url
string | fetch_time
int64 | content_mime_type
string | warc_filename
string | warc_record_offset
int32 | warc_record_length
int32 | text
string | token_count
int32 | char_count
int32 | metadata
string | score
float64 | int_score
int64 | crawl
string | snapshot_type
string | language
string | language_score
float64 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://www.tiwariacademy.in/ncert-solutions/class-10/maths/chapter-13-exercise-13-2/
| 1,721,089,659,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-30/segments/1720763514724.0/warc/CC-MAIN-20240715224905-20240716014905-00494.warc.gz
| 905,884,959 | 10,588 |
NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.2
NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.2 of Surface Areas and Volume in English Medium and Hindi Medium to study online as well as Chose the other exercise like Exercise 13.1 or Exercise 13.3 or Exercise 13.4 or Exercise 13.5 to use it online or view offline. NCERT Solutions for Class 10 other subjects in PDF format is given to download.
NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.2
If you need solutions in Hindi, Click for Hindi Medium solutions of 10 Maths Exercise 13.2
Exercise 13.1
Exercise 13.3
Exercise 13.4
Exercise 13.5
10 Maths Main Page
Class 10 Maths Exercise 13.2 Solutions in Hindi Medium
Exercise 13.1
Exercise 13.3
Exercise 13.4
Exercise 13.5
10 Maths Main Page
To get the solutions in English, Click for English Medium solutions.
Important Questions of Surface Areas and Volumes Exercise 13.2
1. Volume of two spheres is in the ratio 64:125. Find the ratio of their surface areas. [Answer: 16:25]
2. A cylinder and a cone are of same base radius and of same height. Find the ratio of the volumes of cylinder to that of the cone. [Answer: 3:1]
3. If the volume and the surface area of a sphere are numerically equal, then find the radius of the sphere. [Answer: 3 units]
4. If the volume of a cube is 1331 cm³, then find the length of its edge. [Answer: 11 sq. cm]
5. What does the “CAPACITY” for a hollow cylinder means? [Answer: Volume]
6. How many cubes of side 2 cm can be cut from a cuboid measuring (16 cm × 12 cm × 10 cm). [Answer: 240]
7. Find the height of largest right circular cone that can be cut out of a cube whose volume is 729 cm³. [Answer: 9 cm]
8. Twelve solid spheres of the same sizes are made by melting a solid metallic cylinder of base diameter 2 cm and height 16 cm. Find the radius of each sphere. [Answer: 1 cm]
9. The diameters of the two circular ends of the bucket are 44 cm and 24 cm. The height of the bucket is 35 cm. Find the volume of the bucket. [Answer: 32706.6 cm³]
10. The volume of two hemi-sphere are in the ratio 8:27. Find the ratio of their radii. [Answer: 2:3]
11. Find the volume of the largest right circular cone that can be cut out from a cube of edge 4.2 cm? [Answer: 19.4 cm³]
12. A cubical ice cream brick of edge 22 cm is to be distributed among some children by filling ice cream cones of radius 2cm and height 7cm up to its brim. How many children will get ice cream cones? [Answer: 363]
13. Find the volume of the largest right circular cone that can be cut out from a cube of edge 4.9 cm is? [Answer: 30.8 cm³]
14. Three metallic solid cubes whose edges are 3 cm, 4 cm, and 5 cm are melted and converted into a single cube .Find the edge of the cube so formed? [Answer: 6 cm]
15. How many shots each having diameter 4.2 cm can be made from a cuboidal lead solid of dimensions 66 cm × 42 cm × 21 cm? [Answer: 1500]
16. A spherical glass vessel has a cylindrical neck 7 cm long and 4 cm in diameter. The diameter of the spherical part is 21 cm. Find the quantity of water it can hold. [Answer: 4939 cm³]
17. The surface area of a solid metallic sphere is 616 cm2. It is melted and recast into a cone of height 28 cm. Find the diameter of the base of the cone so formed. [Answer: 14 cm]
18. A cylindrical vessel, with internal diameter 10 cm and height 10.5 cm is full of water. A solid cone of base diameter 7 cm and height 6 cm is completely immersed in water. Find the volume of (i) water displaced out of the cylindrical vessel. (ii) water left in the cylindrical vessel. [Answer: (i): 77 cm³, (ii) 748 cm³]
| 1,002 | 3,594 |
{"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-2024-30
|
latest
|
en
| 0.861581 |
https://askgoodquestions.blog/2019/11/11/19-lincoln-and-mandela-part-1/
| 1,680,041,321,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2023-14/segments/1679296948871.42/warc/CC-MAIN-20230328201715-20230328231715-00309.warc.gz
| 158,026,667 | 31,451 |
# #19 Lincoln and Mandela, part 1
Two great leaders will be featured in this post and the next: Abraham Lincoln and Nelson Mandela. Well, to be honest, featured is too strong, but these men provide the background for in-class activities that help students to understand two very important concepts in statistics: random sampling and random assignment.
When I first mention these two terms in class, I suspect that many students only hear random and don’t pay much attention to sampling versus assignment. I admit that I did not make a big deal of this distinction myself when I started teaching. But now I try to emphasize that random sampling and random assignment are very different ideas with very different goals. In a nutshell:
• Random sampling concerns how to select observational units for a sample. Random sampling allows for generalizing the results of a sample to the larger population.
• Random assignment pertains to how observational units come to be in groups to be compared. Random assignment allows for the possibility of drawing a cause-and-effect conclusion.
This post will discuss random sampling with reference to Lincoln, and the next will concern random assignment while mentioning Mandela. Along the way we’ll sneak in a touch of history and also some psychology. As always, questions for students appear in italics.
I begin this activity by asking students to consider the 268 words in this speech as the population of interest:
The natural first question is: What speech is this, and who wrote it? I’m glad that most students recognize this as Lincoln’s Gettysburg Address. Then I give these instructions:
• Circle ten words as a representative sample from this population.
• For each word in your sample, record how many letters are in the word.
• Calculate the average (mean) number of letters per word in your sample.
• Plot your sample average on a dotplot on the board, along with the sample averages of your classmates.
Those who remember post #11 (here) will not be surprised that I next ask students: Identify the observational units and variable, first in your sample and then for the graph on the board. For the students’ samples of ten words, the observational units are words, and the variable is the length of the word, as measured by number of letters. But for the dotplot that students produce on the board, the observational unit are samples of 10 words, and the variable is the average length of a word.
