X is blank.To prove that △LMN ~ △XYZ by the SSS similarity theorem using the information provided in the diagram, it would be enough additional information to know thatLM is 3 units and XZ is 5 units.LM is 4 units and XZ is 6 units.LM is 5 units and XZ is 3 units.LM is 6 units and XZ is 4 units.
Accepted Solution
A:
Answer:LM is 4 units and XZ is 6 units.Step-by-step explanation:Now, first we have to understand that what is SSS similarity theorem. This theorem states that when sides of any two triangles are in proportion, this means that these two triangles are similar.By the data given,we can see that,YZ:MN = 3:1So, there is an assumption that ΔXYZ:ΔLMN = 3:1Now when XY = 12, we need value of LM = 12/3 =4So, XY: LM would become 3:1If value of LN is given as 2, we need value of XZ = (2)*(3) = 6Note that since ΔLMN is smaller triangle by values given, so we need to multiply value of side LN with 3 to get the value of XZ in ratio 3:1So, by the data given in option 2, we would have all lines of both triangles in ratio of 3:1,so,YZ:MN = 3:1XZ:LN = 3:1XY:LM = 3:1Hence, by using SSS postulate for similarity of triangles we would prove thatΔXYZ:ΔLMN = 3:1and also△LMN ~ △XYZ |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are reading an older version of this FlexBook® textbook: CK-12 Algebra I - Second Edition Go to the latest version.
# 6.4: Absolute Value Equations and Inequalities
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Solve an absolute value equation.
• Analyze solutions to absolute value equations.
• Graph absolute value functions.
• Solve absolute value inequalities.
• Rewrite and solve absolute value inequalities as compound inequalities.
• Solve real-world problems using absolute value equations and inequalities.
## Introduction
Timmy is trying out his new roller skates. He’s not allowed to cross the street yet, so he skates back and forth in front of his house. If he skates 20 yards east and then 10 yards west, how far is he from where he started? What if he skates 20 yards west and then 10 yards east?
The absolute value of a number is its distance from zero on a number line. There are always two numbers on the number line that are the same distance from zero. For instance, the numbers 4 and -4 are each a distance of 4 units away from zero.
$|4|$ represents the distance from 4 to zero, which equals 4.
$|-4|$ represents the distance from -4 to zero, which also equals 4.
In fact, for any real number $x$:
$|x| = x$ if $x$ is not negative, and $|x| = -x$ if $x$ is negative.
Absolute value has no effect on a positive number, but changes a negative number into its positive inverse.
Example 1
Evaluate the following absolute values.
a) $|25|$
b) $|-120|$
c) $|-3|$
d) $|55|$
e) $\left | - \frac{5}{4} \right |$
Solution
a) $|25|= 25$ Since 25 is a positive number, the absolute value does not change it.
b) $|-120| = 120$ Since -120 is a negative number, the absolute value makes it positive.
c) $|-3| = 3$ Since -3 is a negative number, the absolute value makes it positive.
d) $|55| = 55$ Since 55 is a positive number, the absolute value does not change it.
e) $\left | -\frac{5}{4} \right | =\frac{5}{4}$ Since $-\frac{5}{4}$ is a negative number, the absolute value makes it positive.
Absolute value is very useful in finding the distance between two points on the number line. The distance between any two points $a$ and $b$ on the number line is $|a - b|$ or $|b - a|$.
For example, the distance from 3 to -1 on the number line is $|3 - (-1)| = |4| = 4$.
We could have also found the distance by subtracting in the opposite order: $|-1 - 3| = |-4| = 4$. This makes sense because the distance is the same whether you are going from 3 to -1 or from -1 to 3.
Example 2
Find the distance between the following points on the number line.
a) 6 and 15
b) -5 and 8
c) -3 and -12
Solution
Distance is the absolute value of the difference between the two points.
a) $\text{distance} = |6 - 15| = |-9| = 9$
b) $\text{distance} = |-5 - 8| = |-13| = 13$
c) $\text{distance} = |-3 - (-12)| = |9| = 9$
Remember: When we computed the change in $x$ and the change in $y$ as part of the slope computation, these values were positive or negative, depending on the direction of movement. In this discussion, “distance” means a positive distance only.
## Solve an Absolute Value Equation
We now want to solve equations involving absolute values. Consider the following equation:
$|x|=8$
This means that the distance from the number $x$ to zero is 8. There are two numbers that satisfy this condition: 8 and -8.
When we solve absolute value equations we always consider two possibilities:
1. The expression inside the absolute value sign is not negative.
2. The expression inside the absolute value sign is negative.
Then we solve each equation separately.
Example 3
Solve the following absolute value equations.
a) $|x| = 3$
b) $|x| = 10$
Solution
a) There are two possibilities: $x = 3$ and $x = -3$.
b) There are two possibilities: $x = 10$ and $x = -10$.
## Analyze Solutions to Absolute Value Equations
Example 4
Solve the equation $|x-4|=5$ and interpret the answers.
Solution
We consider two possibilities: the expression inside the absolute value sign is nonnegative or is negative. Then we solve each equation separately.
$& x-4 = 5 \quad \text{and} \quad x-4=-5\\& \quad \ \ x=9 \qquad \qquad \quad \ x=-1$
$x = 9$ and $x = -1$ are the solutions.
The equation $|x-4|=5$ can be interpreted as “what numbers on the number line are 5 units away from the number 4?” If we draw the number line we see that there are two possibilities: 9 and -1.
Example 5
Solve the equation $|x+3|=2$ and interpret the answers.
Solution
Solve the two equations:
$& x+3 = 2 \quad \text{and} \quad \ \ x+3=-2\\& \quad \ \ x=-1 \qquad \qquad \quad \ x=-5$
$x = -5$ and $x = -1$ are the answers.
The equation $|x+3|=2$ can be re-written as: $|x-(-3)|=2$. We can interpret this as “what numbers on the number line are 2 units away from -3?” There are two possibilities: -5 and -1.
Example 6
Solve the equation $|2x-7|=6$ and interpret the answers.
Solution
Solve the two equations:
$& 2x-7 = 6 \qquad \qquad \quad 2x-7=-6\\& \quad \ \ 2x=13 \qquad \text{and} \qquad \ \ 2x=1\\& \quad \ \ \ x=\frac{13}{2} \qquad \qquad \qquad \ \ x=\frac{1}{2}$
Answer: $x=\frac{13}{2}$ and $x=\frac{1}{2}$.
The interpretation of this problem is clearer if the equation $|2x-7|=6$ is divided by 2 on both sides to get $\frac{1}{2}|2x-7|=3$. Because $\frac{1}{2}$ is nonnegative, we can distribute it over the absolute value sign to get $\left | x-\frac{7}{2} \right |=3$. The question then becomes “What numbers on the number line are 3 units away from $\frac{7}{2}$?” There are two answers: $\frac{13}{2}$ and $\frac{1}{2}$.
## Graph Absolute Value Functions
Now let’s look at how to graph absolute value functions.
Consider the function $y=|x-1|$. We can graph this function by making a table of values:
$x$ $y=|x-1|$
-2 $y=|-2-1|=|-3|=3$
-1 $y=|-1-1|=|-2|=2$
0 $y=|0-1|=|-1|=1$
1 $y=|1-1|=|0|=0$
2 $y=|2-1|=|1|=1$
3 $y=|3-1|=|2|=2$
4 $y=|4-1|=|3|=3$
You can see that the graph of an absolute value function makes a big “V”. It consists of two line rays (or line segments), one with positive slope and one with negative slope, joined at the vertex or cusp.
We’ve already seen that to solve an absolute value equation we need to consider two options:
1. The expression inside the absolute value is not negative.
2. The expression inside the absolute value is negative.
Combining these two options gives us the two parts of the graph.
For instance, in the above example, the expression inside the absolute value sign is $x-1$. By definition, this expression is nonnegative when $x-1 \ge 0$, which is to say when $x \ge 1$. When the expression inside the absolute value sign is nonnegative, we can just drop the absolute value sign. So for all values of $x$ greater than or equal to 1, the equation is just $y=x-1$.
On the other hand, when $x-1 < 0$ — in other words, when $x < 1$ — the expression inside the absolute value sign is negative. That means we have to drop the absolute value sign but also multiply the expression by -1. So for all values of $x$ less than 1, the equation is $y=-(x-1)$, or $y=-x+1$.
These are both graphs of straight lines, as shown above. They meet at the point where $x-1=0$ — that is, at $x=1$.
We can graph absolute value functions by breaking them down algebraically as we just did, or we can graph them using a table of values. However, when the absolute value equation is linear, the easiest way to graph it is to combine those two techniques, as follows:
1. Find the vertex of the graph by setting the expression inside the absolute value equal to zero and solving for $x$.
2. Make a table of values that includes the vertex, a value smaller than the vertex, and a value larger than the vertex. Calculate the corresponding values of $y$ using the equation of the function.
3. Plot the points and connect them with two straight lines that meet at the vertex.
Example 7
Graph the absolute value function $y=|x+5|$.
Solution
Step 1: Find the vertex by solving $x + 5 = 0$. The vertex is at $x = -5$.
Step 2: Make a table of values:
$x$ $y=|x+5|$
-8 $y=|-8+5|=|-3|=3$
-5 $y=|-5+5|=|0|=0$
-2 $y=|-2+5|=|3|=3$
Step 3: Plot the points and draw two straight lines that meet at the vertex:
Example 8
Graph the absolute value function: $y=|3x-12|$
Solution
Step 1: Find the vertex by solving $3x - 12 = 0$. The vertex is at $x = 4$.
Step 2: Make a table of values:
$x$ $y=|3x-12|$
0 $y=|3(0)-12|=|-12|=12$
4 $y=|3(4)-12|=|0|=0$
8 $y=|3(8)-12|=|12|=12$
Step 3: Plot the points and draw two straight lines that meet at the vertex.
## Solve Real-World Problems Using Absolute Value Equations
Example 9
A company packs coffee beans in airtight bags. Each bag should weigh 16 ounces, but it is hard to fill each bag to the exact weight. After being filled, each bag is weighed; if it is more than 0.25 ounces overweight or underweight, it is emptied and repacked. What are the lightest and heaviest acceptable bags?
Solution
The weight of each bag is allowed to be 0.25 ounces away from 16 ounces; in other words, the difference between the bag’s weight and 16 ounces is allowed to be 0.25 ounces. So if $x$ is the weight of a bag in ounces, then the equation that describes this problem is $|x-16|=0.25$.
Now we must consider the positive and negative options and solve each equation separately:
$& x-16 = 0.25 \qquad \text{and} \quad x-16 =-0.25\\& \qquad x=16.25 \qquad \qquad \qquad \ \ x=15.75$
The lightest acceptable bag weighs 15.75 ounces and the heaviest weighs 16.25 ounces.
We see that $16.25 - 16 = 0.25 \ ounces$ and $16 - 15.75 = 0.25 \ ounces$. The answers are 0.25 ounces bigger and smaller than 16 ounces respectively.
The answer you just found describes the lightest and heaviest acceptable bags of coffee beans. But how do we describe the total possible range of acceptable weights? That’s where inequalities become useful once again.
## Absolute Value Inequalities
Absolute value inequalities are solved in a similar way to absolute value equations. In both cases, you must consider the same two options:
1. The expression inside the absolute value is not negative.
2. The expression inside the absolute value is negative.
Then you must solve each inequality separately.
## Solve Absolute Value Inequalities
Consider the inequality $|x| \le 3$. Since the absolute value of $x$ represents the distance from zero, the solutions to this inequality are those numbers whose distance from zero is less than or equal to 3. The following graph shows this solution:
Notice that this is also the graph for the compound inequality $-3 \le x \le 3$.
Now consider the inequality $|x|>2$. Since the absolute value of $x$ represents the distance from zero, the solutions to this inequality are those numbers whose distance from zero are more than 2. The following graph shows this solution.
Notice that this is also the graph for the compound inequality $x < -2$ or $x >2$.
Example 1
Solve the following inequalities and show the solution graph.
a) $|x|<5$
b) $|x| \ge 2.5$
Solution
a) $|x|<5$ represents all numbers whose distance from zero is less than 5.
This answer can be written as “$-5 < x < 5$”.
b) $|x| \ge 2.5$ represents all numbers whose distance from zero is more than or equal to 2.5
This answer can be written as “$x \le -2.5$ or $x \ge 2.5$”.
## Rewrite and Solve Absolute Value Inequalities as Compound Inequalities
In the last section you saw that absolute value inequalities are compound inequalities.
