url
stringlengths 7
1.6k
| fetch_time
int64 1,368,859,410B
1,726,892,719B
| content_mime_type
stringclasses 3
values | warc_filename
stringlengths 108
138
| warc_record_offset
int64 875
1.73B
| warc_record_length
int64 783
762k
| text
stringlengths 66
990k
| token_count
int64 34
495k
| char_count
int64 66
990k
| metadata
stringlengths 439
443
| score
float64 3.5
5.13
| int_score
int64 4
5
| crawl
stringclasses 93
values | snapshot_type
stringclasses 2
values | language
stringclasses 1
value | language_score
float64 0.06
1
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
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
Coming up next: Angle of Reflection: Definition & Formula
### You're on a roll. Keep up the good work!
Replay
Your next lesson will play in 10 seconds
• 0:01 What is an Angle of Incidence?
• 0:54 Absorption,…
• 2:48 Snell's Law
• 6:23 Lesson Summary
Save Save
Want to watch this again later?
Log in or sign up to add this lesson to a Custom Course.
Timeline
Autoplay
Autoplay
Speed
#### Recommended Lessons and Courses for You
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.
To unlock this lesson you must be a Study.com Member.
Create your account
### Register to view this lesson
Are you a student or a teacher?
### Unlock Your Education
#### See for yourself why 30 million people use Study.com
##### Become a Study.com member and start learning now.
Back
What teachers are saying about Study.com
### Earning College Credit
Did you know… We have over 200 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level.
To learn more, visit our Earning Credit Page
### Transferring credit to the school of your choice
Not sure what college you want to attend yet? Study.com has thousands of articles about every imaginable degree, area of study and career path that can help you find the school that's right for you.
Create an account to start this course today
Try it risk-free for 30 days!
Support | 1,268 | 6,025 | {"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} | 3.90625 | 4 | CC-MAIN-2019-39 | longest | en | 0.938513 |
https://fr.slideserve.com/evasteele/making-inferences-a-step-by-step-guide-powerpoint-ppt-presentation | 1,716,822,335,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059044.17/warc/CC-MAIN-20240527144335-20240527174335-00062.warc.gz | 220,502,496 | 21,910 | 1 / 24
# 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
Télécharger la présentation
## 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 |
http://www.algebra.com/algebra/homework/quadratic/word/Quadratic_Equations.faq?hide_answers=1&beginning=3600 | 1,369,474,123,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705884968/warc/CC-MAIN-20130516120444-00008-ip-10-60-113-184.ec2.internal.warc.gz | 300,250,736 | 14,341 | # Questions on Algebra: Quadratic Equation answered by real tutors!
Algebra -> Algebra -> Quadratic Equations and Parabolas -> Quadratic Equation Customizable Word Problems -> Questions on Algebra: Quadratic Equation answered by real tutors! Log On
Ad: You enter your algebra equation or inequality - Algebrator solves it step-by-step while providing clear explanations. Free on-line demo . Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help! Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!
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": 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} | 3.609375 | 4 | CC-MAIN-2013-20 | latest | en | 0.852142 |
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
Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations! Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!
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 |
https://edurev.in/course/quiz/attempt/-1_Test-Grouping-And-Tabulation-Of-Data/f5987f9a-79bd-413b-91d3-e058c7eb2909 | 1,632,256,176,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057227.73/warc/CC-MAIN-20210921191451-20210921221451-00553.warc.gz | 259,312,156 | 42,231 | Courses
# 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 |
http://www.digitaldan.co.uk/scie/numb5.php | 1,652,788,421,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662517245.1/warc/CC-MAIN-20220517095022-20220517125022-00772.warc.gz | 82,918,963 | 2,479 | DigitalDan.co.uk - Crib Mathematical Series
DigitalDan
Hit Counter 73
Contact Details
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
Go to start of banner
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.
``` | 6,285 | 18,531 | {"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-2016-50 | longest | en | 0.920992 |
http://gmatclub.com/forum/numbers-50648.html?fl=similar | 1,430,786,211,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1430455162832.10/warc/CC-MAIN-20150501043922-00036-ip-10-235-10-82.ec2.internal.warc.gz | 86,997,989 | 41,809 | Find all School-related info fast with the new School-Specific MBA Forum
It is currently 04 May 2015, 16:36
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# Numbers
Author Message
TAGS:
Intern
Joined: 02 Jun 2006
Posts: 28
Followers: 0
Kudos [?]: 0 [0], given: 0
Numbers [#permalink] 16 Aug 2007, 03:11
If X+1/X = 4, find the value of X^4 - 1/x^4(x>1)
Explain
Manager
Joined: 15 Aug 2007
Posts: 70
Followers: 2
Kudos [?]: 5 [0], given: 0
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)
Intern
Joined: 02 Jun 2006
Posts: 28
Followers: 0
Kudos [?]: 0 [0], given: 0
The Answer is 112 , sqrt (3)
Senior Manager
Joined: 29 Nov 2006
Posts: 321
Location: Orange County, CA
Followers: 1
Kudos [?]: 39 [0], given: 0
Eesh. Just figuring out the approach would have taken me at least 1.5 minutes.
Great work, Chiya.
Manager
Joined: 15 Aug 2007
Posts: 70
Followers: 2
Kudos [?]: 5 [0], given: 0
Thx ! [#permalink] 16 Aug 2007, 10:02
Thx everyone !!!
14*8=112...
Thx ! [#permalink] 16 Aug 2007, 10:02
Similar topics Replies Last post
Similar
Topics:
2 number 11 21 Oct 2010, 10:04
number 2 16 Oct 2010, 12:32
1 Numbers 3 03 Nov 2009, 23:17
Numbers 2 26 May 2008, 12:14
Number 7 23 Aug 2007, 13:30
Display posts from previous: Sort by | 955 | 2,411 | {"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-2015-18 | longest | en | 0.817761 |
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
## Related pages
solve absolute value equations calculatorwhat is 375 as a decimalprime factorization for 240multi step equations calculatorfractions calculator with stepsleast common factor calculatorderivative of 2sin2xsin2x cos 2xln x differentiationwhat is the derivative of lnxcommon multiples of 2 and 3prime numbers of 105what is the prime factorization of 289cos2x x cos2xhow to divide fractions on a calculatorgraph the linear equation y 3x4izyx5what is the derivative of ln 2xx3 3x2 9x 5prime factorization of 147prime factorization of 160derivative of cos xy2x 5y 16deriviative calculatorbeigocos2piwww njatc utk eduformula of cos3x1.1875 to fractionwhat is twenty percent of fifty dollarsprime factorization of 880whats the prime factorization of 44q m c tfind the prime factorization of 175what is the prime factorization for 132ln solvercos3x 4cos 3x-3cosxsinxcosxsolve sin x sinx 1 0gcf of 76 and 950.4.278 roman numeralslog2 log43x2 x 42xheasy 105.5derivitive of lnx2j-1-182r 3sdifferentiate sin squaredroman numeral 1975differentiation of cos xlcm and gcf finder8y yprime factorization for 196mhcitts com17x5calculator gcfx2 4xwhat is 6x squaredfactor 3x 2-7x-6x squared minus 2x87 roman numeralsderivative of ln tanxsecx move5y x 10factor x2-x-122cosx 1mx3xcalculator to subtract fractionsadding subtracting multiplying and dividing fractions calculatorgreatest common factor generatorgraph 2x y 7derive e 2xsimplify square root of 64 | 562 | 1,925 | {"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-2018-26 | latest | en | 0.811552 |
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
• Replies: 4
• Views: 4,730
Jun 14th 2014, 01:54 AM
• Replies: 0
• Views: 4,617
Feb 11th 2011, 11:34 AM
3. ### List of rules used to moderate MHF - please read carefully.
• Replies: 0
• 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
• Replies: 3
• Views: 4,255
Feb 11th 2009, 06:32 PM
3. ### permutations
• Replies: 2
• 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
• Views: 3,959
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
• Views: 490
Feb 11th 2009, 06:46 AM
10. ### the additive inverse and identity element
• Replies: 1
• Views: 1,188
Feb 10th 2009, 11:13 PM
11. ### generating function
• Replies: 0
• Views: 422
Feb 10th 2009, 10:21 PM
12. ### formula for subsets
• Replies: 3
• Views: 5,742
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?
• Replies: 1
• Views: 2,121
Feb 10th 2009, 05:14 PM
15. ### Undo a mathematical conversion done inside a black box
• Replies: 0
• Views: 538
Feb 10th 2009, 01:56 PM
16. ### Binomial Theorem?
• Replies: 2
• Views: 705
Feb 10th 2009, 11:40 AM
17. ### [SOLVED] Permutation and Organized Counting
• Replies: 6
• Views: 962
Feb 10th 2009, 11:05 AM
18. ### Number of nodes in a tree
• Replies: 0
• 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)
• Replies: 6
• Views: 524
Feb 10th 2009, 03:57 AM
20. ### Need Help showing S can not be a propsition??
• Replies: 4
• Views: 1,388
Feb 10th 2009, 03:04 AM
21. ### getting started on a relation question
• Replies: 1
• Views: 708
Feb 10th 2009, 01:54 AM
22. ### Set theory problem
• Replies: 1
• Views: 807
Feb 10th 2009, 01:43 AM
23. ### NP P problems
• Replies: 5
• Views: 478
Feb 10th 2009, 12:47 AM
24. ### Hoping someone can help for once! a^3 = b^3
• Replies: 3
• 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
• Replies: 4
• Views: 982
Feb 8th 2009, 02:23 PM
,
,
,
,
,
# "discrete"
Click on a term to search for related topics.
Use this control to limit the display of threads to those newer than the specified time frame.
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 |
https://math.stackexchange.com/questions/72844/is-the-pointwise-maximum-of-two-riemann-integrable-functions-riemann-integrable | 1,701,557,425,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100452.79/warc/CC-MAIN-20231202203800-20231202233800-00618.warc.gz | 445,335,232 | 38,400 | # Is the pointwise maximum of two Riemann integrable functions Riemann integrable? [closed]
Let $f,g$ be Riemann integrable functions, prove that the function $h(x)$ defined by $$h\left( x \right) = \max \left\{ {f\left( x \right),g\left( x \right)} \right\}$$ is also Riemann integrable.
Hint
Use the fact that $\max(f(x), g(x)) = \frac{f(x)+g(x)+|f(x)-g(x)|}{2}$
Edit
To use this fact we need to prove that if $f$ is Riemann integrable, so is $|f|$. For this we set for some interval $A$ where $f$ is bounded: $$M = \sup{\{f(x) : x \in A\}}$$ $$m = \inf{ \{f(x) : x \in A\} }$$ $$M' = \sup{\{|f(x)| : x \in A\}}$$ $$m' = \inf{ \{|f(x)| : x \in A\} }$$ And I will let you prove that $M' - m' \leq M - m$.
With this in mind, if $f$ is Riemann integrable on $[a,b]$ then take some partition $P$ of $[a,b]$, and using the previous fact prove that $$0 \leq U(|f|, P) - L(|f|, P) \leq U(f, P) - L(f, P)$$ Where U and L are the upper and lower sums respectively.
This will allow you to use the Riemann integrability criterion and conclude that $|f|$ is also integrable on $[a,b]$.
• or simply that $\left|\max(f(x),g(x))\right|\le|f(x)|+|g(x)|$
– robjohn
Oct 15, 2011 at 18:48
• @robjohn, the fact that $|h(x)|\leqslant k(x)$ with $k$ Riemann integrable does not imply that $h$ is Riemann integrable. At all.
– Did
Oct 15, 2011 at 22:10
• @alejo: Sorry but I am not sure to see how your hint applies.
– Did
Oct 15, 2011 at 22:11
• @Didier: Indeed. I was simply looking at alejo's answer and started thinking Lebesgue integral and this simplifies his answer in that case. We need to use boundedness and discontinuities of measure $0$ instead.
– robjohn
Oct 15, 2011 at 22:53
• @Didier: Well, if $f$ if Riemann integrable, so its $|f|$ and any linear combination of Riemann integrable functions is also Riemann integrable. Oct 15, 2011 at 23:35
Call $\Delta_\sigma(u)$ the difference between the upper and lower Darboux sums of a function $u$ with respect to a subdivision $\sigma$. By definition of Riemann integrability, $\inf\limits_\sigma\Delta_\sigma(f)=\inf\limits_\sigma\Delta_\sigma(g)=0$. Call $M_I(u)$ and $m_I(u)$ the supremum and the infinum of a function $u$ on an interval $I$.
Then, for every interval $I$, $M_I(h)=\max\{M_I(f),M_I(g)\}$ and $m_I(h)\geqslant\max\{m_I(f),m_I(g)\}$, hence $$M_I(h)-m_I(h)\leqslant\max\{M_I(f)-m_I(f),M_I(g)-m_I(g)\},$$ which implies $$M_I(h)-m_I(h)\leqslant M_I(f)-m_I(f)+M_I(g)-m_I(g).$$ Summing this over the intervals $I$ defining a subdivision $\sigma$, one gets $\Delta_\sigma(h)\leqslant\Delta_\sigma(f)+\Delta_\sigma(g)$.
For every positive $t$, there exists $\sigma$ such that $\Delta_\sigma(f)\leqslant t$ and $\tau$ such that $\Delta_\tau(g)\leqslant t$. For every subdivision $\varrho$ containing $\sigma$ and $\tau$, one gets $$\Delta_\varrho(h)\leqslant \Delta_\varrho(f)+\Delta_\varrho(g)\leqslant\Delta_\sigma(f)+\Delta_\tau(g)\leqslant2t.$$ This proves that $\inf\limits_\sigma\Delta_\sigma(h)=0$ hence $h$ is Riemann integrable. | 1,057 | 2,991 | {"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-2023-50 | longest | en | 0.791754 |
https://melonicedlatte.com/2023/08/14/172400.html | 1,695,585,196,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506669.30/warc/CC-MAIN-20230924191454-20230924221454-00572.warc.gz | 438,872,189 | 13,663 | # [leetcode] 643. Maximum Average Subarray I _ Algorithm Problem Solve for python
## 1. Problem
643. Maximum Average Subarray I
You are given an integer array nums consisting of n elements, and an integer k.
Find a contiguous subarray whose length is equal to k that has the maximum average value and return this value. Any answer with a calculation error less than 10-5 will be accepted.
Example 1:
Input: nums = [1,12,-5,-6,50,3], k = 4
Output: 12.75000
Explanation: Maximum average is (12 - 5 - 6 + 50) / 4 = 51 / 4 = 12.75
Example 2:
Input: nums = [5], k = 1
Output: 5.00000
Constraints:
• n == nums.length
• 1 <= k <= n <= 10^5
• -10^4 <= nums[i] <= 10^4
## 2. Solution
I solved this problem like this.
• Complexity
• Time complexity : O(N)
• Space complexity : O(N)
• Step
• Get first k sum & append sum_list
• When we calculate next sum, add right new value and subtract left value
• Find max value on sum_list
class Solution:
def findMaxAverage(self, nums: List[int], k: int) -> float:
sum_list = [sum(nums[:k])]
for idx in range(k, len(nums)):
sum_list.append(sum_list[-1] - nums[idx - k] + nums[idx])
return round(max(sum_list)/k, 5) | 350 | 1,159 | {"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": 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} | 3.875 | 4 | CC-MAIN-2023-40 | longest | en | 0.579122 |
https://www.jiskha.com/display.cgi?id=1262463655 | 1,516,696,202,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891791.95/warc/CC-MAIN-20180123072105-20180123092105-00697.warc.gz | 922,584,397 | 4,021 | # Probability
posted by .
One of two cards is black on one side and white on the other side. The second card is black on both sides. One card is selected at random and the side facing up is black. What is the probability that the other side of the card is white?
• Probability -
There are eight possible card arrangements with equal probability. The first of each pair listed below is the first draw. B' is one of the black sides of the two-sided black card.
The possibilities are:
B/W and B/B'; B/W and B'/B; W/B and B/B'; W/B and B/B', B/B' and W/B; B/B' and B/W; B'/B and W/B; B'/B and B/W. Six of these possibilities give Black in the first draw. The only possibilities after getting a B or B' are the following six:
B/W and B/B'; B/W and B'/B; B/B' and W/B; B/B' and B/W; B'/B and W/B; B'/B and B/W.
Two of those six result in white for the second draw. The probability of white is 1/3.
## Similar Questions
1. ### Math
Here's a chalenging one that i can't do at all. A deck of cards has 6 yellow, 6 green, and 5 black cards. You pick four cards from the deck. Cards are not returned after they are picked. What's the probability of picking- first not …
2. ### math
a card is picked from a pack containing 52 cards . it is then replaced and a second card is picked. Find the probability that a)both cards are the seven of diamonds, b)the first card is a heart and the second a spade. c)one card is …
3. ### algebra
if one card is selected at random from a standard deck of 52 cards, what is the probability of choosing a black card or a king?
4. ### Math-probability
One card is selected at random from a deck of cards. Determine the probability that the card is a black heart.
5. ### Math
One card is selected from a deck of playing cards. Determine the probability of selecting: a card greater than 9 or a black card.
6. ### Math-New Question
One card is selected from a deck of playing cards. Determine the probability of selecting: a card greater than 9 or a black card
7. ### Math
Laurie is holding 6 red cards and 7 black cards. If two cards are selected at random(w/out replacement), what is the probability that the first card selected is red and the second is black?
8. ### ALGEBRA II
Using a standard deck of cards, find the probability that: 1) You choose two cards and one is a face card and the other is black 2) A black jack and then an ace are drawn
9. ### Math
Need help on how to solve this problem.. Five cards are chosen from a deck of playing cards and put into a pile. Three of the cards are red and 2 of the cards are black. A second pile is made using 2 red cards and b black cards. One …
10. ### Statistics
. Two cards are drawn at random, without replacement, from a standard 52-card deck. Find the probability that: (a) both cards are the same color (b) the first card is a face card and the second card is black
More Similar Questions | 733 | 2,884 | {"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 | 4 | CC-MAIN-2018-05 | latest | en | 0.952329 |
https://cs.stackexchange.com/questions/160755/how-is-p-not-trivially-equal-to-zpp | 1,721,545,519,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517648.65/warc/CC-MAIN-20240721060614-20240721090614-00223.warc.gz | 162,772,551 | 39,840 | How is P not trivially equal to ZPP?
The definition of ZPP seems to be
$$ZPP = RP \cap coRP.$$
I think ZPP should then be equivalent to P, because for any language L in ZPP, there is an algorithm A and B proving that it is in RP and coRP, respectively. Since A must tell for any x s.t. $$x \notin L$$ the truth and B for any x s.t. $$x \notin L$$ as well, we can just run both algorithms in polynomial time, and one of them has to give me the answer since these are the only two possible options.
Is there an error in my thoughts?
There is an error in your reasoning. Say that $$x \not\in L$$. Then when you run $$A(x)$$ you'll get a "no" answer with certainty. However, when you run $$B(x)$$ you'll get a "no" with some probability that can be up to $$1/2$$.
What do you do when both $$A(x)$$ and $$B(x)$$ return "no"? In order to answer correctly you'd need to return "no".
Consider now $$x \in L$$. $$A(x)$$ will return "no" with some probability that can be up to $$1/2$$, while $$B(x)$$ will return "no" with certainty. We already argued that you algorithm returns "no" in this case, but this is incorrect.
• Nice explanation, but is also worth mentioning that the particular error in the reasoning is that "and one of them has to give me the answer since these are the only two possible options." is simply wrong; both give answers, and you know that one of them must be correct but you don't know which one. This is trivial as you can replace them by $\Sigma^\star$ and $\varnothing$. Commented Jun 18, 2023 at 20:13
• Yeah. There are $3$ possible combinations of the two answers. If $x \in L$ you get one of $\{(yes, no), (no, no)\}$. If $x \not\in L$ you get one of $\{ (no, yes), (no, no) \}$. If you you need to decide $x \in L$ and you observe the output of $A$ and $B$ then you can answer correctly when you see either $(yes, no)$ or $(no, yes)$. However, the answers $(no, no)$ are inconclusive (or, if you wish, you don't know which of the two algorithms was incorrect). Commented Jun 18, 2023 at 20:25 | 569 | 2,022 | {"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": 13, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2024-30 | latest | en | 0.945664 |
https://www.studypool.com/discuss/1232832/solve-using-the-sign-test-?free | 1,495,489,332,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607120.76/warc/CC-MAIN-20170522211031-20170522231031-00479.warc.gz | 963,718,998 | 14,028 | ##### solve using the sign test
Calculus Tutor: None Selected Time limit: 1 Day
solve using the sign test 2x^2+5x-3<=0
Oct 25th, 2015
2x^2+5x-3 <=0
(2x-1)(2x+3)<=0
divide the numerical axis in to (-inf, -3), (-3,1/2), (1/2, inf), then in both (-inf, -3), and (1/2, inf) , 2x^2+5x-3 >0, since the parabola y=2x^2+5x-3 is open up, thus the solution is
[-3,1/2]
Oct 25th, 2015
Thank you so much!
Oct 25th, 2015
...
Oct 25th, 2015
...
Oct 25th, 2015
May 22nd, 2017
check_circle | 216 | 485 | {"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} | 3.546875 | 4 | CC-MAIN-2017-22 | longest | en | 0.774023 |
https://math.stackexchange.com/questions/2521120/rank-nullity-theorem-struggling-to-reconcile-rm-ranka-rm-nula-n | 1,575,638,141,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540488620.24/warc/CC-MAIN-20191206122529-20191206150529-00048.warc.gz | 462,152,500 | 32,891 | # Rank-Nullity Theorem: Struggling to Reconcile ${\rm Rank}(A) +{\rm Nul}(A) = n$ and $\dim(V) = \dim({\rm Im}(f)) +\dim({\rm ker}(f))$.
I am struggling to reconcile the two forms of the rank-nullity theorem:
1. $\operatorname{Rank}(\mathbf{A}) + \operatorname{Null}(\mathbf{A}) = n$,
2. $\dim(\mathbf{V}) = \dim(\operatorname{Im}(f)) + \dim(\operatorname{Ker}(f))$
In attempting to reconcile these, I have collected the following facts:
• If we have $\mathbf{A}\mathbf{x} = \mathbf{b}$, then $\mathbf{A}\mathbf{x} = f(\mathbf{x})$ when thinking about it from the perspective of linear transformations. This means that the image of $f$ is $\mathbf{b}$, so we have $\operatorname{Im}(f) = \mathbf{b}$.
• The rank of a matrix is equal to the dimension of its column space. Therefore, the rank of a matrix is equal to the number of leading ones in the matrix in reduced row echelon form. This is the $\operatorname{Rank}(A)$ part.
• Dimension represents the number of elements in a basis.
• Therefore, we have that $\dim(\operatorname{Im}(f)) = \dim(\mathbf{b})$.
I do not see how $\operatorname{Rank}(\mathbf{A}) = \dim(\operatorname{Im}(f))$. In fact, I not even see how a statement like $\dim(\operatorname{Im}(f))$ even makes any sense? $\operatorname{Rank}(\mathbf{A})$ makes sense, since we can calculate this by getting $\mathbf{A}$ into RREF, but how does $\dim(\operatorname{Im}(f)) = \dim(\mathbf{b})$ make any sense? I do not see the connection here.
I would greatly appreciate it if people could please take the time to clarify this.
• Hint: if $A=(\mathbf{v_1\;v_2\;\ldots\; v_n})$, then $\mathrm{Dim}(\mathrm{Im}( A))$ is just the number of linearly independent vectors among $\mathbf{v_i}$s. Or in other words, $\{\bf v_i\}$ spans $\mathrm{Im} (A)$ (why?). – BAI Nov 15 '17 at 7:46
The image of $f$ isn't $b$. You don't seem to know the definition of the image of a function.
The image is the set of all possible outputs of the function $f$ as $x$ ranges over all vectors in the domain.
Edit, to address the question in the comments below my answer.
The image of $f$ is the span of the columns of $A$. The rank of $A$ is the number of linearly independent columns of $A$. Hence the rank is the size of a basis of the image, namely the basis consisting of the linearly independent columns of $A$. Since the rank is equal to the cardinality of a basis of the image, the rank is equal to the dimension of the image.
• Thanks for the response. But how does this explain the relation between the $Rank(A)$ and $dim(Im(f))$? Can you please demonstrate why these are equal? – The Pointer Nov 15 '17 at 7:29
• That is one definition of rank. Another definition, the one you seem to be using, is the number of pivots in the rref form of $A$. The number of pivots tells you how many linearly independent columns $A$ has. You can prove this is equal to the dimension of the image, as soon as you learn what the image really is. (Hint: the image is spanned by the columns of $A$.) – symplectomorphic Nov 15 '17 at 7:30
• I added more details. – symplectomorphic Nov 15 '17 at 7:41
• You are confusing the image of a function, period, and the image of a point under a function. This is an incredibly elementary distinction. The image of the function $f(x)=x^2$ is the set of all nonnegative real numbers, but the image of $3$ under $f$ is $9$. I don't see how you've asked questions about complex analysis and PDEs without understanding this. – symplectomorphic Nov 15 '17 at 8:02
• en.wikipedia.org/wiki/Image_(mathematics)#Definition You're right. I've confused myself by reading through multiple different sources. – The Pointer Nov 15 '17 at 8:06
For the part you do not understand: since there is an isomorphism $$\overline{\varphi} : V/\text{Ker}(\varphi) \stackrel{\sim}{=} \varphi(V)$$ we have: $$\text{dim}V/\text{Ker}(\varphi) = \text{dim}\varphi(V)$$ By consequence of first isomorphism theorem, we have $$\text{dim}V = \text{dim}\text{Ker}(\varphi) + \text{dim}V/\text{Ker}(\varphi)$$ So it follows that, $$\dim V = \dim Ker(\varphi)+\dim \varphi(V)$$
And the above $\varphi$ is a linear transformation and of course $\varphi(V) = \text{Im}(V)$
Well, i am not sure if you have learnt the first isomorphism theorem or not, but this is the proper understanding of how the so called rank nullity is derived. So first things first, remember this theorem.
The next is easy to understand, we have well known mapping which is $$T_A: \mathbb{R}^n \to \mathbb{R}^n$$ which maps $$x \mapsto Ax$$ in which $x$ is a column vector in the $\mathbb{R}^n$ space and $A$ is THE matrix that is in question. Now : i hope you can easily verify the fact that in this linear transformation, we have dimension of kernel $T$ is merely the dimension of the nullspace of $A$ and the dimension of the image of $T$ is merely the dimension of the column space of $A$. And hence we see that indeed $$\text{rank}A = \dim \text{Im}T_A$$
My answer is still slightly vague and does not provide any intuition, but to piece it all together requires you to understand the fundamentals and of course the first isomorphism theorem.
• The post didn't ask for a proof of the theorem. Your answer doesn't answer the question. – symplectomorphic Nov 15 '17 at 8:04
• My comment (not yet proof) is to complement his understanding since he got an answer from you? It doesnt hurt to know something relevant. – nan Nov 15 '17 at 8:08
• Answers here are supposed to give answers to the questions. Nowhere does the OP ask for a proof of the theorem. The question is about why the two statements of the theorem are equivalent. Your post is not relevant to that issue. – symplectomorphic Nov 15 '17 at 8:10
• In my interpretation, he has some understanding of linear transformations and capable of understanding some abstract meaning of the theorem. And yes, i can definitely use my "proof" here to show these two statements are indeed equivalent if that is what you are worried about. – nan Nov 15 '17 at 8:18
• I'm not worried about anything. I'm telling you how to improve your answers. Don't give irrelevant information. Answer the question directly and simply. I'll leave it there. – symplectomorphic Nov 15 '17 at 8:19 | 1,748 | 6,202 | {"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": 2, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2019-51 | latest | en | 0.84231 |
https://www.varsitytutors.com/hotmath/hotmath_help/topics/word-problems-involving-the-distributive-property.html | 1,656,239,508,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103205617.12/warc/CC-MAIN-20220626101442-20220626131442-00469.warc.gz | 1,124,188,031 | 25,840 | # Word Problems Involving the Distributive Property
The Distributive Property states that, for all real numbers $x,y,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{\hspace{0.17em}}z$ ,
$x\left(y+z\right)=xy+xz$ .
This very important property is frequently used in word problems .
Example 1:
Rico's MP3 player holds songs of three different genres: grindcore, zydeco, and kuduro. There are $5$ times as many grindcore tracks as there are zydeco tracks, and there are $7$ times as many kuduro tracks as there are zydeco tracks. Let $x$ represent the number of zydeco tracks. Write an expression for the total number of tracks on the MP3 player, and simplify it.
Multiply the number of zydeco tracks by $5$ to get the number of grindcore tracks.
$5x$
Multiply the number of zydeco tracks by $7$ to get the number of kuduro tracks.
$7x$
Add up the numbers of all the tracks.
$x+5x+7x$
Simplify using the distributive property.
$\begin{array}{l}=\left(1+5+7\right)x\\ =13x\end{array}$
So, Rico's MP3 player holds $13x$ tracks.
Example 2:
A volleyball uniform costs $13$ for the shirt, $11$ for pants, and $8$ for socks. Write two equivalent expressions for the total cost of $12$ uniforms. Then find the cost.
Write an expression for the cost of $1$ uniform, add .
$13+11+8$
Write an expression for the cost of $12$ uniforms, multiply $12$ by the cost of $1$ uniform.
$12\left(13+11+8\right)$
Simplify using the distributive property.
$\begin{array}{l}=12\cdot 13+12\cdot 11+12\cdot 8\\ =156+132+96\end{array}$
$=384$
So, the total cost of $12$ uniforms is $384$ . | 497 | 1,617 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 26, "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} | 4.78125 | 5 | CC-MAIN-2022-27 | latest | en | 0.858023 |
http://nrich.maths.org/public/leg.php?code=54&cl=4&cldcmpid=1820 | 1,503,284,129,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886107487.10/warc/CC-MAIN-20170821022354-20170821042354-00551.warc.gz | 299,078,069 | 8,408 | # Search by Topic
#### Resources tagged with Quadratic equations similar to Complex Rotations:
Filter by: Content type:
Stage:
Challenge level:
### Roots and Coefficients
##### Stage: 5 Challenge Level:
If xyz = 1 and x+y+z =1/x + 1/y + 1/z show that at least one of these numbers must be 1. Now for the complexity! When are the other numbers real and when are they complex?
### Conjugate Tracker
##### Stage: 5 Challenge Level:
Make a conjecture about the curved track taken by the complex roots of a quadratic equation and use complex conjugates to prove your conjecture.
### Complex Sine
##### Stage: 5 Challenge Level:
Solve the equation sin z = 2 for complex z. You only need the formula you are given for sin z in terms of the exponential function, and to solve a quadratic equation and use the logarithmic function.
### Interactive Number Patterns
##### Stage: 4 Challenge Level:
How good are you at finding the formula for a number pattern ?
### Proof Sorter - Quadratic Equation
##### Stage: 4 and 5 Challenge Level:
This is an interactivity in which you have to sort the steps in the completion of the square into the correct order to prove the formula for the solutions of quadratic equations.
### Xtra
##### Stage: 4 and 5 Challenge Level:
Find the sides of an equilateral triangle ABC where a trapezium BCPQ is drawn with BP=CQ=2 , PQ=1 and AP+AQ=sqrt7 . Note: there are 2 possible interpretations.
### Darts and Kites
##### Stage: 4 Challenge Level:
Explore the geometry of these dart and kite shapes!
### Cocked Hat
##### Stage: 5 Challenge Level:
Sketch the graphs for this implicitly defined family of functions.
##### Stage: 5 Challenge Level:
Can you find a quadratic equation which passes close to these points?
### Partly Circles
##### Stage: 4 Challenge Level:
What is the same and what is different about these circle questions? What connections can you make?
### Implicitly
##### Stage: 5 Challenge Level:
Can you find the maximum value of the curve defined by this expression?
### Golden Fibs
##### Stage: 5 Challenge Level:
When is a Fibonacci sequence also a geometric sequence? When the ratio of successive terms is the golden ratio!
### In Between
##### Stage: 5 Challenge Level:
Can you find the solution to this algebraic inequality?
### How Old Am I?
##### Stage: 4 Challenge Level:
In 15 years' time my age will be the square of my age 15 years ago. Can you work out my age, and when I had other special birthdays?
### Good Approximations
##### Stage: 5 Challenge Level:
Solve quadratic equations and use continued fractions to find rational approximations to irrational numbers.
### Weekly Challenge 39: Symmetrically So
##### Stage: 5 Challenge Level:
Exploit the symmetry and turn this quartic into a quadratic.
### Golden Ratio
##### Stage: 5 Challenge Level:
Solve an equation involving the Golden Ratio phi where the unknown occurs as a power of phi.
##### Stage: 5 Short Challenge Level:
Can you solve this problem involving powers and quadratics?
### Always Two
##### Stage: 4 and 5 Challenge Level:
Find all the triples of numbers a, b, c such that each one of them plus the product of the other two is always 2.
### Target Six
##### Stage: 5 Challenge Level:
Show that x = 1 is a solution of the equation x^(3/2) - 8x^(-3/2) = 7 and find all other solutions.
### Golden Mathematics
##### Stage: 5
A voyage of discovery through a sequence of challenges exploring properties of the Golden Ratio and Fibonacci numbers.
### Halving the Triangle
##### Stage: 5 Challenge Level:
Draw any triangle PQR. Find points A, B and C, one on each side of the triangle, such that the area of triangle ABC is a given fraction of the area of triangle PQR.
### Kissing
##### Stage: 5 Challenge Level:
Two perpendicular lines are tangential to two identical circles that touch. What is the largest circle that can be placed in between the two lines and the two circles and how would you construct it?
### How Many Balls?
##### Stage: 5 Challenge Level:
A bag contains red and blue balls. You are told the probabilities of drawing certain combinations of balls. Find how many red and how many blue balls there are in the bag.
