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https://norahcolvin.com/tag/counting-coins-to-1/ | 1,685,465,980,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646076.50/warc/CC-MAIN-20230530163210-20230530193210-00793.warc.gz | 475,116,190 | 30,174 | # Lessons about recognising and counting Australian coins – #readilearn
Money is one of those things we all need to understand to be able to participate fully in life as we know it. I think it becomes more difficult for children to understand the value of money as we move towards a cashless society, but maybe that’s because I grew up pre-cards of any kind. Now many children only see transactions made with cards or even phones and watches. Many will have no need to enter a physical bank to deposit or withdraw funds or for any other reason. It is all done online. Perhaps learning about coins and notes will one day be relegated to history lessons, but for now I think it is still important for children to learn about them and their value and they still feature in the Curriculum. For this reason, I have made some resources to support your teaching of young children about our Australian coins and their value. (Lessons about notes will be added later.)
Australian coins helps children learn to recognise, identify and describe the coins according to colour, shape and size, and the identifying icon on the tails side. Additional information is provided about the Australian animals and icons featured on the tails side of each coin. This lesson is ready to teach on the interactive whiteboard.
Australian Coins – Let’s count \$1 gives children practice in counting collections of coins to \$1.
There are three separate sections which can be used over a series of lessons.
1. Count groups of coins of the same value that equal \$1.
2. Count collections of different coins that equal \$1.
3. Make collections of coins to equal \$1.
These lessons are interactive and ready to teach on the interactive whiteboard.
Count Coins to \$1 is a dice game that gives students practice in
• recognising and naming coins
• counting the value of coins to \$1
• comparing the value of coins.
It is a perfect game for maths groups to follow-up lessons with Australian Coins and Australian Coins Let’s count \$1.
Continue reading: Lessons about recognising and counting Australian coins – readilearn | 422 | 2,098 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2023-23 | latest | en | 0.953452 |
https://cs.stackexchange.com/questions/71727/speculating-big-o-for-a-binary-tree | 1,660,742,135,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572908.71/warc/CC-MAIN-20220817122626-20220817152626-00337.warc.gz | 201,340,557 | 65,978 | # Speculating big O for a binary tree
I have a binary tree with $n$ nodes. The tree is not necessarily balanced.
For each node of the tree, I count down the total number of the nodes for its left sub-tree as $n_L$ and then I count down the total number of the nodes for its right sub-tree as $n_R$. Then I calculate $\min(n_L,n_R)$ and then assign the minimum value as a tag to the node. I repeat this process to produce appropriate tags for all the nodes of the tree.
Now I wonder what is the big O for the sum of all the tags. I mean, I wonder if the sum of tags is $O(n\log n)$ or $O(n)$ or $O(n^2)$.
In the most unbalanced case, the binary tree is just a long list like figure below. In this case, I think sum of tags would be 0, therefore the big O would be $O(0)$.
In the case of a perfect binary tree, the tag of the root node might be $2^0\frac{n}{2^{0+1}}$ and at the next level, i.e. root's left and right children, sum of tags might be $2^1\frac{n}{2^{1+1}}$ and for the $i$th level of the perfect binary tree, the sum of tags might be $2^i\frac{n}{2^{i+1}}$.
At the $i$th level, sum of tags might become $2^i\frac{n}{2^{i+1}}=\frac{n}{2}$ therefore looks like that some of tags at each level might be $\frac{n}{2}$ and because a perfect binary three has a total height of $\log_2n$ hence the total sum of tags might be $\frac{n}{2}\log_2n$ and this implies that for a perfect binary tree, the big O for sum of tags might be $O(n\log_2n)$.
• What have you tried so far? Have you tried working through some examples or extreme cases? Have you tried working out what the value will be for a balanced tree? For a maximally unbalanced tree? You should be able to answer your question on your own if you just work through those two cases...
– D.W.
Mar 18, 2017 at 18:58
• Cross-posted: cs.stackexchange.com/q/71727/755, stackoverflow.com/q/42883388/781723. Please do not post the same question on multiple sites. Each community should have an honest shot at answering without anybody's time being wasted.
– D.W.
Mar 29, 2017 at 7:00
Let us denote by $\sigma(T)$ the sum of tags of a tree $T$. As you notice in your worked out examples, $\sigma(T)$ in general depends on the tree structure and not only on the number of nodes. What you are looking for is a tight upper bound on $\sigma(T)$ in terms of the number of nodes.
What is this tight upper bound? For each $n$, we want to know what is the maximal $\sigma$ for a tree having $n$ nodes. Denoting this by $\tau(n)$, this quantity satisfies $\sigma(T) \leq \tau(n)$ for all trees $T$ on $n$ nodes, and furthermore $\tau(n)$ cannot be replaced by any smaller value.
In fact, in practice we often don't care about finding the tight upper bound to this degree of accuracy. It is often enough to find a function $f(n)$ such that for some constant $C>0$, $\sigma(T) \leq Cf(n)$ for all trees $T$ on $n$ nodes, and furthermore for some constant $c>0$, $\sigma(T) \geq cf(n)$ for some tree $T$ on $n$ nodes.
You can prove by induction on $n$ that $\sigma(T) = O(n\log n)$ for all trees on $n$ nodes. Conversely, by taking a nearly complete binary tree $T_n$ on $n$ nodes (that is, a binary tree all of whose leaves are at heights $h,h+1$ for some $h$), you can check that $\sigma(T) = \Omega(n\log n)$. Hence $O(n\log n)$ is a tight upper bound for the sum of tags of binary trees on $n$ nodes.
• Right. Therefore there would be no tree with sum of tags at $O(n^{2})$ because that would violate the tight upper bound. Now I wonder if we can say that there can possibly be a tree with sum of tags at $O(n)$: I'm not sure, maybe that's possible, because $O(n)$ doesn't violate tight upper bound. Mar 19, 2017 at 8:00
• On the contrary, every tree would have sum of tags $O(n^2)$. Don't forget that big O is just an upper bound. Mar 19, 2017 at 8:02 | 1,077 | 3,806 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2022-33 | longest | en | 0.874061 |
https://study.com/academy/lesson/angle-of-incidence-definition-formula-quiz.html | 1,568,858,986,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573415.58/warc/CC-MAIN-20190919015534-20190919041534-00391.warc.gz | 676,679,080 | 46,005 | # Angle of Incidence: Definition & Formula
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Lesson Transcript
Instructor: Damien Howard
Damien has a master's degree in physics and has taught physics lab to college students.
Discover what an angle of incidence is and how it relates to light. Then dive further into learning the related topics of incident angles with the reflection and refraction of light.
## What Is an Angle of Incidence?
If you have ever looked at a white light being shined into a prism, the main thing you probably noticed is it being turned into a rainbow. However, did you notice that the light didn't travel straight through the prism? The light changes direction when it enters the prism, and it changes again when it leaves. This bending of light isn't something that's unique to prisms. It's a common occurrence in translucent and transparent materials. We have even come up with unique names for the angles at which the light enters and travels through the material. The angle at which the light enters is called the angle of incidence. The angle of incidence can be defined as the angle between the oncoming ray of light and the normal vector of the surface of the material it is coming into contact with. The normal vector is a perpendicular vector from a plain or surface of an object.
## Absorption, Reflection and Refraction
There are three possibilities for what can happen to the light after it comes into contact with an object. The first is absorption. In absorption, the light ray travels no further; it is taken into the object and transformed into energy. This most often takes the form of heat. The second is reflection, where the light ray is bounced off the material instead of absorbed. The final possibility is refraction, in which the light penetrates the object, but instead of being turned into internal energy, it changes direction and continues to travel through the material as light.
It's important to note that light does not often do just one of those three possibilities; rather, it travels in a combination of them. For example, we see color because objects reflect the visible light spectrum of the corresponding colors we see back to our eyes, but they absorb the other wavelengths of the visible light spectrum. Another example would be translucent objects. A translucent material, as opposed to a transparent one, only lets some of the light travel through it and absorbs and reflects the rest of it.
With absorption, there are no further relations between the angle of incidence and what happens to the ray of light, but with reflection and refraction, there is more to learn.
## Law of Reflection
The relation between the reflected light ray and the incident light ray is governed by the law of reflection. The law of reflection states that the angle between the reflected ray and the normal vector is equal to the angle of incidence. Mathematically this is expressed by the following simple relation:
thetaI = thetaR
thetaI = angle of incidence
thetaR = angle of reflection
So, if you know either the angle of reflection or the angle of incidence, you automatically know the other as well.
## Snell's Law
Have you ever been spear fishing, or seen it done on television? The fisherman stands in or above shallow water and tries to spear the fish as they swim by. If they try to hit the fish exactly where they see it, they will miss. The fish is not where it seems to be from above the water. This is an effect of refraction. Light bending as it enters the water is what causes the fish to appear in a different spot from where it actually is.
Much like with the angle of reflection, there is also a relation between the angle of incidence and the angle of refraction. When light enters from one medium into another, it can bend to some degree. The calculation of what degree the light will bend is known as the law of refraction, or Snell's Law. The degree to which the light bends is dependent on both the incidence angle entering the second medium and the speed at which the light travels through the two mediums. The relationship between the speed of light and refraction is given in the index of refraction. You will need to understand the index of refraction in order to understand Snell's Law.
n = c / v
n = index of refraction
c = speed of light in a vacuum
v = average speed of light in a medium
While the speed of light in a vacuum is constant, the speed of light in other mediums can vary, and consequently they have different indices of refraction. Some common examples can be seen in this table.
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# Making Inferences: A Step-by-Step Guide
Making Inferences: A Step-by-Step Guide. Arleta High School English Department. What is an Inference?. When you infer , you are making a judgment based on the information you are given You also need to interpret the information that you are given. O = Observe D = Define I = Infer
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## Making Inferences: A Step-by-Step Guide
E N D
### Presentation Transcript
1. Making Inferences:A Step-by-Step Guide Arleta High School English Department
2. What is an Inference? • When you infer, you are making a judgment based on the information you are given • You also need to interpret the information that you are given
3. O = Observe D = Define I = Infer E = Explain The Steps to Making Accurate Inferences
4. First Step: Observe the object or behavior • What can you observe about Odie and Garfield?
5. Step Two: Define the object or behavior • How would you define the behavior? • Is the behavior positive or negative?
6. Step Three: Making the Inference • What judgment can you make about this behavior? • Is it enjoyable? Painful?
7. Step Four: Explain your thinking • In this picture, Odie the dog is tickling Garfield the cat. Garfield is smiling. (Observation) • Tickling is usually fun, and Garfield is smiling, which indicates that he is happy. (Definition) • Garfield enjoys being tickled by Odie. (Inference)
8. The ODIE Method • Remember ODIE! • When you have to make an inference based on something that you read, always: • OBSERVE • DEFINE • INFER • EXPLAIN
9. Now it’s your turn…What can you infer from the following picture? • Observe: What are they doing? • Define: What does it mean? • Infer: What can you tell about their feelings?
10. Now EXPLAIN…Write an explanation of what you were able to infer about Odie and Garfield from the illustration! • In the picture, Odie and Garfield are…
11. It works with Dialogue!Step One: OBSERVE • “Excuse me, Miss, could I…hi, I’m sorry, could I um…I mean, is that seat taken?” • What can you observe about the sentence? • The speaker is talking to a young woman. • What is the punctuation like? • The punctuation has a lot of commas and ellipses. • The sentence ends with a question mark.
12. Step Two: DEFINE • Lots of commas and ellipses means that the speaker is pausing a lot. • The word “um” means that the speaker is hesitating.
13. Step Three: INFER • Pausing a lot and hesitating means that the speaker is uncomfortable speaking to the young woman. He may be nervous.
14. Step Four: EXPLAIN • In this sentence, the author uses many commas and ellipses to show that the speaker is nervous when he is asking the young woman if he can sit near her. The fact that the speaker says “um” a lot probably means that the speaker is shy or hesitant.
15. Odie says Good Job! • Try one more!
16. Step One: OBSERVE • I grabbed LeSange and screamed, “We’re doomed! Doomed, I tell you!” • “Hm,” he said. “We do seem to be in a spot of trouble, don’t we?” • What can you observe about the sentences? • The first speaker uses many exclamation points and negative diction such as the word ‘doomed,’ which is repeated twice. • The first speaker says that he screamed. • The second speaker does not scream. He ends his sentence with a question mark.
17. Step Two: DEFINE • Exclamation points mean that the speaker is very upset or excited. • A question mark means that the speaker is curious. • “Hm” is an expression that means that the speaker is thinking.
18. Step Three: INFER • Many exclamation points and the word “doomed” means that the speaker is frightened and excited. • The “Hm” by the second speaker, and the fact that he does not use any exclamation points means that he is not frightened. He may only be curious, or amused.
19. Step Four: EXPLAIN • In this passage, the author uses exclamation points and the word doomed twice for the first speaker’s sentence. This shows that the speaker is scared and really believes that something terrible is about to happen. • The dialogue for the second speaker is not excited. He says “Hm,” which shows that he is curious, or that he really is not interested. • There is a contrast between the first speaker and the second speaker.
20. One More Time!Step One: OBSERVE • “You have no compassion for my poor nerves,” said Mrs. Bennett. • “You mistake me, my dear,” her husband said. “I have a high respect for your nerves. They are my old friends. I have heard you mention them with consideration these last twenty years at least.” • What can you observe about these sentences? • The first speaker is a wife. • The second speaker is the husband. • There are no exclamation points or question marks. • They are speaking to each other.
21. Step Two: DEFINE • Because there are no exclamation points or question marks, the two characters are speaking calmly. • The wife accuses the husband of not having compassion for her “nerves.” • The husband says that he does, and that he has had compassion for more than twenty years.
22. Step Three: INFER • The wife is complaining that her husband does not pay attention to her problems. • The husband says that he does. He doesn’t fight with her, but he reminds her that he has been listening to her same complaint for twenty years. • They have probably been married for a long time. The wife complains a lot, and the husband is tired of hearing about it.
23. Step Four: EXPLAIN • In this passage, the author presents a husband and wife. Although their voices are calm (they do not yell, because there are no exclamation points), they each have a complaint. The wife feels that the husband never listens to her problems. The husband, a little sarcastically, replies that he has been listening to her problems for at least twenty years.
24. Well Done! • Remember the ODIE method to make inferences! • OBSERVE • DEFINE • INFER • EXPLAIN
More Related | 1,382 | 5,883 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-22 | latest | en | 0.881959 |
https://gateoverflow.in/288374/ugcnet-dec2018-ii-3 | 1,571,280,493,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986672548.33/warc/CC-MAIN-20191017022259-20191017045759-00549.warc.gz | 508,723,344 | 17,891 | 433 views
A box contains six red balls and four green balls. Four balls are selected at random from the box. What is the probability that two of the selected balls will be red and two will be in green?
1. $\frac{1}{14}$
2. $\frac{3}{7}$
3. $\frac{1}{35}$
4. $\frac{1}{9}$
in Others
edited | 433 views
4 balls can be selected from 10 balls in ${10_{C}}_{4}$ ways.
2 red balls can be selected from 6 red balls in ${6_{C}}_{2}$ ways.
2 green balls can be selected from 4 green balls in ${4_{C}}_{2}$ ways.
hence probability= (${6_{C}}_{2}$ *${4_{C}}_{2}$)/${10_{C}}_{4}$ = 3/7
by Active (4.9k points)
+1 vote | 206 | 613 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2019-43 | longest | en | 0.848865 |
https://www.tutorela.com/math/perpendicular-lines/examples-exercises | 1,726,849,688,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701419169.94/warc/CC-MAIN-20240920154713-20240920184713-00620.warc.gz | 964,214,240 | 24,674 | Perpendicular Lines - Examples, Exercises and Solutions
Perpendicular lines are vertical lines that form a right angle between them, that is, an angle of $90°$ degrees.
Perpendicular lines appear in many geometric shapes, such as a rectangle, a square, a right triangle, and others.
Examples with solutions for Perpendicular Lines
Exercise #1
What do the four figures below have in common?
All parallel
Exercise #2
What do the four figures have in common?
Step-by-Step Solution
Remember that perpendicular lines form a 90-degree angle between them.
Parallel lines are lines that never intersect and do not form any angle between them.
To be able to examine the lines, a line will be drawn at each end of the line, as follows:
We notice that from the drawings it seems that all the lines do not form any angle between them, this is because they do not meet each other.
Therefore, all figures show parallel lines.
All show parallel lines.
Exercise #3
Which of the lines are perpendicular to each other?
Step-by-Step Solution
Let's remember that perpendicular lines are lines that form a right angle of 90 degrees between them.
In each of the answers, we will draw the letter T at the point of intersection of the lines.
Let's examine figure A:
We will notice that the lines do not form a right angle, and therefore are not perpendicular.
Let's examine figure B:
The lines do indeed form a 90-degree angle and are therefore perpendicular.
Let's examine figure C:
We notice that the lines do not form a right angle, and therefore are not perpendicular.
Exercise #4
Which lines are perpendicular to each other?
Step-by-Step Solution
Let's remember that perpendicular lines are lines that form an angle of 90 degrees between them.
In each of the answers, we will draw the letter T at the point of intersection of the lines.
Let's examine figure A:
We notice that the lines do form a right angle and the lines are perpendicular to each other.
Let's examine figure B:
We notice that the lines do not meet and do not form any angle; therefore, they are parallel lines and not perpendicular.
Exercise #5
Which lines are perpendicular to each other?
Step-by-Step Solution
Perpendicular lines are lines that form a right angle of 90 degrees between them.
The only drawing where the lines form a right angle of 90 degrees between them is drawing A.
Exercise #6
Which lines are perpendicular to each other?
Step-by-Step Solution
Let's remember that perpendicular lines are lines that form a right angle of 90 degrees between them.
The only drawing where it can be seen that the lines form a right angle of 90 degrees between them is drawing A.
Exercise #7
Which figure(s) show intersecting lines?
Step-by-Step Solution
Lines that intersect each other are lines that divide the side into two equal parts.
The drawings showing that the lines divide the sides into equal parts are drawings 1+3.
In drawing 2, the lines are perpendicular and vertical to each other, and in drawing 4, the lines are parallel to each other.
1 and 3
Exercise #8
Which of the figures show perpendicular lines?
Step-by-Step Solution
Perpendicular lines are lines that form a right angle of 90 degrees between them.
It can be observed that in figures 1 and 3, the angles formed by the lines between them are right angles of 90 degrees.
1 and 3
Exercise #9
Which lines are perpendicular to each other?
Step-by-Step Solution
Let's remember that perpendicular lines form a 90-degree angle with each other.
To check if the lines form a 90-degree angle, we will draw a T at each intersection as follows:
We notice that in each of the four drawings, the lines form a 90-degree angle.
Therefore all are correct as all the lines are perpendicular to each other.
Exercise #10
What do the 4 figures below have in common?
Step-by-Step Solution
Let's think about the different definitions of various lines.
We can see that what is common to all lines is that they intersect with each other, meaning they have a point of intersection.
We'll remember that lines that cross each other are lines that will meet at a certain point.
Therefore, the correct answer is a.
All intersections
Exercise #11
Which of the diagrams shows perpendicular lines?
Step-by-Step Solution
Let's remember that perpendicular lines form a 90-degree angle between them.
To check if the lines form a 90-degree angle, for each of the drawings a T is drawn at the intersecting point of the lines.
Let's examine figure A:
Note that the lines do not form a right angle.
Let's examine figure B:
We notice that the lines form a 90-degree angle and therefore are perpendicular.
Let's examine figure C:
Note that the lines do not form a right angle.
Let's examine figure D:
Note that the lines do not form a right angle.
Exercise #12
Which of these lines are perpendicular to each other?
Step-by-Step Solution
Let's remember that perpendicular lines form a 90-degree angle.
In each of the answers, we will draw the letter T at the intersection point of the lines.
Let's examine figure A:
We notice that the lines do not form a right angle and therefore are not perpendicular.
Let's examine figure B:
We notice that the lines do not form a right angle and therefore are not perpendicular.
Let's examine figure C:
We notice that the lines do not form a right angle and therefore are not perpendicular.