All of this is prelude to the important question: How can we use the dotplot on the board to tell whether this sampling method (my telling students to circle ten words) is any good? Before a student will respond, I often have to add: What additional information would you like to know to help you decide whether this sampling method was good? At this point a student usually responds that they would like to know the average word length in the entire population of 268 words. I reply: Great idea, and before class I calculated this population average to be 4.295 letters per word. Then I draw a vertical line through the dotplot at this value. Here are results from a recent class:
At this point I define sampling bias as a systematic tendency for a sampling method to over-represent some observational units and under-represent others. Then I ask: Would you say that this sampling method (my asking students to circle ten words) is biased? If so, in which direction? How can you tell from the dotplot?
Students recognize that a large majority of the sample averages are greater than the population average. This means that there’s a systematic tendency for this sampling method to over-represent large words and under-represent small words. In other words, this sampling method is biased toward over-estimating the average length of a word in the Gettysburg Address.
I emphasize to students that sampling bias is a property of the sampling method, not of any one sample generated by the method. One illustration of this is to ask: Whose idea was it to select a sample by circling ten words based solely on human judgment? Students reply, somewhat sheepishly, that it was my idea. I respond that this is absolutely right: The sampling bias here is my fault, not theirs, because the sampling method was my idea.
Then I ask: Suggest some reasons for why this sampling method turned out to be biased in this way. Students are quick to suggest good explanations for this sampling bias. They mention that longer words (such as government, battlefield, and consecrate) convey the meaning of the speech better than smaller words (such as a, by, and for). Students also suggest that longer words are more likely to be selected because they are just more interesting than smaller words.
Next I ask whether sample size is the problem: Would asking people to circle twenty words (rather than ten) eliminate, or at least reduce, the sampling bias? Most students realize that taking a larger sample of words would not help with this problem, because people would still be prone to select larger words rather than smaller ones.
Before we conclude this discussion of biased sampling, I ask students to give me a chance to redeem myself by proposing a new sampling method: Suppose that I ask you to close your eyes and point at the page ten times in order to select words for your sample. Would this sampling method be unbiased? (After all, doesn’t closing your eyes guarantee a lack of bias?) Explain. Most students correctly realize that this sampling method is still biased toward longer words. You would be more likely to select longer words than shorter ones, because longer words take up more space on the page.
Finally, I ask: Suggest a different sampling method that would be unbiased. Some students immediately respond with a magic word: random! So I follow up with: What does it mean to select a random sample of words in this situation? This question is harder, but eventually a student says that random sampling gives every word, whether it is an interesting word such as dedicate or a boring word like of, the same chance of being selected.
We then proceed to examine properties of random sampling. Sometimes I ask students to generate their own random samples of words from this population. One option for doing this is to give them a numbered list of the 268 words and then use a random number generator (such as the one at random.org) to select their sample. They can then calculate their sample mean word length and put a dot on a new dotplot on the board, using the same scale as the original dotplot.
Another option is to move directly to using an applet (available here) to select random samples of words. This applet starts by showing the distribution of word lengths in the population, which is skewed to the right:
You can select random samples by first clicking on Show Sampling Options. I ask students to start by selecting one random sample of 5 words, which produces a result such as:
The applet calculates the sample mean word length for this sample and plots that on a graph. Then asking the applet to select 999 more samples results in a graph of sample means that looks like:
Now we’re ready for the key questions: Does this distribution of sample means indicate sampling bias or unbiasedness of this random sampling method? What aspect of the distribution leads you to this conclusion? The shape and variability in this distribution are completely irrelevant to the issue of sampling bias. To address this issue, we focus on the center of the distribution. We see that the center of the distribution of sample means is very close to the population mean. We can quantify this by noting that the mean of the 1000 sample means is 4.336 letters/word, which is quite close to the population mean of 4.295 letters/word. Therefore, this random sampling method appears to be unbiased.
Before moving on, I want to point out how challenging the following statement can be for students:
The mean of the sample means is the population mean.
This sentence contains only ten words, but three of them are the word mean(s)! We can rewrite this statement mathematically, using common notation, as:
Notice that this equation contains only three symbols (in addition to the equals sign), but all three of them describe a mean! It takes considerable time and careful thought for students to recognize and understand what these three means are and how they relate to each other:
• The population mean. For the population of 268 words in the Gettysburg Address, the value of the population mean is 4.295 letters/word.
• The sample mean, which varies from sample to sample. Each student calculated his/her own sample mean and represented it with a dot on the board. The first random sample generated by the applet above had a sample mean of 3.6 letters/word. The applet then generated 999 more random samples and calculated the sample mean number of letters/word for each one.
• The mean of the sample means. We could have calculated this for the students’ sample means in class; we did not bother, but we know from the graph that the mean of the sample means would have been much greater than 4.295. The applet did calculate the mean of the 1000 sample means that it generated; the mean of these sample means turned out to be 4.336 letters/word. If we went on to generate all possible random samples, in the long run the mean of the sample means would be 4.295, the same value as the population mean.
My next question for students: Consider taking random samples of size 20 words per sample, rather than 5 words per sample. How (if at all) would you expect the distribution of sample means to change, in terms of center, variability, and shape? After students think about this, discuss it among themselves, and record their predictions, we use the applet to make this change, which produces a result such as:
We see that the center of this distribution is still close to the population mean of 4.295 letters/word. Most students expect this, because this simply shows that random sampling is still unbiased with a larger sample size. The key finding is that the variability of sample means is smaller with a larger sample size. How can we tell? One way is that the sample means now range from about 3 to 6 letters/word, whereas before (with a smaller sample size of 5) they ranged from about 2 to 8 letters/word. Even better, we can note that the standard deviation of the sample means is now about 0.463, which is much less than its value of 0.945 with the smaller sample size. The shape of the distribution of sample means is a bit more symmetric and normal-looking with the larger sample size than with the smaller sample size, much less skewed than the distribution of the population.
This last point foreshadows the concept of a sampling distribution of a sample mean and the Central Limit Theorem. I think this context and applet provide a great opportunity to study those ideas*, but at this point I prefer to keep the focus on the topics of sampling bias and random sampling.
* One feature that I particularly like about this applet is that it displays three distributions at once, which are crucial (and challenging) for students to keep in mind when studying sampling distributions:
• Population distribution (of word lengths)
• Sample distribution (of word lengths)
• Sampling** distribution (of average word lengths in a sample)
** It’s very unfortunate that the modifier words sample and sampling are so similar, yet the distributions they describe are precisely a key distinction to understand. Perhaps we should avoid using the term sampling distribution and instead say distribution of sample averages. It’s nice to be able to use shorthand when speaking with colleagues who understand the ideas, but in this case the extra words provide clarity for students who are just beginning to consider the ideas.