Inequalities of the type $|x| can be rewritten as “$-a < x < a$”.
Inequalities of the type $|x|>b$ can be rewritten as “$x < -b$ or $x >b$.”
To solve an absolute value inequality, we separate the expression into two inequalities and solve each of them individually.
Example 2
Solve the inequality $|x-3|<7$ and show the solution graph.
Solution
Re-write as a compound inequality: $-7
Write as two separate inequalities: $x-3<7$ and $x-3>-7$
Solve each inequality: $x<10$ and $x > -4$
Re-write solution: $-4 < x < 10$
The solution graph is
We can think of the question being asked here as “What numbers are within 7 units of 3?”; the answer can then be expressed as “All the numbers between -4 and 10.”
Example 3
Solve the inequality $|4x+5| \le 13$ and show the solution graph.
Solution
Re-write as a compound inequality: $-13 \le 4x+5 \le 13$
Write as two separate inequalities: $4x+5 \le 13$ and $4x +5 \ge -13$
Solve each inequality: $4x \le 8$ and $4x \ge -18$
$x \le 2$ and $x \ge -\frac{9}{2}$
Re-write solution: $-\frac{9}{2} \le x \le 2$
The solution graph is
Example 4
Solve the inequality $|x+12|>2$ and show the solution graph.
Solution
Re-write as a compound inequality: $x+12 < -2$ or $x+12 > 2$
Solve each inequality: $x < -14$ or $x > -10$
The solution graph is
Example 5
Solve the inequality $|8x-15| \ge 9$ and show the solution graph.
Solution
Re-write as a compound inequality: $8x-15 \le -9$ or $8x-15 \ge 9$
Solve each inequality: $8x \le 6$ or $8x \ge 24$
$x \le \frac{3}{4}$ or $x \ge 3$
The solution graph is
## Solve Real-World Problems Using Absolute Value Inequalities
Absolute value inequalities are useful in problems where we are dealing with a range of values.
Example 6
The velocity of an object is given by the formula $v=25t-80$, where the time is expressed in seconds and the velocity is expressed in feet per second. Find the times for which the magnitude of the velocity is greater than or equal to 60 feet per second.
Solution
The magnitude of the velocity is the absolute value of the velocity. If the velocity is $25t-80$ feet per second, then its magnitude is $|25t-80|$ feet per second. We want to find out when that magnitude is greater than or equal to 60, so we need to solve $|25t-80| \ge 60$ for $t$.
First we have to split it up: $25t-80 \ge 60$ or $25t-80 \le -60$
Then solve: $25t \ge 140$ or $25t \le 20$
$t \ge 5.6$ or $t \le 0.8$
The magnitude of the velocity is greater than 60 ft/sec for times less than 0.8 seconds and for times greater than 5.6 seconds.
When $t = 0.8 \ seconds, \ v=25(0.8)-80 = -60 \ ft/sec$. The magnitude of the velocity is 60 ft/sec. (The negative sign in the answer means that the object is moving backwards.)
When $t = 5.6 \ seconds, \ v=25(5.6)-80 = 60 \ ft/sec$.
To find where the magnitude of the velocity is greater than 60 ft/sec, check some arbitrary values in each of the following time intervals: $t \le 0.8, \ 0.8 \le t \le 5.6$ and $t \ge 5.6$.
Check $t = 0.5: \ v=25(0.5) -80 = -67.5 \ ft/sec$
Check $t = 2: \ v = 25(2)-80 =-30 \ ft/sec$
Check $t = 6: \ v=25(6)-80 = -70 \ ft/sec$
You can see that the magnitude of the velocity is greater than 60 ft/sec only when $t \ge 5.6$ or when $t \le 0.8$.
## Further Resources
For a multimedia presentation on absolute value equations and inequalities, see http://www.teachertube.com/viewVideo.php?video_id=124516.
## Lesson Summary
• The absolute value of a number is its distance from zero on a number line.
• $|x|=x$ if $x$ is not negative, and $|x|=-x$ if $x$ is negative.
• An equation or inequality with an absolute value in it splits into two equations, one where the expression inside the absolute value sign is positive and one where it is negative. When the expression within the absolute value is positive, then the absolute value signs do nothing and can be omitted. When the expression within the absolute value is negative, then the expression within the absolute value signs must be negated before removing the signs.
• Inequalities of the type $|x| can be rewritten as “$-a < x < a$.”
• Inequalities of the type $|x|>b$ can be rewritten as “$x < -b$ or $x > b$.”
## Review Questions
Evaluate the absolute values.
1. $|250|$
2. $|-12|$
3. $\left | -\frac{2}{5} \right |$
4. $\left | \frac{1}{10} \right |$
Find the distance between the points.
1. 12 and -11
2. 5 and 22
3. -9 and -18
4. -2 and 3
Solve the absolute value equations and interpret the results by graphing the solutions on the number line.
1. $|x-5|=10$
2. $|x+2|=6$
3. $|5x-2|=3$
4. $|x-4|=-3$
Graph the absolute value functions.
1. $y=|x+3|$
2. $y=|x-6|$
3. $y=|4x+2|$
4. $y= \left | \frac{x}{3}-4 \right |$
Solve the following inequalities and show the solution graph.
1. $|x| \le 6$
2. $|x| > 3.5$
3. $|x| < 12$
4. $|x| > 10$
5. $|7x| \ge 21$
6. $|x-5| > 8$
7. $|x+7| < 3$
8. $\left | x-\frac{3}{4} \right | \le \frac{1}{2}$
9. $|2x-5| \ge 13$
10. $|5x+3| < 7$
11. $\left | \frac{x}{3}-4 \right | \le 2$
12. $\left | \frac{2x}{7}+9 \right | > \frac{5}{7}$
1. How many solutions does the inequality $|x| \le 0$ have?
2. How about the inequality $|x| \ge 0$?
13. A company manufactures rulers. Their 12-inch rulers pass quality control if they are within $\frac{1}{32} \ inches$ of the ideal length. What is the longest and shortest ruler that can leave the factory?
14. A three month old baby boy weighs an average of 13 pounds. He is considered healthy if he is at most 2.5 lbs. more or less than the average weight. Find the weight range that is considered healthy for three month old boys.
Feb 23, 2012
Apr 02, 2015 |
The Lemniscate SineSine
# 1 Sine
The sine function, as we learnt in high school, is closely related to the arc length of the circle. We will describe this in a slightly funny way. Consider the circle of unit diameter with center $(0, \frac{1}{2})$:
By some elementary geometry, we find that $r = \sin \theta$. The arc length $s$ is given by
$(\mathrm{d}s)^2 = (\mathrm{d}r)^2 + (r \; \mathrm{d}\theta )^2.$
We have
$\mathrm{d}\theta = \frac{\mathrm{d}r}{\cos \theta } = \frac{\mathrm{d}r}{\sqrt{1 - \sin ^2 \theta }} = \frac{\mathrm{d}r}{\sqrt{1 - r^2}}.$
So we can write the line element as
$\mathrm{d}s = \frac{1}{\sqrt{1 - r^2}}\; \mathrm{d}r$
Thus, the arc length function is given
$s(r_0) = \int _0^{r_0} \frac{1}{\sqrt{1 - r^2}}\; \mathrm{d}r.$
This is, of course, the familiar arcsine function, inverse to the even more familiar sine function.
Note that here we have a square root sitting inside the integrand. For the arc length function, we simply always take the positive square root. However, if we want to extend arcsine and sine to complex functions by replacing the integral with a contour integral, then we cannot always stick with the above choice.
The solution is to instead think about the Riemann surface of the function $\frac{1}{\sqrt{1 - r^2}}$, which is given by the circle $r^2 + t^2 = 1$:
We can then write the integral as
$s(z) = \int _0^z \frac{\mathrm{d}r}{t},$
where $z$ is now viewed as a point in $R = \{ (r, t) \in \mathbb {C}^2: r^2 + t^2 = 1\}$. This integral depends not only on $z$, but also on the path taken from $0$ to $z$. In general, it is well-defined only up to a period, namely the integral of $\frac{\mathrm{d}r}{t}$ around a closed loop. Thus, the inverse function, namely sine, is a singly-periodic function.
It is important to note that this circle is different from the previous circle. When we discuss the Lemniscate sine soon, we will work with two rather different shapes.
Before we end the section, note that to study the sine function, which is a function defined on $R$, it is often convenient to pick an isomorphism between $R$ and $\mathbb {P}^1$ given by stereographic projection.
This is given by the formulae
$r = \frac{2u}{1 + u^2},\quad t = \frac{1 - u^2}{1 + u^2}$
Substituting in, we find
$\frac{\mathrm{d}r}{\sqrt{1 - r^2}} = \frac{1}{\sqrt{1 - \big (\frac{2u}{1 + u^2}\big )^2}} \cdot \frac{2 (1 + u^2) - 4u^2}{(1 + u^2)^2}\; \mathrm{d}u = \frac{1 + u^2}{1 - u^2} \cdot \frac{2 (1 - u^2)}{(1 + u^2)^2}\; \mathrm{d}u = \frac{2\; \mathrm{d}u}{1 + u^2},$
which is the integral of a nice, rational function. |
# Sum of all real solutions for $x$ to the equation $\displaystyle (x^2+4x+6)^{{(x^2+4x+6)}^{\left(x^2+4x+6\right)}}=2014.$
Find the sum of all real solutions for $x$ to the equation $\displaystyle (x^2+4x+6)^{{(x^2+4x+6)}^{\left(x^2+4x+6\right)}}=2014.$
$\bf{My\; Try::}$ Let $y=x^2+4x+6 = (x+2)^2+2\geq 2$.
So our exp. equation convert into $\displaystyle y^{y^{y}} = 2014\;,$ where $y\geq 2$
Now at $y=2\;,$ We Get $\displaystyle 2^{2^{2}} = 16<2014$ and $\displaystyle 3^{3^{3}} = 3^{27}>2014$
So $y$ must be lie between $2$ and $3$.
But I did not Understand How can I calculate it..
Help me
Thanks
• The solution is $-4$! Commented Jul 21, 2014 at 18:01
As you have written if one substitutes $y=x^2 + 4x +6 = (x+2)^2+2\ge 2$, then the resulting equation is $f(y)=2014$ where $f(y)=y^{y^y}$. Due the the fact that $2014>2^{2^2}$, it accepts solutions in ${\mathbb R}$. Now note that $f(y)=y^{y^y}$ is an strictly increasing function on ${\mathbb R}_{\ge 2}$. So for $f(y)=2014$ there exist a unique $y_0\in {\mathbb R}_{\ge 2}$ such that $f(y_0)=2014$. So one have that $$x^2 + 4x +6-y_0 = 0,$$ where the sum of solutions is $-4$. |
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Broad Topics > Calculations and Numerical Methods > Multiplication & division
### The Deca Tree
##### Age 7 to 11 Challenge Level:
Find out what a Deca Tree is and then work out how many leaves there will be after the woodcutter has cut off a trunk, a branch, a twig and a leaf.
### Napier's Bones
##### Age 7 to 11 Challenge Level:
The Scot, John Napier, invented these strips about 400 years ago to help calculate multiplication and division. Can you work out how to use Napier's bones to find the answer to these multiplications?
### The Clockmaker's Birthday Cake
##### Age 7 to 11 Challenge Level:
The clockmaker's wife cut up his birthday cake to look like a clock face. Can you work out who received each piece?
### Sam's Quick Sum
##### Age 7 to 11 Challenge Level:
What is the sum of all the three digit whole numbers?
### The 24 Game
##### Age 7 to 11 Challenge Level:
There are over sixty different ways of making 24 by adding, subtracting, multiplying and dividing all four numbers 4, 6, 6 and 8 (using each number only once). How many can you find?
### Arranging the Tables
##### Age 7 to 11 Challenge Level:
There are 44 people coming to a dinner party. There are 15 square tables that seat 4 people. Find a way to seat the 44 people using all 15 tables, with no empty places.
### Clever Keys
##### Age 7 to 11 Short Challenge Level:
On a calculator, make 15 by using only the 2 key and any of the four operations keys. How many ways can you find to do it?
### The Pied Piper of Hamelin
##### Age 7 to 11 Challenge Level:
This problem is based on the story of the Pied Piper of Hamelin. Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether!
### Function Machines
##### Age 7 to 11 Challenge Level:
If the numbers 5, 7 and 4 go into this function machine, what numbers will come out?