Let's examine figure D:
We notice that the lines do form a right angle therefore are perpendicular lines.
Exercise #13
Which lines are parallel to each other?
Step-by-Step Solution
Let's remember that parallel lines are lines that, if extended, will never intersect.
In diagrams a'+b'+c', all the lines intersect with each other at a certain point, except for diagram d'.
The lines drawn in answer d' will never intersect.
Exercise #14
Which figure shows perpendicular lines?
Step-by-Step Solution
Perpendicular lines are lines that form a right angle between them.
In the drawings A+C+D, you can see that the angles formed are not right angles.
It is possible to point out a right angle in drawing B.
Exercise #15
Which of the figures shows parallel lines?
Step-by-Step Solution
Parallel lines are lines that, if extended, will never meet.
In the drawings A+B+D if we extend the lines we will see that at a certain point they come together.
In drawing C, the lines will never meet, therefore they are parallel lines. | 1,381 | 6,443 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2024-38 | latest | en | 0.930604 |
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Quadratics: solvers Practice! Answers archive Lessons Word Problems In Depth
Question 86665: Solve by using the quadratic formula... 1. 2x^2-7x=5 2. x^2=x+4 3. Find the distance between (7,0) and (-7,0) 4. Solve...x^2+8=17 Click here to see answer by Flake(45)
Question 84581: Steve traveled 600 miles at a certain speed. Had he gone 20mph faster, the trip would have taken 1 hour less. Find the speed of his vehicle. Click here to see answer by Flake(45)
Question 86756: 1.FIND THE CONSTANT TERM THAT SHOULD BE ADDED TO MAKE THE FOLLOWING EXPRESSION A PERFECT SQUARE TRINOMIAL.... X^2+5X 2. SOLVE BY COMPLETING THE SQUARE... X^2-4X+4=0 3. EVALUATE IF POSSIBLE... -3 ROOT OF 64 THANKS IN ADVCANCE FOR ALL OF YOUR HELP!!! Click here to see answer by tutorcecilia(2152)
Question 86765: Hi, I am working on a quadratic equation: x^2 + 4x + 4 = 7. I'm not sure if I've arrived at the correct answer...I have x = 2(plus/minus) sq.rt. 7....am I correct in thinking this is the final answer? Thank you! Click here to see answer by Earlsdon(6287)
Question 86765: Hi, I am working on a quadratic equation: x^2 + 4x + 4 = 7. I'm not sure if I've arrived at the correct answer...I have x = 2(plus/minus) sq.rt. 7....am I correct in thinking this is the final answer? Thank you! Click here to see answer by tutorcecilia(2152)
Question 86792: The base of a ladder is 14 feet away from the wall. The top of the ladder is 17 feet from the floor. Find the length of the ladder to the nearest thousandth. Click here to see answer by checkley75(3666)
Question 86791: simplify: (square root of 7) x (square root of 14) Click here to see answer by checkley75(3666)
Question 86794: Find the altitude of the triangle: length- 13 length- 13 width- 10 Click here to see answer by Flake(45)
Question 86779: Find the distance between (–3, –2) and (1, 4) Click here to see answer by checkley75(3666)
Question 86776: I'm not sure if I'm on the right track to solving this quadratic equation...hopefully you can help! The original equation is: 2(x-5)^2=3. I think that after factoring and re-writing the equation it would look like this: 2x^2-20x+47, is that right? Then, to go from there, I'm not really sure...any help would be greatly appreciated! Thank you! Click here to see answer by Flake(45)
Question 86807: Hi, I've come to an unusual answer on a quadratic equation, and I would greatly appreciate if someone could verify it for me, so I know I've done it correctly. The original equation read like this: 2(x-5)^2=3. After factoring and using the quadratic formula, I came up with two completely different answers, which now has me highly confused as to which one is correct, so any help would be wonderful! My first answer I came to was this: x = 10+/- sq.rt.6 / 2 My second answer I came to was this:x = 5+/- (3/2)^1/2 Now, I am very confused and lost, and am not sure which answer is correct...or if neither one of them are correct. Please help! Thank you! Click here to see answer by jim_thompson5910(28598)
Question 86777: Find the distance between (–4, 0) and (–5, –3) Click here to see answer by jim_thompson5910(28598)
Question 86808: I do not understand quadratic equations. Here's my question from my book. Solve each equation and inequality. Write the solution set for each inequality in interval notation. a.) x^2 - x - 12 = 0 b.) x^2 - x - 12 < 0 c.)x^2 - x - 12 > 0 Click here to see answer by stanbon(57387)
Question 86811: Simplify: square root of 500 Click here to see answer by stanbon(57387)
Question 86811: Simplify: square root of 500 Click here to see answer by bucky(2189)
Question 86802: solve the equation by completeing the square a2(squared)-12a+27=0 Click here to see answer by stanbon(57387)
Question 86802: solve the equation by completeing the square a2(squared)-12a+27=0 Click here to see answer by jim_thompson5910(28598)
Question 86812: HI I have been working this problem over and over but I can't seem to get it right 5/5+3(1-5^2)/ I know the answer is 335 but I am not getting that answer any help will be appriciated Click here to see answer by checkley75(3666)
Question 86790: simplify: square root of 72 Click here to see answer by bucky(2189)
Question 86834: Simplify. Assume x represents a positive number. sqrt45x^4 Click here to see answer by stanbon(57387)
Question 86833: simplify: sqrt75a^2/sqrt5 Click here to see answer by stanbon(57387)
Question 86829: simplify sqrt125/sqrt5 Click here to see answer by Nate(3500)
Question 86914: A garden area is 30 ft long and 20 ft wide. A path of uniform width is set around the edge. If the remaining garden area is 400 square foot, what is the width of the path? Click here to see answer by [email protected](15660)
Question 86816: How can you find the equation of the axis of symmetry for this parabola y=xsquared - 6x + 7 Click here to see answer by scott8148(6628)
Question 86831: Which two expressions are equivalent? Provide evidence for your answer sqrt-49,-7,-sqrt49 Click here to see answer by checkley75(3666)
Question 87000: I still cant figure out the answer to this question even with this website. The question is: Solve by using te quadratic formula. 5x^2+4x-7=0 Click here to see answer by stanbon(57387)
Question 87001: This isnt a quadratic formula problem, I dont know what type of problem it is, I just really need help in it. The problem is: Simplify d^2-2d-3 d^2-10d+24 d+1 d-4 They are like that because the problem is in fraction form Click here to see answer by checkley75(3666)
Question 86795: simplify: square root of 125/square root of 5 Click here to see answer by checkley75(3666)
Question 87033: I need help with a word problem- a radiator contains 10 qts of fluid, 30% of which is antifreeze. How much fluid should be drained and replaced with pure atifreeze so that the new mixture is 40% atifreeze. this is how I set the problem up .30(10-x)+x=.4(10) 3-.30x+x=4 -.30+x=1 .7x=1 1/.7 =1.42 qrts I would appreciate any help because I am not sure if that is correct or not Thank you KRockwell Click here to see answer by scott8148(6628)
Question 86823: simplify 5sqrt6x + 4sqrt6x Click here to see answer by checkley75(3666)
Question 86801: Please help me with this question: sketch the graph of the parabola y=3(x-2)2(squared)-5 I have tried everyway I can. Click here to see answer by Edwin McCravy(8909)
Question 87086: Use the arithmetic sequence of numbers 2, 4, 6, 8, 10… to find the following: a) What is d, the difference between any two consecutive terms? Answer: Show work in this space. b) Using the formula for the nth term of an arithmetic sequence, what is 101st term? Answer: Show work in this space. c) Using the formula for the sum of an arithmetic sequence, what is the sum of the first 20 terms? Answer: Show work in this space. d) Using the formula for the sum of an arithmetic sequence, what is the sum of the first 30 terms? Answer: Show work in this space. e) What observation can you make about the successive partial sums of this sequence (HINT: It would be beneficial to find a few more sums like the sum of the first 2, then the first 3, etc.)? Answer: Click here to see answer by jim_thompson5910(28598)
Question 87089: Please help me with this word problem: A guy wire of length 50 feet is attached to the ground and to the top of an antenna. The height of the antenna is 10 feet larger than the distance from the base of the antenna to the point where the guy wire is attached to the ground. What is the height of the antenna? Click here to see answer by Flake(45)
Question 87213: Would someone please help me with this one: Solve the following quadratic equations by factoring or by any of the quadratic equations available such as square root, completing the square, etc. 4x^2-8x+3=5 Click here to see answer by Nate(3500)
Question 87224: find the quadratic equation when one root is the square of the other Click here to see answer by checkley75(3666)
Question 87224: find the quadratic equation when one root is the square of the other Click here to see answer by Edwin McCravy(8909)
Question 87239: CLASSIC PROBLEM - A traveling salesman (selling shoes) stops at a farm in the Midwest. Before he could knock on the door, he noticed an old truck on fire. He rushed over and pulled a young lady out of the flaming truck. Farmer Crane came out and gratefully thanked the traveling salesman for saving his daughter’s life. Mr. Crane insisted on giving the man an award for his heroism. So, the salesman said, “If you insist, I do not want much. Get your checkerboard and place one grain of wheat on the first square. Then place two grains of wheat on the next square. Then place four grains on the third square. Continue this until all 64 squares are covered with grains of wheat.” As he had just harvested his wheat, Mr. Crane did not consider this much of an award, but he soon realized he made a miscalculation on the amount of wheat involved. a) How much wheat would Mr. Crane have to put on the 24th square? Answer: Show work in this space. b) How much total grain would the traveling salesman receive if the checkerboard only had 24 squares? Answer: Show work in this space. c) Calculate the amount of wheat necessary to fill the whole checkerboard (64 squares). How much wheat would the farmer need to give the salesman? Please provide the answer in either scientific notation, or calculate and show all 20 digits. Answer: Click here to see answer by jim_thompson5910(28598)
Question 87238: Use the geometric sequence of numbers 1, 1/3, 1/9 , 1/27… to find the following: a) What is r, the ratio between 2 consecutive terms? Answer: Show work in this space. b) Using the formula for the sum of the first n terms of a geometric sequence, what is the sum of the first 10 terms? Carry all calculations to 6 decimals on all assignments. Answer: Show work in this space. c) Using the formula for the sum of the first n terms of a geometric sequence, what is the sum of the first 12 terms? Carry all calculations to 6 decimals on all assignments. Answer: Show work in this space. d) What observation can make about the successive partial sums of this sequence? In particular, what number does it appear that the sum will always be smaller than? Answer: Click here to see answer by jim_thompson5910(28598)
Question 87266: Solve the following using the square root method or the quadratic equation. 66x^2 -31x -9 + 0 Not understanding this can you help Please!! Thanks Click here to see answer by jim_thompson5910(28598)
Question 87265: Solve the following absolute value equations. / (stands for a striaght line, not division) /1265z + 26894/ = -36974 I really don't know how to begin this one, any help will be appriciated. Thanks Kim Click here to see answer by kev82(148)
Question 87249: David drew a rectangle is 143 square mm. Derek drew a rectangle inside of David which is 35 square mm. Derek's rectangle has a three mm border between his and David's rectangle. What are the length and width of David rectangle? -Jason Click here to see answer by [email protected](315)
Question 87254: The demand equation for a certain type of printer is given by D = -200p + 35,000 The supply equation is predicted to be S = -p square + 400p - 20,000 Find the equilibraum price Click here to see answer by longjonsilver(2297)
Question 86764: Please help...my original problem is: 2(x-5)^2 = 3. I think the answer is: x= 10 (plus/minus) sq.rt.6 / 2....am I on the right track? Thank you for your help! Click here to see answer by bucky(2189)
Question 87292: Hello, I am having trouble with this word problem involving the use of quadratic equations. Any help would be greatly appreciated! Thank you! The equation h= -16t^2 + 112t gives the height of an arrow, shot upward from the ground with an initial velocity of 112 ft/s, where t is the time after the arrow leaves the ground. Find the time it takes for the arrow to reach a height of 180 feet. Click here to see answer by rapaljer(4667)
Question 87293: Please help! I am having a hard time figuring out this question: The demand equation for a certain type of printer is given by D = -200p +35,000. The supply equation is predicted to be S = -p^2 +400p - 20,000. What is the equilibrium price? Thank you!! Click here to see answer by rapaljer(4667)
Older solutions: 1..45, 46..90, 91..135, 136..180, 181..225, 226..270, 271..315, 316..360, 361..405, 406..450, 451..495, 496..540, 541..585, 586..630, 631..675, 676..720, 721..765, 766..810, 811..855, 856..900, 901..945, 946..990, 991..1035, 1036..1080, 1081..1125, 1126..1170, 1171..1215, 1216..1260, 1261..1305, 1306..1350, 1351..1395, 1396..1440, 1441..1485, 1486..1530, 1531..1575, 1576..1620, 1621..1665, 1666..1710, 1711..1755, 1756..1800, 1801..1845, 1846..1890, 1891..1935, 1936..1980, 1981..2025, 2026..2070, 2071..2115, 2116..2160, 2161..2205, 2206..2250, 2251..2295, 2296..2340, 2341..2385, 2386..2430, 2431..2475, 2476..2520, 2521..2565, 2566..2610, 2611..2655, 2656..2700, 2701..2745, 2746..2790, 2791..2835, 2836..2880, 2881..2925, 2926..2970, 2971..3015, 3016..3060, 3061..3105, 3106..3150, 3151..3195, 3196..3240, 3241..3285, 3286..3330, 3331..3375, 3376..3420, 3421..3465, 3466..3510, 3511..3555, 3556..3600, 3601..3645, 3646..3690, 3691..3735, 3736..3780, 3781..3825, 3826..3870, 3871..3915, 3916..3960, 3961..4005, 4006..4050, 4051..4095, 4096..4140, 4141..4185, 4186..4230, 4231..4275, 4276..4320, 4321..4365, 4366..4410, 4411..4455, 4456..4500, 4501..4545, 4546..4590, 4591..4635, 4636..4680, 4681..4725, 4726..4770, 4771..4815, 4816..4860, 4861..4905, 4906..4950, 4951..4995, 4996..5040, 5041..5085, 5086..5130, 5131..5175, 5176..5220, 5221..5265, 5266..5310, 5311..5355, 5356..5400, 5401..5445, 5446..5490, 5491..5535, 5536..5580, 5581..5625, 5626..5670, 5671..5715, 5716..5760, 5761..5805, 5806..5850, 5851..5895, 5896..5940, 5941..5985, 5986..6030, 6031..6075, 6076..6120, 6121..6165, 6166..6210, 6211..6255, 6256..6300, 6301..6345, 6346..6390, 6391..6435, 6436..6480, 6481..6525, 6526..6570, 6571..6615, 6616..6660, 6661..6705, 6706..6750, 6751..6795, 6796..6840, 6841..6885, 6886..6930, 6931..6975, 6976..7020, 7021..7065, 7066..7110, 7111..7155, 7156..7200, 7201..7245, 7246..7290, 7291..7335, 7336..7380, 7381..7425, 7426..7470, 7471..7515, 7516..7560, 7561..7605, 7606..7650, 7651..7695, 7696..7740, 7741..7785, 7786..7830, 7831..7875, 7876..7920, 7921..7965, 7966..8010, 8011..8055, 8056..8100, 8101..8145, 8146..8190, 8191..8235, 8236..8280, 8281..8325, 8326..8370, 8371..8415, 8416..8460, 8461..8505, 8506..8550, 8551..8595, 8596..8640, 8641..8685, 8686..8730, 8731..8775, 8776..8820, 8821..8865, 8866..8910, 8911..8955, 8956..9000, 9001..9045, 9046..9090, 9091..9135, 9136..9180, 9181..9225, 9226..9270, 9271..9315, 9316..9360, 9361..9405, 9406..9450, 9451..9495, 9496..9540, 9541..9585, 9586..9630, 9631..9675, 9676..9720, 9721..9765, 9766..9810, 9811..9855, 9856..9900, 9901..9945, 9946..9990, 9991..10035, 10036..10080, 10081..10125, 10126..10170, 10171..10215, 10216..10260, 10261..10305, 10306..10350, 10351..10395, 10396..10440, 10441..10485, 10486..10530, 10531..10575, 10576..10620, 10621..10665, 10666..10710, 10711..10755, 10756..10800, 10801..10845, 10846..10890, 10891..10935, 10936..10980, 10981..11025, 11026..11070, 11071..11115, 11116..11160, 11161..11205, 11206..11250, 11251..11295, 11296..11340, 11341..11385, 11386..11430, 11431..11475, 11476..11520, 11521..11565, 11566..11610, 11611..11655, 11656..11700, 11701..11745, 11746..11790, 11791..11835, 11836..11880, 11881..11925, 11926..11970, 11971..12015, 12016..12060, 12061..12105, 12106..12150, 12151..12195, 12196..12240, 12241..12285, 12286..12330, 12331..12375, 12376..12420, 12421..12465, 12466..12510, 12511..12555, 12556..12600, 12601..12645, 12646..12690, 12691..12735, 12736..12780, 12781..12825, 12826..12870, 12871..12915, 12916..12960, 12961..13005, 13006..13050, 13051..13095, 13096..13140, 13141..13185, 13186..13230, 13231..13275, 13276..13320, 13321..13365, 13366..13410, 13411..13455, 13456..13500, 13501..13545, 13546..13590, 13591..13635, 13636..13680, 13681..13725, 13726..13770, 13771..13815, 13816..13860, 13861..13905, 13906..13950, 13951..13995, 13996..14040, 14041..14085, 14086..14130, 14131..14175, 14176..14220, 14221..14265, 14266..14310, 14311..14355, 14356..14400, 14401..14445, 14446..14490, 14491..14535, 14536..14580, 14581..14625, 14626..14670, 14671..14715, 14716..14760, 14761..14805, 14806..14850, 14851..14895, 14896..14940, 14941..14985, 14986..15030, 15031..15075, 15076..15120, 15121..15165, 15166..15210, 15211..15255, 15256..15300, 15301..15345, 15346..15390, 15391..15435, 15436..15480, 15481..15525, 15526..15570, 15571..15615, 15616..15660, 15661..15705, 15706..15750, 15751..15795, 15796..15840, 15841..15885, 15886..15930, 15931..15975, 15976..16020, 16021..16065, 16066..16110, 16111..16155, 16156..16200, 16201..16245, 16246..16290, 16291..16335, 16336..16380, 16381..16425, 16426..16470, 16471..16515, 16516..16560, 16561..16605, 16606..16650, 16651..16695, 16696..16740, 16741..16785, 16786..16830, 16831..16875, 16876..16920, 16921..16965, 16966..17010, 17011..17055, 17056..17100, 17101..17145, 17146..17190, 17191..17235, 17236..17280, 17281..17325, 17326..17370, 17371..17415, 17416..17460, 17461..17505, 17506..17550, 17551..17595, 17596..17640, 17641..17685, 17686..17730, 17731..17775, 17776..17820, 17821..17865, 17866..17910, 17911..17955, 17956..18000, 18001..18045, 18046..18090, 18091..18135, 18136..18180, 18181..18225, 18226..18270, 18271..18315, 18316..18360, 18361..18405, 18406..18450, 18451..18495, 18496..18540, 18541..18585, 18586..18630, 18631..18675, 18676..18720, 18721..18765, 18766..18810, 18811..18855, 18856..18900, 18901..18945, 18946..18990, 18991..19035, 19036..19080, 19081..19125, 19126..19170, 19171..19215, 19216..19260, 19261..19305, 19306..19350, 19351..19395, 19396..19440, 19441..19485, 19486..19530, 19531..19575, 19576..19620, 19621..19665, 19666..19710, 19711..19755, 19756..19800, 19801..19845, 19846..19890, 19891..19935, 19936..19980, 19981..20025, 20026..20070, 20071..20115, 20116..20160, 20161..20205, 20206..20250 | 6,544 | 18,724 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 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http://www.algebra.com/algebra/homework/logarithm/logarithm.faq.question.605412.html | 1,369,434,633,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705195219/warc/CC-MAIN-20130516115315-00014-ip-10-60-113-184.ec2.internal.warc.gz | 307,576,592 | 5,273 | # SOLUTION: log(6x+3)=1+log(x-6)
Algebra -> Algebra -> Logarithm Solvers, Trainers and Word Problems -> SOLUTION: log(6x+3)=1+log(x-6) Log On
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Algebra: Logarithm Solvers Lessons Answers archive Quiz In Depth
Question 605412: log(6x+3)=1+log(x-6)Found 2 solutions by stanbon, bucky:Answer by stanbon(57377) (Show Source): You can put this solution on YOUR website!log(6x+3)=1+log(x-6) ---- log(6x+3) - log(x-6) = 1 ---- log[(6x+3)/(x-6)] = 1 ---- (6x+3)/(x-6) = 10 --- 6x+3 = 10x-60 4x = 63 x = 63/4 ============== cheers, Stan H. ================= Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!Given to solve for x . . Subtract log(x-6) from both sides to get: . . By the rules of logarithms, the subtraction of two logarithms is the same as the logarithm of their division with the divisor being the quantity in the negative logarithm. So we can writ that the above equation is equivalent to: . . Next convert this to exponential form by raising the base (base 10) of the logarithm to the power which is on the right side of the equal sign and setting this equal to the quantity that the logarithm operator is acting on. This results in: . . The 10 to the first power is just 10. So the equation is: . . Get rid of the denominator on the left side by multiplying both sides by x - 6: . . . Do the distributed multiplication on the right side by multiplying 10 times each of the two terms in the parentheses, and the result is: . . Subtract 10x from both sides: . . Subtract 3 from both sides: . . Multiply both sides by -1: . . Solve for x by dividing both sides by 4 and you arrive at the solution: . . . Hope this helps you to understand how this problem can be worked by using the properties of logarithms in a logical way. . | 545 | 1,960 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2013-20 | latest | en | 0.88964 |
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# Test: Grouping And Tabulation Of Data
## 10 Questions MCQ Test Mathematics (Maths) Class 9 | Test: Grouping And Tabulation Of Data
Description
This mock test of Test: Grouping And Tabulation Of Data for Class 9 helps you for every Class 9 entrance exam. This contains 10 Multiple Choice Questions for Class 9 Test: Grouping And Tabulation Of Data (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Grouping And Tabulation Of Data quiz give you a good mix of easy questions and tough questions. Class 9 students definitely take this Test: Grouping And Tabulation Of Data exercise for a better result in the exam. You can find other Test: Grouping And Tabulation Of Data extra questions, long questions & short questions for Class 9 on EduRev as well by searching above.