Before leaving the topic of sampling bias and random sampling, I ask a few more questions of my students, all in the context of selecting a sample of students at our university to complete a survey:
• Would it be easy or hard to select a random sample of 50 Cal Poly students?
It takes a while for some students to realize that selecting such a random sample would be very hard to achieve. It’s unlikely that university administrators would provide a list of all students at the university. Having access to such a list would enable us to select a random sample of students’ names, but we would still face the challenges of contacting them successfully and then, even more problematic, convincing them to respond to our survey.
• Suppose that you select a sample of Cal Poly students by standing in front of the library or recreation center and approaching 50 students who pass by. Would this constitute a random sample of Cal Poly students? What if you stand in front of the recreation center and approach 50 students who pass by?
Most students realize that this sampling method (standing in one location and recruiting passersby) does not constitute random sampling. Some students would be more likely to be selected than others, in part because they are out-and-about on campus more often. It’s also likely that you would be more likely to approach students who appear to be …, well, …, approachable, as opposed to students who look more intimidating or less friendly. Even though the word random is used in an everyday sense to mean anything that is unplanned or unstructured, random sampling has a technical meaning.
• Even though the convenience sampling described above is not random, could it nevertheless result in a sample that is representative of the population of Cal Poly students? Identify a variable for which you would not be willing to consider such a convenience sample (as described above) to be representative of the population of Cal Poly students. Also identify a variable for which you would be willing to consider such a sample (as described above) to be representative of the population of Cal Poly students.
We should certainly not consider a convenience sample, selected from students who pass by the library or recreation center, to be representative of the population for most variables, such as how often a student uses the recreation center per week, and whether or not a student knows where the library is on campus. We should also be wary for variables about the student’s major, or how many hours they study per week, or how much sleep they get per night. But there’s probably no reason to doubt that such a sample is representative of the population for a variable such as blood type.
I have used far more than 268 words to write this post. Clearly I am much less economical with words than Abraham Lincoln in his Gettysburg Address. I look forward to name-dropping Nelson Mandela into the next post, which will feature random assignment and discuss how that is quite different from random sampling.
P.S. Beth Chance and I developed the Gettysburg Address activity based the famous “random rectangles” activity developed by Dick Scheaffer and others. As I told Dick when I interviewed him for the Journal of Statistics Education (here), I suspect that random rectangles is the most widely used activity for teaching statistics of all time, at least among activities that do not involve M&M candies. You can read more about the genesis of the random rectangles activity in this JSE article (here).
P.P.S. This website (here) provides six different versions of the Gettysburg Address, with minor variations (and slightly different numbers of words) among them. The one used above is the Hay copy.
| 3,226 | 15,716 |
{"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-2023-14
|
latest
|
en
| 0.918878 |
https://www.physicsforums.com/threads/continuous-function-on-intervals.279066/
| 1,477,261,626,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2016-44/segments/1476988719437.30/warc/CC-MAIN-20161020183839-00412-ip-10-171-6-4.ec2.internal.warc.gz
| 985,042,129 | 16,068 |
Continuous function on intervals
1. Dec 11, 2008
rapple
1. The problem statement, all variables and given/known data
f:(0,1]->R be a continuous function. Is it possible that f does not have an absolute min or max. Give counter examples
2. Relevant equations
3. The attempt at a solution
Since f is partially bounded, if I break the interval down into smaller sub intervals, each will have an abs min and max. So I will have at least one of the extremes.
2. Dec 11, 2008
Dick
The problem says 'give counterexamples'. Do that. They aren't hard to find.
3. Dec 11, 2008
VeeEight
"if I break the interval down into smaller sub intervals, each will have an abs min and max."
If you break the interval into a union smaller intervals, you will always have one interval of the form (0, a) where a is some real in (0,1]. What is the difference between [0,1] and (0,1]? Does a continuous function on [0,1] attain a min/max?
Think of some functions on (0,1] that may not have a min or max (you might want to think of f(x) as x approaches 0)
4. Dec 12, 2008
rapple
How about f(x) = 1 if x=1 and f(x) = 1/(x-1) if x Not= 1. Then as x->0 , f(x) is increasing and as x-> 1, f(x) is decreasing. Since it is 1 at one, it is not the minimum.
5. Dec 12, 2008
Dick
That f is also not continuous on (0,1]. Keep it simple. How about f(x)=1/x?
6. Dec 12, 2008
rapple
Then f has an abs min at x=1. even though it is continuous
7. Dec 12, 2008
Dick
Oh, I thought you wanted no max or no min. You want neither max NOR min. How about a function that oscillates with increasing increasing frequency and magnitude as x->0? Can you give an example of one of those?
8. Dec 12, 2008
rapple
How about sin(1/x)/x. But I don't think it is continuous because of the sharp peaks. Is it a form of sin function?
9. Dec 12, 2008
Dick
That's perfect. It's continuous on (0,1]. The only place it has a problem is at x=0. But x=0 is not in (0,1].
10. Dec 12, 2008
rapple
Thank you. You seem to have a way of making it happen!
| 607 | 2,016 |
{"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-2016-44
|
longest
|
en
| 0.940319 |
https://brilliant.org/discussions/thread/2-2-4-in-coq/
| 1,627,595,183,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-31/segments/1627046153897.89/warc/CC-MAIN-20210729203133-20210729233133-00532.warc.gz
| 171,603,731 | 13,109 |
# $2 + 2 = 4$ in Coq
Inspiration: Another Test Problem from Kelly's Group.
Coq is an interactive proof assistant. What this means is that Coq helps you prove theorems in a way that can be mechanically checked by Coq itself. This technology could be used to prove mathematical theorems, or develop formally correct software.
A famous example of a notorious theorem whose formal proof was developed in Coq is the four color theorem (download the proof here). Also, it has been used to generate a formally correct C compiler, which is pretty cool, if you think about it.
Behold we prove $2 + 2 = 4$ in Coq as a teaser
To run the proof below, you'll need Coq. You can read more about the Coq Proof Assistant on Wikipedia
We shall begin by defining the natural numbers along with two and four.
1 2 3 4 5 6 Inductive nat : Type := | O : nat | S : nat -> nat. Definition two : nat := S (S O). Definition four : nat := S (S (S (S O))).
