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The figure below shows the front and rear view of a disc, which is shaded with identical patterns. The disc is flipped once with respect to any one of the fixed axes $1 -1$, $2 -2$ or $3 -3$ chosen uniformly at random.
What is the probability that the disc $\text{DOES NOT}$ retain the same front and rear views after the flipping operation?
1. $0$
2. $\frac{1}{3}$
3. $\frac{2}{3}$
4. $1$
Given that, the front and rear view of a disc.
We have a three-axis, and we can flip with respect to $1-1,2-2$ and $3-3:$
• When we flipped disk with axis ${\color{Blue}{1-1:\text{Front View} \Leftrightarrow \text{Rear View}}}$
• When we flipped disk with axis ${\color{Magenta}{2-2:\text{Front View} \nLeftrightarrow \text{Rear View}}}$
• When we flipped disk with axis ${\color{Purple}{3-3:\text{Front View} \nLeftrightarrow \text{Rear View}}}$
Now we have,
• The number of favorable cases $= 2$
• The total number of cases $= 3$
The required probability ${\color{Green}{= \dfrac{\text{The number of favorable cases }}{\text{The total number of cases}}}} = \dfrac{2}{3}.$
$\therefore$ The probability that the disc ${\color{Red}{\text{DOES NOT}}}$ retain the same front and rear views after the flipping operation is $\dfrac{2}{3}.$
Correct Answer $:\text{C}$
by
4.1k points |
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# Mastering Math: Learn How to Divide Fractions Efficiently
by | Dec 16, 2023 | How To
Math can be challenging, and when it comes to fractions, many students find themselves feeling overwhelmed. But fear not! In this tutorial, we will guide you through the process of dividing fractions step by step, making it easier and more manageable. Whether you’re a student looking to improve your skills or a teacher searching for a simplified fraction division tutorial, this article is for you.
Throughout this guide, we will explain various fraction division methods, provide clear examples, and break down the concepts to ensure your understanding. By the end, you’ll have the confidence to tackle any fraction division problem that comes your way.
### Key Takeaways:
• Dividing fractions may seem daunting, but with step-by-step guidance, it can become much easier.
• Understanding the concepts of numerator, denominator, and fraction notation is essential for mastering fraction division.
• Various methods, such as finding common denominators, flipping fractions, and using visual models, can simplify fraction division.
• Addressing common misconceptions and providing a solid foundation in fractions is crucial for success in math.
• Teaching fractions conceptually through real-world stories and visual models can improve students’ understanding and application of fraction division.
## Understanding Denominator and Common Denominators
When working with fractions, it is crucial to have a clear understanding of the denominator and common denominators. The denominator represents the number of equal parts into which a whole is divided. It tells us the size or value of each part. For example, in the fraction 3/5, the denominator is 5, indicating that the whole is divided into 5 equal parts, and each part has a value of 1/5.
Common denominators are important when adding or comparing fractions. In order to add fractions together, they must have the same denominator. When fractions have different denominators, finding a common denominator allows us to simplify and work with the fractions more easily. The least common denominator (LCD) is the smallest multiple of the denominators involved. It ensures that the resulting fractions have the same denominator.
For example, let’s consider adding 1/4 and 2/3. The LCD for 4 and 3 is 12. To make the denominators equal, we need to multiply the numerator and denominator of 1/4 by 3, resulting in 3/12. Similarly, we need to multiply the numerator and denominator of 2/3 by 4, resulting in 8/12. Now that both fractions have a common denominator of 12, we can add the numerators together and keep the common denominator: 3/12 + 8/12 = 11/12.
### Example: Adding Fractions with Common Denominators
Fraction 1 Fraction 2 Common Denominator Sum
1/4 2/3 12 11/12
In the example above, we can see how understanding common denominators is essential for adding fractions. By finding the least common denominator and making the denominators equal, we are able to add the numerators together while keeping the common denominator.
Having a solid grasp of the denominator and common denominators is crucial for working with fractions. It allows us to add, subtract, and compare fractions effectively. By finding the least common denominator, we can simplify fractions and perform operations more easily. So, be sure to understand these concepts thoroughly to excel in working with fractions.
## Adding Fractions with Common Denominators
When adding fractions with common denominators, the process becomes much simpler. The common denominator is the same for both fractions, so you don’t need to find a new one. To add these fractions, you can follow these steps:
2. Keep the common denominator.
Let’s take an example:
“You have 1/4 of a pizza and your friend has 2/4 of a pizza. How much pizza do you have together?”
First, add the numerators together: 1 + 2 = 3. Then, keep the common denominator of 4. So, you have 3/4 of a pizza together.
Adding fractions with common denominators is straightforward and can be easily understood by students. It is a fundamental skill in math and provides a solid foundation for more complex fraction operations.
### Why use common denominators?
Using common denominators ensures that the fractions you are adding are equivalent and can be combined properly. When the denominators are the same, you can simply add the numerators together while keeping the common denominator unchanged. This method simplifies the addition process and allows for accurate calculations.
In summary, adding fractions with common denominators involves adding the numerators while keeping the common denominator. This method is efficient and ensures accurate results.
### Example:
Fraction 1 Fraction 2 Sum
1/3 2/3 3/3
## Adding Fractions without Common Denominators
When adding fractions with different denominators, the process may seem more complex. However, there is an alternative method that can simplify the task. This method involves multiplying the denominators together and cross multiplying the fractions.
To use this method, start by multiplying the denominators of the fractions together. This product will become the new common denominator. Then, cross multiply by multiplying the numerator of the first fraction by the denominator of the second fraction, and vice versa. Write the products as the new numerators.
Once you have the new numerators, add them together and place the sum over the common denominator. This will give you the final result of the addition. Let’s look at an example to better understand the process:
“When adding 1/3 and 1/5 without common denominators, we multiply the denominators 3 and 5 to get a common denominator of 15. Then, we cross multiply: 1 (numerator of 1/3) multiplied by 5 (denominator of 1/5) equals 5, and 1 (numerator of 1/5) multiplied by 3 (denominator of 1/3) equals 3. Next, we add the cross products: 5 + 3 = 8. Finally, we write the sum (8) over the common denominator (15) to get the final result of 8/15.”
Fractions Denominators Cross Multiplication Sum of Cross Products Final Result
1/3 + 1/5 3 and 5 1 * 5 and 1 * 3 5 + 3 8/15
By using this method, you can add fractions with different denominators without having to find a common denominator. It saves time and simplifies the process, especially when dealing with larger fractions. This technique can be particularly useful when working with mixed numbers or complex fractions.
## Dividing Fractions by Flipping
Dividing fractions can be confusing, but there is a handy trick you can use to simplify the process. By flipping the second fraction and changing the division sign to multiplication, you can easily divide two fractions.
To flip a fraction, you need to find its reciprocal. The reciprocal is obtained by interchanging the numerator and denominator. For example, the reciprocal of 2/3 is 3/2.
Once you have the reciprocal, you can multiply the numerators and denominators of the fractions. Multiply the first numerator by the second denominator and the second numerator by the first denominator. Then, simplify the resulting fraction if possible.
Here’s an example to illustrate the process:
Example: Divide 2/3 by 4/5
Flipping the second fraction, we get 2/3 ÷ 5/4
Multiplying the numerators and denominators, we get (2 × 4)/(3 × 5)
Simplifying the fraction, we get 8/15
The result of dividing 2/3 by 4/5 is 8/15.
### Summary
• To divide fractions, flip the second fraction and change the division sign to multiplication.
• The reciprocal of a fraction is obtained by interchanging the numerator and denominator.
• Multiply the numerators and denominators, then simplify the resulting fraction.
Dividend Divisor Quotient
2/3 4/5 8/15
## Using the Circle Method for Improper Fractions
When working with fractions, you may come across improper fractions, which have a numerator that is equal to or greater than the denominator. Converting improper fractions into mixed numbers or whole numbers can help simplify calculations and make them easier to understand. One method to transform improper fractions is the circle method, a visual tool that provides a clear representation of the conversion process.
To use the circle method, begin with the improper fraction and draw a circle. The denominator represents the total number of equal parts the circle is divided into. Starting from the bottom (denominator), multiply it by the top (numerator) and write the product inside the circle. Then, continue moving clockwise around the circle, adding the product of the denominator and numerator at each section.
For example, let’s take the improper fraction 7/4. Draw a circle and divide it into 4 equal sections. Starting from the bottom section, multiply 4 by 1 (numerator) and write the product, 4, inside the circle. Move to the next section and multiply 4 by 1 again, writing 4 in that section. Continue this process until you reach the starting point again. In this case, the resulting circle will have sections filled with the numbers 4, 4, 4, and 3. The numerator of the improper fraction (7) is obtained by adding all the numbers inside the circle, while the denominator remains the same. Therefore, 7/4 as an improper fraction becomes 1 and 3/4 as a mixed number.
## Simplifying Fractions with the Greatest Common Factor
Simplifying fractions is an essential skill in mathematics. It allows us to express fractions in their simplest form, making them easier to work with and compare. One method for simplifying fractions is to find the greatest common factor (GCF) of the numerator and the denominator.
The GCF is the largest number that divides evenly into both the numerator and the denominator. To find the GCF, you can list the factors of both numbers and identify the largest value they have in common. Alternatively, you can use other methods such as prime factorization or a GCF calculator to determine the GCF more efficiently.
Once you have identified the GCF, divide both the numerator and the denominator by this number. This will simplify the fraction by reducing it to its lowest terms. Simplifying fractions not only makes them easier to work with but also helps us compare and order fractions more accurately.
### Example:
“Let’s simplify the fraction 24/36. First, we find the factors of both 24 and 36: 24 (1, 2, 3, 4, 6, 8, 12, 24) and 36 (1, 2, 3, 4, 6, 9, 12, 18, 36). The largest number that appears in both lists is 12. So, we divide both the numerator and the denominator by 12: 24 ÷ 12 = 2 and 36 ÷ 12 = 3. The simplified fraction is 2/3.”
Original Fraction GCF Simplified Fraction
24/36 12 2/3
48/60 12 4/5
15/30 15 1/2
Simplifying fractions with the GCF is a useful technique that simplifies calculations and allows for better understanding and comparison of fractional values. By reducing fractions to their lowest terms, we can work with them more efficiently and accurately in various mathematical operations.
## Teaching Fraction Division with Conceptual Approaches
When it comes to teaching fraction division, adopting conceptual approaches is highly recommended. By using visual models, number sentences, and real-world stories, educators can help students develop a deep understanding of this challenging topic. The 3 Vehicles for Conceptual Math – scale models, number sentences, and stories – can be powerful tools in making fraction division more accessible and meaningful to learners.
Visual models, such as fraction circles or fraction bars, provide a concrete representation of fractions and division. Students can physically manipulate these models to see how a whole is divided into equal parts and how fractions can be divided. This hands-on approach helps build a solid foundation for understanding fraction division.
Number sentences, or numerical expressions, can also aid in conceptualizing fraction division. By breaking down division problems into smaller, manageable steps, students can better grasp the concept. For example, dividing a fraction by a whole number can be visualized as repeated subtraction or as sharing equally among a group.
Using real-world stories and scenarios can bring context and relevance to fraction division. For instance, students can explore how to divide a pizza equally among friends or how to distribute a certain amount of candies among a group of children. By connecting these everyday situations to fraction division, students can see the practical applications of this mathematical skill.
By incorporating these conceptual approaches, educators can foster a deeper understanding of fraction division among students. Instead of relying solely on algorithms and procedures, students develop a conceptual framework that allows them to solve problems flexibly and confidently. This approach equips students with the necessary tools to apply fraction division in various real-life situations, setting them up for success in mathematics and beyond.
## Troubleshooting Common Misconceptions in Fraction Division
When it comes to fraction division, many students encounter common misconceptions that can hinder their understanding of this concept. It is important to address these misconceptions early on and provide students with the necessary tools to overcome them. By focusing on key aspects such as the numerator, denominator, and fraction notation, students can build a solid foundation in fraction division.
One common misconception is that dividing fractions means dividing each part separately. However, it is crucial to understand that when dividing fractions, you are actually multiplying the first fraction by the reciprocal of the second fraction. This confusion can be resolved by emphasizing the relationship between division and multiplication.
Another misconception is that the larger the denominator, the larger the fraction. In reality, the size of a fraction is determined by the ratio between the numerator and the denominator. Students should be encouraged to compare the sizes of fractions by converting them to the same denominator before making any comparisons.
Furthermore, students often struggle with interpreting fraction notation and understanding the relationship between the numerator and the denominator. It is important to reinforce the idea that the numerator represents the number of equal parts being considered, while the denominator represents the total number of equal parts in the whole. Visual aids, such as fraction bars or circles, can be helpful in illustrating this concept.
### Common Misconceptions in Fraction Division:
• Dividing fractions means dividing each part separately
• The larger the denominator, the larger the fraction
• Difficulty interpreting fraction notation and understanding the relationship between the numerator and the denominator
By addressing these misconceptions and providing clear explanations, teachers can help students develop a strong understanding of fraction division. Encouraging hands-on activities, problem-solving exercises, and real-life examples can further enhance students’ comprehension and application of fraction division in various contexts.
Misconception Clarification
Dividing fractions means dividing each part separately When dividing fractions, you are actually multiplying the first fraction by the reciprocal of the second fraction
The larger the denominator, the larger the fraction The size of a fraction is determined by the ratio between the numerator and the denominator, not the value of the denominator alone
Difficulty interpreting fraction notation and understanding the relationship between the numerator and the denominator The numerator represents the number of equal parts being considered, while the denominator represents the total number of equal parts in the whole
By addressing these misconceptions, students can develop a solid foundation in fraction division, enabling them to confidently tackle more complex mathematical concepts in the future.
## The Importance of Teaching Fractions for Success in Math
Teaching fractions is a crucial component of a comprehensive math education. By providing students with a strong foundation in fraction concepts, educators equip them with essential skills for success in higher-level math courses. Fraction mastery is not only about solving specific problems but also about developing critical thinking, problem-solving, and logical reasoning abilities.
When students understand fractions, they are better prepared to tackle advanced mathematical concepts such as ratios, rates, percentages, and algebraic expressions. These topics frequently appear in real-life situations, from calculating discounts during shopping to analyzing data in scientific research. Without a solid grasp of fractions, students may struggle to apply mathematical principles in practical scenarios, hindering their overall mathematical success.
By emphasizing the importance of fractions, teachers can motivate students to engage with this topic and invest time and effort into mastering it. Using various teaching approaches, such as visual models, number sentences, and real-world stories, educators can make fractions more relatable and accessible. These conceptual approaches help students connect abstract concepts to concrete examples, deepening their understanding and fostering a positive attitude towards math.
### Table: The Role of Fractions in Mathematical Success
Mathematical Area Relevance of Fractions
Proportional Relationships Fractions are essential for understanding and solving problems involving proportional relationships, such as scaling, resizing, and comparing quantities.
Probability and Statistics Fractions provide the foundation for understanding probability and statistics, enabling students to interpret data, calculate probabilities, and make informed decisions.
Algebraic Expressions Fractions play a crucial role in algebraic expressions, where they are frequently used in equations, inequalities, and functions. A strong understanding of fractions enhances algebraic problem-solving skills.
Geometry Fractions are integral to geometry, particularly when dealing with measurements, angles, and spatial reasoning. Proficiency in fractions facilitates geometric calculations and analysis.
Teaching fractions goes beyond the curriculum; it equips students with valuable skills that extend into their daily lives. By focusing on fraction foundations, educators empower students to navigate the mathematical challenges they will encounter throughout their academic journey and beyond.
## Conclusion
In conclusion, mastering fraction division is crucial for your mathematical success. Although fractions can be challenging, with the right approach and understanding of key concepts, you can become a fractions master. By focusing on the numerator, denominator, and fraction notation, you can build a solid foundation that will support your progress in math.
Teachers play a vital role in helping you learn fractions effectively and overcome any misconceptions you may have. With their guidance and your dedication to practice, dividing fractions will become easier over time. Remember, fractions are not just a topic to be covered and forgotten. They serve as the building blocks for advanced mathematical concepts, such as ratios, rates, percentages, and algebraic expressions.
By mastering fraction division, you will have the necessary skills to excel in these areas and beyond. So, keep practicing, seeking support when needed, and never underestimate the power of fractions in your mathematical journey. With determination and perseverance, you have the ability to become a true fractions master.
## FAQ
### How do I find the common denominator when adding fractions?
To find the common denominator, you need to multiply the denominators of the fractions you want to add together.
### What is the least common denominator (LCD) and why is it important?
The LCD is the smallest multiple of the denominators. It is important because it simplifies working with fractions and makes addition easier.
### How do I add fractions with common denominators?
To add fractions with common denominators, you simply add the numerators together and keep the common denominator.
### Can I add fractions without common denominators?
Yes, you can add fractions without common denominators by multiplying the denominators together and cross multiplying the fractions.
### How do I divide fractions?
To divide fractions, you flip the second fraction and change the division sign to multiplication. Then, you multiply the numerators and denominators to get the final result.
### What is the circle method for transforming mixed numbers into improper fractions?
The circle method is a visual tool where you start at the bottom number (denominator), multiply it by the top number (numerator), and add the product to the numerator while keeping the denominator the same.
### How do I simplify fractions?
To simplify fractions, you find the greatest common factor (GCF) and divide both the numerator and denominator by it.
### How should I teach fraction division conceptually?
Researchers recommend using visual models, number sentences, and real-world stories to teach fractions conceptually.
### What are some common misconceptions in fraction division?
Common misconceptions in fraction division include misunderstanding the numerator, denominator, and fraction notation.
### Why are fractions important for success in math?
Fractions provide a foundation for advanced math concepts such as ratios, rates, percentages, and algebraic expressions.
### How can I master fraction division?
With the right approach and understanding of key concepts, you can master fraction division. Practice and support are essential. |
# Math (linear algebra)
posted by .
Solve the system of equation.
x- 2y+ 3z = 5
3x + y = -5
-2x+y + z = 8
• Math (linear algebra) -
x- 2y+ 3z = 5
3x + y = -5
-2x+y + z = 8
First thing to check is that we have three equations and three unknowns, which is fine. If there are more unknowns than equations then there is a problem with the question.
x- 2y+ 3z = 5
3x + y = -5
-2x+y + z = 8
we need to eliminate one unknown by combining pairs of equations. Lets say we remove the y terms.
x- 2y+ 3z = 5
3x + y = -5 (multiply by 2)
x- 2y+ 3z = 5
6x + 2y = -10
Add these two and y terms are removed.
7x+3z=-5
Do the same with, but this time subtract the two and the y terms are removed
3x + y = -5
-2x+y + z = 8
5x-z=-13
We now have two equations and two unknwons
7x+3z=-5
5x-z=-13 (multiply by 3)
7x+3z=-5
15x-3z=-39
22x=-44, so x=-2
substitute in here to find y
3x + y = -5
the substitute in one of the above to find z
## Similar Questions
1. ### maths
use the substitution method to solve the linear system. 1. 3x=9 -x+2y=9 3x/3 = 9/3 x = 3 -x(3)+2y = 9 -3+2y = 9 2y = 12 y = 6 is (3,6) correct?
2. ### Algebra
Show equation AX=B represents a linear system of two equations in two unknowns. Solve the system and sustitute into the matrix equation to check results. A=|1,2| |-3,5|, X=|x[1]x[2]|, B=|-4 12|
3. ### Algebra
Show equation AX=B represents a linear system of two equations in two unknowns. Solve the system and substitute into the matrix equation to check results. I have no idea can you help me out A=|1,2| |-3,5|, X=|x[1]x[2]|, B=|-4 12|
4. ### algebra
use method of substitution to solve the system of linear equation 3x+4y=3/2 y=2x-1
5. ### Algebra I
How do you write a system of linear equations in two variables?
6. ### algebra 1
Solve the following system of linear equation using any algebraic method. 2x+5y=3 X-3y=7
7. ### algebra 1
Solve the following system of linear equation using any algebraic method. 2x+5y=3 X-3y=7
8. ### Algebra 1
What is an equation of a line perpendicular to the line with equation y = 1 - 3x a.y=-3x + 5 b. y=3x + 5 c.y=-1/3x + 5 d. y=1/3x + 5 To solve the linear system below, which substitution of unkowns is proper ?
9. ### Algebra
Solve this system of linear equations. * Used the substitution method 3x - y = 8 4x - y = -15 ----------------------------------------- WORK y = 3x - 8 4x - (3x - 8 ) = -15 4x - 3x + 8 = -15 x + 8 = -15 x = -7 3x - y = 8 3(-7) - y …
10. ### Algebra
Explain how you can determine from the graph of a system of two linear equation in two variables whether is an inconsistent system of equation. 2. When you solve a system of equation by the substitution method how do you determine …
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# Quantitative Aptitude for IBPS PO Prelims Exam: 20th September 2018
Dear students,
Numerical Ability or Quantitative Aptitude Section has given heebie-jeebies to the aspirants when they appear for a banking examination. As the level of every other section is only getting complex and convoluted, there is no doubt that this section, too, makes your blood run cold. The questions asked in this section are calculative and very time-consuming. But once dealt with proper strategy, speed, and accuracy, this section can get you the maximum marks in the examination. Following is the Quantitative Aptitude quiz to help you practice with the best of latest pattern questions.
Q1.
360
420
440
380
364
Solution: B
Q2. ∛(25×(169÷13×845) ) + √2704 = ?
111
115
117
119
123
Solution: C
Q3. 52² + 36² + 44² = ? + 64²
1840
1480
1640
2040
1860
Solution: A
Q4. 3434 ÷ 200 + 3256 ÷ 1100 – 3441 ÷ 300 = ?
7.66
6.88
86.6
8.66
0.866
Solution: D
Q5.
2,508
25,080
25,800
24,080
26,080
Solution: B
Q6. An amount is distributed among A, B and C in the ratio 3 : 5 : 8. If part of B is Rs. 480 less than that of C then what is the part of A ?
Rs. 480
Rs. 840
Rs. 680
Rs. 580
Rs. 440
Solution: A
Q7. The ratio of male and female in a village is 4 : 7. If total population of village is 6160 then what will be population of male after one year if it increases at the rate of 20% per annum ?
(
2688
2868
2488
2840
2268
Solution: A
Q8. An amount of Rs. 8000 is invested in a scheme which offers S.I. at the rate of 16 ⅔% per annum for 6 years. Find the total S.I. obtained from scheme after 6 years (in Rs.)
4000
6500
8000
6000
800
Solution: C
Q9. The difference between C.I. and S.I. on a certain sum at a rate of 8% per annum for two years is 1,280 rupees. Find the S.I. for 4 years on the same sum at same rate of interest.
Rs. 32,000
Rs. 64,000
Rs. 6,400
Rs. 3,200
Rs. 64,400
Solution: B
Q10. In a bag there are Rs. 1, 50 paisa and 25 paisa coins in the ratio of 2 : 3 : 5 respectively. If bag contains total amount of Rs.760 in the form of these coins only then find number of coins of 50 paisa.
280
80
360
480
380
Solution: D
Directions (11-15): The following line graph shows the no. of foreigners (in thousand) from three different countries who visited India Gate over 6 different years. Study the chart carefully and answer the following questions.
Q11. Find the average no. of foreigners who visited India gate from Thailand over all the years?
44.16 thousand
48 thousand
37.4 thousand
32.5 thousand
33 thousand
Solution: A
Q12. The foreigners who visited India gate from USA in year 2010 are what percent more or less than the no. of foreigners who visited India gate from UK in the same year?
25% less
20% more
20% less
25% more
None of these
Solution: C
Q13. If 10% more foreigners visited India gate from UK in year 2014 than that in 2013, then find the average no. of foreigners who visited India gate from UK in the years 2012, 2013 and 2014 together. (approximately)
20000
25000
45000
36000
51000
Solution: D
Q14. What is the difference between total no. of foreigners who visited India gate from Thailand and total no. of foreigners who visited India gate from USA overall the years?
50,000
60,000
55,000
65,000
None of these
Solution: C
Q15. From which country, maximum no. of foreigners visited India gate in which year?
2008, UK
2011, UK
2009, USA
2010, Thailand
2012, USA
Solution: B |
# QUADRATIC EQUATIONS AND FUNCTIONS REVIEW
Problem 1 :
The function f(t) = -5t2 + 20t + 60 models the approximate height of an object t seconds after it is launched. How many seconds does it take the object to hit the ground?
Solution:
f(t) = -5t2 + 20t + 60
= 5t2 - 20t - 60
Dividing by 5
t2 - 4t - 12 = 0
t2 - 6t + 2t - 12 = 0
t(t - 6) + 2(t - 6) = 0
(t + 2) (t - 6) = 0
t + 2 = 0t = -2 t - 6 = 0t = 6
As time cannot be negative, the object will hit the ground after 6 seconds.
Problem 2 :
What is the smallest of 3 consecutive positive integers if the product of the smaller two integers is 5 less than 5 times the largest integer?
Solution:
Let the smallest number be x, and the second and third be x + 1 and x + 2.
x(x + 1) = 5(x + 2) - 5
x2 + x = 5x + 10 - 5
x2 - 4x - 5 = 0
(x - 5) (x + 1) = 0
x = 5 and x = -1
The numbers have to be positive, the smallest number is 5.
x + 1 = 5 + 1 = 6
x + 2 = 5 + 2 = 7
So, the 3 consecutive integers are 5, 6 and 7.
Problem 3 :
The larger leg of a right triangle is 3 cm longer than its smaller leg. The hypotenuse is 6 cm longer than the smaller leg. How many centimeters long is the smaller leg?
Solution:
Let x represent the shorter leg.
Then the length of the longer will be x + 3 and
the length of the hypotenuse will be x + 6.
By using Pythagorean Theorem,
x2 + (x + 3)2 = (x + 6)2
x2 + x2 + 6x + 9 = x2 + 12x + 36
2x2 - x2 + 6x - 12x + 9 - 36 = 0
x2 - 6x - 27 = 0
(x - 9) (x + 3) = 0
x = 9 or x = -3
The length can't be negative.
So, x = 9
The length of the shorter leg is 9 centimeters.
Problem 4 :
Which term is a factor of 3a2 + 12a?
A. 3a B. 4a C. 3a2 D. 4a2
Solution:
= 3a² + 12a
= 3a(a + 4)
So, option (A) is correct.
Problem 5 :
Which graph displays function f(x) = (2x + 3) (x - 2)?
Solution:
f(x) = (2x + 3)(x - 2)
x-intercept : put y = 0
2x + 3 = 02x = -3x = -3/2x = -1.5 x - 2 = 0x = 2
Converting into vertex form :
f(x) = (2x + 3)(x - 2)
f(x) = 2x2 - 4x + 3x - 6
f(x) = 2x2 - x - 6
f(x) = 2[x2 - (1/2)x] - 6
f(x) = 2[x2 - (1/2)x] - 6
So, graph (a) is correct.
Problem 6 :
The floor of a rectangular cage has a length 4 feet greater than its width, w. James will increase both dimensions of the floor by 2 feet. Which equation represents the new area, N, of the floor of the cage?
A. N = w2 + 4w B. N = w2 + 6w C. w2 + 6w + 8
D. w2 + 8w + 12
Solution:
Given, the original width = w
The original length = w + 4
Both length and width will be increased by 2 feet
So, new length = w + 6, new width = w + 2
New area N = (w + 6) (w + 2)
= w2 + 2w + 6w + 12
N = w2 + 8w + 12
So, option (D) is correct.
Problem 7 :
Which expression is equivalent to t2 - 36?
A. (t - 6)(t - 6) B. (t + 6)(t - 6)
C. (t - 12)(t - 3) D. (t - 12)(t +3)
Solution:
= t2 - 36
= t2 - 62
(t + 6) (t - 6)
So, option (B) is correct.
Problem 8 :
Draw the graph of the function f(x) = 4x2 - 8x + 7?
Solution:
f(x) = 4x2 - 8x + 7
Let us find vertex and zeroes of the quadratic function.
f(x) = 4[x2 - 2x] + 7
f(x) = 4[(x - 1)2 - 1] + 7
= 4(x - 1)2 - 4 + 7
= 4(x - 1)2 + 3
y-intercept: x = 0
f(x) = 7
y-intercept = (0, 7)
x-intercept: y = 0
4(x - 1)2 + 3 = 0
Will get unreal value, so there is x-intercepts.
So, option (D) is correct.
Problem 9 :
Suppose that the equation V = 20.8x2 - 458.3x + 3,500 represents the value of a car from 1964 to 2002. What year did the car have the least value? (x = 0 in 1964)
A. 1965 B. 1970 C. 1975 D. 1980
Solution:
V = 20.8x2 - 458.3x + 3500
To find the least value, we can figure out the minimum value.
x = -b/2a
x = 458.3/2(20.8)
When V' = 0, it has least value
41.6x - 458.3 = 0
41.6x = 458.3
x = 458.3/41.6
x = 11.01
1964 + 11 = 1975
So, option (C) is correct.
Problem 10 :
The number of bacteria in a culture can be modeled by the function N(t) = 28t2 - 30t + 160, where t is the temperature, in degrees Celsius, the culture is being kept. A scientist wants to have fewer than 200 bacteria in a culture in order to test a medicine effectively. What is the approximate domain of temperatures that will keep the number of bacteria under 200?
A. -1.01°C < t < 2.03°C B. -0.90°C < t < 1.97°C
C. -0.86°C < t < 1.93°C D. -0.77°C < t < 1.85°C
Solution:
N(t) = 28t2 - 30t + 160
200 = 28t2 - 30t + 160
28t2 - 30t - 40 = 0
a = 28
b = -30
c = -40
-0.77°C < t < 1.85°C
So, option (D) is correct
## Recent Articles
1. ### Finding Range of Values Inequality Problems
May 21, 24 08:51 PM
Finding Range of Values Inequality Problems
2. ### Solving Two Step Inequality Word Problems
May 21, 24 08:51 AM
Solving Two Step Inequality Word Problems |
# What are the four types of geometric transformations?
## What are the four types of geometric transformations?
There are four main types of transformations: translation, rotation, reflection and dilation.
## What are the 3 types of transformations in geometry?
Types of transformations:
• Translation happens when we move the image without changing anything in it.
• Rotation is when we rotate the image by a certain degree.
• Reflection is when we flip the image along a line (the mirror line).
• Dilation is when the size of an image is increased or decreased without changing its shape.
What are the basic geometric transformations?
Geometric transformations are needed to give an entity the needed position, orientation, or shape starting from existing position, orientation, or shape. The basic transformations are scaling, rotation, translation, and shear.
### How do you describe shape transformations?
A translation moves a shape up, down or from side to side but it does not change its appearance in any other way. A transformation is a way of changing the size or position of a shape. Every point in the shape is translated the same distance in the same direction.
### What do you mean by geometric transformation?
In mathematics, a geometric transformation is any bijection of a set to itself (or to another such set) with some salient geometrical underpinning. More specifically, it is a function whose domain and range are sets of points — most often both or both. — such that the function is injective so that its inverse exists.
How do you write a transformation in geometry?
The symbol for a composition of transformations (or functions) is an open circle. is read as: “a translation of (x, y) → (x + 1, y + 5) after a reflection in the line y = x”. Composition of transformations is not commutative.
## What are the types of transformations in math?
The four main types of transformations are translations, reflections, rotations, and scaling.
• Translations. A translation moves every point by a fixed distance in the same direction.
• Reflections.
• Rotations.
• Scaling.
• Vertical Translations.
• Horizontal Translations.
• Reflections.
• Learning Objectives.
## Is scaling a geometric transformation?
In Euclidean geometry, uniform scaling (or isotropic scaling) is a linear transformation that enlarges (increases) or shrinks (diminishes) objects by a scale factor that is the same in all directions. When the scale factor is a positive number smaller than 1, scaling is sometimes also called contraction.
How do you write a transformation?
The function translation / transformation rules:
1. f (x) + b shifts the function b units upward.
2. f (x) – b shifts the function b units downward.
3. f (x + b) shifts the function b units to the left.
4. f (x – b) shifts the function b units to the right.
5. –f (x) reflects the function in the x-axis (that is, upside-down). |
Permutations - Formula & Examples
# Permutations
Permutation is the arranging of a given number of things in every possible order of succession. There is a particular rule to write the numbers, such as form the series of numbers 1, 2, 3, 4 etc, up to the number o things to be permuted, and their continued product will be the number of permutations.
For Example: How many different integral numbers may be expressed by writing the 5 significant digits in succession, each figure to be taken once, and once in each number. Its answer is 120 (1.2.3.4.5 = 120).
The order or number of arrangements is known as the permutation. Basically, it is a rearrangement within elements of a given ordered list in one-to-one manner of correspondence. In this section will discuss in detail about permutation.
## Formula
If we have to find permutation of a list of n elements, it is given by n!, read as n factorial. Also, n! can be calculated by finding the product of natural numbers from 1 to n, that is,
n! = n (n – 1) (n – 2) ….. 3 . 2 . 1
For Example: we have three elements in a set say {4, 3, 7} so we have 3! = 6
arrangements which are {3, 4, 7}, {3, 7, 4}, {4, 3, 7}, {4, 7, 3}, {7, 4, 3}, {7, 3, 4}.
If we need to find permutation arrangement of k elements out of a given set of n elements then we have the following formula:
n (n – 1) (n – 2) … (n – k + 1)
It is denoted by various notations. P (n, k), $nP_k$ etc.
The basic formula is given as follows:
$n P_k$ = $\frac{n!}{(n – k)!}$
This formula is equal to zero if $k > n$.
This formula gives the same product as given above. But it is just more simplified one.
But what is the difference between a combination and a permutation? Being precise we can say that if order of things does not matter then we say it is combination and if the order of things makes a difference then we say there is permutation.
For Example: the fruit salad is a combination but the combination of any safe will be a permutation.
So we say that a permutation is nothing but an ordered combination.
In permutation we have two types: one in which repetition is allowed and another without any repetition.
For non repeating permutations we use the formula mentioned above either the factorial form or the k out of n elements form.
For repeating ones we simply multiply n with itself the number of times it is repeating, that is, n^r where n is the number of things to be chosen from and r is the number of items being chosen.
## Examples
Let us see some examples of problems based on permutations.
Example 1: Find out the number of ways in which 4 girls can form a group of 15 in total to be lined up for a click.
Solution: We need to find out P(15,4)
P (15, 4) = $\frac{15!} {(15 - 4)!}$ = $\frac{15!}{11!}$ = $\frac{15 . 14 . 13 . 12 . 11!}{11!}$
= 32760
So the group of 15 girls can form 32760 different lineups as 4.
Example 2: Given a word “MATHS”, find the number of possible arrangements of its letters.
Solution: We can see that there is no repetition in the word MATHS. Also, that there are 5 digits in total in the given word. Also, we have 5 out of 5 choices given.
Hence, P (5, 5) = $\frac{5!} {(5 – 5)!}$ = $\frac{5!}{0!}$ = 5! = 5 * 4 * 3 * 2 * 1 = 120.
So we have 120 numbers of possible arrangements of digits of word MATHS.
Example 3: Sujen wants to draw two cards from a deck of 52 cards, without replacement. What is the probability of drawing a 10 and a jack in that order.
Solution: Sujen wants to draw 10 and a jack from a deck.
In a deck, there are four 10 and four jack
So probability of drawing a 10 and a jack is $P(4, 1) . P(4, 1)$
From a deck of 52 cards Sujen wants to draw two cards i.e. P(52, 2)
=> $\frac{ P(4, 1) .P(4, 1)} {P(52, 2)}$ = $\frac{ 4 . 4} {2652}$ = $\frac{ 16} {2652}$ = $\frac{4}{663}$
The probability of drawing a 10 and a jack is $\frac{4}{663}$. |
# How do you add (7x ^ { 3} - 5x ^ { 2} + 9x + 3) + ( 5x ^ { 3} - 7x ^ { 2} - x + 3)?
Dec 18, 2017
$12 {x}^{3} - 12 {x}^{2} + 8 + 6$
#### Explanation:
$\left(7 {x}^{3} - 5 {x}^{2} + 9 x + 3\right) + \left(5 {x}^{3} - 7 {x}^{2} - x + 3\right)$
This is the same as $7 {x}^{3} - 5 {x}^{2} + 9 x + 3 + 5 {x}^{3} - 7 {x}^{2} - x + 3$.
The first thing we do is look for "like terms" and variables with the same degree. Our only variable here is $x$. We have ${x}^{3}$, ${x}^{2}$, $x$, and numbers.
Let's first combine the ${x}^{3}$ values. We have $7 {x}^{3}$ and $5 {x}^{3}$.
$7 {x}^{3} + 5 {x}^{3} = 12 {x}^{3}$
Now ${x}^{2}$:
$- 5 {x}^{2} - 7 {x}^{2} = - 12 {x}^{2}$
$x$:
$9 x - x = 8 x$
Numbers:
$3 + 3 = 6$
Now that we have simplified/combined each of the terms, let's put them together:
$12 {x}^{3} - 12 {x}^{2} + 8 x + 6$ |
Equation of a Line Worksheet | Problems & Solutions
# Equation of a Line Worksheet
Equation of a Line Worksheet
• Page 1
1.
Which of the following equations does not represent a straight line?
a. $y$ = - $\frac{1}{10}$$x$ + 9 b. $\mathrm{xy}$ = 9 c. 10$x$ - $y$ - 8 = 0 d. ($y$ - 9) = 10($x$ - 8)
#### Solution:
y = - 1 / 10 x + 9 is a straight line.
[Equation of a straight line in slope-intercept form.]
(y - 9) = 10 (x - 8) is a straight line.
[Equation of a straight line in point-slope form.]
10x - y - 8 = 0 is a straight line.
[Equation of a straight line in standard form.]
xy = 9 is not a straight line, because this cannot be written as any one of the above forms of a line.
2.
Which of the following represents the point-slope form of an equation of a line?
a. $y$ = $\mathrm{mx}$ + $b$ b. $\frac{x}{a}$ + $\frac{y}{b}$ = 1 c. $y$ - $y$1 = $m$($x$ - $x$1) d. A$x$ + B$y$ = C
#### Solution:
The point-slope form of an equation of a line is y - y1 = m (x - x1) where m is the slope and (x1, y1) are the coordinates of a given point on the line.
3.
Which of the following lines is perpendicular to the line $y$ = 8$x$ + 7?
a. $y$ = - $x$ + 7 b. $y$ = - $\frac{1}{8}$$x$ c. $y$ = - 7$x$ + 8 d. $y$ = $\frac{1}{7}$$x$
#### Solution:
Two lines are perpendicular if the product of their slopes is - 1.
The slope of the line y = 8x + 7 is 8.
[Find the slope of the line using y = mx + b.]
If x is the slope of the line perpendicular to the line y = 8x + 7, then m × 8 = - 1, m = - 1 / 8
Equation Slope y = - 1 / 8xy = - 7x + 8y = - x + 7y = 1 / 7x - 1 / 8 - 7 - 11 / 7
So, the line y = - 1 / 8x is perpendicular to the line y = 8x + 7.
4.
Determine whether the statement is true or false. The line whose equation is $x$ - 5$y$ + 27 = 0 passes through (- 2, 5) and has a slope $\frac{1}{5}$.
a. False b. True
#### Solution:
x - 5y + 27 = 0
[Equation of the line.]
y = x5 + 27 / 5
[Slope-intercept form of the line.]
Slope = 1 / 5
[Slope of y = mx + c is m.]
Substitute (- 2, 5) in x - 5y + 27 = 0.
- 2 - 5 (5) + 27 = 0
0 = 0, which is always true.
So, the given statement is true.
5.
Which of the following two lines are perpendicular?
I. $y$ = 6$x$
II. $y$ = - 6$x$
III. $y$ = $\frac{1}{6}$$x$
IV. $y$ = - $x$
a. I and II b. II and IV c. II and III d. III and I
#### Solution:
The slope of the line of the form y = mx is m.
Equation Slope y = 6xy = - 6xy = 1 / 6xy = - x 6 - 61 / 6- 1
Two lines are perpendicular if the product of their slopes is - 1.
The product of the slopes of the lines y = - 6x and y = 16x gives - 1.
So, the lines II and III are perpendicular.
6.
Which of the following statements is true for the graph of the function $y$ = 3?
a. The graph of $y$ = 3 in the coordinate plane is a single point (0, 3). b. The graph of $y$ = 3 is a line parallel to the $y$-axis, at a distance of 3 units to the right of it. c. The graph of $y$ = 3 is a line parallel to the $x$-axis, at a distance of 3 units above the $x$-axis. d. The graph of $y$ = 3 is a line parallel to the $x$-axis, at a distance of 3 units below the $x$-axis.
#### Solution:
The graph of y = 3 is a line parallel to the x - axis, at a distance of 3 units above the x - axis.
So, the graph of y = 3 is not a single point (0, 3).
7.
Find the equation of the line perpendicular to 5$x$ - 4$y$ = 9 and having the same $y$ - intercept.
a. 5$x$ - 4$y$ + 9 = 0 b. 16$x$ - 20$y$ + 45 = 0 c. 16$x$ + 20$y$ + 45 = 0 d. 36$x$ + 45$y$ + 20 = 0
#### Solution:
5x - 4y = 9
[Equation of a line.]
y = 5 / 4x - 9 / 4
[Rewrite it in y = mx + c form.]
So, the slope, m = 5 / 4 and y-intercept, c = - 9 / 4
Slope of the perpendicular line = - 1m = - 4 / 5
y - intercept of the required line = - 9 / 4
The equation of the perpendicular line is: y = - 4 / 5x - 9 / 4
[y = mx + c.]
20y = - 16x - 45
[Multiply both sides by 20.]
16x + 20y + 45 = 0
[Simplify.]
8.
Find the equation of the line passing through (1, 0) and parallel to the line passing through (0, 0) and (10, - 5).
a. 4$x$ + 2$y$ - 1 = 0 b. $x$ + 2$y$ - 4 = 0 c. $x$ - 2$y$ + 1 = 0 d. $x$ + 2$y$ - 1 = 0
#### Solution:
The given line is passing through the points (0, 0) and (10, - 5).
Slope = y2-y1x2-x1 = -5-010-0 = - 1 / 2
Slope of the line parallel to the given line = - 1 / 2
[Parallel lines slopes are equal.]
So, equation of the parallel line is, y = - 1 / 2x + c.
[Slope-intercept form.]
This line passes through the point (1, 0).
0 = - 1 / 2(1) + c
c = 12
[Simplify.]
Therefore, the equation of the line is y = - 1 / 2x + 1 / 2.
2y = - x + 1, so x + 2y - 1 = 0
[Multiply both sides by 2.]
9.
Which of the following lines are parallel?
I. $y$ = $\frac{1}{2}$$x$
II. $y$ = $\frac{1}{5}$$x$ + 2
III. $y$ = $\frac{1}{2}$$x$ + $\frac{1}{5}$
IV. $y$ = - 5$x$ + $\frac{1}{2}$
a. II and II b. I, II and III c. II, III and IV d. I and III
#### Solution:
Two lines are parallel if their slopes are equal.
Therefore, the two lines are parallel.
Equation Slope y = 1 / 2x y = 1 / 5x + 2y = 1 / 2x + 1 / 5y = - 5x + 1 / 2 1 / 21 / 51 / 2 - 5
Slopes of the lines y = 1 / 2x and y = 1 / 2x + 1 / 5 are equal.
So, the lines I and III are parallel.
10.
What is the equation of the line passing through the point (4, 5) and perpendicular to the line $y$ = 3$x$ + 8?
a. $x$ + 4 = - 3($y$ - 5) b. $y$ - 5 = - $\frac{1}{3}$($x$ - 4) c. $x$ - 4 = 3($y$ - 5) d. $y$ + 5 = - $\frac{1}{3}$($x$ - 4)
#### Solution:
The slope-intercept form of the equation of a line with slope m and y-intercept b is y = mx + b.
Slope of the line y = 3x + 8 is 3.
[Compare with the equation in step 1.]
Slope of the line perpendicular to y = 3x + 8 = - 1 / 3
[Product of the slopes of perpendicular lines is -1.]
The equation of the line passing through the point (x1, y1) with slope 'm' in point-slope form is y - y1 = m(x - x1).
Point (x1, y1) = (4, 5) and slope m of the perpendicular line = - 1 / 3 .
y - 5 = - 1 / 3(x - 4)
[Substitute x1 = 4, y1 = 5 and m = - 1 / 3 in the equation in step 4.]
The equation of the line passing through the point (4, 5) is y - 5 = - 1 / 3(x - 4). |
# Can both of this vector combine become the following matrix?
• RyozKidz
In summary, the conversation discusses the possibility of combining two vectors, A and B, into a matrix. The question is how to properly arrange the components of the matrix in relation to the original vectors. The answer lies in understanding the definition of matrix multiplication and considering how the resulting matrix should be related to the original vectors.
#### RyozKidz
vector A : ai+bj
vector B : ci+dj
can both of this vector combine become the following matrix?
$$\stackrel{a}{c}$$ $$\stackrel{b}{d}$$ or $$\stackrel{a}{b}$$ $$\stackrel{c}{d}$$
RyozKidz said:
vector A : ai+bj
vector B : ci+dj
can both of this vector combine become the following matrix?
$$\stackrel{a}{c}$$ $$\stackrel{b}{d}$$ or $$\stackrel{a}{b}$$ $$\stackrel{c}{d}$$
What do you mean by combine?
do you mean put vectors together in column or in row?
RyozKidz said:
vector A : ai+bj
vector B : ci+dj
can both of this vector combine become the following matrix?
$$\stackrel{a}{c}$$ $$\stackrel{b}{d}$$ or $$\stackrel{a}{b}$$ $$\stackrel{c}{d}$$
You just need to consider the definition of matrix multiplication. You want to find a matrix A such that
$$\left(\begin{array}{c} a\bold{i} + b\bold{j} \\ c\bold{i} + d\bold{j} \end{array}\right) = A \left(\begin{array}{c} \bold{i} \\ \bold{j} \end{array}\right) = \left(\begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right) \left(\begin{array}{c} \bold{i} \\ \bold{j} \end{array}\right)$$
From this you should be able to work out the required components.
You can put numbers together any way you please and make a matrix. The question is how do you want that matrix to be related to the original vectors and what do you want to do with it?
I think i found the answer by the definition of the mutiplication of matrix...
Thx a lot..~
coz when i read about the article of the transformation matrix i can't get it..hehe thx
## 1. Can you explain what a vector is?
A vector is a mathematical object that represents a quantity with both magnitude and direction. It can be represented graphically as an arrow, with the length of the arrow indicating the magnitude and the direction of the arrow indicating the direction of the vector.
## 2. How can two vectors be combined?
Two vectors can be combined through vector addition, which involves adding the individual components of the vectors. If the vectors are in the same direction, the resulting vector will have a magnitude equal to the sum of the individual magnitudes. If the vectors are in opposite directions, the resulting vector will have a magnitude equal to the difference between the individual magnitudes. The direction of the resulting vector will depend on the direction of the individual vectors.
## 3. What is a matrix and how is it related to vectors?
A matrix is a rectangular array of numbers, symbols or expressions. It can be thought of as a way to organize and manipulate data. Matrices can be used to represent transformations of vectors, where each column of the matrix represents the transformation of a single vector. Matrices can also be used to represent systems of linear equations, where each row represents an equation.
## 4. Can any two vectors be combined to form a matrix?
No, not all combinations of vectors will result in a matrix. In order for two vectors to be combined to form a matrix, they must have the same number of components or dimensions. For example, a 2-dimensional vector and a 3-dimensional vector cannot be combined to form a matrix.
## 5. What are some real-life applications of combining vectors to form a matrix?
Combining vectors to form a matrix is a fundamental concept in linear algebra and has many real-life applications. It is used in computer graphics to represent transformations of 3D objects, in physics to represent forces acting on an object, and in machine learning for data analysis and pattern recognition. It is also used in engineering for structural analysis and optimization problems. |
## Pages
### Converting Logarithms to Exponential Form
Consider the following logarithm equation:
log_{10}(100) = 2
and the following exponential equation:
10^2 =100
We can say that both the above equations are equivalent. They are just different forms of the same equation.
Every logarithmic expression can be converted to an exponential expression by remembering the following rules:
• The base of the logarithm is the base of the exponential expression
• The number/quantity which the logarithmic expression equals is the exponent in the exponential expression
• The argument of the logarithm is the result of the exponential expression
Thus, relating to the above two equations, we can write the following steps:
1. Base of the logarithm log_{10}(100) is 10, hence 10 will be the base of the exponential expression.
2. The logarithmic expression log_{10}(100) equals 2. Thus, 2 is the exponent. That is, 10^2
3. The argument of the logarithmic expression is 100. This becomes the result of the exponential expression. Thus, we can write 10^2 = 100.
Following the above three steps, you can convert any logarithmic expression/equation to its equivalent exponential expression/equation. A more detailed post on this topic is available here.
### Solved Example
Convert the following logarithmic equation to exponential form:
log_{2}(32) = 5
First, identify the base, argument and result of the logarithmic equation:
• Base = 2
• Argument = 32
• Result = 5
Now using the rules described above, we know the following things about the exponential expression:
• The base 2 of the logarithm will become the base of the exponential expression
• The argument 32 of the logarithm will become the result of the exponential expression
• The result 5 of the logarithmic equation will become the exponent
Thus, the exponential equation equivalent to the above equation is,
2^5 = 32
### Worksheet
#### Instructions
Click on Show Answer button to show a particular answer. Click on the answer itself to hide it. Use the buttons at the bottom to show/hide all answers
Question
log_2(3) = 8
log_{10}(1000) = 3
log_{10}(10) = 1
log_{5}(25) = 2
ln(e) = 1 |
# Algebra Help: Solving Exponential Equations
Solve exponential equations with algebra.
Though they may look more intimidating than simple linear equations, exponential equations can be solved using the simple algebraic principle of applying the same operations to both sides of the equation. To solve for an unknown exponent, you need to use logarithms, which are inverses of exponential functions. Think of applying logarithms to both sides like pulling the strings of your shoelaces, pull them all the way out and the knot becomes undone!
This guide will show you how to solve an exponential equation using the example problem given on the right. In case you can't see the image, it's
7.45(1.09)^x = 1.4(2.113)^x
There are also a few more examples worked out.
## Step 1
Let's consider the sample problem 7.45(1.09)^x = 1.4(2.113)^x. The first step is to take the logarithm of both sides of the equation. It does not matter whether you use the natural logarithm function ln(x), or the base-10 logarithm function log(x). In this example we will use the function ln(x). Thus, the original expression becomes
ln[7.45(1.09^x)] = ln[1.4(2.113^x)]
## Step 2
Simplify the logarithm expression using the two basic rules of logs:
log(a*b) = log(a) + log(b)
log(c^d) = d*log(c)
This gives us
ln[7.45] + x*ln[1.09] = ln[1.4] + x*ln[2.113]
which is now a linear algebraic equation in the variable x.
## Step 3
Combine like terms, moving the x terms to one side of the equation and the constant terms on the other.
ln[7.45] - ln[1.4] = x*ln[2.113] - x*ln[1.09]
ln[7.45] - ln[1.4] = x*(ln[2.113] - ln[1.09])
## Step 4
Convert the log values into numbers and solve for x. You can use a scientific calculator to evaluate the natural log function, or tables of logarithms. For the most accurate answer, use a calculator.
1.67174 = x*0.66193
1.67174/0.66193 = x
2.52555 = x
To check your work, you can use the exponential equation solver which solves equations of the form ab^x = cd^x. Just input the values of a, b, c, and d. In this example, a = 7.45, b = 1.09, c = 1.4, and d = 2.113.
Solving Equations (Source: Edu-Stock)
## Another Example
Solve the rational-exponential equation
[2(3^x) + 1]/[3^x - 16] = 5
First multiply both sides of the equation by the denominator 3^x - 16. This creates the simpler and equivalent equation
2(3^x) + 1 = 5(3^x - 16)
2(3^x) + 1 = 5(3^x) - 80
It's important to resist the temptation to incorrectly simplify the expression. 2(3^x) does not equal 6^x, and 5(3^x) does not equal 15^x. Instead, you must treat 2(3^x) and 5(3^x) as like terms and combine them on one side of the equation.
2(3^x) - 5(3^x) = -80 - 1
(2 - 5)(3^x) = -81
-3(3^x) = -81
3^x = 27
At this point you can either use logarithms to solve for x, or you can use basic math skills to deduce that x = 3, since 3^3 = 3*3*3 = 27.
## Last Example
A substance has a half-life of 4 hours. If you start with 9 grams, how long will it be until you have 3 grams? To solve this equation, we start with the expression
Q(t) = 9*e^(rt)
where Q(t) is the quantity of the substance after t hours, e is the mathematical constant 2.71828..., and r is an unknown rate factor. Since the half-life is 4 hours, we know that if t = 4 then Q(t) will equal 4.5. This let's us solve for r:
4.5 = 9*e^(r*4)
4.5/9 = e^(4r)
0.5 = e^(4r)
Ln(0.5) = 4r
Ln(0.5)/4 = r
So the value of r is approximately -0.1732868. Using this we can now solve the original question, find the time t when Q(t) equals 3 grams.
3 = 9*e^(-0.1732868t)
3/9 = e^(-0.1732868t)
1/3 = e^(-0.1732868t)
Ln(1/3) = -0.1732868t
Ln(1/3)/-0.1732868 = t
So t = 6.33985 hours.
## More by this Author
billybuc 2 years ago from Olympia, WA
Oddly, at the high school I attended, we had a choice of taking the math track or the science track. Since science was a mystery to me, I chose math. I always liked algebra because it was a puzzle to solve. These secrets would have come in handy back then. :) |
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Broad Topics > Numbers and the Number System > Place value
### Round the Dice Decimals 1
##### Stage: 2 Challenge Level:
Use two dice to generate two numbers with one decimal place. What happens when you round these numbers to the nearest whole number?
### Round the Three Dice
##### Stage: 2 Challenge Level:
What happens when you round these three-digit numbers to the nearest 100?
### Round the Dice Decimals 2
##### Stage: 2 Challenge Level:
What happens when you round these numbers to the nearest whole number?
### Multiply Multiples 2
##### Stage: 2 Challenge Level:
Can you work out some different ways to balance this equation?
### Multiply Multiples 1
##### Stage: 2 Challenge Level:
Can you complete this calculation by filling in the missing numbers? In how many different ways can you do it?
### Multiply Multiples 3
##### Stage: 2 Challenge Level:
Have a go at balancing this equation. Can you find different ways of doing it?
### Oddly
##### Stage: 2 Challenge Level:
Find the sum of all three-digit numbers each of whose digits is odd.
### (w)holy Numbers
##### Stage: 2 Challenge Level:
A church hymn book contains 700 hymns. The numbers of the hymns are displayed by combining special small single-digit boards. What is the minimum number of small boards that is needed?
### Napier's Bones
##### Stage: 2 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?
### Reach 100
##### Stage: 2 Challenge Level:
Choose four different digits from 1-9 and put one in each box so that the resulting four two-digit numbers add to a total of 100.
### All the Digits
##### Stage: 2 Challenge Level:
This multiplication uses each of the digits 0 - 9 once and once only. Using the information given, can you replace the stars in the calculation with figures?
### Spell by Numbers
##### Stage: 2 Challenge Level:
Can you substitute numbers for the letters in these sums?
### ABC
##### Stage: 2 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?
### Six Is the Sum
##### Stage: 2 Challenge Level:
What do the digits in the number fifteen add up to? How many other numbers have digits with the same total but no zeros?
### The Deca Tree
##### Stage: 2 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.
### Trebling
##### Stage: 2 Challenge Level:
Can you replace the letters with numbers? Is there only one solution in each case?
### Number Detective
##### Stage: 2 Challenge Level:
Follow the clues to find the mystery number.
### Round the Four Dice
##### Stage: 2 Challenge Level:
This activity involves rounding four-digit numbers to the nearest thousand.
### Coded Hundred Square
##### Stage: 2 Challenge Level:
This 100 square jigsaw is written in code. It starts with 1 and ends with 100. Can you build it up?
### Diagonal Sums
##### Stage: 2 Challenge Level:
In this 100 square, look at the green square which contains the numbers 2, 3, 12 and 13. What is the sum of the numbers that are diagonally opposite each other? What do you notice?
### Snail One Hundred
##### Stage: 1 and 2 Challenge Level:
This is a game in which your counters move in a spiral round the snail's shell. It is about understanding tens and units.
### Eleven
##### Stage: 3 Challenge Level:
Replace each letter with a digit to make this addition correct.
### The Thousands Game
##### Stage: 2 Challenge Level:
Each child in Class 3 took four numbers out of the bag. Who had made the highest even number?
### Which Scripts?
##### Stage: 2 Challenge Level:
There are six numbers written in five different scripts. Can you sort out which is which?
### Cayley
##### Stage: 3 Challenge Level:
The letters in the following addition sum represent the digits 1 ... 9. If A=3 and D=2, what number is represented by "CAYLEY"?
### Football Sum
##### Stage: 3 Challenge Level:
Find the values of the nine letters in the sum: FOOT + BALL = GAME
### What Do You Need?
##### Stage: 2 Challenge Level:
Four of these clues are needed to find the chosen number on this grid and four are true but do nothing to help in finding the number. Can you sort out the clues and find the number?
### Some Games That May Be Nice or Nasty
##### Stage: 2 and 3 Challenge Level:
There are nasty versions of this dice game but we'll start with the nice ones...
### Alien Counting
##### Stage: 2 Challenge Level:
Investigate the different ways these aliens count in this challenge. You could start by thinking about how each of them would write our number 7.
### That Number Square!
##### Stage: 1 and 2 Challenge Level:
Exploring the structure of a number square: how quickly can you put the number tiles in the right place on the grid?
### Tis Unique
##### Stage: 3 Challenge Level:
This addition sum uses all ten digits 0, 1, 2...9 exactly once. Find the sum and show that the one you give is the only possibility.
### Four-digit Targets
##### Stage: 2 Challenge Level:
You have two sets of the digits 0 – 9. Can you arrange these in the five boxes to make four-digit numbers as close to the target numbers as possible?
### Two and Two
##### Stage: 2 and 3 Challenge Level:
How many solutions can you find to this sum? Each of the different letters stands for a different number.
### One Million to Seven
##### Stage: 2 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?
### How Many Miles to Go?
##### Stage: 3 Challenge Level:
A car's milometer reads 4631 miles and the trip meter has 173.3 on it. How many more miles must the car travel before the two numbers contain the same digits in the same order?
### Becky's Number Plumber
##### Stage: 2 Challenge Level:
Becky created a number plumber which multiplies by 5 and subtracts 4. What do you notice about the numbers that it produces? Can you explain your findings?
### Three Times Seven
##### Stage: 3 Challenge Level:
A three digit number abc is always divisible by 7 when 2a+3b+c is divisible by 7. Why?
### Which Is Quicker?
##### Stage: 2 Challenge Level:
Which is quicker, counting up to 30 in ones or counting up to 300 in tens? Why?
### Digit Sum
##### Stage: 3 Challenge Level:
What is the sum of all the digits in all the integers from one to one million?
### Arrange the Digits
##### Stage: 3 Challenge Level:
Can you arrange the digits 1,2,3,4,5,6,7,8,9 into three 3-digit numbers such that their total is close to 1500?
### Repeaters
##### Stage: 3 Challenge Level:
Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13.
### Calculator Bingo
##### Stage: 2 Challenge Level:
A game to be played against the computer, or in groups. Pick a 7-digit number. A random digit is generated. What must you subract to remove the digit from your number? the first to zero wins.
### Legs Eleven
##### Stage: 3 Challenge Level:
Take any four digit number. Move the first digit to the 'back of the queue' and move the rest along. Now add your two numbers. What properties do your answers always have?
### Quick Times
##### Stage: 3 Challenge Level:
32 x 38 = 30 x 40 + 2 x 8; 34 x 36 = 30 x 40 + 4 x 6; 56 x 54 = 50 x 60 + 6 x 4; 73 x 77 = 70 x 80 + 3 x 7 Verify and generalise if possible.
### Pupils' Recording or Pupils Recording
##### Stage: 1, 2 and 3
This article, written for teachers, looks at the different kinds of recordings encountered in Primary Mathematics lessons and the importance of not jumping to conclusions!
### X Marks the Spot
##### Stage: 3 Challenge Level:
When the number x 1 x x x is multiplied by 417 this gives the answer 9 x x x 0 5 7. Find the missing digits, each of which is represented by an "x" .
##### Stage: 3 Challenge Level:
Watch our videos of multiplication methods that you may not have met before. Can you make sense of them?
### Basically
##### Stage: 3 Challenge Level:
The number 3723(in base 10) is written as 123 in another base. What is that base?
### Dicey Operations
##### Stage: 2 and 3 Challenge Level:
Who said that adding, subtracting, multiplying and dividing couldn't be fun?
### Chocolate Maths
##### Stage: 3 Challenge Level:
Pick the number of times a week that you eat chocolate. This number must be more than one but less than ten. Multiply this number by 2. Add 5 (for Sunday). Multiply by 50... Can you explain why it. . . . |
# Arithmetic Progression – Selection of Terms, Properties and Formula
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Arithmetic Progression is an important topic for IIT JEE. Multiple Questions can be seen in the JEE Mains and Advanced examination. Various formula based and concept based questions can be seen. JEE aspirants can easily recap the concept and the formulas mentioned in the end.
## Arithmetic Progression (A.P.)
The sequence of numbers is said to be in A.P. when the difference between the number and the previous number is constant. It is also sometimes referred to as Arithmetic Sequence.
Let the sequence be:
Then,
Also, the difference between the successive term and its previous term is denoted by ‘d’.The common difference can be positive, negative and zero. Then the Arithmetic Series can be written as
Where,
Expression to find the nth term of an A.P. is given by:
Expression to find the nth term of an A.P. from the end is given by:
a = first term / initial term
d = common difference
m = the total number of terms in an A.P.
n = term to be found from end
Another formula to find the nth term of an A.P. from the end is:
## Sum of n terms of an AP
Let there be a series whose first term is ‘a’ and the common difference is ‘d’. Then,
Sum of n terms of Arithmetic series is given by:
Or,
Where l is the last term of the sequence.
## Selection of Terms in an Arithmetic Progression(AP)
A different question can be formed based on a certain number of terms in an A.P. Then following are the different ways to select the numbers.
Number of terms Terms Common Difference 3 a – d, a + d d 4 a-3d, a-d, a+d, a+3d 2d 5 a-2d, a-d, a, a+d, a+2d d 6 a-5d, a-3d, a-d, a+d, a+3d, a+5d 2d
Note:
1.Odd (2r +1) number of terms
Middle term – a
Common difference – d
In general,
2. Even ( 2r ) Number of terms
Middle term – a – d, a + d
Common difference – 2d
In general,
## Some Properties of Arithmetic Progression
1. When a fixed non-zero number is added(or subtracted) to each term of an A.P., then the resulting sequence is also an A.P. with the same common difference.
2. When each term of the sequence is multiplied and divided some fixed constant number(non-zero), then the resulting sequence is an A.P. with common difference kd(multiplication) or k/d (division).
3. If there is two arithmetic progression
a1, a2, a3, a4, a5, a6, ………………, an and b1, b2, b3, b4, b5, b6, ………………, b
Then, a1+b1, a2+b2, a3+b3, a4+b4, a5+b5, a6+b6, ………………, an+bn is also an Arithmetic Progression.
1. Let there be three numbers a, b, c. They are in A.P. if and only if 2b=a+c.
2. If an A.P. contains a finite number of terms, then the sum of terms equidistant from beginning and end always remains the same and is equal to the sum of the first term and the last terms.
3. If the nth term of an A.P. is given by the linear expression in n i.e. an = An +B, where A and B are constant, then the common difference is the coefficient of n i.e. A.
4. If the sum of the n terms of an A.P. is given by the quadratic equation i.e. an = An2 +B, where A and B are constants independent of n, then the common difference is 2A i.e. the coefficient of n2e. A.
## Arithmetic Mean(A.M.)
Arithmetic Mean is the average of two numbers. Suppose there are two numbers ‘a’ and ‘c’ which in A.P. Then A.M. of a and c(say b) is the value that can be added in between a and c such that a, b, c is in A.P.
Also, n numbers of Arithmetic Mean A1, A2, A3, A4,……..An can be inserted between two numbers (say a and b). ‘a’ is the first term and b is the (n +2)th term.
Then the sequence obtained is
## Formula
1. To find the common difference of an A.P.
2. To find the nth terms of an Arithmetic progression or to find the total number of terms in an A.P. is given by:
3. To find the last term of an A.P. with n terms.
4. To find the term from the end of an A.P.( m: total number of terms, n: position from the end)
Or
5. To find the sum of n terms
Or
6. To find the Arithmetic Mean
Hope this article on Arithmetic Progression has helped you understand important concepts such as Sum of n terms, Arithmetic Progression formula, Arithmetic series, etc. You can also refer to articles such as Free Body Diagram and Geometric Progression to strengthen your exam preparation.
As we all know, practice is the key to success. Therefore, boost your preparation by starting your practice now.
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# Differentiation
## 1 Differentiation over addition and constant multiple: the linearity
In this chapter, we will be taking a broader look at how we compute the rate of change.
If a function is defined at the nodes of a partition, it is simply a sequence of numbers. And so is its difference quotient. What this means is that this procedure is a special kind of function, a function of functions: $$\newcommand{\ra}[1]{\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{ccccccccccccccc} f & \mapsto & \begin{array}{|c|}\hline\quad \frac{\Delta }{\Delta x} \quad \\ \hline\end{array} & \mapsto & u=\frac{\Delta f}{\Delta x} . \end{array}$$ Furthermore, the derivative is defined as a limit. Unlike the limits we saw prior to derivatives, this one has a parameter, the location $x$. That is why with the input a differentiable function $f$, the output of this limits is another function $f'$. What this means is that this process is a special kind of function too, a function of functions: $$\newcommand{\ra}[1]{\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{ccccccccccccccc} f & \mapsto & \begin{array}{|c|}\hline\quad \frac{d}{dx} \quad \\ \hline\end{array} & \mapsto & f' . \end{array}$$ We need to understand how these two functions operate. We would like to develop shortcuts and algebraic rules for evaluating both difference quotients and the derivatives. The latter will be found without resorting to using limits!
What happens to the output function of differentiation as we perform algebraic operations with the input functions?
The idea of addition of the change is illustrated below:
Here, the bars that represent the change of the output variable are stacked on top of each other, then the heights are added to each other and so are the height differences. The algebra behind this geometry is very simple: $$(A+B)-(a+b)=(A-a)+(B-b).$$ The idea leads to the Sum Rule for Differences from Chapter 1: the difference of the sum of two sequences is the sum of their differences. Below is its analog.
Theorem (Sum Rule). (A) The difference quotient of the sum of two functions is the sum of their difference quotients; i.e., for any two functions $f,g$ defined at the adjacent nodes $x$ and $x+\Delta x$ of a partition, the difference quotients (defined at the corresponding secondary node) satisfy: $$\frac{\Delta(f+g)}{\Delta x}=\frac{\Delta f}{\Delta x}+\frac{\Delta g}{\Delta x}.$$ (B) The sum of two functions differentiable at a point is differentiable at that point and its derivative is equal to the sum of their derivatives; i.e., for any two functions $f,g$ differentiable at $x$, we have at $x$: $$\frac{d(f+g)}{d x}=\frac{d f}{d x}+\frac{d g}{d x}.$$
Proof. Applying the definition to the function $f+g$, we have: $$\begin{array}{lll} \Delta(f+g)(c)&=(f+g)(x+\Delta x)-(f+g)(x)\\ &=f(x+\Delta x)+g(x+\Delta x)-f(x)-g(x)\\ &=\big( f(x+\Delta x)-f(x) \big) +\big(g(x+\Delta x)-g(x) \big)\\ &=\Delta f(c)+\Delta g(c). \end{array}$$ Now, the limit with $c=x$: $$\begin{array}{lll} \frac{\Delta(f+g)}{\Delta x}(x)&=\frac{\Delta f}{\Delta x}(x)+\frac{\Delta g}{\Delta x}(x)&\text{ ...by SR...}\\ &\to\frac{d f}{d x}+\frac{d g}{d x} &\text{ as } \Delta x\to 0.\\ \end{array}$$ $\blacksquare$
In terms of motion, if two runners are running away from each other starting from a common location, then the distance between them is the sum of the distances they have covered.
