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# Algebra: Fraction Problems
In these lessons, we will learn how to solve fraction word problems that deal with fractions i.e. parts of a whole. Remember to read the question carefully to determine the numerator and denominator of the fraction.
We will also learn how to solve word problems that involve comparing fractions, adding mixed numbers, subtracting mixed numbers, multiplying fractions and dividing fractions.
Related Topics: More Algebra Word Problems
## Fraction Word Problems
Example 1:
A class has 20 girls and 30 boys. What part of the class are boys?
Solution:
Step 1: Numerator: boys = 30
Step 2: Denominator: class = 20 + 30 = 50
Step 3: Part or fraction
Example 2:
If John earns \$x in a week and spend \$y, what part of his weekly salary did he save?
Solution:
Step 1: Numerator: amount saved= xy
Step 2: Denominator: salary = x
Step 3: Part or fraction
Example 3:
14 is of what number?
Solution:
Step 1: Assign variables :
Let x = number
Step 2: Solve the equation
Isolate variable x
## Comparing Fractions Word Problems
The following video shows an example of fraction word problem that involves comparing fractions.
On the last math quiz, Tara answered 5/7 of the problems correctly and Julia answered 7/10 of them correctly. If each problem is worth the same amount, who got the higher grade?
## Adding Mixed Numbers Word Problems
Regina rode her bike 2 and 1/4 miles from her house to school and the 1 and 5/8 miles to her friend's house. How many miles did Regina ride in total?
## Subtracting Mixed Numbers Word Problems
If Shelagh describe her age as 15 and 1/2 years old and Luana is 7 and 7/12 years old, what is the difference in their ages? Write the answer as a simplified mixed number.
## Multiplying Fractions Word Problems
The following video is a fraction word problem that involves multiplying fractions.
A recipe for banana oat muffin calls for 3/4 cup of old fashioned oats. You are making have of the recipe. How much oats should you use?
Before leaving for a road trip, Tiffany filled up her gas rank, which holds 16 gallons of gas. After 4 hours, Tiffany notices that the gas tank was 7/8 full. How many gallons of gas were left in the tank?
## Dividing Fractions Word Problems
The following videos are fraction word problem that involve dividing fractions.
A baby t-shirt requires 4/5 yards of fabric. How many t-shirts can be made from 48 yards?
Diya thought that it would be nice to include 2/21 of a pound of chocolate in each of the holiday gift bags she made for her friends and family. How many holiday gift bags could Diya make with 2/3 of a pound of chocolate?
As the swim coach, Minli selects which athletes will participate in the state-wide swim relay. The relay team swims 2/3 of a mile altogether, with each team member responsible for swimming 2/9 of a mile. The team must complete the swim in 2/5 of an hour. How many swimmers does Minli need on the relay team?
You can use the Mathway widget below to practice Algebra or other math topics. Try the given examples, or type in your own problem. Then click "Answer" to check your answer.
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## Tuesday, January 10, 2012
### Math Textbook
Question 6:
Step 1: Find what makes 64, so 8x8=64
Step 2: Find what makes 8, 4x4=8
Step 3: Find what makes 4, 2x2,
Step 4: Circle the prime numbers
64 is a perfect square number because you can multiply 8 by it's self to make 64.
Another way of saying that is...the square root of 64 is 8, it is a whole number, so it is a perfect square.
The area: 64m². You have to remember to draw the "tickeys" to show that it is a "real" square.
11)
12)
15c)
(Blue is your question, Red is your answer.)
16c)
(Blue is your question, Red is your answer.)
19)
''
Step 1: Find the square root of 28 900
Step 2: The square root will tell what the length and width is
Step 3: Add the side together to find the perimeter which is 680
Step4: Then, the teacher told the students to do two laps, so you would multiply 680 by 2
that will give you 1360m
22)
Remember to play MangaHigh! :)
---------------------------------------------------------------------
a² + b² = c²
That formula means, that if you take the legs of the triangle, square them, then add them, you will find (c²) which is the hypotenuse². So if you have a right triangle, the legs² will equal the hypotenuse².
#### 6 comments:
1. Good Job on your scribe Mary. I like how you explained your answers step by step, so we can understand it better, and how it's easy to see. The video was also helpful. On question 15c) it's supposed to be 625 because 25x25=625. Thanks for reminding us to play mangahigh. Keep up the good work:)
2. Hi Mary, good job on your scribe, i like how all your answers were all organized into one or two pictures and Ty for reminding me to play mangahigh, keep up the good work
3. Hi Mary! You did a great job on your post. It had good pictures and a helpful video. There was only one problem though. It was just like what casey said, 265 isn't 25 because 25x25=625. Other than that it was great. Keep up the good work Mary!
4. Hi Mary. Great scribe post! Everything is organized. I like how you explained most if them in steps! I also like how you added borders around your questions and answers. Here is a link: http://www.mathisfun.com/square-root.html
Good Job!
5. HI Mary!! Great job i like how you did the prime factorization and i like the colors of the squares...it's like a rainbow :) .....GREAT JOB Mary :)
6. :(... I Made a typo on the very last question, it's suppose to be 100 is not equal to 121.
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# 7.1: Percent
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When one hears the word “percent,” other words come immediately to mind, words such as “century,” “cents,” or “centimeters.” A century equals 100 years. There are one hundred cents in a dollar and there are 100 centimeters in a meter. Thus, it should come as no surprise that percent means “parts per hundred.”
In the world we live in we are constantly bombarded with phrases that contain the word “percent.” The sales tax in California is 8.25%. An employee is asking his boss for a 5% raise. A union has seen a 6.25% increase in union dues. The population of a town is increasing at a rate of 2.25% per year.
In this chapter we introduce the concept of percent, first addressing how to facilitate writing percents in fraction or decimal form and also performing the reverse operations, changing fractions and decimals to percents. Next we use our expertise in solving equations to solve the more common forms that involve percents, then we apply this ability to solving common applications from the real world that use percents. We’ll tackle applications of commission and sales tax, discount and marked price, percent increase or decrease, and simple interest.
Let’s begin the journey
This page titled 7.1: Percent is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Arnold.
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# Unraveling the Pythagorean Theorem: Formula, Proofs, and Practical Examples
Comprehensive Definition, Description, Examples & RulesÂ
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## Introduction to Pythagoras
Among the many names from the history of mathematics, only a few have the same depth as that of the ancient Greek mathematician and philosopher Pythagoras. Pythagoras was born in about 570 BCE on the island of Samos and founded a religious movement that unified mystic teaching with mathematical accomplishments. The Pythagorean Theorem is one of his lasting contributions to prove that mathematical principles are eternal. This is one of the most important geometrical notions having its application in everyday life. The first exploration in this paper examines the Pythagorean Theorem, its formulas, proofs, and applications.
## Pythagoras Theorem Explained
### Definition and Explanation
Accordingly, the Pythagorean theorem is an important fundamental principle that defines the relationship of the triangle’s sides of the right triangle. The right-angled triangle depicted above encompasses a timeless theorem embodied by the Pythagoras theorem formula a^{2} + b^{2} = c^{2}, where a and b represent the lengths of the smaller sides and c is the hypotenuse’s length.
This very basic formula can be very misleading and has much deeper meaning in the area of geometry. It is a foundation for many mathematical concepts which lead to other advanced topics. At the heart, the theorem states that the square of the hypotenuse is equivalent to the sum of the squares of the other two sides.
### Visual Representation and Basic Geometry
In understanding the Pythagorean Theorem, it would be great to picture its geometrical implications. Suppose a right-angled triangle of ‘a,’ ‘b,’ and ‘c’, of which ‘c’ is a hypotenuse. Squaring ‘a’ and ‘b’ and adding them will result in the square of ‘c.’ In a nutshell, this can be illustrated as squares drawn on the sides of a triangle.
Constructing squares on each side of the triangle yields areas a², b², and c². The Pythagorean theorem states that the sum of the areas of the squares formed by ‘a’ and ‘b’ is equal to the area of the square formed by ‘c.’ This geometric explanation justifies the importance of the theorem and facilitates its use in practice.
## Pythagoras Theorem Examples
The Pythagorean Theorem Examples transcends the classroom and applies in real life. In architecture, navigation, and engineering where precision is necessary, its operational capacity is practical.
### Architecture
The Pythagorean theorem directly impacts the stability and integrity of architectural designs. The dimensions are also necessary for a roof that guides architects and engineers. Applying the theorem in determining the lengths of structural components helps architects come up with good-looking and structurally stable roof designs.
The significance of the theorem even goes as far as affecting navigation, an important area in seafaring and aircraft. The Pythagorean theorem therefore forms an important element of navigaÂtional calculations, including distance measuring and course plotting. It is important to understand the distance relations whether one is sailing in the open sea or developing an air route.
### Engineering
The Pythagorean Theorem is a basic tool that engineers use in designing several structures. Engineers use this theorem in designing bridges and buildings to make sure they are well-balanced and have stability. The theorem also enables one to calculate distances between components on circuit boards which facilitates the operation of electronic devices.
## Proof of Pythagoras Theorem
### Exploring Different Proofs
There are many proofs about the Pythagorean Theorem, each shedding light on its correctness. Often such geometric proofs are associated with ancient scholars like Euclid making use of visual demonstrations using squares and triangles. Thus, by showing that the areas of squares constructed on the sides of a right triangle are equal to the area of the triangle, the Euclidean proof of Pythagoras’ theorem is achieved.
This alternative perspective comes from algebraic proofs which were introduced much later. The Pythagorean Theorem is considered a cornerstone of mathematics because mathematicians have used algebraic expressions representing the sides of the triangle to derive pythagoras theorem proof that validate this theorem. Using these algebraic approaches helps in understanding the inner concept of the theorem.
On the other hand, visual proofs emerged as a more recent phenomenon. They employ diagrams and animations that are continuously used to depict the theorem. This modern perspective promotes ease of use, addressing various learning modalities through multimedia illustration of a proof’s soundness.
### Historical and Modern Perspectives
Proving the Pythagorean theorem, however, has followed a similar route as the evolution of mathematics itself. Every proof is a testament to the smartness of mathematicians at different times, right from ancient geometric demonstrations to algebraic formulations. These proofs are timeless and represent the universality of the Pythagorean Theorem which is valid beyond cultures and epochs.
## Applications and Extensions
The Pythagorean Theorem takes more complicated aspects of mathematics into account, building on its basic use. For example, trigonometry heavily uses the theorem to establish basic identities as well as relations between angles and sides. Trigonometry involves the sine, cosine, and tangent functions rooted in Pythagorean relationships among sides of right-angled triangles.
Modified Pythagorean Theorems expand their use on nontraditional shapes. Pythagorean triples constitute a vast treasure chest full of possibilities to solve new problems as well as an immense opportunity to discover brand-new phenomena in mathematics. It demonstrates the flexibility of the theorem in many mathematical contexts and its longevity.
### Geometry and Beyond
The Pythagorean Theorem is the most fundamental concept in geometry that explains various aspects of space and dimensions. It serves as a base upon which other theorems and conjectures are based. Geometric similarity has its basis in ‘similar right-angled triangles’ where the ratios of the sides correspond with the ‘Pythagorean Theorem’.
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## Key Takeaways
1. Pythagorean Theorem: a2+b2=c2a2+b2=c2.
2. Practical Applications: Architecture, navigation, engineering.
3. Proofs: Geometric, algebraic, visual.
5. Significance: Geometry, basic principles of geometry, widespread usefulness for the real world.
## Quiz
Check your score in the end
Quiz
Check your score in the end
Question of
#### Question comes here
The Pythagorean theorem determines the relationship between the sides of a right-angled triangle. It means that the hypotenuse squared equals the sum of the squares of the other two sides.
The Pythagorean Theorem is used in architecture to construct structures, navigation for estimating lengths, and engineering for equilibrium. It permeates several domains in which accurate dimensions and spatial arrangements matter.
No, the Pythagorean Theorem deals only with right-angled triangles. Nevertheless, different kinds of theorems can be modified in order not to limit them to some mathematical aspects. The law of cosines and the law of sines are more relevant for non-right-angled triangles.
Yes, there are various proofs of the Pythagorean Theorem. The geometric proof, the algebraic proof, and the visual proof all provide different angles on the validity of the theorem. Diversification of proofs speaks volumes about the differentness and multidisciplinary character of mathematical thinking through centuries.
Pythagoras theorem serves as one of the critical principles in the foundation of geometry. It introduces higher mathematics and affects many other ones like trigonometry and number theory. Its wide relevance and applicability prove the theorem’s importance for mathematics in general.
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# 1.2: Special Right Triangles
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Recognize special right triangles.
• Use the special right triangle ratios to solve special right triangles.
## Special Right Triangle #1: Isosceles Right Triangle
An isosceles right triangle is an isosceles triangle and a right triangle. This means that it has two congruent sides and one right angle. Therefore, the two congruent sides must be the legs.
Because the two legs are congruent, we will call them both \begin{align*}a\end{align*} and the hypotenuse \begin{align*}c\end{align*}. Plugging both letters into the Pythagorean Theorem, we get:
\begin{align*}a^2 + a^2 & = c^2\\ 2a^2 & = c^2\\ \sqrt{2a^2} & = \sqrt{c^2}\\ a\sqrt{2} & = c\end{align*}
From this we can conclude that the hypotenuse length is the length of a leg multiplied by \begin{align*}\sqrt{2}\end{align*}. Therefore, we only need one of the three lengths to determine the other two lengths of the sides of an isosceles right triangle. The ratio is usually written \begin{align*}x:x:x\sqrt{2}\end{align*}, where \begin{align*}x\end{align*} is the length of the legs and \begin{align*}x\sqrt{2}\end{align*} is the length of the hypotenuse.
Example 1: Find the lengths of the other two sides of the isosceles right triangles below.
a.
b.
c.
d.
Solution:
a. If a leg has length 8, by the ratio, the other leg is 8 and the hypotenuse is \begin{align*}8\sqrt{2}\end{align*}.
b. If the hypotenuse has length \begin{align*}7\sqrt{2}\end{align*}, then both legs are 7.
c. Because the leg is \begin{align*}10\sqrt{2}\end{align*}, then so is the other leg. The hypotenuse will be \begin{align*}10\sqrt{2}\end{align*} multiplied by an additional \begin{align*}\sqrt{2}\end{align*}.
\begin{align*}10\sqrt{2} \cdot \sqrt{2} = 10 \cdot 2 = 20\end{align*}
d. In this problem set \begin{align*}x\sqrt{2} = 9\sqrt{6}\end{align*} because \begin{align*}x\sqrt{2}\end{align*} is the hypotenuse portion of the ratio.
\begin{align*}x\sqrt{2} & = 9\sqrt{6}\\ x & = \frac{9\sqrt{6}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{9\sqrt{12}}{2} = \frac{18\sqrt{3}}{2} = 9\sqrt{3}\end{align*}
So, the length of each leg is \begin{align*}9\sqrt{3}\end{align*}.
What are the angle measures in an isosceles right triangle? Recall that the sum of the angles in a triangle is \begin{align*}180^\circ\end{align*} and there is one \begin{align*}90^\circ\end{align*} angle. Therefore, the other two angles add up to \begin{align*}90^\circ\end{align*}. Because this is an isosceles triangle, these two angles are equal and \begin{align*}45^\circ\end{align*} each. Sometimes an isosceles right triangle is also referred to as a \begin{align*}45-45-90\end{align*} triangle.
## Special Right Triangle #2: 30-60-90 Triangle
\begin{align*}30-60-90\end{align*} refers to each of the angles in this special right triangle. To understand the ratios of the sides, start with an equilateral triangle with an altitude drawn from one vertex.
Recall from geometry, that an altitude, \begin{align*}h\end{align*}, cuts the opposite side directly in half. So, we know that one side, the hypotenuse, is \begin{align*}2s\end{align*} and the shortest leg is \begin{align*}s\end{align*}. Also, recall that the altitude is a perpendicular and angle bisector, which is why the angle at the top is split in half. To find the length of the longer leg, use the Pythagorean Theorem:
\begin{align*}s^2 + h^2 & = (2s)^2\\ s^2 + h^2 & = 4s^2\\ h^2 & = 3s^2\\ h & = s\sqrt{3}\end{align*}
From this we can conclude that the length of the longer leg is the length of the short leg multiplied by \begin{align*}\sqrt{3}\end{align*} or \begin{align*}s\sqrt{3}\end{align*}. Just like the isosceles right triangle, we now only need one side in order to determine the other two in a \begin{align*}30-60-90\end{align*} triangle. The ratio of the three sides is written \begin{align*}x:x\sqrt{3}:2x\end{align*}, where \begin{align*}x\end{align*} is the shortest leg, \begin{align*}x\sqrt{3}\end{align*} is the longer leg and \begin{align*}2x\end{align*} is the hypotenuse.
Notice, that the shortest side is always opposite the smallest angle and the longest side is always opposite \begin{align*}90^\circ\end{align*}.
If you look back at the Review Questions from the last section we now recognize #8 as a \begin{align*}30-60-90\end{align*} triangle.
Example 2: Find the lengths of the two missing sides in the \begin{align*}30-60-90\end{align*} triangles.
a.
b.
c.
d.
Solution: Determine which side in the \begin{align*}30-60-90\end{align*} ratio is given and solve for the other two.
a. \begin{align*}4\sqrt{3}\end{align*} is the longer leg because it is opposite the \begin{align*}60^\circ\end{align*}. So, in the \begin{align*}x:x\sqrt{3}:2x\end{align*} ratio, \begin{align*}4\sqrt{3} = x\sqrt{3}\end{align*}, therefore \begin{align*}x = 4\end{align*} and \begin{align*}2x = 8\end{align*}. The short leg is 4 and the hypotenuse is 8.
b. 17 is the hypotenuse because it is opposite the right angle. In the \begin{align*}x:x\sqrt{3}:2x\end{align*} ratio, \begin{align*}17 = 2x\end{align*} and so the short leg is \begin{align*}\frac{17}{2}\end{align*} and the long leg is \begin{align*}\frac{17\sqrt{3}}{2}\end{align*}.
c. 15 is the long leg because it is opposite the \begin{align*}60^\circ\end{align*}. Even though 15 does not have a radical after it, we can still set it equal to \begin{align*}x\sqrt{3}\end{align*}.
\begin{align*}x\sqrt{3} & = 15\\ x & = \frac{15}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{15\sqrt{3}}{3} = 5\sqrt{3} \quad \text{So, the short leg is} \ 5\sqrt{3}.\end{align*}
Multiplying \begin{align*}5\sqrt{3}\end{align*} by 2, we get the hypotenuse length, which is \begin{align*}10\sqrt{3}\end{align*}.
d. \begin{align*}24\sqrt{21}\end{align*} is the length of the hypotenuse because it is opposite the right angle. Set it equal to \begin{align*}2x\end{align*} and solve for \begin{align*}x\end{align*} to get the length of the short leg.
\begin{align*}2x & = 24\sqrt{21}\\ x &= 12\sqrt{21}\end{align*}
To find the length of the longer leg, we need to multiply \begin{align*}12\sqrt{21}\end{align*} by \begin{align*}\sqrt{3}\end{align*}.
\begin{align*}12\sqrt{21} \cdot \sqrt{3} = 12\sqrt{3 \cdot 3 \cdot 7} = 36\sqrt{7}\end{align*}
The length of the longer leg is \begin{align*}36\sqrt{7}\end{align*}.
Be careful when doing these problems. You can always check your answers by finding the decimal approximations of each side. For example, in 2d, short leg \begin{align*}= 12\sqrt{21} \approx 54.99\end{align*}, long leg \begin{align*}= 36\sqrt{7} \approx 95.25\end{align*} and the hypotenuse \begin{align*}= 24\sqrt{21} \approx 109.98\end{align*}. This is an easy way to double-check your work and verify that the hypotenuse is the longest side.
## Using Special Right Triangle Ratios
Special right triangles are the basis of trigonometry. The angles \begin{align*}30^\circ, \ 45^\circ, \ 60^\circ\end{align*} and their multiples have special properties and significance in the unit circle (sections 1.5 and 1.6). Students are usually required to memorize these two ratios because of their importance.
First, let’s compare the two ratios, so that we can better distinguish the difference between the two. For a \begin{align*}45-45-90\end{align*} triangle the ratio is \begin{align*}x:x:x\sqrt{2}\end{align*} and for a \begin{align*}30-60-90\end{align*} triangle the ratio is \begin{align*}x:x\sqrt{3}:2x\end{align*}. An easy way to tell the difference between these two ratios is the isosceles right triangle has two congruent sides, so its ratio has the \begin{align*}\sqrt{2}\end{align*}, whereas the \begin{align*}30-60-90\end{align*} angles are all divisible by 3, so that ratio includes the \begin{align*}\sqrt{3}\end{align*}. Also, if you are ever in doubt or forget the ratios, you can always use the Pythagorean Theorem. The ratios are considered a short cut.
Example 3: Determine if the sets of lengths below represent special right triangles. If so, which one?
a. \begin{align*}8\sqrt{3}:24:16\sqrt{3}\end{align*}
b. \begin{align*}\sqrt{5}:\sqrt{5}:\sqrt{10}\end{align*}
c. \begin{align*}6\sqrt{7}:6\sqrt{21}:12\end{align*}
Solution:
a. Yes, this is a \begin{align*}30-60-90\end{align*} triangle. If the short leg is \begin{align*}x = 8\sqrt{3}\end{align*}, then the long leg is \begin{align*}8\sqrt{3} \cdot \sqrt{3} = 8 \cdot 3 = 24\end{align*} and the hypotenuse is \begin{align*}2 \cdot 8\sqrt{3} = 16\sqrt{3}\end{align*}.
b. Yes, this is a \begin{align*}45-45-90\end{align*} triangle. The two legs are equal and \begin{align*}\sqrt{5} \cdot \sqrt{2} = \sqrt{10}\end{align*}, which would be the length of the hypotenuse.
c. No, this is not a special right triangle, nor a right triangle. The hypotenuse should be \begin{align*}12\sqrt{7}\end{align*} in order to be a \begin{align*}30-60-90\end{align*} triangle.
## Points to Consider
• What is the difference between Pythagorean triples and special right triangle ratios?
• Why are these two ratios considered “special”?
## Review Questions
Solve each triangle using the special right triangle ratios.
1. A square window has a diagonal of 6 ft. To the nearest hundredth, what is the height of the window?
2. Pablo has a rectangular yard with dimensions 10 ft by 20 ft. He is decorating the yard for a party and wants to hang lights along both diagonals of his yard. How many feet of lights does he need? Round your answer to the nearest foot.
3. Can \begin{align*}2:2:2\sqrt{3}\end{align*} be the sides of a right triangle? If so, is it a special right triangle?
4. Can \begin{align*}\sqrt{5}:\sqrt{15}:2\sqrt{5}\end{align*} be the sides of a right triangle? If so, is it a special right triangle?
1. Each leg is \begin{align*}\frac{16}{\sqrt{2}} = \frac{16}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{16\sqrt{2}}{2} = 8\sqrt{2}\end{align*}.
2. Short leg is \begin{align*}\frac{\sqrt{6}}{\sqrt{3}} = \sqrt{\frac{6}{3}} = \sqrt{2}\end{align*} and hypotenuse is \begin{align*}2\sqrt{2}\end{align*}.
3. Short leg is \begin{align*}\frac{12}{\sqrt{3}} = \frac{12}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{12\sqrt{3}}{3} = 4\sqrt{3}\end{align*} and hypotenuse is \begin{align*}8\sqrt{3}\end{align*}.
4. The hypotenuse is \begin{align*}4\sqrt{10} \cdot \sqrt{2} = 4\sqrt{20} = 8\sqrt{5}\end{align*}.
5. Each leg is \begin{align*}\frac{5\sqrt{2}}{\sqrt{2}} = 5\end{align*}.
6. The short leg is \begin{align*}\frac{15}{2}\end{align*} and the long leg is \begin{align*}\frac{15\sqrt{3}}{2}\end{align*}.
7. If the diagonal of a square is 6 ft, then each side of the square is \begin{align*}\frac{6}{\sqrt{2}}\end{align*} or \begin{align*}3\sqrt{2} \approx 4.24 \ ft\end{align*}.
8. These are not dimensions for a special right triangle, so to find the diagonal (both are the same length) do the Pythagorean Theorem: \begin{align*}10^2 + 20^2 & = d^2\\ 100 + 400 & = d^2\\ \sqrt{500} & = d\\ 10\sqrt{5} & = d\end{align*} So, if each diagonal is \begin{align*}10\sqrt{5}\end{align*}, two diagonals would be \begin{align*}20\sqrt{5} \approx 45 \ ft.\end{align*} Pablo needs 45 ft of lights for his yard.
9. \begin{align*}2:2:2\sqrt{3}\end{align*} does not fit into either ratio, so it is not a special right triangle. To see if it is a right triangle, plug these values into the Pythagorean Theorem: \begin{align*}2^2 + 2^2 & = \left ( 2\sqrt{3} \right )^2\\ 4 + 4 & = 12\\ 8 & < 12\end{align*} this is not a right triangle, it is an obtuse triangle.
10. \begin{align*}\sqrt{5}:\sqrt{15}:2\sqrt{5}\end{align*} is a \begin{align*}30-60-90\end{align*} triangle. The long leg is \begin{align*}\sqrt{5} \cdot \sqrt{3} = \sqrt{15}\end{align*} and the hypotenuse is \begin{align*}2\sqrt{5}\end{align*}.
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# Kinematics- Acceleration Chapter 5 (pg. 81-103) A Mathematical Model of Motion.
## Presentation on theme: "Kinematics- Acceleration Chapter 5 (pg. 81-103) A Mathematical Model of Motion."— Presentation transcript:
Kinematics- Acceleration Chapter 5 (pg. 81-103) A Mathematical Model of Motion
The Nature of Acceleration Acceleration is the rate of change in _______ with time - the rate at which the velocity changes. If an object is speeding up, it is __________ ! If an object is slowing down, it is ! If an object has constant speed, but is changing direction, it is ! -this is the case when an object travels in a circular path at constant speed ANY VELOCITY CHANGE is an ACCELERATION ! velocity accelerating
Calculating Acceleration Acceleration is a vector, so in a straight line the direction will be represented by +/- sign. Negative acceleration means acceleration in the opposite direction of whatever positive means! Average acceleration is change in velocity divided by change in time. Uniform acceleration means constant acceleration!
a av = v - v o ∆t This is the basic definition of acceleration in the form of an equation. The following equations are basic equations which are derivations of our basic definitions of velocity and acceleration for objects moving with constant acceleration: m/s 2 are the units of acceleration ( think meters per second CHANGE in velocity per second of time) v = v 0 + at ∆d = v 0 t + 1212 at2at2 ∆d = (v 0 + v) 1212
A car accelerates from rest at a rate of 2.00 m/s 2 west for 8.50 s. What is the velocity of the car at this time? What was its displacement during this time? v 0 = 0m/s a = 2.00 m/s 2 west ∆t = 8.50 s v = ? ∆d = ? v = v 0 + at v = 0 + (2.00 m/s 2 west )(8.50 s) v = 17.0 m/s west ∆d = 0 + 0.5(2.00 m/s 2 west)(8.50s) 2 ∆d = 72.3 m west ∆ d = v 0 t + a t 2 1212
3) An object is accelerated to 35.0 m/s at a rate of 2.00 m/s 2 for a time of 7.34 s. How far did it move during this time? 1) How far does an object travel if it accelerates from 12.0 m/s to 45.0 m/s in 15.3 s? 2) An object traveling 48.7 m/s is brought to a stop in 7.61 s. How far did the object move during this time? 4) An object rolled up a hill will stop after having traveled up the hill 18.0 m. If the object is slowed at a rate of 2.50 m/s 2, how long will it take to stop rolling? [203m] [436m] [185m] [3.80s]
Basic Equations: v = v 0 + at ∆d = (v 0 + v) 1212 ∆d = v 0 t + 1212 at2at2
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# Quick Answer: What Are The Six Factors Of 18?
## What is the factor of 10?
The factors of 10 are 1, 2, 5, and 10.
You can also look at this the other way around: if you can multiply two whole numbers to create a third number, those two numbers are factors of the third.
2 x 5 = 10, so 2 and 5 are factors of 10.
1 x 10 = 10, so 1 and 10 are also factors of 10..
## Is 17 a perfect square?
A number is a perfect square (or a square number) if its square root is an integer; that is to say, it is the product of an integer with itself. Here, the square root of 17 is about 4.123. … Anyway, 17 is a prime number, and a prime number cannot be a perfect square.
## What is the smallest factor of 25?
25 = 1 x 25 or 5 x 5. Factors of 25: 1, 5, 25. Prime factorization: 25 = 5 x 5, which can also be written 25 = 5².
## What is the GCF of 24 and 18?
We found the factors and prime factorization of 18 and 24. The biggest common factor number is the GCF number. So the greatest common factor 18 and 24 is 6.
## What is the GCF of 21 35 and 49?
Greatest common factor (GCF) of 21 and 35 is 7. We will now calculate the prime factors of 21 and 35, than find the greatest common factor (greatest common divisor (gcd)) of the numbers by matching the biggest common factor of 21 and 35.
## What is the greatest common factor of 15 and 18?
The factors of 18 are 18, 9, 6, 3, 2, 1. The common factors of 15 and 18 are 3, 1, intersecting the two sets above. In the intersection factors of 15 ∩ factors of 18 the greatest element is 3. Therefore, the greatest common factor of 15 and 18 is 3.
## What is the square of 18?
324 4.243Table of Squares and Square RootsNUMBERSQUARESQUARE ROOT152253.873162564.000172894.123183244.24396 more rows
## What are the factors of 18?
Table of Factors and MultiplesFactorsMultiples1, 2, 4, 8, 16161121, 17171191, 2, 3, 6, 9, 18181261, 191913341 more rows
## What are the factors of 12 and 18?
In terms of numbers, the greatest common factor (gcf) is the largest natural number that exactly divides two or more given natural numbers. Example 1: 6 is the greatest common factor of 12 and 18. Example 2: 2 is a common factor of 12 and 18, but it is NOT the GREATEST common factor.
## How many whole number factors does 18 have?
six different whole number factorsAnswer and Explanation: There are six different whole number factors of 18. The factors are 1, 2, 3, 6, 9, and 18. Whole numbers are numbers beginning with zero that are…
## Is 18 a perfect square?
Is 18 a perfect square number? A number is a perfect square (or a square number) if its square root is an integer; that is to say, it is the product of an integer with itself. Here, the square root of 18 is about 4.243. Thus, the square root of 18 is not an integer, and therefore 18 is not a square number.
## What are the common factors of 18 and 30?
Greatest common factor (GCF) of 18 and 30 is 6. We will now calculate the prime factors of 18 and 30, than find the greatest common factor (greatest common divisor (gcd)) of the numbers by matching the biggest common factor of 18 and 30.
## Whats the HCF of 15 and 20?
5Greatest common factor (GCF) of 15 and 20 is 5. We will now calculate the prime factors of 15 and 20, than find the greatest common factor (greatest common divisor (gcd)) of the numbers by matching the biggest common factor of 15 and 20.
## Is 18 a triangular number?
0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, 231, 253, 276, 300, 325, 351, 378, 406, 435, 465, 496, 528, 561, 595, 630, 666…
## What is the smallest factor of 18?
I write the factors in order from smallest to greatest, so the factors of 18 are: 1, 2, 3, 6, 9, and 18.
## What are the factors of 18 and 3?
18 = 1 x 18, 2 x 9, or 3 x 6. Factors of 18: 1, 2, 3, 6, 9, 18. Prime factorization: 18 = 2 x 3 x 3, which can also be written 18 = 2 x 3².
## What are the factors of 19?
The only factors of 19 are 1 and 19, so 19 is a prime number. That is, 19 is divisible by only 1 and 19, so it is prime.
## What are 5 multiples of 18?
1 Answer. 18,36,54,72,90 are first five multiples of 18 .
## What is the factor of 23?
The factors of 23 are 1 and 23. Since 23 is a prime number, therefore, it has only two factors.
## What is the largest factor of 18?
The factors of 18 are 1, 2, 3, 6, 9, 18. The factors of 27 are 1, 3, 9, 27. The common factors of 18 and 27 are 1, 3 and 9. The greatest common factor of 18 and 27 is 9.
## What is the GCF of 18 and 3?
We found the factors and prime factorization of 3 and 18. The biggest common factor number is the GCF number. So the greatest common factor 3 and 18 is 3.
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# Question Video: Calculating the Scalar Product of Two Vectors Given Their Lengths and the Angle between Them Physics
The diagram shows two vectors, π and π. What is the scalar product of π and π? Give your answer to two significant figures.
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### Video Transcript
The diagram shows two vectors, π and π. What is the scalar product of π and π? Give your answer to two significant figures.
In this question, we are presented with a diagram of two vectors and asked to find their scalar product. We see from the diagram that we have a vector π with a magnitude of 5.4 and that we have a vector π with a magnitude of 6.7. The angle π between these two vectors is 42 degrees. Letβs recall that we can define the scalar product of two vectors π and π as the magnitude of vector π multiplied by the magnitude of vector π multiplied by the cos of the angle π between the two vectors. So letβs substitute in our values for the magnitude of π, magnitude of π, and the value of π into this expression for the scalar product.
We have that the scalar product of π and π is equal to 5.4, the magnitude of π, multiplied by 6.7, the magnitude of π, multiplied by the cos of 42 degrees, the angle between our two vectors. If we evaluate this expression, we get the result 26.88697 and so on with further decimal places. Notice that the question asks us to give our answer to two significant figures. To this precision of two significant figures, our answer to the question of the scalar product of π and π is 27.
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# Find the equation of the circle which passes through the points $(2,3)$ and $(4,5)$ and the centre lies on the straight line $y-4x+3=0$.
$\begin{array}{1 1}x^2+y^2+4x-10y+25=0\\x^2+y^2-4x-10y-38=0\\x^2+y^2-4x-10y+38=0\\x^2+y^2+4x+10y+25=0\end{array}$
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• General equation of the circle is x2+y2+2gx+2fy+c=0" role="presentation" style="position: relative;">x2+y2+2gx+2fy+c=0x2+y2+2gx+2fy+c=0x^2+y^2+2gx+2fy+c=0 where g" role="presentation" style="position: relative;">ggg and f" role="presentation" style="position: relative;">fff are the coordinates of the circle.
Answer : x2+y2−4x−10y+25=0" role="presentation" style="position: relative;">x2+y24x10y+25=0x2+y2−4x−10y+25=0x^2+y^2-4x-10y+25=0
It is given that the circle passes through the points (2,3)" role="presentation" style="position: relative;">(2,3)(2,3)(2,3) and (4,5)" role="presentation" style="position: relative;">(4,5)(4,5)(4,5)
Now substituting these values for x" role="presentation" style="position: relative;">xxx and y" role="presentation" style="position: relative;">yyy in the general equation of the circle we get
x2+y2+2gx+2fy+c=0" role="presentation" style="position: relative;">x2+y2+2gx+2fy+c=0x2+y2+2gx+2fy+c=0x^2+y^2+2gx+2fy+c=0
(2)2+(3)2+2(2)g+2(3)f+c=0" role="presentation" style="position: relative;">(2)2+(3)2+2(2)g+2(3)f+c=0(2)2+(3)2+2(2)g+2(3)f+c=0(2)^2+(3)^2+2(2)g+2(3)f+c=0
⇒4g+6f+c=−13" role="presentation" style="position: relative;">4g+6f+c=13⇒4g+6f+c=−13\Rightarrow 4g+6f+c=-13------(1)
(4)2+(5)2+2(4)g+2(5)f+c=0" role="presentation" style="position: relative;">(4)2+(5)2+2(4)g+2(5)f+c=0(4)2+(5)2+2(4)g+2(5)f+c=0(4)^2+(5)^2+2(4)g+2(5)f+c=0
16+25+8g+10f+c=0" role="presentation" style="position: relative;">16+25+8g+10f+c=016+25+8g+10f+c=016+25+8g+10f+c=0
⇒8g+10f+c=−41" role="presentation" style="position: relative;">8g+10f+c=41⇒8g+10f+c=−41\Rightarrow 8g+10f+c=-41-----(2)
It is also given the centre of the circle lies on the line y−4x−3=0" role="presentation" style="position: relative;">y4x3=0y−4x−3=0y-4x-3=0
⇒f−4g+3=0" role="presentation" style="position: relative;">f4g+3=0⇒f−4g+3=0\Rightarrow f-4g+3=0
Or −4g+f=3" role="presentation" style="position: relative;">4g+f=3−4g+f=3-4g+f=3-----(3)
Now solving the three equations for g,f" role="presentation" style="position: relative;">g,fg,fg,f and c" role="presentation" style="position: relative;">ccc we get,
4g+6f+c=−13" role="presentation" style="position: relative;">4g+6f+c=134g+6f+c=−134g+\;6f+c=-13
8g+10f+c=−41" role="presentation" style="position: relative;">8g+10f+c=418g+10f+c=−418g+10f+c=-41
on subtracting we get,
−4g−4f=28" role="presentation" style="position: relative;">4g4f=28−4g−4f=28-4g-4f=28
−4g+f=3" role="presentation" style="position: relative;">4g+f=3−4g+f=3-4g+\;f=\;\;3
on subtracting we get
−5f=25" role="presentation" style="position: relative;">5f=25−5f=25-5f=25
f=−5" role="presentation" style="position: relative;">f=5f=−5f=-5
−4g−5=3" role="presentation" style="position: relative;">4g5=3−4g−5=3-4g-5=3
−4g=8" role="presentation" style="position: relative;">4g=8−4g=8-4g=8
⇒g=−2" role="presentation" style="position: relative;">g=2⇒g=−2\Rightarrow g=-2
Substituting for g and f in eq(1)
4(−2)+6(−5)+c=0" role="presentation" style="position: relative;">4(2)+6(5)+c=04(−2)+6(−5)+c=04(-2)+6(-5)+c=0
=>−8−30+c=0" role="presentation" style="position: relative;">=>830+c=0=>−8−30+c=0=> -8-30+c=0
c=38" role="presentation" style="position: relative;">c=38c=38c=38
Hence the equation of the circle is x2+y2−4x−10y+38=0" role="presentation" style="position: relative;">x2+y24x10y+38=0x2+y2−4x−10y+38=0x^2+y^2-4x-10y+38=0
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# Polynomials: Factoring by Grouping
Factoring polynomials is an important step in solving polynomial equations. Getting a polynomial into binomials is one of the easiest ways to solve for x-intercepts as you can set each binomial equal to zero and find multiple answers.
We have learned how to factor out a constant and how to factor different types of trinomials into binomials, but what about when you have four terms in a polynomial?
One possibility is factoring by grouping. Sometimes you can factor a number or a variable out of every term in a polynomial. (So, $x^3+3x^2+4x\rightarrow x(3x^2+3x+4)$). When you factor out a term, you group the other terms in parentheses and put the factored term in front, setting up a distributive property expression.
Often, however, you can only factor a number or variable out of some of the terms in a polynomial. In these cases, it can help to "factor by grouping." The associative property says that, in addition problems (and all polynomials are actually long addition problems with positive and negative numbers), you can group the terms however you like. So, try grouping pairs of terms from which you can factor out a number, variable, or term.
Example:
\eqalign{3x^2+6x-8x-16&=0\\(3x^2+6x)+(-8x-16)&=0\quad\text{Break polynomial into groups}}
In this example, you will know that you can pull a 3x out of the first terms and a -8 out of the net two terms, so group those pairs together. (Note, the negative sign stuck to the 8 when we grouped it, if we had left it outside the group, distributive property would have also affected the 16, so best to join your two groups with an addition sign and just keep the signs as they are on the original numbers).
Factoring by grouping gets really interesting when, after you factor a different term out of the different groups, you see that the binomial you have left, in both groups, is the same!
Example
\eqalign{3x^2+6x-8x-16&=0\\(3x^2+6x)+(-8x-16)&=0\quad\text{Break polynomial into groups}\\3x(x+2)-8(x+2)&=0\quad\text{Factor what you can from each group}\\3x\mathbf{(x+2)}-8\mathbf{(x+2)}&=0\quad\text{The binomial is identical}}
As you can see here, both of the groups contain the terms you factored out (3x and -8) and a common binomial (x+2).
The identical binomial in each group means that you could have actually factored out that binomial, you just couldn't recognize it! So, now that you can see it, you can factor the binomial out, put it in the front of the expression (just like you would if you pulled out a single term and wanted to put it in front like a distributive term) and multiply that binomial times the remaining terms. Put those remaining terms together as a binomial and, voila, you have two binomials.
Example
\eqalign{3x^2+6x-8x-16&=0\\(3x^2+6x)+(-8x-16)&=0\quad\text{Break polynomial into groups}\\3x(x+2)-8(x+2)&=0\quad\text{Factor what you can from each group}\\3x\mathbf{(x+2)}-8\mathbf{(x+2)}&=0\quad\text{The binomial is identical, factor it out and put it in front}\\(x+2)(3x-8)&=0\quad\text{Polynomial factored into binomials}}
Not every polynomial will factor nicely by grouping, but it's always worth checking out!
• ## Polynomials: Factoring by Grouping
Factor the following polynomials into binomials by grouping:
1. $2x^2+6x+7x+21$
2. $3x^2-9x-8x+24$
3. $4x^2-28x+12x-84$
4. $2x^2+26x-3x-39$
5. $16x^2+48x-14x-42$
6. $x^3-3x^2-4x+12$
7. $x^3+2x^2+3x+6$
8. $2x^3+2x^2-2x-2$
9. $3x^3+27x^2-x-9$
10. $2x^5-4x^4-3x+6$
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# Not MONEY but GIFT alphametic
There is a well-known alphametic story where a poor college student sends a telegram with the words "SEND MORE MONEY" to his parents, asking for more money, and asking to fill in each letter with a different digit in order to figure out how much he is asking for. In another alphametic story the same poor college student sends a telegram with the words "SEND A GIFT":
SEND = A * GIFT
The parents send him a gift. Although the gift is odd and what the parents send is odd and everything related is odd, the poor college student is happy with it.
Which digit does each letter represent? (Please present the full analysis how these digits can be determined. Every letter stands for a digit in base-10 representation, different letters stand for different digits, and leading digits are always non-zero.)
## 1 Answer
Solution:
There are 15 possible solutions:
$4827 = 3 * 1609$
$5427 = 3 * 1809$
$8463 = 7 * 1209$
$9706 = 2 * 4853$
$3690 = 2 * 1845$
$6970 = 2 * 3485$
$7690 = 2 * 3845$
$9630 = 2 * 4815$
$9670 = 2 * 4835$
$9730 = 2 * 4865$
$9370 = 2 * 4685$
$7852 = 4 * 1963$
$7236 = 4 * 1809$
$8316 = 4 * 2079$
$7854 = 6 * 1309$
Starting point
"The gift is odd". This is the part.
It means the amount is an odd number. So we conclude T has to be odd digit.
$A$ and $T$ obviously can not be $1$
Since $A >=2$ then $G <=4$
Also A cannot be 5 since it will result in $D = 5$
Also A cannot be 9 because it will result in $S = 9$ or go over 4 digits in the result
So A can be one of [2,3,4,6,7,8]
[EDIT]
As Trenin pointed out A can also be even. In my original answer I handled A as odd, that's why the cases are handled in this way..A odd then A even.
now start filling in the blanks using the values we already know
Case 1:
A = 3, T = 7. We immediately see that $G =2$ (otherwise we get to $S = 9$ or go over 4 digits).
The carriage from $3*I + carriage from 3*f + 2$ should be one of [0,2,3] (otherwise we get S = 7 or we go over 4 digits). But 3 cannot be achieved since we cannot get the carriage from 3*f to be 3.
The carriage can be 0 only if $I = 0$ (it cannot be 2 or 3 because they are already taken). We have in this case
20F7 * 3 = 6EN1 F cannot be 4 because it will result in N being 4, F cannot be 5 or 9 since it will result in N being 7 (already taken), F cannot be 8 since it will result in E = 2 (already taken).
So no valid solution here.
The carriage can be 2 only if I is [6,8,9].
Trying every combination for I = 6 and F any available digit gets us to conflicts every time.
Conclusion $A = 3, T = 7$ - not possible
Case 2.
A = 3, T = 9.
We get G is one of [1,2]
Considering G = 1.
1IF9 * 3 = SEN7.
Considering F = 0 we get 1I09 * 3 = SE27.
Possible values for I are [4,5,6,8].
Doing the math for all possible values we get 2 matches. I one of [6,8]
This results in 2 possible solutions:
$4827 = 3 *1609$ and $5427 = 3 * 1809$
For G = 2 we end up with F being one of [6,8].
For F = 6 there is no possible value for I. (we end up with an already used value for E).
For F = 8 There is no possible value for I.
Conclusion: No other possible values that what we found above
Case 3:
A = 7, T = 3. This results in G = 1.
Starting with F = 0 we get no valid value for I.
For F = 2 again no valid value for I.
Going through all the possible values of F ([4,5,6,8,9]) we get the same result. Nothing valid for I.
Conclusion: this doesn't work
Case 4:
A = 7, T = 9. We see immediately that G = 1.
Trying with F one of [0,2,3,4,5,6,8] and following the same process as for the other cases (got lazy here so I won't list all the possible outcomes) we end up with the one solution for F = 0.
$8463 = 7 * 1209$
[EDIT] After Trenin's comment.
In case A and SEND can be even numbers, in addition to the cases above we get:
Case 5.1:
A = 2, T = 3 => G can be only 4 (for 1,2,3 we get duplicate digits).
in this case F can be one of [5,8,9].
IN this case we get only one valid value $9706 = 2 *$4853$Case 5.2: A =2, T = 5 => G can be 1, 3 or 4. otherwise we get conflicts in the digits. Starting with G = 1. We get the possible values for F [3,4,6,7,8,9]. Calculating step by step we end up with these solutions:$3690 = 2 * 1845$continuing with G = 3 we get the possible values for F [4,6,8] Doing the math for all of them we get:$6970 = 2 * 34857690 = 2 * 3845$for G = 4 we get the possible values for F [1,3,6,8]. Doing the math we get$9630 = 2 * 48159670 = 2 * 48359730 = 2 * 48659370 = 2 * 4685$Case 5.3 A = 2, T = 7. Results in G being 1 or 3. Doing the same steps as above, calculating the possible values for F and then filling in the missing value for I we get no valid results. Case 5.4 A = 4, T = 3. Results in G being 1 or 2. Doing the same steps as above, calculating the possible values for F and filling in the missing value for I we get one valid result:$7852 = 4*1963$Case 5.5 to 5.15 (got tired of typing ) A = 4, T = 5 gives no valid result A = 4, T = 7 no results again A = 4, T = 9 gives out$7236 = 4 * 18098316 = 4 * 2079$Starting A = 6 and further is even easier since G can be only 1. A = 6, T = 3 => nothing A = 6, T = 5 => nothing A = 6, T = 7 => nothing A = 6, T = 9 we get:$7854 = 6 * 1309$A = 8, T = 3 => nothing A = 8, T = 5 => nothing A = 8, T = 7 => nothing A = 8, T = 5 => nothing • Why does$A$have to be odd? The clue is only that the gift is odd. If the clue was that the sending of the gift was odd, then I would conclude$SEND$is odd and agree that both$T,A,D$are all odd. But from the clue, I don't think you can infer$A$is odd, but only$GIFT$and thus only$T\$ is odd. Mar 29, 2016 at 12:27
• @Trenin. Hmmm...you might be right on this. I saw the word odd and got all fired up. Anyway, the solutions I found are still valid. I will start digging for the other solutions as well. Thanks for this. Mar 29, 2016 at 12:30
• @Trenin. Fixed it. Thanks again for the tip. I would split the points with you on this if I could :). Mar 29, 2016 at 13:14
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## Definite Integrals
Integration as limit of sum
The summation notation or sigma notation enables us to write a sum with many (finitely or infinitely many) terms in the compact form. It uses the uppercase Greek letter Σ (sigma) to denote various kinds of sums.