We then, define plus
1 2 3 4 5 6 7 8 Fixpoint plus (n : nat) (m : nat) : nat := match n with | O => m | S n' => S (plus n' m) end. Notation "x + y" := (plus x y) (at level 50, left associativity).
Now, we can prove the theorem we desire.
1 2 3 Theorem two_plus_two_four : two + two = four. Proof. simpl. reflexivity. Qed.
And we are done!
Alternately, you could also use this:
1 2 Theorem two_plus_two_four : two + two = four. Proof. trivial. Qed.
Note by Agnishom Chattopadhyay
4 years ago
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
• Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.
MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2
paragraph 1
paragraph 2
[example link](https://brilliant.org)example link
> This is a quote
This is a quote
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$
Sort by:
Another marvelous proof, you couldn't have eludicated any clearer!
I was thinking of proving 2+2=4 using accelerated course, but yours is definitely far superior to mine.
- 4 years ago
| 948 | 3,436 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 11, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "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}
| 3.953125 | 4 |
CC-MAIN-2021-31
|
latest
|
en
| 0.9168 |
http://mathhelpforum.com/calculus/43316-alternating-series.html
| 1,526,924,643,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-22/segments/1526794864461.53/warc/CC-MAIN-20180521161639-20180521181639-00136.warc.gz
| 185,889,773 | 9,642 |
1. ## Alternating Series
Determine the convergence or divergence of the series.
(E = sigma)
infinity
E (-1)^n+1 (2) / e^n - e^-n
n=1
2. i assume it is $\displaystyle \frac{2}{e^n - e^{-n}}$..
notice that it's a sequence of hyperbolic cosecant.. $\displaystyle \frac{2}{e^n - e^{-n}} = \frac{1}{sinh \,\, n} = csch \,\, n$
and from n=1 to infinity, that is monotone and bounded..
3. Originally Posted by element
Determine the convergence or divergence of the series.
(E = sigma)
infinity
E (-1)^n+1 (2) / e^n - e^-n
n=1
Lets put this in math notation:
$\displaystyle \sum_{n=1}^{\infty} (-1)^{n+1} \frac{2}{e^n-e^{-n}}$
Well for large $\displaystyle n$, $\displaystyle \frac{2}{e^n-e^{-n}}$ is strictly decreasing and goes to zero, and so the alternating series test applies and your series is convergent.
RonL
4. yeah, i forgot to mention that $\displaystyle \lim_{n\rightarrow \infty} csch \,\, n = 0$
| 300 | 914 |
{"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}
| 3.65625 | 4 |
CC-MAIN-2018-22
|
latest
|
en
| 0.71741 |
https://gmatclub.com/forum/m02-q14-what-if-the-x-and-y-are-negative-76276.html?sort_by_oldest=true
| 1,495,627,928,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-22/segments/1495463607813.12/warc/CC-MAIN-20170524112717-20170524132717-00106.warc.gz
| 748,570,730 | 59,491 |
Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack
It is currently 24 May 2017, 05:12
### 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
# M02 Q14, what if the X and y are negative?
new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
Intern
Joined: 17 Feb 2008
Posts: 28
Followers: 0
Kudos [?]: 19 [2] , given: 5
M02 Q14, what if the X and y are negative? [#permalink]
### Show Tags
05 Mar 2009, 00:15
2
KUDOS
2
This post was
BOOKMARKED
$$x$$ and $$y$$ are positive integers and $$x > y$$ . What is the value of $$xy^2 + yx^2$$ ?
1. xy = 6
2. x is a prime number
[Reveal] Spoiler: OA
C
Source: GMAT Club Tests - hardest GMAT questions
$$xy^2+ yx^2 = xy(x+y)$$
* From S1, we know that $$xy=6$$ , but we do not know what is $$x+y$$ . Insufficient
* From S2, we know that $$x$$ is a prime number, but there is no information about $$y$$ . Insufficient
Combining the statements, we know that $$xy=6$$ . So $$x$$ is a prime number and $$x>y$$ , therefore $$y=2$$ . The problem can then be solved.
The correct answer is C.
----------------------------------------------------------
But what happens if X & Y are negative as thats a possbility? So the answer must be E
SVP
Joined: 29 Aug 2007
Posts: 2476
Followers: 70
Kudos [?]: 774 [0], given: 19
Re: m02. Q14, what if the X and y are negative? [#permalink]
### Show Tags
05 Mar 2009, 06:20
Forrester300 wrote:
$$x$$ and $$y$$ are positive integers and $$x > y$$ . What is the value of $$xy^2 + yx^2$$ ?
1. xy = 6
2. x is a prime number
(C) 2008 GMAT Club - m02#14
* Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
* Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
* BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
* EACH statement ALONE is sufficient
* Statements (1) and (2) TOGETHER are NOT sufficient
$$xy^2+ yx^2 = xy(x+y)$$
* From S1, we know that $$xy=6$$ , but we do not know what is $$x+y$$ . Insufficient
* From S2, we know that $$x$$ is a prime number, but there is no information about $$y$$ . Insufficient
Combining the statements, we know that $$xy=6$$ . So $$x$$ is a prime number and $$x>y$$ , therefore $$y=2$$ . The problem can then be solved.
The correct answer is C.
----------------------------------------------------------
But what happens if X & Y are negative as thats a possbility? So the answer must be E
2 things to note:
1. x and y are +ves
2. x is a prime, which can never be -ve.
hth
_________________
Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html
GT
Senior Manager
Joined: 18 Aug 2009
Posts: 301
Followers: 4
Kudos [?]: 297 [0], given: 9
Re: M02 Q14, what if the X and y are negative? [#permalink]
### Show Tags
10 Nov 2009, 22:50
The solution can be reached even if X>Y not given (because value of X is not asked, so it doesn't matter as long as we find out that X can be either 2 or 3 and depending on the value of X, Y can be one of 2 or 3).
So unless "X>Y" is meant to be a red herring, it can be removed from the Q.
Senior Manager
Joined: 01 Feb 2010
Posts: 264
Followers: 1
Kudos [?]: 58 [0], given: 2
Re: M02 Q14, what if the X and y are negative? [#permalink]
### Show Tags
18 Mar 2010, 07:26
Forrester300 wrote:
$$x$$ and $$y$$ are positive integers and $$x > y$$ . What is the value of $$xy^2 + yx^2$$ ?