The formula in the Lagrange notation is as follows: $$(f + g)'(x)= f'(x) + g'(x).$$
The same proof applies to subtraction of the change.
Exercise. State the Difference Rule.
In terms of motion, if two runners are running along with each other starting from a common location, then the distance between them is the difference of the distances they have covered.
The idea proportion of the change is illustrated below:
Here, if the heights triple then so do the height differences. The algebra behind this geometry is very simple: $$kA-ka=k(A-a).$$ The idea leads to the Constant Multiple Rule for Differences from Chapter 1: the difference of a multiple of a sequence is the multiple of the sequence's difference. Below is its analog.
Theorem (Constant Multiple Rule). (A) The difference quotient of a multiple of a function is the multiple of the function's difference quotient; i.e., for any function $f$ defined at the adjacent nodes $x$ and $x+\Delta x$ of a partition and any real $k$, the difference quotients (defined at the corresponding secondary node) satisfy: $$\frac{\Delta(kf)}{\Delta x}=k\frac{\Delta f}{\Delta x}.$$ (B) A multiple of a function differentiable at a point is differentiable at that point and its derivative is equal to the multiple of the function's derivative; i.e., for any function $f$ differentiable at $x$ and any real $k$, we have at $x$: $$\frac{d(kf)}{dx}=k\frac{d f}{dx}.$$
Proof. Applying the definition to the function $c\,f$, we have: $$\begin{array}{lll} \Delta(k\cdot f)(c)&=(k\cdot f)(x+\Delta x)-(k\cdot f)(x)\\ &=k\cdot f(x+\Delta x)-k\cdot f(x)\\ &=k\cdot \big( f(x+\Delta x)-f(x) \big)\\ &=k\cdot \Delta f\, (c). \end{array}$$ Now, the limit with $c=x$: $$\begin{array}{lll} \frac{\Delta(kf)}{\Delta x}(x)&=\frac{k\Delta f}{\Delta x}(x)\\ &=k\frac{\Delta f}{\Delta x}(x)&\text{ ...by CMR...}\\ &\to k\frac{d f}{d x}(x)&\text{ as } \Delta x\to 0.\\ \end{array}$$ $\blacksquare$
In terms of motion, if the distance is re-scaled, such as from miles to kilometers, then so is the velocity -- at the same proportion.
The formula in the Lagrange notation is as follows: $$(k\cdot f)'(x) = k\cdot f'(x).$$ Here is another way to write these formulas in the Leibniz notation. This is the Sum Rule: $$\frac{d}{dx}\big( u+v \big) = \frac{du}{dx} + \frac{dv}{dx},$$ and the Constant Multiple Rule: $$\frac{d}{dx}\big( cu \big) = c\frac{du}{dx}.$$
The two theorems can be combined into one. It relies on the following idea: given two functions $f,g$, their linear combination is a new function $pf+qg$, where $p,q$ are two constant numbers.
Theorem (Linearity of Differentiation). (A) The difference quotient of a linear combination of two functions is the linear combination of their difference quotients; i.e., for any two functions $f,g$ defined at the adjacent nodes $x$ and $x+\Delta x$ of a partition, the difference quotients (defined at the corresponding secondary node) satisfy: $$\frac{\Delta(pf+qg)}{\Delta x}=p\frac{\Delta f}{\Delta x}+q\frac{\Delta g}{\Delta x}.$$ (B) A linear combination of two functions differentiable at a point is differentiable at that point and its derivative is equal to the linear combination of their derivatives; i.e., for any two functions $f,g$ differentiable at $x$, we have at $x$: $$\frac{d(pf+qg)}{d x}=p\frac{d f}{d x}+q\frac{d g}{d x}.$$
In other words, our “function of functions” has the same property as a linear polynomial: $$\newcommand{\ra}[1]{\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{ccccccccccccccc} pf+qg & \mapsto & \begin{array}{|c|}\hline\quad \frac{d}{dx} \quad \\ \hline\end{array} & \mapsto & pf' +qg'. \end{array}$$
The hierarchy of polynomials and their derivatives was used in Chapter 7 to model free fall.
• The derivative of a constant polynomial is zero:
$$(c)'=0.$$
• The derivative of a linear polynomial is constant:
$$(mx+b)'=(mx)'+(b)'=m(x)'+0=m\cdot 1=m.$$
• The derivative of a quadratic polynomial is linear:
$$(ax^2+bx+c)'=(ax^2)'+(bx)'+(c)'=a(x^2)'+b(x)'+0=a\cdot 2x+b\cdot 1=2ax+b.$$ And so on: combined with the Power Formula, the two rules above allow us to differentiate all polynomials. Every time, the degree goes down by $1$! The general result is as follows.
Theorem. The derivative of a polynomial of degree $n>0$, $$f(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_{2}x^2+a_{1}x+a_0,\ a_n\ne 0,$$ is a polynomial of degree $n-1$, $$f'(x)=na_nx^{n-1}+(n-1)a_{n-1}x^{n-2}+...+2a_{2}x+a_{1},\ a_n\ne 0.$$
Exercise. Prove the theorem.
## 2 Differentiation over compositions: the Chain Rule
How does one express the derivative of the composition of two functions in terms of their derivatives?
Example. Treating functions as transformations suggest an easy answer.
• If the first transformation is a stretch by a factor of $2$, i.e., the derivative is $2$, and
• the second transformation is a stretch by a factor of $3$, i.e., the derivative is $3$, then
• the composition of the two transformations is a stretch by a factor of $3\cdot 2=6$, i.e., the derivative is $6$:
We multiply the derivatives. $\square$
Example. Let's confirm this idea with a very simple example. Consider two linear polynomials: $$\begin{array}{lllll} x&=qt&\Longrightarrow & \frac{\Delta x}{\Delta t}=\frac{dx}{dt}&=q\\ \quad\quad\circ&&&&\ \ \times\\ y&=mx&\Longrightarrow& \frac{\Delta y}{\Delta x}=\frac{dy}{dx}&=m\\ \hline y&=m(qt)=mqt&\Longrightarrow& \frac{\Delta y}{\Delta t}=\frac{dy}{dt}&=m\cdot q&=\frac{\Delta x}{\Delta t}\cdot\frac{\Delta y}{\Delta x}=\frac{dx}{dt}\cdot\frac{dy}{dx} \end{array}$$ We see their derivatives and, which is the same think for linear polynomials, their difference quotients. In either case, we see how the intermediate variable, whether it is the difference $\Delta x$ or the differential $dx$, is “cancelled”: $$\frac{\tiny{\Delta x}}{\Delta t}\cdot\frac{\Delta y}{\tiny{\Delta x}}=\frac{\Delta y}{\Delta t},\quad \frac{\tiny{dx}}{dt}\cdot\frac{dy}{\tiny{dx}}=\frac{dy}{dt}.$$
$\square$
Example. We pose the following problem. Suppose a car is driven through a mountain terrain. Its location and its speed, as seen on a map, are known. The grade of the road is also known. How fast is the car climbing?
We set up two functions, for the location and the altitude. Then their composition is what we are interested in:
The graph of the second function is literally the profile of the road.
We already know that if the location, $f$, depends on time continuously and the altitude, $g$, depends continuously on location, then the altitude depends on time continuously as well, $g\circ f$. We shall also see that the differentiability of both functions implies the differentiability of the composition.
However, let's first dispose of the “Naive Composition Rule”: $$(f \circ g)' \neq f'\circ g'.$$ We carry out, again, a “unit analysis” to show that such a formula simply cannot be true. Suppose
• $t$ is time measured in $\text{hr}$,
• $x=f(t)$ is the location of the car as a function of time -- measured in $\text{mi}$,
• $y=g(x)$ is the altitude of the road as a function of (horizontal) location -- measured in $\text{ft}$, and
• $y=h(t)=g(f(t))$ is the altitude of the road as a function of time -- measured in $\text{ft}$.
Then,
• $f'(t)$ is the (horizontal) velocity of the car on the road -- measured in $\frac{\text{mi}}{\text{hr}}$, and
• $g'(x)$ is the rate of incline (slope) of the road -- measured in $\frac{\text{ft}}{\text{mi}}$, with the input still measured in $\text{mi}$.
It doesn't even matter now what $h'$ is measured in; just try to compose these two functions... It is impossible because the units of the output of the former and the input of the latter don't match! However, this is possible:
• $f'(t)\cdot g'(x)$ is their product -- measured in $\frac{\text{mi}}{\text{hr}}\cdot \frac{\text{ft}}{\text{mi}}=\frac{\text{ft}}{\text{hr}}$; compare to
• $h'(t)$ is the altitude of the road as a function of time -- measured in $\frac{\text{ft}}{\text{hr}}$.
Why does this make sense?
• 1. How fast you are climbing is proportional to your horizontal speed.
• 2. How fast you are climbing is proportional to the slope of the road.
$\square$
Thus, the derivative of the composition of two linear functions is the product of the two derivatives! Considering the fact that, as far as derivatives at a fixed point are concerned, all functions are linear, we have strong evidence in support of this conjecture.
Unfortunately, derivatives aren't fractions! But difference quotients are: $$\frac{\Delta y}{\Delta x}\cdot\frac{\Delta x}{\Delta t}=\frac{\Delta y}{\Delta t}.$$ The only difference from the other rules we have considered is that there are two partitions and $f$ must map the partition for $t$ to the partition of $x$:
Theorem (Chain Rule). (A) The difference quotient of the composition of two functions is found as the product of the two difference quotients; i.e., for any function $x=f(t)$ defined at two adjacent nodes $t$ and $t+\Delta t$ of a partition and any function $y=g(x)$ defined at the two adjacent nodes $x=f(t)$ and $x+\Delta x=f(t+\Delta t)$ of a partition, we have the difference quotients (defined at the secondary nodes $c$ and $q=f(c)$ within these edges of the two partitions respectively) satisfy, provided $\Delta x\ne 0$: $$\frac{\Delta (g\circ f)}{\Delta t}(c)= \frac{\Delta g}{\Delta x}(q) \cdot \frac{\Delta f}{\Delta t}(c).$$ (B) The composition of a function differentiable at a point and a function differentiable at the image of that point is differentiable at that point and its derivative is found as a product of the two derivatives; specifically, if $x=f(t)$ is differentiable at $t=c$ and $y=g(x)$ is differentiable at $x=q=f(c)$, then we have: $$\frac{d (g\circ f)}{dt}(c)= \frac{dg}{dx}(q) \cdot \frac{df}{dt}(c).$$
Proof. The formula for difference quotients is deduced as follows: $$\begin{array}{lll} \frac{\Delta (g\circ f)}{\Delta t}(c)&=\frac{(g\circ f)(t+\Delta t)-(g\circ f)(t)}{\Delta t}\\ &=\frac{g(f(t+\Delta t))-g(f(t))}{f(t+\Delta t)-f(t)}\frac{f(t+\Delta t)-f(t)}{\Delta t}\\ &=\frac{g(x+\Delta x)-g(x)}{\Delta x}\frac{f(t+\Delta t)-f(t)}{\Delta t}\\ &=\frac{\Delta g}{\Delta x}(q) \cdot \frac{\Delta f}{\Delta t}(c). \end{array}$$ Now we are to take the limit of the formula, with $c=t$, as $$\Delta t \to 0.$$ Now, since $x=x(t)$ is continuous, we conclude that we also have: $\Delta x \to 0$. Therefore, we have: $$\begin{array}{lll} \ \frac{\Delta g}{\Delta t} &=&\ \frac{\Delta g}{\Delta x}(f(t))&\cdot&\ \ \frac{\Delta f}{\Delta t}(t)\\ \quad \downarrow&&\quad \downarrow&&\quad \downarrow\\ \ \frac{dg}{dt} & = &\ \frac{dg}{dx}(f(t))&\cdot&\ \ \frac{df}{dt}(t) \end{array}$$ The idea seems to have worked out... The trouble is, we assumed that $\Delta x \neq 0$! What if $x=f(t)$ is constant in the vicinity of $t$? A complete proof will be provided later. $\blacksquare$
Exercise. Find another, non-constant, example of a function $x=f(t)$ such that $\Delta f$ may be zero even for small values of $\Delta t$.
The formula in the Lagrange notation is as follows: $$(g\circ f)'(t) = g'(f(t))\cdot f'(t).$$
Example. Find the derivative of: $$y = (1 + 2x)^{2}.$$ The function is computed in two consecutive steps (that's how we know this is a composition):
• step 1: from $x$ we compute $1+2x$, and then
• step 2: we square the outcome of the first step.
We then introduce an additional, disposable, variable in order to store the outcome of step 1: $$u=1+2x.$$ Then step 2 becomes: $$y=u^2.$$ This is our decomposition: $x \mapsto u \mapsto y$. Now the derivatives: $$\begin{array}{llll} u & = 1 + 2x &\Longrightarrow&\frac{du}{dx} &= 2 \\ y & = u^{2} &\Longrightarrow&\frac{dy}{du} &= 2u \\ \text{CR } & &\Longrightarrow&\frac{dy}{dx} & = \frac{dy}{du}\cdot\frac{du}{dx} = 2u\cdot 2 = 4u. \end{array}$$ Done. But the answer must be in terms of $x$! Last step: substitute $u = 1 + 2x$. Then the answer is $4(1+2x)$. To verify, expand, $1 + 4x + 4x^{2}$, then use PF. $\square$
Example. Now a very simple example that doesn't allow us to circumvent CR. Let $$y=\sqrt{3x+1}.$$ This is the abbreviated computation (decomposition, the derivatives, CR): $$\begin{array}{llll} x \mapsto u=3x+1 \mapsto y=\sqrt{u}\\ \underbrace{x \mapsto u=3x+1} \\ \qquad \frac{du}{dx} = 3 \\ \qquad\qquad\qquad\underbrace{u \mapsto y=\sqrt{u}}\\ \underbrace{ \qquad\qquad\qquad \frac{dy}{du}= \frac{1}{2\sqrt{u}} } \\ \frac{dy}{dx} = \frac{du}{dx}\cdot\frac{dy}{du} = 3\cdot \frac{1}{2\sqrt{u}}= 3\cdot \frac{1}{2\sqrt{3x+1}}. \end{array}$$ $\square$
Example. Find the derivative of: $$z = e^{\sqrt{3x+1}}$$ Three functions this time: $$x \mapsto u = 3x+1 \ \mapsto y = \sqrt{u} \ \mapsto z = e^{y}.$$ Fortunately, we already know the derivative of the exponent from the last example. We just append that solution with one extra step: $$\begin{array}{llll} x \mapsto u=3x+1 \mapsto y=\sqrt{u} \mapsto z = e^{y}\\ \underbrace{x \mapsto u=3x+1} \\ \qquad \frac{du}{dx} = 3 \\ \qquad\qquad\qquad\underbrace{u \mapsto y=\sqrt{u}}\\ \underbrace{ \qquad\qquad\qquad \frac{dy}{du}= \frac{1}{2\sqrt{u}} } \\ \frac{dy}{dx} = \frac{du}{dx}\cdot\frac{dy}{du} = 3\cdot \frac{1}{2\sqrt{u}} \\ \qquad\qquad\qquad\qquad\qquad\qquad \underbrace{ y \mapsto z = e^{y} }\\ \underbrace{ \qquad\qquad\qquad\qquad\qquad\qquad \frac{dz}{dy}=e^y }\\ \frac{dz}{dx} = \left( \frac{du}{dx}\cdot\frac{dy}{du} \right) \cdot\frac{dz}{dy} =3\cdot \frac{1}{2\sqrt{u}}\cdot e^y=3\frac{1}{2\sqrt{3x+1}} e^{\sqrt{3x+1}}. \end{array}$$ We have applied CR twice! $\square$
The lesson we have learned is: three functions -- three derivatives -- multiply them: $$\begin{array}{rrr} &x &\mapsto u&\mapsto y&\mapsto z \\ \frac{dz}{dx} & = \frac{du}{dx} &\cdot \frac{dy}{du} &\cdot \frac{dz}{dy} \end{array}$$ These “fractions” appear to cancel again... $$\frac{dz}{dx} = \frac{\not{du}}{dx} \cdot \frac{\not{dy}}{\not{du}} \cdot \frac{dz}{\not{dy}}.$$ This is the Generalized Chain Rule about the derivative of the composition (a “chain”!) of $n$ functions.
The short version of the Chain Rule says:
• the derivative of the composition is the product of the derivatives,
as functions.
Example. However, if we fix the location $x=a$, we can make sense of the derivative of the composition as the composition of the derivatives, after all. Indeed, suppose at point $a$ we have the derivative $$\frac{dy}{dx}=m.$$ What if we, again, think of the differentials $dx$ and $dy$ as two new variables -- related to each other by the above equation?
Then we think of the derivative, $m$, not as a number but a linear function: $$dy=m\cdot dx.$$ If now there is another variable with $$\frac{dx}{dt}=q,$$ we think of $q$ as a linear function: $$dx=q\cdot dt.$$ Then, we have to substitute $q$: $$\begin{array}{lllll} x=x(t)&=qt&\Longrightarrow& dx&=q\cdot dt\\ \quad\quad\circ&\quad\circ&&&\quad\quad\circ\\ y=y(x)&=mx&\Longrightarrow& dy&=m\cdot dx\\ \hline y=y(x(t))&=m(qt)&\Longleftrightarrow& dy&=m\cdot (q\cdot dt) \end{array}$$ We have the composition! $\square$
We can use the Chain Rule to find formulas for other important functions.
Theorem. For any $a>0$, we have: $$\left( a^x\right)'=a^x\ln a.$$
Proof. We represent this exponential function in terms of the natural exponential function: $$a^x=e^{\ln a^x}=e^{x\ln a}.$$ Then, $$\left( a^x\right)'=\left( e^{x\ln a} \right)'\ \overset{\text{CR}}{=\! =\! =}\ e^{x\ln a} \cdot (x\ln a)'=a^x\cdot \ln a.$$ $\blacksquare$
Exercise. Use the idea from the proof above to find the derivative of $x^x$.
## 3 Differentiation over multiplication and division
What happens to the output function of differentiation as we perform such an algebraic operation as multiplication with the input functions?
We already know that if the width and the height ($f$ and $g$) of a rectangle are changing continuously then so is its area ($f\cdot g$):
We shall also see that the differentiability of both dimensions implies the differentiability of the area.
However, let's first make sure we avoid the so-called “Naive Product Rule”: $$(f\cdot g)' \neq f'\cdot g'.$$ The formula is extrapolated from the Sum Rule but it simply cannot be true. Let's recast the problem in the terms of motion and take a good look at the units. Suppose
• $x$ is time measured in $\text{sec}$,
• $y=f(x)$ is the location of the first person -- measured in $\text{ft}$, and
• $y=g(x)$ is the location of the second person -- measured in $\text{ft}$.
Then
• $f'(x)$ is the velocity of the first person -- measured in $\frac{\text{ft}}{\text{sec}}$, and
• $g'(x)$ is the velocity of the second person -- measured in $\frac{\text{ft}}{\text{sec}}$.
Suppose they are running in two perpendicular directions (east and north), then
• $y=f(x)\cdot g(x)$ is the area of the rectangle enclosed by the two persons -- measured in $\text{ft}^2$.
Therefore,
• $y=\left( f(x)\cdot g(x) \right)'$ is the rate of change of the area -- measured in $\frac{\text{ft}^2}{\text{sec}}$.
Meanwhile,
• $f(x)'\cdot g(x)'$ is an unknown quantity -- measured in $\frac{\text{ft}}{\text{sec}}\cdot \frac{\text{ft}}{\text{sec}}=\frac{\text{ft}^2}{\text{sec}^2}$!
We do notice now that the product of the location and velocity gives the right units: $$f'f,\ g'g \text{ and also } f'g,\ g'f.$$ Which one(s)?
The correct idea -- cross-multiplication -- is illustrated below:
As the width and the depth are increasing, so is the area of the rectangle. But the increase of the area cannot be expressed entirely in terms of the increases of the width and depth! This increase is split into two parts corresponding to the two terms in the right-hand side of the formula below. It is based on the Product Rule for Differences from Chapter 1: $$\Delta (f \cdot g)(c)=f(x+\Delta x) \cdot \Delta g(c) + \Delta f(c) \cdot g(x).$$
Theorem (Product Rule). (A) The difference quotient of the product of two functions is found as a combination of these functions and their difference quotients. In other words, for any two functions $f,g$ defined at the adjacent nodes $x$ and $x+\Delta x$ of a partition, the difference quotients (defined at the corresponding secondary node $c$) satisfy: $$\frac{\Delta (f\cdot g)}{\Delta x}(c)=f(x+\Delta x) \cdot \frac{\Delta g}{\Delta x}(c) + \frac{\Delta f}{\Delta x}(c) \cdot g(x).$$ (B) The product of two functions differentiable at a point is differentiable at that point and its derivative is found as a combination of these functions and their derivatives; specifically, given two functions $f,g$ differentiable at $x$, we have: $$\frac{d (f\cdot g)}{dx}(x)=f(x) \cdot \frac{dg}{dx}(x) + \frac{df}{dx}(x) \cdot g(x).$$
Proof. $$\begin{array}{lll} \Delta (f \cdot g)(c)&=(f \cdot g)(x+\Delta x)- (f \cdot g)(x)\\ &=f(x+\Delta x) \cdot g(x+\Delta x)- f(x) \cdot g(x)\\ &=f(x+\Delta x) \cdot g(x+\Delta x)- f(x+\Delta x) \cdot g(x) +f(x+\Delta x) \cdot g(x)- f(x) \cdot g(x)\\ &=f(x+\Delta x) \cdot (g(x+\Delta x)- g(x)) +(f(x+\Delta x) - f(x)) \cdot g(x)\\ &=f(x+\Delta x) \cdot \Delta g(c) + \Delta f(c) \cdot g(x). \end{array}$$ Now, the limit with $c=x$: $$\begin{array}{lll} \frac{\Delta (f \cdot g)(x)}{\Delta x}&=f(x+\Delta x) \cdot \frac{\Delta g}{\Delta x} (c)&+ \frac{\Delta f}{\Delta x}(c) \cdot g(x)\\ &\quad\quad \downarrow\quad\quad \quad\ \downarrow&\quad\ \downarrow\quad \quad \quad \\ &\quad\ f(x)\quad \quad \cdot\frac{d g}{d x}(x)&+\ \frac{d f}{d x}(x)\ \cdot g(x)&\text{ as } \Delta x\to 0.\\ \end{array}$$ The first limit is justified by the fact that $f$, as a differentiable function, is continuous. $\blacksquare$
In terms of motion, it is as if two runners are unfurling a flag while running east and north respectively.
The formula in the Lagrange notation is as follows: $$(f \cdot g)'(x) = f(x)\cdot g'(x) + f'(x)\cdot g(x).$$
Example. Let $$y = xe^{x}.$$ Then, $$\begin{array}{lllll} u & = x & \Longrightarrow &\frac{du}{dx} &= (x)' = 1, \\ v & = e^{x} & \Longrightarrow &\frac{dv}{dx} &= (e^{x})' = e^{x}. \end{array}$$ Apply PR via “cross-multiplication”, the idea of which comes from the picture above: $$\frac{dy}{dx} = x\cdot e^{x} + e^{x}\cdot 1 = e^{x}(x + 1).$$ $\square$
Next, the derivatives under division? We already know that if the width and the height ($f$ and $g$) of a triangle are changing continuously then so is the tangent of its base angle ($f/g$):
We shall also see that the differentiability of either dimension implies the differentiability of the tangent.
However, let's first make sure we avoid the so-called “Naive Quotient Rule”: $$(f/ g)' \neq f'/ g'.$$ We can repeat the “unit analysis” to show that such a formula simply cannot be true. The runners still are running in two perpendicular directions, and we have:
• $y=f(x)/ g(x)$ is unitless, and then
• $y=\left( f(x)/ g(x) \right)'$ is measured in $\frac{1}{\text{sec}}$, while
• $f(x)'/ g(x)'$ is unitless!
The following is based on the Quotient Rule for Differences from Chapter 1: $$\Delta (f / g)(c)=\frac{f(x+\Delta x) \cdot \Delta g(c) - \Delta f(c) \cdot g(x)}{g(x)g(x+\Delta x)}.$$
Theorem (Quotient Rule). (A) The difference quotient of the quotient of two functions is found as a combination of these functions and their difference quotients. In other words, for any two functions $f,g$ defined at the adjacent nodes $x$ and $x+\Delta x$ of a partition, the difference quotients (defined at the corresponding secondary node $c$) satisfy: $$\frac{\Delta (f/ g)}{\Delta x}(c)=\frac{f(x+\Delta x) \cdot \frac{\Delta g}{\Delta x}(c) - \frac{\Delta f}{\Delta x}(c) \cdot g(x)}{g(x)g(x+\Delta x)},$$ provided $g(x),g(x+\Delta x) \ne 0$. (B) The quotient of two functions differentiable at a point is differentiable at that point and its derivative is found as a combination of these functions and their derivatives; specifically, given two functions $f,g$ differentiable at $x$, we have: $$\frac{d (f/ g)}{dx}(x)=\frac{f(x) \cdot \frac{dg}{dx}(x) - \frac{df}{dx}(x) \cdot g(x)}{g(x)^2},$$ provided $g(x) \ne 0$.
Proof. We start with the case $f=1$. Then we have: $$\begin{array}{lll} \frac{\Delta (1/g)(x)}{\Delta x}&=\frac{\frac{1}{g(x+\Delta x)}- \frac{1}{g(x)}}{\Delta x}\\ &=\frac{g(x)- g(x+\Delta x)}{\Delta x g(x+\Delta x)g(x)} \\ &=-\frac{g(x+\Delta x)- g(x)}{\Delta x}\cdot \frac{1}{g(x+\Delta x)\cdot g(x)} \\ &=-\frac{\Delta g}{\Delta x}(c)\cdot \frac{1}{g(x+\Delta x)\cdot g(x)} &\text{ with }c=x\\ &\to -\frac{dg}{dx}(x)\cdot\frac{1}{g(x) \cdot g(x)}&\text{ as } \Delta x\to 0. \end{array}$$ The limit of the second fraction is justified by the fact that $g$, as a differentiable function, is continuous. Alternatively, we represent the reciprocal of $g$ as a composition: $$z=\frac{1}{g(x)}\ \Longrightarrow\ z=\frac{1}{y},\ y=g(x)\ \Longrightarrow\ \frac{dz}{dy}=-\frac{1}{y^2},\ \frac{dy}{dx}=g'(x)\ \Longrightarrow\ \frac{dz}{dx}=-\frac{1}{g(x)^2}g'(x),$$ by the Chain Rule. Now the general formula follows from the Product Rule. $\blacksquare$
The formula is similar to the Product Rule in the sense that it also involves cross-multiplication.
The formula in the Lagrange notation is as follows: $$\left( \frac{f(x)}{g(x)} \right)' = \frac{f'(x)\cdot g(x) - f(x)\cdot g'(x)}{g(x)^2},$$
Example. The tangent: \begin{aligned} (\tan x)' & = \left( \frac{\sin x}{\cos x} \right)'\\ & \ \overset{\text{QR}}{=\! =\! =} \frac{(\sin x)' \cos x – \sin x (\cos x)'}{(\cos x)^{2}} \\ & = \frac{\cos x \cos x – \sin x (-\sin x)}{\cos^{2} x} \\ & = \frac{\cos^{2}x + \sin^{2}x}{\cos^{2}x} \quad \text{...use the Pythagorean Theorem...} \\ & = \sec^{2}x. \end{aligned} $\square$
In the Leibniz notation, this is the form of the Product Rule: $$\frac{d}{dx} \left(uv \right) = \dfrac{du}{dx}\cdot v + \dfrac{dv}{dx}\cdot u,$$ and the Quotient Rule: $$\frac{d}{dx} \left(\frac{u}{v}\right) = \dfrac{\dfrac{du}{dx}\cdot v – \dfrac{dv}{dx}\cdot u}{v^{2}}.$$
More examples of differentiation...
Example. Find $$(x^{2} + x^{3})' = \lim_{h \to 0} \frac{(x + h)^{2} +(x + h)^{3} - x^{2} - x^{3}}{h}=...$$ Seems like a lot of work... Instead use SR and PF: $$\begin{array}{lllll} (x^{2} + x^{3})' & = (x^{2})' + (x^{3})' \\ & = 2x +3x^{2}. \end{array}$$ $\square$
Example. We can differentiate any polynomial easily now: $$\begin{array}{lllll} (x^{77} + & 5x^{18} + 6x^{3} - x^{2} + 88)'& \text{ ...try to expand } (x+h)^{77} !\\ & \ \overset{\text{SR}}{=\! =\! =} (x^{77})' + (5x^{18})' + (6x^{3})' - (x^{2})' + (88)' \\ & \ \overset{\text{CMR}}{=\! =\! =}(x^{77})' + (5x^{18})' + (6x^{3})' - (x^{2})' + 0 \\ & \ \overset{\text{PF}}{=\! =\! =} 77x^{77 - 1} + 5\cdot 18x^{13 - 1} + 6\cdot 3x^{3 - 1} - 2x^{2 - 1} \\ & = 77x^{76} + 90x^{17} - 18x^{2} - 2x. \end{array}$$ $\square$
Example. Find $$\left( \frac{\sqrt{x}}{x^{2} + 1} \right)'.$$ Consider: $$\begin{array}{lllll} u & = \sqrt{x} &\Longrightarrow &\frac{du}{dx} &= \frac{1}{2\sqrt{x}}, \\ v & = x^{2} + 1 &\Longrightarrow &\frac{dv}{dx} &= 2x. \end{array}$$ Then, $$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{ \dfrac{1}{2\sqrt{x}} (x^2 + 1) - \sqrt{x}\cdot 2x}{( x^2 + 1)^2}.$$ No need to simplify. $\square$
Example. This is a different kind of example. Evaluate: $$\lim_{x \to 5} \frac{2^{x} - 32}{x - 5}.$$ It's just a limit. But we recognize that this is the derivative of some function. We compare the expression to the formula in the definition: $$f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a},$$ and match. So, we have here: $$a = 5 ,\ f(x) = 2^{x}, \ f(5) = 2^{5} = 32.$$ Therefore, our limit is equal to $f'(5)$ for $f(x) = 2^{x}$. Compute: $$f'(x) = (2^{x})' = 2^{x} \ln 2,$$ so $$f'(5) = 2^{5} \ln 2 = 32 \ln 2.$$ $\square$
This is another interpretation of the formulas. Let's represent the Sum Rule, the Constant Multiple Rule, and the Chain Rule as diagrams: $$\newcommand{\ra}[1]{\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\la}[1]{\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} \newcommand{\ua}[1]{\left\uparrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} \begin{array}{ccc} f,g&\ra{\frac{d}{dx}}&f',g'\\ \ \da{+}&SR &\ \da{+}\\ f+g & \ra{\frac{d}{dx}}&(f+g)'=f'+g' \end{array}\qquad \begin{array}{ccc} f&\ra{\frac{d}{dx}}&f'\\ \ \da{\cdot c}& CMR &\ \da{\cdot c}\\ cf & \ra{\frac{d}{dx}}&(cf)'=cf' \end{array}\qquad \begin{array}{ccc} f,g&\ra{\frac{d}{dx}}&f',g'\\ \ \da{\circ}& CR &\ \da{\circ }\\ f\circ g & \ra{\frac{d}{dx}}&(f\circ g)'=f'\circ g' \end{array}$$ In the first diagram, we start with a pair of functions at the top left and then we proceed in two ways:
• right: differentiate, then down: add the results; or
• down: add them, then right: differentiate the result.
The result is the same! (Neither the Product Rule nor the Quotient Rule has such an interpretation.)
## 4 The rate of change of the rate of change
If a function is known at the nodes of a partition, its difference quotient is also a function -- known at the secondary nodes. Can we treat the latter as a function too? What is the partition then? We saw in Chapter 7 how this idea is implemented in order to derive the acceleration from the velocity.
What can we say about the rate of change of this change? If we know only three values of a function (first line) at the ends of an interval, we compute the difference quotients along the two intervals (second line) and place the results at the corresponding edge: $$\begin{array}{ccccccc} -&f(x_1)&---&f(x_2)&---&f(x_3)&-&\\ -&-\bullet-&\frac{\Delta f}{\Delta x_2}&-\bullet-&\frac{\Delta f}{\Delta x_3}&-\bullet-&-\\ -&-\bullet-&---&\frac{\frac{\Delta f}{\Delta x_3} -\frac{\Delta f}{\Delta x_2}}{c_3-c_2}&---&-\bullet-&-&\\ &x_1&c_2&x_2&c_3&x_3&\\ \end{array}$$ To find the change of this new function, we carry out the same operation and place the result in the middle (third line).
Let's review the construction of the difference quotient.
First, we have an augmented partition of an interval $[a,b]$. We partition it into $n$ intervals with the help of the nodes (the end-points of the intervals): $$a=x_{0},\ x_{1},\ x_{2},\ ... ,\ x_{n-1},\ x_{n}=b;$$ and also provide secondary nodes: $$c_{1} \text{ in } [x_{0},x_{1}], \ c_{2} \text{ in } [x_{1},x_{2}],\ ... ,\ c_{n} \text{ in } [x_{n-1},x_{n}].$$
If a function $y=f(x)$ is defined at the nodes $x_k,\ k=0,1,2,...,n$, the difference quotient of $f$ is defined at the secondary nodes of the partition by: $$\frac{\Delta f}{\Delta x}(c_{k})=\frac{f(x_{k+1})-f(x_k)}{x_{k+1}-x_k},\ k=1,2,...,n.$$
The function represents the slopes of the secant lines over the nodes of the partition. In particular, when the location is represented by a function known only at the nodes of the partition, the velocity is then found in this manner. It is now especially important that we have utilized the secondary nodes as the inputs of the new function. Indeed, we can now carry out a similar construction with this function and find the acceleration!
We have now a new augmented partition, of what? The interval is $$[p,q],\ \text{ with } p=c_0 \text{ and } q=c_n.$$ We partition it into $n-1$ intervals with the help of the nodes that used to be the secondary nodes in the last partition: $$p=c_{1},\ c_{2},\ c_{3},\ ... ,\ c_{n-1},\ c_{n}=b.$$ Then the increments are: $$\Delta c_k=c_{k+1}-c_k.$$ Now, what are the secondary nodes? The primary nodes of the last partition of course! Indeed, we have: $$x_{1} \text{ in } [c_{1},c_{2}], \ x_{2} \text{ in } [c_{2},c_{3}],\ ... ,\ x_{n-1} \text{ in } [c_{n-1},c_{n}].$$
We apply the same construction to this partition to the function $g=\frac{\Delta f}{\Delta x}$. The difference quotient function of $g$ is defined at the secondary nodes of the new partition by: $$\frac{\Delta g}{\Delta x}(x_{k})=\frac{g(c_{k+1})-g(c_k)}{c_{k+1}-c_k},\ k=1,2,...,n.$$
Definition. The second difference quotient of $f$ is defined at the nodes of the partition (denoted) by: $$\frac{\Delta^2 f}{\Delta x^2}(x_{k})=\frac{\frac{\Delta f}{\Delta x}(c_{k+1})-\frac{\Delta f}{\Delta x}(c_k)}{c_{k+1}-c_k},\ k=1,2,...,n.$$
Note that there are:
• $n+1$ values of $f$ (at the nodes),
• $n$ values of $\frac{\Delta f}{\Delta x}$ (at the secondary nodes), and
• $n-1$ values of $\frac{\Delta^2 f}{\Delta x^2}$ (at the nodes except $a$ and $b$).
We will often omit the subscripts for the simplified notation: $$\frac{\Delta^2 f}{\Delta x^2}(x)=\frac{\frac{\Delta f}{\Delta x}(c+\Delta c)-\frac{\Delta f}{\Delta x}(c)}{\Delta c}.$$
Notice that the higher value of the second difference quotient means higher values of the curvature of the graph of $y=f(x)$. As another way to see this, imagine yourself driving along a straight part of the road and seeing the tree ahead to remain the same (no curvature), then, as you start to turn, the trees start to pass your field of vision from right to left (curvature):
This construction will be repeatedly used for approximations and simulations. It will be followed, when necessary, by taking its limit.
Let's differentiate $\sin x$ for the second time. In Chapter 7, we found its difference quotient over a mid-point partition with a single interval. This time we will need at least two intervals:
• three nodes $x$: $a-h$, $a$, and $a+h$, and
• two secondary nodes $c$: $a-h/2$ and $a+h/2$.
We use the two formulas for the difference quotients of $\sin x$ and $\cos x$ from Chapter 7. We write the former for the two secondary nodes, but we re-write the latter for the partition with two nodes $a-h/2,\ a+h/2$ and a single secondary node $x=a$: $$\begin{array}{lllll} \frac{\Delta}{\Delta x}(\sin x)&=\frac{ \sin (h/2)}{h/2}\cdot\cos c,& \frac{\Delta }{\Delta x}(\cos x)=-\frac{ \sin (h/2)}{h/2}\cdot\sin a,\\ \end{array}$$ Therefore, we have at $a$: $$\begin{array}{lllll} \frac{\Delta^2}{\Delta x^2}(\sin x)&=\frac{\Delta }{\Delta x}\left(\frac{\Delta}{\Delta x}( \sin x)\right)(a)\\ &=\frac{\Delta}{\Delta x}\left(\frac{ \sin (h/2)}{h/2}\cdot\cos c\right)&\text{ ...by the first formula... }\\ &=\frac{ \sin (h/2)}{h/2}\frac{\Delta \cos}{\Delta x}(a)&\text{ ...by CMR... }\\ &=\frac{ \sin (h/2)}{h/2}\left(-\frac{ \sin (h/2)}{h/2}\cdot\sin a\right)&\text{ ...by the second formula }\\ &=-\left(\frac{ \sin (h/2)}{h/2}\right)^2\cdot\sin a. \end{array}$$
Similarly, we find: $$\frac{\Delta }{\Delta x}(\cos x)=-\frac{ \sin (h/2)}{h/2}\cdot\sin c\ \Longrightarrow\ \frac{\Delta^2}{\Delta x^2}(\cos x)=-\left(\frac{ \sin (h/2)}{h/2}\right)^2\cdot\cos a.$$
For the exponential function, we need a left-end partition with two intervals:
• three nodes $x$: $a-h$, $a$, and $a+h$, and
• two secondary nodes $c$: $a-h$ and $a$.
Then, we find at $a$: $$\frac{\Delta }{\Delta x}(e^x)=\frac{ e^h-1}{h}\cdot e^{c-h/2}\ \Longrightarrow\ \frac{\Delta^2}{\Delta x^2}(e^x)=\left(\frac{ e^h-1}{h}\right)^2\cdot e^{a-h}.$$
## 5 Repeated differentiation
Example. Let's continue to differentiate the sine: $$\begin{array}{lll} (\sin x)' & = \cos x &\\ (\cos x)' & = -\sin x & \Longrightarrow &(\sin x)' ' &=-\sin x\\ (-\sin x)' & = -\cos x & \Longrightarrow &(\sin x)' ' ' &=-\cos x\\ (-\cos x)' & = \sin x & \Longrightarrow &(\sin x )' ' ' ' &= \sin x. \end{array}$$ And we are back where we started, i.e., the differentiation process for this particular function is cyclic! $\square$
We use the following terminology and notation for the consecutive derivatives of function $f$: $$\begin{array}{|l|l|l|l|} \hline \text{function } & f & f^{(0)}&\\ \text{first derivative } & f' & f^{(1)}&\frac{df}{dx}\\ \text{second derivative } & f' '=(f')' & f^{(2)}=\left(f^{(1)}\right)'&\frac{d^2f}{dx^2}=\frac{d}{dx}\left( \frac{df}{dx} \right)\\ \text{third derivative } & f' ' '=(f' ')'& f^{(3)}=\left(f^{(2)}\right)'&\frac{d^3f}{dx^3}=\frac{d}{dx}\left( \frac{d^2f}{dx^2} \right)\\ ...&&...&...\\ n\text{th derivative } & & f^{(n)}=\left(f^{(n-1)}\right)'&\frac{d^nf}{dx^n}=\frac{d}{dx}\left( \frac{d^{n-1}f}{dx^{n-1}} \right)\\ ...&&...&...\\ \hline \end{array}$$
Thus, a given differentiable function may produce a sequence of functions: $$\newcommand{\ra}[1]{\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{ccc} f & \mapsto & \begin{array}{|c|}\hline\quad \tfrac{d}{dx} \quad \\ \hline\end{array} & \mapsto & f' & \mapsto & \begin{array}{|c|}\hline\quad \tfrac{d}{dx} \quad \\ \hline\end{array} & \mapsto & f' '& \mapsto & ...& \mapsto & f^{(n)}& \mapsto &... \end{array}$$ provided the outcome of each step is differentiable as well. In the abbreviated form the sequence is: $$\newcommand{\ra}[1]{\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\la}[1]{\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} \newcommand{\ua}[1]{\left\uparrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.}\begin{array}{ccc} f &\ra{\frac{d}{dx}} &f' &\ra{\frac{d}{dx}} & f' ' &\ra{\frac{d}{dx}} &...&\ra{\frac{d}{dx}} & f^{(n)} &\ra{\frac{d}{dx}} & ... \end{array}$$
Note that, for a fixed $x$, the sequence of numbers: $$f(x),\ f'(x),\ f' '(x),\ ...,\ f^{(n)}(x),\ ...$$ is just that, a sequence, a concept familiar from Chapter 7. However, as the example of $\sin x$ shows, this sequence doesn't have to converge: $$\left( \sin x \right)^{(n)}\Big|_{x=0},\ n=0,1,2,3,...\ \leadsto\ 0,-1,0,1,0,...$$ We will see in Chapter 15 that some “linear combinations” of the derivatives that produce a sequence convergent to the function...
Let's try to compute as many consecutive derivatives as possible, or even all of them, for the functions below.
Example. The positive integer powers. The PF applies: $$(x^{n})' = nx^{n-1}.$$ The power decreases by $1$ every time. Therefore, $$(x^{n})^{ (n+1)} = 0.$$ Then, it stays $0$: $$(x^{n})^{ (n+1)} = (x^{n})^{ (n+2)}=...=0.$$ The powers in the sequence of derivatives decrease to $0$ and then remain constant. $\square$
Example. The exponent. Since $$(e^{x})' = e^{x},$$ we have: $$(e^{x})^{(n)} = e^{x}.$$ The function remains the same! The sequence of derivatives is constant. $\square$
Example. The trig functions. Same for both sine and cosine: \begin{aligned} (\sin x)^{(4n)} & = \sin x \\ (\cos x)^{(4n)} & = \cos x \end{aligned} The sequence of derivatives is cyclic for both functions. $\square$
Example. The negative integer powers. We apply PF again: \begin{aligned} (x^{-1})' & = -1x^{-2}, \\ (-x^{-2})' & = 2x^{-3},\\ ... \end{aligned} The power goes down by $1$ every time and, as a result, tends to $–\infty$. The sequence doesn't stop. $\square$
Exercise. Show that the same happens with all non-integer powers.
Differentiation creates a certain dynamic among the functions: $$\newcommand{\ra}[1]{\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\la}[1]{\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\lra}[1]{\xleftarrow{\quad\quad#1\quad}\!\to} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} \newcommand{\ua}[1]{\left\uparrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.}\begin{array}{|ccccccccc|} \hline &&&&&&&& \ \curvearrowleft^{\frac{d}{dx}}\\ x^n &\ra{\frac{d}{dx}} &nx^{n-1} &\ra{\frac{d}{dx}} &...&\ra{\frac{d}{dx}} & \text{constant} &\ra{\frac{d}{dx}} & 0\\ \hline \frac{1}{n} &\ra{\frac{d}{dx}} &-\frac{1}{n^2} &\ra{\frac{d}{dx}} & \frac{2}{n^3} &\ra{\frac{d}{dx}} & ...\\ \hline \sin x&\ra{\frac{d}{dx}}&\cos x\\ \ \ua{\frac{d}{dx}}& &\ \da{\frac{d}{dx}}\\ -\cos x & \la{\frac{d}{dx}}& -\sin x\\ \hline e^{-x}&\lra{\frac{d}{dx}}&-e^{-x}\\ \hline \ \curvearrowleft^{\frac{d}{dx}}\\ e^x \\ \hline \end{array}$$
Warning: Starting in Chapter 17, we will see that the function and its derivative are two animals of very different breeds. As a result, the dynamics discussed above will disappear in higher dimensions.
The repeated differentiation process may fail to continue when the $k$th derivative is not differentiable, i.e., when the following limit does not exist: $$f^{(k)}(a)=\lim_{h\to 0} \frac{f^{(k-1)}(a+h)-f^{(k-1)}(a)}{h}.$$
Definition. A function $f$ is called twice, thrice, ..., $n$ times differentiable when $f',f' ',f' ' ',..., f^{(n)}$ exists. When the derivative exists for all $n$, we call the function smooth.
The functions that we have treated above are smooth inside their domains.
Example. This function is differentiable but not twice differentiable: $$f(x)=\begin{cases} -x^2&\text{ if } x<0;\\ x^2&\text{ if } x\ge 0. \end{cases}$$ Its graph looks smooth:
There is no doubt in which direction a beam of light would bounce off such a surface. However, let's compute the derivatives. It is easy for $x\ne 0$ because there is only one formula: $$f(x)=\begin{cases} -2x&\text{ if } x<0;\\ 2x&\text{ if } x> 0. \end{cases}$$ For the case of $x=0$, we consider the two one-sided limits: $$\lim_{h\to 0^-}\frac{f(0+h)-f(0)}{h}=\lim_{h\to 0^-}\frac{f(h)}{h}=\lim_{h\to 0}\frac{-h^2}{h}=\lim_{h\to 0}(-h)=0;$$ $$\lim_{h\to 0^+}\frac{f(0+h)-f(0)}{h}=\lim_{h\to 0^+}\frac{f(h)}{h}=\lim_{h\to 0}\frac{h^2}{h}=\lim_{h\to 0}h=0.$$ They match! Therefore, $$f'(0)=0.$$ We have discovered that $f'(x)=2|x|$. It's not differentiable at $0$! $\square$
Example. More examples of this kind:
• $\sin\frac{1}{x}$ is discontinuous at $x=0$;
• $x\sin\frac{1}{x}$ is continuous at $x=0$ but not differentiable;
• $x^2\sin\frac{1}{x}$ is differentiable at $x=0$ but not twice differentiable.
$\square$
Exercise. Prove the above statements.
Below we visualize the relation between these classes of functions:
What is the geometric meaning of these higher derivatives for a given function?
Let's consider the first derivative. It represents the slopes of the function. Then the second derivative represents the rate of change of these slopes. Notice how changing slopes are seen as rotating tangents:
Specifically, we see:
• decreasing slopes = tangents rotate clockwise;
• increasing slopes = tangents rotate counter-clockwise.
This matches our convention from trigonometry that counter-clockwise is the positive direction for rotations.
Even though we typically have functions with the $n$th derivative for each positive integer $n$, only the first two reveal something visible about the graph of the original function.
Above we compare
• the shapes of the patches of the graph of the function $f$ to the sign of the values of the first derivative $f'$; and
• the shapes of the patches of the graph of the function $f$ to the sign of the values of the second derivative $f' '$.
There are three main levels of analysis of a function:
• Analysis at level $0$: the values of $f$. We ask, how large? The findings are about the values, $x$- and $y$-intercepts, asymptotes and other large-scale behavior, periodicity, etc.
• Analysis at level $1$: the slopes of $f$. We ask, up or down? The findings are about the angles, increasing/decreasing behavior, critical points, etc.
• Analysis at level $2$: the rate of change of the slopes of $f$. We ask, concave up or down? The findings are about the change of steepness, concavity, telling a maximum from a minimum, etc.
We can go on and continue to discover more and more subtle but less and less significant properties of the function...
This three-level analysis also applies to our study of motion, below.
The derivative of the velocity and, therefore, the second derivative of the position, is called the acceleration. The concept allows one to add another level of analysis of motion:
• Analysis at level $0$: the location, where?
• Analysis at level $1$: the velocity, how fast? forward or back?
• Analysis at level $2$: the acceleration, how large is the force?
Suppose $t$ is time and $y$ is the vertical dimension, the height. Now the specific case of free fall... These are the initial conditions:
• $y_0$ is the initial height, $y_0=y\Big|_{t=0}$, and
• $v_y$ is the initial vertical component of velocity, $\frac{dy}{dt}\Big|_{t=0}$.
Then, we have: $$\begin{array}{lll} y&=y_0+v_yt-\tfrac{1}{2}gt^2&\Longrightarrow& \frac{dy}{dt}&=v_y&-gt&\Longrightarrow&\frac{d^2y}{dt^2}&=-g. \end{array}$$ Now, from the point of the physics of the situation, the derivation should go in the opposite direction:
• when there is no force, the velocity is constant;
• when the force is constant, the velocity is linear on time, etc.
However, at this point we still unable to answer these questions:
• How do we know that only the derivatives of constant functions and none others are zero?
• How do we know that only the derivatives of linear functions and none others are constant?
• How do we know that only the derivatives of quadratic functions and none others are linear?
This reversed process is called antidifferentiation. So far, we cannot justify even this, simplest conclusion: $$f'=0 \Longrightarrow f=c,\ \text{ for some real number }c.$$ We will study these and related questions in Chapter 9.
## 6 Change of variables and the derivative
If the distance is measured in miles and time in hours, the velocity is measured in miles per hour. If the distance is measured in kilometers and time in minutes, the velocity is measured in kilometers per minute. In either case, we are dealing with the same functions just measured in different units. If the two distance functions match, do the velocity functions too?
Let's recall that we can interpret every composition as a change of variables. We are especially interested in a change of units because we often measure quantities in multiple ways:
• length and distance: inches, miles, kilometers, light years;
• time: minutes, seconds, hours, years;
• weight: pounds, kilograms, karats;
• temperature: degrees of Celsius, of Fahrenheit,
• etc.
How does such a change affect calculus as we know it?
If $$y=f(x)$$ is a relation between two quantities $x$ and $y$, then either one may be replaced with a new variable. Let's call them $t$ and $z$ respectively and suppose these replacements are given by some functions:
• case 1: $x=g(t)$;
• case 2: $z=h(y)$.
These substitutions create new relations:
• case 1: $y=k(t)=f(g(t))$;
• case 2: $z=k(x)=h(f(x))$.
The Chain Rule gives us the rate of change for each pair:
• case 1:
$$\frac{dk}{dt}=\frac{df}{dx}\frac{dg}{dt};$$
• case 2:
$$\frac{dk}{dx}=\frac{dh}{dy}\frac{df}{dx}.$$
Most often, the conversion formula of a change of units is linear.
This is for Case 1.
Theorem (Linear Chain Rule I). If $$g(t)=mt+b$$ and $y=f(x)$ is differentiable, then the derivative of $y=k(t)=f(g(t))$ is given by: $$k'(t)=mf'(mt+b).$$
Example. What if $x$ is time and we change the moment from which we start measuring time, e.g., the “daylight savings time”? We have: $$g(t)=t+t_0\ \Longrightarrow\ k'(t)=f'(t+t_0).$$ $\square$
Example. Suppose $x$ is time and $y$ is the location, then function $g$ may represent the change of units of time, such as to seconds, $x$, from minutes, $t$: $$x=g(t)=60t.$$ Then, the change of the units won't change a lot about our calculus:
• if $f$ is the location as a function of seconds, $k$ is the location as a function of minutes, and $k(t)=f(60t)$;
• also $f'$ is the velocity as a function of seconds, $k'$ is the velocity as a function of minutes, and $k'(t)=60f'(60t)$;
• also $f' '$ is the acceleration as a function of seconds, then $k' '$ is the acceleration as a function of minutes, and $k' '(t)=60^2f' '(60t)$.
Thus, the graphs of the new quantities describing motion are simply re-scaled versions of the graphs of the old ones. $\square$
This is for Case 2.
Theorem (Linear Chain Rule II). If $$h(y)=my+b,$$ and $y=f(x)$ is differentiable, then the derivative of $z=k(x)=h(f(x))$ is given by: $$k'(x)=mf'(x).$$
Example. What if $y$ is the location and we change the place from which we start measuring, e.g., the Greenwich meridian? We have: $$h(x)=y+y_0\ \Longrightarrow\ k'(x)=f'(x).$$ We can also change the direction of the $x$-axis: $$h(x)=-y\ \Longrightarrow\ k'(x)=-f'(x).$$ $\square$
Example. Suppose $x$ is time and $y$ is the location, then function $h$ may represent the change of units of length, such as from miles, $y$, to kilometers, $x$: $$z=h(y)=1.6y.$$ Then, the change of the units will change very little about the calculus that we have developed; the coefficient, $m=1.6$, is the only adjustment necessary. Furthermore,
• if $f$ is the location in miles, then $k$ is the location in kilometers: $k(x)=1.6f(x)$;
• also $f'$ is the velocity with respect to miles, $k'$ is the velocity with respect to kilometers, and $k'(x)=1.6f'(x)$;
• also $f' '$ is the acceleration with respect to miles, $k' '$ is the acceleration with respect to kilometers, and $k' '(x)=1.6f' '(x)$.
Thus, the quantities describing motion are simply replaced with their multiples. The new graphs are the vertically stretched versions of the old ones. $\square$
Example. Recall the example when we have a function $f$ that records the temperature -- in Fahrenheit -- as a function $f$ of time -- in minutes -- replaced with another to records the temperature in Celsius as a function $g$ of time in seconds:
• $s$ time in seconds;
• $m$ time in minutes;
• $F$ temperature in Fahrenheit;
• $C$ temperature in Celsius.
The conversion formulas are: $$m=s/60,$$ and $$C=(F-32)/1.8.$$
These are the relations between the four quantities: $$g:\quad s \xrightarrow{\quad s/60 \quad} m \xrightarrow{\quad f\quad} F \xrightarrow{\quad (F-32)/1.8\quad} C.$$ And this is the new function: $$F=k(s)=(f(s/60)-32)/1.8.$$ Then, by the Chain Rule, we have: $$\frac{dF}{ds}=\frac{dF}{dC}\frac{dC}{dm}\frac{dm}{ds}=\frac{1}{1.8}\cdot f'(m)\cdot \frac{1}{60}.$$ $\square$
Exercise. Provide a similar analysis for the sizes of shoes and clothing.
Example. The conversion of the number of degrees $y$ to the number of radians $x$ is: $$x=\frac{\pi}{180}y.$$ Then, $$\frac{dx}{dy}=\frac{\pi}{180}.$$ Therefore, the trigonometric differentiation formulas, such as $\left( \sin x \right)'=\cos x$, don't hold anymore! Indeed, let's denote sine and cosine for degrees by $\sin_dy$ and $\cos_dy$ respectively: $$\sin_dy=\sin \left( \frac{\pi}{180}y \right) \text{ and } \cos_dy=\cos \left( \frac{\pi}{180}y \right).$$ Then, $$\begin{array}{lll} \frac{d}{dy}\sin_d y&=\frac{d}{dy}\sin \left( \frac{\pi}{180}y \right)\\ &=\frac{\pi}{180}\cos \left( \frac{\pi}{180}y \right)\\ &=\frac{\pi}{180}\cos_dy. \end{array}$$ $\square$
Example. What if we are to change our unit to a logarithmic scale? For example, $$x=10^t.$$ Then, for any function $y=f(x)$, we have by the Chain Rule: $$\frac{dy}{dt}=\frac{dy}{dx}\Bigg|_{x=10^t}\cdot \left( 10^t \right)'=\frac{dy}{dx}\Bigg|_{x=10^t}10^t\ln 10.$$ The effect on the derivative is not proportional! $\square$
This is the summary of the Chain Rule: $$\newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} \newcommand{\la}[1]{\!\!\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\ua}[1]{\left\uparrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} \begin{array}{ccccccccc} & f(g(x)) &\ra{\frac{d}{dx}} &f'(g(x))g'(x) \\ \small\text{substitution }&\quad \da{u=g(x)} & &\ \ \ua{CR} \\ & f(u) &\ra{\frac{d}{du}} &f'(u) \end{array}$$ The method allows us to get from left to right at the top (differentiation with respect to $x$) by taking a detour. We follow the path around the square: substitution, differentiation with respect to $u$, the Chain Rule formula with back-substitution.
## 7 Implicit differentiation and related rates
We differentiate functions; can we differentiate relations?
Recall from Chapter 2 that relations are represented by equations, but not the kind we are used to: $$\underbrace{x^{2}}_{\text{a number}} - \underbrace{1}_{\text{a number}}=0\quad \leadsto \text{ find a particular number } x.$$ After the substitution, the equation should be true. The equations we are interested in are equations of functions, such as the familiar equation of the circle: $$\underbrace{x^2}_{\text{a function}}+\underbrace{y^{2}}_{\text{a composition of two functions}} =0 \quad \leadsto \text{ find a particular function } y=y(x).$$ After the substitution, the equation should be true for all $x$.
The equation implicitly defines this function. As we have done in the past, we can make the function $y=y(x)$ explicit by solving the equation for $y$: $$y = \sqrt{1 - x^{2}} \text{ or } y=–\sqrt{1 - x^{2}}.$$
However, what if we want only the rate of change of this, unknown, function?
We will rely on the following fact: if two functions are equal, for all nodes $x$, of a partition then so are their difference quotients, for all secondary nodes $c$: $$f(x)=g(x) \text{ for all } x\ \Longrightarrow\ \frac{\Delta f}{\Delta x}(c)=\frac{\Delta g}{\Delta x}(c) \text{ for all } c.$$
Example (circle). Find the secant line through the two points on the circle of radius $1$ centered at $0$: $$(0,1) \text{ and } \left( \tfrac{\sqrt{2}}{2},\tfrac{\sqrt{2}}{2} \right).$$
Typically, a curve has been the graph of a function $y = x^{2}$, $y = \sin x$, etc., given explicitly. This time the equation is: $$x^{2} + y^{2} = 1.$$ To find the slope of the secant line, we need the difference quotient of the function but there is no, explicit, function!
The idea is to consider the above equation as a relation between the two variables. In fact, we think of $y=y(x)$ as a function of $x$, i.e.: $$x^{2} + y(x)^{2} = 1.$$ We will also assume that
• the two $x$-values $x_0 =0$ and $x_1\frac{\sqrt{2}}{2}$ are nodes of a partition of the $x$-axis, and
• the two $y$-values $y_0 =1$ and $y_1= \frac{\sqrt{2}}{2}$ are nodes of a partition of the $y$-axis.
We apply the Chain Rule to both sides of the equation: $$\begin{array}{rll} \frac{\Delta }{\Delta x} \left( x^{2} + y^{2} \right) & = \frac{\Delta }{\Delta x} (1) &\Longrightarrow\\ \frac{\Delta }{\Delta x} x^{2} + \frac{\Delta }{\Delta x}y^{2} &= 0 &\Longrightarrow\\ (x_0+x_1) + (y_0+y_1) \frac{\Delta y}{\Delta x} &= 0 &\Longrightarrow\\ \frac{\Delta y}{\Delta x} &= -\frac{x_0+x_1}{y_0+y_1} &\text{ for } y_0+y_1\ne 0. \end{array}$$ We have found a formula for the difference quotient but it is still implicit -- because we don't have a formula for $y=y(x)$. Fortunately, we don't need the whole function, just those two points on its graph. We substitute these into the formula above to find: $$\frac{\Delta y}{\Delta x}= -\frac{0+\frac{\sqrt{2}}{2}}{1+\frac{\sqrt{2}}{2}}= -\frac{\sqrt{2}}{1+\sqrt{2}}.$$ Finally, from the point-slope formula we obtain the answer: $$y - \frac{\sqrt{2}}{2} = -\frac{\sqrt{2}}{1+\sqrt{2}}\left( x - \frac{\sqrt{2}}{2}\right).$$ We can automate this formula and find more secant lines:
$\square$
What about the derivative? We will rely on the following fact: if the values of two functions are equal for all $x$ then so are the values of their derivatives: $$f(x)=g(x) \text{ for all } x\ \Longrightarrow\ f'(x)=g'(x) \text{ for all } x.$$ We can put it simply as: if two functions are equal then so are their derivatives; i.e., $$\begin{array}{|c|}\hline\quad f=g \ \Longrightarrow\ f'=g' \quad \\ \hline\end{array}$$
Differentiating an equation of functions and finding the derivative of a function defined by this equation is called implicit differentiation.
Let's consider two examples of how this idea may help us with finding tangents to implicit curves.
Example (circle). Find the tangent line for the circle of radius $1$ centered at $0$ at the point $\left( \tfrac{\sqrt{2}}{2},\tfrac{\sqrt{2}}{2} \right)$.
Typically, a curve has been the graph of a function $y = x^{2}$, $y = \sin x$, etc., given explicitly. This time the equation is: $$x^{2} + y^{2} = 1.$$ To find the slope of the tangent line, we need the derivative, but there is no function to differentiate!
Our approach is to differentiate the equation above as a relation between the two variables. As we differentiate, we think of $y=y(x)$ as a function of $x$, i.e.: $$x^{2} + y(x)^{2} = 1.$$ This is the result, via the Chain Rule: $$\begin{array}{rll} \frac{d}{dx} \left( x^{2} + y^{2} \right) & = \frac{d}{dx} (1) &\Longrightarrow\\ \frac{d}{dx} x^{2} + \frac{d}{dx}y^{2} &= 0 &\Longrightarrow\\ 2x + 2y \frac{dy}{dx} &= 0 &\Longrightarrow\\ \frac{dy}{dx} &= -\frac{x}{y} &\text{ for } y\ne 0. \end{array}$$
We have found a formula for the derivative, but it is still implicit -- because we don't have a formula for $y=y(x)$. Fortunately, we don't need the whole function, just a single point on its graph: $$x = \frac{\sqrt{2}}{2},\ y = \frac{\sqrt{2}}{2}$$ We substitute these into the formula above to find: $$\frac{dy}{dx}\Bigg|_{x = \frac{\sqrt{2}}{2},\ y = \frac{\sqrt{2}}{2}}= -\frac{x}{y}\Bigg|_{x = \frac{\sqrt{2}}{2},\ y = \frac{\sqrt{2}}{2}}= -1.$$ Finally, from the point-slope formula we obtain the answer: $$y - \frac{\sqrt{2}}{2} = -1\left( x - \frac{\sqrt{2}}{2}\right).$$
Note that we could use the explicit formula $y = \sqrt{1 - x^{2}}$ with the same result: $$\frac{dy}{dx} \overset{\text{CR}}{=} \frac{-2x}{2\sqrt{1 - x^{2}}} = -\frac{x}{1 - x^{2}},$$ after we substitute $x = \frac{\sqrt{2}}{2}$. However, it's only explicit for the upper half of the circle. For a point below the $x$-axis, we'd need to start over and use the other formula, $y = -\sqrt{1 - x^{2}}$.
Observe also that the derivative $\frac{dy}{dx}$ is undefined at $x= \pm 1$ (implicit or explicit) because the denominator is $0$. How do we find the tangent? From the formula we can proceed in two directions: $$x^{2} + y^{2} = 1 \leadsto \begin{cases} y \text{ depends on } x,\\ x \text{ depends on } y. \end{cases}$$ Then, we can try implicit differentiation of the same equation -- but with respect to $y$ this time. The computation is very similar, and the result is: $$\frac{dx}{dy} = -\frac{y}{x}.$$ The formula is defined for $y = 0$, at the points $(-1,0),\ (1, 0)$. Then, $\frac{dx}{dy} = 0$ at these points. Therefore, the tangent line is $x - 1 = 0 (y-0)$, or $x = 1$. $\square$
Example (Folium of Descartes). This curve is given by: $$x^{3} + y^{3} = 6xy.$$
We differentiate the equation as before: $$\frac{d}{dx} \left( x^{3} + y^{3} \right) = \frac{d}{dx} (6xy).$$ Using CR we notice that every time if we see $y$, the factor $\frac{dy}{dx}$ also appears: $$\begin{array}{rll} \frac{d}{dx} (x^{3}) + \frac{d}{dx} (y^{3}) & = 6\frac{d}{dx} (xy) \\ 3x^{2} + 3y^{2}\cdot \frac{dy}{dx} &= 6 \left(y + x\frac{dy}{dx} \right) . \end{array}$$ Solve for $\frac{dy}{dx}$. $$\begin{array}{rll} 3x^{2} + 3y^{2} \frac{dy}{dx} & = 6y + 6x \frac{dy}{dx} \\ (3y^{2} - 6x) \frac{dy}{dx} & = 6y – 3x^{2} \\ \frac{dy}{dx} & = \underbrace{\frac{6y – 3x^{2}}{3y^{2} - 6x}}_{\text{Fails at } (0,0)!} \end{array}$$ The end result is: if we know the location $(x, y)$, you know the slope of the tangent at that point. For example, at the tip of the curve we have $x=y$. Therefore, the slope is $\frac{dy}{dx}=-1$. $\square$
Note that in either example, we can cut the curve into pieces each of which is the graph of a function:
Now, implicit differentiation also helps with situations when several quantities depend on each other implicitly as well on time. If we differentiate this dependence equation, we get a dependence between their derivatives. The result is related rates.
Example (air balloon). Suppose we have an air balloon, spherical in shape. Air is pumped in it at the rate of $5 {}^{\text{in }^{3}}/_{\text{sec}}$. What is the rate of growth of the radius at different radii?
Step one in word problems: introduce variables; let
• $t$ be time,
• $V$ be the volume, and
• $r$ be the radius.
Next, $V$ depends on $t$ and at that moment we have $$\frac{dV}{dt} = 5,$$ according to the condition. Furthermore, this is a sphere, so $$V = \frac{4}{3}\pi r^{3}.$$ Here we see that $V$ also depends on $r$; altogether, this is the dependencies we face: $$\begin{array}{cccc} t &\to &r\\ &\searrow&\downarrow\\ &&V \end{array}$$ We could reverse the last arrow by finding the inverse: $r = \sqrt[3]{\frac{3}{4\pi}V}$. Instead, we differentiate the equation itself. Thus, if two variables are related (via an equation), then so are their derivatives, i.e., the rates of change (hence, “related rates”).
Keeping in mind that both $V$ and $r$ are functions of time, we differentiate the relation with respect to $t$: $$V= \frac{4}{3} \pi r^{3}.$$ The left-hand side is very simple: $$\frac{d}{dt}V=\frac{dV}{dt},$$ but in the right-hand side $r(t)^{3}$ is a composition: $$\frac{d}{dt}\left( \frac{4}{3} \pi r^{3} \right) = \frac{4}{3} \pi \cdot 3r^{2} \frac{dr}{dt}.$$ Thus, we have: $$\frac{dV}{dt} = \frac{4}{3} \pi \cdot 3r^{2} \frac{dr}{dt}.$$
Recall that the rate of change of $V$ is $5$ (at a given moment), so: $$5 = 4\pi r^{2} \frac{dr}{dt},$$ or $$\frac{dr}{dt} = \frac{5}{4\pi r^{2}}.$$
Next, what's the rate of growth of $r$ when $r = 1,\ r = 2,\ r = 3$? $$\begin{array}{lll} r = 1: & \frac{dr}{dt} = \frac{5}{4\pi}; \\ r = 2: & \frac{dr}{dt} = \frac{5}{16\pi}; \\ r = 3: & \frac{dr}{dt} = \frac{5}{36\pi}. \end{array}$$ Indeed, we see a slow-down. $\square$
Example (sliding ladder). Suppose a $10$ ft. ladder stands against the wall and its bottom is sliding at $2$ ft/sec. How fast is the top moving when it is $6$ ft from the floor?