### ABC
##### Age 7 to 11 Challenge Level:
In the multiplication calculation, some of the digits have been replaced by letters and others by asterisks. Can you reconstruct the original multiplication?
### A-magical Number Maze
##### Age 7 to 11 Challenge Level:
This magic square has operations written in it, to make it into a maze. Start wherever you like, go through every cell and go out a total of 15!
### Buckets of Thinking
##### Age 7 to 11 Challenge Level:
There are three buckets each of which holds a maximum of 5 litres. Use the clues to work out how much liquid there is in each bucket.
### Machines
##### Age 7 to 11 Challenge Level:
What is happening at each box in these machines?
### Clock Face
##### Age 7 to 11 Challenge Level:
Where can you draw a line on a clock face so that the numbers on both sides have the same total?
### One Million to Seven
##### Age 7 to 11 Challenge Level:
Start by putting one million (1 000 000) into the display of your calculator. Can you reduce this to 7 using just the 7 key and add, subtract, multiply, divide and equals as many times as you like?
### Gift Stacks
##### Age 7 to 11 Challenge Level:
Use the information to work out how many gifts there are in each pile.
### Penta Post
##### Age 7 to 11 Challenge Level:
Here are the prices for 1st and 2nd class mail within the UK. You have an unlimited number of each of these stamps. Which stamps would you need to post a parcel weighing 825g?
### Rabbits in the Pen
##### Age 7 to 11 Challenge Level:
Using the statements, can you work out how many of each type of rabbit there are in these pens?
### How Many Eggs?
##### Age 7 to 11 Challenge Level:
Peter, Melanie, Amil and Jack received a total of 38 chocolate eggs. Use the information to work out how many eggs each person had.
### Zargon Glasses
##### Age 7 to 11 Challenge Level:
Zumf makes spectacles for the residents of the planet Zargon, who have either 3 eyes or 4 eyes. How many lenses will Zumf need to make all the different orders for 9 families?
### Oranges and Lemons
##### Age 7 to 11 Challenge Level:
On the table there is a pile of oranges and lemons that weighs exactly one kilogram. Using the information, can you work out how many lemons there are?
### Throw a 100
##### Age 7 to 11 Challenge Level:
Can you score 100 by throwing rings on this board? Is there more than way to do it?
### Rocco's Race
##### Age 7 to 11 Short Challenge Level:
Rocco ran in a 200 m race for his class. Use the information to find out how many runners there were in the race and what Rocco's finishing position was.
### Code Breaker
##### Age 7 to 11 Challenge Level:
This problem is based on a code using two different prime numbers less than 10. You'll need to multiply them together and shift the alphabet forwards by the result. Can you decipher the code?
### The Number Crunching Machine
##### Age 7 to 11 Challenge Level:
Put a number at the top of the machine and collect a number at the bottom. What do you get? Which numbers get back to themselves?
### It Was 2010!
##### Age 5 to 11 Challenge Level:
If the answer's 2010, what could the question be?
### Next Number
##### Age 7 to 11 Short Challenge Level:
Find the next number in this pattern: 3, 7, 19, 55 ...
### How Old?
##### Age 7 to 11 Challenge Level:
Cherri, Saxon, Mel and Paul are friends. They are all different ages. Can you find out the age of each friend using the information?
### Calendar Calculations
##### Age 7 to 11 Challenge Level:
Try adding together the dates of all the days in one week. Now multiply the first date by 7 and add 21. Can you explain what happens?
### Being Resilient - Primary Number
##### Age 5 to 11 Challenge Level:
Number problems at primary level that may require resilience.
### Dividing a Cake
##### Age 7 to 11 Challenge Level:
Annie cut this numbered cake into 3 pieces with 3 cuts so that the numbers on each piece added to the same total. Where were the cuts and what fraction of the whole cake was each piece?
### Sept03 Sept03 Sept03
##### Age 7 to 11 Challenge Level:
This number has 903 digits. What is the sum of all 903 digits?
### X Is 5 Squares
##### Age 7 to 11 Challenge Level:
Can you arrange 5 different digits (from 0 - 9) in the cross in the way described?
### Clever Santa
##### Age 7 to 11 Challenge Level:
All the girls would like a puzzle each for Christmas and all the boys would like a book each. Solve the riddle to find out how many puzzles and books Santa left.
### Sending Cards
##### Age 7 to 11 Challenge Level:
This challenge asks you to investigate the total number of cards that would be sent if four children send one to all three others. How many would be sent if there were five children? Six?
### Route Product
##### Age 7 to 11 Challenge Level:
Find the product of the numbers on the routes from A to B. Which route has the smallest product? Which the largest?
### Book Codes
##### Age 7 to 11 Challenge Level:
Look on the back of any modern book and you will find an ISBN code. Take this code and calculate this sum in the way shown. Can you see what the answers always have in common?
### Forgot the Numbers
##### Age 7 to 11 Challenge Level:
On my calculator I divided one whole number by another whole number and got the answer 3.125. If the numbers are both under 50, what are they?
### Domino Numbers
##### Age 7 to 11 Challenge Level:
Can you see why 2 by 2 could be 5? Can you predict what 2 by 10 will be?
### Looking at Lego
##### Age 7 to 11 Challenge Level:
This task offers an opportunity to explore all sorts of number relationships, but particularly multiplication.
### Difficulties with Division
##### Age 5 to 11
This article for teachers looks at how teachers can use problems from the NRICH site to help them teach division.
### How Much Did it Cost?
##### Age 7 to 11 Challenge Level:
Use your logical-thinking skills to deduce how much Dan's crisps and ice-cream cost altogether.
### Asteroid Blast
##### Age 7 to 11 Challenge Level:
A game for 2 people. Use your skills of addition, subtraction, multiplication and division to blast the asteroids.
### The Money Maze
##### Age 7 to 11 Challenge Level:
Go through the maze, collecting and losing your money as you go. Which route gives you the highest return? And the lowest?
### Cows and Sheep
##### Age 7 to 11 Challenge Level:
Use your logical reasoning to work out how many cows and how many sheep there are in each field.
### Being Resourceful - Primary Number
##### Age 5 to 11 Challenge Level:
Number problems at primary level that require careful consideration.
### Today's Date - 01/06/2009
##### Age 5 to 11 Challenge Level:
What do you notice about the date 03.06.09? Or 08.01.09? This challenge invites you to investigate some interesting dates yourself.
### Fingers and Hands
##### Age 7 to 11 Challenge Level:
How would you count the number of fingers in these pictures?
### Escape from the Castle
##### Age 7 to 11 Challenge Level:
Skippy and Anna are locked in a room in a large castle. The key to that room, and all the other rooms, is a number. The numbers are locked away in a problem. Can you help them to get out?
### Learning Times Tables
##### Age 5 to 11 Challenge Level:
In November, Liz was interviewed for an article on a parents' website about learning times tables. Read the article here. |
# Solve using factoring
It’s important to keep them in mind when trying to figure out how to Solve using factoring. We can solve math problems for you.
## Solving using factoring
This can help the student to understand the problem and how to Solve using factoring. Web math is a type of online math that helps students learn mathematics. Web math can help students learn mathematics by providing interactive tutorials, exercises, and calculators. Web math can also help students learn mathematics by providing online resources, such as video lessons and articles. Web math can also help students learn mathematics by providing online tools, such as graphing calculators and online quizzes. Web math can also help students learn mathematics by providing online tutors who can answer questions and provide feedback. By providing these resources, web math can help students learn mathematics more effectively.
If you're working with continuous data, you'll need to use a slightly different method. First, you'll need to identify the range of the data set - that is, the difference between the highest and lowest values. Then, you'll need to divide this range into a number of intervals (usually around 10). Next, you'll need to count how many data points fall into each interval and choose the interval with the most data points. Finally, you'll need to take the midpoint of this interval as your estimate for the mode. For example, if your data set ranges from 1 to 10 and you use 10 intervals, the first interval would be 1-1.9, the second interval would be 2-2.9, and so on. If you count 5 data points in the 1-1.9 interval, 7 data points in the 2-2.9 interval, and 9 data points in the 3-3.9 interval, then your estimate for the mode would be 3 (the midpoint of the 3-3.9 interval).
How to solve math is a question that has been asked by students for centuries. There is no one answer that will work for everyone, but there are some general tips that can help. First, it is important to understand the problem. Read it carefully and try to identify what information is being asked for. Then, make a list of all the possible steps that could be used to solve the problem. Once you have a plan, it is time to begin working through the steps one by one. If you get stuck, don't be afraid to ask for help from a teacher or tutor. With practice, you will develop problem-solving skills that will last a lifetime.
How to solve for domain: There are many ways to solve for the domain of a function. In algebra, the domain is often defined as the set of all values for which a function produces a real output. However, this definition can be difficult to work with, so it is often useful to think about the domain in terms of graphing. For instance, if a function produces imaginary results for certain input values, then those input values will not be included in the function's domain. Similarly, if a function is undefined for certain input values, those values will also be excluded from the domain. In general, the graphing method is the easiest way to determine the domain of a function. However, it is sometimes necessary to use other methods, such as solving inequalities or using set notation. With practice, you will be able to solve for domain quickly and easily.
Logarithmic functions are a type of math used to calculate an exponent. The log function is the inverse of the exponential function, meaning that it can be used to solve for x when given a number raised to a power. In order to solve logarithmic functions, you need to use a few basic steps. First, identify the base of the logarithm. This is usually either 10 or e. Next, identify the number that is being raised to a power. This number is called the argument. Finally, set up an equation using these two numbers and solve for x. With a little practice, solving logarithmic functions can be easy and even enjoyable!
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Discriminant of a quadratic equation solver Geometry answers free online Solve the equation and check your solution Discrete math solver Algebra 2 help |
# The mean and variance of 7 observations are 8 and 16,
Question:
The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12 and 14. Find the remaining two observations.
Solution:
Let the remaining two observations be x and y.
The observations are 2, 4, 10, 12, 14, xy.
Mean, $\bar{x}=\frac{2+4+10+12+14+x+y}{7}=8$
$\Rightarrow 56=42+x+y$
$\Rightarrow x+y=14$ ...(1)
Variance $=16=\frac{1}{n} \sum_{i=1}^{7}\left(x_{i}-\bar{x}\right)^{2}$
$16=\frac{1}{7}\left[(-6)^{2}+(-4)^{2}+(2)^{2}+(4)^{2}+(6)^{2}+x^{2}+y^{2}-2 \times 8(x+y)+2 \times(8)^{2}\right]$
$16=\frac{1}{7}\left[36+16+4+16+36+x^{2}+y^{2}-16(14)+2(64)\right]$ [Using (1)]
$16=\frac{1}{7}\left[108+x^{2}+y^{2}-224+128\right]$
$\Rightarrow x^{2}+y^{2}=112-12=100$
$x^{2}+y^{2}=100$ $\ldots(2)$
From (1), we obtain
$x^{2}+y^{2}+2 x y=196 \ldots(3)$
$2 x y=196-100$
$\Rightarrow 2 x y=96 \ldots$ (4)
Subtracting (4) from (2), we obtain
$x^{2}+y^{2}-2 x y=100-96$
$\Rightarrow(x-y)^{2}=4$
$\Rightarrow x-y=\pm 2 \ldots$ (5)
Therefore, from (1) and (5), we obtain
x = 8 and y = 6 when x – y = 2
x = 6 and y = 8 when x – y = – 2
Thus, the remaining observations are 6 and 8. |
# An arithmetic sequence is given by 5, 8, 11,... How do you find the value of d?
Jul 18, 2018
$\textcolor{b l u e}{d = 3}$
#### Explanation:
The nth term of an arithmetic sequence is given by:
$a + \left(n - 1\right) d$
Where $\boldsymbol{a}$ is the first term, $\boldsymbol{d}$ is the common difference and $\boldsymbol{n}$ is the nth term.
We have:
$a = 5$
looking at the second element, we have:
$5 + \left(2 - 1\right) d = 8$
$5 + d = 8 \implies d = 3$
Jul 18, 2018
$d = 3$
#### Explanation:
In an arithmetic sequence, $d$ is our common difference, or the value we add/subtract to go to the next term.
To get from $5$ to $8$, we can add $3$.
To get from $8$ to $11$, we can add $3$.
In an arithmetic sequence, we always add or subtract the same amount. We see that to get to the next term, we add $3$.
This is our $d$.