QUESTION: 1
Solution:
QUESTION: 2
### The class mark of class interval 60 – 70 will be
Solution:
class mark = [lower class interval + upper class interval]/2
= (60+70)/2
= 130/2
= 65
(if the intervals are of uniform width)
QUESTION: 3
### The consecutive class marks of a data having continuous class intervals having class width as 5, have a gap of ________ between them.
Solution:
QUESTION: 4
Number of runs scored by a cricket player in 25 innings are as follows:
26, 35, 94, 48, 82, 105, 53, 0, 39, 42, 71, 0, 64, 15, 34, 67, 0, 42, 124, 84, 54, 48, 139, 64, 47.
The number of centuries scored by him is
Solution:
QUESTION: 5
If the class marks in a frequency distribution are 19.5, 26.5, 33.5, 40.5 then the class corresponding to the class mark 33.5 is:
Solution:
QUESTION: 6
A curious 12th class student wants to know and collect the data regarding the percentage of students who got grade A1 in Mathematics during the last 10 years in Board examinations. This collected data is known as
Solution:
Secondary data refers to data that was collected by someone other than the user. Common sources of secondary data for social science include censuses, information collected by government departments, organisational records and data that was originally collected for other research purposes.
QUESTION: 7
Class size for the following distribution: 0 – 0.25, 0.25 – 0.50, 0.50 – 0.75 is
Solution:
QUESTION: 8
Three coins were tossed thirty times. Each time the number of heads occurring was noted as follows:
0, 1, 2, 2, 1, 2, 3, 1, 3, 0
1, 3, 1, 1, 2, 2, 0, 1, 2, 1
3, 0, 0, 1, 1, 2, 3, 2, 2, 0
The frequency distribution table for the given data will be
Solution:
QUESTION: 9
Class mark of a class interval U-L is
Solution:
QUESTION: 10
In the following data, the number of class intervals, (by inclusive method), and taking 4 as the magnitude of class intervals will be
31, 23, 19, 29, 22, 20, 16, 10, 13, 34
38, 33, 28, 21, 15, 18, 36, 24, 18, 15
12, 30, 27, 23, 20, 17, 14, 32, 26, 25
18, 29, 24, 19, 16, 11, 22, 15, 17, 10
Solution: | 885 | 2,901 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2021-39 | latest | en | 0.905239 |
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Fibonacci Series
The Fibonacci Series is a set of numbers that regularly occur in nature. They can be calculated by starting with the numbers 0 and 1. Keep adding the last two numbers to find the next number in the series.
example
0,1 (0 + 1 = 1) so next number is 1
0,1,1 (1 + 1 = 2) so next number is 2
0,1,1,2 (1 + 2 = 3) so next number is 3
0,1,1,2,3 (2 + 3 = 5) so next number is 5
0,1,1,2,3,5 .......
Here are the first few numbers in the Fibonacci Series
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, ....
Lucas Series
There is another similar series of numbers called Lucas Numbers. These are calculated in exactly the same way, but you start with 2,1 (instead of 0,1)
Here are the first few numbers in the Lucas Series
2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843, 1364, 2207, 3571, 5778, 9349, 15127, 24476, 39603, 64079, 103682, 167761, 271443, 439204, 710647, 1149851, 1860498, 3010349, 4870847, ....
Pascal's Triangle
Each number in the Pascal Triangle is calculated by adding together the two numbers directly above it. (Every line starts and ends with 1.)
The numbers in the Pascal Triangle appear in calculations like
(x + 1)n
e.g. (x + 1)6 uses the numbers from the last line of the triangle
(x + 1)6 = 1x6 + 6x5 + 15x4 + 20x3 + 15x2 + 6x + 1
1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1
Triangular Number Series | 647 | 1,588 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2022-21 | longest | en | 0.783125 |
http://kb.eng-software.com/display/ESKB/Static+Head+in+Systems+with+Varying+Fluid+Temperatures | 1,542,390,962,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743110.55/warc/CC-MAIN-20181116173611-20181116195611-00478.warc.gz | 178,452,088 | 13,219 | Page tree
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Q. What is static head and how is the static head calculated for a system that contains a fluid at different temperatures?
A. The static head of a system is that portion of the pump’s Total Head that must be added to the fluid before a single drop of fluid will move in the system. It is the amount of energy, or head, which is used to overcome the pressure and elevation differences in the piping system. It can be calculated using the following equations:
where:
• static, elevation, and pressure head are in units of feet of fluid
• pressure in units of lb/in2 (psi)
• density (ρ) in lb/ft3 of the fluid in the pump
If the system contains fluids at different temperatures the fluid density will vary. So which fluid density should be used in the equation: the density of the hotter fluid, the colder fluid, or an average density? How much difference does it make? Does it affect the differential pressure head or differential elevation head component of static head?
At first glance , based on the equations, it appears that the fluid density will affect the differential pressure head component. But the pressure difference between the two tanks has to be overcome regardless of the densities of the fluid between the two tanks. And since static head is a portion of the Total Head at the pump, the density of the fluid in the pump is used in the pressure head component of the static head calculation.
It is actually the differential elevation head component that is affected by the fluid density. A column of fluid of a given height will equate to a pressure at the bottom of the column, which will be different compared to the same column of fluid with a different density. It is the differential elevation head component that must be adjusted to ensure the terms of elevation head are related to the fluid at the pump.
Consider the following open systems in which water is pumped from a supply tank (bottom elevation = 0 ft, liquid level = 10 ft, and surface pressure = 0 psig) to an elevated product tank (bottom elevation = 150 ft, liquid level = 10 ft, and surface pressure = 10 psig).
The first system is pumping water at 60°F and the second system has a heat exchanger at the 25 ft elevation that heats the water up to 200°F.
The calculation for static head in the first system is:
This value of static head can be confirmed using PIPE-FLO by selecting Graph Resistance Curve from a pump's context menu, and looking at the Head value where the blue line intersects the Y axis.
The Y intercept is this screen shot is 173 ft. It is circled in red and shown at the bottom of the graph by placing the curser on the line.
In the second system, the water from the liquid surface in the supply tank to the inlet of the heat exchanger (at 25 ft elevation) has a density of 62.37 lb/ft3, but the water from the heat exchanger to the liquid surface in the product tank has a density of 60.11 lb/ft3. To calculate the static head requires adjusting the elevation head between the heat exchanger and product tank by the ratio of the fluid densities to put it in terms of head at the pump:
This value of static head can also be confirmed using PIPE-FLO:
The Y intercept in this screen shot is 168 ft. It is circled in red and shown at the bottom of the graph by placing the curser on the line.
The difference in the calculated static head between the two systems may or may not be significant enough to justify calculating the density correction in the system, depending on the degree of accuracy required in the calculations and the magnitude of the difference between the fluid densities.
For reference to older versions of PIPE-FLO see attachments below.
Static Head in Systems with Varying Fluid Temperatures old.pdf
1. Reviewed | 806 | 3,782 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2018-47 | latest | en | 0.9389 |
https://brainly.my/tugasan/235449 | 1,487,745,141,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170914.10/warc/CC-MAIN-20170219104610-00471-ip-10-171-10-108.ec2.internal.warc.gz | 687,521,674 | 10,134 | # Macam mana nak jwab????
2
dari Icepreec
## Jawapan
2016-09-25T23:50:59+08:00
A)perimeter seluruh darjah: 7 + 7 + 7 + ( 60/360 x 2 x22/7 x 7) + 7 = ???
2016-09-26T00:36:59+08:00
PQ=14cm, T = titik tengah PQ, so PT=TQ=14/2=7cm
RS=SP=PU=7cm
π=22/7
a)
lengkok TU = (60/360)×2×(22/7)×7=22/3 cm
lengkok QR = (120/360)×2×(22/7)×14=88/3 cm
perimeter seluruh rajah = RS+SP+PU+TQ+lengkok TU+lengkok QR
= (4×7)+22/3+88/3
= 194/3 cm
b)
luas QRST = luas PQR - luas PST
= (120/360)×(22/7)×14² - (120/360)×(22/7)×7²
= 562/3 cm²
luas PTU
= (60/360)×(22/7)×7²
= 77/3 cm²
luas kaw. berlorek = luas QRST + luas PTU
= 562/3+77/3
= 213 cm² | 363 | 630 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2017-09 | latest | en | 0.164562 |
http://www.gea-web.com/blog/2016/03/04/Slovenian-Nautical-Almanac---Part-3---Astrometric-Coordinates | 1,695,552,666,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506632.31/warc/CC-MAIN-20230924091344-20230924121344-00547.warc.gz | 63,137,849 | 7,196 | # Slovenian Nautical Almanac - Part 3 - Astrometric Coordinates
04.03.2016
Aljoša Kocen
In Part 2 We calculated SSB cetred Rectangular coordinates for planets. We are now able to calculate X, Y, and Z coordinates for any instance in TDB for any planet and Sun. In this post, we will convert these coordinates to Geocentric Equatorial Coordinates, namely:
• Right Ascension, which is an angle, measured along celestial equator, between celestial body and First point of Aries (i.e. point, where plane of Ecliptic crosses the plane of Equator). it is measured in Hours, minutes, and seconds in eastward direction
• Declination, an angle between Body and celestial equator. It is measured in angular degrees, minutes, and seconds, either Noth or South of Celestial Equator
Given that we already know rectangular coordinates on the same plane, the calculation is rather straightforward:
• `RA = Atan2(Y, X)`
• `Dec = Atan(Z / Sqrt(X * X + Y * Y)`
Note that we are using Atan2 function in the calculation of Right Ascension, instead of a simple Atan function. The rational behind this is that Atan2 function automatically registers the correct quadrant, in which the angle is. The formula for computing tangens of an angle in rectangular coordinate system is `tan(φ) = Y / X`. Now let's say that we measure angle in clock-wise direction, with point, say (0,1) being Zero. If we have coordinates (X,Y) = (1,1), formula above would yield 1, and corresponding angle would be 45°. Bit if the coordinates would be (-1,-1), equation would also return 1, and Angle measured would also be 45°. In reality, the angle would be 225°.
Net Framework Math.Atan2 function returns the angle in radians between -π/2 and +π/2. However, because RA is measured in HMS notation, going around full circle, from 0h to 24h, we need to translate this into a positive number. Putting it simply, if RA is negative by using equation above, add 2π to it.
## Reducing SSB centered coordinates to Earth centered coordinates.
The coordinates, given by JPL DE Ephemerides are based on Solar System Barycenter (Solar system Center of Mass). To get RA and Declination, we need to reduce these coordinates to coordinate system, based on center of the Earth. The problem we are facing now is that JPL DE Ephemerides enable us getting XYZ values for Earth - Moon Barycenter. The moon itself does not revolve strictly around the center of the Earth. Instead, it revolves around the center of mass of the Earth - Moon Couple. To get true Earth XYZ Coordinates from tabulated XYZ coordinates of EMB, we need to perform what is called Barycentring.
• From JPL DE Ephemeris, obtain Celestial Body coordinates Xb, Yb, and Zb
• From JPL DE Ephemerides, obtain Earth-Moon-Barycenter coordinates Xemb, Yemb, and Zemb
• From JPL DE Ephemerides, obtain Geocentric rectangular coordinates of the Moon, namely Xm, Ym, and Zm.
All coordinates will be returned in kilometers. We also need to know the Earth-Moon Mass ratio (EMR), which is 81,3005800000000044
To obtain Rectangular coordinates for the Earth (Xe, Ye, and Ze), we look at the solution here
• Xe = Xemb - (Xm / (1 + EMR))
• Ye = Yemb - (Ym / (1 + EMR))
• Ze = Zemb - (Zm / (1 + EMR))
Now we have rectangular coordinates, based on Solar system BaryCenter for Earth. Let us define geocentric coordinates for the Body as XB, YB, and ZB. We proceed by calculating them as follows:
• XB = Xb - Xe
• YB = Yb - Ye
• ZB = Zb - Ze
Now to calculate Right ascension and Declination for the body, we use the two formulas above.
• RA = Atan2(YB / XB)
• Dec = Atan(ZB / Sqrt(YB * YB + XB *XB))
### An Example:
Let us calculate Right Ascension and Declination for Jupiter on January 17, 2006 at 12:00:00 TDB:
Julian Day for the set date and time of day equals 2453753,0. Calculated coordinates for EMB, Jupiter, and Moon for this instance are as follows:
First we will calculate Earth coordinates:
• Xe = Xemb - (Xm / (1 + EMR)) = -66.596.740.39 - (-354.436,33 / (1 + 81,30058)) = -66.592.433,78
• Ye = Yemb - (Ym / (1 + EMR)) = -120.471.863,66 - (172.075.13 / (1 + 81,30058)) = 120.469.772,85
• Ze = Zemb - (Zm / (1 + EMR))= 52.20.9.922.09 - (97.343.28 / (1 + 81,30058)) = 52.208.739,31
To obtain Gocentric coordinates for Jupiter:
• XB = Xb - Xe = -659.559.726,77 - (-66.592.433,78) = - 592.967.292,99
• YB = Yb - Ye = -442.240.720.21 - 120.469.772,85 = - 562.710.493,06
• ZB = Zb - Ze = -173.503.731,00 - 52.208.739,78 = - 225.712.470,31
Calculated Geocentric coordinates for Jupiter would therefore yield:
• RA: -2,38236951 Radians, truncated to positive values equal 3,90081580 or in Hour-Minute-Second (HMS) Notation: 14h 54m 0,07s
• Declination: -0,26939956 Radians, which is -15,43545806 degrees, or in proper notation: S 15° 26' 07,65"
Our VB.Net code to calculate this is shown below:
``` Private Sub CalculatePosition(ByVal TBD As DateTime, ByVal CelBody As CelestialBody)
'PositionCalculation Object
Dim PosCalc As New PositionCalculation
'CartesianCoordinates Objects for Body, EMB and Moon
Dim Body As New CartesianCoordinates
Dim EMBPos As New CartesianCoordinates
Dim Moon As New CartesianCoordinates
'Julian Day Number
Dim JD As Decimal = New DateTime(dtpdateTime.Value.Year,
dtpdateTime.Value.Month,
dtpdateTime.Value.Day,
12, 0, 0).ToJulianDay
'SSB based Body Coordinates
Body = PosCalc.GetBodyPosition(CelBody, JD)
'SSB EMB Coordinates
EMBPos = PosCalc.GetBodyPosition(bodies(2), JD)
'Geocentric position of the moon
Moon = PosCalc.GetBodyPosition(bodies(9), JD)
'SSB based Earth coordinates
Dim Earth As New CartesianCoordinates
Earth.X = EMBPos.X - Moon.X / (1 + Constants.EarthMoonMassRatio)
Earth.Y = EMBPos.Y - Moon.Y / (1 + Constants.EarthMoonMassRatio)
Earth.Z = EMBPos.Z - Moon.Z / (1 + Constants.EarthMoonMassRatio)
'Earth based rectangular coordinates of the body.
Dim BodyEarth As CartesianCoordinates = Body - Earth
Dim X As Decimal = BodyEarth.X
Dim Y As Decimal = BodyEarth.Y
Dim Z As Decimal = BodyEarth.Z
'Astrometric Coordinates
Dim Dec As Decimal = Math.Atan(Z / Math.Sqrt(X ^ 2 + Y ^ 2))
Dim RA As Decimal = Math.Atan2(Y, X)
'Truncate Right Ascension if negative, or if larger than 2*PI.
End Sub```
## Correcting coordinates for Light time
Sun emits light by itself. All other bodies in our solar system don't. What we see is Sun's light reflection from them. The speed of light is roughly 300.000 km/s (299.792,458 km/s to be exact). Above sample can quickly show us, that the distance between Jupiter and Earth equals 848.056.265,11 kilometers. That is some 21200 times Earth circumference, which roughly equals 40.000 km. Light needs to travel from the planet to reach Earth. in the example above, light travel from Jupiter to Earth would last some 2828,81 seconds, or roughly 47 minutes. Even if Earth would be stationary, the calculation above defines coordinates as per the exact instance given. In other words. The position od Jupiter, calculated above, is not the position, which we, on Earth, see at January 17, 2006 at 12:00:00 TBD. It is the position, which we WILL see some 47 minutes later. What we need is XYZ coordinates for the jupiter at some earlier instance. The problem we face is that we do not know in advance, what is the correct Light travel time between the Celestial body and the Earth. We solve this by iteration
1. Calculate XYZ coordinates for Body and the Earth, as shown above.
2. Calculate the distance between Body and the Earth by using formula: Dist = SQRT(X*X + Y*Y + Z*Z)
3. Get Light travel time (lt) for the distance calculated above.
4. Subtract light from Julian day.
5. Recalculate Cordinates XYZ for the Body. (but not Earth)
6. Repeat steps 2 - 5 until a convergence in XYZ positions is achieved
I've noticed that, in practice, only a single pass is usually required, as more than sufficient accuracy is achieved for our purposes. Let us repeat the calculation and include the light time correction:
To calculate the distance between Jupiter and Earth, we will use Geocentric coordinates XB, YB, and ZB, as they were calculated above. The distance between the two bodies is then:
Dist = SQRT(XB * XB + YB * YB + ZB * ZB) = 848.056.265,12 km.
Light travel time for this distance (based on the light speed of 299.792,458 km/s equals 2828,81 seconds. There are 86400 seconds in each day, this would amount to 0,03274087045739 day. Subtracting this from Julian day in question ( 2453753,0) we get the instance at which we need to calculate SSB rectangular coordinates for Jupiter: 2453752,96725913. It is this Julian day we use to calculate new SSB coordinates for Jupiter:
• Xb = - 659.580.860,36
• Yb = - 442.214.938,44
• Zb = - 173.492.165,44
Implementing these values instead of the ones obtained before, our geocentric equatorial coordinates become:
• RA = 3,90077512348832rad or 223,4979514 deg or in proper notation 14h 53m 59,5s
• Declinatination = -0,269387177 rad or -15,43474831 deg, and in the proper notation S 15° 26' 5,09"
These values represent Astrometric Equatorial Coordinates for Jupiter on January 17,2006 at 12:00:00 TDB FOR THE REFERENCE FRAME IERS or J2000.