For example, consider the sum: 2 + 4 + 6 + 8 + 10 + 12, in which each term is of the form '2i', where 'i' is one of the integer from 1 to 6. In sigma notation, this sum can be written as: , and read as "the summation of 2i, where i runs from 1 to 6".
In the notation , the Greek capital letter Σ stands for sum, the expression '2i' is called the summand, the numbers 1 and 6 are called the limits of summation [i.e., 1 is called the lower limit and 6 is called the upper limit of the summation], and 'i' is called the index of summation. The index of summation 'i' tells us that where to begin the sum (at the number below Σ) and where to end (at the number above Σ).
Rules for summation
Rule 1: The sum of the sums of two or more variables is equal to the sum of their summations, i.e., .
Rule 2: The difference of the sums of two or more variables is equal to the difference of their summations, i.e., .
Rule 3: The sum of a constant times the values of a variable is equal to the constant times the sum of the variable, i.e., , where 'p' is any constant.
Rule 4: The sum of a constant taken 'n' times is the constant times 'n', i.e., .
Summation formulae
If 'n' is any positive integer, then
. .
. .
Riemann Sums
Suppose that a function f(x) is continuous on a closed interval [a, b]. The graph of a function f(x) over a closed interval [a, b] is shown below, which may have positive values as well as negative values.
We then partition the interval [a, b] into 'n' sub-intervals by choosing n – 1 points, say x1, x2, x3, x4, x5, . . . . . . . . . . ., xn – 1, between a and b subject only to the condition that a = x0 < x1 < x2 < . . . . . . . . . . . < xn – 1 < xn = b. The set P = {x0, x1, x2, . . ., xn} is called a partition of the closed interval [a, b].
The partition P = {a = x0, x1, x2, x3, . . ., xn = b} divides the closed interval [a, b] into 'n' sub-intervals called [x0, x1], [x1, x2], [x2, x3], . . ., [xi – 1, xi], . . ., [xn – 1, xn] of lengths Δx1, Δx2, Δx3, . . ., Δxi, . . ., Δxn.
In each sub-interval, we select some number. Denote the number chosen from the ith sub-interval by ci. Then, on each sub-interval we stand a vertical rectangle that reaches from the x-axis to touch the curve at (ci, f(ci)). These rectangles could lie either above or below the x-axis.
On each sub-interval, we form the product f(ci) . Δxi. This product can be positive, negative or zero, depending on the value of f(ci). Finally, we take the sum of these products: Sn = f(ci)Δxi. This sum, which depends on the partition P and the choice of the numbers ci, is a Riemann sum for f(x) on the interval [a, b].
If we take the sample points to be right hand endpoints of the sub-intervals, then ci = xi and the Riemann sum = f(xi)Δxi. If we take the sample points to be left hand endpoints of the sub-intervals, then ci = xi – 1 and the Riemann sum = f(xi – 1)Δxi.
Definite Integrals
The definite integral as a limit of Riemann sums: Let f(x) be a function defined on a closed interval [a, b]. For any partition P of [a, b], let the numbers ci be chosen arbitrarily in the sub-intervals [xi – 1, xi]. If there exists a number 'I' such that = I, where "max . Δxi" is the length of the largest sub-interval in the partition P, then 'f(x)' is integrable on [a, b] and I is the definite integral of f(x) over [a, b].
The existence of definite integrals: All continuous functions are integrable. That is, if a function f(x) is continuous on a closed interval [a, b], then its definite integral over [a, b] exists.
The definite integral of a continuous function over [a, b]: Let f(x) be a continuous function defined on a closed interval [a, b] and let [a, b] be partitioned into 'n' sub-intervals of equal length . We let x0 (= a), x1, x2, . . . . . . . ., xn (= b) be the endpoints of these sub-intervals and we choose sample points c1, c2, . . . ., cn in these sub-intervals, so ci lies in the ith sub-interval [xi – 1, xi]. Then the definite integral of f(x) from a to b is: f(x) dx = .
The symbol was introduced by Leibniz and is called an integral sign. It is elongated 'S' and was chosen because an integral is a limit of sums. In the notation: f(x) dx, f(x) is called the integrand, 'a' is called the lower limit of integration, and 'b' is called the upper limit of integration. The symbol 'dx' indicates that x is the variable of integration.
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Equal or Not?
36 teachers like this lesson
Print Lesson
Objective
SWBAT determine if two number sentences are equal.
Big Idea
Students play an interactive game and practice crucial math fluency in this lesson on the meaning of the equals sign.
Setting Up the Learning
5 minutes
The Why:
This standard addresses students' understanding of the equal sign, a concept that is super important for students' later understanding of algebra. A common misconception in equality is that students assume if it "looks wrong", then it can't be true. So 5 = 2+3 is false because 5 is on the "wrong" side.
As students practice equality, they are also getting great practice with math fluency. Fluency is explicitly stated as part of the standards (1.OA.C6), and this lesson is a great example of how to embed number facts throughout the year.
Review:
Yesterday (See the We're Tied! lesson) we played a game where we had to figure out: Are we tied? And we talked about how tied means equal.
Connect:
Objective
Today, our goal is to think about: How can we figure out if number sentence are equal?
Opening Discussion
10 minutes
First, I am going to explain the game. After that, we will practice the game all together.
1. Both partners pull a basketball card.
2. Partners record their number sentences.
3. Each partner figures out how many points he/she got.
4. Partners decide: Did we tie? Are the number sentences equal?
I’ll actually play this game with another student, having that student work out their side on the chart paper while I do mine. Throughout this lesson, we are requiring work for each number sentence as evidence of student thinking. This is aligned with CCSS MP2, “Reason abstractly and quantitatively” , which focuses largely on pushing students to coherently represent a problem.
Step 1: Both partners pull a card.
My card says 3+4. ____’s card says 8 - 1.
Step 2: Partners record their number sentences
Step 3: Each partner figures out how many points he/she got.
I’ll reinforce the importance of proving your answer and both the student and I will draw our strategy.
Step 4: Partners decide: Did we tie? Are the number sentences equal?
• I counted on for my card. I said 3, 4, 5, 6, 7. So 3+4 is the same as 7.(Highlight child’s strategy)-Is 3 + 4 the same as 8 – 1?
• I’ll model writing the number sentence and thinking out loud, “3 + 4 is the same as 8-1, so I can write 3+4=8-1”
I’ll model another round with a different student, this time making sure they are not equal.
Game Practice
10 minutes
Student Work Time:
I’ll present 2 cards to students and have them figure out if they are tied on their own. While students work, I’ll make sure they are proving their thinking. The CCSS emphasizes student independence, this is a time I want the students doing all the heavy lifting!
I had students show their thinking on white boards. Attached you'll see how 2 different students showed their thinking. One student used counting strategies for both. Another drew a base ten model for both equations. This is a great example of having students represent their thinking in writing/pictures. The CCSS MP2, "Reason abstractly and quantitatively". This standard is focused on students being able to show what they are thinking using symbols and pictures.
Student Share:
Guiding Questions:
• Are they tied or not?
• What strategy did you use to PROVE that they are tied? This emphasizes the "Construct viable arguments" portion of MP3. Students have to use math to prove their argument.
• What symbol do we need to show that they each got the same number of points?
If time, we will do one more statement together, such as 4 = 6+2. This number sentence forces students to pay attention to the symbols. If they subtract instead of add they will think it is equal.
Game Time! Partner Work
15 minutes
Students play the game with a partner. To differentiate this activity, I just changed the materials students use. See below for intervention ideas!
Intervention:
Gets card set A. Card set A has all totals under 10. Students get ten frames to record how many they got. This helps support students in deciding if they are equal. Because we have worked so much with the ten frame, they can quickly look at both ten frames and see if each one has the same amount.
See attached Basketball game cards and Recording Sheets!
See attached video of 2 students playing the game. You'll see that they both show their strategies for the equations they pulled.
Closing and Exit Ticket
10 minutes
I'll bring students back together and we will do a set of number cards together and determine if they are equal. This will be brief to allow for time for the exit ticket.
Students complete the exit ticket. See attached exit ticket, 2 per page!
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### by Peter Good
#### Published Sunday December 18 2022 (link)
A plumber had three thin metal pipes with square, rectangular and elliptical cross-sections. In order to fit them into his van, he slid the rectangular pipe inside the elliptical pipe and the elliptical pipe inside the square pipe, before placing the pipe assembly in the van. There are four points where the pipes all touch, as shown in the diagram. The maximum and minimum widths of the elliptical and rectangular pipes and the diagonal width of the square pipe were all even numbers of mm less than 1,000, of which one was a perfect square.
What were the five widths (in increasing order)?
From → Uncategorized
One Comment Leave one →
If the centre of the diagram is placed at the origin (0, 0) of an $$(x, y)$$ graph and we consider its upper right quadrant, we can put the meeting point of three three pipes at $$r$$ and $$s$$. Let the semi-major and semi-minor axes of the ellipse be $$a$$ and $$b$$ respectively, and let the semi-diagonal of the square (which is equal to $$x+y$$) be $$c$$. The ellipse is then given by the function:
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\tag{1}\label{1}$
The slope of the ellipse and the side of the square at the common point are both $$-1$$. Differentiating the equation above gives:$\frac{2x}{a^2}dx+\frac{2y}{b^2}dy=0\tag{2}\label{eq2}$$\frac{dy}{dx}=-\frac{xb^2}{ya^2}=-1\tag{3}\label{3}$and hence $$xb^2=ya^2$$. Using this and the equation of the ellipse at ($$r, s$$) to eliminate $$s$$ now gives:$r=\frac{a^2}{\sqrt{a^2+b^2}}\tag{4}\label{4}$and the equivalent equation for $$s$$ follows in the same way:$s=\frac{b^2}{\sqrt{a^2+b^2}}\tag{5}\label{5}$Since $$c=x+y=\sqrt{a^2+b^2}$$, the five values required are now $$(2a, 2b, 2a^2/c, 2b^2/c, c)$$.
Noting that the triple $$(a, b, c)$$ is a Pythagorean triple, if we restrict ourselves to primitive triples, neither $$a$$ nor $$b$$ are divisible by $$c$$, which means we have to multiply the five values by $$c$$ in order to ensure that they are all integers as required. So the values now become$(2ac, 2bc, 2a^2, 2b^2, 2c^2)$
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# What is a solution to the differential equation dy/dx= y^2? (Please)
Apr 17, 2018
$y \left(x\right) = - \frac{1}{x + C}$
#### Explanation:
This is a separable differential equation. We'll get all $y , \mathrm{dy}$ on the right side and all $x , \mathrm{dx}$ on the left.
$\frac{\mathrm{dy}}{\mathrm{dx}} = {y}^{2}$
$\frac{1}{y} ^ 2 \mathrm{dy} = \mathrm{dx}$
Note that we have no $x$ so all we do is move $\mathrm{dx}$ to the left.
${y}^{-} 2 \mathrm{dy} = \mathrm{dx}$
Integrate both sides:
$\int {y}^{-} 2 \mathrm{dy} = \int \mathrm{dx}$
$- \frac{1}{y} = x + C$
We need an explicit solution in the form $y \left(x\right) :$
$- 1 = y \left(x + C\right)$
$y \left(x\right) = - \frac{1}{x + C}$
Apr 17, 2018
$y = - \frac{1}{x + C}$
#### Explanation:
$\frac{\mathrm{dy}}{\mathrm{dx}} = {y}^{2}$
${y}^{- 2} \mathrm{dy} = \mathrm{dx}$
$\int {y}^{- 2} \mathrm{dy} = \int \mathrm{dx}$
$- {y}^{- 1} = x + C$
${y}^{- 1} = - x - C$
$y = \frac{1}{- x - C}$
$y = - \frac{1}{x + C}$, $C$ is a constant
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# How do you find 20% of an amount?
As finding 10% of a number means to divide by 10, it is common to think that to find 20% of a number you should divide by 20 etc. Remember, to find 10% of a number means dividing by 10 because 10 goes into 100 ten times. Therefore, to find 20% of a number, divide by 5 because 20 goes into 100 five times.
## How do I figure out percentages?
Percentage can be calculated by dividing the value by the total value, and then multiplying the result by 100. The formula used to calculate percentage is: (value/total value)×100%.
## What is the easiest way to calculate percentages?
Generally, the way to figure out any percentage is to multiply the number of items in question, or X, by the decimal form of the percent. To figure out the decimal form of a percent, simply move the decimal two places to the left. For example, the decimal form of 10 percent is 0.1.
## How do I figure out percentage deducted?
To subtract any percentage from a number, simply multiply that number by the percentage you want to remain. In other words, multiply by 100 percent minus the percentage you want to subtract, in decimal form. To subtract 20 percent, multiply by 80 percent (0.8).
## How much is 20% off?
First, convert the percentage discount to a decimal. A 20 percent discount is 0.20 in decimal format. Secondly, multiply the decimal discount by the price of the item to determine the savings in dollars. For example, if the original price of the item equals \$24, you would multiply 0.2 by \$24 to get \$4.80.
## How do you take 20% off in Excel?
If you want to calculate a percentage of a number in Excel, simply multiply the percentage value by the number that you want the percentage of. For example, if you want to calculate 20% of 500, multiply 20% by 500.
## What percent is 125 out of 200?
Solution and how to convert 125 / 200 into a percentage 0.62 times 100 = 62.5. That’s all there is to it!
## What percent is 3 out of 18?
Now we can see that our fraction is 16.666666666667/100, which means that 3/18 as a percentage is 16.6667%.
## What is 30% off?
To take 30 percent off a number: Divide the number by 10. Triple this new number. Subtract your triple from your starting number.
## What is 10 off in money?
While 10 percent of any amount is the amount multiplied by 0.1, an easier way to calculate 10 percent is to divide the amount by 10. So, 10 percent of \$18.40, divided by 10, equates to \$1.84.
## How do I add 20% to a price in Excel?
Increase by Percentage Enter a number in cell A1. Enter a decimal number (0.2) in cell B1 and apply a Percentage format. 2. To increase the number in cell A1 by 20%, multiply the number by 1.2 (1+0.2).
## How do you take 10% off in Excel?
For example, if you type the formula =10/100 in cell A2, Excel will display the result as 0.1. If you then format that decimal as a percentage, the number will be displayed as 10%, as you ‘d expect.
## What percentage is 35 out of 210?
Note that you can reverse steps 1 and 2 and still come to the same solution. If you multiply 35 by 100 and then divide the result by 210, you will still come to 16.67!
## What is a 25 out of 50 grade?
Now we can see that our fraction is 50/100, which means that 25/50 as a percentage is 50%. And there you have it! Two different ways to convert 25/50 to a percentage.
## How do you find 20% of a number?
Multiply the original price by 0.2 to find the amount of a 20 percent markup, or multiply it by 1.2 to find the total price (including markup). If you have the final price (including markup) and want to know what the original price was, divide by 1.2.
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Inverse functions
Exponential and logarithmic functions
Exponential and logarithmic functions are mutually inverse functions
Inverse functions
The inverse function, usually written f -1, is the function whose domain and the range are respectively the range and domain of a given function f, that is
f -1(x) = y if and only if ƒ (y) = x .
Thus, the composition of the inverse function and the given function returns x, which is called the identity function, i.e.,
f -1(ƒ (x)) = x and ƒ (f -1 (x)) = x.
The inverse of a function undoes the procedure (or function) of the given function.
A pair of inverse functions is in inverse relation.
Example: If given ƒ(x) = log2 x then f -1(x) = 2x since,
Therefore, to obtain the inverse of a function y = ƒ (x), exchange the variables x and y, i.e., write x = ƒ (y) and solve for y. Or form the composition ƒ (f -1(x)) = x and solve for f -1.
Example: Given y = ƒ (x) = log2 determine f -1(x).
Solution: a) Rewrite y = ƒ (x) = log2 x to x = log2 y and solve for y, which gives y = f -1(x) = 2x.
b) Form ƒ (f -1(x)) = x that is, log2 (f -1(x)) = x and solve for f -1, which gives f -1(x) = 2x.
The graphs of a pair of inverse functions are symmetrical with respect to the line yx.
Exponential functions
- Exponential function y = ex <=> x = ln y, e = 2.718281828...the base of the natural logarithm.
The exponential function is inverse of the natural logarithm function, so that eln x = x.
- Exponential function y = ax <=> x = loga y, where a > 0 and a is not 1.
The exponential function with base a is inverse of the logarithmic function, so that
The graph of the exponential function y = ax = ebxa > 0 and b = ln a
The exponential function is inverse of the logarithmic function since its domain and the range are respectively the range and domain of the logarithmic function, so that
ƒ (f -1(x)) = x that is, ƒ (f -1(x)) = ƒ (ax) = loga (ax) = x.
The domain of ƒ (x) = ax is the set of all real numbers.
The range of the ƒ (x) = ax is the set of all positive real numbers.
If a > 1 then ƒ is an increasing function and if 0 < a < 1 then ƒ is a decreasing function.
The graph of the exponential function passes through the point (0, 1). The x-axis is the horizontal asymptote to the graph, as shows the above picture.
Translated logarithmic and exponential functions
Example: Given translated logarithmic function y - 2 = log3 (x + 1), find its inverse and draw their graphs.
Solution: Exchange the variables and solve for y, that is x - 2 = log3 (y + 1) which gives y + 1 = 3x - 2.
Note that y - y0 = loga (x - x0) represents translated logarithmic function to base a
and y - y0 = a (x - x0) represents translated exponential function with base a
where, x0 and y0 are the coordinates of translations of the graph in the direction of the coordinate axes.
Functions contents D
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How many different words, each containing 2 vowels and 3 consonants can be formed with 5 vowels and 17 consonants?
Asked by Aaryan | 1 year ago | 129
##### Solution :-
Given:
Total number of vowels = 5
Total number of consonants = 17
Number of ways = (No. of ways of choosing 2 vowels from 5 vowels) × (No. of ways of choosing 3 consonants from 17 consonants)
= (5C2) × (17C3)
By using the formula,
nCr = $$\dfrac{ n!}{r!(n – r)!}$$
= 10 × (17×8×5)
= 10 × 680
= 6800
Now we need to find the no. of words that can be formed by 2 vowels and 3 consonants.
The arrangement is similar to that of arranging n people in n places which are n! Ways to arrange. So, the total no. of words that can be formed is 5!
So, 6800 × 5! = 6800 × (5×4×3×2×1)
= 6800 × 120
= 816000
The no. of words that can be formed containing 2 vowels and 3 consonants are 816000.
Answered by Sakshi | 1 year ago
### Related Questions
#### How many words each of 3 vowels and 2 consonants can be formed from the letters of the word INVOLUTE?
How many words each of 3 vowels and 2 consonants can be formed from the letters of the word INVOLUTE?
#### Find the number of permutations of n distinct things taken r together, in which 3 particular things must occur together.
Find the number of permutations of n distinct things taken r together, in which 3 particular things must occur together.
#### How many words, with or without meaning can be formed from the letters of the word ‘MONDAY’,
How many words, with or without meaning can be formed from the letters of the word ‘MONDAY’, assuming that no letter is repeated, if
(i) 4 letters are used at a time
(ii) all letters are used at a time
(iii) all letters are used but first letter is a vowel ?
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Lesson - The Real Number System
There are 5 different classifications of numbersyou need to know: The Real Number Systemnatural a.k.a counting numbers whole numbersintegersrational numbersirrational numbers The natural numbers are also known as the counting numbers. They begin with 1, 2, 3,...and they continue indefinitely.So, the number 254,673,843 is also a natural number.Natural Numbers The set of whole numbers includes all of the natural numbers and ONE other number: ZERO!So, if we wanted to list the whole numbersin ascending order we would start:0, 1, 2, 3, 4, 5, 6, 7...Whole Numbers The set of numbers called integers (this word comes from the same root as the word integrity) include all of the whole numbers (and, thus, allof the natural numbers), as well as,their opposites (negatives).So, we say that integers are all of the positiveand negative whole numbers and zero....-3, -2, -1, 0, 1, 2, 3,...Integers The root word of rational is ratio, so all rationalnumbers can be written as ratios (a.k.a. fractions).This set of numbers includes all integers, wholenumbers, and natural numbers as well as allnumbers that can be written as a fraction of twointegers, or a terminating (ending) or repeatingdecimal.Examples: 3/4 -8/2 0.342 -0.75 √64-15/3, -3.5, -1, 0, 1.2, 6/9, 32/8, 65/6Rational Numbers Irrational numbers are a set of numbers thatcannot be written as fractions or decimals thatterminate or repeat. They must be written usingspecial symbols because, if we tried to write a decimal equivalent, we would never be ableto stop writing.The most commonly known irrational number isπ (pi). All square roots of numbers that are notperfect squares are irrational (i.e. √7, √94, √2).Irrational Numbers -56Rational NumbersWhole Numbers-7.5 = -15/2The Real Number SystemNatural Numbers0.75 = 3/40.6 = 2/3Integers0103√9-√4Irrational Numbers√60.1010010001...√2√17π√5√3e Let's check what you know:How would you classify 1,345,752?IrrationalRational onlyRational and integer onlyRational, integer, and whole onlyRational, integer, whole, and natural 1,345,752 is rational, integer, whole, and naturalHow would you classify 1/3?IrrationalRational onlyRational and integer onlyRational, integer, and whole onlyRational, integer, whole, and natural 1/3 = 0.3 and is rational onlyHow would you classify 8/2?IrrationalRational onlyRational and integer onlyRational, integer, and whole onlyRational, integer, whole, and natural 8/2 = 4 is rational, integer, whole, and naturalHow would you classify -10?IrrationalRational onlyRational and integer onlyRational, integer, and whole onlyRational, integer, whole, and natural -10 is rational and integer onlyHow would you classify -2/4?IrrationalRational onlyRational and integer onlyRational, integer, and whole onlyRational, integer, whole, and natural -2/4 = -1/2 = -0.5 and is rational onlyHow would you classify √11?IrrationalRational onlyRational and integer onlyRational, integer, and whole onlyRational, integer, whole, and natural IrrationalRational onlyRational and integer onlyRational, integer, and whole onlyRational, integer, whole, and natural√11 is between 3 and 4 and can only beapproximated and is irrationalHow would you classify √81? √81 = 9 and is rational, integer, whole, and natural
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# Multiplication by 3
Let us now learn the multiplication tables of 3.
Multiplication by 3.
Count the number of petals.
We write: 3 + 3 + 3 + 3 + 3 = 15
or 5 × 3 = 15
We read: 5 threes are fifteen.
So, there are 15 petals in all.
Multiplication Tables of 3
1 × 3 = 3
2 × 3 = 6
3 × 3 = 9
4 × 3 = 12
5 × 3 = 15
6 × 3 = 18
7 × 3 = 21
8 × 3 = 24
9 × 3 = 36
10 × 3 = 30
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# Trigonometric Ratios with a Calculator
## Solving for values when triangles aren't special right triangles.
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Practice Trigonometric Ratios with a Calculator
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Trigonometric Ratios with a Calculator
What if you wanted to find the missing sides of a right triangle with angles of \begin{align*}20^\circ\end{align*} and \begin{align*}70^\circ\end{align*} and a hypotenuse length of 10 inches? How could you use trigonometry to help you? After completing this Concept, you'll be able to solve problems like this one.
### Guidance
The trigonometric ratios are not dependent on the exact side lengths, but the angles. There is one fixed value for every angle, from \begin{align*}0^\circ\end{align*} to \begin{align*}90^\circ\end{align*}. Your scientific (or graphing) calculator knows the values of the sine, cosine and tangent of all of these angles. Depending on your calculator, you should have [SIN], [COS], and [TAN] buttons. Use these to find the sine, cosine, and tangent of any acute angle. One application of the trigonometric ratios is to use them to find the missing sides of a right triangle. All you need is one angle, other than the right angle, and one side.
#### Example A
Find the trigonometric value, using your calculator. Round to 4 decimal places.
a) \begin{align*}\sin 78^\circ\end{align*}
b) \begin{align*}\cos 60^\circ\end{align*}
c) \begin{align*}\tan 15^\circ\end{align*}
Depending on your calculator, you enter the degree and then press the trig button or the other way around. Also, make sure the mode of your calculator is in DEGREES.
a) \begin{align*}\sin 78^\circ = 0.97815\end{align*}
b) \begin{align*}\cos 60^\circ = 0.5\end{align*}
c) \begin{align*}\tan 15^\circ = 0.26795\end{align*}
#### Example B
Find the value of each variable. Round your answer to the nearest tenth.
We are given the hypotenuse. Use sine to find \begin{align*}b\end{align*}, and cosine to find \begin{align*}a\end{align*}. Use your calculator to evaluate the sine and cosine of the angles.
#### Example C
Find the value of each variable. Round your answer to the nearest tenth.
We are given the adjacent leg to \begin{align*}42^\circ\end{align*}. To find \begin{align*}c\end{align*}, use cosine and use tangent to find \begin{align*}d\end{align*}.
Any time you use trigonometric ratios, use only the information that you are given. This will result in the most accurate answers.
Watch this video for help with the Examples above.
#### Concept Problem Revisited
Use trigonometric ratios to find the missing sides. Round to the nearest tenth.
Find the length of \begin{align*}a\end{align*} and \begin{align*}b\end{align*} using sine or cosine ratios:
### Vocabulary
Trigonometry is the study of the relationships between the sides and angles of right triangles. The legs are called adjacent or opposite depending on which acute angle is being used. The three trigonometric (or trig) ratios are sine, cosine, and tangent.
### Guided Practice
1. What is \begin{align*}\tan 45^\circ\end{align*}?
2. Find the length of the missing sides and round your answers to the nearest tenth: .
3. Find the length of the missing sides and round your answers to the nearest tenth: .
1. Using your calculator, you should find that \begin{align*}\tan 45^\circ=1\end{align*}?
2. Use tangent for \begin{align*}x\end{align*} and cosine for \begin{align*}y\end{align*}.
3. Use tangent for \begin{align*}y\end{align*} and cosine for \begin{align*}x\end{align*}.
### Practice
Use your calculator to find the value of each trig function below. Round to four decimal places.
1. \begin{align*}\sin 24^\circ\end{align*}
2. \begin{align*}\cos 45^\circ\end{align*}
3. \begin{align*}\tan 88^\circ\end{align*}
4. \begin{align*}\sin 43^\circ\end{align*}
5. \begin{align*}\tan 12^\circ\end{align*}
6. \begin{align*}\cos 79^\circ\end{align*}
7. \begin{align*}\sin 82^\circ\end{align*}
Find the length of the missing sides. Round your answers to the nearest tenth.
1. Find \begin{align*}\sin 80^\circ\end{align*} and \begin{align*}\cos 10^\circ\end{align*}.
2. Use your knowledge of where the trigonometric ratios come from to explain your result to the previous question.
3. Generalize your result to the previous two questions. If \begin{align*}\sin\theta=x\end{align*}, then \begin{align*}\cos ?=x\end{align*}.
4. How are \begin{align*}\tan\theta \end{align*} and \begin{align*}\tan(90-\theta)\end{align*} related? Explain.
### Vocabulary Language: English
Hypotenuse
Hypotenuse
The hypotenuse of a right triangle is the longest side of the right triangle. It is across from the right angle.
Legs of a Right Triangle
Legs of a Right Triangle
The legs of a right triangle are the two shorter sides of the right triangle. Legs are adjacent to the right angle.
Trigonometric Ratios
Trigonometric Ratios
Ratios that help us to understand the relationships between sides and angles of right triangles.
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# Calculating with polynomials
For a full and interactive version of this page with examples and exercises, check out the Pass Your Math platform!
If $$f(x)$$ and $$g(x)$$ are two polynomials and $$c$$ is a real number, then the following expressions also are polynomials:
• $$c\cdot {f(x)}$$
• $$f(x)+g(x)$$
• $$f(x)-g(x)$$
• $$f(x)\cdot g(x)$$
Rules of calculation for polynomials
• Multiplying a polynomial by a constant is equivalent to multiplying each term of the polynomial by that constant
• Addition of two polynomials in $$x$$ is equivalent to adding the coefficients of terms with the same power of $$x$$.
• Subtraction of polynomial $$g(x)$$ by a polynomial $$f(x)$$ is the same as subtracting the coefficients of terms in $$g(x)$$ by the coefficients of the same power of $$x$$ in $$f(x)$$.
• Multiplication of two polynomials is obtained by multiplying each term of one polynomial by each term of the other polynomial and adding all the products.
The rules specify how we can add, subtract and multiply polynomials. The quotient $$\frac{f(x)}{g(x)}$$ of two polynomials is not always a polynomial, but does result in a rational function.
Degree of polynomials that result from arithmetic operations
Let $$f(x)$$ and $$g(x)$$ be polynomials of degree respectively $$m$$ and $$n$$, and let $$c$$ be a real number.
• The degree of $$c\cdot f(x)$$ is the degree of $$f(x)$$ if $$c\ne0$$.
• The degree of $$f(x)\cdot g(x)$$ is the sum of the degrees $$f(x)$$ and $$g(x)$$.
• if $$m\gt n$$, then the degree of $$f(x)+g(x)$$ is equal to the degree of $$f(x)$$.
• If $$m=n$$, then the degree of $$f(x)+g(x)$$ is less than or equal to the degree of $$f(x)$$.
Let’s practice!
What is the product of polynomial $$f(x)=x^3+x^2-4\cdot x-4$$ and the constant $$2$$?
In order to compute the product of $$f(x)$$ and $$2$$ we multiply the coefficient of each power of $$x$$ in $$f(x)$$ with $$2$$:
$$\begin{array}{rcl}2\cdot f(x)&=& 2\cdot \left(x^3+x^2-4\cdot x-4\right)\\ &=&{ -8+{ (2\cdot1)\cdot x^2 }+{ (2\cdot-4)\cdot x }+{ (2\cdot-4) }} \\ &=& 2\cdot x^3+2\cdot x^2-8\cdot x-8\end{array}$$
$$2\cdot x^3+2\cdot x^2-8\cdot x-8$$
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# Activities to Introduce Students to Partial Sums
Partial sums is a mathematical concept that is often studied in elementary math classrooms. It is the sum of all the partial numbers in a series. This concept is important in mathematics because it is used to add numbers quickly. However, some students might find the concept of partial sums challenging at first. It is essential to provide activities that will help them grasp the idea of partial sums. In this article, we will discuss some of the activities that teachers can use to introduce students to partial sums.
1. The Count and Circle Game
This activity is a fun way of introducing students to partial sums. The teacher should write some numbers on the board and ask the students to count them. The students should then circle all the odd numbers, and the teacher will add them to get the partial sum. The process is repeated several times with different numbers until the students understand the concept of partial sums.
2. The Partial Sums Tower
In this activity, the teacher can provide the students with building blocks and ask them to construct a tower by building one block at a time. Each block will have a number on it, and the students will add them as they go. For example, the students will add 1+2+3+4+5 to get the partial sum of 15.
3. The Partial Sums Bingo
Bingo is a popular game that can be used to teach partial sums to students. The teacher should prepare bingo cards with different numbers written on them. The students will then be asked to add the numbers on their cards to get the partial sum. The first student to complete the partial sum wins the game.
4. The Partial Sum Maze
This activity involves creating a maze with numbers written on it. The students will navigate the maze and add the numbers they come across to get the partial sum. The maze can be designed with different levels of difficulty, depending on the grade level of the students.
5. The Partial Sums Chart
The teacher can introduce the concept of partial sums using a chart. The chart will have different numbers written on it in rows and columns. The students will be asked to add the numbers in each row to get the partial sum. The teacher can also challenge the students by asking them to find the partial sum of the diagonal numbers.
In conclusion, there are several fun and engaging activities that teachers can use to introduce students to partial sums. These activities will help the students understand the concept and apply it in real-life situations. The key to teaching partial sums is to make it fun and interactive and provide opportunities to practice repeatedly until the students master the skill.
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GREATEST COMMON FACTOR WORD PROBLEMS
Problem 1 :
Find the greatest possible length that can be used to measure exactly the lengths 84 feet, 98 feet and 126 feet.
Solution :
The greatest possible length that can be used to measure the given lengths is the greatest common factor of 84, 63 and 97.
Find the greatest common factor of 84, 98 and 126.
Product of common factors of 84, 98 and 126 is
= 2 x 7
= 14
GCF (84, 98, 126) = 14.
The greatest possible length that can be used to measure the given lengths is 14 feet.
Problem 2 :
A warehouse has three shelves that can hold 8, 12, or 16 skateboards. Each shelf has sections holding the same number of skateboards. What is the greatest number of skateboards that can be put in a section? Explain
Solution :
Write the factors of 8, 12 and 16.
8 = 1, 2, 4, 8
12 = 1, 2, 3, 4, 6, 12
16 = 1, 2, 4, 8, 16
The greatest common factor of 8, 12 and 16 is 4.
So, the greatest number of skateboards can be put in a section is 4.
Problem 3 :
Lily has 15 oranges, 9 peaches, and 18 pears. She wants to put all of the fruit into decorative baskets. Each basket must have the same number of pieces of fruit in it. Without mixing fruits, what is the greatest number of pieces of fruit Lily can put in each basket? Explain.
Solution :
Write the factors of 15, 9 and 18.
15 = 1, 2, 3, 5, 15
9 = 1, 3, 9
18 = 1, 2, 3, 6, 9, 18
The greatest common factor of 15, 9 and 18 is 3.
So, the greatest number of pieces of fruit Lily can put in each basket is 3.
Problem 4 :
A store has 120 lbs and 250 lbs of two kinds of meat. The store wants to sell the meat by filling the two kinds in boxes of equal volumes and no meat left over. Find the greatest volume of such a box.
Solution :
The given two quantities 120 and 250 can be divided by 2, 5, 10,... exactly. That is, both kinds of meat can be sold in boxes of equal volume of 2, 5, 10,...
But, the volume of meat filled in boxes must be greatest.
So, we have to find the largest number which exactly divides 120 and 250. That is the highest common factor (HCF) of 120 and 250.
HCF (120, 250) = 10
The 1st kind 120 lbs is sold in 12 boxes of volume 10 lbs in each box.
The 2nd kind 250 lbs is sold in 25 boxes of volume 10 lbs in each box.
Hence, the greatest volume of the box is 10 lbs.
Problem 5 :
Two numbers are in the ratio 2 : 3. If the second number is 27, find their greatest common factor.
Solution :
Because the two numbers are in the ratio 2 : 3, the numbers can be assumed as 2x and 3x.
Find the greatest common multiple of 2x and 3x.
The common factor of 2x and 3x is x.
Greatest common factor (2x, 3x) = x
But, it is given that the second number is 27.
Then,
3x = 27
Divide each side by 3.
x = 9
So, the greatest factor of the two numbers is 9.
Problem 6 :
Two numbers are in the ratio 3 : 5 and their greatest common factor is 18. Find the numbers.
Solution :
Because the two numbers are in the ratio 3 : 5, the numbers can be assumed as 3x and 5x.
Greatest common factor (3x, 5x) = x
But, it is given that the greatest common factor of the two numbers is 18.
Then,
x = 18
Substitute x = 18 in 3x and 5x.
3x = 3(18) = 54
5x = 5(18) = 90
So, the two numbers are 54 and 90.
Kindly mail your feedback to v4formath@gmail.com
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# SAT Math : How to find the solution to an equation
## Example Questions
### Example Question #62 : Algebra
If 6h – 2g = 4g + 3h
In terms of g, h = ?
2g
3g
4g
5g
g
2g
Explanation:
If we solve the equation for b, we add 2g to, and subtract 3h from, both sides, leaving 3h = 6g. Solving for h we find that h = 2g.
### Example Question #341 : Algebra
If 2x + y = 9 and y – z = 4 then 2x + z = ?
Cannot be determined
29
21
13
5
5
Explanation:
If we solve the first equation for 2x we find that 2x = 9 – y. If we solve the second equation for z we find z = –4 + y. Adding these two manipulated equations together we see (2x) + (y) = (9 – y)+(–4 + y).
The y’s cancel leaving us with an answer of 5.
### Example Question #61 : How To Find The Solution To An Equation
11/(x – 7) + 4/(7 – x) = ?
15/(7 – x)
15
(–7)/(7 – x)
7/(7 – x)
15/(x – 7)
(–7)/(7 – x)
Explanation:
We must find a common denominator and here they changed the first fraction by removing a negative from the numerator and denominator, leaving –11/(7 – x). We add the numerators and keep the same denominator to find the answer.
Jack has 14 coins consisting of nickels and dimes that total $0.90. How many nickels does Jack have? Possible Answers: 8 10 12 6 4 Correct answer: 10 Explanation: In order to solve this question we must first set up two equations. We know the number of nickels and the number of dimes equals 14 (n + d = 14). We also know the value of nickels and dimes. For the second equation we simply multiply the number of nickels we have by their value, added to the number of dimes we have by their value to get the total (0.05n + 0.10d = 0.90). Solve the first equation for n giving us n = 14 – d. We can then substitute 14 – d into the second equation wherever there is an “n”. Giving us 0.05 (14 – d) + 0.10d = 0.90. When we solve the equation we find the number of dimes is d = 4; therefore the remaining 10 coins must be nickels. ### Example Question #31 : Linear / Rational / Variable Equations If a = 1/3b and b = 4c, then in terms of c, a – b + c = ? Possible Answers: c 5/3c –5/3c –11/3c Correct answer: –5/3c Explanation: To begin we must find how a and c relate to each other. Using the second equation we know that we can plug in 4c everywhere there is a b in the first equation, giving us a = 4/3c. Now we can plug into the last equation. We plug in 4/3c for a, 4c for b, and leave c as it is. We must find a common denominator (4/3c – 12/3c + 3/3c) and add the numerators to find that our equation equals –5/3c. ### Example Question #11 : How To Find The Solution To An Equation If x= 8, then x2(4/(3 – x))(2/(4 – x)) – (4/x2) = ? Possible Answers: 22 35 16 15 0 Correct answer: 15 Explanation: There is really no need to alter this equation using algebra. Simply find that x = 2 and plug in. We see that 4(4)(1) – (1)=15. Remember to use correct order of operations here (parentheses, exponents, multiplication, division, addition, subtraction). ### Example Question #71 : Algebra Find the intersection of the following two equations: 3x + 4y = 6 15x - 4y = 3 Possible Answers: (0.5, 1.125) (1, 0.5) (18, 0) (0.2, 0) (3, 4) Correct answer: (0.5, 1.125) Explanation: The point of intersection for two lines is the same as the values of x and y that mutually solve each equation. Although you could solve for one variable and replace it in the other equation, use elementary row operations to add the two equations since you have a 4y and -4y: 3x + 4y = 6 15x - 4y = 3 18x = 9; x = 0.5 You can now plug x into the first equation: 3 * 0.5 + 4y = 6; 1.5 +4y = 6; 4y = 4.5; y = 1.125 Therefore, our point of intersection is (0.5, 1.125) ### Example Question #32 : Linear / Rational / Variable Equations Two cars start 25 mile apart and drive away from each other in opposite directions at speeds of 50 and 70 miles per hour. In approximately how many minutes will they be 400 miles apart? Possible Answers: None of the other answers 3.33 200 3.125 187.5 Correct answer: 187.5 Explanation: The cars have a distance from each other of 25 + 120t miles, where t is the number of hours, 25 is their initial distance and 120 is 50 + 70, or their combined speeds. Solve this equation for 400: 25 + 120t = 400; 120t = 375; t = 3.125 However, the question asked for minutes, so we must multiply this by 60: 3.125 * 60 = 187.5 minutes. ### Example Question #33 : Linear / Rational / Variable Equations A given university has an average professor pay of$40,000 a year and an average administrator pay of $45,000 per year. If the ratio of professors to administrators is 4 to 3, and the total pay for professors and administrators in a year is$40,415,000, how many professors does the college have?
548
475
411
500
375
548
Explanation:
Set up a system of linear equations based on our data:
40,000P + 45,000A = 40,415,000
P/A = 4/3
To make things easiest, solve the second equation for A in terms of P:
A = (3/4) P
Replace this value into the first equation:
40,000P + 45,000 * (3/4)P = 40,415,000
Simplify:
40,000P + 33,750P = 40,415,000
73,750P = 40,415,000
P = 548 (The number of professors)
### Example Question #19 : How To Find The Solution To An Equation
Abby works at a car dealership and receives a commission "c" which is a percent of the profit the dealership makes from the sale, which is the difference between the price "p" of the car and the value "v" of the car. How much, in dollars, does the dealership earn per transaction?
pv(0.01c)
(p – v)(1 – 0.01c)
p(v – 0.01c)
(p – v)(1 – c)
(p – v)(0.01c)
(p – v)(1 – 0.01c)
Explanation:
To show that c is of the profit of the transaction, we must represent the profit as the difference between the price and the value of the car, or "(p – v)"
To show that Abby's commission in dollars is a percentage of the profit, we use 0.01 * c to convert the commission she earns to a percent.
To shift the earnings from Abby to the dealership (which is what the question requires of us), we must take 1 – 0.01c since this will accommodate for the remaining percentage. For example, it shifts 75% (0.75) to 25% (1 – 0.75 or 0.25).
Putting this all together, we get a final expression of:
(p – v)(1 – 0.01c) = dealership earnings
Check answer with arbitrary values: letting p = 300, v = 200, and c = 20, we get a value of 80 which makes sense as the $100 profit must be distributed evenly between Abby ($20) and the dealership (\$80).
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# Given that the slope of a line is -1/5, what is the slope of a line that is perpendicular to it?
Jan 28, 2017
Slope is 5
#### Explanation:
The perpendicular to a given slope is its negative reciprocal. This means the fraction is flipped and multiplied by $- 1$. So perpendicular to $- \frac{1}{5}$ is $5$
Jan 28, 2017
$m = 5$
#### Explanation:
The perpendicular slope of any original slope is derived by negating the original slope and then "flipping" the fraction. By "flipping" the fraction, I mean find the inverse of the original slope. So for example:
Original slope: ${m}_{\text{orig}} = - \frac{1}{5}$
Step 1. Negate the original slope. Remember that a negative of a negative is a positive.
$- \left(- \frac{1}{5}\right) = \frac{1}{5}$
Step 2. "Flip" the fraction, finding it's inverse. Remember that whole numbers can be turned automatically into fractions by placing them over a 1.
$\frac{5}{1} = 5 = {m}_{\text{perp}}$
More generally, you can always find the perpendicular slope using this formula:
${m}_{\text{perp")=-1/m_("orig}}$
Jan 28, 2017
All you have to do is remember and follow the method in the first bit of the explanation.
The rest is supportive expansion and contains the actual solution.
#### Explanation:
$\textcolor{b l u e}{\text{THE REALLY IMPORTANT BIT: The basic rule}}$
Let the slope (gradient) of the first line be $m$
Then the gradient of the perpendicular line is $- \frac{1}{m}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Comment}}$
This s true for any straight or curved line graph.
The only difference is that for a straight line it is a constant value but for a curved line it changes to suit the gradient at each and every point
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{The calculation}}$
The given slope is $- \frac{1}{5}$ which is a constant. Thus the graph is that of a straight line.
The gradient of the perpendicular is: $\left(- 1\right) \times \left(- \frac{5}{1}\right) = + \frac{5}{1}$
$\text{ } \textcolor{red}{\uparrow}$
" "color(red)("Inverting the given gradient")
.................................................................................
$\textcolor{b l u e}{\text{Foot note}}$
I left the answer in the format of $\frac{5}{1}$ as it represents a ratio.
For every 1 along you go up 5
The teacher will expect you to write the answer gradient as 5 and not $\frac{5}{1}$
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# Chapter2.6
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### Chapter2.6
1. 1. Warm Up California Standards Lesson Presentation Preview
2. 2. Warm Up Add or subtract. 1 + – 1. 2 5 2. 7 12 3 – 2 3. 6.5 + (–1.2) 4. 3.4 – 0.9 5.3 2.5 3 5 11 12 1 5 1 1 3 1
3. 3. NS1.2 Add, subtract , multiply, and divide rational numbers (integers, fractions , and terminating decimals) and take positive rational numbers to whole-number powers. NS2.2 Add and subtract fractions by using factoring to find a common denominator. California Standards
4. 4. To add and subtract fractions with unlike denominators, first find a common denominator using one of these methods: Method 1 Find a common denominator by multiplying the denominators. Method 2 Find the least common denominator (LCD).
5. 5. Find a common denominator: 8(7)=56. 2 7 + 1 8 Multiply by fractions equal to 1. Rewrite with a common denominator. Additional Example 1A: Adding and Subtracting Fractions with Unlike Denominators Add or subtract. Method 1: Add numerators. Keep the denominator. = 23 56 2 7 + 1 8 7 7 8 8 = 16 56 + 7 56 = 7 + 16 56 =
6. 6. 1 6 1 – 1 5 8 List the multiples of each denominator and find the LCD. Write as improper fractions. Rewrite with the LCD. Multiply by fractions equal to 1. Subtract numerators. Keep the denominator. Additional Example 1B: Adding and Subtracting Fractions with Unlike Denominators Add or subtract. Method 2: Multiples of 6: 6, 12, 18, 24, 30 Multiples of 8: 8, 16, 24, 32 11 24 = – – 7 6 13 8 = 13 8 – 7 6 4 4 3 3 = 39 24 – 28 24 = 28 – 39 24 =
7. 7. Find a common denominator: 3(8)=24. 5 8 + 1 3 Multiply by fractions equal to 1. Rewrite with the LCD. Check It Out! Example 1A Add or subtract. Method 1: Add numerators. Keep the denominator. 23 24 = 5 8 + 1 3 8 8 3 3 = 15 24 + 8 24 = = 8 + 15 24
8. 8. 1 6 2 + 3 4 List the multiples of each denominator and find the LCD Write as an improper fraction. Rewrite with the LCD. Multiply by fractions equal to 1. Check It Out! Example 1B Add or subtract. Method 2: Add numerators. Keep the denominator. Multiples of 6: 6, 12, 18, 24, 30 Multiples of 4: 4, 8, 12, 16 11 12 = 2 35 12 = 3 4 + 13 6 = 3 4 + 13 6 2 2 33 = 9 12 + 26 12 = 26 + 9 12 =
9. 9. 25 56 Write the prime factorization of each denominator. Circle the common factors. List all the prime factors of the denominators, using the circled factors only once. Multiply. Additional Example 2: Using Factoring to Find the LCD Find + . Write the answer in simplest form. 37 84 2, 2, 2, 7, 3 2 ∙ 2 ∙ 2 ∙ 7 ∙ 3 = 168 The LCD is 168. Factors of 56: 2 ∙ 2 ∙ 2 ∙ 7 Factors of 84: 2 ∙ 2 ∙ 3 ∙ 7
10. 10. 25 56 Multiply by fractions equal to 1 to get a common denominator. Rewrite using the LCD. Additional Example 2 Continued Find + . Write the answer in simplest form. 37 84 168 ÷ 56 = 3 168 ÷ 84 = 2 Add numerators. Keep the denominator. = + 25 56 37 84 3 3 2 2 = + 75 168 74 168 = 149 168
11. 11. Write the prime factorization of each denominator. Circle the common factors. List all the prime factors of the denominators, using the circled factors only once. Multiply. Check It Out! Example 2 Find + . Write the answer in simplest form. 2, 2, 2, 2, 2, 5 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 5 = 160 The LCD is 160. Factors of 32: 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 Factors of 80: 2 ∙ 2 ∙ 2 ∙ 2 ∙ 5 19 32 9 80
12. 12. Multiply by fractions that equal to 1 to get a common denominator. Rewrite using the LCD. 160 ÷ 32 = 5 160 ÷ 80 = 2 Add numerators. Keep the denominator. Check It Out! Example 2 Continued Find + . Write the answer in simplest form. = + 19 32 9 80 5 5 2 2 = + 95 160 18 160 = 113 160 19 32 9 80
13. 13. Additional Example 3: Consumer Application Subtract both amounts from 36 to find the amount of ribbon left. Write as improper fractions. The LCD is 8. Two dancers are making necklaces from ribbon for their costumes. They need pieces measuring 13 inches and 12 inches. How much ribbon will be left over after the pieces are cut from a 36-inch length? Simplify. 36 – 12 – 13 7 8 3 4 There will be 9 inches left. 3 8 7 8 3 4 103 8 55 4 36 1 – – = 103 8 288 8 – – 110 8 = , or 9 3 8 75 8 =
14. 14. Check It Out! Example 3 Subtract both amounts from 12 to find the amount of board left. Write as improper fractions. The LCD is 12. Simplify. Fred and Jose are building a tree house. They need to cut a 6 foot piece of wood and a 4 foot piece of wood from a 12 foot board. How much of the board will be left? 12 – 6 – 4 3 4 5 12 There will be foot left. 5 6 5 12 3 4 27 4 53 12 12 1 – – = 81 12 144 12 – – 53 12 = , or 5 6 10 12 =
15. 15. Evaluate t – for t = . Additional Example 4: Evaluating Expressions with Rational Numbers Multiply by fractions equal to 1. 4 5 5 6 Rewrite with a common denominator: 6(5) = 30. Simplify. 4 5 – 5 6 24 30 – 25 30 = Substitute for t. 5 6 t – = 4 5 4 5 – 5 6 5 5 6 6 = 1 30 =
16. 16. Check It Out! Example 4 Multiply by fractions equal to 1. Evaluate – h for h =– . 5 9 Rewrite with the LCD. Simplify. 7 12 Substitute – for h. 7 12
17. 17. Add or subtract. 1. + 5 14 1 7 2. 12 2 3 8 1 – 3. – 2 + 3 5 2 3 4. Evaluate – n for n = . 1 38 9 16 Lesson Quiz Robert is 5 feet 6 inches tall. Judy is 5 feet 3 inches tall. How much taller is Robert than Judy? 5. 3 4 12 13 16 7 1 6 – 2 1 15 12 2 in. 3 4
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Strategies to Add & Subtract within 20 | Grade K-2 Lesson Plan & PDF Worksheet | Generation Genius
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WHAT IS A STRATEGY?