Hope this helps! |
Instasolv
IIT-JEE NEET CBSE NCERT Q&A
4.5/5
# NCERT Exemplar Class 11 Maths Chapter 11 Solutions: Conic Sections
NCERT Exemplar Solutions for Class 11 Maths Chapter 11 will help you study the equations of curved shapes such as circles, parabola, hyperbola, and ellipses. The NCERT Exemplar Class 11 Maths will prove to be essential for learning and building the confidence that is required to face the exams.
In NCERT Exemplar class 11 maths chapter 11, you will understand that the conic sections can be achieved by the intersection of a plane with a right circular cone. The main topics of chapter 11 are sections of a cone, parabola, hyperbola, ellipse, the focal distance of hyperbola, focal Distance of an ellipse, circle, equation of a circle, standard equations of parabola, the focal distance of a point of the parabola, directrix of the parabola, and parametric equation of conics.
There are 5 exercises with a total of 87 questions in Chapter 11 of conic sections. These problems will help you to understand the types of questions that will be asked in the exams. Also, the solutions provided by Instasolv will help you develop a mindset that is required in cracking tough exams.
## Important Topics for NCERT Maths Exemplar Solutions Class 11 Chapter 11 – Conic Sections
Circles
The set of all points on a surface that is located at an equal distance from a fixed point in the plane is known as a circle.
• The equation of a circle with center (h, k) and the radius r is denoted by (p – h)2 + (q – k)2 = r2
• The equation of a circle with center (0, 0) and the radius r is denoted by p2 + q2 = r2.
• Features of the equation of a circle:
1. Second degree in p and q
2. Coefficient of p2 = coefficient of q2
3. Coefficient of pq = 0
Parabola
The set of all points on a surface that is located at an equal distance from a fixed line and a fixed point on the surface is called Parabola. In geometric, Parabola is a locus of the point which keeps on moving so that its distance from a fixed point is similar to the distance from moving point to a fixed straight line.
The standard equation of the parabola with focus at (a, 0) a > 0 and directrix p = – a is q2 = 4ax.
Ellipse
The set of all points in a surface where the sum of whose distances from two fixed points in the surface is a constant. If the surface cuts completely across one nappe of the cone and Ɵ < α < 90 degrees, then the curve of intersection of cone and surface is an Ellipse.
An ellipse with foci on the x-axis can be represented in the form of p2/a2+ q2/b2 = 1
Hyperbola
The set of all points in a surface where the difference of whose distances from two fixed points in the surface is a constant is termed as a hyperbola.
A hyperbola with foci on the x-axis can be written in the form of the equation as follows: x2/a2 – y2/b2 = 1
## Exercise-wise Discussion of NCERT Maths Exemplar Solutions Class 11 Chapter 11 – Conic Sections
Exercise 11.1 – Short Answer Type Questions
In this exercise of Chapter 11, there are a total of 22 questions that are based on the concepts of parabola, circles, and section of a cone.
Exercise 11.2 – Long Answer Type Questions
NCERT Exemplar Class 11 for Chapter 11 Exercise 11.2 contains a total of 10 questions. They are long answer type questions based on the ellipse, the focal distance of hyperbola, the focal Distance of an ellipse, circle, equation of a circle, standard equations of parabola, and focal distance of a point of the parabola.
Exercise 11.3 – Objective Type Questions
This exercise consists of a total of 27 questions which are objective type questions based on the important topics related to the conic section. In these questions, you will be provided with 4 options and you have to choose the right answer.
Exercises 4 and 5 consists of very short answer type questions including match the following, fill in the blanks and state true or false. Solving these questions will help you have prepared for the competitive exams as well.
## Why Use NCERT Maths Exemplar Solutions Class 11 Chapter 11 – Conic Sections by Instasolv?
• These NCERT exemplar solutions are prepared by the subject matter experts at Instasolv in order to guide you in your academics.
• The exercise-wise solutions of the important topics of Chapter 11 – Conic Sections are available here on our web page, which you can easily follow at any time and from anywhere.
• These solutions are very useful for your conceptual knowledge and to build up problem-solving ability in yourself.
Referring to the NCERT exemplar class 11 maths solutions for chapter 11 will provide you with the basics and the advanced level of questions and answers that are mostly asked in various competitive exams.
More Chapters from Class 11 |
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# The Ultimate Q51 Guide [Expert Level]
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Joined: 16 Aug 2015
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
### Show Tags
07 Jan 2020, 17:21
[GMAT math practice question]
(Geometry) The figure shows right triangle ABC and AB = 5, BC = 4 and CA = 3. What is the area of triangle BDE?
1) AC = AD
2) AB is perpendicular to DE.
Attachment:
1.3ds.png [ 22.3 KiB | Viewed 416 times ]
=>
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.
Since we have 2 variables (AD and EC) in the original condition, C is most likely the answer. We can figure out where D and E are by AD and EC. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.
Conditions 1) & 2)
Attachment:
1.3ds(a).png [ 12.99 KiB | Viewed 416 times ]
Since triangles AED and AEC are congruent, we have DE = CE and DB = 5 - 3 = 2.
The area of triangle ABC = (1/2)*4*3 = 6, which is also the sum of the areas of triangle ABE and triangle AEC.
Then we have (1/2)*5*DE + (1/2)*3*CE = (5/2)DE + (3/2)DE = (8/2)DE = 4DE = 6.
Thus, we have DE = 6/4, or DE = 3/2.
The area of triangle DBE is (1/2)*DB*DE = (1/2)*2*(3/2) = 3/2.
Since both conditions together yield a unique solution, they are sufficient.
Therefore, C is the answer.
Answer: C
In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2) when taken together. Obviously, there may be occasions on which the answer is A, B, C, or D.
_________________
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
### Show Tags
08 Jan 2020, 23:44
[GMAT math practice question]
(Geometry) The figure shows right triangle ABC and AB = 5, BC = 4 and CA = 3. What is the area of triangle BDE?
1) AC = AD.
2) AB is perpendicular to DE.
Attachment:
1.3ds.png [ 22.3 KiB | Viewed 386 times ]
=>
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.
Since we have 2 variables (AD and EC) in the original condition, C is most likely the answer. We can figure out where D and E are by AD and EC. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.
Conditions 1) & 2)
Attachment:
1.3ds(a).png [ 12.99 KiB | Viewed 386 times ]
Since triangles AED and AEC are congruent, we have DE = CE and DB = 5 - 3 = 2.
The area of triangle ABC = (1/2)*4*3 = 6, which is also the sum of the areas of triangle ABE and triangle AEC.
Then we have (1/2)*5*DE + (1/2)*3*CE = (5/2)DE + (3/2)DE = (8/2)DE = 4DE = 6.
Thus, we have DE = 6/4, or DE = 3/2.
The area of triangle DBE is (1/2)*DB*DE = (1/2)*2*(3/2) = 3/2.
Since both conditions together yield a unique solution, they are sufficient.
Therefore, C is the answer.
Answer: C
In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2) when taken together. Obviously, there may be occasions on which the answer is A, B, C, or D.
_________________
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Math Revolution GMAT Instructor
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
### Show Tags
09 Jan 2020, 19:33
[GMAT math practice question]
(Inequalities) a, b and c are positive and a < 5. a, b and c satisfy a2 - a - 2b - 2c = 0, and a + 2b - 2c + 3 = 0. What is the order among a, b and c?
A. c > a > b
B. a > c > b
C. a > b > c
D. c > b > a
E. b > c > a
=>
When we add the two equations a^2 - a - 2b - 2c = 0 and a + 2b - 2c + 3 = 0, we have a^2 - a - 2b - 2c + a + 2b - 2c + 3 = 0 + 0, a^2 – 4c + 3 = 0 or 4c = a^2 + 3.
When we subtract 4a from the both sides of the last equation, we have 4c - 4a = a^2 - 4a + 3, 4(c - a) = (a - 1)(a - 3).
Rearranging the equations given in the question gives us a^2 - a = 2(b + c) and a + 3 = 2(c - b) given in the question. When we subtract a + 3 = 2(c - b) from a^2 - a = 2(b + c), we have a^2 - a - (a + 3) = 2(b + c) - 2(c - b), a^2 - a - a - 3 = 2b + 2c - 2c +2b, 4b = a^2 – 2a – 3, 4b = (a - 3)(a + 1) > 0 and a > 3.
We have 4(c - a) = (a - 1)(a - 3) > 0 or c > a since a > 3 and (a - 1)(a - 3) > 0.
We have 4(b - a) = 4b – 4a and 4b – 4a = a^2 – 2a – 3 – 4a (because 4b = a^2 – 2a – 3). Then 4b - 4a = a^2 – 6a – 3 = a(a - 3) – 3(a - 3) – 12 = (a - 3)2 - 12 < 0 or b < a since 3 < a < 5.
Thus, we have c > a > b.
Therefore, A is the answer.
Answer: A
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
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12 Jan 2020, 17:51
[GMAT math practice question]
(Geometry) As the figure below shows, △ABC is a right triangle. What is the measure of ∠B?
1) DM is a perpendicular bisector of segment AB.
2) ∠MAD = ∠CAD
Attachment:
1.6ds.png [ 14.59 KiB | Viewed 361 times ]
=>
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.
The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.
Condition 1) tells us that triangle ABD is isosceles with BD = AC and ∠B = ∠MBD = ∠MAD. Condition 2) tells us that ∠MAD = ∠CAD.
Since ∠C = 90, we have ∠MBD + ∠MAD + ∠CAD + ∠C = 180, ∠MBD + ∠MAD + ∠CAD + 90 = 180, or ∠MBD + ∠MAD + ∠CAD = 90 and ∠MBD = ∠MAD = ∠CAD. As each angle is the same and all three add up to 90, each angle must be 30.
Thus we have ∠B = ∠MBD = 30.
Since both conditions together yield a unique solution, they are sufficient.
Therefore, C is the answer.
Answer: C
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
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13 Jan 2020, 17:40
[GMAT math practice question]
(Inequalities) x is an integer. y and z are real numbers with x < 2y < 3z. What is the value of x?
1) x + 2y + 3z = 4
2) 2x + 3y + 4z = 5
=>
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.
The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.
Since we have 3 variables (x, y, and z) and 0 equations, E is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.
Conditions 1) & 2)
We have x + 2y + 3z = 4 and 2x + 3y + 4z = 5.
When we subtract twice the second equation from three times the first equation, we have
3(x + 2y + 3z) - 2(2x + 3y + 4z) = 3(4) - 2(5)
3x + 6y + 9z - 4x - 6y - 8z = 12 - 10
-x + z = 2
z = x + 2.
Substituting z = x + 2 into the first equation gives us:
x + 2y + 3z = 4
x + 2y + 3(x + 2) = 4
x + 2y + 3x + 6 = 4
4x + 2y = -2
2y = -4x - 2
y = -2x – 1
x < 2y < 3z is equivalent to x < 2(-2x – 1) < 3(x + 2), or x < -4x – 2 < 3x + 6.
Then we have x < -2/5 from x < -4x – 2, because:
x < -4x – 2
5x < -2
x < -2/5.
We also have x > -8/7 from -4x - 2 < 3x + 6, because:
-4x - 2 < 3x + 6
-7x < 8
x > -8/7 (the inequality sign changes direction since we divided by a negative)
Thus -8/7 < x < -2/5 and x = -1 since x is an integer.
Since both conditions together yield a unique solution, they are sufficient.
Since this question is an integer question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.
Condition 1)
x = -2, y = 1, z = 4/3 and x = -3, y = 0, z = 1 are solutions.
Condition 2)
x = -2, y = 0, z = 9/4 and x = -1, y = 0, z = 7/4 are solutions.
Since condition 2) does not yield a unique solution, it is not sufficient.
Therefore, C is the answer.
Answer: C
In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C, or D.
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
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14 Jan 2020, 20:10
[GMAT math practice question]
(Function) A function f(x) is defined as x *<x>. If <x> is the greatest factor of x except x, what is f(500)?
A. 1000
B. 5000
C. 15000
D. 120000
E. 125000
=>
All the factors of 500 are 1, 2, 4, 5, 10, 20, 25, 50, 100, 125, 250 and 500.
Thus <500> = 250.
Then f(500) = 500*250 = 125000.
Therefore, E is the answer.
Answer: E
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
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15 Jan 2020, 17:24
[GMAT math practice question]
(Coordinate Geometry) The line y = ax - 1 passes through (1/3, 0). On the line, there is a point whose x-coordinate and y-coordinate are the same. What is this point?