The complete code, which this is shown below:
``` Private Sub CalculatePosition(ByVal TBD As DateTime, ByVal CelBody As CelestialBody)
'PositionCalculation Object
Dim PosCalc As New PositionCalculation
'CartesianCoordinates Objects for Body, EMB and Moon
Dim Body As New CartesianCoordinates
Dim EMBPos As New CartesianCoordinates
Dim Moon As New CartesianCoordinates
'Julian Day Number
Dim JD As Decimal = New DateTime(TBD.Year,
TBD.Month,
TBD.Day,
12, 0, 0).ToJulianDay
'SSB based Body Coordinates
Body = PosCalc.GetBodyPosition(CelBody, JD)
'SSB EMB Coordinates
EMBPos = PosCalc.GetBodyPosition(bodies(2), JD)
'Geocentric position of the moon
Moon = PosCalc.GetBodyPosition(bodies(9), JD)
'SSB based Earth coordinates
Dim Earth As New CartesianCoordinates
Earth.X = EMBPos.X - Moon.X / (1 + Constants.EarthMoonMassRatio)
Earth.Y = EMBPos.Y - Moon.Y / (1 + Constants.EarthMoonMassRatio)
Earth.Z = EMBPos.Z - Moon.Z / (1 + Constants.EarthMoonMassRatio)
'Earth based rectangular coordinates of the body.
Dim BodyEarth As CartesianCoordinates = Body - Earth
Dim X As Decimal = BodyEarth.X
Dim Y As Decimal = BodyEarth.Y
Dim Z As Decimal = BodyEarth.Z
'Astrometric Coordinates
Dim Dec As Decimal = Math.Atan(Z / Math.Sqrt(X ^ 2 + Y ^ 2))
Dim RA As Decimal = Math.Atan2(Y, X)
'Truncate Right Ascension if negative, or if larger than 2*PI.
'In order to correct for light travel time, we add the following:
'Distance between earth and Jupiter in kilometers
Dim Distance As Decimal = BodyEarth.Distance
'Light travel time:
Dim lt As Decimal = Distance / (299792.458 * 86400)
'Get a modified Julian Day (MJD)
Dim MJD As Decimal= JD - lt
'Get New Body position for Modified Julian Day
Body=PosCalc.GetBodyPosition(CelBody,MJD)
'Geocentric XYZ coordinates:
BodyEarth= Body - Earth
X=BodyEarth.X
Y=BodyEarth.Y
Z=BodyEarth.Z
'Astrometric geocentric coordinates (in radians)
Dec = Math.Atan(Z / Math.Sqrt(X ^ 2 + Y ^ 2))
RA = MathFunctions.TruncateRad(Math.Atan2(Y, X))
End Sub```
## WHat the hell is TDB, and on which clock is measured.
Up till now, we used Terrestial Dynamic Barycentric Time as an input to our calculations. This is dynamic time scale, and is not shown on any clock. We will discuss different time scales and their relations in my next post.
Best regards. | 3,186 | 11,419 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2023-40 | latest | en | 0.895619 |
http://slidegur.com/doc/20888/so-machines-2014---marshall-middle-school | 1,481,332,505,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542932.99/warc/CC-MAIN-20161202170902-00069-ip-10-31-129-80.ec2.internal.warc.gz | 247,404,112 | 13,588 | ### SO-Machines 2014 - Marshall Middle School
```What do Machines do?
Do they allow one to do more work?
Not really, at best they make completing a task easier.
So then what do Machines do?
• Multiply the force.
• Multiply the distance.
• Change the direction of the force.
Work = Force x Distance an object moves
while the force is applied.
W=Fxd
In SI Units:
Force is measured in newtons (N)
distance is measured in meters (m)
Work in N.m which is a joule (J).
Named after James Prescott Joule
What does work do?
Work causes a change in Energy. In
other words, it can do any of the
following:
• Make something move faster.
• Lift something up.
• Move something against friction.
• A combination of the above.
Examples of Work:
A cart is pushed to the right as illustrated.
50.0 N
distance cart is moved 4.00 m.
How much work is done on the cart?
W = F x d = (50.0 N)(4.00 m) = 200. J
Examples of Work:
W=Fxd
W = (80.0 N)(20. m)
W = 1600 J
80.0 N
How much work is done on
the box?
distance box is moved 20. m.
A box is lifted as illustrated.
Now let’s apply this to
some of our machines.
The simplest is most
likely levers.
Class 1 Lever: load on one side of the
fulcrum and the effort on the other side.
Effort
The Fulcrum is the pivot point.
The Load is what we are trying to
lift or the output of the machine.
The Effort is the force that is applied to
lift the load or the input of the machine.
Load
Fulcrum
Class 1 Lever: Load on one side of the
fulcrum and the Effort on the other side.
Effort
Load
Fulcrum
Class 1 Lever: More terminology
Note that in lifting the load the Effort moved
much farther than the Load.
Effort
With a smaller Effort we could lift a
Load that is heavier.
Load
Fulcrum
Class 1 Lever: More terminology
Fulcrum
Load = 900N
Effort distance, dE = 60 cm
Effort = 300 N The Effort moved 60 cm
while the Load moved only
20 cm. We moved 3 times
farther than the LOAD.
Load distance, dL = 20 cm
With an Effort of 300 N we were able to lift a Load of
900 N. We multiplied the input force by 3.
Class 1 Lever: More terminology
Work, W = F x d
WIN = E x dE = (300 N)(0.60 m) = 180 J
Fulcrum
Load = 900N
Note: WorkIN = WorkOUT
We didn’t do more work,
we just did it with less
Effort than if I tried to lift
it without the lever.
Load distance, dL = 20 cm
Effort distance, dE = 60 cm
WOUT = L x dL = (900 N)(0.20 m) = 180 J
Effort = 300 N
Class 1 Lever: More terminology
MA = Load/Effort
MA = (900 N)/(300 N)
MA = 3
MA = dE/dL
MA = (60 cm)/(20 cm)
MA = 3
Fulcrum
Load = 900N
Effort distance, dE = 60 cm
Effort = 300 N
Load distance, dL = 20 cm
We say that we have a Mechanical Advantage, MA.
We can lift 3 times more than our input Effort.
Class 1 Lever: More terminology
MA = Load/Effort
MA = (900 N)/(300 N)
MA = 3
MA = dE/dL
MA = (60 cm)/(20 cm)
MA = 3
Fulcrum
Load = 900N
Effort distance, dE = 60 cm
Effort = 300 N
Load distance, dL = 20 cm
We triple our Effort (input force) at the expense of
moving the Load ⅓ as much.
Class 1 Lever: More terminology
We can also analyze the lever by measuring the
distance from the Effort to the fulcrum (pivot point)
and the distance from the Load to the fulcrum.
This is called the lever arm or just arm and is often
given the variable name “x.”
Effort arm, xE = 3.0 m
Load arm, xL = 1.0 m
Fulcrum
Class 1 Lever: More terminology
This can also be used to calculate the
Mechanical Advantage.
MA = xE/xL = (3.00 m)/(1.00 m) = 3
Looks familiar doesn’t it.
Effort arm, xE = 3.0 m
Load arm, xL = 1.0 m
Fulcrum
More terminology:
Often we use the terms, Ideal Mechanical
Advantage, IMA and Actual Mechanical
Advantage, AMA
IMA = xE/xL or dE/dL
AMA = L/E with the load being just what you
ultimately wanted to move, excluding anything
else that may have to be moved with it.
This will become clearer when we look at a 2nd class lever.
More terminology:
Often we use the terms, Ideal Mechanical
Advantage, IMA and Actual Mechanical
Advantage, AMA
We want to find out how well the particular
machine does its work. This is called
Efficiency, Eff.
Efficiency, Eff = (AMA/IMA) x 100%
2nd class lever:
Notice that the Effort and the Load
are on the same side of the Fulcrum
and the Load is between the Effort
and the Fulcrum.
Effort
Load
Fulcrum
2nd class lever:
Again the Effort moves much farther than the Load. We
are getting more force out than what we put in, but the
load only moves a short distance. We are also lifting
the lever along with the load.
Fulcrum
2nd class lever:
IMA = xE/xL = (2.8 m)/(0.35 m) = 8
The load is 900 N and since the Effort has
to lift the Load and the lever, let’s say that
the Effort is 150 N.
Effort
Effort arm, xE = 2.8 m
Load arm, xL = 0.35 m
Load
Fulcrum
2nd class lever:
IMA = xE/xL = (2.8 m)/(0.35 m) = 8
AMA = L/E = (900 N)/(150 N) = 6
Eff = AMA/IMA = 6/8 x 100% = 75%
E = 150N
L = 900 N
Effort arm, xE = 2.8 m
Load arm, xL = 0.35 m
Fulcrum
3rd class lever:
Notice that the Effort and the Load
are on the same side of the Fulcrum
and the Effort is between the Load
and the Fulcrum.
Load
Effort
Fulcrum
3rd class lever:
IMA = xE/xL = (0.50 m)/(2.0 m) = ¼ = 0.25
AMA = L/E = (200 N)/(1000 N) = ⅕ = 0.2
Eff = AMA/IMA = (0.2/0.25) x 100% = 80%
L = 200 N
E = 1000 N
xL = 2.0 m
xE = 50. cm
Fulcrum
There is a lab part of the competition.
Let’s look at some of the basic concepts
for a lever that is in static equilibrium.
The easiest lever to analyze is the first class
lever (seesaw), that is balanced by itself.
The center of gravity of the lever is on the
fulcrum.
c.g.
If a lever is not moving (rotating) then it is said to
be at static equilibrium. When an object is at
static equilibrium the following is true:
ΣF = 0, that is netF = 0, no unbalanced forces.
Στ = 0, that is there are no unbalanced torques.
If you place a seesaw so that its center of gravity
is on the fulcrum, it will balance. That is, the left
side balances the right side.
Torque is the tendency of a force to
cause an object to rotate around an
axis. In the case of a lever, the axis is
the fulcrum.
Force
In this case the force would make the left
side of the lever go down or rotate the
lever counterclockwise, ccw.
Torque is the tendency of a force to
cause an object to rotate around an
axis. In the case of a lever, the axis is
the fulcrum.
Force
In this case the force would make the left
side of the lever go down or rotate the
lever counterclockwise, CCW.
What if the force is 24 N, what torque is applied?
Earlier we talked about the lever arm or arm being
the distance from the fulcrum (axis) to the force. We
will use the letter “x” as the symbol for lever arm.
F = 24 N
The symbol for torque is the Greek letter tau, τ
x = 1.2 m
τ = (F)(x) = (24 N)(1.2 m)
τ = 28.8 N.m = 29 N.m
A torque of 29 N.m will rotate the lever CCW.
The weight of the seesaw on the left creates a torque
that tries to rotate the seesaw counter-clockwise, CCW,
so that the left side would go down.
The weight of the seesaw on the right creates a torque
that tries to make it rotate clockwise,CW, so that the right
side would go down.
F
F
The two balance each other
and it does not rotate.
Another way to look at this is that we can place all
the weight of the seesaw ( FL ) at its center of
gravity.
The center of gravity of the seesaw is at the axis of
rotation (fulcrum) so the value of the lever arm is
zero and the force creates no torque.
c.g.
Note: The center of
gravity may not be at
the geometric center.
FL
Especially when
using wooden
meter sticks!
Sample problem: Two identical 40.0 kg twin girls
are sitting on opposite ends of a seesaw that is 4.0
m long and weighs 700 N, so that the center of
gravity of the seesaw is on the fulcrum.
How do we analyze this situation?
c.g.
First, we need to draw a torque diagram of the seesaw.
This is a free body diagram which includes the lever arms.
FN = 1500 N
Next, we define the axis of
rotation (circle with a dot
in the middle) and the
lever arms.
We place all the forces at
their proper location.
x2 = 2.0 m
x1 = 2.0 m
c.g.
.
F1 = 400 N
FL = 700 N
F2 = 400 N
Στ = 0 or ΣτCCW = ΣτCW
F1x1 = F2x2
FN = 1500 N
(400 N)(2.0 m) = (400 N)(2.0 m)
FL and FN both act through
the axis of rotation, so their
lever arm is zero, making
their torque 0.
The torques balance so the
seesaw can be in static
equilibrium.
800 Nm = 800 Nm
x2 = 2.0 m
x1 = 2.0 m
c.g.
.
F1 = 400 N
FL = 700 N
F2 = 400 N
It is important that the seesaw be
level, so that the force applied by each
of the girls is acting downward and is
perpendicular to the lever arm. If the
Force and the Lever Arm are not
perpendicular, then the equation for
the torque becomes complex. It is
better that we avoid that situation.
So, what do you do to balance the seesaw if the two
people are not the same weight (mass)?
One 400 N girl sits on one end of a seesaw that is
centered on the fulcrum, is 4.0 m long, and weighs 700 N.
Where must her 650 N brother sit in order for the seesaw
to be in static equilibrium?
?
c.g.
Option #1, move the
heavier person closer to
the fulcrum.
FL and FN both act through
the axis of rotation, so their
lever arms are zero.
Στ = 0 or ΣτCCW = ΣτCW
FGxG = FBxB
FN = 1750 N
xB = (800 Nm)/(650 N)
(400 N)(2.0 m) = (650 N)xB
xB = 1.23 m
800 = 650x
B
xG = 2.0 m
xB = ?? m
c.g.
.
FG = 400 N
FL = 700 N
FB = 650 N
Option #2, move the center of gravity of the seesaw so that
more of the seesaw is on the side of the lighter person,
One 400 N girl sits on one end of a 4.0 m long seesaw
weighing 700 N That has moved the center of gravity of
the lever 0.2 meters towards her. Where must her 650 N
brother sit in order for the seesaw to be in static
equilibrium?
?
c.g.
Now the weight of the
seesaw creates a torque
helping the girl.
FN acts through the axis of
rotation, so its lever arm is zero.
Στ = 0 or ΣτCCW = ΣτCW
FGxG + FLxL = FBxB
FN = 1750 N
(400)(2.2) + (700)(0.2) = (650 N)xB
xB = (1020 Nm)/(650 N)
880 + 140 = 650xB
xB = 1.57 m
xG = 2.2 m
xB = ?? m
c.g.
.
xL = 0.2 m
FG = 400 N
FL = 700 N
FB = 650 N
For the Middle School
(Division B) Competition,
you will need to build a
simple first class lever
system. The lever may not
be longer than 1.00 meter.
Simple Machines. (Simple case.)
Given small mass placed on one side.
Given unknown large mass on the other.
Unless the values are too extreme, you may be able to
move the large mass close enough to the fulcrum.
c.g.
If this is the setup, you
don’t have to worry about
the weight of the lever.
Simple Machines. (Simple case.)
c.g.
Big Mass
Small Mass
FN
xS
xB
c.g.
.
FS
FL
FB
Simple Machines. (Simple case.)
In this case the torque equation is: τCCW = τCW
(FS)(xS) = (FB)(xB) and you can solve for any value.
FN
xS
xB
c.g.
.
FS
FL
FB
So far we have been dealing with the force applied by the
hanging mass. This force is known as the weight of the
object or the force of gravity (Fg) acting on the object.
The force of gravity acting on an object is the product of
the mass of the object multiplied by the gravity constant
on the planet Earth (9.8 N/kg).
Fg = mg = m(9.8 N/kg) so, mass, m = Fg/(9.8 N/kg)
This gets quite confusing because weight is measured in
newtons and mass is measured in grams or kilograms (kg).
You may have been told to weigh something but you
actually measured its mass in grams.
Simple Machines. (Simple case.)
Knowing that Fg = mg
(FS)(xS) = (FB)(xB) This equation can be written:
(msg)(xS) = (mBg)(xB) Dividing by g we get:
(msg)(xS)/g = (mBg)(xB)/g
FN
xS
(ms)(xS) = (mB)(xB)
xB
c.g.
.
FS = mSg
FL
F B = m Bg
Simple Machines. (Simple case.)
We can now solve for a mass using this equation
and modify our torque diagram as shown:
(ms)(xS) = (mB)(xB)
FN
xS
xB
c.g.
.
mS
FL
mB
Simple Machines. (Simple case.)
Suppose that you were given a small mass of 125
grams and an unknown large mass. You set up
your lever so it balances as shown:
FN
xB = 10.0 cm
xS = 47.6 cm
c.g.
.
mS = 125 g
FL
mB = ??
Simple Machines. (Simple case.)
(ms)(xS) = (mB)(xB)
(125 g)(47.6 cm) = (mB)(10.0 cm)
mB = (125 g)(47.6 cm)/(10.0 cm) = 595 grams
FN
xB = 10.0 cm
xS = 47.6 cm
c.g.
.
mS = 125 g
FL
mB = ??
Simple Machines (More realistic case)
If the difference between the unknown mass and
the known mass is large, move the fulcrum near
one end of the lever.
Place the unknown large mass on the short side.
Then place the small mass on the long side.
c.g.
Now the lever helps
balance the large weight.
FN acts through the axis of
rotation, so its lever arm is zero.
Στ = 0 or ΣτCCW = ΣτCW
FSxS + FLxL = FBxB
FN
xB = cm
xS = cm
c.g.
.
xL = cm
FS =
N
FL = N
FB = ?? N
FN acts through the axis of
rotation, so its lever arm is zero.
Στ = 0 or ΣτCCW = ΣτCW
FSxS + FLxL = FBxB
FN
(mSg)xS + (mLg)xL = (mBg)xB
If you divide through by “g” you get:
xS = cm
xB = cm
c.g.
.
mSxS + mLxL = mBxB
FS = mSg
FL = mLg
xL = cm
FB = mBg
Sample: Suppose that you have set up
your 1.00 meter long lever of mass 83.4
grams so that the center of gravity is
FN
20.0 cm from the fulcrum. You have also
determined that the big unknown mass
will be placed 10.0 cm from the fulcrum.
xB = 10.0 cm
xS = cm
c.g.
.
xL = 20.0 cm
FS = mSg
FL = mLg
FB = mBg
You are given a known mass of 76.2
grams and a big mass. Putting the
masses on the lever, you find that it
balances when the little mass is 62.8
centimeters from the fulcrum.
FN
xB = 10.0 cm
xS = 62.8 cm
c.g.
.
xL = 20.0 cm
mS = 76.2 g
mL = 83.4 g
mB = ??
mSxS + mLxL = mBxB
(76.2 g)(62.8 cm) + (83.4 g)(20.0 cm) = mB(10.0 cm)
FN
mB = [(76.2 g)(62.8 cm) + (83.4 g)(20.0 cm)]/(10.0 cm)
mB = 645 grams
xS = 62.8 cm
xB = 10.0 cm
c.g.
.
xL = 20.0 cm
mS = 76.2 g
mL = 83.4 g
mB = ??
As long as the levers are horizontal
and in static equilibrium, you can
use the equations with mass
instead of force or weight. Your
Physics teacher probably will not
be too happy, but the equation is
mathematically correct for the
situation.
c.g.
c.g.
B
First Class Lever
Second Class Lever
For the High School Competition you will need to
build a compound lever system made up of a first
class lever connected to a second class lever. Each
lever may not be longer than 50.0 cm.
c.g.
First part of lever system
mSxS + mL1xL1 = ExE1
xE1 = cm
xS = cm
c.g.
.
xL1 = cm
mS =
g
mL1 =
g
E = ?? g
Second part of lever system
c.g.
ExE2 = mL2xL2 + mBxB
B
xL2 = cm
xE2 = cm
c.g.
.
E=
g
mL2 =
g
xB = cm
mB = ?? g
Sample:
c.g.
xE1 = 5.0 cm
xE2 = 35.0 cm
c.g.
xB = 10.0 cm
B
Suppose that you built your lever system so that
the fulcrum in the class 1 lever was 5.0 cm from
the string connecting the levers.