A strategy is a way you can tackle a problem. In math, there are often many strategies you use to solve a problem. These strategies can all give you the correct answer! Just pick the one that makes the most sense to you.
To better understand strategies…
WHAT IS A STRATEGY?. A strategy is a way you can tackle a problem. In math, there are often many strategies you use to solve a problem. These strategies can all give you the correct answer! Just pick the one that makes the most sense to you. To better understand strategies…
## LET’S BREAK IT DOWN!
### Counting all to find a sum.
You have 9 toy cars and 7 toy cars. How many cars do you have in all? 9 + 7 = ? You can count all of the cars to find the sum. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16. There are 16 cars in all.
Counting all to find a sum. You have 9 toy cars and 7 toy cars. How many cars do you have in all? 9 + 7 = ? You can count all of the cars to find the sum. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16. There are 16 cars in all.
### Make a ten to find how many cars.
Counting all of the cars will get you the correct answer but it takes a long time. Especially when numbers get bigger! You can make a ten to find how many in all. 9 + 1 = 10, so take 1 from 7 and add it to the 9. That leaves 6 left. So, 9 + 7 = 10 + 6. There are 16 cars in all.
Make a ten to find how many cars. Counting all of the cars will get you the correct answer but it takes a long time. Especially when numbers get bigger! You can make a ten to find how many in all. 9 + 1 = 10, so take 1 from 7 and add it to the 9. That leaves 6 left. So, 9 + 7 = 10 + 6. There are 16 cars in all.
### Use doubles facts to add spots.
A Dalmatian has 8 spots and 9 spots. How many spots in all? 8 and 9 are close to a double. 8 + 8 = 16. 9 is 1 more than 8. So 8 + 9 is the same as 8 + 8 + 1, or 16 + 1 = 17. There are 17 spots in all.
Use doubles facts to add spots. A Dalmatian has 8 spots and 9 spots. How many spots in all? 8 and 9 are close to a double. 8 + 8 = 16. 9 is 1 more than 8. So 8 + 9 is the same as 8 + 8 + 1, or 16 + 1 = 17. There are 17 spots in all.
### Count to compare cherries.
A jar has 15 cherries. You use 9 cherries. How many cherries are left? You can count back from 15 to 9 find the difference. 14, 13, 12, 11, 10, 9. You can also count on to find a difference. Count on from 9 to 15. 10, 11, 12, 13, 14, 15. You counted 6 numbers both ways. There are 6 cherries left.
Count to compare cherries. A jar has 15 cherries. You use 9 cherries. How many cherries are left? You can count back from 15 to 9 find the difference. 14, 13, 12, 11, 10, 9. You can also count on to find a difference. Count on from 9 to 15. 10, 11, 12, 13, 14, 15. You counted 6 numbers both ways. There are 6 cherries left.
### Make a ten to subtract.
You can make a ten to subtract. Start with 15 cherries. 15 is 10 and 5 more, so break 9 into 5 and 4. Subtract 5: 15 – 5 = 10. Then subtract 4 more. 10 – 4 = 6. So, 15 – 9 = 6.
Make a ten to subtract. You can make a ten to subtract. Start with 15 cherries. 15 is 10 and 5 more, so break 9 into 5 and 4. Subtract 5: 15 – 5 = 10. Then subtract 4 more. 10 – 4 = 6. So, 15 – 9 = 6.
## STRATEGIES TO ADD & SUBTRACT WITHIN 20 VOCABULARY
Finding the total, or sum, by combining two or more numbers.
Sum
Subtraction
Taking one number away from another.
Difference
The result of subtracting.
Decompose
Breaking a number into two or more parts.
When adding two numbers, break one number into two parts, so that one part added to the first number makes 10.
When adding two numbers, break one number into two parts so that two of the numbers make a doubles fact.
When subtracting two numbers, count on from the smaller number to the bigger number. How many you counted on is the difference.
When subtracting two numbers, count back from the bigger number the amount given by the smaller number. Where you stop counting is the difference.
## STRATEGIES TO ADD & SUBTRACT WITHIN 20 DISCUSSION QUESTIONS
### What strategy would you use to find 6 + 7?
Answers will vary. Students may make a 10 by adding 6 + 4 + 3. Students may use a doubles fact: 6 + 6 + 1.
### What strategy would you use to find 8 + 4?
Answers will vary. Students may make a 10 by adding 8 + 2 + 2. Students may use doubles facts and add 4 + 4 + 4.
### How do you decide which addition strategy to use?
Answers will vary. Students may say they always use one strategy or that they choose the doubles strategy if the addends are the same or 1 apart.
### What strategy would you use to find 14 – 7?
Answers will vary. Students may describe a counting back strategy: 14 – 4 = 10, 10 – 3 = 7. They may count on 7 + 3 = 10, 10 + 4 = 14. 3 + 4 = 7. Students may also remember that 14 and 7 are part of a doubles fact, so the difference must be 7.
### How do you decide which subtraction strategy to use?
Answers will vary. Students may say they always use one strategy or that they choose the count on strategy when they have to 'cross 10' when subtracting.
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# PP 11.5 - Conic Sections Conic sections, or conics, get...
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Conic Sections
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Conic sections, or conics, get their name because they result from intersecting a cone with a plane.
PARABOLA A parabola is the set of points in a plane that are equidistant from a fixed point F (the focus) and a fixed line (the directrix). The vertex is the point halfway between the focus and the directrix.
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The line through the focus perpendicular to the directrix is called the axis of symmetry of the parabola.
Equation of a Parabola To obtain an equation for a parabola, we place its vertex at the origin O and its directrix parallel to the x -axis. We will use the definition of parabola to get its equation.
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The defining property of a parabola is that the distance to the focus equals the distance to the directrix. 2 2 ( ) x y p y p + - = +
Hence, 2 2 2 2 2 2 2 2 2 2 ( ) ( ) 2 2 4 x y p y p y p x y py p y py p x py + - = + = + + - + = + + = 2 2 ( ) x y p y p + - = +
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Similar arguments yield the equations of the other three possibilities. Vertical Parabolas
Horizontal Parabolas; notice that these curves do NOT represent functions.
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Example Find the focus and directrix of the parabola y 2 + 10 x = 0 and sketch the graph. 2 5 2 5 2 ) ( 4 10 - = - = - = p x x y 2 5 2 5 : ), 0 , ( = - = x D F Horizontal parabola
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ELLIPSE An ellipse is the set of points in a plane the sum of whose distances from two fixed points F 1 and F 2 is a constant. These points are called the foci. Q
To obtain the equation for an ellipse, we place the foci on the x -axis at the points (– c , 0) and ( c , 0) so that the origin is halfway between the foci.
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P ( x , y ) is a point on the ellipse, the definition says that |PF 1 | + | PF 2 | = 2 a Let the sum of the distances from a point on the ellipse to the foci be 2 a > 0. 2
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## This note was uploaded on 01/21/2011 for the course PHYS 4A 60865 taught by Professor L. oldewurtel during the Fall '09 term at Irvine Valley College.
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PP 11.5 - Conic Sections Conic sections, or conics, get...
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# 8.6: Partial Fraction Decomposition
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This section uses systems of linear equations to rewrite rational functions in a form more palatable to Calculus students. In College Algebra, the function
$\label{falg} f(x) = \dfrac{x^2-x-6}{x^4+x^2} \tag{1}$
is written in the best form possible to construct a sign diagram and to find zeros and asymptotes, but certain applications in Calculus require us to rewrite $$f(x)$$ as
$\label{fcalc} f(x) = \dfrac{x+7}{x^2+1} - \dfrac{1}{x} - \dfrac{6}{x^2} \tag{2}$
If we are given the form of $$f(x)$$ in (2), it is a matter of Intermediate Algebra to determine a common denominator to obtain the form of $$f(x)$$ given in (1). The focus of this section is to develop a method by which we start with $$f(x)$$ in the form of (1) and ‘resolve it into partial fractions’ to obtain the form in (2). Essentially, we need to reverse the least common denominator process. Starting with the form of $$f(x)$$ in (1), we begin by factoring the denominator
$\dfrac{x^2-x-6}{x^4+x^2} = \dfrac{x^2-x-6}{x^2 \left(x^2+1\right)}\nonumber$
We now think about which individual denominators could contribute to obtain $$x^2 \left(x^2+1\right)$$ as the least common denominator. Certainly $$x^2$$ and $$x^2+1$$, but are there any other factors? Since $$x^2+1$$ is an irreducible quadratic1 there are no factors of it that have real coefficients which can contribute to the denominator. The factor $$x^2$$, however, is not irreducible, since we can think of it as $$x^2 = xx = (x-0)(x-0)$$, a so-called ‘repeated’ linear factor.2 This means it’s possible that a term with a denominator of just $$x$$ contributed to the expression as well. What about something like $$x \left(x^2+1\right)$$? This, too, could contribute, but we would then wish to break down that denominator into $$x$$ and $$\left(x^2+1\right)$$, so we leave out a term of that form. At this stage, we have guessed
$\dfrac{x^2-x-6}{x^4+x^2} = \dfrac{x^2-x-6}{x^2 \left(x^2+1\right)} = \dfrac{?}{x} + \dfrac{?}{x^2} + \dfrac{?}{x^2+1}\nonumber$
Our next task is to determine what form the unknown numerators take. It stands to reason that since the expression $$\frac{x^2-x-6}{x^4+x^2}$$ is ‘proper’ in the sense that the degree of the numerator is less than the degree of the denominator, we are safe to make the that all of the partial fraction resolvents are also. This means that the numerator of the fraction with $$x$$ as its denominator is just a constant and the numerators on the terms involving the denominators $$x^2$$ and $$x^2+1$$ are at most linear polynomials. That is, we guess that there are real numbers $$A$$, $$B$$, $$C$$, $$D$$ and $$E$$ so that
$\dfrac{x^2-x-6}{x^4+x^2} = \dfrac{x^2-x-6}{x^2 \left(x^2+1\right)} = \dfrac{A}{x} + \dfrac{Bx+C}{x^2} + \dfrac{Dx+E}{x^2+1}\nonumber$
However, if we look more closely at the term $$\frac{Bx+C}{x^2}$$, we see that $$\frac{Bx+C}{x^2} = \frac{Bx}{x^2} + \frac{C}{x^2} = \frac{B}{x} + \frac{C}{x^2}$$. The term $$\frac{B}{x}$$ has the same form as the term $$\frac{A}{x}$$ which means it contributes nothing new to our expansion. Hence, we drop it and, after re-labeling, we find ourselves with our new guess:
$\dfrac{x^2-x-6}{x^4+x^2} = \dfrac{x^2-x-6}{x^2 \left(x^2+1\right)} = \dfrac{A}{x} + \dfrac{B}{x^2} + \dfrac{Cx+D}{x^2+1}\nonumber$
Our next task is to determine the values of our unknowns. Clearing denominators gives
$x^2 - x- 6 = Ax\left(x^2+1\right) + B\left(x^2+1\right) + (Cx+D)x^2\nonumber$
Gathering the like powers of $$x$$ we have
$x^2 - x - 6 = (A+C)x^3+(B+D)x^2+Ax + B\nonumber$
In order for this to hold for all values of $$x$$ in the domain of $$f$$, we equate the coefficients of corresponding powers of $$x$$ on each side of the equation3 and obtain the system of linear equations
$\left\{ \begin{array}{lrcrl} (E1) & A+C & = & 0 & \text{From equating coefficients of x^{3}} \\ (E2) & B+D & = & 1 & \text{From equating coefficients of x^{2}} \\ (E3) & A & = & -1 & \text{From equating coefficients of x} \\ (E4) & B & = & -6 & \text{From equating the constant terms} \\ \end{array} \right.\nonumber$
To solve this system of equations, we could use any of the methods presented in Sections 8.1 through 8.5, but none of these methods are as efficient as the good old-fashioned substitution you learned in Intermediate Algebra. From $$E3$$, we have $$A=-1$$ and we substitute this into $$E1$$ to get $$C = 1$$. Similarly, since $$E4$$ gives us $$B=-6$$, we have from $$E2$$ that $$D = 7$$. We get
$\dfrac{x^2-x-6}{x^4+x^2} = \dfrac{x^2-x-6}{x^2 \left(x^2+1\right)} = -\dfrac{1}{x} - \dfrac{6}{x^2} + \dfrac{x+7}{x^2+1}\nonumber$
which matches the formula given in (2). As we have seen in this opening example, resolving a rational function into partial fractions takes two steps: first, we need to determine the form of the decomposition, and then we need to determine the unknown coefficients which appear in said form. Theorem 3.16 guarantees that any polynomial with real coefficients can be factored over the real numbers as a product of linear factors and irreducible quadratic factors. Once we have this factorization of the denominator of a rational function, the next theorem tells us the form the decomposition takes. The reader is encouraged to review the Factor Theorem (Theorem 3.6) and its connection to the role of multiplicity to fully appreciate the statement of the following theorem.
###### Theorem 8.10.
Suppose $$R(x) = \dfrac{N(x)}{D(x)}$$ is a rational function where the degree of $$N(x)$$ less than the degree of $$D(x)$$ and $$N(x)$$ and $$D(x)$$ have no common factors.a
• If $$\alpha$$ is a real zero of $$D$$ of multiplicity $$m$$ which corresponds to the linear factor $$ax+b$$, the partial fraction decomposition includes
$\dfrac{A_1}{ax+b} + \dfrac{A_2}{(ax+b)^2} + \ldots + \dfrac{A_{m}}{(ax+b)^m}\nonumber$
for real numbers $$A_{1}$$, $$A_{2}$$, …$$A_{m}$$.
• If $$\alpha$$ is a non-real zero of $$D$$ of multiplicity $$m$$ which corresponds to the irreducible quadratic $$ax^2+bx+c$$, the partial fraction decomposition includes
$\dfrac{B_1x + C_1}{ax^2+bx+c} + \dfrac{B_2x + C_2}{\left(ax^2+bx+c\right)^2} + \ldots +\dfrac{B_{m}x + C_{m}}{\left(ax^2+bx+c\right)^m}\nonumber$
for real numbers $$B_{1}$$, $$B_{2}$$, …$$B_{m}$$ and $$C_{1}$$, $$C_{2}$$, …$$C_{m}$$.
a In other words, $$R(x)$$ is a proper rational function which has been fully reduced.
The proof of Theorem 8.10 is best left to a course in Abstract Algebra. Notice that the theorem provides for the general case, so we need to use subscripts, $$A_{1}$$, $$A_{2}$$, etc., to denote different unknown coefficients as opposed to the usual convention of $$A$$, $$B$$, etc.. The stress on multiplicities is to help us correctly group factors in the denominator. For example, consider the rational function
$\dfrac{3x-1}{\left(x^2-1\right)\left(2-x-x^2\right)}\nonumber$
Factoring the denominator to find the zeros, we get $$(x+1)(x-1)(1-x)(2+x)$$. We find $$x = -1$$ and $$x=-2$$ are zeros of multiplicity one but that $$x=1$$ is a zero of multiplicity two due to the two different factors $$(x-1)$$ and $$(1-x)$$. One way to handle this is to note that $$(1-x) = -(x-1)$$ so
$\dfrac{3x-1}{(x+1)(x-1)(1-x)(2+x)} = \dfrac{3x-1}{-(x-1)^2(x+1)(x+2)} = \dfrac{1-3x}{(x-1)^2(x+1)(x+2)}\nonumber$
from which we proceed with the partial fraction decomposition
$\dfrac{1-3x}{(x-1)^2(x+1)(x+2)} = \dfrac{A}{x-1} + \dfrac{B}{(x-1)^2} + \dfrac{C}{x+1} + \dfrac{D}{x+2}\nonumber$
Turning our attention to non-real zeros, we note that the tool of choice to determine the irreducibility of a quadratic $$ax^2+bx+c$$ is the discriminant, $$b^2-4ac$$. If $$b^2 - 4ac < 0$$, the quadratic admits a pair of non-real complex conjugate zeros. Even though one irreducible quadratic gives two distinct non-real zeros, we list the terms with denominators involving a given irreducible quadratic only once to avoid duplication in the form of the decomposition. The trick, of course, is factoring the denominator or otherwise finding the zeros and their multiplicities in order to apply Theorem 8.10. We recommend that the reader review the techniques set forth in Sections 3.3 and 3.4. Next, we state a theorem that if two polynomials are equal, the corresponding coefficients of the like powers of $$x$$ are equal. This is the principal by which we shall determine the unknown coefficients in our partial fraction decomposition.
###### Theorem 8.11.
Suppose $a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{2} x^{2}+a_{1} x+a_{0}=b_{m} x^{m}+m_{m-1} x^{m-1}+\cdots+b_{2} x^{2}+b_{1} x+b_{0}\nonumber$
for all $$x$$ in an open interval $$I$$. Then $$n=m$$ and $$a_{i} = b_{i}$$ for all $$i = 1 \ldots n$$.
Believe it or not, the proof of Theorem 8.11 is a consequence of Theorem 3.14. Define $$p(x)$$ to be the difference of the left hand side of the equation in Theorem 8.11 and the right hand side. Then $$p(x) = 0$$ for all $$x$$ in the open interval $$I$$. If $$p(x)$$ were a nonzero polynomial of degree $$k$$, then, by Theorem 3.14, $$p$$ could have at most $$k$$ zeros in $$I$$, and $$k$$ is a finite number. Since $$p(x) = 0$$ for all the $$x$$ in $$I$$, $$p$$ has infinitely many zeros, and hence, $$p$$ is the zero polynomial. This means there can be no nonzero terms in $$p(x)$$ and the theorem follows. Arguably, the best way to make sense of either of the two preceding theorems is to work some examples.
##### Example 8.6.1
Resolve the following rational functions into partial fractions.
1. $$R(x)=\frac{x+5}{2 x^{2}-x-1}$$
2. $$R(x)=\frac{3}{x^{3}-2 x^{2}+x}$$
3. $$R(x)=\frac{3}{x^{3}-x^{2}+x}$$
4. $$R(x)=\frac{4 x^{3}}{x^{2}-2}$$
5. $$R(x)=\frac{x^{3}+5 x-1}{x^{4}+6 x^{2}+9}$$
6. $$R(x)=\frac{8 x^{2}}{x^{4}+16}$$
###### Solution
1. We begin by factoring the denominator to find $$2 x^{2}-x-1=(2 x+1)(x-1)$$. We get $$x=-\frac{1}{2}$$ and $$x = 1$$ are both zeros of multiplicity one and thus we know $\frac{x+5}{2 x^{2}-x-1}=\frac{x+5}{(2 x+1)(x-1)}=\frac{A}{2 x+1}+\frac{B}{x-1}\nonumber$ Clearing denominators, we get $$x+5=A(x-1)+B(2 x+1)$$ so that $$x+5=(A+2 B) x+B-A$$. Equating coefficients, we get the system $\left\{\begin{array}{r} A+2 B=1 \\ -A+B=5 \end{array}\right.\nonumber$ This system is readily handled using the Addition Method from Section 8.1, and after adding both equations, we get $$3B = 6$$ so $$B = 2$$. Using back substitution, we find $$A = −3$$. Our answer is easily checked by getting a common denominator and adding the fractions. $\frac{x+5}{2 x^{2}-x-1}=\frac{2}{x-1}-\frac{3}{2 x+1}\nonumber$
2. Factoring the denominator gives $$x^{3}-2 x^{2}+x=x\left(x^{2}-2 x+1\right)=x(x-1)^{2}$$ which gives $$x = 0$$ as a zero of multiplicity one and $$x = 1$$ as a zero of multiplicity two. We have $\\frac{3}{x^{3}-2 x^{2}+x}=\frac{3}{x(x-1)^{2}}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{(x-1)^{2}}\nonumber$ Clearing denominators, we get $$3=A(x-1)^{2}+B x(x-1)+C x$$, which, after gathering up the like terms become $$3=(A+B) x^{2}+(-2 A-B+C) x+A$$. Our system is \left\{\begin{aligned} A+B &=0 \\ -2 A-B+C &=0 \\ A &=3 \end{aligned}\right.\nonumber Substituting $$A = 3$$ into $$A + B = 0$$ gives $$B = −3$$, and substituting both for $$A$$ and $$B$$ in $$−2A − B + C = 0$$ gives $$C = 3$$. Our final answer is $\frac{3}{x^{3}-2 x^{2}+x}=\frac{3}{x}-\frac{3}{x-1}+\frac{3}{(x-1)^{2}}\nonumber$
3. The denominator factors as $$x\left(x^{2}-x+1\right)$$. We see immediately that $$x = 0$$ is a zero of multiplicity one, but the zeros of $$x^{2}-x+1$$ aren’t as easy to discern. The quadratic doesn’t factor easily, so we check the discriminant and find it to be $$(-1)^{2}-4(1)(1)=-3<0$$. We find its zeros are not real so it is an irreducible quadratic. The form of the partial fraction decomposition is then $\frac{3}{x^{3}-x^{2}+x}=\frac{3}{x\left(x^{2}-x+1\right)}=\frac{A}{x}+\frac{B x+C}{x^{2}-x+1}\nonumber$ Proceeding as usual, we clear denominators and get $$3=A\left(x^{2}-x+1\right)+(B x+C) x$$ or $$3=(A+B) x^{2}+(-A+C) x+A$$. We get \left\{\begin{aligned} A+B &=0 \\ -A+C &=0 \\ A &=3 \end{aligned}\right.\nonumber From $$A = 3$$ and $$A + B = 0$$, we get $$B = −3$$. From $$−A + C = 0$$, we get $$C = A = 3$$. We get $\frac{3}{x^{3}-x^{2}+x}=\frac{3}{x}+\frac{3-3 x}{x^{2}-x+1}\nonumber$
4. Since $$\frac{4 x^{3}}{x^{2}-2}$$ isn’t proper, we use long division and we get a quotient of $$4x$$ with a remainder of $$8x$$. That is, $$\frac{4 x^{3}}{x^{2}-2}=4 x+\frac{8 x}{x^{2}-2}$$ so we now work on resolving $$\frac{8 x}{x^{2}-2}$$ into partial fractions. The quadratic $$x^{2}-2$$, though it doesn’t factor nicely, is, nevertheless, reducible. Solving $$x^{2}-2=0$$ gives us $$x=\pm \sqrt{2}$$, and each of these zeros must be of multiplicity one since Theorem 3.14 enables us to now factor $$x^{2}-2=(x-\sqrt{2})(x+\sqrt{2})$$. Hence, $\frac{8 x}{x^{2}-2}=\frac{8 x}{(x-\sqrt{2})(x+\sqrt{2})}=\frac{A}{x-\sqrt{2}}+\frac{B}{x+\sqrt{2}}\nonumber$ Clearing fractions, we get $$8 x=A(x+\sqrt{2})+B(x-\sqrt{2})$$ or $$8 x=(A+B) x+(A-B) \sqrt{2}$$. We get the system $\left\{\begin{array}{r} A+B=8 \\ (A-B) \sqrt{2}=0 \end{array}\right.\nonumber$ From $$(A-B) \sqrt{2}=0$$, we get $$A = B$$, which, when substituted into $$A + B = 8$$ gives $$B = 4$$. Hence, $$A = B = 4$$ and we get $\frac{4 x^{3}}{x^{2}-2}=4 x+\frac{8 x}{x^{2}-2}=4 x+\frac{4}{x+\sqrt{2}}+\frac{4}{x-\sqrt{2}}\nonumber$
5. At first glance, the denominator $$D(x)=x^{4}+6 x^{2}+9$$ appears irreducible. However, $$D(x)$$ has three terms, and the exponent on the first term is exactly twice that of the second. Rewriting $$D(x)=\left(x^{2}\right)^{2}+6 x^{2}+9$$, we see it is a quadratic in disguise and factor $$D(x)=\left(x^{2}+3\right)^{2}$$. Since $$x^{2}+3$$ clearly has no real zeros, it is irreducible and the form of the decomposition is $\frac{x^{3}+5 x-1}{x^{4}+6 x^{2}+9}=\frac{x^{3}+5 x-1}{\left(x^{2}+3\right)^{2}}=\frac{A x+B}{x^{2}+3}+\frac{C x+D}{\left(x^{2}+3\right)^{2}}\nonumber$ When we clear denominators, we find $$x^{3}+5 x-1=(A x+B)\left(x^{2}+3\right)+C x+D$$ which yields $$x^{3}+5 x-1=A x^{3}+B x^{2}+(3 A+C) x+3 B+D$$. Our system is \left\{\begin{aligned} A &=1 \\ B &=0 \\ 3 A+C &=5 \\ 3 B+D &=-1 \end{aligned}\right. We have $$A=1$$ and $$B=0$$ from which we get $$C = 2$$ and $$D = −1$$. Our final answer is $\frac{x^{3}+5 x-1}{x^{4}+6 x^{2}+9}=\frac{x}{x^{2}+3}+\frac{2 x-1}{\left(x^{2}+3\right)^{2}}\nonumber$
6. Once again, the difficulty in our last example is factoring the denominator. In an attempt to get a quadratic in disguise, we write $x^{4}+16=\left(x^{2}\right)^{2}+4^{2}=\left(x^{2}\right)^{2}+8 x^{2}+4^{2}-8 x^{2}=\left(x^{2}+4\right)^{2}-8 x^{2}\nonumber$ and obtain a difference of two squares: $$\left(x^{2}+4\right)^{2}$$ and $$8 x^{2}=(2 x \sqrt{2})^{2}$$. Hence, $x^{4}+16=\left(x^{2}+4-2 x \sqrt{2}\right)\left(x^{2}+4+2 x \sqrt{2}\right)=\left(x^{2}-2 x \sqrt{2}+4\right)\left(x^{2}+2 x \sqrt{2}+4\right)\nonumber$ The discrimant of both of these quadratics works out to be −8 < 0, which means they are irreducible. We leave it to the reader to verify that, despite having the same discriminant, these quadratics have different zeros. The partial fraction decomposition takes the form $\frac{8 x^{2}}{x^{4}+16}=\frac{8 x^{2}}{\left(x^{2}-2 x \sqrt{2}+4\right)\left(x^{2}+2 x \sqrt{2}+4\right)}=\frac{A x+B}{x^{2}-2 x \sqrt{2}+4}+\frac{C x+D}{x^{2}+2 x \sqrt{2}+4}\nonumber$ We get $8 x^{2}=(A+C) x^{3}+(2 A \sqrt{2}+B-2 C \sqrt{2}+D) x^{2}+(4 A+2 B \sqrt{2}+4 C-2 D \sqrt{2}) x+4 B+4 D\nonumber$ which gives the system \left\{\begin{aligned} A+C &=0 \\ 2 A \sqrt{2}+B-2 C \sqrt{2}+D &=8 \\ 4 A+2 B \sqrt{2}+4 C-2 D \sqrt{2} &=0 \\ 4 B+4 D &=0 \end{aligned}\right.\nonumber We choose substitution as the weapon of choice to solve this system. From $$A + C = 0$$, we get $$A = −C$$; from $$4B + 4D = 0$$, we get $$B = −D$$. Substituting these into the remaining two equations, we get $\left\{\begin{array}{r} -2 C \sqrt{2}-D 2 C \sqrt{2}+D=8 \\ -4 C-2 D \sqrt{2}+4 C-2 D \sqrt{2}=0 \end{array}\right.\nonumber$ or $\left\{\begin{array}{l} -4 C \sqrt{2}=8 \\ -4 D \sqrt{2}=0 \end{array}\right.\nonumber$ We get $$C=-\sqrt{2}$$ so that $$A=-C=\sqrt{2}$$ and $$D = 0$$ which means $$B = −D = 0$$. We get $\frac{8 x^{2}}{x^{4}+16}=\frac{x \sqrt{2}}{x^{2}-2 x \sqrt{2}+4}-\frac{x \sqrt{2}}{x^{2}+2 x \sqrt{2}+4}\nonumber$
## 8.6.1. Exercises
In Exercises 1 - 6, find only the form needed to begin the process of partial fraction decomposition. Do not create the system of linear equations or attempt to find the actual decomposition.
1. $$\frac{7}{(x-3)(x+5)}$$
2. $$\frac{5 x+4}{x(x-2)(2-x)}$$
3. $$\frac{m}{(7 x-6)\left(x^{2}+9\right)}$$
4. $$\frac{a x^{2}+b x+c}{x^{3}(5 x+9)\left(3 x^{2}+7 x+9\right)}$$
5. $$\frac{\text { A polynomial of degree }<9}{(x+4)^{5}\left(x^{2}+1\right)^{2}}$$
6. $$\frac{\text { A polynomial of degree }<7}{x(4 x-1)^{2}\left(x^{2}+5\right)\left(9 x^{2}+16\right)}$$
In Exercises 7 - 18, find the partial fraction decomposition of the following rational expressions.
1. $$\frac{2 x}{x^{2}-1}$$
2. $$\frac{-7 x+43}{3 x^{2}+19 x-14}$$
3. $$\frac{11 x^{2}-5 x-10}{5 x^{3}-5 x^{2}}$$
4. $$\frac{-2 x^{2}+20 x-68}{x^{3}+4 x^{2}+4 x+16}$$
5. $$\frac{-x^{2}+15}{4 x^{4}+40 x^{2}+36}$$
6. $$\frac{-21 x^{2}+x-16}{3 x^{3}+4 x^{2}-3 x+2}$$
7. $$\frac{5 x^{4}-34 x^{3}+70 x^{2}-33 x-19}{(x-3)^{2}}$$
8. $$\frac{x^{6}+5 x^{5}+16 x^{4}+80 x^{3}-2 x^{2}+6 x-43}{x^{3}+5 x^{2}+16 x+80}$$
9. $$\frac{-7 x^{2}-76 x-208}{x^{3}+18 x^{2}+108 x+216}$$
10. $$\frac{-10 x^{4}+x^{3}-19 x^{2}+x-10}{x^{5}+2 x^{3}+x}$$
11. $$\frac{4 x^{3}-9 x^{2}+12 x+12}{x^{4}-4 x^{3}+8 x^{2}-16 x+16}$$
12. $$\frac{2 x^{2}+3 x+14}{\left(x^{2}+2 x+9\right)\left(x^{2}+x+5\right)}$$
13. As we stated at the beginning of this section, the technique of resolving a rational function into partial fractions is a skill needed for Calculus. However, we hope to have shown you that it is worth doing if, for no other reason, it reinforces a hefty amount of algebra. One of the common algebraic errors the authors find students make is something along the lines of $\frac{8}{x^{2}-9} \neq \frac{8}{x^{2}}-\frac{8}{9}\nonumber$ Think about why if the above were true, this section would have no need to exist.
1. $$\frac{A}{x-3}+\frac{B}{x+5}$$
2. $$\frac{A}{x}+\frac{B}{x-2}+\frac{C}{(x-2)^{2}}$$
3. $$\frac{A}{7 x-6}+\frac{B x+C}{x^{2}+9}$$
4. $$\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x^{3}}+\frac{D}{5 x+9}+\frac{E x+F}{3 x^{2}+7 x+9}$$
5. $$\frac{A}{x+4}+\frac{B}{(x+4)^{2}}+\frac{C}{(x+4)^{3}}+\frac{D}{(x+4)^{4}}+\frac{E}{(x+4)^{5}}+\frac{F x+G}{x^{2}+1}+\frac{H x+I}{\left(x^{2}+1\right)^{2}}$$
6. $$\frac{A}{x}+\frac{B}{4 x-1}+\frac{C}{(4 x-1)^{2}}+\frac{D x+E}{x^{2}+5}+\frac{F x+G}{9 x^{2}+16}$$
7. $$\frac{2 x}{x^{2}-1}=\frac{1}{x+1}+\frac{1}{x-1}$$
8. $$\frac{-7 x+43}{3 x^{2}+19 x-14}=\frac{5}{3 x-2}-\frac{4}{x+7}$$
9. $$\frac{11 x^{2}-5 x-10}{5 x^{3}-5 x^{2}}=\frac{3}{x}+\frac{2}{x^{2}}-\frac{4}{5(x-1)}$$
10. $$\frac{-2 x^{2}+20 x-68}{x^{3}+4 x^{2}+4 x+16}=-\frac{9}{x+4}+\frac{7 x-8}{x^{2}+4}$$
11. $$\frac{-x^{2}+15}{4 x^{4}+40 x^{2}+36}=\frac{1}{2\left(x^{2}+1\right)}-\frac{3}{4\left(x^{2}+9\right)}$$
12. $$\frac{-21 x^{2}+x-16}{3 x^{3}+4 x^{2}-3 x+2}=-\frac{6}{x+2}-\frac{3 x+5}{3 x^{2}-2 x+1}$$
13. $$\frac{5 x^{4}-34 x^{3}+70 x^{2}-33 x-19}{(x-3)^{2}}=5 x^{2}-4 x+1+\frac{9}{x-3}-\frac{1}{(x-3)^{2}}$$
14. $$\frac{x^{6}+5 x^{5}+16 x^{4}+80 x^{3}-2 x^{2}+6 x-43}{x^{3}+5 x^{2}+16 x+80}=x^{3}+\frac{x+1}{x^{2}+16}-\frac{3}{x+5}$$
15. $$\frac{-7 x^{2}-76 x-208}{x^{3}+18 x^{2}+108 x+216}=-\frac{7}{x+6}+\frac{8}{(x+6)^{2}}-\frac{4}{(x+6)^{3}}$$
16. $$\frac{-10 x^{4}+x^{3}-19 x^{2}+x-10}{x^{5}+2 x^{3}+x}=-\frac{10}{x}+\frac{1}{x^{2}+1}+\frac{x}{\left(x^{2}+1\right)^{2}}$$
17. $$\frac{4 x^{3}-9 x^{2}+12 x+12}{x^{4}-4 x^{3}+8 x^{2}-16 x+16}=\frac{1}{x-2}+\frac{4}{(x-2)^{2}}+\frac{3 x+1}{x^{2}+4}$$
18. $$\frac{2 x^{2}+3 x+14}{\left(x^{2}+2 x+9\right)\left(x^{2}+x+5\right)}=\frac{1}{x^{2}+2 x+9}+\frac{1}{x^{2}+x+5}$$
## Reference
1 Recall this means it has no real zeros; see Section 3.4.
2 Recall this means $$x = 0$$ is a zero of multiplicity.
3 We will justify this shortly.
This page titled 8.6: Partial Fraction Decomposition is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Carl Stitz & Jeff Zeager via source content that was edited to the style and standards of the LibreTexts platform.
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## Image Moments
An Image moment is a number calculated using a certain formula. Understand what that formula means might be hard at first. In fact, I got a lot of questions about moments from the tracking tutorial I did long back. So, here it is - an explanation of what moments area!
## The math of moments
In pure math, the nth order moment about the point c is defined as:
This definition holds for a function that has just one independent variable. We're interested in images - they have two dimensions. So we need two independent variables. So the formula becomes:
Here, the f(x, y) is the actual image and is assumed to be continuous. For our purposes, we need a discrete way (think pixels) to describe moments:
The intergrals has been replaced by summations. The order of the moment is m + n. Usually, we calculate the moments about (0, 0). So you can simply ignore the constants cx and cy.
Now with the math part out of the way, let's have a look at what you can calculate with this thing.
## Calculating area
To calculate the area of a binary image, you need to calculate its zeroth moment:
The x0 and y0 don't have any effect and can be removed.
Now, in a binary image, a pixel is either 0 or 1. So for every white pixel, a '1' is added to the moment - effectively calculating the area of the binary image! Another thing to note is that there is only one zeroth order moment.
## Centroid
To calculate the centroid of a binary image you need to calculate two coordinates -
How did I get that? Here's a quick explanation. Consider the first moment:
The two summations are like a for loop. The x coordinate of all white pixels (where f(x, y) = 1) is added up.
Similarly, we can calculate the sum of y coordinates of all white pixels:
Now we have the sum of several pixels' x and y coordinates. To get the average, you need to divide each by the number of pixels. The number of pixels is the area of the image - the zeroth moment. So you get:
and
One interesting thing about this technique is that it is not very sensitive to noise. The centroid might move a little bit but not much.
Also, from the math it's clear this technique holds only for single blobs. If you have two white blobs in your image, the centroid will be somewhere in between. You'll have to extract each blob separately to get their centroids.
## Central moments
In fact, this kind of division is very common - dividing a moment by the zeroth order moment. It's so common that it has a name of its own - central moments.
So to calculate the centroid, you need to calculate the first order central moments.
## Higher order moments
Going onto higher order moments, things get complicated really fast. You have three 2nd order moments, four 3rd order moments, etc. You can combine several of these moments so that they are translation invariant, scale invariant and even rotation invariant.
While reading about moments, I found an entire book dedicated to pattern recognition with moments. In fact, there are terms called skewness and kurtosis. These refer to third and fourth order moments. They measure how skewed an image is and whether an image is tall and thin or short and fat. Clearly, there's a LOT that can be learned about these mathematical tools.
Utkarsh Sinha created AI Shack in 2010 and has since been working on computer vision and related fields. He is currently at Microsoft working on computer vision.
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When solving trigonometry equations, more than one method will usually work — although one method is often quicker or easier than another. With practice, you’ll get good at choosing the better of the ways to solve the equation.
And then you’ll come across a trig equation that defies your finest attempts. Two methods that you can use when solving these more difficult trig equations are: to square both sides of the equation, or to multiply each term through by a trig function that you’ve carefully selected.
Examples of equations that respond well to squaring both sides include
1. Square both sides of the equation.
When squaring a binomial, be sure not to forget the middle term.
2. Use the Pythagorean identity to replace sin2 x + cos2 x.
3. Subtract 1 from each side. Then replace the expression on the left with the sine double-angle formula.
4. Solve for the value of 2x by using the inverse function. Then write a few angle solutions to determine a pattern.
2x = sin–1(1) = 90°, 450°, 810°, . . .
Because you’re supposed to find all the possible solutions, you’re not bound by only two rotations.
5. Divide every term by 2.
2x = 90°, 450°, 810°, . . .
x = 45°, 225°, 405°, . . .
6. Write an expression for all the solutions.
x = 45° + 180°n
In the next example, you need to do a little shifting at first.
equation by itself. Otherwise, when you square both sides, you end up with a radical factor in one of the terms. That situation isn’t always bad, but dealing with it is usually a little more awkward than not.
Solve the equation for all the possible angles from 0 to 360 degrees.
1. Add the radical term to both sides and subtract 1 from both sides.
2. Square both sides.
3. Replace sin2 x with 1 – cos2 x from the Pythagorean identity.
Doing so creates an equation with terms that have all the same functions, cos x, in them.
cos2 x – 2cos x + 1 = 3(1 – cos2 x)
4. Simplify the equation by distributing the 3 on the right and then bringing all the terms to the left to set the equation equal to 0.
5. Divide every term by 2.
2cos2 x – cos x – 1 = 0
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# 4 Perfect Lesson Plan Objectives In Math Solutions
Lesson Plan Objectives In Math - Requirements and objectives your state core curriculum/pupil achievement preferred(s): wide variety and operations – fractions: 3 provide an explanation for equivalence of fractions in special cases, and compare fractions with the aid of reasoning about their length. Range and operations – fractions: 3a recognize fractions as equal (equal) if they may be the same length, or the equal point on more than a few line. Number and operations – fractions: 3b apprehend and generate easy equal fractions, e.G., ?=2/four, four/6=2/3). Give an explanation for why the fractions are equal, e.G., By the usage of a visible fraction model.
## Lesson Plan Objectives In Math Top Modules 18/22: Designing, Differentiating Instruction, Ppt Images
2. Presentation of latest statistics or modeling (10-15 min): begin by way of retaining up 6 pencils. Ask what number of pencils we've got within the set. (6) ask how many pencils might be half of of the set. (3) ask how we would represent this as a fraction. (3/6). Write the fraction at the board. We simply determined that one-1/2 of the 6 pencils is three pencils, or three/6 (factor to fraction). We also recognise that one-half can be written as ½. (Write this at the board). Draw two rectangles on the board. Divide one in half. Divide the alternative one into 6. Colour in one 1/2 of each rectangle. Ask the elegance to provide an explanation for what they be aware approximately those fractions. 3/6 and ½ are the identical quantity of the entire so they are same. Write an same signal among the 2 fractions. Provide an explanation for that when two fractions are equal, they may be referred to as equivalent fractions. Have the students write the phrases “equal fractions” in their math journals. Write the definition: “fractions that constitute the equal part of the complete.?? have the scholars repeat and write it in their journals. Role play activity: king fraction start through studying a brief introductory tale: once upon a time, there has been a country positioned in a small valley in the center of the treturous mountains. The population of this nation were very glad because they lived at the best inhabitable land for a hundred miles. The kindgom’s king, king fraction, changed into a fair and generous king. King fraction’s favorite hobby become to ensure all and sundry in his state cherished fractions as a great deal as he did. King fraction tested his topics’ love of fractions as a minimum 4 instances a day by means of calling the whole nation collectively and having them play his fraction game. For the king it was thoroughly enjoyable to look at his subjects scramble round trying to observe his orders. Those subjects who participated had been handsomely rewarded every month with a large cargo of meals. Turn around and put on home made crown. .
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# 2004 AMC 8 Problems/Problem 25
## Problem
Two $4 \times 4$ squares intersect at right angles, bisecting their intersecting sides, as shown. The circle's diameter is the segment between the two points of intersection. What is the area of the shaded region created by removing the circle from the squares?
$[asy] unitsize(6mm); draw(unitcircle); filldraw((0,1)--(1,2)--(3,0)--(1,-2)--(0,-1)--(-1,-2)--(-3,0)--(-1,2)--cycle,lightgray,black); filldraw(unitcircle,white,black); [/asy]$
$\text{(A)}\ 16-4\pi\qquad \text{(B)}\ 16-2\pi \qquad \text{(C)}\ 28-4\pi \qquad \text{(D)}\ 28-2\pi \qquad \text{(E)}\ 32-2\pi$
## Solution 1
If the circle was shaded in, the intersection of the two squares would be a smaller square with half the sidelength, $2$. The area of this region would be the two larger squares minus the area of the intersection, the smaller square. This is $4^2 + 4^2 - 2^2 = 28$.
The diagonal of this smaller square created by connecting the two points of intersection of the squares is the diameter of the circle. This value can be found with Pythagorean or a $45^\circ - 45^\circ - 90^\circ$ circle to be $2\sqrt{2}$. The radius is half the diameter, $\sqrt{2}$. The area of the circle is $\pi r^2 = \pi (\sqrt{2})^2 = 2\pi$.
The area of the shaded region is the area of the two squares minus the area of the circle which is $\boxed{\textbf{(D)}\ 28-2\pi}$.
## Solution 2
Find the area of the overlapping squares. $2 \cdot 4^2 - 2^2=28$. Now, we find the chord of the circle to be 2, so then the radius of the circle would be $\sqrt{2}$. Now, the area of the circle is $\pi r^2$, so we put in the values to get $2\pi$. Now, we can subtract the two values. Now, we can find the answer to be $\boxed{\textbf{(D)}\ 28-2\pi}$.
~ pi_is_3.14
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Illustration: Chelsea Miller
The Rule of 72
Calculating compound interest can be tricky and intimidating. However, there are some shortcuts you can take to get a quick estimate of growth rates. A common shortcut is known as the Rule of 72.
The Basics
The Rule of 72 is a tool used to estimate how long it will take an investment to double at a given interest rate, assuming a fixed annual rate of interest.
If you know the interest rate, you can easily determine how long it will take you to double your money or what interest rate you should look for in order to double your money in a given period of time.
To figure out how long it will take to double your money, take the fixed annual interest rate and divide that number into 72. Let’s say your interest rate is 8%. 72 ∕ 8 = 9, so it will take about 9 years to double your money. A 10% interest rate will double your investment in about 7 years (72 ∕ 10 = 7.2); an amount invested at a 12% interest rate will double in about 6 years (72 ∕ 12 = 6).
Using the Rule of 72, you can easily determine how long it will take to double your money.
To figure out what interest rate to look for, use the same basic formula, but run it backward: divide 72 by the number of years. So if you want to double your money in about 6 years, look for an interest rate of 12%.
The basic algebraic formula looks like this, where Y is the number of years and r is the interest rate:
Y = 72 ∕ r and r = 72 ∕ Y
This rule works for interest rates from about 4% up to about 20%; after that, the error becomes significant and more straightforward math is required. However, for 4%–20%, the rule works remarkably well.
Illustration: Chelsea Miller
Why 72?
Here, we merely scrape the surface of that “more straightforward math.” To really dive deep into why the rule works, check out this article.
The Rule of 72 is itself an estimation. It uses a concept called natural logarithms to estimate compounding periods. In mathematics, the natural logarithm is the amount of time needed to reach a particular level of growth using continuous compounding.
For math enthusiasts out there: it is easiest to see how this works through continuously compounded interest. (The Rule of 72 addresses annually compounded interest, but we’ll get there in a minute.)
When dealing with continuously compounding interest, you can work out the exact time it takes an investment to double by using the time value of money formula (TVM) and simplifying the equation until eventually, you are left with something like this:
ln(2)= rY
The natural log (ln) of 2 is about 0.693. Solve for interest rate (r) or number of years (Y), and then multiply by 100 to express as a percentage or year, respectively.
Wait...
If our new formula is based on the number 69.3 (0.693 × 100), that begs the question: Why isn’t it called the Rule of 69.3?
First, that just doesn’t sound quite as good as “The Rule of 72.” Second, there are two points to remember:
1. The “Rule of 69.3” is not an estimation. It is the actual amount of time that it will take money to double, and works for any range of interest rates.
2. The Rule of 69.3 works for continuously compounded interest. The Rule of 72 works for a fixed annual rate of interest.
The math equation for fixed annual interest is slightly more complex, and simplifying it leaves us with approximately 72.7.
Normally, we would round up to 73. However, 72 is much easier to work with, as it is readily divisible by 2, 3, 4, 6, 8, 9, and 12. As we are already estimating, convenience wins out, and we are left with the Rule of 72.
History
The Rule of 72 was first introduced in the late fifteenth century by the Franciscan friar and Italian mathematician Luca Pacioli. A contemporary of Leonardo da Vinci, Pacioli is considered by many to be the father of accounting. The Rule of 72 was introduced in his book Summa de arithmetica, geometria, proportioni et proportionalita, published in 1494 for use as a textbook for schools in what is now northern Italy.