A. (1/5 , 1/5)
B. (1/3 , 1/3)
C. (1/2 , 1/2)
D. (- 1/4 , - 1/4)
E. (- 1/5 , - 1/5)
=>
Since the line y = ax – 1 passes through (1/3, 0), we have 0 = a/3 – 1 or a = 3.
To figure out the point with the same x-coordinate and y-coordinate on the line y = 3x – 1, we make y = x, which gives us x = 3x – 1 or 2x = 1.
Thus we have x = 1/2, y = 1/2, and the point is (1/2, 1/2).
Therefore, C is the answer.
Answer: C
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The Ultimate Q51 Guide [Expert Level] [#permalink]
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16 Jan 2020, 17:25
[GMAT math practice question]
(Coordinate Geometry) The following figure shows line l: y = 2x. If the area of BOC is 8, what is the area of triangle AOB?
A. 4
B. 6
C. 8
D. 11
E. 13
Attachment:
1.8ps.png [ 7.31 KiB | Viewed 280 times ]
=>
Since the area of the triangle OBC is 8 and the length of the base OC is 4, and the area of a triangle = 1/2bh, we have:
8 = 1/2(4)h
8 = 2h
h = 4.
The height of the triangle is 4, and the y-coordinate of point B is 4.
Since point B is on the line y = 2x, we have 4 = 2x or x = 2.
Then we have point B(2, 4).
We can then determine the equation of the line passing through B (2, 4) and C (4, 0). We first determine the slope of the line using the equation
(y1 - y2) / (x1 - x2)
(4 - 0) / (2 - 4)
4 / -2
Slope = m = -2.
Then, using the point (4, 0) and m = 4 we get:
y = mx + b
0 = -2(4) + b
b = 8
The line passing through B and C is y = -2x + 8.
Point A has an x-coordinate of 0, so y = -2(0) + 9, or y = 8.
Then we have point A(0,8).
The triangle AOB has base OA = 8, and the height equals 2 since the x-coordinate of B is 2. Then the area of triangle OAB is (1/2)*8*2 = 8.
Therefore, C is the answer.
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The Ultimate Q51 Guide [Expert Level] [#permalink]
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19 Jan 2020, 17:52
[GMAT math practice question]
(Number Properties) x, y, and z are integers with 3 ≤ x < y < z ≤ 30 and y is a prime number. What is the value of x + y + z?
1) 1/x + 1/y = 1/2 + 1/z
2) 2xy = z
=>
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.
The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.
Condition 2)
x = 3, y = 5, z = 2*3*5 = 30 are unique solutions as it is the only combination of numbers that works within the given conditions of 3 ≤ x < y < z ≤ 30 and y is a prime number. If x and y are larger numbers than z is greater than 30. We then have x + y + z = 3 + 5 + 30 = 38.
Since condition 2) yields a unique solution, it is sufficient.
Condition 1)
Since 3 ≤ x < y < z ≤ 30, we have 1/30 ≤ 1/z < 1/y < 1/x ≤ 1/3 when we take reciprocals.
Since we have 1/x + 1/y = 1/2 + 1/z, we have 1/2 < 1/x + 1/y < 1/x + 1/x = 2/x or 1/2 = 2/4 < 1/x.
Thus x < 4 and we have x = 3.
Since we have 1/2 = 1/3 + 1/y, we have 1/6 < 1/y or y < 6.
Since 3 < y < 6 and y is a prime number, we have y = 5.
1/z = 1/x + 1/y – 1/2 = 1/3 + 1/5 – 1/2 = 10/30 + 6/30 – 15/30 = 1/30 or z = 30.
Then, x + y + z = 3 + 5 + 30 = 38.
Since condition 1) yields a unique solution, it is sufficient.
Therefore, D is the answer.
Answer: D
This question is a CMT 4(B) question: condition 2) is easy to work with, and condition 1) is difficult to work with. For CMT 4(B) questions, D is most likely the answer.
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
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20 Jan 2020, 17:36
[GMAT math practice question]
(Algebra) What is k?
1) 3x + 5y = k + 1 and 2x + 3y = k
2) x + y = 2
=>
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.
The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.
Since we have 3 variables (x, y, and k) and 0 equations, E is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.
Conditions 1) & 2)
Since x + y = 2, we have y = 2 – x.
Substituting y = 2 - x into 3x + 5y = k + 1 gives us 3x + 5(2 - x) = k + 1, 3x + 10 - 5x = k + 1, -2x + 10 = k + 1 or 2x + k = 9.
Substituting y = 2 - x into 2x + 3y = k gives us 2x + 3(2 - x) = k, 2x + 6 = 3x = k, -x + 6 = k or x + k = 6.
We now have 2 equations: 2x + k = 9 and x + k = 6. Rewriting the first equation gives us k = 9 - 2x. Substituting this into the second equation gives us x + 9 - 2x = 6, -x = -3, and x = 3. Then x + k = 6 becomes 3 + k = 6, and k = 3.
Then we have x = 3 and k = 3.
Therefore, C is the answer.
Answer: C
In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2) when taken together. Obviously, there may be occasions on which the answer is A, B, C, or D.
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
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21 Jan 2020, 17:43
[GMAT math practice question]
(Coordinate Geometry) The figure shows that OABC is a parallelogram. What is the equation of the line passing through A and C?
A. y = 1/10x + 3/10
B. y = 5/11x + 20/11
C. y = 5/12x + 3/12
D. y = 1/13x + 3/13
E. y = 5/14x + 3/14
Attachment:
1.13PS.png [ 12.51 KiB | Viewed 209 times ]
=>
Since AB = CO = 4, we have point C(-4, 0).
Then the slope of the line AC is (5 - 0) / (7 - (-4)) = 5/11.
The equation of the line AC is y - 0 = (5/11)(x - (-4)) or y = (5/11)x + 20/11.
Therefore, B is the answer.
Answer: B
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
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27 Jan 2020, 06:05
[GMAT math practice question]
(Probability) On each face of a cube, one of 1, 2 or 3 is written. The number of 1’s on a face is a, the number of 2’s is b, and the number of 3’s is c. What is c?
1) a = 2 and b = 3.
2) The probability of throwing the two identical cubes and getting a sum of 3 is 1/3.
=>
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.
The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.
We have 3 variables and 1 equation. However, we should check condition 1) alone first, since it has 2 equations.
Condition 1)
Since we have a + b + c = 6, a = 2 and b = 3, we have 2 + 3 + c = 6, 5 + c = 6, and c = 1.
Since condition 1) yields a unique solution, it is sufficient.
Condition 2)
Condition 2) tells us that c/6 + c/6 = 1/3, (2c)/6 = 1/3, c/3 = 1/3, c = 3/3. Then we have c = 1.
Since condition 2) yields a unique solution, it is sufficient.
Therefore, D is the answer.
Answer: D
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
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27 Jan 2020, 18:23
[GMAT math practice question]
(Probability) What is the probability that a three-digit integer corresponding with the three numbers thrown on three different dice is a multiple of 11?
A. 1/3
B. 2/27
C. 4/27
D. 5/27
E. 2/9
=>
In order for a three-digit integer abc to be a multiple of 11, a – b + c must be a multiple of 11 since we have 100a + 10b + c = 99a + a + 11b – b + c = (99a + 11b) + a – b + c = 11(9a+b) + a – b +c.
If a – b + c = 0, then the possible cases for (a, b, c) are
(1, 2, 1), (1, 3, 2), (1, 4, 3), (1, 5, 4), (1, 6, 5), (2, 3, 1), (2, 4, 2), (2, 5, 3), (2, 6, 4), (3, 4, 1), (3, 5, 2), (3, 6, 3), (4, 5, 1), (4, 6, 2) and (5, 6, 1), and we have 15 cases.
If a – b + c = 11, then we have only the case (a, b, c) = (6, 1, 6).
Thus the probability is 16/(6^3) = 16/216 = 2/27.
Therefore, the answer is B.
Answer: B
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
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28 Jan 2020, 17:47
[GMAT math practice question]
(Geometry) Is triangle ADE an isosceles triangle?
1) AB = AC
2) BD = CE
Attachment:
1.21DS.png [ 7.09 KiB | Viewed 164 times ]
=>
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.
Since triangle ADE has three sides, we have 3 variables and 0 equations, and E is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.
Conditions 1) & 2)
When we consider condition 1), triangle ABC is an isosceles triangle, and ∠B and ∠C are congruent.
Since BD = EC from condition 2) and we have AB = AC, and ∠B = ∠C, triangles ABD and ACE are congruent to each other using the SAS property.
Thus, we have AD = AE, and the triangle ADE is isosceles.
Since both conditions together yield a unique solution, they are sufficient.
Therefore, C is the answer.
Answer: C
In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2) when taken together. Obviously, there may be occasions on which the answer is A, B, C, or D.
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
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29 Jan 2020, 17:27
[GMAT math practice question]
(Geometry) The figure shows that ∠ABC is 80. What is ∠ADC?
1) Point O is the circumcenter of △ABC.
2) Point O is the circumcenter of △ACD.
Attachment:
1.24DS.png [ 9.51 KiB | Viewed 148 times ]
=>
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.
Since we have 4 angles of a quadrilateral, we have 4 variables (∠A, ∠B, ∠C, and ∠D) and 2 equations (∠B = 80 and ∠A + ∠B + ∠C + ∠D = 360), and E is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.
Conditions 1) & 2)
When we consider both conditions together, we have a quadrilateral inscribed by a circle and ∠B + ∠D = 180.
Since we have ∠B = 80, we have ∠D = 100.
Since both conditions together yield a unique solution, they are sufficient.
Therefore, C is the answer.
Answer: C
Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B, or D is the answer. This occurs in Common Mistake Types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D, or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
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30 Jan 2020, 17:19
[GMAT math practice question]
(Probability) There are six cards numbered 0, 0, 1, 2, 3, and 4. How many 3-digit even numbers are possible when picking 3 cards?
A. 30
B. 32
C. 34
D. 36
E. 38
=>
Case 1: X00 (We have two zero digits.)
The number of possible cases is 4.
Case 2: X0Y / XY0 (We have one zero digit.)
X0Y: The number of possible values of y is 2 since y is an even number.
Then the number of possible values of x is 3.
Thus we have 2 * 3 = 6 cases.
XY0: The number of possible cases is 4 * 3 = 12
When we have one zero digit, we have 6 + 12 = 18 cases.
Case 3: XYZ (We don’t have any zero digits.)
The number of possible values of z is 2 since z is an even number.
Then we have 3 * 2 = 6 cases for each even integer z.
Thus we have 6 * 2 = 12 cases.
Then we have 4 + 18 + 12 = 34 cases.
Therefore, the answer is C.
Answer: C
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
### Show Tags
02 Feb 2020, 18:32
[GMAT math practice question]
(Geometry) What is the measure of ∠BOC in the figure?
1) Point O is the circumcenter of triangle ABC.
2) ∠OAC = 23° and ∠OBA = 48°
Attachment:
1.27DS.png [ 13.67 KiB | Viewed 110 times ]
=>
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.
Since we have 4 variables (∠BOC, ∠ABC, ∠BCA, and ∠CAB) and 1 equation (∠ABC + ∠BCA + ∠CAB = 180°), C is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.
Conditions 1) & 2)
Since point O is the circumcenter of triangle ABC from condition 1), triangles OAB, OBC, and OCA are isosceles and OA, OB, OC are congruent.
Since ∠OAC = 23°, we have ∠OCA = 23° (because it is an isosceles triangle) and ∠AOC = 134° (180° - 23° - 23°).
Since ∠OBA = 48°, we have ∠OAB = 48° and ∠AOB = 84°.
Then we have ∠AOB + ∠BOC + ∠COA = 360°, 134° + 84° + ∠BOC = 360°, 218° + ∠BOC = 360° or ∠BOC = 142°.
Since both conditions together yield a unique solution, they are sufficient.
Therefore, C is the answer.
Answer: C
In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B, or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2) when taken together. Obviously, there may be occasions in which the answer is A, B, C, or D.
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
### Show Tags
03 Feb 2020, 18:14
[GMAT math practice question]
(Geometry) The figure shows that ∠BAD is 30°, and ∠CAE is 40°. What is the measure of ∠ADE?
1) Point O is the circumcenter of triangle ABC.
2) Point I is the incenter of triangle ABC.
Attachment:
1.30DS.png [ 7.84 KiB | Viewed 91 times ]
=>
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.