Introduce variables:
• $x$ the distance of the bottom from the wall,
• $y$ the distance of the top from the floor, both functions of
• $t$ the time.
Translate the information, as well as the question, about the variables into equations: $$\begin{array}{ll|l} &\text{quantities:}&\text{functions:}\\ \hline \text{always}& x^{2} + y^{2} = 10^{2}&(x(t))^{2} + (y(t))^{2} = 10^{2}\\ \text{now}&\frac{dx}{dt} = 2&x'(t_0)=2\\ \text{now}&y = 6&y(t_0)=6 \\ \text{now}&\frac{dy}{dt} = ?& y'(t_0)=? \end{array}$$ That's the purely mathematical problem to be solved.
We differentiate the equation with respect to the independent variable, $t$: $$\begin{array}{rlll} \frac{d}{dt}\left( x^{2} + y^{2} \right) & = \frac{d}{dt}\left(100\right) \\ 2x\frac{dx}{dt} + 2y\frac{dy}{dt} & = 0,& \text{ solve for } \frac{dy}{dt} \\ \frac{dy}{dt} &= - \frac{x}{y}\frac{dx}{dt},& \text{ substitute } \\ &= -\frac{x}{6} 2, & \text{ now } x=8 \text{ comes from } x^{2} + y^{2} = 100, \\ &= -\frac{8}{6} 2 \\ & = -\frac{8}{3}. \end{array}$$ It is going down! $\square$
Exercise. Solve the problem for the moment when the ladder hits the floor.
## 8 Radar gun: the math
Problem. Suppose you are driving at a speed $80$ mph when you see a police car positioned off $40$ feet the road. What is the radar gun's reading?
How does the radar gun work? In fact, how does a radar work? A signal is sent, it bounces off an object, and, when it comes back, the time lapse is recorded. Then, the distance to the object is computed as: $$S = \underbrace{\text{ signal's speed }}_{\text{ known }} \cdot \underbrace{\text{ time passed }}_{\text{ measured }}.$$
A radar run does this twice.
A signal is sent, it comes back, the time is measured. Then the second time:
• $S_{1} =$ speed $\cdot$ time, at time $t=t_1$,
• $S_{2} =$ speed $\cdot$ time, at time $t=t_2$.
Then, the reading is computed as: $$\text{ estimated speed }= \frac{\text{ change of distance }}{\text{ time between signals }}.$$ No radar gun can do better than that!
To summarize: $$\frac{dS}{dt}\approx \frac{\Delta S}{\Delta t},$$ where
• $\Delta S=S_{2} - S_{1}$ is the change of distance between the two cars,
• $\Delta t=t_2-t_1$ is the time between signals.
Now the question, is the reading of the radar gun $80$ m/h?
To get an idea of what can happen, consider this extreme example: what if you are just passing in front of the police car, like this?
It is conceivable that at time $t_1$ your car is the same distance from the intersection as it is past the intersection at time $t_2$. Then the $\Delta S=0$! So, the reading can be off by a lot...
These are the variables:
• $S$, the distance between the police car to yours.
• $P$, the distance between your car to the intersection.
• $t$, the time, the independent variable, also
• $D=40$, distance from the police car to the road.
Since $80$ m/h is your speed, $\frac{dP}{dt} = 80$. That's what the radar gun is meant to detect. But what does the radar measure in reality is $\frac{dS}{dt}$!
How good an approximation of the real velocity $\frac{dP}{dt}$ is the perceived velocity $\frac{dS}{dt}$? The spreadsheet contains a column of locations $P$ of your car (distances to the intersection), the next one is for the distance $S$ to the police car (plotted first), and finally the average rate of change of $S$.
As we can see, the approximation is best away from the intersection. But, within $75$ feet from the intersection, the reading will be less than $70$ mph!
Next, we establish a functional relation between the two via the Pythagorean Theorem: $$P^{2} + D^{2} = S^{2} \gets \text{These aren't numbers, but variables, i.e., functions.}$$ This connects $P$ and $S$, but not $\frac{dP}{dt}$ and $\frac{dS}{dt}$ yet. We differentiate equation with respect to $t$ to get there: \begin{aligned} \frac{d}{dt}\left(P^{2} + D^{2}\right) & = \frac{d}{dt}\left(S^{2}\right)\ \Longrightarrow \\ 2P\cdot \frac{dP}{dt} + 2D\underbrace{\frac{dD}{dt}}_{=0} &= 2S\cdot \frac{dS}{dt}\ \Longrightarrow \\ P\cdot\frac{dP}{dt} &= S\cdot \frac{dS}{dt}\ \Longrightarrow \\ \frac{dS}{dt} &= \frac{P}{S}\frac{dP}{dt}. \end{aligned} Thus, we finally have a relation between these functions. In fact, this is what the radar gun shows: $$\frac{dS}{dt} = \frac{P}{\sqrt{P^2+D^2}}\cdot 80.$$ We plot this relation below, to confirm our earlier conclusions:
Furthermore, we can simplify this relation: $$\cos \alpha = \frac{P}{S},$$ where $\alpha$ is the angle between the road ahead of you and the direction to the police car.
How does $\alpha$ change as you drive?
• Early: $\alpha$ is close to $0$, so $\cos \alpha$ close to $1$, and, therefore, $\frac{dS}{dt}$ is close to $80$.
• Then, as $\alpha$ increases, $\cos \alpha$ decreases toward $0$, and so does $\frac{dS}{dt}$.
• In the middle, we have $\alpha = \frac{\pi}{2}$, $\cos \alpha = 0$, $\frac{dS}{dt} = 0$.
• As $\alpha$ passes $\frac{\pi}{2}$, $\cos \alpha$ decreases to negative values, and so does $\frac{dS}{dt}$.
• Late: $\alpha$ approaches $\pi$, and $\cos \alpha$ approaches $1$, and, therefore, $\frac{dS}{dt}$ approaches $80$.
Conclusion: The radar gun always underestimates your speed: $$\left| \frac{dS}{dt} \right| < 80.$$ Unless, the police car is on the road! In that case, what can you do to “improve” the reading? What do you want $\alpha$ to be -- as large as possible!
## 9 The derivative of the inverse function
Let's recall from Chapter 3 that for a given one-to-one and onto function $y=f(x)$, its inverse is the function, $x=f^{-1}(y)$, that satisfies $$f^{-1}(y)=x \text{ if and only if }f(x)=y.$$ An idea is that a function and its inverse represent the same relation:
• $x$ and $y$ are related when $y=F(x)$, or
• $x$ and $y$ are related when $x=F^{-1}(y)$.
For example, these are pairs of functions inverse to each other: $$\begin{array}{rl} y=x+2& \text{ vs. } & x=y-2,\\ y=3x&\text{ vs. } & x=\frac{1}{3}y,\\ y=x^2&\text{ vs. } & x=\sqrt{y} \quad\text{ for }x,y\ge 0,\\ y=e^x&\text{ vs. } & x=\ln y \quad\text{ for }y > 0. \end{array}$$
Can we express the derivative of the inverse of a function in terms of the derivative of the function?
Let's recall that the inverse “undoes” the effect of the function. The idea applies especially well to the interpretation of the functions as transformations. Indeed, what is the meaning of the derivative of a transformation? It is its stretch ratio. Now, if $f$ stretches the $x$-axis at the rate of $k$ (at x=a$), then$f^{-1}$shrinks the$y$-axis at the rate of$k$(at$b=f(a)$); i.e., $$\frac{df^{-1}}{dy}(b)=\frac{1}{\frac{df}{dx}(a)}.$$ It's the reciprocal! We can also guess this relation from the following simple picture: As the$xy-plane is flipped about the diagonal, this is what happens: $$\text{slope of }f=\frac{\text{ change of }y}{\text{ change of } x}=\frac{A}{B}=\frac{1}{B/A}=\frac{1}{\text{slope of }f^{-1}}.$$ Now the algebra. We will need the following, algebraic, property of the inverse presented in Chapter 3: $$\begin{array}{lll} f^{-1}(f(x))=x\quad\text{for all }x;\\ f\left(f^{-1}(y) \right) = y \quad\text{for all }y. \end{array}$$ Here is a flowchart representation of this idea: $$\begin{array}{ccccccccccccccc} x & \mapsto & \begin{array}{|c|}\hline\quad f \quad \\ \hline\end{array} & \mapsto & y & \mapsto & \begin{array}{|c|}\hline\quad f^{-1} \quad \\ \hline\end{array} & \mapsto & x &(\text{same }x);\\ y & \mapsto & \begin{array}{|c|}\hline\quad f^{-1} \quad \\ \hline\end{array} & \mapsto & x & \mapsto & \begin{array}{|c|}\hline\quad f \quad \\ \hline\end{array} & \mapsto & y &(\text{same }y). \end{array}$$ Example. Suppose we want to find the derivative of the logarithm. We'll use only its definition via the exponential function, as follows. We differentiate this equation (that amounts to the definition of the logarithm) of functions: $$e^{\ln x} = x.$$ The flow chart below shows the dependencies: $$x \mapsto u = \ln x \mapsto y =e^{u}$$ By CR, we have: \begin{aligned} (e^{\ln x})' & = (\ln x)'\cdot e^{u} \\ & = (\ln x)' e^{\ln x} \\ & = (\ln x)' x &=(x)'=1. \end{aligned} Therefore, $$(\ln x)' = \frac{1}{x},$$ wheneverx > 0$.$\square$Similarly, we can find the derivatives of$\sin^{-1} x$,$\cos^{-1} x$, etc. Let's find the general formula instead. We differentiate the equation: $$f^{-1}(f(x))=x.$$ Then, by CR, we have $$\frac{\Delta f^{-1}}{\Delta y}\frac{\Delta f}{\Delta x}=1 \text{ and } \frac{d f^{-1}}{dy}\frac{df}{dx} = 1.$$ The other equation produces the same result! We have proven the theorem below. We just need to be careful with the variables, as follows. Theorem (Inverse Rule). (A) The difference quotient of the inverse of a function is found as the reciprocal of the its difference quotient; i.e., for any function$f$defined at two adjacent nodes$x$and$x+\Delta x$of a partition with$f(x)\ne f(x+\Delta x)$so that its inverse function$f^{-1}$is defined at the two adjacent nodes$f(x)$and$f(x+\Delta x)$of a partition, we have the difference quotients (defined at the secondary nodes$c$and$q$within these edges of the two partitions respectively) satisfy: $$\frac{\Delta f^{-1}}{\Delta y}(q)= \frac{1}{\frac{\Delta f}{\Delta x}(c)}.$$ (B) For any one-to-one function$y=f(x)$differentiable at$x=a$, its inverse$x=f^{-1}(y)$is differentiable at$b=f(a)$and we have: $$\frac{df^{-1}}{dy}(f(a))= \frac{1}{\frac{df}{dx}(a)},$$ or $$\frac{df^{-1}}{dy}(b)= \frac{1}{\frac{df}{dx}(f^{-1}(b))}.$$ The formulas in the Lagrange notation are as follows: $$\left( f^{-1} (f(a)) \right)' =\frac{1}{f'(a)}$$ and $$\left( f^{-1} (b) \right)' =\frac{1}{f'(f^{-1}(b))}.$$ Example. Find$(\sin^{-1} y)'$. There is no formula for this function, but its meaning is (for$-\pi/2\le x\le \pi/2$) as follows: $$y = \sin x, \text{ or } x = \sin^{-1}y.$$ Since$(\sin x)'=\cos x$, we conclude: $$(\sin^{-1} y)' =\frac{1}{\cos x}= \frac{1}{\cos(\sin^{-1}y)}.$$ That may be the answer, but it's too cumbersome and should be simplified. We need express$\cos x$in terms of$\sin x$, which is$y. By the Pythagorean Theorem $$\sin^{2} x + \cos^{2} x = 1,$$ we have \begin{aligned} \cos x & = \sqrt{1 - \sin^{2} x} \\ & = \sqrt{ 1 - y^{2}}. \end{aligned} Therefore, $$(\sin^{-1} y)'=\frac{1}{\sqrt{ 1 - y^{2}}}.$$\squareWe can apply the theorem to other trigonometric functions. These are the results: \begin{aligned} (\sin^{-1}x)' & = \frac{1}{\sqrt{1 - x^{2}}}; \\ (\cos^{-1}x)' & = -\frac{1}{\sqrt{1 - x^{2}}}; \\ (\tan^{-1}x)' & = \frac{1}{1+x^{2}}. \end{aligned} Exercise. Prove the formulas. Exercise. Since(\sin^{-1}x)' = -(\cos^{-1}x)'$, does it mean that$\sin^{-1}x = -\cos^{-1}x$? We can re-write the Inverse Rule in the Leibniz notation: $$\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}\text{ or }\frac{dx}{dy}\frac{dy}{dx}=1.$$ Then the derivatives of inverses are the reciprocals of each other. Even though the derivatives aren't fractions, the difference quotients, i.e., the slopes of the secant lines, are: Then the formula follows from these limits: $$\begin{array}{lll} \frac{\Delta x}{\Delta y}&\cdot& \frac{\Delta y}{\Delta x} &=1\\ \quad \downarrow&&\quad \downarrow\\ \ \frac{dx}{dy}&\cdot&\ \frac{dy}{dx}&=1& \text{ as } \Delta x\to 0,\ \Delta y\to 0. \end{array}$$ The fact that $$\Delta x\to 0 \Longrightarrow\ \Delta y\to 0$$ follows from the continuity of$f$. Furthermore, if we concentrate on a single point$(a,b)$, where$b=f(a)$, on the graph of$y=f(x)$and its tangent line, the derivatives $$\frac{dy}{dx}\Bigg|_{x=a} \text{ and } \frac{dx}{dy}\Bigg|_{y=b}$$ are indeed fractions and the reciprocals of each other: Theorem (Reciprocal Powers). For any positive integer$n$we have: $$\frac{ d y^{\frac{1}{n}} }{dy}=\frac{1}{n} y^{ \frac{1}{n}-1 }.$$ Proof. The inverse of$x=y^{\frac{1}{n}}$is$y=x^n$. Therefore, $$\begin{array}{lll} \frac{ d y^{\frac{1}{n}} }{dy}&=\frac{1}{\frac{d x^{n}}{dx}} &=\frac{1}{nx^{n-1}} &=\frac{1}{n\left( y^{\frac{1}{n}} \right)^{n-1}}&=\frac{1}{n y^{\frac{n-1}{n}} } &=\frac{1}{n}y^{\frac{1}{n}-1}. \end{array}$$$\blacksquare$Theorem (Rational Powers). For any positive integers$n$and$m$we have: $$\frac{ d y^{\frac{m}{n}} }{dy}=\frac{m}{n} y^{ \frac{m}{n}-1 }.$$ Exercise. Prove the theorem. ## 10 Reversing differentiation We have encountered the following question several times by now: when we know the velocity at every moment of time, how do we find the location? The question applies equally to the velocity acquired from the location as its difference quotient or as its derivative. We need to “reverse” the effect of differentiation on a function. But let's start with an even simpler problem: if we know the displacements during each of the time periods, can we find our location? Just add them together to find the total displacement! This is about the difference. Suppose we have a function$y=p(x)$defined at the secondary nodes,$c$, of a partition. How do we find a function$y=f(x)$defined at the nodes,$x$, of the partition so that$g$is its difference: $$\Delta f(c)=p(c)?$$ In other words, we need to solve this equation for$f$: $$\Delta f=p.$$ Suppose this function$g$is known but$f$isn't except for one, initial, value:$y_0=f(a)$. Then the above equation becomes: $$\Delta f(c_1)=f(x_0+\Delta x_1)-f(x_0)=p(c_1),$$ and we can solve it: $$f(x_1)= f(x_0+\Delta x_1)=f(x_0)+p(c_1).$$ We continue in this manner and find the rest of the values of$f$: $$f(x_{k+1})= f(x_k+\Delta x_k)=f(x_k)+p(c_k).$$ This formula is recursive: we need to know the last value of$f$in order to find the next. Though not an explicit formula, the solution is very simple! Now, the difference quotient. Suppose we have a function$y=g(x)$defined at the secondary nodes,$c$, of a partition. How do we find a function$y=f(x)$defined at the nodes,$x$, of the partition so that$g$is its difference quotient: $$\frac{\Delta f}{\Delta x}(c)=g(c)?$$ In other words, we need to solve this equation for$f$: $$\frac{\Delta f}{\Delta x}=g.$$ We follow exactly the process above. Suppose this function$g$is known but$f$isn't except for one, initial, value:$y_0=f(a)$. Then the above equation becomes: $$\frac{\Delta f}{\Delta x}(c_1)=\frac{f(x_0+\Delta x_1)-f(x_0)}{\Delta x_1}=g(c_1),$$ and we can solve it: $$f(x_1)= f(x_0+\Delta x_1)=f(x_0)+g(c_1)\Delta x_1.$$ We continue in this manner and find the rest of the values of$f$: $$f(x_{k+1})= f(x_k+\Delta x_k)=f(x_k)+g(c_k)\Delta x_k.$$ This formula is also recursive, but, within this limitation, the problem of reversing the effect of the difference quotient is solved! These two problems are similar to the one of finding the inverse of a function. This is how inverse functions appear in algebra; they come from solving equations, for$x$: $$\begin{array}{llllll} x^{2} & = 4 & \Longrightarrow& x = 2 & \text{ via } \sqrt{\ \cdot\ }; \\ 2^{x} & = 8 & \Longrightarrow& x = 3 & \text{ via } \log_{2}(\cdot ); \\ \sin x & = 0 & \Longrightarrow& x = 0 & \text{ via } \sin^{-1}(\cdot ), \text{ etc.} \end{array}$$ Now, what if we know the result of differentiation and want to know where it came from? We have just discovered that the inverse of the difference is the sum, no surprise! There may be also some explicit formulas. For example, we can solve this equation, for$f$: $$\Delta f=(e^h-1)\cdot e^c.$$ This is the solution: $$f=e^x.$$ Similarly, we solve the equation: $$\frac{\Delta }{\Delta x}(f)=\frac{ \sin (h/2)}{h/2}\cdot\cos c.$$ by $$f=\sin x.$$ This is called anti-differentiation. What about the derivative? Because it not a fraction but a limit of a fraction, there is no formula, even recursive. Some particular cases are considered in Chapter 9. The above approach still applies. We solve equations with respect to a variable function; for example: $$\begin{array}{llllll} f' & = 2x & \Longrightarrow& f &= x^{2}; \\ f' & = \cos x & \Longrightarrow& f &= \sin x ;\\ f' & =e^x & \Longrightarrow& f &= e^x. \end{array}$$ Example. The importance of this “inverse” problem stems from the need to find location from velocity or velocity from acceleration. For example, this is what we derive from our experience with differentiation. For the motion of a free fall, we have the following for the horizontal component: • the acceleration is zero$\Longrightarrow$the velocity is constant$\Longrightarrow$the location is a linear function. And for the vertical component, we have: • the acceleration is constant$\Longrightarrow$the velocity is a linear function$\Longrightarrow$the location is a quadratic function.$\square$We illustrate the idea with a diagram: $$\newcommand{\ra}[1]{\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{ccccccccccccccc} x^2 & \mapsto & \begin{array}{|c|}\hline\quad \tfrac{d}{dx} \quad \\ \hline\end{array} & \mapsto & 2x; \\ 2x & \mapsto & \begin{array}{|c|}\hline\quad \left( \tfrac{d}{dx}\right)^{-1} \quad \\ \hline\end{array} & \mapsto & x^2 ... \end{array}$$ ... are there any others? Yes,$(x^{2} + 1)' = 2x $. And more: $$\begin{array}{cclc} &&x^2+1\\ &\nearrow&\\ 2x&\to&x^2\\ &\searrow\\ &&x^2-1 \end{array}$$ As a function -- a function of functions --$\tfrac{d}{dx}\$ isn't one-to-one!
Exercise. We can make any function one-to-one by restricting its domain. How? |
# Multiples of 11
Created by: Team Maths - Examples.com, Last Updated: May 21, 2024
## Multiples of 11
Multiples of 11 are the result of multiplying 11 by any integer, producing a sequence of numbers like 11, 22, 33, and so on. In mathematics, these numbers are significant because they share common factors and divisors with 11. Understanding multiples helps in recognizing patterns and relationships in multiplication. Factors and divisors play a key role in identifying whether a number is a multiple of another. Multiples of 11 are integral to various mathematical concepts and problem-solving strategies.
## What are Multiples of 11?
Multiples of 11 are numbers obtained by multiplying 11 with any integer, such as 11, 22, 33, and so on. These numbers share common factors and divisors with 11, highlighting important mathematical patterns.
Prime factorization of 11: 11 = 1 × 11 First 10 multiples of 11: 11, 22, 33, 44, 55, 66, 77, 88, 99, 111.
Table of 11
## Important Notes
### Step 1: Understanding Multiples
Definition: A multiple of a number is the product of that number and any integer. For instance, multiples of 11 are obtained by multiplying 11 with integers.
### Step 2: Identifying the Pattern
Pattern Recognition: Multiples of 11 follow a simple arithmetic progression. The sequence starts at 11 and increases by 11 each time. Here are the first few multiples of 11:
• 11×1 = 11
• 11×2 = 22
• 11×3 = 33
• 11×4 = 44
• 11×5 = 55
### Step 3: List of Initial Multiples of 11
First 10 Multiples of 11:
• 11
• 22
• 33
• 44
• 55
• 66
• 77
• 88
• 99
• 110
### Step 4: Recognizing Properties
Properties of Multiples of 11:
• Alternating Sum Rule: A number is a multiple of 11 if the difference between the sum of its digits in odd positions and the sum of its digits in even positions is a multiple of 11 (including 0).
• Evenly Spaced: The difference between any two consecutive multiples of 11 is always 11.
• Divisibility: A number is divisible by 11 if it fits the alternating sum rule.
### Step 5: Practical Applications
Uses of Multiples of 11:
• Mathematical Problems: Frequently used in problems involving sequences and series.
• Daily Life: Used in scenarios involving periodic events or repeating patterns, such as time schedules (e.g., buses arriving every 11 minutes).
• Coding and Algorithms: Helpful in creating checksums or hash functions that ensure data integrity.
## Examples on Multiples of 11
### Example 1: Checking if a Number is a Multiple of 11
Problem: Determine if 121 is a multiple of 11.
Solution:
• Apply the alternating sum rule. For 121:
• Sum of digits in odd positions: 1+1 = 2
• Sum of digits in even positions: 2
• Difference: 2−2 = 0
Since the difference is 0, 121 is a multiple of 11.
Verification: 121 = 11×11
### Example 2: Finding the 15th Multiple of 11
Problem: Calculate the 15th multiple of 11.
Solution:
• Use the formula: nth multiple 11 = 11×n
• For n=15: 11×15 = 165
So, the 15th multiple of 11 is 165.
### Example 3: Using Multiples of 11 in Real Life
Scenario: A bus arrives at a stop every 11 minutes. If the first bus arrives at 7:00 AM, when will the 8th bus arrive?
Solution:
The time interval between buses is 11 minutes.
To find the arrival time of the 8th bus:
Multiply the number of intervals by 11: 11×(8−1) = 77 minutes.
Add 77 minutes to 7:00 AM.
7:00 AM+77 minutes = 8:17 AM
So, the 8th bus arrives at 8:17 AM.
## What are multiples of 11?
Multiples of 11 are numbers that can be expressed as 11 times an integer. For example, the first few multiples of 11 are 11, 22, 33, 44, and so on.
## How do you determine if a number is a multiple of 11?
To determine if a number is a multiple of 11, you can use the alternating sum rule. Subtract the sum of the digits in the odd positions from the sum of the digits in the even positions. If the result is a multiple of 11 (including 0), then the number is a multiple of 11.
## What is the 10th multiple of 11?
The 10th multiple of 11 is 110. This is calculated as 11 multiplied by 10 (11 x 10 = 110).
## Are all multiples of 11 also multiples of other numbers?
Yes, multiples of 11 can also be multiples of other numbers. For example, 22 is a multiple of both 11 and 2. However, not all multiples of other numbers are multiples of 11.
## Can negative numbers be multiples of 11?
Yes, negative numbers can be multiples of 11. For example, -11, -22, and -33 are all multiples of 11.
## What is the least common multiple (LCM) of 11 and another number?
The least common multiple (LCM) of 11 and another number is the smallest number that is a multiple of both 11 and that number. For instance, the LCM of 11 and 5 is 55.
## How are multiples of 11 used in real life?
Multiples of 11 are used in various real-life scenarios such as scheduling events, packaging items, financial planning, and arranging objects in rows. For example, a bus arriving every 11 minutes creates a schedule based on multiples of 11.
## Are multiples of 11 always odd or even?
Multiples of 11 can be both odd and even. For example, 11 and 33 are odd multiples, while 22 and 44 are even multiples.
## What is the sum of the first five multiples of 11?
The sum of the first five multiples of 11 (11, 22, 33, 44, and 55) is 165. This is calculated as 11+22+33+44+55 = 165.
## How can you generate a sequence of multiples of 11?
You can generate a sequence of multiples of 11 by starting with 11 and repeatedly adding 11 to the previous number. For example, starting with 11, the sequence is 11, 22, 33, 44, and so on.
Text prompt |
## Basic College Mathematics (10th Edition)
-4 (t + 2) = 12 Step 1 : Use the distributive property on the left side of the equation. -4(t + 2) = -4*t + -4*2 or -4t -8 Now the equation looks like this. -4t - 8 = 12 Step 2: Combine like terms. Check the left side of the equation. There are no like terms. Check the right side. No like terms there either, so go on to Step 3. Step 3: add 8 to both sides to get the variable term by itself on the right side -4t - 8 + 8 = 12 + 8 -4t = 20 Step 4 : Divide both sides by -4. $\frac{-4t}{-4}$ = $\frac{20}{-4}$ t= -5 Step 5: Check. Go back to the original equation -4t - 8 = 12 Replace t with -5 -4*-5 - 8 = 12 20 - 8 = 12 12 = 12 True, so -5 is the correct solution |
Prime and Composite Numbers
# Prime and Composite Numbers
## Prime Numbers
Definition: An integer $p > 1$ is said to be Prime if the only divisors of $p$ are $\pm 1$ and $\pm p$.
For a long time it was debated whether the number $1$ should be classified as a prime number or not. The importance of $1$ being classified as a prime number was uninteresting and so the first prime number is officially $2$.
If $\mathbb{P}$ is the set of all prime numbers, then the first few primes in $\mathbb{P}$ are:
(1)
\begin{align} \quad \mathbb{P} = \{2, 3, 5, 7, 11, 13, ... \} \end{align}
We will now look at a very simple theorem that says the greatest common divisor between an integer $a$ and any prime $p$ is either $1$ or $p$ depending on whether $p$ divides $a$.
Theorem 1: If $a, p \in \mathbb{Z}$ and $p$ is a prime number then $\gcd (a, p) = 1$ if $p \not \mid a$ or $\gcd (a, p) = p$ if $p \mid a$.
• Proof: Since $p$ is prime, the only positive divisors of $p$ are $1$ and $p$ so $\gcd (a, p) = 1$ if $p \not \mid a$ or $\gcd (a, p) = p$ if $p \mid a$. $\blacksquare$
One important aspect of prime numbers is that every integer $n > 1$ is divisible by a prime number $p$ which we prove in the following lemma.
Lemma 1: If $n \in \mathbb{Z}$ and $n > 1$ then there exists a prime $p \in \mathbb{P}$ such that $p \mid n$.
Lemma 1 also holds if $n \in \mathbb{Z}$ and $n < 1$ and can be proven in an analogous manner. Furthermore, if $n = 0$ then every number divides $0$ and by extension, every prime $p \in \mathbb{P}$ divides $0$.
• Proof: Let $S$ be the set of positive divisors of $n$, that is:
(2)
\begin{align} \quad S = \{ d \in \mathbb{Z} : d \mid n \} \end{align}
• The set $S$ is nonempty since $1, n \in S$. If $1$ and $n$ are the only elements in $S$ then this implies that $n$ is prime and so $p = n$ divides $n$. If there exists more elements in $S$, then since $S \setminus \{1 \}$ is a nonempty subset of integers we have that by The Well-Ordering Principle of the Natural Numbers that a least element $d$ exists in $S \setminus \{ 1 \}$.
• There are two subcases to consider. If $d_1$ is prime then $p =d_1$ and $d_1 \mid n$. If $d_1$ is not prime then there exists a positive divisor $d_2$ of $d_1$, so $1 < d_2 < d_1 < n$. But this implies that $d_2 \in S \setminus \{ 1 \}$ and $d_2 < d_1$, which is a contradiction since $d_1$ is the least element in $S \setminus \{ 1 \}$. Therefore $d_1$ is actually prime and again, $d_1 \mid n$.
• Therefore if $n > 1$ then there exists a prime $p \in \mathbb{P}$ such that $p \mid n$. $\blacksquare$
## Composite Numbers
Definition: An integer $n \geq 1$ is said to be Composite if it is not prime, that is, there exists a $d \in \mathbb{Z}$, $1 < d < n$ such that $d \mid n$ OR $n = 1$.
For example, the numbers in the following set are composite:
(3)
\begin{align} \quad \{4, 6, 8, 9, 10, 12, ... \} \end{align}
One important aspect of composite numbers is that every nonnegative even number (except $2$) is composite since all of such numbers has $2$ as a nontrivial divisor. |
## Linear Dependence of Vectors
In this page linear dependence of vectors we are going to see how to check whether the given vectors are linearly dependent or independent.
Definition :
A system of vectors X₁,X₂,.........Xn are said to be linearly dependent,if at least one of the vectors is a linear combination of remaining vectors.Otherwise it is called linearly independent.
In other words we can say the system of vectors X₁,X₂,.........Xn are said to be linearly dependent,if there exists numbers λ₁, λ₂,............. λn in which at least one of them is non zero satisfying the equation λ₁ X₁ + λ₂ X₂ + ................ + λn Xn = 0.
There are three methods to test linear dependence or independence of vectors in matrix.
Procedure for Method I
Definition can be directly used to test linear dependence or independence of vectors in matrix.
Procedure for Method II
• First we have to write the given vectors as row vectors in the form of matrix.
• Next we have to use elementary row operations on this matrix in which all the element in the nth column below the nth element are zero.
• The row which is having every element zero should be below the non zero row.
• Now we have to count the number of non zero vectors in the reduced form. If number of non zero vectors = number of given vectors,then we can decide that the vectors are linearly independent. Otherwise we can say it is linearly dependent.
Procedure for Method III
• If the matrix formed by the given vectors as row vectors is the square matrix,then we have to find rank.
• If the rank of the matrix = number of given vectors,then the vectors are said to be linearly independent otherwise we can say it is linearly dependent.
1. Test whether the vectors (1,-1,1), (2,1,1) and (3,0,2) are linearly dependent.If so write the relationship for the vectors
Method 2 Method 3
2. Test whether the vectors (1,3,1), (-1,1,1) and (3,1,-1) are linearly dependent.If so write the relationship for the vectors
Method 1 Method 2 Method 3
3. Test whether the vectors (1, 3, 1), (-1, 1, 1) and (2, 6, 2) are linearly dependent.If so write the relationship for the vectors
Method 1 Method 2 Method 3
4. Test whether the vectors (1,1,1), (1,0,1) and (0,2,0) are linearly dependent.If so write the relationship for the vectors
Method 1 Method 2 Method 3
5. Test whether the vectors (1,1,1), (1,2,3) and (2,-1,1) are linearly dependent.If so write the relationship for the vectors
Method 1 Method 2 Method 3 |
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# Evaluate $\int {\dfrac{{{e^x}}}{{{e^x} + 1}}dx}$ ?
Last updated date: 29th Feb 2024
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Hint: First of all, we will add and subtract a number in the numerator, so that they get separated. After that, we will let the denominator as any variable (say t) and will convert dx into dt. Then after solving further we will re-put the values that we have left.
Complete Step by Step Solution:
Let us consider $\int {\dfrac{{{e^x}}}{{{e^x} + 1}}dx}$ as y
$\Rightarrow y = \int {\dfrac{{{e^x}}}{{{e^x} + 1}}dx}$
Now, we will add and subtract 1 in the numerator
$\Rightarrow y = \int {\dfrac{{{e^x} + 1 - 1}}{{{e^x} + 1}}dx}$
Now, we will separate it into two parts (first one will be the $\int {\dfrac{{{e^x} + 1}}{{{e^x} + 1}}dx}$ and the other one is $\int {\dfrac{{ - 1}}{{{e^x} + 1}}dx}$ )
$\Rightarrow y = \int {\dfrac{{{e^x} + 1}}{{{e^x} + 1}}dx + \int {\dfrac{{ - 1}}{{{e^x} + 1}}} } dx$
We can also write it as
$\Rightarrow y = \int {\dfrac{{{e^x} + 1}}{{{e^x} + 1}}dx} - \int {\dfrac{1}{{{e^x} + 1}}dx}$
$\Rightarrow y = \int {1dx} - \int {\dfrac{1}{{{e^x} + 1}}dx}$
As we know that $\int {1dx = x}$
Therefore, $\Rightarrow y = x - \int {\dfrac{1}{{{e^x} + 1}}dx}$
Let us consider $- \int {\dfrac{1}{{{e^x} + 1}}dx}$ as I
$\Rightarrow y = x + I$
And $I = - \int {\dfrac{1}{{{e^x} + 1}}dx}$ ……(i)
Let ${e^x} + 1 = t$ ……(ii)
Differentiating both sides of the above equation with respect to t
$\Rightarrow \dfrac{{d\left( {{e^x}} \right)}}{{dt}} + \dfrac{{d\left( 1 \right)}}{{dt}} = \dfrac{{d\left( t \right)}}{{dt}}$
On further simplification,
$\Rightarrow {e^x}\left( {\dfrac{{dx}}{{dt}}} \right) + 0 = 1$
$\Rightarrow {e^x}\left( {dx} \right) = dt$
$\Rightarrow dx = \dfrac{{dt}}{{{e^x}}}$ ……(iii)
Now, by putting the value of dx and ${e^x} + 1$ from (ii) and (iii) in (i), we get
$\Rightarrow I = - \int {\dfrac{1}{{t(t - 1)}}dt}$
Now, adding and subtracting t in the numerator
$\Rightarrow I = - \int {\dfrac{{1 + t - t}}{{t(t - 1)}}dt}$
We can rewrite the above equation as
$\Rightarrow I = - \int {\dfrac{{t - \left( {t - 1} \right)}}{{t(t - 1)}}dt}$
Now, we will separate the above equation into two parts (first one will be the $- \int {\dfrac{t}{{t\left( {t - 1} \right)}}dt}$ and the other one is $- \int {\dfrac{{ - \left( {t - 1} \right)}}{{t\left( {t - 1} \right)}}dt}$ )
$\Rightarrow I = - \int {\dfrac{t}{{t\left( {t - 1} \right)}}dt - \int {\dfrac{{ - \left( {t - 1} \right)}}{{t\left( {t - 1} \right)}}dt} }$
$\Rightarrow I = - \int {\dfrac{1}{{t - 1}}dt - \int {\dfrac{{ - 1}}{t}dt} }$
As we know that $\int {\dfrac{1}{x}dx = \ln \left( x \right)}$
Hence $I = - \ln \left( {t - 1} \right) + \ln \left( t \right)$
Now, putting the value of t from (ii)
$\Rightarrow I = - \ln \left( {{e^x} + 1 - 1} \right) + \ln \left( {{e^x} + 1} \right)$
We can also rewrite the above equation as
$\Rightarrow I = \ln \left( {{e^x} + 1} \right) - \ln \left( {{e^x}} \right)$
As $y = x + I$
As we know $\ln \left( {{e^x}} \right) = x$
Therefore, $y = x + \ln ({e^x} + 1) - x$
$\Rightarrow y = \ln ({e^x} + 1)$
Note:
While doing these types of problems, strictly take care of dx and dt. When you let a variable (say x) to another variable (say t) then take care that you do not forget to change the dx into dt. And in the last, do not forget to put the given variables. Do not leave the answer in those variables which you have taken (let). |
# How do you divide a polynomial by a binomial of the form $ax^2+b$, where $a$ and $b$ are greater than one?
I came across a question that asked me to divide $-2x^3+4x^2-3x+5$ by $4x^2+5$. Can anyone help me?
#### Solutions Collecting From Web of "How do you divide a polynomial by a binomial of the form $ax^2+b$, where $a$ and $b$ are greater than one?"
As mentioned, long division works fine. But general modular techniques also work quickly, and require no specialized knowledge. For example, here modulo $\rm\:4\:x^2 + 5\:,\:$ we have $\rm\ {\color{red}{4\:x^2}} \equiv -5\$ so $$\rm\:2\ f(x)\ =\ (-{\color{red}{4\:x^2}}-6)\ x+2\cdot {\color{red}{\:4\: x^2}}+10\ \equiv\ (5-6)\ x -10 + 10\ \equiv\:\: -x$$
Therefore $\rm\ 2\ f(x)\ =\:\: -x + (a\:x+b)\ (4\:x^2+5)\$ so $\rm\ a = -1,\ b = 2\$ by comparing coef’s of $\rm\:x^3,\ x^0\:.$
Note $\$ I multipled $\rm f(x)$ by $2$ only to simplify the arithmetic (eliminate fractions).
The answer in no way depends on whether $a$ and $b$ are more than $1$ or less; the only fact about those that you need is that the polynomial you’re dividing by is not zero. You have this:
$$\begin{array}{ccccccccccccccccc} \\ 4x^2+5 & \big) & -2x^3 & + & 4x^2 & – & 3x & + & 5 \\ & & \end{array}$$
So ask what you need to multiply $4$ by to get $-2$. In other words, divide $-2$ by $4$. You get $-2/4=-1/2$. So write
$$\begin{array}{ccccccccccccccccc} & & \frac{-1}{2} x \\ \hline 4x^2+5 & \big) & -2x^3 & + & 4x^2 & – & 3x & + & 5 \\ & & \end{array}$$
Then multiply:
$$\begin{array}{ccccccccccccccccc} & & \frac{-1}{2} x \\ \hline 4x^2+5 & \big) & -2x^3 & + & 4x^2 & – & 3x & + & 5 \\ & & -2x^3 & & & -& \frac 52 x \end{array}$$
Then subtract:
$$\begin{array}{ccccccccccccccccc} & & \frac{-1}{2} x \\ \hline 4x^2+5 & \big) & -2x^3 & + & 4x^2 & – & 3x & + & 5 \\ & & -2x^3 & & & -& \frac 52 x \\ \hline & & & & 4x^2 & – & \frac 12 x & + 5 \\ \hline \end{array}$$
The ask what you have to multiply $4x^2$ by to get $4x^2$, then multiply, then subtract…..
You can do long division with polynomials almost the same way you would for integers. For example, $4x^2+5$ can be multiplied by $-\frac{1}{2}x$ to get a leading term of $-2x^3$, so we might say that $4x^2+5$ goes into $-2x^3 + 4x^2 – 3x + 5$ about $-\frac{1}{2}x$ times. As with long division, we then multiply $4x^2+5$ by $-\frac{1}{2}x$ and subtract the result from $-2x^3 + 4x^2 – 3x + 5$; then repeat the process until we can’t anymore. If there’s a remainder at the end, we divide it by the divisor and add that to the end of the result (again, similar to long division).
In this case you should verify that the result is $-\frac{1}{2}x + 1 + \frac{-x}{8x^2+10}$.
Nobody seems to have posted the synthetic division route; I’ll include it here for completeness.
We first monicize the divisor (i.e., set things up such that the divisor has leading coefficient $1$); thus, consider
$$\frac{-\frac12x^3+x^2-\frac34x+\frac54}{x^2+\frac54}$$
and set up the array
$$\begin{array}{r|cc|cc} &-\frac12&1&-\frac34&\frac54\\\hline 0&\times& & &\times\\ -\frac54&\times&\times& & \\\hline & & & & \end{array}$$
Note that the coefficients in the leftmost column are the negated coefficients of the terms with degree less than the degree of the polynomial. The second dividing line (can’t do dashed lines in an array environment, sorry) indicates that we expect the quotient to be linear ($3-2=1$). The $\times$ entries indicate the sections of the array that shouldn’t be filled.
This generalized synthetic division proceeds like this:
$$\begin{array}{r|cc|cc} &-\frac12&1&-\frac34&\frac54\\\hline 0&\times& & &\times\\ -\frac54&\times&\times& & \\\hline &-\frac12& & & \end{array}$$
$$\begin{array}{r|cc|cc} &-\frac12&1&-\frac34&\frac54\\\hline 0&\times&0& &\times\\ -\frac54&\times&\times&\frac58& \\\hline &-\frac12& & & \end{array}$$
(multiply the bottom of the first column by the coefficients in the leftmost column, and fill the array diagonally)
$$\begin{array}{r|cc|cc} &-\frac12&1&-\frac34&\frac54\\\hline 0&\times&0& &\times\\ -\frac54&\times&\times&\frac58& \\\hline &-\frac12&1& & \end{array}$$
$$\begin{array}{r|cc|cc} &-\frac12&1&-\frac34&\frac54\\\hline 0&\times&0&0&\times\\ -\frac54&\times&\times&\frac58&-\frac54\\\hline &-\frac12&1& & \end{array}$$
(multiply bottom of the second column by the coefficients in the leftmost column and fill diagonally)
$$\begin{array}{r|cc|cc} &-\frac12&1&-\frac34&\frac54\\\hline 0&\times&0&0&\times\\ -\frac54&\times&\times&\frac58&-\frac54\\\hline &-\frac12&1&-\frac18&0 \end{array}$$
(add numbers in third and fourth columns)
We thus have the result
$$\frac{-\frac12x^3+x^2-\frac34x+\frac54}{x^2+\frac54}=-\frac12 x+1+\frac{-\frac18x+0}{x^2+\frac54}=-\frac12 x+1+\frac{-\frac12x}{4x^2+5}$$
I will try this way:
Since you are dividing a 3rd degree polynomial by a 2nd degree polynomial, WLOG, we may assume
$$-2x^3+4x^2-3x+5=(4x^2+5)(ax+b)+cx+d\quad(1)$$
Now, comparing the coefficients of $x^3$ and $x^2$ readily give $a=-\frac{1}{2}$ and $b=1$. Comparing coefficients of $x$, we have $5a+c=-3\Rightarrow c=-\frac{1}{2}$. Finally comparing the constant term, we get $5b+d=5\Rightarrow d=0$. |
# Vector components
##### Intros
###### Lessons
1. Introduction to vector components:
• What are x and y components?
• How to break a vector into its components
##### Examples
###### Lessons
1. Break a displacement into vector components
A boat sails 33.0 km at 55.0° north of west. What are the north and west components of its displacement?
1. Use components of velocity for calculations
A plane that is taking off travels at 67.0 m/s at a 15.0° angle above the horizontal.
1. How long does it take the plane to reach its cruising altitude of 12.0 km?
2. If the plane is headed east, how far east has it travelled in this time?
1. Add two 2D vectors using components
Solve the equation $\vec{v}_{1} + \vec{v}_{2} = \vec{v}_{res}$ by breaking $\vec{v}_{1}$ and $\vec{v}_{2}$ into their x and y components.
###### Topic Notes
In this lesson, we will learn:
• What are x and y components?
• How to break a vector into its components
• Problem solving with vector components
Notes:
• Components of a vector are other vectors that add up tip-to-tail give you the original vector.
• The x and y components of a vector are the components that are pointed directly in the x and y directions, respectively, and are useful for solving problems.
• The x and y components can be found with trigonometry, since they always form a right triangle with the original vector.
Right Triangle Trigonometric Equations
$\sin(\theta) = \frac{opp.}{hyp.}$
$\cos(\theta) = \frac{adj.}{hyp.}$
$\tan(\theta) = \frac{opp.}{adj.}$
$a^{2}+b^{2}=c^{2}$ (Pythagorean theorem)
$\theta$: angle, in degrees (°)
$opp.$: side opposite angle
$adj.$: side adjacent angle
$hyp.$: hypotenuse of triangle (longest side, side opposite 90° angle)
$a$ and $b$: non-hypotenuse sides of triangle
$c$: hypotenuse of triangle |
TWO VERTICES OF AN ISOSCELES TRIANGLE IS (2,0) AND (2,5).FIND IT’S THIRD VERTEX.. NO SPAMMING SPAMMING MAY LEAD TO DELETION OF UR ACCOUNT..
Question
TWO VERTICES OF AN ISOSCELES TRIANGLE IS (2,0) AND (2,5).FIND IT’S THIRD VERTEX.. NO SPAMMING SPAMMING MAY LEAD TO DELETION OF UR ACCOUNT.. THANK YOU!!
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1 month 2021-10-27T02:44:49+00:00 2 Answers 0 views 0
1. Step-by-step explanation:
hola sis here is ur ans
Let ΔABC be the isosceles triangle, the third vertex be C(a,b).
and A(2,0) and B(2,5)
Let AC and BC be equal sides of the triangle
By distance formula we have,answr
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Two vertices of an isoscele…
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Two vertices of an isosceles triangle are (2,0) and (2,5). Find the third vertex if the length of the equal sides is 3.
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Let ΔABC be the isosceles triangle, the third vertex be C(a,b).
and A(2,0) and B(2,5)
Let AC and BC be equal sides of the triangle
By distance formula we have,
D=
(x
2
−x
1
)+(y
2
−y
1
)
∴AB=
(2−2)
2
+(5−0)
2
=
25
=5
Now,
AC=3 [length of equal sides=3 ]
AC
2
=3
2
=9……….(a)
but by distance formula,
AC
2
=
(a−2)
2
+b
2
………..(b)
Combining (a) and (b)
⇒(a−2)
2
+b
2
=9
⇒a
2
+4−4a+b
2
=9
⇒a
2
+b
2
−4a=5 ———- (1)
Consider BC
BC
2
=
(a−2)
2
+(b−5)
2
but BC=3
∴(a−2)
2
+(b−5)
2
=9
⇒a
2
+4−4a+b
2
+25−10b=9
⇒a
2
+b
2
−4a−10b+20=0 ——- (2)
⇒−10b+20+5=0 ———(from 1 )
⇒10b=25
⇒b=
10
25
=
2
5
Now, a
2
+
4
25
−4a=5
⇒a
2
−4a=
4
−5
⇒a
2
−4a+
4
5
=0
⇒a=
2
16−4∗1∗
4
5
=
2
11
=
2
2+
11
,
2
2−
11
∴ Coordinate of third vertex (a,b) = (
2
2+
11
,
2
5
) and (
2
2−
11
,
2
5
) |
AlgOneLectOne
# When we carry out that division incompletely to form
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larger numerator by a smaller denominator. When we carry out that division incompletely to form a whole number quotient plus a remainder portion of the numerator (yet to be divided) we have produced the mixed number (whole number plus a proper fraction). Saxon Lesson Learning Concept Statement Number Concept Name Description (answers what is this concept?) Question (among many) Correct Answer 1 4 Mixed number notation m l/k means m + l/k What is a mixed number? A mixed number is a whole number plus a fraction with the plus sign "understood" Click me for video > QuickTimeª and a decompressor are needed to see this picture.
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9 Adding Mixed Numbers To add mixed numbers we keep in mind that each mixed number is itself two numbers: a whole number and a fraction to which it is added. Thus adding mixed numbers generally means we are adding four numbers: two are whole numbers and two are fractions. Since addition can proceed in any order we can separately add the fractions together and the whole numbers together to give us the procedure for adding mixed numbers. The result is then a mixed number. Sometimes the fraction is an improper fraction whose value exceeds one in which case further simplification is required. We cover that next. Saxon Lesson Learning Concept Statement Number Concept Name Description (answers what is this concept?) Question (among many) Correct Answer 1 5 Addition of mixed numbers Add fractions and add whole numbers. Then simplify fraction. m l/k + n i/j = ? (m + n) + (jl + ki)/kj _____________________________________________________________________________________________________ Saxon Lesson Learning Concept Statement Number Concept Name Description (answers what is this concept?) Question (among many) Correct Answer 1 6 Simplify fraction Convert to mixed number if fraction exceeds 1 m/n = ? If m > n Int + j/n where j < n A mixed number that has an improper fraction is not really following the rule for a mixed number in which the fractional part is preferred to be a proper fraction- less than one. When the mixed number resulting from the addition of mixed numbers produces an improper fraction, we convert the improper fraction to its own mixed notation. In that conversion a whole number and a proper fraction result. The whole number generated from the improper fraction is then added to the original whole number to produce the sought after mixed number result. Click me for video > QuickTimeª and a decompressor are needed to see this picture.
10 More on Mixed Numbers So for example, let’s consider the sum of: 2 7/8 + 3 5/9 =. After writing these with common denominators, we have instead 2 63/72 + 3 40/72 = Performing the indicated additions gives us: 5 103/72 = Then reducing the improper fraction to its own mixed form yields, 5 + 1 31/72 = 6 31/72, where we have shown the final mixed result containing the whole number 6 and the proper fraction 31/72. ___________________________________________________________________________________ Click me for video ^ QuickTimeª and a decompressor are needed to see this picture.
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11 Subtracting Mixed Numbers Subtracting mixed numbers is very similar to the addition of them. Here we subtract the fractional parts and the whole parts separately. If the two numbers each had proper fractional parts, the resulting fractional part will also be a proper fraction.
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# ICSE Number Line Class 6 Chapter 18 Data Handling Exercise 18E Solutions
## ICSE Number Line Class 6 Chapter 18 Data Handling Exercise 18E Solutions
This post is created to help all the ICSE Class 6 students for the Solutions of Number Line Class 6 Mathematics Book, Chapter 18 Data Handling Exercise 18E. Here students can easily find step by step solutions of all the problems for Data Handling Exercise 18E. Here students all find exercise wise proper solutions for every problems. Here all the problem are solved with easy to understand methods so that all students can understand easily.
Chapter 18 – Data Handling
Data Handling – Exercise 18E all Questions Solution
Question no – (1)
Solution :
Arranging in ascending order
82, 85, 85, 87, 87, 87, 90, 90, 90
No of observation = 9
(9 + 1/2)th position
Median = 10/2th position
= 5th position
= 87
Question no – (2)
Solution :
In ascending order
8, 10, 10, 12, 12, 15
No of observation = 6 (even)
So,
Median = x/2th position + (x/2 + 1) position/2
= 6/2 position + (6/2 + 1) position/2
= 3rd position + 4th position/2
= 10 + 12/2
= 11
Question no – (3)
Solution :
In ascending order
5000, 8000, 8000, 9000, 9000, 10000, 10000, 10000, 50000
No of observation = 9 (odd)
So, median = (x + 1/2)th position
= 9 + 1/2th position
= 5th position
= Rs 9000
Question no – (4)
Solution :
Mean = 7 + 6 + 5 + 6 + 5 + 7 + 6 + 5/8
= 47/8
= 5.875
In ascending order
5, 5, 5, 6, 6, 6, 7, 7
No of observation = 8
Median = (x/2th position + (x/2 + 1)th position)/2
= (4th position + (4 + 1)th position)/2
= 4th position + 5th position/2
= 6 + 6/2
= 6
No of mean and median are not same.
Multiple Choice Question
(i) On which day were the maximum number of parcels delivered?
Solution :
Maximum number of parcels delivered – (d) Thursday
(ii) On which day were the minimum number of parcels delivered?
Solution :
Minimum number of parcels delivered – (a) Saturday
(iii) How many more parcels were delivered on Tuesday than on Monday?
Solution :
Difference between Tuesday and Monday per parcels
= (9 – 7) × 10
= 2 × 10
= 20
So, The correct option is – (a) 20
(iv) What was the total number of parcels delivered over the whole week?
Solution :
Total parcels delivered over a week
(5 + 7 + 4 + 9 + 6 + 3) × 10
= 34 × 10
= 340
So, The correct option is – (c) 340
Updated: January 31, 2023 — 9:36 pm |
# Class 9 NCERT Solutions- Chapter 11 Constructions – Exercise 11.1
Last Updated : 15 Mar, 2022
### Question 1. Construct an angle of 90° at the initial point of a given ray and justify the construction.
Solution:
Steps of construction
1. Take a ray with initial point A.
2. Taking care center and same radius draw an Arc of a circle which intersect AB at C.
3. With C as Centre and the same radius, draw an arc intersecting the previous arc at E.
4. With E as Centre and the same radius draw an arc which intersects the arc drawn in step 2 at F.
5. With E as Centre and the same radius, draw another arc, intersecting the previous arc at G.
6. Draw the ray AG.
7. Then ∠BAG is the required angle 90°
Justification:
Join AE, CE, EF and AE, AF
AC = CE = AE [ by construction]
∴ ACE is an equilateral Triangle
⇒ ∠CAE = 60° —————–1
Similarly, AE = EF = AF
∴Triangle AEF is an equilateral Triangle
⇒ ∠EAF = 60°
Because AG bisects ⇒ ∠EAF
∴∠GAE = 1\2 = 30° = 30°————2
1+2
∴∠CAE + ∠GAE = 60°+30°
∠GAB=30°
### Question 2. Construct an angle of 45° at the initial point of a given ray and justify the construction.
Solution:
Step of Construction:
1. Take a ray AB with initial point A
2. Draw ∠BAF=90°
3. Taking C as Centre and radius more than draw an arc.
4. Taking G as Centre and the same radius as before, draw another arc.
5. Taking G as Centre and the same radius as before, draw another arc. Intersecting previous arc at H.
6. Draw the ray AH.
7. Then ∠BAH is the required angle of 45°
Justification:
Join GH and HC (construct)
In ∆ AHG and ∆ AHC
AH=AH………………[common]
AHG≅AHC [S.S.S]
∠HAG=∠HAC [C.P.C.T]
But ∠HAG+∠HAC=90
∠HAG=∠HAC=90\2=45
∴∠BAH=45
### Question 3. Draw the angles of the following measurement
i) 30°
Solution:
Step of construction
1. Draw a ray AB with initial point A.
2. With A as centre, draw an arc intersecting AB at c.
3. With c as centre and the same radius, draw another arc, intersecting the previously drawn arc at D.
4. Taking C and D as centre and with the radius more than 1\2 DC draws arcs to intersect each other at E.
5. Draw ray AE. ∠EAB is the required angle of 30.
ii) 22 ½°
Solution:
Steps of construction
1. Take a ray AB
2. Draw an angle ∠AB=90° on point A.
3. Bisect ∠CAB and draw ∠DAB=45°
4. Bisect ∠DAB and draw ∠EAB
5. ∠EAB is required angle of 22 ½°
iii) 15°
Solution:
Steps of construction
1. Take a ray AB.
2. Draw an arc on AB, by taking A a center, which intersect AB at c.
3. From C with the same radius draw another re which intersect the previous arc at D.
4. Join DA.
5. ∠DAB =60°
6. Bisect ∠DAB and draw angle EAB=30°
7. Bisect ∠EAB and draw ∠FAB
8. ∠FAB is the required angle.
### (i) 75°
Solution:
Steps of construction
1. Draw a ray AB with initial point A.
2. At point A draw an angle ∠CAB=90°
3. At point A draw ∠DAB=60°
5. ∠EAB=75° {∠EAB=∠EAD+∠DAB=15°+60°=75°}
### (ii) 105°
Solution:
Steps of construction
1. Draw a ray AB with initial point A.
2. At point A draw an angle ∠CAB=90°
3. At point A draw ∠DAB=120°
5. ∠EAB=75° {∠EAB=∠EAC+∠CAB=15°+90°=105°}
### (iii) 135°
Solution:
Steps of construction
1. Draw a ray AB with initial point A.
2. At point A draw an angle ∠CAB=120°
3. At point A draw ∠DAB=150°
4. Bisect ∠CAD, now ∠EAC=15°
5. ∠EAB=135° {∠EAB=∠EAC+∠CAB=15°+120°=135°}
### Question 5. Construct an equilateral triangle, given its side and justify the construction.
Solution:
Steps of construction
1. Draw a line segment of AB of a given length.
2. With A and B as centre and radius equal to AB draw arcs to intersect each other at c.
3. Join AC and BC.
Then ABC is the required equilateral triangle.
Justification:
AB=AC ……………. [by construction]
AB=BC ……………..[by construction]
AB=AC=BC
Hence, ∆ABC is required equilateral triangle.
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# How do you write 2 divided by 6?
## Introduction
Writing fractions can be a daunting task, especially when the fractions are not in their simplest form. Knowing how to convert fractions into their simplest form and then write them in a way that makes sense is essential for anyone who wants to understand fractions. In this blog post, we will look at the process of writing 2 divided by 6, which is an example of a fraction that is not in its simplest form. We will look at how to convert the fraction to its simplest form and then how to write it in an understandable way. By the end of this blog post, you will have a better understanding of how to write fractions, and you will be able to apply this knowledge to other fractions.
## Simplifying Fractions
When writing fractions, it is important to make sure that the fraction is in its simplest form. This means that the numerator and denominator should have no common factors and that the fraction should be written in a way that makes it easier to understand. To simplify a fraction, you need to find the Greatest Common Factor (GCF) of the numerator and denominator. The GCF is the largest number that can be divided into both the numerator and denominator. In the case of 2 divided by 6, the GCF is 2. This means that the fraction can be simplified by dividing both the numerator and denominator by 2. The simplified fraction is 1 divided by 3.
### Writing the Fraction
Now that we have the fraction in its simplest form, it is time to write it in an understandable way. There are two main ways to write fractions: in words and in symbols. When writing a fraction in words, you should use the word “over” to indicate division. For example, the fraction 2 divided by 6 can be written as “two over six.” When writing a fraction in symbols, you should use a slash (/) to indicate division. For example, the fraction 2 divided by 6 can be written as “2/6.”
### Improper Fractions
In some cases, a fraction may be written in an “improper” form. An improper fraction is a fraction where the numerator is larger than the denominator. For example, the fraction 6 divided by 2 is an improper fraction. To write an improper fraction in an understandable way, you should convert it to a “mixed number.” A mixed number is a combination of a whole number and a fraction. To convert an improper fraction to a mixed number, you need to divide the numerator by the denominator and then write the quotient as the whole number and the remainder as the fraction. For example, the fraction 6 divided by 2 can be written as 3 and 1/2.
### Decimals
In some cases, you may want to convert a fraction into a decimal. To convert a fraction into a decimal, you need to divide the numerator by the denominator. For example, the fraction 2 divided by 6 can be written as a decimal by dividing 2 by 6. The answer is 0.3333.
#### Conclusion
Writing fractions can be a difficult task, especially when the fractions are not in their simplest form. By following the steps outlined in this blog post, you should have a better understanding of how to write fractions, and you should be able to apply this knowledge to other fractions. Good luck!
How do you write 2 divided by 6?
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# Difference between revisions of "Limit"
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A limit is the behavior of a function as its inputs approach arbitrarily close to a given value.
Limits are written in the following form:
$\lim_{x \to a}f(x) = L$
The expression above states that when $x\,$ approaches arbitrarily close to $a\,$, the function $f(x)\,$ becomes arbitrarily close to the value $L\,$, which is called the limit.
## Informal Definition
We can examine the limit of a simple continuous function, $f(x)=x^2 \,$.
The limit at x=0 of $f(x) = x^2\,$ would pretty clearly be 0, since $f(0)=0$.
Indeed for this function, $\lim_{x \to a}f(x) = f(a) \,$. But this is a special case, in the majority of limits cannot be solved in this manner.
For a very different example; given
$f(x)=\left\{\begin{matrix} {x}^2, & \mbox{if }x\ne 0 \\ \\ 1, & \mbox{if }x=0. \end{matrix}\right.$ (as pictured below)
The limit of $f(x)\,$ because x approaches 0 is 0 (just as in $f(x)\,$), but $\lim_{x\to 0}f(x)\neq f(0)$; $f(x)$ is not continuous at $x = 0$ (as shown on the right).
In other cases a limit can fail to exist, as approaching the limit from different sides produces conflicting values.
Here we look at one such case:
$f(x)=\left\{\begin{matrix} {x}^2, & \mbox{if }x\ne 0 \\ \\ 1, & \mbox{if }x=0. \end{matrix}\right.$ (as pictured above)
## Rigorous Definition of Limit
This definition is more appropriate for 2nd year calculus students and higher.
The original statement $\lim_{x \to a}f(x) = L$ now means that given any $\varepsilon > 0$, a $\delta > 0 \,$ exists such that if $0 < |x-a| < \delta \,$, then $|f(x)-L|< \varepsilon$.
What this means to us is that for every
We use the definition as follows to prove for the function $f(x) = 3x-1 \,$, $\lim_{x \to 2}f(x) = 5$
Given any $\delta\,$, we choose a $\varepsilon\,$ such that $\delta \leq \varepsilon/3$
If $|x-2|<\delta\,$, we can derive $|f(x)-5|<3\delta\,$.
$3|x-2|<3\delta\,$
$|3x-6|<3\delta\,$
$|f(x)-5|<3\delta\,$
$3\delta \leq 3*\varepsilon/3 = \varepsilon\,$
Thus $\lim_{x \to 2}f(x) = 5$
## Properties of Limits
The limit of a sum of two functions is equal to the sum of the limits of the functions.
$\lim_{x \to a} (f(x)+g(x)) = \lim_{x \to a} f(x) + \lim_{x \to a} g(x)$
The limit of a difference between two functions is equal to the difference between the limits of the functions.
$\lim_{x \to a} (f(x)-g(x)) = \lim_{x \to a} f(x) - \lim_{x \to a} g(x)$
The limit of a product of two functions is equal to the product of the limits of the functions.
$\lim_{x \to a} (f(x)*g(x)) = \lim_{x \to a} f(x) * \lim_{x \to a} g(x)$
The limit of a quotient of two functions is equal to the quotient of the limits of the functions (assuming a non-zero denominator).
$\lim_{x \to a} (\frac{f(x)}{g(x)}) = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}$
## Ideas for the Future
- an interactive diagram in which changing the size of epsilon shows a corresponding delta, or something. |
# Recursive Formula
If t1, t2, t3,….,tn,… is a set of series or a sequence. Then a recursive formula for this sequence will be needed to compute all the previous terms and find the value of tn.
tn = tn-1
This formula can also be defined as Arithmetic Sequence Recursive Formula. As you can observe from the sequence itself, it is an arithmetic sequence, which includes the first term followed by other terms and a common difference, d between each term is the number you add or subtract to them.
A recursive function can also be defined for a geometric sequence, where the terms in the sequence have a common factor or common ratio between them. And it can be written as;
tn = r x tn-1
## Recursive Formula Examples
Example 1:
Let t1=10 and tn= 2tn-1+1
So the series becomes;
t1=10
t2=2t1+1=21
t3=2t2+1= 43
And so on…
Example 2: Find the recursive formula which can be defined for the following sequence for n > 1.
65, 50, 35, 20,….
Solution:
Given sequence is 65, 50, 35, 20,….
a1 = 65
a2 = 50
a3 = 35
a2 – a1 = 50 – 65 = -15
a3 – a2 = 35 – 50 = -15
Thus, a2 = a1 – 15
Similarly, a3 – a2 = -15
From this we can write the recursive formula as: an = an-1 – 15
Example 3: Calculate f(9) for the recursive series f(x) = 3. f(x – 2) + 4 which has a seed value of f(3) = 9.
Solution:
Given,
f(3) = 9
f(x) = 3.f(x – 2) + 4
f(9) = 3.f(9-2) + 4
= 3.f(7) + 5
f(7) = 3.f(7-2) + 4
= 3.f(5) + 4
f(5) = 3.f(5-2) + 4
= 3.f(3) + 4
Substituting f(3) = 9,
f(5) = 3(9) + 4 = 27 + 4 = 31
f(7) = 3(31) + 4 = 93 + 4 = 97
f(9) = 3(97) + 4 = 291 + 4 = 295
Register at BYJU’S to learn other mathematical topics in an interesting way. |
# The Mysterious Quotient: Understanding the Result of 100 Divided by 3
## The History of the Quotient
The concept of quotient dates back to ancient times when people had a need to 100 divided by 3 objects into equal portions. The word ‘quotient’ comes from the Latin term ‘quotiens,’ meaning “how many times.” In mathematics, the concept refers to obtaining an exact number of times one quantity is contained in another.
In ancient Greece and India, mathematicians used different methods for division, including multiplication by reciprocals, repeated subtraction, and using tables with pre-calculated results. However, these methods were often cumbersome and imprecise.
division notation was introduced by John Pell and William Oughtred. They used fractions to represent quotients in equations. Later on, Rene Descartes popularized this method further.
Today’s standard long division algorithm was developed in Europe during the Renaissance period. It allowed for more precise calculations without relying on memorization or pre-calculated results.
Understanding the history behind mathematical concepts like quotient can give us insight into how we arrived at our current level of understanding and appreciation for math as a whole!
## The Different Types of Quotients
When we talk about quotients, there are various types that exist. Understanding these different types is essential to ensure that we accurately calculate the correct result.
One type of quotient is the exact quotient. This means that when dividing two numbers, there will be no remainder left. For example, 10 divided by 2 equals exactly 5 with no remainder.
Another type of quotient is the rounded or approximate quotient. As the name suggests, this type involves rounding up or down to get an estimate of what the answer might be.
There’s also a mixed number quotient where fractions are involved in division. Here, we need to convert any improper fractions into mixed numbers and then proceed with division as usual.
There’s also a decimal quotient where instead of getting an answer in fraction form, we get it as a decimal value instead.
Knowing these different types of quotients can help us choose which method to use depending on our needs and requirements while solving problems involving division.
## How to Calculate the Quotient
Calculating the quotient is a fundamental operation in mathematics that involves dividing one number by another. To calculate the quotient, you need to understand the basic principles of division and be familiar with some simple techniques.
Firstly, you should identify the numbers that are being divided. The first number is called the dividend, while the second number is known as the divisor.
Lastly but most importantly keep practicing! Calculating quotients might seem daunting at first glance with practice and experience.
## The Results of 100 Divided by 3
When we divide 100 by 3, the result is a decimal number with an infinite amount of digits after the decimal point. This type of result is called a “repeating decimal” and is represented by placing a bar over the numbers that repeat.
The quotient of 100 divided by 3 can be expressed as 33.333… or simply as 33 with a remainder of one. The remainder indicates that there is still one more unit left to make another group of three.
This result may seem mysterious at first glance, but it’s actually quite common in mathematics. In fact, many fractions cannot be expressed as whole numbers and instead produce repeating decimals when divided.
In some cases, repeating decimals can be converted into fractions using algebraic techniques such as long division or simplification. However, this process can become complicated for longer sequences of repeating digits.
## Conclusion
Understanding the quotient is essential for anyone who wants to be proficient in mathematics. One example is dividing 100 by 3 which gives us a result that appears mysterious at first glance but can easily be understood with the right knowledge.
By understanding how division works and the concept of remainders, we can see that the result of 100 divided by 3 is not actually a mystery but rather a simple calculation with an interesting remainder. Moreover, this exercise has shown us how numbers can tell stories and reveal hidden patterns when examined more closely.
Therefore, let’s continue exploring the fascinating world of mathematics and keep learning about its mysteries together! |
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# Alg1 8.2 Substitution Method
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## Alg1 8.2 Substitution MethodPresentation Transcript
• Algebra 1 Lesson 8.2 Substitution Method
• Whadda ya know 'bout... Substitution Method
• how much does each item cost? \$3.50 fries = \$1.50 burger= \$1 \$7.50
• Goal: Use substitution to solve a system of linear equations.