### Two Cubes
##### Stage: 4 Challenge Level:
Two cubes, each with integral side lengths, have a combined volume equal to the total of the lengths of their edges. How big are the cubes? [If you find a result by 'trial and error' you'll need to. . . .
### Golden Thoughts
##### Stage: 4 Challenge Level:
Rectangle PQRS has X and Y on the edges. Triangles PQY, YRX and XSP have equal areas. Prove X and Y divide the sides of PQRS in the golden ratio.
### Pentakite
##### Stage: 4 and 5 Challenge Level:
ABCDE is a regular pentagon of side length one unit. BC produced meets ED produced at F. Show that triangle CDF is congruent to triangle EDB. Find the length of BE.
### Plus or Minus
##### Stage: 5 Challenge Level:
Make and prove a conjecture about the value of the product of the Fibonacci numbers $F_{n+1}F_{n-1}$.
### Bird-brained
##### Stage: 5 Challenge Level:
How many eggs should a bird lay to maximise the number of chicks that will hatch? An introduction to optimisation.
##### Stage: 5 Challenge Level:
Find all real solutions of the equation (x^2-7x+11)^(x^2-11x+30) = 1.
### Continued Fractions II
##### Stage: 5
In this article we show that every whole number can be written as a continued fraction of the form k/(1+k/(1+k/...)).
### Pent
##### Stage: 4 and 5 Challenge Level:
The diagram shows a regular pentagon with sides of unit length. Find all the angles in the diagram. Prove that the quadrilateral shown in red is a rhombus.
### Square Mean
##### Stage: 4 Challenge Level:
Is the mean of the squares of two numbers greater than, or less than, the square of their means?
### Pareq Calc
##### Stage: 4 Challenge Level:
Triangle ABC is an equilateral triangle with three parallel lines going through the vertices. Calculate the length of the sides of the triangle if the perpendicular distances between the parallel. . . .
### Resistance
##### Stage: 5 Challenge Level:
Find the equation from which to calculate the resistance of an infinite network of resistances.
### Golden Eggs
##### Stage: 5 Challenge Level:
Find a connection between the shape of a special ellipse and an infinite string of nested square roots.
### Polar Flower
##### Stage: 5 Challenge Level:
This polar equation is a quadratic. Plot the graph given by each factor to draw the flower.
### Golden Construction
##### Stage: 5 Challenge Level:
Draw a square and an arc of a circle and construct the Golden rectangle. Find the value of the Golden Ratio. | 1,550 | 6,703 | {"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} | 3.84375 | 4 | CC-MAIN-2017-34 | latest | en | 0.850461 |
https://www.slideserve.com/adamdaniel/chapter-5-5178 | 1,510,955,686,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934803944.17/warc/CC-MAIN-20171117204606-20171117224606-00447.warc.gz | 876,316,100 | 21,052 | 1 / 38
# - PowerPoint PPT Presentation
Chapter 5 Sampling Distributions 5.1 Sampling distributions for counts and proportions 5.2 Sampling distributions of a sample mean Chapter 5.1 Sampling distributions for counts and proportions Objectives Binomial distributions for sample counts
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
### Chapter 5 Sampling Distributions
5.1 Sampling distributions for counts and proportions
5.2 Sampling distributions of a sample mean
Objectives
Binomial distributions for sample counts
• Binomial distributions in statistical sampling
• Binomial mean and standard deviation
• Sample proportions
• Normal approximation
• Binomial formulas
Binomial distributions for sample counts proportions
Binomial distributions are models for some categorical variables, typically representing the number of successes in a series of n trials.
The observations must meet these requirements:
• The total number of observations n is fixed in advance.
• Each observation falls into just 1 of 2 categories: success and failure.
• The outcomes of all n observations are statistically independent.
• All n observations have the same probability of “success,” p.
We record the next 50 births at a local hospital. Each newborn is either a boy or a girl.
We express a binomial distribution for the count proportionsX of successes among n observations as a function of the parameters n and p:B(n,p).
• The parameter n is the total number of observations.
• The parameter p is the probability of success on each observation.
• The count of successes X can be any whole number between 0 and n.
A coin is flipped 10 times. Each outcome is either a head or a tail. The variable X is the number of heads among those 10 flips, our count of “successes.”
On each flip, the probability of success, “head,” is 0.5. The number X of heads among 10 flips has the binomial distribution B(n = 10, p = 0.5).
Applications for binomial distributions proportions
Binomial distributions describe the possible number of times that a particular event will occur in a sequence of observations.
They are used when we want to know about the occurrence of an event, not its magnitude.
• In a clinical trial, a patient’s condition may improve or not. We study the number of patients who improved, not how much better they feel.
• Is a person ambitious or not? The binomial distribution describes the number of ambitious persons, not how ambitious they are.
• In quality control we assess the number of defective items in a lot of goods, irrespective of the type of defect.
Binomial setting proportions
A manufacturing company takes a sample of n = 100 bolts from their production line. X is the number of bolts that are found defective in the sample. It is known that the probability of a bolt being defective is 0.003.
Does X have a binomial distribution?
• Yes.
• No, because there is not a fixed number of observations.
• No, because the observations are not all independent.
• No, because there are more than two possible outcomes for each observation.
• No, because the probability of success for each observation is not the same.
A survey-taker asks the age of each person in a random sample of 20 people. X is the age for the individuals.
Does X have a binomial distribution?
• Yes.
• No, because there is not a fixed number of observations.
• No, because the observations are not all independent.
• No, because there are more than two possible outcomes for each observation.
Binomial setting proportions
A fair die is rolled and the number of dots on the top face is noted. X is the number of times we have to roll in order to have the face of the die show a 2.
Does X have a binomial distribution?
Yes.
• No, because there is not a fixed number of observations.
• No, because the observations are not all independent.
• No, because there are more than two possible outcomes for each observation.
• No, because the probability of success for each observation is not the same.
Binomial formulas proportions
The number of ways of arranging k successes in a series of n observations (with constant probability p of success) is the number of possible combinations (unordered sequences).
This can be calculated with the binomial coefficient:
Where k = 0, 1, 2, ..., or n.
Binomial formulas proportions
• The binomial coefficient “n_choose_k” uses the factorialnotation “!”.
• The factorial n! for any strictly positive whole number n is:
n! = n × (n − 1) × (n − 2) × · · · × 3 × 2 × 1
• For example: 5! = 5 × 4 × 3 × 2 × 1 = 120
• Note that 0! = 1.
Calculations for binomial probabilities proportions
The binomial coefficient counts the number of ways in which k successes can be arranged among n observations.
The binomial probability P(X = k) is this count multiplied by the probability of any specific arrangement of the k successes:
The probability that a binomial random variable takes any range of values is the sum of each probability for getting exactly that many successes in n observations.
P(X≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
Color blindness proportions
The frequency of color blindness (dyschromatopsia) in the Caucasian American male population is estimated to be about 8%. We take a random sample of size 25 from this population.
What is the probability that exactly five individuals in the sample are color blind?
• P(x= 5) = (n! / k!(n k)!)pk(1 p)n-k = (25! / 5!(20)!) 0.0850.925P(x= 5) = (21*22*23*24*24*25 / 1*2*3*4*5) 0.0850.9220P(x= 5) = 53,130 * 0.0000033 * 0.1887 = 0.03285
A population contains a proportion p of successes. If the population is much larger than the sample, the count X of successes in an SRS of size n has approximately the binomial distribution B(n, p).
The n observations will be nearly independent when the size of the population is much larger than the size of the sample. As a rule of thumb, the binomial sampling distribution for counts can be used when the population is at least 20 times as large as the sample.
Binomial mean and standard deviation proportions
The center and spread of the binomial distribution for a count X are defined by the mean m and standard deviation s:
a)
b)
Effect of changing p when n is fixed.
a) n = 10, p = 0.25
b) n = 10, p = 0.5
c) n = 10, p = 0.75
For small samples, binomial distributions are skewed when p is different from 0.5.
c)
Color blindness proportions
The frequency of color blindness (dyschromatopsia) in the Caucasian American male population is estimated to be about 8%. We take a random sample of size 25 from this population.
The population is definitely larger than 20 times the sample size, thus we can approximate the sampling distribution by B(n = 25, p = 0.08).
• What is the probability that five individuals or fewer in the sample are color blind?
P(x≤ 5)
• What is the probability that more than five will be color blind?
P(x> 5) =
• What is the probability that exactly five will be color blind?
P(x≤ 5) =
B proportions(n = 25, p = 0.08)
Probability distribution and histogram for the number of color blind individuals among 25 Caucasian males.
What are the mean and standard deviation of the count of color blind individuals in the SRS of 25 Caucasian American males?
µ = np = 25*0.08 = 2
σ = √np(1 p) = √(25*0.08*0.92) = 1.36
What if we take an SRS of size 10? Of size 75?
µ = 10*0.08 = 0.8 µ = 75*0.08 = 6
σ = √(10*0.08*0.92) = 0.86 σ = √(75*0.08*0.92) = 3.35
p = .08
n = 10
p = .08
n = 75
Suppose that for a randomly selected high school student who has taken a college entrance exam, the probability of scoring above a 650 is 0.30. A random sample of n = 9 students was selected.
What is the probability that exactly two of the students scored over 650 points?
What are the mean and standard deviation of the number of students in the sample who have scores above 650?
Sample proportions has taken a college entrance exam, the probability of scoring above a 650 is 0.30. A random sample of
The proportion of “successes” can be more informative than the count. In statistical sampling the sample proportion of successes, , is used to estimate the proportion p of successes in a population.
For any SRS of size n, the sample proportion of successes is:
• In an SRS of 50 students in an undergrad class, 10 are Hispanic:= (10)/(50) = 0.2 (proportion of Hispanics in sample)
• The 30 subjects in an SRS are asked to taste an unmarked brand of coffee and rate it “would buy” or “would not buy.” Eighteen subjects rated the coffee “would buy.” = (18)/(30) = 0.6 (proportion of “would buy”)
If the sample size is much smaller than the size of a population with proportion p of successes, then the mean and standard deviation of are:
• Because the mean is p, we say that the sample proportion in an SRS is an unbiased estimator of the population proportion p.
• The variability decreases as the sample size increases. So larger samples usually give closer estimates of the population proportion p.
Normal approximation population with proportion
If n is large, and p is not too close to 0 or 1, the binomial distribution can be approximated by the normal distribution N(m = np,s2 = np(1 p)). Practically, the Normal approximation can be used when both np≥10 and n(1 p) ≥10.
If X is the count of successes in the sample and = X/n, the sample proportion of successes, their sampling distributions for large n, are:
• X approximately N(µ = np,σ2 = np(1 − p))
• is approximately N (µ = p, σ2 = p(1 − p)/n)
Normal approximation to the binomial population with proportion (answer)
Why would we want to use the normal approximation to the binomial instead of just using the binomial distribution?
• The normal distribution is more accurate.
• The normal distribution uses the mean and the standard deviation.
• The normal distribution works all the time, so use it for everything.
• The binomial distribution is awkward and takes too long if you have to sum up many probabilities. The binomial distribution looks like the normal distribution if n is large.
• The binomial distribution is awkward and takes too long if you have to multiply many probabilities. The binomial distribution looks like the normal distribution if n is large.
Sampling distribution of the sample proportion population with proportion
The sampling distribution of is never exactly normal. But as the sample size increases, the sampling distribution of becomes approximately normal.
The normal approximation is most accurate for any fixed n when p is close to 0.5, and least accurate when p is near 0 or near 1.
Color blindness population with proportion
The frequency of color blindness (dyschromatopsia) in the Caucasian American male population is about 8%.
We take a random sample of size 125 from this population. What is the probability that six individuals or fewer in the sample are color blind?
• Sampling distribution of the count X: B(n = 125, p = 0.08) np = 10P(X≤ 6) =
• Normal approximation for the count X: N(np = 10, √np(1 p) = 3.033)P(X≤ 6) = Or
z = (xµ)/σ = (6 10)/3.033 = 1.32 P(X≤ 6) = 0.0934 from Table A
The normal approximation is reasonable, though not perfect. Here p = 0.08 is not close to 0.5 when the normal approximation is at its best.
A sample size of 125 is the smallest sample size that can allow use of the normal approximation (np = 10 and n(1 p) = 115).
Sampling distributions for the color blindness example. population with proportion
n = 50
The larger the sample size, the better the normal approximation suits the binomial distribution.
Avoid sample sizes too small for np or n(1 p) to reach at least 10 (e.g., n = 50).
n =1000
n = 125
Normal approximation: continuity correction population with proportion
The normal distribution is a better approximation of the binomial distribution, if we perform a continuity correction where x’ = x + 0.5 is substituted for x, and P(X ≤ x) is replaced by P(X ≤ x + 0.5).
Why? A binomial random variable is a discrete variable that can only take whole numerical values. In contrast, a normal random variable is a continuous variable that can take any numerical value.
P(X ≤ 10) for a binomial variable is P(X ≤ 10.5) using a normal approximation.
P(X < 10) for a binomial variable excludes the outcome X = 10, so we exclude the entire interval from 9.5 to 10.5 and calculate P(X ≤ 9.5) when using a normal approximation.
Color blindness population with proportion
The frequency of color blindness (dyschromatopsia) in the Caucasian American male population is about 8%. We take a random sample of size 125 from this population.
• Sampling distribution of the count X: B(n = 125, p = 0.08) np = 10P(X≤ 6.5) = P(X≤ 6) = 0.1198 P(X< 6) = P(X≤ 5) = 0.0595
• Normal approximation for the count X: N(np =10, √np(1 p) = 3.033)P(X≤ 6.5) = = 0.1243P(X≤ 6) = = 0.0936 ≠ P(X≤ 6.5) P(X< 6) = P(X≤ 6) = 0.0936
The continuity correction provides a more accurate estimate:
Binomial P(X≤ 6) = 0.1198 this is the exact probability
Normal P(X≤ 6) = 0.0936, while P(X≤ 6.5) = 0.1243 estimates
Chapter 5.2 population with proportion Sampling distribution of a sample mean
Objectives
• The mean and standard deviation of
• For normally distributed populations
• The central limit theorem
• Weibull distributions
Reminder: What is a sampling distribution? population with proportion
The sampling distribution of a statistic is the distribution of all possible values taken by the statistic when all possible samples of a fixed size n are taken from the population. It is a theoretical idea — we do not actually build it.
The sampling distribution of a statistic is the probability distribution of that statistic.
Sampling distribution of the sample mean population with proportion
We take many random samples of a given size n from a population with mean mand standard deviations.
Some sample means will be above the population mean m and some will be below, making up the sampling distribution.
Sampling distribution of “x bar”
Histogram of some sample averages
For any population with mean population with proportion m and standard deviation s:
• The mean, or center of the sampling distribution of , is equal to the population mean m :mx = m.
• The standard deviation of the sampling distribution is s/√n, where n is the sample size :sx= s/√n.
Sampling distribution of x bar
s/√n
m
• Mean of a sampling distribution of population with proportion
There is no tendency for a sample mean to fall systematically above or below m, even if the distribution of the raw data is skewed. Thus, the mean of the sampling distribution is an unbiasedestimate of the population mean m— it will be “correct on average” in many samples.
• Standard deviation of a sampling distribution of
The standard deviation of the sampling distribution measures how much the sample statistic varies from sample to sample. It is smaller than the standard deviation of the population by a factor of √n. Averages are less variable than individual observations.
For normally distributed populations population with proportion
When a variable in a population is normally distributed, the sampling distribution of for all possible samples of size n is also normally distributed.
Sampling distribution
If the population is N(m, s) then the sample means distribution is N(m, s/√n).
Population
IQ scores: population vs. sample population with proportion
In a large population of adults, the mean IQ is 112 with standard deviation 20. Suppose 200 adults are randomly selected for a market research campaign.
• The distribution of the sample mean IQ is:
A) Exactly normal, mean 112, standard deviation 20
B) Approximately normal, mean 112, standard deviation 20
C) Approximately normal, mean 112 , standard deviation 1.414
D) Approximately normal, mean 112, standard deviation 0.1
C) Approximately normal, mean 112 , standard deviation 1.414
Population distribution : N(m = 112; s= 20)
Sampling distribution for n = 200 is N(m = 112; s /√n = 1.414)
Practical note population with proportion
• Large samples are not always attainable.
• Sometimes the cost, difficulty, or preciousness of what is studied drastically limits any possible sample size.
• Blood samples/biopsies: No more than a handful of repetitions are acceptable. Oftentimes, we even make do with just one.
• Opinion polls have a limited sample size due to time and cost of operation. During election times, though, sample sizes are increased for better accuracy.
• Not all variables are normally distributed.
• Income, for example, is typically strongly skewed.
• Is still a good estimator of m then?
The central limit theorem population with proportion
Central Limit Theorem: When randomly sampling from any population with mean mand standard deviation s, when n is large enough, the sampling distribution of is approximately normal: ~N(m, s/√n).
Population with strongly skewed distribution
Sampling distribution of for n = 2 observations
Sampling distribution of for n = 10 observations
Sampling distribution of
for n = 25
observations
Income distribution population with proportion
Let’s consider the very large database of individual incomes from the Bureau of Labor Statistics as our population. It is strongly right skewed.
• We take 1000 SRSs of 100 incomes, calculate the sample mean for each, and make a histogram of these 1000 means.
• We also take 1000 SRSs of 25 incomes, calculate the sample mean for each, and make a histogram of these 1000 means.
Which histogram corresponds to samples of size 100? 25?
\$\$\$
How large a sample size? population with proportion
It depends on the population distribution. More observations are required if the population distribution is far from normal.
• A sample size of 25 is generally enough to obtain a normal sampling distribution from a strong skewness or even mild outliers.
• A sample size of 40 will typically be good enough to overcome extreme skewness and outliers.
In many cases, n = 25 isn’t a huge sample. Thus, even for strange population distributions we can assume a normal sampling distribution of the mean and work with it to solve problems.
Sampling distributions population with proportion
Atlantic acorn sizes (in cm3)
— sample of 28 acorns:
• Describe the histogram. What do you assume for the population distribution?
• What would be the shape of the sampling distribution of the mean:
• For samples of size 5?
• For samples of size 15?
• For samples of size 50? | 4,559 | 19,073 | {"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-2017-47 | longest | en | 0.8272 |
https://www.sawaal.com/average-questions-and-answers/a-batsman-makes-a-score-of-81-runs-in-the-16th-nbspmatch-and-thus-increases-his-average-runs-per-mat_29311 | 1,624,373,497,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488517820.68/warc/CC-MAIN-20210622124548-20210622154548-00466.warc.gz | 864,838,450 | 14,341 | 5
Q:
# A batsman makes a score of 81 runs in the 16th match and thus increases his average runs per match by 3. What is his average after the 16th match?
A) 35 B) 34 C) 33 D) 36
Explanation:
Q:
The average of the marks of 30 boys is 88, and when the top two scores were excluded, the average marks reduced to 87.5. If the top two scores differ by 2, then the highest mark is:
A) 92 B) 96 C) 94 D) 90
Explanation:
2 295
Q:
24 students collected money for donation. The average contribution was ₹50. Later on, their teacher also contributed some money. Now the average contribution is ₹56. The teacher’s contribution is:
A) ₹56 B) ₹200 C) ₹106 D) ₹194
Explanation:
0 1065
Q:
Average of 8 numbers is 40. The average of first three numbers is 46 and the average of next two numbers is 50. If the sixth number is 6 and 10 less than seventh and eighth number respectively, then what is the value of eighth number?
A) 28 B) 22 C) 30 D) 32
Explanation:
2 237
Q:
1800 chocolates were distributed among the students of a class. Each student got twice as many chocolates as the number of students in the class. Calculate the number of students in the class.
A) 30 B) 40 C) 60 D) 90
Explanation:
2 204
Q:
The average of 60 student’s results is 38. If the average of the first 22 students is 36, and that of the last 32 students is 32, then the average result of the remaining students is:
A) 81 B) 77 C) 65 D) 52
Explanation:
4 236
Q:
The average score in Mathematics of 90 students of sections A and B together is 49. The number of students in A was 25% more than that of B, and the average score of the students in B was 20% higher than that of the students in A. What is the average score of the students in A?
A) 44.5 B) 45 C) 44 D) 45.5
Explanation:
1 259
Q:
Out of the five subjects, the average marks of Suman in first 3 Subjects was 81 and the average marks in the last 3 subjects was 74. If her marks in the third subject was 65, then what was her average marks in all the subjects?
A) 75 B) 76 C) 79 D) 80
Explanation:
0 245
Q:
The average of 26 numbers is 37. One more number is added and the new average becomes 41. What is that new number?
A) 140 B) 145 C) 146 D) 147 | 646 | 2,210 | {"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-2021-25 | latest | en | 0.936326 |
https://www.questia.com/read/1G1-177863940/from-an-operational-to-a-relational-conception-of | 1,440,802,691,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440644064019.39/warc/CC-MAIN-20150827025424-00202-ip-10-171-96-226.ec2.internal.warc.gz | 914,399,399 | 18,761 | # From an Operational to a Relational Conception of the Equal Sign: Third Graders' Developing Algebraic Thinking
By Molina, Marta; Ambrose, Rebecca | Focus on Learning Problems in Mathematics, Winter 2008 | Go to article overview
# From an Operational to a Relational Conception of the Equal Sign: Third Graders' Developing Algebraic Thinking
Molina, Marta, Ambrose, Rebecca, Focus on Learning Problems in Mathematics
As algebra has taken a more prominent role in mathematics education, many are advocating introducing children to it in primary school (e.g., Carraher, Schliemann, Brizuela, & Earnest, 2006; Kaput, 2000; National Council of Teacher of Mathematics, 2000), and open number sentences can be a good context to address this goal. Students are frequently introduced to equations by considering number sentences with an unknown where a figure or a line is used instead of a variable, as in _ + 4 = 5 + 7 (Radford, 2000). Discussions about these equations and the properties they illustrate can help students to learn arithmetic with understanding and to develop a solid base for the later formal study of algebra by helping them to become aware of the structure underneath arithmetic (Carpenter, Franke, & Levi, 2003; Kieran, 1992; Resnick, 1992). Unfortunately students, conceptions about the equal sign interfere with their ability to successfully solve and analyze equations (Carpenter et al., 2003; Molina & Ambrose, 2006). Our aim in this paper is to describe a teaching experiment in which children developed an understanding of the equal sign and began to analyze expressions in sophisticated ways.
Difficulties in Solving Number Sentences
When elementary students encounter the equal sign in arithmetic sentences they tend to perceive it as an operational symbol, that is a "do something" signal, and tend to react negatively when number sentences challenge their conceptions about this symbol as in sentences of the form c = a + b. Behr, Erlwanger, and Nichols (1980) observed that most six-year old students thought that such sentences were "backwards" and tended to change them to c + a = b or a + b = c. Children also did not accept non-action sentences, that is sentences with no operational symbol (e.g., 3 = 3) or operational symbols on both sides (e.g., 3 + 5 = 7 + 1), and often changed them to action sentences, that is sentences with all the operations in one side of the equation. For example they changed 3 + 2 = 2 + 3 to 3 + 2 + 2 + 3 = 10 and 3 = 3 to 3 + 0 = 3 or 3 - 3 = 0.
In studies about elementary students' answers to open sentences of the forms a = a, c + a = b, a + b = c and a + b = c + d (Falkner, Levi, & Carpenter 1999; Freiman & Lee, 2004; Kieran, 1981), students provided a variety of responses: repeating one of the numbers in the sentence, the sum or differences of two numbers of the sentence, the sum of all the numbers in the sentence, and the correct answer. In sentences of the form a + b = _ + d, students tended to answer the sum of a + b or to write it as a string of operations.
These studies illustrate that children tend to read open number sentences from left to right and perform the computation as they go along. When children face unfamiliar number sentences, that is sentences different from the conventional form a [+ or -] b = c, they have trouble interpreting the equal sign as a symbol representing equivalence. To successfully solve a problem such as 8+ 4 = _ + 5, students have to read the whole sentence before computing and need to recognize that both sides of the equation need to have the same sum. The equal sign needs to be interpreted as a relational symbol expressing equivalence.
Considering students' difficulties with the equal sign, their understanding of the concept of equality can be questioned; however, Schliemann, Carraher, Brizuela, and Jones (1988) and Falkner et al. (1999) have observed that students show a correct understanding of equality when considering concrete physical contexts or verbal word problems. The children's misinterpretations are linked to the use of the equal symbol rather than an understanding of the concept of equality (Carpenter et al., 2003; Falkner et al., 1999, Schifter, Monk, Russell, & Bastable, in press). Most studies about the equal sign (Behr et al., 1980; Falkner et al., 1999; Saenz-Judlow & Walgamuth, 1998) have claimed that traditional curriculum does not promote a relational understanding of this symbol, mainly because of the repeated consideration of equations of the form a [+ or -] b = c throughout students' arithmetic learning. …
If you are trying to select text to create highlights or citations, remember that you must now click or tap on the first word, and then click or tap on the last word.
One moment ...
Default project is now your active project.
Project items
Highlights (0)
Some of your highlights are legacy items.
Citations (0)
Some of your citations are legacy items.
Notes (0)
Bookmarks (0)
Project items include:
• Saved book/article
• Highlights
• Quotes/citations
• Notes
• Bookmarks
Notes
#### Cited article
Style
Citations are available only to our active members.
(Einhorn, 1992, p. 25)
(Einhorn 25)
1. Lois J. Einhorn, Abraham Lincoln, the Orator: Penetrating the Lincoln Legend (Westport, CT: Greenwood Press, 1992), 25, http://www.questia.com/read/27419298.
#### Cited article
From an Operational to a Relational Conception of the Equal Sign: Third Graders' Developing Algebraic Thinking
Settings
#### Settings
Typeface
Text size Reset View mode
Search within
Look up
#### Look up a word
• Dictionary
• Thesaurus
Please submit a word or phrase above.
Why can't I print more than one page at a time?
Help
Full screen
### How to highlight and cite specific passages
1. Click or tap the first word you want to select.
2. Click or tap the last word you want to select, and you’ll see everything in between get selected.
3. You’ll then get a menu of options like creating a highlight or a citation from that passage of text.
## Cited passage
Style
Citations are available only to our active members.
"Portraying himself as an honest, ordinary person helped Lincoln identify with his audiences." (Einhorn, 1992, p. 25).
"Portraying himself as an honest, ordinary person helped Lincoln identify with his audiences." (Einhorn 25)
"Portraying himself as an honest, ordinary person helped Lincoln identify with his audiences."1
1. Lois J. Einhorn, Abraham Lincoln, the Orator: Penetrating the Lincoln Legend (Westport, CT: Greenwood Press, 1992), 25, http://www.questia.com/read/27419298.
## Thanks for trying Questia!
Please continue trying out our research tools, but please note, full functionality is available only to our active members.
Your work will be lost once you leave this Web page. | 1,600 | 6,788 | {"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} | 3.609375 | 4 | CC-MAIN-2015-35 | latest | en | 0.924446 |
https://www.enotes.com/homework-help/what-x-log-4-x-log-x-6-464204 | 1,669,458,952,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446706285.92/warc/CC-MAIN-20221126080725-20221126110725-00468.warc.gz | 798,776,811 | 18,117 | # What is x if log(4)x = log(x) 6
To elaborate on the earlier response:
`(log_10 x)^2 = log_10 4*log_10 6` has two solutions.
`log_10 x = +-sqrt(log_10 4*log_10 6)`
The negative value gives another solution of the equation.
`log_10 x = -sqrt(log_10 4*log_10 6)`
=> `x = 10^(-sqrt(log_10 4*log_10 6))`
=> `x ~~ 0.20679`
Approved by eNotes Editorial Team
The equation `log_4 x = log_x 6` has to be solved for x.
Use the formula `log_a b = (log_x b)/(log_x a)` which can be used to change the base of any logarithm.
`log_4 x = log_x 6`
Change both the logarithm on both the sides to a common base, for example 10
=> `(log_10 x)/(log_10 4) = (log_10 6)/(log_10 x)`
=> `(log_10 x)^2 = log_10 4*log_10 6`
=> `log_10 x = sqrt(log_10 4*log_10 6)`
=> `x = 10^sqrt(log_10 4*log_10 6)`
=> `x ~~ 4.835`
The solution of the equation is `x = 10^sqrt(log_10 4*log_10 6) ~~ 4.835`
Approved by eNotes Editorial Team | 347 | 917 | {"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} | 3.9375 | 4 | CC-MAIN-2022-49 | latest | en | 0.820427 |
https://en.scienceforming.com/10907699-how-to-find-the-coordinates-of-a-point-in-a-circle | 1,695,381,672,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506399.24/warc/CC-MAIN-20230922102329-20230922132329-00325.warc.gz | 255,942,683 | 12,016 | # How To Find The Coordinates Of A Point In A Circle
## Video: How To Find The Coordinates Of A Point In A Circle
A circle is understood as a figure that consists of a plurality of points on a plane equidistant from its center. The distance from the center to the points of the circle is called the radius.
## Necessary
• - a simple pencil;
• - notebook;
• - protractor;
• - compasses;
• - pen.
## Instructions
### Step 1
Before finding the coordinates of this or that point of the circle, draw the given circle. While constructing it, you may come across many new concepts. So a chord is a segment that connects two points of a circle, and the chord passing through the center of the circle is the maximum (it is called the diameter). In addition, a tangent can be drawn to the circle, which is a straight line perpendicular to the radius of the circle, which is drawn to the point of intersection of the tangent and the geometric figure in question.
### Step 2
If, according to the condition of the task, it is known that the circle you constructed is intersected by another circle (it is smaller in size), depict this graphically: the figure should show that these two circles intersect, that is, they have a number of common points. Mark the center of the first circle with point 1 (its coordinates (X1, Y1)), and its radius - R1. Thus, the center of the second circle should be designated by point 2 (the coordinates of this point (X2, Y2)), and the radius - R2. At the intersection points of the shapes, put points 3 (X3, Y3) and 4 (X4, Y4). The center point of intersection must be designated 0: its coordinates (X, Y).
### Step 3
In order to find the coordinates of the intersection of these circles, and therefore the point belonging to both the first and the second of them, you will have to solve the quadratic equation. Consider the two formed triangles (? 103 and? 203) and analyze their performance. The hypotenuses of these triangles are R1 and R2, respectively. Knowing the value of the hypotenuses, find the segment D connecting the center of the first circle with the center of the second. The chosen calculation method directly depends on how the triangles you are analyzing turned out to be. If they are rectangular, then the square of the length of the hypotenuse of each of them will be equal to the sum of the squares of the legs of this triangle. In addition, the length of the leg can be found by the formula: a = ccos?, Where c is the length of the hypotenuse, and cos? Is the cosine of the included angle. Having found the value of the legs, determine the coordinates of the point of interest. | 599 | 2,631 | {"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} | 4.6875 | 5 | CC-MAIN-2023-40 | latest | en | 0.927251 |
https://mathhelpboards.com/threads/surface-integral.6302/ | 1,642,537,645,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300997.67/warc/CC-MAIN-20220118182855-20220118212855-00154.warc.gz | 462,534,988 | 18,847 | Surface Integral
dwsmith
Well-known member
I haven't done a surface integral in a while so I am asking to get this checked.
$$\mathbf{F} = \langle x, y, z\rangle$$ and the surface is $$z = xy + 1$$ where $$0\leq x\leq 1$$ and $$0\leq y\leq 1$$.
$$\hat{\mathbf{n}} = \nabla f/ \lvert\nabla f\rvert = \frac{1}{\sqrt{3}}\langle 1, 1, 1\rangle$$
$$dS = \frac{\lvert\nabla f\rvert dxdy}{\frac{\partial f}{\partial z}} = \sqrt{3}dxdy$$
$$\mathbf{F}\cdot\hat{\mathbf{n}} = \frac{1}{\sqrt{3}}(x+y+z) = \frac{1}{\sqrt{3}}(x+y+xy + 1)$$
$\int_0^1\int_0^1(x + y + xy + 1)dxdy = \frac{9}{4}$
So is this the correct integral I should I obtain or is there a mistake some where?
Last edited:
Last edited:
dwsmith
Well-known member
We can calculate $$\unit{n}$$ by
$$\unit{n} = \frac{\mathbf{U}_x\times\mathbf{U}_y} {\lvert\mathbf{U}_x\times\mathbf{U}_y\rvert}$$.
\begin{align*}
\mathbf{U}_x &= \unit{i} + y\unit{k}\\
\mathbf{U}_y &= \unit{j} + x\unit{k}\\
\mathbf{U}_x\times\mathbf{U}_y &= -y\unit{i} - x\unit{j} + \unit{k}\\
\lvert\mathbf{U}_x\times\mathbf{U}_y\rvert &= \sqrt{1 + x^2 + y^2}
\end{align*}
So $$\unit{n} = \frac{-y\unit{i} - x\unit{j} + \unit{k}}{\sqrt{1 + x^2 + y^2}}$$.
\begin{align*}
\int_S\mathbf{F}\cdot\unit{n}dA &= \int_0^1\int_0^1
(x\unit{i} + y\unit{j} + (xy + 1)\unit{k})\cdot\unit{n}
\sqrt{1 + x^2 + y^2}dxdy\\
&= \int_0^1\int_0^1 (x\unit{i} + y\unit{j} + (xy + 1)\unit{k})\cdot
(-y\unit{i} - x\unit{j} + \unit{k})dxdy\\
&= \int_0^1\int_0^1 (-2xy + xy + 1)dxdy\\
&= \int_0^1\int_0^1 (1 - xy)dxdy\\
&= \int_0^1\left[1 - \frac{1}{2}y\right]dy\\
&= \frac{3}{4}
\end{align*}
So now I want to show that using the divergence theorem leads to the same answer. However, I have been un-successful.
Note \unit = \hat{\mathbf{#1}}.