1. xy = 6
2. x is a prime number
[Reveal] Spoiler: OA
C
Source: GMAT Club Tests - hardest GMAT questions
$$xy^2+ yx^2 = xy(x+y)$$
* From S1, we know that $$xy=6$$ , but we do not know what is $$x+y$$ . Insufficient
* From S2, we know that $$x$$ is a prime number, but there is no information about $$y$$ . Insufficient
Combining the statements, we know that $$xy=6$$ . So $$x$$ is a prime number and $$x>y$$ , therefore $$y=2$$ . The problem can then be solved.
The correct answer is C.
----------------------------------------------------------
But what happens if X & Y are negative as thats a possbility? So the answer must be E
Its C. explanation given above.
It is given that x & y are (+)ve.
Intern
Joined: 30 Jul 2009
Posts: 20
Followers: 0
Kudos [?]: 4 [0], given: 3
Re: M02 Q14, what if the X and y are negative? [#permalink]
### Show Tags
18 Mar 2010, 10:24
Well.. Q is saying, X and Y is +ve and they are both integers and X>Y.
ST 1 is XY=6, X could be either 6 and Y=1 OR x=3 and Y=2
so its insuff
ST 2... we dont know anything about Y or any value for any of X or Y... insuff
Together x has to be 3 and Y is 2... suff
Ans C
Senior Manager
Joined: 13 Dec 2009
Posts: 263
Followers: 10
Kudos [?]: 193 [0], given: 13
Re: M02 Q14, what if the X and y are negative? [#permalink]
### Show Tags
19 Mar 2010, 07:29
Question is saying that both the numbers are positive integers, so there is no question of considering the case when they are negative.
_________________
My debrief: done-and-dusted-730-q49-v40
Manager
Joined: 03 Aug 2010
Posts: 105
GMAT Date: 08-08-2011
Followers: 1
Kudos [?]: 73 [0], given: 63
Re: M02 Q14, what if the X and y are negative? [#permalink]
### Show Tags
22 Mar 2011, 05:57
Question stem reworded: What's the value of XY (Y + X) when x & Y are positive integers and x > y?
1) xy = 6. We still need to determine (Y+X) (insufficient)
2) X is prime. X can be 2, 3, 5, 7, etc and we know nothing of Y (insufficient)
1 and 2) Together, xy=6. Therefore x or y has to be an even integer. Since x is prime and larger than Y, X has to be 3 and Y 2. Both statements together provide that Xy(Y+X) = 30
Ans: C
Senior Manager
Joined: 01 Nov 2010
Posts: 289
Location: India
Concentration: Technology, Marketing
GMAT Date: 08-27-2012
GPA: 3.8
WE: Marketing (Manufacturing)
Followers: 10
Kudos [?]: 80 [0], given: 44
Re: M02 Q14, what if the X and y are negative? [#permalink]
### Show Tags
22 Mar 2011, 06:02
ANS IS C.
XY(X+Y)=?
ON COMBINING TWO STATEMENTS;
XY=6
AND X IS PRIME AND X>Y; ONLY 3 SATISFIES BOTH CONDITION.
SO, X=3, Y=2
ONE CAN FIND THE ANS AS 30.
_________________
kudos me if you like my post.
Attitude determine everything.
all the best and God bless you.
SVP
Joined: 16 Nov 2010
Posts: 1666
Location: United States (IN)
Concentration: Strategy, Technology
Followers: 34
Kudos [?]: 533 [0], given: 36
Re: M02 Q14, what if the X and y are negative? [#permalink]
### Show Tags
22 Mar 2011, 07:10
xy^2 + yx^2
= xy(X+y)
(1) xy = 6, so x = 6 and y = 1, or x = 3 and y = 2, (x+y) is different, so not sufficient.
(2) x is prime, but no information about y, so not sufficient.
From (1) and (2), x = 3, y = 2, so sufficient.
_________________
Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)
GMAT Club Premium Membership - big benefits and savings
Manager
Status: Seeking new horizons...
Joined: 03 Sep 2010
Posts: 64
Location: Taiwan
Concentration: Strategy, Technology
Schools: ESADE '15 (A)
Followers: 2
Kudos [?]: 26 [0], given: 29
Re: M02 Q14, what if the X and y are negative? [#permalink]
### Show Tags
22 Mar 2011, 21:43
If they state that x and y are negative integers, x cannot be a prime number.
Thus the question would become invalid.
~Nilay
_________________
Learn to walk before you run.
Intern
Joined: 13 Mar 2012
Posts: 2
Followers: 0
Kudos [?]: 0 [0], given: 0
Re: M02 Q14, what if the X and y are negative? [#permalink]
### Show Tags
26 Mar 2012, 09:50
The correct answer is C.
if notice that that statement one gives you (x)(y)=6
then you can see that the only factors for 6 is (1)(6) and (2)(3)
for X to be greater than Y. you will need x to be either 6 or 3
Now the second statement tells you that x is prime leaving us with the possibility of 3.
Intern
Joined: 20 Dec 2011
Posts: 19
Followers: 0
Kudos [?]: 2 [0], given: 2
Re: M02 Q14, what if the X and y are negative? [#permalink]
### Show Tags
26 Mar 2012, 12:18
Easy one from the GMATCLUB which rarely happens. Such questions really motivate you if face such questions after doing many wrong.
Manager
Joined: 15 Aug 2012
Posts: 110
Location: India
Concentration: Technology, Strategy
Schools: Merage '15 (A)
GPA: 3.6
WE: Consulting (Computer Software)
Followers: 6
Kudos [?]: 47 [0], given: 22
Re: M02 Q14, what if the X and y are negative? [#permalink]
### Show Tags
27 Mar 2013, 05:13
x and y are positive integers and x>y . What is the value of xy2 + yx2 ?
1. xy = 6
2. x is a prime number
For Case 1:
xy = 6 and x>y, has two solution x,y = {3,2 and 6,1}
for 3,2 the value is 12 + 18 = 30
for 6,1 the value is 6 + 36 = 42
For Case 2:
x is prime=> x can be 2,3,5
and y can have any value so the equation has infinite values.
Combining case 1 and 2:
The only possible solution for x,y is 3,2 and the equation solves to 30.
Hence C.
Intern
Joined: 11 Mar 2013
Posts: 8
Followers: 0
Kudos [?]: 6 [1] , given: 11
Re: M02 Q14, what if the X and y are negative? [#permalink]
### Show Tags
27 Mar 2013, 05:29
1
KUDOS
Can anyone tell the aproximate level of this question?