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Students can use the Spectrum Math Grade 3 Answer Key Chapter 7 Posttest as a quick guide to resolve any of their doubts.
Check What You Learned
Question 1.
a. About how much does a paper clip weigh?
a. 1 gram
b. 100 grams
c. 1,000 grams
a. 1 gram,
Explanation:
Generally the mass of a regular paper clip is about 1 gram. The paper clip weighs about 1 gram respectively.
b. About how much juice can a baby bottle hold?
a. 3 liters
b. 30 milliliters
c. 300 liters
30 milliliters,
Explanation:
Generally about 30 milliliters of juice can be a baby bottle can hold.
Solve.
Question 2.
Kennedy popped 24 cups of popcorn in 3 days. If she popped the same number of cups each day, how many cups did she pop each day?
8 cups,
Explanation:
Kennedy popped 24 cups of popcorn in 3 days, Number of cups of popcorn did Kennedy popped in 3 days = 24 cups, Number of cups of popcorn did Kennedy popped each day = 24 cups / 3 = 8 cups , Therefore 8 cups did Kennedy pop each day.
Complete the graphs.
Question 3.
a.
Key __________ = 20 miles
Key is 20 miles for Team 1,
Explanation:
Completed the graph Key is 20 miles which comes for Team 1.
b.
Explanation:
Completed the graph X-axis favorite pets and Y-axis as number of favorite pets.
Find the area of the figure.
Question 4.
a.
____________ sq. units
6 sq. units,
Explanation:
In given rectangle length as 6 units and breadth as 1 unit. So, area of rectangle is (length X breadth) = 6 units X 1 unit = 6 sq. units respectively.
b.
____________ sq. units
9 sq. units,
Explanation:
Given from figure each side represents 3 units so, area of square is (side X side) sq. units = 3 units x 3 units = 9 sq. units, therefore area of square is 9 sq. units.
Question 5.
Draw the square units to show the area of the rectangle.
A = ____________ sq. units
24 sq. units,
Explanation;
From the given figure rectangle, length is 8 units and breadth is 3 units, so therefore the area of rectangle is l X b that is 8 X 3 = 24 sq. units .
Question 6.
Multiply to find the area.
A = _____________ sq. in.
12 sq. in,
Explanation:
Given, from figure length is 4 in. and breadth is 3 in. so, area is 4 X 3 = 12 sq. in.
Question 7.
Find the area.
A = _____________ sq. units
24 sq. units,
Explanation:
Given, In the figure ABCDEF ,draw a line from D to G then DG = 3 units so, area of a square DEFG each side is 3 units that is 3 X 3 = 9 sq. units, Area of rectangle ABCG length is 5 units and breadth is 3 units that is 5 X 3 = 15 sq. units. So, therefore Area of square DEFG + Area of rectangle ABCG = 9 sq. units + 15 sq. units = 24 sq. units.
Question 8.
Solve.
An equilateral triangle has one side that measures 9 cm. How many centimeters is the perimeter of the triangle?
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# Question Video: Finding the Transpose of a Matrix Mathematics
Given the matrix 𝐴 = [−8, 4, 3 and 4, 1, −1], find (𝐴^(𝑇))^(𝑇).
02:28
### Video Transcript
Given the matrix 𝐴 equals the two-by-three matrix negative eight, four, three, four, one, negative one, find the transpose of 𝐴 transpose.
In this question, we are given a two-by-three matrix 𝐴 and asked to find the transpose of the transpose of this matrix. We can do this in two ways. First, we recall that we find the transpose of a matrix by switching the rows with the corresponding columns of the matrix. We can use this to find the transpose of matrix 𝐴. We can start by writing the first row of matrix 𝐴 as the first column in its transpose. This gives us a first column of negative eight, four, three. We can follow the same process for the second row of 𝐴. We write this as the second column in the transpose of 𝐴 to obtain a second column of four, one, negative one.
We want to find the transpose of 𝐴 transpose, so we need to apply this process once again to this new matrix 𝐴 transpose. We start by writing the first row of this matrix as the first column in its transpose. This gives us a first column of negative eight, four. We then write the second row of this matrix as the second column of its transpose. We obtain a second column of four, one. We follow this process one final time by writing the final row of the matrix as the final column of the new matrix. The final column is three, negative one. Therefore, we have shown that the transpose of 𝐴 transpose is the two-by-three matrix negative eight, four, three, four, one, negative one.
However, this is not the only way we can answer this question. We can note that the transpose of 𝐴 transpose is actually equal to 𝐴. We can show why this is true by noting that taking the transpose of 𝐴 transpose will switch the rows with the columns and then switch them back. So for any matrix 𝑀, taking the transpose of 𝑀 transpose will leave the matrix unchanged. We can apply this result with matrix 𝐴 instead of 𝑀 to get that the transpose of 𝐴 transpose is equal to 𝐴, the two-by-three matrix negative eight, four, three, four, one, negative one.
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# How do you solve 3|p-5|=2p?
Jul 18, 2017
See a solution process below:
#### Explanation:
First, divide each side of the equation by $\textcolor{red}{3}$ to isolate the absolute value function while keeping the equation balanced:
$\frac{3 \left\mid p - 5 \right\mid}{\textcolor{red}{3}} = \frac{2 p}{\textcolor{red}{3}}$
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} \left\mid p - 5 \right\mid}{\cancel{\textcolor{red}{3}}} = \frac{2 p}{3}$
$\left\mid p - 5 \right\mid = \frac{2 p}{3}$
The absolute value function takes any negative or positive term and transforms it to its positive form. Therefore, we must solve the term within the absolute value function for both its negative and positive equivalent.
Solution 1
$p - 5 = - \frac{2 p}{3}$
$- \textcolor{red}{p} + p - 5 = - \textcolor{red}{p} - \frac{2 p}{3}$
$0 - 5 = - \left(\frac{3}{3} \times \textcolor{red}{p}\right) - \frac{2 p}{3}$
$- 5 = - \frac{3 p}{3} - \frac{2 p}{3}$
$- 5 = \frac{- 5 p}{3}$
$- 5 \times \frac{\textcolor{red}{3}}{\textcolor{b l u e}{- 5}} = \frac{- 5 p}{3} \times \frac{\textcolor{red}{3}}{\textcolor{b l u e}{- 5}}$
$\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{- 5}}} \times \frac{\textcolor{red}{3}}{\cancel{\textcolor{b l u e}{- 5}}} = \frac{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{- 5}}} p}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}} \times \frac{\cancel{\textcolor{red}{3}}}{\cancel{\textcolor{b l u e}{- 5}}}$
$3 = p$
$p = 3$
Solution 2
$p - 5 = \frac{2 p}{3}$
$- \textcolor{red}{p} + p - 5 = - \textcolor{red}{p} + \frac{2 p}{3}$
$0 - 5 = - \left(\frac{3}{3} \times \textcolor{red}{p}\right) + \frac{2 p}{3}$
$- 5 = - \frac{3 p}{3} + \frac{2 p}{3}$
$- 5 = \frac{- 1 p}{3}$
$- 5 \times \frac{\textcolor{red}{3}}{\textcolor{b l u e}{- 1}} = \frac{- 1 p}{3} \times \frac{\textcolor{red}{3}}{\textcolor{b l u e}{- 1}}$
$\frac{- 15}{-} 1 = \frac{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{- 1}}} p}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}} \times \frac{\cancel{\textcolor{red}{3}}}{\cancel{\textcolor{b l u e}{- 1}}}$
$15 = p$
$p = 15$
The Solutions Are: $p = 3$ and $p = 15$
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# Solve 2 equations
There are a variety of methods that can be used to Solve 2 equations. Math can be difficult for some students, but with the right tools, it can be conquered.
## Solving 2 equations
As a student, there are times when you need to Solve 2 equations. It can be incredibly satisfying to solve a hard math problem - the feeling of elation and accomplishment is like nothing else. When you're stuck on a problem, it can seem impossible, but when you finally crack the code and find the solution, it's all worth it.
Hard problems are those that are difficult to solve. The best way to describe a hard problem is as a challenge, or an obstacle that must be overcome. A hard problem can be something as simple as learning how to ski for the first time, or as complex as curing cancer. Hard problems are typically more difficult than they have any right to be. Sometimes it’s even impossible to solve them. But if you stick with it, eventually you will find a solution. There are two types of hard problems: those that can be solved and those that cannot be solved. Seemingly impossible problems often turn out to have solutions after all. The trick is finding them. It doesn’t matter whether your problem is big or small, complicated or simple. If you’re willing to put in the work, you can solve almost any problem you encounter.
Quadratic equations are a type of mathematical problem that can be difficult to solve. However, there are Quadratic equation solvers that can provide step by step solutions to these types of problems. Quadratic equation solvers typically take a Quadratic equation and break it down into smaller pieces that can be solved more easily. These Quadratic equation solvers will then provide the steps necessary to solve the problem, as well as the final answer. Quadratic equation solvers can be found online and in many mathematical textbooks.
To solve a trinomial, first find the coefficients of all of the terms in the expression. In this example, we have ("3x + 2"). Now you can start solving for each variable one at a time using algebraic equations. For example, if you know that x = 0, y = 9 and z = -2 then you can solve for y with an equation like "y = (0)(9)/(-2)" After you've figured out all of the variables, use addition or subtraction to combine them into one final answer.
There are a few different ways to solve logarithmic equations, but one of the most common methods is by using logs. To do this, you first take the log of both sides of the equation, and then solve for the variable. This is usually a fairly simple process, but it can be tricky if you're not familiar with logs. Another method that can be used is to rewrite the equation in exponential form, and then solve from there. This can be a bit more difficult
## More than just an app
This really helped me complete my homework. I was having lots of trouble concentrating in online school. Especially in math. When I found this app by my friend it really helped me. I didn’t just copy the answers. I was more focused on learning the solutions. In the solutions box it gives lots and lots of detail in order for you to understand. If you are having trouble with math, I really suggest you to get this app. Overall I really liked it!!!
### Nova Sanders
It is very much reliable and understandable and it also provides accurate answers not just for a person who doesn't understand a math concept but it is recommended for even marking your answers
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# Find the probability distribution of the number of heads in two tosses of a coin.
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Hint: Firstly, we need to find the sample space for tossing of a coin, after that we can find the probability of getting heads for each event. This data can then be combined to form the required probability distribution table.
Formula to be used:
$P = \dfrac{f}{T}$ where, P is the probability, f and T are favorable and total outcomes respectively.
We are tossing two coins at a time, so the sample space i.e. possible outcomes every time are:
S = { HH, TT, TH, HT }
Where, H and T denotes heads and tails respectively.
If X denotes number of heads, then the value of X for each event can be given as:
X (HH) = 2 [2 heads in this event]
X (TH) = 1 [1 head in this event]
X (HT) = 1 [1 heads in this event]
X (TT) = 0 [0 heads in this event]
Thus, there are 3 possible values of X i.e. 0, 1, 2 and 4 total outcomes.
The total outcomes will be 4 and the favorable outcomes will be the occurrence of that value of X in the above data.
Then probability of each event using the formula $P = \dfrac{f}{T}$ can be given as:
When X = 2,
$\Rightarrow P\left( X \right) = \dfrac{1}{4}$ as 2 appears only 1 time in the data.
When X = 1,
$\Rightarrow P\left( X \right) = \dfrac{2}{4} \to \dfrac{1}{2}$ as 1 appears 2 times in the data.
When X = 0,
$\Rightarrow P\left( X \right) = \dfrac{1}{4}$ as 0 appears only 1 time in the data.
Now, making the probability distribution using this information:
X 0 1 2 P (X) $\dfrac{1}{4}$ $\dfrac{1}{2}$ $\dfrac{1}{4}$
This is the required probability distribution of the number of heads in two tosses of a coin.
Note: We generally take variable X to denote the required outcome for calculation of probability distribution. We can also denote the probability of X along with its value like P (X = 1) for the value of X as 1.
We can also include the number of occurrences in the table, then the table would look like:
X 0 1 2 Number of occurrences of heads 1 2 1 P (X) $\dfrac{1}{4}$ $\dfrac{1}{2}$ $\dfrac{1}{4}$
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# AP Class 10 Maths Chapter 9 Tangents and Secants
The word tangent was introduced by Danish mathematician Thomas Fineke in 1583 and it means to touch in Latin. Let us look at the definition of tangent and secant from AP Class 10 Maths Chapter 9 Tangents and Secants.
• Tangent – It is a line that intersects the circle at one point only
• Secant – It is a line that intersects the circle in two different points
Referring to the AP Board 10th Class Maths Chapter 9 Tangents and Secants notes and solutions are the best way to study for the board exams, understand the concepts, revise the key topics and to ace the Maths paper.
## Tangents of a Circle
• The common point at which the tangent and the circle meet is known as the point of contact.
• The tangent at any point of a circle is perpendicular to the radius through the point of contact.
• The length of tangents drawn from an external point to a circle are equal.
• The area of the segment of a circle can be calculated using the formula
$\frac{x^{\circ}}{360^{\circ}}\times \Pi r^{2}-area\, of \,the \,triangle$
In the next section, let us look at a few solved chapter questions to better understand the concepts discussed in the chapter.
### Class 10 Maths Chapter 9 Tangents and Secants Questions
1. The tangent BC at a point B of a circle of 6 cm meets a line through the centre O at a point C such that AC =14cm. Find the length of BC.
Solution:
From the figure, we notice that $AB\perp BC$,
According to Pythagoras theorem,
In $\Delta ABC$,
$AC^2=AB^2+BC^2$
Substituting the values in the equation, we get
$14^2=6^2+BC^2$ $BC^2=14^2-6^2$ $BC^2=160$ $BC=\sqrt{160}=12.65 cm$
The length of BC is 12.65 cm.
Stay tuned to BYJUS to get the latest notification on SSC exam along with AP SSC model papers, exam pattern, marking scheme and more.
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# Dress a Snowman! Multiplication and Division Game {3.OA.C.7}
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Have fun dressing a snowman in a crazy outfit while practicing multiplication and division fact fluency! After answering a multiplication or division problem correctly, students will choose a "Clothing Card" that will dictate how they will "dress" their snowman! This could be a great early finisher activity or math center activity!
Includes:
Title Page
Thank you page
How to Play
Multiplication Recording Sheet
Division Recording Sheet
Playing Card Mat
24 Clothing Cards
24 Multiplication Cards
24 Division Cards
Abby Sandlin
Check out the Dress Santa version of this activity!
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Find all factor pairs for a whole number in the range 1-100. Recognize that a whole number is a multiple of each of its factors. Determine whether a given whole number in the range 1-100 is a multiple of a given one-digit number. Determine whether a given whole number in the range 1-100 is prime or composite.
Interpret a multiplication equation as a comparison, e.g., interpret 35 = 5 × 7 as a statement that 35 is 5 times as many as 7 and 7 times as many as 5. Represent verbal statements of multiplicative comparisons as multiplication equations.
Fluently multiply and divide within 100, using strategies such as the relationship between multiplication and division (e.g., knowing that 8 × 5 = 40, one knows 40 ÷ 5 = 8) or properties of operations. By the end of Grade 3, know from memory all products of two one-digit numbers.
Apply properties of operations as strategies to multiply and divide. Examples: If 6 × 4 = 24 is known, then 4 × 6 = 24 is also known. (Commutative property of multiplication.) 3 × 5 × 2 can be found by 3 × 5 = 15, then 15 × 2 = 30, or by 5 × 2 = 10, then 3 × 10 = 30. (Associative property of multiplication.) Knowing that 8 × 5 = 40 and 8 × 2 = 16, one can find 8 × 7 as 8 × (5 + 2) = (8 × 5) + (8 × 2) = 40 + 16 = 56. (Distributive property.)
Use multiplication and division within 100 to solve word problems in situations involving equal groups, arrays, and measurement quantities, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem.
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## Maxima and Minima of Functions of Two Variables
The problem of determining the maximum or minimum of function is encountered in geometry, mechanics, physics, and other fields, and was one of the motivating factors in the development of the calculus in the seventeenth century.
Let us recall the procedure for the case of a function of one variable y=f(x). First, we determine points x_c where f'(x)=0. These points are called critical points. At critical points the tangent line is horizontal. This is shown in the figure below.
The second derivative test is employed to determine if a critical point is a relative maximum or a relative minimum. If f''(x_c)>0, then x_c is a relative minimum. If f''(x_c)<0, then x_c is a maximum. If f''(x_c)=0, then the test gives no information.
The notions of critical points and the second derivative test carry over to functions of two variables. Let z=f(x,y). Critical points are points in the xy-plane where the tangent plane is horizontal.
Since the normal vector of the tangent plane at (x,y) is given by
The tangent plane is horizontal if its normal vector points in the z direction. Hence, critical points are solutions of the equations:
because horizontal planes have normal vector parallel to z-axis. The two equations above must be solved simultaneously.
Example
Let us find the critical points of
The partial derivatives are
f_x=0 if 1-x^2=0 or the exponential term is 0. f_y=0 if -2y=0 or the exponential term is 0. The exponential term is not 0 except in the degenerate case. Hence we require 1-x^2=0 and -2y=0, implying x=1 or x=-1 and y=0. There are two critical points (-1,0) and (1,0).
The Second Derivative Test for Functions of Two Variables
How can we determine if the critical points found above are relative maxima or minima? We apply a second derivative test for functions of two variables.
Let (x_c,y_c) be a critical point and define
We have the following cases:
• If D>0 and f_xx(x_c,y_c)<0, then f(x,y) has a relative maximum at (x_c,y_c).
• If D>0 and f_xx(x_c,y_c)>0, then f(x,y) has a relative minimum at (x_c,y_c).
• If D<0, then f(x,y) has a saddle point at (x_c,y_c).
• If D=0, the second derivative test is inconclusive.
An example of a saddle point is shown in the example below.
Example: Continued
For the example above, we have
For x=1 and y=0, we have D(1,0)=4exp(4/3)>0 with f_xx(1,0)=-2exp(2/3)<0. Hence, (1,0) is a relative maximum. For x=-1 and y=0, we have D(-1,0)=-4exp(-4/3)<0. Hence, (-1,0) is a saddle point.
The figure below plots the surface z=f(x,y).
Notice the relative maximum at (x=1,y=0). (x=-1,y=0) is a relative maximum if one travels in the y direction and a relative minimum if one travels in the x-direction. Near (-1,0) the surface looks like a saddle, hence the name.
Maxima and Minima in a Bounded Region
Suppose that our goal is to find the global maximum and minimum of our model function above in the square -2<=x<=2 and -2<=y<=2? There are three types of points that can potentially be global maxima or minima:
1. Relative extrema in the interior of the square.
2. Relative extrema on the boundary of the square.
3. Corner Points.
We have already done step 1. There are extrema at (1,0) and (-1,0). The boundary of square consists of 4 parts. Side 1 is y=-2 and -2<=x<=2. On this side, we have
The original function of 2 variables is now a function of x only. We set g'(x)=0 to determine relative extrema on Side 1. It can be shown that x=1 and x=-1 are the relative extrema. Since y=-2, the relative extrema on Side 1 are at (1,-2) and (-1,-2).
On Side 2 (x=-2 and -2<=y<=2)
We set h'(y)=0 to determine the relative extrema. It can be shown that y=0 is the only critical point, corresponding to (-2,0).
We play the same game to determine the relative extrema on the other 2 sides. It can be shown that they are (2,0), (1,2), and (-1,2).
Finally, we must include the 4 corners (-2,-2), (-2,2), (2,-2), and (2,2). In summary, the candidates for global maximum and minimum are (-1,0), (1,0), (1,-2), (-1,-2), (-2,0), (2,0), (1,2), (-1,2), (-2,-2), (-2,2), (2,-2), and (2,2). We evaluate f(x,y) at each of these points to determine the global max and min in the square. The global maximum occurs (-2,0) and (1,0). This can be seen in the figure above. The global minimum occurs at 4 points: (-1,2), (-1,-2), (2,2), and (2,-2).
Another example of a bounded region is the disk of radius 2 centered at the origin. We proceed as in the previous example, determining in the 3 classes above. (1,0) and (-1,0) lie in the interior of the disk.
The boundary of the disk is the circle x^2+y^2=4. To find extreme points on the disk we parameterize the circle. A natural parameterization is x=2cos(t) and y=2sin(t) for 0<=t<=2*pi. We substitute these expressions into z=f(x,y) and obtain
On the circle, the original functions of 2 variables is reduced to a function of 1 variable. We can determine the extrema on the circle using techniques from calculus of on variable.
In this problem there are not any corners. Hence, we determine the global max and min by considering points in the interior of the disk and on the circle. An alternative method for finding the maximum and minimum on the circle is the method of Lagrange multipliers.
Maxima and Minima for Functions of More than 2 Variables
The notion of extreme points can be extended to functions of more than 2 variables. Suppose z=f(x_1,x_2,...,x_n). (a_1,a_2,...,a_n) is extreme point if it satisfies the n equations
There is not a general second derivative test to determine if a point is a relative maximum or minimum for functions of more than two variables.
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# What is a formula of area of triangle?
## What is a formula of area of triangle?
So, the area A of a triangle is given by the formula A=12bh where b is the base and h is the height of the triangle. Example: Find the area of the triangle. The area A of a triangle is given by the formula A=12bh where b is the base and h is the height of the triangle.
What is a formula triangle?
Formula triangles are a tool to help students use equations without needing to rearrange them. In the example here it’s the volume: cover up V to see the required equation is moles divided by concentration. Or if you want the number of moles, cover up the n and then, as c and V are side by side, multiply them together.
Who gave the formula for area of triangle?
In geometry, Heron’s formula (sometimes called Hero’s formula), named after Hero of Alexandria, gives the area of a triangle when the length of all three sides are known. Unlike other triangle area formulae, there is no need to calculate angles or other distances in the triangle first.
### How find the area of a right triangle?
Area of one right triangle = 1/2 × l × w. We usually represent the legs of the right-angled triangle as base and height. Thus, the formula for the area of a right triangle is, Area of a right triangle = 1/2 × base × height.
How do I calculate volume of a triangle?
Calculating volume
1. Remember the formula for calculating volume is: Volume = Area by height. V = A X h.
2. For a triangle the area is calculated using the formula: Area = half of base by altitude. A = 0.5 X b X a.
3. So to calculate the volume of a triangular prism, the formula is: V = 0.5 X b X a X h.
What is the formula for area of all shapes?
Perimeter, Area, and Volume
Table 2. Area Formulas
Shape Formula Variables
Square A=s2 s is the length of the side of the square.
Rectangle A=LW L and W are the lengths of the rectangle’s sides (length and width).
Triangle A=12bh b and h are the base and height
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# How do you find the slope of the graph of 3x+5y=10?
May 5, 2018
$- \frac{3}{5}$ is the slope of the line.
#### Explanation:
Rearrange the equation into slope-intercept form to find the slope:
$y = m x + b$
$\text{m = slope}$
$\text{b = y-intercept}$
So rearrange the problem so that the $y$ is by itself on the left of the equal sign:
$3 x + 5 y = 10$
$5 y = - 3 x + 10$
$y = - \frac{3}{5} x + \frac{10}{5}$
$y = - \frac{3}{5} x + 2$
The slope is next to the $x$, so it is called $m$ in the slope-intercept form:
$y = \textcolor{b l u e}{m} x + b$
$y = \textcolor{b l u e}{- \frac{3}{5}} x + 2$
$- \frac{3}{5}$ is the slope of the line.
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A parabola is a conic section. It is a slice of a right cone parallel to one side (a generating line) of the cone. Like the circle, the parabola is a quadratic relation, but unlike the circle, either x will be squared or y will be squared, but not both. You worked with parabolas in Algebra 1 when you graphed quadratic equations. We will now be investigating the conic form of the parabola equation to learn more about the parabola's graph.
A parabola is defined as the set (locus) of points that are equidistant from both the directrix (a fixed straight line) and the focus (a fixed point).
This definition may be hard to visualize. Let's take a look. For ANY point on a parabola, the distance from that point to the focus is the same as the distance from that point to the directrix. (Looks like hairy spider legs!) Notice that the "distance" being measured to the directrix is always the shortest distance (the perpendicular distance). The specific distance from the vertex (the turning point of the parabola) to the focus is traditionally labeled "p". Thus, the distance from the vertex to the directrix is also "p". The focus is a point which lies "inside" the parabola on the axis of symmetry. The directrix is a line that is ⊥ to the axis of symmetry and lies "outside" the parabola (not intersecting with the parabola).
Conic Equations of Parabolas:
You recognize the equation of a parabola as being y = x2 or
y = ax2 + bx + c from your study of quadratics. And, of course, these remain popular equation forms of a parabola. But, if we examine a parabola in relation to its focal point (focus) and directrix, we can determine more information about the parabola. We are now going to look more closely at the coefficient of the x2 term to see what additional information it can tell us about the graph of the parabola. Keep in mind that all information you already know about parabolas remains true!
Parabola with Vertex at Origin (0,0)
(axis of symmetry parallel to the y-axis)
Conic Forms of Parabola Equations: with the vertex at (0,0), focus at (0, p) and directrix y = -p
In the example at the right, the coefficient of x² is 1, so , making p = ¼.
The vertex is (0,0), the focus is (0,¼), and the directrix is y = -¼.
The distance from the vertex
(in this case the origin) to the
focus is traditionally labeled as "p".
The leading coefficient in y = ax2 + bx + c is labeled "a". So when examining the coefficient of x2, we are examining a. p is the distance from the vertex to the focus.
You remember the vertex form of a parabola as being y = a(x - h)2 + k where (h, k) is the vertex of the parabola. If we let the coefficient of x2 (or a) = and perform some algebraic maneuvering, we can get the next equation.
Parabola with Vertex at (h, k)
(axis of symmetry parallel to the y-axis)
(Known as "standard form".)
Conic Form of Parabola Equation: (x - h)2 = 4p(y - k) with the vertex at (h, k), the focus at (h, k+p) and the directrix y = k - p
Since the example at the right is a translation of the previous graph, the relationship between the parabola and its focus and directrix remains the same (p = ¼). So with a vertex of (2,-3), we have:
(x - 2)2 = 4(¼) (y - (-3))
(x - 2)2 = y + 3
The focus is at (2,-3+¼) or (2,-2¾) and
the directrix is y = -3-¼ or y = -3¼
This "new" equation is just another form
of the old "vertex form" of a parabola.
OLD: y = (x - 2)² - 3
NEW: (x - 2)² = y + 3
S O M E T H I N G N E W ! ! !
Up to this point, all of your parabolas have been either opening upward or opening downward, depending upon whether the leading coefficient was positive or negative respectively. The axis of symmetry of those parabolas is parallel to the y-axis. We will now be looking at a parabola that opens to the right or to the left ("sideways"), with its axis of symmetry parallel to the x-axis.
"Sideways" Parabola Vertex (0,0)
(axis of symmetry parallel to x-axis)
Conic Forms of Parabola Equations: with the vertex at (0,0), focus at (p, 0), and directrix x = -p
We will now be examining the coefficient of y², instead of x².
In the example at the right, the coefficient of y² is 1, so , making p = ¼.
The vertex is (0,0), the focus is (¼,0), and the directrix is x = -¼.
For parabolas opening to the right or
to the left, the y-variable is being
squared (instead of the x² we are
used to seeing for parabolas).
"Sideways" Parabolas Vertex (h,k)
(axis of symmetry parallel to x-axis)
Conic Form of Parabola Equation: (y - k)2 = 4p(x - h) with the vertex at (h, k), the focus at (h+p, k) and the directrix x = h - p
Sideways Equation in Standard Vertex Form: x = a(y - k)2 + h with the vertex at (h, k).
Since the example at the right is a translation of the previous graph, the relationship between the parabola and its focus and directrix remains the same.
In the example at the right,
(y - 1)2 = 4(¼) (x - (-2))
(y - 1)2 = x + 2
The vertex is (-2,1), the focus is (-1¾,1)
and the directrix is y = -2¼.
Notice that parabolas that open right
or left, are NOT functions. They fail
the vertical line test for functions.
These parabolas are considered relations.
The Standard "Vertex Form"
and the Conic Form are the same:
VERTEX FORM: x = (y - 1)² - 2
CONIC FORM: (y - 1)² = x + 2
S u m m i n g U p: (p is distance from vertex to focus)
Vertical Parabola (up/down) Vertex (0,0): Vertex (h,k):
Horizontal Parabola (left-right) Vertex (0,0): Vertex (h,k):
Deriving the Conic Parabola Equation:
Deriving the Parabola Equation Start by placing the parabola's vertex at the origin, for ease of computation. By definition, the distance, p, from the origin to the focus will equal the distance from the origin to the directrix (which will be y = -p). The focus is point F and FA = AB by definition. Using the Distance Formula, we know Since we know FA = AB, we have The distance from the vertex (origin) to the focus is traditionally labeled as "p". ("p" is also the distance from the vertex to the directrix.)
Examples:
Given x2 = 16y, state whether the parabola opens upward, downward, right or left, and state the coordinates of the focus.
ANSWER: Form: x2 = 4py 4p = 16 p = 4 The focal length is 4. Since this "form" squares x, and the value of 4p is positive, the parabola opens upward. This form of parabola has its vertex at the origin, (0,0). The focal length (distance from vertex to focus) is 4 units. The focus is located at (0,4).
Given the parabola, (x - 3)2 = -8(y - 2), state whether the parabola opens upward, downward, right or left, and state the coordinates of the vertex, the focus, and the equation of the directrix.
ANSWER: Form: (x - h)2 = 4p(y - k) Vertex: (h,k) = (3,2) 4p = 8 p = 2 The focal length is 2. Since this "form" squares x, and the value of 4p is negative, the parabola opens downward. This form of parabola has its vertex at (h,k) = (3,2). The focal length (distance from vertex to focus) is 2 units. The focus is located at (3,0). The directrix is y = 4.
Write the equation of a parabola with a vertex at the origin and a focus of (0,-3).
ANSWER: Make a sketch. Remember that the parabola opens "around" the focus. Vertex: (0,0) and Focus: (0,-3) Focal length p = 3. Opening downward means negative. Form of Equation: x2 = 4py EQUATION: x2 = 4(-3)y x2 = -12y
Find the focus and directrix of the parabola whose equation
is x2 - 6x + 3y + 18 = 0.
ANSWER: You need to complete the square so the vertex, focus and directrix information will be visible. • The vertex is (3,-3). • The x-squared term indicates the parabola opens upward or downward. • The negative value indicates the parabola opens downward. • The focal length, p, is: 4p = 3; p = ¾ • The focus is at (3, -3¾) • The directrix is y = -2¼
Write the equation of a parabola whose focus (-2,1) and whose directrix is x = -6.
ANSWER: Make a sketch. Remember that the parabola opens "around" the focus, and the vertex is halfway between the focus and the directrix. • Vertex: (-4,1) = (h,k) • Opens to the right (around the focus) • Focal length, p = 2 • Form of Equation: (y - k)2 = 4p(x - h) EQUATION: (y - 1)2 = 4(2)(x - (-4)) (y - 1)2 = 8(x + 4)
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# Limits and asymptotes : 12th grade math lesson
Limits (sum, product, quotient) in a 12th grade math lesson with the study of indeterminate forms. In this lesson, we will conduct a study of horizontal, vertical and oblique asymptotes in the 12th grade for the compulsory education.
Knowledge needed for this chapter:
Determine the possible limit of a geometric sequence.
Study the limit of a sum, product or quotient
of two suites.
Use a comparison or framing theorem
to determine a limit of a sequence.
Establish (by derivation or not) the variations of a function.
## I.Limit of a function at infinity
In this section, is the representative curve of the function f in any plane.
### 1. Finite limit in infinity
Definition:
Let f be a function defined at least on an interval of of the type .
The function f has limit ℓ in if any open interval containing ℓ contains all
values of f (x) for x large enough. Then we note: .
Example:
Let f be the function defined on by . We have .
Indeed, the inverse of x approaches 0 as x increases.
Let be an open interval I such that . Then f (x) will always be in I for x large enough.
Graphically, as narrow as a band parallel to the line of equation y = 1 may be, and which
contains, there is always a value of x beyond which does not leave this band.
Horizontal asymptote.
The line with equation y = ℓ is horizontal asymptote to at if .
Remark:
Analogously, we define which characterizes a horizontal asymptote at in of equation y = ℓ.
Example:
We have seen previously that . We also have .
Therefore, the line with equation y = 1 is horizontal asymptote to the curve in and in .
Property (admitted): finite limits of usual functions in ± .
Let n be a non-zero natural number.
and .
## II. Infinite limit in infinity
Definition:
The function f has limit in if any interval of of the type contains
all values of f (x) for x large enough. Then we note: .
Example:
Let f be the square root function. We have.
Indeed, becomes as large as we want as x increases.
Let be an open interval . Then f (x) will always be in I for x large enough.
Graphically, if we consider the upper boundary half-plane a line of equation
y = a, there is always a value of a beyond which does not leave this half-plane.
Property (admitted): infinite limits of usual functions in ±.
Let n be a non-zero natural number.
and
### 2. Infinite limit in a real
Definition:
Let f be a function defined on an open interval of of type or .
The function f has limit in if any interval of of type contains all
the values of f (x) for x close enough to . Then we note: .
Definition: vertical asymptote.
The line of equation is vertically asymptotic to if or .
Property (admitted) : finite limits of usual functions in 0.
Let n be a non-zero natural number.
and
## III. Operations on the limits.
Property: limit of a sum, product and quotient of two functions.
## IV. Limit of a composite function
### 1. Compound function
Definition:
Let f be a function defined on E and having values in F, and let g be a function defined on F.
The compound of f followed by g is the function defined on E by .
Remark:
Do not confuse and which are, in general, different.
### 2. Theorem of composition of limits
Theorem:
Let h be the composite of the function f followed by g and a, b and c three real or ± .
If and , then .
## V. Limitations and comparison
Theorem:
### 2. The so-called “gendarmes” or “sandwich” framing theorem.
Theorem:
Let there be two real a and ℓ and three functions f , g and h such that, for x > a, we have .
If , then .
Remark:
As for the previous comparison theorem, we have two theorems
analogous when x tends to – and when x tends to a real .
Example:
Let’s determine the limit in – of .
The limit of cos x in – is indeterminate. Therefore the one of f (x) too.
However, for any strictly negative real x, so .
And dividing member by member by we have :
.
For ,.
Now, .
So, according to the gendarme theorem,.
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# 3.2 Solving Systems Algebraically. Solving System Algebraically Substitution y = 2x + 5 x = -y + 14.
## Presentation on theme: "3.2 Solving Systems Algebraically. Solving System Algebraically Substitution y = 2x + 5 x = -y + 14."— Presentation transcript:
3.2 Solving Systems Algebraically
Solving System Algebraically Substitution y = 2x + 5 x = -y + 14
Solving System Algebraically Substitution y = 4x – 7 y = ½ x + 7
Solving System Algebraically Elimination x + 6y = 10 2x + 5y = 6
Solving System Algebraically Elimination 2x + 5y = -1 3x + 4y = -5
When to use substitution? 1)A variable in an equation is isolated 2)Both equations are in y = mx +b form
When to use elimination? 1)Equations are in standard form ax + by = c
Special Case #1 x + 3y = 10 2x + 6y = 19 The solution to they system is false because 0 = -1. There is no solution because the lines are parallel.
Special Case #2 2x – 5y = 8 -4x + 10y = -16 The solution to they system is always true because 0 = 0. There is an infinite number of solutions is because they are the same line.
Parametric Equations Parametric Equations are equations that express the coordinates of x and y as separate functions of a common third variable, called the parameter. You can use parametric equations to determine the position of an object over time.
Parametric Example Starting from a birdbath 3 feet above the ground, a bird takes flight. Let t equal time in seconds, x equal horizontal distance in feet, and y equal vertical distance in feet. The equation x(t)= 5t and y(t)=8t+3 model the bird’s distance from the base of the birdbath. Using a graphing calculator, describe the position of the bird at time t=3.
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# Math Help: Creating a Number Line
When you're adding and subtracting positive and negative numbers, it's helpful to use a number line to visualize the operations. Read on to learn about different types of number lines and how to use them!
## How to Make a Number Line
Number lines usually include both positive and negative numbers. You can just make a short number line to help you solve a problem with lower numbers, or you can make a long number line that wraps all the way around the walls of your classroom. Either way, the process is the same:
1. Gather a piece of paper and a pencil, and then mark a dot in the center of the paper. This point will be the zero on your number line. Next, draw a horizontal line through the zero point.
2. Working from zero and moving to the right, divide the line into equal sections and label them with the positive numbers (1, 2, 3…).
3. Do the same thing on the left side of zero, this time labeling the marks with the negative numbers.
### Variations
If you'd like to make your number line a bit more interesting, there are lots of creative options. Here are a few ideas:
1. Make your number line vertical instead of horizontal, like a thermometer. The positive numbers go on the top half of the line (above zero), and the negative ones go on the bottom (below zero).
2. Use colored pens or markers to decorate your number line. For example, positive numbers could be red, and negative numbers could be blue.
3. Tailor your number line to the problem you're working on. For instance, if your math problem is about the distance a train is traveling, you could label your number line in miles and create a legend to show the directions.
Now that you've made yourself a number line, you might need some instructions on how to use it. First, keep in mind that number lines are most helpful for adding and subtracting positive and negative numbers. Here's how to use your number line to solve this kind of problem:
1. The first number in the problem is your starting point on the number line. For instance, if your problem is -5 + 7, then your starting point is negative five.
2. The sign in the middle of your problem tells you which direction you need to go from the starting point. In the problem -5 + 7, the '+' sign tells you to go to the right. If it were a '-' sign, you'd move to the left.
3. The second number in the problem tells you how many spaces on the number line you should go in the chosen direction. In our example problem, you go seven spaces to the right.
4. The number that you land on is your solution. Since positive two (2) is seven spaces to the right of negative five (-5), the answer is two (-5 + 7 = 2).
Did you find this useful? If so, please let others know!
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# Introduction to Asymptotic Analysis and Big O
In this lesson, we will learn about asymptotic notation, an important tool applied to the analysis of algorithms.
We'll cover the following
We have seen that the time complexity of an algorithm can be expressed as a polynomial. To compare two algorithms, we can compare the respective polynomials. However, the analysis performed in the previous lessons is a bit cumbersome and would become intractable for bigger algorithms that we tend to encounter in practice.
## Asymptotic Analysis
One observation that helps us is that we want to worry about large input sizes only. If the input size is really small, how bad can a poorly designed algorithm get, right? Mathematicians have a tool for this sort of analysis called the asymptotic notation. The asymptotic notation compares two functions, say, $f(n)$ and $g(n)$ for very large values of $n$. This fits in nicely with our need for comparing algorithms for very large input sizes.
## Big O Notation
One of the asymptotic notations is the Big O notation. A function $f(n)$ is considered $O(g(n))$, read as big oh of $g(n)$, if there exists some positive real constant $c$ and an integer $n_0 > 0$, such that the following inequality holds for all $n \geq n_0$:
$f(n) \leq cg(n)$
The following graph shows a plot of a function $f(n)$ and $cg(n)$ that demonstrates this inequality.
Note that the above inequality does not have to hold for all $n$. For $n < n_0$, $f(n)$ is allowed to exceed $cg(n)$, but for all values of $n$ beyond some value $n_0$, $f(n)$ is not allowed to exceed $cg(n)$.
What good is this? It tells us that for very large values of $n$, $f(n)$ will be at most within a constant factor of $g(n)$. In other words, $f(n)$ will grow no faster than a constant multiple of $g(n)$. Put yet another way, the rate of growth of $f(n)$ is within constant factors of that of $g(n)$.
People tend to write $f(n)$ = $O(g(n))$, which isn’t technically accurate. A whole lot of functions can satisfy the $O(g(n))$ conditions. Thus, $O(g(n))$ is a set of functions. It is OK to say that $f(n)$ belongs to $O(g(n))$.
### Example
Let’s consider an algorithm whose running time is given by $f(n) = 3n^3 + 4n + 2$. Let us try to verify that this algorithm’s time complexity is in $O(n^3)$. To do this, we need to find a positive constant $c$ and an integer $n_0 > 0$, such that for all $n \geq n_0$:
$3n^3 + 4n + 2 \leq cn^3$
The above inequality would still be true if we re-wrote it while replacing $cn^3$ with $3n^3 + 4n^3 + 2n^3$. What we have done is the replacement of the variable part in all terms with $n^3$, the variable-part of the highest order term. This gives us:
$3n^3 + 4n + 2 \leq 3n^3 + 4n^3 + 2n^3 = 9n^3$
This does not violate the inequality because each term on the right-hand side is greater than the corresponding term on the left-hand side. Now, comparing it with the definition of Big-O, we can pick c = 9.
That leaves $n_0$. For what values of $n$ is the inequality $9n^3 \leq cn^3$ satisfied? All of them, actually! So, we can pick $n_0 = 1$.
The above solution $(c=9, n_0 = 1)$ is not unique. We could have picked any value for $c$ that exceeds the coefficient of the highest power of $n$ in $f(n)$. Suppose, we decided to pick $c = 4$. The reader can verify that the inequality $3n^3 + 4n + 2 \leq cn^3$ still holds for $n_0 = 3$ or higher.
Note that it is not possible to find a constant $c$ and $n_0$ to show that $f(n) = 3n^3 + 4n + 2$ is $O(n^2)$ or $O(n)$. It is possible to show that $f(n)$ is $O(n^4)$ or $O(n^5)$ or any higher power of $n$. Mathematically, it is correct to say that $3n^3 + 4n + 2$ is $O(n^4)$. It gives us a loose bound on the asymptotic running time of the algorithm. When dealing with time and space complexities, we are generally interested in the tightest possible bound when it comes to the asymptotic notation.
Note: All challenges in this chapter should be answered with the lowest order big oh.
Suppose algorithms A and B have running time of $O(n)$ and $O(n^2)$, respectively. For sufficiently large input sizes, algorithm A will run faster than algorithm B. That does not mean that algorithm A will always run faster than algorithm B.
Algorithm A and B both have running time $O(n)$. The execution time for these algorithms, for very large input sizes, will be within constant factors of each other. For all practical purposes, they are considered equally good.
## Simplified Asymptotic Analysis
Once we have obtained the time complexity of an algorithm by counting the number of primitive operations as discussed in the previous two lessons, we can arrive at the Big O notation for the algorithm simply by:
• Dropping the multiplicative constants with all terms
• Dropping all but the highest order term
Thus, $n^2 + 2n + 1$ is $O(n^2)$ while $n^5 + 4n^3 + 2n + 43$ is $O(n^5)$.
Notice how the constant coefficients have become insignificant in the Big-Oh notation. Recall that these constants represent the number of primitive operations on a given line of code. This means that while analyzing code, counting a line of code as contributing $4$ primitive operations is as good as counting it as $1$ primitive operation. What matters is correctly counting the number of times each line of code is repeated.
Moving forward, we will simplify the analysis by just counting the number of executions of each line of code, instead of the number of operations.
Sometimes the constant coefficients do become important. For example, consider algorithms A and B that have a worst-case running time of 100000n + 4 and 10n + 6, respectively. Asymptotically, both are $O(n)$. However, the worst-case running time for algorithm B is numerically better than A.
## A Comparison of Some Common Functions
It is easy to work with simple polynomials in $n$, but when the time complexity involves other types of functions like $log()$, you may find it hard to identify the “highest order term”. The following table lists some commonly encountered functions in ascending order of rate of growth. Any function can be given as Big O of any other function that appears later in this table.
Function Name Function Name
1. Any constant Constant 7. $n^2$ Quadratic
2. $log n$ Logarithmic 8. $n^3$ Cubic
3. $log^2 n$ Log-square 9. $n^4$ Quartic
4. $\sqrt n$ Root-n 10. $2^n$ Exponential
5. $n$ Linear 11. $e^n$ Exponential
6. $nlogn$ Linearithmic 12. $n!$ n-Factorial
The following graph visually shows some of the functions from the above table.
Quick quiz on Big O!
1
$e^{3n}$ is in $O(e^n)$
A)
True
B)
False
Question 1 of 20 attempted
Now that you are familiar with the Big O notation, let’s discuss other notations in the next lesson.
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# Perpendicular Axis Theorem
Last updated date: 23rd May 2024
Total views: 72.9k
Views today: 1.72k
## What is the Perpendicular Axis Theorem?
The parallel and perpendicular axis theorem deals with the moment of inertia. So, before studying the theorems, let's know about the moment of inertia. It is the property of a body by virtue of which a body resists angular acceleration.
Angular acceleration is the sum of the product of masses of particles of the body with the square of the distance from the axis of rotation.
Moment of inertia $I_{i} = \sum m_{i}r_{i}^{2}$
The equation is applicable only in-plane lamina.
This theorem is used for symmetric objects, i.e. when two out of three axes are symmetric. The moment of inertia of the third axis can be calculated by this equation when the moment of inertia of the other two axes is known.
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### State and Prove Theorem of Perpendicular Axis
Perpendicular axis theorem statement - The perpendicular axis theorem states that the moment of inertia, for any axis which is perpendicular to the plane, is equal to the sum of any two perpendicular axes of the body which intersects with the first axis.
Let us consider a plane lamina made up of a large number of particles in the x-y plane as represented by the figure. Consider a particle having mass 'm' at point P.
From P, draw PN and PN' perpendicular to the x and y-axis, respectively.
The moment of inertia about the x-axis = my².
The moment of inertia of the whole lamina about the x-axis is given by
Ix = ∑my²-----(1)
The moment of inertia of the whole lamina about why the axis is given by
Iy = ∑mx²-------(2)
Similarly, the moment of inertia of the whole lamina about the z-axis is given by,
Iz = ∑mr²
But r² = x² + y²
Therefore,
Iz = ∑m (x² + y²)
From eq(1) and (2), we get:
i.e., Iz = ∑mx² + ∑my²
(or)
Iz = Ix + Iy.
The perpendicular axis theorem helps calculate the moment of inertia of a body where it's difficult to access one vital axis of the body.
### Parallel Axis Theorem and Perpendicular Axis Theorem
The parallel and perpendicular axis theorem is mentioned below:
1. Parallel Axis Theorem
The parallel axis theorem states that the moment of inertia of a body about an axis that is parallel to the axis of the body and passing through its centre is equal to the sum of moments of inertia of the body about the axis passing through the centre of the product of the mass of the body and the square of the distance between two axes.
Statement
The parallel axis theorem is represented by the following equation:
I = Ic + Mh2
Where,
I = moment of inertia of the body.
Ic = moment of inertia about the centre of the body.
M = mass of the body.
h2 = square of the distance between the two axes.
1. Perpendicular Axis Theorem
Let’s see an example for this theorem:
Let us suppose that we want to calculate the moment of inertia of a uniform ring about its diameter.
Let the centre of the ring be MR2/2,
Where,
M = mass of the ring
R = radius of the ring
According to the perpendicular axis theorem,
Iz = Ix + Iy
Since it is a uniform ring; so all its diameter is equal
Therefore, $I_{x} = I_{y}$ $I_{z} = 2 I_{x}$
I= Iy
∴ Iz = 2 Ix
Iz = (MR²)/2
Hence, the moment of inertia of a ring about its diameter is MR2/2.
### Parallel Axis Theorem Derivation
Let Ic be the moment of inertia of an object with the axis passing through the centre of mass of the object. Let, I will be the moment of inertia about the axis A'B' (from figure AB) at a distance of h.
Let us consider a particle having mass ‘m’ and located at a distance ‘r’ from the centre of gravity of the body.