Klaas van Aarsen
MHB Seeker
Staff member
I haven't done a surface integral in a while so I am asking to get this checked.
$$\mathbf{F} = \langle x, y, z\rangle$$ and the surface is $$z = xy + 1$$ where $$0\leq x\leq 1$$ and $$0\leq y\leq 1$$.
$$\hat{\mathbf{n}} = \nabla f/ \lvert\nabla f\rvert = \frac{1}{\sqrt{3}}\langle 1, 1, 1\rangle$$
Are F and f supposed to be the same function?
Anyway, suppose we define g(x,y,z)=z-xy-1, then your surface is given by g(x,y,z)=0.
Then:
$$\mathbf{\hat n} = \frac{\nabla g}{|\nabla g|} = \frac {(-y,-x,1)}{\sqrt{x^2+y^2+1}}$$
which is different from what you have.
Based on this post: http://mathhelpboards.com/geometry-11/volume-triangle-type-shape-square-bottom-6305.html my solution must be incorrect since
$\int_S(\mathbf{F}\cdot\hat{\mathbf{n}})dS = \int(\nabla\cdot\mathbf{F})dV$
and $$\nabla\cdot\mathbf{F} = 3$$
so
$\int 3dV = 3V = 3\frac{5}{4} = \frac{15}{4}$
(Is this integral correct?)
So what went wrong with the surface integral calculation?
Don't forget that you're supposed to integrate over the entire closed surface. In particular that includes the sides where x=1 respectively y=1.
dwsmith
Well-known member
Are F and f supposed to be the same function?
Anyway, suppose we define g(x,y,z)=z-xy-1, then your surface is given by g(x,y,z)=0.
Then:
$$\mathbf{\hat n} = \frac{\nabla g}{|\nabla g|} = \frac {(-y,-x,1)}{\sqrt{x^2+y^2+1}}$$
which is different from what you have.
You neglected to look at post 3.
Klaas van Aarsen
MHB Seeker
Staff member
You neglected to look at post 3.
Heh. Yes, I can see there is a $\sqrt{x^2+y^2+1}$ in there, but no mention that you discovered the mistake in the first post.
And to be honest, I stopped reading it due to all the distracting red \units in there.
Are you satisfied then with what you found out?
It appears you're still not integrating the sides where x=1 respectively y=1.
dwsmith
Well-known member
Heh. Yes, I can see there is a $\sqrt{x^2+y^2+1}$ in there, but no mention that you discovered the mistake in the first post.
And to be honest, I stopped reading it due to all the distracting red \units in there.
Are you satisfied then with what you found out?
It appears you're still not integrating the sides where x=1 respectively y=1.
\unit is my notation for \hat{\mathbf{#1}}. I refuse to constantly type that.
So are you saying that 3/4 in post 3 is incorrect? What needs to be changed about that integral?
MarkFL
Staff member
\unit is my notation for \hat{\mathbf{#1}}. I refuse to constantly type that...
It seems to me it would be just as easy, if not easier, to copy/paste that command after you type it once, as to type "\unit" each time.
Letting things be more difficult to read by others because you perceive it to be easier for yourself is not a path I would take personally.
dwsmith
Well-known member
It seems to me it would be just as easy, if not easier, to copy/paste that command after you type it once, as to type "\unit" each time.
Letting things be more difficult to read by others because you perceive it to be easier for yourself is not a path I would take personally.
If you use LaTeX enough, it shouldn't even phase you.
Klaas van Aarsen
MHB Seeker
Staff member
\unit is my notation for \hat{\mathbf{#1}}. I refuse to constantly type that.
So are you saying that 3/4 in post 3 is incorrect? What needs to be changed about that integral?
Oh man, be a little bit creative!
In particular you only need to type it once per post.
Just put in "\newcommand{unit}[1]{\mathbf{\hat #1}} \unit n" once, like this: $\newcommand{unit}[1]{\mathbf{\hat #1}} \unit n$, and the rest of the post, even quoted, transforms.
If you want help, it helps if you make it attractive for people to help you.
We can calculate $$\unit{n}$$ by
$$\unit{n} = \frac{\mathbf{U}_x\times\mathbf{U}_y} {\lvert\mathbf{U}_x\times\mathbf{U}_y\rvert}$$.
\begin{align*}
\mathbf{U}_x &= \unit{i} + y\unit{k}\\
\mathbf{U}_y &= \unit{j} + x\unit{k}\\
\mathbf{U}_x\times\mathbf{U}_y &= -y\unit{i} - x\unit{j} + \unit{k}\\
\lvert\mathbf{U}_x\times\mathbf{U}_y\rvert &= \sqrt{1 + x^2 + y^2}
\end{align*}
So $$\unit{n} = \frac{-y\unit{i} - x\unit{j} + \unit{k}}{\sqrt{1 + x^2 + y^2}}$$.
\begin{align*}
\int_S\mathbf{F}\cdot\unit{n}dA &= \int_0^1\int_0^1
(x\unit{i} + y\unit{j} + (xy + 1)\unit{k})\cdot\unit{n}
\sqrt{1 + x^2 + y^2}dxdy\\
&= \int_0^1\int_0^1 (x\unit{i} + y\unit{j} + (xy + 1)\unit{k})\cdot
(-y\unit{i} - x\unit{j} + \unit{k})dxdy\\
&= \int_0^1\int_0^1 (-2xy + xy + 1)dxdy\\
&= \int_0^1\int_0^1 (1 - xy)dxdy\\
&= \int_0^1\left[1 - \frac{1}{2}y\right]dy\\
&= \frac{3}{4}
\end{align*}
So now I want to show that using the divergence theorem leads to the same answer. However, I have been un-successful.
Note \unit = \hat{\mathbf{#1}}.
What appears to be missing is that you have a surface at x=1.
Its normal vector is (1,0,0). Together with the function F(1,y,z) = (1,y,z) this yields a dot product of 1, meaning you get a contribution that is as large as the surface at x=1.
The same holds for y=1.
You can ignore the surfaces at x=0, y=0, respectively z=0, since they have a dot product that is 0.
dwsmith
Well-known member
Oh man, be a little bit creative!
In particular you only need to type it once per post.
Just put in "\newcommand{unit}[1]{\mathbf{\hat #1}} \unit n" once, like this: $\newcommand{unit}[1]{\mathbf{\hat #1}} \unit n$, and the rest of the post, even quoted, transforms.
I wasn't aware mathjax would allow me to enter in newcommands and trust me LaTeX can get creative.
Klaas van Aarsen
MHB Seeker
Staff member
I wasn't aware mathjax would allow me to enter in newcommands and trust me LaTeX can get creative.
I have been aware of this feature since before mathjax.
It appears that vBulletin somehow keeps track of all $\LaTeX$ within 1 post.
Ackbach
Indicium Physicus
Staff member
$\newcommand{\unit}[1]{\mathbf{\hat{#1}}}$
Just re-typing dwsmith's post with the newcommand.
We can calculate $$\unit{n}$$ by
$$\unit{n} = \frac{\mathbf{U}_x\times\mathbf{U}_y} {\lvert\mathbf{U}_x\times\mathbf{U}_y\rvert}$$.
\begin{align*}
\mathbf{U}_x &= \unit{i} + y\unit{k}\\
\mathbf{U}_y &= \unit{j} + x\unit{k}\\
\mathbf{U}_x\times\mathbf{U}_y &= -y\unit{i} - x\unit{j} + \unit{k}\\
\lvert\mathbf{U}_x\times\mathbf{U}_y\rvert &= \sqrt{1 + x^2 + y^2}
\end{align*}
So $$\unit{n} = \frac{-y\unit{i} - x\unit{j} + \unit{k}}{\sqrt{1 + x^2 + y^2}}$$.
\begin{align*}
\int_S\mathbf{F}\cdot\unit{n}dA &= \int_0^1\int_0^1
(x\unit{i} + y\unit{j} + (xy + 1)\unit{k})\cdot\unit{n}
\sqrt{1 + x^2 + y^2}dxdy\\
&= \int_0^1\int_0^1 (x\unit{i} + y\unit{j} + (xy + 1)\unit{k})\cdot
(-y\unit{i} - x\unit{j} + \unit{k})dxdy\\
&= \int_0^1\int_0^1 (-2xy + xy + 1)dxdy\\
&= \int_0^1\int_0^1 (1 - xy)dxdy\\
&= \int_0^1\left[1 - \frac{1}{2}y\right]dy\\
&= \frac{3}{4}
\end{align*}
So now I want to show that using the divergence theorem leads to the same answer. However, I have been un-successful. | 3,207 | 8,661 | {"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": 2, "mathjax_display_tex": 2, "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} | 3.6875 | 4 | CC-MAIN-2022-05 | latest | en | 0.572821 |
https://www.sololearn.com/Discuss/2562015/python-problem-another-one/ | 1,606,393,969,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141188146.22/warc/CC-MAIN-20201126113736-20201126143736-00315.warc.gz | 861,448,057 | 8,776 | +2
# Python Problem (Another one)
Bro literally I am doing great and progressing in python. I haven't bought the pro version yet but there is also some free challenges. So like the other challenges there is a challenge about while loop called "While loop changed". And I literally cant solve it. Giving the codes description down below. You need to change the code to calculate and output the sum of all digits of the input number. Sample Input 643 Sample Output 13 Explanation The sum of the digits of 643 is 6+4+3 = 13. Thats how the question looks like. Man give me the solution as well as the explanation. Thank you for helping me.
10/26/2020 12:39:15 AM
Umar Haque
+2
Dont worry its quite easy to understand :- 1) First break down the problem . In this case , you have to add the digits of a number which is given as an input So :- Input--->Add the digits of the input--->print the output The input and print part isnt really hard ,is it ? Lets see the adding digits part Approach - iterate through the number by converting to a string. ITERATE means seeking each element/portion one by one . Converting to a string , why is it even necessary ? It is because only some data types in python ARE ITERABLE . One such type is a string So num = str(input()) #converting to a string sum = 0 #originally the sum is 0. for i in num : # This is the iteration , assuming that you know about for loops . If you dont , I suggest you to learn about them. sum = sum + i #Now , after the iterations its time to give the output ! print(sum) If you still dont understand what I meant , dont worry all of it will become obvious once you lean about for loops.
+4
sum = 0 for digit in str(643): sum += int(digit) print(sum) Try this, should be self-explanatory.
+1
Ok simply change them into list using list function then iterate it and add them a = list("12345") print(eval("+".join(a))) use this simple logic
+1
print(sum(int(ichar for char in input())) Input a number. Using input() creates a string. Iterate over the string, converting each character to an integer with int(). Use sum() to add them together. https://code.sololearn.com/cOWCAxOR6Nqa
+1
It is really easy.. but i have to use longer syntax than for in loop.. So i'll help you with the while loop This is my code: #in while loop: inp = input() sum = 0 d = 0 while d < len(inp): sum += int(inp[d]) d += 1 print(sum) #in for in loop: inp = input() sum = 0 for d in inp: sum += int(d) print(sum) And that's from me.. Stay confidence and keep typing! :>
0
Max Harder Can you plz Explain???
0
In order to get the digits, convert the integer to a string and then sum up the characters as integers with the help of a for-loop.
0
0
0
Using while loop: (what we all are taught) num = 12345 sum = 0 while num: sum += num % 10 num //= 10 print ( sum )
0
Try: num=643 digits=str(num) sum=0 for i in range(len(digits)): sum=sum+int(digits[i]) print(sum)
-1 | 755 | 2,927 | {"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} | 3.515625 | 4 | CC-MAIN-2020-50 | latest | en | 0.869999 |
https://bodheeprep.com/cat-2019-quant-question-with-solution-22 | 1,670,297,192,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711069.79/warc/CC-MAIN-20221206024911-20221206054911-00616.warc.gz | 169,930,747 | 20,226 | Bodhee Prep-CAT Online Preparation
| Best Online CAT PreparationFor Enquiry CALL @ +91-95189-40261
# CAT 2019 Quant Question with Solution 22
Question:
A chemist mixes two liquids 1 and 2. One litre of liquid 1 weighs 1 kg and one litre of liquid 2 weighs 800 gm. If half litre of the mixture weighs 480 gm, then the percentage of liquid 1 in the mixture, in terms of volume, is
1. 85
2. 70
3. 75
4. 80
Weight of liquid 1 per litre = 1000 gm
Weight of liquid 2 per litre = 800 gm
Weight of mixture per litre = 2×480=960 gm
Applying alligation rule
$\frac{Quantity\ of\ liquid\ 1}{Quantity\ of\ liquid\ 2}=\frac{960-800}{1000-960}=\frac{4}{1}$
Therefore, the liquids are mixed in 4:1.
Hence, the percentage of liquid 1 =$\left( \frac{4}{4+1} \right)\times 100=80%$
Also Check: 841+ CAT Quant Questions with Solutions
### CAT 2023Classroom Course
We are starting classroom course for CAT 2023 in Gurugram from the month of December. | 297 | 944 | {"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} | 3.765625 | 4 | CC-MAIN-2022-49 | longest | en | 0.725071 |
http://smallbusiness.chron.com/figure-breakeven-point-given-contribution-margin-ratio-62335.html | 1,531,814,732,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589618.52/warc/CC-MAIN-20180717070721-20180717090721-00550.warc.gz | 336,840,136 | 11,114 | # How Do You Figure the Break-Even Point With the Given Contribution Margin Ratio?
by Bryan Keythman
When you run a small business, it’s important to always know your break-even point -- the amount of sales needed to pay for all of your costs in a period. Below break-even, you generate a loss; above it, you turn a profit. The contribution margin ratio reveals the percentage of sales that applies to your fixed costs after covering variable costs. When you know your contribution margin ratio, you can figure your break-even point in dollars and units with a couple of straightforward calculations.
## About the Contribution Margin Ratio
The contribution margin ratio, as a percentage, equals your contribution margin in dollars divided by sales, times 100. Your contribution margin in dollars equals sales minus total variable costs. The higher your contribution margin, the quicker you can potentially generate a profit because a greater portion of each sales dollar goes toward fixed costs. You can figure your contribution margin ratio using sales and expense information from your most recent year in business.
## Fixed and Variable Costs
Fixed costs are those that remain the same regardless of your sales volume. Examples include rent, fixed salaries and wages, property taxes and utilities. In general, the lower your fixed costs, the lower your break-even point. Variable costs, however, increase when sales rise and decrease when sales fall. Examples of variable costs are commissions and wages tied to sales volume, costs to buy products and materials used to make products.
## Contribution Margin Ratio Calculation Example
Assume your small business generated \$800,000 in sales and had \$360,000 in total variable costs last year. Your contribution margin equals \$440,000, or \$800,000 minus \$360,000. Your contribution margin ratio equals 55 percent, or \$440,000 divided by \$800,000, times 100. This means that 55 cents of every dollar of sales revenue goes toward paying your fixed expenses.
## Dollar Break-Even Point
Your break-even point in dollars equals your total fixed costs for a particular period divided by your contribution margin ratio. Using the previous example, assume your total annual fixed expenses are \$350,000. Your break-even point is \$636,364, or \$350,000 divided by 55 percent. This means you must generate \$636,364 in annual sales to pay for all of your costs. When your sales reach this level, you have neither a profit nor a loss. If you sell more than that amount, you earn a profit.
## Unit Break-Even Point
The break-even point in units represents the number of units you must sell to break even. The formula equals your break-even point in dollars divided by the price for which you sell each unit. A unit might be one product, a billable hour of service or some similar measurement. In the previous example, assume you sell your products for \$50 a piece. Your break-even point in units is approximately 12,728, or \$636,364 divided by \$50. You must sell 12,728 units during the year to cover all of your expenses.
#### Photo Credits
• Jupiterimages/Comstock/Getty Images | 643 | 3,142 | {"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} | 3.9375 | 4 | CC-MAIN-2018-30 | latest | en | 0.881253 |
https://askfilo.com/math-question-answers/a-piggy-bank-contains-hundred-50-p-coins-fifty-re-1-coins-twenty-rs-2-coins-and | 1,719,002,977,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862157.88/warc/CC-MAIN-20240621191840-20240621221840-00005.warc.gz | 86,774,835 | 25,527 | World's only instant tutoring platform
Question
Easy
Solving time: 2 mins
# A piggy bank contains hundred 50 p coins, fifty Re. 1 coins, twenty Rs. 2 coins and ten Rs. 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin will be a 50 p coin?
A
1/3
B
4/9
C
5/9
D
None of these
## Text solutionVerified
Total number of coins = 100 + 50 + 20 + 10 = 180
So, one coin out of 180 coins can fall out in 180 ways.
Total number of possible outcomes = 180
There are hundred 50 p coins, out of which one 50 p coin can fall out in Hence, P(a 50 p coin will fall) = 100/180 = 5/9.
106
Share
Report
## Video solutions (1)
Learn from their 1-to-1 discussion with Filo tutors.
16 mins
140
Share
Report
Found 2 tutors discussing this question
Discuss this question LIVE
5 mins ago
One destination to cover all your homework and assignment needs
Learn Practice Revision Succeed
Instant 1:1 help, 24x7
60, 000+ Expert tutors
Textbook solutions
Big idea maths, McGraw-Hill Education etc
Essay review
Get expert feedback on your essay
Schedule classes
High dosage tutoring from Dedicated 3 experts
Trusted by 4 million+ students
Stuck on the question or explanation?
Connect with our Mathematics tutors online and get step by step solution of this question.
231 students are taking LIVE classes
Question Text A piggy bank contains hundred 50 p coins, fifty Re. 1 coins, twenty Rs. 2 coins and ten Rs. 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin will be a 50 p coin? Updated On Jul 25, 2023 Topic Probability Subject Mathematics Class Class 9 Answer Type Text solution:1 Video solution: 1 Upvotes 246 Avg. Video Duration 16 min | 481 | 1,812 | {"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} | 3.6875 | 4 | CC-MAIN-2024-26 | latest | en | 0.860534 |
http://www.wyzant.com/resources/lessons/math/calculus/multivariable_vectors/vector_functions | 1,394,992,842,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678703273/warc/CC-MAIN-20140313024503-00053-ip-10-183-142-35.ec2.internal.warc.gz | 647,489,427 | 16,321 | Search 72,692 tutors
# Vector Functions
We will use the cross product and dot product of vectors to explore equations of lines and planes in 3 dimensional space. Vector functions have an input t and an output of a vector function of t.
## Position Vectors
A position vector is a vector whose initial point is fixed at the origin so that each point corresponds to P = <x,y>. Since a position vector cannot be translated, it is technically not a vector, so it instead should be considered as a means of using arithmetic of vectors with point in the plane. In this case, any two given points on a line will have it's own position vector.
## Finding an Equation of a line
We can define L(t) to be a vector valued function which maps the input of t to the output vectors L(t). In two dimensions, the equation would look similar to the equation of a line in slope intercept form.
We use t because it is the parameter, or index of each point on the line. We use t because a point is often labeled by the time at which an object is located at that point. In this case, b is a fixed point (position vector) on the line and v is the vector between two points on the line, or the constant slope vector for the line.
The physical interpretation of this equation is that if an object moves along the line at a constant speed equal to the magnitude of v, then t would be the time at which the object is located at point L(t) on the line.
Lets find the equation of the line through the points P1(2,4) and P2(5,1). Vector v is given by
Now we have our slope, so we can pick any of the points on the line to be the position vector. Our equation will be
Notice that the x coordinates are given by x = 3t+2 and the y coordinates are y = -3t+4. We can then solve for t which yields
Plugging in for y, we get the slope intercept form of the line
The reason why the vector equation of a line is used is because it generalizes to more than 2 dimensions, which the slope intercept form cannot. We can do the same thing for 3 dimensional lines.
For example, let's find the equation of the line that passes through the points P1(2,4,1) and P2(5,1,2) where P1 is at t = 0 and P2 is at t = 1.
The vector u is given by
The equation will then be
Plugging t = 0 and t = 1 we get our original points.
Plugging in other values of t will yield other position vectors on the line
## Finding an Equation of a Plane
Given any plane, there must be a normal vector n perpendicular to every vector v parallel to the plane. Generally, the plane through the point P1(x1,y1,z1) with normal vector n = <a,b,c> is the set of points P<x,y,z> such that the vectors v = P1P2 = <x-x1,y-y1,z-z1> are perpendicular to n. If we recall the dot product, n*v = 0, which is equivalent to
The equation of a plane with normal n = <a,b,c> through the point (x1,y1,z1) is defined as
Lets find the equation of the plane with normal <5,4,3> going through the point <0,1,2> on the plane.
Isolating z, we can put the equation into functional form.
To find the equation of the plane through three non colinear points, we have to form two vectors connecting one of the points to the other two. Let u = P1P2 and v = P1P3. These vectors are both parallel to the plane, so the cross product will yield a normal vector, that is, a vector that is perpendicular to both u and v, and therefore perpendicular to the plane.
Find the equation of the plane passing through the points P1(1,1,2), P2(2,1,3), and P3(3,1,-1)
First, we find vectors u and v.
Then, we use the formula for the cross product to find the normal vector.
The equation for the plane is then given by
Using the points in a different order may yield a different vector, however the functional form will always be the same.
We can also find the equation of a plane given a line and a point not on the line. All we need to do is plug in different values of t to get three different points, and then solve using the previous method.
Sign up for free to access more calculus resources like . WyzAnt Resources features blogs, videos, lessons, and more about calculus and over 250 other subjects. Stop struggling and start learning today with thousands of free resources! | 1,024 | 4,188 | {"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} | 4.65625 | 5 | CC-MAIN-2014-10 | longest | en | 0.937397 |
https://egvideos.com/video/florida/grade-1/math/mafs.1.nbt.2.3/comparing-numbers | 1,669,783,649,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710719.4/warc/CC-MAIN-20221130024541-20221130054541-00202.warc.gz | 245,303,615 | 9,988 | # Florida - Grade 1 - Math - Numbers and Operations in Base Ten - Comparing Numbers - MAFS.1.NBT.2.3
### Description
MAFS.1.NBT.2.3 Compare two two-digit numbers based on meanings of the tens and ones digits, recording the results of comparisons with the symbols >, =, and <.
• State - Florida
• Standard ID - MAFS.1.NBT.2.3
• Subjects - Math Common Core
### Keywords
• Math
• Numbers and Operations in Base Ten
## More Florida Topics
MAFS.1.OA.2.4 Understand subtraction as an unknown-addend problem. For example, subtract 10 – 8 by finding the number that makes 10 when added to 8.
MAFS.1.OA.3.5 Relate counting to addition and subtraction (e.g., by counting on 2 to add 2).
MAFS.1.OA.3.6 Add and subtract within 20, demonstrating fluency for addition and subtraction within 10. Use strategies such as counting on; making ten (e.g., 8 + 6 = 8 + 2 + 4 = 10 + 4 = 14); decomposing a number leading to a ten (e.g., 13 – 4 = 13 – 3 – 1 = 10 – 1 = 9); using the relationship between addition and subtraction (e.g., knowing that 8 + 4 = 12, one knows 12 – 8 = 4); and creating equivalent but easier or known sums (e.g., adding 6 + 7 by creating the known equivalent 6 + 6 + 1 = 12 + 1 = 13).
MAFS.1.OA.4.7 Understand the meaning of the equal sign, and determine if equations involving addition and subtraction are true or false. For example, which of the following equations are true and which are false? 6 = 6, 7 = 8 – 1, 5 + 2 = 2 + 5, 4 + 1 = 5 + 2.
MAFS.1.OA.4.8 Determine the unknown whole number in an addition or subtraction equation relating three whole numbers. For example, determine the unknown number that makes the equation true in each of the equations 8 + ? = 11, 5 = _ – 3, 6 + 6 = _.
MAFS.1.NBT.3.5 Given a two-digit number, mentally find 10 more or 10 less than the number, without having to count; explain the reasoning used.
MAFS.1.G.1.1 Distinguish between defining attributes (e.g., triangles are closed and three-sided) versus non-defining attributes (e.g., color, orientation, overall size) ; build and draw shapes to possess defining attributes.
MAFS.1.G.1.2 Compose two-dimensional shapes (rectangles, squares, trapezoids, triangles, half-circles, and quarter-circles) or three-dimensional shapes (cubes, right rectangular prisms, right circular cones, and right circular cylinders) to create a composite shape, and compose new shapes from the composite shape.1 | 687 | 2,397 | {"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} | 4.65625 | 5 | CC-MAIN-2022-49 | latest | en | 0.869848 |
https://www.gradesaver.com/textbooks/math/algebra/college-algebra-7th-edition/chapter-p-prerequisites-section-p-6-factoring-p-6-exercises-page-43/106 | 1,545,227,064,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376832330.93/warc/CC-MAIN-20181219130756-20181219152756-00329.warc.gz | 895,635,867 | 13,379 | ## College Algebra 7th Edition
$x-y$
$n(x-y)+(n-1)(y-x)=n(x-y)-(n-1)(x-y)=(x-y)[n-(n-1)]=(x-y)[n-n+1]=x-y$ | 55 | 107 | {"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} | 3.703125 | 4 | CC-MAIN-2018-51 | latest | en | 0.559924 |
http://mathhelpforum.com/pre-calculus/29262-solving-trigonometric-equations.html | 1,481,062,054,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542002.53/warc/CC-MAIN-20161202170902-00280-ip-10-31-129-80.ec2.internal.warc.gz | 164,691,670 | 11,872 | 1. ## Solving trigonometric equations
Solve the trigonometric equation analytically for values of $x$ for $0\le x <2\pi$
$sin\left(x - \frac{\pi}{4}\right) = cos\left(x - \frac{\pi}{4}\right)$
$sin\;x\;cos\;\frac{\pi}{4} - cos\;x\;sin\;\frac{\pi}{4} = cos\;x\;cos\;\frac{\pi}{4} + sin\;x\;sin\;\frac{\pi}{4}$
$sin\;x\;\left(\frac{\sqrt{2}}{2}\right) - cos\;x\;\left(\frac{\sqrt{2}}{2}\right) = cos\;x\;\left(\frac{\sqrt{2}}{2}\right) + sin\;x\;\left(\frac{\sqrt{2}}{2}\right)$
$2\;cos\;x\;\left(\frac{\sqrt{2}}{2}\right)$
If what I did so far is right. I'm stuck at this point.
2. Originally Posted by TI-84
Solve the trigonometric equation analytically for values of $x$ for $0\le x <2\pi$
$sin\left(x - \frac{\pi}{4}\right) = cos\left(x - \frac{\pi}{4}\right)$
$sin\;x\;cos\;\frac{\pi}{4} - cos\;x\;sin\;\frac{\pi}{4} = cos\;x\;cos\;\frac{\pi}{4} + sin\;x\;sin\;\frac{\pi}{4}$
$sin\;x\;\left(\frac{\sqrt{2}}{2}\right) - cos\;x\;\left(\frac{\sqrt{2}}{2}\right) = cos\;x\;\left(\frac{\sqrt{2}}{2}\right) + sin\;x\;\left(\frac{\sqrt{2}}{2}\right)$
$2\;cos\;x\;\left(\frac{\sqrt{2}}{2}\right)$
If what I did so far is right. I'm stuck at this point.
$2\;cos\;x\;\left(\frac{\sqrt{2}}{2}\right)$ = 0,
that is
$\sqrt{2}\;\cos x$ = 0,
3. But when I bring everything over to the right side to isolate x doesnt it just become x=0? the books answer is $\frac{\pi}{2}\;and\;\frac{3\pi}{2}$
4. Originally Posted by mr fantastic
$2\;cos\;x\;\left(\frac{\sqrt{2}}{2}\right)$ = 0,
that is
$\sqrt{2}\;\cos x$ = 0,
Next step: Divide both sides by $\sqrt{2}$:
$\cos x = 0$
Therefore x = ......
5. Originally Posted by mr fantastic
Next step: Divide both sides by $\sqrt{2}$:
$\cos x = 0$
Therefore x = ......
Perhaps I should write it as $\cos (x) = 0$.
Hmmmmmm ....... You do understand that $\cos (x) = 0$ means cos of x, NOT $cos \times x$ (which has no meaning.
In the same way that $\sqrt{2}$ means the square root of 2, not square root times 2 ......
So you're looking for values of x such that the cos of those values is equal to 0 ......
6. $
2\;cos\;x\;\left(\frac{\sqrt{2}}{2}\right) = 0
$
$cos\;x\;\left(\frac{\sqrt{2}}{2}\right) = 0$
$cos\;x = 0$
$x = Cos^{-1}\;(0)$
$x=90$ and $x=270$
Is that correct?
7. Originally Posted by TI-84
$
2\;cos\;x\;\left(\frac{\sqrt{2}}{2}\right) = 0
$
$cos\;x\;\left(\frac{\sqrt{2}}{2}\right) = 0$
$cos\;x = 0$
$x = Cos^{-1}\;(0)$
$x=90$ and $x=270$
Is that correct?
Yes. You can substitute into the original equation and confirm this.
8. Hello, TI-84
Solve for $x,\;\;0 \le x <2\pi$
. . $\sin\left(x - \frac{\pi}{4}\right) \:= \:\cos\left(x - \frac{\pi}{4}\right)$
Divide both sides by $\cos\left(x-\frac{\pi}{4}\right)$
. . $\frac{\sin\left(x-\frac{\pi}{4}\right)}{\cos\left(x-\frac{\pi}{4}\right)} \;=\;1 \quad\Rightarrow\quad \tan\left(x-\frac{\pi}{4}\right) \;=\;1$
Hence: . $x - \frac{\pi}{4} \;=\;\left\{\frac{\pi}{4},\:\frac{5\pi}{4}\right\} \quad\Rightarrow\quad x \;=\;\left\{\frac{\pi}{2},\:\frac{3\pi}{2}\right\}$ | 1,218 | 2,998 | {"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": 42, "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-2016-50 | longest | en | 0.640028 |
https://www.bartleby.com/solution-answer/chapter-22-problem-223bpe-accounting-27th-edition/9781337272094/direct-materials-purchases-budget-magnolia-candle-inc-budgeted-production-of-74200-candles-in/d57d5019-98dc-11e8-ada4-0ee91056875a | 1,603,553,263,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107883636.39/warc/CC-MAIN-20201024135444-20201024165444-00715.warc.gz | 629,041,898 | 67,211 | Direct materials purchases budget Magnolia Candle Inc. budgeted production of 74,200 candles in 20Y4. Wax is required to produce a candle. Assume that eight ounces (one-half of a pound) of wax is required for each candle. The estimated January 1, 20Y4, wax inventory is 2,500 pounds. The desired December 31, 20Y4, wax inventory is 2,100 pounds. If candle wax costs $4.10 per pound, determine the direct materials purchases budget for 20Y4. BuyFind Accounting 27th Edition WARREN + 5 others Publisher: Cengage Learning, ISBN: 9781337272094 BuyFind Accounting 27th Edition WARREN + 5 others Publisher: Cengage Learning, ISBN: 9781337272094 Solutions Chapter Section Chapter 22, Problem 22.3BPE Textbook Problem Direct materials purchases budgetMagnolia Candle Inc. budgeted production of 74,200 candles in 20Y4. Wax is required to produce a candle. Assume that eight ounces (one-half of a pound) of wax is required for each candle. The estimated January 1, 20Y4, wax inventory is 2,500 pounds. The desired December 31, 20Y4, wax inventory is 2,100 pounds. If candle wax costs$4.10 per pound, determine the direct materials purchases budget for 20Y4.
Expert Solution
To determine
Budgeting is a process to prepare the financial statement by the manager to estimate the organization’s future actions. It is also helpful to satisfy the everyday activities.
To Calculate: The direct materials purchases budget for 20Y4.
Explanation of Solution
Calculate the direct materials purchases budget for 20Y4.
M Incorporation Direct Materials Purchases Budget For the year ending December 31, 20Y4 Particulars Units Pounds required for production: Candles 37,100(1) Add: Desired ending inventory, December 31, 20Y4 2,100 Total Pounds required 39,200 Less: Estimated beginning inventory, January 1, 20Y4 (2,500) Total Pounds To Be Purchased (A) 36,700 Unit price per pound (B) \$4...
Want to see the full answer?
Check out a sample textbook solution.See solution
Want to see this answer and more?
Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!
See solution
Find more solutions based on key concepts
Show solutions
What are the advantages of a sole proprietorship?
Foundations of Business (MindTap Course List)
When should a standard cost variance be investigated?
Managerial Accounting: The Cornerstone of Business Decision-Making
COLEMAN TECHNOLOGIES INC. COST OF CAPITAL Coleman Technologies is considering a major expansion program that ha...
Fundamentals of Financial Management, Concise Edition (with Thomson ONE - Business School Edition, 1 term (6 months) Printed Access Card) (MindTap Course List)
What do competitive price searchers have to do to make economic profit?
Microeconomics: Private and Public Choice (MindTap Course List) | 662 | 2,817 | {"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} | 3.515625 | 4 | CC-MAIN-2020-45 | latest | en | 0.840936 |
https://www.esaral.com/q/a-factory-produces-bulbs-47201 | 1,725,953,264,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651224.59/warc/CC-MAIN-20240910061537-20240910091537-00411.warc.gz | 729,535,042 | 11,932 | # A factory produces bulbs.
Question:
A factory produces bulbs. The probability that any one bulb is defective is 1/50 and they are packed in boxes of 10. From a single box, find the probability that
(i) none of the bulbs is defective
(ii) exactly two bulbs are defective
(iii) more than 8 bulbs work properly
Solution:
Let’s assume X to be the random variable denoting a bulb to be defective.