Intern
Joined: 27 Oct 2011
Posts: 15
Concentration: Finance, Technology
GMAT Date: 04-10-2012
Followers: 0
Kudos [?]: 22 [0], given: 7
Re: M02 Q14, what if the X and y are negative? [#permalink]
### Show Tags
27 Mar 2013, 05:45
Solution C
1) xy = 6
Considering x>y, below two scenarios can come up
a) x = 6
y = 1
Putting the above x & Y values in the equuation xy2+ yx2 = 6+36 = 42
b) x = 3
y = 2
Putting the above x & Y values in the equuation xy2+ yx2 = (3*4)+(2*9) = 40
Two solutions so 1 itself is not sufficient.
2) x is a prime number
x can be anything 1,3,5, 7.. so as y. not sufficient
Taking 1 & 2 together
XY = 6, X is a prime number & X >Y we get unique values for X & Y
i.e. X=3 & Y = 2, this will get you the unique solutin of the equation xy2+ yx2
Manager
Joined: 12 Dec 2012
Posts: 230
GMAT 1: 540 Q36 V28
GMAT 2: 550 Q39 V27
GMAT 3: 620 Q42 V33
GPA: 2.82
WE: Human Resources (Health Care)
Followers: 4
Kudos [?]: 79 [0], given: 181
Re: M02 Q14, what if the X and y are negative? [#permalink]
### Show Tags
27 Mar 2013, 06:36
chose A ... x+y is not known but has a fixed value .... therefore sufficient . Can anybody please explain why this cannot be correct?
_________________
My RC Recipe
http://gmatclub.com/forum/the-rc-recipe-149577.html
My Problem Takeaway Template
http://gmatclub.com/forum/the-simplest-problem-takeaway-template-150646.html
Math Expert
Joined: 02 Sep 2009
Posts: 38853
Followers: 7722
Kudos [?]: 105972 [2] , given: 11602
Re: M02 Q14, what if the X and y are negative? [#permalink]
### Show Tags
27 Mar 2013, 06:50
2
KUDOS
Expert's post
TheNona wrote:
chose A ... x+y is not known but has a fixed value .... therefore sufficient . Can anybody please explain why this cannot be correct?
When a DS question asks about the value of some variable (or an expression with variable(a)), then the statement(s) is sufficient ONLY if you can get the single numerical value of this variable (expression).
x and y are positive integers and x > y . What is the value of xy^2 + yx^2 ?
What is the value of $$xy^2 + yx^2=xy(y+x)$$?
(1) xy = 6 --> $$xy(y+x)=6(y+x)$$. Now, since x and y are positive integers and x > y, then from xy = 6 it follows that x=6 and y=1 OR x=3 and y=2, thus 6(y+x)=42 or 6(y+x)=30. Two different values, thus this statement is NOT sufficient.
(2) x is a prime number. Clearly insufficient.
(1)+(2) Since from (2) we know that x is a prime number, then x=3=prime and y=2, thus 6(y+x)=42. Sufficient.
Hope it's clear.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 38853
Followers: 7722
Kudos [?]: 105972 [0], given: 11602
Re: M02 Q14, what if the X and y are negative? [#permalink]
### Show Tags
27 Mar 2013, 06:54
Recobita wrote:
Can anyone tell the aproximate level of this question?
I'd say the difficulty level is ~600.
_________________
Manager
Joined: 12 Dec 2012
Posts: 230
GMAT 1: 540 Q36 V28
GMAT 2: 550 Q39 V27
GMAT 3: 620 Q42 V33
GPA: 2.82
WE: Human Resources (Health Care)
Followers: 4
Kudos [?]: 79 [0], given: 181
Re: M02 Q14, what if the X and y are negative? [#permalink]
### Show Tags
27 Mar 2013, 07:15
Bunuel wrote:
TheNona wrote:
chose A ... x+y is not known but has a fixed value .... therefore sufficient . Can anybody please explain why this cannot be correct?
When a DS question asks about the value of some variable (or an expression with variable(a)), then the statement(s) is sufficient ONLY if you can get the single numerical value of this variable (expression).
x and y are positive integers and x > y . What is the value of xy^2 + yx^2 ?
What is the value of $$xy^2 + yx^2=xy(y+x)$$?
(1) xy = 6 --> $$xy(y+x)=6(y+x)$$. Now, since x and y are positive integers and x > y, then from xy = 6 it follows that x=6 and y=1 OR x=3 and y=2, thus 6(y+x)=42 or 6(y+x)=30. Two different values, thus this statement is NOT sufficient.
(2) x is a prime number. Clearly insufficient.
(1)+(2) Since from (2) we know that x is a prime number, then x=3=prime and y=2, thus 6(y+x)=42. Sufficient.
Hope it's clear.
yessssssssssssssssss ... I did not consider the option of 6& 1... only 2&3 . Thanks Bunuel
_________________
My RC Recipe
http://gmatclub.com/forum/the-rc-recipe-149577.html
My Problem Takeaway Template
http://gmatclub.com/forum/the-simplest-problem-takeaway-template-150646.html
Intern
Joined: 02 Nov 2012
Posts: 2
Followers: 0
Kudos [?]: 0 [0], given: 5
Re: M02 Q14, what if the X and y are negative? [#permalink]
### Show Tags
27 Mar 2013, 11:38
saratchandra wrote:
Solution C
1) xy = 6
Considering x>y, below two scenarios can come up
a) x = 6
y = 1
Putting the above x & Y values in the equuation xy2+ yx2 = 6+36 = 42
b) x = 3
y = 2
Putting the above x & Y values in the equuation xy2+ yx2 = (3*4)+(2*9) = 40
Two solutions so 1 itself is not sufficient.
2) x is a prime number
x can be anything 1,3,5, 7.. so as y. not sufficient
Taking 1 & 2 together
XY = 6, X is a prime number & X >Y we get unique values for X & Y
i.e. X=3 & Y = 2, this will get you the unique solutin of the equation xy2+ yx2
So far the best explanation-Thanks Dude.