Since we have 6 variables (∠ABC, ∠BCA, ∠CAB, ∠ADE, ∠DAE, and ∠DEA) and 2 equations (∠ABC + ∠BCA + ∠CAB = 180°, ∠ADE + ∠DAE + ∠DEA = 180°), C is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.
Conditions 1) & 2)
Since point I is the incenter of triangle ABC, we have ∠IAB = ∠IAC = 40° and ∠DAE = ∠IAB - ∠BAD, ∠DAE = 40° - 30°, ∠DAE = 10° .
Since point O is the circumcenter of triangle ABC, we have ∠BAO = ∠ABO = 30° and ∠OBC = ∠OCB = (1/2)(180° – (2*30° + 2*50°)) = 10°.
Then ∠ABC = ∠ABO + ∠OBC = 30° + 10° = 40°.
Since ∠ADE is an exterior angle of triangle ABD, we have ∠ADE = ∠DAB + ∠ABD = 30° + 40° = 70°.
Since both conditions together yield a unique solution, they are sufficient.
Therefore, C is the answer.
Answer: C
In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B, or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2) when taken together. Obviously, there may be occasions in which the answer is A, B, C, or D.
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Re: The Ultimate Q51 Guide [Expert Level] [#permalink]
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04 Feb 2020, 17:18
[GMAT math practice question]
(Geometry) Point I is the incenter of triangle ABC in the figure. The figure shows AB = 10, AC = 8, ∠ABI = 22° and ∠ACI = 30°. Line DE is parallel to line BC. What is the length of the perimeter of triangle ADE?
Attachment:
1.27DS.png [ 13.67 KiB | Viewed 78 times ]
A. 14
B. 16
C. 18
D. 20
E. 22
=>
Since I is the incenter of the triangle ABC, IB and IC bisect ∠ABC and ∠ACB, respectively. We know ∠IBC = 22° and ∠ICB = 30°. Since ∠DIB and ∠CBI are alternate interior angles, we have ∠BID = 22° and triangle DBI is an isosceles. Since ∠EIC and ∠BCI are alternate interior angles, we have ∠EIC = 30° and triangle CEI is an isosceles. Thus we have the perimeter of triangle ADE = AD + DE + EA = AD + DI + IE + EA = AD + DB + CE + EA = AB + AC = 10 + 8 = 18.
Therefore, C is the answer.
Answer: C
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The Ultimate Q51 Guide [Expert Level] [#permalink]
### Show Tags
05 Feb 2020, 17:20
[GMAT math practice question]
(Geometry) Point O is the circumcenter of triangle ABC, and the length of AC is 7. The length of the perimeter of triangle ABC is 19. What is the area of the circumscribed circle of triangle ABC?
A. 30π
B. 32π
C. 34π
D. 36π
E. 38π
Attachment:
1.28ps.png [ 8.69 KiB | Viewed 49 times ]
=>
Since point O is the circumcenter of triangle ABC, OA = OB = OC is the radius of the circumscribed circle of the triangle.
Since the perimeter of the triangle ABC is 19, we have OA + OC + 7 = 19, OA + OA + 7 = 19, 2(OA) + 7 = 19, 2(OA) = 12, or the radius OA = 6.
Thus, the area of the circumscribed circle of the triangle is (62)π = 36π.
Therefore, the answer is D.
Answer: D
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The Ultimate Q51 Guide [Expert Level] [#permalink] 05 Feb 2020, 17:20
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# HSPT Quantitative : How to make geometric comparisons
## Example Questions
### Example Question #21 : Geometric Comparison
Examine (a), (b), and (c) and find the best answer.
a) The area of a square with a side length of
b) The area of a square with a side length of
c) The area of a circle with a radius of
c > b > a
a = b < c
c < b < a
a = c < b
c > b > a
Explanation:
a) The area of a square with a side length of
To find the area of a square, square the side length:
b) The area of a square with a side length of
c) The area of a circle with a radius of
To find the area of a circle, multiply the radius by .
(Here, we rounded to , because an exact number isn't necessary to answer the question.)
Therefore (c) is larger than (b) which is larger than (a).
### Example Question #22 : Geometric Comparison
Examine (a), (b), and (c) to find the best answer:
a) the interior angle of an equilateral triangle
b) the interior angle of a square
c) the interior angle of a regular pentagon
Explanation:
Since the interior angles of a triangle add up to , each angle of an equilateral triangle is degrees.
Each of the interior angles of a square is degrees.
The interior angles of a pentagon add up to , so each angle in a regular pentagon is degrees.
### Example Question #23 : Geometric Comparison
Examine (a), (b), and (c) to find the best answer:
a) half the volume of a cube with dimensions inches by inches by inches
b) the volume of a cube with dimensions inches by inches by inches
c) the volume of a cube with dimensions inches by inches by inches
(a) is equal to (b) but not (c)
(a) is equal to (c) but not (b)
(a), (b), and (c) are all equal
(a), (b), and (c) are all unequal
(a) is equal to (b) but not (c)
Explanation:
Find the three volume by multiplying height by length by width:
a)
Half of this volume is .
b)
c)
Remember that we are only looking at half of the volume in a).
Therefore (a) and (b) are equal but (c) is not.
### Example Question #21 : Geometric Comparison
Examine (a), (b), and (c) to find the best answer:
a) area of a rectangle with side lengths and
b) area of a rectangle with side lengths and
c) area of a square with side length
Explanation:
Area is calculated by multiplying the side lengths:
a) area of a rectangle with side lengths and
b) area of a rectangle with side lengths and
c) area of a square with side length
Therefore (b) is less than (a), which is less than (c).
### Example Question #21 : How To Make Geometric Comparisons
Examine (a), (b), and (c) to find the best answer:
a) side length of a cube with a volume of inches cubed
b) side length of a square with an area of inches squared
c) side length of a square with an area of inches squared
Explanation:
To find the side length of a cube from its volume, find the cube root:
To find the side length of a square from its area, find the square root:
b)
c)
(a) is smaller than (c), which is smaller than (b)
### Example Question #26 : Geometric Comparison
What are the relationships between the areas of these shapes?
b. A square with side
c. A rectangle with side lengths of and
Explanation:
First, we find the areas of a, b, and c.
Now we put them in order of size.
### Example Question #27 : Geometric Comparison
Find the relationship between the perimeters of these shapes.
a. A square with area
b. A circle with diameter
c. A pentagon with side length
Explanation:
First, find the perimeter of the shapes.
Since the area of is , its side length is , giving it a perimeter of .
The perimeter of is .
The perimeter of is .
Since , .
Therefore, .
### Example Question #28 : Geometric Comparison
Find the relationship between these lengths.
a. Side of a square of area
b. Side of a square with perimeter
c. Diameter of a circle with area
Explanation:
First, find the lengths given.
Since the square in has area , the side length is
All sides of a square have equal lengths, so gives us a length of .
The area of a circle is , and diameter is . Since area is , , giving us .
The three lengths are equal, so .
### Example Question #29 : Geometric Comparison
Find the relationship between the areas of the following shapes.
a. Square with perimeter
b. Triangle with base and height
c. Circle with circumference
Explanation:
First, find the areas.
Since the perimeter of is , its side length is , making the area .
For triangles, , so the area of is .
For , circumference is , making , giving us an area of .
Putting them in order, we get .
### Example Question #30 : Geometric Comparison
Find the relationship between the perimeters of the following shapes.
a. Regular hexagon with side length
b. Square with side length
c. Equilateral triangle with side length |
Question
## Introduction
Page Contents
What is the fractional of hundred? In this post I will explain what is the fractional of hundred and how to write it as a decimal.
## The fractional of hundred is the number equal to one hundred and a whole number.
It can be written as a decimal, fraction or mixed number. For example, if you have \$100 in your bank account, then this is considered to be a fractional of hundred because it has two whole numbers: 100 and 1 (in this case).
## In other words, it is the part of 100 before the decimal separator.
In other words, it is the part of 100 before the decimal separator. For example, if you have \$100 and you want to express it as a fraction, then you would say that your amount is 1/2 or 50%.
Let’s take another example: If someone says “I want to buy 5 apples for \$5 each”, then he/she means that there are 5 apples in total and each apple costs \$1. So if we add up all these prices together, we get 5*(\$1+\$1+\$1+\$1+\$1), which equals 25 cents (or 1/4).
## A fractional of hundred can be written as a decimal, as long as its denominator is not greater than 9.
A fractional of hundred can be written as a decimal, as long as its denominator is not greater than 9. For example, 0.1, 0.2 and so on.
## A portion of one hundred
A hundred is a whole number, and it’s divisible by 1, 2, 5, 10 and 100.
100 can be written as a decimal with a denominator between 1 and 9. The most common fractional of hundred is 1/100
In conclusion, a fractional of hundred is the number equal to one hundred and a whole number. It can be written as a decimal as long as its denominator is not greater than 9. In other words, a fractional of hundred can be written as long as it has a denominator that ends in zero or five. For example: 0.75
1. # WHAT IS THE FRACTIONAL OF HUNDRED: What is the fractional of hundred?
In mathematics, the fractional of a number is a number that is less than or equal to 100. Fractions are important because they allow us to handle decimals, which are a type of number that we use all the time. In this blog post, we will explore what fractional of hundred is and how it works. We will also explore how to convert between fractions and decimals. Finally, we will show you an example problem that uses fractional of hundred.
## What is the fractional of hundred?
The fractional of hundred is one hundredth. The fractional of a thousand is one thousandth.
## How to calculate fractional of hundred?
The fractional of hundred is a calculation that is used to find the percentage of a number. The fractional of hundred is calculated by dividing 100 by the number that you are looking to calculate the fraction for.
For example, if someone wants to find out how much 20% of 300 is, they would divide 300 by 20 and get 60. This would then give them the answer that 20% of 300 is 60/300 or 0.8. |
Shared11 - Function Analysis Part 2 Assignment / Function Analysis Part 2 Notes.sagewsOpen in CoCalc
This material was developed by Aaron Tresham at the University of Hawaii at Hilo and is
### Prerequisites:
• Intro to Sage
• Graphing and Solving Equations
• Differentiation
• Function Analysis, Part 1
# Function Analysis, Part 2
We continue to try to understand the key features of a function using calculus, rather than depending on a graph alone. In part 1, we only had functions with domain . In this lab, we'll consider functions where the function (or its derivative) has domain restrictions.
## Example 1
Analyze and graph .
Step 1 Find the domain. Discuss vertical asymptotes and holes.
For a rational function, the domain is all real numbers such that the denominator is not . So let's find out when the denominator is .
solve(x^4 + 9*x^3 + 18*x^2 - 32*x - 96==0,x)
[x == -3, x == 2, x == -4]
The denominator is when , , or . So the domain is all real numbers except , , and , in other words, .
We need to determine what happens near these holes in the domain, so we will take limits.
f(x)=(x^3 - x^2 - 14*x + 24)/(x^4 + 9*x^3 + 18*x^2 - 32*x - 96)
limit(f(x),x=-4)
limit(f(x),x=-3)
limit(f(x),x=2)
Infinity Infinity -1/30
Based on these limits, we know that has vertical asymptotes at and , and has a hole at the point .
Step 2 Find the derivative.
df(x)=derivative(f,x);show(df(x))
Step 3 Find the critical points of .
We need to know where the derivative is or undefined. Notice that the derivative has the same denominator as , so it has the same gaps in the domain (, , and ).
Now let's find where the derivative is .
solve(df(x)==0,x)
[x == -sqrt(42) + 3, x == sqrt(42) + 3]
N(-sqrt(42) + 3)
N(sqrt(42) + 3)
-3.48074069840786 9.48074069840786
So is at and .
Step 4 See if the sign of actually changes at the critical points of , and determine whether has a local maximum or local minimum at these points.
First, we'll check x-values near .
df(-3.5)
df(-3.4)
-4.00000000000000 18.0555555555558
Since changes sign from negative to positive, has a local min at .
Now check x-values near .
df(9)
df(10)
1/4056 -1/4732
Since changes sign from positive to negative, has a local max at .
Step 5 Find .
d2f(x)=derivative(f,x,2);show(d2f(x))
Step 6 Find the critical points of .
Once again, is undefined at the same x-values as and (, , and ).