Also suppose that in the second class lever the
string connecting the levers is 35.0 cm from its
fulcrum and you set up the lever so that the
unknown big mass is 10.0 cm from the fulcrum.
Sample
xE1 = 5.0 cm
c.g.
xE2 = 35.0 cm
c.g.
xS = 32.7 cm
xB = 10.0 cm
B
You are given a small mass of 86.0 grams and an
unknown large mass that you place at the 10.0
cm mark.
You slide the small mass along the class 1 lever
and manage to get the levers to balance when
the small mass is 32.7 cm from its fulcrum.
xE1 = 5.0 cm
c.g.
xS = 32.7 cm
First part of lever system
You also measured the mass of the
lever as 27.4 g and arranged the
lever so that its center of gravity is
16.0 cm from the fulcrum.
xE1 = 5.0 cm
xS = 32.7 cm
c.g.
.
xL1 = 16.0 cm
mS = 86.0 g
mL1 = 27.4 g
mSxS + mL1xL1 = ExE1
E = ?? g
First part of lever system
mSxS + mL1xL1 = ExE1
(86.0 g)(32.7 cm) + (27.4 g)(16.0 cm) = E(5.0 cm)
E = [(86.0 g)(32.7 cm) + (27.4 g)(16.0 cm)]/(5.0 cm)
E = 650 g
xE1 = 5.0 cm
x = 32.7 cm
S
c.g.
.
xL1 = 16.0 cm
mS = 86.0 g
mL1 = 27.4 g
E = ?? g
xE2 = 35.0 cm
Suppose the mass of the lever is
31.4 g and its center of gravity is
18.0 cm from the fulcrum.
c.g.
xB = 10.0 cm
Second part of lever system
B
xE2 = 35.0 cm
xL2 = 18.0 cm
c.g.
.
xB = 10.0 cm
E = 650 g
mL2 = 31.4 g
ExE2 = mL2xL2 + mBxB
mB = ?? g
Second part of lever system
ExE2 = mL2xL2 + mBxB
(650 g)(35.0 cm) = (31.4 g)(18.0 cm) + mB(10.0 cm)
mB = [(650 g)(35.0 cm) - (31.4 g)(18.0 cm)]/(10.0 cm)
mB = 2218 g = 2.218 kg
xL2 = 18.0 cm
xE2 = 35.0 cm
c.g.
.
xB = 10.0 cm
E = 650 g
mL2 = 31.4 g
mB = ?? g
Things to note:
• You must build your own lever system.
• You may want to have two set places to have
your fulcrum depending on the given masses.
• You may want to have the unknown mass at a
predetermined spot and thus notching the lever
at that point.
• Make sure that you know the mass of your lever
and have marked the location of its center of
gravity.
Fixed Pulley
A fixed pulley is basically a First
Class Lever that can rotate. The
mechanical advantage is 1. All a
fixed pulley does is change the
direction of the force.
Effort
Load
Fulcrum
Effort
Load
Movable Pulley
A movable pulley is basically a
Second Class Lever that can rotate.
The mechanical advantage is 2.
Effort
Effort
Load
Load
Fulcrum
Block & Tackle
A Block and Tackle is a combination
of a movable pulley connected to a
fixed pulley. In this case the
mechanical advantage of the
movable pulley is 2 and the MA of
the fixed pulley is 1. Combined the
mechanical advantage is 2.
In order to calculate the Ideal
Mechanical Advantage, IMA, of a
Block and Tackle, you count the
number of supporting ropes.
Effort
Load
Block & Tackle
A Block and Tackle is a combination
of a movable pulley connected to a
fixed pulley. In this case the
mechanical advantage of the
movable pulley is 2 and the MA of
the fixed pulley is 1. Combined the
mechanical advantage is 2.
Effort
Load
Work is done to lift the Load, but you also
must lift the movable pulley with it. The
AMA will be less, not only because of
friction in the system but because the
weight of the movable pulley also has to
be lifted by the Effort.
dE
Wheel and Axle
dL
A Wheel and Axle is two different
diameter cylinders on the same shaft. This
also is a first class lever that can rotate.
The Ideal Mechanical Advantage is the
ratio of the diameters. IMA = dE/dL.
Load
Effort
Effort
Load
Fulcrum
All these are examples of a wheel and Axle.
Inclined Plane
x
h
An inclined plane is basically a ramp that is stationary.
A load is pushed up the ramp instead of being lifted
straight up. The Ideal Mechanical Advantage of a
ramp is the ratio of the length of the ramp (x) to the
height of the ramp (h). IMA = x/h
Wedge
A wedge is like an inclined plane, but instead of being
stationary the wedge is driven into something or
between things.
All these are examples of wedges.
LEVERS.
Class 1 Levers.
Class 1 Levers. One of these can also be used
as a class 2 Lever. Which one? How?
COMPOUND
MACHINES.
This is actually a compound machine.
The teeth are wedges that cut into the wood.
The whole blade is a wheel and axle.
These too are compound machines.
The teeth are wedges that cut the wires.
The handles make a class 1 lever.
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# Numbers
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Numbers [#permalink] 16 Aug 2007, 03:11
If X+1/X = 4, find the value of X^4 - 1/x^4(x>1)
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X^4 - 1/x^4 = ? Solution [#permalink] 16 Aug 2007, 06:57
X^4 - 1/x^4 = ?
we can write it like
(X^2 + 1/x^2)(X^2 - 1/x^2)
1st root (X^2 + 1/x^2)
2nd root (X^2 - 1/x^2)
Soln of 1st root
x + 1/x =4
Now square it
(x + 1/x)^2 =4^2
=>
x^2 + 1/x^2 + 2 = 16
=>
x^2 + 1/x^2 = 14
Soln of 2nd root
(X^2 - 1/x^2)
square it
(X^2 - 1/x^2)^2 =x^4 + 1/x^4 -2 ....... let this is euation 1
Now square the soln of 1st root
(x^2 + 1/x^2)^2 =14^2
=>
x^4 + 1/x^4 + 2 = 196
=>
x^4 + 1/x^4 = 194
Now keep this value into equation 1
(X^2 - 1/x^2)^2 =x^4 + 1/x^4 -2 = 194 - 2 = 192
=> (X^2 - 1/x^2) =sqrt(192) =8 sqrt(3)
Now we got both root
multiple both
(X^2 + 1/x^2)(X^2 - 1/x^2) = 14 * 8 * sqrt(3)
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The Answer is 112 , sqrt (3)
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Eesh. Just figuring out the approach would have taken me at least 1.5 minutes.
Great work, Chiya.
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Thx ! [#permalink] 16 Aug 2007, 10:02
Thx everyone !!!
14*8=112...
Thx ! [#permalink] 16 Aug 2007, 10:02
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https://metanumbers.com/1005809 | 1,642,526,868,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300934.87/warc/CC-MAIN-20220118152809-20220118182809-00646.warc.gz | 452,016,556 | 7,412 | 1005809 (number)
1,005,809 (one million five thousand eight hundred nine) is an odd seven-digits composite number following 1005808 and preceding 1005810. In scientific notation, it is written as 1.005809 × 106. The sum of its digits is 23. It has a total of 2 prime factors and 4 positive divisors. There are 862,116 positive integers (up to 1005809) that are relatively prime to 1005809.
Basic properties
• Is Prime? No
• Number parity Odd
• Number length 7
• Sum of Digits 23
• Digital Root 5
Name
Short name 1 million 5 thousand 809 one million five thousand eight hundred nine
Notation
Scientific notation 1.005809 × 106 1.005809 × 106
Prime Factorization of 1005809
Prime Factorization 7 × 143687
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 1005809 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 1,005,809 is 7 × 143687. Since it has a total of 2 prime factors, 1,005,809 is a composite number.
Divisors of 1005809
4 divisors
Even divisors 0 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 1.1495e+06 Sum of all the positive divisors of n s(n) 143695 Sum of the proper positive divisors of n A(n) 287376 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1002.9 Returns the nth root of the product of n divisors H(n) 3.49998 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 1,005,809 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 1,005,809) is 1,149,504, the average is 287,376.
Other Arithmetic Functions (n = 1005809)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 862116 Total number of positive integers not greater than n that are coprime to n λ(n) 431058 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 78765 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 862,116 positive integers (less than 1,005,809) that are coprime with 1,005,809. And there are approximately 78,765 prime numbers less than or equal to 1,005,809.
Divisibility of 1005809
m n mod m 2 3 4 5 6 7 8 9 1 2 1 4 5 0 1 5
The number 1,005,809 is divisible by 7.
Classification of 1005809
• Arithmetic
• Semiprime
• Deficient
Expressible via specific sums
• Polite
• Non-hypotenuse
• Square Free
Other numbers
• LucasCarmichael
Base conversion (1005809)
Base System Value
2 Binary 11110101100011110001
3 Ternary 1220002201012
4 Quaternary 3311203301
5 Quinary 224141214
6 Senary 33320305
8 Octal 3654361
10 Decimal 1005809
12 Duodecimal 406095
16 Hexadecimal f58f1
20 Vigesimal 65ea9
36 Base36 lk35
Basic calculations (n = 1005809)
Multiplication
n×y
n×2 2011618 3017427 4023236 5029045
Division
n÷y
n÷2 502904 335270 251452 201162
Exponentiation
ny
n2 1011651744481 1017528429464690129 1023439252111450513959361 1029384410726965929994950928049
Nth Root
y√n
2√n 1002.9 100.193 31.6686 15.8673
1005809 as geometric shapes
Circle
Radius = n
Diameter 2.01162e+06 6.31968e+06 3.1782e+12
Sphere
Radius = n
Volume 4.26221e+18 1.27128e+13 6.31968e+06
Square
Length = n
Perimeter 4.02324e+06 1.01165e+12 1.42243e+06
Cube
Length = n
Surface area 6.06991e+12 1.01753e+18 1.74211e+06
Equilateral Triangle
Length = n
Perimeter 3.01743e+06 4.38058e+11 871056
Triangular Pyramid
Length = n
Surface area 1.75223e+12 1.19917e+17 821240
Cryptographic Hash Functions
md5 188733a96b1b87639af5b93e3943bb5b 27a9245fc2446df2c5ba81edd810796fc826c6e4 28e9c7d35851f01ba295b2671ec14b8bb42d07d4eb65217fedb5b4f2b26d7d0a 6160cdbfc3bea65ef6f5526cd3e679077446b89ad5b55179d4f12f065e604e30821e42cadcc7ed873b9f585afe5b7bc2c2e8d05010a81781fba1bd27a9f3192a c6100adeeca8cab6005d323a6b693dd2e46d05a5 | 1,533 | 4,299 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2022-05 | latest | en | 0.814718 |
msd0707.blogspot.com | 1,539,920,854,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512268.20/warc/CC-MAIN-20181019020142-20181019041642-00425.warc.gz | 231,268,665 | 11,028 | ## SPOJ "The Bulk" Hint
### Basic Steps :
1. Remove faces non perpendicular to Z axis
2. Remove Unnecessary Points from all faces
3. Sort faces on the basis of height along Z axis
4. Assign sign to each face whether it contributes positive volumes or negative volume
5. Find Area of each face
6. volume contributed by each face = sign*height along Z axis*area
7. Answer is sum of volume contributed by each face.
### 1. Remove faces non perpendicular to Z axis
we need only faces which are perpendicular to a particular axis lets take it as z axis since if you know all faces of a cube perpendicular to z axis then you can predict the other faces so that is just an extra information.
### 2. Remove Unnecessary Points from all faces
If there are three points (x1,y,z),(x2,y,z) and (x3,y,z) given in order as shown in fig. then x2 is an unnecessary point removing that point will still result in the same line so remove such points. Similarly apply for y as shown in fig.
### 3. Sort faces on the basis of height along Z axis
Now to find empty cubes we need to decide the sign of faces . if the volume below some face is empty its sign is negative. Now to decide which face is below which face we have to short faces on the basis of there heights along z axis.
### 4. Assign sign to each face whether it contributes positive volumes or negative volume
Now to decide sign of a face its sign would be opposite to that of one above it. Since we have already sorted the faces start with the top face giving it a positive sign. Now for each face find a face above it which covers it or in other words overlaps it and give it a sign opposite to that face. Now to find such face for each face steps are
1. Among the points with highest Y value in a face find one with least x value let this point be (X,Y,Z).
2. Now for each face above it starting from the one just above it in terms of height with respect to Z axis find the no of horizontal lines(parallel to x axis) passing through the point(X,y,Z) such that y >= Y.
3. If the count is odd it means the face is covering current face .
4. Give current face opposite sign of this face and move to next face
### 5. Find Area of each face
To find area of a face using co-ordinates (xi,yi) iterate over cordinates .Take the point (x1,y1) and next point (x2,y2) and add up to the answer (x1*y2 - y1*x2).do it for all the points and (sum/2) is the area of face.
### 7. Answer is sum of volume contributed by each face.
To get the c++ code for reference request me for the same via my email [email protected] i will share link to my code with you. | 646 | 2,667 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2018-43 | longest | en | 0.882365 |
https://www.physicsforums.com/threads/tricky-definite-integral.677340/ | 1,532,273,247,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593302.74/warc/CC-MAIN-20180722135607-20180722155607-00634.warc.gz | 965,576,654 | 18,442 | # Homework Help: Tricky Definite Integral
1. Mar 9, 2013
### pierce15
1. The problem, the whole problem, and nothing but the problem
$$\int_0^\pi \frac{x \cdot sin(x)}{1+cos^2(x)} \, dx$$
2. Relevant equations
Integration by parts
trig substitution
3. The attempt at a solution
My first idea was to break up the integral by letting $u=x$ and $dv=sin(x)/(1+cos^2 x)$. I will omit the work, but it got me here:
$$\int_0^\pi \frac{x \cdot sin(x)}{1+cos^2(x)} \, dx = (x \cdot tan^{-1} (cos(x)) \big|_0^\pi - \int_0^\pi tan^{-1}(cos(x)) \, dx$$
I am fairly sure that I arrived at the second integral correctly. Is it integrateable, if that's a word?
Last edited: Mar 9, 2013
2. Mar 9, 2013
$works for inline. For the integral, it is a trick. Try substituting u=pi-x. Finding an indefinite integral is really pretty hopeless. Last edited: Mar 9, 2013 3. Mar 9, 2013 ### pierce15 $$- \int_0^\pi tan^{-1} cos(x) \, dx$$ $$u= \pi-x, du=- dx$$ $$\int_\pi^0 tan^{-1} cos(\pi-u) \, du$$ $$- \int_0^\pi tan^{-1}(cos(\pi)cos(u) + sin(\pi)sin(u)) \, du$$ $$- \int_0^\pi tan^{-1}(-cos(u)) \, du$$ $$\int_0^\pi tan^{-1}cos(u) \, du$$ Hmm... does that mean that since the top equation is equal to the bottom, and they are negatives, the integral must be 0? Last edited: Mar 9, 2013 4. Mar 9, 2013 ### Dick I meant you to use that substitution in the original integral. You should get the original integral back with a different sign plus something you can integrate. Equate it to the original integral. 5. Mar 9, 2013 ### pierce15 $$\int_0^\pi \frac{x \cdot sin(x)}{1+cos^2x} \, dx$$ $$u=\pi-x, du=-1$$ $$- \int_\pi^0 \frac{(\pi-u)(sin(\pi-u))}{1+cos^2(\pi-u)} \, du$$ $$\int_0^\pi \frac{(\pi-u)(-sin(u))}{1+ cos^2(\pi-u)} \, du$$ $$-\int_0^\pi \frac{sin(u)(\pi-u)}{1+ cos^2(\pi-u)} \, du$$ $$-\int_0^\pi \frac{sin(u)(\pi-u)}{1+ (-cos(u))^2} \, du$$ $$-\int_0^\pi \frac{sin(u)(\pi-u)}{1+ cos^2u} \, du$$ $$-\pi \int_0^\pi \frac{sin(u)}{1+ cos^2u} \, du + \int_0^\pi \frac{u \cdot sin(u)}{1+ cos^2u}$$ $$v=cos(u), dv=-sin(u) \, du$$ $$\pi \int_1^{-1} \frac{dv}{1+v^2} + \int_0^\pi \frac{u \cdot sin(u)}{1+ cos^2u}$$ Going back to the beginning: $$\int_0^\pi \frac{x \cdot sin(x)}{1+cos^2x} \, dx = -\pi \int_{-1}^1 \frac{dv}{1+v^2} + \int_0^\pi \frac{u \cdot sin(u)}{1+ cos^2u}$$ $$0=\int_{-1}^1 \frac{dv}{1+v^2}$$ Am I on the right track? Last edited: Mar 9, 2013 6. Mar 9, 2013 ### Dick First fix an error. sin(pi-u)=sin(u). Then break it into two integrals. 7. Mar 9, 2013 ### pierce15 $$\int_0^\pi \frac{x \cdot sin(x)}{1+cos^2x} \, dx$$ $$u=\pi-x, du=-1$$ $$- \int_\pi^0 \frac{(\pi-u)(sin(\pi-u))}{1+cos^2(\pi-u)} \, du$$ $$\int_0^\pi \frac{(\pi-u)(sin(u))}{1+ cos^2(\pi-u)} \, du$$ $$\int_0^\pi \frac{sin(u)(\pi-u)}{1+ cos^2(\pi-u)} \, du$$ $$\int_0^\pi \frac{sin(u)(\pi-u)}{1+ (-cos(u))^2} \, du$$ $$\int_0^\pi \frac{sin(u)(\pi-u)}{1+ cos^2u} \, du$$ $$\pi \int_0^\pi \frac{sin(u)}{1+ cos^2u} \, du - \int_0^\pi \frac{u \cdot sin(u)}{1+ cos^2u}$$ $$v=cos(u), dv=-sin(u) \, du$$ $$\pi \int_1^{-1} \frac{dv}{1+v^2} - \int_0^\pi \frac{u \cdot sin(u)}{1+ cos^2u} \, du$$ Is the following OK: $$I = \int_0^\pi \frac{x \cdot sin(x)}{1+cos^2x} \, dx = \pi \int_{-1}^1 \frac{dv}{1+v^2} - \int_0^\pi \frac{u \cdot sin(u)}{1+ cos^2u}$$ $$2I = \pi \int_{-1}^1 \frac{dv}{1+v^2}$$ If so, how can I evaluate this integral? 8. Mar 9, 2013 ### Dick You know, having to constantly go back and check whether you have reedited a past post really sucks. Use preview to see whether you have what you want, then just LEAVE IT. If you change your mind, just post the correction. Don't go back and change the past. And yes, your result is fine. 9. Mar 9, 2013 ### pierce15 Sorry 10. Mar 9, 2013 ### Dick S'ok. Just advice for the future. But you got the right answer. 11. Mar 9, 2013 ### pierce15 So the answer is just 0? 12. Mar 9, 2013 ### Dick It's an arctan, isn't it? Why do you think it's zero? 13. Mar 9, 2013 ### Dick Now I forgot to quote the integral. Doesn't look like zero to me. 14. Mar 9, 2013 ### pierce15 Oh, right. I'm pretty tired right now. Just did out the trig substitution and verified that it's 0. How did you know to make the substitution u=pi-x? 15. Mar 9, 2013 ### pierce15 [itex] 1/(1+v^2)$ is also an even function which is probably enough to show that the answer is 0, considering that the upper limit is equal to the lower limit
16. Mar 9, 2013
### Dick
Tired, ok. But it's NOT zero. Check it in the morning if that works better. I guessed to do the substitution because I've seen problems like this before. Just experience.
17. Mar 9, 2013
### Dick
Wrong. Get some sleep if you need to.
18. Mar 9, 2013
### pierce15
$$\int_{-1}^1 \frac{dv}{v^2+1}$$
$$v= tan\theta, dv=sec^2\theta d\theta$$
$$\int_{-\pi/4}^{\pi/4} \frac{sec^2 \theta}{sec^2 \theta} d\theta$$
$$\int_{-\pi/4}^{\pi/4} \theta d\theta$$
$$\frac{\theta^2}{2} \big|_{-\pi/4}^{\pi/4}$$
$$(\pi/4)^2/2 - (-\pi/4)^2/2$$
Alright, I see that I'm wrong by looking at the graph. What's the problem here?