• What does it mean to... "solve a system of linear equations?" answer: find the solution(s) y = 4x - 4 y = -2x + 8 the point that is a solution to both lines in the point of intersection, (2,4)
• Figure out the solution by using the guess and check method! System of equations: 2x + y = 20 x + y = 16 the only point that is true for both equations is (4.12) x 0 1 2 3 4 y 20 18 16 14 12
• Figure out the solution by using the guess and check method! System of equations: 3x - 4y = 17 x = 2y + 5
• A.E. I.N. SUBSTITUTION P. METHOD, FTW! L.F.
• Types of solutions: One Solution No Solutions Different slope Same slope Infinitely Many Solutions Different y-inter. Different y-inter. Same slope ordered pair false statement Same y-inter. identity
• How to subsitute System of equations: 3x - 4y = 17 x = 2y + 5 substitute find y: SMOTD 3(2y + 5) - 4y = 17 6y + 15 - 4y = 17 2y + 15 = 17 2y = 2 y=1 find x: x = 2y + 5 x = 2(1) + 5 x=2+5 x=7 answer: (7,1)
• How to substitute substitute SMOTD Isolate one variable and substitute into... Equation ivse-solvse Equation eev-say-sol-vee-say Solve for unknown variable and substitute into...
• System of equations: One solution (ordered pair) 3x = 2y - 6 x - y = -1 3(y-1) = 2y - 6 3y - 3 = 2y - 6 3x = 2y - 6 y - 3 = -6 3x = 2(-3) - 6 y = -3 3x = -6 - 6 3x = -12 answer: (-4,-3) x = -4
• System of equations: No Solution (false statement) 8x - 4y = 6 2x - y = 9 8x - 4y = 6 8x - 4(2x-9) = 9 8x - 8x + 36 = 9 36 = 9 FALSE STATEMENT- no solution
• System of equations: Infinitely Many (identity) 3y - 6x = 24 8 + 2x = y 3(8 + 2x) - 6x = 24 24 + 6x - 6x = 24 24 + 0 = 24 24 = 24 IDENTITY- infinitely many solutions |
# Video: Pack 4 • Paper 3 • Question 14
Pack 4 • Paper 3 • Question 14
05:38
### Video Transcript
𝐴𝐵, 𝐵𝐶, and 𝐶𝐷 are sides of a regular 10-sided polygon. And 𝐶𝐷 and 𝐶𝐸 are sides of a regular octagon. Work out the size of angle 𝐶𝐸𝐵. You must show all of your working out.
First of all, we’re actually gonna mark on to the diagram some of the things that we know. First of all, because we actually have 𝐶𝐷 as a shared side — so it’s a side of the regular 10-sided polygon and it’s also a side of the regular octagon — so therefore we can actually say that the sides 𝐵𝐶 and 𝐶𝐸 must also be the same, because we’ve already said, like I said, that 𝐶𝐷 is a shared side in both of the polygons.
Okay, great! because of this, we actually know that the triangle 𝐵𝐶𝐸 is actually going to be an isosceles triangle. And this is gonna be really helpful a bit later on in the question. To enable us to actually find the angle 𝐶𝐸𝐵, which is what the question is looking for, we need to first of all find out the angle 𝐵𝐶𝐸. And the way we can actually do that is by splitting it in two, because actually what we’ve got here are two exterior angles. So what we actually have is the exterior angle of our 10-sided polygon, which is called a decagon, and also the exterior angle of the regular octagon, so our eight-sided shape.
Okay, so how do we find out what the exterior angle is going to be? Well, there are actually a couple of ways we can solve this using different relationships and formula. So first of all, we know that the exterior angle of a regular polygon is equal to 360 degrees divided by 𝑛, where 𝑛 is the number of sides of the polygon.
The key word that allows us to use this is actually regular because we do know that the exterior angle of any polygon add up to 360 degrees. However, we won’t be able to calculate the individual values of those exterior angles if it wasn’t a regular polygon. But because it’s a regular polygon, we can actually do that, because we can just divide 360 degrees by 𝑛, the number of sides.
Okay, so that’s one method. The next method is actually to calculate one of the interior angles. And you can calculate an interior angle by 180 multiplied by 𝑛 minus two over 𝑛. And that’s where 𝑛 again is the size of our polygon. And we can use that to solve this kind of problem, because actually we know another relationship and that’s the exterior plus interior angles add up to 180 degrees. And this is because their angles are on a straight line.
Okay, great! So we’ve got some different ways to solve this. First of all, I’m gonna start using the first formula to solve it. So I’m gonna start with the exterior angle of our decagon. And I’ve actually used 𝐸𝐷 to represent that. So we can say that 𝐸𝐷 is equal to 360 divided by 10, because it’s 10 sides, which gives us an angle of 36 degrees.
And now we’re gonna move on to the exterior angle of the octagon. And to calculate the exterior angle of our octagon, it’s gonna be 360 degrees divided by eight, which gives an angle of 45 degrees.
Now as I said before, we can actually check this by using the other method. So if we do, we’ve got the interior angle of 𝐷, so the interior angle of our decagon, is equal to 180 multiplied by 10 minus two over 10, which is equal to 1440 divided by 10, which gives us 144 degrees. So that’s our interior angle.
So therefore, using our final relationship that an exterior plus an interior angle equals 180 degrees, we can state that the exterior angle of the decagon is gonna be equal to 180 minus 144, which gives us 36 degrees.
So first of all, we can say that the interior angle of our octagon is equal to 180 multiplied by eight minus two divided by eight, which gives us 1080 divided by eight, which is equal to 135 degrees. Okay, great! So again now we can move on and find the exterior angle of the octagon. And again, we use our final relationship that exterior plus interior equals 180 degrees. So we can say that the exterior angle of our octagon is equal to 180 minus 135, which gives us an angle of 45 degrees, which again, great, fits the angle that we got using the first method.
Okay, so now let’s move on and actually work out the size of angle 𝐶𝐸𝐵. So what I’ve actually done is drawn a sketch of the triangle that we’re using here. So we’re trying to find the angle 𝐶𝐸𝐵. But we’ve just found out the angle 𝐵𝐶𝐸, because the angle 𝐵𝐶𝐸 is equal to 36 degrees plus 45 degrees, cause they are exterior angles that we calculated. So this is gonna give us an angle of 81 degrees.
Okay, great! So now what’s the next step? Well, the next step is to actually use the relationship that we found very early on. And that’s that this is actually an isosceles triangle. So therefore, 𝐶𝐸𝐵 and 𝐶𝐵𝐸 are actually the same. So they’re equal to each other. So therefore, we can say that 𝐶𝐸𝐵 plus 𝐶𝐵𝐸 is gonna be equal to 180 minus 81, cause that’s just the angles in a triangle adding up to 180 degrees. And this gives an answer of 99 degrees.
So great! We know that if we add those two angles together, it’s gonna give us 99 degrees. So therefore, angle 𝐶𝐸𝐵 is gonna be equal to 99 divided by two. That’s because as it’s an isosceles triangle, we’ve already said 𝐶𝐸𝐵 and 𝐶𝐵𝐸 are gonna be the same. So then we get our final answer. And that’s that 𝐶𝐸𝐵 is equal to 49.5 degrees, cause 99 divided by two gives us 49.5 degrees. |
# Permutation or Combination?
Students often struggle with deciding whether the problem required permutations or combinations.
The key is to decide whether order must be considered or not when counting the possibilities. If order matters and selections may not be repeated, then you should use permutations (via the permutation formula). If order does not matter and selections may not be repeated, then you should use combinations (via the combinations formula). If order matters and selection may be repeated, then you have permutations but you will need a slot diagram to count the possibilities
To demonstrate the difference between permutations and combinations, Let’s examine two similar problems.
Professional baseball team have 25 members. Of these 25 members, nine are on the field at any time when the opposition is at bat. The nine players occupy certain positions such as pitcher, catcher, first basemen, ect.
Question 1: How many ways is there to select 9 players from the 25 members to take the field?
Question 2: How many ways is there to select 9 players for the different positions on the field from the 25 members?
For both questions, we cannot repeat the selection since you need nine different player. Additionally, we are selecting 9 players from the 25 man roster. The only question we really need to consider is whether order makes a difference.
In Question 1, we are simply choosing nine players from the entire team. They are NOT assigned to any particular position so reordering is not counted differently. SO the number of ways to choose the 9 players is
25C9 = 2,042,975
In Question 2, we are choosing 9 players and also assigning them to positions on the field. This means that if we rearrange any nine players on the field we would count them differently since there would be a different player pitching, catching, ect. The number of permutations of 9 players chosen from 25 is
25P9 = 7.41354768 x 1011
Because order makes a difference, there are many more possibilities. |
8.3: Arcs, Semi-Circles, and Central Angles
Difficulty Level: At Grade Created by: CK-12
Learning Objectives
• Measure central angles.
• Measure arcs of circles.
• Find relationships between minor arcs, semicircles, and major arcs.
Arc, Central Angle
In a circle, the central angle is formed by two radii of the circle with its vertex at the center of the circle. An arc is a section of the circle.
A circle’s central angle is an angle with a vertex at the _____________________ of the circle.
Each side of a central angle is also a _______________________ of the circle.
An ________________ is a section of the outer edge of the circle.
Minor and Major Arcs, Semicircle
A semicircle is half of a circle.
A major arc is longer than a semicircle and a minor arc is shorter than a semicircle.
A ______________________________ is half of a circle.
A __________________ arc is longer than a semicircle and a __________________ arc is shorter than a semicircle.
An arc can be measured in degrees or in a linear measure (cm, ft, etc.). In this lesson we will concentrate on degree measure.
The measure of the minor arc is the same as the measure of the central angle that corresponds to it.
This means that the measure of the arc in between the sides of the central angle (from point A\begin{align*}A\end{align*} to point B\begin{align*}B\end{align*} below) is the same degree measure as the central angle, like in the picture below:
The measure of the major arc is equal to 360\begin{align*}360^\circ\end{align*} minus the measure of the minor arc.
In the example above, the larger arc from point A\begin{align*}A\end{align*} to point B\begin{align*}B\end{align*} (in a counter-clockwise direction) is 36085=275\begin{align*}360^\circ - 85^\circ = 275^\circ\end{align*}. Minor arcs are named with two letters—the letters that denote the endpoints of the arc. Below, the minor arc corresponding to the central angle AOC\begin{align*}\angle AOC\end{align*} is called ACˆ\begin{align*}\widehat{AC}\end{align*}.
In order to prevent confusion, semicircles and major arcs are named with three letters—the letters that denote the endpoints of the arc and any other point on the major arc. In the figure above, the major arc corresponding to the central angle AOC\begin{align*}\angle AOC\end{align*} is called ABCˆ\begin{align*}\widehat{ABC}\end{align*} because point B\begin{align*}B\end{align*} is on the major arc.
Minor arcs are named with ________ letters, which represent the __________________ of the arc.
Major arcs are named with ________ letters, which represent the endpoints of the arc and a _________________ on the major arc.
Congruent Arcs
Two arcs that correspond to congruent central angles will also be congruent.
In the figure below, AOCBOD\begin{align*}\angle AOC \cong \angle BOD\end{align*} because they are vertical angles. This also means that ACˆDBˆ\begin{align*}\widehat{AC} \cong \widehat{DB}\end{align*}:
If central angles are congruent, their corresponding arcs are also __________________.
The measure of the arc formed by two adjacent arcs (or two arcs next to each other) is the sum of the measures of the two arcs.
In other words, mRQˆ=mRPˆ+mPQˆ\begin{align*}m \widehat{RQ} = m \widehat{RP} + m \widehat{PQ}\end{align*}:
1. True/False: A central angle in a circle creates two arcs along its circumference: the minor arc is the smaller arc and the major arc is the larger arc.
2. True/False: If two central angles in a circle are congruent, then their corresponding arcs are supplementary.
3. For question #2 above, give an example that proves why this statement is either true or false. Explain your reasoning in words.
\begin{align*}{\;}\end{align*}
\begin{align*}{\;}\end{align*}
\begin{align*}{\;}\end{align*}
\begin{align*}{\;}\end{align*}
Chord, Diameter, Secant
A chord is defined as a line segment starting at one point on the circle and ending at another point on the circle.
A chord that goes through the center of the circle is called the diameter of the circle. Notice that the diameter is twice as long as the radius of the circle.
A secant is a line that cuts through the circle and continues infinitely in both directions.
• A line segment from one point on a circle to another point on the circle is called a _____________________.
• When a chord goes through the center of a circle, it is a ___________________.
• A chord that is a line extending in both directions is a _____________________.
Congruent Chords Have Congruent Minor Arcs
In the same circle or congruent circles, congruent chords have congruent minor arcs.
Proof. Draw the diagram above, where the chords DB¯¯¯¯¯¯¯¯\begin{align*}\overline{DB}\end{align*} and AC¯¯¯¯¯¯¯¯\begin{align*}\overline{AC}\end{align*} are congruent. Construct ΔDOB\begin{align*}\Delta DOB\end{align*} and ΔAOC\begin{align*}\Delta AOC\end{align*} by drawing the 4 radii from the center O\begin{align*}O\end{align*} to points A,B,C,\begin{align*}A, B, C,\end{align*} and D\begin{align*}D\end{align*}.
Then, ΔAOCΔBOD\begin{align*}\Delta AOC \cong \Delta BOD\end{align*} by the SSS Postulate.
This means that central angles AOCBOD\begin{align*}\angle AOC \cong \angle BOD\end{align*} (by CPCTC), which leads to the conclusion that ACˆDBˆ\begin{align*}\widehat{AC} \cong \widehat{DB}\end{align*}.
Congruent Minor Arcs Have Congruent Chords and Congruent Central Angles
In the same circle or congruent circles, congruent minor arcs have congruent chords and congruent central angles.
Proof. Draw the following diagram, in which ACˆDBˆ\begin{align*}\widehat{AC} \cong \widehat{DB}\end{align*}. In the diagram, DO¯¯¯¯¯¯¯¯,OB¯¯¯¯¯¯¯¯,AO¯¯¯¯¯¯¯¯\begin{align*}\overline{DO}, \overline{OB}, \overline{AO}\end{align*}, and OC¯¯¯¯¯¯¯¯\begin{align*}\overline{OC}\end{align*} are each a radius of the circle.
Since ACˆDBˆ\begin{align*}\widehat{AC} \cong \widehat{DB}\end{align*}, this means that the corresponding central angles are also congruent:
AOCBOD\begin{align*}\angle AOC \cong \angle BOD\end{align*}
Therefore, ΔAOCΔBOD\begin{align*}\Delta AOC \cong \Delta BOD\end{align*} by the SAS Postulate.
We conclude that DB¯¯¯¯¯¯¯¯AC¯¯¯¯¯¯¯¯\begin{align*}\overline{DB} \cong \overline{AC}\end{align*}.
Here are some examples in which we apply the concepts and theorems we discussed in this lesson.
Example 1
Find the measure of each arc:
A. mMLˆ\begin{align*}m \widehat{ML}\end{align*}
B. mPMˆ\begin{align*}m \widehat{PM}\end{align*}
C. mLMQˆ\begin{align*}m \widehat{LMQ}\end{align*}
Solutions:
A. mMLˆ\begin{align*}m \widehat{ML}\end{align*} is equal to mLOM\begin{align*}m \angle LOM\end{align*} (the central angle) =60\begin{align*}= 60^\circ\end{align*}
B. mPMˆ\begin{align*}m \widehat{PM}\end{align*} is equal to mPOM\begin{align*}m \angle POM\end{align*} (which is supplementary to the angle LOM\begin{align*}\angle LOM\end{align*})
18060=120\begin{align*}180^\circ - 60^\circ = 120^\circ\end{align*}
C. mLMQˆ=mMLˆ+mPMˆ+mPQˆ=60+120+60=240\begin{align*}m \widehat{LMQ} = m \widehat{ML} + m \widehat{PM} + m \widehat{PQ}\!\\ = 60^\circ + 120^\circ + 60^\circ = 240^\circ\end{align*}
Example 2
Find mABˆ\begin{align*}m \widehat{AB}\end{align*} in circle \begin{align*}O\end{align*}. The measures of all three arcs shown must add to \begin{align*}360^\circ\end{align*}.
All three arcs must add to \begin{align*}360^\circ\end{align*} because all three central angles add to \begin{align*}360^\circ\end{align*} (since they complete a circle.)
Fill in the arc measurements based on the picture above:
\begin{align*}m \angle AOB = m \widehat{AB} =\end{align*} _________________
\begin{align*}m \angle BOC = m \widehat{BC} =\end{align*} _________________
\begin{align*}m \angle AOC = m \widehat{AC} =\end{align*} _________________
We can add all three arcs together:
\begin{align*}m \widehat{AB}{\;\;} + m \widehat{BC}{\;\;} + m \widehat{AC}{\;\;\;} &= 360^\circ\\ x + 20 + 4x + 5 + 3x + 15 &= 360\\ 8x + 40 &= 360\\ 8x &= 320\\ x &= 40\end{align*}
We are looking for \begin{align*}m \widehat{AB}\end{align*} so we substitute \begin{align*}x = 40\end{align*} back into the arc measurement:
\begin{align*}m \widehat{AB} = x + 20 = 40 +20 = 60^\circ\end{align*}
What other postulate is similar to the Arc Addition Postulate? Describe.
\begin{align*}{\;}\end{align*}
\begin{align*}{\;}\end{align*}
\begin{align*}{\;}\end{align*}
\begin{align*}{\;}\end{align*}
Graphic Organizer for Lessons 1 – 2: PARTS OF A CIRCLE
Circle Part Draw a Picture What is it? Is there a formula I need to know for this part?
Center
Diameter
Circumference
Area
Central Angle
Semicircle
Minor Arc
Major Arc
Chord
Secant
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# Numbers, words and symbols (tenths/hundredths)
Lesson
If you had an apple and you cut it into $2$2 equal pieces to share with a friend, how much would each of you get? You would each get half an apple. One half is an example of a number that is less than one whole. It can be written as $\frac{1}{2}$12 as a fraction or $0.5$0.5 as a decimal.
We can express a part of a whole in words, or we can use numbers and symbols (e.g. fractions or decimals).
Watch this video to see how.
#### Worked examples
##### Question 1
How do we write the number $0.89$0.89 in words?
1. eighty nine
A
eighty nine tenths
B
eighty nine hundredths
C
eighty nine thousandths
D
##### Question 2
Look at the diagram.
1. How many squares are shaded in total?
2. What part of the grid is coloured in yellow?
Written Expression $\editable{}$ out of $100$100 $\frac{\editable{}}{100}$100 $0.$0.$\editable{}$
3. What part of the grid is coloured in blue?
Written Expression $\editable{}$ out of $100$100 $\frac{\editable{}}{100}$100 $0.$0.$\editable{}$
4. What part of the grid is coloured in red?
Written Expression $\editable{}$ out of $100$100 $\frac{\editable{}}{100}$100 $0.$0.$\editable{}$
5. How many squares are shaded in total?
Written Expression $\editable{}$ out of $100$100 $\frac{\editable{}}{100}$100 $\editable{}$
##### Question 3
We want to write "$690$690 hundredths" as a decimal.
1. Which of the following statements is true?
$6.9$6.9 and $6.90$6.90 are different values
A
$6.9$6.9 is a simpler way of writing $6.90$6.90
B
$6.90$6.90 is a simpler way of writing $6.9$6.9
C
### Outcomes
#### 5.NN1.07
Demonstrate and explain equivalent representations of a decimal number, using concrete materials and drawings |
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• Level: GCSE
• Subject: Maths
• Word count: 4280
For this piece of work I am investigating Pythagoras. Pythagoras was a Greek mathematician.
Extracts from this document...
Introduction
--
Sarah Williams
Introduction
For this piece of work I am investigating Pythagoras. Pythagoras was a Greek mathematician. Pythagoras lived on the island of Samos and was born around 569BC. He did not write anything but he is regarded as one of the world’s most important characters in maths. His most famous theorem is named after him and is called the Pythagoras Theorem. It is basically a²+b²=c². This is what the coursework is based on.
I am going to look at the patterns, which surround this theorem and look at the different sequences that can be formed.
The coursework
The numbers 3, 4 and 5 satisfy the condition 3²+4²=5² because
3²=3x3=9
4²=4x4=16
5²=5x5=25
And so
3²+4²=9+16=25=5²
I will now check that the following sets of numbers also satisfy the similar condition of (smallest number) ²+(middle number) ²=(largest number) ²
a) 5, 12, 13
5²=5x5=25
12²=12x12=144
25+144=169
√169 = 13
This satisfies the condition as
5²+12²=25+144=169=13²
b) 7, 24, 25
7²=7x7=49
24²=24x24=576
49+576=625
√625=25
This satisfies the condition as
7²+24²=49+576=625=25²
The numbers 3,4 and 5 can be the lengths – in appropriate units – of the side of a right-angled triangle.
5
3
The perimeter and area of this triangle are:
Perimeter = 3+4+5=12 units
Area = ½x3x4=6 units ²
The numbers 5,12,13 can also be the lengths – in appropriate units – of a right-angled triangle.
Perimeter = 5+12+13=30
Area=½x5x12=30
This is also true for the numbers 7,24,25
Area=½x7x24=84
I have put these results into a table to see if I can work out any patterns.
Length of shortest side Length of middle side Length of longest side Perimeter Area 3 4 5 12 6 5 12 13 30 30 7 24 25 56 84
I have noticed that there seems to be a recurring pattern between the length of the middle side and the length of the longest side. They are always consecutive numbers. I shall now investigate this.
I will assume that the hypotenuse has length b + 1 where b is the length of the middle side.
Middle
2256
25944
24
49
1200
1201
2450
29400
25
51
1300
1301
2652
33150
Another pattern that I have noticed is that there must be a connection between the sides of the triangle because the middle side and the longest side have a difference of one. I will now try and work out the formula. I will basically use trial and error and try out different formulae that I think of.
(Shortest side) ²+(middle side) ²=(longest side) ²
I already know that this is correct because it is Pythagoras theorem. However I will still check it using the Pythagorean triple 3,4,5.
3²+4²=5²
I know that this is correct because
3²=9
4²=16
9 + 16 = 25, which is also 5²
(Middle side) ²+(longest side) ²=(shortest side) ²
To see if this statement is true I shall use the Pythagorean triple 9,40,41.
40²+41²=9²
This is not true because
40² = 1600
41² = 1681
which add together to make 3281
and
9²=81
This means that this is not a correct formula.
(Longest side) ²/(shortest side) ² = (middle side)
To check this I will use the Pythagorean triple 15,112,113.
113²/15²=112²
This is wrong as
113²=12769
15²=225
and
12769/225 = 56.751111
and as
The middle side = 113 this formula has to be wrong.
Shortest side x longest side = (middle side) ²
To check this I will use the Pythagorean triple 7,24,25
7 x 25 = 175
This is wrong because
24²= 576 and this is very different
(Longest side) ²-(middle side) ²=shortest side
To check this formula I will use the Pythagorean triple 5,12,13
13²-12²=5²
This is right as
13²=169
12²=144
169-144=25 which is also 5².
So this formula is correct
(Middle side) ²+Smallest side=Largest side
To check this I will use the Pythagorean triple 5, 12,13.
12²+5=13
This is silly and I shouldn’t really have tried it, as I should have been able to work out that this formula is incorrect just by looking at the numbers involved. 144 is bigger than 13 without adding the 5.
Largest side + Middle side = (shortest side) ²
To check this I will use the Pythagorean triple 3,4,5
5+4=3²
This is correct because 5 + 4 = 9 which is also 9.
Just to check that this formula works with all triples I shall pick 3 random numbers from the N column of the table and use the formula with them. I shall use (i) 23 (ii) 10 (iii) 16
1. 1105+1104=47² which is also 2209. This is correct
2. 221+220=21² which is also 441. This is correct
3. 545+544=33² which is also 1089. This is correct
This seems to be a correct formula.
So the formulae I have worked out in this section are:
• (Shortest side)²+(middle side)²=(longest side)²
• (Longest side) ²-(middle side) ²=shortest side
• Largest side + Middle side = (shortest side) ²
All these are based on the assumption that the difference between the longest side and the middle side is 1.
However, I am interested about what happens when the difference between the longest side and the middle side is different from 1. I shall investigate this going up to c=b+10.
In the next few investigations I was required to work out the square roots of numbers. However, although it is possible to work out a square root of every number I chose only to investigate the perfect squares. Those numbers which are whole numbers. This is because I am only investigating Pythagorean triples with positive integers.
Investigating c=b+2
I will now investigate when longest side – middle side= 2. (c-b=2)
a²+b²=c²
therefore
a²=c²-b²
factorising
a²=(c-b)(c+b)
= 2(2b+2)
=4b+4
I can now produce a table for different values of b. Perfect squares are coloured red.
B a²=4b+4 √a² Triples (a,b,c) 1 8 2 12 3 16 4 This becomes 4,3,5 4 20 5 24 6 28 7 32 8 36 6 This becomes 6,8,10 9 40 10 44 11 48 12 52 13 56 14 60 15 64 8 This becomes 8,15,17 16 68 17 72 18 76 19 80 20 84 21 88 22 92 23 96 24 100 10 This becomes 10, 24, 26, which is a multiple of 5,12,13. 25 104
Investigating c=b+3
I will now investigate when longest side – middle side=3. (c-b=3)
a²+b²=c²
therefore
a²=c²-b²
factorising
a²=(c-b)(c+b)
= 3(2b+3)
=6b+9
I can now produce a table for different values of b. Perfect squares are coloured red.
B a²=6b+9 √a² Triples (a,b,c) 1 15 2 21 3 27 4 33 5 39 6 45 7 51 8 57 9 63 10 69 11 75 12 81 9 This becomes 9,12,15, which is a multiple of 3,4,5 13 87 14 93 15 99 16 105 17 111 18 117 19 123 20 129 21 135 22 141 23 147 24 153 25 159
Investigating c=b+4
I will now investigate when longest side – middle side=4. (c-b=4)
a²+b²=c²
therefore
a²=c²-b²
factorising
a²=(c-b)(c+b)
= 4(2b+4)
=8b+16
I can now produce a table for different values of b. Perfect squares are coloured red.
B a²=8b+16 √a² Triples (a,b,c) 1 24 2 32 3 40 4 48 5 56 6 64 8 This becomes 8,6,10 which is a multiple of 4,3,5 7 72 8 80 9 88 10 96 11 104 12 112 13 120 14 128 15 136 16 144 12 This becomes 12,16,20 which is a multiple of 3,4,5 17 152 18 160 19 168 20 176 21 184 22 192 23 200 24 208 25 216
Investigating c=b+5
I will now investigate when longest side – middle side=5. (c-b=5)
a²+b²=c²
therefore
a²=c²-b²
factorising
a²=(c-b)(c+b)
= 5(2b+5)
=10b+25
I can now produce a table for different values of b. Perfect squares are coloured red.
B a²=10b+25 √a² Triples (a,b,c) 1 35 2 45 3 55 4 65 5 75 6 85 7 95 8 105 9 115 10 125 11 135 12 145 13 155 14 165 15 175 16 185 17 195 18 205 19 215 20 225 15 This becomes 15,20.25 which is a multiple of 3,4,5 21 235 22 245 23 255 24 265 25 275
Conclusion
Internet
An interesting thing I found on the Internet, which I had never heard of before, was Pythagorean Quads (quadruples).
Pythagorean quads are the 3D version of the triples and can all be generated by the following formula, where a, b, c and d are whole numbers.
If p = a
2 + b2 - c2 - d2
r = 2(ac + bd)
and s = a2 + b2 + c2 + d2
then p
2 + q2 + r2 = s2.
I shall now try and work out a few of these to check if this formula is correct
I shall choose the numbers a=4, b=1, c=1 and d=1.
This means that p=16+1-1-1 = 15
q=2(4-1)=6
r=2(4+1)=10
s=16+1+1+1=19
From looking further on the Internet for confirmation I conclude that this formula does work because the quad that I worked out above is correct.
Conclusion
I started this investigation by noticing that there was a difference of 1 between the middle side and the longest side. I then went on to work out formulae relating to this. All these formulae were based on the assumption that c-b+1. however after investigating further I realised that the difference between the middle side and the longest side could be any number. So I investigated all the differences up to a difference of 5. I decided that I would stop each table at the value of b=25. I realised that this investigation could go on forever and you could investigate unlimited values of b and there are unlimited differences that I could have investigated but I had to set a limit. I also decided that I would only investigate up to c=b+10 as I felt that was a sufficient number of investigations. I wrote the extension because I thought it was interesting to see what else other people had discovered
This student written piece of work is one of many that can be found in our GCSE Pythagorean Triples section.
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# How do you write the standard form of a line given (1/2 , -2/3 ); perpendicular to the line 4x − 8y = 1?
Feb 16, 2018
$6 x + 3 y = 1$
#### Explanation:
$\text{the equation of a line in "color(blue)"standard form}$ is.
$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{A x + B y = C} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
$\text{where A is a positive integer and B, C are integers}$
$\text{the equation of a line in "color(blue)"slope-intercept form}$ is.
•color(white)(x)y=mx+b
$\text{where m is the slope and b the y-intercept}$
$\text{rearrange "4x-8y=1" into this form}$
$- 8 y = - 4 x + 1 \Rightarrow y = \frac{1}{2} x - \frac{1}{8} \to m = \frac{1}{2}$
$\text{given a line with slope m then the slope of a line}$
$\text{perpendicular to it is }$
•color(white)(x)m_(color(red)"perpendicular")=-1/m
$\Rightarrow {m}_{\textcolor{red}{\text{perpendicular}}} = - \frac{1}{\frac{1}{2}} = - 2$
$\Rightarrow y = - 2 x + b \leftarrow \textcolor{b l u e}{\text{is the partial equation}}$
$\text{to find b substitute "(1/2,-2/3)" into the partial}$
$\text{equation}$
$- \frac{2}{3} = - 1 + b \Rightarrow b = - \frac{2}{3} + 1 = \frac{1}{3}$
$\Rightarrow y = - 2 x + \frac{1}{3} \leftarrow \textcolor{red}{\text{in slope-intercept form}}$
$\text{multiply through by 3 to eliminate fraction}$
$\Rightarrow 3 y = - 6 x + 1$
$\Rightarrow 6 x + 3 y = 1 \leftarrow \textcolor{red}{\text{in standard form}}$ |
# What are Mathematical Combinations and How To Calculate Them
If you’re looking to understand how many unique combinations can be created from a set of objects, you’ve come to the right place. Combinations are a powerful tool utilized in a variety of ways, such as in mathematics, cryptography, and even in business. With the right understanding of how combinations work and the different kinds of combinations, you can explore a seemingly infinite number of possibilities. In this blog post, we’ll discuss the different types of combinations and the process of calculating the total number of combinations in a set of objects. We’ll also provide a few examples to help you better understand how to calculate the combinations of any set of objects. So, read on to learn more about how many combinations are possible and the process of calculating them.
## Formula for finding how many combinations you have
The number of ways to create a non-repeating arrangement of items (r) from a larger set of distinct items (n) is determined by the formula C(n, r) = (n!) / [(r!) x (n – r)!] The factorials, which are the products of all positive integers equal to and less than the number you are computing, must also be computed in order to use the formula for calculating combinations. The factorial in the combination formula is indicated by the symbol (! ), which stands for the factorial functions you must use in your calculation.
## What is a combination?
Combinations are the various configurations that can be produced when a sample of values or items from a larger set is taken. How many subsets you can create from the entire set of items is demonstrated by the combinations you can make. In a mathematical combination, the order of items is unimportant. This means that although some combinations produce an ordered sequence, resulting in a permutation, combinations can be formed in any order. The combinations (0, 0, 1, 1, 2, 3) and (0, 1, 2, 3, 4) are examples of non-repeating combinations.
## How to calculate combinations
To determine how many combinations you can get from a sample set, use the following steps and the formula C(n, r) = (n!) / [(r!) x (n – r)!]
### 1. Determine your r and n values
By selecting a smaller set of items from a larger set, you can determine the values of r and n.
Assume, for instance, that you want to select four books from a shelf of eight to read. The (r) denotes the sample of the four books you select, and the (n) denotes the larger set of eight books in the formula. When you arrive at these numbers, replace the (r) and (n) variables in the formula with them:
C(n, r) = (8!) / [(4!) x (8 – 4)!]
Subtract these two numbers when determining your sample set and larger set for the r and n variables in the formula. Find the difference using the n = 30 books and r = 10 books from the previous step as an example:
C(n, r) = (8!) / [(4!) x (8 – 4)!] =
C(n, r) = (8!) / [(4!) x (4)!]
### 3. Expand on the factorials
After simplifying the formula’s expressions, you can begin computing each of the problem’s factorials. Expand each factorial in the following formula using the larger set of eight books, the sample set of four books, and the difference you discover when you subtract these values:
8! = 8 7 6 5 4 3 2 1
= 4 x 3 x 2 x 1 x 4 x 3 x 2 x 1 = 4! x 4!
### 4. Cancel like terms and divide
Since these values are connected through division, you can eliminate the like terms between your factorials in the formula before dividing them. The common terms on the top and bottom of the fractional part of the equation are 4, 3, 2, 1 between 8 and (4! x 4!) Canceling these terms out results in:
C(n, r) = (8x7x6x5) / (4x3x2x1) =
C(n, r) = (1,680) / (24) =
C(n, r) = 70
When four books are selected from a set of eight, the result is 70, which is the number of combinations possible when n r 0. This means that your (n) value must be greater than or equal to your (r) value, and that all values you work with must be greater than or equal to zero. Additionally, your values must not repeat when calculating how many combinations you can get using this formula, and the order is irrelevant.
## Combinations vs. permutations
Combinations can have both repeating and non-repeating arrangements. But it doesn’t matter in what order you find these arrangements. However, since permutations must occur in an ordered set, the order is crucial when dealing with them. The distinction between these two ideas can be remembered by recalling that a permutation is an ordered combination.
Permutations, like combinations, can have repeating or non-repeating values inside a set. Assume, for instance, that you need to discover a lock’s combination. The lock’s combination is an example of a permutation where the order of the numbers is crucial to the combination’s functionality. However, you might have a non-repeating permutation (such as the combination 3, 4, 5) or a repeating permutation (such as the combination 4, 4, 4).
## Examples
You can use two formulas to calculate combinations when the order matters, as it does with permutations:
You can determine the permutations in the following example issue by using these formulas:
Let’s say you need to figure out the combination to a lock, and you have a choice of six numbers. Six represents your (n) value in both formulas. Three becomes your (r) value, or the smaller subset you select from the entire set, if the lock requires three numbers as part of the combination. Use the formula nr to determine how many combinations you can create using repeating values:
n^r = (6) ^ (3) =
*6 x 6 x 6 = 216
As a result, you can find 216 combinations that have repeating values. To determine the possible combinations for each of the three items you select from the set of six values, however, if the lock combination has non-repeating values, use the formula (n!) / (n – r)!
(n!) / (n – r)! =
(6!) / (6 – 3)! =
(6 x 5 x 4 x 3 x 2 x 1) / (3)! =
(3 x 2 x 1) / (6 x 5 x 4 x 3 x 2 x 1) equals
6 x 5 x 4 / 1 =
120 / 1 = 120
This finding shows that a lock with values arranged in permutations has 120 possible combinations. When calculating non-repeating permutations, you can simply reduce the factorials that are shared by each factor you divide in the formula, similar to when calculating combinations where the order is irrelevant.
## FAQ
How do you calculate the number of possible combinations?
We will use the formula nCr = n! / r! * (n – r)! to calculate combinations, where n denotes the total number of items and r denotes the number of items being chosen at a time. You must compute a factorial in order to calculate a combination.
How many combinations of 4 items are there?
The answer to this question (which you got right) is 24 if you meant to say “permutations.” If so, then you were likely asking “how many different ways can I arrange the order of four numbers?” Here’s how to observe this: 1.
How many combinations of 2 numbers are there?
Additionally, I am aware that there are 10 combinations of a single digit from 0–9 and 100 combinations of two digits from 0–9. I know this because 10 10 equals 100 two-digit combinations. |
# In a ∆ABC, right angled at A,
Question:
In a $\Delta \mathrm{ABC}$, right angled at $\mathrm{A}$, if $\tan \mathrm{C}=\sqrt{3}$, find the value of $\sin \mathrm{B} \cos \mathrm{C}+\cos \mathrm{B} \sin \mathrm{C}$.
Solution:
Given:
$\tan C=\sqrt{3}$
To find: $\sin B \cos C+\cos B \sin C$
The given $\triangle A B C$ is as shown in figure below
Side BC is unknown and can be found using Pythagoras theorem
Therefore,
$B C^{2}=A B^{2}+A C^{2}$
Now by substituting the value of known sides from figure (a)
We get,
$B C^{2}=(\sqrt{3})^{2}+1^{2}$
$=3+1$
$=4$
Now by taking square root on both sides
We get,
$B C=\sqrt{4}$
$=2$
Therefore Hypotenuse side BC = 2 …… (1)
Now $\sin B=\frac{\text { Perpendicular side opposite to } \angle B}{\text { Hypotenuse }}$
Therefore,
$\sin B=\frac{A C}{B C}$
Now by substituting the values from equation (1) and figure (a)
We get,
$\sin B=\frac{1}{2}$.....(2)
Now $\cos B=\frac{\text { Base side adjacent to } \angle B}{\text { Hypotenuse }}$
Therefore,
$\cos B=\frac{A B}{B C}$
Now by substituting the values from equation (1) and figure (a)
We get,
$\cos B=\frac{\sqrt{3}}{2}$....(3)
Now $\sin C=\frac{\text { Perpendicular side opposite to } \angle C}{\text { Hypotenuse }}$
Therefore,
$\sin C=\frac{A B}{B C}$
Now by substituting the values from equation (1) and figure (a)
We get,
$\sin C=\frac{\sqrt{3}}{2} \ldots \ldots$(4)
Now by definition,
$\tan C=\frac{\sin C}{\cos C}$
Therefore,
$\cos C=\frac{\sin C}{\tan C}$
Now by substituting the value of $\sin C$ and $\tan C$ from equation (4) and given data respectively
We get,
$\cos C=\frac{\frac{\sqrt{3}}{2}}{\sqrt{3}}$
$\cos C=\frac{\sqrt{3}}{2 \sqrt{3}}$
Now $\sqrt{3}$ gets cancelled as it is present in both numerator and denominator
Therefore,
$\cos C=\frac{1}{2}$.....(5)
Now by substituting the value of $\sin B, \cos B, \sin C$ and $\cos C$ from equation $(2),(3),(4)$ and (5) respectively in $\sin B \cos C+\cos B \sin C$
We get,
$\sin B \cos C+\cos B \sin C=\frac{1}{2} \times \frac{1}{2}+\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}$
$=\frac{1}{4}+\frac{3}{4}$
$=\frac{4}{4}$
$=1$
$\sin B \cos C+\cos B \sin C=1$ |
Solving Lesson 18 Page 80 SBT Math 10 – Kite>
Topic
A person walks along the beach from the position A to the location REMOVE and observe a ship OLD moored offshore. That person took measurements and got the results: $$AB = 30m,\widehat {CAB} = {60^0},\widehat {CBA} = {50^0}$$ (Figure 23). Calculate distance from location A to the ship OLD(round the result to tenths of a meter)?
Solution method – See details
Step 1: Calculate the angle measure $$\widehat {ACB}$$
Step 2: Use the sine theorem to calculate the length AC byABC then conclude
Detailed explanation
We have: $$\widehat {ACB} = {180^0} – (\widehat {CBA} + \widehat {CAB}) = {70^0}$$
Apply the sine theorem toABC we have: $$\frac{{AC}}{{\sin \widehat {CBA}}} = \frac{{AB}}{{\sin \widehat {ACB}}} \Rightarrow AC = \frac{{ AB.\sin \widehat {CBA}}}{{\sin \widehat {ACB}}} = \frac{{30.\sin {{50}^0}}}{{\sin {{70}^0 }}} \approx 24.5$$
So the distance from the position A to the ship OLD is 24.5 m |
Question
# Today a father deposits $12,500 in a bank that pays 8% annual interest. Additionally, make annual contributions due of$2,000 annually for 3 years. The fund is for your son to receive an annuity and pay for his studies for 5 years. If the child starts college after 4 years, how much is the value of the annuity? solve how well it is for an exam
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## Answer to a math question Today a father deposits $12,500 in a bank that pays 8% annual interest. Additionally, make annual contributions due of$2,000 annually for 3 years. The fund is for your son to receive an annuity and pay for his studies for 5 years. If the child starts college after 4 years, how much is the value of the annuity? solve how well it is for an exam
Bud
4.6
To find the value of the annuity, we need to calculate the future value of the initial deposit and the annual contributions.
Step 1: Calculate the future value of the initial deposit. The formula for calculating the future value compounded annually is:
FV = PV$1+r$^n
Where:
FV = Future value
PV = Present value $initial deposit$
r = Interest rate
n = Number of years
In this case, the present value $PV$ is $12,500, the interest rate $r$ is 8%, and the number of years $n$ is 4 since the child starts college after 4 years. FV = 12,500$1+0.08$^4 FV = 12,500$1.08$^4 FV \approx 12,500 \times 1.36049 FV \approx 17,006.13 Step 2: Calculate the future value of the annual contributions. Since the annual contributions are received for 3 years, we can calculate the future value using the formula: FV = P [$1 + r$^n - 1] / r Where: FV = Future value of the annuity P = Annual contribution r = Interest rate n = Number of years In this case, the annual contribution $P$ is$2,000, the interest rate $r$ is 8%, and the number of years $n$ is 3.
FV = 2,000 [$1 + 0.08$^3 - 1] / 0.08
FV = 2,000 [$1.08$^3 - 1] / 0.08
FV = 2,000 [1.259712 - 1] / 0.08
FV = 2,000 \times 0.259712 / 0.08
FV\approx6,492.8
Step 3: Add the future value of the initial deposit and the future value of the annual contributions to find the total future value of the annuity.
TotalFutureValue=17,006.13+6,492.8
TotalFutureValue\approx23,498.93
Answer: The value of the annuity is approximately \$23,498.93.
Frequently asked questions $FAQs$
What is the radius of a circle represented by the equation x^2 + y^2 = 16?
+
What is the period and amplitude of the sine function f$x$ = sin$x$?
+
What is the value of f$x$ if f$x$ is a constant function with f$x$=c, where c is a fixed real number?
+ |
# How Do You Calculate The Average Of Two Percentages?
## How do you average 3 percentages?
The average of a set of numbers is simply the sum of the numbers divided by the total number of values in the set.
For example, suppose we want the average of 24 , 55 , 17 , 87 and 100 .
Simply find the sum of the numbers: 24 + 55 + 17 + 87 + 100 = 283 and divide by 5 to get 56.6 ..
## How can I find the percentage of a number?
To find the percentage of a number when it is in decimal form, you just need to multiply the decimal number by 100. For example, to convert 0.5 to a percentage, 0.5 x 100 = 25% The second case involves a fraction. If the given number is in fractional form, first convert it to a decimal value and multiply by 100.
## What is the formula for the average?
Average, which is the arithmetic mean, and is calculated by adding a group of numbers and then dividing by the count of those numbers. For example, the average of 2, 3, 3, 5, 7, and 10 is 30 divided by 6, which is 5.
## How do you calculate average percentage?
Calculate the average percentage by dividing the total items represented by percentages by the overall total of items. In the example, a total of 200 pencils were removed out of a total of 500 pencils. Divide 200 by 500, which is equal to 0.40.
## How do I calculate a percentage of a percentage?
To calculate a percentage of a percentage, convert both percentages to fractions of 100, or to decimals, and multiply them. For example, 50% of 40% is calculated; (50/100) x (40/100) = 0.50 x 0.40 = 0.20 = 20/100 = 20%.
## How do you compare two percentages?
First: work out the difference (increase) between the two numbers you are comparing. Then: divide the increase by the original number and multiply the answer by 100. % increase = Increase ÷ Original Number × 100. If your answer is a negative number, then this is a percentage decrease.
## How do you calculate weighted average of two percentages?
To find a weighted average, multiply each number by its weight, then add the results….In a data set of four test scores where the final test is more heavily weighted than the others:50(. 15) = 7.5.76(. 20) = 15.2.80(. 20) = 16.98(. 45) = 44.1.Mar 22, 2021
## How do you calculate the average percentage in Excel?
By constructing an Excel spreadsheet, this calculation becomes a simple matter of data entry.Open Microsoft Excel.Enter the data to be averaged in column A. … Enter the corresponding percentages in column B. … Enter “=A1*B1” without quotes in cell C1.More items…
## How do I calculate percentages?
1. How to calculate percentage of a number. Use the percentage formula: P% * X = YConvert the problem to an equation using the percentage formula: P% * X = Y.P is 10%, X is 150, so the equation is 10% * 150 = Y.Convert 10% to a decimal by removing the percent sign and dividing by 100: 10/100 = 0.10.More items…
## Why You Should not average percentages?
Therefore, the temptation of averaging percentages can provide inaccurate results. As previously mentioned, there is one exception where the average of percentages agrees with the accurate percentage calculation. This occurs when the sample size in both groups are the same.
## Can you sum percentages?
Percents can be added directly together if they are taken from the same whole, which means they have the same base amount. … You would add the two percentages to find the total amount. You can use Microsoft Excel to to perform this operation. |
# Polar Equations
A polar rose (Rhodonea Curve)
This polar rose is created with the polar equation: $r = cos(\pi\theta)$.
# Basic Description
Polar equations are used to create interesting curves, and in most cases they are periodic like sine waves. Other types of curves can also be created using polar equations besides roses, such as Archimedean spirals and limaçons. See the Polar Coordinates page for some background information.
# A More Mathematical Explanation
Note: understanding of this explanation requires: *calculus, trigonometry
## Rose
The general polar equations form to create a rose is UNIQ1db4ec0833600791-math-00000001-Q [...]
## Rose
The general polar equations form to create a rose is $r = a \sin(n \theta)$ or $r = a \cos(n \theta)$. Note that the difference between sine and cosine is $\sin(\theta) = \cos(\theta-\frac{\pi}{2})$, so choosing between sine and cosine affects where the curve starts and ends. $a$ represents the maximum value $r$ can be, i.e. the maximum radius of the rose. $n$ affects the number of petals on the graph:
• If $n$ is an odd integer, then there would be $n$ petals, and the curve repeats itself every $\pi$.
Examples:
• If $n$ is an even integer, then there would be $2n$ petals, and the curve repeats itself every $2 \pi$.
Examples:
• If $n$ is a rational fraction ($p/q$ where $p$ and $q$ are integers), then the curve repeats at the $\theta = \pi q k$, where $k = 1$ if $pq$ is odd, and $k = 2$ if $pq$ is even.
Examples:
$r = \cos(\frac{1}{2}\theta)$ The angle coefficient is $\frac{1}{2} = 0.5$. $1 \times 2 = 2$, which is even. Therefore, the curve repeats itself every $\pi \times 2 \times 2 \approx 12.566.$ $r = \cos(\frac{1}{3}\theta)$ The angle coefficient is $\frac{1}{3} \approx 0.33333$. $1 \times 3 = 3$, which is odd. Therefore, the curve repeats itself every $\pi \times 3 \times 1 \approx 9.425.$
• If $n$ is irrational, then there are an infinite number of petals.
Examples:
$r = \cos(e \theta)$
$\theta \text{ from } 0 \text{ to...}$
...$10$ ...$50$ ...$100$
$\text{Note: }e \approx 2.71828$
Below is an applet to graph polar roses, which is used to graph the examples above:
If you can see this message, you do not have the Java software required to view the applet.
Source code: Rose graphing applet
## Other Polar Curves
Archimedean Spirals
Archimedes' Spiral $r = a\theta$ The spiral can be used to square a circle and trisect an angle. Fermat's Spiral $r = \pm a\sqrt\theta$ This spiral's pattern can be seen in disc phyllotaxis. Hyperbolic spiral$r = \frac{a}{\theta}$ It begins at an infinite distance from the pole, and winds faster as it approaches closer to the pole. Lituus $r^2 \theta = a^2$It is asymptotic at the $x$ axis as the distance increases from the pole.
Limaçon[1]
The word "limaçon" derives from the Latin word "limax," meaning snail. The general equation for a limaçon is $r = b + a\cos(\theta)$.
• If $b = a/2$, then it is a trisectrix (see figure 2).
• If $b = a$, then it becomes a cardioid (see figure 3).
• If $2a > b > a$, then it is dimpled (see figure 4).
• If $b \geq 2a$, then the curve is convex (see figure 5).
$r = \cos(\theta)$ 1 $r = 0.5 + \cos(\theta)$ 2 Cardioid $r = 1 + \cos(\theta)$3 $r = 1.5 + \cos(\theta)$4 $r = 2 + \cos(\theta)$5
## Finding Derivatives[2]
A derivative gives the slope of any point in a function.
Consider the polar curve $r = f(\theta)$. If we turn it into parametric equations, we would get:
• $x = r \cos(\theta) = f(\theta) \cos(\theta)$
• $y = r \sin(\theta) = f(\theta) \sin(\theta)$
Using the method of finding the derivative of parametric equations and the product rule, we would get:
$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{\frac{dr}{d\theta} \sin(\theta) + r \cos(\theta)}{\frac{dr}{d\theta} \cos(\theta) - r \sin(\theta)}$
Note: It is not necessary to turn the polar equation to parametric equations to find derivatives. You can simply use the formula above.
Examples:
Find the derivative of $r = 1 + \sin(\theta)$ at $\theta = \frac{\pi}{3}$.
$\frac{dr}{d\theta} = \cos(\theta)$
$\frac{dy}{dx} = \frac{\frac{dr}{d\theta} \sin(\theta) + r \cos(\theta)}{\frac{dr}{d\theta} \cos(\theta) - r \sin(\theta)}$
$= \frac{\cos(\theta) \sin(\theta) + (1 + \sin(\theta) ) \cos(\theta)}{\cos(\theta)\cos(\theta) - (1 + \sin(\theta) ) \sin(\theta)}$
$= \frac{\cos(\theta)\sin(\theta) + \cos(\theta) + \cos(\theta)\sin(\theta)}{\cos^2(\theta) - \sin(\theta) - \sin^2(\theta)}$
Note: Using the double-angle formula, we get $\cos^2(\theta) - \sin^2(\theta) = 1 - 2\sin^2(\theta)$
$= \frac{\cos(1+2\sin(\theta))}{1-2\sin^2(\theta)-\sin(\theta)}$
$= \frac{\cos(1+2\sin(\theta))}{(1+\sin(\theta))(1-2\sin(\theta))}$
$\frac{dy}{dx} \Big |_{\theta=\pi/3} = \frac{\cos(\pi/3)(1+2\sin(\pi/3))}{(1+\sin(\pi/3))(1-2\sin(\pi/3))}$
## Finding Areas and Arc Lengths[2]
Area of a sector of a circle.
To find the area of a sector of a circle, where $r$ is the radius, you would use $A = \frac{1}{2} r^2 \theta$.
$A = \int_{-\frac{\pi}{4}}^\frac{\pi}{4}\! \frac{1}{2} \cos^2(2\theta) d\theta$
Therefore, for $r = f(\theta)$, the formula for the area of a polar region is:
$A = \int\limits_a^b\! \frac{1}{2} r^2 d\theta$
The formula to find the arc length for $r = f(\theta)$ and assuming $r$ is continuous is:
$L = \int\limits_a^b\! \sqrt{r^2 + {\bigg(\frac{dr}{d\theta}\bigg)} ^2}$ $d\theta$
# Why It's Interesting
Polar coordinates are often used in navigation, such as aircrafts. They are also used to plot gravitational fields and point sources. Furthermore, polar patterns are seen in the directionality of microphones, which is the direction at which the microphone picks up sound. A well-known pattern is a cardioid.
## Possible Future work
• More details can be written about the different curves, maybe they can get their own pages.
• Applets can be made to draw these different curves, like the one on the page for roses.
# About the Creator of this Image
Polar Coordinates
Cardioid
Source code: Rose graphing applet
# References
Wolfram MathWorld: Rose, Limacon, Archimedean Spiral
Wikipedia: Polar Coordinate System, Archimedean Spiral, Fermat's Spiral
1. Weisstein, Eric W. (2011). http://mathworld.wolfram.com/Limacon.html. Wolfram:MathWorld.
2. 2.0 2.1 Stewert, James. (2009). Calculus Early Transcendentals. Ohio:Cengage Learning. |
# Relationship between Temperature Scales
## Relationship between Temperature Scales
### Relationship between Fahrenheit and Celsius
##### To determine the relationship between Fahrenheit and Celsius, we could draw a graph of F (temperature in °F) versus C (temperature in °C), for C in the range 0 ≤ C ≤ 100.
Using the general equation for a straight line:
#### y = m x + c
using the general straight line equation and above graph, the following equation obtained
We can now also express Celsius in terms of Fahrenheit. From Equation -1:
Now we solved the basic formulas for conversion of Celsius to Fahrenheit or vice versa. Using the same above procedure we can also calculate the relationship between different temperature scales.
### Quickly Convert Between Temperatures:
• Celsius to Fahrenheit: Multiply the temperature by 2 and then add 30
• Fahrenheit to Celsius: subtract 30 from the temp. and then divide by 2
• Kelvin to Fahrenheit: Subtract 273.15, multiply by 1.8, then add 32
• Fahrenheit to Kelvin: Subtract 32, multiply by 5, divide by 9, then add 273.15
• Kelvin to Celsius: Add 273
• Celsius to Kelvin: Subtract 273
Example -1
#### Convert the absolute zero of temperature (–273.15 °C), to degrees Fahrenheit.
F = (9/5)C + 32 = (9/5)×(-273.15) + 32 = -459.67 °F ≈ -460 °F
Example -2
#### Convert 86 °F into degrees Celsius, Rankine and Kelvin.
C = (5/9)(F – 32) = (5/9)×(86 – 32) = 30 °C
R = F + 460 = 86 + 460 = 546 R
K = C + 273 = 30 + 273 = 303 K
Example -3
#### Determine the temperature in degrees Fahrenheit that would translate into the same value in degrees Celsius.
In Equation -2 (or Equation -1), set C = F:
F = (5/9)(F – 32) ⇒ 9F = 5F – 160 ⇒ 4F = -160 ⇒ F = -40 °F
Example -4
#### Convert 0 °F to degrees Celsius.
C = (5/9)(F – 32) = (5/9)(0 – 32) = -17.78 °C |
#### Basic Rules of Algebra
The real numbers form an abstract set, denoted R which contains two special distinct numbers 0 and 1 and for which we have two operations, addition and multiplication that combine two given real numbers to make a third real number. Addition of a and b is denoted a+b and multiplication of a and b is denoted ab. Moreover, the following basic rules are satisfied:
BR1
a + b = b + a , for all a, b in R
BR2
(a+b) + c = a + (b+c) , for all a, b, c in R
BR3
a + 0 = a , for all a in R
BR4
For any a in R, there exists a number -a in R so that a + (-a) = 0
BR5
ab = ba , for all a, b in R
BR6
(ab)c = a(bc) , for all a, b, c in R
BR7
a1 = a , for all a in R
BR8
If a in R and a not equal to 0, then there exists a number b in R so that ba = 1 (Usually this b is denoted 1/a)
BR9
a(b+c) = ab + ac , for all a, b, c in R
This rule is called expanding. If you are given an expression of the form (a + b)(c + d) and are told to expand it, you essentially do the same thing. Informally, it is called "FOILing" the expression. "FOIL" stands for First, Outside, Inside and Last. This means that you multiply the first term in each bracket together (i.e. ac), then multiply the outside terms together (i.e. ad), then multiply the inside terms together (i.e. bc) and finally multiply the last terms together (i.e. bd).The result is:
(a + b)(c + d) = ac + ad + bc + bd, for all a,b,c,d in R
Notation: We usually write a - b instead of a + (-b).
The real numbers are usually pictured as a number line, which is a horizontal straight line with 0 and 1 marked on it with 1 to the right of 0.
``` ________________________________________________
0 1
```
There are order properties that the real number line has that originate in the choice of 1 being to the right of 0. If a is a point on the number line on the same side of 0 as 1 is, we say a is greater than 0 and denote this by a > 0. If a and b are two points on the number line such that a - b > 0, then we say a is greater than b and denote this by a > b. We may also write b < a if this is more convenient. That is, a > b and b < a mean exactly the same thing. You recognize it on the number line as a being to the right of b. We read b < a as "b is less than a".
There are three more basic rules of the real numbers that involve order. We list them here formally so that they can be easily refered to later.
BR10
For any a in R, exactly one of the following statements is true:
(i)
a > 0
(ii)
a = 0
(iii)
a < 0
BR11
For a,b in R, if a > 0 and b > 0, then a + b > 0
BR12
For a,b in R, if a > 0 and b > 0, then ab > 0 |
# What is a Centroid?
Almost every closed geometrical shape has a specific point where all its weight lies. That point is normally called the Centroid of that figure.
Mathematically, we can write the centroid definition as “The point where a specific figure is balanced is called a Centroid”.
This specific point lies inside that figure and may be its central point. So, we can also say that it is the center of the figure where its weight lies, not the actual central point.
## What is the Centroid theorem?
Like other parameters, a centroid also has some specific properties that can be summarized in the form of a theorem. The centroid theorem states that the lengths of lines on both sides of the centroid should be in a ratio of 2:1.
It means that one side of the line of the centroid will be double to the other side. In simple words, the centroid doesn’t cut the medians from the center but it cuts it from one of its sides. Mathematically, we can show the centroid theorem using the following example.
Suppose we have a triangle with vertices, “A”, “B”, and “C” with the midpoints “D”, “E”, and “F” while the centroid of that triangle is represented by point “P”. So, we can write the general theorem of centroid as
AP : PD = BP : PE = CP : PF = 2 : 1
The above statement shows that the sides written on the left side of the ratio will be double as compared to the sides written on the right side of the ratio. This is the actual statement of the centroid theorem that summarizes all its properties in a mathematical relation.
## What is a Centroid of a triangle?
Generally, most of the students assume that a triangle only involves three sides with three angles. It is completely wrong! Because multiple other terms and lengths are also involved in this figure. One of those terms/parameters is a centroid where the whole triangle is considered to be balanced.
The centroid of a triangle is defined as the point where the medians of all its sides meet. It is considered the central point of any triangle too. It doesn’t matter whether you are dealing with a right triangle or a scalene triangle, you can find its centroid using some simple steps that we will discuss further.
## How to find the Centroid of a Triangle?
The centroid of a triangle is the point that you can find practically as well as mathematically too. To find this point mathematically, you can use the centroid of a triangle formula given below.
Centroid of a Triangle = [(x1 + x2 + x33) ,(y1 + y2 + y33)]
In this formula, “x” and “y” are the coordinates of three points of a triangle respectively. You only have to put the values of the points and solve them using basic mathematical operations.
## Where is the Centroid of a right triangle located?
The centroid of a right triangle is located inside this figure and it is the point where the medians of the triangle meet. As it is a specific type of triangle with a particular shape, it might be hard to find this point.
To make the process easier and faster, you can use our Centroid Calculator. This online maths calculator has been programmed for this calculation. You can locate the centroid of a right triangle within seconds using this advanced tool. Just enter the values and you will get the answer quickly with 100% accuracy.
## Solved Example of Calculating the Centroid
If you are struggling with the concept of the centroid of a triangle and are unable to understand the calculation, you should check the following example. We have solved it just for your better understanding
example:
Find the centroid of a triangle if its vertices are: A(2, -1), B(4, 7), and C(6, 9).
Solution:
By using the above formula, we can easily find the centroid of this triangle.
Centroid of a Triangle = [(x1 + x2 + x33) ,(y1 + y2 + y33)]
= [(2 + 4 + 63) ,(-1 + 7 + 93)]
After simplifying the equation, we get:
Centroid of a triangle = (4, 5)
## Conclusion
You can easily find the centroid of any triangle using the above formula. It might be possible that you have cleared all your doubts after viewing the solved example. We have also shared the comprehensive overview of the centroid of a triangle to let you understand this parameter.
But if you are still looking for assistance, you should use the centroid of a triangle calculator. This tool will help you in solving your assignments as well as let you understand the solution properly.
### FAQ
How do you find the centroid?
To find the centroid of any figure, we need to use different formulas. We can also find this point practically using a compass, scale, and pencil.
Is the centroid always 2/3?
Yes, the centroid of a triangle will always be at a distance of 2/3 of the distance between a vertex and the midpoint of the sides
Is the centroid of a triangle 2/3 or 1/3?
The centroid of a triangle is at a distance of 2/3, not 1/3.
Where is the centroid of an acute triangle?
The centroid of an acute angle lies inside the figure. Doesn’t matter which type of triangle you are dealing with, the centroid point will be in its inside.
How do you find the centroid of a right triangle?
To find the centroid of a right triangle, we can use a specific formula or draw the medians. The point where all medians will meet is called the centroid of that triangle.
What are the Properties of the Centroid of a triangle?
The centroid of a triangle has specific properties given below:
• It lies inside the figure.
• There is only one centroid of one triangle.
• The centroid of a triangle lies at a distance of 2/3 of the total distance between a vertex and a midpoint of the sides.
What is the Relationship Between the Orthocentre, Circumcentre, and Centroid of the Triangle?
In a triangle, the orthocenter, circumcenter, and centroid will be collinear. It means all these points lie on the same line |
What is length and weight? We all come up with a question “What is your height"? And we all answer it accordingly. For example my height is 185 cm. But again the question arises what is height actually? Let’s find out the answer to all these questions.
Length is actually the measurement of how long is something.
• As mentioned above my height is 185 cm or I can say that I am 185 cm long.
• Here my measurement of height is my length.
Now all of you will ask this question to each other and we all get to know each other’s height. Isn’t it interesting?
So let’s explore more about the topic.
• There are instruments to measure the length of any body like rulers (scales), tapes etc.
• The length is always measured in mm (millimeter), cm (centimeters), m (meters), km (kilometers) etc. These are the units of length.
• The standard unit of length is Meter.
• For shorter length like of a pencil we use the unit cm, for a bigger length like of a person we use the unit m, and for a larger length like of a road we use the unit km.
Conversions
The changing of unit into the other using some mathematical formulas are known as conversions.
• The smallest unit to measure a distance is millimeter (mm).
1 mm = 0.1 cm = 0.001 m
Example 1:- The length of tip of your pencil is 5mm or 0.5 cm.
• The next unit to measure length is centimeter (cm).
1 cm = 0.01 m and 1 cm = 10 mm
Example 2:- The length of your pencil is 20 cm.
Therefore in mm it would be 20 x 10 = 200mm.
• The next unit to measure length is meter (m).
1 m = 100 cm = 1000mm.
Example 3:- The length of your classroom is 7 m.
Therefore it will be 700 cm and 7000 mm.
• The largest unit to measure length is km.
1 km = 1000 m = 100000 cm = 1000000 mm.
Example 4:- The length of Narmada River is 1312 km.
It equals to 1312 x 1000 m = 1312000 m.
Let us now find where and how we are going use these concepts which we have learned till now.
Example 5:- Convert 3 m into cm.
Solution:- As we know that 1 m = 100 cm.
Therefore 3 m = (3 x 100) cm = 300 cm.
Example 6:- Convert 5.32 m into cm.
Solution:- As we know that 1 m = 100 cm.
Therefore 5.32 m = (5.32 x 100) cm = 532 cm.
Example 7:- A shopkeeper has 128 m of wire. He sold 23 m wire to a customer. How much length of wire is he left with?
Solution:- Length of wire before selling = 128 m
Length of wire sold = 23 m
Length of wire remaining is = (128-23) m = 105 m.
Example 8:- Identify the length of pencil with the help of scale.
Here the length of pencil can be identified as 9 cm.
Practice questions
Question 1:- Convert the following as directed.
1. 5 cm into mm.
2. 32 cm into mm.
3. 123 m into mm.
4. 52 km into m.
Question 2:- Ajay has ropes of 3 colors Red, Blue and Black. The Red rope is 23 m long; the Black rope is 12m long. Ajay has to measure a length of 69 m with these ropes. How much length of Blue rope is required?
Question 3:- State True or False:
1. The smallest unit of length is cm.
2. 1 cm = 100 m
3. 1 km = 1000 cm.
4. 3 m = 330mm
Weight
We all again come up with a question “What is your weight". And we answer it accordingly. I will answer it by saying “My weight is 71 Kg." But still the question arises what actually weight is? Let’s just find out the answers.
• Weight is actually the measurement of how heavy is something.
• As mentioned above I am 71 kg heavy or my weight is 71 kg.
Let’s do a activity and ask each other how heavy they are.
And let’s explore more about the topic weight.
• The instruments used to measure the weight are Weighing machines, balance scales etc.
• The units of weight are g (Gram) and kg (Kilogram).
• The standard unit of weight is Kilogram.
For lighter objects like for weighing of a mobile phone we use the unit Gram and for weighing heavier objects like for weighing of an Elephant we use the unit Kilogram.
Conversions
The changing of units into one another using some mathematical formulae are known as conversions.
• 1 g = 0.001 Kg or 1/1000 gm.
Example 1:- The mass of an object is 250 gm. What will it be in Kilograms?
Solution:- As we know that 1 g = 0.001 kg
Therefore 250 g = (250 x 0.001) kg = 0.250 kg.
• 1 Kg = 1000 g.
Example 2:- The weight of an object is 12 kg. Convert it into grams.
Solution:- As we know that 1 kg = 1000 g.
Therefore 12 kg = (12 x 1000) g = 12000 g.
Example 3:- Ram is going to a market to purchase sugar and Flour. Ram has a carry bag which can sustain a max weight of 14 kg and 300 g. If Ram has to buy 6 kg and 200 g of sugar then how much flour can he purchase so that he can take both of them inside the carry bag back to home?
Solution:- Weight of sugar = 6 kg and 200 g.
Maximum weight that can be kept inside the carry bag
= 14 kg and 300 g.
Therefore without damaging the carry bag the weight of flour purchased must be =
(Max weight) – (Weight of Sugar)
= 14 kg – 6 kg and 300 g – 200 g
= 8 kg and 100 g.
Hence Ram can purchase maximum of 8 kg and 100 g of flour.
Practice Questions
Question 1:- Convert 23.23 kg into g .
Question 2:- Convert 6300 g into Kg.
Question 3:- State True or False:
1. 1 kg = 1000 g
2. 200 g =0.2 kg
3. 2 kg = 42000 g
Question 4:- The weight of Ajay is 30 kg and the weight of Raman is 28 kg. Who is heavier and by how much?
Recap
• Length and Weight are the measurement quantities.
• The standard unit of length is meter and weight is Kg.
• 1 km = 1000 m and 1 m = 100 cm.
• 1 kg = 1000 g.
• For shorter distance we use the unit m or cm and for longer distance we use the unit km.
For lighter objects we use the unit g and for the heavier objects we use the unit Kg. |
I. Fundamentals
Chelsey Hamm and Mark Gotham
Key Takeaways
• are meters in which the beat divides into three and then further subdivides into six.
• have groupings of two beats, have groupings of three beats, and have groupings of four beats. You can determine these groupings aurally by listening carefully and tapping along to the beat.
• There are different conducting patterns for duple, triple, and quadruple meters; these are the same in both compound and simple meters.
• in compound meters express two things: how many divisions are contained in each measure (the top number), and the —which note gets the division (the bottom number).
• Rhythms in compound meters get different counts based upon their division unit. Beats that are not articulated (because they contain more than one beat or because of ties, rests, or dots) receive parentheses around their counts.
In the previous chapter, Simple Meter and Time Signatures, we explored rhythm and time signatures in —meters in which the beat divides into two and further subdivides into four. In this chapter, we will learn about —meters in which the beat divides into three and further subdivides into six.
# Listening to and Conducting Compound Meters
Compound meters can be , , or , just like simple meters. In other words, the beats of compound meters group into sets of either two, three, or four. The difference is that each beat divides into three divisions instead of two, as you can hear by listening carefully to the following examples:
• “End of the Road” (1992) by Boyz II Men is in a —the beats group into a two pattern. Tap along to the beat and notice how it divides into three parts instead of two. If you further divide the beat (by tapping twice as fast), you will feel that the beat subdivides into six parts.