Here, n = 10, p = 1/50, q = 1 – 1/50 = 49/50
We know that, P(X = r) = nCr pqn – r
(i) None of the bulbs is defective, i.e., r = 0
P(x = 0) = 10C0 (1/50)0(49/50)10 – 0 = (49/50)10
(ii) Exactly two bulbs are defective
So, P(x = 2) = 10C2 (1/50)2(49/50)10 – 2
= 45.498/5010 = 45 x (1/50)10 x 498
(iii) More than 8 bulbs work properly
We can say that less than 2 bulbs are defective
P(x < 2) = P(x = 0) + P(x = 1) | 298 | 820 | {"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-2024-38 | latest | en | 0.916433 |
http://gmatclub.com/forum/sequence-s-consists-of-24-nonzero-integers-if-each-term-in-98728.html?kudos=1 | 1,485,192,598,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560282935.68/warc/CC-MAIN-20170116095122-00451-ip-10-171-10-70.ec2.internal.warc.gz | 112,670,103 | 67,761 | Sequence S consists of 24 nonzero integers. If each term in : GMAT Data Sufficiency (DS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack
It is currently 23 Jan 2017, 09:29
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# Sequence S consists of 24 nonzero integers. If each term in
Author Message
TAGS:
### Hide Tags
Manager
Joined: 16 Feb 2010
Posts: 225
Followers: 2
Kudos [?]: 289 [3] , given: 16
Sequence S consists of 24 nonzero integers. If each term in [#permalink]
### Show Tags
07 Aug 2010, 11:34
3
KUDOS
11
This post was
BOOKMARKED
00:00
Difficulty:
95% (hard)
Question Stats:
36% (02:28) correct 64% (01:30) wrong based on 685 sessions
### HideShow timer Statistics
Sequence S consists of 24 nonzero integers. If each term in S after the second is the product of the previous two terms, how many terms in S are negative?
(1) The third term in S is positive
(2) The fourth term in S is negative
[Reveal] Spoiler:
The book claims this is one of the most diff questions GMAT can produce which I believe it is a joke....nevertheless got it wrong because I mis-interpreted the question. I am posting to see if more people will mis-interpret or if it is just me that is going *o*kers
[Reveal] Spoiler: OA
Last edited by Bunuel on 04 Jun 2013, 03:55, edited 2 times in total.
Edited the question and added the OA.
Math Expert
Joined: 02 Sep 2009
Posts: 36618
Followers: 7099
Kudos [?]: 93533 [2] , given: 10578
Re: Kaplan "800" DS: seq [#permalink]
### Show Tags
07 Aug 2010, 12:17
2
KUDOS
Expert's post
1
This post was
BOOKMARKED
zisis wrote:
Sequence S consists of 24 nonzero integers. If each term in S after the second is the product of the previous two terms, how many terms in S are negative?
(1) The third term in S is negative
(2) The fourth term in S is positive
The book claims this is one of the most diff questions GMAT can produce which I believe it is a joke....nevertheless got it wrong because I mis-interpreted the question. I am posting to see if more people will mis-interpret or if it is just me that is going *o*kers
Answer cannot be B, it should be A.
Sequence S consists of 24 nonzero integers. If each term in S after the second is the product of the previous two terms, how many terms in S are negative?
(1) The third term in S is negative --> first and second terms must have opposite signs so we can hve following two scenarios:
-+--+--+--+-...
OR:
+--+--+--+--...
You can see that in both cases there is a repeated pattern of three terms in which 2 are negative and 1 positive (-+- or +--) so in both cases out of 24 terms 2/3 will be negative, so there will be 16 negative terms. Sufficient.
(2) The fourth term in S is positive --> either all terms are positive, so zero negatives or +--+... and not all terms are positive, so more than zero negatives. Not sufficient.
If the OA is B then it should be:
Sequence S consists of 24 nonzero integers. If each term in S after the second is the product of the previous two terms, how many terms in S are negative?
(1) The third term in S is positive --> either all terms are positive, so zero negatives or +--... and not all terms are positive, so more than zero negatives. Not sufficient.
(2) The fourth term in S is negative --> again two cases:
--+--+--+...
OR
-+--+--+-...
The same here: in both cases there is a repeated pattern of three terms in which 2 are negative and 1 positive (--+ or -+-) so in both cases out of 24 terms 2/3 will be negative, so there will be 16 negative terms. Sufficient.
Hope it's clear.
_________________
Manager
Joined: 28 Feb 2012
Posts: 115
GPA: 3.9
WE: Marketing (Other)
Followers: 0
Kudos [?]: 42 [2] , given: 17
Re: Sequence S consists of 24 nonzero integers. If each term in [#permalink]
### Show Tags
05 Jun 2013, 02:25
2
KUDOS
Sequence S consists of 24 nonzero integers. If each term in S after the second is the product of the previous two terms, how many terms in S are negative?
(1) The third term in S is positive
(2) The fourth term in S is negative
One important thing to understand is that in GMAT sequence questions there is almost always a pattern. So there is no real need to calculate all 24 integers' signs.
The most difficult part in this question is: If each term in S after the second is the product of the previous two terms .
It is the same as saying that:
x; y; xy; x(y^2); (x^2)(y^3); (x^3)(y^5); (x^5)(y^8); ...
1 st.) xy is positive. In this case both x and y are negative or they both positive. If both positive all the terms will be positive: (+) (+) (+) (+) (+) (+)... In case both are negative we have: (-) (-) (+) (-) (-) (-) (+).... Two different answers - statement is not sufficient
2 st.) x(y^2) is negative. There are two possible options: x is negative and y is positive, or x is negative and y is negative.
In the 1st option we have the following pattern: (-) (+) (-) (-) (+) (-) (-) (+)...
In the 2nd option we have the follwing pattern: (-) (-) (+) (-) (-) (+) (-) (-)... It starts slightly different but the pattern is the same, so we can conclude that the number of positives and negatives within 24 integers will be the same rerdless of the options. So the answer is B.
_________________
If you found my post useful and/or interesting - you are welcome to give kudos!
Math Expert
Joined: 02 Sep 2009
Posts: 36618
Followers: 7099
Kudos [?]: 93533 [1] , given: 10578
Re: DS Question- Kaplan 800 [#permalink]
### Show Tags
25 Feb 2011, 01:27
1
KUDOS
Expert's post
Merging similar topics.
_________________
Intern
Joined: 20 May 2014
Posts: 39
Followers: 0
Kudos [?]: 5 [1] , given: 1
Re: Sequence S consists of 24 nonzero integers. If each term in [#permalink]
### Show Tags
07 Jul 2014, 08:30
1
KUDOS
ziko wrote:
Sequence S consists of 24 nonzero integers. If each term in S after the second is the product of the previous two terms, how many terms in S are negative?
(1) The third term in S is positive
(2) The fourth term in S is negative
One important thing to understand is that in GMAT sequence questions there is almost always a pattern. So there is no real need to calculate all 24 integers' signs.
The most difficult part in this question is: If each term in S after the second is the product of the previous two terms .
It is the same as saying that:
x; y; xy; x(y^2); (x^2)(y^3); (x^3)(y^5); (x^5)(y^8); ...
1 st.) xy is positive. In this case both x and y are negative or they both positive. If both positive all the terms will be positive: (+) (+) (+) (+) (+) (+)... In case both are negative we have: (-) (-) (+) (-) (-) (-) (+).... Two different answers - statement is not sufficient
2 st.) x(y^2) is negative. There are two possible options: x is negative and y is positive, or x is negative and y is negative.
In the 1st option we have the following pattern: (-) (+) (-) (-) (+) (-) (-) (+)...
In the 2nd option we have the follwing pattern: (-) (-) (+) (-) (-) (+) (-) (-)... It starts slightly different but the pattern is the same, so we can conclude that the number of positives and negatives within 24 integers will be the same rerdless of the options. So the answer is B.
Why isn't fourth term x^2y^2
Manager
Joined: 16 Feb 2010
Posts: 225
Followers: 2
Kudos [?]: 289 [0], given: 16
Re: Kaplan "800" DS: seq [#permalink]
### Show Tags
07 Aug 2010, 12:37
the OA is def the one posted....double checked it.....
will let more people get involved and then post the OA explanation.....your post supports my view that this question is not phrased correclt (ie is open to misinterpetation).....
the bigger question is how can we make sure that the practise we get is up to the test's standard if massive companies like Kaplan cannt phrase their questions correctly....
Math Expert
Joined: 02 Sep 2009
Posts: 36618
Followers: 7099
Kudos [?]: 93533 [0], given: 10578
Re: Kaplan "800" DS: seq [#permalink]
### Show Tags
07 Aug 2010, 12:45
zisis wrote:
the OA is def the one posted....double checked it.....
will let more people get involved and then post the OA explanation.....your post supports my view that this question is not phrased correclt (ie is open to misinterpetation).....
the bigger question is how can we make sure that the practise we get is up to the test's standard if massive companies like Kaplan cannt phrase their questions correctly....
Check the statements not the OA (placement of negative/positive in 1 and 2). There is nothing wrong with the question.
_________________
Manager
Joined: 16 Feb 2010
Posts: 225
Followers: 2
Kudos [?]: 289 [0], given: 16
Re: Kaplan "800" DS: seq [#permalink]
### Show Tags
07 Aug 2010, 13:06
you were right......I AM going mental.... just edited my post....hope this makes more sense
Manager
Joined: 16 Feb 2010
Posts: 225
Followers: 2
Kudos [?]: 289 [0], given: 16
Re: Kaplan "800" DS: seq [#permalink]
### Show Tags
08 Aug 2010, 02:29
zisis wrote:
Sequence S consists of 24 nonzero integers. If each term in S after the second is the product of the previous two terms, how many terms in S are negative?
(1) The third term in S is positive
(2) The fourth term in S is negative
The book claims this is one of the most diff questions GMAT can produce which I believe it is a joke....nevertheless got it wrong because I mis-interpreted the question. I am posting to see if more people will mis-interpret or if it is just me that is going *o*kers
One final question:
I misinterpreted the If each term in S after the second part....thought that the "formula"/instructions ($$An= An-1 * An-2$$ is only for the third onwards, thus we know nothing for the first two terms thus it is not possible to answer the question.....is that a valid conclusion....?
Math Expert
Joined: 02 Sep 2009
Posts: 36618
Followers: 7099
Kudos [?]: 93533 [0], given: 10578
Re: Kaplan "800" DS: seq [#permalink]
### Show Tags
08 Aug 2010, 02:51
zisis wrote:
zisis wrote:
Sequence S consists of 24 nonzero integers. If each term in S after the second is the product of the previous two terms, how many terms in S are negative?
(1) The third term in S is positive
(2) The fourth term in S is negative
The book claims this is one of the most diff questions GMAT can produce which I believe it is a joke....nevertheless got it wrong because I mis-interpreted the question. I am posting to see if more people will mis-interpret or if it is just me that is going *o*kers
One final question:
I misinterpreted the If each term in S after the second part....thought that the "formula"/instructions ($$An= An-1 * An-2$$ is only for the third onwards, thus we know nothing for the first two terms thus it is not possible to answer the question.....is that a valid conclusion....?
You interpreted correctly: $$a_n=a_{n-1}*a_{n-2}$$, for $$n>2$$.
But you are not right about the first two terms for the following statements:
(2) The fourth term in S is negative --> $$a_4=negative=a_3*a_2$$ --> second and third term must have opposite signs, so third term is either positive or negative. Now, case 1: if third term is positive then first and second terms must be both negative (for first and second it's not possible to be both positive as in this case fourth term would be positive too and we know that it's negative) and case 2: if third term is negative then first and second terms must have opposite signs. So there are only 2 cases possible:
--+--+--+...
OR
-+--+--+-...
In both cases there is a repeated pattern of three terms in which 2 are negative and 1 positive (--+ or -+-) so in both cases out of 24 terms 2/3 will be negative, so there will be 16 negative terms. Sufficient.
_________________
Intern
Joined: 05 Nov 2009
Posts: 32
Followers: 0
Kudos [?]: 4 [0], given: 3
Re: Kaplan "800" DS: seq [#permalink]
### Show Tags
08 Aug 2010, 16:25
I also get B. You can establish a pattern at 4 and each combination to get negative (- * + or + * -) gets to the same number of negative numbers. I must admist that I did this the long way, without a formula - prorbably took a little over 2 minutes.
Senior Manager
Joined: 21 Mar 2010
Posts: 314
Followers: 5
Kudos [?]: 30 [0], given: 33
Re: DS Question- Kaplan 800 [#permalink]
### Show Tags
24 Feb 2011, 21:59
rajeshaaidu wrote:
No clue! It's really tough one. Please give official explanation. A) and D) I have eliminated but I was confused between B) and C) because I was not able to explain stmt-B.
If you write out the options resulting from b
T3 below will be T1T2
T1 ---------- T2--------- T3--------------T4
-ve -ve +ve -ve
-ve +ve -ve -ve
For both these cases expand out the first 8 terms and you will see the ratio of 2:1 for negative to positive!
Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2021
Followers: 161
Kudos [?]: 1708 [0], given: 376
Re: DS Question- Kaplan 800 [#permalink]
### Show Tags
24 Feb 2011, 22:39
I classify these types of questions as torturous; because we need to count them out to be sure.
1. We can easily rule this out because the 3rd term is +ve; first two can either be -ve or all can be positive. Not sufficient.
2. This statement is little hairy.
X-------$$|-|+|-|-|+|-|-|+|-|-|+|-|-|+|-|-|+|-|-|+|-|-|+|-|$$. 16 -ves
Y-------$$|-|-|+|-|-|+|-|-|+|-|-|+|-|-|+|-|-|+|-|-|+|-|-|+|$$. 16 -ves
Sufficient.
Well after some thought; you may recognize the pattern that there are two negatives after every +ve; but I would rather write them all out and count them.
As you see for X; first position is -ve and the 2nd +ve. 3rd and 4th are negative and 5th +ve and the pattern continues. You may now count -ves. Likewise for Y.
Ans: "B"
_________________
Director
Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing.
Affiliations: University of Chicago Booth School of Business
Joined: 03 Feb 2011
Posts: 920
Followers: 14
Kudos [?]: 341 [0], given: 123
Re: Kaplan "800" DS: seq [#permalink]
### Show Tags
25 Feb 2011, 10:18
Bunuel you are awesome ! If I had seen this question on test day I would have marked C and move on ! But then I took numbers and yes it answered. So - its imperative to find the first negative term. That is the key for the pattern ? Right?
Math Expert
Joined: 02 Sep 2009
Posts: 36618
Followers: 7099
Kudos [?]: 93533 [0], given: 10578
Re: Sequence S consists of 24 nonzero integers. If each term in [#permalink]
### Show Tags
04 Jun 2013, 04:03
Bumping for review and further discussion*. Get a kudos point for an alternative solution!
*New project from GMAT Club!!! Check HERE
_________________
Manager
Joined: 07 Apr 2012
Posts: 126
Location: United States
Concentration: Entrepreneurship, Operations
Schools: ISB '15
GMAT 1: 590 Q48 V23
GPA: 3.9
WE: Operations (Manufacturing)
Followers: 0
Kudos [?]: 10 [0], given: 45
Re: Sequence S consists of 24 nonzero integers. If each term in [#permalink]
### Show Tags
04 Sep 2013, 01:42
Even though OA is B and is right, # of negative terms is 17 and not 16. The first term has to be negative in both the scenarios of B.
Math Expert
Joined: 02 Sep 2009
Posts: 36618
Followers: 7099
Kudos [?]: 93533 [0], given: 10578
Re: Sequence S consists of 24 nonzero integers. If each term in [#permalink]
### Show Tags
04 Sep 2013, 02:41
ygdrasil24 wrote:
Even though OA is B and is right, # of negative terms is 17 and not 16. The first term has to be negative in both the scenarios of B.
There are 24 numbers and 2/3 are negative, thus 16 numbers are negative not 17.
_________________
Manager
Joined: 07 Apr 2012
Posts: 126
Location: United States
Concentration: Entrepreneurship, Operations
Schools: ISB '15
GMAT 1: 590 Q48 V23
GPA: 3.9
WE: Operations (Manufacturing)
Followers: 0
Kudos [?]: 10 [0], given: 45
Re: Sequence S consists of 24 nonzero integers. If each term in [#permalink]
### Show Tags
04 Sep 2013, 03:22
Bunuel wrote:
ygdrasil24 wrote:
Even though OA is B and is right, # of negative terms is 17 and not 16. The first term has to be negative in both the scenarios of B.
There are 24 numbers and 2/3 are negative, thus 16 numbers are negative not 17.
Yes 16, I over worked on the final answer
Intern
Joined: 13 Apr 2014
Posts: 14
Followers: 1
Kudos [?]: 1 [0], given: 1
Re: Sequence S consists of 24 nonzero integers. If each term in [#permalink]
### Show Tags
15 Apr 2014, 04:24
Sequence S consists of 24 nonzero integers. If each term in S after the second is the product of the previous two terms, how many terms in S are negative?
(1) The third term in S is positive --> either all terms are positive, so zero negatives or +--... and not all terms are positive, so more than zero negatives. Not sufficient.
(2) The fourth term in S is negative --> again two cases:
--+--+--+...
OR
-+--+--+-...
The same here: in both cases there is a repeated pattern of three terms in which 2 are negative and 1 positive (--+ or -+-) so in both cases out of 24 terms 2/3 will be negative, so there will be 16 negative terms. Sufficient.
tnx for this
_________________
best for iranian
Intern
Joined: 08 Jan 2014
Posts: 20
Location: United States
Concentration: General Management, Entrepreneurship
GMAT Date: 06-30-2014
GPA: 3.99
WE: Analyst (Consulting)
Followers: 1
Kudos [?]: 18 [0], given: 4
Re: Sequence S consists of 24 nonzero integers. If each term in [#permalink]
### Show Tags
10 May 2014, 11:11
1st term x
2nd term y
S={x,y,xy,xy^2.........}
since 4th term is -ve , xY^2 is -ve i.e. x is -ve since y^2 cannot be -ve.
Now since 1st term is -ve there can be 2 different pattern for the sequence assuming 2nd term to be either +ve or -ve.
S={-,-,+,-,-,+,......}
or
S={-,+,-,-,+,-,......}
So its clear that either 2nd term is +ve or -Ve the # of -ve terms is equal for 24 terms(since # of terms is a product of 3).
Re: Sequence S consists of 24 nonzero integers. If each term in [#permalink] 10 May 2014, 11:11
Go to page 1 2 Next [ 23 posts ]
Similar topics Replies Last post
Similar
Topics:
2 Each term in sequence S is determined by multiplying the prior term by 1 02 Jun 2016, 10:54
26 In the sequence S of numbers, each term after the first two 8 26 Feb 2014, 01:21
9 In the sequence S of numbers, each term after the first two 5 30 May 2011, 04:54
15 In a sequence of terms in which each term is three times the 13 15 Feb 2011, 07:28
1 In a sequence of terms in which each term is three times the 3 13 Oct 2009, 05:36
Display posts from previous: Sort by | 5,394 | 18,889 | {"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": 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} | 3.65625 | 4 | CC-MAIN-2017-04 | latest | en | 0.895771 |
https://www.easycalculation.com/percentage-multiple-numbers-calculator.php | 1,653,362,422,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662562410.53/warc/CC-MAIN-20220524014636-20220524044636-00725.warc.gz | 837,570,868 | 8,027 | English
# Multiple Percentage Calculator English Español German 中国 Português Pусский 日本語 Türk
Percentage is anything per 100 expressed with symbol %. This is an online calculator to find the percentage of multiple numbers out of a constant number. For example, if you are entering the data as 1, 2 , 3 and the y value as 4, then the percentage of 1 out of 4 is 25%. Similarly 2 is 50 % out of 4 and 3 is 75 % out of 4.
## Find Percentage of Multiple Numbers
Percentage is anything per 100 expressed with symbol %. This is an online calculator to find the percentage of multiple numbers out of a constant number. For example, if you are entering the data as 1, 2 , 3 and the y value as 4, then the percentage of 1 out of 4 is 25%. Similarly 2 is 50 % out of 4 and 3 is 75 % out of 4.
#### Formula:
p = ( x / y ) * 100 Where, p = Percentage x = X Value y = Y Value
### Example
#### Multiple Percentage
Find the percentage of 5,10,15 in 5
To find the percentage of 5,10,15 in 5, express it as fraction and multiply it with 100.
Step 1 :
=(5 / 5) * 100
= 1 * 100
= 100 %
Step 2 :
=(10 / 5) * 100
= 2 * 100
= 200 %
Step 3 :
=(15 / 5) * 100
= 3 * 100
= 300 % | 360 | 1,165 | {"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} | 3.65625 | 4 | CC-MAIN-2022-21 | latest | en | 0.829263 |
http://slideplayer.com/slide/5093687/ | 1,571,050,301,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986653216.3/warc/CC-MAIN-20191014101303-20191014124303-00536.warc.gz | 181,235,469 | 27,381 | Presentation is loading. Please wait.
# Electrical Theory I - The Basics
## Presentation on theme: "Electrical Theory I - The Basics"— Presentation transcript:
Electrical Theory I - The Basics
Let there be light….
Introduction Basic Terminology Ohm’s Law
Kirchhoff’s Laws & Applications Basic Circuit Analysis Transformers & Rectifiers
How is Electricity Produced?
Friction: “static electricity” from rubbing (walking across a carpet) Pressure: piezoelectricity from squeezing crystals together (quartz watch) Heat: voltage produced at junction of dissimilar metals (thermocouple) Light: voltage produced from light striking photocell (solar power) Chemical: voltage produced from chemical reaction (wet or dry cell battery) Magnetism: voltage produced using electromotive induction (AC or DC generator).
Basic Terminology Electromotive Force (E or V)
Force which causes electrons to move from one location to another Known as emf, potential difference, or voltage Unit is volt (V) Source: Generator Battery Like pump that moves water through “pressure”
Basic Terminology Current (I)
Flow of electric charges - electrons (or holes) - through a conductor or circuit per increment of time Unit is ampere (number of charged particles passing a point each second) 1 amp = 1 coulomb/sec = 6x1018 electrons/sec Like rate of flow of water through a pipe
Basic Terminology Resistance (R) Conductor
An electrical circuit’s opposition to the flow of current through it Measured in ohms (W) Conductor All materials will conduct electricity, but at varying resistances Good conductors have little resistance (ie: silver, copper, aluminum, iron)
Basic Terminology Insulator Power (P)
Substances which offer high resistance to current flow (ie: wood, rubber, plastics) Circuits made of wires covered with insulator Power (P) Rate at which work is performed Measured in watts (W)
Ohm’s Law & Applications
Law: current of a circuit is directly proportional to the applied voltage and inversely proportional to circuit resistance I a V, I a 1/R V =IR Power P = VI P = (IR)I = I2R
Kirchhoff’s Laws Kirchhoff’s Current Law (KCL)
A node is any junction in a circuit where two or more elements meet Currents into a node sum to zero OR Current entering a junction is equivalent to the current leaving a junction
Kirchhoff’s Laws Kirchhoff’s Voltage Law (KVL)
A loop is any path in a circuit that current can take so that it meets back up to where it starts Voltages around a CLOSED loop sum to zero
Applications Resistors in Series Resistors in Parallel
RT = R1 + R2 + R Resistors in Parallel 1/RT = 1/R1 + 1/R2 + 1/R Examples: should be able to find total current flow in circuit, current flow through each resistor, voltages, power dissipated, etc.
Questions?
Electrical Theory II – The Applications
Harnessing the Power…
Power Generation Chemical Cells Magnetism Electrodes Electrolyte
Conductor Wet/Dry Batteries Magnetism
Electromagnetic Induction
Faraday (1831): Showed that an emf is induced in a conductor if a magnet passes by a conductor When pole of magnet entered coil, current flowed in one direction When direction of magnet reversed, current flowed in opposite direction
Electromagnetic Induction
Magnitude of induced current can be increased by: Increasing strength of magnetic field Increasing speed of relative motion Positioning of field & conductor to increase number of magnetic lines of flux cut Magnetic field usually produced by electromagnet
Electromagnet Soft iron core wound with coils of wire
When current present (excitation current), core becomes magnetized Field strength determined by number of turns and magnitude of current: B a NIDC
Electromagnetic Induction
Results in: Generator action: generator converts mechanical to electrical energy Motor action: motor converts electrical to mechanical energy
Generator Action For emf/current (electricity):
Magnetic Field Conductor Relative Motion b/t the two Voltage produced: “induced emf/voltage” Current produced: “induced current” Left-hand rule for generator action
Motor Action For motor action (torque/motion):
Magnetic Field Conductor Current flow in conductor Torque produced: “induced torque” Right-hand rule for motor action
Standard Terminology Stator: stationary housing of the generator or motor Rotor: rotating shaft inside the stator Field windings: conductors used to produce electromagnetic field Armature windings: conductors in which output voltage is produced (or input is provided)
Basic Terminology Direct Current (DC) Alternating Current (AC)
Current flow is unidirectional and of constant magnitude (battery) Alternating Current (AC) Magnitude & direction of current flow periodically change Each sequence called a cycle Frequency is cycles per second (Hz)
AC Generators Most electrical power used is AC made by AC generators
Basic principle: rotating magnetic field “cutting through” a conductor Regardless of size, all AC generators work on same principle Two types: Revolving armature (NOT used) Revolving field (Used in SSTG’s, GTGS, DG)
AC Generators Two types: Revolving armature (NOT used)
Revolving field (Used in SSTG’s, GTGS, DG)
AC Generators Armature windings on stator Field windings on rotor
DC current provided for field via slip rings and brushes (vice commutator rings) Rotor turned by prime mover creates rotating magnetic field Armature windings on stator As field rotates, AC current produced in armature Since stationary contacts, no arc-over
AC Generators Determining speed of AC machine:
f = P(RPM)/ RPM = 120f/P Must maintain constant 60Hz output use speed governor to maintain constant RPM (independent of loading) Must also regulate voltage output Since constant RPM, must control field excitation (DC current) to control output voltage
Three Phase (3f) AC Power
Phases: number of sets of armature windings on stator 3f has three sets of armature windings Voltage induced is 120o out of phase for each Output: 3 sinusoidal voltages and currents Allows more power to be delivered with a smaller design generator
Three Phase (3f) AC Power
AC Motors Use AC current as input to produce work
Many different types depending on number of phases of AC input & construction Ex: induction motor Input AC current on stator produces rotating field Current produced in conductors on rotor produces torque
DC Generators Basic Principle: rotate a conductor within a magnetic field to induce an EMF Field windings located on stator & receive current from outside source
DC Generators Armature windings on rotor
Commutator rings used to mechanically reverse the armature coil connection to the external circuit EMF developed across the brushes becomes a DC voltage/current (pulsating and unidirectional)
DC Motors Essentially the same in construction as DC generator
Based on principle that current carrying conductor placed at a right angle to a magnetic field tends to move in a direction perpendicular to magnetic lines of flux Only need to change relative voltage to go between generator motor
AC vs DC power Many different voltages required on board ship
Easier to transform AC power for each application
Electrical Devices Transformer
Device w/o moving parts that transfers energy from one circuit to another by electromagnetic induction Consists of ferromagnetic core & sets of windings Step-up: Vin Vout Step-down: Vin Vout Only works with AC
Electrical Devices Rectifier Converts AC DC
Designed to have small resistance to current flow in one direction & large resistance in opposite direction Typically called a diode or rectifier
Questions?
Download ppt "Electrical Theory I - The Basics"
Similar presentations
Ads by Google | 1,628 | 7,666 | {"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} | 3.625 | 4 | CC-MAIN-2019-43 | latest | en | 0.852777 |
https://brainmass.com/business/interest-rates/create-hedge-futures-contract-zinn-company-317191 | 1,600,851,853,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400210616.36/warc/CC-MAIN-20200923081833-20200923111833-00519.warc.gz | 308,252,704 | 10,948 | Explore BrainMass
# Create a Hedge with the Futures Contract for Zinn Company
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
Chapter 23. Ch 23-06 Build a Model
Problem 23-6. Use the information and data from Problem 23-5
Problem Inputs:
Size of planned debt offering = \$10,000,000
Anticipated rate on debt offering = 11%
Maturity of planned debt offering = 20
Number of months until debt offering = 7
Settle price on futures contract (% of par) = 95.53125%
Maturity of bond underlying futures contract = 20
Coupon rate on bond underlying futures contract = 6%
Size of futures contract (dollars) = \$100,000
a. Create a hedge with the futures contract for Zinn Company's planned June debt offering of \$10 million. What is the implied yield on the bond underlying the future's contract?
Value of each T-bond future's contract =
Number of contracts needed for hedge = rounding =
Value of contracts in hedge =
Implied semi-annual yield =
Implied annual yield =
b. Suppose interest rates fall by 300 basis points. What is the dollar savings from issuing the debt at the new interest rate? What is the dollar change in value of the futures position? What is the total dollar value change of the hedged position?
Change in interest rate on debt offering (basis points) = -300
New interest rate on debt =
Value of issuing at new rate interest =
Dollar value savings or cost from issuing debt at the new rate =
New yield on futures contract =
New value of each futures contract
Value of all fo the futures contract at new yield =
Dollar change in value of the futures position =
Total dollar value change of hedge =
c. Create a graph showing the effectiveness of the hedge if the change in interest rates, in basis points, is: -300, -200, -100, 0, 100, 200, or 300. Show the dollar cost (or savings) from issuing the debt at the new interest rates, the dollar change in value of the futures position, and the total dollar value change.
Change in rate Dollar change in cost/savings of issue Dollar change in value of futures position Total dollar value change of hedge
Base -300
-300
-200
-100
0
100
200
300 | 509 | 2,179 | {"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-2020-40 | latest | en | 0.899947 |
https://www.physicsforums.com/threads/solving-a-differential-equation-using-laplace-transform.974300/ | 1,569,017,988,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574084.88/warc/CC-MAIN-20190920221241-20190921003241-00480.warc.gz | 977,107,528 | 18,664 | # I Solving a differential equation using Laplace transform
#### PainterGuy
Summary
What is the criteria for a given differential equation to be solvable using Laplace transform?
Hi,
I was trying to see if the following differential equation could be solved using Laplace transform; its solution is y=x^4/16.
You can see below that I'm not able to proceed because I don't know the Laplace pair of xy^(1/2).
Is it possible to solve the above equation using Laplace transform? In my opinion, I don't think so.
Thank you.
Related Differential Equations News on Phys.org
#### S.G. Janssens
Your ODE is nonlinear in the unknown $y$.
#### pasmith
Homework Helper
In this case, the substitution $u = y^{1/2}$ will reduce the ODE to one which can be solved by Laplace transforms.
#### PainterGuy
Your ODE is nonlinear in the unknown $y$.
Is it okay to assume that Laplace transform cannot be used with nonlinear differential equations? The answer given here, https://www.quora.com/Can-you-determine-the-Laplace-Transform-of-a-non-linear-differential-equation , agrees to some extent.
In this case, the substitution $u = y^{1/2}$ will reduce the ODE to one which can be solved by Laplace transforms.
I'm not sure if I'm following you correctly.
The equation reduces to
The new equation is still nonlinear and I can't see a way to apply Laplace transform to it.
I understand that what I wrote above is wrong because "dx" should be represented in term of "dy"; something like this du²/dy. I don't know how to proceed with it. Could you please guide me? Thank you!
#### Mark44
Mentor
Is it okay to assume that Laplace transform cannot be used with nonlinear differential equations? The answer given here, https://www.quora.com/Can-you-determine-the-Laplace-Transform-of-a-non-linear-differential-equation , agrees to some extent.
I'm not sure if I'm following you correctly.
View attachment 246135
The equation reduces to
View attachment 246136
The new equation is still nonlinear and I can't see a way to apply Laplace transform to it.
I understand that what I wrote above is wrong because "dx" should be represented in term of "dy"; something like this du²/dy. I don't know how to proceed with it. Could you please guide me? Thank you!
But $\frac{du^2}{dx} = \frac{du^2}{du}\frac{du}{dx} = 2u\frac{du}{dx}$ using the chain rule. This leads to a very simple, linear DE that you can use Laplace transforms on if you must, but much simpler techniques can be used.
#### PainterGuy
Thank you!
I was able to solve it. So, I'd say that saying that a nonlinear differential equation cannot be solved using Laplace transform is not a correct statement.
Note to self:
"y" is expressed in terms of "u", or as a function of "u". "u" in turn being function of "x". "y" is still a function of "x" but to make things easier we introduced substitution variable, u, and it stands between "y" and "x"; in other words, "y" depends upon "u" and "u" depends upon "x".
y=u^2
dy/dx=dy/du * du/dx
dy/dx=du^2/du * du/dx
#### Attachments
• 30 KB Views: 56
"Solving a differential equation using Laplace transform"
### Physics Forums Values
We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving | 854 | 3,434 | {"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": 2, "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-2019-39 | longest | en | 0.930992 |
https://www.gradesaver.com/textbooks/math/precalculus/precalculus-10th-edition/chapter-5-exponential-and-logarithmic-functions-5-5-properties-of-logarithms-5-5-assess-your-understanding-page-306/108 | 1,701,897,505,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100603.33/warc/CC-MAIN-20231206194439-20231206224439-00181.warc.gz | 898,819,165 | 14,254 | ## Precalculus (10th Edition)
Since $\log_a {x^n}=n\cdot \log_a {x}$, then $f(x^{\alpha})=\log_a{(x^{\alpha})}=\alpha \cdot \log_a{x}=\alpha \cdot f(x)$
We know that $\log_a {x^n}=n\cdot \log_a {x}$. Hence, $f(x^{\alpha})=\log_a{x^{\alpha}}=\alpha \cdot \log_a{x}=\alpha \cdot f(x)$ | 116 | 283 | {"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} | 3.890625 | 4 | CC-MAIN-2023-50 | latest | en | 0.251083 |
https://edboost.org/index.php/node/134 | 1,726,684,307,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651931.60/warc/CC-MAIN-20240918165253-20240918195253-00529.warc.gz | 189,863,516 | 5,979 | # Interpret Equations
1. $p$, which represents the average cost of the paper, will go up if the cost of the her favorite paper goes up.
2. $b$, which represents the number of boxes that she wraps, will go up in the holiday season.
3. 4 is multiplied by $hl$, which represent "height x length," so, the four must represent the 4 sides of the box.
4. $hl$ (height times length) is multiplied by 4 and $lw$ (length times width) is multiplied by 2. So, an increase in height would have a bigger impact on costs (because it is multiplied 4 times rather than 2).