19. Mar 9, 2013
### Dick
Now you are making me sleepy. sec(θ)^2/sec(θ)^2 isn't θ. It's 1.
20. Mar 9, 2013
### pierce15
$$\theta \big|_{-\pi/4}^{\pi/4} = \pi/2$$
$$2I= \pi/2$$
$$I = \pi/4$$
Were there any arithmetic errors in there? haha
21. Mar 9, 2013
### Dick
Sadly, yes. You dropped the factor of pi you had correct in post 7. You would do it easily with the hint I gave you if you were less tired.
22. Mar 9, 2013
### pierce15
Alright, so $I = \pi^2/4$. Thank you very much for the help, now I'm going to get some beauty sleep
23. Mar 9, 2013
### Dick
Very welcome, and good idea! | 2,194 | 5,574 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2018-30 | latest | en | 0.689096 |
https://www.sbank.in/2019/12/ibps-clerks-reasoning.html | 1,653,360,216,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662562410.53/warc/CC-MAIN-20220524014636-20220524044636-00162.warc.gz | 1,146,208,581 | 58,081 | Test-III : Reasoning Ability
given questions:
Point G is 17m to the north of Point H. Point G
is 4m to the east of Point J. Point K is 25m to the south of Point J. Point K
is 12m to the west of Point L. Point M is 8m to the north of Point L.
98. How far and in which direction is point M
with respect to point H?
1) 23m towards South 2) 8m towards East 3) 25m towards West 4) 12m towards East
5) 8m towards West
99. If Point Z is 8m to the west of Point L,
then which of the following points will form a straight line?
1) ZKM 2)
HZM 3) ZLG 4) GHZ
5) JHZ
100. In which direction is Point G with respect
to Point L?
1) North-East
2) South-East 3) South 4) North-West 5) North
Directions (Q. 92-96): Study the given information carefully to
Ten boxes are kept above one another in a
stack. Only three boxes are kept between A and S. Q is kept immediately below
S. Only two boxes are kept between S and M. S is kept at one of the positions
above M. Only one box is kept between M and B. Only one box is kept below B. C
is kept immediately above K but not at the topmost position. R is kept one of
the positions above W but below G.
92. As per the given arrangement, four of the
following five are alike in a certain way and hence form a group. Which one of
the following does not belong to the group?
1) SG 2)
MW 3) CA 4) KR
5) BR
93. As per the given arrangement, as many boxes
are kept above G as below
1) C 2)
W 3) M
4) R 5) Q
94. Which box(es) is/are kept exactly between R
and A?
1) Only M
2) Both B and W 3) Only C 4) Both S and Q 5) Only B
95. Which box is kept third from the top of the
stack?
1) A 2)
K 3) R
4) G 5) C
96. How many boxes are kept between B and W?
1) Five
2) Four 3) None 4) Three
5) Two
97. How many such pairs of letters are there in
the word NATIONALS each of which has as many letters between them in the word
(in both forward and backward directions) as they have between them in the
English alphabetical order?
1) One 2)
Four 3) Two 4) More than four 5) Three
Directions (Q. 66-69) : Study the given information carefully to
In a certain code language, ‘roam around the
hills’ is coded as ‘mu pa jl ko’, ‘hills on left side’ is coded as ‘el rf pa nt’,
‘left roam alone’ is coded as ‘jl dg el hv’, ‘alone for the journey’ is coded
as ‘bx ko oy dg’.
(Note :
All codes are two-letter codes only.)
66. As per the given code language, what is
‘hills’ coded as?
1) mu 2)
nt 3) pa 4) oy
5) jl
67. If in the given code language, ‘for alone
discovery’ is coded as ‘dg sz oy’, then how will ‘discovery to journey’ be
coded in the same code language?
1) dg co sz
2) el jib x 3) bx sz hv 4) hv bx dg
5) sz hv el
68. As per the given code language, _______ may
be coded as ‘el ko’.
1) the left
2) to the 3) roam alone 4) alone roam 5) left around
69. In the given code language, what is ‘side’
coded as?
1) mu 2)
either ‘nt’ or ‘rf’ 3) either ‘el’ or
‘pa’ 4) dg 5) pa
Directions (Q. 76-80): Study the given information carefully to
Eight people are sitting in a straight line.
Some of them are facing north and some are facing south. B is sitting second
from one of the extreme ends of the line. E sits fourth to the right of B. Only
two people sit between E and H. The immediate neighbours of H face opposite
directions. A sits on the immediate left of D. A is neither an immediate
neighbor of E nor of H. D and A face opposite directions. People sitting at the
extreme ends of the line face the same direction. F sits fourth to the left of
C. G sits at one of the positions to the left of both H and E. E faces north.
76. Who is sitting second to the left of H?
1) D 2)
F 3) A
4) C 5) E
77. Who amongst the following is an immediate
neighbor of D?
1) E 2)
C 3) F
4) B 5) G
78. How many people are sitting between G and
B?
1) More than three 2) None
3) Three 4) Two 5) One
79. Which of the following statements is TRUE with respect to the given
information?
1) Only two people sit between B and C. 2) None of the given statements is true.
3) A faces south. 4) D sits at one of the extreme ends of the line.
5) F and H face the same direction.
80. Four of the following five are alike in a
certain way based on their directions as per the given arrangement and so form
a group. Which one of them does not belong to the group?
1) A 2)
F 3) C
4) D 5) B
Directions (Q. 81-84): In this question three statements followed by two conclusions I and II have been given. You have to take the given statements to be true even if they seem to be at variance with the commonly known facts and then decide which of the given conclusions logically follows from the given statements. Give answer
1) if only conclusion I follows. 2) if only conclusion II follows.
3) if either conclusion I or II follows. 4) if both conclusion I and II follow.
5) if neither conclusion I nor II follows.
81. Statements: All
watches are rings.
Only
a few rings are carrots.
Only a few carrots are phones.
Conclusions: I. No ring is a phone.
II.
All watches can never be carrots.
82. Statements: No test is a paper.
Only
a few papers are registers.
No
register is a book.
Conclusions: I. All tests being registers is a possibility.
II.
No paper is a book.
83. Statements: Only a few breads are juices.
Only
a few juices are toasters.
All
toasters are pans.
Conclusions: I. All breads being toasters is a possibility.
II.
At least some juices are pans.
84. Statements: Only a few wires are loops.
All
loops are circles.
No
circle is a rectangles.
Conclusions: I. At least some wires are circles.
II. All wires can never be
rectangles
given questions.
Each of the six bags, viz A, B, C, D, E and F,
is of a different weight. B is heavier than E but lighter than both F and C. C
is lighter than D. D is not the heaviest. E is not the lightest. The weight of
the second lightest bag is 5 kg. D weighs 21 kg. (Note: All the given weights are in whole numbers only.)
89. With respect to the weights of the given
bags, if F + E = 29 kg and F + B = 34 kg, then B + D = _________?
1) 45 kg
2) 15 kg 3) Cannot be
determined 4) 31 kg 5) 26 kg
90. Which is the third heaviest bag?
1) C 2)
D 3) A
4) B 5) E
91. What is the probable weight of bag C(in
kg)?
1) 4 2)
Either 2 or 3 3) 15 4) 24
5) Either 22 or 25
@@@@@@@@@@@@@@@@@@@@@@
ANSWERS GIVEN IN DIFFERENT ORDER -- A FEW SIGNS ARE NOT APPEARING PROPERLY IN BLOGGER.COM
REASONING
ABILITY
(66-69): Given,
Roam around the
hills
mu pa jl ko ….(i)
Hills on left side
jl dg el hv …
(ii)
Left to roam alone
jl dg el hv ….
(iii)
Alone for the
journey
bx ko oy dg ….
(iv)
From (i) and (ii),
hills
pa …
(v)
From (i) and
(iii), roam
jl ….
(vi)
From (ii) and
(iii), left
el ….
(vii)
From (i) and (iv),
the
ko ...
(viii)
From (i), (v),
(vi) and (viii), around
mu …
(ix)
From (iii) and
(iv), alone
dg ….
(x)
From (ii), (v) and
(vii), on/side
rf/nt …
(xi)
From (iii), (vi),
(vii) and (x), to
hv …
(xii)
From (iv), (viii)
and (x), for/journey
bx/oy ….
(xiii)
66.(3)
67.(3) for alone
discovery
dg sz oy ….
(a)
From (xiii) and
(a), for
oy ….
(xiv)
From (x) and (a),
Alone dg and journey bx ….
(xv)
Now, from (a),
(xiv) and (xv), discovery
sz ….
(xvi)
discovery to journey sz hv bx
68.(1) el ko left the
69.(2)
(
(76-80):
76.(4) 77.(1) 78.(5) 79.(1) 80.(4) (f)
(b) h g
(c) ( e) d
(a)
81.(5) All watches
are rings (A) + Only a few rings are carrots (I) = A + I = No conclusion. Hence
conclusion II does not follow.
From statement
second + third (I + I =) No conclusion. Hence conclusion I does not follow.
Thus, neither
conclusion I nor II follows.
82.(1) No test is
a paper (E) + Only a few papers are registers (I) = E + I = O* = Some registers
are not tests. But the possibility in I exists. Hence conclusion I follows.
Again, Only a few
papers are registers (I) + No register is a book (E) = I + E = O = Some papers
are not books. Hence conclusion II does not follow.
83.(4) Only a few
breads are juices (I) + Only a few juices are toasters (I) = I + I = No
conclusion. But the possibility in I exists. Hence conclusion I follows.
Again, Only a few
juices are toasters (I) + All toasters are pans (A) = I + A = I = Some juices
are pans. Hence conclusion II follows.
84.(4) Only a few
wires are loops (I) + All loops are circles (A) = I + A = I = Some wires are
circles (I). Hence conclusion I follows.
Again, Some wires
are circles (I) + No circle is a rectangle (E) = I + E = O = Some wires are not
rectangles. Hence conclusion II follows.
(89-91)
:
F > D > C > B > E > A
21 5 kg
89.(4) F + E = 29 F + 5 = 29
F = 29 – 5 = 24 kg
Again, F + B = 34 B = 34 – 24 = 10 kg
Now, B + D = 10 +
21 = 31 kg
90.(1)
91.(3) 21 > C
> 5
Thus, probable
weight of C will be 15 kg.
(92-96)
:
Box number Box 10 G 9 C 8 K 7 S 6 Q 5 R 4 M 3 A 2 B 1 W
92.(3) 93.(2) 94.(1) 95.(2) 96.(3)
97.(2)
(98-100)
:
98.(2) MH = KL –
JG = 12 – 4 = 8m
Hence, Point M is
8m towards east of Point H.
99.(4) GHZ will
make a straight line.
100.(4)
@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ | 3,018 | 9,522 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2022-21 | latest | en | 0.922292 |
http://mathhelpforum.com/calculus/75894-differenciation.html | 1,481,168,626,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542323.80/warc/CC-MAIN-20161202170902-00173-ip-10-31-129-80.ec2.internal.warc.gz | 187,351,563 | 11,366 | 1. ## differenciation
1)
$
(x + \frac {1}{x})' = \frac {1}{2}x^{-0.5} - \frac {x^{0.5}}{x}= \frac {1}{2\sqrt{x}} - \frac {\sqrt{x}}{x} = \frac{\sqrt{x}}{2x}
$
The solution is supposed to be $\frac{\sqrt{x}(x-1)}{2x^2}$
2)
$
(ln \sqrt{\frac{a+x}{a-x}})' = \frac{a-x}{2(a+x)}
$
The solution is supposed to be $\frac{a}{a^2-x^2}$
2. #1) Think of it as $x+x^{-1}$ Use the power rule like normal.
3. if i use that i get $\frac{1}{2}x^{-\frac{1}{2}} - \frac{1}{2}x^{-\frac{3}{2}}$
4. Originally Posted by metlx
if i use that i get $\frac{1}{2}x^{-\frac{1}{2}} - \frac{1}{2}x^{-\frac{3}{2}}$
You shouldn't. d/dx x = 1 and d/dx $x^{-1}=-1(x)^{-2}$
5. Originally Posted by metlx
if i use that i get $\frac{1}{2}x^{-\frac{1}{2}} - \frac{1}{2}x^{-\frac{3}{2}}$
HOW do you get that? Please show your work. The power rule: $(x^n)'= nx^{n-1}$ won't give fractional powers if you don't start with them!
6. aww. i just realized I forgot the square root.
1) is $(\sqrt{x} + \frac{1}{\sqrt{x}})'$
7. Originally Posted by metlx
aww. i just realized I forgot the square root.
1) is $(\sqrt{x} + \frac{1}{\sqrt{x}})'$
Please read the replies you're getting. There should be NO SQUARE ROOTS unless you simplify after taking the derivative.
8. no i forgot to add the square root into the initial derivative..
I wrote $(x + \frac {1}{x})'$ instead of $(\sqrt{x} + \frac{1}{\sqrt{x}})'$
9. by the way, 2) is no longer needed. Managed to solve it
10. Then your first part if right. $\frac{d}{dx} \frac{1}{\sqrt{x}}= \frac{d}{dx} x^{-\frac{1}{2}}=-\frac{1}{2}x^{-\frac{3}{2}}$
11. I'm guessing that the solution $\frac{\sqrt{x}(x-1)}{2x^2}$ that is in the book is wrong then?
12. Originally Posted by metlx
I'm guessing that the solution $\frac{\sqrt{x}(x-1)}{2x^2}$ that is in the book is wrong then?
$\frac{1}{2\sqrt{x}}-\frac{1}{2}x^{-\frac{3}{2}}$
I am busy right now so I can't do this for you, but try to get these two terms into one expression. Then see if the book's answer makes more sense. | 735 | 1,992 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 18, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2016-50 | longest | en | 0.774376 |
https://www.physicsforums.com/threads/rotational-inertia-problem.134306/ | 1,544,918,841,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827137.61/warc/CC-MAIN-20181215222234-20181216004234-00370.warc.gz | 1,035,230,457 | 12,371 | # Homework Help: Rotational Inertia problem
1. Oct 1, 2006
### John O' Meara
A lawn roller in the form of a hollow cylinder of mass M is pulled horizontally with a constant force F applied by a handle attached to the axle. If it rolls without slipping, find the acceleration and the frictional force.
Let R1 be the radius of the hollow and R2 the outer radius, and (alpha) the angular acceleration.
Sum Fy= n - M*g
Sum Fx= F - f.
Where n = normal reaction and f = friction force. Applying Sum F = M*a, we get n=M*g ....as ay=0.
F - f = M*ax ....the equation for the translational motion of the center of mass. And where ax and ay are the accelerations in the x and y directions resprctively.
And the equation of rotational motion about the axis through the center of mass is:
f*R2 = I*(alpha)=.5*M*(R1^2+R2^2)
f*R2 = .5*M*(R1^2+R2^2)*ax/R2: let a=ax
f = .5*M*(R1^2+R2^2)*a/R2^2 .......(ii)
F-f = M*a...(i) After adding i and ii
a=F/(M+.5*M*(R1^2+R2^2)/R2^2.
The question is, is this correct or not? And where is it wrong?
2. Oct 1, 2006
### Staff: Mentor
Looks OK to me. | 347 | 1,076 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2018-51 | latest | en | 0.827952 |
https://www.mkamimura.com/2019/11/Mathematics-Python-Algebra-Integer-Calculation-Integer-Division-Divisors-Factors-Multiples-when-Divided.html | 1,726,709,782,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651944.55/warc/CC-MAIN-20240918233405-20240919023405-00808.warc.gz | 804,607,317 | 24,394 | ## 2019年11月23日土曜日
### 数学 - Python - 代数学 - 整式の計算 - 整式の除法 - 割り切れる場合、約数、因数、倍数
1. $\begin{array}{l}\frac{2{x}^{2}+7x-4}{2x-1}\\ =\frac{\left(2x-1\right)\left(x+4\right)}{2x-1}\\ =x+4\end{array}$
2. $\begin{array}{l}\frac{{x}^{3}-38x-12}{x+6}\\ =\frac{\left(x+6\right)\left({x}^{2}-6x-2\right)}{x+6}\\ ={x}^{2}-6x-2\end{array}$
3. $\begin{array}{l}\frac{{a}^{4}+{a}^{2}+1}{{a}^{2}-a+1}\\ =\frac{\left({a}^{2}-a+1\right)\left({a}^{2}+a+1\right)}{{a}^{2}-a+1}\\ ={a}^{2}+a+1\end{array}$
4. $\begin{array}{l}\frac{6{y}^{4}+3{y}^{3}+2{y}^{2}-2y-4}{2{y}^{2}+y+2}\\ =\frac{\left(2{y}^{2}+y+2\right)\left(3{y}^{2}-2\right)}{2{y}^{2}+y+2}\\ =3{y}^{2}-2\end{array}$
コード
#!/usr/bin/env python3
from sympy import symbols, pprint
from unittest import TestCase, main
print('17.')
x = symbols('x')
class MyTest(TestCase):
def test(self):
spam = [(2 * x ** 2 + 7 * x - 4, 2 * x - 1),
(x ** 3 - 38 * x - 12, x + 6),
(x ** 4 + x ** 2 + 1, x ** 2 - x + 1),
(6 * x ** 4 + 3 * x ** 3 + 2 * x ** 2 - 2 * x - 4, 2 * x ** 2 + x + 2)]
egg = [x + 4,
x ** 2 - 6 * x - 2,
x ** 2 + x + 1,
3 * x ** 2 - 2]
for (num, den), t in zip(spam, egg):
self.assertEqual((num / den).simplify(), t)
if __name__ == '__main__':
main()
% ./sample17.py -v
17.
test (__main__.MyTest) ... ok
----------------------------------------------------------------------
Ran 1 test in 0.322s
OK
% | 659 | 1,352 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 4, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2024-38 | latest | en | 0.231004 |
https://bestlovevashikaranspecialists.com/qa/quick-answer-what-can-equal-23.html | 1,624,205,660,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488249738.50/warc/CC-MAIN-20210620144819-20210620174819-00553.warc.gz | 128,185,136 | 10,804 | Quick Answer: What Can Equal 23?
What are the first 4 multiple of 6?
Here again are the multiples of 6, 0, 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96,….
What are the factors of 42?
The factors of 42 are 1, 2, 3, 6, 7, 14, 21, and 42.
Are all factors of 16 even?
The Factors of 16 are all the integers (positive and negative whole numbers) that you can evenly divide into 16. 16 divided by a Factor of 16 will equal another Factor of 16. The Postive Factors of 16 are therefore all the numbers we used to divide (divisors) above to get an even number.
Which is the smallest prime number?
The first 1000 prime numbers121–202321–40737941–6017918161–8028329315 more rows
What can equal 33?
33 = 1 x 33 or 3 x 11. Factors of 33: 1, 3, 11, 33. Prime factorization: 33 = 3 x 11. When 33 is a clue in the FIND THE FACTORS 1 – 12 puzzles, always use 3 and 11 as the factors.
What are the first 5 factors of 16?
The factors of 16 are 1, 2, 4, 8, 16.
What can equal 48?
The factor pairs of the number 48 are: 1 x 48, 2 x 24, 3 x 16, 4 x 12, and 6 x 8.
What two numbers make 23?
The factors of 23 are 1 and 23. Since 23 is a prime number, therefore, it has only two factors.
15+8 = 23.
What is the prime factorization for 23?
Factors of 23 by Prime Factorization It has only two factors, i.e., the number 1 and the number 23 itself. So the prime factorization of the number 23 is 23.
Is 23 a prime number Yes or no?
List of prime numbers up to 100: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, …
What are the factors of 20?
Factors of 20Factors of 20: 1, 2, 4, 5, 10 and 20.Negative Factors of 20: -1, -2, -4, -5, -10 and -20.Prime Factors of 20: 2, 5.Prime Factorization of 20: 2 × 2 × 5 = 22 × 5.Sum of Factors of 20: 42.
What can equal 22?