Re: M02 Q14, what if the X and y are negative? [#permalink] 27 Mar 2013, 11:38
Go to page 1 2 Next [ 22 posts ]
Similar topics Replies Last post
Similar
Topics:
4 M26-25 If x and y are negative numbers 14 12 Aug 2014, 01:58
1 M02 Q18 Is the total number of divisors of x^3 a multiple 3 18 Jan 2013, 15:26
29 What is the number of integers from 1 to 1000 (m07q14) 18 09 Apr 2013, 08:48
40 m02#24 14 17 Jul 2013, 05:55
22 m02#8 16 16 Dec 2013, 20:48
Display posts from previous: Sort by
# M02 Q14, what if the X and y are negative?
new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Moderator: Bunuel
Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.
| 5,063 | 15,535 |
{"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}
| 4.34375 | 4 |
CC-MAIN-2017-22
|
latest
|
en
| 0.846171 |
http://www.enotes.com/homework-help/what-value-1-x-sqrt-x-2-2x-limit-x-equal-inf-243589
| 1,477,649,582,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2016-44/segments/1476988721606.94/warc/CC-MAIN-20161020183841-00530-ip-10-171-6-4.ec2.internal.warc.gz
| 420,168,481 | 10,243 |
# What is the value of 1/(x - sqrt(x^2 + 2x)) for limit x equal to inf.
Asked on by xetaalpha2
justaguide | College Teacher | (Level 2) Distinguished Educator
Posted on
We have to find lim x--> inf, [1/(x - sqrt(x^2 + 2x))].
[1/(x - sqrt(x^2 + 2x))]
=> (x + sqrt(x^2 + 2x)/ (x - sqrt(x^2 + 2x)* (x + sqrt(x^2 + 2x)
=> (x + sqrt(x^2 + 2x)/ x^2 - x^2 - 2x
=> (x + sqrt(x^2 + 2x)/-2x
=> (x/-2x) – sqrt (x^2/4x^2 + 2x/4x^2)
=> -1/2 – sqrt (1/4 + 1/2x)
The required limit is now:
lim x--> +inf [-1/2 – sqrt (1/4 + 1/2x)]
substitute x = + inf, 1/x = 0
=> -1/2 – sqrt (1/4)
=> -1/2 – 1/2
=> -1
The required result of the limit is -1.
giorgiana1976 | College Teacher | (Level 3) Valedictorian
Posted on
To determine the limit of the fraction, we'll substitute infinite into the expression and we'll get, at denominator, the indeterminacy case: infinite - infinite.
We'll multiply the fraction by the conjugate of denominator, that is x + sqrt(x^2 + 2x).
We'll get to the denominator the difference of squares:
[x - sqrt(x^2 + 2x)][x + sqrt(x^2 + 2x)] = x^2 - x^2 - 2x
We'll eliminate like terms and we'll get:
[x - sqrt(x^2 + 2x)][x + sqrt(x^2 + 2x)] = -2x
We'll re-write the fraction:
1/[x - sqrt(x^2 + 2x)] = - [x + sqrt(x^2 + 2x)]/2x
We'll apply limit both sides:
lim - [x + sqrt(x^2 + 2x)]/2x
We'll factorize by x the numerator:
lim - x[1+ sqrt(1 + 2/x)]/2x
We'll simplify and we'll get:
lim - [1+ sqrt(1 + 2/x)]/2 = -2/2 = -1
lim 1/[x - sqrt(x^2 + 2x)] = -1, x -> infinite
We’ve answered 317,954 questions. We can answer yours, too.
| 626 | 1,564 |
{"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-2016-44
|
latest
|
en
| 0.774812 |
https://mathforums.com/threads/calculations-of-protons.348169/
| 1,585,637,494,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2020-16/segments/1585370500331.13/warc/CC-MAIN-20200331053639-20200331083639-00054.warc.gz
| 525,000,754 | 16,792 |
# calculations of protons
#### JeffMv2
Information:
Mass of proton:$$\displaystyle 1.67×10^−27kg$$
Mass of electron: $$\displaystyle 9.11×10^−31kg$$
Proton charge $$\displaystyle +1.60×10^-19 C$$
Question:
A proton is accelerated through a voltage of 248 kV.
a) Calculate the energy of the proton in electron volt and in Joule.
b) Assuming that the mass of the proton remains the same as its rest mass, calculate the speed of the photon.
c)The proton is curved by a magnetic field of 1.5 Telsa. Calculate the radius of the curvature of its path.
My workings:
a) is 248keV.
$$\displaystyle 3.97*10^{−14 }J.$$
b) v = √(2K/m)
$$\displaystyle √(2*3.97*10^{-14})/1.67*10^{-27}) = 6.89354 * 10^6 ms^-1$$
c) r = mv/(eB) r=$$\displaystyle 1.67*10^{-19}* 6.89354 *10^6/(1.60*10^{-19}*1.5) = 4.8 * 10^6 m$$
Have I done these correctly?
#### skeeter
Math Team
(a) and (b) are ok
check the mass value of the proton you used in part (c)
#### JeffMv2
(a) and (b) are ok
check the mass value of the proton you used in part (c)
$$\displaystyle (1.67∗10^-27∗6.89354∗10^6)/(1.60∗10^−19∗1.5)=0.048$$ is it nm or m?
#### skeeter
Math Team
$$\displaystyle (1.67∗10^-27∗6.89354∗10^6)/(1.60∗10^−19∗1.5)=0.048$$ is it nm or m?
do a dimensional analysis
$\dfrac{mv}{qB} = \dfrac{(kq)(m/s)}{(C)(T)}$
what are the SI units for a Tesla?
topsquark
#### JeffMv2
do a dimensional analysis
$\dfrac{mv}{qB} = \dfrac{(kq)(m/s)}{(C)(T)}$
what are the SI units for a Tesla?
right so 0.048m
#### JeffMv2
my numerical value is correct?
Math Team
| 578 | 1,539 |
{"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}
| 3.578125 | 4 |
CC-MAIN-2020-16
|
latest
|
en
| 0.640694 |
https://teacher.7generationgames.com/category/educational-resources/lessons/10-minute-lesson/page/3/
| 1,718,841,229,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-26/segments/1718198861853.72/warc/CC-MAIN-20240619220908-20240620010908-00043.warc.gz
| 498,152,616 | 12,270 |
## Ten Math PowerPoints for Grade 7
Check out these middle school lesson plans by math teacher Zeb Hammond. Featuring data analysis, statistics, probability, ratios, rational numbers, proportional relationships, and number sense.