Distance from A’B’ = r + h
I = ∑m (r + h)2
I = ∑m (r2 + h2 + 2rh)
I = ∑mr2 + ∑mh2 + ∑2rh
I = Ic + h2∑m + 2h∑mr
I = Ic + Mh2 + 0
I = Ic + Mh2
### Difference Between Parallel and Perpendicular Axis Theorem
Parallel Axis Theorem Perpendicular Axis Theorem The MOI around any axis is equal to the sum of moments of inertia about an axis parallel to the axis passing through the centre of mass (COM) of the object and the product of the mass of the object with the square of the perpendicular distance from the axis in consideration and the COM axis parallel to it. The moment of inertia of an object about an axis perpendicular to it is equal to the sum of moments of inertia of the object about two mutually perpendicular axes lying in the plane of the object. Io = Ic + md2Here,Io = M.O.I of the object about the point O.Ic = M.O.I of the object about the centroid C.md2 = added M.O.I due to the distance between O and C. Izz = Ixx + IyyIZZ = MOI of the object about the 3D plane along the z-axisIxx = MOI of the object about the 3D plane along the x-axisIyy = MOI of the object about the 3D plane along the y-axis This theorem is applicable for any object, there is no such restriction. This theorem applies to planar bodies, which means two-dimensional bodies.
## FAQs on Perpendicular Axis Theorem
1. Why is the Perpendicular Axis Theorem not applicable for 3-D objects?
Perpendicular axis theorem is not applicable for 3D objects in the case of a planar object in the x-y plane.
However, the perpendicular axis theorem doesn't work for three-dimensional objects because the equation is derived from the assumption that the object is planar.
2. What are the applications of the Perpendicular Axis Theorem?
Some of the most important application of the perpendicular axis theorem is outlined here below:
• With the help of the perpendicular axis theorem, the moment of inertia about a third axis can be calculated.
• By using the perpendicular axis theorem, the moment of inertia for a three-dimensional object can be calculated.
3. The MOI of a thin Uniform Rod with Mass M and Length L about an axis perpendicular to the rod, through its centre, is I. What is the MOI of a rod on an axis perpendicular to the rod at its endpoint?
a. 4I
b. 2I
c. I/4
d. I/2
a. $I_{center} = \frac{ML^{2}}{12}$and
Ipendpoint = $\frac{ML^{2}}{3}$
4. When do we use a Parallel Axis Theorem?
It is used to calculate the moment of inertia of the area of a rigid object, whose axis is parallel to the axis of the object, and passes through the centre of gravity of the object.
5. How to manage time while writing a Physics Exam?
Many students say that though they went prepared to give their Physics exam, due to the lengthy and tricky questions, it became difficult. To avoid such an experience students are recommended to practice writing faster and memorizing the keywords. Also working out objective type questions beforehand saves time in analyzing and answering them. Lastly, students are advised to divide their time into sections namely A, B, C and D and attempt all the questions without fail.
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# Cube Root of 1728
The cube root of 1728, expressed as 3√1728, is equal to a value that when multiplied three times by itself, will give the original number. To find the value of 3√1728, we can use two methods, i.e. prime factorisation method and the estimation method, without using any calculator. Also, learn cube root of numbers here.
Cube root of 1728, 3√1728 = 12
Since 1728 is a four-digit number, so finding its prime factors can be a little lengthy. Thus, we can use an estimation method for fast calculation. For this method, we have to learn the value of cubes of natural numbers from 1 to 10, which is provided here in the later part. Now, let us look at both methods one by one.
Also, check:
## Cube Root of 1728 by Prime Factorisation Method
Using prime factorisation, we will find prime factors of 1728, since it is a perfect cube and then will pair them in a group of three. Let us understand it in a step-by-step procedure.
Step 1: Find the prime factors of 1728
1728 = 2x2x2x2x2x2x3x3x3
Step 2: Group the factors in a pair of three and write in the form of cubes.
1728 = (2x2x2)x(2x2x2)x(3x3x3)
1728 = 23x23x33
Step 3: Apply the cube root on both sides and take out the terms in cubes out of the cube root.
3√1728 = 3√(23x23x33) = 2 x 2 x 3 = 12
Hence, 3√1728 =12
## Cube Root of 1728 By Estimation Method
Now, we will learn here to find the cube root of the four-digit number 1728 using the estimation method. Before we get to know about this method, we need to memorise the value of cubes from 1 to 10. Here is the table for reference.
Number Cubes 1 1 2 8 3 27 4 64 5 125 6 216 7 343 8 512 9 729 10 1000
Now we will proceed to find the value of 3√1728 following these steps:
Step 1:
We will firstly take the digit at the unit’s place.
So, here we get 8.
Step 2:
Now we will check the cube table, the cube of which number has 8 at its unit digit place.
We see, 23 = 8.
That means the cube root of 1728 will have 2 at the unit place.
Step 3:
Now ignore the last 3 digits of 1728, i.e. 728
Taking 1 as a reference number, we know the cube of 1 is equal to 1.
Therefore, we get the two-digit answer.
Hence, 3√1728 = 12
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COMPARE FRACTIONS INSTRUCTIONS
When Compare Fractions starts, you will be given two fractions to compare, as in the example below:
You are to decide if the fraction on the left is less than, equal to, or greater than the fraction on the right. You will choose < for less than, = for equal to, or > for greater than. Shown in the dialog box are your choices of <, >, or =.
If the denominators are the same, the fraction with the larger numerator is larger and if the numerators are the same, the fraction with the larger denominator is smaller.
The fractions 3/4 and 2/3 are pictured with number lines below:
Let's make the denominators the same so that we can compare the numerators. Fractions with the same denominators are like fractions.
Here, we will introduce the idea of the least common denominator or LCDLCD is an idea that will be used in comparing fractions, and adding and subtracting fractions. The LCD is the smallest number that both 4 and 3 will divide into evenly. The LCD for the fractions 3/4 and 2/3 is 12 because both denominators 4 and 3 divide evenly into 12.
Then, write each fraction with the common denominator 12 to make them like. The illustration shows that 3/4 is equal to 9/12 and 2/3 is equal to 8/12. Once each fraction is renamed with a common denominator, you can compare the numerators - the larger the numerator the larger the fraction.
Since 3/4 is greater than 2/3, you will select the > symbol.
See the program RENAME IN HIGHER TERMS for more information on renaming fractions.
One way to find the LCD is to see if the smaller denominator 3 will divide evenly into the larger denominator 4. If not, multiply the larger denominator 4 by 2 to get 8. Will the smaller denominator 3 divide into 8? No, so multiply the larger denominator by 3 to get 12. Will 3 divide evenly into 12? Yes, so 12 is the LCD for the denominators 3 and 4. If not, then multiply by 4, then 5, etc. until the smaller denominator divides into the product.
Also, thinking of the pictures of the fractions will help you decide which is the larger.
Choose the < (less than) button if you think the first fraction is smaller than the second. Choose the = (equal button) if you think the two fractions are the same size. Choose the > (greater than) if you think the fraction is larger. If correct, circles showing the comparative sizes of the two fractions will appear. Also, the number lines will show the two fractions with the LCD.
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# Weighted mean
The weighted mean is similar to an arithmetic mean (the most common type of average), where instead of each of the data points contributing equally to the final average, some data points contribute more than others. The notion of weighted mean plays a role in descriptive statistics and also occurs in a more general form in several other areas of mathematics.
If all the weights are equal, then the weighted mean is the same as the arithmetic mean. While weighted means generally behave in a similar fashion to arithmetic means, they do have a few counterintuitive properties, as captured for instance in Simpson's paradox.
The term weighted average usually refers to a weighted arithmetic mean, but weighted versions of other means can also be calculated, such as the weighted geometric mean and the weighted harmonic mean.
## Examples
The following examples are designed to introduce this topic to those with only basic math skills.
### Basic example
Given two school classes, one with 20 students, and one with 30 students, the grades in each class on a test were:
Morning class = 62, 67, 71, 74, 76, 77, 78, 79, 79, 80, 80, 81, 81, 82, 83, 84, 86, 89, 93, 98
Afternoon class = 81, 82, 83, 84, 85, 86, 87, 87, 88, 88, 89, 89, 89, 90, 90, 90, 90, 91, 91, 91, 92, 92, 93, 93, 94, 95, 96, 97, 98, 99
The straight average for the morning class is 80 and the straight average of the afternoon class is 90. The straight average of 80 and 90 is 85, the mean of the two class means. However, this does not account for the difference in number of students in each class, hence the value of 85 does not reflect the average student grade (independent of class). The average student grade can be obtained by averaging all the grades, without regard to classes (add all the grades up and divide by the total number of students):
${\displaystyle {\bar {x}}={\frac {4300}{50}}=86.}$
Or, this can be accomplished by weighting the class means by the number of students in each class (using a weighted mean of the class means):
${\displaystyle {\bar {x}}={\frac {(20\times 80)+(30\times 90)}{20+30}}=86.}$
Thus, the weighted mean makes it possible to find the average student grade in the case where only the class means and the number of students in each class are available.
### Convex combination example
Since only the relative weights are relevant, any weighted mean can be expressed using coefficients that sum to one. Such a linear combination is called a convex combination.
Using the previous example, we would get the following:
${\displaystyle {\frac {20}{20+30}}=0.4\,}$
${\displaystyle {\frac {30}{20+30}}=0.6\,}$
${\displaystyle {\bar {x}}={\frac {(0.4\times 80)+(0.6\times 90)}{0.4+0.6}}=86.}$
This simplifies to:
${\displaystyle {\bar {x}}=(0.4\times 80)+(0.6\times 90)=86.}$
## Mathematical definition
Formally, the weighted mean of a non-empty set of data
${\displaystyle \{x_{1},x_{2},\dots ,x_{n}\},}$
with non-negative weights
${\displaystyle \{w_{1},w_{2},\dots ,w_{n}\},}$
is the quantity
${\displaystyle {\bar {x}}={\frac {\sum _{i=1}^{n}w_{i}x_{i}}{\sum _{i=1}^{n}w_{i}}},}$
which means:
${\displaystyle {\bar {x}}={\frac {w_{1}x_{1}+w_{2}x_{2}+\cdots +w_{n}x_{n}}{w_{1}+w_{2}+\cdots +w_{n}}}.}$
Therefore data elements with a high weight contribute more to the weighted mean than do elements with a low weight. The weights cannot be negative. Some may be zero, but not all of them (since division by zero is not allowed).
The formulas are simplified when the weights are normalized such that they sum up to ${\displaystyle 1}$, i.e. ${\displaystyle \sum _{i=1}^{n}{w_{i}}=1}$. For such normalized weights the weighted mean is simply ${\displaystyle {\bar {x}}=\sum _{i=1}^{n}{w_{i}x_{i}}}$.
Note that one can always normalize the weights by making the following transformation on the weights ${\displaystyle w_{i}'={\frac {w_{i}}{\sum _{i=1}^{n}{w_{i}}}}}$. Using the normalized weight yields the same results as when using the original weights. Indeed,
${\displaystyle {\bar {x}}=\sum _{i=1}^{n}w'_{i}x_{i}=\sum _{i=1}^{n}{\frac {w_{i}}{\sum _{i=1}^{n}w_{i}}}x_{i}={\frac {\sum _{i=1}^{n}w_{i}x_{i}}{\sum _{i=1}^{n}w_{i}}}.}$
The common mean ${\displaystyle {\frac {1}{n}}\sum _{i=1}^{n}{x_{i}}}$ is a special case of the weighted mean where all data have equal weights, ${\displaystyle w_{i}=w}$. When the weights are normalized then ${\displaystyle w_{i}'={\frac {1}{n}}.}$
## Statistical properties
The weighted sample mean, ${\displaystyle {\bar {x}}}$, with normalized weights (weights summing to one) is itself a random variable. Its expected value and standard deviation are related to the expected values and standard deviations of the observations as follows,
If the observations have expected values
${\displaystyle E(x_{i})={\bar {x_{i}}},}$
then the weighted sample mean has expectation
${\displaystyle E({\bar {x}})=\sum _{i=1}^{n}{w_{i}{\bar {x_{i}}}}.}$
Particularly, if the expectations of all observations are equal, ${\displaystyle {\bar {x_{i}}}=c}$, then the expectation of the weighted sample mean will be the same,
${\displaystyle E({\bar {x}})=c.\,}$
For uncorrelated observations with standard deviations ${\displaystyle \sigma _{i}}$, the weighted sample mean has standard deviation
${\displaystyle \sigma ({\bar {x}})={\sqrt {\sum _{i=1}^{n}{w_{i}^{2}\sigma _{i}^{2}}}}.}$
Consequently, when the standard deviations of all observations are equal, ${\displaystyle \sigma _{i}=d}$, the weighted sample mean will have standard deviation ${\displaystyle \sigma ({\bar {x}})=d{\sqrt {V_{2}}}}$. Here ${\displaystyle V_{2}}$ is the quantity
${\displaystyle V_{2}=\sum _{i=1}^{n}{w_{i}^{2}},}$
such that ${\displaystyle 1/n\leq V_{2}\leq 1}$. It attains its minimum value for equal weights, and its maximum when all weights except one are zero. In the former case we have ${\displaystyle \sigma ({\bar {x}})=d/{\sqrt {n}}}$, which is related to the central limit theorem.
Note that due to the fact that one can always transform non-normalized weights to normalized weights all formula in this section can be adapted to non-normalized weights by replacing all ${\displaystyle w_{i}}$ by ${\displaystyle w_{i}'={\frac {w_{i}}{\sum _{i=1}^{n}{w_{i}}}}}$.
## Dealing with variance
For the weighted mean of a list of data for which each element ${\displaystyle x_{i}\,\!}$ comes from a different probability distribution with known variance ${\displaystyle {\sigma _{i}}^{2}\,}$, one possible choice for the weights is given by:
${\displaystyle w_{i}={\frac {1}{\sigma _{i}^{2}}}.}$
The weighted mean in this case is:
${\displaystyle {\bar {x}}={\frac {\sum _{i=1}^{n}(x_{i}/{\sigma _{i}}^{2})}{\sum _{i=1}^{n}(1/{\sigma _{i}}^{2})}},}$
and the variance of the weighted mean is:
${\displaystyle \sigma _{\bar {x}}^{2}={\frac {1}{\sum _{i=1}^{n}(1/{\sigma _{i}}^{2})}},}$
The significance of this choice is that this weighted mean is the maximum likelihood estimator of the mean of the probability distributions under the assumption that they are independent and normally distributed with the same mean.
### Correcting for over- or under-dispersion
Weighted means are typically used to find the weighted mean of experimental data, rather than theoretically generated data. In this case, there will be some error in the variance of each data point. Typically experimental errors may be underestimated due to the experimenter not taking into account all sources of error in calculating the variance of each data point. In this event, the variance in the weighted mean must be corrected to account for the fact that ${\displaystyle \chi ^{2}}$ is too large. The correction that must be made is
${\displaystyle \sigma _{\bar {x}}^{2}\rightarrow \sigma _{\bar {x}}^{2}\chi _{\nu }^{2}\,}$
where ${\displaystyle \chi _{\nu }^{2}}$ is ${\displaystyle \chi ^{2}}$ divided by the number of degrees of freedom, in this case n − 1. This gives the variance in the weighted mean as:
${\displaystyle \sigma _{\bar {x}}^{2}={\frac {1}{\sum _{i=1}^{n}1/{\sigma _{i}}^{2}}}\times {\frac {1}{(n-1)}}\sum _{i=1}^{n}{\frac {(x_{i}-{\bar {x}})^{2}}{\sigma _{i}^{2}}};}$
when all data variances are equal, ${\displaystyle \sigma _{i}=\sigma _{0}}$, they cancel out in the weighted mean variance, ${\displaystyle \sigma _{\bar {x}}^{2}}$, which then reduces to the standard error of the mean (squared), ${\displaystyle \sigma _{\bar {x}}^{2}=\sigma ^{2}/n}$, in terms of the sample standard deviation (squared), ${\displaystyle \sigma ^{2}=\sum _{i=1}^{n}(x_{i}-{\bar {x}})^{2}/(n-1)}$.
## Weighted sample variance
Typically when a mean is calculated it is important to know the variance and standard deviation about that mean. When a weighted mean ${\displaystyle \mu ^{*}}$ is used, the variance of the weighted sample is different from the variance of the unweighted sample. The biased weighted sample variance is defined similarly to the normal biased sample variance:
${\displaystyle \sigma ^{2}\ ={\frac {\sum _{i=1}^{N}{\left(x_{i}-\mu \right)^{2}}}{N}}}$
${\displaystyle \sigma _{\mathrm {weighted} }^{2}={\frac {\sum _{i=1}^{N}w_{i}\left(x_{i}-\mu ^{*}\right)^{2}}{V_{1}}}}$
where ${\displaystyle V_{1}=\sum _{i=1}^{n}w_{i}}$, which is 1 for normalized weights.
For small samples, it is customary to use an unbiased estimator for the population variance. In normal unweighted samples, the N in the denominator (corresponding to the sample size) is changed to N − 1. While this is simple in unweighted samples, it is not straightforward when the sample is weighted. The unbiased estimator of a weighted population variance (assuming each ${\displaystyle x_{i}}$ is drawn from a Gaussian distribution with variance ${\displaystyle 1/w_{i}}$) is given by [1]:
${\displaystyle s^{2}\ ={\frac {V_{1}}{V_{1}^{2}-V_{2}}}\sum _{i=1}^{N}w_{i}\left(x_{i}-\mu ^{*}\right)^{2},}$
where ${\displaystyle V_{2}=\sum _{i=1}^{n}{w_{i}^{2}}}$ as introduced previously. The degrees of freedom of the weighted, unbiased sample variance vary accordingly from N − 1 down to 0.
The standard deviation is simply the square root of the variance above.
If all of the ${\displaystyle x_{i}}$ are drawn from the same distribution and the integer weights ${\displaystyle w_{i}}$ indicate frequency of occurrence in the sample, then the unbiased estimator of the weighted population variance is given by
${\displaystyle s^{2}\ ={\frac {1}{V_{1}-1}}\sum _{i=1}^{N}w_{i}\left(x_{i}-\mu ^{*}\right)^{2},}$
If all ${\displaystyle x_{i}}$ are unique, then ${\displaystyle N}$ counts the number of unique values, and ${\displaystyle V_{1}}$ counts the number of samples.
For example, if values ${\displaystyle \{2,2,4,5,5,5\}}$ are drawn from the same distribution, then we can treat this set as an unweighted sample, or we can treat it as the weighted sample ${\displaystyle \{2,4,5\}}$ with corresponding weights ${\displaystyle \{2,1,3\}}$, and we should get the same results.
## Vector-valued estimates
The above generalizes easily to the case of taking the mean of vector-valued estimates. For example, estimates of position on a plane may have less certainty in one direction than another. As in the scalar case, the weighted mean of multiple estimates can provide a maximum likelihood estimate. We simply replace ${\displaystyle \sigma ^{2}}$ by the covariance matrix[2]:
${\displaystyle W_{i}=\Sigma _{i}^{-1}.}$
The weighted mean in this case is:
${\displaystyle {\bar {\mathbf {x} }}=\left(\sum _{i=1}^{n}\Sigma _{i}^{-1}\right)^{-1}\left(\sum _{i=1}^{n}\Sigma _{i}^{-1}\mathbf {x} _{i}\right),}$
and the covariance of the weighted mean is:
${\displaystyle \Sigma _{\bar {\mathbf {x} }}=\left(\sum _{i=1}^{n}\Sigma _{i}^{-1}\right)^{-1},}$
For example, consider the weighted mean of the point [1 0] with high variance in the second component and [0 1] with high variance in the first component. Then
${\displaystyle \mathbf {x} _{1}:=[10]^{\top },\qquad \Sigma _{1}:={\begin{bmatrix}1&0\\0&100\end{bmatrix}}}$
${\displaystyle \mathbf {x} _{2}:=[01]^{\top },\qquad \Sigma _{2}:={\begin{bmatrix}100&0\\0&1\end{bmatrix}}}$
then the weighted mean is:
${\displaystyle {\bar {\mathbf {x} }}=\left(\Sigma _{1}^{-1}+\Sigma _{2}^{-1}\right)^{-1}\left(\Sigma _{1}^{-1}\mathbf {x} _{1}+\Sigma _{2}^{-1}\mathbf {x} _{2}\right)}$
${\displaystyle ={\begin{bmatrix}0.9901&0\\0&0.9901\end{bmatrix}}{\begin{bmatrix}1\\1\end{bmatrix}}={\begin{bmatrix}0.9901\\0.9901\end{bmatrix}}}$
which makes sense: the [1 0] estimate is "compliant" in the second component and the [0 1] estimate is compliant in the first component, so the weighted mean is nearly [1 1].
## Accounting for correlations
In the general case, suppose that ${\displaystyle \mathbf {X} =[x_{1},\dots ,x_{n}]}$, ${\displaystyle \mathbf {C} }$ is the covariance matrix relating the quantities ${\displaystyle x_{i}}$, ${\displaystyle {\bar {x}}}$ is the common mean to be estimated, and ${\displaystyle \mathbf {W} }$ is the design matrix [1, ..., 1] (of length ${\displaystyle n}$). The Gauss–Markov theorem states that the estimate of the mean having minimum variance is given by:
${\displaystyle \sigma _{\bar {x}}^{2}=(\mathbf {W} ^{T}\mathbf {C} ^{-1}\mathbf {W} )^{-1},}$
and
${\displaystyle {\bar {x}}=\sigma _{\bar {x}}^{2}(\mathbf {W} ^{T}\mathbf {C} ^{-1}\mathbf {X} ).}$
## Decreasing strength of interactions
Consider the time series of an independent variable ${\displaystyle x}$ and a dependent variable ${\displaystyle y}$, with ${\displaystyle n}$ observations sampled at discrete times ${\displaystyle t_{i}}$. In many common situations, the value of ${\displaystyle y}$ at time ${\displaystyle t_{i}}$ depends not only on ${\displaystyle x_{i}}$ but also on its past values. Commonly, the strength of this dependence decreases as the separation of observations in time increases. To model this situation, one may replace the independent variable by its sliding mean ${\displaystyle z}$ for a window size ${\displaystyle m}$.
${\displaystyle z_{k}=\sum _{i=1}^{m}w_{i}x_{k+1-i}.}$
Range weighted mean interpretation
Range (1–5) Weighted mean equivalence
3.34–5.00 Strong
1.67–3.33 Satisfactory
0.00–1.66 Weak
## Exponentially decreasing weights
In the scenario described in the previous section, most frequently the decrease in interaction strength obeys a negative exponential law. If the observations are sampled at equidistant times, then exponential decrease is equivalent to decrease by a constant fraction ${\displaystyle 0<\Delta <1}$ at each time step. Setting ${\displaystyle w=1-\Delta }$ we can define ${\displaystyle m}$ normalized weights by
${\displaystyle w_{i}={\frac {w^{i-1}}{V_{1}}}}$,
where ${\displaystyle V_{1}}$ is the sum of the unnormalized weights. In this case ${\displaystyle V_{1}}$ is simply
${\displaystyle V_{1}=\sum _{i=1}^{m}{w^{i-1}}={\frac {1-w^{m}}{1-w}}}$,
approaching ${\displaystyle V_{1}=1/(1-w)}$ for large values of ${\displaystyle m}$.
The damping constant ${\displaystyle w}$ must correspond to the actual decrease of interaction strength. If this cannot be determined from theoretical considerations, then the following properties of exponentially decreasing weights are useful in making a suitable choice: at step ${\displaystyle (1-w)^{-1}}$, the weight approximately equals ${\displaystyle {e^{-1}}(1-w)=0.39(1-w)}$, the tail area the value ${\displaystyle e^{-1}}$, the head area ${\displaystyle {1-e^{-1}}=0.61}$. The tail area at step ${\displaystyle n}$ is ${\displaystyle \leq {e^{-n(1-w)}}}$. Where primarily the closest ${\displaystyle n}$ observations matter and the effect of the remaining observations can be ignored safely, then choose ${\displaystyle w}$ such that the tail area is sufficiently small.
## Weighted averages of functions
The concept of weighted average can be extended to functions.[3] Weighted averages of functions play an important role in the systems of weighted differential and integral calculus.[4]
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Unit 8 Section 3 : Plotting Scatter Diagrams
# Unit 8 Section 3 : Plotting Scatter Diagrams
In this section we review plotting scatter diagrams and discuss the different types of correlation that you can expect to see on these diagrams.
Strong positive correlation between x and y. The points lie close to a straight line with
y increasing as x increases.
Weak, positive correlation between x and y. The trend shown is that
y increases as x increases
but the points are not close to a straight line.
No correlation between x and y; the points are distributed randomly on the graph.
Weak, negative correlation between x and y. The trend shown is that
y decreases as x increases
but the points do not lie close to a straight line.
Strong, negative correlation. The points lie close to a straight line, with
y decreasing as x increases.
If the points plotted were all on a straight line we would have perfect correlation, but it could be positive or negative as shown in the diagrams above.
## Example 1
The following table lists values of x and y.
x 2 3 5 6 9 11 12 15 y 10 7 8 5 6 2 5 2
(a)
Use the data to draw a scatter graph.
(b)
Describe the type of correlation that you observe.
It shows weak, negative correlation.
## Example 2
What sort of correlation would you expect to find between:
(a)
a person's age and their house number,
No correlation, because these two quantities are not linked in any way.
(b)
a child's age and their height,
Positive correlation, because children get taller as they get older.
(c)
an adult's age and their height ?
No correlation, because the height of adults does not change with their age.
## Exercises
Question 1
Consider the following scatter graphs:
(a) Which graph shows strong correlation? Which graphs show positive correlation? Which graph shows negative correlation? Which graph shows a weak, positive correlation?
Question 2
The following table lists values of x and y.
x 2 4 6 7 8 9 10 11 12 y 3 5 8 5 9 6 9 9 11
(a)
Plot a scatter graph for this data.
(b)
Describe the correlation between x and y.
Question 3
Complete the table below for 10 people in your class.
(a)
Plot a scatter graph for your data.
(b)
Describe the type of correlation that there is between these two quantities.
Question 4
A driver keeps a record of the distance travelled and the amount of fuel in his tank on a long journey.
Distance Travelled (km) Fuel in Tank (litres) 0 50 100 150 200 250 300 80 73 67 61 52 46 37
(a)
Illustrate this data with a scatter plot.
(b)
Describe the type of correlation that is present.
Question 5
What type of correlation would you expect to find between each of the following quantities:
(a)
Age and pocket money
(b)
IQ and height,
(c)
Price of house and number of bedrooms,
Possibly strong positive correlation in a single, smallish geographical area. For wider areas with greater mix of housing, little or no correlation.
(d)
Person's height and shoe size ?
Question 6
In a class 10 pupils took a Science test and an English test. Their scores are listed in the following table:
Pupil English Score Science Score A B C D E F G H I J 2 10 18 4 9 7 18 19 3 10 18 12 6 3 11 20 4 17 7 2
(a)
Draw a scatter graph for this data.
(b)
Describe the correlation between the two scores.
Question 7
Chris carries out an experiment in which he measure the extension of a spring when he hangs different masses on it. The following table lists his results:
Mass (grams) Extension (cm) 20 50 100 120 200 1.2 3 6 7.2 12
(a)
Draw a scatter graph for this data.
(b)
Describe the correlation between the mass and the extension.
Perfect positive correlation
Question 8
Every day Peter picks the ripe tomatoes in his greenhouse. He keeps a record of their mass and the number that he picks. His results are listed in the following table:
Number of Tomatoes Picked Total Mass (grams) 1 3 2 5 8 6 7 4 40 180 60 270 390 220 420 210
(a)
Draw a scatter graph for this data.
(b)
Describe the correlation between the number of tomatoes picked and their total mass.
Question 9
A competition has 3 different games.
(a)
Jeff plays 2 of the games.
Score Game A Game B Game C 62 53
To win, Jeff needs a mean score of 60. How many points does he need to score in Game C?
Mean score ≥ 60 ⇒ total score ≥ 3 × 60 = 180 ⇒ score in Game C ≥ 180 – 62 – 53 = 65;
so he needs to score at least 65 in Game C.
(b)
Imran and Nia play the 3 games. Their scores have the same mean.
The range of Imran's score is twice the range of Nia's scores.
Fill in the missing scores in the following table.
Imran's Scores Nia's Scores 40 35 40 45
The scatter diagrams show the scores of everyone who plays all 3 games.
(c)
Look at the scatter diagrams. Choose a statement which most closely describes the relationship between the games.
Game A and Game B:
Game A and Game C:
(d)
What can you tell about the relationship between the scores on Game B and the scores on Game C? Write down the statement below which most closely describes the relationship.
Game B and Game C:
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# Mathematics for Technology I (Math 1131)
Durham College, Mathematics
Free
• 55 lessons
• 1 quizzes
• 10 week duration
• ##### Numerical Computation
Here you'll be introduced to the bare basics of mathematics. Topics include commonly used words and phrases, symbols, and how to follow the order of operations.
• ##### Measurements
An introduction to numerical computation. Emphasis is placed on scientific and engineering notation, the rule of significant figures, and converting between SI and Imperial units.
• ##### Trigonometry with Right Triangles
Here we focus on right angle triangles within quadrant I of an x-y plane. None of the angles we evaluate here are greater than 90°. A unit on trigonometry with oblique triangles is covered later.
• ##### Trigonometry with Oblique Triangles
This unit is a continuation of trigonometry with right triangles except we'll extend our understanding to deal with angles *greater* than 90°. Resolving and combining vectors will be covered at the end of this unit.
• ##### Geometry
This unit focuses on analyzing and understand the characteristics of various shapes, both 2D and 3D.
## Mathematics for Technology I (Math 1131)
### Identify, measure, and calculate different types of straight lines and angles
Much of this section can be summarized using the diagram:
Notice how lines 1 and 2 (denoted L1 and L2) are parallel (they never intersect), and that’s indicated by red arrows. Line T is called the transversal, it simply cuts through L1 and L2. How each of these angles are related is explained below.
# Opposite Angles
Angles that are across from each other are equal (× pattern).
• Using the diagram above:
• ∠A = ∠D
• ∠C = ∠B
• ∠E = ∠H
• ∠G = ∠F
• Generally:
# Supplementary Angles
Any two angles whose measures sum to 180°. In the original diagram, ∠A + ∠B = 180° (though other examples also exist).
• Generally:
# Complementary Angles
Angle pairs whose measures sum to one right angle (90°). This is not illustrated in the example above.
• Generally:
# Corresponding Angles
When two parallel lines are crossed by the transversal, the angles in matching corners are equal (F pattern). In the original diagram, ∠A are ∠E are equal (though other examples also exist).
• Generally:
# Alternate Interior Angles
Two interior angles which lie on different parallel lines and on opposite sides of a transversal are equal (Z pattern). In the original diagram, ∠E are ∠D are equal (though other examples also exist).
• Generally:
# Co-interior Angles
Angles on the same side of the transversal and inside the parallel lines are supplementary (both angles add up to 180°).
• Using the same diagram we started with, ∠C and ∠E summed up should equal to 180°. Algebraically, this can be represented as ∠C + ∠E = 180°.
A summary on parallel lines is shown below:
Examples where we apply these rules are shown below:
• Part 2, Part 3, and Part 4 can be accessed by clicking each link. Parts 3 and 4 incorporate algebraic expressions into the mix. It’s just another example displaying the versatility of algebra found across many fields of mathematics.
# Corresponding Segments
When a number of parallel lines are cut by two transversals, the portions of the transversals lying between the same parallels are called corresponding segments.
Let’s apply this to a real-life situation:
Question: A portion of a street map is shown. Find the distances PQ and QR.
Solution: The transversal lines are Avenue A and B. Start by comparing these two sides via a fraction:
$\frac{355}{402}\phantom{\rule{0ex}{0ex}}$
This fraction is then made equal to 172 over PQ:
$\frac{355}{402}=\frac{172}{PQ}\phantom{\rule{0ex}{0ex}}$
To solve, you can cross-multiply to get 195 feet. Similarly, to find QR:
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How do you solve u(u - 5) + 8u = u(u + 2) - 4 ?
May 27, 2017
See a solution process below:
Explanation:
First, expand the terms in parenthesis on each side of the equation by multiplying each term within the parenthesis by the term outside the parenthesis:
$\textcolor{red}{u} \left(u - 5\right) + 8 u = \textcolor{b l u e}{u} \left(u + 2\right) - 4$
$\left(\textcolor{red}{u} \times u\right) - \left(\textcolor{red}{u} \times 5\right) + 8 u = \left(\textcolor{b l u e}{u} \times u\right) + \left(\textcolor{b l u e}{u} \times 2\right) - 4$
${u}^{2} - 5 u + 8 u = {u}^{2} + 2 u - 4$
${u}^{2} + \left(- 5 + 8\right) u = {u}^{2} + 2 u - 4$
${u}^{2} + 3 u = {u}^{2} + 2 u - 4$
Next, subtract $\textcolor{red}{{u}^{2}}$ from each side of the equation to eliminate this term while keeping the equation balanced:
$- \textcolor{red}{{u}^{2}} + {u}^{2} + 3 u = - \textcolor{red}{{u}^{2}} + {u}^{2} + 2 u - 4$
$0 + 3 u = 0 + 2 u - 4$
$3 u = 2 u - 4$
Now, subtract $\textcolor{red}{2 u}$ from each side of the equation to solve for $u$ while keeping the equation balanced:
$- \textcolor{red}{2 u} + 3 u = - \textcolor{red}{2 u} + 2 u - 4$
$\left(- \textcolor{red}{2} + 3\right) u = 0 - 4$
$1 u = - 4$
$u = - 4$
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# TS 10th Class Maths Model Paper Set 2 with Solutions
The strategic use of TS 10th Class Maths Model Papers Set 2 can significantly enhance a student’s problem-solving skills.
## TS SSC Maths Model Paper Set 2 with Solutions
Time: 3 Hours
Maximum Marks: 80
General Instructions:
1. Answer all the questions under Part – A on a separate answer book.
2. Write the answers to the questions under Part – B on the question paper itself and attach it to the answer book of Part – A.
Part – A (60 Marks)
Section – I (6 × 2 = 12 Marks)
Note :
1. Answer ALL the following questions.
2. Each question carries 2 marks.
Question 1.
Expand log a3b2c5.
Solution:
log a3b2c5 = log a3 + log b2 + log c5
= 3 log a + 2 log b + 5 log c
Question 2.
If p(x) = x2 + 3x + 4, then find the values of p(0) and p(1).
Solution:
p(x) = x2 + 3x + 4
p(0) = (0)2 + 3(0) + 4
= 0 + 0 + 4 = 4
∴ p(0) = 4
p(1) = (1)2 + 3(1) + 4
= 1 + 3 + 4 = 8
∴ p(1) = 8
Question 3.
Find the 10th term of the arithmetic progression 3, 5, 7, ……………
Solution:
3, 5, 7 ……………. A.P
a = 3, d = 2, n = 10
an = a + (n – 1)d
a10 = 3 + (10 – 1)2
= 3 + (9 × 2)
= 3 + 18 = 21
Question 4.
Express tan θ in terms of sin θ.
Solution:
tan θ = $$\frac{\sin \theta}{\cos \theta}$$
sin2 θ + cos2 θ = 1
cos2 θ = 1 – sin2θ
cos θ = $$\sqrt{1-\sin ^2 \theta}$$
∴ tan θ = $$\frac{\sin \theta}{\sqrt{1-\sin ^2 \theta}}$$.
Question 5.
If a dice is rolled once, then find the probability of getting an odd number.
Solution:
Total outcomes = {1, 2, 3, 4, 5, 6)
Number of Total outcomes = 6
Favourable outcomes = {1, 3, 5}
Number of favourable outcomes = 3
P (an odd number)
= $$\frac{\text { Number of favourable outcomes }}{\text { Number of Total outcomes }}$$
= $$\frac{3}{6}$$ = $$\frac{1}{2}$$
Question 6.
“The top of a tower is observed at an angle of elevation 45° and the foot of ‘ the tower is at a distance of 30 metres from the observer”. Draw a suitable diagram for this data.
Solution:
C – Point of observation
AB – Tower
A – Top of the tower
Section – II (6 × 3 = 18 Marks)
Note :
1. Answer ALL the following questions.
2. Each question carries 3 marks.
Question 7.
Solve : 2x + y = 5 and 5x + 3y = 11.
Solution:
2x + y = 5 ……………. (1)
5x + 3y = 11 …………… (2)
Substituting the value of x in equation (1)
⇒ 2(4) + y = 5
⇒ 8 + y = 5 ⇒ y = 5 – 8 ∴ y = -3
∴ x = 4, y = – 3 are the solution of the equations.
Question 8.
5, 8, 11, 14, ………. is an arithmetic progression. Find the sum of first 20 terms of it.
Solution:
5, 8, 11, 14, ……………. A.P
a = 5, d = 8 – 5 = 3, n = 20
Sn = $$\frac{n}{2}$$[2a + (n – 1) d]
S20 = $$\frac{20}{2}$$[(2 × 5) + (20 – 1)3]
= 10[10 + 57]
= 10 × 67 = 670
Question 9.
Write a Quadratic equation, whose roots are 3 + √5 and 3 – √5 .
Solution:
Roots are 3 + √5 and 3 – √5 .
Sum of the roots = (3 + √5) + (3 – √5) = 6
Product of roots = (3 + √5) (3 – √5)
= 9 – 5 = 4
∴ Required Quadratic equation = x2 – x (sum of the roots) + Product of roots = 0
∴ x2 – 6x + 4 = 0
Question 10.
A box contains 20 cards which are numbered from 1 to 20. If one card is selected at random from the box, find the probability that it bears (i) a prime number, (ii) an even number.
Solution:
Total possibilities
= {1, 2, 3, 4, 5, 6, 7, 8, 9, ……………,20}
Number of Total possibilities = 20
i) Prime numbers
= {2, 3, 5, 7, 11, 13, 17, 19}
Number of favourable outcomes = 8
P(Prime Number)
= $$\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}$$
= $$\frac{8}{20}$$ = $$\frac{2}{5}$$
∴ P(Prime Number) = $$\frac{2}{5}$$
ii) Even numbers
= {2,4, 6, 8, 10,12, 14, 16, 18,20}
Number of favourable outcomes = 10
P (Even Number)
$$\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}$$
= $$\frac{10}{20}$$ = $$\frac{1}{2}$$
∴ P(Prime Number) = $$\frac{1}{2}$$
Question 11.
If two persons standing on either side of a tower of height 100 metres observes the top of it with angles of elevation of 60° and 45° respectively, then find the distance between the two persons.
[Note : Consider the two persons and the tower are on the same line.]
Solution:
Tower height = 100 m
Angles of elevation = 60° and 45°
In ΔABP tan 60° = $$\frac{\mathrm{AB}}{\mathrm{PB}}$$
$$\frac{\sqrt{3}}{1}=\frac{100}{x}$$
√3 = x = 100 ⇒ x = $$\frac{100}{\sqrt{3}}$$ m
In ΔABQ tan 45° = $$\frac{\mathrm{AB}}{\mathrm{BQ}}$$
$$\frac{1}{1}=\frac{100}{y}$$ ⇒ y = 100 m
∴ The distance between Two persons = x + y
$$\frac{100}{\sqrt{3}}$$ + 100 = $$\frac{100(\sqrt{3}+1)}{\sqrt{3}}$$ m
Question 12.
In a trapezium ABCD, AB || DC. If diagonals intersect each other at point ‘O’, then show that $$\frac{\mathrm{AO}}{\mathrm{BO}}$$ = $$\frac{\mathrm{CO}}{\mathrm{DO}}$$
Solution:
Given: In trapezium oABCD, AB // CD.
Diagonals AC, BD intersect at O.
R.T.P : $$\frac{\mathrm{AO}}{\mathrm{BO}}$$ = $$\frac{\mathrm{CO}}{\mathrm{DO}}$$
Construction : Draw a line EF passing through the point ‘O’ and parallel to CD and AB.
Proof: In ΔACD, EO//CD
∴ $$\frac{\mathrm{AO}}{\mathrm{CO}}$$ = $$\frac{\mathrm{AE}}{\mathrm{DE}}$$
[∵ line drawn parallel to one side of a triangle divides other two sides in the same ratio by Basic proportionality theorem]
In ΔABD, EO //AB
Hence, $$\frac{\mathrm{DE}}{\mathrm{AE}}$$ = $$\frac{\mathrm{DO}}{\mathrm{BO}}$$
[∵ Basic proportionality theorem]
$$\frac{\mathrm{BO}}{\mathrm{DO}}$$ = $$\frac{\mathrm{AE}}{\mathrm{ED}}$$ ………… (2)
From (1) and (2)
$$\frac{\mathrm{AO}}{\mathrm{CO}}$$ = $$\frac{\mathrm{BO}}{\mathrm{DO}}$$ [∵ Alternendo]
Section – III (6 × 5 = 30 Marks)
Note :
1. Answer all the following questions.
2. In this section every question has internal choice. Answer any one alternative.
3. Each question carries 5 marks.
Question 13.
A) If sec θ + tan θ = P, then prove that sin θ = $$\frac{\mathbf{P}^2-1}{\mathbf{P}^2+1}$$
Solution:
If sec θ + tan θ = P
Then sec θ – tan θ = $$\frac{1}{\mathrm{P}}$$
[∵ sec2 θ – tan2 θ = 1]
by adding (+) ⇒ 2 sec θ = P + $$\frac{1}{\mathrm{P}}$$
sec θ = $$\frac{\mathrm{P}^2+1}{2 \mathrm{P}}$$ ……………… (1)
(OR)
B) Find the median for the following data.
Solution:
Median = l + ($$\frac{\frac{N}{2}-c f}{f}$$) × h
l = 20, $$\frac{N}{2}$$ = $$\frac{44}{2}$$ = 22, cf = 16
f = 12, h = 10
Median = 20 + ($$\frac{22-16}{12}$$) × 10
20 + ($$\frac{1}{12}$$ × 10)
= 20 + ($$\frac{60}{12}$$) = 20 + 5 = 25
∴ Median =25
Question 14.
A) Prove that √5 + √7 Is an irrational number.
Solution:
Suppose √5 + √7 is not an irrational number.
√5 + √7 = $$\frac{p}{q}$$ ; p, q ∈ Z, q ≠ 0
Squaring on both sides
(√5 + √7)2 = ($$\frac{p}{q}$$)2
⇒ 5 + 7 + 2$$\sqrt{35}$$ = $$\frac{p^2}{q^2}$$
⇒ 2$$\sqrt{35}$$ = $$\frac{p^2}{q^2}$$ – 12
⇒ $$\sqrt{35}$$ = $$\frac{p^2-12 q^2}{2 q^2}$$
[p2 – 12q2, 2q2∈ Z and 2q2 ≠ 0]
∴ $$\frac{p^2-12 q^2}{2 q^2}$$ is a rational number.
But $$\sqrt{35}$$ is an irrational number.
An irrational number never be-comes equal to a rational number.
So our supposition that √5 + √7 is not an irrational number is false.
∴ √5 + √7 is an irrational number.
(OR)
B) The angle of elevation of the top of a hill from the foot of a tower is 60° and the angle of elevation of the top of the tower from the foot of the hill is 30°. If the tower is 50 m high. Find the height of the hill.
Solution:
Given height of the tower = AB = 50 m
Let height of hill be CD = h m
and distance between their feet be AC = x m
∠ACB = 30°, ∠CAD = 60°
From right angled Δ ABC, tan 30° = $$\frac{\mathrm{AB}}{\mathrm{AC}}$$
$$\frac{1}{3}$$ = $$\frac{50}{x}$$ ⇒ x = 50√3 m
From right angled Δ ACD, tan 60° = $$\frac{\mathrm{CD}}{\mathrm{AC}}$$
√3 = $$\frac{h}{x}$$ ⇒ h = x√3 m
h = 50√3 .√3 (∵ x = 50√3m)
h = 50 × 3
⇒ h = 150 m.
Question 15.
A) Show that the distance of the points (5, 12), (7, 24) and (35, 12) from the origin are arranged in ascending order forms an arithmetic progression. Find the common difference of the progression.
Solution:
The distance of the point (x, y) from the origin = $$\sqrt{x^2+y^2}$$
The distance of the point (5, 12) from the origin = $$\sqrt{(5)^2+(12)^2}$$ = $$\sqrt{25+144}$$ = $$\sqrt{169}$$ = 13 units;
The distance of the point (7, 24) from the origin = $$\sqrt{(7)^2+(24)^2}$$ = $$\sqrt{49+576}$$ = $$\sqrt{625}$$ = 25 units.
The distance of the point (35, 12) from the origin = $$\sqrt{(35)^2+(12)^2}$$ = $$\sqrt{1225+144}$$ = $$\sqrt{1369}$$ = 37 units.,
The ascending order of the distances is 13, 25, 37
a1 a2 a3
a2 – a1 = 25 – 13 = 12
a3 – a2 = 37 – 25 = 12
a2 – a1 = a3 – a2
a1, a2, a3 are in A.R
Common difference (d) = 12
(OR)
B)
The sum of the radius of base and height of a solid right circular cylinder is 37 cm. If it’s total surface area is 1628 square centimeters (cm2), then find the volume of the cylinder
(use π = $$\frac{22}{7}$$)
Solution:
In the right circular cylinder
Height = h
r + h = 37 cm
Total surface area = 1628 sq.cm
Volume = ?
Total surface area of the cylinder = 2πr (r + h)
2πr(r + h) = 1628
2 × $$\frac{22}{7}$$ × r × 37 = 1628
r = $$\frac{1628 \times 7}{2 \times 22 \times 37}$$ = 7 cm
7 + h = 37 ⇒ h = 37 – 7 = 30 cm
∴ Volume = πr2h
= $$\frac{22}{7}$$ × 7 × 7 × 30 = 4620 cm3
Question 16.
A) From the given Venn diagram, write the sets A∪B, A∩B, A – B and B – A.
Solution:
A = {1, 2, 3, 4, 5}, B = {2, 4, 6, 8}
A∪B = {1, 2, 3, 4, 5, 6, 8}
A∩B = (2, 4}
A- B = {1, 3, 5}
B – A = {6, 8}
(OR)
B) Sum of the areas of two squares is 850 m2. If the difference of their perimeters is 40 m, find the sides of the two squares.
Solution:
Let the side of first square be ‘a’
the side of second square be ‘b’
Area of first square = ‘a2‘ m2 ,
Area of second square = ‘b2‘ m2
Sum of the area of two squares = 850 m2
a2 + b2 = 850 ………… (1)
Perimeter of first square = ‘4a’m
Perimeter of second square = ‘4b’m
Difference of their perimeters = 40 m
4a – 4b = 40
a – b = 10 …………. (2)
a = b + 10
substitute this in eqn (1)
(b + 10)2 + b2 = 850
b2 + 20b + 100 + b2 = 850
2b2 + 20b – 750 = 0
b2 + 10b – 375 = 0
b2 + 25b – 15b – 375 = 0
b(b + 25) – 15(b + 25) = 0
(b – 15) (b + 25) = 0
b = 15, b = – 25
from (2) a – 15 = 10 ⇒ a = 25
∴ The sides of two squares are 25m, 15m.
Question 17.
A) Draw the graph of the polynomial p(x) x2 + 2x – 3 and find the zeroes of the polynomial from the graph.
Solution:
y = p(x) = x2 + 2x – 3
Zeros of the polynomial are – 3, 1
(OR)
B) Solve the equations graphically 3x + 4y = 10 and 4x – 3y = 5.
Solution:
3x + 4y = 10 ……………… (1)
x 2 -2 6 y 1 4 -2 (x, y) (2, 1) (-2, 4) (6, -2)
4x – 3y = 5. ……………… (2)
x 2 -1 5 y 1 -3 5 (x, y) (2, 1) (-1, -3) (5, 5)
Intersecting point is (2, 1)
∴ x = 2, y = 1
Question 18.
A) Draw a circle of radius 4 cm. from a point 9 cm away from it’s centre, construct a pair of tangents to the circle.
Solution:
Steps of Construction :
1. Draw a circle of radius 4 cm.
2. Locate the point ‘P’ at a distance of 9 cm from the centre of the circle ‘O’ and join O, P.
3. Draw a perpendicular bisector for the line segment OP
4. Name the point of intersection of OP and its peipendicular bisector as M.
5. Consider M as centre and MO as the radius draw a circle.
6. Name the points of intersection of the two circles as A and B.
7. Draw PA and PB.
8. PA and PB are required tangents.
(OR)
B) Draw less than Ogive for the following frequency distribution. Find the median from obtained curve.