22 = 1 x 22 or 2 x 11. Factors of 22: 1, 2, 11, 22. Prime factorization: 22 = 2 x 11. When 22 is a clue in the FIND THE FACTORS 1 – 12 puzzles, use 2 and 11 as the factors.
What can equal 25?
Solution: The factors of 25 are 1, 5, and 25 but when we divide 25 by 6, 8, and 9, it leaves a reminder.
What can equal to 24?
The factor pairs of 24 are: 1 x 24, 2 x 12, 3 x 8, and 4 x 6.
What are the 5 factors of 16?
Factors of 16 are numbers that, when multiplied in pairs give the product as 16. There are 5 factors of 16, which are 1, 2, 4, 8, and 16. Here, 16 is the biggest factor. The Prime Factors of 16 are 1, 2, 4, 8, 16 and its Factors in Pairs are (1, 16), (2, 8), and (4, 4).
What are the first 5 multiples of 10?
Multiples of 10 are numbers like 10, 20, 30, 40, 50, 60, and so on.
What are the factors of 21?
The sum of all factors of 21 is 32 and its factors in Pairs are (1, 21) and (3, 7).Factors of 21: 1, 3, 7 and 21.Negative Factors of 21: -1, -3, -7 and -21.Prime Factors of 21: 3, 7.Prime Factorization of 21: 3 × 7 = 3 × 7.Sum of Factors of 21: 32. | 1,114 | 2,900 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2021-25 | latest | en | 0.885342 |
http://openstudy.com/updates/507c21cfe4b040c161a228ef | 1,411,091,258,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657129431.12/warc/CC-MAIN-20140914011209-00154-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 203,460,085 | 36,398 | ## shubhamsrg Group Title what is the ratio of area of a traingle ABC to the area of the triangle whose sides are equal to the medians of ABC ? one year ago one year ago
1. shubhamsrg
or,,prove that this ratio always equals 4/3
2. mukushla
im watching :)
3. estudier
|dw:1350313254678:dw|
4. estudier
x^2 + h^2 = c^2 and y^2 +h^2 = b^2, x+y = a x^2-y^2 = c^2-b^2 ......hmm might be better off with Cosine Rule:-(
5. estudier
Yukky algebra....think, think...
6. estudier
|dw:1350318976709:dw|
7. estudier
|dw:1350319309051:dw|
8. estudier
|dw:1350319508249:dw|
9. estudier
152 is a quarter 123 -> 123/235 is 4/3 -> 143/362 is 4/3 as well
10. estudier
Think that's right...
11. shubhamsrg
why is 352 half of 362 ?
12. estudier
Median divides triangle area in half?
13. shubhamsrg
how is 35 a median ?
14. ganeshie8
|dw:1350323581495:dw|
15. ganeshie8
72 || 34 coz 72 is the midsegment
16. ganeshie8
|dw:1350323706822:dw|
17. ganeshie8
7824 is a parallelog gram, now imagine we translate it down
18. ganeshie8
|dw:1350323784808:dw|
19. estudier
Yes, sorry, I didn't make the construction very clear.......
20. ganeshie8
it translates to 1672
21. shubhamsrg
why will 87 meet 6 ?
22. ganeshie8
diagonals bisect each other in ||gram
23. ganeshie8
so 65=52
24. ganeshie8
87 meets at 6, thats right, but before that we should prove this : 326 is the triangle made by medians
25. ganeshie8
87 meets at 6 becoz, 47 is congruent to 26, 42 is congruent to 21
26. ganeshie8
the only tricky part in this proof i see is proving this : 326 is the triangle made by medians
27. shubhamsrg
i still dont get 876 thing..
28. ganeshie8
ok did u get estudier diagram
29. shubhamsrg
yep..i followed every bit till now..
30. ganeshie8
he constructed, 26 \(\cong\) 47
31. shubhamsrg
yes..
32. ganeshie8
26 is just translation of 47, so they are || one more construction he did is, he translated other median also, the translated medians intersect at 6. (we can prove this, later)
33. shubhamsrg
ahh alright,,so 36 isnt a median yet..okay,,
34. ganeshie8
assume it is a median so that we prove the thing at hand, :)
35. shubhamsrg
alright,,
36. ganeshie8
72 || 84, agree ?
37. shubhamsrg
yes..
38. ganeshie8
becoz 72 is the midsegment
39. shubhamsrg
yes,,i could figure out..
40. ganeshie8
7284 is a ||gram i have constructed
41. shubhamsrg
7248*
42. shubhamsrg
yes,,clear till now..
43. ganeshie8
also, 74 congruent to 62, by construction
44. shubhamsrg
okay,,7426 is a //gm..
45. ganeshie8
now imagine we slide the ||gram 7248 along the side 41
46. shubhamsrg
so you're saying those 2 //gms are congruent..
47. ganeshie8
the diagonal in the ||gram 7248 translated to where ?
48. shubhamsrg
coincides with 62
49. ganeshie8
|dw:1350324847401:dw|
50. ganeshie8
the diagonal 74 coincides wid 62 the side 42 coincides wid 21
51. shubhamsrg
yes,,the 2 //gms are congruent..
52. ganeshie8
|dw:1350324939445:dw|
53. ganeshie8
7248 congruent 6127
54. ganeshie8
we just use this : 6127 is a ||gram that gives 5 is midpoint of diagonal 62
55. shubhamsrg
that one we already knew right ?
56. ganeshie8
which one ? i thought we're trying to prove 5 is midpoint of 62 lol
57. ganeshie8
whats ur question
58. shubhamsrg
we knew 5 was the mid point quite earlier..after that assumption, everything fitted..those 2 became congruent //gms and everything else..
59. ganeshie8
no
60. shubhamsrg
??
61. ganeshie8
those 2 became congruent || grams by construction, 5 got nothing to do wid it
62. ganeshie8
let me put everything in one reply
63. shubhamsrg
we assumed 36 to be the median..that gave 16 // 27 ..by which it was a //gm,, and thus 5 was the midpoint ( 2 diagnols intersected)
64. ganeshie8
ok if that make sense good :)
65. ganeshie8
so only assumption we need to prove is 36 is a median, right ?
66. shubhamsrg
yep..
67. sauravshakya
68. shubhamsrg
excellency yet again..!! thank you @sauravshakya
69. shubhamsrg
and big thank you to @ganeshie8 ..your idea was used here after all..
70. shubhamsrg
+ @estudier you sparked the figure, ofcourse ! :P
71. estudier
I managed to get an outline of this just with algebra --- what a mess!! | 1,454 | 4,239 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2014-41 | longest | en | 0.752708 |
http://experiment-ufa.ru/what-number-is-10-less-than-2681 | 1,529,597,006,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864191.74/warc/CC-MAIN-20180621153153-20180621173153-00362.warc.gz | 105,639,133 | 6,441 | # What number is 10 less than 2681
## What number is 10 less than 2681 - solution with step by step explanation
If it's not what You are looking for, type integer numbers into the boxes below and see the solution.
## What is 10 less than 2681?
We are looking for a new number which is 10 less than 2681.
We will get the new number by subtracting 10 from 2681.
We write it down as:
2681-10=2671
And finally the solution for: What number is 10 less than 2681?
is 2671
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https://logosconcarne.com/2021/02/18/sideband-72-trig-is-easy/ | 1,679,383,349,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943637.3/warc/CC-MAIN-20230321064400-20230321094400-00120.warc.gz | 441,597,939 | 43,794 | # Sideband #72: Trig Is Easy!
Trigonometry is infamously something most normal people fear and loath. Or at least don’t understand and don’t particularly want to deal with. (In fairness, it doesn’t pop up much in regular life.) As with matrix math, trig often remains opaque even for those who do have a basic grasp of other parts of math.
Excellent and thorough tutorials exist for those interested in digging into either topic, but (as with matrix math) I thought a high-altitude flyover might be helpful in pointing out important concepts.
The irony, as it turns out, is that trig is actually pretty easy!
At least the basics are. The thing to keep in mind about trigonometry is that it’s just triangles. And right-angle triangles at that.
[I’m reminded of an old calm down and carry on saying among computer geeks: “Calm down, it’s just ones and zeros.” So for trig: Calm down, it’s just triangles.]
But the other thing about trig (again, as with matrix math) is that it’s often taught by teachers without a fundamental understanding or who even share that vague fear and loathing. Math, in general, often suffers from rote teaching by teachers who learned it the same way.
If I may be forgiven for one last comparison to matrix math, in the defense of teachers these may not be the easiest subjects to teach. As with any subject, there are gestalts that provide a “big picture” view. Without those, one has little else to resort to than those rote rules and procedures which, at best, are boring and, at worst, are utterly opaque and off-putting.
And, to repeat, most people just don’t run into trigonometry very often in life. Who among us has ever needed to calculate the height of a flagpole using trigonometry? It does seem to fall under the frequent student complaint: Why must I learn this? What’s it good for?
The intent here is to try to provide those basic gestalts — a basic understanding of trigonometry on a fundamental level. And, perhaps, to demonstrate what it’s good for (besides figuring out how tall a flagpole is).
§
Figure 1. The Trig Triangle
As mentioned, it’s all just right-angle triangles.
Figure 1 shows the canonical trigonometry right-angle triangle. It stands for all triangles with the following two properties:
• One of the three corners has a 90° angle.
• The long side is set to a length of 1 (no matter how long it actually is).
Such triangles can be tall and skinny, or squat and broad, or (as in Figure 1) somewhere in the middle.
A primary characteristic of triangles in flat space is that the sum of their three angles is always 180°. (In fact, a test of a space to see if it’s flat is to see if this is true. If it isn’t, the space isn’t flat.) Since a trig triangle has a 90° angle, the other two angles must sum to 90°.
Effectively that means we only care about one of the angles (which we call theta, θ), since the other angle is always 90-θ. For example, if theta is 30° then the other angle is 90-30=60°.
Setting the long side (the hypotenuse) to 1 makes the math easier. For real-world triangles with real-world lengths, we use the actual value of the hypotenuse as a scaling factor. For example, if the hypotenuse is actually 4.5, then, as you’ll see, we scale our various trig values for the sides by that much.
Figure 2.
A key point is that the 90° angle along with locking the hypotenuse to 1 constrains the a and b sides such that, if the triangle is tall — side a is long — then side b is forced to be short, making the triangle skinny.
Conversely, if side b is long, side a is forced to be short, and the triangle is squat and broad.
Note that, as shown in Figure 2, the orientation of the triangle isn’t important, only that it has a 90° angle.
Note also that in Figure 2 I kept the hypotenuses the same length, even though that length (per the background grid) is obviously not 1. In fact the actual length (per the grid) is 4.1231… (it’s the square root of 17).
Which brings us to another important aspect of right-angle triangles and trigonometry: the Pythagorean theorem. It says the hypotenuse is the square root of the sum of the squares of the sides. Which always sounds torturous expressed in words. It’s much simpler in numbers:
$\sqrt{a^2+b^2}=hypotenuse$
Since we’re setting the hypotenuse to 1, we have:
$\sqrt{a^2+b^2}=1$
If we take the square root of both sides:
$a^2+b^2=1$
Which, incidentally, is the equality that defines a circle (the unit circle, but as with the triangle, it can be scaled to any real-world size).
And now the locking of the side lengths is obvious, because:
$a^2=1-b^2,\;\;\;b^2=1-a^2$
Making one side longer makes the other side shorter, and vice versa. This proportional locking is fundamental to trigonometry — it’s why it works.
§
So trig boils down to one angle and one side of a right-angle triangle. How easy is that?
Not only are the two sides locked, so is the angle, theta, and this is where the trigonometry kicks in. Given that angle, and the length of one side, we can determine all the sides and angles of a right-angle triangle.
We can also determine the angle by knowing the lengths of at least two of the sides.
Essentially, given two pieces of information about the triangle, we can determine all the information about it. That’s the value of trig.
§ §
The trick is to not let words like sine and cosine (or arctangent!) intimidate. They’re attached to simple concepts.
[That said, as with most math topics, one can get deep into the weeds where things get very involved and there’s lots to learn. We won’t need any of that here.]
Figure 3. Sine & Cosine
To explain, consider Figure 3.
To be canonical, we need to switch from calling them side a and side b to calling them side y and side x, respectively. This places trigonometry on the Cartesian plane.
The sine and cosine are nothing more than the proportions of the sides given some angle, theta (θ). They are functions that take an angle and return the length of the side.
Note that the length returned assumes the hypotenuse is set to 1. As mentioned above, when that’s not true, the actual hypotenuse length becomes a scaling factor we multiply the returned sine or cosine value by to get the actual side length.
Again, all sine and cosine do is, given an angle, give us the length of a side assuming the hypotenuse is set to 1. That’s their whole deal.
In particular, the sine gives the length of the side opposite the angle, and the cosine gives the length of the side next to (“co”) the angle (see Figure 1).
§
You may have noticed that, while sine has its own Wiki page, cosine (and all other trig functions) link to the trigonometric functions page (or the inverse trigonometric functions page for ones like arctangent).
That’s because sine is the fundamental function. The cosine, due to how the sides are proportionally locked, is just sine’s mirror image. In fact, by simply switching to the other angle, the sine and cosine swap.
If you’ve ever perused a trig table, you’ve noted both contain the same numbers in reversed order. (Actually, not reversed but 90° out of phase with each other.)
[For the more mathematically inclined, the cosine is the derivative of the sine and vice versa, modulo some sign changes.]
At root, the sine is nothing more than the length of the opposite side divided by the length of the hypotenuse. Cosine is the length of the adjacent side divided by the hypotenuse. At heart, they’re just fractions that, because of the proportional locking, are tied to the angle. (See the Wiki trig functions page.)
A sine wave is what we get if we feed the sine function a progression of increasing angles:
Generating a red sine (and blue cosine) wave. [from Wikipedia]
The above animation nicely demonstrates how a sine wave comes from circular motion. In fact, the A.C. electricity that comes into our home follows a sine wave because generators turn in circles.
The animation also illustrates how the sine and cosine are 90° out of phase.
§
That’s pretty much the deal. The rest is just elaboration, details, and building on these basic tools.
Another big trig function is tangent, which is just the opposite side’s length divided by the adjacent side’s length. That ends up making the tangent the slope of the hypotenuse, which is why it isn’t defined for 90° (or 270°) — the slope is infinite on a vertical line.
[To be precise, the slope isn’t defined for a vertical line because slope is defined as Δy/Δx, and Δx is zero in a vertical line. The Δ (delta) just means change or difference.]
The other common trig function is arctangent, which is the inverse function to tangent. It takes the slope and returns the angle. (The inverse functions take a length and return the angle, whereas the normal functions take an angle and return a length.)
Arctangent is handy when you know the size of the triangle and want to figure out the angle. For example, the green triangle in Figure 2 has lengths of x=1.0 and y=4.0 and, thus, the slope of the hypotenuse is 4.0 (y/x).
My calculator says the arctangent of 4 is 75.9+ which means the angle (in the lower left) is nearly 76°. We can check this by using the angle and a known side:
$\sin(76)=0.97\ldots$
Remember this assumes the hypotenuse is 1.0, but (per the background grid), that triangle’s actual hypotenuse is:
$\sqrt{1^2+4^2}=\sqrt{1+16}=\sqrt{17}=4.1231\ldots$
So that’s our scaling factor:
$\sin(76)=0.97\times{4.1231}={3.999}\ldots$
That vertical side is 4.0, so (given the rounding) exactly right.
I cannot stress enough that the only way to really learn these things is to work with them. Just reading about them won’t do it.
[As a side note to programmers, the common atan2 function provided by many code libraries is the arctangent function plus it does the slope calculation. The atan2 function takes two parameters, the two sides, calculates the slope and then returns the arctangent of that slope. It’s such a common operation the library just does it for you.]
§ §
This was a hard post to write. I’ve had a note about a “Trig Is So Easy” post for years, but I’ve been working with trig so long it’s hard to judge what needs to be explained and what is obvious.
And it just now occurs to me to wonder if that’s why Rachel Maddow often seems to over-explain things on her MSNBC show. She’s so smart and well-read that she misjudges what her viewers need explained and what is well-known to them? (One time she spent about five minutes explaining a “dead man switch.” I kind of assume anyone smart enough to watch Maddow in the first place already knows what that is, but maybe that’s my own blind spot speaking.)
Well, if that’s what’s going on there, I can sure relate. It ain’t easy!
Stay trigonometric, my friends! Go forth and spread beauty and light.
The canonical fool on the hill watching the sunset and the rotation of the planet and thinking what he imagines are large thoughts. View all posts by Wyrd Smythe
#### 30 responses to “Sideband #72: Trig Is Easy!”
• Wyrd Smythe
The height of flag poles aside, in physics trigonometry gives us a map between the angular location of a point and its x and y coordinates in space. It also gives us the distance (magnitude) of the point through the Pythagorean theorem.
If nothing else, associate cosine with the x coordinate and sine with the y coordinate.
There is also a strong connection with sine waves, and recall that any wave form is a combination of sine waves, so trig is also a tool for wave analysis.
• Wyrd Smythe
[sigh] Yet another reason to hate the WordPress Reader: It “grays out” older posts, a UI technique that usually means “disabled” but to the WP Reader apparently just means “old”.
Note you can still click into these posts and read them even though the graying out suggests you wouldn’t be able to. I’ve complained about this, but it’s apparently intended behavior.
Add that to how the Reader sometimes makes your paragraphs merge into one block of text if you use the Classic editor. If you use the Reader to comment, you might notice it does that to comments at first, but if you exit and come back, your paragraphs are usually intact.
The damn Reader also strips out style information, even if the (Classical) editor put it there in the first place. Did you center something? Well, it won’t be centered in the Reader. And forget about any font color or font style or indent you set, because it won’t be there in the Reader.
The only virtue I see in the Reader is that it also strips out ads for blogs that haven’t upgraded to having no ads (you’re welcome). There are a couple of blogs I follow that I won’t visit the actual site because of the ads, but I can read the posts ad-free in the Reader.
• Wyrd Smythe
Heh. I just noticed this is post #1100. 🙂
• World of Lies | Logos con carne
[…] also Pi Day, so either do some trigonometry (it’s easy) or eat some pie. (But put me down for Tau Day when I ken has double the pizza […]
• The Paltry Sum
I have never been good with numbers, all this homeschooling has me now throwing my hands in the air and making sad faces. Ill just slink off and feel…intimidated by triangles.
• Wyrd Smythe
It’s all about one’s early training, and for most that training hasn’t been good so it all does seem pretty intimidating. The hard part is getting started and facing those first foothills. Most just don’t see the need or point to it. A question I’ve heard a lot is “Why?!” If I can ever help, answer any questions, feel free to ask!
• The Paltry Sum
Bit of proud mom moments – kiddo pulled his math grade up from a C to an A – I paid for homeschool…found him watching math lectures and up early slogging at it. He doesn’t bother asking me anymore. Thank you for the offer, Ill let him know if he gets stuck you were kind enough to offer! Math boggles me entirely.
• Wyrd Smythe
Well, you’re certainly not alone in that — a lot of people see it as a foreign language. Which is unfortunate and far more due to the teachers than the students. Anyone can learn it given a teacher who knows and loves the subject (although, sadly, learning anything gets harder as one ages — I’m trying to learn quantum mechanics, and it’s a challenge).
That said, I took German classes for four years in high school and it never took, like, at all, so I think people are just better at some things than others.
Kudos to the kiddo! Math fundamentals open a lot of doors, whether it be the sciences, baseball stats, or computer animation (which is where I started getting into trig). Learning math definitely falls under the heading of “teaching someone to fish.”
• The Paltry Sum
…I went to a very nice school post 16. I was told to not bother attending math, just accept I would fail it. I got an A. which I am not wholly blaming on speed…but everything certainly made more sense for five minutes. As soon as I could, I dumped it and went where I belonged – liberal arts degree at a nice university. I used to stare at the stem guys in awe.