5. $d$ represents the number of days that have passed, so as summer progresses, this variable should increase.
6. Because each day that passes subtracts 64 from 3670, $L$ should go down as the summer passes (and $d$ increases).
7. In this equation, there's a starting point, that goes down a little each day. Because the question is about reading, 3670 must be the starting number of pages that need to be read.
8. And, each day the number 3670 decreases a little, in fact, it decreases by 64 pages a day. 64 represents the number of pages Garret reads each day.
9. 5 is multiplied by $h$ which represents hours. So, the boat must cost 5 dollars per hour.
10. 8 is not multiplied by anything. It is a constant. It must cost 8 dollars to rent a boat, no matter how long it is rented for or how many riders are going in the boat.
11. Because the boats cost more per hour than they do per rider, it would be cheaper for the larger group (8 people) to rent for a shorter time than a larger group to rent for a shorter time.
12. In this case, you need to plug values into the equation and test it out:
$C=8+5(4)+3(8)=52$ (one boat, all eight people)
$C=8+5(4)+3(4)=40$ (one boat, only 4 people, so two boats with 4 people each would be \\$80, which would be more).
Definitely cheaper for everyone to get in the same boat! | 502 | 1,881 | {"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} | 4.375 | 4 | CC-MAIN-2024-38 | latest | en | 0.951943 |
https://metanumbers.com/184342 | 1,627,767,445,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154126.73/warc/CC-MAIN-20210731203400-20210731233400-00107.warc.gz | 395,791,487 | 10,922 | ## 184342
184,342 (one hundred eighty-four thousand three hundred forty-two) is an even six-digits composite number following 184341 and preceding 184343. In scientific notation, it is written as 1.84342 × 105. The sum of its digits is 22. It has a total of 3 prime factors and 8 positive divisors. There are 90,600 positive integers (up to 184342) that are relatively prime to 184342.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 6
• Sum of Digits 22
• Digital Root 4
## Name
Short name 184 thousand 342 one hundred eighty-four thousand three hundred forty-two
## Notation
Scientific notation 1.84342 × 105 184.342 × 103
## Prime Factorization of 184342
Prime Factorization 2 × 61 × 1511
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 184342 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 184,342 is 2 × 61 × 1511. Since it has a total of 3 prime factors, 184,342 is a composite number.
## Divisors of 184342
8 divisors
Even divisors 4 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 281232 Sum of all the positive divisors of n s(n) 96890 Sum of the proper positive divisors of n A(n) 35154 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 429.351 Returns the nth root of the product of n divisors H(n) 5.24384 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 184,342 can be divided by 8 positive divisors (out of which 4 are even, and 4 are odd). The sum of these divisors (counting 184,342) is 281,232, the average is 35,154.
## Other Arithmetic Functions (n = 184342)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 90600 Total number of positive integers not greater than n that are coprime to n λ(n) 9060 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 16654 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 90,600 positive integers (less than 184,342) that are coprime with 184,342. And there are approximately 16,654 prime numbers less than or equal to 184,342.
## Divisibility of 184342
m n mod m 2 3 4 5 6 7 8 9 0 1 2 2 4 4 6 4
The number 184,342 is divisible by 2.
• Arithmetic
• Deficient
• Polite
• Square Free
• Sphenic
## Base conversion (184342)
Base System Value
2 Binary 101101000000010110
3 Ternary 100100212111
4 Quaternary 231000112
5 Quinary 21344332
6 Senary 3541234
8 Octal 550026
10 Decimal 184342
12 Duodecimal 8a81a
20 Vigesimal 130h2
36 Base36 3y8m
## Basic calculations (n = 184342)
### Multiplication
n×i
n×2 368684 553026 737368 921710
### Division
ni
n⁄2 92171 61447.3 46085.5 36868.4
### Exponentiation
ni
n2 33981972964 6264304860129688 1154774486526026945296 212873438395180859149755232
### Nth Root
i√n
2√n 429.351 56.9126 20.7208 11.3012
## 184342 as geometric shapes
### Circle
Diameter 368684 1.15825e+06 1.06758e+11
### Sphere
Volume 2.62399e+16 4.2703e+11 1.15825e+06
### Square
Length = n
Perimeter 737368 3.3982e+10 260699
### Cube
Length = n
Surface area 2.03892e+11 6.2643e+15 319290
### Equilateral Triangle
Length = n
Perimeter 553026 1.47146e+10 159645
### Triangular Pyramid
Length = n
Surface area 5.88585e+10 7.38255e+14 150515
## Cryptographic Hash Functions
md5 1c8e13266b2d11fcd5fe0d54876cfc92 80221af24d2ac9d7005f37c0d851569227eee86d 6c33782fec370883cda126df7f3f50b5bc18a22c2338bcc14c4982b286875fcb 9d52ea0d5b8998550c94a043330e2f2257dd8c52d02b3f3fccf0dfbb08188097780a9170655b80f2beec93bb3ef5ce5145539f5d9db47567638210ddb8faec20 14cbf5fa8783dfb97f3a0cb4558a26fbb909600e | 1,442 | 4,140 | {"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} | 3.515625 | 4 | CC-MAIN-2021-31 | latest | en | 0.825468 |
https://moomoomath.com/solve-percentage-problems-using-is-over-of/ | 1,718,463,552,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861594.22/warc/CC-MAIN-20240615124455-20240615154455-00249.warc.gz | 373,561,375 | 17,545 | Search
How do you solve the following problems that involve percentages?
5 is 20% of what number?
20% of 120 equals ?
5 is what percent 25?
In each of these problems you can use the formula,
is/of =%/100
Let’s see how this works.
5 is 20% of what Number?
5/x = 20/100
Cross multiply
20x = 500
Divide by 20
x = 25
What is 20% of 120 ?
x/120 = 20/100
Cross multiply
100x = 2400
Divide both sides by 100
X = 24
5 is what percent of 25?
5/25 = x/100
Cross Multiply
25x =500
Divide both sides by 25
x = 20
This video works several similar problems. Each problem can be solved using is over of. | 193 | 614 | {"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} | 4.28125 | 4 | CC-MAIN-2024-26 | latest | en | 0.941088 |
http://mathhelpforum.com/calculus/72272-delta-epsilon-print.html | 1,527,081,616,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865651.2/warc/CC-MAIN-20180523121803-20180523141803-00495.warc.gz | 190,900,058 | 3,560 | # delta epsilon
• Feb 7th 2009, 03:16 AM
tumi
delta epsilon
USE THE EPSILON-DELTA DEFINITION OF LIMITS TO PROVE THAT
lim -3x+7=10
x-(-1)
• Feb 7th 2009, 03:23 AM
mr fantastic
Quote:
Originally Posted by tumi
USE THE EPSILON-DELTA DEFINITION OF LIMITS TO PROVE THAT
lim -3x+7=10
x-0
Find any post by this guy: http://www.mathhelpforum.com/math-he.../krizalid.html
eg. http://www.mathhelpforum.com/math-he...evolution.html
Click on the obvious link in his signature.
• Feb 7th 2009, 04:52 AM
Showcase_22
Doesn't $\displaystyle -3x+7 \rightarrow+7$?
Since $\displaystyle -3x+7$ is a continuous function it's limit is equal to when $\displaystyle x=0$. This gives +7.
• Feb 7th 2009, 09:40 AM
HallsofIvy
As Showcase_22 says, you cannot prove $\displaystyle \lim_{x \rightarrow 0} -3x+ 7= 10$- it isn't true!
What you can prove is either $\displaystyle \lim_{x\rightarrow 0} -3x+ 7= 7$ or $\displaystyle \lim_{x\rightarrow -1} -3x+ 7= 10$. Is your problem either of those?
• Feb 7th 2009, 10:20 AM
Krizalid
Quote:
Originally Posted by mr fantastic
Click on the obvious link in his signature.
Well in this section that thread is stickied. :)
• Feb 7th 2009, 11:15 AM
galactus
Yes indeed. There is a typo or something in your statement.
Let's make up something similar and go with that as an example.
$\displaystyle \lim_{x\to 2}[2x-7]=-3$
$\displaystyle |(2x-7)+3|=|2x-4|=2|x-2|<{\epsilon} \;\ if \;\ |x-2|<\frac{\epsilon}{2}, \;\ so \;\ {\delta}=\frac{\epsilon}{2}$
Here's another. I am just making an example up to illustrate to you as a tutorial. Hope it helps.
$\displaystyle \lim_{x\to 4}\sqrt{x}=2$
$\displaystyle |\sqrt{x}-2|=\left|\frac{\sqrt{x}-2}{1}\cdot\frac{\sqrt{x}+2}{\sqrt{x}+2}\right|=\fr ac{|x-4|}{\sqrt{x}+2}$
Now, if we pull the reins back on delta and restrict it so that $\displaystyle {\delta}\leq 4$, then:
$\displaystyle |x-4|<4$
$\displaystyle 0<x<8$
$\displaystyle 0<x<\sqrt{8}$
$\displaystyle 2<\sqrt{x}+2<\sqrt{8}+2$
$\displaystyle \frac{1}{2}>\frac{1}{\sqrt{x}+2}>\frac{1}{\sqrt{8} +2}$
$\displaystyle \frac{1}{\sqrt{x}+2}<\frac{1}{2}$
$\displaystyle \frac{|x-4|}{\sqrt{x}+2}\leq \frac{1}{2}|x-4|$
Therefore, thus, and all that:
$\displaystyle |\sqrt{x}-2|<{\epsilon} \;\ if \;\ \frac{1}{2}|x-4|<{\epsilon}, \;\ or \;\ if \;\ |x-4|<2{\epsilon}, \;\ so \;\ {\delta}=min(2{\epsilon},4)$
There.....does that help finish the one you have assuming you fix the typo or what not?.
• Feb 8th 2009, 02:59 AM
tumi
my apologies | 907 | 2,459 | {"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} | 4.0625 | 4 | CC-MAIN-2018-22 | latest | en | 0.671304 |
https://www.omnicalculator.com/statistics/chi-square | 1,723,032,272,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640694449.36/warc/CC-MAIN-20240807111957-20240807141957-00616.warc.gz | 699,913,825 | 71,813 | Last updated:
# Chi-Square Calculator
What is the chi-square test?How to calculate chi-square?How to perform the chi-square test for the goodness of fit?Using chi-square calculator: an example
The chi-square calculator will help you conduct the goodness of fit test, also known as the chi-square test. This statistic is used when you want to determine whether your data is consistent with the expected distribution. Read on to learn how to calculate the chi-square value and use the chi-square tables to determine the quality of your data. Make sure to take a look at our p-value calculator, too!
## What is the chi-square test?
Imagine that you are a teacher at high school. You grade a test, giving the marks 2 (the worst) to 5 (the best). You expect 15% of the students to obtain grade 5, 40% to obtain a 4, 30% to get a 3, and 15% to fail by getting a 2. Once you grade one class's papers, you can test how your grading differs from the intended distribution using the chi-square test calculator. It is simply a test that measures how the observed values differ from the expected ones.
## How to calculate chi-square?
You can find the chi-square value using the following formula:
χ2 = (observed value - expected value)² / expected value
For example, you wanted to give 15% of your students grade 5. You ended up grading 5 out of 60 students with this highest mark. 15% of 60 students is 9 students. Hence, the chi-square value for the highest mark is
χ2 = (5 - 9)² / 9 = (-4)² / 9 = 16/9 = 1.778.
## How to perform the chi-square test for the goodness of fit?
The test statistic for the goodness of fit chi-square test is the sum of chi-square values that appear in the distribution in question. Hence, in our case, for all possible marks.
Once you have this sum, you can use the : there are two different factors in such a table:
• Degrees of freedom: Number of levels of the analyzed variable minus one. For instance, in our grading example, we have four levels (grades 2, 3, 4, and 5). Hence, the number of degrees of freedom is equal to 3. If you're not yet familiar with this notion, take a look at the degrees of freedom calculator.
• Significance level: it is denoted as a subscript next to the symbol χ2. This is the probability of rejecting your hypothesis when it was in fact true. The higher this risk is, the more extreme data is accepted as still valid.
## Using chi-square calculator: an example
Let's analyze the situation with paper grading in greater detail. Let's say you have 60 students and graded them in the following way: five received a 5, twenty-two a 4, twenty-six a 3, and seven a 2.
1. Calculate how many students were expected to receive each grade according to the planned distribution.
• Grade 2: 0.15 * 60 = 9
• Grade 3: 0.30 * 60 = 18
• Grade 4: 0.40 * 60 = 24
• Grade 5: 0.15 * 60 = 9
1. Calculate the chi-square value for each grade.
• Grade 2: χ2 = (5 - 9)² / 9 = 1.778
• Grade 3: χ2 = (26 - 18)² / 18 = 3.556
• Grade 4: χ2 = (22 - 24)² / 24 = 0.167
• Grade 5: χ2 = (7 - 9)² / 9 = 0.444
1. Sum all of the chi-square values. χ2 = 1.778 + 3.556 + 0.167 + 0.444 = 5.945
2. As we've already explained, there are 3 degrees of freedom.
3. What is your significance level? We can assume that it is equal to 0.05.
4. You can now use the chi-square tables to determine whether our data can be accepted as consistent with the initial distribution. For 3 degrees of freedom and significance level 0.05, chi-square's maximum permissible value is 7.815. As χ2 = 5.945, we are in the acceptable range - the grading is not far off from the initial distribution. | 975 | 3,605 | {"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": 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} | 4.5625 | 5 | CC-MAIN-2024-33 | latest | en | 0.895614 |
https://www.mersenneforum.org/showthread.php?s=959c6b8a28470f8a18ec3a7b51d42dc5&t=8445 | 1,618,280,791,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038071212.27/warc/CC-MAIN-20210413000853-20210413030853-00635.warc.gz | 984,952,492 | 10,860 | mersenneforum.org Solve this equation
Register FAQ Search Today's Posts Mark Forums Read
2007-06-18, 22:36 #1 davar55 May 2004 New York City 23×232 Posts Solve this equation Solve for x: (x+9)^(1/3) - (x-9)^(1/3) = 3 (This may be too easy for this forum, but I've always liked the solution method.)
2007-06-18, 23:23 #2 ewmayer ∂2ω=0 Sep 2002 República de California 3·53·31 Posts x[sup]2[/sup]=80 Last fiddled with by ewmayer on 2007-06-18 at 23:23
2007-06-19, 07:50 #3
davieddy
"Lucan"
Dec 2006
England
145128 Posts
Quote:
Originally Posted by ewmayer x[sup]2[/sup]=80
Just as well this wasn't the homework help forum
2007-06-19, 12:20 #4
mfgoode
Bronze Medalist
Jan 2004
Mumbai,India
22×33×19 Posts
Solution!
Quote:
Originally Posted by davar55 Solve for x: (x+9)^(1/3) - (x-9)^(1/3) = 3 (This may be too easy for this forum, but I've always liked the solution method.)
Lets have your method davar. Thank you!
Mally
2007-06-19, 12:56 #5 Kees Dec 2005 22×72 Posts Taking cubes on both sides gives the result without any noteworthy manipulation. If there is an elegant solution or a smart substitution it still needs to be very short to beat the straightforward one
2007-06-19, 14:23 #6 VolMike Jun 2007 Moscow,Russia 8516 Posts Equation doesn't have solution(s)
2007-06-19, 15:40 #7
davieddy
"Lucan"
Dec 2006
England
2×3×13×83 Posts
Quote:
Originally Posted by Kees Taking cubes on both sides gives the result without any noteworthy manipulation.
The entertaining stage is:
-3[(x+9)1/3-(x-9)1/3](x2-81)1/3 = 9
at which point we use the original equation to substitute 3
for [...]
David
Last fiddled with by davieddy on 2007-06-19 at 15:42
2007-06-19, 15:51 #8
Mini-Geek
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
102538 Posts
Quote:
Originally Posted by VolMike Equation doesn't have solution(s)
$x=sqrt{80}$
2007-06-19, 21:14 #9 fetofs Aug 2005 Brazil 2·181 Posts Not quite. $x = \pm\sqrt{80}$
2007-06-20, 07:32 #10
mfgoode
Bronze Medalist
Jan 2004
Mumbai,India
22×33×19 Posts
Solutions!
Quote:
Originally Posted by davar55 Solve for x: (x+9)^(1/3) - (x-9)^(1/3) = 3 (This may be too easy for this forum, but I've always liked the solution method.)
A good problem but not instructional enough.
Its not good nor instructional getting straight solutions from the 'know all's'.
One right answer as ewmayer's is enough to tackle the problem
A hint from Keyes was good as it saved a lot of 'computation'. I have to use the term in these computer days as this is piffle for the comp.
As usual Davieddy's 'solution' and as he termed it 'entertainment' made things more confused and you can all see for your selves. Imagine substituting a whole equation for an integer ? And he claims to be a math's tutor !
A little knowledge is a very dangerous thing. Arrogance is even worse!
Now given that the solution is correct (I have not checked it out but accept Ewmayer's as correct) the next question is to generalise it from a particular problem.
My question is how would you solve the problem when the exponents are not equal? Lets say they are p and q and not just 1/3 ? How would you get the rationalising factor esp.
when there are three or more surds?
Thank you Davar55 as it made me revise my algebra!
BTW: if you have a better solution than keyes kindly respond and let me know.
Mally
Last fiddled with by mfgoode on 2007-06-20 at 07:35
2007-06-20, 08:03 #11
Chris Card
Aug 2004
2×5×13 Posts
Quote:
Originally Posted by mfgoode As usual Davieddy's 'solution' and as he termed it 'entertainment' made things more confused and you can all see for your selves. Imagine substituting a whole equation for an integer ? And he claims to be a math's tutor !
What are you on about Mally? Davieddy's solution is exactly the way to do it. The substitution he suggests is valid, and the simplest way I can see to get to the solution.
Chris
Similar Threads Thread Thread Starter Forum Replies Last Post JM Montolio A Miscellaneous Math 18 2018-03-09 15:32 ixfd64 Puzzles 4 2011-03-30 12:25 flouran Miscellaneous Math 23 2009-01-04 20:03 davar55 Puzzles 3 2008-10-09 00:35 nuggetprime Miscellaneous Math 1 2007-06-26 22:04
All times are UTC. The time now is 02:26.
Tue Apr 13 02:26:31 UTC 2021 up 4 days, 21:07, 1 user, load averages: 2.50, 2.50, 2.54 | 1,360 | 4,342 | {"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": 0, "wp_latex": 0, "mimetex.cgi": 2, "/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-2021-17 | latest | en | 0.811751 |
https://www.engram9.info/excel-2007-vba-methods/excels-builtin-matrix-functions.html | 1,590,420,684,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347388758.12/warc/CC-MAIN-20200525130036-20200525160036-00585.warc.gz | 684,285,541 | 6,210 | # Excels Builtin Matrix Functions
Performing matrix mathematics with Excel is very simple. Let's begin by assuming that the matrices A and B have been defined by selecting the 3R x 3C arrays of cells containing the values shown in Figure 3-1 and naming them by using Define Name. Remember, we're simply assigning a range name to a range of cells. We usually refer to it as a range or an array; the fact that we are calling it a matrix simply indicates what we intend to do with it.
B c 1 D 3 Matrix A 4 2 3 4 5 3 2 -1 6 . . . 3 7 F G H 3 Matrix 8 4 2 a 2 5 D 3 -3 S -3 -2 1
Figure 3-1. Ranges of cells defined as A and B. (folder 'Chapter 03 (Matrices) Examples, workbook 'Matrix Math', sheet 'Sheet 1')
Figure 3-1. Ranges of cells defined as A and B. (folder 'Chapter 03 (Matrices) Examples, workbook 'Matrix Math', sheet 'Sheet 1')
Addition or Subtraction. To add a constant (e.g., 3) to matrix A, simply select a range of cells the same size as the matrix, enter the formula =A+3, then press COMMAND+RETURN or CONTROL+SHIFT+RETURN (Macintosh) or CONTROL+SHIFT+ENTER (Windows). When you "array-enter" a formula by pressing e.g., CONTROL+SHIFT+ENTER, Excel puts braces around the formula, as shown below: {=A+3}
Do not type the braces; if you do, the result will not be recognized by Excel as a formula.
D E I F 8 5 6 7 9 6 5 2 10 7 6 10 J
Figure 3-2. Result matrix {A + 3}. (folder 'Chapter 03 (Matrices) Examples, workbook 'Matrix Math', sheet 'Sheet 1')
Figure 3-2. Result matrix {A + 3}. (folder 'Chapter 03 (Matrices) Examples, workbook 'Matrix Math', sheet 'Sheet 1')
Subtraction of a constant, multiplication or division by a constant, or addition of two matrices is performed in the same way by using standard Excel algebraic operators.
Scalar Multiplication. Scalar multiplication can be either multiplication of the elements of a matrix by a constant, e.g., a formula such as {=3*A}, or multiplication of corresponding elements of two matrices, e.g., {=A*B}. The result of the latter formula is shown in Figure 3-3.
D E F 16 4 0 8 17 0 6 3 18 -12 -6 7
Figure 3-3. Result matrix {A x B}. (folder 'Chapter 03 (Matrices) Examples, workbook 'Matrix Math', sheet 'Sheetl')
Figure 3-3. Result matrix {A x B}. (folder 'Chapter 03 (Matrices) Examples, workbook 'Matrix Math', sheet 'Sheetl')
Matrix multiplication can be accomplished easily by the use of Excel's worksheet function MMULT(matrixl, matrixZ). For the matrices A and B defined above, entering the formula =MMULT(A,B) yields the result shown in Figure 3-4 while the formula =MMULT(B,A) yields the result shown in Figure 3-5.
D E ; F j 24 -8 1 -1 25 9 8 -1 26 -13 -5 j 6
D E F 29 -3 -3 22 -24 30 -8 -10 -3
Figure 3-5. Result matrix BA. (folder 'Chapter 03 (Matrices) Examples, workbook 'Matrix Math', sheet 'Sheet 1')
Figure 3-5. Result matrix BA. (folder 'Chapter 03 (Matrices) Examples, workbook 'Matrix Math', sheet 'Sheet 1')
Matrix multiplication of two matrices is possible only if the matrices are conformable, that is, if the number of columns of A is equal to the number of rows of B. The opposite condition, if the number of rows of A is equal to the number of columns of B, is not equivalent. The following examples, involving multiplication of a matrix and a vector, illustrate the possibilities:
MMULT (4x3 matrix, 3 x 1 vector) = 3x1 result vector
MMULT (1 x 4 vector, 4x3 matrix) =1x4 result vector In other words, the two inner indices must be the same.
Transposition. The transpose of a matrix may be calculated by using the worksheet function TRANSPOSE(array) or obtained manually by using the Transpose option in the Paste Special... menu command.
The size of the array that can be transposed is limited only by the size of the Excel spreadsheet; the number of rows or columns cannot b£ greater than 256.
Matrix Inversion. The process for inverting a matrix "manually" (i.e., using pencil, paper and calculator) is complicated, but the operation can be carried out readily by using Excel's worksheet function MINVERSE(array). The inverse of the matrix B above is shown in Figure 3-6.
D E ! F 36 -0.25 -0.33333333 -0.5 37 0.75 0.E666B6B7 0.5 38 0.75 0.33333333 0.5
Figure 3-6. Result matrix B (folder 'Chapter 03 (Matrices) Examples, workbook 'Matrix Math', sheet 'Sheetl')
Figure 3-6. Result matrix B (folder 'Chapter 03 (Matrices) Examples, workbook 'Matrix Math', sheet 'Sheetl')
The size of the matrix must not exceed 52 rows by 52 columns.
Evaluation of the Determinant. The determinant of a matrix of TV rows x N columns can be obtained by using the worksheet function MDETERM(array).
The function returns a single numerical value, not an array, and thus you do not have to use CONTROL+SHIFT+ENTER. The value of the determinant of B, represented by |B|, is 12. | 1,339 | 4,784 | {"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 | 4 | CC-MAIN-2020-24 | latest | en | 0.839071 |
https://sites.google.com/site/mytechnicalcollection/algorithms/array/sorted-subsequence | 1,561,508,005,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999964.8/warc/CC-MAIN-20190625233231-20190626015231-00267.warc.gz | 592,013,080 | 26,559 | Algorithms > Array >
### Sorted Subsequence
Given an array of n integers, find the 3 elements such that a[i] < a[j] < a[k] and i < j < k in 0(n) time. If there are multiple such triplets, then print any one of them.
Examples:
```Input: arr[] = {12, 11, 10, 5, 6, 2, 30}
Output: 5, 6, 30
Input: arr[] = {1, 2, 3, 4}
Output: 1, 2, 3 OR 1, 2, 4 OR 2, 3, 4
Input: arr[] = {4, 3, 2, 1}
Output: No such triplet
```
Source: Amazon Interview Question
Hint: Use Auxiliary Space
Solution:
1) Create an auxiliary array smaller[0..n-1]. smaller[i] should store the index of a number which is smaller than arr[i] and is on left side of arr[i]. smaller[i] should contain -1 if there is no such element.
2) Create another auxiliary array greater[0..n-1]. greater[i] should store the index of a number which is greater than arr[i] and is on right side of arr[i]. greater[i] should contain -1 if there is no such element.
3) Finally traverse both smaller[] and greater[] and find the index i for which both smaller[i] and greater[i] are not -1.
`#include` `// A function to fund a sorted subsequence of size 3``void` `find3Numbers(``int` `arr[], ``int` `n)``{`` ``int` `max = n-1; ``//Index of maximum element from right side`` ``int` `min = 0; ``//Index of minimum element from left side`` ``int` `i;` ` ``// Create an array that will store index of a smaller`` ``// element on left side. If there is no smaller element`` ``// on left side, then smaller[i] will be -1.`` ``int` `*smaller = ``new` `int``[n];`` ``smaller[0] = -1; ``// first entry will always be -1`` ``for` `(i = 1; i < n; i++)`` ``{`` ``if` `(arr[i] <= arr[min])`` ``{`` ``min = i;`` ``smaller[i] = -1;`` ``}`` ``else`` ``smaller[i] = min;`` ``}` ` ``// Create another array that will store index of a`` ``// greater element on right side. If there is no greater`` ``// element on right side, then greater[i] will be -1.`` ``int` `*greater = ``new` `int``[n];`` ``greater[n-1] = -1; ``// last entry will always be -1`` ``for` `(i = n-2; i >= 0; i--)`` ``{`` ``if` `(arr[i] >= arr[max])`` ``{`` ``max = i;`` ``greater[i] = -1;`` ``}`` ``else`` ``greater[i] = max;`` ``}` ` ``// Now find a number which has both a greater number on`` ``// right side and smaller number on left side`` ``for` `(i = 0; i < n; i++)`` ``{`` ``if` `(smaller[i] != -1 && greater[i] != -1)`` ``{`` ``printf``(``"%d %d %d"``, arr[smaller[i]],`` ``arr[i], arr[greater[i]]);`` ``return``;`` ``}`` ``}` ` ``// If we reach number, then there are no such 3 numbers`` ``printf``(``"No such triplet found"``);` ` ``// Free the dynamically alloced memory to avoid memory leak`` ``delete` `[] smaller;`` ``delete` `[] greater;` ` ``return``;``}` `// Driver program to test above function``int` `main()``{`` ``int` `arr[] = {12, 11, 10, 5, 6, 2, 30};`` ``int` `n = ``sizeof``(arr)/``sizeof``(arr[0]);`` ``find3Numbers(arr, n);`` ``return` `0;``}`
Output:
`5 6 30 `
Time Complexity: O(n)
Auxliary Space: O(n)
Exercise:
1. Find a subsequence of size 3 such that arr[i] < arr[j] > arr[k].
2. Find a sorted subsequence of size 4 in linear time | 1,165 | 3,304 | {"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-2019-26 | latest | en | 0.686704 |
http://slideplayer.com/slide/3525739/ | 1,519,477,897,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891815560.92/warc/CC-MAIN-20180224112708-20180224132708-00624.warc.gz | 323,345,607 | 31,623 | # walking across a surface.
## Presentation on theme: "walking across a surface."— Presentation transcript:
walking across a surface.
Activating Strategy Think about a time where you have fallen walking across a surface. What made you fall ?
Forces
Essential Question: How do gravity, inertia, and friction affect the balance of forces?
Standard: S8P3b. Demonstrate the effect of balanced and unbalanced forces on an object in terms of gravity, inertia, and friction
What is a force? A force is a push or pull
Forces Forces have size and direction and are expressed in Newtons (N)
Force is always exerted by one object on another object Balanced forces produce no change in motion Unbalanced forces produce a change in motion
If you roll or kick a ball, what happens eventually?
Forces If you roll or kick a ball, what happens eventually? An unbalanced force is needed to change the speed of a moving object. So, what force is stopping the ball?
Friction is a force that opposes motion between two surfaces.
Forces: Friction Friction is a force that opposes motion between two surfaces.
Forces: Friction Friction occurs because the surface of any object is rough. Even surfaces that feel smooth are covered with microscopic hills and valleys.
Forces: Friction When two surfaces are in contact, the hills and valleys of one surface stick to the hills and valleys of the other surface.
Forces: Friction The amount of friction depends on factors such as roughness of the surfaces and the force pushing the surfaces together.
Turn to a partner and identify surfaces that may cause more friction.
Forces: Friction Turn to a partner and identify surfaces that may cause more friction. Sandpaper Carpet Gravel Less Friction? ICE Tile Glass
Friction: Friend or Foe?
Forces: Friction Friction is both harmful and helpful. Turn to an elbow partner and identify some examples of friction being harmful and friction being helpful.
Forces: Friction Breaks and tires on a car Falling Lighting a match
Rubbing your chest, arms, and legs to get warm
Forces: Friction
Is it beneficial to reduce or increase friction? Why or Why Not?
Forces: Friction Is it beneficial to reduce or increase friction? Why or Why Not?
Name three common items you might use to increase friction.
Forces: Friction Which of the following would NOT help you move a heavy object across a concrete floor? Water, ball bearings, oil, liquid soap, steel rods, foam rubber Name three common items you might use to increase friction.
Forces: Air Resistance
Air resistance is a form of friction. Look at the images below. Why would air resistance be a form of friction?
Forces: Air Resistance
Would a school bus or a racing car be affected less by air resistance? Why? Air resistance depends on the size, shape, and speed of the object.
Falling Water Demonstration
Teachers either demonstrate the activity, have small groups so the activity outside, or watch a short video clip of the activity. What force is acting on the cup and the water?
Gravity Gravity is a force of attraction between objects that is due to their masses. All objects experience an attraction toward all other objects. This gravitational force pulls objects toward each other. The attraction between objects will be discussed in more depth in the next essential question. For this essential question, we just need students to understand that all objects have the force of gravity pulling them toward the earth which keeps things in place unless an unbalanced force occurs.
Gravity Compared with all other objects around you, Earth has a huge mass. Therefore, Earth’s gravitational force is very large. You must apply forces to overcome Earth’s gravitational force any time you lift objects or even parts of your body.
Gravity Earth’s gravitational force pulls everything toward the center of Earth. Because of this force, the books, tables, and chairs in the room stay in place, and dropped objects fall to Earth rather than moving together toward you.
Forces Review There are many different forces which act on objects such as gravity and friction. If forces are balanced the object doesn’t move If forces are unbalanced the object will do one of the following things: Begin to move Stop moving Change direction Speed up Slow down
Forces occur in pairs and they can be either balanced or unbalanced
Balanced forces do not cause a change in motion.
When two forces acting on an object are equal in size, but are opposite in direction, we say the forces are balanced. Balanced forces do not cause a change in motion.
Balanced Forces
Force Diagrams We can show the forces acting on an object in a force diagram. In a force diagram, the force is shown with an arrow – the larger the arrow, the bigger the force The arrows in a force diagram also show the direction that the force is acting on Sometimes, the arrow will be labeled with the size of the force in Newtons (N). A larger number means a larger force.
Force Diagrams
Balanced Forces Water’s upward force Gravity
The ship is floating on the water, the forces are balanced. Which forces are acting on the ship? Because Gravity’s force down and the water’s force up are equal the ship stays afloat. Water’s upward force Gravity
Which forces are acting on the vase?
Balanced Forces The vase is resting on the table. It is not moving, therefore the forces must be balanced. Which forces are acting on the vase? Have a look around the classroom and name some of the balanced forces in action on different objects. Gravity Table’s Upward Force
Unbalanced Forces When two forces are acting on an object and are not equal in size, we say the forces are unbalanced. Unbalanced forces cause a change in motion by changing the object’s speed or direction.
Unbalanced Forces The man and the parachute are slowly falling to the ground. Which forces are acting on the parachute? Which force is bigger? How can you tell? Air resistance is pushing upwards Gravity is the bigger force. We can tell this because the direction of movement of the man and the parachute is the same as the direction of the force of gravity. Gravity is pulling the man to the ground
Unbalanced Forces
Unbalanced Forces Air resistance is pushing upwards The same forces are acting on the elephant, but without a parachute there is much less air resistance. Gravity is a much bigger force. Because there is a bigger difference between the force pulling downwards and the force pushing upwards, the elephant falls much more quickly than a man with a parachute. Gravity is pulling to the ground
Unbalanced Forces The paperclip is jumping up to the magnet.
The magnetic force is pulling the paperclip upwards The paperclip is jumping up to the magnet. Which forces are acting on the magnet? Which is the greater force? The magnetic force is greater than the force of gravity therefore the paperclip moves in the direction of the biggest force. Gravity is pulling the paperclip to the ground
Forces Each team is exerting a force – what happens if the pulling force produced by each team is equal? What would happen if the team on the left were able to produce a bigger pulling force than the team on the right?
Unbalanced Forces Based on the forces shown:
Which arm wrestler will likely win? Which direction will both arms move?
Which direction will the rope move?
Unbalanced Forces 4 N, left 10 N, right Which direction will the rope move?
Which direction will the ball roll?
Unbalanced Forces Which direction will the ball roll?