Solution:
n = 103 ⇒ $$\frac{n}{2}$$ = $$\frac{103}{2}$$ = 51.5
Form graph, median = 100
Part – B (20 Marks)
Note :
2. Each question carries 1 mark.
3. Answers are to be written in the Question paper only.
4. Marks will not be awarded in any case of over writing, rewriting or erased answers.
Note : Write the capital letters (A, B, C, D) showing the correct answer for the following questions in the brackets provided against them. (Marks; 20 × 1 = 20)
Question 1.
If A⊂B, then A∩B is …………….
A) A
B) B
C) μ
D) Φ
A) A
Question 2.
The coefficient of x3 in the polynomial 2x4 – 5x3 + 6x2 + 5 is ……………..
A) -5
B) 5
C) 6
D) 2
A) -5
Question 3.
If the slope of the line joining the points (2, 5) and (x, 3) is 2, then the value of ‘x’ is ……………
A) 0
B) 1
C) -1
D) 2
B) 1
Question 4.
The product of prime factors of 108 is ……………
A) 23 × 32
B) 22 × 32
C) 22 × 33
D) 23 × 33
C) 22 × 33
Question 5.
The number of solutions of the pair of linear equations
3x + 2y = 6 and 6x + 4y = 18 is ……………..
A) 0
B) 1
C) 2
D) infinite
A) 0
Question 6.
“The total cost of 2 pens and 3 books is Rs. 110”.
Linear equation representing this data is ……………..
A) x + y = 110
B) 5x = 110
C) x2 + y3 = 110
D) 2x + 3y = 110
D) 2x + 3y = 110
Question 7.
If the nth term of an arithmetic progression is 4n – 2, then its 10th term is ……………..
A) 38
B) 28
C) 42
D) 24
A) 38
Question 8.
If one root of the Quadratic equation x2 – kx + 36 = 0 is 4, then the value of ‘k’ is ……………
A) 12
B) 17
C) 18
D) 13
D) 13
Question 9.
The nature of roots of the Quadratic equation x2 + 6x + 9 = 0 is …………………
A) Real and distinct.
B) Real and equal.
C) No real roots.
D) One is positive and the other is negative.
B) Real and equal.
Question 10.
The sum of first 10 natural numbers is …………….
A) 45
B) 65
C) 55
D) 35
C) 55
Question 11.
From the given Ogive curve, the value of the median of the data is ……………..
A) 20
B) 25
C) 15
D) 30
D) 30
Question 12.
In the formula of volume of right circular cylinder V = πr2h, the letter ‘r’ represents ……………..
A) Diameter
B) Height
C) Volume
Question 13.
If E and $$\overline{\mathbf{E}}$$ are complementary events in a random experiment and P($$\overline{\mathbf{E}}$$) = 0.3, the value of P(E) is ……………….
A) 0.3
B) 0.7
C) 1
D) 0
B) 0.7
Question 14.
If one letter is selected randomly from the letters of the word “COVID”, then the probability of getting a vowel is ……………..
A) $$\frac{4}{5}$$
B) $$\frac{3}{5}$$
C) $$\frac{2}{5}$$
D) $$\frac{1}{5}$$
C) $$\frac{2}{5}$$
Question 15.
ΔABC ~ ΔDEF, if ∠A = 45° and ∠E = 75°, then ∠C is ……………….
A) 90°
B) 120°
C) 30°
D) 60°
D) 60°
Question 16.
ΔABC is a right triangle, right angled at B. If AB = 9 cm, BC = 12 cm, then AC is …………….
A) 13 cm
B) 14 cm
C) 15 cm
D) 16 cm
C) 15 cm
Question 17.
In the given figure OA and OB are radii. PA and PB are tangents to the circle at points A and B. If ∠AOB = 130°, then ∠APB = ?
A) 40°
B) 50°
C) 60°
D) 70°
B) 50°
Question 18.
If sin θ = $$\frac{3}{5}$$, then the value of cos θ is (θ is acute angle) ……………..
A) $$\frac{1}{5}$$
B) $$\frac{5}{3}$$
C) $$\frac{4}{5}$$
D) $$\frac{2}{5}$$
C) $$\frac{4}{5}$$
Question 19.
The maximum number of tangents can be drawn from an external point to a circle is ……………..
A) 1
B) 2
C) 3
D) 4
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# 2006 AMC 12A Problems/Problem 6
The following problem is from both the 2006 AMC 12A #6 and 2006 AMC 10A #7, so both problems redirect to this page.
## Problem
The $8\times18$ rectangle $ABCD$ is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square. What is $y$?
$[asy] unitsize(3mm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=4; draw((0,4)--(18,4)--(18,-4)--(0,-4)--cycle); draw((6,4)--(6,0)--(12,0)--(12,-4)); label("A",(0,4),NW); label("B",(18,4),NE); label("C",(18,-4),SE); label("D",(0,-4),SW); label("y",(3,4),S); label("y",(15,-4),N); label("18",(9,4),N); label("18",(9,-4),S); label("8",(0,0),W); label("8",(18,0),E); dot((0,4)); dot((18,4)); dot((18,-4)); dot((0,-4));[/asy]$
$\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 7\qquad\mathrm{(C)}\ 8\qquad\mathrm{(D)}\ 9\qquad\mathrm{(E)}\ 10$
## Solution 1
Since the two hexagons are going to be repositioned to form a square without overlap, the area will remain the same. The rectangle's area is $18\cdot8=144$. This means the square will have four sides of length 12. The only way to do this is shown below.
$[asy] size(175); pair A,B,C,D,E,F,G,H; A=(0,8); B=(12,12); C=(12,4); D=(0,0); E=(0,12); F=(12,0); G=(6,4); H=(6,8); draw(A--E--B--C--G--H--A--D--F--C); label("A",A,W); label("B",B,NE); label("C",(12.6,4)); label("D",D,SW); label("12",E--B,N); label("12",D--F,S); label("4",E--A,W); label("4",(12.4,-1.75),E); label("8",A--D,W); label("8",(12.4,4),E); label("y",A--H,S); label("y",G--C,N); [/asy]$
As you can see from the diagram, the line segment denoted as $y$ is half the length of the side of the square, which leads to $y = \frac{12}{2} = \boxed{\textbf{(A) }6}$.
### Solution 1 (Shortcut)
As solution 1 says, the two hexagons are going to be repositioned to form a square without overlap. Thus we create this square out of the original rectangle.
$[asy] size(175); pair A,B,C,D,E,F,G,H; A=(0,8); B=(12,12); C=(12,4); D=(0,0); E=(0,12); F=(12,0); G=(6,4); H=(6,8); draw(A--E--B--C--G--H--A--D--F--C); label("y",A--H,S); label("y",G--C,N); [/asy]$
As you can see from the diagram, the length $y$ fits into the previously blank side, so we know that it is equal to $y$.
$[asy] unitsize(3mm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=4; draw((0,4)--(18,4)--(18,-4)--(0,-4)--cycle); draw((6,4)--(6,0)--(12,0)--(12,-4)); label("y",(9,-2),NW); label("A",(0,4),NW); label("B",(18,4),NE); label("C",(18,-4),SE); label("D",(0,-4),SW); label("y",(3,4),S); label("y",(15,-4),N); label("18",(9,4),N); label("18",(9,-4),S); label("8",(0,0),W); label("8",(18,0),E); dot((0,4)); dot((18,4)); dot((18,-4)); dot((0,-4));[/asy]$
From there we can say $3y = 18$ so $y = \frac{18}{3} = \boxed{\textbf{(A) }6}$.
~Ezraft
## Solution 2 (Cheap)
Because the two hexagons are congruent, we know that the perpendicular line to $A$ is half of $BC$, or $4$. Next, we plug the answer choices in to see which one works. Trying $A$, we get the area of one hexagon is $72$ , as desired, so the answer is $\boxed{\textbf{(A) }6}$ .
~coolmath2017
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# An Interesting Right Triangle Property You’ll Need to Know for the GMAT
In a, we discussed medians, altitudes and angle bisectors of isosceles and equilateral triangles. Today, we will discuss an interesting property of perpendicular bisectors and circumcenter of right triangles.
Property: The circumcenter of a right triangle is the mid point of the hypotenuse.
Let’s prove this first and then we will see its application.
Say, we have a right triangle ABC right angled at B. Let’s draw the perpendicular bisector of AB which intersects AB at its mid point M. Say this line intersects the hypotenuse AC at N. We need to prove that AN = CN. Note that triangle AMN and triangle ABC are similar triangles using the AA property (angle AMN = angle ABC = 90 degrees and angle A is common to both triangles). So the ratio of the sides of the two triangle is the same. Since MN is the perpendicular bisector of line AB, AM = MB which means that AM is half of AB.
So AM/AB = 1/2 = AN/AC
Hence AN = NC
So N is the mid point of AC.
Using the exact same logic for side BC, we will see that its perpendicular bisector also bisects the hypotenuse. So N would be the circumcenter of triangle ABC and the mid point of AC.
Using an official question, let’s see how this property can be useful to us:
In the rectangular coordinate system shown above, points O, P, and Q represent the sites of three proposed housing developments. If a fire station can be built at any point in the coordinate system, at which point would it be equidistant from all three developments?
(A) (3,1)
(B) (1,3)
(C) (3,2)
(D) (2,2)
(E) (2,3)
First, let’s see how we will solve this question without knowing this property and using co-ordinate geometry instead.
Method 1:
Points O and Q lie on the X axis and are 4 units apart. We need a point equidistant from both O and Q. All such points will lie on the line lying in the middle of O and Q and perpendicular to the X axis. The equation of such a line will be x = 2. The fire station should be somewhere on this line.
Points O and P lie on the Y axis and are 6 units apart. We need a point equidistant from both O and P. All such points will lie on the line lying in the middle of O and P and perpendicular to the Y axis. The equation of such a line will be y = 3. The fire station should be somewhere on this line too.
Any two lines on the XY plane intersect at most at one point (if they are not overlapping). Since the fire station must lie on both these lines, it must be on their intersection i.e. at (2, 3).
This point (2,3) will be equidistant from O, Q and P. Therefore, the answer is E.
Method 2:
Think of the question in terms of the perpendicular bisectors of triangle OPQ. Their point of intersection will be equidistant from all three vertices.
We know that the circumcenter lies on the mid point of the hypotenuse. The end points of the hypotenuse are (4, 0) and (0, 6). The mid point will be
x = (4 + 0)/2 = 2
y = (0 + 6)/2 = 3
As in Method 1, the point (2, 3) will be equidistant from all three points, O, P and Q. Again, the answer is E.
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## What is an example of adding decimals?
This addition of decimals can be understood with the help of a simple example: 12.5 + 4.9. We can separate the decimals from the whole numbers. 12.5 + 4.9 = (12 + 0.5) + (4 + 0.9) = (12 + 4) + (0.5 + 0.9) = 16 + 1.4 = 16 + (1 + 0.4) = 17.4.
How do you add decimals math is fun?
Adding and Subtracting Decimals To add decimals, follow these steps: Write down the numbers, one under the other, with the decimal points lined up. Put in zeros so the numbers have the same length (see below for why that is OK) Then add, using column addition, remembering to put the decimal point in the answer.
### What are the steps in adding or subtracting decimals?
TO ADD OR SUBTRACT DECIMALS: 1) Line up the decimal points vertically. Fill in any 0’s where necessary. 2) Add or subtract the numbers as if they were whole numbers. 3) Place the decimal point in the sum or difference so that it lines up vertically with the numbers being added or subtracted.
When adding decimals by hand you must?
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# What is the derivative of x=y^2?
Dec 3, 2014
We can solve this problem in a few steps using Implicit Differentiation.
Step 1) Take the derivative of both sides with respect to x.
• $\frac{\Delta}{\Delta x} \left({y}^{2}\right) = \frac{\Delta}{\Delta x} \left(x\right)$
Step 2) To find $\frac{\Delta}{\Delta x} \left({y}^{2}\right)$ we have to use the chain rule because the variables are different.
• Chain rule: $\frac{\Delta}{\Delta x} \left({u}^{n}\right) = \left(n \cdot {u}^{n - 1}\right) \cdot \left(u '\right)$
• Plugging in our problem: $\frac{\Delta}{\Delta x} \left({y}^{2}\right) = \left(2 \cdot y\right) \cdot \frac{\Delta y}{\Delta x}$
Step 3) Find $\frac{\Delta}{\Delta x} \left(x\right)$ with the simple power rule since the variables are the same.
• Power rule: $\frac{\Delta}{\Delta x} \left({x}^{n}\right) = \left(n \cdot {x}^{n - 1}\right)$
• Plugging in our problem: $\frac{\Delta}{\Delta x} \left(x\right) = 1$
Step 4) Plugging in the values found in steps 2 and 3 back into the original equation ( $\frac{\Delta}{\Delta x} \left({y}^{2}\right) = \frac{\Delta}{\Delta x} \left(x\right)$ ) we can finally solve for $\frac{\Delta y}{\Delta x}$.
• $\left(2 \cdot y\right) \cdot \frac{\Delta y}{\Delta x} = 1$
Divide both sides by $2 y$ to get $\frac{\Delta y}{\Delta x}$ by itself
• $\frac{\Delta y}{\Delta x} = \frac{1}{2 \cdot y}$
This is the solution
Notice: the chain rule and power rule are very similar, the only differences are:
-chain rule: $u \ne x$ "variables are different" and
-power rule: $x = x$ "variables are the same"
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# NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 24 Homework 24
Name Date
1. Solve using mental math. If you cannot solve mentally, use your place value chart
and place value disks.
## b. 130 20 = ______ 130 30 = ______ 130 40 = ______
2. Solve using your place value chart and place value disks. Unbundle the hundred or
ten when necessary. Circle what you did to model each problem.
a. b.
115 50 = ______ 125 57 = ______
## I unbundled the hundred. Yes No I unbundled the hundred. Yes No
I unbundled a ten. Yes No I unbundled a ten. Yes No
c. d.
88 39 = ______ 186 39 = ______
## I unbundled the hundred. Yes No I unbundled the hundred. Yes No
I unbundled a ten. Yes No I unbundled a ten. Yes No
e. f.
162 85 = ______ 172 76 = ______
## I unbundled the hundred. Yes No I unbundled the hundred. Yes No
I unbundled a ten. Yes No I unbundled a ten. Yes No
## Lesson 24: Use manipulatives to represent subtraction with decompositions of
1 hundred as 10 tens and 1 ten as 10 ones. 310
NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 24 Homework 24
g. h.
121 89 = ______ 131 98 = ______
## I unbundled the hundred. Yes No I unbundled the hundred. Yes No
I unbundled a ten. Yes No I unbundled a ten. Yes No
i. j.
140 65 = ______ 150 56 = ______
## I unbundled the hundred. Yes No I unbundled the hundred. Yes No
I unbundled a ten. Yes No I unbundled a ten. Yes No
k. l.
163 78 = ______ 136 87 = ______
## I unbundled the hundred. Yes No I unbundled the hundred. Yes No
I unbundled a ten. Yes No I unbundled a ten. Yes No
3. 96 crayons in the basket are broken. The basket has 182 crayons. How many
crayons are not broken?
## Lesson 24: Use manipulatives to represent subtraction with decompositions of
1 hundred as 10 tens and 1 ten as 10 ones. 311
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# Division Calculator
## Calculate the result of dividing two integers
### How does this work?
Using this calculator is simple! Enter two integers, the numerator (X) and the denominator (Y), to calculate the result of dividing X by Y. After entering the values, click on "Calculate" to see the result.
### Understanding Division
Division is one of the four basic arithmetic operations, along with addition, subtraction, and multiplication. It can be thought of as the inverse of multiplication, in the sense that it undoes the operation of multiplication. When we divide a number (called the dividend) by another number (called the divisor), we are essentially finding out how many times the divisor can fit into the dividend.
In mathematical terms, we express division as:
Dividend ÷ Divisor = Quotient
Division helps us solve problems related to sharing or distributing objects, resources, or values equally among a certain number of groups. For example, if you have 20 apples and want to divide them equally among 4 friends, you would perform the operation 20 ÷ 4, which gives you the quotient 5. This means that each friend would receive 5 apples.
### Division in Everyday Life
Division plays a significant role in various aspects of our daily lives. Here are a few examples where division is useful:
1. Budgeting: When managing your finances, you often need to divide your monthly income by the number of weeks or days in a month to determine how much you can spend per week or per day.
2. Cooking: If you're preparing a meal for a specific number of people, you might need to divide the ingredients' quantities to adjust the recipe to serve the desired number of portions.
3. Sharing expenses: When splitting the cost of a group activity, like a dinner bill or the rent for a shared apartment, division helps you determine how much each person owes.
4. Time management: You may need to divide the total time available for a project by the number of tasks to allocate time effectively and ensure that everything gets completed on schedule.
5. Traveling: When planning a trip, you can use division to estimate the number of days required for the journey by dividing the total distance to be covered by the average distance you expect to travel per day.
As you can see, division is a crucial mathematical concept that helps us solve various everyday problems and make informed decisions.
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If you have a line segment with endpoints A and B, and point C is between points A and B, then AC + CB = AB. The Angle Addition Postulate: This postulates states that if you divide one angle into two smaller angles, then the sum of those two angles must be equal to the measure of the original angle.
Keeping this in consideration, What is perpendicular line?
In elementary geometry, the property of being perpendicular (perpendicularity) is the relationship between two lines which meet at a right angle (90 degrees). A line is said to be perpendicular to another line if the two lines intersect at a right angle.
Also know, How is addition related to angle measurement? Students should be able to recognize when an angle is decomposed into non-overlapping parts, the angle measure of the whole is the sum of the angle measures of the parts. Students should be able to find the angle measure in degrees using addition or subtraction.
Contents
## How do you bisect an angle?
Construction: bisect ∠ABC.
1. STEPS:
2. Place compass point on the vertex of the angle (point B).
3. Stretch the compass to any length that will stay ON the angle.
4. Swing an arc so the pencil crosses both sides (rays) of the given angle.
5. Place the compass point on one of these new intersection points on the sides of the angle.
Two angles are Adjacent when they have a common side and a common vertex (corner point) and don’t overlap.
## How do you solve postulates?
If you have a line segment with endpoints A and B, and point C is between points A and B, then AC + CB = AB. The Angle Addition Postulate: This postulates states that if you divide one angle into two smaller angles, then the sum of those two angles must be equal to the measure of the original angle.
## What is the segment addition postulate formula?
The segment addition postulate states that if we are given two points on a line segment, A and C, a third point B lies on the line segment AC if and only if the distances between the points meet the requirements of the equation AB + BC = AC.
## What does it mean to be congruent?
The adjective congruent fits when two shapes are the same in shape and size. If you lay two congruent triangles on each other, they would match up exactly. Congruent comes from the Latin verb congruere “to come together, correspond with.” Figuratively, the word describes something that is similar in character or type.
## What is perpendicular line?
In elementary geometry, the property of being perpendicular (perpendicularity) is the relationship between two lines which meet at a right angle (90 degrees). A line is said to be perpendicular to another line if the two lines intersect at a right angle.
## Are vertical angles congruent?
When two lines intersect to make an X, angles on opposite sides of the X are called vertical angles. These angles are equal, and here’s the official theorem that tells you so. Vertical angles are congruent: If two angles are vertical angles, then they’re congruent (see the above figure).
## How is addition related to angle measurement?
Recognize angle measure as additive. When an angle is decomposed into non-overlapping parts, the angle measure of the whole is the sum of the angle measures of the parts.
## How do you define angles?
In plane geometry, an angle is the figure formed by two rays, called the sides of the angle, sharing a common endpoint, called the vertex of the angle. Angles formed by two rays lie in a plane, but this plane does not have to be a Euclidean plane.
## Is a segment a defined term?
Definition of segment. (Entry 1 of 2) 1 : a portion cut off from a geometric figure by one or more points, lines, or planes: such as. a : the area of a circle bounded by a chord and an arc of that circle. b : the part of a sphere cut off by a plane or included between two parallel planes.
## What are the angle properties?
Each angle is opposite to another and form a pair of what are called opposite angles. Angles a and c are opposite angles. Opposite angles are equal. Opposite angles are sometimes called vertical angles or vertically opposite angles.
## How do you solve congruent segments?
Indicating Congruent Segments in Writing
We indicate a line segment by drawing a line over its two end points. We indicate that two line segments are congruent by using the congruence symbol. Use of the congruence symbol indicates that line segment AB is equal in length to line segment CD.
## What are the 5 postulates in geometry?
Geometry/Five Postulates of Euclidean Geometry
• A straight line segment may be drawn from any given point to any other.
• A straight line may be extended to any finite length.
• A circle may be described with any given point as its center and any distance as its radius.
• All right angles are congruent.
## What is the highest number of degrees that an angle can have?
Congruent. Angles are congruent when they are the same size (in degrees or radians). Sides are congruent when they are the same length.
## What is the highest number of degrees that an angle can have?
When two lines intersect to make an X, angles on opposite sides of the X are called vertical angles. These angles are equal, and here’s the official theorem that tells you so. Vertical angles are congruent: If two angles are vertical angles, then they’re congruent (see the above figure).
## What is the ruler postulate?
Ruler Postulate. Ruler Postulate: The points on a line can be put into a one-to-one correspondence (paired) with the real numbers. The distance between any two points is represented by the absolute value of the difference between the numbers.
## Are parallel lines congruent?
If two parallel lines are cut by a transversal, the corresponding angles are congruent. If two lines are cut by a transversal and the corresponding angles are congruent, the lines are parallel. Interior Angles on the Same Side of the Transversal: The name is a description of the “location” of the these angles.
## What are the different types of theorems?
A
• AF+BG theorem (algebraic geometry)
• ATS theorem (number theory)
• Abel’s binomial theorem (combinatorics)
• Abel’s curve theorem (mathematical analysis)
• Abel’s theorem (mathematical analysis)
• Abelian and tauberian theorems (mathematical analysis)
• Abel–Jacobi theorem (algebraic geometry)
## What is ARC addition postulate?
semicircleAn arc that measures . Arc Addition PostulateArc addition postulate states that the measure of the arc formed by two adjacent arcs is the sum of the measures of the two arcs.
## Which angles are congruent?
The vertical angles theorem is about angles that are opposite each other. These vertical angles are formed when two lines cross each other as you can see in the following drawing. Theorem: Vertical angles are congruent. Congruent is quite a fancy word. Put simply, it means that vertical angles are equal.
## What is an angle postulate?
The Corresponding Angles Postulate states that, when two parallel lines are cut by a transversal , the resulting corresponding angles are congruent . The converse is also true; that is, if two lines l and m are cut by a transversal in such a way that the corresponding angles formed are congruent , then l∥m .
## How do you read a protractor?
How to measure an angle with a protractor:
1. Place the midpoint of the protractor on the VERTEX of the angle.
2. Line up one side of the angle with the zero line of the protractor (where you see the number 0).
3. Read the degrees where the other side crosses the number scale.
180
## Does angle bisector bisect opposite side?
The “Angle Bisector” Theorem says that an angle bisector of a triangle will divide the opposite side into two segments that are proportional to the other two sides of the triangle. An angle bisector is a ray in the interior of an angle forming two congruent angles.
## Which angles are congruent?
Congruent angles are two or more angles that have the same measure. In simple words, they have the same number of degrees. It’s important to note that the length of the angles‘ edges or the direction of the angles has no effect on their congruency. As long as their measure is equal, the angles are considered congruent.
## How many postulates are in geometry?
A postulate is a statement that is assumed true without proof. A theorem is a true statement that can be proven. Listed below are six postulates and the theorems that can be proven from these postulates.
## Is there an angle subtraction postulate?
There are four subtraction theorems you can use in geometry proofs: two are for segments and two are for angles. Each of these corresponds to one of the addition theorems. Here are the subtraction theorems for three segments and three angles (abbreviated as segment subtraction, angle subtraction, or just subtraction):
## What is vertical angle theorem?
Congruent angles are two or more angles that have the same measure. In simple words, they have the same number of degrees. It’s important to note that the length of the angles‘ edges or the direction of the angles has no effect on their congruency. As long as their measure is equal, the angles are considered congruent.
## How do you read a protractor?
How to measure an angle with a protractor:
1. Place the midpoint of the protractor on the VERTEX of the angle.
2. Line up one side of the angle with the zero line of the protractor (where you see the number 0).
3. Read the degrees where the other side crosses the number scale.
## What is ARC addition postulate?
The vertical angles theorem is about angles that are opposite each other. These vertical angles are formed when two lines cross each other as you can see in the following drawing. Theorem: Vertical angles are congruent. Congruent is quite a fancy word. Put simply, it means that vertical angles are equal.
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# Differentiation [ (7x^3+4)^(1/x) ] / Integration [ x^x(1+ln(x)) ]
1. Apr 20, 2007
1. The problem statement, all variables and given/known data
Differentiate the following expression leaving in the simplest form:
$$\left( {7x^3 + 4} \right)^{\frac{1}{x}}$$
Integrate the following leaving in the simplest form:
$$x^x \left( {1 + \ln x} \right)$$
2. The attempt at a solution
Here is my worked solution for the differentiation problem:
$$$\begin{array}{l} \frac{d}{{dx}}\left[ {\left( {7x^3 + 4} \right)^{\frac{1}{x}} } \right] \\ y = \left( {7x^3 + 4} \right)^{\frac{1}{x}} \\ \ln y = \left( {\frac{1}{x}} \right)\ln \left( {7x^3 + 4} \right) \\ = \frac{{\ln \left( {7x^3 + 4} \right)}}{x} \\ \frac{d}{{dx}}\left[ {\ln \left( {7x^3 + 4} \right)} \right] = \frac{1}{{7x^3 + 4}}.\frac{d}{{dx}}\left[ {7x^3 + 4} \right] \\ = \frac{{21x^2 }}{{7x^3 + 4}} \\ \frac{d}{{dx}}\left[ {\frac{{\ln \left( {7x^3 + 4} \right)}}{x}} \right] = \frac{{x\left( {\frac{{21x^2 }}{{7x^3 + 4}}} \right) + 1\left( {\ln \left( {7x^3 + 4} \right)} \right)}}{{x^2 }} \\ = \frac{{21x^3 + \left( {7x^3 + 4} \right)\ln \left( {7x^3 + 4} \right)}}{{x^2 }} \\ \\ \frac{1}{y} = \\ y' = y.\frac{{21x^3 + \left( {7x^3 + 4} \right)\ln \left( {7x^3 + 4} \right)}}{{x^2 \left( {7x^3 + 4} \right)}} \\ = \frac{{\left( {7x^3 + 4} \right)^{\frac{1}{x}} \left( {21x^3 + \left( {7x^3 + 4} \right)\ln \left( {7x^3 + 4} \right)} \right)}}{{x^2 \left( {7x^3 + 4} \right)}} \\ = \frac{{\left( {7x^3 + 4} \right)^{\frac{1}{x} - 1} \left( {21x^3 + \left( {7x^3 + 4} \right)\ln \left( {7x^3 + 4} \right)} \right)}}{{x^2 }} \\ \end{array}$$$
______________________________________
For the integration problem I am not quite certain on how to integrate the expression given. I know from previous experience that the expression x^x when differentiated will give x^x(1+ln(x)).
______________________________________
many thanks for all suggestions and help
2. Apr 20, 2007
### HallsofIvy
Staff Emeritus
For the integration, my first thought was to make a substitution like u= xx. Of course, to do that I would need a derivative for xx. Find the derivative of xx and think you will realize that the integral is surprizingly easy.
3. Apr 20, 2007
### lalbatros
I got this result, with a small difference:
(4 + 7*x^3)**(-1 + 1/x)*(21*x^3 - (4+7*x^3)*Log(4 + 7*x^3)))/x^2
The mistake appeared probably in the 7th line of your calculation.
You probably forgot the minus sign from the rule:
(a/b)' = (a'b-ab')/b²
Last edited: Apr 20, 2007
4. Apr 20, 2007
### Gib Z
Since from previous experience you know the integral is x^x, and x^x happens to appear in that integrand, the easiest way out will always be the substitution which gives the whole integrand! Same goes for any integral like that.
5. Apr 20, 2007
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# A question using arithmetic progression: How long does it take to fill a rectangular tank?
Water fills a tank at a rate of $$150$$ litres during the $$1$$st hour, $$350$$ litres during the $$2$$nd hour, $$550$$ litres during the third hour and so on.
Find the number of hours necessary to fill a rectangular tank $$16m \times 9m \times 9m$$.
So far I have tried this... $$16 \times 9\times 9= 1296 m^3$$ .
Converted is $$1296000$$ litres.
The common difference is $$200$$ litres an hour as it show arithmetic progression $$(a_1=150, a_2=350,a_3=550) a_3-a_2=200, a_2-a_1=200$$.
Using the formula $$S_n=\frac{n}2(2a+(n-1)d)$$
I get $$1296000=\frac{n}2(150 \times 150(n-1)200)$$ which is then $$2,592,000= n(22,500+200n-200)$$
then $$2,592,000=n(200n-22,500) = 2,592,000=200n^2-22,500n-2,592,000=0$$.
I feel I am going wrong some where though. I'm unsure of the next step.
$a=150, d=200, S_n=1,296,000, n=?$.
Now, $S_n=\frac{n}{2}[2a+(n-1)d] \iff 1,296,000=\frac{n}{2}[2(150)+(n-1)200].$
Expanding and simplifying, we get: $$100n^2+50n-1,296,000=0.$$ You now have a quadratic in $n$.
Use the quadratic formula (Try it as an exercise!) to give: $$\underbrace{n\approx-114.09 \ \text{hours}}_{\text{silly}}, \underbrace{\boxed{n\approx113.59 \ \ \text{hours}}}_{\text{sensible}}.$$
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# Lesson: Fractions: Equivalent
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### Lesson Objective
SWBAT identify equivalent fractions
### Lesson Plan
Materials Needed: scrap paper for DN, white board, dry erase markers, pencils, GP/IND Worksheet.
Vocabulary: whole (or ONE or unit), denominator, numerator, equivalent
……….
Do Now (2 -3 min): One the board the teacher writes the following: 1. Are ½ and 2/4 the same value? 2. Are 2/4 and ¾ the same value? 3. Are 4/5 and 8/10 the same value? 4. Are ¾ and 2/3 the same value? Students should write answer to DN on scrap paper.
Opening (2 -3 min): The teacher should state the objective, “Yesterday, we modeled model halves, thirds, fourths, sixths and twelfths on a clock face. Today, we are going to focus equivalent fractions. By the end of this lesson, you will be able to identify equivalent fractions.”
Direct Instruction (10 min): The teacher explains, “We are going to learn about equivalent fractions today. Lets start by making sure we understand what equivalent means? Does anything sound familiar about the word equivalent? [equal]. Yes, that is right, I sort of hear the word equal in equivalent. So knowing that we sort of hear the word equal in equivalent, who wants to take a guess at what equivalent means? [ It means when two different fractions, with different numerators and denominators, equal the same amount]. Oh my gosh! That is a perfect answer. I loved how you used your vocabulary words. Yes, equivalent fractions are fractions that have the same value. Let’s look at our DN answers now.” The teacher reviews the answer to the DN.
1. Yes, they are equivalent
2. No, they have the same denominator
3. Yes, they are equivalent
4. No, different denominators and different numerators, but something tells me they aren’t equivalent.
Then the teacher continues, “Ok, so I gave you the answers, now let me explain why # 1 and # 3 are equivalent. The first thing we have to do is set them up across from each other, like this. Then I look at the relationship between the two fractions, and ask myself is there a mathematical relationship between these two fractions. Well, YES there is, I need to look at the numerators and the denominators both separately and together. If I can multiply the numerator and get the new numerator, I am half way there. I have to be able to multiply the denominator BY THE SAME NUMBER, say it with me “Whatever I do to the bottom, I do to the top” and get the new denominator. Is that the case with these two fractions? YES, I multiply the numerator by 2 and get 2, I multiply the denominator by 2 and get 4, so what does that mean?!?! IT MEANS THESE ARE Equivalent fractions. If I gave [INSERT STUDENT NAME] a circle divided in 2 with one part shaded, and I gave [INSERT STUDENT NAME] a circle divided in 4 parts with 2 parts shaded, they would have the same amounts shaded!! This is the Equivalent Fractions Rule: if both the numerator and denominator of a fraction are multiplied by the same non-zero number, the result is a fraction that is equivalent to the original fraction.”
½ 2/4
The teacher then demonstrates another equivalent pair of fractions and one pair that is not equivalent.
Guided Practice (15 min): The teacher passes out the GP Worksheet. The students will be working on identifying/listing all the equivalent fractions for 13 fractions. The teacher will be working with the students on items 1-8. The rest will be completed for IND Practice.
The teacher should give the student’s scrap paper to work on making equivalent fractions. For ½ the teacher should write options on the board and the students determine if it is equivalent or not. For 2/2 the students should multiple the numerator and the denominator by 2, 3, 4, and 5. This should continue for items 3-8.
Independent (10 min): The teacher then tells the students that they are to work on the remainder of the Worksheet by independently, using the skills that they were just taught. They will be required to hand in the worksheet at the end of the lesson. The teacher should circulate to assist students that may need it.
Closing (2-3 min): Teacher calls the attention of the students back toward the front of the class to quickly review the answers to the Independent Practice worksheet/ ask what we learned about.
### Lesson Resources
IND Practice Lesson 8 Classwork
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Decimal places
Here you will learn about decimal places, including how to round numbers to the tenths place, hundredths place and thousandths place.
Students will first learn about decimal places as part of number and operations in base ten, in elementary school.
What are decimal places?
Decimal places refer to the position of the digits that fall to the right of a decimal point, or period.
The first decimal place to the right of the decimal point is the tenths place.
The second decimal place is the hundredths place.
The third decimal place is the thousandths place.
The decimal system continues up to the ten-thousandths place, the hundred-thousandths place, the millionths place and beyond.
• Using a place value chart will help you to identify the value of a specific digit.
For example,
The decimal 0.2 is represented on the place value chart.
Notice that there is a 0 in the ones place and the 2 is in the tenths place.
So the value of the digit 2 is two-tenths.
Using additional zeros in the place value chart won’t change the value of the decimal, but would change the decimal place and how you read the decimal.
0.20 can be read as twenty hundredths.
0.200 can be read as two-hundred thousandths
These values are all equivalent.
• You can use place value to round a number to given decimal places such as tenths, hundredths, and thousandths.
To round a decimal, you find the decimal place you wish to round to and look at the value after that number. If the value after the chosen place value is a 5 or above you round up, if it is less than 5, you round down.
It is important to give the number of digits required after the decimal point, even if the last digit would be a zero.
For example,
Round 19.237 to the nearest tenth.
19.{\color{orange}2¦}{\color{blue}3}7
3 is less than 5, so we’d leave the 2 as it is.
19.237 rounded to the nearest tenth would be 19.20.
Common Core State Standards
How does this relate to 4th grade math and 5th grade math?
• Grade 5: Number and Operations in Base Ten (5.NBT.A3)
Read, write, and compare decimals to the thousandths
• Grade 5: Number and Operations in Base Ten (5.NBT.A4)
Use place value understanding to round decimals to any place
How to round to decimal places
In order to round to a decimal place with the rule:
1. Find the rounding place and look at the digit immediately to the right of the rounding place.
2. When that digit is \bf{5} or greater, add \bf{1} to the rounding digit. When it is less than \bf{5} , leave the rounding digit alone.
3. Drop all the digits to the right of the rounded digit.
4. Write the rounded number.
Decimal place examples
Example 1: rounding to the tenths place
Round 36.248 to the nearest tenth.
1. Find the rounding place and look at the digit immediately to the right of the rounding place.
You need to locate the tenths place, the first digit to the right of the decimal point.
2When that digit is \bf{5} or greater, add \bf{1} to the rounding digit. When it is less than \bf{5} , leave the rounding digit alone.
4 is less than 5.
As 4 is less than 5, you round down.
3Drop all the digits to the right of the rounded digit.
4Write the rounded number.
36.248 is \bf{36.2} to the nearest tenth.
Example 2: rounding to the tenths place
Round 7.97 to the nearest tenth.
Find the rounding place and look at the digit immediately to the right of the rounding place.
When that digit is \bf{5} or greater, add \bf{1} to the rounding digit. When it is less than \bf{5} , leave the rounding digit alone.
Drop all the digits to the right of the rounded digit.
Write the rounded number.
Example 3: rounding to the hundredths place
Round 97.071 to the nearest hundredth.
Find the rounding place and look at the digit immediately to the right of the rounding place.
When that digit is \bf{5} or greater, add \bf{1} to the rounding digit. When it is less than \bf{5} , leave the rounding digit alone.
Drop all the digits to the right of the rounded digit.
Write the rounded number.
Example 4: rounding to the hundredths place
A calculator displayed an answer to a calculation as 53.298. What is this value rounded to the nearest hundredths place?
Find the rounding place and look at the digit immediately to the right of the rounding place.
When that digit is \bf{5} or greater, add \bf{1} to the rounding digit. When it is less than \bf{5} , leave the rounding digit alone.
Drop all the digits to the right of the rounded digit.
Write the rounded number.
Example 5: rounding to the thousandths place
Round 14.9792 to the nearest thousandth.
Find the rounding place and look at the digit immediately to the right of the rounding place.
When that digit is \bf{5} or greater, add \bf{1} to the rounding digit. When it is less than \bf{5} , leave the rounding digit alone.
Drop all the digits to the right of the rounded digit.
Write the rounded number.
Example 6: rounding to the thousandths place
Round 0.0715 to the nearest thousandth.
Find the rounding place and look at the digit immediately to the right of the rounding place.
When that digit is \bf{5} or greater, add \bf{1} to the rounding digit. When it is less than \bf{5} , leave the rounding digit alone.
Drop all the digits to the right of the rounded digit.
Write the rounded number.
Teaching tips for decimal places
• Allowing students to use a place value chart when first studying decimal places, allows them to visualize how each number relates with the place value names and positions.
• Hands-on decimal activities, including using manipulatives, allows students to explore the decimal system in an impactful way.
• Number lines are a great place to start students when first learning how to round. The rule makes more sense once they’ve had the opportunity to see how rounding happens. Not giving students the opportunity to use the number line can lead to misconceptions and mistakes later down the line.
Easy mistakes to make
• Adding \bf{1} to all of the previous digits when rounding up
A common error when rounding up is to increase all of the digits to the left by 1.
For example, if 53.683 was rounded to the tenths place, a student may write 64.7 instead of the correct answer 53.7.
• Decreasing the value of the rounding digit when rounding down
When rounding down, the value of the rounding digit remains the same, it does not decrease. For example, round 5.24 to the nearest tenth. The 4 tells us to round down, so the rounding number would stay the same, 5.2, not decrease to 5.1.
• Leaving out zeros when rounding a \bf{9} up
It is important to remember to write the number of decimals to the place value given. For example, if 23.997 is rounded to the hundredths place, it would be written as 24.00.
• Mixing up whole numbers and decimals
It is confusing to some when trying to determine the difference between decimal place value: thousandths, hundredths and tenths with whole numbers place value: thousands, hundreds and tens. Consistent practice with these terms will help with familiarity.
Decimal places practice questions
1. Round 5.162 to the nearest tenth.
5.1
5.2
6.2
5.0
The digit in the tenths place is 1. The value to the right of this is 6, which is greater than 5, so round up.
The answer is 5.2.
2. Round 36.715 to the nearest tenth.
40
36.8
36.7
37.7
The digit in the tenths place is 7. The value to the right of this is 1, which is less than 5, so round down.
The answer is 36.7.
3. Round 21.734 to the nearest hundredth.
21.74
21.73
21.83
21.80
The digit in the hundredths place is 3. The value to the right of this is 4, which is less than 5, so round down.
The answer is 21.73.
4. Round 1.7368 to the nearest hundredth.
1.70
0.74
1.74
1.84
The digit in the hundredths place is 3. The value to the right of this is 6, which is greater than 5, so round up.
The answer is 1.74.
5. Round 0.6998 to the thousandths place.
0.700
0.699
0.7
0.600
The digit in the thousandths place is 9 and the value to the right of this is 8. \, 8 is greater than 5 so round up – increasing 9 by 1 gives us 10, so you need to increase the previous 9 by 1 which also gives us 10.
Therefore you need to increase the 6 by 1 to give us 7.
You need to include the zeros in our answer so that it is written to the thousandths place.
The answer is 0.700.
6. Round 0.03197 to the nearest thousandth.
0.031
0.040
0.142
0.032
The digit in the thousandths place is 1. The value to the right of this is 9, which is greater than 5, so round up.
The answer is 0.032.
Decimal places FAQs
What decimals should I expect my students to know and work with?
Following the Common Core recommendations, in 5th grade, students should work with tenths place, hundredths place and thousandths place. Check your state’s specific standards for further clarification.
Why is learning about decimal places important?
Decimals are used in everyday life, including money and measurements. Decimals provide a more precise number that are needed in some situations.
What is “decimal place value”?
Decimal place values are all the digits in a decimal number. A tenth is the first digit to the right of a decimal point. The second digit to the right of a decimal point is the hundredths place and the third is the thousandths place. Decimals can also include ten thousandths and even millionths.
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Prepare for math tests in your state with these Grade 3 to Grade 6 practice assessments for Common Core and state equivalents.
40 multiple choice questions and detailed answers to support test prep, created by US math experts covering a range of topics!
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The length of an elastic string is $a$ metre when the longitudinal tension is $4\,N$ and $b$ metre when the longitudinal tension is $5\,N$. The length of the string in metre when longitudinal tension is $9\,N$ is:A. $a - b$B. $5b - 4a$C. $2b - \dfrac{1}{4}a$D. $4a - 3b$
Last updated date: 13th Jul 2024
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Hint:In order to answer this question, you must be aware about the concept of Tension and force constant. Force constant is a proportionality constant. The greater the constant of force, the greater the restored force.
Let $L$ be the original length of the wire and $k$ be the force constant of the wire.
Final length = Original length + Elongation
${L^1} = L + \dfrac{F}{k}$
In the first case where tension is $4\,N$, $a = L + \dfrac{4}{k}$ ……. (1)
In the second case where tension is $5\,N$, $b = L + \dfrac{5}{k}$ ……. (2)
By solving eq. (1) and (2), we get
$L = 5a - 4b$ and $k = \dfrac{1}{{b - a}}$
Now, when the longitudinal tension is 9N, length of the string will be
${L^1} = L + \dfrac{9}{k}$
$\Rightarrow {L^1} = 5a - 4b + 9(b - a)$
$\therefore {L^1} = 5b - 4a$
Hence, option B is correct.
Note:Tension is described as the pulling force transmitted axially by the means of a string. It can also be defined as the action – reaction pair of forces acting at each end of the said elements.While considering a rope, the tension force is felt by every section of the rope in both the directions, apart from the endpoints.
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# Modulus of x-a is continuous at x=a but not differentiable
This page will discuss the continuity and differentiability of the absolute value of $x-a$, that is, of the function $|x-a|$, at the point $x=a$.
Note that the function $|x-a|$ is defined as follows:
$|x-a| = x-a$ if $x\geq a$
$=-(x-a)$ if $x<a$
## Continuity of |x-a| at x=a
Let us now discuss the continuity of the modulus function $|x-a|$ at $x=a$.
Write $f(x)=|x-a|$
Note that $\lim\limits_{x \to a-} f(x)=\lim\limits_{x \to a-} (-(x-a))=0$
$\lim\limits_{x \to a+} f(x)=\lim\limits_{x \to a+} (x-a)=0$
and $f(a)=0$
So we get that
$\lim\limits_{x \to a-} f(x)=\lim\limits_{x \to a+} f(x)=f(a)$
Thus the function f(x)=|x-a| is continuous at x=a.
## Differentiability of |x-a| at x=a
We will now show the modulus of $x-a$ is not differentiable at $x=a$.
Let $f(x)=|x-a|.$
Note that the function $|x-a|$ will be differentiable at $x=a$ if the following limit $L$ exists.
$L=\lim\limits_{h \to 0} \dfrac{f(a+h)-f(a)}{h}.$
The left-hand limit $=\lim\limits_{h \to 0-} \dfrac{f(a+h)-f(a)}{h}$
$=\lim\limits_{h \to 0-} \dfrac{-(a+h-a)-0}{h}$
$=\lim\limits_{h \to 0-} \dfrac{-h}{h}$ $=-1$
The right-hand limit $=\lim\limits_{h \to 0+} \dfrac{f(a+h)-f(a)}{h}$
$=\lim\limits_{h \to 0+} \dfrac{(a+h-a)-0}{h}$
$=\lim\limits_{h \to 0-} \dfrac{h}{h}$ $=1$
As both the left-hand and the right-hand limit exist and are not equal, we conclude that the limit L does not exist. Thus the function f(x)=|x-a| is not differentiable at x=a.
Epsilon-Delta definition of Continuity
Sinx is continuous with proof
## FAQs
Q1: Discuss the continuity and differentiability of |x-a| at x=a.
Answer: The function f(x)=|x-a| is continuous at x=a but not differentiable at x=a.
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# How do you find the area between f(x)=x^2+2x+1,g(x)=3x+3?
Nov 16, 2016
4.5 square units
#### Explanation:
The points of intersection of the two functions are at:
$\left(- 1 , 0\right)$ and $\left(2 , 9\right)$.
So, we have to find the area under the curve of $g \left(x\right)$, then subtract the area under the curve of $f \left(x\right)$ because $g \left(x\right)$ is above $f \left(x\right)$ in the interval $\left[- 1 , 2\right]$.
${\int}_{-} {1}^{2} g \left(x\right) \mathrm{dx} - {\int}_{-} {1}^{2} f \left(x\right) \mathrm{dx}$
${\int}_{-} {1}^{2} \left(3 x + 3\right) \mathrm{dx} - {\int}_{-} {1}^{2} \left({x}^{2} + 2 x + 1\right) \mathrm{dx}$
Simplify the are under the curve of the linear equation using geometry, and the area under the curve of the parabola using a fundamental theorem of calculus.
Let $F \left(x\right)$ be the antiderivative of $f \left(x\right)$
$\frac{1}{2} \left(3\right) \left(9\right) - \left[F \left(2\right) - F \left(- 1\right)\right]$
$\frac{27}{2} - \left[F \left(2\right) - F \left(- 1\right)\right]$
To simplify this, we need to know $F \left(x\right)$. Integrate $f \left(x\right)$:
$F \left(x\right) = \int {x}^{2} + 2 x + 1$
$F \left(x\right) = \frac{1}{3} {x}^{3} + {x}^{2} + x + c$
Solve $F \left(2\right)$ and $F \left(- 1\right)$:
$F \left(2\right) = \frac{1}{3} {\left(2\right)}^{3} + {2}^{2} + 2 + c = \frac{26}{3} + c$
$F \left(- 1\right) = \frac{1}{3} {\left(- 1\right)}^{3} + {\left(- 1\right)}^{2} - 1 + c = - \frac{1}{3} + c$
Now continue to simplify the expression above:
$\frac{27}{2} - \left[F \left(2\right) - F \left(- 1\right)\right]$
$13.5 - \left[\left(\frac{26}{3} \cancel{+ c}\right) - \left(- \frac{1}{3} \cancel{+ c}\right)\right]$
$= 13.5 - 9$
$= 4.5$ square units
Nov 16, 2016
Start by finding the intersection points of the two curves.
$\left\{\begin{matrix}y = {x}^{2} + 2 x + 1 \\ y = 3 x + 3\end{matrix}\right.$
$3 x + 3 = {x}^{2} + 2 x + 1$
$0 = {x}^{2} - x - 2$
$0 = \left(x - 2\right) \left(x + 1\right)$
$x = 2 \mathmr{and} - 1$
$y = 3 x + 3$
$y = 9 \mathmr{and} 0$
Hence, the points of intersection are $\left\{2 , 9\right\}$ and $\left\{- 1 , 0\right\}$
We now do the graph of each function (on the same grid). We find that $g \left(x\right)$ is above $f \left(x\right)$ in the area that they share, so we find the area under $g \left(x\right)$, and will subtract the area of $f \left(x\right)$ from that.