• Wyrd Smythe
Yeah, one of the really fucked up things is that whole “girls can’t do math” crap. Tell it to Emmy Noether or any of the other brilliant female mathematicians. Then watch out for the left hook…
I was a science nerd until I got into theatre in high school and discovered my artistic side, so I went from STEM to STEAM. My college years (as a film student) were firmly liberal arts, and one of my key phrases for what’s wrong with the world is: “The Death of a Liberal Arts Education” (or, really, just the death of a good education in general).
• The Paltry Sum
The Boy has never been to conventional school. He might go for high school next year. He is testing a year ahead, and is steaming through the work. I am with you about the death of a good education. I’m old enough that I was still subject to a more rigorous schooling. I actually think I am ‘math dyslexic;’ – not the teacher’s fault. I had tutor after lovely tutor, I just can’t see numbers. It was, curiously, a girl’s only school that banned me from math. I had come from a non feepaying school and was so behind them that she was irritated by me. Film, huh! That sounds fun. English Lit here.
• Wyrd Smythe
It does seem like our brains can be wired, or not wired, for certain things. I’m a retired software designer, and I’m fluent in over a dozen programming languages, but I just can’t seem to wrap my head around foreign languages. And I’ve met others who are the same when it comes to numbers. I get that way with my taxes and complicated forms in general. My mind just swims and drowns.
Theatre in high school, filmmaking in college. I lived in Los Angeles then, and I was gonna be the next George Lucas, but that thing about having to know someone in the industry was really true. (How do you get a Union Card so you can get work? You need 22 hours of work experience. How do you get those 22 hours? You need a Union Card. [D’Oh!]) I decided I just wasn’t willing to work my way up the ladder so maybe some day I might direct a film. Meanwhile, I fell into a job that spoke to my technical side, did well, and the rest, as they say, is history.
English Lit! Big reader? (From your blog it looks like you have some interest in science fiction.) I’m currently trying to read everything Agatha Christie wrote. My reading tastes are decidedly low-brow; mostly science fiction with a side of detective stories and murder mysteries. I’m a long-time fan of Christie’s Hercule Poirot, but only recently have I figured out what an amazing writer she was. I’m working my way through the Miss Marple series at the moment.
• The Paltry Sum
I used to be a reader, but the degree destroyed it. Now my reading interests are niche – sci fi, the Beat poets, non fiction, Japanese literature. Im currently reading Basho, which is more grist to the writing mill. I am tending towards writing at this point in life, over reading. I love Christie! I was just discussing with the Boy how there needs to be a really great Sherlock Holmes game for playstation. THey totally wasted the ones they did do, total borefests.
So you are an effects guy? Im sorry you didn’t get to direct…but it sounds like you found your work fulfilling.
I’m just watching Concrete Cowboy. I loved Elba in The Wire, but am just not feeling him in this. The young man playing his son is great. I am going to battle through and hope that it speeds up! The lighting is awful, I can barely see anything….
• Wyrd Smythe
I generally had a pretty good career, although I was never a great fit in Corporate America. The one upside was a pension, although things got bad enough late in my career that I retired early which meant a reduced pension. Social Security started for me this year, and that’s been a big help. (And Medicare, so I no longer have to buy medical insurance; that’s helped, too. Things were starting to look a bit dicey.)
I would have been far more a fit in theatre or film, but success in those is seriously iffy. I had (and still have) some interest in special effects, and my theatre activities were certainly as a backstage guy (lighting mostly).
I’m not familiar with Concrete Cowboy, although I do like Elba’s work. (I did very much enjoy The Wire.) Luther always sounded kind of interesting. Those British shows are often a lot less violent and dark than our American shows. (I’ve had a growing dislike of the casual violence of media ever since Luke blew up the Death Star.)
• The Paltry Sum
Luther is terrifying, not gory but scary as heck. Really recommend it if you get the time. It sounds like you had an interesting career. Spiderman you say. DC or Marvel?
• Wyrd Smythe
With the exception of, despite many tries, never being successful in love, my life has much to be thankful for, and I’ve really enjoyed just about every minute. I do count myself very fortunate.
Spiderman and, as I mentioned, Superman and friends. As a kid it was mostly DC, but when I got older the darker aspects of Marvel held some attraction. Quite some time ago I got seriously into graphic novels (“gnovels” I call them), but that seems to have faded. I generally don’t care for the big budget movies. Too much casual destruction, fight scenes always seem to involve people hitting each other, and the stories really bother my logical “yeah, but” side. The ones I’ve liked (Deadpool, for instance) are the ones that don’t take themselves too seriously and have some fun. I especially like the deconstruction of the Deadpool movies.
• The Paltry Sum
Deadpool is cool, especially the deconstruction of the third wall. I am always disappointed by the movies.
Love is overrated. Good friends is where it is at!
• Wyrd Smythe
That certainly has been my experience!
• The Paltry Sum
It’s on Netflix. I really struggle to find things on netflix to watch and it is my only TV.
I just realized the Giants are playing the Padres, and I cant fucking watch it. I hate MLB.
• Wyrd Smythe
Netflix, okay, I’ll add it to my queue (although it has a lot in front of it).
Netflix, let’s see… So much depends on taste… Things I’ve really enjoyed: The Good Place, Community, Russian Doll, Happy!, Love, Death & Robots, Black Mirror, Disenchantment, The IT Crowd, Lucifer, Unbreakable Kimmy Schmidt, Dear White People, Black AF, Grace and Frankie, The Kominsky Method, Medical Police, Space Force,… But obviously those are all just my tastes, and I have no idea what you may have seen already.
• The Paltry Sum
I really loved Russian Doll. Black Mirror is always fun, but they lost me on one of the episodes that I found too ‘edgy’. I have been enjoying Longmire…yeah..I know…but it is a really good modern western, with a passable fake Clint!
• Wyrd Smythe
Do you know about the Netflix “special codes”? Ways to dig deeper into their catalog. Google for [netflix special codes]
• “For the Third Time…” | Logos con carne
[…] — for example, the three corners pick out a unique plane and a unique circle in that plane. (Triangles also ground trigonometry.) Tripods allow a stable platform on uneven or rough […]
• Wyrd Smythe
The other irony is that, while being good at math can make one seem less human in popular vision, doing math is something that is especially human. We invented the stuff.
As I’ve said many times, being bad at math should be viewed the same way we view being bad at reading or writing. Don’t forget: Math was the third R! It’s a key way you don’t get fooled again. | 5,275 | 22,446 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 7, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2023-14 | latest | en | 0.951158 |
http://cbse-notes.blogspot.com/2016/12/cbse-class-8-mathematics-direct-and.html?utm_source=feedburner&utm_medium=feed&utm_campaign=Feed%3A+CbsePapersQuestionsAnswersMcq+%28CBSE+Papers%2C+Questions%2C+Answers%2C+MCQ+...%29 | 1,485,279,747,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560285001.96/warc/CC-MAIN-20170116095125-00135-ip-10-171-10-70.ec2.internal.warc.gz | 46,790,559 | 23,783 | ## Saturday, December 31, 2016
### CBSE Class 8 - Mathematics - Direct and Inverse Proportions
Direct and Inverse Proportions
CBSE Class 8 - Mathematics
NCERT Exemplar Problems and Solutions
Q1: If x and y are directly proportional and when x = 13, y = 39, which of the following is not a possible pair of corresponding values of x and y ?
(a) 1 and 3
(b) 17 and 51
(c) 30 and 10
(d) 6 and 18
For direct variation,
x1 = x2 y1 y2
Thus x/y = 13/39 = 1/3. Options (a), (b) and (d) have the same ratio while option (c) does not.
Q2: Which of the following is in direct proportion?
(a) One side of a cuboid and its volume.
(b) Speed of a vehicle and the distance travelled in a fixed time interval.
(c) Change in weight and height among individuals.
(d) Number of pipes to fill a tank and the time required to fill the same tank.
Answer: (b) Speed of a vehicle and the distance travelled in a fixed time interval.
∵ Speed / distance = fixed ratio.
Q3: Both u and v vary directly with each other. When u is 10, v is 15, which of the following is not a possible pair of corresponding values of u and v?
(a) 2 and 3
(b) 8 and 12
(c) 15 and 20
(d) 25 and 37.5
u/v = 2/3
Option (C) does not maintain the same ratio.
Q4: Assuming land to be uniformly fertile, the area of land and the yield on it vary
(a) directly with each other.
(b) inversely with each other.
(c) neither directly nor inversely with each other.
(d) sometimes directly and sometimes inversely with each other.
Answer: (d) sometimes directly and sometimes inversely with each other.
Q5: A truck needs 54 litres of diesel for covering a distance of 297 km. The diesel required by the truck to cover a distance of 550 km is
(a) 100 litres
(b) 50 litres
(c) 25.16 litres
(d) 25 litres
More diesel means more distance covered. It is a case of direct variation.
∴ 54/297 = x/550
⇒ x = 54 × 550 / 297 = 100
Q6: By travelling at a speed of 48 kilometres per hour, a car can finish a certain journey in 10 hours. To cover the same distance in 8 hours, the speed of the car should be
(a) 60 km/h
(b) 80 km/h
(c) 30 km/h
(d) 40 km/h
Higher the speed, less time will it take to finish the journey. It is a case of inverse propotion.
∴ 48 × 10 = s × 8
⇒ s = 60
Q7: Which of the following vary inversely with each other?
(a) speed and distance covered.
(b) distance covered and taxi fare.
(c) distance travelled and time taken.
(d) speed and time taken.
Answer: (d) speed and time taken.
Q8: Both x and y are in direct proportion, then 1/x and 1/y are
(a) in indirect proportion.
(b) in inverse proportion.
(c) neither in direct nor in inverse proportion.
(d) sometimes in direct and sometimes in inverse proportion.
Q9: 100 persons had food provision for 24 days. If 20 persons left the place, the provision will last for
(a) 30 days
(b) 96/5 days
(c) 120 days
(d) 40 days
Less people, food will last more days. ⇒ It is an inverse proportion.
∴ 100 × 24 = 80 × days
⇒ days = 30
Q10: (Fill in the blanks)
(a) Amrita takes 18 hours to travel 720 kilometres. Time taken by her to travel 360 kilometres is _______.
(b) If x and y are inversely proportional then _____ = k where k is positive constant.
(c) If x = 5y, then x and y vary ______ with each other.
(d) If xy = 10, then x and y vary ______ with each other.
(e) x and y are said to vary directly with each other if for some positive number k, ______ =k.
(f) If 12 pumps can empty a reservoir in 20 hours, then time required by 45 such pumps to empty the same reservoir is ______ hours. | 1,009 | 3,523 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2017-04 | latest | en | 0.861846 |
http://mathhelpforum.com/discrete-math/index276.html | 1,529,832,921,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267866926.96/warc/CC-MAIN-20180624083011-20180624103011-00626.warc.gz | 199,733,935 | 15,160 | # Discrete Math Forum
Discrete Math Help Forum: Discrete mathematics, logic, set theory
1. ### Difference equation tutorial: draft of part I
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Jun 14th 2014, 01:54 AM
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Feb 11th 2011, 11:34 AM
3. ### List of rules used to moderate MHF - please read carefully.
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• Views: 3,095
Jul 19th 2010, 10:33 PM
1. ### Discrete Proof
• Replies: 7
• Views: 411
Feb 11th 2009, 07:17 PM
2. ### One-to-one and Onto Functions
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• Views: 4,255
Feb 11th 2009, 06:32 PM
3. ### permutations
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• Views: 1,036
Feb 11th 2009, 05:42 PM
4. ### Cardinality
• Replies: 4
• Views: 751
Feb 11th 2009, 05:36 PM
5. ### [SOLVED] Proof involing power sets
• Replies: 4
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Feb 11th 2009, 12:52 PM
6. ### [SOLVED] Organized Counting
• Replies: 1
• Views: 525
Feb 11th 2009, 11:16 AM
7. ### The natural number p is prime
• Replies: 1
• Views: 529
Feb 11th 2009, 08:16 AM
8. ### True or False Statement?
• Replies: 1
• Views: 617
Feb 11th 2009, 08:04 AM
9. ### [SOLVED] Beginner Type Proof: subsets
• Replies: 1
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Feb 11th 2009, 06:46 AM
10. ### the additive inverse and identity element
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Feb 10th 2009, 11:13 PM
11. ### generating function
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• Views: 422
Feb 10th 2009, 10:21 PM
12. ### formula for subsets
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Feb 10th 2009, 06:42 PM
13. ### Help with a few propositional logic proofs
• Replies: 5
• Views: 990
Feb 10th 2009, 06:27 PM
14. ### Recursive and Explicit Equations...How to Write Them?
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• Views: 2,121
Feb 10th 2009, 05:14 PM
15. ### Undo a mathematical conversion done inside a black box
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Feb 10th 2009, 01:56 PM
16. ### Binomial Theorem?
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• Views: 705
Feb 10th 2009, 11:40 AM
17. ### [SOLVED] Permutation and Organized Counting
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• Views: 962
Feb 10th 2009, 11:05 AM
18. ### Number of nodes in a tree
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• Views: 439
Feb 10th 2009, 06:02 AM
19. ### NEED Help ASAP (Due in 2 hour) - Just tell if true or false dealing with empty set)
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Feb 10th 2009, 03:57 AM
20. ### Need Help showing S can not be a propsition??
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• Views: 1,388
Feb 10th 2009, 03:04 AM
21. ### getting started on a relation question
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• Views: 708
Feb 10th 2009, 01:54 AM
22. ### Set theory problem
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• Views: 807
Feb 10th 2009, 01:43 AM
23. ### NP P problems
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• Views: 478
Feb 10th 2009, 12:47 AM
24. ### Hoping someone can help for once! a^3 = b^3
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• Views: 613
Feb 9th 2009, 03:10 PM
25. ### Double summation - Binomial Theorem?
• Replies: 1
• Views: 3,883
Feb 9th 2009, 10:20 AM
26. ### Modulo Classes Problems
• Replies: 1
• Views: 524
Feb 9th 2009, 06:50 AM
• Replies: 0
• Views: 441
Feb 8th 2009, 04:42 PM
28. ### Cardinality
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• Views: 982
Feb 8th 2009, 02:23 PM
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# "discrete"
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Allows you to choose the data by which the thread list will be sorted. | 1,223 | 3,125 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2018-26 | latest | en | 0.710325 |
https://cs.stackexchange.com/questions/83928/sqrtn-omega3-logn-is-not-true-why | 1,718,509,540,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861640.68/warc/CC-MAIN-20240616012706-20240616042706-00243.warc.gz | 167,563,108 | 35,491 | # $\sqrt{n} = \Omega(3^{\log{}n})$ is not true? Why? [duplicate]
Had this question on an algorithm exam. Can any one explain why $\sqrt{n} = \Omega(3^{\log n})$ is false?
• It really depends on the base of the logarithm. Commented Nov 14, 2017 at 22:44
We'll start with some arithmetic, and assume the base of the logarithm to be $x$. First, remember the log law: $$\log_{a}(b) = \frac{\log_{c}(b)}{\log_{c}(a)}$$ So $$\begin{array}{rl} 3^{\log_{x}(n)} &= 3^{\frac{\log_{3}(n)}{\log_{3}(x)}}\\ &=(3^{\log_{3}(n)})^{1/\log_{3}(x)} &=n^{1/\log_{3}(x)} \end{array}$$ Whether $n^{\frac{1}{2}}$ is in big-omega of this then depends on what $x$ is.
If we make the right and proper computer science assumption that the base 2 logarithm is the only logarithm (that was a joke of course), then $1/\log_{3}(x)=1/\log_{3}(2)\approx 1.6$, which is certainly bigger than $1/2$.
You'd need $x\geq 9$ for the statement to be true.
• The problem actually states log base is 2. I forgot to add it. Thank you Commented Nov 14, 2017 at 23:27 | 359 | 1,028 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2024-26 | latest | en | 0.849179 |
http://www.enotes.com/homework-help/2sin3x-1-0-184067 | 1,462,050,650,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860112727.96/warc/CC-MAIN-20160428161512-00172-ip-10-239-7-51.ec2.internal.warc.gz | 498,165,139 | 11,847 | 2sin3x -1 = 0
Posted on
2sin3x -1 =0
==> 2sin3x = 1
Divide by 2:
==> sin3x = 1/2
But we know that sin pi/6 = 1/2
==> 3x = pi/6
Also we know that sinx is positive in 1st and 2nd quadrants:
==> 3x1 = pi/6 ==> x1= pi/18
==> 3x2 = pi-pi/6 = 5pi/6 ==> x2= 5pi/18
Posted on
To solve 2sin3x - 1 = 0
2sin3x -1 +1 = 0+1
2sin3x = 1
sin3x =1/2 =sin 30 = sin (n*180+(-1)^n* 30) deg, n =0,1,2,...
3x = (180n+(-1)^n*30)/3 deg, n= 0, 1,2,...
x= (180n+(-1)^n30)/3 deg, n=0,1,2...
x = (60n+(-1)^n*10) deg, n= 0,1,2,3..
x= 10 , 50, 130, 170, 250, 290, 370 in deg etc | 317 | 573 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2016-18 | longest | en | 0.616421 |
http://www.tennisabstract.com/blog/2012/08/20/the-implications-of-the-10-point-tiebreak/ | 1,519,418,177,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814833.62/warc/CC-MAIN-20180223194145-20180223214145-00754.warc.gz | 570,633,275 | 11,873 | # The Implications of the 10-Point Tiebreak
I’m not sure how we got here, but we now live in a world where a lot of people consider a 10-point tiebreak equivalent to a set. Apparently it’s more fan-friendly and better for television. And of course it’s faster.
Whatever its practical uses, it’s obvious that the first-to-10 breaker isn’t the same as a set. I’ll leave the moral debate to others; let’s take a statistical approach.
In general, the more points (or games, or sets) required to win a match, the more likely it is that the better player wins. Some commentators have taken to calling the 10-point breakers “shootouts,” and for good reason. Reduce the number of points required to win, and you increase the role played by luck.
Of course, sometimes a shootout is the best idea. You’ve got to end a match somehow, and when players end up equal after two sets, four sets, or four sets and twelve games, it’s all the more likely that luck will have to intervene. But the structure of the match determines just how much luck is permitted to play a part.
To compare a 10-point tiebreak with the set it replaces, we need to know how much more luck it introduces into the game. For that, we need an example to work with.
Take two players: Player A wins 70% of points on serve, and Player B wins 67% of points on serve. Playing best of three tiebreak sets, Player A has a 63.9% chance of winning the match.
If A and B split sets, A’s probability of winning falls to 59.3%. In other words, the shorter time frame makes it more likely that B gets lucky, or is able to put together an unusually good run of play long enough to win the match.
If the match is decided by a 10-point tiebreak, however, A’s probability of winning falls all the way to 56.0%, erasing more than one-third of the favorite’s edge in the third set. In fact, the 10-point breaker is barely more favorable to A than a typical 7-pointer, in which A would have a 55.1% chance.
(If you like playing around with this stuff, see my python code to calculate tiebreak odds.)
Somehow I don’t think anyone would advocate replacing the deciding set with a 7-point tiebreak. Yet a 10-point tiebreak is much closer to its 7-point cousin than it is to a full set.
Adding a few more points doesn’t resolve the discrepancy, either. To maintain Player A’s 59.3% chance of winning, the third set would have to be replaced by a 26-point tiebreak. But that, I’m sure, wouldn’t attract many new advertisers.
## 6 thoughts on “The Implications of the 10-Point Tiebreak”
1. Always with the facts and figures and sober analyses! Can’t you just be driven by fear and politics like the rest of us?
1. Who said I was sober?
2. Nancy says:
Wow! Thanks for doing the math for those of us not so gifted! This is fascinating. The numbers tell the story.
3. Jeff – Good post on a topic that needed revisiting. I recall reading that before the SuperBreaker (10 points) was introduced instead of 3rd sets in doubles, the Bryans and most other top seeds were opposed to it on the grounds that they would lose more often than they would in a 3 settter. Their acceptance after a year was influenced by the results of a study of the first 12 months’ results of all matches between seeds and unseeded teams, indicating that seeded teams had won almost the exact same percentage of the time.
If this was this case then, I wonder if it’s continued to be the case, and if so, what might explain it.
Rick | 846 | 3,464 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2018-09 | latest | en | 0.948793 |
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