• The second movement (Minuet) of Franz Joseph Haydn’s Sonata no. 42 in G Major (1784) is in a compound . Listen for the groupings of three beats, each of which divides into three.
• Finally, a compound contains four beats, each of which divides into three. Listen to “Exogenesis Symphony Part III” (2010) by the alternative rock band Muse. This is in a compound quadruple meter; in other words, the beats are grouped into a four pattern.
In general, it is less common for music to be written in compound meters. Nonetheless, you must learn how to read music and perform in these meters in order to master Western musical notation.
Review the conducting patterns for simple meters in the previous chapter, as they are the same for compound meters.
# Time Signatures
in compound meters are equivalent to one beat grouping (duple, triple, or quadruple), just as they are in simple meters. However, the two numbers in the express different information for compound meters. The top number of a time signature in compound meter expresses the number of divisions in a measure, while the bottom number expresses the —which note value is the division. Example 1 shows a common compound-meter time signature.
Just like in simple meter, compound-meter time signatures are not fractions (and there is no line between the two numbers), and they are placed after the clef on the staff. In Example 1, the top number (6) means that each measure will contain six divisions; the bottom number (8) means that the eighth note is the division. This means that each measure in this time signature will contain six eighth notes, as you can verify by examining Example 1.
In compound meters, the top number is always a multiple of three. Divide this number by three to find the corresponding number of beats in simple meter: top numbers of 6, 9, and 12 correspond to duple, triple, and quadruple meters respectively. In compound meters, the bottom number is usually one of the following:
• 8, which means the eighth note receives the division.
• 4, which means the quarter note receives the division.
• 16, which means the sixteenth note receives the division.
The following table summarizes the six categories of meters that we have covered so far:
[table id=36 /]
Example 2. Categories of meters.
# Counting in Compound Meter
While counting compound meter rhythms, it is recommended that you conduct in order to keep a steady tempo. Because beats in compound meter divide into three, they are always dotted. Beats in compound meter are as follows:
• If 8 is the bottom number, the beat is a dotted quarter note (equivalent to three eighth notes).
• If 4 is the bottom number, the beat is a dotted half note (equivalent to three quarter notes).
• If 16 is the bottom number, the beat is a dotted eighth note (equivalent to three sixteenth notes).
In simple meters, the beat divides into two parts, the first accented and the second non-accented. In compound meters, the beat divides into three parts, the first accented and the second and third non-accented. The counts for compound meter are different from simple meter, as demonstrated in Example 3, which is in $\mathbf{^6_8}$.
Example 3. Counting in a compound duple meter.
In this time signature, each measure has two beats (6÷3=2), indicating duple meter. Each dotted quarter note (the beat) gets a count, which is expressed in Arabic numerals, like in simple meter. For notes that are longer than one beat (such as the dotted half note in the fourth measure of Example 3), the beats that are not counted out loud are still written in parentheses. Divisions are counted using the syllables “la” (first division) and “li” (second division). As the final measure of Example 3 shows, further subdivisions at the sixteenth-note level are counted as “ta,” with the “la” and “li” syllables on the eighth-note subdivisions remaining consistent.
The third measure of Example 3 presents two of the most common compound-meter rhythms with divisions, so make sure to review this measure carefully if you are not familiar with compound meter.
Please note that your instructor may employ a different counting system. Open Music Theory privileges American traditional counting, but this is not the only method.
Example 4 gives examples of rhythms in (a) duple, (b) triple, and (c) quadruple meter. Just as with simple meters, compound duple meters have only two beats, compound triple meters have three beats, and compound quadruple meters have four beats.
Example 4. (a) Compound duple has two beats, (b) compound triple has three beats, and (c) compound quadruple has four beats.
Like in simple meters, beats that are not articulated because of rests and ties are written in parentheses and not counted out loud, as shown in Example 5. However, because dotted notes receive the beat in compound meters, dotted rhythms do not cause beats to be written in parentheses the way they do in simple meters.
Example 5. Beats that are not counted out loud are put in parentheses.
# Counting with Division Units of 4 and 16
So far, we have focused on meters with a dotted-quarter beat. In compound meters with other beat units (shown in the bottom number of the time signature), the same counting patterns are used for the beats and subdivisions, but they correspond to different note values (Example 6).
Example 6. The same rhythm written with three different beat units: (a) dotted quarter, (b) dotted half, and (c) dotted eighth.
Each of these rhythms sounds the same and is counted the same. They are also all considered compound triple meters. The difference in each example is the bottom number—which note gets the division unit (eighth, quarter, or sixteenth), which then determines the beat unit.
# Beaming, Stems, and Flags
In compound meters, still connect notes together by beat; beaming therefore changes in different time signatures. In the first measure of Example 7, sixteenth notes are grouped into sets of six, because six sixteenth notes in a $\mathbf{^6_8}$ time signature are equivalent to one beat. In the second measure of Example 7, sixteenth notes are grouped into sets of three, because three sixteenth notes in a $\mathbf{^{\:6}_{16}}$ time signature are equivalent to one beat.
Example 7. Beaming in two different meters.
When the music involves note values smaller than a quarter note, you should always clarify the meter with beams, regardless of whether the time signature is simple or compound. Example 8 shows twelve sixteenth notes beamed properly in two different meters. The first measure is in simple meter, so the notes are grouped by beat into sets of four; in the second measure, the compound meter requires the notes to be grouped by beat into sets of six.
The same rules of stemming and flagging that applied in simple meter still apply in compound meter. For notes above the middle line, stems and flags point downward on the left side of the note, and for notes below the middle line, stems and flags point upward on the right side of the note. Stems and flags on notes on the middle line can point in either direction, depending on the surrounding notes.
Like in simple meters, partial beams can be used for mixed rhythmic groupings. If you aren’t yet familiar with these conventions, pay special attention to how the notes in Example 9 are beamed.
Example 9. The most common partially beamed variations with a division unit of the eighth note.
Online Resources
Assignments from the Internet
1. Meter Identification (Simple and Compound) (.pdf,), and with Bar Lines (.pdf)
2. Meter Beaming (Simple and Compound) (.pdf), and pp. 4 and 5 (.pdf)
3. Time Signatures (Simple and Compound) (.pdf)
4. Counting in 6/8 (.pdf.pdf.pdf)
5. Time Signatures (.pdf.pdf, .pdf)
6. Bar Lines (.pdf), and p. 2 (.pdf)
Assignments
1. Notes, Rests, Bar Lines (.pdf, .docx)
2. Re-beaming (.pdf, .musx) |
# Changing Numbers To Percents Worksheet Page 78 77 6th Pre-algebra
A Realistic Figures Worksheet will help your youngster be more informed about the concepts behind this rate of integers. With this worksheet, pupils should be able to fix 12 various troubles linked to rational expression. They will learn to increase a couple of phone numbers, class them in couples, and figure out their goods. They will likely also process simplifying realistic expression. As soon as they have mastered these methods, this worksheet is a valuable device for continuing their research. Changing Numbers To Percents Worksheet Page 78 77 6th Pre-algebra.
## Realistic Figures really are a percentage of integers
There are 2 kinds of phone numbers: rational and irrational. Realistic numbers are defined as whole figures, whilst irrational phone numbers usually do not perform repeatedly, and possess an infinite quantity of numbers. Irrational figures are non-zero, no-terminating decimals, and square beginnings that are not best squares. They are often used in math applications, even though these types of numbers are not used often in everyday life.
To define a logical amount, you need to understand such a reasonable number is. An integer can be a total quantity, as well as a reasonable quantity is really a proportion of two integers. The rate of two integers may be the variety ahead split through the quantity on the bottom. If two integers are two and five, this would be an integer, for example. There are also many floating point numbers, such as pi, which cannot be expressed as a fraction.
## They could be manufactured into a small percentage
A reasonable amount includes a numerator and denominator that are not zero. Consequently they can be expressed being a fraction. In addition to their integer numerators and denominators, logical figures can furthermore have a negative worth. The unfavorable importance must be put on the left of along with its definite benefit is its distance from absolutely nothing. To easily simplify this case in point, we will point out that .0333333 is really a portion that may be created being a 1/3.
As well as adverse integers, a logical amount can also be made right into a small fraction. As an example, /18,572 is actually a rational number, whilst -1/ is just not. Any fraction consisting of integers is logical, given that the denominator is not going to consist of a and will be published for an integer. Likewise, a decimal that leads to a level is also a realistic variety.
## They make sensation
Regardless of their label, realistic amounts don’t make a lot feeling. In math, they may be single entities with a special span around the variety range. This means that when we count up anything, we are able to order the size by its ratio to its authentic number. This contains real even if there are infinite rational amounts in between two distinct figures. In other words, numbers should make sense only if they are ordered. So, if you’re counting the length of an ant’s tail, a square root of pi is an integer.
In real life, if we want to know the length of a string of pearls, we can use a rational number. To obtain the period of a pearl, by way of example, we could add up its size. One particular pearl weighs in at 10 kgs, which is actually a rational variety. In addition, a pound’s excess weight is equal to twenty kilograms. As a result, we should certainly break down a pound by ten, without worry about the duration of an individual pearl.
## They may be expressed as being a decimal
If you’ve ever tried to convert a number to its decimal form, you’ve most likely seen a problem that involves a repeated fraction. A decimal variety may be composed like a several of two integers, so 4 times 5 is equal to 8-10. A comparable problem involves the frequent fraction 2/1, and either side needs to be divided up by 99 to obtain the proper answer. But how will you create the conversion? Here are a few examples.
A rational quantity can also be printed in various forms, which include fractions and a decimal. One way to symbolize a rational quantity within a decimal is to divide it into its fractional equal. You will find 3 ways to split a reasonable amount, and each of these methods produces its decimal comparable. One of these brilliant techniques would be to separate it into its fractional comparable, and that’s what’s called a terminating decimal. |
# Blast Off
## Problem:
We would like to find out the accuracy using a foam rocket, by aiming to get the foam rocket into a bucket at certain distances.
## Research:
We have found that foam rockets can shoot up to distances of thirty feet. We know that our rocket is 120 centimeters cubed, and the bucket that we are shooting into has a volume of 17,710 centimeters cubed. This means that it is 148 times smaller than the bucket.
## Hypothesis:
If you shoot a rocket from 0 to 20 feet, then you will be able to make it in the bucket with out missing by more than 45cm.
## Materials:
• One bucket (preferably 35cm wide, 22 cm long, and 23 cm tall)
• One foam rocket
• A long empty space (indoors (30 feet long)
• A person who can shoot a foam rocket
## Procedures:
1. Place your bucket at the end of a long hall.
2. Create a data table with four trials, and 5, 10, 20, 30, and average distance away options.
3. Mark zones of 5, 10, 20, and 30 feet.
4. Stand at 5 feet away and try to shoot the foam rocket into the bucket.
5. Record Data (If misses write distance away)
6. Repeat #4 and 5 with all of the distances.
7. After this, find the distance away you were from the bucket. of times that you made the rocket in from each distance.
## Data and Calculations:
:::::::::::::::5 feet :10 feet :20 feet :30 feet
Trial 1 ::::Yes :::::Yes ::::::85cm :::90cm
Trial 2 ::::Yes :::::Yes ::::::23cm :::33cm
Trial 3 ::::10cm ::14cm :::Yes ::::::93cm
Average*:3.3cm:4.6cm:::36cm::::72cm
(Average is average away)
## Conclusion:
We now know that when shooting a rocket from different distances, you shall have your rocket land further away when you are further from the target. Our hypothesis was wrong, because we had 85 cm away when at 20 feet from the target. We support this by our results, showing in our averages that the further away you are from a target the further it shall probably be from the target. We believe this, because when you shoot mass over a longer distance it will be pulled down by gravity. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 4.4: Slope and Rate of Change
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Find positive and negative slopes.
• Recognize and find slopes for horizontal and vertical lines.
• Understand rates of change.
• Interpret graphs and compare rates of change.
## Introduction
We come across many examples of slope in everyday life. For example, a slope is in the pitch of a roof, the grade or incline of a road, and the slant of a ladder leaning on a wall. In math, we use the word slope to define steepness in a particular way.
Slope= distance moved verticallydistance moved horizontally\begin{align*}\text{Slope} =\ \frac{\text{distance moved vertically} }{ \text{distance moved horizontally}}\end{align*}
This is often reworded to be easier to remember:
Slope= riserun\begin{align*}\text{Slope} = \ \frac{\text{rise}}{\text{run}}\end{align*}
Essentially, slope is the change in y\begin{align*}y\end{align*} if x\begin{align*}x\end{align*} increases by 1.
In the picture to the right, the slope would be the ratio of the height of the hill (the rise) to the horizontal length of the hill (the run).
Slope=34=0.75\begin{align*}\text{Slope} = \frac{3} {4} = 0.75\end{align*}
If the car were driving to the right it would climb the hill. We say this is a positive slope. Anytime you see the graph of a line that goes up as you move to the right, the slope is positive.
If the car were to keep driving after it reached the top of the hill, it may come down again. If the car is driving to the right and descending, then we would say that the slope is negative. The picture at right has a negative slope of -0.75.
Do not get confused! If the car turns around and drives back down the hill shown, we would still classify the slope as positive. This is because the rise would be -3, but the run would be -4 (think of the x\begin{align*}x-\end{align*}axis - if you move from right to left you are moving in the negative x\begin{align*}x-\end{align*}direction). Our ratio for moving left is:
Slope=34=0.75A negative divided by a negative is a positive.\begin{align*}\text{Slope} = \frac{-3} {-4} = 0.75 && \text{A negative divided by a negative is a positive.}\end{align*}
So as we move from left to right, positive slopes increase while negative slopes decrease.
## Find a Positive Slope
We have seen that a function with a positive slope increases in y\begin{align*}y\end{align*} as we increase x\begin{align*}x\end{align*}. A simple way to find a value for the slope is to draw a right angled triangle whose hypotenuse runs along the line. It is then a simple matter of measuring the distances on the triangle that correspond to the rise (the vertical dimension) and the run (the horizontal dimension).
Example 1
Find the slopes for the three graphs shown right.
There are already right-triangles drawn for each of the lines. In practice, you would have to do this yourself. Note that it is easiest to make triangles whose vertices are lattice points (i.e. the coordinates are all integers).
a. The rise shown in this triangle is 4 units, the run is 2 units.
Slope=42=2\begin{align*}\text{Slope} = \frac{4} {2} = 2\end{align*}
b. The rise shown in this triangle is 4 units, the run is also 4 units.
Slope=44=1\begin{align*}\text{Slope} = \frac{4} {4} = 1\end{align*}
c. The rise shown in this triangle is 2 units, the run is 4 units.
Slope=24=12\begin{align*}\text{Slope} = \frac{2} {4} = \frac{1} {2}\end{align*}
Example 2
Find the slope of the line that passes through the points (1, 2) and (4, 7).
We already know how to graph a line if we are given two points. We simply plot the points and connect them with a line. Look at the graph shown at right.
Since we already have coordinates for our right triangle, we can quickly work out that the rise would be 5 and the run would be 3 (see diagram). Here is our slope.
Slope=7241=53\begin{align*}\text{Slope} = \frac{7-2} {4-1} = \frac{5} {3}\end{align*}
If you look closely at the calculations for the slope you will notice that the 7 and 2 are the y\begin{align*}y-\end{align*}coordinates of the two points and the 4 and 1 are the x\begin{align*}x-\end{align*}coordinates. This suggests a pattern we can follow to get a general formula for the slope between two points (x1,y1)\begin{align*}(x_1, y_1)\end{align*} and (x2,y2)\begin{align*}(x_2, y_2)\end{align*}.
Slope between (x1,y1)\begin{align*} (x_1, y_1)\end{align*} and (x2,y2)=y2y1x2x1\begin{align*} (x_2, y_2) = \frac{y_2 - y_1} {x_2 - x_1}\end{align*} or m=yx\begin{align*} m = \frac{\triangle y} {\triangle x}\end{align*}
In the second equation, the letter m\begin{align*}m\end{align*} denotes the slope (you will see this a lot in this chapter) and the Greek letter delta (Δ)\begin{align*}(\Delta)\end{align*} means change. So another way to express slope is change in y\begin{align*}y\end{align*} divided by change in x\begin{align*}x\end{align*}. In the next section, you will see that it does not matter which point you choose as point 1 and which you choose as point 2.
## Find a Negative Slope
Any function with a negative slope is simply a function that decreases as we increase x\begin{align*}x\end{align*}. If you think of the function as the incline of a road a negative slope is a road that goes downhill as you drive to the right.
Example 3
Find the slopes of the lines on the graph to the right.
Look at the lines. Both functions fall (or decrease) as we move from left to right. Both of these lines have a negative slope.
Neither line passes through a great number of lattice points, but by looking carefully you can see a few points that look to have integer coordinates. These points have been identified (with rings) and we will use these to determine the slope. We will also do our calculations twice, to show that we get the same slope whichever way we choose point 1 and point 2.
For line A\begin{align*}A\end{align*}:
(x1,y1)m=(6,3) (x2,y2)=(5,1)=y2y1x2x1=(1)(3)(5)(6)=4110.364(x1,y1)=(5,1) (x2,y2)=(6,3)m=y2y1x2x1=(3)(1)(6)(5)=4110.364\begin{align*}(x_1, y_1) &= (-6, 3)\ (x_2, y_2) = (5, -1) && (x_1, y_1) = (5, -1)\ (x_2, y_2) = (-6, 3) \\ m &= \frac{y_2 - y_1} {x_2 - x_1} = \frac{(-1) - (3)} {(5) - (-6)} = \frac{-4} {11} \approx -0.364 && m = \frac{y_2 - y_1} {x_2 - x_1} = \frac{(3) - (-1)} {(-6) - (-5)} = \frac{-4} {11} \approx -0.364\end{align*}
For line B\begin{align*}B\end{align*}:
(x1,y1)m=(4,6) (x2,y2)=(4,5)=y2y1x2x1=(5)(6)(4)(4)=118=1.375(x1,y1)=(4,5) (x2,y2)=(4,6)m=y2y1x2x1=(6)(5)(4)(4)=118=1.375\begin{align*}(x_1, y_1) &= (-4, 6)\ (x_2, y_2) = (4, -5) && (x_1, y_1) = (4, -5)\ (x_2, y_2) = (-4, 6) \\ m &= \frac{y_2 - y_1} {x_2 - x_1} = \frac{(-5) - (6)} {(4) - (-4)} = \frac{-11} {8} = -1.375 && m = \frac{y_2 - y_1} {x_2 - x_1} = \frac{(6) - (-5)} {(-4) - (4)} = \frac{11} {-8} = -1.375\end{align*}
You can see that whichever way you select the points, the answers are the same!
Solution
Line A\begin{align*}A\end{align*} has slope -0.364. Line B\begin{align*}B\end{align*} has slope -1.375.
Multimedia Link The series of videos starting at Khan Academy Slope (8:28) models several more examples of finding the slope of a line given two points.
## Find the Slopes of Horizontal and Vertical lines
Example 4
Determine the slopes of the two lines on the graph at the right.
There are two lines on the graph. A(y=3)\begin{align*}A (y = 3)\end{align*} and B(x=5)\begin{align*}B(x= 5)\end{align*}.
Let's pick two points on line A\begin{align*}A\end{align*}. say, (x1,y1)=(4,3)\begin{align*}(x_1, y_1 )=(-4, 3)\end{align*} and (x2,y2)=(5,3)\begin{align*}(x_2, y_2 )=(5, 3)\end{align*} and use our equation for slope.
m=y2y1x2x1=(3)(3)(5)(4)=09=0\begin{align*} m = \frac{y_2 - y_1} {x_2 - x_1} = \frac{(3) - (3)} {(5) - (-4)} = \frac{0} {9} = 0\end{align*}
If you think about it, this makes sense. If there is no change in \begin{align*}y\end{align*} as we increase \begin{align*}x\end{align*} then there is no slope, or to be correct, a slope of zero. You can see that this must be true for all horizontal lines.
Horizontal lines \begin{align*}(y = \text{constant})\end{align*} all have a slope of 0.
Now consider line \begin{align*}B\end{align*}. Pick two distinct points on this line and plug them in to the slope equation.
\begin{align*}(x_1, y_1 )=(5, -3)\end{align*} and \begin{align*}(x_2, y_2 )=(5, 4)\end{align*}.
\begin{align*} m = \frac{y_2 - y_1} {x_2 - x_1} = \frac{(4) - (-3)} {(5) - (5)} = \frac{7} {0} && \text{A division by zero!}\end{align*}
Divisions by zero lead to infinities. In math we often use the term undefined for any division by zero.
Vertical lines \begin{align*}(x = \text{constant})\end{align*} all have an infinite (or undefined) slope.
## Find a Rate of Change
The slope of a function that describes real, measurable quantities is often called a rate of change. In that case, the slope refers to a change in one quantity \begin{align*}(y)\end{align*} per unit change in another quantity \begin{align*}(x)\end{align*}.
Example 5
Andrea has a part time job at the local grocery store. She saves for her vacation at a rate of \$15 every week. Express this rate as money saved per day and money saved per year.
Converting rates of change is fairly straight forward so long as you remember the equations for rate (i.e. the equations for slope) and know the conversions. In this case \begin{align*}1\ \text{week} =\ 7\ \text{days}\end{align*} and \begin{align*}52\ \text{weeks} = 1\ \text{year}\end{align*}.
\begin{align*}\text{rate} & = \frac{\15} {1\ \text{week}} \cdot \frac{1\ \text{week}} {7\ \text{days}} = \frac{\ 15} {7\ \text{days}} = \frac{15} {7}\ \text{dollars per day} \approx \ \ 2.14\ \text{per day} \\ \text{rate} & = \frac{\ 15} {1\ \text{week}} \cdot \frac{52\ \text{week}} {1\ \text{year}} = \15 \cdot \frac{52} {\text{year}} = \ 780\ \text{per year}\end{align*}
Example 6
A candle has a starting length of 10 inches. Thirty minutes after lighting it, the length is 7 inches. Determine the rate of change in length of the candle as it burns. Determine how long the candle takes to completely burn to nothing.
In this case, we will graph the function to visualize what is happening.
We have two points to start with. We know that at the moment the candle is lit \begin{align*}(\text{time} = 0)\end{align*} the length of the candle is 10 inches. After thirty minutes \begin{align*}(\text{time} = 30)\end{align*} the length is 7 inches. Since the candle length is a function of time we will plot time on the horizontal axis, and candle length on the vertical axis. Here is a graph showing this information.
The rate of change of the candle is simply the slope. Since we have our two points \begin{align*}(x_1, y_1 )=(0,10)\end{align*} and \begin{align*}(x_2, y_2 )=(30, 7)\end{align*} we can move straight to the formula.
\begin{align*}\text{Rate of change} = \frac{y_2 - y_1} {x_2 - x_1}= \frac{(7\ inches)- (10\ inches)} {(30\ minutes) - (0\ minutes)} = \frac{-3 \ inches} {30\ minutes} = -0.1\ inches \ per \ minute\end{align*}
The slope is negative. A negative rate of change means that the quantity is decreasing with time.
We can also convert our rate to inches per hour.
\begin{align*}\text{rate} = \frac{- 0.1\ inches} {1\ minute} \cdot \frac{60\ minutes} {1\ hour} = \frac{-6\ inches} {1\ hour} = -6\ inches \ per \ hour\end{align*}
To find the point when the candle reaches zero length we can simply read off the graph (100 minutes). We can use the rate equation to verify this algebraically.
\begin{align*}\text{Length burned } & = \text{rate} \times \text{time}\\ 0.1 \times 100 & = 10\end{align*}
Since the candle length was originally 10 inches this confirms that 100 minutes is the correct amount of time.
## Interpret a Graph to Compare Rates of Change
Example 7
Examine the graph below. It represents a journey made by a large delivery truck on a particular day. During the day, the truck made two deliveries, each one taking one hour. The driver also took a one hour break for lunch. Identify what is happening at each stage of the journey (stages A through E)
Here is the driver's journey.
A. The truck sets off and travels 80 miles in 2 hours.
B. The truck covers no distance for 1 hours.
C. The truck covers \begin{align*}(120 - 80) = 40\end{align*} miles in 1 hours
D. the truck covers no distance for 2 hours.
E. The truck covers -120 miles in 2 hours.
Lets look at the rates of change for each section.
A. \begin{align*}\text{Rate of change} = \frac{\Delta y} {\Delta x} = \frac{80 \ miles} {2 \ hours} = 40 \ miles \ per \ hour\end{align*}
• The rate of change is a velocity! This is a very important concept and one that deserves a special note!
The slope (or rate of change) of a distance-time graph is a velocity.
You may be more familiar with calling miles per hour a speed. Speed is the magnitude of a velocity, or, put another way, velocity has a direction, speed does not. This is best illustrated by working through the example.
On the first part of the journey sees the truck travel at a constant velocity of 40 mph for 2 hours covering a distance of 80 miles.
B. Slope = 0 so rate of change = 0 mph. The truck is stationary for one hour. This could be a lunch break, but as it is only 2 hours since the truck set off it is likely to be the first delivery stop.
C. \begin{align*}\text{Rate of change} = \frac{\Delta y} {\Delta x} = \frac{(120 -80)\ miles} {(4 - 3)\ hours} = 40 \ miles\ per\ hour\end{align*}. The truck is traveling at 40 mph.
D. Slope = 0 so rate of change = 0 mph. The truck is stationary for two hours. It is likely that the driver used these 2 hours for a lunch break plus the second delivery stop. At this point the truck is 120 miles from the start position.
E. \begin{align*}\text{Rate of change} = \frac{\triangle y} {\triangle x} = \frac{(0 - 120)\ miles} {(8 - 6)\ hours} = \frac{-120\ miles} {2\ hours} = -60 \ miles \ per \ hour\end{align*}. The truck is traveling at negative 60 mph.
Wait, a negative velocity? Does this mean that the truck is reversing? Well, probably not. What it means is that the distance (and don’t forget that is the distance measured from the starting position) is decreasing with time. The truck is simply driving in the opposite direction. In this case, back to where it started from. So, the speed of the truck would be 60 mph, but the velocity (which includes direction) is negative because the truck is getting closer to where it started from. The fact that it no longer has two heavy loads means that it travels faster (60 mph as opposed to 40 mph) covering the 120 mile return trip in 2 hours.
## Lesson Summary
• Slope is a measure of change in the vertical direction for each step in the horizontal direction. Slope is often represented as “\begin{align*}m\end{align*}”.
• \begin{align*} \text{Slope} = \frac{rise} {run}\end{align*} or \begin{align*} m = \frac{\triangle y} {\triangle x}\end{align*}
• The slope between two points \begin{align*}(x_1 , y_1 )\end{align*} and \begin{align*} (x_2 , y_2 ) = \frac{y_2 - y_1} {x_2 - x_1}\end{align*}
• Horizontal lines \begin{align*}(y =\text{constant})\end{align*} all have a slope of 0.
• Vertical lines \begin{align*}(x = \text{constant})\end{align*} all have an infinite (or undefined) slope.
• The slope (or rate of change) of a distance-time graph is a velocity.
## Review Questions
1. Use the slope formula to find the slope of the line that passes through each pair of points.
1. (-5, 7) and (0, 0)
2. (-3, -5) and (3, 11)
3. (3, -5) and (-2, 9)
4. (-5, 7) and (-5, 11)
5. (9, 9) and (-9, -9)
6. (3, 5) and (-2, 7)
2. Use the points indicated on each line of the graphs to determine the slopes of the following lines.
3. The graph below is a distance-time graph for Mark’s three and a half mile cycle ride to school. During this ride, he rode on cycle paths but the terrain was hilly. He rode slower up hills and faster down them. He stopped once at a traffic light and at one point he stopped to mend a tire puncture. Identify each section of the graph accordingly.
1. -1.4
2. 2.67
3. -2.8
4. undefined
5. 1
6. -0.4
1. 3
2. 0.5
3. -2
4. 1
5. undefined
6. \begin{align*}\frac{1}{3}\end{align*}
1. uphill
2. stopped (traffic light)
3. uphill
4. downhill
5. stopped (puncture)
6. uphill
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# Prim’s Algorithm
By BYJU'S Exam Prep
Updated on: September 25th, 2023
Prim’s Algorithm is developed to discover a graph’s minimum spanning tree. The algorithm is popular and follows different steps to solve the same problem. The prim’s algorithm selects the root vertex initially and then traverses from vertex to vertex adjacently.
Prim’s algorithm is widely used as a Greedy algorithm that helps discover the most miniature spanning tree for a weighted undirected graph. This algorithm tends to search the subgroup of the edges that can construct a tree, and the complete poundage of all the edges in the tree should be minimal. Here, we will discuss what is Prim’s algorithm along with its pseudocode, algorithm, and other essential details.
Download Formulas for GATE Computer Science Engineering – Programming & Data Structures
## What is Prim’s Algorithm?
Prim’s algorithm has the property that the edges in the set T, which contains the edges of the minimum spanning tree when the algorithm proceeds step-by-step, always form a single tree. That is, at each step, there is only one connected component.
### Prim’s Algorithm Steps
The following steps are as follows:
• Begin with one starting vertex (say v) of a given graph G(V, E).
• Then, choose a minimum weight edge (u, v) connecting vertex v in set A to the vertices in the set (V-A) in each iteration. We always find an edge (u, v) of minimum weight, such as v ∈ A and u ∈ V-A. Then we modify the set A by adding u, A ← A ∪ {u}.
• The process is repeated until ≠V, that is, until all the vertices are not in the set A.
## Pseudocode of Prim’s Algorithm
The pseudocode for prim’s algorithm shows how we create two sets of vertices U and V-U. U contains the list of vertices that have been visited, and V-U is the list of vertices that haven’t. One by one, we move vertices from set V-U to set U by connecting the least weight edge. The following pseudocode is used to construct the MCST(minimum cost spanning tree), using prim’s algorithm that is as follows:
PRIMS MCST(G, w)
/* Input: An undirected connected weighted graph G = (V, E).
/* Output: A minimum cost spanning tree T(V, E’) of G
{
Т ← ф // T contains the edges of the MST
A ← (Any arbitrary member of V)
while (A≠V)
{
find an edge (u, v) of minimum weight such that u ∈ V – A and v ∈ A
A ← A U {(u, v)}
B ← B U(u)
}
return T
}
## Working on Prim’s Algorithm
Prim’s algorithm finds the subset of edges that includes every vertex of the graph such that the sum of the weights of the edges can be minimized. Prim’s algorithm starts with the single node and explores all the adjacent nodes with all the connecting edges at every step. The working of prim’s algorithm is as follows:
1. Initially, the set A of nodes contains a single arbitrary node (starting vertex), and the set T of edges are empty.
2. Prim’s algorithm searches for the shortest possible edge (u, v) at each step such that u ∈ V-A and V ∈
3. In this way, the edges in T form a minimal spanning tree for the nodes in A. Repeat this process until A ≠ V.
## Analysis of Prim’s Algorithm (By using min-heap)
Prim’s algorithm (also known as Jarník’s algorithm) is a greedy algorithm that finds a minimum spanning tree for a weighted undirected graph. This means it finds a subset of the edges that forms a tree that includes every vertex, where the total weight of all the edges in the tree is minimized. The analysis of prim’s algorithm by using min-heap is as follows.
For constructing heap: O(V)
The loop executes |V| times, and extracting the minimum element from the min-heap takes log V time, so while loop takes VlogV.
Total ‘E’ decrease key operations. Hence takes E log V
(Vlog V+E log V) = (V+ E) log V ≈ Elog V
## Time Complexity of Prim’s Algorithm
Assume we are given V vertices and E edges in a graph and need to find an MST. To complete one iteration, we delete the min node from the Min-Heap and add several edge weights to the Min-Heap.
We delete the V vertex from the Min-Heap because we have V vertices in the graph, and each iteration deletes 1 edge, for a total of V-1 edges in MST, with a complexity of O(log(V)). And we add up to E edges altogether, with each addition having a complexity of O(log(V)). As a result, the total complexity is O((V+E)Log(V)).
• Using Fibonacci heap
The prim’s algorithm time complexity using the Fibonacci heap is O(E + V log V).
• Using binomial heap
The prim’s algorithm time complexity using the binomial heap is O(V + E).
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# Concept: Average
INTRODUCTION
The concept of averages is important since questions based on averages regularly appear in Mathematics and Data-interpretation sections of various MBA entrance examinations. Average is a very simple but an effective way of representing an entire group by a single value.
The sum of all observations divided by the number of observations is referred to as average of those observations.
Average =$\frac{\mathrm{Sum of all quantities}}{\mathrm{Number of quantities}}$
It is also referred to as the Arithmetic Mean.
AVERAGE
• Let the n quantities be x1, x2, x3…xn, then the formula for the average a is
• If out of n quantities, x1 occurs f1 times, x2 occurs f2 times, …, xk occurs fk times, then their average a is given by
• For example, if the wages of five workers are 80, 120, 70, 100, 110, then the average wage per worker is
a = = $\frac{80+70+120+100+110}{5}$$\frac{480}{5}$ = Rs.96
• If a is the average of n quantities, then the sum of all the quantities S is given by S = a × n
• If the average and Sum of certain quantities is a and S respectively, then the number of quantities is given by n = $\frac{S}{a}$
Remember:
• If the value of each item is increased by the same value p, then the average of the group of items will also get increased by p.
• If the value of each item is decreased by the same value p, then the average of the group of items will also get decreased by p.
• If the value of each item is multiplied by the same value p, then the average of the group of items will also get multiplied by p.
• If the value of each item is divided by the same value p (p ≠ 0), then the average of group of items will also be divided by p.
• The average of a group of item will always lie between the smallest value in the group and largest value in the group.
• If all the observations form an Arithmetic Progression, then the middle term (Median) will be the average of the set of observations.
Example : Find the average of the following numbers: 351, 244, 479, 588, 105, 78, 483, 536.
(a) 479
(b) 356
(c) 424
(d) 358
(e) None of these
Solution :
Average = = 2864/8 = 358
Example : There are six numbers 30, 72, 53, 68, X and 87, out of which X is unknown. The average of the numbers is 60. What is the value of X?
(a) 40
(b) 60
(c) 70
(d) 30
(e) None of these
Solution
Sum of all 6 numbers = 60 × 6 = 360
So, 30 + 72 + 53 + 68 + X + 87 = 360
⇒ 310 + X = 360
⇒ x = 50
Example : The average age of a woman and her daughter is 16 years. The ratio of their ages is 7 : 1 respectively. What is the woman’s age?
(a) 4 years
(b) 28 years
(c) 32 years
(d) 6 years
(e) None of these
Solution
Let the woman’s age be 7x years and her daughter’s age be x years.
According to the question, 7x + x = 2 × 16
⇒ 8x = 32 ⇒ x = 4
⇒ Woman’s age = 7x = 7 × 4 = 28 years.
Example : The average of 4 consecutive even numbers is 103. What is the product of the smallest and the largest number?
(a) 10400
(b) 10504
(c) 10605
(d) 10600
(e) None of these
Solution
Sum of the four consecutive even numbers = 4 × 103 = 412.
⇒ x + (x + 2) + (x + 4) + (x + 6) = 412
4x + 12 = 412 ⇒ 4x = 400 ⇒ x = 100
Thus, product of the smallest and the largest number = 100 × (100 + 6) = 10600.
Example : The average of 6 consecutive odd numbers is 24. What is the largest number?
(a) 25
(b) 27
(c) 29
(d) 31
(e) None of these
Solution
Let the first odd number be x.
As per the question, x + (x + 2) + (x + 4) + (x + 6) + (x + 8) + (x + 10) = 24 × 6 = 144
⇒ 6x + 30 = 144 ⇒ 6x = 114 ⇒ x = 19
Thus, the largest number = 19 + 10 = 29
Example : If 36a + 36b = 576, then what is the average of a and b?
(a) 16
(b) 8
(c) 12
(d) 6
(e) 10
Solution
36a + 36b = 576
⇒ 36(a + b) = 576 ⇒ (a + b) = 16
Thus, average of a and b = (a + b)/2 = 8
Example : The average monthly expenditure of a family was Rs. 2,200 during first 3 months, Rs. 2,550 during next 4 months and Rs. 3,120 during last 5 months of the year. If the total saving during the year was Rs. 1,260, find average monthly income.
Solution
Total yearly income
= yearly expenditure + yearly saving
= [2200 × 3 + 2550 × 4 + 3120 × 5] + [1260]
= Rs. 33,660
∴ Average monthly income = 33660/12 = Rs.2805
Example : The average age of 5 children is 8 years. If the age of the father be included, the average is increased by 7 years, find the age of the father.
Solution
The total age of 5 children = 5 × 8 = 40 years
Total ages of (5 children + the father) = 6 × (8 + 7) = 90 years
∴ Age of the father = 90 years – 40 years = 50 years
Example : The average weight of 6 men is 65 kg. Two of them weigh 70 and 48 kg respectively, what is the average weight of the remaining four?
Solution
Total weight of six = 6 × 65 = 390 kg
The weight of the remaining four = 390 – 70 – 48 = 272
∴ Average weight of the remaining four = 272/4 = 68 kgs
Example : The average temperature for Sunday, Monday & Tuesday is 37.3° C, whereas the average temperature for Monday, Tuesday & Wednesday is 38.7°C. If Sunday’s temperature recorded is 40°C, find Wednesday’s temperature.
Solution
Total temperature of Sunday, Monday & Tuesday = 37.3 × 3 = 111.9° C
∴ Monday + Tuesday = 111.9 – 40 = 71.9° C
Total temperature of Monday, Tuesday & Wednesday = 38.7 × 3 = 116.1° C
Wednesday’s temperature = 116.1° C – 71.9° C = 44.2° C
Example : There are three numbers. The second is half of the first and the third is half of the second. If the average of the three numbers is 28, find the first number.
Solution
Sum of the three numbers = 3 × 28 = 84
Let, the first number be 4x, so the second number is 2x and the third number is x.
∴ 4x + 2x + x = 84 ⇒ x = 12
∴ First number is 4x = 48
Example : 30 horses were purchased for Rs. 12000. The average cost of 12 horses out of them is Rs. 250. Find the average cost of the remaining horses.
Solution :
Cost of 30 horses = Rs. 12000
Cost of 12 horses out of them = 12 × 250 = 3000
∴ Average cost of remaining horses =(12000-3000)/18 = Rs. 500
ASSUMED AVERAGE METHOD
In some cases, we can avoid tedious calculations of adding all the numbers and dividing by the number of observations. The calculation of averages can be simplified using the assumed average method. Take some arbitrary average (A), which lies between the smallest and the largest value. Then, calculate the deviations (differences) of the given values from A and find the average of all these deviations. Add this to A to find the average.
∴ Required average = Assumed average + Average of deviations
Example : Find the average of 101, 98, 105, 96, 103.
Solution
Here, let’s take the assumed average as 100.
Now, the required average = assumed average + average of deviations
∴ Average = 100 + $\frac{\left(101-100\right)+\left(98-100\right)+\left(105-100\right)+\left(96-100\right)+\left(103-100\right)}{5}$ = 100 + $\frac{1-2+5-4+3}{5}$ = 100 + $\frac{3}{5}$ = 100.6
Remember:
If the average of ‘n’ numbers is ‘a‘ and
• a new number equal to ‘a’ is added to the group, the average of the new group will remain same.
• a new number greater than ‘a’ is added to the group, the average of the new group will increase.
• a new number less than ‘a’ is added to the group, the average of the new group will decrease.
• a new number equal to ‘a’ is removed from the group, the average of the new group will remain same.
• a number greater than ‘a’ is removed from the group, the average of the new group will decrease.
• a number less than ‘a’ is removed from the group, the average of the new group will increase.
• a existing number is replaced with a greater number, the average of the group will increase.
• a existing number is replaced with a smaller number, the average of the group will decrease.
Let’s discuss these in detail.
A PERSON JOINING THE GROUP
When a number is added/removed from a group of numbers, we can calculate how much more/less, the number being added/removed, is from the earlier average by multiplying the remaining no. of numbers with the change in average.
Example : The average weight of a class of 24 students is 36 kg. When the weight of the teacher is also included, the average weight increases by 1kg. What is the weight of the teacher?
(a) 60 kg
(b) 61 kg
(c) 37 kg
(d) 39 kg
Solution
The average weight of a class of 24 students = 36 kgs.
Therefore, the total weight of the class = 24 × 36 = 864 kgs
When the weight of the teacher is included, the average weight increases by 1 kg i.e. the new average weight = 37 kgs.
Therefore, the total weight of the 24 students plus the teacher = 25 × 37 = 925 kgs
∴ 864 + Teacher’s weight = 925
∴ Weight of the teacher = 925 - 864 = 61 kgs.
Now, the same question can also be solved is a direct manner
Example : The average weight of a class of 24 students is 36 kg. When the weight of the teacher is also included, the average weight increases by 1 kg. What is the weight of the teacher?
(a) 60 kg
(b) 61 kg
(c) 37 kg
(d) 39 kg
Solution
The average weight of a class of 24 students = 36 kgs.
Had Teacher’s weight been 36 kgs, the average of the new group would not change. But since, the average of new group increases, it means Teacher weights more than 36 kgs.
Now, since the average of now 25 people increases by 1 kg, that means Teacher brought in 25 × 1 = 25 kg weight more than the earlier average of the group.
i.e. Teacher’s weight – 36 = 25
⇒ Weight of the Teacher = 36 + 25 = 61 kgs.
Example : The average weight of a class of 24 students is 36 kg. When a new student joins the class, the average weight decreases by 0.2 kg. What is the weight of the new student?
(a) 30 kg
(b) 31 kg
(c) 37 kg
(d) 29 kg
Solution
The average weight of a class of 24 students = 36 kgs.
Had new student’s weight been 36 kgs, the average of the new group would not change. But since, the average of new group decreases, it means new student weights less than 36 kgs.
Now, since the average of now 25 people decreases by 1 kg, that means new student brought in 25 × 0.2 = 5 kg weight less than the earlier average of the group.
i.e. 36 – New student’s weight = 5
⇒ Weight of the new student = 36 - 5 = 31 kgs.
Example : The average weight of class of 20 students is 30 kgs. If a new student whose weight is 51 kgs is added to the group, what will be the new average of the group.
Solution
Had the new student weighed 30 kgs, the new average would not change.
But, the new student brought in 51 – 30 = 21 kgs weight more than the average.
To calculate the average, this extra weight will be divided equally among the 21 students.
∴ Each student will gain 21/21 = 1 kg.
Hence, the new average of the class will be 30 + 1 = 31 kgs
A PERSON LEAVING THE GROUP
Example : The average weight of a class of 24 students is 36 kg. When one of the students leaves the group, the average weight decreases by 1 kg. What is the weight of the student leaving?
(a) 60 kg
(b) 61 kg
(c) 57 kg
(d) 59 kg
Solution
The average weight of a class of 24 students = 36 kgs.
Therefore, the total weight of the class = 24 × 36 = 864 kgs
When the student leaves the group, the average weight decreases by 1 kg i.e. the new average weight = 35 kgs.
Therefore, the total weight of the remaining 23 students = 23 × 35 = 805 kgs
∴ 805 + Leaving student’s weight = 864
∴ Weight of the student leaving = 964 - 805 = 59 kgs.
Alternately,
The average weight of a class of 24 students = 36 kgs.
If the weight of the student leaving was 36 kgs, the average of the new group would not change. But since, the average of new group decreases, it means the student leaving weights more than 36 kgs.
Now, since the average of now 23 people decreases by 1 kg, that means the student leaving took 23 × 1 = 23 kgs weight more than the earlier average of the group.
i.e. Weight of the student leaving – 36 = 23
⇒ Weight of the student leaving = 36 + 23 = 59 kgs.
Example : The average weight of a class of 24 students is 36 kg. When one of the students leaves the group, the average weight increases by 0.2 kg. What is the weight of the student leaving?
(a) 30.5 kg
(b) 31.4 kg
(c) 32 kg
(d) 31.6 kg
Solution
The average weight of a class of 24 students = 36 kgs.
If the weight of the student leaving was 36 kgs, the average of the new group would not change. But since, the average of new group increases, it means the student leaving weights less than 36 kgs.
Now, since the average of now 23 people increases by 0.2 kg, that means the student leaving took 23 × 0.2 = 4.6 kgs weight less than the earlier average of the group.
i.e. 36 - Weight of the student leaving = 4.6
⇒ Weight of the student leaving = 36 – 4.6 = 31.4 kgs.
Example : The average weight of class of 21 students is 50 kgs. If a student whose weight is 30 kgs is removed to the group, what will be the new average of the group.
Solution
Had the student leaving weighed 50 kgs, the new average would not change.
But, the student leaving took 50 – 30 = 21 kgs weight less than the average.
To calculate the average, this extra weight will be divided equally among the remaining 20 students.
∴ Each student will gain 20/20 = 1 kg.
Hence, the new average of the class will be 50 + 1 = 51 kgs.
Example : When a student weighing 45 kg left a class, the average weight of the remaining 59 students increased by 200g. What is the average weight of the original group of students?
(a) 57
(b) 56.8
(c) 58.2
(d) 52.2
Solution
Let the average weight of the original group be a kg.
∴ new average weight of the class will be = (a + 0.2) kg
Now, the average weight of 59 students increases by 0.2 kg. It means the student leaving the group took 59 × 0.2 = 11.8 kgs less than the earlier average.
⇒ Weight of the student leaving = (a - 11.8) kgs
∴ a – 11.8 = 45
⇒ a = 56.8 kgs
A PERSON BEING REPLACED IN THE GROUP
Example : The average weight of a class of 25 students is 36 kg. When one of the students whose weight is 30 kgs is replaced by another in the group, the average weight of the group increases by 0.2 kg. What is the weight of the student joining?
(a) 35 kg
(b) 31 kg
(c) 33.2 kg
(d) 35 kg
Solution
The average weight of a class of 25 students = 36 kgs.
Therefore, the total weight of the class = 25 × 36 = 900 kgs
When the student weighing 30 kgs leaves the group and another students weighing x kgs joins the group, the average weight increases by 0.2 kg i.e. the new average weight = 36.2 kgs.
Therefore, the total weight of the group = 25 × 36.2 = 905 kgs
∴ 900 – 30 + x = 905
∴ Weight of the student joining = 905 – 900 + 30 = 35 kgs.
Alternately
The average weight of a class of 25 students = 36 kgs.
If the weight of the student leaving and joining were same, the average of the new group would not change. But since, the average of new group increases, it means the student leaving weighs less than the student coming in.
Now, since the average of total 25 people increases by 0.2 kg, that means the student leaving weighs 25 × 0.2 = 5 kgs less than the students coming in.
i.e. Weight of the student joining - Weight of the student leaving = 5
⇒ Weight of the student joining = 30 + 5 = 35 kgs.
Example : The average age of 10 men in a play is increased by 1.5 years when two men aged 31 years and 49 years are replaced by two females. Find the average age of females.
Solution :
Overall increase in the total age = 10 × 1.5 years = 15 years
∴ Total age of two females = 31 + 49 + 15 = 95 years.
Average age of two females = 95/2 = 47.5 years
WEIGHTED AVERAGE
Let W1, W2, … ,Wn be the weights assigned to the quantities X1, X2, … , Xn respectively, then their weighted average (XW ) is defined as :
Weighted averages are used in connection with average speed, average price, average consumption etc.
Combined Average
If the number of items and their averages are given for two or more groups, then the formula for their combined average is
Combined average
$\stackrel{-}{X}=\frac{\sum {n}_{k}\stackrel{-}{{X}_{k}}}{\sum {n}_{k}}$
Example : The average age of the students in section A of 40 students is 40 years and the average of students in section B of 30 students is 12 years. Find the average age of students in both sections taken together.
Solution :
Average = $\frac{40×40+30×12}{40+30}$ = $\frac{1960}{70}$ = 28 years
∴ Average age of all the students is 28 years.
Example : A man bought 13 shirts of Rs. 50 each, 15 pairs of trousers of Rs. 60 each and 12 pairs of shoes at Rs. 65 each. Find the average value of each article.
Solution
Average = $\frac{13×50+15×60+12×65}{13+15+12}$ = Rs. 58.25
Example : The average of marks obtained by 60 students in a certain examination is 35. If the average marks of students who passed are 39 and that of the students who failed are 15, how many students passed in the examination?
Solution :
Let the number of passing students be x, then 39x + 15 (60 - x) = 60 × 35
⇒ 24x = 1200 ⇒ x = 50
∴ Number of passing students = 50
Example : The average monthly salary of 8 workers and 1 supervisor in a factory was Rs. 195. The supervisor,who used to get a monthly salary of Rs. 300, retired and a new supervisor was appointed in his place as a result of which the average monthly salary of the factory workers reduced to Rs. 185. Calculate the monthly salary of new supervisor.
Solution :
Average monthly salary of 8 workers and 1 supervisor = Rs. 195
Total salary of these 9 persons = 9 × 195 = Rs. 1,755
Total salary after the retirement of the supervisor = 1,755 – 300 = Rs. 1,455
Let the salary of new supervisor be Rs. x
∴ Total salary after the entry of new supervisor = Rs. (1,455 + x)
Average salary after the new supervisor joined = Rs. 185and total salary = 9 × 185 = Rs. 1,665
Hence, 1,455 + x = 1,665
⇒ x = 1,665 – 1,455 = Rs. 210
Example : 30 persons live in a hotel. On an increase of 10 persons, expenses of food increased by Rs. 60 but average expenses are reduced by Rs. 1. Find out what was the earlier total monthly expenditure of hotel for 30 persons?
Solution :
Let the monthly average expenditure of 30 persons = $\overline{x}$
∴ Total monthly expenditure of 30 persons = 30$\overline{x}$
Average monthly expenditure of 40 (= 30 + 10) persons = $\overline{x}$ – 1
∴ Total monthly expenditure of these 40 persons = 40 ($\overline{x}$ – 1)
According to the question, 40($\overline{x}$ - 1) - 30$\overline{x}$ = 60
⇒ 40$\overline{x}$ - 40 - 30$\overline{x}$ = 60
⇒ 10$\overline{x}$ = 100
∴ x ̅ = 10
Hence, Earlier total expenditure = 30 × 10 = Rs. 300
Example : Visitors to a show were charged Rs. 15 each on the first day, Rs. 7.5 each on the second, Rs. 2.5 each on the third and the total attendance on three days were in the ratio 2 : 5 : 13 respectively. Find the average charge per person for the whole show.
Solution :
Suppose the attendance on the first, second and third day were 2x, 5x and 13x respectively
∴ Total attendance = 2x + 5x + 13x = 20x
Money collected on the first day = 15 × 2x = 30x
Money collected on the second day = 7.5 × 5x = 37.5x
Money collected on the third day = 2.5 × 13x = 32.5 x
∴ Total money collected on all the three days = Rs. (30x + 37.50x + 32.50x) = Rs. 100x
∴ The average charge per person for the whole show = $\frac{100x}{20x}$ = Rs. 5
Example : The average weight of boys in a class is 43 kg. Later, four boys, whose weights are respectively 42 kg, 36.5 kg, 39 kg and 42.5 kg, join them. The average now becomes 42.5 kg. Find the initial number of boys in the class.
Solution :
Suppose initial number of boys in the class be x.
As their average weight = 43 kg
Their total weight = 43x kg
∴ Total weight of (x + 4) boys
= (43 x + 42 + 36.5 + 39 + 42.5) kg
= (43x + 160) kg
∴ Average weight of (x + 4) boys = $\frac{43x+160}{x+4}$ kg
According to the question, $\frac{43x+160}{x+4}$ = 42.5
⇒ 43x + 160 = 42.5x + 170
⇒ 43x – 42.5 x = 170 – 160
⇒ 0.5x = 10 ⇒ x = 10/0.5 = 20
Hence, initial number of boys in class = 20
Example : The average of 8 results is 20. The average of first two results is 15 1/2, The average of next three results is 21 1/3. The sixth result is 4 less than seventh and 7 less than eighth. Find the result.
Solution :
As the average of results is 20, the total of 8 results = 20 × 8 = 160.
Also, average of first two results =15 1/2
⇒ Total of first two result = $15\frac{1}{2}$ × 2 = 31
And, average of next three results = $21\frac{1}{3}$
⇒ Total of next three results = $21\frac{1}{3}$ × 3 = 64
∴ Total of the remaining three results = 160 – (31 + 64) = 160 – 95 = 65
Suppose 6th result = x, then 7th result = x + 4 and 8th result = x + 7
∴ x + (x + 4) + (x + 7) = 65
⇒ 3x + 11 = 65
⇒ 3x = 54 ⇒ x = 18
∴ Eighth result = 18 + 7 = 25
Example : In each row of a hand written book of 180 pages, there are 12 words on an average and each page consists of 16 rows on an average. After printing, in each row of the book there are 18 words on average and each page contains 20 rows on average. Find the number of pages of the printed book.
Solution :
No. of words in hand-written book = 180 × 12 × 16
Let the number of printed pages be x. Then, number of words in printed book = 18 × 20 × x
∴ 18 × 20 × x = 180 × 12 × 16
⇒ x = $\frac{180×12×16}{18×20}$ = 96
∴ Required number of pages = 96
Example : There were 35 students in a hostel. If the number of students increased by 7, the expenses of the mess were increased by Rs. 42 per day while the average expenditure per head diminished by Rs. 1. Find the initial expenditure of the mess.
Solution :
Let the initial expenditure be x. Then the initial average expenditure =x/35
New average expenditure = $\frac{x+42}{42}$
$\frac{x}{35}-\frac{x+42}{42}$ = 1 ⇒ x = 420
∴ Initial expenditure = Rs. 420
Example : Three years ago, the average of a family of 5 members was 17 years. A baby having been born, the average of the family is the same today, find the age of the baby.
Solution :
Present age of 5 members of the family + new born baby = 17 × 6 = 102
Present age of 5 members of the family = 17 × 5 + 3 × 5 = 100 years
Age of the baby = (102 - 100) years i.e., 2 years
Example : 5 years ago, the average age of Ram and Shyam was 20 years. At present, the average age of Ram, Shyam and Mohan is 30 years. What will be Mohan’s age 10 years hence?
Solution :
Total age of Ram & Shyam 5 years ago = (2×20) = 40
∴ Total age of Ram and Shyam now = (40 + 5 + 5) = 50
Total age of Ram, Shyam and Mohan now = (3×30) = 90
Mohan’ age now = 90 – 50 = 40
Mohan’s age 10 years hence = 40 + 10 = 50 years
Example : A batsman has a certain average runs for 16 innings. In the 17th inning, he made a score of 87 runs thereby increasing his average by 3. What is his average after 17th inning?
Solution :
Let the average for 16 innings be x runs
Total runs in 16 innings = 16x
Total runs in 17 innings = 16x + 87
Average of 17 innings = $\frac{16x+87}{17}$
$\frac{16x+87}{17}$ = x + 3
⇒ x = 36
Thus, average of 17 innings = 36 + 3 = 39
Example : The average age of the husband and wife who got married 7 years ago was 25 years then.The average age of the family, including a child born afterwards, is 22 years now. How old is the child now?
Solution :
Total age of Husband and wife, 7 years ago = (2 × 25) = 50
Total age of husband & wife now = (50 + 7 + 7) = 64
Total age of husband, wife & child now = (3 × 22) = 66
Age of child now = 66 – 64 = 2 years
Example : Out of 24 students of a class, 6 are 1 m 15 cm in height; 8 are 1 m 5 cm and the rest are 1 m 11 cm. What is the average height of all the students?
Solution :
Sum of the heights of all the 24 students = (6 × 115 + 8 × 105 + 10 × 111) cm = 2640 cm.
∴ Average height =2640/24=110 cm = 1 m 10 cm.
Example : The average age of a family of 6 members is 22 years. If the age of the youngest member be 7 years, then what was the average age of the family just before the birth of the youngest member?
Solution :
Total age of family = 6 × 22 = 132 years
7 years ago, total sum of ages = 132 – (6 × 7) = 90 years
But at that time there are 5 members in the family
∴ Average at that time =90/5=18 years
Example : 10 years ago the average age of a family of 4 members was 24 years. 2 children having been born, the average age of the family is same today. Find the present age of the two children, assuming that the children’s ages differ by 2 years.
Solution :
10 years ago, the total age of 4 members = 4 × 24 = 96 years
Total age of these 4 members at present = 96 + 10 × 4 = 136 years
∵ The average age of family is same after the birth of 2 children.
∴ Total age of 6 (= 4 + 2) members
= 24 × 6 = 144 years
∴ Total age of the 2 children = 144 – 136 = 8 years
Let the age of children be ‘x’ years and ‘x + 2’ years
Hence, x + (x + 2) = 8
⇒ 2x = 8 – 2 = 6 ⇒ x = 6/2 = 3 years
Hence, ages of the children would be 3 years and 5 years.
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# Difference between revisions of "2001 AMC 12 Problems/Problem 18"
## Problem
A circle centered at $A$ with a radius of 1 and a circle centered at $B$ with a radius of 4 are externally tangent. A third circle is tangent to the first two and to one of their common external tangents as shown. The radius of the third circle is
$[asy] unitsize(0.75cm); pair A=(0,1), B=(4,4); dot(A); dot(B); draw( circle(A,1) ); draw( circle(B,4) ); draw( (-1.5,0)--(8.5,0) ); draw( A -- (A+(-1,0)) ); label("1", A -- (A+(-1,0)), N ); draw( B -- (B+(4,0)) ); label("4", B -- (B+(4,0)), N ); label("A",A,E); label("B",B,W); filldraw( circle( (12/9,4/9), 4/9 ), lightgray, black ); dot( (12/9,4/9) ); [/asy]$
$\text{(A) }\frac {1}{3} \qquad \text{(B) }\frac {2}{5} \qquad \text{(C) }\frac {5}{12} \qquad \text{(D) }\frac {4}{9} \qquad \text{(E) }\frac {1}{2}$
## Solution
### Solution 1
$[asy] unitsize(1cm); pair A=(0,1), B=(4,4), C=(4,1); dot(A); dot(B); draw( circle(A,1) ); draw( circle(B,4) ); draw( (-1.5,0)--(8.5,0) ); draw( (A+(4,0)) -- A -- (A+(0,-1)) ); draw( A -- B -- (B+(0,-4)) ); label("A",A,N); label("B",B,N); label("C",C,E); filldraw( circle( (12/9,4/9), 4/9 ), lightgray, black ); dot( (12/9,4/9) ); draw( rightanglemark(A,C,B) ); [/asy]$
In the triangle $ABC$ we have $AB = 1+4 = 5$ and $BC=4-1 = 3$, thus by the Pythagorean theorem we have $AC=4$.
We can now pick a coordinate system where the common tangent is the $x$ axis and $A$ lies on the $y$ axis. In this coordinate system we have $A=(0,1)$ and $B=(4,4)$.
Let $r$ be the radius of the small circle, and let $s$ be the $x$-coordinate of its center $S$. We then know that $S=(s,r)$, as the circle is tangent to the $x$ axis. Moreover, the small circle is tangent to both other circles, hence we have $SA=1+r$ and $SB=4+r$.
We have $SA = \sqrt{s^2 + (1-r)^2}$ and $SB=\sqrt{(4-s)^2 + (4-r)^2}$. Hence we get the following two equations:
\begin{align*} s^2 + (1-r)^2 & = (1+r)^2 \\ (4-s)^2 + (4-r)^2 & = (4+r)^2 \end{align*}
Simplifying both, we get
\begin{align*} s^2 & = 4r \\ (4-s)^2 & = 16r \end{align*}
As in our case both $r$ and $s$ are positive, we can divide the second one by the first one to get $\left( \frac{4-s}s \right)^2 = 4$.
Now there are two possibilities: either $\frac{4-s}s=-2$, or $\frac{4-s}s=2$. In the first case clearly $s<0$, hence this is not the correct case. (Note: This case corresponds to the other circle that is tangent to both given circles and the $x$ axis - a large circle whose center is somewhere to the left of $A$.) The second case solves to $s=\frac 43$. We then have $4r = s^2 = \frac {16}9$, hence $r = \boxed{\frac 49}$.
### Solution 2
The horizontal line is the equivalent of a circle of curvature $0$, thus we can apply Descartes' Circle Formula.
The four circles have curvatures $0, 1, \frac 14$, and $\frac 1r$.
We have $2(0^2+1^2+\frac {1}{4^2}+\frac{1}{r^2})=(0+1+\frac 14+\frac 1r)^2$
Simplifying, we get $\frac{34}{16}+\frac{2}{r^2}=\frac{25}{16}+\frac{5}{2r}+\frac{1}{r^2}$
$\frac{1}{r^2}-\frac{5}{2r}+\frac{9}{16}=0$
$\frac{16}{r^2}-\frac{40}{r}+9=0$
$(\frac{4}{r}-9)(\frac{4}{r}-1)=0$
Obviously $r$ cannot equal $4$, therefore $r = \boxed{\frac 49}$. |
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# Warm-up Solve. 1. 8x – 9 = x + 12 = 4x x – 8 = 7x + 3
## Presentation on theme: "Warm-up Solve. 1. 8x – 9 = x + 12 = 4x x – 8 = 7x + 3"— Presentation transcript:
Warm-up Solve. 1. 8x – 9 = 23 2. 9x + 12 = 4x + 37 3. 6x – 8 = 7x + 3
x = – , or –5 3
More practice 5. 25 = -5x -10 6. -4x – 8 = -2x + 24
Solve. 5. 25 = -5x -10 6. -4x – 8 = -2x + 24 7. 6x – 14 = -2(x + 3) x = -7 x = -16 x = 1
Homework Review
How do I use order of operations to solve for a specified value?
GSE Algebra I UNIT QUESTION: Why is it important to understand the relationship between quantities? Standard: MCC9-12.N.Q.1-3, MCC9-12.A.SSE.1, MCC9-12.A.CED.1-4 Today’s Question: How do I use order of operations to solve for a specified value? Standard: MCC9-12.A.CED.4
Solving Equations and Formulas
Objectives: Solve literal equations (formulas) for a specified variable Apply these skills to solve practical problems
Definitions: An equation is a mathematical sentence that contains an equal sign ( = ). Ex: y = 3x A formula is an equation that states a rule for the relationship between certain quantities. Ex: A = lw What do A, l, and w stand for?
What it means to solve: To solve for x would mean to get x by itself on one side of the equation, with no x’s on the other side. (x = __ ) Similarly, to solve for y would mean to get y by itself on one side of the equation, with no y’s on the other side. (y = __ )
The DO-UNDO chart 1) Solve the equation -5x + y = -56 for x.
Ask yourself: What is the first thing being done to x, the variable being solved for? x is being multiplied by DO UNDO · y What is being done next? y ÷(-5) y is being added to -5x.
Show all of your work! First, subtract y from both sides of the equation. Next, divide by -5. This process actually requires LESS WORK than solving equations in one variable Ex: -5x + y = -56 - y -y -5x = y x = y = 56 + y
Let’s try another: Ex: Solve 2x - 4y = 7 for x.
Complete the do-undo chart. DO UNDO +4y + 4y 2x = 7 + 4y + 4y ÷ 2 x 2 - 4y x = 7 + 4y 2 This fraction cannot be simplified unless both terms in the numerator are divisible by 2.
Another example: Solve a(y + 1) = b for y. Complete the do-undo chart.
a a ÷ a - 1 y + 1 = b a + 1 x a y = b - 1 a
Try a few on your own. Solve P = 1.2W for W. H2
Solve P = 2l + 2w for l. Solve F = 9C + 32 for C. 5
The answers: DO UNDO · 1.2 · H2 ÷ H2 ÷ 1.2 W = PH2 1.2 DO UNDO · 2 -2w
l = P - 2w 2
The answers: DO UNDO x 9/5 - 32 + 32 ÷9/5 or times 5/9 C = 5 (F – 32)
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# How do you find the critical points for f(x)= -(sinx)/ (2+cosx) and the local max and min?
Sep 14, 2016
The critical points are at:
$\left(\frac{2 \pi}{3} , \frac{\sqrt{3}}{3}\right)$is a minimum point
$\left(\left(4 \frac{\pi}{3}\right) , \frac{\sqrt{3}}{3}\right)$ is the maximum point.
#### Explanation:
To find the critical points we have to find $f ' \left(x\right)$
then solve for $f ' \left(x\right) = 0$
$f ' \left(x\right) = - \frac{\left(\sin x\right) ' \left(2 + \cos x\right) - \left(2 + \cos x\right) ' \sin x}{2 + \cos x} ^ 2$
$f ' \left(x\right) = - \frac{\cos x \left(2 + \cos x\right) - \left(- \sin x\right) \sin x}{2 + \cos x} ^ 2$
$f ' \left(x\right) = - \frac{2 \cos x + {\cos}^{2} \left(x\right) + {\sin}^{2} \left(x\right)}{2 + \cos x} ^ 2$
Since ${\cos}^{2} \left(x\right) + {\sin}^{2} \left(x\right) = 1$ we have:
$f ' \left(x\right) = - \frac{2 \cos x + 1}{2 + \cos x} ^ 2$
Let us dolce for $f ' \left(x\right) = 0$to find the critical points:
$f ' \left(x\right) = 0$
$\Rightarrow - \frac{2 \cos x + 1}{2 + \cos x} ^ 2 = 0$
$\Rightarrow - \left(2 \cos x + 1\right) = 0$
$\Rightarrow \left(2 \cos x + 1\right) = 0$
$\Rightarrow 2 \cos x = - 1$
$\Rightarrow \cos x = - \frac{1}{2}$
$\cos \left(\pi - \left(\frac{\pi}{3}\right)\right) = - \frac{1}{2}$
or
$\cos \left(\pi + \left(\frac{\pi}{3}\right)\right) = - \frac{1}{2}$
Therefore,
$x = \pi - \left(\frac{\pi}{3}\right) = \frac{2 \pi}{3}$
or $x = \pi + \left(\frac{\pi}{3}\right) = \frac{4 \pi}{3}$
Let's compute f((2pi)/3)=-sin((2pi)/3)/(2+cos((2pi)/3)
$f \left(\frac{2 \pi}{3}\right) = - \frac{\frac{\sqrt{3}}{2}}{2 - \frac{1}{2}}$
$f \left(\frac{2 \pi}{3}\right) = - \frac{\frac{\sqrt{3}}{2}}{\frac{3}{2}}$
$f \left(\frac{2 \pi}{3}\right) = - \left(\frac{\sqrt{3}}{3}\right)$
Since$f \left(x\right)$ is decreasing on $\left(0 , \frac{2 \pi}{3}\right)$
Then$\left(\left(\frac{2 \pi}{3}\right) , - \frac{\sqrt{3}}{3}\right)$ is minimum point
Since then the function increases till $x = \left(4 \frac{\pi}{3}\right)$ then the point
$\left(\left(4 \frac{\pi}{3}\right) , \frac{\sqrt{3}}{3}\right)$ is the maximum point. |
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# Similarity Part C: Trigonometry (35 minutes)
## Session 8, Part C
### In this part
• Right Triangle Ratios
• Trigonometric Functions
### Right Triangles Ratios
The word “trigonometry” is enough to strike fear into the hearts of many high school students. But it simply means “triangle measuring.” A trigon is another name for a triangle — think pentagon and hexagon, for example — and “-metry” means “measuring,” just as it does in “geometry,” or “earth measuring.” Trigonometry is about measuring similar right triangles.
Set up a proportion between the corresponding sides of the two triangles, the original one and the new one with unknown sides. There are many different proportions you might set up depending on what you’re looking for (e.g., a1/c1 = a2/c2, or, a1/a2 = c1/c2, etc.). In each proportion try to use three known side lengths, so the only unknown is the side you’re looking for.
### Problem C1
Here’s a right triangle:
The triangles below are all similar to the original triangle above. Find the length a for each triangle. Explain how you did it. Note 7
### Problem C2
For each triangle above, find the length b. Explain how you did it.