Now we'll find where is .
solve(d2f(x)==0,x)
[x == 1/294*294^(2/3)*(21*I*sqrt(3) - 21) - 1/2*294^(1/3)*(I*sqrt(3) + 1) + 3, x == 1/294*294^(2/3)*(-21*I*sqrt(3) - 21) - 1/2*294^(1/3)*(-I*sqrt(3) + 1) + 3, x == 1/7*294^(2/3) + 294^(1/3) + 3]
N(1/294*294^(2/3)*(21*I*sqrt(3) - 21) - 1/2*294^(1/3)*(I*sqrt(3) + 1) + 3)
N(1/294*294^(2/3)*(-21*I*sqrt(3) - 21) - 1/2*294^(1/3)*(-I*sqrt(3) + 1) + 3)
N(1/7*294^(2/3) + 294^(1/3) + 3)
-3.48287967940368 - 0.288421242066844*I -3.48287967940368 + 0.288421242066844*I 15.9657593588074
We have one real solution, so is at .
Step 7 See if the sign of actually changes at the critical points of , and determine whether has an inflection point at these points.
Check the sign of near .
d2f(15)
d2f(16)
-55/3333474 13/27436000
Since changes sign from negative to postive, changes from concave down to concave up at , so has an inflection point here.
Step 8 Find the x- and y-intercepts.
solve(f(x)==0,x) #find x-intercepts
[x == 3]
f(0) #find y-intercept
-1/4
Step 9 Determine the end behavior.
limit(f(x),x=-Infinity);limit(f(x),x=+Infinity)
0 0
So the x-axis is a horizontal asymptote for this function.
Step 10 Make an informed graph. Mark any - and -intercepts, relative maxima and minima, and inflection points.
First, we'll calculate the function values at the critical points.
f(-3.48074069840786);f(9.48074069840786);f(15.9657593588074)
25.9614813968190 0.0385186031842795 0.0342406411926452
So, we'll need a graph that covers these three points. We also need to remember the hole, vertical asymptotes, and intercepts.
Here is a first attempt:
f(x)=(x^3 - x^2 - 14*x + 24)/(x^4 + 9*x^3 + 18*x^2 - 32*x - 96)
plot(f,xmin=-5,xmax=20,ymin=-10,ymax=30)
So this picture is not very clear. We need to make xmin smaller so we can see more of the branch on the left. It would also be nice to extend the range on the y-axis. The vertical asymptotes at and are marked with solid lines, but it would be nice to make them dashed instead. We also should mark the hole at .
We can see the local minimum at , but the local maximum at is not clear from the graph. We also can't see the inflection point at . The x-intercept at is also not obvious.
If we want to see all these features, we'll need a really big picture (which won't fit on our screen).
So we will have to produce a second graph that reveals the local max, inflection point, and x-intercept.
Here's another graph of the overall picture, and then we'll do a second plot that zooms in on the positive x-axis.
plot(f,xmin=-15,xmax=20,ymin=-20,ymax=40,detect_poles=True,figsize=5)+point((2,-1/30),size=15,faceted=True,color='white',markeredgecolor='blue')+line([(-4,-25),(-4,45)],linestyle='dashed')+line([(-3,-25),(-3,45)],linestyle='dashed')+point([(-3.48,25.96),(9.48,0.0385),(15.97,0.0342),(3,0)],color='black',size=20)
This is a good picture of the overall shape. Note: I do not expect you to add dashed lines for the asymptotes or open circles for the holes.
Here is a close up so we can see the local max and inflection point:
plot(f,xmin=0,xmax=40,ymin=-.05,ymax=.05)+point((2,-1/30),size=15,faceted=True,color='white',markeredgecolor='blue')+point([(9.48,0.0385),(15.97,0.0342),(3,0)],color='black',size=20)
This is another good example of why we bother analyzing the function with calculus. A single graph may not show all the important features of a function.
Step 11 Discuss absolute max/min, increasing/decreasing, concave up/down.
There is no absolute max or min.
The function is increasing on , , and .
The function is decreasing on , , and .
The function is concave up on and .
The function is concave down on , , and .
Comment: Notice where all the endpoints come from: critical points and domain restrictions (don't forget the hole at ).
## Example 2
Analyze and graph .
Step 1 Find the domain. Discuss vertical asymptotes and holes.
The domain of is .
There is a vertical asymptote or hole at . We'll use a limit to determine which it is.
f(x)=ln(x)/x
limit(f(x),x=0)
-Infinity
Since the limit is infinite, there is a vertical asymptote at .
Step 2 Find the derivative.
df(x)=derivative(f,x);show(df(x))
Step 3 Find the critical points of .
The derivative has the same domain as the original , so we just need to find where the derivative is 0.
solve(df(x)==0,x)
[e == x]
Our one critical point is .
Step 4 See if the sign of actually changes at the critical points of , and determine whether has a local maximum or local minimum at these points.
df(2.5);df(3.0)
0.0133934829001352 -0.0109569209631233
Since changes sign from positive to negative, has a local max at .
Step 5 Find .
d2f(x)=derivative(f,x,2);show(d2f(x))
Step 6 Find the critical points of .
has the same domain as and , so we need to find where is 0.
solve(d2f(x)==0,x)
[x == e^(3/2)]
N(e^(3/2))
4.48168907033806
We have at .
Step 7 See if the sign of actually changes at the critical points of , and determine whether has an inflection point at these points.
d2f(4.4);d2f(4.5)
-0.000431899396031749 0.0000894901898770770
Since changes sign, there is an inflection point at .
Step 8 Find the x- and y-intercepts.
solve(f(x)==0,x) #find x-intercept
[x == 1]
There is no y-intercept, since 0 is not in the domain.
Step 9 Determine the end behavior.
limit(f(x),x=0,dir='right');limit(f(x),x=infinity)
-Infinity 0
We have a vertical asymptote at (y-axis) and a horizontal asymptote at (x-axis).
Step 10 Make an informed graph. Mark any - and -intercepts, relative maxima and minima, and inflection points.
Our graph needs to include and their corresponding y-values.
N(f(e));N(f(e^(3/2)))
0.367879441171442 0.334695240222645
Here is a first attempt. Notice that the inflection point and the horizontal asymptote are not clear.
plot(f,xmin=0,xmax=5,ymin=-.5,ymax=.5)
Here is a second attempt, with the important points marked.
plot(f,xmin=0,xmax=25,ymin=-.5,ymax=.5)+point([(1,0),(e,f(e)),(e^1.5,f(e^1.5))],color='black',size=20)
Step 11 Discuss absolute max/min, increasing/decreasing, concave up/down.
There is no absolute min. The absolute max is at .
The function is increasing on and decreasing on .
The function is concave up on and concave down on .
## Example 3
Analyze and graph
Step 1 Find the domain. Discuss vertical asymptotes and holes.
The only potential domain issue is dividing by 0, but since is never 0, the domain is .
Step 2 Find the derivative.
f(x)=(x^2-6*x+9)^(1/3)/(x^2+1)^(1/5)
df(x)=derivative(f,x)
show(df(x))
Step 3 Find the critical points of .
First, we'll think about where the derivative is undefined.
Looking back at the formula for , we see we get division by 0 when .
solve(x^2-6*x+9==0,x)
[x == 3]
So the derivative is undefined at , so this is a critical point.
Now we'll find where the derivative is 0.
solve(df(x)==0,x)
[x == -1/4*sqrt(41) - 9/4, x == 1/4*sqrt(41) - 9/4, x == 3]
Notice that Sage gives as a solution, even though we know does not exist. This is actually a mistake in Sage's solve command.
But the other two numbers are places where the derivative is 0, so they are critical points. Let's convert them to decimals.
N(-1/4*sqrt(41) - 9/4)
N(1/4*sqrt(41) - 9/4)
-3.85078105935821 -0.649218940641788
We have three critical points: , and .
Step 4 See if the sign of actually changes at the critical points, and determine whether has a local maximum or local mininum at these points.
df(2.9)
df(3.1)
-0.934289205592418 0.879868999227662
Since changes sign from negative to positive, has a local min at .
df(-3.9)
df(-3.8)
-0.000792013772198064 0.000843740680385702
Since changes sign from negative to positive, has a local min at .
df(-.7)
df(-.6)
0.0170948233809796 -0.0192489860586108
Since changes sign from postive to negative, has a local max at .
Step 5 Find .
d2f(x)=derivative(f,x,2)
show(d2f(x))
Step 6 Find the critical points of .
Looking at the formula for , we see the same problem with division by 0 that we saw for , so does not exist.
Now we need to find where .
solve(d2f(x)==0,x)
[x == -1/132*sqrt((4356*(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(2/3) + 142593*(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3) + 550564)/(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3)) - 1/2*sqrt(-(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3) - 137641/1089/(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3) + 40905/2/sqrt((4356*(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(2/3) + 142593*(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3) + 550564)/(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3)) + 4321/66) - 9/4, x == -1/132*sqrt((4356*(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(2/3) + 142593*(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3) + 550564)/(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3)) + 1/2*sqrt(-(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3) - 137641/1089/(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3) + 40905/2/sqrt((4356*(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(2/3) + 142593*(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3) + 550564)/(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3)) + 4321/66) - 9/4, x == 1/132*sqrt((4356*(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(2/3) + 142593*(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3) + 550564)/(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3)) - 1/2*sqrt(-(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3) - 137641/1089/(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3) - 40905/2/sqrt((4356*(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(2/3) + 142593*(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3) + 550564)/(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3)) + 4321/66) - 9/4, x == 1/132*sqrt((4356*(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(2/3) + 142593*(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3) + 550564)/(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3)) + 1/2*sqrt(-(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3) - 137641/1089/(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3) - 40905/2/sqrt((4356*(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(2/3) + 142593*(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3) + 550564)/(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3)) + 4321/66) - 9/4, x == 3]
Notice that Sage incorrectly gives us again.
Let's convert the four remaining solutions to decimals.
N(-1/132*sqrt((4356*(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(2/3) + 142593*(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3) + 550564)/(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3)) - 1/2*sqrt(-(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3) - 137641/1089/(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3) + 40905/2/sqrt((4356*(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(2/3) + 142593*(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3) + 550564)/(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3)) + 4321/66) - 9/4)
N(-1/132*sqrt((4356*(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(2/3) + 142593*(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3) + 550564)/(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3)) + 1/2*sqrt(-(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3) - 137641/1089/(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3) + 40905/2/sqrt((4356*(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(2/3) + 142593*(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3) + 550564)/(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3)) + 4321/66) - 9/4)
N(1/132*sqrt((4356*(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(2/3) + 142593*(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3) + 550564)/(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3)) - 1/2*sqrt(-(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3) - 137641/1089/(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3) - 40905/2/sqrt((4356*(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(2/3) + 142593*(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3) + 550564)/(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3)) + 4321/66) - 9/4)
N(1/132*sqrt((4356*(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(2/3) + 142593*(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3) + 550564)/(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3)) + 1/2*sqrt(-(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3) - 137641/1089/(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3) - 40905/2/sqrt((4356*(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(2/3) + 142593*(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3) + 550564)/(25/1331*I*sqrt(404272455)*sqrt(11) + 24113314/35937)^(1/3)) + 4321/66) - 9/4)
-10.5718681539223 + 5.47924906323833e-17*I -1.26168389059917 + 1.72972993360071e-16*I 0.680639364999719 - 4.18499764306598e-17*I 2.15291267952170 - 1.85915507561795e-16*I
Notice that all four of these solutions have "" in the answer. This is Sage's notation for , and it usually indicates a complex number (which we will ignore). However, as we can see from the decimal approximation, the imaginary part of these solutions actually cancels out in the end, and all four solutions are real numbers (Notice the scientific notation in each of the decimals above).
We conclude that has five critical points: , , , , and .
Step 7 See if the sign of actually changes at the critical points of , and determine whether has an inflection point at these points.
d2f(2.9)
d2f(3.1)
-2.82489609113212 -3.18861678816105
Since does not change sign at , does not have an inflection point here.
d2f(-10.6)
d2f(-10.5)
-6.87833320245357e-6 0.0000179922253477384
Since the sign changes, has an inflection point at .
d2f(-1.3)
d2f(-1.2)
0.00770424683905568 -0.0147822036307242
Since the sign changes, has an inflection point at .
d2f(.6)
d2f(.7)
-0.0805864668319887 0.0166625176968990
Since the sign changes, has an inflection point at .
d2f(2.1)
d2f(2.2)
0.0175507707028564 -0.0171586030217367
Since the sign changes, has an inflection point at .
Step 8 Find the x- and y-intercepts.
solve(f(x)==0,x) #find x-intercept
[x == 3]
f(0) #find y-intercept
9^(1/3)
N(_)
2.08008382305190
The x-intercept is , and the y-intercept is .