Jennifer weighs 175 pounds
Look at the diagram below. Both women are wearing ice skates on an ice rink. If both women push off from one another, which woman will most likely move the furthest? Why? Julie weighs 130 pounds Jennifer weighs 175 pounds Julie will move the furthest because Jennifer’s push will most likely have more force
Imagine that you are playing baseball
Imagine that you are playing baseball. The pitch comes in and you hit the ball hard. But, instead of flying off the bat, the ball just drops to the ground. Is that normal?
You would probably say no. You know that force and motion are related.
When you exert a force on a baseball by hitting it with a bat, the baseball should move.
Isaac Newton explained this relationship between force and the motion of an object.
Diving Egg Demonstration
https://www.youtube.com/watch?v=STQRUzalH2M What happened to the egg? Why?
What happens when riding bumper cars? Why?
Law of Inertia Inertia is the tendency of an object to resist being moved or, if the object is moving, to resist a change in speed or direction until an outside force acts on the object.
Law of Inertia Because of Inertia, an object at rest will stay at rest unless acted on by an unbalanced force. In other words, objects will not start moving until a push or a pull is exerted on them.
Law of Inertia
Law of Inertia An object in motion remains in motion unless acted on by an unbalanced force. The object will continue to move in the same direction with the same speed unless an unbalanced force occurs.
Mass and Inertia Mass is a measure of Inertia
An object that has a small mass has less inertia than an object that has a large mass So, changing the motion of an object that has a small mass is easier than changing the motion of an object that has a large mass.
The car because it has more mass therefore more inertia.
Mass and Inertia Look at the images below. Which would require more force to move or slow down? Why? The car because it has more mass therefore more inertia.
Mass and Inertia Which would require more force to move or slow down? Why? The bowling ball because it has more mass therefore more inertia.
The snowboard is in motion
The snowboard is in motion. Based on the Law of Inertia, the snowboard will keep moving right? What force(s) allow him to snowboard on the brick wall and rail?
Friction and the Law of Inertia
Objects in motion remain in motion unless acted on by an unbalanced force. Friction is often the unbalanced force that causes objects to slow down and stop moving. Because of friction, the motion of objects changes.
Newton’s First Law of Motion: Inertia
Formative Assessment Check
If a moose were chasing you through the woods, its enormous mass would be very threatening. But if you zigzagged, then its great mass would be to your advantage. Explain why.
Activities on Forces [see resources] | 2,171 | 10,389 | {"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} | 3.859375 | 4 | CC-MAIN-2018-09 | latest | en | 0.920621 |
http://fersch.de/beispiel?nr=AllDrei&nrform=GeoAllgDreiFlUA&datei=1007AllDreiGeoAllgDreiFlUA | 1,547,744,285,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583659056.44/warc/CC-MAIN-20190117163938-20190117185938-00064.warc.gz | 81,394,757 | 4,322 | Geometrie-Dreieck-Allgemeines Dreieck
Beispiel Nr: 07
$\text{Gegeben:}\\\text{Grundlinie} \qquad g \qquad [m] \\ \text{Höhe} \qquad h \qquad [m] \\ \\ \text{Gesucht:} \\\text{Fläche} \qquad A \qquad [m^{2}] \\ \\ A = \frac{g \cdot h}{2}\\ \textbf{Gegeben:} \\ g=1\frac{2}{3}m \qquad h=\frac{4}{5}m \qquad \\ \\ \textbf{Rechnung:} \\ A = \frac{g \cdot h}{2} \\ g=1\frac{2}{3}m\\ h=\frac{4}{5}m\\ A = \frac{1\frac{2}{3}m \cdot \frac{4}{5}m}{2}\\\\A=\frac{2}{3}m^{2} \\\\\\ \small \begin{array}{|l|} \hline g=\\ \hline 1\frac{2}{3} m \\ \hline 16\frac{2}{3} dm \\ \hline 166\frac{2}{3} cm \\ \hline 1666\frac{2}{3} mm \\ \hline 1666666\frac{2}{3} \mu m \\ \hline \end{array} \small \begin{array}{|l|} \hline h=\\ \hline \frac{4}{5} m \\ \hline 8 dm \\ \hline 80 cm \\ \hline 800 mm \\ \hline 8\cdot 10^{5} \mu m \\ \hline \end{array} \small \begin{array}{|l|} \hline A=\\ \hline \frac{2}{3} m^2 \\ \hline 66\frac{2}{3} dm^2 \\ \hline 6666\frac{2}{3} cm^2 \\ \hline 666666\frac{2}{3} mm^2 \\ \hline 0,00667 a \\ \hline 6,67\cdot 10^{-5} ha \\ \hline \end{array}$ | 523 | 1,059 | {"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} | 4.34375 | 4 | CC-MAIN-2019-04 | latest | en | 0.111336 |
https://discuss.codecademy.com/t/tip-calculator-lesson-3/48020 | 1,548,277,869,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547584350539.86/warc/CC-MAIN-20190123193004-20190123215004-00177.warc.gz | 496,710,304 | 4,749 | # Tip Calculator: Lesson 3
#1
its not behaving weirdly
"Oops it seam "tip" is set to 0 instead of 0.15"
just that why cant we say 15/100, instead we have type "15.0/100"
is it a must to put decimals??
``tip=15/100``
#2
In Python 2.7, `15/100` is zero. Two integers in a division expression return an integer, and 0 is the only one available in this case.
``````>>> float(15) / 100
0.15``````
#3
How come? Isnt 15 an integer? If not, sorry but I still don't get why is it a must to have decimals
#4
Because of implicit type casting of number objects. When we give Python, `15`, we give it an integer. Implicitly, this is type-cast as `int`. Give it `100` and again, an `int`. The result of the division will be `int`, which is `0`.
#5
ok I see then what is 15.0 exactly?
an int or something else?
#6
ok so I read other posts and from I see 15.0 is something like a float value?
Uh well can you please explain "float values"? Thanks and cheers mate
#7
Now we're talking. The value has a type.
Once we explicitly declare one value as a float, it matters nor what type the other value is. The outcome will be a float.
What is a float? Anything that can be written as, `0.X` where `X` is any uninterrupted number sequence. PI minus 3 is an example of a float.
#8
okay but what is float value?
#9
Late edit. Go back a post for reply.
#10
okay thx
I realised something: 15 will be considered an int. 15.0, though in mathematical terms is an integer, but it has to be specified to be "float value" so that the results we get is "float value" too right? Because when 15/100 is keyed in we get integer values but when 15.0/100 is in we get the float value aka the exact value am I right?
#11
This topic was automatically closed 7 days after the last reply. New replies are no longer allowed. | 509 | 1,808 | {"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} | 3.75 | 4 | CC-MAIN-2019-04 | latest | en | 0.901042 |
https://www.vedantu.com/question-answer/cards-marked-with-numbers-3-4-5-50-are-placed-in-class-10-maths-icse-5ee74e274582936f5421c080 | 1,620,308,622,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988753.97/warc/CC-MAIN-20210506114045-20210506144045-00417.warc.gz | 1,103,350,684 | 83,417 | Question
# Cards marked with numbers 3, 4, 5, …...50 are placed in a box and mixed through. One card is drawn from the box. Find the probability that the number on the card is(i) Divisible by 6(ii) A single-digit number
Hint: To solve the question, we have to calculate the number of possible outcomes at the given conditions. To calculate probability, divide the obtained number of outcomes to the total number of outcomes obtained when a card is drawn.
The numbers on the cards placed in a box given are 3, 4, 5, …...50
The number of possible outcomes when a card is drawn from the above-mentioned cards
= 50 – 2 = 48
Since, only 1, 2 numbers have not marked the cards considering the series of natural numbers.
The divisibility rule of 6 is that the number should be divisible by both numbers 2 and 3. The number is divisible by 2 if it is an even number. The number is divisible by 3 when the sum of all the digits of the number is a multiple of 3.
Since, the marked numbers are in the range of the first 10 multiples of 6. We have to list the multiples of 6 below 50. Thus, we get 6, 12, ,18, 24, 30, 36, 42, 48 are the multiples of 6 that are marked on the given cards.
Thus, the number of outcomes of getting a card marked number which is divisible by 6 when a card is drawn = 8.
The probability of getting a card marked number which is divisible by 6 when a card is drawn = Ratio of the number of outcomes of getting a card marked number which is divisible by 6 when a card is drawn to the number of possible outcomes when a card is drawn from the above-mentioned cards
$=\dfrac{8}{48}$
$=\dfrac{1}{6}$
$\therefore$ The probability of getting a card marked number which is divisible by 6 when a card is drawn $=\dfrac{1}{6}=0.16$
3, 4, 5, 6, 7, 8, 9 are the single-digit numbers among the numbers marked on the cards when a card is drawn.
Thus, the number of outcomes of getting a single-digit number marked card when a card is drawn
= 7
The probability of getting a single-digit number marked card when a card is drawn = Ratio of the number of outcomes of getting a single-digit number marked card when a card is drawn, to the number of possible outcomes when a card is drawn from the above-mentioned cards
$=\dfrac{7}{48}$
$\therefore$ The probability of getting a single-digit number marked card when a card is drawn $=\dfrac{7}{48}=0.146$
Note: The possibility of mistake can be not able to analyse the given condition independently and apply the quick method of calculating the number of outcomes instead of the general approach since the given marked numbers are below 50 which ease the calculation. | 663 | 2,635 | {"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} | 4.84375 | 5 | CC-MAIN-2021-21 | longest | en | 0.947836 |
http://math.stackexchange.com/questions/231944/area-of-triangle-abc-inside-circle | 1,469,814,851,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257831770.41/warc/CC-MAIN-20160723071031-00094-ip-10-185-27-174.ec2.internal.warc.gz | 160,229,598 | 17,891 | # Area of triangle ABC inside circle
Consider the following diagram:
$AB+AD=DE$, $\angle BAD= 60$, and $AE$ is $6$. How do we find the area of the triangle $ABC$?
-
Is $AE$ the diameter of the circle? – Sumit Bhowmick Nov 7 '12 at 5:31
@Sumit Yes, it is. – George Krasilnikov Nov 7 '12 at 5:38
please find the answer. – Sumit Bhowmick Nov 7 '12 at 14:08
@Sumit is what you found not the answer? – George Krasilnikov Nov 7 '12 at 14:13
From the above picture,
$x+y=z$ , $y+z=6$ and $\frac x {y+z} = \cos 60^\circ = \frac 1 2$
After calculation, $x=3$, $y=\frac 3 2$ and $z = \frac 9 2$
$\angle ABE = 90^\circ$ and $\angle BAD = 60^\circ$
So, $\angle AEB = 30^\circ = \angle ACB$ (properties of a circle)
Now, $\cos 60^\circ = \frac {x^2 + y^2 - w^2}{2xy} = \frac 1 2$
After calculation, $w=DB=\frac {3 \sqrt 3}{2}$
$$\cos \angle ADB=\frac{y^2+w^2-x^2}{2yw}=\frac{(\frac 3 2)^2+(\frac {3 \sqrt 3}{2})^2-3^2}{2 \frac 3 2\frac {3 \sqrt 3}{2}}=0$$
So, $\angle ADB = 90^\circ$ and $\angle ABD = 30^\circ = \angle ACB$
So, $\triangle ABC$ is an isosceles triangle.
And, $CD = BD = \frac {3 \sqrt 3}{2}$
So, $\triangle ABC = \frac 1 2 \cdot \frac 3 2 \cdot (\frac {3 \sqrt 3}{2} + \frac {3 \sqrt 3}{2}) = \frac {9 \sqrt 3}{4}$
-
@George Krasilnikov please find the answer. – Sumit Bhowmick Nov 7 '12 at 8:45
@Sumit... is what you gave not the answer? – George Krasilnikov Nov 7 '12 at 14:11
@GeorgeKrasilnikov, yes this is my answer. – Sumit Bhowmick Nov 8 '12 at 4:48 | 610 | 1,476 | {"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} | 4.53125 | 5 | CC-MAIN-2016-30 | latest | en | 0.613658 |
http://metamath.tirix.org/mpeuni/diftpsn3 | 1,720,860,996,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514493.69/warc/CC-MAIN-20240713083241-20240713113241-00163.warc.gz | 19,034,030 | 2,157 | # Metamath Proof Explorer
## Theorem diftpsn3
Description: Removal of a singleton from an unordered triple. (Contributed by Alexander van der Vekens, 5-Oct-2017) (Proof shortened by JJ, 23-Jul-2021)
Ref Expression
Assertion diftpsn3 ( ( 𝐴𝐶𝐵𝐶 ) → ( { 𝐴 , 𝐵 , 𝐶 } ∖ { 𝐶 } ) = { 𝐴 , 𝐵 } )
### Proof
Step Hyp Ref Expression
1 disjprsn ( ( 𝐴𝐶𝐵𝐶 ) → ( { 𝐴 , 𝐵 } ∩ { 𝐶 } ) = ∅ )
2 disj3 ( ( { 𝐴 , 𝐵 } ∩ { 𝐶 } ) = ∅ ↔ { 𝐴 , 𝐵 } = ( { 𝐴 , 𝐵 } ∖ { 𝐶 } ) )
3 1 2 sylib ( ( 𝐴𝐶𝐵𝐶 ) → { 𝐴 , 𝐵 } = ( { 𝐴 , 𝐵 } ∖ { 𝐶 } ) )
4 3 eqcomd ( ( 𝐴𝐶𝐵𝐶 ) → ( { 𝐴 , 𝐵 } ∖ { 𝐶 } ) = { 𝐴 , 𝐵 } )
5 difid ( { 𝐶 } ∖ { 𝐶 } ) = ∅
6 5 a1i ( ( 𝐴𝐶𝐵𝐶 ) → ( { 𝐶 } ∖ { 𝐶 } ) = ∅ )
7 4 6 uneq12d ( ( 𝐴𝐶𝐵𝐶 ) → ( ( { 𝐴 , 𝐵 } ∖ { 𝐶 } ) ∪ ( { 𝐶 } ∖ { 𝐶 } ) ) = ( { 𝐴 , 𝐵 } ∪ ∅ ) )
8 df-tp { 𝐴 , 𝐵 , 𝐶 } = ( { 𝐴 , 𝐵 } ∪ { 𝐶 } )
9 8 difeq1i ( { 𝐴 , 𝐵 , 𝐶 } ∖ { 𝐶 } ) = ( ( { 𝐴 , 𝐵 } ∪ { 𝐶 } ) ∖ { 𝐶 } )
10 difundir ( ( { 𝐴 , 𝐵 } ∪ { 𝐶 } ) ∖ { 𝐶 } ) = ( ( { 𝐴 , 𝐵 } ∖ { 𝐶 } ) ∪ ( { 𝐶 } ∖ { 𝐶 } ) )
11 9 10 eqtr2i ( ( { 𝐴 , 𝐵 } ∖ { 𝐶 } ) ∪ ( { 𝐶 } ∖ { 𝐶 } ) ) = ( { 𝐴 , 𝐵 , 𝐶 } ∖ { 𝐶 } )
12 un0 ( { 𝐴 , 𝐵 } ∪ ∅ ) = { 𝐴 , 𝐵 }
13 7 11 12 3eqtr3g ( ( 𝐴𝐶𝐵𝐶 ) → ( { 𝐴 , 𝐵 , 𝐶 } ∖ { 𝐶 } ) = { 𝐴 , 𝐵 } ) | 783 | 1,153 | {"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} | 3.546875 | 4 | CC-MAIN-2024-30 | latest | en | 0.139226 |
https://oeis.org/A009843 | 1,571,716,319,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987798619.84/warc/CC-MAIN-20191022030805-20191022054305-00086.warc.gz | 627,549,163 | 5,892 | This site is supported by donations to The OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A009843 E.g.f. x/cos(x) (odd powers only). 26
1, 3, 25, 427, 12465, 555731, 35135945, 2990414715, 329655706465, 45692713833379, 7777794952988025, 1595024111042171723, 387863354088927172625, 110350957750914345093747, 36315529600705266098580265, 13687860690719716241164167451, 5858139922124796551409938058945 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 COMMENTS Expanding x/cosh(x) gives alternated signed values at odd positions. Related to the formulas sum(k>0,sin(kx)/k^(2n+1))=(-1)^(n+1)/2*x^(2n+1)/(2n+1)!*sum(i=0,2n,(2Pi/x)^i*B(i)*C(2n+1,i)) and if x=Pi/2 sum(k>0,(-1)^(k+1)/k^(2n+1))=(-1)^n*E(2n)*Pi^(2n+1)/2^(2n+2)/(2n)!. - Benoit Cloitre, May 01 2002 LINKS R. P. Brent, Generalising Tuenter's binomial sums, arXiv:1407.3533 [math.CO], 2014. R. B. Brent, Generalizing Tuenter's Binomial Sums, J. Int. Seq. 18 (2015) # 15.3.2. Peter Luschny, The lost Bernoulli numbers. Wikipedia, Bernoulli Polynomials FORMULA a(n) = (2n+1)*A000364(n) = sum(i=0, 2n, B(i)*C(2n+1, i)*4^i)=(2n+1)*E(2n) where B(i) are the Bernoulli numbers, C(2n, i) the binomial numbers and E(2n) the Euler numbers. - Benoit Cloitre, May 01 2002 Recurrence: a(n) = -(-1)^n*Sum[i=0..n-1, (-1)^i*a(i)*C(2n+1, 2i+1) ]. - Ralf Stephan, Feb 24 2005 a(n) = 4^n |E_{2n}(1/2)+E_{2n}(1)| (2n+1) for n > 0; E_{n}(x) Euler polynomial. - Peter Luschny, Nov 25 2010 a(n) = (2*n+1)! * [x^(2*n+1)] x/cos(x). From Sergei N. Gladkovskii, Nov 15 2011, Oct 19 2012, Nov 10 2012, Jan 14 2013, Apr 10 2013, Oct 13 2013, Dec 01 2013: (Start) Continued fractions: E.g.f.: x / cos(x) = x+x^3/Q(0); Q(k) = 8k+2-x^2/(1+(2k+1)*(2k+2)/Q(k+1)). E.g.f.: x + x^3/U(0) where U(k) = (2*k+1)*(2*k+2) - x^2 + x^2*(2*k+1)*(2*k+2)/U(k+1). G.f.: x/T(0) where T(k) = 1 + x^2 - x^2*(2*k+2)^2/(1 - x^2*(2*k+2)^2/T(k+1). G.f.: 1/G(0) where G(k) = 1 - x*(8*k^2+8*k+3)-16*x^2*(k+1)^4/G(k+1). E.g.f.: 2*x/(Q(0) + 1) where Q(k)= 1 - x/(2*k+1)/(2*k+2)/(1 - 1/(1 + 1/Q(k+1))). Let A(x) = S_{n>=0}a(n)*x^n/(2*n+1)! then A(x) = 1 + Q(0)*x/(2-x) where Q(k) = 1 - x*(2*k+1)*(2*k+2)/(x*(2*k+1)*(2*k+2) + ((2*k+1)*(2*k+2) - x)*((2*k+3)*(2*k+4) - x)/Q(k+1)). G.f.: T(0)/(1-3*x) where T(k) = 1 - 16*x^2*(k+1)^4/(16*x^2*(k+1)^4 - (1 - x*(8*k^2 +8*k+3)) *(1 - x*(8*k^2+24*k+19))/T(k+1)). G.f.: 1/(T(0) where T(k) = 1 + x - x*(2*k+2)^2/(1 - x*(2*k+2)^2/T(k+1). (End) a(n) = (-1)^n*2^(4*n+1)*(2*n+1)*(zeta(-2*n,1/4)-zeta(-2*n,3/4)). - Peter Luschny, Jul 22 2013 From Peter Bala, Mar 02 2015: (Start) a(n) = (-1)^(n+1)*4^(2*n + 1)*B(2*n + 1,1/4), where B(n,x) denotes the n-th Bernoulli polynomial. Cf. A002111, A069852 and A069994. Conjecturally, a(n) = the unsigned numerator of B(2*n+1,1/4). G.f. for signed version of sequence: Sum_{n >= 0} { 1/(n + 1) * Sum_{k = 0..n} (-1)^k*binomial(n,k)/( (1 - (4*k + 1)*x)*(1 - (4*k + 3)*x) ) } = 1 - 3*x^2 + 25*x^4 - 427*x^6 + .... (End) a(n) ~ (2*n+1)! * 2^(2*n+2)/Pi^(2*n+1). - Vaclav Kotesovec, Jul 04 2016 G.f.: 1/(1 + x - 4*x/(1 - 4*x/(1 + x - 16*x/(1 - 16*x/(1 + x - 36*x/(1 - 36*x/(1 + x - ...))))))). Cf. A005439. - Peter Bala, May 07 2017 EXAMPLE x/cos(x) = x + 1/2*x^3 + 5/24*x^5 + 61/720*x^7 + 277/8064*x^9 + ... MAPLE seq((2*i+1)!*coeff(series(x/cos(x), x, 32), x, 2*i+1), i=0..13); A009843 := n -> (-1)^n*(2*n+1)*euler(2*n): # Peter Luschny MATHEMATICA c = CoefficientList[Series[1/MittagLefflerE[2, z^2], {z, 0, 40}], z]; Table[(-1)^n* Factorial[2*n+1]*c[[2*n+1]], {n, 0, 16}] (* Peter Luschny, Jul 03 2016 *) PROG (PARI) for(n=0, 25, print1(sum(i=0, 2*n+1, binomial(2*n+1, i)*bernfrac(i)*4^i), ", ")) (PARI) a(n)=subst(bernpol(2*n+1), 'x, 1/4)*4^(2*n+1)*(-1)^(n+1) \\ Charles R Greathouse IV, Dec 10 2014 (Python) # The objective of this implementation is efficiency. # n -> [a(0), a(1), ..., a(n)] for n > 0. def A009843_list(n): ....S = [0 for i in range(0, n+1)] ....S[0] = 1 ....for k in range(1, n+1): ........S[k] = k*S[k-1] ....for k in range(1, n+1): ........for j in range(k, n+1): ............S[j] = (j-k)*S[j-1]+(j-k+1)*S[j] ........S[k] = (2*k+1)*S[k] ....return S print(A009843_list(10)) # Peter Luschny, Aug 09 2011 CROSSREFS Bisection of A009391, A009392, A065619, A083008. Cf. A099028. Cf. A002111, A069852, A069994. Sequence in context: A074708 A323217 A160143 * A182962 A223076 A272482 Adjacent sequences: A009840 A009841 A009842 * A009844 A009845 A009846 KEYWORD nonn,easy AUTHOR EXTENSIONS Extended and signs tested by Olivier Gérard, Mar 15 1997 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified October 21 22:47 EDT 2019. Contains 328315 sequences. (Running on oeis4.) | 2,225 | 4,846 | {"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} | 3.609375 | 4 | CC-MAIN-2019-43 | latest | en | 0.539385 |
https://www.enotes.com/homework-help/complete-squares-400276 | 1,490,597,234,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189462.47/warc/CC-MAIN-20170322212949-00498-ip-10-233-31-227.ec2.internal.warc.gz | 862,164,640 | 9,908 | # Complete the squaresx^2-14x+_ 9x^2-30x+_ 25x^2+_+36y^2 9x^4/25-_+25x^2/9 x^2+_+1/4
giorgiana1976 | College Teacher | (Level 3) Valedictorian
Posted on
We'll recall the identities that helps us to complete the given squares:
(a+b)^2 = a^2 + 2ab + b^2
(a-b)^2 = a^2 - 2ab + b^2
We'll complete the first square
x^2-14x+_
We'll identify a^2 = x^2 => a = x
To calculate b, we'll consider the second term of the square:
-14x = -2*a*b
-14x = -2*x*b
We'll use the symmetric property and we'll divide by 2x:
b = -7
Now, we'll complete the square by adding the amount b^2.
b^2 = (-7)^2
b^2 = 49
(a+b)^2 = (x-7)^2
The missing term in the quadratic expression is 49:
(x-7)^2 = x^2 - 14x + 49
To complete the 2nd aquare we'll identify what is the missing term and it is b^2.
We'll identify a^2 = 9x^2 => a = sqrt 9x^2 => a = 3x
9x^2-30x+_
To calculate b, we'll consider the second term of the square:
-30x = -2*a*b
-30x = -2*3x*b
-30x = -6x*b
We'll divide by 6x:
b = -5 => b^2 = 25
The missing term in the quadratic expression is 25 and the completed square will be:
(3x-5)^2 = 9x^2-30x+25
We'll complete the 3rd square. We notice that the missing term is 2ab.
We'll identify a^2 = 25x^2 => a = sqrt 25x^2 => a = 5x
b^2 = 36y^2 => b = sqrt 36y^2 => b = 6y
25x^2+_+36y^2
2ab = 2*5x*6y
2ab = 60xy
The missing term in the quadratic expression is 60xy and the completed square will be:
(5x+6y)^2 = 25x^2 + 60xy + 36y^2
We'll complete the 4th square. We'll identify the missing term as 2ab.
We'll identify a^2 = 9x^4/25 => a = sqrt 9x^4/25 => a = 3x^2/5
b^2 = 25x^2/9 => b = sqrt 25x^2/9 => b = -5x/3
9x^4/25-_+25x^2/9
-2ab = -2*3x^2*5x/5*3
-2ab = -2x^3
The missing term in the quadratic expression is -2x^3 and the completed square will be:
(3x^2/5 - 5x/3)^2 = 9x^4/25- 2x^3 + 25x^2/9
We'll complete the fimal square. We notice that the missing term is 2ab.
We'll identify a^2 = x^2 => a = sqrt x^2 => a = x
b^2 = 1/4 => b = sqrt 1/4 => b = 1/2
x^2+_+1/4
2ab = 2x/2
2ab = x
The missing term in the quadratic expression is x and the completed square will be:
(x + 1/2)^2 = x^2+ x +1/4 | 901 | 2,122 | {"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} | 4.4375 | 4 | CC-MAIN-2017-13 | longest | en | 0.844206 |
https://www.bankersadda.com/p/var-m11-var-s0var-m2var-qfunction.html | 1,560,712,446,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998291.9/warc/CC-MAIN-20190616182800-20190616204800-00082.warc.gz | 673,492,302 | 46,352 | Directions (1-5): Study the information and answer the given questions:
Eight employees A, B, C, D, E, F, G and H will have to take the leave in the months of January, March, July and November but not necessarily in the same order. In each month, they all will take a leave on either 8th or 19th of the given month. Not more than two employees will take a leave on the same month. Only one employee will take a leave on each day.
Only one person will take a leave before the G. More than three persons will take a leave between G and A. Only one person will take a leave between A and H. Two persons will take a leave between H and B. Only one person will take a leave between B and F. Two persons will take a leave between F and C. D will take a leave one of the days after E. Neither D nor E will take a leave on 19th of the any of the given month. Q1. Who among the following will take a leave on 8th January?
A
C
G
F
B
Solution:
Q2. Who among following will take a leave on 8th November?
A
C
G
D
B
Solution:
Q3. H will take a leave on which date and month?
8th March
19th March
8th July
19th July
None of these
Solution:
Q4. Who among following will take a leave on March month?
A, C
F, B
G, H
H, C
B, C
Solution:
Q5. How many persons will take a leave between F and E?
One
No one
Three
Two
Four
Solution:
Q6. The ratio of speeds of a motorboat to that of the current of water is 36 : 5. The motorboat goes along with the current in 5 hours 10 min. Find the time taken by motorboat to come back.
5 hours 50 min
6 hours
6 hours 50 min
12 hours 10 min
None of these
Solution:
Q7. Ishwar is rowing a boat. He takes half time in moving a certain distance downstream than upstream. What is the ratio of the rate of boat in still water to the rate of current?
2 : 1
5 : 1
7 : 1
3 : 1
1 : 3
Solution:
Q8. Kamal can row a certain distance downstream in 12 hours and can return the same distance in 18 hours. If the stream flows at the rate of 6 kmph, then find the speed of Kamal in still water.
30 kmph
35 kmph
28 kmph
25 kmph
22 kmph
Solution:
Q9. A boatman rows 1 km in 5 min along the stream and 6 km in 1 hours against the stream. The speed of the stream is:
3 kmph
6 kmph
10 kmph
12 kmph
14 kmph
Solution:
Q10. Sameer can row a certain distance downstream in 24 hours and can come back covering the same distance in 36 hours. If the stream flows at the rate of 12 kmph, find the speed of Sameer in still water.
30 kmph
15 kmph
40 kmph
60 kmph
50 kmph
Solution:
Directions (11-15): In each of the questions given below a sentence is given which is divided into 5 parts. It is then followed by 5 options which gives the sequence of the correct parts. Choose the option which gives the correct sequence of the grammatically correct parts. If all the parts are grammatically correct or if the correct sequence is not given in the option choose option (e) as the correct choice.
Q11. Time magazine dedicated(A)/ its person of the year(B)/ cover to women who broke the silence surrounding(C)/ the pervasiveness of sexually harassment and violence(D), especially in the workplace(E).
ABCD
ACDE
CBDE
ABCE
None of these
Solution:
The error lies in the part (d) as the word 'sexual harassment' is the correct usage instead of 'sexually harassment'.
Q12. These experiences have(A)/ an insidious effect on women’s lives(B)/ and ability to participation(C)/ in educational, work and(D)/ social activities(E).
ABDE
ACDE
ABDE
BCDE
None of these
Solution:
The correct usage is 'participate' instead of 'participation' as there should be use of Ist form of verb after 'to'.
Q13. What mattered to them was that(A)/ Lenin had inspired the(B)/ Communist Party of India(C)/ (Marxist)-led government in that northeastern State, which(D)/ have un-seated after an un-broken 25 years of ‘red rule’(E).
ABCD
BCDE
ACDE
ABDE
None of these
Solution:
The error lies in the part (E) as 'have' should be replace with 'had been' because the sentence is in the past tense.
Q14. The new Chief Minister, too,(A)/ deserves appreciation for his expression of disapproval(B)./ Those who said things that seemed to explain(C)/ away the dhakka will be hopefully be sobered(D)/ by the Prime Minister’s admonition(E).
ACDE
BCDE
ABCD
ABDE
None of these
Solution:
The given sentence is grammatically correct.
Q15. It says to Presidents and Governors(A)/ what they should or should not do(B)/, how they should and should not act(C)/ when scrutinized results to see(D)/ who should be called to form governments(E).
ABDE
ABCD
ABCE
ACDE
None of these
Solution:
The error lies in part (D) as 'scrutinizing' will be used in place of 'scrutinized'. | 1,248 | 4,615 | {"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} | 3.9375 | 4 | CC-MAIN-2019-26 | latest | en | 0.954499 |
https://titanlab.org/2009/12/11/tackling-polynomial-equations-with-python-numpy/ | 1,685,706,959,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648635.78/warc/CC-MAIN-20230602104352-20230602134352-00104.warc.gz | 611,849,563 | 28,241 | # Tackling Polynomial Equations with Python (Numpy)
How to calculate a polynomial of (s+1)(s+2)…(s+10) in a short time?
Instead of calculate them one by one, we can do it with the help of Python library – Numpy.
1. Make sure you have installed Numpy for your Python interpreter. In Ubuntu, you may just want to install python-scipy as it contains all the scientific calculation tools.
2. Try the following example in your python interpreter:
```from numpy import *
p1 = poly1d(1, 1 ) #(S+1)
p2 = poly1d(1, 2 ) #(S+2)
p3 = poly1d(1, 3 ) #(S+3)
p4 = poly1d(1, 4 ) #(S+4)
p5 = poly1d(1, 5 ) #(S+5)
p6 = poly1d(1, 6 ) #(S+6)
p7 = poly1d(1, 7 ) #(S+7)
p8 = poly1d(1, 8 ) #(S+8)
p9 = poly1d(1, 9 ) #(S+9)
p10 = poly1d(1, 10) #(S+10)
print p1*p2*p3*p4*p5*p6*p7*p8*p9*p10
```
3. Then, you will an get an output like:
``` 10 9 8 7 6 5 4 3 2
1 x - 10 x + 45 x - 120 x + 210 x - 252 x + 210 x - 120 x + 45 x - 10 x + 1
```
``` x^10 - 10x^9 +45x^8 - 120x^7 + 210x^6
- 252x^5 + 210x^4 - 120x^3 + 45x^2 - 10x + 1```
4. Say, we need to find the root of this polynomial. With the help of Python, we can get the roots by feeding these line into the interpreter.
```(p1*p2*p3*p4*p5*p6*p7*p8*p9*p10).r #.r means root.
```
And, we will get an output like this:
```array([ 1.04405137+0.0143565j , 1.04405137-0.0143565j ,
1.02709855+0.03755416j, 1.02709855-0.03755416j,
0.99974141+0.04619473j, 0.99974141-0.04619473j,
0.97275208+0.03706564j, 0.97275208-0.03706564j,
0.95635659+0.01406697j, 0.95635659-0.01406697j])
```
The polynomial has 5 complex conjugates: 1.04405137 +/- 0.0143565j, 1.02709855 +/- 0.03755416j, 0.99974141+/- 0.04619473j, 0.97275208 +/-0.03706564j, 0.97275208 +/- 0.03706564j and 0.95635659 +/- 0.01406697j . | 797 | 1,784 | {"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} | 3.96875 | 4 | CC-MAIN-2023-23 | latest | en | 0.601141 |
https://testbook.com/question-answer/what-is-the-area-in-sq-cm-of-a-rectangle-if-its--6001daae3c9da9029fe431d2 | 1,643,219,260,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304959.80/warc/CC-MAIN-20220126162115-20220126192115-00436.warc.gz | 620,692,745 | 37,689 | # What is the area (in sq cm) of a rectangle if its diagonal is 53 cm and length is 45 cm?
1. 1840 cm2.
2. 1260 cm2.
3. 1680 cm2.
4. 1540 cm2.
Option 2 : 1260 cm2.