${\int}_{- 1}^{2} \left(3 x + 3\right) \mathrm{dx}$
$\implies \frac{3}{2} {x}^{2} + 3 x {|}_{- 1}^{2}$
$\implies \frac{3}{2} {\left(2\right)}^{2} + 3 \left(2\right) - \left(\frac{3}{2} {\left(- 1\right)}^{2} + 3 \left(- 1\right)\right)$
$\implies \frac{3}{2} \left(4\right) + 6 - \frac{3}{2} + 3$
$\implies 6 + 6 - \frac{3}{2} + 3$
$\implies \frac{27}{2}$
Now for $f \left(x\right)$:
${\int}_{- 1}^{2} \left({x}^{2} + 2 x + 1\right)$
$\implies \frac{1}{3} {x}^{3} + {x}^{2} + x {|}_{- 1}^{2}$
$\implies \frac{1}{3} {\left(2\right)}^{3} + {2}^{2} + 2 - \left(\frac{1}{3} {\left(- 1\right)}^{3} + {\left(- 1\right)}^{2} - 1\right)$
$\implies \frac{8}{3} + 4 + 2 + \frac{1}{3} - 1 + 1$
$\implies 9$
Now subtract the two areas.
$\implies \frac{27}{2} - 9 = \frac{9}{2} = 4.5$
Hence, the area between the graphs of $f \left(x\right) = {x}^{2} + 2 x + 1$ and $g \left(x\right) = 3 x + 3$ is $4.5 {\text{ units}}^{2}$.
Hopefully this helps!
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# Idaho - Grade 1 - Math - Numbers and Operations in Base Ten - Tens & Ones - 1.NBT.2
### Description
Understand that the two digits of a two-digit number represent amounts of tens and ones. Understand the following as special cases: A. 10 can be thought of as a bundle of ten ones — called a “ten.” B. The numbers from 11 to 19 are composed of a ten and one, two, three, four, five, six, seven, eight, or nine ones. C. The numbers 10, 20, 30, 40, 50, 60, 70, 80, 90 refer to one, two, three, four, five, six, seven, eight, or nine tens (and 0 ones).
### Additional Info
• State - Idaho
• Standard ID - 1.NBT.2
• Subjects - Math Common Core
• Grade - 1
### Keywords
• Math
• Idaho grade 1
• Numbers and Operations in Base Ten
## More Idaho Topics
Distinguish between defining attributes (e.g., triangles are closed and three-sided) versus non-defining attributes (e.g., color, orientation, overall size) ; build and draw shapes to possess defining attributes.
Use addition and subtraction within 20 to solve word problems involving situations of adding to, taking from, putting together, taking apart, and comparing, with unknowns in all positions, e.g., by using objects, drawings, and equations with a symbol for the unknown number to represent the problem.
Solve word problems that call for addition of three whole numbers whose sum is less than or equal to 20, e.g., by using objects, drawings, and equations with a symbol for the unknown number to represent the problem.
Count to 120, starting at any number less than 120. In this range, read and write numerals and represent a number of objects with a written numeral.
Add within 100, including adding a two-digit number and a one-digit number, and adding a two-digit number and a multiple of 10, using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a written method and explain the reasoning used. Understand that in adding two-digit numbers, one adds tens and tens, ones and ones; and sometimes it is necessary to compose a ten.
## Here is the skill that Idaho requires you to master
• Grade Level 1
• State Test The ISAT by Smarter Balanced
• State Standards Idaho Core Standards
• Subject Math
• Topic Name Tens & Ones
• Standard ID 1.NBT.2
• Description
Understand that the two digits of a two-digit number represent amounts of tens and ones. Understand the following as special cases: A. 10 can be thought of as a bundle of ten ones — called a “ten.” B. The numbers from 11 to 19 are composed of a ten and one, two, three, four, five, six, seven, eight, or nine ones. C. The numbers 10, 20, 30, 40, 50, 60, 70, 80, 90 refer to one, two, three, four, five, six, seven, eight, or nine tens (and 0 ones).
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### Idaho has adopted the Idaho Core Standards
Idaho has adopted the Idaho Core Standards and will administer the ISAT by Smarter Balanced to measure student proficiency of Idaho standards. Education Galaxy’s ISAT by Smarter Balanced Assessments preparation program provides online assessment and practice for students in Grades 2-8 to help build mastery towards the Idaho Core Standards. Our unique online program is easy to use and enjoyable for both teachers and students. Students work on their Study Plans practicing important concepts while teachers pull formative assessment reports to identify the strengths and weaknesses of their classroom and individual students.
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# 5.4.4 - Checking Normality
5.4.4 - Checking Normality
## Using Normal Probability Plot to Check Normality
If the sample size is less than 30, one needs to use a Normal Probability Plot to check whether the assumption that the data come from a normal distribution is valid.
Normal Probability Plot
The Normal Probability Plot is a graph that allows us to assess whether or not the data comes from a normal distribution.
Note! This plot should be used as a guide for us to assess if the assumption that the data come from a normal distribution is valid or not. It should not be used to “test” an assumption.
## Example 5-7: Rattlesnake Lengths
It is very time consuming to find rattlesnakes and nerve racking to measure them (for obvious reasons). A scientist randomly finds 12 snakes from the central Pennsylvania area and measures their length. The following twelve measurements in inches are obtained:
40.2, 43.1, 45.5, 44.5, 39.5, 38.5, 40.2, 41.0, 41.6, 43.1, 44.9, 42.8
Using the above data, find a 90% confidence interval for the mean length of rattlesnakes in the central Pennsylvania area.
#### Step 1 Check Conditions
Think about what conditions you need to check. The sample size is only 12. The scenario does not give us an indication that the lengths follow a normal distribution. Therefore, let's do a normal probability plot to check whether the assumption that the data come from a normal distribution is valid.
##### Minitab: Creating a normal probability plot
To create a normal probability plot in Minitab:
1. Enter the 12 measurements into one column (name it length for this example) or upload the snakes.txt file.
2. Type or upload the data in the first column in Minitab.
3. Choose Graph > Probability Plot
Here is the normal probability plot for the rattlesnake data. What do you conclude about whether they may come from a normal distribution?
Since the points all fall within the confidence limits, it is reasonable to suggest that the data come from a normal distribution.
#### Step 2 Construct the CI
Now, we can proceed to find the 90% t-interval for the mean length of rattlesnakes in the central Pennsylvania area since even though the sample size is less than 30, the normality plot shows that the data may come from a normal distribution.
##### Minitab: Find the t-interval using Minitab
1. Enter the 12 measurements into one column (name it length for this example)
2. Choose Stat > Basic Statistics > 1-Sample t
3. Click on the variable (length for this example) and change to the desired confidence level
The Minitab output will provide the confidence interval. We get the following:
N Mean StDev SE Mean 90% CI
12 42.075 2.257 0.652 (40.905, 43.245)
View the video to see these steps within Minitab.
#### Step 3 Interpret the Interval
We are 90% confidence that the population mean lengths of rattlesnakes is between 40.905 and 43.245 inches.
[1] Link ↥ Has Tooltip/Popover Toggleable Visibility
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# Lesson 10Getting CenteredSolidify Understanding
## Jump Start
Find the missing term that makes each expression a perfect square trinomial, and then write an equivalent expression. (You may use a diagram to help.)
## Learning Focus
Write and graph the equation of a circle.
Find the center and radius of a circle in general form.
Do translations work on circles like they do on functions?
How are different forms of the equation of a circle related?
## Open Up the Math: Launch, Explore, Discuss
Malik’s family has decided to put in a new sprinkler system in their yard. Malik has volunteered to plan and draw the system out on a grid. Sprinklers are available at the hardware store that spray a full circle with a maximum radius of . The sprinklers can be adjusted so that they spray a smaller radius. Malik needs to be sure that the entire yard gets watered, which he knows will require that some of the circular water patterns overlap. He gets out a piece of graph paper and begins with a scale diagram of the yard. In this diagram, the length of the side of each square represents .
### 1.
As he begins to think about locating sprinklers on the lawn, his parents tell him to try to cover the whole lawn with the fewest number of sprinklers possible so that they can save some money. The equation of the first circle that Malik draws to represent the area watered by the sprinkler is:
1. Draw this circle on the diagram using a compass.
2. Help Malik lay out a possible configuration for the sprinkling system by adding two more sprinklers, and the area that they cover, to the diagram that includes the first sprinkler pattern that you drew.
3. Find the equation of each of the circles that you have drawn.
Malik wrote the equation of one of the circles and, just because he likes messing with the algebra, he did this:
Original equation:
Malik thought, “That’s pretty cool. It’s like a different form of the equation. I guess that there could be different forms of the equation of a circle like there are different forms of the equation of a parabola or the equation of a line.” He showed his equation to his sister, Sapana, and she thought he was nuts.
Sapana said, “That’s a crazy equation. I can’t even tell where the center is or the length of the radius anymore.” Malik said, “Now it’s like a puzzle for you. I’ll give you an equation in the new form. I’ll bet you can’t figure out where the center is.”
Sapana said, “Of course I can. I’ll just do the same thing you did, but work backwards.”
### 2.
Malik gave Sapana this equation of a circle:
Help Sapana find the center and the length of the radius of the circle.
### 3.
Sapana said, “OK. I made one for you. What’s the center and length of the radius for this circle?”
### 4.
Sapana said, “I still don’t know why this form of the equation might be useful. When we had different forms for other equations like lines and parabolas, each of the various forms highlighted different features of the relationship.” Why might this form of the equation of a circle be useful?
How might you use the general form of the equation of a circle, , to find the equation of the circle that contains the following points: , , ?
Try it. See if you can find values for , , and to write the equation of the circle.
## Takeaways
General form of the equation of a circle:
Standard form of the equation of a circle:
Changing from general form to standard form:
Example:
Description of steps:
## Lesson Summary
In this lesson, we learned to write equations of circles in both standard form and general form. We used the process of completing the square to change an equation from standard form to general form.
## Retrieval
### 1.
The graph of is translated to the right units and down units and undergoes a vertical stretch by a factor of . Write the equation for the transformed function.
Solve for .
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# Solving a Simple Cubic Equation
Alignments to Content Standards: A-REI A-APR.B.3
1. Find all the values of $x$ for which the equation $9x=x^3$ is true.
2. Use graphing technology to graph $f(x)=x^3-9x$. Explain where you can see the answers from part (a) in this graph, and why.
3. Someone attempts to solve $9x=x^3$ by dividing both sides by $x$, yielding $9=x^2$, and going from there. Does this approach work? Why or why not?
## IM Commentary
The purpose of this task is twofold. First, it prompts students to notice and explain a connection between the factored form of a polynomial and the location of its zeroes when graphed. Second, it highlights a complication that results from a seemingly innocent move that students might be tempted to make: "dividing both sides by x." When students solve equations in earlier grades they tend to be limited to linear equations in one variable, where "doing the same thing to both sides" is encouraged and never causes problems. This is because moves like "add one to both sides" are invertible. Once students try to solve more complicated equations, though, "doing the same thing to both sides" isn't necessarily always okay. In this case, the problem comes from dividing both sides by something that could take the value zero. Once students understand the Zero Product Property, and are able to rewrite expressions in factored form, they have another strategy for solving equations besides "performing operations on both sides." This task suggests that checking the graph of an associated function for locations of x-intercepts is a way to avoid inadvertently "losing" solutions.
A technology note: the default graphing window for many graphing calculators is from -10 to 10 both vertically and horizontally. Strictly speaking, that's fine for this task, because all of the zeroes of the polynomial will be visible. However, the maximum and minimum values will be outside the viewing window, and it's a good habit for students to consider all the important features of a graph when analyzing it. Some students default to "zooming out" to change the window, which is often too coarse a move to be effective, which it is in this case. This could be an opportunity to help students understand the benefit of strategically selecting a viewing window in a graphing calculator's settings. A suggested viewing window is x:[-10,10] and y:[-15,15].
## Solution
1. The values of $x$ making the equation true are 0, -3, and 3. A student could find these by reasoning about equality and operations. Or, one could write the equivalent equation $x^3-9x=0$, rewrite the left side as $x(x^2-9)=0$, use the Zero Product Property to reason that if that is true, then $x=0$ or $x^2-9=0$. So one solution is $x=0$. We can deal with $x^2-9=0$ by either factoring one more time to get $(x+3)(x-3)=0$ (hence $x=3$ or $x=-3$) or reasoning directly that if $x^2=9$ then $x$ could equal $3$ or $-3$.
2. This graph intercepts the $x$-axis when $x$ is $-3$, $0$, and $3$, which matches the solutions to the equation given in part (a). The equation was equivalent to $0=x^3-9x$. In the $xy$-plane we graphed $y=x^3-9x$. The $x$-intercepts are the values of $x$ when $y$ or $x^3 - 9x$ is $0$. From part (a), we know that this is true when $x$ is $0$, $-3$, or $3$.
3. If you first divide both sides by $x$ and then solve $x^2=9$, you get only two solutions: $x = -3$ and $x = 3$. So something is obviously wrong with this approach, because there were three solutions to the original equation: $x = 0$,$x = -3$, and $x = 3$. When the person divided by $x$, they neglected the possibility that $x$ could be zero. So solving $x^2 = 9$ gives the same non-zero solutions as the original equation $x^3 = 9x$ but the solution $x = 0$ is lost because, in this case, we performed the illegal move of dividing both sides of the equation $x^3 = 9x$ by $0$.
The operation ''dividing by $x$'' can be undone (namely by ''multiplying by $x$'') as long as $x$ is non-zero so we do not lose the non-zero solutions to the equation $x^3 = 9x$ when we divide both sides by $x$. Dividing both sides by $x$ will not produce spurious solutions but it does remove a valid solution. In a similar way, multiplying both sides by $x-1$, for example, will preserve all of the solutions to the equation but will introduce an additional solution, namely $x = 1$.
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# In Figure, AD is a median of a triangle ABC and AM $$\perp$$ BC. Prove that : $$(i) AC^2 = AD^2 + BC.DM + 2 (\frac{BC}{2} )^2$$ $$(ii) AB^2 = AD^2 – BC.DM + 2 (\frac{BC}{2} )^2$$ $$(iii) AC^2 + AB^2 = 2 AD^2 + {{1} \over {2}} BC^2$$
(i) By applying Pythagoras Theorem in ?AMD, we get,
$$AM^2 + MD^2 = AD^2$$ ………………. (i)
Again, by applying Pythagoras Theorem in $$\triangle$$ AMC, we get,
$$\Rightarrow AM^2 + MC^2 = AC^2$$
$$\Rightarrow AM^2 + (MD + DC)^ 2 = AC^2$$
$$\Rightarrow (AM^2 + MD^2 ) + DC^2 + 2MD.DC = AC^2$$
From equation(i), we get,
$$AD^2 + DC^2 + 2MD.DC = AC^2$$
Since, DC=$$\frac{BC}{2}$$ , thus, we get,
$$\Rightarrow AD^2 + (BC/2)^ 2 + 2MD.(BC/2)^ 2 = AC^2$$
$$\Rightarrow AD^2 + (BC/2)^ 2 + 2MD × BC = AC^2$$
Hence, proved.
(ii) By applying Pythagoras Theorem in -ABM, we get;
$$\Rightarrow AB^2 = AM^2 + MB^2$$
$$\Rightarrow (AD^2 - DM^2 ) + MB^2$$
$$\Rightarrow (AD^2 - DM^2 ) + (BD - MD)^2$$
$$\Rightarrow AD^2 - DM^2 + BD^2 + MD^2 - 2BD × MD$$
$$\Rightarrow AD^2 + BD^2 - 2BD × MD$$
$$\Rightarrow AD^2 + (\frac{BC}{2} )^2 – 2 \frac{BC}{2} MD$$
$$\Rightarrow AD^2 + (\frac{BC}{2} )^2 – BC.MD$$
Hence, proved.
(iii) By applying Pythagoras Theorem in $$\triangle$$ ABM, we get,
$$AM^2 + MB^2 = AB^2$$ ………………….… (i)
By applying Pythagoras Theorem in $$\triangle$$ AMC, we get,
$$\Rightarrow AM^2 + MC^2 = AC^2$$ …………………..… (ii)
Adding both the equations (i) and (ii), we get,
$$\Rightarrow 2AM^2 + MB^2 + MC^2 = AB^2 + AC^2$$
$$\Rightarrow 2AM^2 + (BD - DM)^2 + (MD + DC)^2 = AB^2 + AC^2$$
$$\Rightarrow 2AM^2+BD^2 + DM^2 - 2BD.DM + MD^2 + DC^2 + 2MD.DC = AB^2 + AC^2$$
$$\Rightarrow 2AM^2 + 2MD^2 + BD^2 + DC62 + 2MD (- BD + DC) = AB^2 + AC^2$$
$$\Rightarrow 2(AM^2+ MD^2) + (\frac{BC}{2} )^2 + (\frac{BC}{2} )^2+ 2MD(\frac{-BC}{2} + \frac{BC}{2} )^ 2 = AB^2 + AC^2$$
$$\Rightarrow 2AD^2 + (\frac{BC}{2} )^2 = AB^2 + AC^2$$
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# 5.2 Definite Integrals In this section we move beyond finite sums to see what happens in the limit, as the terms become infinitely small and their number.
## Presentation on theme: "5.2 Definite Integrals In this section we move beyond finite sums to see what happens in the limit, as the terms become infinitely small and their number."— Presentation transcript:
5.2 Definite Integrals In this section we move beyond finite sums to see what happens in the limit, as the terms become infinitely small and their number infinitely large. Sigma notation enables us to express a large sum in compact form:
Definite Integrals The Greek capital letter, sigma, stands for “sum”.
The index k tells us where to begin the sum (at the number below the sigma) and where to end (at the number above the sigma). If the symbol, infinity, appears above the sigma, it indicates that the terms go on indefinitely. These sums are called Riemann sums. LRAM, MRAM, and RRAM are examples of Riemann sums – not because they estimated area, but because they were constructed in a particular way.
Definite Integrals Figure 5.12 is a continuous function f(x) defined on a closed integral [a , b]. It may have negative values as well as positive values.
Definite Integrals To make the notation consistent, we denote a by x0 and b by xn. The set P = {x0, x1, x2, …, xn} is called a partition of [a , b]. The partition P determines n closed subintervals. The kth subinterval is [xk – 1 , xk], which has length xk = xk – xk – 1.
Definite Integrals The value of the definite integral of a function over any particular interval depends on the function and not on the letter we choose to represent its independent variable. If we decide to use t or u instead of x, we simply write the integral as: No matter how we represent the integral, it is the same number, defined as a limit of Riemann sums. Since it does not matter what letter we use to run from a to b, the variable of integration is called a dummy variable.
Using the Notation The interval [-1 , 3] is partitioned into n subintervals of equal length Let mk denote the midpoint of the kth subinterval. Express the limit as an integral.
Revisiting Area Under a Curve
Evaluate the integral
If an integrable function y = f(x) is nonpositive, the nonzero terms in the Riemann sums for f over an interval [a , b] are negatives of rectangle areas. The limit of the sums, the integral of f from a to b, is therefore the negative of the area of the region between the graph of f and the x-axis.
If an integrable function y = f(x) has both positive and negative values on an interval [a , b], then the Riemann sums for f on [a , b] add areas of rectangles that lie above the x-axis to the negatives of areas of rectangles that lie below the x-axis. The resulting cancellations mean that the limiting value is a number whose magnitude is less than the total area between the curve and the x-axis. The value of the integral is the area above the x-axis minus the area below.
Constant Functions Integrals of constant functions are easy to evaluate. Over a closed interval, they are simply the constant times the length of the interval (Figure 5.21).
Revisiting the Train Problem
A train moves along a track at a steady 75 miles per hour from 7:00 A.M. to 9:00 A.M. Express its total distance traveled as an integral. Evaluate the integral using Theorem 2.
Using NINT Evaluate the following integrals numerically.
More Practice!!!!! Homework – Textbook p. 282 – 283 #1 – 22 ALL.
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# Lesson 2Root VariationSolidify Understanding
## Learning Focus
Examine how changes in the quantities of a context transform the graph of the square root function that models the context.
How can we model the motion of swinging on a swing?
What quantities might affect the length of time it takes to swing forward and back, returning to the starting position?
## Open Up the Math: Launch, Explore, Discuss
Tehani, Taska, and Trevor have entered an engineering competition in which students have to redesign an Earth-based amusement park full of rides so the rides will feel and behave the same when installed on other planets, specifically Mars or Jupiter. Because the pull of gravity on each of these three planets is different, there are a lot of mathematical challenges involved in recreating the amusement park rides. Tehani’s team has been assigned to redesign the swings that will appear throughout the park, some smaller swings in the children’s portion of the park, a large swing in the adult section of the park, and some swings that replicate swinging vines in the Tarzan: Escape from the Jungle ride.
Trevor has found that that the period, (one forward and back motion of a swing), is related to the length, , of the swing by the formula . In this formula, the constant depends on the acceleration of gravity acting on the weight of the object carried by the swing, and therefore the value of will change from planet to planet. On Earth, when the length of the swing is measured in feet and the period is measured in seconds.
### 1.
Calculate the period of a swing that is long.
### 2.
If you double the length of the swing, will you double the length of time it takes for one period?
Why or why not (that is, provide evidence for your claim)?
### 3.
How much longer do you need to make the long swing to double its period?
### 4.
How many times longer do you need to make the long swing to double its period?
### 5.
How many times longer do you need to make the long swing to triple its period?
### 6.
In general, if you want to extend the period of the long swing by a factor of , how much longer do you need to make the swing?
Taska is working on the redesign of the swings for the park on Jupiter. The formula for the period of a swing on Jupiter is .
### 7.
Describe how a ride on a swing transported from Earth to Jupiter without adjustment would compare to the ride experienced on the same swing on Earth, where the period is given by .
### 8.
Trevor has found that included in their instructions they were given a graph of the period of a swing as a function of its length on the planet Earth. That is, this is a graph of . Draw the graph of the period of a swing for the planet Jupiter on this same coordinate grid. Remember that the formula for the period of a swing on Jupiter is .
### 9.
Add the graph of the period of a swing for Mars to this graph of the period of a swing on the Earth, given that the formula for the period of a swing on Mars relative to its length is .
### 10.
Tehani has inspected Taska’s work and has several changes she wants made to various swings around the playground. Since each swing that needs to be adjusted is of a different length, she has written her instructions using to represent the original length of the swing before the desired adjustment is made. Interpret the following notation in terms of what Taska is being asked to adjust. Be specific, in terms of the context, by describing what quantity needs to be changed and by how much, including units.
### 11.
The graph and a table for the period of a swing of length on Jupiter is given below. Complete either the table or the graph for each of the following related functions. Use descriptions of the contextual changes to the swing (see problem 10) to justify the coordinates you list in the table or the shape and location of the new graphs.
The Language of Proportionality Relationships (or Variation)
You studied proportionality relationships in Grade 7. One way to tell if two quantities are related by a proportionality relationship is to see how a change in the input affects the change in the output. In general, if doubling the input doubles the output, tripling the input triples the output, cutting the input in half cuts the output in half, etc., then the input and output quantities are related by a direct proportionality relationship. Mathematicians use language, such as, “The output quantity is proportional to the input quantity,” or, “the output quantity varies directly with the input quantity” to describe such relationships.
### 12.
If you are running at a rate of , your distance as a function of time is given by . Is this relationship a direct variation relationship?
Why or why not?
### 13.
Is the period of a swing relationship , a direct variation relationship?
Why or why not?
While the period of a swing equation is not a direct variation, scientists use the language of proportionality to describe this type of relationship using the language, “The period of a swing is proportional to the square root of the length.” You might wonder why this is called a proportionality relationship since doubling the length of the swing does not double the period, etc. However, this statement is not about the relationship between the quantities period and length, but rather the quantities period and square root of the length.
Table 1:
Length Period $2$ $4$ $6$ $8$ $10$ $12$ $14$ $16$ $18$ $20$ $22$ $24$ $1.697$ $2.4$ $2.939$ $3.394$ $3.794$ $4.157$ $4.490$ $4.8$ $5.091$ $5.367$ $5.628$ $5.879$
Graph 1:
Table 2:
Graph 2:
### 14.
Examine the two tables and graphs given. Which table illustrates a direct variation?
How do you know?
### 15.
Which graph illustrates a direct variation?
How do you know?
Based on the information in the last part of the task, is there a way that you can state and justify a direct variation relationship for the equation for the area of a circle: ?
That is, varies directly with .
Use a table or a graph to justify your direct variation statement.
## Takeaways
The square root function can be transformed in the same way as other functions we have seen.
For example:
• .
• .
• .
• .
The following tests can be used to determine if two quantities and are proportional to each other (or in other words, are in a direct variation relationship):
An equation like is not a direct variation in the quantities and , but is a direct variation
## Lesson Summary
In this lesson, we learned the graph of a square root function can be transformed in the same way as linear, exponential and quadratic functions are transformed, including horizontal and vertical translations and dilations. We also reviewed conditions that could be used to verify if two quantities are proportional to each other. In this lesson, we called these relationships direct variations and found that if we define the quantities carefully, we can find direct variations between such quantities as the “square root of the length of a pendulum” and “the period of a pendulum.”
## Retrieval
### 1.
An astronomer, Dr. Astral, wrote a system of two equations to see if he could predict when the paths of the objects he was studying would intersect. He solved the system by setting the equations equal to each other and solving for . He then used substitution to find . After solving the systems, he looked at the graphs of the equations to check if he was right.
Set the equations equal to each other and solve for .
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### Part I. Section I. Chapter 2. “Explanation of the Signs + Plus and - Minus.”
8 When we have to add one given number to another, this is indicated by the sign +, which is placed before the second number, and is read plus. Thus 5 + 3 signifies that we must add 3 to the number 5, in which case, every one knows that the result is 8; in the same manner 12 + 7 make 19; 25 + 10 make 41; the sum of 25 + 41 is 66, etc.
9 We also make use of the same sign + plus, to connect several numbers together for example, 7 + 5 + 9 signifies that to the number 7 we must add 5, and also 9, which make 21. The reader will therefore understand what is meant by
8 + 5 + 13 + 11 + 1 + 3 + 10,
namely, the sum of all these numbers, which is 51.
10 All this is evident; and we have only to mention, that in Algebra, in order to generalise numbers, we represent them by letters, as $$a$$, $$b$$, $$c$$, $$d$$, etc. Thus, the expression $$a + b$$, signifies the sum of two numbers, which we express by $$a$$ and $$b$$, and these numbers may be either very great, or very small. In the same manner, $$f+m+b+x$$, signifies the sum of the numbers represented by these four letters.
If we know therefore the numbers that are represented by letters, we shall at all times be able to find, by arithmetic, the sum or value of such expressions.
11 When it is required, on the contrary, to subtract one given number from another, this operation is denoted by the sign -, which signifies minus, and is placed before the number to be subtracted: thus, 8 - 5 signifies that the number 5 is to be taken from the number 8; which being done, there remain 3. In like manner 12 - 7 is the same as 5; and 20 - 14 is the same as 6, etc.
12 Sometimes also we may have several numbers to subtract from a single one ; as, for instance,
50 - 1 - 3 - 5 - 7 - 9.
This signifies, first, take 1 from 50, and there remain 49; take 3 from that remainder, and there will remain 46; take away 5, and 41 remain; take away 7, and 34 remain; lastly, from that take 9, and there remain 25: this last remainder is the value of the expression. But as the numbers 1, 3, 5, 7, 9, are all to be subtracted, it is the same thing if we subtract their sum, which is 25, at once from 50, and the remainder will be 25 as before.
13 It is also easy to determine the value of similar expressions, in which both the signs + plus and - minus are found. For example; 12 - 3 - 5 + 2 - 1 is the same as 5.
We have only to collect separately the sum of the numbers that have the sign + before them, and subtract from it the sum of those that have the sign -. Thus, the sum of 12 and 2 is 14; and that of 3, 5, and 1, is 9; hence 9 being taken from 14, there remain 5.
14 It will be perceived, from these examples, that the order in which we write the numbers is perfectly indifferent and arbitrary, provided the proper sign of each be preserved. We might with equal propriety have arranged the expression in the preceding article thus;
12 + 2 - 5 - 3 - 1,
or
2 - 1 - 3 - 5 + 12,
or
2 + 12 -3 -1 -5,
or in still different orders; where it must be observed, that in the arrangement first proposed, the sign + is supposed to be placed before the number 12.
15 It will not be attended with any more difficulty if, in order to generalise these operations, we make use of letters instead of real numbers. It is evident, for example, that
$a - b - c - d - e,$
signifies that we have numbers expressed by $$a$$ and $$d$$, and that from these numbers, or from their sum, we must subtract the numbers expressed by the letters $$b$$, $$c$$, $$e$$, which have before them the sign -.
16 Hence it is absolutely necessary to consider what sign is prefixed to each number: for in Algebra, simple quantities are numbers considered with regard to the signs which precede, or affect them. Farther, we call those positive quantities, before which the sign + is found; and those are called negative quantities, which are affected by the sign -.
17 The manner in which we generally calculate a person’s property, is an apt illustration of what has just been said. For we denote what a man really possesses by positive numbers, using, or understanding the sign +; whereas his debts are represented by negative numbers, or by using the sign -. Thus, when it is said of any one that he has 100 crowns, but owes 50, this means that his real possession amounts to 100 - 50; or, which is the same thing, +100-50, that is to say, 50.
18 Since negative numbers may be considered as debts, because positive numbers represent real possessions, we may say that negative numbers are less than nothing. Thus, when a man has nothing of his own, and owes 50 crowns, it is certain that he has 50 crowns less than nothing; for if any one were to make him a present of 50 crowns to pay his debts, he would still be only at the point nothing, though really richer than before.
19 In the same manner, therefore, as positive numbers are incontestably greater than nothing, negative numbers are less than nothing. Now, we obtain positive numbers by adding 1 to 0, that is to say, 1 to nothing; and by continuing always to increase thus from unity. This is the origin of the series of numbers called natural numbers; the following being the leading terms of this series:
0, +1, +2, +3, +4, +5, +6, +7, +8, +9, +10,
and so on to infinity.
But if, instead of continuing this series by successive additions, we continued it in the opposite direction, by perpetually subtracting unity, we should have the following series of negative numbers:
0, -1, -2, -3, -4, -5, -6, -7, -8, -9, -10,
and so on to infinity.
20 All these numbers, whether positive or negative, have the known appellation of whole numbers, or integers, which consequently are either greater or less than nothing. We call them integers, to distinguish them from fractions, and from several other kinds of numbers, of which we shall hereafter speak. For instance, 50 being greater by an entire unit than 49, it is easy to comprehend that there may be, between 49 and 50, an infinity of intermediate numbers, all greater than 49, and yet all less than 50. We need only imagine two lines, one 50 feet, the other 49 feet long, and it is evident that an infinite number of lines may be drawn, all longer than 49 feet, and yet shorter than 50.
21 It is of the utmost importance through the whole of Algebra, that a precise idea should be formed of those negative quantities, about which we have been speaking. I shall, however, content myself with remarking here, that all such expressions as
+1 - 1, +2 - 2, +3 - 3, +4 - 4, etc.
are equal to 0, or nothing. And that +2 - 5 is equal to -3: for if a person has 2 crowns, and owes 5, he has not only nothing, but still owes 3 crowns. In the same manner, 7 - 12 is equal to -5, and 25 - 40 is equal to -15.
22 The same observations hold true, when, to make the expression more general, letters are used instead of numbers; thus 0, or nothing, will always be the value of $$+a - a$$; but if we wish to know the value of $$+a - b$$, two cases are to be considered.
The first is when $$a$$ is greater than $$b$$; $$b$$ must then be subtracted from $$a$$, and the remainder (before which is placed, or understood to be placed, the sign +) shows the value sought.
The second case is that in which $$a$$ is less than $$b$$: here $$a$$ is to be subtracted from $$b$$, and the remainder being made negative, by placing before it the sign -, will be the value sought.
#### Editions
1. Leonhard Euler. Elements of Algebra. Translated by Rev. John Hewlett. Third Edition. Longmans, Hurst, Rees, Orme, and Co. London. 1822.
2. Leonhard Euler. Vollständige Anleitung zur Algebra. Mit den Zusätzen von Joseph Louis Lagrange. Herausgegeben von Heinrich Weber. B. G. Teubner. Leipzig and Berlin. 1911. Leonhardi Euleri Opera omnia. Series prima. Opera mathematica. Volumen primum.
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# Surface Area Of A Cylinder- Formula, Derivation, Volume, And Examples
In this blog, we will discuss the surface area of a cylinder. Do you know what a cylinder is? Yes, of course, water bottles, pens, jars, bamboo, etc are examples of the cylinder.
But I want to introduce the right circular cylinder. If we take a number of circular sheets of paper and stack them up what would we get? Here, if the stack is kept vertically up, we get what is called a right circular cylinder, since it has been kept at right angles to the base, and the base is circular.
Take a rectangular sheet. Let us roll it. It turns into a cylindrical shape. The line joining the center of the circle is called the axis of the cylinder. The area of the sheet gives us the curved surface area of the cylinder. Observe that the length of the sheet is equal to the circumference of the circular base which is equal to 2πr.
So, the curved surface area of the cylinder = area of the rectangular sheet = length × breadth
= perimeter of the base of the cylinder × h = 2πr × h = 2rπh, where r is the radius of the base of the cylinder and h is the height of the cylinder
Here we haven’t considered the top and the base of the cylinder. Look at this bottle a closed cylinder. Let us open it. On opening, we can observe that we have a rectangle and 2 circles. If the top and the bottom of the cylinder are also to be covered, then we need two circles (in fact, circular regions) to do that, each of radius r.
Thus having an area of πr2 each giving us the total surface area as curved surface area + area of 2 circles = 2πrh + 2πr2 = 2πr(r + h). So, Total Surface Area of a Cylinder = 2πr(r + h) where h is the height of the cylinder and r its radius.
Examples of Surface Area Of A Cylinder
Savitri had to make a model of a cylindrical kaleidoscope for her science
project. She wanted to use chart paper to make the curved surface of the kaleidoscope. What would be the area of chart paper required by her, if she wanted to make a kaleidoscope of length 25 cm with a 3.5 cm radius? You may take π= 22/7
Solution: Radius of the base of the cylindrical kaleidoscope (r) = 3.5 cm.
Height (length) of kaleidoscope (h) = 25 cm.
Area of chart paper required = curved surface area of the kaleidoscope
= 2πrh
= 2x 22/7 x3.5×25 cm square
= 550 cm square
Let’s understand the volume of cylinder. Just as a cuboid is built up with rectangles of the same size, a right circular cylinder can be built using circles of the same size. So, using the same argument as for a cuboid, we can see that the volume of a cylinder can be obtained as :
base area × height = area of circular base × height = πr2h
The volume of a Cylinder = πr2h, where r is the base radius and h is the height of the cylinder.
Volume of a Cylinder = πr2h
1. Let us take an example The pillars of a temple are cylindrically shaped. If each pillar has a circular base of a radius of 20 cm and a height of 10 m, how much concrete mixture would be required to build 14 such pillars?
Solution: Since the concrete mixture that is to be used to build up the pillars is going to occupy the entire space of the pillar, what we need to find here is the volume of the cylinders.
The radius of the base of a cylinder = 20 cm.
Height of the cylindrical pillar = 10 m = 1000 cm
So, volume of each cylinder = πr2h that is 22/7 into20 into 20 into 1000cm. On solving it gives 8800000cmcube or = 8.8/ 7m3
Therefore, the volume of 14 pillars = volume of each cylinder × 14
=8.8/7 into 14m cube
= 17.6 m cube
So, 14 pillars would need a 17.6 m cube of a concrete mixture.
2. At a Ramzan Mela, a stall keeper in one of the food stalls has a large cylindrical vessel of base radius 15 cm filled up to a height of 32 cm with orange juice. The juice is filled in small cylindrical glasses of radius 3 cm up to a height of 8 cm and sold for Rs 3 each. How much money does the stall keeper receive by selling the juice completely?
Solution: The volume of juice in the vessel = volume of the cylindrical vessel = πR2H (where R and H are taken as the radius and height respectively of the vessel). = π× 15 × 15 × 32 cm3
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# How many squares and rectangles are there in this picture | All Math Tricks
In this article discussed about formulas to find number of squares and rectangles in given figure of of ‘n’ number of rows and ‘m’ is the number of columns
## Count number of squares and number of rectangles in a given N x M Grid | How to count square in reasoning
### How many squares are there in the given figure
How many squares are there in the figure of same number of rows and columns
( i.e Number of squares in square grid )
Example – 1 :
How many squares are there in an 4 x 4 grid
Solution : There are 4 rows and 4 columns in the above figure. So let n =4
Here we using two types of formulas for finding number of squares in an n x n grid as follows
Formula – 1
n2 + (n -1 )2 + (n-2)2 + – – – – – + (n – n)2
Now substitute n = 4 in the above formula
= 42 + ( 4-1)2 + (4- 2 )2 + (4 – 3)2 + (4 – 4)2
= 16 + 9 + 4 + 1 + 0 = 30
Formula – 2
Apply the formula
Substitute n = 4 in above formula
= 4 x 5 x 9 / 6 = 30
So number of squares in an 4 x 4 grid is 30
Example -2
How many squares are there in an 5 x 5 grid
Solution : There are 5 rows and 5 columns in the above figure. Hence n=5
Formula
n2 + (n -1 )2 + (n-2)2 + – – – – – + (n – n)2
So according to above substitute n = 5
= 52 + ( 5-1)2 + (5- 2 )2 + (5 – 3)2 + (5 – 4)2+ (5 – 5)2
= 25 + 16 + 9 + 4 + 1 + 0 = 55
Number of squares in an 5 x 5 grid is 55
How many squares are there in the figure of’n’ number of rows and ‘m’ number of columns
( i.e Number of squares in rectangle grid )
Example – 3 How many squares are there in an 3 x 4 grid
Solution : There are 4 rows and 5 columns in the above figure.
Let number of rows ( n)= 4 & number of columns (m) = 5
Here we using simple formulas as follows
Formula- 1
( n x m ) + (n -1 ) (m – 1) + (n-2 ) ( m- 2) + – – – – – + (n – n ) or (m – m)
Now substitute n = 5 and m = 4
= ( 4 x 5) + ( 4 – 1) (5 – 1 ) + ( 4 – 2) (5 – 2 ) + ( 4 – 3) (5 – 3 ) + ( 4 – 4) (5 – 4 )
= 20 + 12 + 6 + 2 +0 = 40
Formula- 2
Here consider large value is n
i.e n = 5 and m = 4, Now substitute these values in above formulas
= [ 4 x 1 x 5 / 2 ] + [ 4 x 5 x 9 / 6 ]
= 10 + 30 = 40
Number of squares in given figure = 40
### How many rectangles are there in the given figure
Count number of rectangles in the figure of same number of rows and columns grid
( i.e Counting rectangles within a square )
Example – 4 How many rectangles are there in an 5 x 5 grid
Solution: There are 5 rows and 5 columns in the above figure.
Let number of rows or columns ( n)=5
Here for finding the rectangles there are having two methods
Formula – 1
[n + ( n -1 ) + ( n -2 ) + ( n -3 ) + – – – – – + ( n -n )]2
Now substitute the values of’ n‘ in the above formula
= [ 5 + (5 – 1 ) + (5 – 2 ) + (5 – 3 ) + (5 – 4 ) + (5 – 5) ]2
= [ 5 + 4 + 3+ 2 + 1 +0 ]2
= 15 x 15 = 225
Formula – 2
To count the number of rectangles within a square by using formula of
Now substitute the values of’ n‘ in the above formula
= [ 5 ( 5 + 1 ) /2 ] 2 =[ 5 x 6 / 2 ]2
= 15 2 = 225
Total number of rectangles in given figure = 225
Count number of rectangles in the figure of ‘n’ number of rows and ‘m’ number of columns
( i.e Counting rectangles within a rectangle )
Example – 5 How many rectangles are there in an 4 x 5 grid
Solution: There are 4 rows and 5 columns in the above figure.
Let number of rows ( n)=4 & number of columns (m) = 5
Here for finding the rectangles there having two methods
Formula – 1 ( Formula for finding number of rectangles in figure of ‘n’ number of rows and ‘m’ is the number of columns)
[n + ( n -1 ) + ( n -2 ) + ( n -3 ) + – – – – – + ( n -n )] x [m + ( m -1 ) + (m -2 ) + (m-3 ) + – – – – – + ( m -m )]
Now substitute the values of’ n‘ and ‘ m‘ in the above formula
= [ 4 + (4 – 1 ) + (4 – 2 ) + (4 – 3 ) + (4 – 4 )] x [ 5 + (5 – 1 ) + (5 – 2 ) + (5 – 3 ) + (5 – 4 ) + (5 – 5) ]
= [ 4 +3 +2 +1 +0 ] x [ 5 + 4 + 3+ 2 + 1 +0 ]
= 10 x 15 = 150
Formula – 2
Now substitute the values of’ n‘ and ‘ m‘ in the above formula
= 4 x 5 x 5 x 6 / 4 = 150
Total number of rectangles in given figure = 150
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## 2 thoughts on “How many squares and rectangles are there in this picture | All Math Tricks”
#### arun
(December 13, 2020 - 8:04 am)
good but very difficult
#### Deepak
(July 25, 2022 - 10:11 am)
Very nice
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Difference between revisions of "2020 AMC 12A Problems/Problem 17"
Problem 17
The vertices of a quadrilateral lie on the graph of $y=\ln{x}$, and the $x$-coordinates of these vertices are consecutive positive integers. The area of the quadrilateral is $\ln{\frac{91}{90}}$. What is the $x$-coordinate of the leftmost vertex?
$\textbf{(A) } 6 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 13$
Solution 1
Let the left-most $x$-coordinate be $n.$
Recall that, by the shoelace formula, the area of the quadrilateral must be $-\ln{(n)}+\ln{(n+1)}+\ln{(n+2)}-\ln{(n+3)}.$ That equals to $\ln\frac{(n+1)(n+2)}{n(n+3)}.$
$\ln\frac{(n+1)(n+2)}{n(n+3)} = \ln\frac{n^{2}+3n+2}{n^{2}+3n}$
$\ln\frac{n^{2}+3n+2}{n^{2}+3n} = \ln\frac{91}{90}$
$\ln\frac{n^{2}+3n+2}{n^{2}+3n} = \ln\frac{182}{180}$
$n^{2}+3n = 180$
$n^{2}+3n-180 = 0$
$(n-12)(n+15) = 0$
The $x$-coordinate is, therefore, $\boxed{\textbf{(D) } 12.}$~lopkiloinm.
Solution 2
Like above, use the shoelace formula to find that the area of the triangle is equal to $\ln\frac{(n+1)(n+2)}{n(n+3)}$. Because the final area we are looking for is $\ln\frac{91}{90}$, the numerator factors into $13$ and $7$, which one of $n+1$ and $n+2$ has to be a multiple of $13$ and the other has to be a multiple of $7$. Clearly, the only choice for that is $\boxed{12}$
~Solution by IronicNinja
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• Level: GCSE
• Subject: Maths
• Word count: 1850
# Investigate the relationship between the T-total and the T-number
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Introduction
Maths Coursework
Aim: Investigate the relationship between the T-total and the T-number. Use grids of different sizes and translate the T-shape to different positions. Investigate relationships between the T-total and the T-number and the grid size.
Method:In order to find this rule we will have to go through a trial and error process. To secure an algebraic rule expressed in the nth term we will have to test it in many variables. These variables consist of random placement of the T-shapes; another is the shape being placed at different degrees. This will provide us with numerous results, which will help us, gain a conclusive overall algebraic term.
9x9 Number grid (Across)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81
T-Number T-Total Difference 20 1+2+3+11+20=37 - 21 2+3+4+12+21=42 5 22 3+4+5+13+22=47 5 23 4+5+6+14+23=52 5 24 5+6+7+15+24=57 5 25 6+7+8+16+25=62 5
As the T-Shape translates across by 1, the T-Number increases by 1. The difference in T-Total is 5 each time so far. Therefore the T-Total for the T-Number 24 should be 57 if the pattern follows. The T-Total for the T-Number 25 should be 62.
The results show the predictions made earlier are correct proving that the pattern follows.
9x9 Square Grid Patterns down
T-Number T-Total Difference 20 1+2+3+11+20=37 - 29 10+11+12+20+29= 45 38 19+20+21+29+38= 45 47 28+29+30+38+47= 45 56 37+38+39+47+56=217 45 65 40+41+42+50+59=262 45
I have noticed patterns within my table of results.
Middle
T-Total=41+42+43+51+60=237
T-Number=60
As the other example shows, my formula is correct as the same T-Total was found for each T-Number when using both methods.
8x8 Number grid (Across)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64
T-Number T-Total Difference 18 1+2+3+10+18= - 19 2+3+4+11+19= 5 20 3+4+5+12+20= 5 21 4+5+6+13+21= 5 22 5+6+7+14+22=54 5 23 6+7+8+15+23=59 5
As the T-Shape translates across by 1, the T-Number increases by 1. The difference in T-Total is 5 each time so far. Therefore the T-Total for the T-Number 22 should be 54 if the pattern follows. The T-Total for the T-Number 23 should be 59.
The results show the predictions made earlier are correct proving that the pattern follows
8x8 Square Grid patterns down
T-Number T-Total Difference 18 1+2+3+10+18=34 - 26 9+10+11+18+26=74 40 34 17+18+19+26+34=114 40 42 25+26+27+34+42=154 40 50 33+34+35+42+50=194 40 58 41+42+43+50+59=234 40
I have noticed patterns within my table of results. One of the patterns I have noticed is, as the T-Shape translates down by 1, the T-Number increases by 8, which is also the size of the grid. Another noticeable pattern is the T-Total difference is 40 each time.
To prove that the patterns work anywhere within the grid I am testing it at a different point within the grid. The T-numbers I will test it with is 50 and 58. If my patterns are work at any point in the grid, than the difference between both results should be 45.
Conclusion
The results above of the predictions made earlier as to what the T-Total should be if the
T-Number was 62 and 72 prove that the pattern follows anywhere within the grid.
The values inside the T of any 10x10 grid can be expressed in an algebraic way. This is as follows with n being the value of the T-Number.
Algebraic expression for a 10x10 grid:
Using the algebraic expressions a formula can be found so the
T-Total for any T-Number inside an 10x10 grid could be found straight away. The rule is:
T-Total = n+(n-10)+(n-21)+(n-20)+(n-19)
This can be simplified to:
T-total = 5n – 70
For example if n=82:
T-Total=(5x82)-70
T-Total=410-70
T-Total=340
To see if the formula is correct for this I will find the T-Total using the original method.
61 62 63 72 82
T-Total=82+72+61+62+63=340
T-Number=82
To establish that this expression is correct I will present further examples to show that it is correct
If n=92:
T-Total=(5x92)-70
T-Total=460-70
T-Total=390
71 72 73 82 92
T-Total=71+72+73+82+92=390
T-Number=92
As the other example shows, my formula is correct as the same T-Total was found for each T-Number when using both methods.
9x9 Grid Rotations:
This student written piece of work is one of many that can be found in our GCSE T-Total section.
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# Related GCSE T-Total essays
1. ## T Total and T Number Coursework
There will be three separate formulas for +90 degrees, -90 degrees and 180 degrees. I will use a 9x9 grid for this part of the investigation. The shape on the left is the original and the new shape is on the right.
2. ## To prove that out of town shopping is becoming increasingly popular with shoppers, and ...
For example, parking. A family with children visits the town centre to do some shopping and they are forced either to spend money to park at a car park, which can take a long time itself when looking for free spaces, or spend half an hour or so trying to
1. ## Maths GCSE Coursework &amp;#150; T-Total
17 18 19 20 21 22 23 24 25 26 27 28 29 30 T = (5x16) - (2 x 6) T = 80 - 12 T = 68 In addition, the T-Total by hand is; T = 9 + 10 + 11 + 16 + 22 T = 68
2. ## T-totals. I am going to investigate the relationship between the t-total, T, and ...