If you solved the problems above, you probably used the sine and cosine functions, even if you didn’t know you were doing it. To find the missing value of a, you need to multiply the length of the hypotenuse by 3/5. To find the length b, you multiply the hypotenuse by 4/5.
Those ratios — (side a)/(hypotenuse) = 3/5 and (side b)/(hypotenuse) = 4/5 — will hold for any right triangle that has another angle the same as A. All such right triangles will be similar to each other, so the ratios must be the same as in our original triangle.
### Trigonometric Functions
As you’ve seen, the trigonometric functions such as sine, cosine, and tangent are nothing more than ratios of particular sides in right-angle triangles. Note 8 These ratios depend only on the measure of an angle and have special names:
In the triangle above, the sine of angle A is defined as BC/AB. It is abbreviated sin A.
The cosine of angle A is defined as AC/AB. It is abbreviated cos A.
The tangent of angle A is defined as BC/AC. It is abbreviated as tan A. Note 9
Use the Interactive Activity to answer Problem C3 and C4 and to experiment with the trigonometric functions. For example, how big can sine get? What about cosine? How do you make them as big as possible? What about tangent? Does it seem to have a maximum value? (This interactive has been disabled due to Flash content.)
### Problem C3
For the original triangle:
a. Find sin A and cos A. b. Find sin B and cos B.
### Problem C4
For the following triangle:
a. Find sin A and cos A. b. Find sin B and cos B. c. Find tan A and tan B.
### Problem C5
a. Find sin 30°, cos 30°, and tan 30° in the triangle below. Then find sin 60°, cos 60°, and tan 60°. b. Find sin 45°, cos 45°, and tan 45° in the triangle below.
### Problem C6
Suppose you have any right triangle ABC (with the right angle at C). Explain why it must be true that sin A = cos B.
### Problem C7
Most public buildings were built before wheelchair-access ramps became widespread. When it came time to design the ramps, the doors of buildings were already in place. Suppose a particular building has a door two feet off the ground. How long must a ramp be to reach the door if the ramp is to make a 10° angle with the ground? (This is why so many access ramps must make one or more turns!)
sin 10° 0.17 and cos 10° 0.98 Note 10
Problem C7 taken from Connected Geometry, developed by Educational Development Center, Inc. p. 341. © 2000 Glencoe/McGraw-Hill. Used with permission. www.glencoe.com/sec/math
### Note 7
If you’re working with a group, compare different methods other participants have come up with. Think about how these methods connect to yours.
### Note 8
These ratios have been immensely useful in many fields, such as engineering, surveying, astronomy, and architecture. In addition to the three trigonometric functions mentioned in this session (sine, cosine, and tangent), there are three additional functions not covered in this course (secant, cosecant, and cotangent).
### Note 9
In mathematics, if we have an idea that works in some particular cases, we often look for ways to extend that idea to a more general situation. For example, the definitions of the trigonometric functions on the right triangle are valid only when dealing with angles between 0° and 90° — no other angle can appear in a right triangle!
So, in order to make sense of and compute the values of trigonometric functions of any angle, we need to extend this definition. According to the extended definition, sin and cos of angle (theta) are defined to be y and x coordinates, respectively, of a point on the unit circle. (The unit circle is a circle centered at the origin of the coordinate system, with a radius equal to 1.)
Notice that as we move the point P along the circle to create angles between 0° and 360°, some coordinates will be positive and some will be negative.
### Note 10
The values for sine and cosine have been calculated for many triangles. You can find them in a trigonometry table, or you can calculate them yourself using a scientific calculator.
### Problem C1
The triangle with hypotenuse 4: By similarity with the original triangle, we can say:
5/4 = 3/a, so a = 12/5.
The triangle with hypotenuse 10: By similarity with the original triangle, we can say:
5/10 = 3/a, so a = 30/5 = 6.
The triangle with hypotenuse 1: By similarity with the original triangle, we can say:
5/1 = 3/a, so a = 3/5.
### Problem C2
The triangle with hypotenuse 4: By similarity with the original triangle, we can say:
5/4 = 4/b, so b = 16/5.
The triangle with hypotenuse 10: By similarity with the original triangle, we can say:
5/10 = 4/b, so b = 40/5 = 8.
The triangle with hypotenuse 1: By similarity with the original triangle, we can say:
5/1 = 4/b, so b = 4/5.
### Problem C3
a. sin A = 3/5 and cos A = 4/5
b. sin B = 4/5 and cos B = 3/5
Problem C4
We calculate the hypotenuse with the Pythagorean theorem and find that it is 13.
a. sin A = 12/13, cos A = 5/13 b. sin B = 5/13, cos B = 12/13 c. tan A = 12/5, tan B = 5/12
Problem C5
a. sin 30° = cos 60° = 1/2 sin 60° = cos 30° = √3/2 tan 30° = 1/√3 tan 60° = These values are constant for any triangle with angles 30°-60°-90°. b. sin 45° = cos 45° = 1/√2 tan 45° = 1These values are constant for any triangle with angles 45°-45°-90°.
### Problem C6
Angles A and B are adjacent, so sin A = cos B. The hypotenuse is the same for both angles, but the roles of “adjacent side” and “opposite side” switch. The side opposite angle B is adjacent to angle A, and vice versa.
### Problem C7
Let x be the length of the ramp. Then we have a right triangle with hypotenuse x, shorter leg 2, and the angle opposite to the shorter leg of 10°. Since sin 10° = 2/x, we have x = 2/sin 10° = 2/0.17 11.765 feet. |
# Video: AQA GCSE Mathematics Higher Tier Pack 5 β’ Paper 2 β’ Question 27
πΏππππ is a rectangular pyramid. The horizontal base πΏπππ has side lengths 8 cm and 12 cm and centre πΆ. Angle ππΆπΏ is 90Β°. Angle πΏππΆ is 34Β°. Remember that the volume of a pyramid = (1/3) Γ area of base Γ perpendicular height. Work out the volume of the pyramid.
05:54
### Video Transcript
πΏππππ is a rectangular pyramid. The horizontal base πΏπππ has side lengths eight centimeters and 12 centimeters and centre πΆ. Angle ππΆπΏ is 90 degrees. Angle πΏππΆ is 34 degrees. Remember that the volume of a pyramid is equal to one-third times the area of the base times the perpendicular height. Work out the volume of the pyramid.
Letβs begin by going through the given information. Weβre told this is a rectangular pyramid. So the horizontal base πΏπππ would be a rectangle with side lengths eight centimeters, which means over here would be eight centimeters and since itβs a rectangle and opposite sides of a rectangle are equal, and 12 centimeters, which means this had to be 12 centimeters.
And weβre also told that the centre of this rectangle is πΆ. And this is important because when combined with the next given piece of information, the fact the angle ππΆπΏ is 90 degrees and πΆ is the centre of the rectangle, then ππΆ will be the perpendicular height of this pyramid. And we donβt know this length and we need it for the volume of the pyramid. So letβs go ahead and call it π₯. And lastly, angle πΏππΆ is 34 degrees.
So for the volume of the pyramid, we need to take one-third times the area of the base times the perpendicular height. Since our base is a rectangle, the area of our base will be length times width: eight centimeters times 12 centimeters giving us 96 centimeters squared. And then last, weβll multiply by the perpendicular height, which we donβt know. So weβll call it π₯. So this perpendicular height is the only piece of information that weβre missing to find this volume.
And here, we have a right-angled triangle and π₯ is the length of one of its sides. And we know an angle of this triangle: angle πΏππΆ is equal to 34 degrees. So if we could find one of these side lengths, we would be able to solve for the perpendicular height ππΆ, which weβre calling π₯ using trigonometry. Well, since the base is a rectangle, triangle πΏππ would be a right triangle because angles in a rectangle are 90 degrees.
So if we found the length of the hypotenuse β the longest side of this triangle β and then found half of that, that would be a length of the side of the pink triangle that we could use to help us find π₯ and this length would be half of πΏπ. Thatβs what we need to find. So letβs begin by first finding the length of πΏπ. We can find the length of πΏπ by using Pythagorasβs theorem. The length of the hypotenuse squared is equal to, so πΏπ squared is equal to the sum of the other two side lengths squared. Eight squared is 64 and 12 squared is 144. Adding these together, we get 208.
So if we have that πΏπ squared is equal to 208, to solve for πΏπ, we simply square root both sides. 208 is 16 times 13. And weβre doing this to simplify the square root. 16 is a perfect square. Itβs four times four. So this means we find that πΏπ is equal to four square root 13. So if we would like to find half of πΏπ for the pink triangle, we need to take one-half times four square root of 13. Two goes into four twice. So we are left with two square root of 13.
Notice we only took one-half times four. Thatβs because both of these numbers are not inside of a square root. So we cannot take one-half times the square root of 13 because the square root of 13 is underneath the square root, where one-half is not. So now we can use this length to help us find the perpendicular height π₯.
So here is our pink triangle. From our angle of 34 degrees, two square root of 13 will be considered our opposite side and π₯ would be considered the adjacent side. So out of sine, cosine, and tangent, itβs the tangent function that uses the opposite and adjacent sides. The tangent of an angle is equal to the opposite side divided by the adjacent side. So we have that the tangent of 34 degrees is equal to two square root of 13 divided by π₯.
And since we want to solve for π₯, we can multiply both sides of the equation by π₯ β this way itβs moved up to the numerator β and then divide both sides of the equation by tangent 34. So we are left with π₯ is equal to two square root of 13 divided by the tangent of four. So weβll plug this into our calculator. So this means π₯ is about 10.69089 and so on. Now to avoid rounding error, we want to keep this entire number on our calculator and then multiply by 96 and then multiply by one-third or dividing by three. And remember this perpendicular height is in centimeters. So when we have centimeters squared times centimeters, our volume should be in centimeters cubed.
And after multiplying these together and we get 342.108773822 cubic centimeters. Now this is not specify how many decimal places to round. Normally, itβs one or two. Letβs go ahead and round one. But either way would be fine. So we need to decide whether to round the one up or round it down. So we look at the number to right, zero. Since zero is less than five, we round down. So weβll keep the one a one.
Therefore, the volume of this rectangular pyramid would be 342.1 cubic centimeters. |
Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Differentiation Ex 1.1 Questions and Answers.
## Maharashtra State Board 12th Maths Solutions Chapter 1 Differentiation Ex 1.1
Question 1.
Differentiate the following w.r.t. x :
(i) (x3 – 2x – 1)5
Solution:
Method 1:
Let y = (x3 – 2x – 1)5
Put u = x3 – 2x – 1. Then y = u5
Method 2:
Let y = (x3 – 2x – 1)5
Differentiating w.r.t. x, we get
(ii) $$\left(2 x^{\frac{3}{2}}-3 x^{\frac{4}{3}}-5\right)^{\frac{5}{2}}$$
Solution:
Let y = $$\left(2 x^{\frac{3}{2}}-3 x^{\frac{4}{3}}-5\right)^{\frac{5}{2}}$$
Differentiating w.r.t. x, we get
(iii) $$\sqrt{x^{2}+4 x-7}$$
Solution:
(iv) $$\sqrt{x^{2}+\sqrt{x^{2}+1}}$$
Solution:
Let y = $$\sqrt{x^{2}+\sqrt{x^{2}+1}}$$
Differentiating w.r.t. x, we get
(v) $$\frac{3}{5 \sqrt[3]{\left(2 x^{2}-7 x-5\right)^{5}}}$$
Solution:
Let y = $$\frac{3}{5 \sqrt[3]{\left(2 x^{2}-7 x-5\right)^{5}}}$$
Differentiating w.r.t. x, we get
(vi) $$\left(\sqrt{3 x-5}-\frac{1}{\sqrt{3 x-5}}\right)^{5}$$
Solution:
Let y = $$\left(\sqrt{3 x-5}-\frac{1}{\sqrt{3 x-5}}\right)^{5}$$
Differentiating w.r.t. x, we get
Question 2.
Diffrentiate the following w.r.t. x
(i) cos(x2 + a2)
Solution:
Let y = cos(x2 + a2)
Differentiating w.r.t. x, we get
$$\frac{d y}{d x}$$ = $$\frac{d}{d x}$$[cos(x2 + a2)]
= -sin(x2 + a2)∙$$\frac{d}{d x}$$x2 + a2)
= -sin(x2 + a2)∙(2x + 0)
= -2xsin(x2 + a2)
(ii) $$\sqrt{e^{(3 x+2)}+5}$$
Solution:
Let y = $$\sqrt{e^{(3 x+2)}+5}$$
Differentiating w.r.t. x, we get
(iii) log[tan($$\frac{x}{2}$$)]
Solution:
Let y = log[tan($$\frac{x}{2}$$)]
Differentiating w.r.t. x, we get
(iv) $$\sqrt{\tan \sqrt{x}}$$
Solution:
Let y = $$\sqrt{\tan \sqrt{x}}$$
Differentiating w.r.t. x, we get
(v) cot3[log (x3)]
Solution:
Let y = cot3[log (x3)]
Differentiating w.r.t. x, we get
(vi) 5sin3x+ 3
Solution:
Let y = 5sin3x+ 3
Differentiating w.r.t. x, we get
(vii) cosec ($$\sqrt{\cos X}$$)
Solution:
Let y = cosec ($$\sqrt{\cos X}$$)
Differentiating w.r.t. x, we get
(viii) log[cos (x3 – 5)]
Solution:
Let y = log[cos (x3 – 5)]
Differentiating w.r.t. x, we get
(ix) e3 sin2x – 2 cos2x
Solution:
Let y = e3 sin2x – 2 cos2x
Differentiating w.r.t. x, we get
(x) cos2[log (x2+ 7)]
Solution:
Let y = cos2[log (x2+ 7)]
Differentiating w.r.t. x, we get
(xi) tan[cos (sinx)]
Solution:
Let y = tan[cos (sinx)]
Differentiating w.r.t. x, we get
(xii) sec[tan (x4 + 4)]
Solution:
Let y = sec[tan (x4 + 4)]
Differentiating w.r.t. x, we get
= sec[tan(x4 + 4)]∙tan[tan(x4 + 4)]∙sec2(x4 + 4)(4x3 + 0)
= 4x3sec2(x4 + 4)∙sec[tan(x4 + 4)]∙tan[tan(x4 + 4)].
(xiii) elog[(logx)2 – logx2]
Solution:
Let y = elog[(logx)2 – logx2]
= (log x)2 – log x2 …[∵ elog x = x]
Differentiating w.r.t. x, we get
(xiv) sin$$\sqrt{\sin \sqrt{x}}$$
Solution:
Let y = sin$$\sqrt{\sin \sqrt{x}}$$
Differentiating w.r.t. x, we get
(xv) log[sec(ex2)]
Solution:
Let y = log[sec(ex2)]
Differentiating w.r.t. x, we get
(xvi) loge2(logx)
Solution:
Let y = loge2(logx) = $$\frac{\log (\log x)}{\log e^{2}}$$
(xvii) [log{log(logx)}]2
Solution:
let y = [log{log(logx)}]2
Differentiating w.r.t. x, we get
(xviii) sin2x2 – cos2x2
Solution:
Let y = sin2x2 – cos2x2
Differentiating w.r.t. x, we get
= 2sinx2∙cosx2 × 2x + 2sinx2∙cosx2 × 2x
= 4x(2sinx2∙cosx2)
= 4xsin(2x2).
Question 3.
Diffrentiate the following w.r.t. x
(i) (x2 + 4x + 1)3 + (x3 – 5x – 2)4
Solution:
Let y = (x2 + 4x + 1)3 + (x3 – 5x – 2)4
Differentiating w.r.t. x, we get
$$\frac{d y}{d x}$$ = $$\frac{d}{d x}$$[(x2 + 4x + 1)3 + (x3 – 5x – 2)4]
= $$\frac{d}{d x}$$ = (x2 + 4x + 1)3 + $$\frac{d}{d x}$$(x3 – 5x – 2)4
= 3(x2 + 4x + 1)2∙$$\frac{d}{d x}$$(x2 + 4x + 1) + 4(x3 – 5x – 2)4∙$$\frac{d}{d x}$$(x3 – 5x – 2)
= 3(x2 + 4x + 1)3∙(2x + 4 × 1 + 0) + 4(x3 – 5x – 2)3∙(3x2 – 5 × 1 – 0)
= 6 (x + 2)(x2 + 4x + 1)2 + 4 (3x2 – 5)(x3 – 5x – 2)3.
(ii) (1 + 4x)5(3 + x − x2)8
Solution:
Let y = (1 + 4x)5(3 + x − x2)8
Differentiating w.r.t. x, we get
= 8 (1 + 4x)5 (3 + x – x2)7∙(0 + 1 – 2x) + 5 (1 + 4x)4 (3 + x – x2)8∙(0 + 4 × 1)
= 8 (1 – 2x)(1 + 4x)5(3 + x – x2)7 + 20(1 + 4x)4(3 + x – x2)8.
(iii) $$\frac{x}{\sqrt{7-3 x}}$$
Solution:
Let y = $$\frac{x}{\sqrt{7-3 x}}$$
Differentiating w.r.t. x, we get
(iv) $$\frac{\left(x^{3}-5\right)^{5}}{\left(x^{3}+3\right)^{3}}$$
Solution:
Let y = $$\frac{\left(x^{3}-5\right)^{5}}{\left(x^{3}+3\right)^{3}}$$
Differentiating w.r.t. x, we get
$$\frac{d y}{d x}$$ = $$\frac{d}{d x}\left[\frac{\left(x^{3}-5\right)^{5}}{\left(x^{3}+3\right)^{3}}\right]$$
(v) (1 + sin2x)2(1 + cos2x)3
Solution:
Let y = (1 + sin2x)2(1 + cos2x)3
Differentiating w.r.t. x, we get
= 3(1 + sin2x)2 (1 + cos2x)2∙[2cosx(-sinx)] + 2 (1 + sin2x)(1 + cos2x)3∙[2sinx-cosx]
= 3 (1 + sin2x)2 (1 + cos2x)2 (-sin 2x) + 2(1 + sin2x)(1 + cos2x)3(sin 2x)
= sin2x (1 + sin2x) (1 + cos2x)2 [-3(1 + sin2x) + 2(1 + cos2x)]
= sin2x (1 + sin2x)(1 + cos2x)2(-3 – 3sin2x + 2 + 2cos2x)
= sin2x (1 + sin2x)(1 + cos2x)2 [-1 – 3 sin2x + 2 (1 – sin2x)]
= sin 2x(1 + sin2x)(1 + cos2x)2 (-1 – 3 sin2x + 2 – 2 sin2x)
= sin2x (1 + sin2x)(1 + cos2x)2(1 – 5 sin2x).
(vi) $$\sqrt{\cos x}+\sqrt{\cos \sqrt{x}}$$
Solution:
Let y = $$\sqrt{\cos x}+\sqrt{\cos \sqrt{x}}$$
Differentiating w.r.t. x, we get
$$\frac{d y}{d x}$$ = $$\frac{d}{d x}[\sqrt{\cos x}+\sqrt{\cos \sqrt{x}}]$$
(vii) log(sec 3x+ tan 3x)
Solution:
Let y = log(sec 3x+ tan 3x)
Differentiating w.r.t. x, we get
(viii) $$\frac{1+\sin x^{\circ}}{1-\sin x^{\circ}}$$
Solution:
(ix) cot$$\left(\frac{\log x}{2}\right)$$ – log$$\left(\frac{\cot x}{2}\right)$$
Solution:
Let y = cot$$\left(\frac{\log x}{2}\right)$$ – log$$\left(\frac{\cot x}{2}\right)$$
Differentiating w.r.t. x, we get
(x) $$\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}$$
Solution:
(xi) $$\frac{e^{\sqrt{x}}+1}{e^{\sqrt{x}}-1}$$
Solution:
let y = $$\frac{e^{\sqrt{x}}+1}{e^{\sqrt{x}}-1}$$
Differentiating w.r.t. x, we get
(xii) log[tan3x·sin4x·(x2 + 7)7]
Solution:
Let y = log [tan3x·sin4x·(x2 + 7)7]
= log tan3x + log sin4x + log (x2 + 7)7
= 3 log tan x + 4 log sin x + 7 log (x2 + 7)
Differentiating w.r.t. x, we get
= 6cosec2x + 4 cotx + $$\frac{14 x}{x^{2}+7}$$
(xiii) log$$\left(\sqrt{\frac{1-\cos 3 x}{1+\cos 3 x}}\right)$$
Solution:
(xiv) log$$\left(\sqrt{\left.\frac{1+\cos \left(\frac{5 x}{2}\right)}{1-\cos \left(\frac{5 x}{2}\right)}\right)}\right.$$
Solution:
Using log$$\left(\frac{a}{b}\right)$$ = log a – log b
log ab = b log a
$$-\frac{5}{2}$$cosec$$\left(\frac{5 x}{2}\right)$$
(xv) log$$\left(\sqrt{\frac{1-\sin x}{1+\sin x}}\right)$$
Solution:
(xvi) log$$\left[4^{2 x}\left(\frac{x^{2}+5}{\sqrt{2 x^{3}-4}}\right)^{\frac{3}{2}}\right]$$
Solution:
(xvii) log$$\left[\frac{e^{x^{2}}(5-4 x)^{\frac{3}{2}}}{\sqrt[3]{7-6 x}}\right]$$
Solution:
(xviii) log$$\left[\frac{a^{\cos x}}{\left(x^{2}-3\right)^{3} \log x}\right]$$
Solution:
(xix) y= (25)log5(secx) − (16)log4(tanx)
Solution:
y = (25)log5(secx) − (16)log4(tanx)
= 52log5(secx) – 42log4(tanx)
= 5log5(sec5x) – 4log4(tan2x)
= sec2x – tan2x … [∵ = x]
∴ y = 1
Differentiating w.r.t. x, we get
$$\frac{d y}{d x}$$ = $$\frac{d}{d x}$$(1) = 0
(xx) $$\frac{\left(x^{2}+2\right)^{4}}{\sqrt{x^{2}+5}}$$
Solution:
Let y = $$\frac{\left(x^{2}+2\right)^{4}}{\sqrt{x^{2}+5}}$$
Differentiating w.r.t. x, we get
Question 4.
A table of values of f, g, f ‘ and g’ is given
(i) If r(x) = f [g(x)] find r’ (2).
Solution:
r(x) = f[g(x)]
∴ r'(x) = $$\frac{d}{d x}$$f[g(x)]
= f'[g(x)]∙$$\frac{d}{d x}$$[g(x)]
= f'[g(x)∙[g'(x)]
∴ r'(2) = f'[g(2)]∙g'(2)
= f'(6)∙g'(2) … [∵ g(x) = 6, when x = 2]
= -4 × 4 … [From the table]
= -16.
(ii) If R(x) = g[3 + f(x)] find R’ (4).
Solution:
R(x) = g[3 + f(x)]
∴ R'(x) = $$\frac{d}{d x}$${g[3+f(x)]}
= g'[3 + f(x)]∙$$\frac{d}{d x}$$[3 + f(x)]
= g'[3 +f(x)]∙[0 + f'(x)]
= g'[3 + f(x)]∙f'(x)
∴ R'(4) = g'[3 + f(4)]∙f'(4)
= g'[3 + 3]∙f'(4) … [∵ f(x) = 3, when x = 4]
= g'(6)∙f'(4)
= 7 × 5 … [From the table]
= 35.
(iii) If s(x) = f[9− f(x)] find s’ (4).
Solution:
s(x) = f[9− f(x)]
∴ s'(x) = $$\frac{d}{d x}$${f[9 – f(x)]}
= f'[9 – f(x)]∙$$\frac{d}{d x}$$[0 – f(x)]
= f'[9 – f(x)]∙[0 – f'(x)]
= -f'[9 – f(x)] – f'(x)
∴ s'(4) = -f'[9 – f(4)] – f'(4)
= -f'[9 – 3] – f'(4) … [∵ f(x) = 3, when x = 4]
= -f'(6) – f'(4)
= -(-4)(5) … [From the table]
= 20.
(iv) If S(x) = g[g(x)] find S’ (6)
Solution:
S(x) = g[g(x)]
∴ S'(x) = $$\frac{d}{d x}$$g[g(x)]
= g'[g(x)]∙$$\frac{d}{d x}$$[g(x)]
= g'[g(x)]∙g'(x)
∴ S ‘(6) = g'[g'(6)]∙g'(6)
= g'(2)∙g'(6) … [∵ g (x) = 2, when x = 6]
= 4 × 7 … [From the table]
= 28.
Question 5.
Assume that f ‘(3) = -1, g'(2) = 5, g(2) = 3 and y = f[g(x)] then $$\left[\frac{d y}{d x}\right]_{x=2}$$ = ?
Solution:
y = f[g(x)]
∴ $$\frac{d y}{d x}$$ = $$\frac{d}{d x}$${[g(x)]}
Question 6.
If h(x) = $$\sqrt{4 f(x)+3 g(x)}$$, f(1) = 4, g(1) = 3, f ‘(1) = 3, g'(1) = 4 find h'(1).
Solution:
Given f(1) = 4, g(1) = 3, f ‘(1) = 3, g'(1) = 4 …..(1)
Question 7.
Find the x co-ordinates of all the points on the curve y = sin 2x – 2 sin x, 0 ≤ x < 2π where $$\frac{d y}{d x}$$ = 0.
Solution:
y = sin 2x – 2 sin x, 0 ≤ x < 2π
= cos2x × 2 – 2cosx
= 2 (2 cos2x – 1) – 2 cosx
= 4 cos2x – 2 – 2 cos x
= 4 cos2x – 2 cos x – 2
If $$\frac{d y}{d x}$$ = 0, then 4 cos2x – 2 cos x – 2 = 0
∴ 4cos2x – 4cosx + 2cosx – 2 = 0
∴ 4 cosx (cosx – 1) + 2 (cosx – 1) = 0
∴ (cosx – 1)(4cosx + 2) = 0
∴ cosx – 1 = 0 or 4cosx + 2 = 0
∴ cos x = 1 or cos x = $$-\frac{1}{2}$$
∴ cos x = cos 0
Question 8.
Select the appropriate hint from the hint basket and fill up the blank spaces in the following paragraph. [Activity]
“Let f (x) = x2 + 5 and g(x) = ex + 3 then
f [g(x)] = _ _ _ _ _ _ _ _ and g [f(x)] =_ _ _ _ _ _ _ _.
Now f ‘(x) = _ _ _ _ _ _ _ _ and g'(x) = _ _ _ _ _ _ _ _.
The derivative off [g (x)] w. r. t. x in terms of f and g is _ _ _ _ _ _ _ _.
Therefore $$\frac{d}{d x}$$[f[g(x)]] = _ _ _ _ _ _ _ _ _ and [$$\frac{d}{d x}$$[f[g(x)]]]x = 0 = _ _ _ _ _ _ _ _ _ _ _.
The derivative of g[f(x)] w. r. t. x in terms of f and g is _ _ _ _ _ _ __ _ _ _ _.
Therefore $$\frac{d}{d x}$$[g[f(x)]] = _ _ _ _ _ _ _ _ _ and [$$\frac{d}{d x}$$[g[f(x)]]]x = 1 = _ _ _ _ _ _ _ _ _ _ _.”
Hint basket : { f ‘[g(x)]·g'(x), 2e2x + 6ex, 8, g'[f(x)]·f ‘(x), 2xex2 + 5, -2e6, e2x + 6ex + 14, ex2 + 5 + 3, 2x, ex}
Solution:
f[g(x)] = e2x + 6ex + 14
g[f(x)] = ex2 + 5 + 3
f'(x) = 2x, g’f(x) = ex
The derivative of f[g(x)] w.r.t. x in terms of and g is f'[g(x)]∙g'(x).
∴ $$\frac{d}{d x}$${f[g(x)]} = 2e2x + 6ex and $$\frac{d}{d x}$${f[g(x)]}x = 0 = 8
The derivative of g[f(x)] w.r.t. x in terms of f and g is g’f(x)]∙f'(x).
∴ $$\frac{d}{d x}$${g[(f(x)]} = 2xex2 + 5 and
$$\frac{d}{d x}$${g[(f(x)]}x = -1 = -2e6. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 9.9: Factor Polynomials Using Special Products
Difficulty Level: Basic Created by: CK-12
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Practice Special Products of Polynomials
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What if the area of a square playground were 10,000 square feet? Instead of taking the square root of the area to find the length of one of the playground's sides, you could set up the equation \begin{align*}s^2 = 10,000\end{align*}, subtract 10,000 from both sides to get \begin{align*}s^2 - 10,000 = 0\end{align*}, and solve for \begin{align*}s\end{align*} by factoring \begin{align*}s^2 -10,000\end{align*} on the left side of the equation. But how would you factor this? After completing this Concept, you'll know how to factor expressions like this one by using special products.
### Guidance
When we learned how to multiply binomials, we talked about two special products: the Sum and Difference Formula and the Square of a Binomial Formula. In this Concept, we will learn how to recognize and factor these special products.
Factoring the Difference of Two Squares
We use the Sum and Difference Formula to factor a difference of two squares. A difference of two squares can be a quadratic polynomial in this form: \begin{align*}a^2-b^2\end{align*}. Both terms in the polynomial are perfect squares. In a case like this, the polynomial factors into the sum and difference of the square root of each term.
\begin{align*}a^2-b^2=(a+b)(a-b)\end{align*}
In these problems, the key is figuring out what the \begin{align*}a\end{align*} and \begin{align*}b\end{align*} terms are. Let’s do some examples of this type.
#### Example A
Factor the difference of squares.
(a) \begin{align*}x^2-9\end{align*}
(b) \begin{align*}x^2y^2-1\end{align*}
Solution:
(a) Rewrite \begin{align*}x^2-9\end{align*} as \begin{align*}x^2-3^2\end{align*}. Now it is obvious that it is a difference of squares.
We substitute the values of \begin{align*}a\end{align*} and \begin{align*}b\end{align*} in the Sum and Difference Formula:
\begin{align*}(x+3)(x-3)\end{align*}
The answer is \begin{align*}x^2-9=(x+3)(x-3)\end{align*}.
(b) Rewrite \begin{align*}x^2y^2-1\end{align*} as \begin{align*}(xy)^2-1^2\end{align*}. This factors as \begin{align*}(xy+1)(xy-1)\end{align*}.
Factoring Perfect Square Trinomials
A perfect square trinomial has the form:
\begin{align*}a^2+2ab+b^2 \qquad \text{or} \qquad a^2-2ab+b^2\end{align*}
The factored form of a perfect square trinomial has the form:
\begin{align*}&&(a+b)^2 \ if \ a^2+2(ab)+b^2\\ \text{And}\\ &&(a-b)^2 \ if \ a^2-2(ab)+b^2\end{align*}
In these problems, the key is figuring out what the \begin{align*}a\end{align*} and \begin{align*}b\end{align*} terms are. Let’s do some examples of this type.
#### Example B
Factor \begin{align*}x^2+8x+16\end{align*}.
Solution:
Check that the first term and the last term are perfect squares.
\begin{align*}x^2+8x+16 \qquad \text{as} \qquad x^2+8x+4^2.\end{align*}
Check that the middle term is twice the product of the square roots of the first and the last terms. This is true also since we can rewrite them.
\begin{align*}x^2+8x+16 \qquad \text{as} \qquad x^2+2 \cdot 4 \cdot x+4^2\end{align*}
This means we can factor \begin{align*}x^2+8x+16\end{align*} as \begin{align*}(x+4)^2\end{align*}.
#### Example C
Factor \begin{align*}x^2-4x+4\end{align*}.
Solution:
Rewrite \begin{align*}x^2-4x+4\end{align*} as \begin{align*}x^2+2 \cdot (-2) \cdot x+(-2)^2\end{align*}.
We notice that this is a perfect square trinomial and we can factor it as \begin{align*}(x-2)^2\end{align*}.
Solving Polynomial Equations Involving Special Products
We have learned how to factor quadratic polynomials that are helpful in solving polynomial equations like \begin{align*}ax^2+bx+c=0\end{align*}. Remember that to solve polynomials in expanded form, we use the following steps:
Step 1: Rewrite the equation in standard form such that: Polynomial expression = 0.
Step 2: Factor the polynomial completely.
Step 3: Use the Zero Product Property to set each factor equal to zero.
Step 4: Solve each equation from step 3.
### Guided Practice
Solve the following polynomial equations.
\begin{align*}x^2+7x+6=0\end{align*}
Solution: No need to rewrite because it is already in the correct form.
Factor: We write 6 as a product of the following numbers:
\begin{align*}6& =6 \times 1 && \text{and} && 6+1=7\\ x^2+7x+6 & = 0 && \text{factors as} && (x+1)(x+6)=0\end{align*}
Set each factor equal to zero:
\begin{align*}x+1=0 \qquad \text{or} \qquad x+6=0\end{align*}
Solve:
\begin{align*}x=-1 \qquad \text{or} \qquad x=-6\end{align*}
Check: Substitute each solution back into the original equation.
\begin{align*}(-1)^2+7(-1)+6&=1+(-7)+6=0\\ (-6)^2+7(-6)+6&=36+(-42)+6=0\end{align*}
### Practice
Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Factoring Special Products (10:08)
Factor the following perfect square trinomials.
1. \begin{align*}x^2+8x+16\end{align*}
2. \begin{align*}x^2-18x+81\end{align*}
3. \begin{align*}-x^2+24x-144\end{align*}
4. \begin{align*}x^2+14x+49\end{align*}
5. \begin{align*}4x^2-4x+1\end{align*}
6. \begin{align*}25x^2+60x+36\end{align*}
7. \begin{align*}4x^2-12xy+9y^2\end{align*}
8. \begin{align*}x^4+22x^2+121\end{align*}
Factor the following differences of squares.
1. \begin{align*}x^2-4\end{align*}
2. \begin{align*}x^2-36\end{align*}
3. \begin{align*}-x^2+100\end{align*}
4. \begin{align*}x^2-400\end{align*}
5. \begin{align*}9x^2-4\end{align*}
6. \begin{align*}25x^2-49\end{align*}
7. \begin{align*}-36x^2+25\end{align*}
8. \begin{align*}16x^2-81y^2\end{align*}
Solve the following quadratic equations using factoring.
1. \begin{align*}x^2-11x+30=0\end{align*}
2. \begin{align*}x^2+4x=21\end{align*}
3. \begin{align*}x^2+49=14x\end{align*}
4. \begin{align*}x^2-64=0\end{align*}
5. \begin{align*}x^2-24x+144=0\end{align*}
6. \begin{align*}4x^2-25=0\end{align*}
7. \begin{align*}x^2+26x=-169\end{align*}
8. \begin{align*}-x^2-16x-60=0\end{align*}
Mixed Review
1. Find the value for \begin{align*}k\end{align*} that creates an infinite number of solutions to the system \begin{align*}\begin{cases} 3x+7y=1\\ kx-14y=-2 \end{cases}\end{align*}.
2. A restaurant has two kinds of rice, three choices of mein, and four kinds of sauce. How many plate combinations can be created if you choose one of each?
3. Graph \begin{align*}y-5= \frac{1}{3}(x+4)\end{align*}. Identify its slope.
4. \$600 was deposited into an account earning 8% interest compounded annually.
1. Write the exponential model to represent this situation.
2. How much money will be in the account after six years?
1. Divide: \begin{align*}4 \frac{8}{9} \div -3\frac{1}{5}\end{align*}.
2. Identify an integer than is even and not a natural number.
### Notes/Highlights Having trouble? Report an issue.
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### Vocabulary Language: English Spanish
TermDefinition
difference of two squares The difference of two squares has the form $a^2-b^2=(a+b)(a-b)$.
Difference of Squares A difference of squares is a quadratic equation in the form $a^2-b^2$.
Perfect Square Trinomial A perfect square trinomial is a quadratic expression of the form $a^2+2ab+b^2$ (which can be rewritten as $(a+b)^2$) or $a^2-2ab+b^2$ (which can be rewritten as $(a-b)^2$).
Quadratic form A polynomial in quadratic form looks like a trinomial or binomial and can be factored like a quadratic expression.
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The beauty of Montessori Math continues with the Golden Bead Static Addition lesson.
Each child needs a tray. Ask one child to get the numeral “1341” from the small numeral cards. Ask him to read the cards and slide them together. Ask him to get the quantity from the bank. Give a second child the numeral “2435” and proceed with the same approach.
When each child returns from the bank with their quantities, ask them to tell you what they’ve brought you.
Have the first child place his quantity in the correct columns at the top of the mat, and his numerals to the right. The second child then should place his quantity and symbols a few inches below the first child’s.
“Now we have 1341 and 2435 beads here. I am going to put them together in one place. When we put things together it is called addition because we are adding two numbers together. We always add the units first.
Slide the top group of numerals down to the bottom group and both groups to the bottom edge of the mat, below the equal sign.
Ask one of the children to count the units and retrieve a large numeral card, which tells us how many beads we have after adding them together. Place the large numeral card below the units column of the small numeral cards.
Ask the second child to scoop the ten bars together and then to the bottom of the mat. Ask him to count the ten bars, retrieve a large numeral card that corresponds to the number of ten bars, and place the large numeral card next to the units numeral card.
Continue with the approach. One child adds 100s. The other child adds 1000s.
“Now look what happened when we added 1341 and 2435…we got 3776. When we add, we use a special sign to show what we’re doing. It is called a plus sign, and it goes here.
We also need another sign to show that we are writing the new number that we got by adding. It is called the equal sign and it goes here. So now we have 1341 + 2435 = 3776.”
Review the names for the plus and equal signs with a 3-period lesson.
Invite the children to put the quantity and symbols away. Then give them a new number. Completing three problems is a great start for the first time with this work. With each problem, withdraw more and more.
Related Read: Learn the Teen Board Montessori Math Lesson
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# Trigonometry Part 5: Advance math questions.
Trigonometry Part 5: Advance math questions.
Q161. The length of the shadow of a vertical tower on level ground increases by 10 metres when the altitude of the sun changes from 45° to 30°. Then the height of the tower is
(a)5( √3+ 1) metres
(b) 5( √3- 1) metres
(c) 5 √3 metres
(d) 5/√3 metres
Q162. If a pole of 12 m height casts a shadow of 4√3 m long on the ground, then the sun’s angle of elevation at that instant is
(a) 30°
(b) 60°
(c) 45°
(d) 90°
Q163. The angle of elevation of the top of a tower from a point on the ground is 30° and moving 70 metres towards the tower it becomes 60°. The height of the tower is
(a)10 metre
(b)10/√3 metre
(c) 10√3 metre
(d) 35√3 metre
Q164. In circular measure, the value of the angle 11°15’ is
(a) π°/16
(b) π°/8
(c) π°/4
(d) π°/12
Q165. In a triangle ABC, ΔABC = 75° and ΔABC =π°/4. The circular measure of ΔBAC is
Q166. The circular measure of an angle of an isosceles triangle is 5π/9. Circular measure of one of the other angles must be
(a) 5π/8
(b) 5π/9
(c) 2π/9
(d) 4π/9
Q167. The degree measure of 1 radian (taking π = 22/7) is
(a)57°61’22” (approx.)
(b) 57°16’22” (approx.)
(c) 57°22’16” (approx.)
(d) 57°32’16” (approx.)
Q168. 3π/5 radians is equal to
(a)100°
(b) 120°
(c) 108°
(d) 180°
Q169. If the sum of two angles is 135° And their difference is π/12 , then the circular measure of the greater angle is
(a) 2π/3
(b) 3π/5
(c) 5π/12
(d) π/3
Q170. There are two vertical posts, one on each side of a road. Just opposite to each other. One post is 108 metre high. From the top of this post, the angle of depression of the top and foot of the other post are 30° and 60° respectively. The height of the other post (in metre) is
(a) 36
(b) 72
(c) 108
(d) 110
Q171. There are two temples, one on each bank of a river, just opposite to each other. One temple is 54 m high. From the top of this temple, the angles of depression of the top and the foot of the other temple are 30° and 60° respectively. The length of the temple is:
(a) 18 m
(b) 36 m
(c) 36√3 m
(d) 18√3 m
Q172. The top of two poles of height 24 m are 36m are connected by a wire. If the wire makes an angle of 60° with the horizontal. Then the length of the wire is
(a) 6 m
(b) 8√3 m
(c) 8 m
(d) 6√3 m
Q173. From the top of a hill 200 m high, the angle of depression of the top and the bottom of a tower are observed to be 30° and 60°. The height of the tower is (in m):
(a) 400√3/3
(b) 166
(c) 133
(d) 200√3
Q174. From a tower 125 metres high, the angle of depression of two objects, which are in horizontal line through the base of the tower, are 45° and 30° and they are on the same side of the tower. The distance (in metres) between the objects is
(a) 125√3
(b) 125(√3- 1)
(c) 125/(√3 – 1)
(d) 125( √3+ 1)
Q175. From the top of a tower of height 180 m the angles of depression of two objects on either sides of the tower are 30° and 45°. Then the distance between the objects are
(a) 180 (3 +√3) m
(b) 180 (3 -√3) m
(c) 180 ( √3– 1) m
(d) 180 ( √3+ 1) m
Q176. From the peak of a hill which is 300 m high, the angle of depression of two sides of a bridge lying on a ground are 45° and 30° (both ends of the bridge are on the same side of the hill). Then the length of the bridge is
(a) 300 ( √3– 1) m
(b) 300 ( √3+ 1) m
(c) 300√3 m
(d) 300/√3 m
Q177. From an aeroplane just over a river, the angle of depression of two palm trees on the opposite bank of the river are found to be 60° and 30° respectively. If the breadth of the river is 400 metres, then the height of the aeroplane above the river at that instant is
(Assume √3 = 1.732)
(a) 173.2 metres
(b) 346.4 metres
(c) 519.6 metres
(d) 692.8 metres
Q178. If tan θ = 1, then the value of
(a) 2
(b) 2
(c) 3
(d) 4/5
Q179. If θ is a positive acute angle and cos2θ + cos4 θ = 1 then the value of tan2 θ + tan4 θ is
(a) 3/2
(b) 1
(c) ½
(d) 0
Q180. What is the value of tan 4°.tan 43°.tan 47°.tan 86° ?
(a) 0
(b) 1
(c) √3
(d) 1/√3
Q181. If then the value of is
(a) 0.5
(b) – 0.5
(c) 3.0
(d) -3.0
Q182. If tan 15°=2-√3 then tan 15° cot75°+ tan75° cot15° is
(a) 14
(b) 12
(c) 10
(d) 8
Q183. If θ is an acute angle and tanθ+cotθ=2 then tan 5 θ cot 10 θ is
(a) 1
(b) 2
(c) 3
(d) 4
Q184. What will be the value of (sin21°+sin23°+sin25°+……………..+sin285°+sin287°+sin289°)?
(a) 21
(b) 22
(c) 22
(d) 23
Q185. If sin θ – cos θ = 7/13 and 0< θ < 90° then what is the value of sin θ + cos θ ?
(a) 17/13
(b) 13/17
(c) 1/13
(d) 1/17
Q186. If sinα + cosβ =2 (0° ≤ a ≤ 90°) then what will be the value of sin
(a) sin a/2
(b) cos a/3
(c) sin a/3
(d) cos 2a/3
Q187. If cosθ – sinθ then what will be the value of 2 cos2 θ -1
(a) 0
(b) 1
(c) 2/3
(d) 3/2
Q188. What is the value of cot 10° . cot 20° . cot 60° . cot 70° . cot 80°.
(a) 1
(b) – 1
(c) √3
(d) 1/√3
Q189. If sin a sec ( 30° + a ) = 1 (0< a < 60° ), the value of sin a + cos 2a is
(a) 1
(b)
(c) 0
(d) √2
Q190. What will be simplified value of (secA – cosA)2 + (cosec A- sinA)2 – (cot A – tanA)2 ?
(a) 0
(b) 1/2
(c) 1
(d) 2
Q191. If θ is an acute angle and 7 sin2 θ+ 3cos2θ then what is the value of tan θ ?
(a) √3
(b) 1/√3
(c) 1
(d) 0
Q192. What is the value of sin2 1°+ sin2 5° + sin2 9° + …….. + sin2 89° ?
(a) 11
(b) 11 √2
(c) 11
(d) 11/ √2
Q193. The numerical value of cot 18°(cot 72° cos2 22° + ) is
(a) 1
(b) √2
(c) 3
(d) 1/√3
Q194. If A = sinθ + cos2 θ for any value of then what is the value of A ?
(a)
(b)
(c)
(d)
Q195. If sinθ + cosecθ = 2, then in the situation of 0° ≤ θ ≤ 90°. What will be the value of sin5 θ + cosec5 θ ?
(a) 0
(b) 1
(c) 10
(d) 2
Q196. sin2 5° sin210° + sin2 15°+ ………. + sin2 85° + sin2 90° will be equal to
(a) 7
(b) 8
(c) 9
(d) 9
Q197. If tan 2θ.tan4θ =1, then the value of tan3θ is
(a) √3
(b) 0
(c) 1
(d) 1/√3
Q198. If 2 cosθ – sinθ = 1/ √2 , (0° < θ < 90°) then the value of 2 sin θ + cos θ is
(a) 1/√2
(b) √2
(c) 3/√2
(d) √2/3
Q199. If then the value of sin4 θ – cos4 θ is
(a) 1/5
(b) 2/5
(c) 3/5
(d) 4/5
Q200. The value of tan1°.tan2°.tan3°.tan4°………………tan87°.tan88°.tan89° is
(a) 1/√3
(b) √3
(c) 1
(d) Not defined
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# Time and Work Quiz Set 003
### Question 1
A can finish a work in 5 days. B can do the same work in 7 days. They work together for 2 days. The fraction of work that is left is?
A
\${11/35}\$.
B
\$1{11/35}\$.
C
\${11/34}\$.
D
\${1/3}\$.
Soln.
Ans: a
They together finish \$1/5 + 1/7\$ work in a day. In 2 days they finish 2 × \$(1/5 + 1/7)\$ work, which is \$24/35\$. So the un-finished work is 1 - \$24/35\$ = \${11/35}\$.
### Question 2
A can finish a work in 6 days. B can do the same work in 15 days. They work together for 4 days. The fraction of work that is left is?
A
\${1/15}\$.
B
\$1{1/15}\$.
C
\${1/14}\$.
D
\${1/8}\$.
Soln.
Ans: a
They together finish \$1/6 + 1/15\$ work in a day. In 4 days they finish 4 × \$(1/6 + 1/15)\$ work, which is \$84/90\$. So the un-finished work is 1 - \$84/90\$ = \${1/15}\$.
### Question 3
A can do a piece of work in 39 days. B is 30% more efficient than A. In how many days will they complete the work if they work together?
A
\$16{22/23}\$ days.
B
17 days.
C
\$17{1/23}\$ days.
D
\$17{2/23}\$ days.
Soln.
Ans: a
Let us first calculate the one day work of B. One day work of A is given as \$1/39\$. If B is 30% efficient, then one day work of B is \$1/39\$ × \$130/100\$ = \$1/30\$. Putting x = 39 and y = 30 in the shortcut method, we get \${xy}/{x + y}\$ = \${390/23}\$, which is same as: \$16{22/23}\$.
### Question 4
If 70 men can do a task in 17 days, how many men are required to complete the task in 14 days?
A
85.
B
15.
C
13.
D
17.
Soln.
Ans: a
If m1 men can do a task in d1 days, and m2 in d2, then we must have m1 × d1 = m2 × d2. Putting m1 = 70, d1 = 17 and d2 = 14, we get m2 = 85.
### Question 5
A new tub can be filled by a tap in 13 minutes. But the tub is worn out, and there is a leakage that can empty the tub in 17 minutes. In how many minutes will the tap be able to fill the tub?
A
\$55{1/4}\$ mins.
B
\$56{1/4}\$ mins.
C
\$57{1/4}\$ mins.
D
\$58{1/4}\$ mins.
Soln.
Ans: a
Putting x = 13 and y = 17 in the shortcut method, we get \${xy}/{y - x}\$ = \${221/4}\$, which is same as: \$55{1/4}\$. |
# Difference between revisions of "1986 AIME Problems/Problem 11"
## Problem
The polynomial $1-x+x^2-x^3+\cdots+x^{16}-x^{17}$ may be written in the form $a_0+a_1y+a_2y^2+\cdots +a_{16}y^{16}+a_{17}y^{17}$, where $y=x+1$ and the $a_i$'s are constants. Find the value of $a_2$.
## Solution
### Solution 1
Using the geometric series formula, $1 - x + x^2 + \cdots - x^{17} = \frac {1 - x^{18}}{1 + x} = \frac {1-x^{18}}{y}$. Since $x = y - 1$, this becomes $\frac {1-(y - 1)^{18}}{y}$. We want $a_2$, which is the coefficient of the $y^3$ term in $-(y - 1)^{18}$ (because the $y$ in the denominator reduces the degrees in the numerator by $1$). By the Binomial Theorem, this is $(-1) \cdot (-1)^{15}{18 \choose 3} = \boxed{816}$.
### Solution 2
Again, notice $x = y - 1$. So
\begin{align*}1 - x + x^2 + \cdots - x^{17} & = 1 - (y - 1) + (y - 1)^2 - (y - 1)^3 + \cdots - (y - 1)^{17} \\ & = 1 + (1 - y) + (1 - y)^2 + (1 - y)^3 \cdots + (1 - y)^{17}\end{align*}.
We want the coefficient of the $y^2$ term of each power of each binomial, which by the binomial theorem is ${2\choose 2} + {3\choose 2} + \cdots + {17\choose 2}$. The Hockey Stick Identity tells us that this quantity is equal to ${18\choose 3} = \boxed{816}$.
### Solution 3
Again, notice $x=y-1$. Substituting $y-1$ for $x$ in $f(x)$ gives: \begin{align*}1 - x + x^2 + \cdots - x^{17} & = 1 - (y - 1) + (y - 1)^2 - (y - 1)^3 + \cdots - (y - 1)^{17} \\ & = 1 + (1 - y) + (1 - y)^2 + (1 - y)^3 \cdots + (1 - y)^{17}\end{align*}. From binomial theorem, the coefficient of the $y^2$ term is ${2\choose 0} + {3\choose 1} + \cdots + {17\choose 15}$. This is actually the sum of the first 16 triangular numbers, which evaluates to $\frac{(16)(17)(18)}{6} = \boxed{816}$.
### Solution 4(calculus)
Let $f(x)=1-x+x^2-x^3+\cdots+x^{16}-x^{17}$ and $g(y)=a_0+a_1y+a_2y^2+\cdots +a_{16}y^{16}+a_{17}y^{17}$.
Then, since $f(x)=g(y)$, $$\frac{d^2f}{dx^2}=\frac{d^2g}{dy^2}$$ $\frac{d^2f}{dx^2} = 2\cdot 1 - 3\cdot 2x+\cdots-17\cdot 16x^{15}$ by the power rule.
Similarly, $\frac{d^2g}{dy^2} = a_2(2\cdot 1) + a_3(3\cdot 2y)+\cdots+a_{17}(17\cdot 16y^{15})$
Now, notice that if $x = -1$, then $y = 0$, so $f^{''}(-1) = g^{''}(0)$
$g^{''}(0)= 2a_2$, and $f^{''}(-1) = 2\cdot 1 + 3\cdot 2 +\cdots + 16\cdot 17$.
Now, we can use the hockey stick theorem to see that $2\cdot 1 + 3\cdot 2 +\cdots + 16\cdot 17 = 2\binom{18}{3}$
Thus, $2a_2 = 2\binom{18}{3}\rightarrow a_2 = \binom{18}{3}=\boxed{816}$
-AOPS81619
## See also
1986 AIME (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 12 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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## How many prime numbers are there between 1 and 300?
List of Prime Numbers From 1 to 500
Range of Numbers List of Prime Numbers Total
201- 300 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293 16
301 – 400 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397 16
What are the co-prime numbers from 1 to 100?
Co-prime Numbers from 1 to 100 Some of the co-prime number pairs that exist from 1 to 100 are (1, 2), (3, 67), (2, 7), (99, 100), (34, 79), (54, 67), (10, 11), etc.
### What are all the prime numbers of 300?
Thus, the prime factorization of 300 is 2 × 2 × 3 × 5 × 5 = 22 × 3 × 52 where 2, 3, and 5 are the prime numbers.
What are the co-prime numbers between 1 to 50?
What Is Co Prime Number?
1,2 2,3 3,5
1,49 2,97 13,67
1,50 2,99 13,71
17,19 19,23 13,73
17,23 19,29 13,79
## How many prime numbers are there between 1 and 100000?
Table 2. π(x) verse x/ln x
x π(x) x/ln x
1000 168 145
10000 1229 1086
100000 9592 8686
1000000 78498 72382
Why 1 is not a prime number?
1 can only be divided by one number, 1 itself, so with this definition 1 is not a prime number. It is important to remember that mathematical definitions develop and evolve. Throughout history, many mathematicians considered 1 to be a prime number although that is not now a commonly held view.
### Is 7 and 15 Coprime numbers?
Factors of 7 are 1 and 7. Factors of 15 are 1, 3, 5 and 15. . since there is only one common factor the given numbers are co-prime.
Is 8 and 15 are Coprime numbers?
So, by this definition, 15 and 8 are co-prime, but 15 and 9 are not. There are 4 positive integers less than 12 and co- prime with 12. There are 8 positive integers less than 15 and co- prime with 15. They are 1, 2, 4, 7, 8, 11, 13 and 14.
## What is the prime factorization for 144?
The factors of composite number 144 are 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, and 144. 144 is a square number. The prime factorization of 144 is 2 × 2 × 2 × 2 × 3 × 3.
What is the prime factor of 42?
Prime factors of 42 are 1, 2, 3, and 7. Factors of 60 are the numbers which divide 60 exactly without any remainder.
### Is 17 and 68 a Coprime number?
17 and 68 are not co-prime because 1 is not the only common factor of these numbers. Example, 17 is another common factor of 17 and 68. 215 and 216 are co-prime because 1 is the only common factor of these numbers.
Are there any prime numbers from 1 to 300?
List of Prime Numbers from 1 to 300. 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293. Sequence. Prime Number. 1. 2.
## Why are 11 and 30 considered coprime numbers?
11 and 30 are coprime numbers because they both are commonly divisible only by 1 and no other factor. 11 = 1 x 11 30 = 1 x 2 x 3 x 5 As, we can see here, there is no common factor other than 1 between 11 and 30.
Are there any coprime numbers from 1 to 100?
There are several pairs of co-primes from 1 to 100 which follow the above properties. Some of them are: (75, 41) and so on. Also, we can write any number with the combination of 1 as a coprime pair such as (22, 1), (31, 1), (4, 1), (90, 1), (1, 100). In this way, many coprime numbers are defined from 1 to 100.
### Are there any numbers that are always co prime?
Any two successive numbers/ integers are always co-prime: Take any consecutive number such as 2, 3 or 3, 4 or 5, 6 and so on; they have 1 as their HCF. The sum of any two co-prime numbers are always co-prime with their product: 2 and 3 are co-prime and have 5 as their sum (2+3) and 6 as the product (2×3). |
## First principle of calculus
A secant line is a straight line that connects two points on a curve. A tangent line is a straight line that “just touches” the curve at a single point.
The slope of the secant line gives us the average rate of change of the function on the interval between the two points. The slope of the tangent line gives us the instantaneous rate of change.
Finding the slope of the secant line is easy:
m_"sec" = (Delta y)/(Delta x) = (y_2 - y_1)/(x_2 - x_1) = (3 - 4)/(4 - 2) = -1/2.
We can generalize this to give us the average rate on the interval [x_1, x_2] for any function f with
m_"sec" = (f(x_2) - f(x_1))/(x_2 - x_1),
but what about the tangent line? We could draw it with a ruler and pick two points that is passes through, but this method is tedious and inaccurate. No, instead we can use our newly-acquired limit skills.
The key thing to realize here is that the secant line and the tangent line are related. A tangent line is just a secant line whose interval happens to be very, very small—infinitesimal, to be exact (this means indefinitely small). In other words, the slope of the tangent line at x=a is equal to the slope of the secant line on the interval [a, a+h] as h approaches zero:
m_"tan" = lim_(h->0)(f(a+h) - f(a))/h.
This is the first principle of calculus. As with most other things, graphing will help us understand it:
As we move the second point closer to the first point, thereby making h approach zero, the slope of the secant line will converge on the slope of the tangent line at a.
Suppose we want to find the slope of the tangent line at 3 on f(x) = x^2. We simply plug it into our first principle of calculus, giving us
m_"tan" = lim_(h->0)(f(3+h) - f(3))/h = lim_(h->0)((3+h)^2 - 3^2)/h,
which simplifies to
lim_(h->0)(9+6h+h^2 - 9)/h = lim_(h->0)(6h+h^2)/h = lim_(h->0)6+h = 6+0 = 6,
therefore the instantaneous rate of change at x = 3 is 6. |
# What Is Median Average?
The median average (or median) is the middle number in a set of numbers that has been arranged in order.
# How to Find the Median Average
Question: What is the median of 8, -15, 14, 54, and 8?
Step 1
Sort the list from smallest to largest.
Step 2
Select the middle number. That's the median.
# How to Find the Median Average with an Even Amount of Numbers
When a set has an odd amount of numbers, the median is easily found. (It's the one in the middle with an equal amount of numbers either side of it). However, if the amount of numbers is even, finding the median is a little trickier as there is no middle number. For example:
Question: What is the median of 16, 9, 24, and 33?
Step 1
Sort the list from smallest to largest.
Step 2
Select the middle 2 numbers.
Step 3
Add those 2 numbers and divide by 2. That's the median.
show
# FINDING THE MIDDLE NUMBER QUICKLY
To find the middle number quickly, add 1 to the amount of numbers in the set and then divide by 2. For example, if there are 9 numbers in the set, add 1 (giving you 10), then divide by 2 (giving you 5). The 5th number in the set is the median.
With a set with an even amount of numbers, do the same. For example, if there are 10 numbers in the set, add 1 (giving you 11), then divide by 2 (giving you 5.5). This means we have to take the 5th and 6th numbers in the set. (You then have to add them and divide by 2 for your median.
This tip is useful with large sets of numbers. For example, if there are 75 numbers in the set, add 1 (giving you 76), then divide by 2 (giving you 38). This means we have to take the 5th and 6th numbers in the set. The 38th number in the set is the median.
WHEN IS THE MEDIAN USEFUL?
Median is useful when the set has numbers outside the expected range which would skew the mean.
For example, if we were to look at the average age of a school class, the mean will be skewed by the age of the teacher. The median, however, would give us a better idea of the age of the children. |
# Algebra formula solver
This Algebra formula solver supplies step-by-step instructions for solving all math troubles. We can solve math problems for you.
## The Best Algebra formula solver
Algebra formula solver is a mathematical tool that helps to solve math equations. Algebra is a branch of mathematics that uses arithmetical and geometrical methods to solve equations. Algebra is the mathematics of equations and variables, which means that algebra unsolved for x is incomplete. Algebraic equations are equations that have one or more variable terms, such as x + 3 = 5. The variable x represents an unknown quantity, and solving for x means finding the value of the variable that makes the equation true. In this case, solving for x would give us the answer 2, because 2 + 3 = 5. Algebra can be used to solve for other unknowns in equations as well, making it a powerful tool for mathematical problem-solving. Thanks to algebra, we can unlock the solutions to many mysteries hidden in equations.
A rational function is any function which can be expressed as the quotient of two polynomials. In other words, it is a fraction whose numerator and denominator are both polynomials. The simplest example of a rational function is a linear function, which has the form f(x)=mx+b. More generally, a rational function can have any degree; that is, the highest power of x in the numerator and denominator can be any number. To solve a rational function, we must first determine its roots. A root is a value of x for which the numerator equals zero. Therefore, to solve a rational function, we set the numerator equal to zero and solve for x. Once we have determined the roots of the function, we can use them to find its asymptotes. An asymptote is a line which the graph of the function approaches but never crosses. A rational function can have horizontal, vertical, or slant asymptotes, depending on its roots. To find a horizontal asymptote, we take the limit of the function as x approaches infinity; that is, we let x get very large and see what happens to the value of the function. Similarly, to find a vertical asymptote, we take the limit of the function as x approaches zero. Finally, to find a slant asymptote, we take the limit of the function as x approaches one of its roots. Once we have determined all of these features of the graph, we can sketch it on a coordinate plane.
Word math problems can be tricky, but there are a few tips that can help you solve them more quickly and easily. First, read the problem carefully and identify the key information. Then, determine what operation you need to perform in order to solve the problem. Next, write out the equation using numbers and symbols. Finally, solve the equation and check your work to make sure you've found the correct answer. By following these steps, you can approach word math problems with confidence and avoid making common mistakes. With a little practice, you'll be solving them like a pro in no time!
When dealing with data, there are typically three different types of averages that can be used in order to summarize the information: the mean, the median, and the mode. Of these, the mode is often the most difficult to calculate. However, once you understand the definition of mode and how it is used, solving for it becomes a relatively straightforward process. Mode is simply the value that appears most frequently in a data set. In order to calculate it, first identify all of the unique values in your data set and then count how many times each one occurs. The value that occurs most often is the mode. In some cases, there may be more than one mode, or no mode at all. When this happens, it is said to be bimodal or multimodal if there are two or more modes, respectively, and unimodal if there is only one.
In a right triangle, the longest side is called the hypotenuse, and the other two sides are called legs. To solve for x in a right triangle, you will need to use the Pythagorean theorem. This theorem states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. In other words, if you know the lengths of all three sides of a right triangle, you can solve for any one of them using this equation. To solve for x specifically, you will need to square both sides of the equation and then take the square root of each side. This will give you the length of side x. You can then use this information to calculate the other two sides if needed.
## Help with math
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# It's Not Magic, It's Mathematics
I have been teaching mathematics in an Australian High School since 1982, and I am a contributing author to mathematics text books.
## Math Magic Tricks
Entertainers such as magicians and mentalists incorporate numbers into their staged illusions. I am referring not to sleight of hand card tricks or other such manipulations but a display of mathematics camouflaged by razzle-dazzle and cries of “abracadabra".
Although we know it is not real magic, it still seems like they are doing the impossible, just like creating impossible mathematics shapes such as the ones shown below.
## Understanding Number Magic
This article will hopefully go some way to demystify so-called number magic and encourage you to explore the fascinating world of number patterns and algebra.
## Magic 1: Is That a Zebra Crossing?
Let’s begin with one where I predict the outcome regardless of your initial choice of number.
Carry out these steps in turn, keeping track of your answer each time.
1. Think of any number.
2. Square it. That means multiply it with itself, such as 3 x 3, 8 x 8.
3. Add the result to your original number.
4. Divide the answer by your original number.
6. Subtract from the answer the number you started with.
7. Divide by 10.
8. Now add 16.
9. If A = 1, B = 2, C = 3, D = 4, etc, work out the letter that corresponds to your final answer.
10. Think of a 4-legged animal whose name starts with the letter you found.
I’m sure the animal you came up with has stripes and looks like a donkey!
Try this again using a different number. What can you conclude?
Now let’s see mathematically what is happening.
We will use the letter N to represent the start number and perform each of the 10 steps using this letter. The solution is shown alongside each step.
1. Think of any number. [ N]
2. Square it. [N x N]
3. Add the result to your original number. [N + N x N]
4. Divide the answer by your original number. [(N + N x N) ÷ N = 1 + N]
5. Add 99. [100 + N]
6. Subtract from the answer the number you started with. [100 + N – N = 100]
7. Divide by 10. [100 ÷ 10 =10]
8. Now add 16. [10 + 16 = 26]
9. If A = 1, B = 2, C = 3, D = 4, etc, work out the letter that corresponds to your final answer. [26]
10. Think of a 4-legged animal whose name starts with the letter you found.
[Zebra is the animal everyone would immediately come up with]
Scroll to Continue
## Read More From Owlcation
We conclude that the number we start with has no effect on the final number, which is always 26.
## Magic 2: I Know Your Age
Here is one where you can precisely determine a person’s age even though their choice of the start number is completely random.
Let’s assume it is currently January 1, 2018, the person was born on 14/8/1995, and he chooses 4 as his start number. The solution is shown alongside each step.
1. Ask them to think of a number from 2 to 9. [ 4]
2. Multiply the result by 2. [ 8]
3. Add 5 to the answer. [ 13]
4. Now multiply by 50. [ 650]
5. If the person has had their birthday, add 1767. If the person is yet to have their birthday, add 1768. [ 2418]
6. Ask them to subtract from their answer the year they were born. [ 423]
The last 2 digits of the answer is their age. [ 23]
We can now show why this method works by letting N be the start number and writing down the result of each step in terms of N.
1. Ask them to think of a number from 2 to 10.
[ N]
2. Multiply the result by 2.
[ 2xN]
[ 2xN + 5]
4. Now multiply by 50.
[ 100xN + 250]
5. If the person has had their birthday, add 1767.
[ 100xN + 2017]
If the person is yet to have their birthday, add 1768.
[ 100xN + 2018]
6. Ask them to subtract from their answer the year they were born.
[ 100xN + (2018 – year of birth)] or [ 100xN + (2017 – year of birth)]
100xN can only have the values 200, 300, …, 900. This can be ignored in the final answer. Then (2018 – year of birth) or (2017 – year of birth) is the person’s birth year, which is obtained from the last 2 digits of the answer.
## Magic 3: Hieroglyphics Prediction
This one is both interesting and easy to explain. We will use 46 as our initial number.
1. Think of a number from 10 to 99. [ 46]
2. Add its two digits together. [ 4 + 6 = 10]
3. Subtract the total from the original number. [46 – 10 = 36]
4. Find the shape next to your answer. [ a circle inside another circle]
It turns out that the answer will always correspond to a number with a circle next to it.
Let’s see why by reworking and explaining each step.
1. Suppose our 2-digit number is AB. This can be written as 10xA + B. For example, 46 = 10x4 + 6.
2. Add the two digits together to get A + B.
3. To subtract the total from the original number, we write 10xA + B – (A + B).
This is the same as 10xA + B – A – B, which simplifies to 9xA.
Now, A is the first digit, which can be any of the digits 1, 2, 3, 4, 5, 6, 7, 8, 9.
Therefore, 9xA are the first 9 multiples of 9.
Hence the only possible answers for choosing an initial number from 10 to 99 are 9, 18, 27, 36, 45, 54, 63, 72, 81 or 90.
If you look again at the diagram above, you will notice that the symbol next to each of these multiples of 9 is the same; a circle inside another circle.
## Magic 4: Symbols Galore
This one is an interesting variation of Magic 3.
1. Choose two different digits and make a number from 10 to 99. Suppose we choose 5 and 7 to form the number 57.
2. Reverse the two digits to get another number. 75
3. Subtract the smaller number from the larger number. 75 – 57 = 18
4. Find the symbol under your answer.
The shape is a box.
The following provides a proof that the result is always the same.
1. Suppose our two digits are A and B and we form the 2-digit number is AB.
This can be written as 10xA + B.
2. We reverse AB to get BA. This can be written as 10xB + A.
3. Let’s assume 10xA + B is the smaller of the two numbers.
Subtracting the smaller number from the larger number gives
(10xB + A) – (10xA + B)
This is the same as 10xB + A – 10xA – B.
This simplifies to 9B – 9A which is the same as 9x(B – A)
Now, the possible values for the difference, B – A, are 1, 2, 3, 4, 5,6 ,7 ,8, 9.
Therefore, 9x(B – A) are the first 9 multiples of 9.
Again, if you look at the diagram above, you will see that each multiple of 9 has a box shape adjacent to it.
As our final exploration, let’s look at an extension of Magic 3.
## Magic 5: It’s All Smiles and Smooth Sailing
1. Pick any number between 100 and 999 with its first digit greater than its last digit. Suppose we choose 453.
2. Reverse the digits and subtract the smaller answer from the larger answer. The reverse of 453 is 354. Subtracting 354 from 453 gives 99.