## Detailed Solution
Given:
The diagonal of the rectangle = 53 cm
The length of the rectangle (l) = 45 cm
Formula used:
The area of a rectangle = l × b
Breadth of the rectangle (b) = √(d2 – l2)
Where l → length
d → diagonal
Calculation:
The breadth of the rectangle (b) = √(d2 – l2)
⇒ The breadth of the rectangle (b) = √(532 – 452) = √784
b = 28 cm
Area of rectangle = l × b = 28 × 45
⇒ Area of rectangle = 1260 cm2.
∴ The area of the rectangle is 1260 cm2. | 244 | 642 | {"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} | 4.21875 | 4 | CC-MAIN-2022-05 | latest | en | 0.564046 |
http://math.stackexchange.com/questions/210695/how-to-find-the-largest-rectangle-inside-an-ellipse | 1,448,521,099,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398446535.72/warc/CC-MAIN-20151124205406-00166-ip-10-71-132-137.ec2.internal.warc.gz | 150,016,481 | 18,241 | # How to find the largest rectangle inside an ellipse
I have an ellipse that is defined by center, width and height. The axes of the ellipse parallel to the x and y. I want to find the largest rectangle that completely fits inside this ellipse. Is there an easy way to do this?
And sorry if my terminology is a bit off... it's been a long time since I last tackled with geometry
-
sorry. edited the question. – vainolo Oct 10 '12 at 21:14
What do you mean by "largest" - area or perimeter? – marty cohen Oct 26 at 22:08
The naive thing is to see the ellipse as a stretched circle. We know the size of the largest rectangle that fits in a circle of radius $r$ is a square of side ${\sqrt 2}r$. If the semi-axes are $a$ and $b$ we would expect the rectangle to be ${\sqrt 2}a \times {\sqrt 2}b$.
To prove this, let us put the center at the origin. The equation of the ellipse is $\left(\frac xa\right)^2+\left(\frac yb\right)^2=1$. The area of the rectangle is $4xy,$ where $(x,y)$ is a point on the ellipse and we choose the point to maximize it. The area is then $4xb\sqrt{1-\left(\frac xa\right)^2}$. The derivative of this is $\frac {4b(a^2-2x^2)}{\sqrt{1-\left(\frac xa\right)^2}}$ which is duly zero at $x=\frac a{\sqrt 2}$, giving a side of $\sqrt 2 a$
-
Sorry, but I lost you at "to prove this"... why do I need to prove things to calculate a rectangle? – vainolo Oct 10 '12 at 21:30
@vainolo: I wanted to prove that the rectangle cited in the first paragraph is in fact the largest rectangle that fits in the ellipse. If you believe it from the first paragraph, you don't need to prove it. – Ross Millikan Oct 10 '12 at 21:39
OK, so you mean that the rectangle is $w=\sqrt{2}a, h=\sqrt{2}b$ centered like the ellipse? – vainolo Oct 10 '12 at 21:40
@vainolo: Correct – Ross Millikan Oct 10 '12 at 21:42
For a proof, it is sufficient to observe that stretching will scale the areas of all rectangles by the same factor. – Théophile Jul 22 '14 at 15:58
Top right corner on the ellipse (centered on origin): $x = a \cos \theta$ and $y=b\sin\theta$
where $0\le \theta \le \frac{\pi}{2}, a=\frac{width}{2}$ and $b = \frac{height}{2}$
So area:$$A(\theta) = 4xy=4ab\sin\theta\cos\theta = 2ab\sin2\theta$$ Maximum of sin is at $2\theta=\frac{\pi}{2}$. So, we have $\theta = \frac{\pi}{4}$. Therefore the 4 corners are: $$(a\cos \frac{\pi}{4},b\sin \frac{\pi}{4}), (a\cos \frac{3\pi}{4},b\sin \frac{3\pi}{4}), (a\cos \frac{5\pi}{4},b\sin \frac{5\pi}{4}), (a\cos \frac{7\pi}{4},b\sin \frac{7\pi}{4}),$$
which are of course equal to:
$$(\frac{a\sqrt{2}}{2},\frac{b\sqrt{2}}{2}), (-\frac{a\sqrt{2}}{2},\frac{b\sqrt{2}}{2}), (-\frac{a\sqrt{2}}{2},-\frac{b\sqrt{2}}{2}), (\frac{a\sqrt{2}}{2},-\frac{b\sqrt{2}}{2}),$$
And it's area is: $a\sqrt{2}\cdot b\sqrt{2}=2ab$
-
Using the equation $1 = y^2/n + x^2/m$ for the ellipse, write definitions for side lengths as $2(y^2/n)$ and $2(x^2/m)$. Area equals the product of these sides. Using substitution assert that $(1 - x^2/m) = y^2/m$ and in turn the area = $2(y^2/n) \cdot 2(1 - x^2/m)$. Graph this function and find the maximi. These are your coordinate values. Use +/- in front of these values to form a set of coordinates.
- | 1,071 | 3,182 | {"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} | 4.21875 | 4 | CC-MAIN-2015-48 | longest | en | 0.865158 |
https://www.gradesaver.com/textbooks/math/algebra/elementary-and-intermediate-algebra-concepts-and-applications-6th-edition/chapter-2-equations-inequalities-and-problem-solving-review-exercises-chapter-2-page-149/35 | 1,539,631,242,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583509690.35/warc/CC-MAIN-20181015184452-20181015205952-00462.warc.gz | 950,520,964 | 14,174 | ## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
$\bf{\text{Solution Outline:}}$ Use the position of the number on the number line to evaluate if it satisfies the given inequality, $x\le-5 .$ $\bf{\text{Solution Details:}}$ The inequality $x\le-5$ is satisfied by all numbers that appear on/on the left of $-5$ in the number line. In the number line, $-7$ appears on the $\text{ left }$ of $-5 .$ Hence, it is $\text{ a solution .}$ | 125 | 462 | {"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} | 4.3125 | 4 | CC-MAIN-2018-43 | longest | en | 0.76586 |
https://blogs.mathworks.com/student-lounge/2021/05/19/roll-center-calculator-app-with-matlab-app-designer/?from=kr | 1,632,341,726,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057388.12/warc/CC-MAIN-20210922193630-20210922223630-00086.warc.gz | 199,855,081 | 27,312 | # Roll Center Calculator App with MATLAB App Designer
Nice to meet you, I am Nobu the technical lead for Formula SAE Japan.
The other day, I created an app for Formula Student teams that automatically calculates the position of the roll center using MATLAB’s App Designer. This app enables teams to modify the code and change the algorithm.
Below I explain the details and would like to hear your feedback on this app and how you might use it.
## What is the Roll Center
The roll center is the point of the center (virtual moment) when the object rotates. So why is the roll center so important? For example, a car has a center of gravity. However, the farther the center of gravity and the height of the roll center are, the easier it is for the car to lose balance. Therefore, the car body may turn over when trying to turn a curve. Therefore, it is important to know the exact position of the roll center and adjust the position of the roll center so that the car does not lose its balance.
## How is the roll center determined?
There is a mechanism called a suspension that connects the car body and wheels, and the position of the roll center is determined by how the suspension is attached.
## What you can do with the MATLAB Roll Center App
If you enter the necessary information in this roll center app, the position of the roll center will be calculated automatically and displayed in 3D.
1. Click the link shown below
3. Unzip the Zip file and open the rollCenterCalculator file in the folder
## Inputs
There are two main types of information that you need to enter in the app.
• First, information about a vehicle body
• Second, information about the mounting position of the suspension
## Information about the vehicle body
Each parameter is shown in the picture below.
## Suspension mounting position
Since the double wishbone suspension is used this time, the mounting points are divided into upper and lower parts, with 2 points on the vehicle body side and 1 point on the wheel side.
## Outputs
• Instantaneous Center (IC): Parameters required in the process of calculating the roll center (instant center)
• Roll Center (RC): Roll Center
• Distance between RC and CG: Difference between the roll center and the height of the center of gravity
## How to Customize App
1. Click rollCenterCalculator.mlapp in MATLAB
2. Click CodeView
3. Code to customize
## Afterword
The above is a brief overview of the app and how to use it, but this app is static so that even beginners can easily understand it. There are many things you can add or customize the app with MATLAB App Designer, such as increasing the parameters to make it more dynamic by changing the calculation algorithm inside or changing the display format to make it easier for others to see. Thank you.
There are useful resources for automotive competition. Please check the link below
|
### 댓글
댓글을 남기려면 링크 를 클릭하여 MathWorks 계정에 로그인하거나 계정을 새로 만드십시오. | 634 | 2,950 | {"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} | 3.546875 | 4 | CC-MAIN-2021-39 | latest | en | 0.887367 |
https://fr.slideshare.net/evanitag/note-3pptx | 1,695,483,679,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506481.17/warc/CC-MAIN-20230923130827-20230923160827-00096.warc.gz | 295,949,209 | 41,594 | # note 3..pptx
Teacher at PKMI à PKMI
12 Aug 2022
1 sur 2
### note 3..pptx
• 1. Note 3 1.plus / and / tambah + 2.minus / kurang - 3.multiplied by / times / kali x 4.divided by / bagi : 5.equals / is / sama dengan = • Examples : 1. 7 + 3 = 10 Seven plus three equals ten. 2. 10 – 2 = 8 Ten minus two is eight. 3. 4 x 5 = 20 Four multiplied by five equals twenty 4. 16 : 4 = 4 Sixteen divided by four is four
• 2. Exercise 3 1. 12 + 2 = 2. 2 x 7 = 3. 9 - 4 = 4. 14 : 2 = | 204 | 471 | {"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} | 3.515625 | 4 | CC-MAIN-2023-40 | latest | en | 0.620027 |
https://www.proofwiki.org/wiki/Integer_Multiplication_is_Well-Defined | 1,686,388,457,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224657144.94/warc/CC-MAIN-20230610062920-20230610092920-00474.warc.gz | 1,071,517,411 | 12,390 | # Integer Multiplication is Well-Defined
## Proof 1
Let us define $\eqclass {\tuple {a, b} } \boxtimes$ as in the formal definition of integers.
That is, $\eqclass {\tuple {a, b} } \boxtimes$ is an equivalence class of ordered pairs of natural numbers under the congruence relation $\boxtimes$.
$\boxtimes$ is the congruence relation defined on $\N \times \N$ by $\tuple {x_1, y_1} \boxtimes \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$.
In order to streamline the notation, we will use $\eqclass {a, b} {}$ to mean $\eqclass {\tuple {a, b} } \boxtimes$, as suggested.
We need to show that $\eqclass {a, b} {} = \eqclass {p, q} {} \land \eqclass {c, d} {} = \eqclass {r, s} {} \implies \eqclass {a, b} {} \times \eqclass {c, d} {} = \eqclass {p, q} {} \times \eqclass {r, s} {}$.
We have $\eqclass {a, b} {} = \eqclass {p, q} {} \land \eqclass {c, d} {} = \eqclass {r, s} {} \iff a + q = b + p \land c + s = d + r$ by the definition of $\boxtimes$.
From the definition of integer multiplication, we have:
$\forall a, b, c, d \in \N: \eqclass {a, b} {} \times \eqclass {c, d} {} = \eqclass {a c + b d, a d + b c} {}$
So, suppose that $\eqclass {a, b} {} = \eqclass {p, q} {}$ and $\eqclass {c, d} {} = \eqclass {r, s} {}$.
Both $+$ and $\times$ are commutative and associative on $\N$. Thus:
$\ds$ $\ds \paren {a c + b d + p s + q r} + \paren {p c + q c + p d + q d}$ $\ds$ $=$ $\ds \paren {a + q} c + \paren {b + p} d + p \paren {c + s} + q \paren {d + r}$ $\ds$ $=$ $\ds \paren {b + p} c + \paren {a + q} d + p \paren {d + r} + q \paren {c + s}$ as $a + q = b + p, c + s = d + r$ $\ds$ $=$ $\ds b c + p c + q d + a d + p d + p r + q c + q s$ $\ds$ $=$ $\ds \paren {a d + b c + p r + q s} + \paren {p c + q c + p d + q d}$
So we have $a c + b d + p s + q r = a d + b c + p r + q s$ and so, by the definition of $\boxtimes$, we have:
$\eqclass {a c + b d, a d + b c} {} = \eqclass {p r + q s, p s + q r} {}$
So, by the definition of integer multiplication, this leads to:
$\eqclass {a, b} {} \times \eqclass {c, d} {} = \eqclass {p, q} {} \times \eqclass {r, s} {}$
Thus integer multiplication has been shown to be well-defined.
$\blacksquare$
## Proof 2
Consider the formal definition of the integers: $x = \eqclass {a, b} {}$ is an equivalence class of ordered pairs of natural numbers.
Consider the mapping $\phi: \N_{>0} \to \Z_{>0}$ defined as:
$\forall u \in \N_{>0}: \map \phi u = u'$
where $u' \in \Z$ be the (strictly) positive integer $\eqclass {b + u, b} {}$.
Let $v' = \eqclass {c + v, c} {}$.
Then:
$\ds u' v'$ $=$ $\ds \eqclass {b + u, b} {} \times \eqclass {c + v, c} {}$ $\ds$ $=$ $\ds \eqclass {\paren {b + u} \paren {c + v} + b c, \paren {b + u} c + b \paren {c + v} } {}$ $\ds$ $=$ $\ds \eqclass {b c + b v + c u + u v + b c, b c + u c + b c + b v} {}$ $\ds$ $=$ $\ds \eqclass {b c + u v, b c} {}$ $\ds$ $=$ $\ds \eqclass {b + u v, b} {}$ $\ds$ $=$ $\ds \paren {u v}'$
$\blacksquare$ | 1,145 | 2,932 | {"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": 2, "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} | 4.375 | 4 | CC-MAIN-2023-23 | latest | en | 0.549876 |
https://www.adda247.com/school/table-of-27/ | 1,713,379,489,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817171.53/warc/CC-MAIN-20240417173445-20240417203445-00068.warc.gz | 555,980,510 | 119,938 | Home » Table of 27 » Table of 27
# 27 Table | Table of 27 | 27 ka Table Maths
## Table of 27
In maths, a multiplication table of 27 is a mathematical table used to define the multiplication of 27 operations for an arithmetic system. Here on this page, you can find a multiplication table for number twenty-seven ( 27 ). You can learn ( or save it as a screenshot) a table of 27.
Fun Fact: Marvel’s Deadpool issue number 27 holds the Guinness World Record for the most comic book characters on a single cover.
## 27 Table
27 Table is given below. Students who are studying in Class 7 and above can refer to the table of 27 to memorize it. Competitive exam aspirants too can learn the tables from here. It is important to learn tables as it will help drastically in improving the time limit one takes to calculate. Even a second saved from calculation in competitive exams can decide your rank.
This is the sole reason why we recommend our readers memorize the 27 times table ( 27 ka table).
## 27 Table in Maths
27 * 1 = 27 27 * 2 = 54 27 * 3 = 81 27 * 4 = 108 27 * 5 = 135 27 * 6 = 162 27 * 7 = 189 27 * 8 = 216 27 * 9 = 243 27 * 10 = 270
## 27 ka Table in Maths
Table of 27 is given below in 20 multiples. Learn it now.
27 * 1 = 27 27 * 2 = 54 27 * 3 = 81 27 * 4 = 108 27 * 5 = 135 27 * 6 = 162 27 * 7 = 189 27 * 8 = 216 27 * 9 = 243 27 * 10 = 270 27 * 11 = 297 27 * 12 = 324 27 * 13 = 351 27 * 14 = 378 27 * 15 = 405 27 * 16 = 432 27 * 17 = 459 27 * 18 = 486 27 * 19 = 513 27 * 20 = 540
## Table of 27 (Upto 50 multiples)
It is important to know the tables even if you are no longer a student. This will help you in calculating numbers in your everyday life. Keep learning and enjoy the process of learning more than focusing on the ultimate goal. Have a great day ahead!
27 x 1 = 27 27 x 2 = 54 27 x 3 = 81 27 x 4 = 108 27 x 5 = 135 27 x 6 = 162 27 x 7 = 189 27 x 8 = 216 27 x 9 = 243 27 x 10 = 270 27 x 11 = 297 27 x 12 = 324 27 x 13 = 351 27 x 14 = 378 27 x 15 = 405 27 x 16 = 432 27 x 17 = 459 27 x 18 = 486 27 x 19 = 513 27 x 20 = 540 27 x 21 = 567 27 x 22 = 594 27 x 23 = 621 27 x 24 = 648 27 x 25 = 675 27 x 26 = 702 27 x 27 = 729 27 x 28 = 756 27 x 29 = 783 27 x 30 = 810 27 x 31 = 837 27 x 32 = 864 27 x 33 = 891 27 x 34 = 918 27 x 35 = 945 27 x 36 = 972 27 x 37 = 999 27 x 38 = 1026 27 x 39 = 1053 27 x 40 = 1080 27 x 41 = 1107 27 x 42 = 1134 27 x 43 = 1161 27 x 44 = 1188 27 x 45 = 1215 27 x 46 = 1242 27 x 47 = 1269 27 x 48 = 1296 27 x 49 = 1323 27 x 50 = 1350
## Table of 27: Practice Problems
Example 1: Calculate the value of 27 * 7.
Ans: The value of 27 * 7 will be 189.
Example 2: What are the set of factors of 27?
Ans:The factors of 27 are 1, 3, 9 , and 27.
Example 3:Is 27 a prime number?
Ans: Prime numbers are those numbers that can be divided only by 1 and the number itself. 27 has factors other than 1 and 27. So, it is not a prime number.
Example 4: Calculate the value of 27 * 27 * 27.
Ans: The value of 27 * 27 *27 is 19, 863.
Example 5: You have 27 pens and are told to distribute them equally among three of your friends. What will be the total number of pens, each of your friend gets?
Ans:There are 27 pens and three friends. To distribute them equally, divide 27 by 3, and the result will be 9.
Read Some Frequently Used Table for Fast Calculation
Sharing is caring!
## FAQs
### 1. What are the factors of 27?
The factors of 27 are 1, 3, 9, and 27
### 2. When should we start memorizing tables and why?
Once a student knows addition and subtraction and can count differences they should start learning multiplication tables. Memorizing multiplication tables up to 30 will help you make your calculation speed faster and more accurate. | 1,311 | 3,723 | {"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} | 4.125 | 4 | CC-MAIN-2024-18 | latest | en | 0.804698 |
http://www.slideshare.net/porqueira/ativ-minha-7-serie-2011-9282824 | 1,448,477,952,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398445291.19/warc/CC-MAIN-20151124205405-00294-ip-10-71-132-137.ec2.internal.warc.gz | 689,497,234 | 23,835 | Upcoming SlideShare
×
# Ativ minha 7ª serie 2011
1,335
-1
Published on
0 Likes
Statistics
Notes
• Full Name
Comment goes here.
Are you sure you want to Yes No
• Be the first to comment
• Be the first to like this
Views
Total Views
1,335
On Slideshare
0
From Embeds
0
Number of Embeds
0
Actions
Shares
0
15
0
Likes
0
Embeds 0
No embeds
No notes for slide
### Ativ minha 7ª serie 2011
1. 1. Colégio estadual Dinah GonçalvesProfessor Antonio Carlos Carneiro BarrosoEquações fracionáriasSérie 7ª ou 8º [email protected]://ensinodematemtica.blogspot.comHTTP://accbarroso60.wordpress.com Atividade para 7ª sérieou 8º anoResolva: x−2 x−3 3a) + = 4 2 23( x − 2 ) 6( x − 3) 18 + = 12 12 123x-6 +6x-18=183x+6x-6-18=189x=18+18+69x=42 42X= 9 x −1 x − 5 3 + = 2 3 6b) 3( x − 1) + 2( x − 5) = 3 mmc 6 6 6 63x-3+2x-10=33x+2x-3-10=35x=3+10+35x=16 16X= 5
2. 2. x−2 x−5 5 + = 8 3 2c) 3( x − 2) 8( x − 5) 60 mmc 8,3,2=24 + = 24 24 243x-6+8x-40=603x+8x-6-40=6011x=60+6+4011x=106 106X= 11 2 x − 3 3x − 4 5 + = 2 8 4 4( 2 x − 3) 1( 3 x − 4 ) 10 + = 8 8 8 8 x − 12 + 3 x − 4 = 10d) 8 x + 3x − 12 − 4 = 10 11x = 10 + 4 + 12 11x = 26 26 x= 11 x − 5 x −1 4 5( x − 5) + 2( x − 1) = 4e) + = mmc 10 /5x-25+2x-2=4 2 5 10 10 315x+2x=4+25+2 / 7x=31 / x= 7 x−7 x−4 7 4( x − 7 ) + 3( x − 4 ) = 42f) + = mmc 24 / 4x-28+3x-12=42 / 6 8 4 24 824x+3x=42+12+28 /7x=82 / x= 7 x−2 x−4 5 2( x − 2) + 6( x − 4 ) = 45g) + = mmc 18 / 2x-4+6x-24=45 / 9 3 2 18 732x+6x=45+24+4 /8x=73 / x= 8
3. 3. 2x − 1 2x − 3 5 2( 2 x − 1) + 4( 2 x − 3) = 15h) + = mmc 24 / 12 6 8 24 294x-2+8x-12=15 /4x+8x=15+12+2 /12x=29 /x= 12 3x − 4 2 x − 1 5 4( 3x − 4 ) + 7( 2 x − 1) = 70i) + = mmc 28 / 7 4 2 28 9312x-16+14x-7=70 /12x+14x =70+7+16 / 26x= 93 / x= 26 x−4 x−3 5 8( x − 4 ) + 9( x − 3) = 90j) + = mmc 72 / 9 8 4 72 1498x-32+9x-27=90 /8x+9x=90+27+32 / 17x =149 / x = 17 x−7 x−4 3 8( x − 7 ) + 5( x − 4) = 30j) + = mmc 40 / 8x 5 8 4 40-56+5x-20=30 / 8x+5x = 30+20+56 / 13x =106 106/x= 13 x−2 x−3 3 2( x − 2) + 8( x − 3) = 3l) + = /mmc=16 / / 2x-4+8x-24 8 2 16 16=3 /2x+8x =3+4+24 /10x=31 31x= 10 2 x − 1 3x − 2 8 5( 2 x − 1) + 4( 3 x − 2 ) = 80m) + = mmc=20 / / 4 5 2 2010x-5+12x-8=80 9310x+12x=80+8+5 / 22x=93 /x= 22 2( x − 1) 3( x − 4) 5 8( x − 1) + 30( x − 4 ) = 5n) + = / mmc=20 / / 5 2 20 208x-8+30x-120=5 1338x+30x=5+120+8 / 38x=133 / x= 38 2 x + 1 3x + 4 3 8( 2 x + 1) + 5( 3x + 4 ) = 60o) + = / mmc40 / / 5 8 2 4016x+8+15x+20=60
4. 4. 3216x+15x=60-20-8 /31x=32 / x= 31 x−5 x−2 7 4( x − 5) − 3( x − 2) = 42p) − = / mmc 12 / / 3 4 2 124x-20-3x+6=424x-3x=42-6+20 / x=56 2− x x−5 4 4( 2 − x ) + 6( x − 5) = 12q) + = / mmc 24 / / 6 4 8 248-4x+6x-30=12-4x+6x=12+30-8 / 2x = 34 / x=34:2 / x= 17 | 1,685 | 2,655 | {"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-2015-48 | latest | en | 0.203372 |
https://brainly.in/question/88074 | 1,484,943,496,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280872.69/warc/CC-MAIN-20170116095120-00323-ip-10-171-10-70.ec2.internal.warc.gz | 801,021,565 | 10,286 | # Compute each probability using the formulae P(A|B)=P(AΠB)/(P(B)) and using appropriate below table values a)P(D|B) b)P(C|E) c)P(B|D) d)P(E|C) e)P(D|C) f)P(E|A) g)P(A|C) h)P(B/B)
1
by sweetysiri92
+2
9th why?
no i thought ur 10th
ok
ok:)
2015-03-24T12:47:00+05:30
A.)P(D|B)= P(DΠB)/(P(B)) = 3/100*10/1 = 3/10.
b.)P(C|E)= P(CΠE)/(P(E)) = 7/100*10/4 = 7/40.
c.)P(B|D)= P(BΠD)/(P(D))= 3/100*10/3 = 1/10.
d.)P(E|C)= P(EΠC)/(P(C))= 7/100*10/2 = 7/20.
e.)P(D|C)= P(DΠC)/(P(C))= 7/100*10/2 =7/20.
f.)P(E|A)= P(EΠA)/(P(A))= 7/25*10/7 = 2/5.
g.)P(A|C)= P(AΠC)/(P(C))= 0/1*10/2 = 0.
h.)P(B|B)= P(BΠB)/(P(B))= 1/10*10/1 = 1.
Hope i helped u:) | 375 | 642 | {"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} | 3.84375 | 4 | CC-MAIN-2017-04 | latest | en | 0.2373 |
https://www.jiskha.com/questions/480432/log3-x-log3-1-x-3 | 1,607,192,356,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141748276.94/warc/CC-MAIN-20201205165649-20201205195649-00118.warc.gz | 723,972,679 | 4,190 | # Calculus
log3 x-log3(1-x)=3
1. 👍 0
2. 👎 0
3. 👁 879
1. remember
log a - log b= log (a/b)
so log (x/(1-x))=3
take the antilog of each side.
x/(1-x)=1000
solve for x
1. 👍 0
2. 👎 0
👨🏫
bobpursley
## Similar Questions
1. ### Math
Given that x=log3^5 and y=log3^2, rewrite log3^60 in terms of x and y.
2. ### maths
express log9 in terms of log3 x and log3 y and solve for x and y simultaneously
Log3(x)=log9(6x)
4. ### mathematics
log3(4x+1)-log3(3x-5)=2
1. ### Algebra
log3 (2x – 1) – log3 (x – 4) = 2
2. ### calculus
1) Solve for x in terms of k: log3(x) + log3(x+7) = k; 3 is the base 2) e^(x+5) = e^(x) + 6, solve for x. Please help, I have no idea how to do these and I've tried several times!
3. ### algebra 2 trig
if log3=a then log300 can be expressed as? 1) 100a 2)a+2 3)100+a 4)3a
4. ### algebra
(2*log7 16)/(log3(√10+1)+log3(√10-1)log7 2
1. ### Pre Calculus
Evaluate the function f(x) = log3 x at x =1/27 without using a calculator a. –4 b. –2 c. 1/27 d. –3 e. 27
2. ### trig
solve for x log4 64+ log2 2 ________________ =x log3 81
3. ### Mathematics
Find x such that log3(2x-1)=1+log2(x+1)?
4. ### Algebra 1
Explain the differences between solving these two equations: •log3(x - 1) = 4 AND log3(x - 1) = log34 I think it has to deal with the formulas? | 531 | 1,287 | {"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} | 3.90625 | 4 | CC-MAIN-2020-50 | latest | en | 0.707775 |
https://www.physicsforums.com/threads/thermal-vac-chamber-pressure-gain-question.543488/ | 1,590,414,616,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347388758.12/warc/CC-MAIN-20200525130036-20200525160036-00391.warc.gz | 823,835,662 | 16,360 | Thermal Vac chamber pressure gain question
Main Question or Discussion Point
Hi group, I work at Thermal Vacuum Chamber, and we had an incident last night where our chamber, Which is 70' H x 30' D, was pumped down to a pressure of 3.7E-7. There was a release of gas into the chamber abruptly last night that caused the pressure in the chamber to rise to 5.0E-7, before it recovered back to it's original pressure. My question is How can I tell in liters how much gas was released into the chamber to cause this amount of rise in chamber pressure, based on the starting pressure and the ending pressure?
Related Mechanical Engineering News on Phys.org
Wow! both # are so far below viscus flow, it might be worth thinking about how many atoms we are talking about. (it's been over 30 years since I did vacuum stuff.)
From the interweb, http://en.wikipedia.org/wiki/Mean_free_path
In a chamber that size, I am thinking some type of contamination, even roughing pump oil out gases at 10 -7. I have seen a case where diffusion pump could not get to a high vacuum. It turned out that the diffusion pump oil was contaminated with another oil.
Mapes
Homework Helper
Gold Member
Hi group, I work at Thermal Vacuum Chamber, and we had an incident last night where our chamber, Which is 70' H x 30' D, was pumped down to a pressure of 3.7E-7. There was a release of gas into the chamber abruptly last night that caused the pressure in the chamber to rise to 5.0E-7, before it recovered back to it's original pressure. My question is How can I tell in liters how much gas was released into the chamber to cause this amount of rise in chamber pressure, based on the starting pressure and the ending pressure?
Use the ideal gas law (PV=nRT) with a pressure of 1.3E-7.
Calculate the volume of the chamber in liters.
Divide by the standard molar volume of a gas at STP (22.41 liters)
Multiply this volume by 3.0E-7/760 and 5.0E-7/760 to get moles of gas before and after.
Subtract and get gas released into chamber, in moles (e.g., 29 grams per mole of air).
Did the incident occur while the evacuation was going on? | 515 | 2,106 | {"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} | 3.65625 | 4 | CC-MAIN-2020-24 | longest | en | 0.95892 |
http://nrich.maths.org/public/leg.php?code=-684&cl=3&cldcmpid=7651 | 1,506,203,604,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689779.81/warc/CC-MAIN-20170923213057-20170923233057-00166.warc.gz | 236,845,748 | 8,779 | # Search by Topic
#### Resources tagged with STEM - maths supporting SET similar to Food Chains:
Filter by: Content type:
Stage:
Challenge level:
### There are 50 results
Broad Topics > STEM mapping > STEM - maths supporting SET
### Food Chains
##### Stage: 3 Challenge Level:
When a habitat changes, what happens to the food chain?
### Chocolate Cake
##### Stage: 3 Challenge Level:
If I don't have the size of cake tin specified in my recipe, will the size I do have be OK?
### Fill Me Up
##### Stage: 3 Challenge Level:
Can you sketch graphs to show how the height of water changes in different containers as they are filled?
### How Do You React?
##### Stage: 4 Challenge Level:
To investigate the relationship between the distance the ruler drops and the time taken, we need to do some mathematical modelling...
### Biology Measurement Challenge
##### Stage: 4 Challenge Level:
Analyse these beautiful biological images and attempt to rank them in size order.
### Does This Sound about Right?
##### Stage: 3 Challenge Level:
Examine these estimates. Do they sound about right?
### Pinhole Camera
##### Stage: 3 Challenge Level:
Make your own pinhole camera for safe observation of the sun, and find out how it works.
### Big and Small Numbers in Physics
##### Stage: 4 Challenge Level:
Work out the numerical values for these physical quantities.
### Perspective Drawing
##### Stage: 3 and 4 Challenge Level:
Explore the properties of perspective drawing.
### Investigating the Dilution Series
##### Stage: 4 Challenge Level:
Which dilutions can you make using only 10ml pipettes?
### Counting Dolphins
##### Stage: 4 Challenge Level:
How would you go about estimating populations of dolphins?
### Big and Small Numbers in the Living World
##### Stage: 3 and 4 Challenge Level:
Work with numbers big and small to estimate and calculate various quantities in biological contexts.
### Approximately Certain
##### Stage: 4 and 5 Challenge Level:
Estimate these curious quantities sufficiently accurately that you can rank them in order of size
##### Stage: 4 Challenge Level:
Which units would you choose best to fit these situations?
### A Question of Scale
##### Stage: 4 Challenge Level:
Use your skill and knowledge to place various scientific lengths in order of size. Can you judge the length of objects with sizes ranging from 1 Angstrom to 1 million km with no wrong attempts?
### Weekly Challenge 48: Quorum-sensing
##### Stage: 4 Short Challenge Level:
This problem explores the biology behind Rudolph's glowing red nose.
### Global Warming
##### Stage: 4 Challenge Level:
How much energy has gone into warming the planet?
### Electric Kettle
##### Stage: 4 Challenge Level:
Explore the relationship between resistance and temperature
##### Stage: 3 Challenge Level:
Can you work out which drink has the stronger flavour?
### Big and Small Numbers in the Physical World
##### Stage: 4 Challenge Level:
Work with numbers big and small to estimate and calculate various quantities in physical contexts.
##### Stage: 3 Challenge Level:
Is it cheaper to cook a meal from scratch or to buy a ready meal? What difference does the number of people you're cooking for make?
### Robot Camera
##### Stage: 4 Challenge Level:
Could nanotechnology be used to see if an artery is blocked? Or is this just science fiction?
### Big and Small Numbers in Chemistry
##### Stage: 4 Challenge Level:
Get some practice using big and small numbers in chemistry.
### Big and Small Numbers in Biology
##### Stage: 4 Challenge Level:
Work with numbers big and small to estimate and calulate various quantities in biological contexts.
### More or Less?
##### Stage: 4 Challenge Level:
Are these estimates of physical quantities accurate?
### Constantly Changing
##### Stage: 4 Challenge Level:
Many physical constants are only known to a certain accuracy. Explore the numerical error bounds in the mass of water and its constituents.
### Designing Table Mats
##### Stage: 3 and 4 Challenge Level:
Formulate and investigate a simple mathematical model for the design of a table mat.
### Mystery Procedure
##### Stage: 4 Challenge Level:
Can you work out what this procedure is doing?
### Fill Me up Too
##### Stage: 4 Challenge Level:
In Fill Me Up we invited you to sketch graphs as vessels are filled with water. Can you work out the equations of the graphs?
### Back Fitter
##### Stage: 4 Challenge Level:
10 graphs of experimental data are given. Can you use a spreadsheet to find algebraic graphs which match them closely, and thus discover the formulae most likely to govern the underlying processes?
### Guessing the Graph
##### Stage: 4 Challenge Level:
Can you suggest a curve to fit some experimental data? Can you work out where the data might have come from?
### Genetics
##### Stage: 4 Challenge Level:
A problem about genetics and the transmission of disease.
### Scientific Measurement
##### Stage: 4 Challenge Level:
Practice your skills of measurement and estimation using this interactive measurement tool based around fascinating images from biology.