Step 9 Determine the end behavior.
limit(f(x),x=+infinity)
limit(f(x),x=-infinity)
+Infinity +Infinity
This function has no horizontal asymptote.
Step 10 Make an informed graph. Mark any - and -intercepts, relative maxima and minima, and inflection points.
Our graph needs to include the following x-values with their corresponding y-values:
f(3);f(-3.85);f(-.65);f(-10.57);f(-1.26);f(.68);f(2.15)
0 2.07621357743516 2.20928352836990 2.21130811676451 2.17283611390427 1.62422212052417 0.635272502740562
We don't want the inflection point at to be right on the edge of the graph, so let's try and .
plot((x^2-6*x+9)^(1/3)/(x^2+1)^(1/5),xmin=-20,xmax=5,ymin=0,ymax=3)+point([(3,0),(-3.85,2.08),(-.65,2.21),(-10.57,2.21),(-1.26,2.17),(.68,1.62),(2.15,.64),(0,2.08)],color='black',size=20)
It's hard to see that the function is concave down on the left side, but if we went far enough in the negative direction to see that, we would obscure the rest of the graph.
Step 11 Discuss absolute max/min, increasing/decreasing, concave up/down.
There is no absolute maximum.
The absolute minimum is 0 at .
The function is increasing on and .
The function is decreasing on and .
The function is concave up on and
The function is concave down on , , , and . |
Question
Two alternate exterior angles are formed by a transversal through two parallel lines. If one of these alternate exterior angles has an angle measure of , and the other angle has a measure of , then what is the sum of their angle measures
1. The sum of the angle measures of the exterior angles is 180.
### What is an exterior angle?
The exterior angle refers to the angle between a side of a polygon and an extended adjacent side.
Now, given that 5x-21 and 2x+12 are two alternate exterior angles. We know that the sum of the two exterior angles is 180° because the two exterior angles are lying on a straight line and straight line is 180°.
Thus, (5x-21)+(2x+12) = 180°
=> 5x-21+2x+12 = 180
=> 5x+ 2x -21 +12 = 180
=> 7x – 9 = 180
=>7x = 189
=> x = 189/7 = 27
Hence the angles are:
1. 5x-21 = (5*27) – 21 = 135 – 21 = 114
1. 2x+12 = 2 * 27 + 12 = 54+12 = 66
And sum = 114+66= 180 |
# Algebra II Chapter 5 Notes and Practice- Solving Quadratic Equations
Document Sample
``` Algebra II Chapter 5 Notes and Practice – Solving Quadratic Equations
Recall that a quadratic equation can take one of two forms. Those forms are Standard
form, which is, y ax 2 bx c or vertex form, which is y a ( x h) 2 k . We are now going to
solve quadratic equations. If a quadratic equation starts in standard form and “ b 0 ” we will
use two different methods to solve it: Factoring and The Quadratic Formula. If the quadratic
equation is in vertex form or if "b 0" we will solve it using the square root method. Your
answers can be Real Numbers or they could be Imaginary Numbers.
Solving a Quadratic Using the Square Root Principle.
Solving a quadratic of the form a( x h)2 k 0 or ax 2 c 0
To solve a quadratic equation in vertex form or to solve a quadratic whose “b” value = 0
we can use the square root principle.
How to Use the Square Root Method to solve a quadratic
2. Take the square root of each side of the equation
x2 c
Recall that if x 2 c and “c” is a positive number then
so x c
3. If, after taking the square root of the quadratic expression the “x” is
still not isolated continue solving until it is.
Examples of Solving a Quadratic Equation Using Square Root Method.
Example 1 Solve: x 2 20 0 Example 2. Solve: 3x 2 27 0
Solution :
Solution :
Step1: 3 x 2 27 27 0 27
Step1: x 20 20 0 20
2
3 x 2 27
x 10
2
3 3
Step 2 : x 2 20 x 9
2
x 20 Step 2 : x 2 9
x 45 x i 9 3i
x 2 5
Example 3: Solve: 4( x 3) 2 6 18 Example 4: Solve: 2( x 5) 2 4 20
Solution: Solution:
2( x 5) 2 4 4 20 4
4( x 3) 2 6 6 18 6 2( x 5) 2 24
4( x 3) 2 12 2 2
( x 5) 12
2
4 4
( x 3) 3
2
( x 5) 2 12
( x 3) 2 3 x 5 i 4 3
x3 3 x 5 2i 3
x 3 3 3 3 x 5 5 5 2i 3
x 3 3 x 5 2i 3
Solving Quadratics in Standard form Using Factoring
If a quadratic equation starts in Standard form that is ax 2 bx c =0 form then one
method for solving these is factoring. Note: The quadratic must be equal to zero before
checking for factorability. First you will need to check to see if the quadratic is factorable by
multiplying “a x c” and seeing if there are factors of that product that add to “b”. If there are
then you know the quadratic is factorable and you can use the process below to solve.
2. Find the “zero” of each factor. What ever the zeros are represent the
solutions to the equation.
Solve the following using factoring
Ex 1. x 2 5 x 14 0 Ex: 2 4 x 2 12 x 9 0
Step 1 Factor Step 1 factoring
( x 7)( x 2) 0 4 x2 6 x 6 x 9 0
2 x( x 3) 3(2 x 3) 0
Step 2 Find the zeros of each factor (2 x 3)(2 x 3) 0
x 7, x 2 Step 2 Find the zero
2x 3 0
3
x double root
2
Unfortunately not all quadratics in standard form are factorable so we need to have an
additional method to solve these. This method requires you to use the Quadratic Formula. The
quadratic formula is a formula that you substitute the “a” , “b”, and “c” values into to find the
two solutions.
b b 2 4ac
2a
Part of this formula contains a value underneath the radical “ b2 4ac ”. This value underneath
the radical is called the discriminant of the formula. The value of the discriminant will tell you
how many solutions you have and what type of solutions they are.
Determining What types of solutions you have from the discriminant.
Discriminant > 0 (Positive)
If the value of the discriminant "b2 4ac 0" This tells you the equation has two real
solutions.
Discriminant = 0
If the value of the discriminant "b2 4ac 0" This tells you the equation has one real
b
solution and the solution = . Note: Quadratic is factorable if discriminant = 0.
2a
Discriminant < 0 (Negative)
If the value of the discriminant "b2 4ac 0" This tells you the equation has no real
solutions but it does have two complex conjugate solutions. Note: IF you are asked to solve
over the set of real numbers only and the discriminant is negative, then you would state
there are no real solutions and you would be finished.
1. Identify the “a”, “b”, and “c” value from the quadratic equation. Make sure equation
= 0.
2. Find the value of the discriminant to determine what types of solutions you have.
Note: If solving over the set of Real numbers and the discriminant is negative
then you would state there are no real solutions and you would be finished. If
you are solving for solutions over the set of complex numbers you would
continue to step three.
b
3. If the discriminant is zero find the solution by determining the value of . If the
2a
discriminant is not zero then plug the value of the discriminant and the other values
into the quadratic formula and solve for the solutions.
4. Make sure the final answer is simplified
Examples:
Solve over the set of complex numbers
Ex 1. 2 x2 4 x 3 0
Step 1: Identify a, b, and c values
a = 2, b = -4, and c=-3
Step 2: Find the value of the discriminant " b2 4ac "
(4)2 4(2)(3)
16 24 40 (Two real solutions)
Step 3: Since the discriminant is not zero plug it in with other values into quadratic
formula.
(4) 40 4 2 10 2(2 10)
x
2(2) 4 4
2 10
x
2
Ex 2: x2 5x 7 0
Step 1: a = 1, b= 5, c=7
(5)2 4(1)(7)
Step 2:
25 28 3 (two complex conjugate solutions)
(5) 3
x
2(1)
Step 3:
5 i 3
x
2
Solve the following Problems using the Square root principle.
1. 3( x 3) 2 3 0 2. 2( x 4) 2 10 3. 4( x 1) 2 5 5
Solve the following Problems using Factoring
4. x2 7 x 8 0 5. 5 x 2 5 0 6. 4 x2 x 5 0
Solve the following Problems using the quadratic formula over the set of complex numbers
7. 3x 2 5 x 1 0 8. 2 x 2 4 x 1 0 9. 2 x 2 4 x 5
Find the value of the discriminant for the given quadratic. From the value of the
discriminant determine how many solutions there are and the type of solutions.
10. 5 x 2 3x 4 0 11. 6 x 2 2 x 1 0 12. x 2 2 x 1 0
Discriminant=________ Discriminant=________ Discriminant=________
# of Solutions = ________ # of Solutions = ________ # of Solutions = ________
Solve using any method over the set of real numbers
13. 4 x 2 9 0 14. x 2 4 x 12 0 15. 2 x2 4 x 3 0
```
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# Division Theorem
## Theorem
For every pair of integers $a, b$ where $b \ne 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $0 \le r < \size b$:
$\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < \size b$
In the above equation:
$a$ is the dividend
$b$ is the divisor
$q$ is the quotient
$r$ is the remainder.
### Half Remainder Version
For every pair of integers $a, b$ where $b \ne 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $-\dfrac {\size b} 2 \le r < \dfrac {\size b} 2$:
$\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, -\dfrac {\size b} 2 \le r < \dfrac {\size b} 2$
## Proof 1
$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
That is, the result holds for positive $b$.
$\Box$
It remains to show that the result also holds for negative values of $b$.
Let $b < 0$.
Consider:
$\size b = -b > 0$
where $\size b$ denotes the absolute value of $b$: by definition $\size b > 0$.
From Division Theorem: Positive Divisor, we have the existence of $\tilde q, \tilde r \in \Z$ such that:
$a = \tilde q \size b + \tilde r, 0 \le \tilde r < \size b$
Since $\size b = -b$:
$a = \tilde q \paren {-b} + \paren {\tilde r} = \paren {-\tilde q} b + \tilde r$
Taking:
$q = -\tilde q, r = \tilde r$
the existence has been proved of integers $q$ and $r$ that satisfy the requirements.
The proof that they are unique is the same as that for the proof for positive $b$, but with $\size b$ replacing $b$.
$\blacksquare$
## Proof 2
Consider the set of integer multiples $x \size b$ of $\size b$ less than or equal to $a$:
$M := \set {k \in \Z: \exists x \in \Z: k = x \size b, k \le a}$
We have that:
$-\size a \size b \le -\size a \le a$
and so $M \ne \O$.
From Set of Integers Bounded Above by Integer has Greatest Element, $M$ has a greatest element $h \size b$.
Then $h \size b \le a$ and so:
$a = h \size b + r$
where $r \ge 0$.
On the other hand:
$\paren {h + 1} \size b = h \size b + \size b > h \size b$
So:
$\paren {h + 1} \size b > a$
and:
$h \size b + \size b > h \size b + r$
Thus:
$r \le b$
Setting:
$q = h$ when $b > 0$
$q = -h$ when $b < 0$
it follows that:
$h \size b = q b$
and so:
$a = q b + r$
as required.
$\blacksquare$
## Informal Proof
### Existence
Consider the arithmetic sequence:
$\ldots, a - 3 b, a - 2 b, a - b, a, a + b, a + 2 b, a + 3 b, \ldots$
which extends in both directions.
Then by the Well-Ordering Principle, there must exist a smallest non-negative element, denoted by $r$.
So $r = a - q b$ for some $q \in \Z$.
$r$ must be in the interval $\hointr 0 b$ because otherwise $r - b$ would be smaller than $r$ and a non-negative element in the sequence.
$\Box$
### Uniqueness
Suppose we have another pair $q_0$ and $r_0$ such that $a = b q_0 + r_0$, with $0 \le r_0 < b$.
Then:
$b q + r = b q_0 + r_0$
Factoring we see that:
$r - r_0 = b \paren {q_0 - q}$
and so:
$b \divides \paren {r - r_0}$
Since $0 \le r < b$ and $0 \le r_0 < b$, we have that:
$-b < r - r_0 < b$
Hence:
$r - r_0 = 0 \implies r = r_0$
So now:
$r - r_0 = 0 = b \paren {q_0 - q}$
which implies that:
$q = q_0$
Therefore the solution is unique.
$\blacksquare$
## Also known as
Otherwise known as the Quotient Theorem, or (more specifically) the Quotient-Remainder Theorem (as there are several other "quotient theorems" around).
Some sources call this the division algorithm but it is preferable not to offer up a possible source of confusion between this and the Euclidean Algorithm to which it is closely related.
It is also known by some as Euclid's Division Lemma, and by others as the Euclidean Division Property. |
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