3. Find your answer in the grid below.
A smiley face.
## Practice It Yourself
Do you think you can go solo in proving that the answer is always going to be a multiple of 99? Try it before looking at the solution given below.
Suppose our 3-digit number between 100 and 999 is ABC.
This can be written as 100xA + 10xB + C.
The reverse of ABC is CBA, which we can write as 100OC + 10xB + A.
Let’s assume 100xA + 10xB + C is the smaller of the two numbers.
Subtracting the smaller number from the larger number gives
(100xC + 10xB + A) – (100xA + 10xB + C).
This is the same as writing 100xC + 10xB + A – 100xA – 10xB – C, which simplifies to 99xC – 99xA. This can also be written as 99x(C – A).
The possible values for the difference, C – A, are 1, 2, 3, 4, 5,6 ,7 ,8, 9.
Therefore, 99x(C – A) are multiples of 99.
Examining the diagram above confirms that each multiple of 99 has a type of smiley face underneath it.
## Now You Know the Magician's Secrets
So, the next time you see a magician’s amazing number crunching or a mind-reader’s apparent probing of your mind, you will gently smile and say to yourself, “Yep, I know how it is done!” |
# Fun 5th Grade Math Activities to Do at Home
In 5th grade, students learn to measure capacity, or volume, and work with fractions. If these concepts are giving your child difficulty, try using the fun activities below to help boost his or her understanding.
Among other skills, your 5th grader will begin learning to measure volume and recognize volume as a characteristic of a solid figure. When discussing volume with your child, feel free to use the terms volume and capacity interchangeably, as they both refer to the amount a figure can hold.
Fractions with unlike denominators will consistently be found within your 5th grader's mathematical curriculum. Solving problems containing fractions with unlike denominators is often difficult because of the multiple steps required to reach an answer. Consider using the hands-on activities below to help your child understand these difficult mathematical skills.
### Volume Activities
#### Scavenger Hunt
Before beginning the scavenger hunt, set a few guidelines for what you expect your child to locate. For instance, you could ask your child to bring back containers that he thinks will hold an equal amount.
Send your child on a search for containers around your house that fit within the guidelines you provided. After your child has located the containers, ask her to arrange them in order from smallest capacity to largest capacity. Fill the containers with water, recording the amount each container holds. If any containers need to be rearranged, discuss the reasoning with your child.
#### Cooking up Math
Using a measuring cup, have your child measure out a specific amount of rice. Ask your child to make a prediction about how the measurement of the rice may change after it has been cooked. Boil the rice and have your child measure it again. Ask your child questions like: What has changed regarding the measurement? How much rice is there now and why do you think this happened? Feel free to substitute pasta noodles for the rice in this activity.
### Fraction Activities
#### Picture This Fraction
For this activity, draw two rectangles and divide each of them into a different number of sections. Have your child use construction paper strips to model specific fractions. For instance, for 2/3, your child would divide a rectangle into three sections and cover two of them with paper strips.
Use this activity to help your child compare fractions with unlike denominators. Compare 2/3 with 4/5 by repeating the activity above. Your child will find that 4/5 is greater than 2/3.
#### Pizza Fractions
Cook two frozen pizzas to help your child visualize fractions with unlike denominators. After the pizzas are cooked, slice one of them into four pieces and the other one into eight pieces. Have your child create a fraction for each pizza. Point out to your child that two of the smaller pieces will 'fit' on top of one of the larger pieces. Create addition and subtraction problems for your child to solve based on the pizza fractions.
Did you find this useful? If so, please let others know!
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Imagine a math teaching tool so effective that it need only be employed twice per week for less than an hour to result in huge proficiency gains. Impossible, you say? Not so...and MIND Research Institute has the virtual penguin to prove it.
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Should kindergartners put away the building blocks and open the math books? According to recent research, earlier is better when it comes to learning mathematical concepts. But that could put undue pressure on kids, parents and even teachers.
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Inverse Properties of Logarithms
Simplify expressions using two properties of inverse logs
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Practice Inverse Properties of Logarithms
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Inverse Properties of Logarithmic Functions
If you continue to study mathematics into college, you may take a course called Differential Equations. There you will learn that the solution to the differential equation $y' = y$ is the general function $y = Ce^x$ . What is the inverse of this function?
Guidance
By the definition of a logarithm, it is the inverse of an exponent. Therefore, a logarithmic function is the inverse of an exponential function. Recall what it means to be an inverse of a function. When two inverses are composed (see the Inverse Functions concept), they equal $x$ . Therefore, if $f(x)=b^x$ and $g(x)=\log_b x$ , then:
$f \circ g=b^{\log_b x}=x$ and $g \circ f =\log_b b^x=x$
These are called the Inverse Properties of Logarithms.
Example A
Find:
a) $10^{\log 56}$
b) $e^{\ln6} \cdot e^{\ln2}$
Solution: For each of these examples, we will use the Inverse Properties.
a) Using the first property, we see that the bases cancel each other out. $10^{\log 56}=56$
b) Here, $e$ and the natural log cancel out and we are left with $6 \cdot 2=12$ .
Example B
Find $\log_4 16^x$
Solution: We will use the second property here. Also, rewrite 16 as $4^2$ .
$\log_4 16^x=\log_4 (4^2)^x=\log_4 4^{2x}=2x$
Example C
Find the inverse of $f(x)=2e^{x-1}$ .
Solution: See the Finding the Inverse concept for the steps on how to find the inverse.
Change $f(x)$ to $y$ . Then, switch $x$ and $y$ .
$& y=2e^{x-1} \\& x=2e^{y-1}$
Now, we need to isolate the exponent and take the logarithm of both sides. First divide by 2.
$& \frac{x}{2}=e^{y-1} \\& \ln \left(\frac{x}{2}\right)= \ln e^{y-1}$
Recall the Inverse Properties from earlier in this concept. $\log_b b^x=x$ ; applying this to the right side of our equation, we have $\ln e^{y-1}=y-1$ . Solve for $y$ .
$& \ln \left(\frac{x}{2}\right)=y-1 \\& \ln \left(\frac{x}{2}\right)+1=y$
Therefore, $\ln \left(\frac{x}{2}\right)+1$ is the inverse of $2e^{y-1}$ .
Intro Problem Revisit Switch x and y in the function $y = Ce^x$ and then solve for y .
$x = Ce^y\\\frac{x}{C} = e^y\\ln \frac{x}{C} = ln (e^y)\\ln \frac{x}{C} = y$
Therefore, the inverse of $y = Ce^x$ is $y = ln \frac{x}{C}$ .
Guided Practice
1. Simplify $5^{\log_5 6x}$ .
2. Simplify $\log_9 81^{x+2}$ .
3. Find the inverse of $f(x)=4^{x+2}-5$ .
1. Using the first inverse property, the log and the base cancel out, leaving $6x$ as the answer.
$5^{\log_5 6x}=6x$
2. Using the second inverse property and changing 81 into $9^2$ we have:
$\log_9 81^{x+2} &= \log_9 9^{2(x+2)} \\&= 2(x+2) \\&= 2x+4$
3. Follow the steps from Example C to find the inverse.
$f(x) &= 4^{x+2}-5 \\y &= 4^{x+2}-5 \\x &= 4^{y+2}-5 \\x+5 &= 4^{y+2} \\\log_4 (x+5) &= y+2 \\\log_4 (x+5)-2 &= y$
Vocabulary
Inverse Properties of Logarithms
$\log_b b^x=x$ and $b^{\log_b x}=x, b \ne 1$
Practice
Use the Inverse Properties of Logarithms to simplify the following expressions.
1. $\log_3 27^x$
2. $\log_5 \left(\frac{1}{5}\right)^x$
3. $\log_2 \left(\frac{1}{32}\right)^x$
4. $10^{\log(x+3)}$
5. $\log_6 36^{(x-1)}$
6. $9^{\log_9(3x)}$
7. $e^{\ln(x-7)}$
8. $\log \left(\frac{1}{100}\right)^{3x}$
9. $\ln e^{(5x-3)}$
Find the inverse of each of the following exponential functions.
1. $y=3e^{x+2}$
2. $f(x)=\frac{1}{5}e^\frac{x}{7}$
3. $y=2+e^{2x-3}$
4. $f(x)=7^{\frac{3}{x}+1-5}$
5. $y=2(6)^\frac{x-5}{2}$
6. $f(x)=\frac{1}{3}(8)^{\frac{x}{2}-5}$
Vocabulary Language: English
Inverse Properties of Logarithms
Inverse Properties of Logarithms
The inverse properties of logarithms are $\log_b b^x=x$ and $b^{\log_b x}=x, b \ne 1$. |
## Tuesday, June 7, 2011
### Problem 615: Right Triangle with a Square, Diagonals, Center, Area of Quadrilateral
Geometry Problem
Level: Mathematics Education, High School, College.
Click the figure below to see the complete problem 615.
1. OABC is a cyclic quadrilateral
By Ptolemy's Theorem:
AB.OC+BC.OA=OB.AC
(c+a)(b/√2)=6b, c+a=6√2
Required area
=(ABC)+(COA)
=(1/2)ca + (b²/4)
=(1/4)(2ca+c²+a²)by Pythagoras
=(1/4)(c + a)²
=18 sq units
2. http://s30.postimg.org/qiktgwn6p/pro_615.png
Solution of this problem depend on value b , length of side of square ACDE.
Case 1: b< 6, we have no solution
Case 2: 6< b< 6.sqrt(2) we have 2 solutions. Area of ABCO is 18
Case 3: b> 6.sqrt(2) , we have 2 solutions. Point B is inside the square ACDE
And the area of ACBO is undefined.
3. Area • ac/2 + b^2/4 = (a+c)^2 / 4 using Pythagoras
Applying Ptolemy to cyclic quad ABCO
ab/sqrt2 + cb/sqrt2 = 6b
Hence a+c = 6aqrt2 and so area = 36 X 2 /4 = 18
Sumith Peiris
Moratuwa
Sri Lanka
4. Append triangles congruent to ABC to the other sides of the square, thus creating a new square called BFGH. Connect BO up through G. Since 0 is the center of both squares, then BO is 12. The area of the BFGH is 72 = 12^2 / 2. Since quadrilateral AOCB is 1/4 of the area of the square BFGH, then the area of AOCB is 18.
5. https://photos.app.goo.gl/UGeJgNArp2tU6dzJ8
6. Let the length of the square be x
<AOC=90=<ABC |
# Algebra 2. Graph the quadratic equation. Vertex: (-3, 4) Axis of symmetry: x = -3.
## Presentation on theme: "Algebra 2. Graph the quadratic equation. Vertex: (-3, 4) Axis of symmetry: x = -3."— Presentation transcript:
Algebra 2
Graph the quadratic equation. Vertex: (-3, 4) Axis of symmetry: x = -3
Simplify radical expressions Rationalize the denominator Solve quadratic equations
Product Property: Quotient Property:
1.No fractions in square roots 2.No perfect squares in the radical 3.No radicals in the denominator Did you know…? That it is against the law in 32 states (including Illinois) to leave a radical in the denominator of a fraction.
Steps: 1.Use a “factor tree” to factor completely 2.Circle pairs of the same number (they go out) 3.Any other factors not part of a pair stay in
Steps: 1.Use a “factor tree” to factor completely 2.Circle pairs of the same number (they go out) 3.Any other factors not part of a pair stay in 1) 2) 3)
Steps: 1.Split up into two square root problems 2.Simplify the numerator and denominator 3.Remember the “NO-NO” rules
Steps: 1.Split up into two square root problems 2.Simplify the numerator and denominator 3.Remember the “NO-NO” rules
Steps: 1.Split up into two square root problems 2.Simplify the numerator and denominator 3.Remember the “NO-NO” rules 1) 2)
Steps: 1.Isolate the variable 2.Include both answers 1) 2) Note: This method only works when there is no “b” term!
Steps: 1.Isolate the variable 2.Include both answers 3) 4)
pg. 267 #32-34, 38, 41, 49, 54-57, 60 Quiz on Tuesday, November 20 th
Download ppt "Algebra 2. Graph the quadratic equation. Vertex: (-3, 4) Axis of symmetry: x = -3."
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Home » NCERT Solutions » NCERT Solutions class 12 Maths Exercise 10.4 (Ex 10.4) Chapter 10 Vector Algebra
# NCERT Solutions class 12 Maths Exercise 10.4 (Ex 10.4) Chapter 10 Vector Algebra
## NCERT Solutions for Class 12 Maths Exercise 10.4 Chapter 10 Vector Algebra – FREE PDF Download
Free PDF download of NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 (Ex 10.4) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 12 Maths Chapter 10 Vector Algebra Exercise 10.4 Questions with Solutions to help you to revise complete Syllabus and Score More marks.
# NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra (Ex 10.4) Exercise 10.4
1. Find if and
Ans. Given: and
Expanding along first row,
=
### 2. Find a unit vector perpendicular to each of the vectors and where and
Ans. Given: and
On Subtracting = =
Therefore,
Expanding along first row =
=
Therefore, a unit vector perpendicular to both and is
= =
±23iˆ23jˆ23kˆ±23i^∓23j^∓23k^
### 3. If a unit vector makes an angle with with and an acute angle with then find and hence, the components of .
Ans. Let be a unit vector. ……….(i)
Squaring both sides, ……….(ii)
Given: Angle between vectors and iˆi^ is
……….(iii)
Again, given Angel between vectors and jˆj^ is
……….(iv)
Again, given Angel between vectors and kˆk^ is where is acute angle.
……….(v)
Putting the values of and in eq. (ii),
Since is acute angle, therefore is positive and hence
From eq. (v),
Putting values of and in eq. (i),
Components of are coefficients of in
and angle
### 4. Show that (a→−b→)×(a→+b→)=2(a→×b→)(a→−b→)×(a→+b→)=2(a→×b→)
Ans. L.H.S. = =
= 2(a×b)2(a→×b→) = R.H.S.
### 5. Find and if
Ans. Given:
Expanding along first row,
=
Comparing the coefficients of on both sides, we have
……….(i)
……….(ii)
And ……….(iii)
From eq. (ii),
From eq. (iii),
Putting the values of and in eq. (i),
0 = 0
Therefore, and λ=3.λ=3.
### 6. Given that and What can you conclude about the vectors and
Ans. Given:
or or
or or vector is perpendicular to …..(i)
Again, given
or or
or or vector and are collinear or parallel. …..(ii)
Since, vectors & are perpendicular to each other as well as parallel are not possible. ..(iii)
Therefore, from eq. (i), (ii) and (iii), either or
and
### 7. Let the vectors be given as then show that
Ans. Given: Vector and
Now L.H.S. =
+
[By Property of Determinants]
= R.H.S.
### 8. If either and then Is the converse true? Justify your answer with an example.
Ans. Given: Either or
or ……….(i)
[Using eq. (i)]
[By definition of zero vector]
But the converse is not true.
Let
is a non-zero vector.
Let
is a non-zero vector.
But
Taking 2 common from R3 = [ R2 and R3 are identical]
### 9. Find the area of the triangle with vertices A (1, 1, 2), B (2, 3, 5) and C (1, 5, 5).
Ans. Vertices of are A (1, 1, 2), B (2, 3, 5) and C (1, 5, 5).
Position vector of point A = (1, 1, 2) =
Position vector of point B = (2, 3, 5) =
Position vector of point C = (1, 5, 5) =
Now = Position vector of point B – Position vector of point A
=
And = Position vector of point C – Position vector of point A
=
x =
= 6i3j+4k−6i→−3j→+4k→
Now Area of triangle ABC =
sq. units
### 10. Find the area of the parallelogram whose adjacent sides are determined by the vectors and
Ans. Given: Vectors representing two adjacent sides of a parallelogram are
and
= =
Now Area of parallelogram =
sq. units
### 11. Let the vectors and such that then is a unit vector, if the angle between and is:
(A)
(B)
(C)
(D)
Ans. Given: and is a unit vector.
, where is the angle between and
Therefore, option (B) is correct.
### 12. Area of a rectangle having vertices A, B, C and D with position vectors and respectively is:
(A)
(B) 1
(C) 2
(D) 4
Ans. Given: ABCD is a rectangle.
Now = Position vector of point B – Position vector of point A
=
AB =
And = Position vector of point D – Position vector of point A
= |
# Dividing Decimals
## Related calculator: Long Division Calculator
It is pretty easy to divide decimals after learning dividing decimals by whole numbers and dividing whole numbers by decimals.
To use techniques from these notes, simultaneously move decimal points to the right, until you get at least one whole number.
Example 1. Calculate ${11.5}\div{4.6}$.
Simultaneously move decimal point one position to the right: ${115}\div{46}$.
We obtained two whole numbers, so can perform long division as always: ${115}\div{46}={2.5}$.
Thus, ${\color{purple}{{{11.5}\div{4.6}={2.5}}}}$.
Now, let's see an example which leads to dividing decimal by a whole number.
Example 2. Calculate ${7.68}\div{2.4}$.
Simultaneously move decimal point one position to the right: ${76.8}\div{24}$.
We obtained one whole number and now can divide decimal by a whole number: ${76.8}\div{24}={3.2}$.
So, ${\color{purple}{{{7.68}\div{2.4}={3.2}}}}$.
Final example leads to dividing whole number by a decimal.
Example 3. Calculate ${16.54}\div{3.308}$.
Simultaneously move decimal point two positions to the right: ${1654}\div{330.8}$.
We obtained one whole number, so can divide whole number by a decimal: ${1654}\div{330.8}={5}$.
Thus, ${\color{purple}{{{16.54}\div{3.308}={5}}}}$.
Now, it is time to practice:
Exercise 1. Find ${3.8}\div{1.9}$.
Exercise 2. Calculate ${32.4}\div{0.16}$.
Exercise 3. Find $-{0.025}\div{2.5}$.
Exercise 4. Find $-{5.95}\div{\left(-{2.8}\right)}$.
Exercise 5. Find ${3.4}\div{0.008}$. |
# Maths
## Count by 3 | Skip Count by 3 | Skip Counting | Skip Counting Song | Jack Hartmann
Count by 3 engages kids with exercise and movement as they learn to skip county by 3's. I give kids simple, easy - to - follow exercises to do as they get ready to skip county by 3's. Kids follow my exercises and skip county by 3's out loud up to 30.
## Count by 2 | Dancing 2's | Skip Counting by 2 | Count to 100 | Educational Songs | Jack Hartmann
Dancing 2's is a fun way for kids to learn to skip count by 2. Children follow me doing different dances as we count to 100 by 2. Every time we hit a place holder of tens we change dances. Children sing-a-long with me as we dance to this fun skip counting song.
During this term will be focusing on:
Multiplication and Division:
• Year 1: solve one-step problems involving multiplication and division, by calculating the answer using concrete objects, pictorial representations and arrays with the support of the teacher.
• Year 2: recall and use multiplication and division facts for the 2, 5 and 10 multiplication tables, including recognising odd and even numbers, calculate mathematical statements for multiplication and division within the multiplication tables and write them using the multiplication (×), division (÷) and equals (=) signs show that multiplication of two numbers can be done in any order (commutative) and division of one number by another cannot solve problems involving multiplication and division, using materials, arrays, repeated addition, mental methods, and multiplication and division facts, including problems in contexts.
Fractions (including decimals and percentages):
• Year 1: recognise, find and name a half as one of two equal parts of an object, shape or quantity, recognise, find and name a quarter as one of four equal parts of an object, shape or quantity.
• Year 2: recognise, find, name and write fractions ⅓, ¼, 2⁄4 and ¾ of a length, shape, set of objects or quantity write simple fractions e.g. ½ of 6 = 3 and recognise the equivalence of two quarters and one half.
Top |
Learn how to count money | Grade 1 | Learning Concepts
Money
# Counting Money for Class 1 Math
This concept will help them the students in counting tmoney from coins and notes. Also, they will learn about the denominations.
Also, the students will learn to
• Identify the Indian coins
• Make total of the denominations
• Also find difference between two denominations
Each concept is explained to class 1 maths students with examples and illustrations, and a concept map is given at the end to summarise the idea. At the end of the page, two printable Counting Money worksheets with solutions are attached for students to practice. Download the worksheets and assess your knowledge.
How to Count Money?
We have learned about Indian currency – ‘Indian rupee’.
## Coins:
All coins are made up of metals. In India, ₹1, ₹2, ₹5, ₹10 and ₹20 coins are in circulation.
• From the above, we can clearly see the denomination of coins is written on it. Therefore, we can easily identify the coins.
## Notes:
In India, ₹5, ₹10, ₹20, ₹50, ₹100, ₹200, ₹500, ₹2000 notes are in circulation.
• On each and every note, the denomination is printed on it. So, we can easily identify the value of the note.
## Counting Money
Now, let’s learn how to count the money.
Here, there is only one coin of ₹1. So, the amount is ₹1.
Here, there are two coins of ₹1. Add the number 1 two times, we get:
1 + 1 = 2. So, the amount is ₹2.
Here, there is only one coin of ₹2. So, the amount is ₹2.
Here, there are two coins of ₹2. Add the number 2 two times, we get:
2 + 2 = 4. So, the amount is ₹4.
Adding ₹1 and ₹2, we get:
1 + 2 = 3. So, the total amount is ₹3.
Here, add the numbers 5 and 10, we get:
5 + 10 = 15
So, the total amount is ₹15.
## How to count the money for buying something?
• We use the money for buying something that we want.
• So, we have to count the money equal to the cost of that thing. But we have a fixed value of currency coins and notes, so sometimes we can’t count the money exactly equal to the cost of that thing.
• To count the money, basically we use addition and subtraction.
### Example 1:
A boy wants to buy a pencil that cost is ₹5.
How can he give the amount of ₹5?
### Solution:
He can count ₹5 in different ways.
### Example 2:
Mini has Rs. 10 note, she wants to buy an ice-cream. The cost of ice-cream is Rs. 15. Can she buy the ice-cream with the money she has?
### Solution:
Mini has = ₹10
Cost of the ice-cream = ₹15
10 is less than 15. So, she has less amount of money.
### Example 3:
Jenny wants to buy a balloon. She has a ₹20 note. When she gives that note to the shopkeeper, he returns ₹10. What is the cost of that balloon?
### Solution:
Jenny has spent some money buying a balloon which means she gave some amount of money to the shopkeeper. So, here we have to subtract that money.
We will do subtraction here.
She has now ₹10. She spent ₹10. So, the cost of a balloon is ₹10.
• When we spent some amount of money or take away some amount of money, to find the left amount we need to do subtraction.
• We use here, give and take money method.
• - |
# What is a Product in Math?
A product in math is the result we obtain when we multiply two or more numbers together.
Author
Michelle Griczika
Published
September 13, 2023
# What is a product in math?
A product in math is the result we obtain when we multiply two or more numbers together.
Author
Michelle Griczika
Published
September 13, 2023
# What is a Product in Math?
A product in math is the result we obtain when we multiply two or more numbers together.
Author
Michelle Griczika
Published
September 13, 2023
Key takeaways
• The product meaning in math refers to the result achieved when two or more numbers, known as factors, are multiplied together. It is a fundamental concept in many areas of mathematics.
• Calculating a product is straightforward, involving multiplying the numbers together, whether they are whole numbers, fractions, decimals, or even include variables.
• The understanding of the concept of the product extends beyond math assignments. It is practical in daily life, from calculating the prices of multiple items to figuring out the area of a room.
Understanding the language of mathematics can sometimes feel like learning a new language altogether for both students and parents alike. The term product often comes up in math lessons and homework. But what is a product in math? We’ll explain what a product is, show how to find it, and provide practice problems to help reinforce the concept.
## Definition of product in math
So, what is “product” in math? In mathematics, product means the result you get after multiplying numbers together. The numbers that are being multiplied are known as factors. So, the final result is called the product when we multiply factors, whether they’re whole numbers, fractions, or even decimals. This concept is at the heart of many areas in mathematics, which is why it’s so important to understand the products definition.
## How to find the product in math
Figuring out the product in math is simple: you multiply the numbers. The process can vary from elementary multiplication problems, such as 2 x 3, to more complex ones involving bigger numbers or decimals. Regardless of the complexity, the answer derived from the multiplication operation is called a product – and yes, this holds even when variables are in the equation.
But what happens when there are more than two numbers or factors involved? Do not fret! The method remains the same when wondering, “what’s product in math?”.
Consider a simple example: To find the product of 2, 3, and 4, multiply them in any order. You can multiply 2 and 3 first to get 6, then multiply 6 by 4 to get 24. Alternatively, you can multiply 4 and 2 first to get 8, then multiply 8 by 3 to get 24. This flexibility is due to the commutative property of multiplication, which states that the order of numbers does not change the product.
It’s also important to remember that the math product of any number and zero is always zero. This is known as the zero property of multiplication.
When dealing with fractions or decimals, the process remains essentially the same. You multiply the fractions or the decimals directly. However, calculating these might require additional steps or a good grasp of fraction and decimal operations.
Lastly, multiplication and its product have a wide range of applications in real-world scenarios, from calculating the prices of multiple items to figuring out the area of a room. Therefore, understanding the concept of a product is not just crucial for math assignments but also practical in daily life.
Why is understanding the concept of 'product' important?
Understanding the concept of ‘product’ is a foundational skill in mathematics. Being familiar with it can make more advanced math topics easier to understand.
## Practice problems
If your child is having difficulty, using our math practice app to help reinforce these concepts might be beneficial. Consistent practice makes perfect!
## FAQs about products in math
A little blurb of text about why we crafted these really helpful FAQs for you!
In mathematics, ‘product’ refers to the result of multiplying numbers together. For example, if you multiply 3 and 4, the product is 12.
So, what is the product in math? To find the product in math, you perform multiplication. This could involve two or more numbers. For example, the product of 5 and 6 is 30 because 5 multiplied by 6 equals 30.
Yes, the term ‘product’ is specifically used to denote the result of a multiplication operation. It is not used in addition, subtraction, or division.
In a multiplication operation, the numbers being multiplied are called ‘factors,’ while the result of the multiplication is called the ‘product.’ For example, in 6 * 4 = 24, 6 and 4 are factors, and 24 is the product.
Lesson credits
Michelle Griczika
Michelle Griczika is a seasoned educator and experienced freelance writer. Her years teaching first and fifth grades coupled with her double certification in elementary and early childhood education lend depth to her understanding of diverse learning stages. Michelle enjoys running in her free time and undertaking home projects.
Michelle Griczika
Michelle Griczika is a seasoned educator and experienced freelance writer. Her years teaching first and fifth grades coupled with her double certification in elementary and early childhood education lend depth to her understanding of diverse learning stages. Michelle enjoys running in her free time and undertaking home projects.
# Are you a parent, teacher or student?
Are you a parent or teacher?
## Hi there!
Book a chat with our team |
# Mathematical Induction Proof for the Sum of Cubes | Summary and Q&A
7.6K views
April 27, 2022
by
The Math Sorcerer
Mathematical Induction Proof for the Sum of Cubes
## TL;DR
The video explains how to prove the formula for the sum of cubes using the principle of mathematical induction.
## Key Insights
• 👍 Mathematical induction is a powerful proof technique used to prove statements about integers.
• 🔺 The base case is the starting point of an induction proof, where the statement is shown to be true for the smallest value.
• 👍 The induction hypothesis assumes the truth of the statement for a specific value and is used to prove the truth for the next value.
• 🥹 The induction step involves showing that if the statement holds for a value, it also holds for the next value.
• ⚾ By combining the base case, induction hypothesis, and induction step, the principle of mathematical induction guarantees the truth of the statement for all positive integers.
• ❓ Summation notation can be used to represent the problem more succinctly and with greater clarity.
## Transcript
hello in this problem we're going to prove that 1 cubed plus 2 cubed plus dot dot dot plus n cubed is equal to n squared times m plus 1 squared divided by 4 for all positive integers n and we're going to do it via the principle of mathematical induction so let's go ahead and start proof so before we jump into the proof i wanted to show you that you... Read More
### Q: What is the purpose of using summation notation in the proof?
Summation notation provides clarity and a more concise representation of the problem by clearly indicating the range of values being summed.
### Q: Why is it important to understand what you're trying to prove in an induction proof?
Understanding the statement you're trying to prove allows you to structure your proof correctly and ensures that you are addressing the specific claim.
### Q: How is the induction hypothesis used in the proof?
The induction hypothesis assumes that the formula is true for a specific value (k) and uses that assumption to prove that it is also true for the next value (k + 1).
### Q: Why is it necessary to specify the use of the induction hypothesis in the proof?
Specifying the use of the induction hypothesis adds clarity to the proof and helps the reader understand where the assumption is being applied.
## Summary & Key Takeaways
• The video introduces the problem of proving the formula for the sum of cubes using mathematical induction.
• The base case, where the formula is shown to be true for the smallest positive integer, is demonstrated.
• The induction hypothesis is stated and used to show that the formula holds true for any positive integer.
• The video concludes by stating that the principle of mathematical induction guarantees that the formula is true for all positive integers. |
Integration by Parts
Let $f(x)$ and $g(x)$ be differentiable functions. Then the product rule
$$(f(x)g(x))’=f'(x)g(x)+f(x)g'(x)$$
\label{eq:intpart}
\int f(x)g'(x)dx=f(x)g(x)-\int f'(x)g(x)dx.
The formula \eqref{eq:intpart} is called integration by parts. If we set $u=f(x)$ and $v=g(x)$, then \eqref{eq:intpart} can be also written as
\label{eq:intpart2}
\int udv=uv-\int vdu.
Example. Evaluate $\int x\cos xdx$.
Solution. Let $u=x$ and $dv=\cos xdx$. Then $du=dx$ and $v=\sin x$. So,
\begin{align*}
\int x\cos xdx&=x\sin x-\int\sin xdx\\
&=x\sin x+\cos x+C,
\end{align*}
where $C$ is a constant.
Example. Evaluate $\int\ln xdx$.
Solution. Let $u=\ln x$ and $dv=dx$. Then $du=\frac{1}{x}dx$ and $v=x$. So,
\begin{align*}
\int\ln xdx&=x\ln x-\int x\cdot\frac{1}{x}dx\\
&=x\ln x-x+C,
\end{align*}
where $C$ is a constant.
Often it is required to apply integration by parts more than once to evaluate a given integral. In that case, it is convenient to use tabular integral (which is equivalent to integration by parts) as shown in the following example.
Example. Evaluate $\int x^2e^xdx$
Solution. In the following table, the first column represents $x^2$ and its derivatives, and the second column represents $e^x$ and its integrals.
$$\begin{array}{ccc} x^2 & & e^x\\ &\stackrel{+}{\searrow}&\\ 2x & & e^x\\ &\stackrel{-}{\searrow}&\\ 2 & & e^x\\ &\stackrel{+}{\searrow}&\\ 0 & & e^x. \end{array}$$
This table shows the repeated application of integration by parts. Following the table, the final answer is given by
$$\int x^2e^xdx=x^2e^x-2xe^x+2e^x+C,$$
where $C$ is a constant.
In order to understand why this works and why this is equivalent to integration by parts, assume that we are to evaluate the integral $\int fgdx$. Let $u=f$ and $dv=gdx$. Then by \eqref{eq:intpart2} $$\int fgdx=f\int gdx-\int f’\left(\int gdx\right)dx$$ If $\int f’\left(\int gdx\right)dx$ requires integration by parts, we would have $$\int fgdx=f\int gdx-f’\int\left(\int gdx\right)dx+\int f^{\prime\prime}\left(\int\left(\int gdx\right)dx\right)dx$$ If $\int f^{\prime\prime}\left(\int\left(\int gdx\right)dx\right)dx$ still requires integration by parts, we would have \begin{align*}\int fgdx&=f\int gdx-f’\int\left(\int gdx\right)dx+f^{\prime\prime}\int\left(\int\left(\int gdx\right)dx\right)dx\\&-\int f^{\prime\prime\prime}\left(\int\left(\int\left(\int gdx\right)dx\right)\right)dx\end{align*} This illustrates tabular integral and the process continues until the last integral no longer requires integration by parts. Sometimes the last integral becomes the same type of integral as the one in the LHS up to a constant. When that happens the last integral can be combined with the one in the LHS.
Example. Evaluate $\int x^3\sin xdx$.
Solution. In the following table, the first column represents $x^3$ and its derivatives, and the second column represents $\sin x$ and its integrals.
$$\begin{array}{ccc} x^3 & & \sin x\\ &\stackrel{+}{\searrow}&\\ 3x^2 & & -\cos x\\ &\stackrel{-}{\searrow}&\\ 6x & & -\sin x\\ &\stackrel{+}{\searrow}&\\ 6 & & \cos x\\ &\stackrel{-}{\searrow}&\\ 0 & & \sin x. \end{array}$$
Following the table, the final answer is given by
$$\int x^3\sin xdx=-x^3\cos x+3x^2\sin x+6x\cos x-6\sin x+C,$$
where $C$ is a constant.
Example. Evaluate $\int e^x\cos xdx$.
Solution. In the following table, the first column represents $e^x$ and its derivatives, and the second column represents $\cos x$ and its integrals.
$$\begin{array}{ccc} e^x & & \cos x\\ &\stackrel{+}{\searrow}&\\ e^x & & \sin x\\ &\stackrel{-}{\searrow}&\\ e^x & & -\cos x. \end{array}$$
Now, this is different from the previous two examples. While the first column repeats the same function $e^x$, the functions second column changes from $\cos x$ to $\sin x$ and to $\cos x$ again up to sign. In this case, we stop there and write the answer as we have done in the previous two examples and add to it $\int e^x(-\cos x)dx$. Notice that the integrand is the product of functions in the last row and this is the last integral we obtain from a multiple applications of integration by parts, which becomes the same type of integral as the one you began with. Hence,
$$\int e^x\cos xdx=e^x\sin x+e^x\cos x-\int e^x\cos xdx.$$
For now we do not worry about the constant of integration. Solving this for $\int e^x\cos xdx$, we obtain the final answer
$$\int e^x\cos xdx=\frac{1}{2}e^x\sin x+\frac{1}{2}e^x\cos x+C,$$
where $C$ is a constant.
Example. Evaluate $\int e^x\sin xdx$.
Solution. In the following table, the first column represents $e^x$ and its derivatives, and the second column represents $\sin x$ and its integrals.
$$\begin{array}{ccc} e^x & & \sin x\\ &\stackrel{+}{\searrow}&\\ e^x & & -\cos x\\ &\stackrel{-}{\searrow}&\\ e^x & & -\sin x. \end{array}$$
This is similar to the above example. The first columns repeats the same function $e^x$, and the functions in the second column changes from $\sin x$ to $\cos x$ and to $\sin x$ again up to sign. So we stop there and write
$$\int e^x\sin xdx=-e^x\cos x+e^x\sin x-\int e^x\sin xdx.$$
Solving this for $\int e^x\sin xdx$, we obtain
$$\int e^x\sin xdx=-\frac{1}{2}e^x\cos x+\frac{1}{2}e^x\sin x+C,$$
where $C$ is a constant.
Example. Evaluate $\int e^{5x}\cos 8xdx$.
Solution. In the following table, the first column represents $e^{5x}$ and its derivatives, and the second column represents $\cos 8x$ and its integrals.
$$\begin{array}{ccc} e^{5x} & & \cos 8x\\ &\stackrel{+}{\searrow}&\\ 5e^{5x} & & \frac{1}{8}\sin 8x\\ &\stackrel{-}{\searrow}&\\ 25e^{5x} & & -\frac{1}{64}\cos 8x. \end{array}$$
The first columns repeats the same function $e^{5x}$ up to constant multiple, and the functions in the second column changes from $\cos 8x$ to $\sin 8x$ and to $\cos 8x$ again to constant multiple. This case also we do the same.
$$\int e^{5x}\cos 8xdx=\frac{1}{8}e^{5x}\sin 8x+\frac{5}{64}e^{5x}\cos 8x-\frac{25}{64}\int e^{5x}\cos 8xdx.$$
Solving this for $\int e^{5x}\cos 8xdx$, we obtain
$$\int e^{5x}\cos 8xdx=\frac{8}{89}e^{5x}\sin 8x+\frac{5}{89}e^{5x}\cos 8x+C,$$
where $C$ is a constant.
The evaluation of a definite integral by parts can be done as
\label{eq:intpart3}
\int_a^b f(x)g'(x)dx=[f(x)g(x)]_a^b-\int_a^b f'(x)g(x)dx.
Example. Find the area of the region bounded by $y=xe^{-x}$ and the x-axis from $x=0$ to $x=4$.
The graph of y=xexp(-x), x=0..4
Solution. Let $u=x$ and $dv=e^{-x}dx$. Then $du=dx$ and $v=-e^{-x}$. Hence,
\begin{align*}
A&=\int_0^4 xe^{-x}dx\\
&=[-xe^{-x}]0^4+\int_0^4 e^{-x}dx\\
&=-4e^{-4}+[-e^{-x}]_0^4\\
&=1-5e^{-4}.
\end{align*}
One thought on “Integration by Parts”
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# Your question: Why is it important to understand transformational geometry?
Contents
## Why is transformation in geometry important?
Geometric transformations are needed to give an entity the needed position, orientation, or shape starting from existing position, orientation, or shape. The basic transformations are scaling, rotation, translation, and shear. Other important types of transformations are projections and mappings.
## Why is it important to know about transformations?
Now, the way transformations are taught gives students the ability to manipulate figures in the plane freely, which sets the foundation for other areas of study, such as the verification of perpendicular segments, the derivation of the equation of a circle, and perhaps most notably, congruence and similarity.
## What do you understand by geometric transformation?
A geometric object is represented by its vertices (as position vectors) A geometric transformation is an operation that modifies its shape, size, position, orientation etc with respect to its current configuration operating on the vertices (position vectors).
## What do you know about transformations?
A transformation is a process that manipulates a polygon or other two-dimensional object on a plane or coordinate system. Mathematical transformations describe how two-dimensional figures move around a plane or coordinate system. A preimage or inverse image is the two-dimensional shape before any transformation.
## What is the result of a transformation?
A transformation can be a translation, reflection, or rotation. A transformation is a change in the position, size, or shape of a geometric figure. The given figure is called the preimage (original) and the resulting figure is called the new image.
THIS IS IMPORTANT: What is a generator step up unit?
## Where are geometric transformations used?
Abstract Geometric transformations are widely used for image registration and the removal of geometric distortion. Common applications include construction of mosaics, geographical mapping, stereo and video. A spatial transformation of an image is a geometric transformation of the image coordinate system.
## How do you explain transformations in math?
A transformation is a way of changing the size or position of a shape. Every point in the image is the same distance from the mirror line as the original shape. The line joining a point on the original shape to the same point on the image is perpendicular to the mirror line.
## Why are translations in math important?
When you translate something in geometry, you’re simply moving it around. You don’t distort it in any way. … Similarly, if you translate an angle, the measure of the angle doesn’t change. These properties may seem obvious, but they’re important to keep in mind later on when we do proofs.
## What are some real life examples of translations?
Real life examples of translations are:
• the movement of an aircraft as it moves across the sky.
• the lever action of a tap (faucet)
• sewing with a sewing machine.
• punching decorative studs into belts.
• throwing a shot-put.
• making pasta such as spaghetti. |
# Algebra of Limit of Sequence
Consider two sequences ${x}_{{n}}$ and ${y}_{{n}}$. When we talk about sum of these sequences, we talk about sequence ${x}_{{n}}+{y}_{{n}}$, whose elements are ${x}_{{1}}+{y}_{{1}},{x}_{{2}}+{y}_{{2}},{x}_{{3}}+{y}_{{3}}\ldots.$. Same can be said about other arithmetic operations. In other words sum of sequences is sequence with elements that are sum of corresponding elements of initial two sequences.
For example, consider sequence ${x}_{{n}}={\left\{{1},{3},{5},{7},{9},\ldots\right\}}$ and sequence ${y}_{{n}}={\left\{{2},{4},{6},{8},{10},\ldots\right\}}$ then ${x}_{{n}}+{y}_{{n}}={\left\{{1}+{2},{3}+{4},{5}+{6},{7}+{8},{9}+{10},\ldots\right\}}={\left\{{3},{7},{11},{15},{19},\ldots\right\}}$.
Following facts are important because with their help in many cases we can easily find limit without using definition.
Limit of sum (difference) equals sum (difference) of limits. If sequences ${x}_{{n}}$ and ${y}_{{n}}$ have finite limits: $\lim{x}_{{n}}={a}$ and $\lim{y}_{{n}}={b}$, then their sum and difference also have finite limits, and $\lim{\left({x}_{{n}}+{y}_{{n}}\right)}={a}+{b}$, $\lim{\left({x}_{{n}}-{y}_{{n}}\right)}={a}-{b}$.
This fact holds for any finite number of summands.
Example 1. Let ${x}_{{n}}=\frac{{1}}{{n}}$, ${y}_{{n}}={1}+\frac{{1}}{{{n}}^{{2}}}$, ${z}_{{n}}=\frac{{1}}{{{n}}^{{3}}}$.
Then
$\lim{\left({x}_{{n}}+{y}_{{n}}-{z}_{{n}}\right)}=\lim{x}_{{n}}+\lim{y}_{{n}}-\lim{z}_{{n}}=\lim\frac{{1}}{{n}}+\lim{\left({1}+\frac{{1}}{{{n}}^{{2}}}\right)}-\lim\frac{{1}}{{{n}}^{{3}}}=$
$={0}+{1}-{0}={1}$.
Limit of product equals product of limits. If sequences ${x}_{{n}}$ and ${y}_{{n}}$ have finite limits: $\lim{x}_{{n}}={a}$ and $\lim{y}_{{n}}={b}$, then their product also has finite limit and $\lim{\left({x}_{{n}}{y}_{{n}}\right)}={a}{b}$.
This fact holds for any finite number of factors.
Example 2. Let ${x}_{{n}}=\frac{{1}}{{n}}$, ${y}_{{n}}=\frac{{1}}{{n}}$.
Then $\lim\frac{{1}}{{{n}}^{{2}}}=\lim{\left({x}_{{n}}{y}_{{n}}\right)}=\lim{\left(\frac{{1}}{{n}}\cdot\frac{{1}}{{n}}\right)}=\lim\frac{{1}}{{n}}\cdot\lim\frac{{1}}{{n}}={0}\cdot{0}={0}$.
So, $\lim\frac{{1}}{{{n}}^{{2}}}={0}$.
Limit of quotient equals quotient of limits. If sequences ${x}_{{n}}$ and ${y}_{{n}}$ have finite limits: $\lim{x}_{{n}}={a}$ and $\lim{y}_{{n}}={b}$ $\left({b}\ne{0}\right)$ then their quotient also has finite limit and $\lim\frac{{{x}_{{n}}}}{{{y}_{{n}}}}=\frac{{a}}{{b}}$.
Example 3. Let ${x}_{{n}}=\frac{{1}}{{n}}$, ${y}_{{n}}={1}+\frac{{1}}{{n}}$.
Then $\lim\frac{{{x}_{{n}}}}{{{y}_{{n}}}}=\lim\frac{{\frac{{1}}{{n}}}}{{{1}+\frac{{1}}{{n}}}}=\frac{{\lim\frac{{1}}{{n}}}}{{\lim{\left({1}+\frac{{1}}{{n}}\right)}}}=\frac{{0}}{{1}}={0}$. |
# Limits, flipping approaching direction
• plutonium
In summary, a limit is a mathematical concept that represents the value a function approaches as its input gets closer to a certain value or goes to infinity. To calculate a limit, the given value is plugged into the function and the resulting value is observed. Limits can approach from the left or right, which can affect the final value. In calculus, limits are important for analyzing function behavior and determining continuity and differentiability. They are also used in the definition of derivatives and integrals. In real-life, limits have applications in physics, economics, engineering, and computer science.
They let $$t = -x$$
Thank you for sharing your discovery of the proof for lim x-> 0 sinx/x = 1. This is a fundamental limit in calculus and it's great that you were able to find the proof for it.
Regarding the step you mentioned, it is common in mathematics to approach limits from both sides. In this case, we are looking for the limit as x approaches 0, but we can approach 0 from the positive side (x>0) and the negative side (x<0). In the proof, they first approach 0 from the positive side and then flip the direction to approach from the negative side. This is a valid approach because the limit should be the same regardless of the direction we approach from.
I hope this helps clarify the step for you. Keep up the great work in exploring and understanding mathematical concepts.
## What is a limit?
A limit is a mathematical concept that represents the value that a function approaches as its input approaches a certain value or goes to infinity. It is denoted by the symbol "lim" and is an important tool in calculus.
## How do you calculate a limit?
To calculate a limit, you must plug in the value that the function is approaching into the function and see what value it approaches. This can be done algebraically or graphically. If the function does not approach a specific value and instead goes to infinity, the limit is said to be undefined.
## What does it mean for a limit to approach from the left or right?
When a limit approaches from the left, it means that the input values are approaching the given value from the left side of the number line. Similarly, approaching from the right means that the input values are approaching from the right side. This is important because the limit value can be different depending on which direction it is approached from.
## Why are limits important in calculus?
Limits are important in calculus because they allow us to analyze the behavior of functions as they approach certain values. They also help us determine the continuity, differentiability, and other properties of functions. Limits are also used in the definition of derivatives and integrals, which are fundamental concepts in calculus.
## Are there any real-life applications of limits?
Yes, there are many real-life applications of limits. For example, in physics, limits are used to calculate instantaneous velocity and acceleration. In economics, limits are used to analyze the behavior of demand and supply functions. Limits are also used in engineering, computer science, and other fields to model and solve real-world problems.
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How do you solve 1/2(x+8)^2=14?
Jan 28, 2017
$x = 2 \sqrt{7} - 8$
Explanation:
$\frac{1}{\text{2}} {\left(x + 8\right)}^{2} = 14$
Multiplying by 2 on both sides
cancel(2) × 1/cancel("2")(x + 8)^2 = 2 × 14
${\left(x + 8\right)}^{2} = 28$
Square root both side
sqrt((x + 8)^2 = $\sqrt{28}$
$x + 8 = \pm 2 \sqrt{7}$
$x = 2 \sqrt{7} - 8 \mathmr{and} x = - 2 \sqrt{7} - 8$
$x = - 8 \pm 2 \sqrt{7}$
Explanation:
$\frac{1}{2} {\left(x + 8\right)}^{2} = 14$
I want to expand the squared term, so to do that I'll first multiply both sides by 2:
$\frac{1}{2} \textcolor{red}{\times 2} {\left(x + 8\right)}^{2} = 14 \textcolor{red}{\times 2}$
${\left(x + 8\right)}^{2} = 28$
${x}^{2} + 16 x + 64 = 28$
${x}^{2} + 16 x + 64 \textcolor{red}{- 28} = 28 \textcolor{red}{- 28}$
${x}^{2} + 16 x + 36 = 0$
I don't see any easy factors, so let's use the quadratic formula. The general formula is:
$x = \frac{- \textcolor{g r e e n}{b} \pm \sqrt{{\textcolor{g r e e n}{b}}^{2} - 4 \textcolor{b l u e}{a} \textcolor{b r o w n}{c}}}{2 \textcolor{b l u e}{a}}$
and relates to a trinomial this way:
$\textcolor{b l u e}{a} {x}^{2} + \textcolor{g r e e n}{b} x + \textcolor{b r o w n}{c}$
$x = \frac{- 16 \pm \sqrt{{16}^{2} - 4 \left(1\right) \left(36\right)}}{2 \left(1\right)}$
$x = \frac{- 16 \pm \sqrt{256 - 144}}{2}$
$x = \frac{- 16 \pm \sqrt{112}}{2} = \frac{- 16 \pm \sqrt{16 \times 7}}{2} = \frac{- 16 \pm 4 \sqrt{7}}{2} = - 8 \pm 2 \sqrt{7}$
And we can see these two solutions in the graphing of both sides of the original equation:
graph{(y-(1/2)(x+8)^2)(y-0x-14)=0 [-16.69, 3.31, 9.14, 19.136]} |
# 2000 AIME I Problems/Problem 6
## Problem
For how many ordered pairs $(x,y)$ of integers is it true that $0 < x < y < 10^6$ and that the arithmetic mean of $x$ and $y$ is exactly $2$ more than the geometric mean of $x$ and $y$?
## Solutions
### Solution 1
$\begin{eqnarray*} \frac{x+y}{2} &=& \sqrt{xy} + 2\\ x+y-4 &=& 2\sqrt{xy}\\ y - 2\sqrt{xy} + x &=& 4\\ \sqrt{y} - \sqrt{x} &=& \pm 2\end{eqnarray*}$
Because $y > x$, we only consider $+2$.
For simplicity, we can count how many valid pairs of $(\sqrt{x},\sqrt{y})$ that satisfy our equation.
The maximum that $\sqrt{y}$ can be is $\sqrt{10^6} - 1 = 999$ because $\sqrt{y}$ must be an integer (this is because $\sqrt{y} - \sqrt{x} = 2$, an integer). Then $\sqrt{x} = 997$, and we continue this downward until $\sqrt{y} = 3$, in which case $\sqrt{x} = 1$. The number of pairs of $(\sqrt{x},\sqrt{y})$, and so $(x,y)$ is then $\boxed{997}$.
### Solution 2
Let $a^2$ = $x$ and $b^2$ = $y$, where $a$ and $b$ are positive.
Then $$\frac{a^2 + b^2}{2} = \sqrt{{a^2}{b^2}} +2$$ $$a^2 + b^2 = 2ab + 4$$ $$(a-b)^2 = 4$$ $$(a-b) = \pm 2$$
This makes counting a lot easier since now we just have to find all pairs $(a,b)$ that differ by 2.
Because $\sqrt{10^6} = 10^3$, then we can use all positive integers less than 1000 for $a$ and $b$.
We know that because $x < y$, we get $a < b$.
We can count even and odd pairs separately to make things easier*:
Odd: $$(1,3) , (3,5) , (5,7) . . . (997,999)$$
Even: $$(2,4) , (4,6) , (6,8) . . . (996,998)$$
This makes 499 odd pairs and 498 even pairs, for a total of $\boxed{997}$ pairs.
$*$Note: We are counting the pairs for the values of $a$ and $b$, which, when squared, translate to the pairs of $(x,y)$ we are trying to find.
### Solution 3
Since the arithmetic mean is 2 more than the geometric mean, $\frac{x+y}{2} = 2 + \sqrt{xy}$. We can multiply by 2 to get $x + y = 4 + 2\sqrt{xy}$. Subtracting 4 and squaring gives $$((x+y)-4)^2 = 4xy$$ $$((x^2 + 2xy + y^2) + 16 - 2(4)(x+y)) = 4xy$$ $$x^2 - 2xy + y^2 + 16 - 8x - 8y = 0$$
Notice that $((x-y)-4)^2 = x^2 - 2xy + y^2 + 16 - 8x +8y$, so the problem asks for solutions of $$(x-y-4)^2 = 16y$$ Since the left hand side is a perfect square, and 16 is a perfect square, $y$ must also be a perfect square. Since $0 < y < (1000)^2$, $y$ must be from $1^2$ to $999^2$, giving at most 999 options for $y$.
However if $y = 1^2$, you get $(x-5)^2 = 16$, which has solutions $x = 9$ and $x = 1$. Both of those solutions are not less than $y$, so $y$ cannot be equal to 1. If $y = 2^2 = 4$, you get $(x - 8)^2 = 64$, which has 2 solutions, $x = 16$, and $x = 0$. 16 is not less than 4, and $x$ cannot be 0, so $y$ cannot be 4. However, for all other $y$, you get exactly 1 solution for $x$, and that gives a total of $999 - 2 = \boxed{997}$ pairs.
- asbodke
### Solution 4 (Similar to Solution 3)
Rearranging our conditions to
$$x^2-2xy+y^2+16-8x-8y=0 \implies$$ $$(y-x)^2=8(x+y-2).$$
Thus, $4|y-x.$
Now, let $y = 4k+x.$ Plugging this back into our expression, we get
$$(k-1)^2=x-1.$$
There, a unique value of $x, y$ is formed for every value of $k$. However, we must have
$$y<10^6 \implies (k+1)^2< 10^6-1$$
and
$$x=(k-1)^2+1>0.$$
Therefore, there are only $997$ pairs of $(x,y).$
Solution by Williamgolly
### Solution 5
First we see that our condition is $\frac{x+y}{2} = 2 + \sqrt{xy}$. Then we can see that $x+y = 4 + 2\sqrt{xy}$. From trying a simple example to figure out conditions for $x,y$, we want to find $x-y$ so we can isolate for $x$. From doing the example we can note that we can square both sides and subtract $4xy$: $(x-y)^2 = 16 + 16\sqrt{xy} \implies x-y = -2( \sqrt{1+\sqrt{xy}})$ (note it is negative because $y > x$. Clearly the square root must be an integer, so now let $\sqrt{xy} = a^2-1$. Thus $x-y = -2a$. Thus $x = 2 + \sqrt{xy} - a = 2 + a^2 - 1 -2a$. We can then find $y$, and use the quadratic formula on $x,y$ to ensure they are $>0$ and $<10^6$ respectively. Thus we get that $y$ can go up to 999 and $x$ can go down to $3$, leaving $997$ possibilities for $x,y$. |
# TS Inter 2nd Year Maths 2A Question Paper May 2017
Thoroughly analyzing TS Inter 2nd Year Maths 2A Model Papers and TS Inter 2nd Year Maths 2A Question Paper May 2017 helps students identify their strengths and weaknesses.
## TS Inter 2nd Year Maths 2A Question Paper May 2017
Time: 3 Hours
Maximum Marks: 75
Note: This question paper consists of three Sections A, B, and C.
Section – A
(10 × 2 = 20 Marks)
I. Very Short Answer Type Questions.
• Each question carries two marks.
Question 1.
Find the complex conjugate of (3 + 4i) (2 – 3i).
Solution:
Let Z = (3 + 4i) (2 – 3i)
= 6 – 9i + 8i – 12i2
= 6 – i – 12(-1)
= 6 – i + 12
= 18 – i
∴ $$\bar{Z}$$ = 18 + i
∴ The complex conjugate of (3 + 4i) (2 – 3i) is 18 + i.
Question 2.
If the Arg $$\bar{Z}_1$$ and Arg Z2 are $$\frac{\pi}{5}$$ and $$\frac{\pi}{3}$$ respectively then find Arg Z1 + Arg Z2.
Solution:
Question 3.
If 1, w, w2 are the cube roots of unity, then prove that $$\frac{1}{2+w}+\frac{1}{1+2 w}=\frac{1}{1+w}$$.
Solution:
Given 1, w, w2 are the cube roots of unity.
∴ 1 + w + w2 = 0 and w3 = 1
Question 4.
Find the value of m for the equation having equal roots x2 – 15 – m(2x – 8) = 0.
Solution:
Given equation is x2 – 15 – m(2x – 8) = 0
⇒ x2 – 15 – 2mx + 8m = 0
⇒ x2 – 2mx + 8m – 15 = 0
Since it has equal roots
∴ (-2m)2 – 4(1)(8m – 15) = 0
⇒ 4m2 – 32m + 60 = 0
⇒ m2 – 8m + 15 = 0
⇒ m2 – 3m – 5m + 15 = 0
⇒ m(m – 3) – 5(m – 3) = 0
⇒ (m – 3) (m – 5) = 0
⇒ m = 3, 5
Question 5.
If α, β, 1 are the roots of x3 – 2x2 – 5x + 6 = 0, then find α, β.
Solution:
Given α, β, 1 are the roots of x3 – 2x2 – 5x + 6 = 0
∴ Sum of the roots = $$\frac{-(-2)}{1}$$
⇒ α + β + 1 = 2
⇒ α + β = 1 ………(1)
Product of the roots = $$\frac{-(6)}{1}$$
⇒ α . β . 1 = -6
∴ (α – β)2 = (α + β)2 – 4αβ
= 12 – 4(-6)
= 1 + 24
= 25
⇒ α – β = 5 ………(2)
(1) + (2) ⇒ 2α = 6
⇒ α = 3
From (1), 3 + β = 1
⇒ β = -2
∴ α = 3, β = -2
Question 6.
Find the number of terms in the expansion of (2x + 3y + z)7.
Solution:
The number of terms in the expansion of (2x + 3y + z)7 = $$\frac{(7+1)(7+2)}{2}$$ = 36
Question 7.
Find the mean deviation about the mean for data 3, 6, 10, 4, 9, 10.
Solution:
The arithmetic mean of the given data is
$$\bar{x}=\frac{3+6+10+4+9+10}{6}$$
= $$\frac{42}{6}$$
= 7
The absolute values of the deviations are |3 – 7|, |6 – 7|, |10 – 7|, |4 – 7|, |9 – 7|, |10 – 7| = 1, 1, 3, 3, 2, 3
∴ The mean deviation from the mean = $$\frac{4+1+3+3+2+3}{6}$$
= $$\frac{16}{6}$$
= 2.666
Question 8.
Find the number of injections of set A with 5 elements to set B with 7 elements.
Solution:
Here n(A) = 5 and n(B) = 7
∴ The number of injections of set A to a set B is $$n(B)_{p_n(A)}={ }^7 p_5$$
= 7 × 6 × 5 × 4 × 3
= 2,520
Question 9.
If 9C3 + 9C4 = 10Cr then find r.
Solution:
Given 9C3 + 9C4 = 10Cr
we know nCr + nCr-1 = n+1Cr
10C4 = 10Cr
⇒ r = 4 (or) 10 = 4 + r
⇒ r = 4 (or) r = 6
Question 10.
If the mean and variance of a binomial variable x are 2.4 and 1.44, then find n.
Solution:
Let X follow a binomial distribution with parameters n and p.
∴ Mean np = 2.4 and variance npq = 1.44
$$\frac{\mathrm{npq}}{\mathrm{np}}=\frac{1.44}{2.4}$$
⇒ q = 0.6
∴ p = 1 – q
= 1 – 0.6
= 0.4
∴ np = 2.4
⇒ n(0.4) = 2.4
⇒ n = 6
Section – B
(5 × 4 = 20 Marks)
• Attempt any five questions.
• Each question carries four marks.
Question 11.
Show that the four points in the Argand plane represented by the complex numbers 2 + i, 4 + 3i, 2 + 5i, and 3i are the vertices of a square.
Solution:
Let A, B, C, and D be the points in the Argand plane.
∴ A = (2, 1), B = (4, 3), C = (2, 5), D = (0, 3)
∴ AB = BC = CD = DA and AC = BD.
∴ A, B, C, D are the vertices of a square.
Question 12.
Determine the range of the expression $$\frac{x^2+x+1}{x^2-x+1}$$.
Solution:
Let y = $$\frac{x^2+x+1}{x^2-x+1}$$
⇒ x2y – xy + y = x2 + x + 1
⇒ (y – 1)x2 – x(y + 1) + y – 1 = 0
x is real ⇒ b2 – 4ac ≥ 0
⇒ [-(y + 1)]2 – 4(y – 1) (y – 1) ≥ 0
⇒ (y + 1)2 – 4(y – 1)2 ≥ 0
⇒ y2 + 2y + 1 – 4(y2 – 2y + 1) ≥ 0
⇒ y2 + 2y + 1 – 4y2 + 8y – 4 ≥ 0
⇒ -3y2 + 10y – 3 ≥ 0
⇒ 3y2 – 10y + 3 ≤ 0
⇒ 3y2 – 9y – y + 3 ≤ 0
⇒ 3y(y – 3) – 1(y – 3) ≤ 0
⇒ (3y – 1) (y – 3) ≤ 0
⇒ y ∈ [$$\frac{-1}{3}$$, 3]
∴ The range of y is [$$\frac{-1}{3}$$, 3]
Question 13.
Resolve $$\frac{3 x^3-2 x^2-1}{x^4+x^2+1}$$ into partial fractions,
Solution:
x4 + x2 + 1 = x4 + 2x2 + 1 – x2
= (x2 + 1)2 – x2
= (x2 + 1 + x) (x2 + 1 – x)
= (x2 + x + 1) (x2 – x + 1)
Let $$\frac{3 x^3-2 x^2-1}{x^4+x^2+1}=\frac{A x+B}{\dot{x}^2+x+1}+\frac{C x+D}{x^2-x+1}$$
⇒ 3x3 – 2x2 – 1 = (Ax + B) (x2 – x + 1) + (Cx + D) (x2 + x + 1)
Equating the coefficients of like terms, we have
A + C = 3 ……..(1)
⇒ C = 3 – A
-A + B + C + D = -2 …….(2)
A – B + C + D = 0 ……….(3)
B + D = -1 ………(4)
⇒ D = -1 – B
Substitute C, D in (2)
-A + B + 3 – A – 1 – B = -2
⇒ -2A = -4
⇒ A = 2
∴ C = 3 – 2
⇒ C = 1
Substitute C, D in (3)
A – B + 3 – A – 1 – B = 0
⇒ -2B = -2
⇒ B = 1
∴ D = -1 – 1
⇒ D = -2
∴ A = 2, B = 1, C = 1, D = -2
∴ $$\frac{3 x^3-2 x^2-1}{x^4+x^2+1}=\frac{2 x+1}{x^2+x+1}+\frac{x-2}{x^2-x+1}$$
Question 14.
A, B, C are 3 newspapers from a city. 20% of the population read A, 16% read B, 14% read C, 8% both A and B, 5% both A and C, 4% both B and C and 2% all the three. Find the percentage of the population who read atleast one newspaper.
Solution:
Given P(A) = $$\frac{20}{100}$$
P(B) = $$\frac{16}{100}$$
P(C) = $$\frac{14}{100}$$
P(A ∩ B) = $$\frac{8}{100}$$
P(B ∩ C) = $$\frac{4}{100}$$
P(A ∩ C) = $$\frac{5}{100}$$
P(A ∩ B ∩ C) = $$\frac{2}{100}$$
We know P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(C ∩ A) + P(A ∩ B ∩ C)
= $$\frac{20}{100}+\frac{16}{100}+\frac{14}{100}-\frac{8}{100}-\frac{4}{100}-\frac{5}{100}+\frac{2}{100}$$
= $$\frac{35}{100}$$
∴ Percentage of population who read atleast one news paper = $$\frac{35}{100}$$ × 100 = 35%.
Question 15.
Define conditional probability. State and prove the multiplication theorem of probability.
Solution:
Conditional Event: Suppose A and B are two events of a random experiment. If event B occurs after the occurrence of the event A. The event happening of B after the happening of A is called a conditional event and is denoted by $$\frac{B}{A}$$.
Conditional Probability: If A and B are two events of a sample space S and P(A) ≠ 0 then the probability of B after the event A has occured is called the conditional probability given A and is denoted by P($$\frac{B}{A}$$).
$$P\left(\frac{B}{A}\right)=\frac{P(B \cap A)}{P(A)}$$, P(A) ≠ 0
Multiplication Theorem of Probability: If A and B are two events of a random experiment with P(A) > 0 and P(B) > 0 then P(A ∩ B) = P(A) P($$\frac{B}{A}$$) = P(B) P($$\frac{A}{B}$$)
Proof: Let S be the sample space associated with the random experiment.
Let A and B be two events of S such that P(A) > 0 and P(B) > 0.
By the definition of conditional probability,
$$P\left(\frac{B}{A}\right)=\frac{P(B \cap A)}{P(A)}$$
∴ P(B ∩ A) = P(A) P($$\frac{B}{A}$$)
Since P(B) > 0, we interchange A and B in the above.
∴ P(A ∩ B) = P(B ∩ A) = P(B) P($$\frac{A}{B}$$)
Question 16.
If the letters of the word PRISON are permuted in all possible ways and the words thus formed are arranged in dictionary order, find the rank of the word PRISON.
Solution:
The letters of the given word in dictionary order are I, N, O, P, R, S
The number of words that begin with
I – – – – – = 5!
N – – – – – = 5!
O – – – – – = 5!
P I – – – – = 4!
P N – – – – = 4!
P O – – – – = 4!
P R I N – – = 2!
P R I O – – = 2!
P R I S N – = 1!
∴ The next word is P R I S O N = 1
The rank of the word PRISON = 3(5!) + 3(4!) + 2(2!) + 1! + 1
= 3(120) + 3(24) + 2(2) + 1 + 1
= 360 + 72 + 4 + 1 + 1
= 438
Question 17.
Simplify $${ }^{34} C_5+\sum_{r=0}^4{ }^{(38-r)} C_4$$.
Solution:
Section – C
(5 × 7 = 35 Marks)
• Attempt any five questions.
• Each question carries seven marks.
Question 18.
If n is a positive integer show that (1 + i)n + (1 – i)n = $$2^{\frac{n+2}{2}} \cos \left(\frac{n \pi}{4}\right)$$.
Solution:
Question 19.
Solve the equation 2x5 + x4 – 12x3 – 12x2 + x + 2 = 0.
Question 20.
If the coefficients of 4 consecutive terms in the expansion of (1 + x)n are a1, a2, a3, a4 respectively then show that $$\frac{a_1}{a_1+a_2}+\frac{a_3}{a_3+a_4}=\frac{2 a_2}{a_2+a_3}$$.
Solution:
Given a1, a2, a3, and a4 are the coefficients of 4 consecutive terms in (1 + x)n respectively.
Question 21.
Find the sum of the infinite series $$1-\frac{4}{5}+\frac{4.7}{5.10}-\frac{4.7 .10}{5 \cdot 10.15}$$ + ……….
Solution:
Question 22.
Find the mean deviation from the mean of the following data using the step deviation method:
Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 Number of Students 6 5 8 15 7 6 3
Question 23.
Three Boxes B1, B2, and B3 contain balls with different colours as shown below:
White Black Red B1 2 1 2 B2 3 2 4 B3 4 3 2
A die is thrown. B1 is chosen if either 1 or 2 turns up. B2 is chosen if 3 or 4 turns up and B3 if 5 or 6 turns up. Having chosen a box in this way, a ball is chosen at random from this box, if the ball drawn is found to be red, find the probability that it is drawn from box B2.
Solution:
Let E1, E2, E3 be the events of choosing boxes B1, B2, and B3 respectively.
Question 24.
The range of a random variable X is {0, 1, 2}. Given that P(X = 0) = 3C3, P(X = 1) = 4C – 10C2, P(X = 2) = 5C – 1.
(i) Find the value of C.
(ii) P(X < 1), P(1 < X ≤ 2) and P(0 < X ≤ 3).
Solution:
Range or random variable X is {0, 1, 2}
(i) P(X = 0) + P(X = 1) + P(X = 2) = 1
⇒ 3C3 + 4C – 10C2 + 5C – 1 = 1
⇒ 3C3 – 10C2 + 9C – 2 = 0
⇒ (C – 1) (3C2 – 7C + 2) = 0
⇒ (C – 1) (C – 2) (3C – 1) = 0
⇒ C = 1 (or) 2 (or) $$\frac{1}{3}$$
If C = 1 (or) C = 2 then P(X = 0) > 1
∴ C ≠ 1 and C ≠ 2
(i) C = $$\frac{1}{3}$$
(ii) P(X < 1) = P(X = 0)
= 3C3
= $$3\left(\frac{1}{3}\right)^3$$
= 3($$\frac{1}{27}$$)
= $$\frac{1}{9}$$
(iii) P(1 < x ≤ 2) = P(X = 2)
= 5C – 1
= 5($$\frac{1}{3}$$) – 1
= $$\frac{5-3}{3}$$
= $$\frac{2}{3}$$
(iv) P(0 < x ≤ 3) = P(X = 1) + P(X = 2)
= 4C – 10C2 + 5C – 1
= -10C2 + 9C – 1
= $$-10\left(\frac{1}{9}\right)+9\left(\frac{1}{3}\right)-1$$
= $$\frac{-10+27-9}{9}$$
= $$\frac{8}{9}$$ |
# The Box or Area Method: an alternative to traditional long division
Long division is often considered one of the most challenging topics to teach. Luckily, there are strategies that we can teach to make multi-digit division easier to understand and perform.