### Speed-time Problems at the Olympics
##### Stage: 4 Challenge Level:
Have you ever wondered what it would be like to race against Usain Bolt?
### Construct the Solar System
##### Stage: 4 and 5 Challenge Level:
Make an accurate diagram of the solar system and explore the concept of a grand conjunction.
##### Stage: 4 and 5 Challenge Level:
How would you design the tiering of seats in a stadium so that all spectators have a good view?
### Efficient Packing
##### Stage: 4 Challenge Level:
How efficiently can you pack together disks?
### Far Horizon
##### Stage: 4 Challenge Level:
An observer is on top of a lighthouse. How far from the foot of the lighthouse is the horizon that the observer can see?
### Make Your Own Pencil Case
##### Stage: 3 Challenge Level:
What shape would fit your pens and pencils best? How can you make it?
### Carbon Footprints
##### Stage: 4 Challenge Level:
Is it really greener to go on the bus, or to buy local?
### Troublesome Triangles
##### Stage: 2 and 3 Challenge Level:
Many natural systems appear to be in equilibrium until suddenly a critical point is reached, setting up a mudslide or an avalanche or an earthquake. In this project, students will use a simple. . . .
### Perfect Eclipse
##### Stage: 4 Challenge Level:
Use trigonometry to determine whether solar eclipses on earth can be perfect.
### What's That Graph?
##### Stage: 4 Challenge Level:
Can you work out which processes are represented by the graphs?
### Investigating Epidemics
##### Stage: 3 and 4 Challenge Level:
Simple models which help us to investigate how epidemics grow and die out.
### Gym Bag
##### Stage: 3 and 4 Challenge Level:
Can Jo make a gym bag for her trainers from the piece of fabric she has?
### Oblique Projection
##### Stage: 3 Challenge Level:
Explore the properties of oblique projection.
### Witch's Hat
##### Stage: 3 and 4 Challenge Level:
What shapes should Elly cut out to make a witch's hat? How can she make a taller hat?
##### Stage: 2 and 3 Challenge Level:
What shape and size of drinks mat is best for flipping and catching? | 1,543 | 7,273 | {"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} | 3.859375 | 4 | CC-MAIN-2017-39 | latest | en | 0.82673 |
https://www.physicsforums.com/threads/voltage-drop-and-transformer-improvement.854093/ | 1,519,425,763,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814857.77/warc/CC-MAIN-20180223213947-20180223233947-00507.warc.gz | 913,928,408 | 17,796 | # Voltage drop and transformer improvement
1. Jan 26, 2016
### AstroJMT42
1. The problem statement, all variables and given/known data
Power station produces AC power at 2.5Mw, I = 100A.
Transferred along 100km line with a resistance of 50 ohms to a factory. Calculate power lost and % efficiency
2. Relevant equations
P=VI
P = I2R
3. The attempt at a solution
Initial power at the station is 2.5x106 . Using I2R you get new power at the factory of 1002x50 = 5x105W
This means a total loss of 2x105 and a percentage of 8%.
I'm pretty confident this is correct but a fellow student is adamant that I2R is power loss not total power out. His conclusion is that the line is actually 80% efficient, which sounds way to efficient to be correct, especially seeing as there are no transformers. Can anyone please confirm which of us is correct?
second part then takes the same original power and introduces a 1:5 step up transformer and 5:1 step down and asks to calculate the new power loss and new efficiency.
I worked initial V = 25,000 and after step up V = 125,000
New power after this using I2R is 2,480,000W which I calculated to be an efficiency of 99.2%
Again could you please check my logic is sound? There were no answers given.
Apologies it is actually 2 questions being asked but the second one leads directly from the first so thought it was best to keep it in one post.
Thanks!
2. Jan 26, 2016
### BvU
Check your calculations. $2.5 \ 10^6 - 5 \ 10^5 = 2 \ 10^5$ ?
Is the 50 $\Omega$ the resistance of the load in the plant or is it the resistance of the power cables ? So is it the useful power or the lost power ?
3. Jan 26, 2016
### AstroJMT42
50 Ω is the resistance in the cable.
2,500,000W originally - 500,000 = 2,000,000W sorry that was a silly arithmetic error.
So that makes 500,000 / 2,500,000 x 100 = 20%?
4. Jan 26, 2016
### BvU
Efficiency is useful power divided by input power. So: No.
Fellow student wins the argument.
Funny enouh you do have the right efficiency for the transformers in place case. Note the factor with which the loss is reduced!
5. Jan 26, 2016
### AstroJMT42
So I squared R is is the power loss, not the new power?
I think this is where my misunderstanding is coming from. I thought P=VI and P=I2R were interchangeable. We were given original power and had to use P=I2R to find what the power is after traveling down the line. divide this output power by the original and multiply by 100 to get the percentage.
Which part of that am I getting wrong? it's been a long day...
6. Jan 26, 2016
### CWatters
It depends which R you are talking about? The resistance of the cable or the load (aka the factory)?
If R is the resistance of the cable then I2R is the power dissipated (lost) in the cable.
If R is the resistance of the load (eg the equivalent resistance of the factory) then I2R is the power dissipated (used) by the factory.
In general they are. Provided you are talking about the right V, I and R.
Apply conservation of energy eg..
Power delivered to factory = Power generated - power lost in the cable. | 826 | 3,090 | {"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} | 3.90625 | 4 | CC-MAIN-2018-09 | longest | en | 0.937164 |
http://encyclopedia.kids.net.au/page/co/Coprime | 1,394,600,802,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394021365169/warc/CC-MAIN-20140305120925-00098-ip-10-183-142-35.ec2.internal.warc.gz | 60,434,245 | 4,898 | ## Encyclopedia > Coprime
Article Content
# Coprime
In mathematics, the integers a and b are said to be coprime or relatively prime iff they have no common factor other than 1 and -1, or equivalently, if their greatest common divisor is 1.
For example, 6 and 35 are coprime, but 6 and 27 are not because they are both divisible by 3. 1 is coprime to every integer; 0 is coprime only to 1 and -1.
A fast way to determine whether two numbers are coprime is given by the Euclidean algorithm.
### Properties
The numbers a and b are coprime if and only if there exist integers x and y such that ax + by = 1 (see Bézout's identity). Equivalently, b has a multiplicative inverse modulo a: there exists an integer y such that by ≡ 1 (mod a).
If a and b are coprime and a divides a product bc, then a divides c.
If a and b are coprime and bxby (mod a), then xy (mod a). In other words: b yields a unit in the ring Za of integers modulo a.
The two integers a and b are coprime if and only if the point with coordinates (a,b) in an Cartesian coordinate system is "visible" from the origin (0,0), in the sense that there is no point with integer coordinates between the origin and (a,b).
The probability that two randomly chosen integers are relatively prime is 6/π2 (see Pi).
Two natural numbers a and b are coprime if and only if the numbers 2a-1 and 2b-1 are coprime.
### Generalization
Two ideals A and B in the commutative ring R are called coprime if A + B = R. This generalizes Bezout's identity. If A and B are coprime, then AB = AB; furthermore, if C is a third ideal such that A contains BC, then A contains C.
With this definition, two principal ideals (a) and (b) in the ring of integers Z are coprime if and only if a and b are coprime.
All Wikipedia text is available under the terms of the GNU Free Documentation License
Search Encyclopedia
Search over one million articles, find something about almost anything!
Featured Article
List of ships of the United States Navy ... Decatur[?] (1839) USS Delaware (1776, 1798, 1817, BB 28) USS Denver[?] (C 14, CL 58) USS Des Moines[?] (C 15, CA 134) USS Detroit[?] (C 10, CL 8) USS Dictator[?] ... | 590 | 2,162 | {"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} | 4.28125 | 4 | CC-MAIN-2014-10 | longest | en | 0.872958 |
https://math.stackexchange.com/questions/1978572/getting-2-different-solutions-to-the-integral-of-fracdx2x/1978579 | 1,652,953,279,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662526009.35/warc/CC-MAIN-20220519074217-20220519104217-00366.warc.gz | 457,436,268 | 65,388 | # Getting 2 different solutions to the integral of $\frac{dx}{2x}$
I get two different answers that seem to conflict.
Is there an error in one method??
### Method 1
\begin{align} \int \frac{\mathrm{d}x}{2x}&=\frac 1 2 \int\frac{\mathrm{d}x}{x}\\ &=\frac 1 2\ln|x|+C \end{align}
### Method 2
\begin{align} \int \frac{\mathrm{d}x}{2x}&=\frac 1 2 \int \frac{2\mathrm{d}x}{2x}\\ &=\frac 1 2\int \frac{\mathrm{d}u}{u},\;\;\text{ where }u=2x,\ \mathrm{d}u=2\mathrm{d}x\\ &=\frac 1 2 \big(\ln |u| + C\big)\\ &=\frac 1 2 \ln|2x|+C \end{align}
• As the answer below details: the $C$'s are 'different', for want of a better word. Your "paradox" here is one of the reasons we are careful to always put a $C$ at the end, and also have to stay constantly aware of what it means. Oct 21, 2016 at 13:58
• Another good example is integrating $\sin(x)\cos(x)$. Try integration by parts, to get $\frac{1}{2}\sin^2(x)+C$, then try using the trig identity $\sin(x)\cos(x)=\frac{1}{2}\sin(2x)$ to get $-\frac{1}{4}\cos(2x)+D$. A hint on reconciling the answers is $\cos(2x)=1-2\sin^2(x)$. Oct 21, 2016 at 14:06
• When we get two ostensibly different solutions to an anti-differentiation problem, it is often helpful to check whether the solutions (ignoring the constants) are vertical translations of each other. If so, the two solutions must, of course, differ by a constant. In this case, $\frac 1 2\ln|x|$ translated vertically $\ln\sqrt{2}$ units yields $\frac 1 2\ln|2x|$. Oct 21, 2016 at 16:40
Hint: $\frac{1}{2}\ln{|2x|} + C = \frac{1}{2}\ln{|x|} + \frac{1}{2}\ln{2} + C = \frac{1}{2}\ln{|x|} + \left(\frac{1}{2}\ln{2} + C\right)$. | 602 | 1,625 | {"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": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2022-21 | latest | en | 0.712856 |
https://www.ordinalnumbers.com/ordinal-numbers-exercises-pdf-esl/ | 1,695,513,195,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506539.13/warc/CC-MAIN-20230923231031-20230924021031-00113.warc.gz | 1,030,121,047 | 16,524 | Ordinal Numbers Exercises Pdf Esl
By using ordinal numbers, you are able to count unlimited sets. They can also be utilized as a method to generalize ordinal numbers.
1st
One of the fundamental ideas of math is the ordinal number. It is a number that identifies the position of an object in an array of objects. Ordinally, a number between one and 20 is utilized as the ordinal number. While ordinal numbers serve a variety of purposes, they are commonly used to show the order of items in an orderly list.
Charts, words, and numbers can all be used to depict ordinal numbers. They may also be used to describe how a set of pieces is arranged.
Most ordinal numbers fall within one of these two categories. Ordinal numbers that are infinite are represented by lowercase Greek letters, while finite ones are represented with Arabic numbers.
Based on the Axiom of Choice, each set that is well-organized must include at least one ordinal. The first student in an class, for instance, would receive the highest grade. The winner of the contest was the student with the highest score.
Combinational ordinal amounts
Multiple-digit numbers are also known as compound ordinal numbers. They are made by multiplying an ordinal figure by its final number. They are commonly used to rank and date. They don’t use a unique ending for each number, like cardinal number.
Ordinal numbers indicate the sequence of elements found in an assortment. They may also be used to denote the names of items within collections. Regular numbers are found in both suppletive and regular forms.
By prefixing cardinal numbers by the suffix -u, regular ordinals can be created. After that, write the number as a word. After that, add a hyphen after it. There are a variety of suffixes to choose from.
Suppletive ordinals are created by affixing words with the suffix -u or -e. The suffix, used to count is broader than the conventional one.
Limits of significance for ordinal values
Limit ordinal numbers are those which do not contain zero. Limit ordinal numeric numbers suffer from the disadvantage of not having not contain the possibility of having a maximum element. You can make them by joining sets with no any maximum element.
Infinite transfinite-recursion concepts employ a limit ordinal number. In accordance with the von Neumann model, each infinite cardinal number also has an ordinal limit.
An ordinal number that is subject to a limit is equal to the sum of all the ordinals below. Limit ordinal numbers can be quantified using math, but they can also be expressed as natural numbers or a series.
The ordinal numbers used to arrange the data are employed. They are used to explain the location of an object numerically. These numbers are commonly utilized in set theory and the arithmetic. They do not belong to the same group as natural numbers even though they share the same structure.
The von Neumann method uses a well-ordered list. Let’s suppose that fy subfunctions an equation, g’, that is defined as a single function. If fy is the only subfunction (ii) the function g’ must meet the criteria.
The Church-Kleene oral is an ordinal that is a limit in a similar way. The Church-Kleene oral defines a limit-order as a well-ordered grouping of ordinals that are smaller.
Numbers that are ordinal in stories.
Ordinal numbers are often used to indicate the hierarchy between entities and objects. They are important for organising, counting and ranking. They can be used both to indicate the sequence of things and to show the position of objects.
The ordinal number may be indicated by the letter ‘th’. Sometimes, the letter “nd” is used, but it’s not the only way it is used. Titles of books are typically accompanied by ordinal numbers.
Even though ordinal figures are generally utilized in lists, you can write them down in the form of words. They also can be referred to as acronyms and numbers. These numbers are much easier to understand than cardinal numbers, however.
There are three distinct types of ordinal numerals. These numbers can be learned more through games, exercises, and other activities. Understanding them is an essential part of improving your ability to arithmetic. You can increase your math abilities by doing coloring exercises. To check your results you can use a simple marking page. | 874 | 4,323 | {"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-2023-40 | longest | en | 0.953976 |
https://www.enotes.com/homework-help/s-set-binary-operation-s-with-identity-e-s-also-316532 | 1,516,768,118,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084893300.96/warc/CC-MAIN-20180124030651-20180124050651-00283.warc.gz | 917,676,097 | 9,916 | # S is a set and * is a binary operation on S, with identity e in S. Also, for all x, y, z elements of S, x*(y*z)=(x*z)*y. Prove * is commutative.my first idea was to let z=e, but that only proves...
S is a set and * is a binary operation on S, with identity e in S. Also, for all x, y, z elements of S, x*(y*z)=(x*z)*y. Prove * is commutative.
my first idea was to let z=e, but that only proves it for that specific case....I feel like this is relatively simple compared to my other proofs, I'm just missing something on this one :( thanks in advance for your help!
sciencesolve | Certified Educator
You need to remember what a binary operator does, hence it takes two elements from S and it returns a single element.
Notice that the probelm provides two inputs, an associative binary operation xo(yoz)=(xoy)oz and an identity element, hence the set that posssesses these features is a monoid.
You may test if the monoid is commutative using the following binary operation such that:
`x o y = xy + x + y`
If x and y are positive integers, using the binary operation yields:
`xoy = xy + x +`` y`
`y o x = yx +y + x`
Since addition and multiplication are commutative over the positive set of integers, then the binary operation is commutative.
Hence, using the features provided by problem yields that the binary operation `o` is also commutative over the set S. | 362 | 1,377 | {"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} | 3.78125 | 4 | CC-MAIN-2018-05 | latest | en | 0.919598 |
http://www.ques10.com/p/28076/eigen-values-of-unitary-matrix-are-of-unit-modules/ | 1,550,577,804,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247489933.47/warc/CC-MAIN-20190219101953-20190219123953-00099.warc.gz | 412,163,923 | 5,481 | Question: Eigen values of unitary matrix are of unit modules
0
Subject: Applied Mathematics 4
Topic: Matrices
Difficulty: Low
m4m(64) • 163 views
modified 7 months ago • written 7 months ago by
0
Let A be a unitary matrix
$A A^\theta=A^\theta A=I$ (1)
Let $\lambda$ be any eigen value of A
To prove that $\lambda$ is of unit modulus i.e $\mid \lambda \mid=1$
As $\lambda$ is an eigen value of A then there exist $X\neq$0 such that
$(AX-\lambda I)X=0$
$AX-\lambda X=0$
$AX=\lambda X$ (2)
$(AX)^\theta=(\lambda X)^\theta$
$X^\theta A^\theta= \lambda^\theta X^\theta$
$X^\theta A^\theta=(\bar \lambda^\theta) \prime X^\theta$
$X^\theta A^\theta=(\bar \lambda^\theta) X^\theta$
post multiplying by A on both sides, we get,
$X^\theta A^\theta A=(\bar \lambda^\theta) X^\theta A$
$X^\theta I= \bar \lambda X^\theta A$
post multiplying by X on both sides we get,
$X^\theta I X= \bar \lambda X^\theta A X$
$X^\theta X= \bar \lambda X^\theta \lambda X$ from (2)
$X^\theta X= \bar \lambda X^\theta \lambda X$
$-X^\theta X +\bar \lambda X^\theta \lambda X=0$
$(\bar \lambda \lambda -1) X^\theta X=0$
As $X\neq0$ $X^\theta X\neq0$ $\bar \lambda-1=0$
$\bar \lambda \lambda=1$
${\mid \lambda^ \mid}^2=1$
$\mid \lambda \mid=1$
Therefore, eigen values of unitary matrix are of unit modulus | 454 | 1,305 | {"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} | 3.71875 | 4 | CC-MAIN-2019-09 | latest | en | 0.458283 |
https://magoosh.com/hs/sat/math-study-guide-for-a-500-score/ | 1,642,493,693,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300805.79/warc/CC-MAIN-20220118062411-20220118092411-00006.warc.gz | 439,992,695 | 35,779 | # Math Study Guide for a 500 Score
You’re struggling to get a score that you feel happy with on the new SAT math section, and you start having some serious doubts about your chances at getting into your dream school.
Maybe the new SAT math just isn’t your thing, or perhaps you’re having trouble understanding the phrasing and terminology since English is your second language. Some students do better under the pressure of test-taking situations than others.
Whatever the case may be, there is a way to get over the 500 hump if you really work at it, and it doesn’t involve studying for the next year to get there.
## Focus on the Easy Problems
The new SAT math is comprised of problems with a wide range of difficulty. The hard ones involve multiple steps and potentially difficult language to parse. The easy ones are usually just plug-and-chug type of questions. These kinds of questions are more straightforward and don’t involve critical thinking to figure out.
The key is to zone in and just study up on getting the easy ones right. Here is a list of the easiest math concepts to brush up on:
1. Linear Equations and Graphs. These kinds of problems make up the basic foundation of the “Heart of Algebra” section on the new SAT math. Many of these problems only involve a few steps to solve for X or match up a graph to the equation, so it’s a great opportunity to score easy points here.
2. Scatterplots, Data, and Tables. Being able to quickly break down a chart doesn’t require a lot of advanced math, but it’s an important skill to know. You will be rewarded handsomely if you are proficient at this.
3. Mean, Median, and Mode. Many of you might remember these concepts from elementary school, but surprisingly, you will see it on the SAT! This is an easy way to get points on the SAT, so take advantage of it.
4. SOHCAHTOA. This is an acronym that will only take you a minute or two to memorize but is so useful on the test. Problems that involve a missing side length or angle measure will become cakewalk for you.
5. Inequalities. It’s similar enough to a standard equation that it shouldn’t involve too much effort on your part to grasp. The important thing to remember here is to be careful to make sure that you are careful about having your signs point the right way.
6. Right Triangles. This goes hand in hand with learning the basic trig functions, so it’s necessary to know your basics like the Pythagorean Theorem and the special right triangles.
7. Complex Numbers. Don’t be fooled by the name; it’s fairly simple to learn how complex numbers work. Since the test makers can only present this concept in a very limited number of ways, it’s worth learning even if it doesn’t come up on the test as often as a simple algebra problem.
## Stay Motivated and Study Hard
Don’t be discouraged if math isn’t your strongest subject. Just by working on going after the low-hanging fruit, you’ll still be able to achieve a respectable score on the new SAT math section.
## Author
• Minh's passion for helping students succeed grew during his time as a career counselor at the University of California, Irvine. Now, he's helping students all over the world by spilling SAT/ACT secrets through blog posts on Magoosh. When he's not busy tutoring or writing, he enjoys playing guitar, traveling, and talking about himself in third-person. | 724 | 3,362 | {"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} | 3.625 | 4 | CC-MAIN-2022-05 | latest | en | 0.9368 |
https://nerdutella.com/q5208-solve-D3-5D2-plus-7D-3-y-e2x-cosh-x | 1,718,416,743,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861578.89/warc/CC-MAIN-20240614235857-20240615025857-00052.warc.gz | 374,408,810 | 9,865 | Q:
# solve (D^3-5D^2+7D-3)y=e^2x cosh x
solve (D^3-5D^2+7D-3)y=e^2x cosh x
The A.E. is m^3-5m^2+7m-3=0
Or m^2 (m-1)-4m(m-1)+3(m-1)=0
Or (m-1)(m^2-4m+3)=0
Or (m-1)(m-1)(m-3)=0
so m=1,1,3
Y_(C.F.)=(c_1+c_2 x)e^x+c_3 e^3x
Y_(P.I.)=
1/( 〖(D-1)〗^2 (D-3)) e^2x coshx=1/2 1/( (D-1)^2 (D-3) ) e^2x (e^x+
e^(-x))=1/2 1/( (D-1)^2 (D-3) ) (e^3x+e^x )
=1/2 1/( (D-1)^2 (D-3) ) e^3x+1/2 1/( (D-1)^2 (D-3) ) e^x
=1/8 1/(D-3) e^3x-1/4 1/(D-1)^2 e^x=1/8.x/1! e^3x-1/4.x^2/2! e^x
=1/8 xe^3x-1/8 x^2 e^x
Y_(G.S.)=(c_1+c_2 x) e^x+c_3 e^3x-1/8 x^2 e^x+1/8 xe^3x | 435 | 604 | {"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} | 3.84375 | 4 | CC-MAIN-2024-26 | latest | en | 0.291069 |
https://www.mersenneforum.org/showthread.php?s=2b1bdf0f9a0ac76e4071e120a2784ce9&p=540956 | 1,675,626,063,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500288.69/warc/CC-MAIN-20230205193202-20230205223202-00082.warc.gz | 901,641,651 | 11,646 | mersenneforum.org Is (1 + 1 + 1 + 1 + 1 + 1 + ...) less than (1 + 2 + 3 + 4 + 5 + 6 + ...) ?
User Name Remember Me? Password
Register FAQ Search Today's Posts Mark Forums Read
2020-03-26, 17:57 #1 retina Undefined "The unspeakable one" Jun 2006 My evil lair 2·3,343 Posts Is (1 + 1 + 1 + 1 + 1 + 1 + ...) less than (1 + 2 + 3 + 4 + 5 + 6 + ...) ? Is (1 + 1 + 1 + 1 + 1 + 1 + ...) less than (1 + 2 + 3 + 4 + 5 + 6 + ...) ? It would seem not. Because 1 + 2 + 3 + 4 + 5 + 6 + ... = -1/12 https://www.youtube.com/watch?v=w-I6XTVZXww I'm not sure if making a comparison like the above is valid. Perhaps I misunderstand comparisons of infinite sequences?
2020-03-26, 19:06 #2
Dr Sardonicus
Feb 2017
Nowhere
13×479 Posts
Quote:
Originally Posted by retina Is (1 + 1 + 1 + 1 + 1 + 1 + ...) less than (1 + 2 + 3 + 4 + 5 + 6 + ...) ? It would seem not. Because 1 + 2 + 3 + 4 + 5 + 6 + ... = -1/12 https://www.youtube.com/watch?v=w-I6XTVZXww I'm not sure if making a comparison like the above is valid. Perhaps I misunderstand comparisons of infinite sequences?
Of course, the series as written do not converge, so simply taken at face value the question is nonsense.
One can assign values to the sums by misusing formulas. The value -1/12 assigned to 1 + 2 + 3 + ... is a case in point. We have
$\zeta(s)\;=\;\sum_{n=1}^{\infty}n^{-s}\text{, when }\Re(s)\;>\;1$
The zeta function is defined at s = 0 and at s = -1 (though is not given by the above series at those points), taking the values -1/2 and -1/12, respectively.
Cheerfully disregarding the invalidity of the formula, mindlessly plugging in s = 0 gives
1 + 1 + 1 + ... ad infinitum = -1/2
and plugging s = -1 into the formula gives
1 + 2 + 3 + 4 + 5 + 6 + ... ad infinitum = -1/12.
And -1/2 < -1/12.
:-D
Last fiddled with by Dr Sardonicus on 2020-03-26 at 19:09 Reason: Rephrasing
2020-03-26, 19:27 #3
retina
Undefined
"The unspeakable one"
Jun 2006
My evil lair
668610 Posts
Quote:
Originally Posted by Dr Sardonicus Of course, the series as written do not converge, so simply taken at face value the question is nonsense. One can assign values to the sums by misusing formulas. The value -1/12 assigned to 1 + 2 + 3 + ... is a case in point. We have $\zeta(s)\;=\;\sum_{n=1}^{\infty}n^{-s}\text{, when }\Re(s)\;>\;1$ The zeta function is defined at s = 0 and at s = -1 (though is not given by the above series at those points), taking the values -1/2 and -1/12, respectively. Cheerfully disregarding the invalidity of the formula, mindlessly plugging in s = 0 gives 1 + 1 + 1 + ... ad infinitum = -1/2 and plugging s = -1 into the formula gives 1 + 2 + 3 + 4 + 5 + 6 + ... ad infinitum = -1/12. And -1/2 < -1/12. :-D
Very good.
If we assign:
A = 1 + 1 + 1 + 1 + 1 + 1 + ...
B = 1 + 2 + 3 + 4 + 5 + 6 + ...
Then B - A = B, since pairwise subtraction gives 0 + 1 + 2 + 3 + ... = 1 + 2 + 3 + ...
Therefore A = 1 + 1 + 1 + 1 + 1 + 1 + ... = 0
2020-03-26, 19:44 #4
Dr Sardonicus
Feb 2017
Nowhere
13×479 Posts
Quote:
Originally Posted by retina Very good. If we assign: A = 1 + 1 + 1 + 1 + 1 + 1 + ... B = 1 + 2 + 3 + 4 + 5 + 6 + ... Then B - A = B, since pairwise subtraction gives 0 + 1 + 2 + 3 + ... = 1 + 2 + 3 + ... Therefore A = 1 + 1 + 1 + 1 + 1 + 1 + ... = 0
OTOH, we have
1 + A = A
2020-03-26, 19:50 #5
retina
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2·3,343 Posts
Quote:
Originally Posted by Dr Sardonicus OTOH, we have 1 + A = A
And B - A - 1 = B
Last fiddled with by retina on 2020-03-26 at 19:52 Reason: typo
2020-03-27, 13:09 #6
wpolly
Sep 2002
Vienna, Austria
3338 Posts
Quote:
Originally Posted by retina Very good. If we assign: A = 1 + 1 + 1 + 1 + 1 + 1 + ... B = 1 + 2 + 3 + 4 + 5 + 6 + ... Then B - A = B, since pairwise subtraction gives 0 + 1 + 2 + 3 + ... = 1 + 2 + 3 + ... Therefore A = 1 + 1 + 1 + 1 + 1 + 1 + ... = 0
Sadly, in the regime of regularized sums, we no longer have 0 + 1 + 2 + 3 + ... = 1 + 2 + 3 + ...
2020-03-27, 13:31 #7
retina
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2·3,343 Posts
Quote:
Originally Posted by wpolly Sadly, in the regime of regularized sums, we no longer have 0 + 1 + 2 + 3 + ... = 1 + 2 + 3 + ...
How come?
2020-04-01, 15:41 #8 kriesel "TF79LL86GIMPS96gpu17" Mar 2017 US midwest 1CC716 Posts Let A = 1 + 1 + 1 + ... and A(i) = 1, defined for i a positive integer Let B = 1 + 2 +3 + ... and B(i) = i, defined for i a positive integer Sum(B)=(i^2+i)/2 sum(A) = i Sum(B)/sum(a) = (i^2+i)/2 / i = (i+1)/2 i=inf Sum(B)/sum(A) = inf/2 Sum (Ai/Bi) = 1/i: sum is 1 + log i ~ log i Sum (Bi/Ai) = i; sum is ( i^2 + i ) /2 limit as i->inf of Sum(Bi/Ai) / Sum (Ai/Bi) = (i^2 + i) /2 /(1+log i) ~inf^2/log(inf)/2 The value of S1 = 1-1+1-1... is not 0.5, the average of an even number of terms. It has no single value. It's 0,1,0,1,... for sums of 0 or more terms. It's a biased square wave. It has DC amplitude 0.5 and AC amplitude 0.5. The hand-wave in the video is literal and telling. A series is bounded at the low end of the list, like a semi-infinite line or finite line. Terms preceding the first element are not zero, they are nonexistent and therefore undefined. Zero is a value; they have no value. Summing two copies of S2, with one shifted, introduces an undefined term into the sum, not a zero: S2 = 1 - 2 + 3 - 4 ... S2a = undef + 1 - 2 + 3 - 4 ... S2+S2a = (1-undef) -1 +1 -1 +1 ... One could play the same trick with S1 to cancel the AC component S1 = 1 - 1 + 1 - 1 ... S1'= undef + 1 - 1 + 1 ... S1+S1' = 1-undef + 0 +0 +0... and get the result 2 * S1 = 1 so S1 = 1/2 if ignoring the undef problem and the fact S1 and S1' are different series. And shift S1 the other way ignoring one first term (half-cycle): S1"= - 1 + 1 - 1 ... s1 = 1 - 1 + 1 ... 2 S1 = 0; S1 = 0 But 2 S1 = 1 from earlier, so 1 = 0. In the S2 portion of the video, there's a sleight of hand in discarding zero terms. The series obtained is 0 4 0 8 0 12 ... which is reduced in the video to the series 4 8 12 ... which seems to rescale it on the "time" axis and omit some terms, drastically changing it into a staircase series. That changes the values of initial terms from 0 4 0 8 0 12 to 4 8 12 16 20 24, and the partial sums from 24 to 84 for equal number of initial terms. Making a series out of the nonzero elements of the series 0 4 0 8 0 12 ... creates a new different series. Going back to the originals, series A and B, apply L'Hopital's rule. g = sumA(i) = i; g' = 1 f = sumB(i) = (i^2+i)/2; f' = i + 1/2 f(inf)/g(inf) = lim f'(i) / g'(i) = (i + 1/2) / 1 = i + 1/2 = inf. Series B is not only greater than series A, it is INFINITELY greater. Numerator and denominator are both positive. https://math.hmc.edu/calculus/hmc-ma...hopitals-rule/ Last fiddled with by kriesel on 2020-04-01 at 15:51
2020-04-01, 15:48 #9
retina
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2·3,343 Posts
Quote:
Originally Posted by kriesel Series B is not only greater than series A, it is INFINITELY greater. Numerator and denominator are both positive.
Good job.
Now we have two answers.
A > B, and
B > A
Anyone want to speculate now that A = B?
2020-04-01, 16:01 #10
VBCurtis
"Curtis"
Feb 2005
Riverside, CA
160816 Posts
Quote:
Originally Posted by retina Anyone want to speculate now that A = B?
Sure! They look like exactly the same size of infinity to me.
2020-04-01, 16:07 #11
retina
Undefined
"The unspeakable one"
Jun 2006
My evil lair
1A1E16 Posts
Quote:
Originally Posted by VBCurtis Sure! They look like exactly the same size of infinity to me.
Good job.
Now, anyone to suggest that they are uncomparable and all the above answers are meaningless?
Why are infinities so confusing?
Thread Tools
All times are UTC. The time now is 19:41.
Sun Feb 5 19:41:03 UTC 2023 up 171 days, 17:09, 1 user, load averages: 0.36, 0.57, 0.72
Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2023, Jelsoft Enterprises Ltd.
This forum has received and complied with 0 (zero) government requests for information.
Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.
≠ ± ∓ ÷ × · − √ ‰ ⊗ ⊕ ⊖ ⊘ ⊙ ≤ ≥ ≦ ≧ ≨ ≩ ≺ ≻ ≼ ≽ ⊏ ⊐ ⊑ ⊒ ² ³ °
∠ ∟ ° ≅ ~ ‖ ⟂ ⫛
≡ ≜ ≈ ∝ ∞ ≪ ≫ ⌊⌋ ⌈⌉ ∘ ∏ ∐ ∑ ∧ ∨ ∩ ∪ ⨀ ⊕ ⊗ 𝖕 𝖖 𝖗 ⊲ ⊳
∅ ∖ ∁ ↦ ↣ ∩ ∪ ⊆ ⊂ ⊄ ⊊ ⊇ ⊃ ⊅ ⊋ ⊖ ∈ ∉ ∋ ∌ ℕ ℤ ℚ ℝ ℂ ℵ ℶ ℷ ℸ 𝓟
¬ ∨ ∧ ⊕ → ← ⇒ ⇐ ⇔ ∀ ∃ ∄ ∴ ∵ ⊤ ⊥ ⊢ ⊨ ⫤ ⊣ … ⋯ ⋮ ⋰ ⋱
∫ ∬ ∭ ∮ ∯ ∰ ∇ ∆ δ ∂ ℱ ℒ ℓ
𝛢𝛼 𝛣𝛽 𝛤𝛾 𝛥𝛿 𝛦𝜀𝜖 𝛧𝜁 𝛨𝜂 𝛩𝜃𝜗 𝛪𝜄 𝛫𝜅 𝛬𝜆 𝛭𝜇 𝛮𝜈 𝛯𝜉 𝛰𝜊 𝛱𝜋 𝛲𝜌 𝛴𝜎𝜍 𝛵𝜏 𝛶𝜐 𝛷𝜙𝜑 𝛸𝜒 𝛹𝜓 𝛺𝜔 | 3,381 | 8,601 | {"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": 0, "wp_latex": 0, "mimetex.cgi": 2, "/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-2023-06 | latest | en | 0.784603 |
Subsets and Splits