} -7 = 258 The t-total for the rotated t-shape agrees with the results from the formula. We can therefore say that our formulae are consistent. Combined Transformations Rotation followed by translation n+c-dg-d-cg+a-bg+2-g a b n+c-dg-d-cg+a-bg n+c-dg-d-cg+a-bg+1 n+c-dg-d-cg+a-bg+2 n+c-dg-d-cg+a-bg+2+g n-2g-1 n-2g n-2g+1 centre of rotation n-g n The magnitude and
1. ## Objectives Investigate the relationship between ...
11 12 13 19 20 21 22 28 29 30 31 * T21 2 3 4 11 12 13 20 21 22 2+3+4+12+21=42 * T19 270� Rotation 10 11 12 19 20 21 28 29 30 10+19+28+20+21=98 T-shape T-total Increment T19 42 T19 (270�)
2. ## T-Shapes Coursework
13 65 99 164 14 70 102 172 15 75 105 180 16 80 108 188 17 85 111 196 c) Here are the results of the 5 calculations for a 7x4 "T" on Width 11 Grid: Middle Number Sum of Wing Sum of Tail Total Sum (Wing + Tail)
1. ## In this section there is an investigation between the t-total and the t-number.
51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 Red t-shape 5tn- (7*G)+7= t-total 5*41 - 63+7 = 149# Blue t-shape 5tn- (7*G)
2. ## For my investigation, I will be investigating if there is a relationship between t-total ...
+ (N+1) + (N-4) + (N+2) + (N+8) This can be simplified to: = 5N + 7 I then tested this formula on another shape in the 6 by 6 table, using the numbers 8, 9, 10, 4 and 16. 4 8 9 10 16 The calculated T-Total for this shape is 32, with the T-Number being 8.
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2 Vector Algebra
2 Vector Algebra - 46 2 VECTOR ALGEBRA 2.1 Introduction We...
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Unformatted text preview: 46 2. VECTOR ALGEBRA 2.1 Introduction : We have already studied two operations ‘addition’ and ‘subtraction’ on vectors in class XI. In this chapter we will study the notion of another operation, namely product of two vectors. The product of two vectors results in two different ways, viz., a scalar product and a vector product. Before defining these products we shall define the angle between two vectors. 2.2 Angle between two vectors : Let two vectors a → and b → be represented by OA → and OB → respectively. Then the angle between a → and b → is the angle between their directions when these directions both converge or both diverge from their point of intersection. Fig. 2. 1 Fig. 2. 2 It is evident that if θ is the numerical measure of the angle between two vectors, then 0 ≤ θ ≤ π . 2.3 The Scalar product or Dot product Let a → and b → be two non zero vectors inclined at an angle θ . Then the scalar product of a → and b → is denoted by a → . b → and is defined as the scalar | | a → | | b → cos θ . Thus a → . b → = | | a → | | b → cos θ = ab cos θ Note : Clearly the scalar product of two vectors is a scalar quantity. Therefore the product is called scalar product. Since we are putting dot between a → and b → , it is also called dot product. O A B a b O A B a b a b θ a b θ 47 Geometrical Interpretation of Scalar Product Let OA → = a → , OB → = b → Let θ be the angle between a → and b → . From B draw BL ⊥ r to OA . OL is called the projection of b → on a → . From ∆ OLB , cos θ = OL OB Fig. 2.3 ⇒ OL = ( OB ) (cos θ ) ⇒ OL = | | b → (cos θ ) … (1) Now by definition a → . b → = | | a → | | b → cos θ = | | a → ( OL ) [ ‡ using (1)] ∴ a → . b → = | | a → [ ] projection of b → on a → Projection of b → on a → = a → . b → | | a → = a → | | a → . b → = a ∧ . b → Projection of a → on b → = a → . b → | | b → = a → . b → | | b → = a → . b ∧ 2.3.1 Properties of Scalar Product : Property 1 : The scalar product of two vectors is commutative (i.e.,) a → . b → = b → . a → for any two vectors a → and b → Proof : Let a → and b → be two vectors and θ the angle between them. a → . b → = | | a → | | b → cos θ … (1) ∴ b → . a → = | | b → | | a → cos θ a b θ O L A B a b θ O L A B 48 b → . a → = | | a → | | b → cos θ … (2) From (1) and (2) a → . b → = b → . a → Thus dot product is commutative. Property 2 : Scalar Product of Collinear Vectors : (i) When the vectors a → and b → are collinear and are in the same direction, then θ = 0 Thus a → . b → = | | a → | | b → cos θ = | | a → | | b → (1) = ab … (1) (ii) When the vectors a → and b → are collinear and are in the opposite direction, then θ = π Thus a → . b → = | | a → | | b → cos θ = | | a → | | b → (cos π ) … (1) = | | a → | | b → ( − 1) = − ab Property 3 : Sign of Dot Product...
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# NCERT solutions for Class 9 Maths chapter 7 - Triangles [Latest edition]
## Chapter 7: Triangles
Exercise 7.1Exercise 7.2Exercise 7.3Exercise 7.4Exercise 7.5
Exercise 7.1 [Pages 118 - 120]
### NCERT solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.1 [Pages 118 - 120]
Exercise 7.1 | Q 1 | Page 118
In quadrilateral ACBD, AC = AD and AB bisects ∠A (See the given figure). Show that ΔABC ≅ ΔABD. What can you say about BC and BD?
Exercise 7.1 | Q 2 | Page 119
ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (See the given figure). Prove that
(i) ΔABD ≅ ΔBAC
(ii) BD = AC
(iii) ∠ABD = ∠BAC.
Exercise 7.1 | Q 3 | Page 119
AD and BC are equal perpendiculars to a line segment AB (See the given figure). Show that CD bisects AB.
Exercise 7.1 | Q 4 | Page 119
l and m are two parallel lines intersected by another pair of parallel lines p and q (see the given figure). Show that ΔABC ≅ ΔCDA.
Exercise 7.1 | Q 5 | Page 119
Line l is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A (see the given figure). Show that:-
(i) ΔAPB ≅ ΔAQB
(ii) BP = BQ or B is equidistant from the arms of ∠A.
Exercise 7.1 | Q 6 | Page 120
In the given figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.
Exercise 7.1 | Q 7 | Page 120
AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB (See the given figure). Show that
(i) ΔDAP ≅ ΔEBP
Exercise 7.1 | Q 8 | Page 120
In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see the given figure). Show that:
(i) ΔAMC ≅ ΔBMD
(ii) ∠DBC is a right angle.
(iii) ΔDBC ≅ ΔACB
(iv) CM = 1/2AB
Exercise 7.2 [Pages 123 - 124]
### NCERT solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.2 [Pages 123 - 124]
Exercise 7.2 | Q 1 | Page 123
In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O. Show that: (i) OB = OC (ii) AO bisects ∠A
Exercise 7.2 | Q 2 | Page 123
In ΔABC, AD is the perpendicular bisector of BC (see the given figure). Show that ΔABC is an isosceles triangle in which AB = AC.
Exercise 7.2 | Q 3 | Page 124
ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see the given figure). Show that these altitudes are equal.
Exercise 7.2 | Q 4 | Page 124
ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see the given figure). Show that
(i) ΔABE ≅ ΔACF
(ii) AB = AC, i.e., ABC is an isosceles triangle.
Exercise 7.2 | Q 5 | Page 124
ABC and DBC are two isosceles triangles on the same base BC (see the given figure). Show that ∠ABD = ∠ACD.
Exercise 7.2 | Q 6 | Page 124
ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see the given figure). Show that ∠BCD is a right angle.
Exercise 7.2 | Q 7 | Page 124
ABC is a right angled triangle in which ∠A = 90º and AB = AC. Find ∠B and ∠C.
Exercise 7.2 | Q 8 | Page 124
Show that the angles of an equilateral triangle are 60º each.
Exercise 7.3 [Page 128]
### NCERT solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.3 [Page 128]
Exercise 7.3 | Q 1 | Page 128
ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see the given figure). If AD is extended to intersect BC at P, show that
(i) ΔABD ≅ ΔACD
(ii) ΔABP ≅ ΔACP
(iii) AP bisects ∠A as well as ∠D.
(iv) AP is the perpendicular bisector of BC.
Exercise 7.3 | Q 2 | Page 128
AD is an altitude of an isosceles triangles ABC in which AB = AC. Show that
Exercise 7.3 | Q 3 | Page 128
Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ΔPQR (see the given figure). Show that:
(i) ΔABM ≅ ΔPQN
(ii) ΔABC ≅ ΔPQR
Exercise 7.3 | Q 4 | Page 128
BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
Exercise 7.3 | Q 5 | Page 128
ABC is an isosceles triangle with AB = AC. Drawn AP ⊥ BC to show that ∠B = ∠C.
Exercise 7.4 [Pages 132 - 133]
### NCERT solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.4 [Pages 132 - 133]
Exercise 7.4 | Q 1 | Page 132
Show that in a right angled triangle, the hypotenuse is the longest side.
Exercise 7.4 | Q 2 | Page 132
In the given figure sides AB and AC of ΔABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.
Exercise 7.4 | Q 3 | Page 132
In the given figure, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.
Exercise 7.4 | Q 4 | Page 132
AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see the given figure). Show that ∠A > ∠C and ∠B > ∠D.
Exercise 7.4 | Q 5 | Page 132
In the given figure, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR >∠PSQ.
Exercise 7.4 | Q 6 | Page 133
Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
Exercise 7.5 [Page 133]
### NCERT solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.5 [Page 133]
Exercise 7.5 | Q 1 | Page 133
ABC is a triangle. Locate a point in the interior of ΔABC which is equidistant from all the vertices of ΔABC.
Exercise 7.5 | Q 2 | Page 133
In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.
Exercise 7.5 | Q 3 | Page 133
In a huge park people are concentrated at three points (see the given figure):
A: where there are different slides and swings for children,
B: near which a man-made lake is situated,
C: which is near to a large parking and exit.
Where should an ice-cream parlour be set up so that maximum number of persons can approach it?
(Hint: The parlor should be equidistant from A, B and C)
Exercise 7.5 | Q 4 | Page 133
Complete the hexagonal and star shaped rangolies (see the given figures) by filling them with as many equilateral triangles of side 1 cm as you can. Count the number of triangles in each case. Which has more triangles?
## Chapter 7: Triangles
Exercise 7.1Exercise 7.2Exercise 7.3Exercise 7.4Exercise 7.5
## NCERT solutions for Class 9 Maths chapter 7 - Triangles
NCERT solutions for Class 9 Maths chapter 7 (Triangles) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CBSE Class 9 Maths solutions in a manner that help students grasp basic concepts better and faster.
Further, we at Shaalaa.com provide such solutions so that students can prepare for written exams. NCERT textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.
Concepts covered in Class 9 Maths chapter 7 Triangles are Concept of Triangles - Sides, Angles, Vertices, Interior and Exterior of Triangle, Properties of a Triangle, Some More Criteria for Congruence of Triangles, Inequalities in a Triangle, Criteria for Congruence of Triangles, Congruence of Triangles.
Using NCERT Class 9 solutions Triangles exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in NCERT Solutions are important questions that can be asked in the final exam. Maximum students of CBSE Class 9 prefer NCERT Textbook Solutions to score more in exam.
Get the free view of chapter 7 Triangles Class 9 extra questions for Class 9 Maths and can use Shaalaa.com to keep it handy for your exam preparation
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# Physics Snap
## A very large sheet of a conductor carries a uniform charge density of 4.00 pC/mm2 on its surfaces. What is the electric field strength 3.00 mm outside the surface of the conductor? (8.85 x 10-12 C2/N m²) Select one: a. 0.226 N/C 5 b. 4.52x10 N/C 5 c. 2.26x10 N/C d. 0.452 N/C •
#### STEP 1
List of assumptions necessary for solving the problem: 1. The conductor sheet is infinite in extent. 2. The charge density is uniform across the surface of the conductor. 3. The electric field due to an infinite sheet of charge is perpendicular to the surface and has the same magnitude at all points at the same distance from the sheet. 4. The electric field inside the conductor is zero.
#### STEP 2
The strategy to solve for the electric field strength outside the surface of the conductor involves using Gauss's law for an infinite sheet of charge. The relevant equation derived from Gauss's law for an infinite sheet of charge is: $E = \frac{\sigma}{2\epsilon_0}$ where $E$ is the electric field strength, $\sigma$ is the surface charge density, and $\epsilon_0$ is the permittivity of free space.
#### STEP 3
First, we need to convert the given surface charge density from pC/mm² to C/m² to use in the formula. The conversion factors are $1 \text{ pC} = 10^{-12} \text{ C}$ and $1 \text{ mm}^2 = 10^{-6} \text{ m}^2$.
#### STEP 4
Perform the conversion for the charge density: $\sigma = 4.00 \text{ pC/mm}^2 = 4.00 \times 10^{-12} \text{ C/mm}^2 \times \frac{10^6 \text{ mm}^2}{1 \text{ m}^2} = 4.00 \times 10^{-6} \text{ C/m}^2$
#### STEP 5
Now, we can use the formula for the electric field strength due to an infinite sheet of charge: $E = \frac{\sigma}{2\epsilon_0}$
#### STEP 6
Substitute the given value for the permittivity of free space, $\epsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/\text{N m}^2$, and the converted charge density into the formula: $E = \frac{4.00 \times 10^{-6} \text{ C/m}^2}{2 \times 8.85 \times 10^{-12} \text{ C}^2/\text{N m}^2}$
#### STEP 7
Perform the calculation to find the electric field strength: $E = \frac{4.00 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} \text{ N/C}$
#### STEP 8
Calculate the numerator and denominator separately: $\text{Numerator} = 4.00 \times 10^{-6}$ $\text{Denominator} = 2 \times 8.85 \times 10^{-12} = 1.77 \times 10^{-11}$
#### STEP 9
Divide the numerator by the denominator to get the electric field strength: $E = \frac{4.00 \times 10^{-6}}{1.77 \times 10^{-11}} \text{ N/C}$
#### STEP 10
Perform the division to find the electric field strength: $E = \frac{4.00}{1.77} \times 10^{5} \text{ N/C}$
##### SOLUTION
Calculate the value: $E = 2.26 \times 10^{5} \text{ N/C}$ The electric field strength 3.00 mm outside the surface of the conductor is $2.26 \times 10^{5} \text{ N/C}$, which corresponds to option c.
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# Understanding Fractals, Part 3
On Monday we covered the real, imaginary, and complex numbers. Yesterday we went over the complex plane. Only one thing left…
Constructing the Mandelbrot Set
Let’s play a game.
You think of any number. I’ll take that number, square it, and add the original number. Then I’ll take that number, square it, and add the original number. And so on.
Let’s try it with the number 2:
It’s easy to see that the numbers will just keep getting higher and higher forever.
But what if we played the game with a different number? Let’s say, -1.
No matter how long we play this game, we’ll never get anything besides -1 and 0.
This “game” is called an iterative algorithm – “iterative” meaning “doing the same thing over and over,” and “algorithm” meaning “a series of well-defined steps.” The formal mathematical definition for our game would be:
That is, the next number in the game equals the current number squared, plus our original number.
But enough formality. Let’s get back to the game.
We’ve only played this game twice, but already we can see there are two different classes of numbers. Some numbers, like 2, shoot off to infinity. We’ll call those the “Superman Numbers,” since they are up, up, and away. Other numbers, like -1, just circle around in one spot no matter how long we wait. We’ll call those the “Tornado Numbers.”
So far, though, we’ve only played the game with real numbers. Let’s try it with i.
By the way, if you want a quick way to do these calculations, try Google. The main search bar actually does all kinds of calculations, even with complex numbers. For instance, to figure out Step 3, you’d just type in the following, and hit Enter: (i – 1)^2 + i
Try it!
Anyway – we can see that i is another Tornado Number. Let’s try just one more example, with another complex number, i + 1. And by the way, I’m totally cheating and using Google for these calculations.
Actually, when I tried this just now, Google got confused on Step 4 and, for some unfathomable reason, tried to correct my spelling (and refused to do the calculation). Very helpful. I did it by hand, but I’m pretty sure it’s right.
Anyway, i + 1 is another Superman Number.
Let’s take the four numbers we’ve done so far and plot them on the complex plane. We’ll put the Superman Numbers in blue, and the Tornado Numbers in green. (Click image to enlarge.)
Suppose we got a computer to play this game, over and over, for millions of points on the complex plane, still plotting the Tornado Numbers in green and the Superman Numbers in blue. What would that look like?
By now, you can probably guess…
Ta-da! The Mandelbrot Set!
I’m out of time, so I’ll stop here for today. Let me know if you have any questions. Tomorrow we’ll dive a little further into this monster we’ve just created!
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# 13.3 - Smooth-by-Eye Method
13.3 - Smooth-by-Eye Method
## Motivation of the Smooth-by-Eye Method
Recall that in 17.2, one main limitation of the Narrow-Strip Method is that it assumes the detectability of objects is perfect within the strip. But, in reality, the detectability decreases as the distance to the transect line increases.
In this section, the main idea of the Smooth-by-Eye Method is to use the histogram for distance x from the transect line to approximate the density of the detectability function f (x). We will approximate f (x) as a decreasing function, which agrees with the situation in real life.
## How to Approximate the Detectability Function f (x)
The first step is to construct a histogram for the distance x from the transect line.
The formula for the height of the histogram for a given distance x is
$$\hat{f}(x)=\dfrac{y_x}{y \times w_x}$$
• $$y_x$$ - number of observations in the interval containing x
• $$y$$ - total number of observations
• $$w_x$$ - interval width
## Example 13-2:
( Reference: Section 17.2 page 231 of the text)
y = 18 (total number of observations)
• 5 birds were seen within 10 meters
• 7 birds were seen between 10 and 20 meters
Choose the interval width of the histogram to be 10 meters.
Thus,
• the height for the 1st interval is $$5/[18(10)] = 0.0028$$
• the height for the 2nd interval is $$7/[18(10)] = 0.039$$
Similarly,
• the height for the 3rd interval is 0.017
• the height for the 4th interval is 0.011
• the height for the 5th interval is 0.006
Knowing the heights of each interval, we got the following histogram:
Notice that the smooth-by-eye curve intersects the vertical axis at 0.048
Notice from the histogram that the detectability actually increases a little bit in the second interval compared with that in the first interval. However, the true density of detectability decreases smoothly with distance. We regard the increase in the irregularities in the histogram due to random chance and the small number of observations.
We still fit a smooth, decreasing curve irrespective of the irregularity. The curve is fitted by eyeballing and each person may have a slightly different fit.
According to the “height of histogram” formula given before, we know:
$$f_0=\dfrac{y_0}{y \times w_0}$$
Combining with the formula in 17.2:
$$\hat{D}=\dfrac{y_0}{2w_{0}L}$$
We got a new formula to estimate the density:
$$\hat{D}=\dfrac{\hat{f}(0) \times y}{2L}$$
$$\hat{f}(0)=0.048$$ (the smooth-by-eye curve intersects the vertical axis at 0.048)
Thus,
$$\hat{D}=\dfrac{\hat{f}(0) \times y}{2L}=\dfrac{0.048 \times 18}{2 \times 100}=0.00432$$
So, the estimated bird population density is 0.00432 birds per square meter or 43.2 birds per hectare.
[1] Link ↥ Has Tooltip/Popover Toggleable Visibility
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+0
# need help!
0
262
1
+4
The perimeter of the rectangle is 596 yards. The length of it is 2 yards less than quadruple the width. What are the dimensions of the rectangle?
This is all the info given
Apr 11, 2018
edited by herrstein1010 Apr 11, 2018
#1
+7612
+1
Let's call the length of the rectangle, in yards, L .
Let's call the width of the rectangle, in yards, W .
The perimeter of the rectangle is 596 yards, so....
L + L + W + W = 596
Let's solve this equation for L . Comine like terms.
2L + 2W = 596
Subtract 2W from both sides of the equation.
2L = 596 - 2W
Divide through by 2 .
L = 298 - W
Its length is 2 yards less than 4 times its width, so...
L = 4W - 2
Since L = 298 - W we can replace L with 298 - W .
298 - W = 4W - 2
298 + 2 = 4W + W
Combine like terms.
300 = 5W
Divide both sides by 5 .
60 = W
Now we can use this value for the width to find the length.
L = 298 - W
L = 298 - 60
L = 238
So the width is 60 yards and the length is 238 yards.
Apr 11, 2018
#1
+7612
+1
Let's call the length of the rectangle, in yards, L .
Let's call the width of the rectangle, in yards, W .
The perimeter of the rectangle is 596 yards, so....
L + L + W + W = 596
Let's solve this equation for L . Comine like terms.
2L + 2W = 596
Subtract 2W from both sides of the equation.
2L = 596 - 2W
Divide through by 2 .
L = 298 - W
Its length is 2 yards less than 4 times its width, so...
L = 4W - 2
Since L = 298 - W we can replace L with 298 - W .
298 - W = 4W - 2
298 + 2 = 4W + W
Combine like terms.
300 = 5W
Divide both sides by 5 .
60 = W
Now we can use this value for the width to find the length.
L = 298 - W
L = 298 - 60
L = 238
So the width is 60 yards and the length is 238 yards.
hectictar Apr 11, 2018
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Question Video: Determining Whether a Triangle of Given Side Lengths Can Exist | Nagwa Question Video: Determining Whether a Triangle of Given Side Lengths Can Exist | Nagwa
# Question Video: Determining Whether a Triangle of Given Side Lengths Can Exist Mathematics • Second Year of Preparatory School
## Join Nagwa Classes
Does the triangle with side lengths 8, 32, and 16 exist?
01:47
### Video Transcript
Does the triangle with side lengths eight, 32, and 16 exist?
To answer this question, letโs recall the triangle inequality. The triangle inequality helps us to determine whether given lengths can be sides of a triangle. And it states, โThe sum of the lengths of any two sides of a triangle must be greater than the length of the third side.โ This means that if a triangle ๐ด๐ต๐ถ exists, then ๐ด๐ต plus ๐ต๐ถ is greater than ๐ด๐ถ, ๐ด๐ต plus ๐ด๐ถ is greater than ๐ต๐ถ, and ๐ต๐ถ plus ๐ด๐ถ is greater than ๐ด๐ต. For a triangle with side lengths eight, 32, and 16 to exist, those numbers must make all three triangle inequalities true.
First, we will ask ourselves if eight plus 32 is greater than 16, and that is true. Moving on to the second inequality, we ask ourselves if eight plus 16 is greater than 32. This statement is false because 24 is not greater than 32. As soon as we find one of the triangle inequalities is not true, we know it is not possible to form a triangle. So thereโs no need to check the third inequality.
Because it was not possible to verify all three triangle inequalities were true, we can say, โNo, a triangle with side lengths eight, 32, and 16 does not exist.โ
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## Want to keep learning?
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4.2
## UNSW Sydney
One polynomial, seven zeroes
# Factors of quadratic polynomials and zeroes
Quadratic functions are examples of polynomials, which have a pleasant arithmetic much like that of numbers. The question of factoring polynomials is particularly interesting. Descartes found an important relation between the zeroes of a polynomial, and linear factors of that polynomial.
In this step, we will
• review basics of polynomial arithmetic
• relate zeroes and factors of quadratics via Descartes theorem
• give a proof of Descartes theorem.
## Polynomials and their arithmetic
A polynomial is a general algebraic expression made from powers of $$\normalsize{x}$$ mixed with arbitrary coefficients, such as
$\normalsize{p(x)=x^2-4x+3} \quad\text{ or } \quad \normalsize{q(x)=x+2}.$
The degree of a polynomial is the highest power of $$\normalsize{x}$$ that appears, so that $$\normalsize{p}$$ has degree $$\normalsize{2}$$ and $$\normalsize{q}$$ has degree $$\normalsize{1}$$. The coefficients of $$\normalsize{p}$$ are $$\normalsize{1,-4}$$ and $$\normalsize{3}$$ corresponding to $$\normalsize{x^2,x}$$ and $$\normalsize{1}$$ respectively.
There is an arithmetic with polynomials: addition or subtraction is just addition or subtraction of the corresponding coefficients. For example
$\Large{(p+q)(x)=x^2-3x+5}$ $\Large{(p-q)(x)=x^2-5x+1}.$
Multiplication by a number is also performed on all of the coefficients independently
$\Large{(2p)(x) = 2(x^2 - 4x + 3) = 2x^2-8x+6}.$
Multiplication by $$\normalsize{x}$$ increases the exponent of all the $$\normalsize{x}$$’s by one.
$\Large{(xp)(x) = x(x^2 - 4x + 3) = x^3-4x^2+3x}.$
Finally, multiplication by a polynomial combines these operations.
\Large{\begin{align*} (qp)(x) & =(x+2)p(x) \\ & = (xp)(x) + (2p)(x) \\ & = (x^3-4x^2+3x) + (2x^2-8x+6) \\ & = x^3-2x^2-5x+6.\end{align*}}
Q1 (E): If $$\normalsize{r(x)=x+3}$$ and $$\normalsize{s(x)=2x-1}$$ then determine the polynomials i) $$\normalsize{r+s}$$ ii) $$\normalsize{r-s}$$ and iii) $$\normalsize{rs}$$.
Q2 (E): If $$\normalsize{u(x)=x^2+x+1}$$ and $$\normalsize{v(x)=x^3-1}$$ then determine the polynomials i) $$\normalsize{u+v}$$ ii) $$\normalsize{u-v}$$ and iii) $$\normalsize{uv}$$.
## Factoring polynomials
Sometimes polynomials can also be divided: just like numbers, this occurs most smoothly when one polynomial is a multiple of the other.
For example you will all know the factorization
$\Large{x^2-1=(x+1)(x-1)}.$
In this case we say that $$\normalsize{x^2-1}$$ is a multiple of $$\normalsize{x+1}$$, and that equivalently $$\normalsize{x+1}$$ is a factor of $$\normalsize{x^2-1}$$.
Natural numbers factor uniquely (up to rearrangement) into primes, for example
$\Large{1428=2\times2\times3\times7\times17}.$
So too a polynomial factors uniquely (up to rearrangement) into irreducible polynomials that do not have proper factors (factors aside from the trivial factors of $$1$$ and the polynomial itself). If we consider only polynomials with real coefficients, then the irreducible factos can be linear or quadratic.
Given any two linear polynomials $$\normalsize{u(x)=ax+b}$$ and $$\normalsize{v(x)=cx+d}$$ their product is the quadratic polynomial
$\Large{(uv)(x)=(ax+b)(cx+d)=acx^2+(ad+bc)x+bd.}$
Please check that you understand completely how the arithmetic works here! A rather deep and interesting problem is the reverse: given a quadratic polynomial, how can you write it as a product of linear factors.
Q3 (M): I have multiplied two linear factors together to get the quadratic polynomial
$\Large{30x^2-28x-240}.$
What were my two linear polynomials?
## Factoring challenge
Please take two linear factors, and multiply them, and challenge your fellow participants to find the factors. And also try to find the factors of at least one of your fellow course mates quadratic.
## Zeroes of a quadratic polynomial
If $$\normalsize{p(x)=x^2-12x+35}$$ is a polynomial, then a zero of $$\normalsize{p(x)}$$ is a number $$\normalsize{r}$$ with the property that $$\normalsize{p(r)=0}$$. In other words, we are solving the equation $$\normalsize{r^2-12r+35=0}$$. The basic technique for solving this kind of quadratic equation goes back to the ancient Hindus.
If we look at a table of values for $$\normalsize{p(x)}$$, we can find such zeros:
$$\normalsize{x}$$ $$\normalsize{p(x)}$$
-3 80
-2 63
-1 48
0 35
1 24
2 15
3 8
4 3
5 0
6 -1
7 0
8 3
9 8
10 15
We see that the zeroes of $$\normalsize{p(x)}$$ are $$\normalsize{5}$$ and $$\normalsize{7}$$. However we have to consider ourselves lucky here: not every quadratic has zeroes that can be found so easily.
We could also find the zeroes graphically, at least approximately, by plotting the corresponding polynomial function
$\Large{y=p(x)=x^2-12x+35}:$
## The connection with factors
We say that the polynomial $$\normalsize{p(x)}$$ has a linear factor $$\normalsize{(x-r)}$$ when $$\normalsize{p(x)=(x-r)q(x)}$$ for some other polynomial $$\normalsize{q(x)}$$. Descartes realized that the zeroes of $$\normalsize{p(x)}$$ were intimately connected with its linear factors. Note that we can write
$\Large{x^2-12x+35=(x-5)(x-7)}$
and this explains why this polynomial has zeroes $$\normalsize{5}$$ and $$\normalsize{7}$$.
Q4 (E): What are the linear factors of $$\normalsize{x^2+8x+15}$$?
Now we come to a major theoretical insight of Descartes.
Theorem (Descartes’ Factor Theorem) If $$\normalsize{p(x)}$$ is a polynomial, then $$\normalsize{p(x)}$$ has a zero $$\normalsize{r}$$ precisely when $$\normalsize{(x-r)}$$ is a factor of $$\normalsize{p(x)}$$.
Note that we are using the term ‘precisely when’ to mean that there is a two-way implication here:
• if $$\normalsize{p(x)}$$ has a zero $$\normalsize{r}$$ then $$\normalsize{(x-r)}$$ is a factor, and
• if $$\normalsize{(x-r)}$$ is a factor then $$\normalsize{p(x)}$$ has a zero $$\normalsize{r}$$.
We are getting two statements for the price of one!
## Examples with Descartes’ Factor Theorem
To illustrate the Theorem, let’s look at some examples. The polynomial $$\normalsize{x^2-12x+35}$$ factors has zeroes $$\normalsize{5}$$ and $$\normalsize{7}$$, and factors as
$\Large{x^2-12x+35=(x-5)(x-7)}.$
Another example: the polynomial $$\normalsize{x^2+3x-28}$$ factors as
$\Large{x^2+3x-28=(x-4)(x+7)}$
and so it has zeroes $$\normalsize{4}$$ and $$\normalsize{-7}$$.
Although we like to give examples with integers, fractions are also possible.
Q5 (M): What are the linear factors of $$\normalsize{x^2+\frac{7}{2}x+3}$$?
It is not always the case that a polynomial has zeroes, and hence factors. The equation $$\normalsize{x^2+2=0}$$ has no real solutions. Hence the polynomial $$\normalsize{x^2+2}$$ has no linear factors.
Q6 (C): How many linear factors does the polynomial have whose graph is shown in the image to this step?
## A proof of Descartes theorem (advanced)
Why is Descartes theorem true? If you have some stronger background, we invite you to follow the argument for quadratic polynomials. In fact the same proof applies more generally.
Proof: If $$\normalsize{(x-r)}$$ is a factor of $$\normalsize{p(x)=ax^2+bx+c}$$, it means that $$\normalsize{p(x)=(x-r)q(x)}$$ for some other polynomial $$\normalsize{q(x)}$$. In that case if we substitute $$\normalsize{r}$$ we have $$\normalsize{p(r)=(r-r)q(r)}$$ which means that $$\normalsize{p(r)=0}$$ so that yes, $$\normalsize{r}$$ is a zero of $$\normalsize{p(x)}$$.
Now suppose that $$\normalsize{r}$$ is a zero of $$\normalsize{p(x)}$$, so that $$\normalsize{p(r)=ar^2+br+c=0}$$ then
\Large{\begin{align*} p(x) &= p(x) - p(r)\\ & = ax^2+bx+c-(ar^2+br+c)\\ &=a(x^2-r^2)+b(x-r)\\ &=a(x-r)(x+r) + b(x-r) \\ & = (x-r) (a(x+r) + b) \end{align*}}
So $$\normalsize{p(x) = (x-r) q(x) }$$, which means that $$\normalsize{(x-r)}$$ is a factor of $$\normalsize{p(x)}$$. QED.
A1.
i) The sum is $$\normalsize{(r+s)(x)=x+3+2x-1=3x+2}$$.
ii) The difference is $$\normalsize{(r-s)(x)=x+3-(2x-1)=-x+4}$$.
iii) The product is $$\normalsize{(rs)(x)=(x+3)\times(2x-1)=2x^2+5x-3}$$.
A2.
i) The sum is $$\normalsize{(u+v)(x)=x^2+x+1+x^3-1=x^3+x^2+x}$$.
ii) The difference is $$\normalsize{(u-v)(x)=x^2+x+1-(x^3-1)=-x^3+x^2+x+2}$$.
iii) The product is \normalsize{\begin{align}(uv)(x)&=(x^2+x+1)\times(x^3-1)\\&=x^5-x^2+x^4-x+x^3-1 \\&=x^5+x^4+x^3-x^2-x-1. \end{align}}
A3. The factorization is:
$\Large{30x^2-28x-240=(5x+12)(6x-20)}.$
How could we get that? We need to find two numbers that multiply to $$\normalsize{30}$$, and two other numbers that multiply to $$\normalsize{-240}$$ such that the sum of products of two of them with the other two is $$\normalsize{-28}$$. It certainly appears that trial and error are needed!
A4. The linear factors of $$\normalsize{x^2+8x+15}$$ are $$(x+3)$$ and $$(x+5)$$.
A5. The linear factors of $$\normalsize{x^2+\frac{7}{2} x+3}$$ are $$\normalsize{(x+\frac{3}{2})}$$ and $$\normalsize{(x+2)}$$.
A6. The polynomial visually has zeros at $$\normalsize{x=-3,-2,-1,0,1,2}$$ and $$\normalsize{3}$$. So according to Descartes’ theorem, it will have factors $$\normalsize{(x+3),(x+2),(x+1),x,(x-1),(x-2)}$$ and $$\normalsize{(x-3)}$$. So a good guess for the polynomial might be:
$\Large{p(x)= (x+3)(x+2)(x+1)x(x-1)(x-2)(x-3)}.$
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## How do you convert to fractional notation?
How to Find a Fractional Notation?
1. Write the number divided by 1 i.e., [(Decimal Number) / 1]
2. Multiply both the numerator and the denominator by 100.
3. Finally, simplify the obtained fraction.
What does convert to fractional notation mean?
Fractional notation simply means that a number is written in fraction form. It is written as a/b where neither a nor b is equal to 0. A fraction has two parts, the numerator and denominator.
### What is 0.238 written as a fraction?
How to Write 0.238 or 23.8% as a Fraction?
Decimal Fraction Percentage
0.238 119/500 23.8%
0.236 118/500 23.6%
0.23944 119/497 23.944%
0.23896 119/498 23.896%
What is the fractional notation of 45%?
Explanation: 45% can be written as . 45, which is 45100 in the non-reduced fractional form. This cannot be reduced any further, therefore, 45% is 920 in fractional form.
#### How do you find the fractional notation of a percent?
To convert a percent to a fraction, we have to remove the percent sign and divide the given number by 100. And, then we express the fractional form of the percentage in the simplest form. For example, 1% is 1/100, 2% is 2/100 which can be reduced to 1/50.
What is the fractional number of 100?
Example Values
Percent Decimal Fraction
100% 1
125% 1.25 5/4
150% 1.5 3/2
200% 2
## What Is percent notation?
In mathematics, a percentage (from Latin per centum “by a hundred”) is a number or ratio expressed as a fraction of 100. It is often denoted using the percent sign, “%”, although the abbreviations “pct.”, “pct” and sometimes “pc” are also used.
What is .8 as a fraction?
4/5
Answer: 0.8 as a fraction is expressed as 4/5.
### What is 125 as a decimal?
So 125% = 125/100. That can be reduced to 5/4. To convert to a decimal, you can do long division, dividing 5 into 4. Or, you can realize 5/4 = 1 and 1/4, = 1.25.
What is 45 as a percentage?
Now we can see that our fraction is 45/100, which means that 45/100 as a percentage is 45%. We can also work this out in a simpler way by first converting the fraction 45/100 to a decimal.
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# Learning How To Factor?
## Learning How To Factor?
The four main types of factoring are the Greatest common factor (GCF), the Grouping method, the difference in two squares, and the sum or difference in cubes.
## What are the 4 methods of factoring?
The four main types of factoring are the Greatest common factor (GCF), the Grouping method, the difference in two squares, and the sum or difference in cubes.
## What are the 7 factoring techniques?
What are the 7 factoring techniques?
• Factoring out the GCF.
• The sum-product pattern.
• The grouping method.
• The perfect square trinomial pattern.
• The difference of squares pattern.
## How do you introduce students to factors?
A “friendly” way of introducing factors to 4th grade students, is to have them work out factor pairs using only words they are familiar with. For example, Two numbers multiply to 24, and sum to 11.
This question makes students work out all the factor pairs of 24:
1. x 24.
2. x 12.
3. x 8.
4. x 6.
## How do you explain factors to students?
Definition of Factors
Factors are numbers which you can multiply together to get another number. For Example : The numbers 2 and 3 are factors of 6 because 2 x 3 = 6. A number can have many factors!
## How do you explain factors?
factor, in mathematics, a number or algebraic expression that divides another number or expression evenly—i.e., with no remainder. For example, 3 and 6 are factors of 12 because 12 ÷ 3 = 4 exactly and 12 ÷ 6 = 2 exactly. The other factors of 12 are 1, 2, 4, and 12.
## What is the factoring method?
Factoring is the process by which we go about determining what we multiplied to get the given quantity. … A common method of factoring numbers is to completely factor the number into positive prime factors. A prime number is a number whose only positive factors are 1 and itself.
3
## Why do we teach factors?
Traveling with Factors
Understanding factoring allows you to easily navigate number relationships in the real world without relying on your calculator or phone to do the work for you.
## What are factors for beginners?
Factors are what we can multiply to get the number. Multiples are what we get after multiplying the number by an integer (not a fraction).
Example: 12
• 3 × 4 = 12, so 3 and 4 are factors of 12.
• Also 2 × 6 = 12, so 2 and 6 are also factors of 12,
• And 1 × 12 = 12, so 1 and 12 are factors of 12 as well.
## How do you explain common factors to kids?
Common factors are numbers that are in the list of factors for two different numbers. To find common factors, simply factor each number then compare them. If they share any of the same number, then they have common factors.
## How can I be a good factor?
Here are some basic tips that will help you to factor faster.
1. Always start with real numbers: Students are more familiar with calculations with real number than variables, so working with real number will reduced the the amount of calculation and chance of making mistakes. …
2. Recognize common terms: …
3. cross multiplication.
## Which is the smallest factor?
number 1
The number 1 is the smallest factor of every number.
## Which number has only one factor?
1 is the only number which has only single factor, that is, 1 itself.
## What is factor of every number?
1 is the factor of every number as one divides every number exactly, without leaving any remainder behind and gives the quotient as the number itself.
## What is a factor lesson?
Lesson Summary
When we multiply numbers to get a final answer, the numbers we use are called factors. A factor is a number that we multiply by another number to get a product (answer).
## What is the importance of factoring?
Factoring reduces your bookkeeping costs and your overhead expenses. Factoring allows you to make cash payments to your suppliers, which means you can take advantage of discounts and reduce your production costs. Factoring makes it possible for a business to finance its operations from its own receivables.
## What is a factor pair of 18?
Pair factors of 18 are (1,18), (2,9), and (3,6). 1 is a factor of every number. The factor of a number is always less than or equal to the given number. Prime factorization is expressing the number as a product of its prime factors.
## How do you teach 5th grade factors?
See more articles in category: Education
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# How do you solve (x-1)(x-2)(x-3)>=0 using a sign chart?
Jan 2, 2017
The answer is x in [1 ,2 ] uu [3, +oo [
#### Explanation:
Let $f \left(x\right) = \left(x - 1\right) \left(x - 2\right) \left(x - 3\right)$
Now, we can establish the sign chart
$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a}$$1$$\textcolor{w h i t e}{a a a a a a}$$2$$\textcolor{w h i t e}{a a a a a a a}$$3$$\textcolor{w h i t e}{a a a a a a}$$- \infty$
$\textcolor{w h i t e}{a a a a}$$x - 1$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a a}$$+$$\textcolor{w h i t e}{a a a a a}$$+$$\textcolor{w h i t e}{a a a a a}$$+$
$\textcolor{w h i t e}{a a a a}$$x - 2$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a a}$$+$$\textcolor{w h i t e}{a a a a a}$$+$
$\textcolor{w h i t e}{a a a a}$$x - 3$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a a}$$+$
$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a a a}$$+$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a a}$$+$
Therefore,
$f \left(x\right) \ge 0$ when x in [1 ,2 ] uu [3, +oo [
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# Y-Intercept(KS2, Year 6)
homesitemaplinear equationsthe y-intercept
The y-intercept is the y-coordinate of the point where a line, curve or surface crosses the y-axis.
## What Is the Y-Intercept of a Line?
The y-intercept of a line is the y-coordinate of the point where the line crosses the y-axis. The line below crosses the y-axis at y = 3. The y-intercept of the line is 3.
finding the y-intercept of a line
## The Y-Axis Crosses the X-Axis When X = 0
The y-axis crosses the x-axis when x = 0. This means that all points on the y-axis have x = 0. This is useful, because it let's us find the y-intercept of a line or a curve when we know its equation.
• Finding the Y-Intercept of a Line from a Linear Equation The y-intercept of a line can be found from the equation of a line (called a linear equation). The slope-intercept form of a linear equation is:
In this equation, y and x are variables. m and c are letters that stand in for numbers. An example of a linear equation would be y = 2x + 1 (the m = 2 and c = 1). The y-intercept occurs when the line crosses the y-axis. On the y-axis, x = 0. The y-intercept of this equation is c.
• Finding the Y-Intercept of a Line from a Function The linear equation is an example of a function. For any value of x, it tells you y. We can find the y-intercept of any function by substituting x = 0. Learn more about finding the y-intercept of a function
## Positive, Zero and Negative Y-Intercepts
A positive y-intercept means the line crosses the y-axis above the x-axis:
A zero y-intercept means the line crosses the y-axis at the origin:
A negative y-intercept means the line crosses the y-axis below the x-axis:
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# How to construct a 3x3 magic square
In this section we will be constructing the basic magic square with the dimension 3x3 starting value 1 and common difference 1. If you do not know what a magic square is please refer to the Magic Squares: Introduction section. It is also important to note that this method works with all magic squares that have an odd
dimension. i.e. 5x5 7x7 9x9.......
1. ) First we will begin with an empty 3x3 array.
2.) Next we place the starting value 1 in the middle of the top row.
3.) Now we move right one space and up one space,
but as we can see highlighted in yellow this places the 2
outside the bounds of the 3x3 square.When this happens we
simply bring the 2 down to the bottom square of the column it is positioned over.
4.) In this step we start with the 2 and once again we go right
one space and up one space. And once again this leaves us
out side the bounds of the 3x3 square so we place the 3 at the
beginning of the row it is outside of.
5.) Here we start with the 3 and once again move right
one space and up one space, but this time it puts us in an
occupied space. when that happens we simply place the 4
underneath the 3.
6.) In this step we start with the 4 and again move right one
space and up one space. This space is within the bounds of the
square and unoccupied so we can simply leave the 5 here.
7. ) Here we start with the 5 and go one space right and one
space up and as we can see we can once again simply just
place the 6 in this unoccupied space
8.) This time when we move right one space and up one
space. we are not only outside the bounds of the square
but we are also on a diagonal In case this happens we
place the 7 underneath the 6.
9.) Here we start with the 7 and move right one space
and up one space. Here we are again out side the bounds
of the square so we simply take the 8 and place it at the
beginning of the row it is beside.
10.) now in the final step we only have one space left where we could go ahead and place the 9 there how ever I am still going to illustrate that the pattern still applies when there is only one space. so we start with the 8 move right one space and up one space. and again we are out side the bounds of the square, so we will move the 9 to the bottom of the column it is over.
Next in the series we will be looking deeper into the math behind these squares.
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## Elementary Linear Algebra 7th Edition
The first row remained unchanged. The second row is transformed: The first row multiplied by $2$ is added to it. The third row is transformed: The first row multiplied by $5$ is added to it.
The first row remained unchanged. The second and the third rows were transformed $$\text{Row 1 Original} = \left[\begin{matrix}-1&-2&3&-2\end{matrix}\right]\Longrightarrow \text{Row 1 Transformed} = \left[\begin{matrix}-1&-2&3&-2\end{matrix}\right];$$ $$\text{Row 2 Original} = \left[\begin{matrix}2&-5&1&-7\end{matrix}\right]\Longrightarrow \text{Row 2 Transformed} = \left[\begin{matrix}0&-9&7&-11\end{matrix}\right].$$ $$\text{Row 3 Original} = \left[\begin{matrix}5&4&-7&6\end{matrix}\right]\Longrightarrow \text{Row 3 Transformed} = \left[\begin{matrix}0&-6&8&-4\end{matrix}\right].$$ To transform a row we can only multiply each element in it by some number and add the element in the same column of another row multiplied by some number to it. This can be achieved here by Step 1: Adding to the second row the first one multiplied by $2$. Indeed, we will have checking this for each element: $$2+2\times(-1) = 0\quad -5+2(-2) = -9\quad 1+2\times3 = 7\quad -7+2\times(-2) = -11$$ Step 2: Adding the first row multiplied by $5$ to the third row. Again, checking for each element we have: $$5+5\times(-1) = 0\quad 4+5\times(-2) = -6\quad -7+5\times3 = 8\quad 6+5\times(-2) = -4.$$
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# How do you evaluate \frac { 12} { 7} \div ( - \frac { 6} { 11} )?
Mar 18, 2018
See a solution process below:
#### Explanation:
First, we can rewrite this expression as:
$\frac{\frac{12}{7}}{\frac{- 6}{11}}$
Now, we can use this rule for dividing fractions to evaluate the expression:
$\frac{\frac{\textcolor{red}{a}}{\textcolor{b l u e}{b}}}{\frac{\textcolor{g r e e n}{c}}{\textcolor{p u r p \le}{d}}} = \frac{\textcolor{red}{a} \times \textcolor{p u r p \le}{d}}{\textcolor{b l u e}{b} \times \textcolor{g r e e n}{c}}$
$\frac{\frac{\textcolor{red}{12}}{\textcolor{b l u e}{7}}}{\frac{\textcolor{g r e e n}{- 6}}{\textcolor{p u r p \le}{11}}} \implies$
$\frac{\textcolor{red}{12} \times \textcolor{p u r p \le}{11}}{\textcolor{b l u e}{7} \times \textcolor{g r e e n}{- 6}} \implies$
$\frac{\cancel{\textcolor{red}{12}} \textcolor{red}{2} \times \textcolor{p u r p \le}{11}}{\textcolor{b l u e}{7} \times \cancel{\textcolor{g r e e n}{- 6}} \textcolor{g r e e n}{- 1}} \implies$
$\frac{22}{-} 7 \implies$
$- \frac{22}{7}$
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## Intermediate Algebra (12th Edition)
$t=\left\{ \dfrac{2-\sqrt{3}}{5},\dfrac{2+\sqrt{3}}{5} \right\}$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $(2-5t)^2=12 ,$ take the square root of both sides (Square Root Property) and simplify the radical. Then use the properties of equality to isolate the variable. $\bf{\text{Solution Details:}}$ Taking the square root of both sides (Square Root Property), the equation above is equivalent to \begin{array}{l}\require{cancel} 2-5t=\pm\sqrt{12} .\end{array} Simplifying the radical and then using the properties of equality to isolate the variable, the equation above is equivalent to \begin{array}{l}\require{cancel} 2-5t=\pm\sqrt{4\cdot3} \\\\ 2-5t=\pm\sqrt{(2)^2\cdot3} \\\\ 2-5t=\pm2\sqrt{3} \\\\ -5t=-2\pm2\sqrt{3} \\\\ t=\dfrac{-2\pm2\sqrt{3}}{-5} \\\\ t=\dfrac{2\pm2\sqrt{3}}{5} .\end{array} The solutions are \begin{array}{l}\require{cancel} t=\dfrac{2-\sqrt{3}}{5} \\\\\text{OR}\\\\ t=\dfrac{2+2\sqrt{3}}{5} .\end{array} Hence, $t=\left\{ \dfrac{2-\sqrt{3}}{5},\dfrac{2+\sqrt{3}}{5} \right\} .$
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