The Box Method, also referred to as the Area Model, is one of these strategies. It is a mental math based approach that will enhance number sense understanding. Students solve the equation by subtracting multiples until they get down to 0, or as close to 0 as possible.
If you plan on teaching the partial quotients strategy in your classroom (which I highly recommend) the Box Method is a great way to get started. It uses the same steps as partial quotients, but is organized a bit differently.
Let’s learn how to perform the Box Method/Area Model for long division!
Below, I have included both a video tutorial and step-by-step instructions.
VIDEO TUTORIAL
STEP-BY-STEP INSTRUCTIONS
Suppose that we want to solve the equation 324÷2.
Step 1:
First we draw a box. We write the dividend inside the box, and the divisor on the left side.
Step 2:
We want to figure out how many groups of 2 can be made from 324. We will do this in parts to make it easier. We could start by making 100 groups of 2, since we know that we have at least this many groups. So we multiply 100×2 to make 200, and then take that 200 away from 324. Now we have 124 left.
Step 3:
We make another box and carry the 124 over to it. Now let’s take away another easy multiply of 2. How about 50 groups of 2? We know that we can take out another 50 groups of 2 from 124. 50×2=100, so we take 100 from 124. Now we have 24 left.
Step 4:
We make another box and carry the 24 over to it. We know that 12 groups of 2 makes 24, so let’s write a 12 on top and take away 24 from the 24. Now we end up with 0, so we know that we are finished our equation.
Step 5:
Now we add the “parts” from the top of the boxes to find our quotient. 100+50+12=162, so we know that 324÷2=162.
ONE MORE EXAMPLE (WITH A REMAINDER)
Let’s take a look at one more example. In this example, we will solve 453÷4.
1. First we wrote our dividend inside the box, and our divisor on the left side.
2. We took out 100 groups of 4 first. This made 400. We subtracted 400 from 453 and were left with 53.
3. We carried the 53 to the next box, and then took out another 10 groups of 4 to make 40. We took the 40 away from the 53 and were left with 13.
4. We carried the 13 over to the next box, and then took out 3 groups of 4 to make 12. We took the 12 away from the 13 and were left with 1.
5. We cannot take any more groups of 4 out, so our remainder is 1. To find our final quotient, we add 100+10+3+remainder 1 to make 113 R1.
HELPFUL RESOURCES FOR THE BOX STRATEGY/AREA MODEL FOR DIVISION
I would love to help you teach the box strategy for long division in your classroom. You may find the following resources helpful:
FREE MINI COURSE
Register here for free Multi-Digit Multiplication and Division Mini Course. It will only take you about an hour to complete and you’ll leave with tons of new ideas, a strategic plan of action, free resources, a PD certificate, and more!
These task cards give students the opportunity to practice the box method/area model for long division in a variety of different ways. Students will calculate quotients, solve division problems, figure out missing dividends and divisors, think about how to efficiently solve an equation using the box method, and more. See the Box Method Task Cards HERE or the Big Bundle of Long Division Task Cards HERE.
THE LONG DIVISION STATION
The Long Division Station is a self-paced, student-centered math station for long division. Students gradually learn a variety of strategies for long division, the box method being one of them. One of the greatest advantages to this Math Station is that is allows you to target every student and their unique abilities so that everyone is appropriately challenged. See The Long Division Station HERE.
OR SEE ALL RESOURCES |
RD Sharma 2017 Solutions for Class 9 Math Chapter 7 Introduction To Euclid's Geometry are provided here with simple step-by-step explanations. These solutions for Introduction To Euclid's Geometry are extremely popular among class 9 students for Math Introduction To Euclid's Geometry Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma 2017 Book of class 9 Math Chapter 7 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma 2017 Solutions. All RD Sharma 2017 Solutions for class 9 Math are prepared by experts and are 100% accurate.
#### Question 1:
Define the following terms:
(i) Line segment
(ii) Collinear points
(iii) Parallel lines
(iv) Intersecting lines
(v) Concurrent lines
(vi) Ray
(vii) Half-line.
#### Answer:
The given problem requires definitions of various terms.
(i) Line segment:
A line segment AB can be defined as the part of the line with end points A and B, whereA and B are the two points of the line.
It is denoted by
Let us take a line with two points A and B
This is a line AB
While,
This is a line segment AB.
(ii) Collinear points:
When three or more points lie on the same line; they are said to be collinear.
Example:
Let us take a line l. P, Q, R points lie on it.
So,
Here, P, Q and R are collinear points.
(iii) Parallel lines:
Two or more lines are said to be parallel to each other if there is no point of intersection between them.
For Example:
Since, there is no point of intersection between l and m, they are parallel.
(iv) Intersecting lines:
Two or more lines are said to be intersecting lines if they meet each other at a point or they have a common point.
For Example:
l and m are the two lines both passing through point O. Hence, they are intersecting lines.
(v) Concurrent lines:
Two or more lines are said to be concurrent if they all pass through a common point or there exist a point common to all of them.
For Example:
m, n, o and p are concurrent as they all have a common point O.
(vi) Ray:
A ray is defined as the part of the line with one end point such that it can be extended infinitely in the other direction.
It is represented by
For Example:
Here, is a ray as it has one end point A and it can be extended indefinitely in other direction.
(vii) Half-line:
A half-line can be defined as a part of the line which has one end point and extends indefinitely in the other direction. It is different from ray as the end point is not included in the half-line.
For example,
When A is included in the part, then it is called a ray AB, but when A is not included then is called a half-line AB.
#### Question 2:
(i) How many lines can pass through a given point?
(ii) In how many points can two distinct lines at the most intersect?
#### Answer:
(i) Let us take a point A.
If we try to draw lines passing through this point A, we can see that we can draw many lines.
Therefore, infinite number of lines can pass through a given point.
(ii) Let us take two lines l and m, and intersect them.
As, we can see here the two lines have only one point in common that is O.
Therefore, there is only one point where two distinct lines can intersect.
#### Question 3:
(i) Given two points P and Q, find how many line segments do they deter-mine.
(ii) Name the line segments determined by the three collinear points P, Q and R.
#### Answer:
(i) In this problem we are given two points P and Q.
If we try to join these two points through a line segment, we can see that there can be only one such line segment PQ.
Therefore, given two points, only one line segment is determined by them.
(ii) In the given problem, we are given three collinear points P, Q and R. Collinear points lie on the same line, so they can be represented as
So, the various line segments determined here are PQ, QR and PR.
#### Question 4:
Write the turth value (T/F) of each of the following statements:
(i) Two lines intersect in a point.
(ii) Two lines may intersect in two points.
(iii) A segment has no length.
(iv) Two distinct points always determine a line.
(v) Every ray has a finite length.
(vi) A ray has one end-point only.
(vii) A segment has one end-point only.
(viii) The ray AB is same as ray BA.
(ix) Only a single line may pass through a given point.
(x) Two lines are coincident if they have only one point in common.
#### Answer:
(i) False
The given statement is false, as it is not necessary that the two lines always intersect. In the case of parallel lines, the two lines never intersect or there is no common point between them.
(ii) False
The given statement is false, as there is only one point common between two intersecting lines. So, two lines cannot intersect at two points.
(iii) False
A line segment is a part of line defined by two end points, so it is always of a definite length. Hence, the given statement is false.
(iv) True
The given statement is true as a unique line can be determined by minimum of two distinct points.
(v) False
A ray is defined as a part of the line with one end point, where the other end can be stretched indefinitely. Therefore, a ray cannot have a finite length.
(vi) True
The given statement is true as a ray is defined as a part of the line with one end point, where the other end can be stretched indefinitely.
(vii) False
A line segment is a part of line defined by two end points. So the given statement is false as a line segment has two end points.
(viii) False
While denoting a ray, the first letter denotes the end point and the second one denotes the end which can be extended. So, for ray AB, A is the end point and B is the end which can be stretched while for ray BA, B is the end point and A the end which can be stretched. Therefore, ray AB is not the same as ray BA.
(ix) False
The given statement is false as infinite number of lines can pass through a given point.
(x) False
Two lines are said to be coincident if they lie exactly on top of each other, which means that they have all the points in common. So the given statement is false.
#### Question 5:
In Fig. 7.17, name the following:
(i) five line segments.
(ii) Five rays.
(iii) Four collinear points.
(iv) Two pairs of non-intersecting line segments.
#### Answer:
(i) A line segment is a part of line defined by two end points. So in the given figure 7.17, five line segments are:
(1) AC
(2) CD
(3) AP
(4) PQ
(5) RS
(ii) A ray is the part of line with one end point and one end which can be extended. So in the given figure 7.17, five rays are:
(1) Ray RB
(2) Ray RS
(3) Ray PQ
(4) Ray DS
(5) Ray AB
(iii) Collinear points are the points which are present on the same line. In the present figure 7.17, there are two sets of four collinear points.
(1) A, P, R, B
(2) C, D, Q, S
(iv) In the given figure 7.17, two pairs of non intersecting line segments are:
(1) AB and CS
(2) AC and PQ
#### Question 6:
Fill in the blanks so as to make the following statements true:
(i) Two distinct points in a plane determine a ........line.
(ii) Two distinct ........... in a plane cannot have more than one point in common.
(iii) Given a line and a point, not on the line, there is one and only ......... line which passes through the given point and is ......... to the given line.
(iv) A line separates a plane into ......... parts namely the ...........and the ........... itself.
#### Answer:
(i) Two distinct points in a plane determine a unique line.
(ii) Two distinct lines in a plane cannot have more than one point in common.
(iii) Given a line and a point, not on the line, there is one and only perpendicular line which passes through the given point and is perpendicular to the given line.
(iv) A line separates a plain into three parts namely the two half planes and the line itself.
#### Question 1:
How many least number of distinct points determine a unique line?
#### Answer:
If we have a single point A then we can draw infinite number of lines through it, while if we have two points A and B; then only one unique line passes through both of them.
When we have one point,
When we have two points,
Therefore, a minimum of two distinct points are required to determine a unique line.
#### Question 2:
How many lines can be drawn through both of the given points?
#### Answer:
Given two distinct points, only one unique line can be drawn passing through both of them.
For example, if we have two distinct points A and B, only one line is there which passes through both of them.
#### Question 3:
How many lines can be drawn through a given point.
#### Answer:
Given a single point, then we can draw infinite number of lines through that point.
For example: If we have a point A, then there are infinite numbers of lines passing through it.
Here, lines m, n, o and p all pass through point A.
#### Question 4:
In how many points two distinct lines can intersect?
#### Answer:
Two distinct lines can intersect at only point, as there is only one common point between two intersecting lines.
For example, if l and m are two intersecting lines then there is only one common point O between them. This is the point of intersection.
#### Question 5:
In how many points a line, not in a plane, can intersect the plane?
#### Answer:
A line which is not present in the plane can intersect the plane at only one single point.
#### Question 6:
In how many points two distinct planes can intersect?
#### Answer:
Two distinct planes can intersect at only one unique line, which is common to both of them and one line is made up of infinite points.
Therefore, two distinct planes can intersect at infinite points.
#### Question 7:
In how many lines two distinct planes can intersect?
#### Answer:
If we look at two intersecting plane, we can see that there is only one unique line at which the two planes intersect.
Therefore, two distinct planes can intersect each other at a single unique line as only a single line is common between two intersecting planes.
#### Question 8:
How many least number of distinct points determine a unique plane?
#### Answer:
If we have two distinct points, then we can draw infinite number of planes passing through those two points. While if we have three distinct non collinear points, only a single unique plane can be drawn passing through those three points.
Therefore, a minimum of three distinct non collinear points are required to get a unique plane.
#### Question 9:
Given three distinct points in a plane, how many lines can be drawn by joining them?
#### Answer:
Given three distinct points A, B and C in a plane, they can either be collinear or non collinear.
If they are collinear, then there can be only one line joining them.
If they are non collinear, then there can be three lines joining them.
For example, if we have three distinct non collinear points P, Q and R. Then we can draw three lines l, m and n joining them.
#### Question 10:
How many planes can be made to pass through a line and a point not on the line?
#### Answer:
Given a line and a distinct point not lying on the line, only a single plane can be drawn through both of them as there can be only plane which can accommodate both the line and the point together.
Let us take a line l and a point A, as we can see there can be only plane which pass through both of them.
#### Question 11:
How many planes can be made to pass through two points?
#### Answer:
Given two distinct points, we can draw many planes passing through them. Therefore, infinite number of planes can be drawn passing through two distinct points or two points can be common to infinite number of planes.
#### Question 12:
How many planes can be made to pass through three distinct points?
#### Answer:
The number of planes that can pass through three distinct points is dependent on the arrangement of the points.
• If the points are collinear, then infinite number of planes may pass through the three distinct points.
• If the points are non collinear, then only one unique plane can pass through the three distinct points.
View NCERT Solutions for all chapters of Class 9 |
## Intermediate Algebra (12th Edition)
$(5x-6)(2x+3)$
$\bf{\text{Solution Outline:}}$ To factor the quadratic expression $ax^2+bx+c,$ find two numbers whose product is $ac$ and whose sum is $b$. Use these $2$ numbers to decompose the middle term of the quadratic expression and then use factoring by grouping. $\bf{\text{Solution Details:}}$ In the given expression, $10x^2+3x-18 ,$ the value of $ac$ is $10(-18)=-180$ and the value of $b$ is $3 .$ The possible pairs of integers whose product is $ac$ are \begin{array}{l}\require{cancel} \{1,-180\}, \{2,-90\}, \{3,-60\}, \{4,-45\}, \{5,-36\}, \{6,-30\}, \{9,-20\}, \{10,-18\}, \{12,-15\}, \{-1,180\}, \{-2,90\}, \{-3,60\}, \{-4,45\}, \{-5,36\}, \{-6,30\}, \{-9,20\}, \{-10,18\}, \{-12,15\} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ -12,15 \}.$ Using these $2$ numbers to decompose the middle term of the given expression results to \begin{array}{l}\require{cancel} 10x^2-12x+15x-18 .\end{array} Grouping the first and third terms and the second and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (10x^2-12x)+(15x-18) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 2x(5x-6)+3(5x-6) .\end{array} Factoring the $GCF= (5x-6)$ of the entire expression above results to \begin{array}{l}\require{cancel} (5x-6)(2x+3) .\end{array} |
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# The sum of two numbers is 8 and 15 times the sum of their reciprocals is also 8. Find the numbers.
Last updated date: 09th Aug 2024
Total views: 459.9k
Views today: 5.59k
Verified
459.9k+ views
Hint: - Use the formula ${\left( {x - y} \right)^2} = {\left( {x + y} \right)^2} - 4xy$
Let the numbers be $x$ and $y$.
According to the question , the sum of two numbers is 8.
$\Rightarrow x + y = 8...............\left( 1 \right)$
Reciprocals of the numbers be$\dfrac{1}{x}$and $\dfrac{1}{y}$
According to question 15 times the sum of reciprocals is also 8
$\Rightarrow 15\left( {\dfrac{1}{x} + \dfrac{1}{y}} \right) = 8$
$\Rightarrow 15\left( {\dfrac{{x + y}}{{xy}}} \right) = 8$
From equation 1
$\Rightarrow 15\left( {\dfrac{8}{{xy}}} \right) = 8 \\ \Rightarrow xy = 15.........\left( 2 \right) \\$
Now it is known fact that ${\left( {x - y} \right)^2} = {\left( {x + y} \right)^2} - 4xy$
Now from equation (1) and (2)
${\left( {x - y} \right)^2} = {8^2} - 4 \times 15 = 64 - 60 = 4 = {2^2} \\ \Rightarrow x - y = 2...........\left( 3 \right) \\$
Now add equation (1) and (3)
$\Rightarrow x + y + x - y = 8 + 2 \\ \Rightarrow 2x = 10 \\ \Rightarrow x = 5 \\$
From equation 1
$\Rightarrow x + y = 8 \\ \Rightarrow 5 + y = 8 \\ \Rightarrow y = 3 \\$
So, the required numbers are 5 and 3.
Note: - Whenever we face such types of problems the key concept we have to remember is that always remember the formula ${\left( {x - y} \right)^2} = {\left( {x + y} \right)^2} - 4xy$. It will give us a simple approach to solve this kind of problem, then simplify we will get the required answer. |
1. ## the House Problem
Consider this series: 1, 2, 3,..., N-1, N, N+1,..., M.
Given 1 + 2 + 3 + ... + (N-1) = (N+1) + (N+2) +...+ M. Find N.
(hint: N is a 4 digits number)
2. Originally Posted by john_n82
Consider this series: 1, 2, 3,..., N-1, N, N+1,..., M.
Given 1 + 2 + 3 + ... + (N-1) = (N+1) + (N+2) +...+ M. Find N.
(hint: N is a 4 digits number)
Note, $\displaystyle 1+2+...+(N-1) = \tfrac{1}{2}N(N-1) \text{ and }(N+1)+(N+2)+...+M = $$\displaystyle \tfrac{1}{2}(M-N)(M-N+1) + N(M-N) Therefore, \displaystyle N^2 = \tfrac{1}{2}M(M+1), therefore \displaystyle M^2 + M - 2N^2 = 0. We need the discriminant to be a square, so we need \displaystyle 1 + 8N^2 = x^2 . This is a Pellian equation. 3. Originally Posted by ThePerfectHacker Note, \displaystyle 1+2+...+(N-1) = \tfrac{1}{2}N(N-1) \text{ and }(N+1)+(N+2)+...+M =$$\displaystyle \tfrac{1}{2}(M-N)(M-N+1) + N(M-N)$
Therefore, $\displaystyle N^2 = \tfrac{1}{2}M(M+1)$, therefore $\displaystyle M^2 + M - 2N^2 = 0$.
We need the discriminant to be a square, so we need $\displaystyle 1 + 8N^2 = x^2$.
This is a Pellian equation.
I found M = 8, so N = 6.
4. Originally Posted by john_n82
Consider this series: 1, 2, 3,..., N-1, N, N+1,..., M.
Given 1 + 2 + 3 + ... + (N-1) = (N+1) + (N+2) +...+ M. Find N.
(hint: N is a 4 digits number)
i think you mean N^2 is a four-digit number
in this case M = 49, N = 35
5. i found 2 more values for M
M = 288, N = 204
M 9800, N = 6930
What is the exact answer for this problem?
6. Originally Posted by john_n82
i found 2 more values for M
M = 288, N = 204
M 9800, N = 6930
What is the exact answer for this problem?
There are many answers to this problem.
You are told that you need N to be a 4 digit number so N=6930 |
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# How to Add and Subtract Unlike Fractions?
Last updated date: 17th Apr 2024
Total views: 156k
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## Introduction to Fractions
Fractions are simply numbers in which the numerator is divided by the denominator. Fractions can get a little complicated when it comes to simple operations like addition and subtraction. When the denominators of given fractions are the same, we can add the numerators directly. However, when the denominators are different, such fractions need to be solved using a different method. Don’t worry, we’ve got you covered. In this chapter we’ll learn about adding and subtracting unlike fractions, addition and subtraction of fractions with unlike denominators, steps for addition and subtraction of unlike fractions, and addition and subtraction of like and unlike fractions.
Additions and Subtraction of Unlike Fractions
## What are Like Fractions?
Like fractions are fractions that have the same denominators. For example, $\dfrac{10}{19}, \dfrac{11}{19}, \dfrac{13}{19}, and \dfrac{15}{19}$ are like fractions.
## What are Unlike Fractions?
Fractions that have different denominators are known as, unlike fractions. For example,
$\left\{\dfrac{2}{3}, \dfrac{5}{6}\right\},\left\{\dfrac{4}{9}, \dfrac{8}{7}\right\}$ are two sets of unlike fractions. The addition and subtraction operations of such fractions are different from those with the same denominator.
Let us see how to perform these operations on such fractions.
## Adding and Subtracting Like Fractions
Let us add the fractions with like denominators in numerical terms. In this case, we need to add $\dfrac{1}{5} + \dfrac{2}{5}$. Let us use the following steps to understand the addition.
Step 1: Add the numerators of the given fractions. Here, the numerators are 1 and 2, so it will be 1 + 2 = 3
Step 2: Retain the same denominator. Here, the denominator is 5.
Step 3: Therefore, the sum of $\dfrac{1}{5} + \dfrac{2}{5} = \dfrac{1 + 2}{5}= \dfrac{3}{5}$.
Now, let us subtract the fractions with like denominators in numerical terms. In this case, we need to subtract $\dfrac{2}{5} - \dfrac{1}{5}$. Let us understand the procedure using the following steps.
Step 1: We will subtract the numerators of the given fractions. Here, the numerators are 2 and 1, so it will be 2 - 1 = 1
Step 2: Retain the same denominator. Here, the denominator is 5.
Step 3: Therefore, the difference of $\dfrac{2}{5} - \dfrac{1}{5} = \dfrac{2 - 1}{5} = \dfrac{1}{5}$
## How to Add or Subtract Fractions with Unlike Denominators?
When the given fractions have a different denominator, we cannot add the numerators directly without considering the denominators. Below are the steps for the addition and subtraction of unlike fractions. Firstly we will see the steps for addition.
## Steps for Addition of Unlike Fractions
Example: Add $\dfrac{1}{5}+ \dfrac{1}{3}$
Ans: For adding unlike fractions, we need to use the following steps
Step 1: Find the denominators Least Common Multiple (LCM).
Here, the LCM of 5 and 3 is 15.
Step 2: Convert the given fractions to like fractions by writing the equivalent fractions for the respective fractions such that their denominators remain the same. Here, it will be $\dfrac{1}{5} \times \dfrac{3}{3}=\dfrac{3}{15}$
Step 3: Similarly, an equivalent fraction of $\dfrac{1}{3}$ with denominator 15 is $\dfrac{1}{3} \times \dfrac{5}{5}=\dfrac{5}{15}$
Step 4: Now that we have converted the given fractions to like fractions, we can add the numerators and retain the same denominator. This will be $\dfrac{3}{15}+ \dfrac{5}{15}=\dfrac{8}{15}$
## Steps to Subtract Fractions with Unlike Denominators
Let’s understand it, with an example.
Example: Subtract $\dfrac{5}{6}-\dfrac{1}{3}$
Solution: For subtracting unlike fractions, we need to use the following steps.
Step 1: Find the denominators Least Common Multiple (LCM). Here, the LCM of 6 and 3 is 6.
Step 2: Convert the given fractions to like fractions by writing the equivalent fractions for the respective fractions such that their denominators remain the same. Here, it will be $\dfrac{5}{6} \times \dfrac{1}{1}=\dfrac{5}{6}$
Step 3: Similarly, an equivalent fraction of $\dfrac{1}{3}$ with denominator 6 is $\dfrac{1}{3}$ $\times \dfrac{2}{2}=\dfrac{2}{6}$
Step 4: Now that we have converted the given fractions to like fractions, we can subtract the numerators and retain the same denominator. This will be $\dfrac{5}{6}-\dfrac{2}{6}=\dfrac{3}{6}$. This can be further reduced to $\dfrac{1}{2}$
## Solved Examples
Below is an example of some of the problems based on the addition and subtraction of fractions with unlike denominators
Q 1. $\operatorname{Add} \dfrac{5}{9}+\dfrac{3}{2}$
Ans: The fractions $\dfrac{5}{9}$ and $\dfrac{3}{2}$ have different denominators.
LCM of 9 and $2=18$
Multiply $\dfrac{5}{9}$ by $\dfrac{2}{2}$
$\dfrac{5}{9} \times \dfrac{2}{2}=\dfrac{10}{18}$
Multiply $\dfrac{3}{2}$ by $\dfrac{9}{9}$
$\dfrac{3}{2} \times \dfrac{9}{9}=\dfrac{27}{18}$
Now,
$\dfrac{10}{18}+\dfrac{27}{18}=\dfrac{37}{18}$
Thus, upon adding $\dfrac{5}{9}$ and $\dfrac{3}{2}$ we get $\dfrac{37}{18}$ as the result.
Q 2. $\operatorname{Add} \dfrac{1}{9}$ and $\dfrac{7}{3}$
Ans: The fractions $\dfrac{1}{9}$ and $\dfrac{7}{3}$ have different denominators.
LCM of 9 and $3=27$
Multiply $\dfrac{1}{9}$ by $\dfrac{3}{3}$
$\dfrac{1}{9} \times \dfrac{3}{3}=\dfrac{3}{27}$
Multiply $\dfrac{7}{3}$ by $\dfrac{9}{9}$
$\dfrac{7}{3} \times \dfrac{9}{9}=\dfrac{63}{27}$
Now,
$\dfrac{3}{27}+\dfrac{63}{27}=\dfrac{66}{27}=\dfrac{22}{9}$
Thus, upon adding $\dfrac{1}{9}$ and $\dfrac{7}{3}$ we get $\dfrac{22}{9}$ as the result.
Q 3. Subtract $\dfrac{1}{3}$ from $\dfrac{5}{7}$
Ans: $\operatorname{LCM}(3,7)=21$
Multiply $\dfrac{1}{3}$ by $\dfrac{7}{7}$
$\dfrac{1}{3} \times \dfrac{7}{7}=\dfrac{7}{21}$
Multiply $\dfrac{5}{7}$ by $\dfrac{3}{3}$
$\dfrac{5}{7} \times \dfrac{3}{3}=\dfrac{15}{21}$
Hence,$\dfrac{15}{21}-\dfrac{7}{21}=\dfrac{8}{21}$
Thus, the required answer is $\dfrac{8}{21}$.
## Worksheet for Addition and Subtraction of Unlike fractions
Below are some of the questions based on adding and subtracting unlike fractions and like fractions.
Q 1. $\dfrac{1}{3}+\dfrac{9}{3}$ (Ans. $\dfrac{10}{3}$)
Q 2. $\dfrac{2}{7}+\dfrac{1}{9}$ (Ans. $\dfrac{25}{63}$)
Q 3. $\dfrac{9}{7}-\dfrac{5}{3}$ (Ans. $\dfrac{-8}{21}$)
Q 4. $\dfrac{8}{3}-\dfrac{1}{3}$ (Ans. $\dfrac{7}{3}$)
## Summary
Just like our counting numbers, fractions can also be easily subtracted and added. The rules for addition and subtraction of like and unlike fractions are quite simple and it involves three steps to do so: finding the same common denominator if it's an unlike fraction, then adding or subtracting the numerators. And if the answer is an improper form you have to reduce the fraction into a mixed number. Solving these unlike fractions using the given steps is easy once you grasp the important step for finding the LCM and rationalizing the denominators. You can also download the addition and subtraction of unlike fractions worksheets PDF.
## FAQs on How to Add and Subtract Unlike Fractions?
1. Do like fractions and unlike fractions have the same numerator?
Like fractions are fractions having the same denominators; the numerators can be the same or different. Unlike fractions are fractions have different denominators. The numerators in unlike fractions also can be the same or different.
2. How to Convert Unlike Fractions to Like Fractions?
Unlike fractions can be converted to like fractions by finding the lowest common multiple (LCM) of the denominators first and then calculating their equivalent fractions with the same denominator.
3. What are the seven types of fractions?
Seven types of fractions are as follows:
• Proper fraction
• Improper fraction
• Mixed fraction
• Like fraction
• Unlike fraction
• Equivalent fraction |
# The Square Root Property Video
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• 1:44 Using the Square Root Property
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Lesson Transcript
Instructor: Laura Pennington
Laura has taught collegiate mathematics and holds a master's degree in pure mathematics.
In this lesson, we'll learn what the square root property is. We'll also look at how to use the square root property to solve quadratic equations and for which types of quadratic equations it can be used.
## Square Root Property
Have you ever moved into an unfurnished apartment or home? If so, then you probably know that one of the best parts of moving is decorating your new living area! Imagine that you've just moved into a new home that has a large square dining room with hardwood floors. You decide that you want to get a nice area rug for this room.
You take the measurements of the room and find that you will want a square area rug with sides 12 feet in length to cover the floor adequately.
You go online to look at area rugs and find a few that you really like. However, the rugs are listed by area, not by dimensions, so you're not sure which ones will fit the space appropriately. You narrow it down to three different rugs with the areas of 100 square feet, 121 square feet, and 144 square feet.
You know that to find the area of a square rug, you would use the formula A = s2, where A is the area of the square, and s is the length of a side of the square. You realize that if you plug each of the areas into this formula for A, you can solve for s. If you get 12 for s, then you know it will fit the space perfectly! In other words, we want to solve three different equations.
s2 = 100
s2 = 121
s2 = 144
Okay, we know what you need to do. Now, we just need to figure out how to do it! It just so happens that there is a property we can use to solve these specific types of equations, and that property is called the square root property.
The square root property can be used to solve certain quadratic equations, and it states that if x2 = c, then x = √c or x = -√c, where c is a number.
## Using the Square Root Property
In words, the square root property states that if we have an equation with a perfect square on one side and a number on the other side, then we can take the square root of both sides and add a plus or minus sign to the side with the number and solve the equation.
Let's use this to see which of the area rugs you've found will fit your new dining room! First, let's consider the rug with area 100 ft2. We plug A = 100 into the area formula and use the square root property to solve for s.
We write:
A = s2
We plug in 100 for A. Now our equation reads:
100 = s2
We use the square root property and have two equations:
s = √100, or
s = - √100
When we simplify, we have:
s = 10 or s = -10
Since we're talking about the length, our answer will be a positive number. So, our answer is:
s = 10 feet
We see that this rug with area 100 ft.2 has sides of length 10 feet. This rug is too small! Let's look at the rug with area 121 ft2. We plug A = 121 into our formula and solve for s again.
We write:
A = s2
We plug in 121 for A. Now our equation reads:
121 = s2
We use the square root property and have two equations:
s = √121, or
s = - √121
When we simplify, we have:
s = 11 or s = -11
Since we're talking about the length, our answer will be a positive number. So, our answer is:
s = 11 feet
We see that this rug with area 121 ft2 has sides of length 11 feet. This one is also too small. Let's hope the third one will be a good fit! To check the third one, we plug A = 144 into the formula and solve for s.
We write:
A = s2
We plug in 144 for A. Now our equation reads:
144 = s2
We use the square root property and have two equations:
s = √144, or
s = - √144
When we simplify, we have:
s = 12 or s = -12
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NCERT Solutions for Exercise 1.6 Class 9 Maths Chapter 1 - Number Systems
# NCERT Solutions for Exercise 1.6 Class 9 Maths Chapter 1 - Number Systems
Edited By Vishal kumar | Updated on Oct 03, 2023 09:23 AM IST
## NCERT Solutions for Class 9 Maths Chapter 1: Number Systems Exercise 1.6- Download Free PDF
NCERT Solutions for Class 9 Maths Chapter 1 - Number Systems Exercise 1.6- NCERT Solutions for Class 9 Maths Exercise 1.6 deals with the idea of simplification of the real number with power. Before exercise 1.6 Class 9 Maths we learnt about how to solve exponents with positive power when they are multiplied, and divided in a different condition when bases are the same or exponents is the same but in exercise 1.6 Class 9 Maths we are dealing with rational number as power along with the rational number power we are also introduced to the negative power and its application.
In NCERT solutions for Class 9 Maths exercise 1.6, there are four basic rules which have been used. These principles are the basis of these types of questions. There is only one way of solving them by applying these rules. The following exercises are also present along with Class 9 Maths Chapter 1 exercise 1.6.
9th class maths exercise 1.6 answers have been expertly crafted by Careers360 subject experts. They are in simple language for easy understanding. This class 9 maths chapter 1 exercise 1.6 contains three questions with multiple parts. A PDF version is available for free download, enabling offline use.
## Access Number Systems Class 9 Chapter 1 Exercise: 1.6
Given number is $64^{\frac{1}{2}}$
Now, on simplifying it we will get
$\Rightarrow 64^{\frac{1}{2}} = (8^2)^\frac{1}{2} = 8$
Given number is $32^{\frac{1}{5}}$
Now, on simplifying it we will get
$\Rightarrow 32^{\frac{1}{5}} = (2^5)^\frac{1}{5} = 2$
Given number is $125^{\frac{1}{3}}$
Now, on simplifying it we will get
$\Rightarrow 125^{\frac{1}{3}} = (5^3)^\frac{1}{3} = 5$
Given number is $9^{\frac{3}{2}}$
Now, on simplifying it we will get
$\Rightarrow 9^{\frac{3}{2}} = (3^2)^\frac{3}{2} = 3^3 = 27$
Given number is $32^{\frac{2}{5}}$
Now, on simplifying it we will get
$\Rightarrow 32^{\frac{2}{5}} = (2^5)^\frac{2}{5} = 2^2 = 4$
Q2 (iii) Find : $16^{\frac{3}{4}}$
Given number is $16^{\frac{3}{4}}$
Now, on simplifying it we will get
$\Rightarrow 16^{\frac{3}{4}} = (2^4)^\frac{3}{4} = 2^3 = 8$
Given number is $125^{\frac{-1}{3}}$
Now, on simplifying it we will get
$\Rightarrow 125^{\frac{-1}{3}} = (5^3)^\frac{-1}{3} = 5^{-1} = \frac{1}{5}$
Therefore, the answer is $\frac{1}{5}$
Given number is $2^{\frac{2}{3}}.2^{\frac{1}{5}}$
Now, on simplifying it we will get
$\Rightarrow 2^{\frac{2}{3}}.2^{\frac{1}{5}} = 2^{\frac{2}{3}+\frac{1}{5}} = 2^{\frac{10+3}{15}} = 2^\frac{13}{15}$ $\left ( \because a^n.a^m = a^{n+m} \right )$
Therefore, the answer is $2^{\frac{13}{15}}$
Given number is $\left (\frac{1}{3^{3}} \right )^{7}$
Now, on simplifying it we will get
$\Rightarrow \left ( \frac{1}{3^3} \right )^7= \frac{1^7}{3^{3\times7}} = \frac{1}{3^{21}} = 3^{-21}$ $\left ( \because (a^n)^m = a^{n.m} \ and \ \frac{1}{a^m}= a^{-m}\right )$
Therefore, the answer is $3^{-21}$
Given number is $\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}$
Now, on simplifying it we will get
$\Rightarrow \frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}} = 11^{\frac{1}{2}-\frac{1}{4}}= 11^{\frac{2-1}{4}} = 11^\frac{1}{4}$ $\left ( \because \frac{a^n}{a^m}= a^{n-m}\right )$
Therefore, the answer is $11^{\frac{1}{4}}$
Given number is $7^{\frac{1}{2}}.8^{\frac{1}{2}}$
Now, on simplifying it we will get
$\Rightarrow 7^{\frac{1}{2}}.8^{\frac{1}{2}}= (7\times8)^{\frac{1}{2}} = 56^{\frac{1}{2}}$ $\left ( \because a^n.b^n=(a.b)^n\right )$
Therefore, the answer is $56^{\frac{1}{2}}$
## More About NCERT Solutions for Class 9 Maths Exercise 1.6: Laws of Exponents for Real Numbers
The NCERT book Class 9 Maths Exercise 1.6 is about applying the Laws of Exponents on numbers with rational number exponents. there are different ways of representing the same exponent. is the most common way which is used in the NCERT solutions for Class 9 Maths exercise 1.6. In problems, 'a' is a real number that is greater than zero and m and n are the integers and both m and n have no common factor other than one and the n which is the denominator of the power is not zero because fraction cannot exist with denominator zero.
Also Read| Number Systems Class 9 Notes
## Benefits of NCERT Solutions for Class 9 Maths Exercise 1.6:
• NCERT solutions for Class 9 Maths exercise 1.6 help us establish our fundamental understanding of numbers and their properties.
• By solving the NCERT solutions for Class 9 Maths chapter 1 exercise 1.6 builds a strong foundation of mathematical understanding as well as confidence in approaching new topics in higher grades.
• NCERT syllabus Class 9 Maths exercise 1.6 will assist us in understanding how to conduct operations on real numbers, which will aid us in future courses.
## key Features of Exercise 1.6 Class 9 Maths
1. Subject Expert Solutions: The 9th class maths exercise 1.6 answers are crafted by subject matter experts. This ensures accuracy and clarity in explanations.
2. Detailed Explanations: Each class 9 maths chapter 1 exercise 1.6 solution is accompanied by detailed explanations, making it easier for students to understand the concepts and problem-solving techniques.
3. Step-by-Step Format: The exercise 1.6 class 9 maths solutions are presented in a step-by-step format, helping students follow the solution process.
4. Multiple Questions: Class 9 maths ex 1.6 contains a variety of questions with multiple parts, providing ample practice for students.
5. PDF Availability: Class 9 ex 1.6 solutions are often available in PDF format, allowing students to download and access them offline for convenient studying.
6. Conceptual Clarity: The ex 1.6 class 9 exercise focuses on reinforcing the understanding of fundamental mathematical concepts related to real numbers and their properties.
7. Syllabus Alignment: Exercise 1.6 aligns with the prescribed syllabus, covering all relevant topics and concepts.
8. Free Access: These solutions are typically provided free of charge, ensuring accessibility to all students.
Also see-
## Subject Wise NCERT Exemplar Solutions
1. In Class 9 Mathematics chapter 1 exercise 1.6, how many solved examples are there?
One solved example is there in Class 9 Maths chapter 1 exercise 1.6
2. What number of questions are there in chapter 1 exercise 1.6 of Class 9 Maths?
Three questions are there in chapter 1 exercise 1.6 of Class 9 Maths. The question one has three subparts, question two has four sub-parts and question three has four subparts.
3. What kinds of questions may be found in NCERT Solutions for Class 9 Maths chapter 1 Exercise 1.6?
Two types of questions are there in chapter 1 exercise 1.6 of Class 9 Math. Firstly we have a question in which we have to find a solution to the problem using the law of exponent and secondly we have a question in which we have to simplify.
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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is
Option 1) Option 2) Option 3) Option 4)
A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up ? Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms−2 :
Option 1) 2.45×10−3 kg Option 2) 6.45×10−3 kg Option 3) 9.89×10−3 kg Option 4) 12.89×10−3 kg
An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range
Option 1) Option 2) Option 3) Option 4)
A particle is projected at 600 to the horizontal with a kinetic energy . The kinetic energy at the highest point
Option 1) Option 2) Option 3) Option 4)
In the reaction,
Option 1) at STP is produced for every mole consumed Option 2) is consumed for ever produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .
How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?
Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2
If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will
Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.
With increase of temperature, which of these changes?
Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.
Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is
Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023
A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is
Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9 |
## Learning and Applying Mathematics using Computing
### Theory: Lines
This post is a bit of background on lines, which we’ll be using in the next post to finish off our bus scenario.
Mathematics typically only deals with lines that are infinite in length. However, in programming we more often deal with lines that are finite, stretching from one endpoint to another. There are several different ways to represent infinite and finite lines, which we’ll look at in turn.
#### Infinite line: y = mx + c
You may well have come across the equation “$y = m \times x + c$” before. You can look at it in two ways. One is that it is a way to calculate y, given a value for x (and thus is a way to draw the line). The other is that it is a relation that holds true for points on the line, and is false otherwise. So let’s say you have $y = 3 \times x + 2$. If you plug in $x = 1, y = 5$ then you will find the equation is true: the point (1, 5) is on the line. But if you plug in $x = 1, y = 6$ then you will find the equation is false: the point (1, 6) is not on the line.
This form is usually used for straight lines on graphs because it expresses the relation between x and y. However, it has one major weakness: you can’t express a vertical line in this form, which only exists at one value of x. This makes it unsuitable for programming purposes, because you don’t want to have to program a special case for when a line happens to be vertical.
#### Infinite line: point and direction
Another way to express a line is as a combination of a point on the line (as a vector), and a direction. For example, a point on the line might be (0, 2), and the direction might be (1, 3). In mathematics it is often useful to combine these two into an expression which represents all points on the line. This is done by introducing a variable, which I’ll call scalar, to represent a form of distance along the line:
$\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 2 \end{pmatrix} + \text{scalar} \times \begin{pmatrix} 1 \\ 3 \end{pmatrix}$
The idea here is that all points on the line have a value for the scalar variable. So for example, (1.5, 6.5) is on the line: use $\text{scalar}=1.5$ and you get:
$\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 2 \end{pmatrix} + 1.5 \times \begin{pmatrix} 1 \\ 3 \end{pmatrix}$
$\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 2 \end{pmatrix} + \begin{pmatrix} 1.5 \\ 4.5 \end{pmatrix}$
$\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 1.5 \\ 6.5 \end{pmatrix}$
However, there’s no such value of $\text{scalar}$ for the point (2, 9), because it’s not on the line. Two slightly unexpected features of this representation are that it doesn’t matter which point on the line you pick for the expression, and you can multiply the direction vector by any constant without it changing the line. So the above line is equivalent to either of the following:
$\begin{pmatrix} 1 \\ 5 \end{pmatrix} + \text{scalarB} \times \begin{pmatrix} -1 \\ -3 \end{pmatrix}$
$\begin{pmatrix} -10 \\ -28 \end{pmatrix} + \text{scalarC} \times \begin{pmatrix} 6 \\ 18 \end{pmatrix}$
#### Finite line: two endpoints
The simplest way to represent a finite line is to simply write down its two endpoints. So you might have a line from (1, 3) to (6, 7). That’s a good, simple, unambiguous definition of the line. However, to do any calculations involving the line, you usually need to convert it to another representation, such as the last on our list:
#### Finite line: point and direction
You can take the two endpoints of a line and turn it into a form that looks identical to our earlier point and direction form. So using our (1, 3) to (6, 7) example, you get:
$\begin{pmatrix} 1 \\ 3 \end{pmatrix} + \text{scalar} \times \begin{pmatrix} 5 \\ 4 \end{pmatrix}$
The difference now is that the line extends from $\text{scalar}=0$ (the start-point) to $\text{scalar}=1$ (the end-point). Any values for the scalar variable outside the 0–1 range would be on the line if it’s infinite, but are not when it’s finite.
## End of the Line
Today’s post was a lot of theory without an actual application, but we’ll be using it in our next post when we’ll look at finding out if two finite lines intersect. |
# The Cube – B Cube Formula: A Comprehensive Guide
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When it comes to mathematics, formulas play a crucial role in solving complex problems. One such formula that often comes up in algebraic equations is the cube – b cube formula. In this article, we will explore the cube – b cube formula in detail, understand its significance, and learn how to apply it effectively.
## What is the Cube – B Cube Formula?
The cube – b cube formula is a mathematical expression used to simplify the difference of cubes. It is derived from the algebraic identity (a – b)(a^2 + ab + b^2) = a^3 – b^3. By factoring the difference of cubes, we can simplify complex expressions and solve equations more efficiently.
## Understanding the Components of the Formula
Before diving into the applications of the cube – b cube formula, let’s break down its components:
• a: Represents the first term or number in the equation.
• b: Represents the second term or number in the equation.
• a^3: Represents the cube of the first term.
• b^3: Represents the cube of the second term.
## Applications of the Cube – B Cube Formula
The cube – b cube formula finds its applications in various mathematical problems, including algebraic equations, simplification of expressions, and factorization. Let’s explore some practical examples to understand its usage:
### Example 1: Simplifying Expressions
Suppose we have the expression 8^3 – 2^3. By applying the cube – b cube formula, we can simplify it as follows:
(8 – 2)(8^2 + 8 * 2 + 2^2) = 6(64 + 16 + 4) = 6(84) = 504
Therefore, 8^3 – 2^3 simplifies to 504.
### Example 2: Solving Equations
Consider the equation x^3 – 27 = 0. To solve this equation, we can use the cube – b cube formula:
x^3 – 3^3 = 0
(x – 3)(x^2 + 3x + 9) = 0
Now, we can set each factor equal to zero:
x – 3 = 0 or x^2 + 3x + 9 = 0
Solving these equations, we find that x = 3 or x = -1.5 + 2.598i or x = -1.5 – 2.598i.
## Advantages of Using the Cube – B Cube Formula
The cube – b cube formula offers several advantages when it comes to solving mathematical problems:
• Simplification: By factoring the difference of cubes, complex expressions can be simplified, making calculations easier and more manageable.
• Efficiency: The formula allows for quicker and more efficient solving of equations, saving time and effort.
• Generalization: The cube – b cube formula can be applied to a wide range of problems, making it a versatile tool in algebraic manipulations.
## Common Mistakes to Avoid
While using the cube – b cube formula, it is essential to be aware of common mistakes that can lead to incorrect results. Here are a few mistakes to avoid:
• Incorrect Sign: Ensure that the signs are correctly applied when factoring the difference of cubes. A simple sign error can lead to an incorrect solution.
• Missing Terms: Double-check that all terms are included when applying the formula. Missing a term can result in an incomplete or inaccurate solution.
• Incorrect Order: Pay attention to the order of terms when factoring the difference of cubes. Switching the order can lead to incorrect results.
## Summary
The cube – b cube formula is a powerful tool in algebraic manipulations, allowing for the simplification of complex expressions and efficient solving of equations. By understanding its components and applications, you can enhance your problem-solving skills and tackle mathematical problems with confidence.
## Q&A
### Q1: Can the cube – b cube formula be applied to any two numbers?
A1: Yes, the cube – b cube formula can be applied to any two numbers. However, it is important to note that the formula is specifically designed for the difference of cubes.
A2: Yes, apart from the cube – b cube formula, there are other formulas related to cubes, such as the sum of cubes formula (a^3 + b^3 = (a + b)(a^2 – ab + b^2)) and the cube root formula (a^(1/3)). These formulas have their own applications and uses in mathematics.
### Q3: Can the cube – b cube formula be used in calculus?
A3: While the cube – b cube formula is primarily used in algebraic manipulations, it can also be applied in calculus when dealing with limits and derivatives involving cubes.
### Q4: Are there any real-world applications of the cube – b cube formula?
A4: The cube – b cube formula may not have direct real-world applications, but its underlying principles and concepts are used in various fields such as engineering, physics, and computer science. Understanding the formula can help in solving practical problems in these domains.
### Q5: Can the cube – b cube formula be extended to higher powers?
A5: No, the cube – b cube formula is specific to cubes and cannot be extended to higher powers. However, similar formulas exist for higher powers, such as the difference of fourth powers formula.
Navya Menon
Navya Mеnon is a tеch bloggеr and cybеrsеcurity analyst spеcializing in thrеat intеlligеncе and digital forеnsics. With еxpеrtisе in cybеr thrеat analysis and incidеnt rеsponsе, Navya has contributеd to strеngthеning cybеrsеcurity mеasurеs. |
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Angular Speed Calculator
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Angular speed is the distance covered in the circular path in the given time. In short it is the scalar measure of rotation rate.It differs by linear speed as it travels in a circular path instead of linear path.It is given by
$\omega$ = 2 $\pi$ f
Where $\omega$ = Angular speed and
f = frequency
The frequency is the number of rotations taken in the given time. Hence it is given by
f = $\frac{1}{T}$
Where T = time taken.
Angular speed Calculator
helps to calculate the frequency if angular speed is given and vice-versa. It also calculates the linear speed, angular or radius of circular path if any of the two quantities are given.
## Steps for Angular Speed Calculator
Step 1 :
Go through the problem and observe the given parameters.
Step 2 :
If frequency is given
Use the formula:
$\omega$ = 2 $\pi$ f
where $\omega$ = angular speed and
f = frequency taken
and if linear speed and radius of circular path is given
$\omega$ = $\frac{V}{r}$
where V = Linear speed and
r = radius of circular path
Substituting the given values we get the desired parameter.
## Problems on Angular Speed Calculator
1. ### A wheel of 0.60m in radius is moving with a speed of 10m/s. Find the angular speed.
Step 1 :
Given:
Linear speed V = 10 m/s
radius of wheel r = 0.6 m
Step 2 :
Using the formula:
$\omega$ = $\frac{V}{r}$
$\omega$ = $\frac{10 m/s}{0.6 m}$
The angular speed $\omega$ = 16.67 rad/s
2. ### The angular speed of a giant wheel ia 20 rad/s. Calculate its frequency.
Step 1 :
given: $\omega$ = 20 rad/s,
frequency f = ?
Step 2 :
Use the formula:
$\omega$ = 2 $\pi$ f
$\therefore$ f = $\frac{\omega}{2 \pi}$
f = $\frac{20 rad/s}{2 \pi}$ |
# Factors of 728 in pair
Factors of 728 in pair are (1, 728) , (2, 364) , (4, 182) , (7, 104) , (8, 91) , (13, 56) , (14, 52) and (26, 28)
#### How to find factors of a number in pair
1. Steps to find factors of 728 in pair 2. What is factors of a number in pair? 3. What are Factors? 4. Frequently Asked Questions 5. Examples of factors in pair
### Example: Find factors of 728 in pair
Factor Pair Pair Factorization
1 and 728 1 x 728 = 728
2 and 364 2 x 364 = 728
4 and 182 4 x 182 = 728
7 and 104 7 x 104 = 728
8 and 91 8 x 91 = 728
13 and 56 13 x 56 = 728
14 and 52 14 x 52 = 728
26 and 28 26 x 28 = 728
Since the product of two negative numbers gives a positive number, the product of the negative values of both the numbers in a pair factor will also give 728. They are called negative pair factors.
Hence, the negative pairs of 728 would be ( -1 , -728 ) .
#### What does factor pairs in mathematics mean?
In mathematics, factor pair of a number are all those possible combination which when multiplied together give the original number in return. Every natural number is a product of atleast one factor pair. Eg- Factors of 728 are 1 , 2 , 4 , 7 , 8 , 13 , 14 , 26 , 28 , 52 , 56 , 91 , 104 , 182 , 364 , 728. So, factors of 728 in pair are (1,728), (2,364), (4,182), (7,104), (8,91), (13,56), (14,52), (26,28).
#### What is the definition of factors?
In mathematics, factors are number, algebraic expressions which when multiplied together produce desired product. A factor of a number can be positive or negative.
#### Properties of Factors
• Each number is a factor of itself. Eg. 728 is a factor of itself.
• 1 is a factor of every number. Eg. 1 is a factor of 728.
• Every number is a factor of zero (0), since 728 x 0 = 0.
• Every number other than 1 has at least two factors, namely the number itself and 1.
• Every factor of a number is an exact divisor of that number, example 1, 2, 4, 7, 8, 13, 14, 26, 28, 52, 56, 91, 104, 182, 364, 728 are exact divisors of 728.
• Factors of 728 are 1, 2, 4, 7, 8, 13, 14, 26, 28, 52, 56, 91, 104, 182, 364, 728. Each factor divides 728 without leaving a remainder.
• Every factor of a number is less than or equal to the number, eg. 1, 2, 4, 7, 8, 13, 14, 26, 28, 52, 56, 91, 104, 182, 364, 728 are all less than or equal to 728.
#### Steps to find Factors of 728
• Step 1. Find all the numbers that would divide 728 without leaving any remainder. Starting with the number 1 upto 364 (half of 728). The number 1 and the number itself are always factors of the given number.
728 ÷ 1 : Remainder = 0
728 ÷ 2 : Remainder = 0
728 ÷ 4 : Remainder = 0
728 ÷ 7 : Remainder = 0
728 ÷ 8 : Remainder = 0
728 ÷ 13 : Remainder = 0
728 ÷ 14 : Remainder = 0
728 ÷ 26 : Remainder = 0
728 ÷ 28 : Remainder = 0
728 ÷ 52 : Remainder = 0
728 ÷ 56 : Remainder = 0
728 ÷ 91 : Remainder = 0
728 ÷ 104 : Remainder = 0
728 ÷ 182 : Remainder = 0
728 ÷ 364 : Remainder = 0
728 ÷ 728 : Remainder = 0
Hence, Factors of 728 are 1, 2, 4, 7, 8, 13, 14, 26, 28, 52, 56, 91, 104, 182, 364, and 728
#### Frequently Asked Questions
• Is 728 a composite number?
Yes 728 is a composite number.
• Is 728 a prime number?
No 728 is not a prime number.
• Is 728 a perfect square?
No 728 is not a perfect square.
• Write five multiples of 728.
Five multiples of 728 are 1456, 2184, 2912, 3640, 4368.
• Write all odd factors of 728?
The factors of 728 are 1, 2, 4, 7, 8, 13, 14, 26, 28, 52, 56, 91, 104, 182, 364, 728.
Odd factors of 728 are 1 , 7 , 13 , 91.
#### Examples of Factors
Ariel has been asked to write all factor pairs of 728 but she is finding it difficult. Can you help her find out?
Factors of 728 are 1, 2, 4, 7, 8, 13, 14, 26, 28, 52, 56, 91, 104, 182, 364, 728. So, factors of 728 in pair are (1,728), (2,364), (4,182), (7,104), (8,91), (13,56), (14,52), (26,28).
Sammy wants to write all the negative factors of 728 in pair, but don't know how to start. Help Sammy in writing all the factor pairs.
Negative factors of 728 are -1, -2, -4, -7, -8, -13, -14, -26, -28, -52, -56, -91, -104, -182, -364, -728. Hence, factors of 728 in pair are (-1,-728), (-2,-364), (-4,-182), (-7,-104), (-8,-91), (-13,-56), (-14,-52), (-26,-28).
Help Deep in writing the positive factors of 728 in pair and negative factor of 728 in pair.
Factors of 728 are 1, 2, 4, 7, 8, 13, 14, 26, 28, 52, 56, 91, 104, 182, 364, 728. Positive factors of 728 in pair are (1,728), (2,364), (4,182), (7,104), (8,91), (13,56), (14,52), (26,28). Negative factors of 728 in pair are (-1,-728), (-2,-364), (-4,-182), (-7,-104), (-8,-91), (-13,-56), (-14,-52), (-26,-28).
Find the product of all factors of 728.
Factors of 728 are 1, 2, 4, 7, 8, 13, 14, 26, 28, 52, 56, 91, 104, 182, 364, 728. So the product of all factors of 728 would be 1 x 2 x 4 x 7 x 8 x 13 x 14 x 26 x 28 x 52 x 56 x 91 x 104 x 182 x 364 x 728 = 7.889528230345372e+22.
Find the product of all prime factors of 728.
Factors of 728 are 1, 2, 4, 7, 8, 13, 14, 26, 28, 52, 56, 91, 104, 182, 364, 728. Prime factors are 2, 2, 2, 7, 13. So, the product of all prime factors of 728 would be 2 x 2 x 2 x 7 x 13 = 728.
Can you help Sammy list the factors of 728 and also find the factor pairs?
Factors of 728 are 1, 2, 4, 7, 8, 13, 14, 26, 28, 52, 56, 91, 104, 182, 364, 728.
Factors of 728 in pair are (1,728), (2,364), (4,182), (7,104), (8,91), (13,56), (14,52), (26,28).
Sammy has 728 blocks and he wants to arrange them in all possible ways to form a rectangle but he doesn't know the technique for doing that, help Sammy in arrangements.
To arrange 728 blocks in all possible ways to form a rectangle, we need to calculate factors of 728 in pair. Therefore, factors of 728 in pair are (1,728), (2,364), (4,182), (7,104), (8,91), (13,56), (14,52), (26,28)
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# Division with Money: Calculating Unit Prices
In this worksheet, students calculate the price of a single item, correctly using the pounds and pence signs.
Key stage: KS 2
Curriculum topic: Measurement
Curriculum subtopic: Solve Four Operations Problems
Difficulty level:
### QUESTION 1 of 10
Sometimes we see signs like this:
We can work out the price of one can by working out £2.88 ÷ 6.
So the price for one can or the UNIT PRICE is 48 p.
6 cans for £2.58
Work out the unit price of a can.
(Don't forget the p sign)
6 bottles for £5.04
Work out the unit price of a bottle.
(Don't forget the p sign)
8 bottles for £7.84
Work out the unit price of a bottle.
(Don't forget the p sign)
12 cakes for £7.92
Work out the unit price of a cake.
(Don't forget the p sign)
4 cakes for £2.92
Work out the unit price of a cake.
(Don't forget the p sign)
4 apples for £1.48
Work out the unit price of an apple.
(Don't forget the p sign)
6 eggs for £1.44
Work out the unit price of an egg.
(Don't forget the p sign)
6 cartons for £7.44
Work out the unit price of a carton.
(Don't forget the £ sign)
3 CDs for £26.97
Work out the unit price of a CD.
(Don't forget the £ sign)
7 bottles for £34.02
Work out the unit price of a bottle.
(Don't forget the £ sign)
• Question 1
6 cans for £2.58
Work out the unit price of a can.
(Don't forget the p sign)
43p
EDDIE SAYS
£2.58 ÷ 6 = 43p
• Question 2
6 bottles for £5.04
Work out the unit price of a bottle.
(Don't forget the p sign)
84p
EDDIE SAYS
£5.04 ÷ 6 = 84p
• Question 3
8 bottles for £7.84
Work out the unit price of a bottle.
(Don't forget the p sign)
98p
EDDIE SAYS
£7.84 ÷ 8 = 98p
• Question 4
12 cakes for £7.92
Work out the unit price of a cake.
(Don't forget the p sign)
66p
EDDIE SAYS
£7.92 ÷ 12 = 66p
• Question 5
4 cakes for £2.92
Work out the unit price of a cake.
(Don't forget the p sign)
73p
EDDIE SAYS
£2.92 ÷ 4 = 73p
• Question 6
4 apples for £1.48
Work out the unit price of an apple.
(Don't forget the p sign)
37p
EDDIE SAYS
£1.48 ÷ 4 = 37p
• Question 7
6 eggs for £1.44
Work out the unit price of an egg.
(Don't forget the p sign)
24p
EDDIE SAYS
£1.44 ÷ 6 = 24p
• Question 8
6 cartons for £7.44
Work out the unit price of a carton.
(Don't forget the £ sign)
£1.24
EDDIE SAYS
£7.44 ÷ 6 = £1.24
• Question 9
3 CDs for £26.97
Work out the unit price of a CD.
(Don't forget the £ sign)
£8.99
EDDIE SAYS
£26.97 ÷ 3 = £8.99
• Question 10
7 bottles for £34.02
Work out the unit price of a bottle.
(Don't forget the £ sign)
£4.86
EDDIE SAYS
£34.02 ÷ 7 = £4.86
---- OR ----
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# 2.10: Identify and Apply Number Properties in Fraction Operations
Difficulty Level: At Grade Created by: CK-12
Estimated8 minsto complete
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Progress
Practice Fractions
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Do you know how to simplify an expression with fractions by using properties? Take a look at this dilemma.
Simplify: 23×(27×32)\begin{align*}\frac{2}{3}\times\left(\frac{2}{7}\times\frac{3}{2}\right)\end{align*}
To simplify this expression, you will need to know how to work with number properties and fractions. This Concept will show you how to accomplish this task successfully.
### Guidance
Now that we are working with fractions, you will have a chance to investigate how the different properties of addition and subtraction can help us when we work with fractions in expressions.
Here are a few properties.
The sum of any number and zero is that number: 311+0=311\begin{align*}\frac{3}{11}+0=\frac{3}{11}\end{align*}
The sum of any number and its inverse is zero: 34+(34)=0\begin{align*}\frac{3}{4}+ \left( - \frac{3}{4}\right)=0\end{align*}
Which of the following shows the Additive Inverse Property?
a. xy+0=0\begin{align*}\frac{x}{y}+0=0\end{align*}
b. xy+0=xy\begin{align*}\frac{x}{y}+0=\frac{x}{y}\end{align*}
c. xy+(xy)=0\begin{align*}\frac{x}{y}+\left(-\frac{x}{y}\right)=0\end{align*}
Consider choice a.
This equation states that a number added to zero is equal to zero. This is not necessarily correct, unless xy\begin{align*}\frac{x}{y}\end{align*} is also equal to zero.
Consider choice b.
This equation states that the sum of a number and zero is equal to that number. This is correct, but illustrates the additive identity, not the additive inverse property.
Consider choice c.
This equation states that the sum of a number and its inverse is equal to zero. This illustrates the additive inverse property, so this is the correct equation.
We can also use these properties to help us when we simplify a numerical expression. Remember that a numerical expression is a group of numbers and operations. Because we are working with fractions, the numerical expressions in this section will be made up of fractions.
Simplify: 18+(14+38)\begin{align*}\frac{1}{8}+\left(\frac{1}{4}+\frac{3}{8}\right)\end{align*}
You can use addition properties to reorganize this expression to make it easier to simplify.
First apply the commutative property: 18+(14+38)=18+(38+14)\begin{align*}\frac{1}{8}+\left(\frac{1}{4} + \frac{3}{8}\right)=\frac{1}{8}+\left(\frac{3}{8}+\frac{1}{4}\right)\end{align*}
Then apply the associate property: 18+(38+14)=(18+38)+14\begin{align*}\frac{1}{8}+\left(\frac{3}{8} + \frac{1}{4}\right)=\left ( \frac{1}{8}+\frac{3}{8}\right)+\frac{1}{4}\end{align*}
Now you can easily simplify to find the sum.
(18+38)+14=48+14=24+14=34\begin{align*}\left(\frac{1}{8}+\frac{3}{8}\right)+\frac{1}{4}=\frac{4}{8}+\frac{1}{4}=\frac{2}{4}+\frac{1}{4}=\frac{3}{4}\end{align*}
The answer is 34\begin{align*}\frac{3}{4}\end{align*}.
Using properties to reorganize fractions can help us to work with these fractions. Notice that we reorganized the common denominators together and this simplified our work.
Now let’s look at how the properties of multiplication and division can help us when working with fractions. You have already learned the Commutative Property, the Associative Property, and the Distributive Property.
Multiplicative Identity
The product of any number and one is that number: 311×1=311\begin{align*}\frac{3}{11}\times 1=\frac{3}{11}\end{align*}
Zero Property
The product of any number and zero is zero: 47×0=0\begin{align*}\frac{4}{7}\times 0=0\end{align*}
Multiplicative Inverse
The product of any number and its reciprocal is one: 34×43=1\begin{align*}\frac{3}{4}\times\frac{4}{3}=1\end{align*}
Which of the following shows the Multiplicative Inverse Property?
a. xy×0=xy\begin{align*}\frac{x}{y}\times 0=\frac{x}{y}\end{align*}
b. xy×yx=0\begin{align*}\frac{x}{y}\times \frac{y}{x}=0\end{align*}
c. xy×yx=1\begin{align*}\frac{x}{y}\times \frac{y}{x}=1\end{align*}
Consider choice a.
This equation states that the product of a number and zero is equal to that number. This is not correct.
Consider choice b.
This equation states that the product of a number and its reciprocal is equal to zero. This is not correct.
Consider choice c.
This equation states that the product of a number and its reciprocal is equal to one. This illustrates the multiplicative inverse property, so this is the correct equation.
Take a few minutes to write these properties down in your notebooks. Be sure to include an example of each.
We could also use properties when working with a variable. Take a look at this one.
23×(a×32)\begin{align*}\frac{2}{3}\times\left(a \times \frac{3}{2}\right)\end{align*}
First, we can apply the commutative property: 23×32×a\begin{align*}\frac{2}{3} \times \frac{3}{2} \times a\end{align*}
Now we apply the multiplication inverse property: 23×32=1\begin{align*}\frac{2}{3} \times \frac{3}{2} =1\end{align*}
Our simplified expression is a\begin{align*}a\end{align*}.
If we had a value substituted in for a\begin{align*}a\end{align*}, then that would be our answer.
#### Example A
Name the property: 38×0\begin{align*}\frac{3}{8} \times 0\end{align*}
Solution: Zero Property
#### Example B
Name the property: 56×65\begin{align*}\frac{5}{6} \times \frac{6}{5}\end{align*}
Solution: Multiplicative Inverse
#### Example C
Simplify: 34(b×43)\begin{align*}\frac{3}{4} (b \times \frac{4}{3})\end{align*}
Solution: a\begin{align*}a\end{align*}
Now let's go back to the dilemma from the beginning of the Concept.
Simplify: 23×(27×32)\begin{align*}\frac{2}{3}\times\left(\frac{2}{7}\times\frac{3}{2}\right)\end{align*}
You can use multiplication properties to reorganize this expression to make it easier to simplify.
First apply the commutative property: 23×(27×32)=23×(32×27)\begin{align*}\frac{2}{3}\times\left(\frac{2}{7}\times\frac{3}{2}\right)=\frac{2}{3} \times \left(\frac{3}{2} \times \frac{2}{7}\right)\end{align*}
Then apply the associate property: 23×(32×27)=(23×32)×27\begin{align*}\frac{2}{3}\times\left(\frac{3}{2}\times\frac{2}{7}\right)=\left(\frac{2}{3} \times \frac{3}{2}\right) \times \frac{2}{7}\end{align*}
Then apply the multiplicative inverse property: (23×32)×27=1×27\begin{align*}\left(\frac{2}{3}\times\frac{3}{2}\right)\times\frac{2}{7}=1\times\frac{2}{7}\end{align*}
Finally, apply the multiplicative identity property: 1×27=27\begin{align*}1 \times \frac{2}{7}=\frac{2}{7}\end{align*}
### Vocabulary
any number plus zero is still that number.
any number plus it’s opposite or inverse is equal to 0.
Multiplicative Identity
any number times 1 is the same number.
Zero Property
any number times 0 is zero.
Multiplicative Inverse
any number times it’s reciprocal is 1.
### Guided Practice
Here is one for you to try on your own.
Simplify: 45+12+x\begin{align*}\frac{4}{5} + \frac{1}{2} + x\end{align*}
Solution
First, we find a common denominator so that we can add the fractions. The lowest common denominator for 5 and 2 is 10. Let's rename both fractions in terms of tenths.
810+510+x\begin{align*}\frac{8}{10} + \frac{5}{10} + x\end{align*}
Now we can add the fractions.
1310+x\begin{align*}\frac{13}{10} + x\end{align*}
Change the improper fraction to a mixed number.
1310+x\begin{align*}1 \frac{3}{10} + x\end{align*}
This is our simplified expression.
### Practice
Directions: Identify each property shown below.
1. 34+0=34\begin{align*}\frac{3}{4} + 0 = \frac{3}{4}\end{align*}
2. 34+34=0\begin{align*}\frac{3}{4} + -\frac{3}{4} = 0\end{align*}
3. 67×0=0\begin{align*}\frac{6}{7} \times 0 = 0\end{align*}
4. 58×0=0\begin{align*}\frac{5}{8} \times 0 = 0\end{align*}
5. 67×76=1\begin{align*}\frac{6}{7} \times \frac{7}{6} = 1\end{align*}
6. 34+x=x+34\begin{align*}\frac{3}{4} + x = x + \frac{3}{4}\end{align*}
7. 14+y=y+14\begin{align*}\frac{1}{4} + y = y + \frac{1}{4}\end{align*}
8. 12×(x+3)=12x+12(3)\begin{align*}\frac{1}{2} \times (x + 3)= \frac{1}{2}x + \frac{1}{2}(3)\end{align*}
Directions: Simplify each expression.
9. 34+14+x\begin{align*}\frac{3}{4} + \frac{1}{4} + x\end{align*}
10. 67×13×x\begin{align*}\frac{6}{7} \times \frac{1}{3} \times x\end{align*}
11. 2×48x\begin{align*}2 \times \frac{4}{8}x\end{align*}
12. 3x×68\begin{align*}3x \times \frac{6}{8}\end{align*}
13. 45+12+610\begin{align*}\frac{4}{5} + \frac{1}{2} + \frac{6}{10}\end{align*}
14. 61013\begin{align*}\frac{6}{10} - \frac{1}{3}\end{align*}
15. \begin{align*}\frac{1}{2} \times 3x\end{align*}
### Vocabulary Language: English
The sum of any number and zero is the number itself.
The additive inverse or opposite of a number x is -1(x). A number and its additive inverse always sum to zero.
Multiplicative Identity
Multiplicative Identity
The multiplicative identity for multiplication of real numbers is one.
Multiplicative Inverse
Multiplicative Inverse
The multiplicative inverse of a number is the reciprocal of the number. The product of a real number and its multiplicative inverse will always be equal to 1 (which is the multiplicative identity for real numbers).
Zero Property
Zero Property
The zero property of multiplication says that the product of any number and zero is zero. The zero property of addition states that the sum of any number and zero is the number.
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Date Created:
Dec 19, 2012 |
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