## Lesson Plans: Multiplying by N
Standards CCSS.MATH.CONTENT.3.OA.A.1 – Interpret products of whole numbers, e.g., interpret 5 × 7 as the total number of objects in 5 groups of 7 objects each. CCSS.MATH.CONTENT.3.OA.A.3– Use multiplication and division within 100 to solve word problems in situations involving equal groups, arrays, and measurement quantities. CCSS.MATH.CONTENT.3.OA.A.4-Determine the unknown whole number in a multiplication or
## Video: Ojibwe Clan Lessons
Lesson and Video Resources: Clans of the Ojibwe People Learn Ojibwe History and Culture with Making Camp Premium from 7 Generation Games 7 Generation Games Clan Lesson Open your 7 Generation Games app, Making Camp. If you need a download, visit our site to purchase a download (1 user license) for \$1.99. Go to the
## Teaching with primary sources: Buffalo hunting
10-Minute Lesson Plan on Primary Sources
## Introducing Fractions
Standards CCSS.MATH.CONTENT.4.NF.B.3Understand a fraction a/b with a > 1 as a sum of fractions 1/b. CCSS.MATH.CONTENT.3.NF.A.1Understand a fraction 1/b as the quantity formed by 1 part when a whole is partitioned into b equal parts; understand a fraction a/b as the quantity formed by a parts of size 1/b. Summary Before you jump into teaching how to add, subtract, multiply or divide
| 381 | 1,532 |
{"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-2024-26
|
latest
|
en
| 0.75831 |
https://math.answers.com/math-and-arithmetic/How_do_you_turn_.535_into_a_whole_number
| 1,723,116,095,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-33/segments/1722640726723.42/warc/CC-MAIN-20240808093647-20240808123647-00060.warc.gz
| 316,349,570 | 47,892 |
0
# How do you turn .535 into a whole number?
Updated: 9/27/2023
Wiki User
6y ago
0.535 is a fractional number in decimal form and there is no sensible way to turn it into a whole number.
Wiki User
6y ago
Earn +20 pts
Q: How do you turn .535 into a whole number?
Submit
Still have questions?
Related questions
### Round 535 to nearest tenth?
Since 535 is a whole number - it's already been rounded !
### Is 535 an even or odd number?
535 is an odd number.
### What is divisible by 535?
The factors of 535 are: 1, 5, 107 & 535.
### How do you turn a whole number into a percentage?
To turn a whole number into a percentage: Simply multiply it by 100%.
### How do you turn 710 into a whole number?
710 is a whole number.
### How do you turn a negative number to a whole number?
You do nothing. A negative number is a whole number.
### How do you turn a whole number to a percent?
Multiply the whole number to 100%.
### How Do You Turn 152 Into A Whole Number?
Step 1: Close you eyes.Step 2: Open them. Its done! 152 is a whole number so you do not need to do anything to "turn it" into a whole number!
1, 5, 107, 535.
### How do you turn 25 in a whole number?
25 IS a whole number: you need do nothing.
### How do you turn 4.53 into whole number?
It rounds up to 5 as a whole number
### How do you change a decimal to a whole number?
you turn the decimal into a fraction then you turn the fraction into a whole number and add a percent sign
| 392 | 1,473 |
{"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}
| 3.921875 | 4 |
CC-MAIN-2024-33
|
latest
|
en
| 0.858217 |
https://www.dataunitconverter.com/kibibit-to-yottabit/6
| 1,679,502,900,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2023-14/segments/1679296943845.78/warc/CC-MAIN-20230322145537-20230322175537-00773.warc.gz
| 805,428,889 | 11,745 |
# Kibibit to Yottabit - 6 Kibit to Ybit Conversion
## Conversion History (Last 6)
Input Kibibit - and press Enter
Kibit
RESULT ( Kibibit → Yottabit ) :
6 Kibit = 0.000000000000000000006144 Ybit
Copy
Calculated as → 6 x 1024 / 10008...view detailed steps
## Kibit to Ybit - Conversion Formula and Steps
Kibibit and Yottabit are units of digital information used to measure storage capacity and data transfer rate. Kibibit is a binary standard unit where as Yottabit is decimal. One Kibibit is equal to 1024 bits. One Yottabit is equal to 1000^8 bits. There are 976,562,500,000,000,000,000 Kibibits in one Yottabit.
Source Data UnitTarget Data Unit
Kibibit (Kibit)
Equal to 1024 bits
(Binary Unit)
Yottabit (Ybit)
Equal to 1000^8 bits
(Decimal Unit)
The formula of converting the Kibibit to Yottabit is represented as follows :
Ybit = Kibit x 1024 / 10008
Note : Here we are converting the units between different standards. The source unit Kibibit is Binary where as the target unit Yottabit is Decimal. In such scenario, first we need to convert the source unit to the basic unit - Bit - multiply with 1024, and then convert to target unit by dividing with 1000^8 .
Now let us apply the above formula and, write down the steps to convert from Kibibit (Kibit) to Yottabit (Ybit).
1. STEP 1 → Yottabit = Kibibit x 1024 / 10008
2. STEP 2 → Yottabit = Kibibit x 1024 / (1000x1000x1000x1000x1000x1000x1000x1000)
3. STEP 3 → Yottabit = Kibibit x 1024 / 1000000000000000000000000
4. STEP 4 → Yottabit = Kibibit x 0.000000000000000000001024
If we apply the above steps, conversion from 6 Kibit to Ybit, will be processed as below.
1. = 6 x 1024 / 10008
2. = 6 x 1024 / (1000x1000x1000x1000x1000x1000x1000x1000)
3. = 6 x 1024 / 1000000000000000000000000
4. = 6 x 0.000000000000000000001024
5. = 0.000000000000000000006144
6. i.e. 6 Kibit is equal to 0.000000000000000000006144 Ybit.
(Result rounded off to 40 decimal positions.)
#### Definition : Kibibit
A Kibibit (Kib or Kibit) is a unit of digital information that is equal to 1024 bits. It is defined by the International Electro technical Commission(IEC) and is used to measure the amount of digital data. The prefix "kibi" is derived from the binary number system, it is used to distinguish it from the decimal-based "kilobit" (Kb) and it is widely used in the field of computing as it more accurately represents the amount of data storage and data transfer in computer systems.
#### Definition : Yottabit
A Yottabit (Yb or Ybit) is a unit of measurement for digital information transfer rate. It is equal to 1,000,000,000,000,000,000,000,000 (one septillion) bits. It is used to measure the speed of extremely high-speed data transfer over communication networks, such as high-speed internet backbones and advanced computer networks.
### Excel Formula to convert from Kibit to Ybit
Apply the formula as shown below to convert from 6 Kibibit to Yottabit.
ABC
1Kibibit (Kibit)Yottabit (Ybit)
26=A2 * 0.000000000000000000001024
3
| 948 | 2,995 |
{"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}
| 3.765625 | 4 |
CC-MAIN-2023-14
|
latest
|
en
| 0.819626 |
http://en.wikibooks.org/wiki/Circuit_Theory/1Source_Excitement/Example_7/code
| 1,432,554,937,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2015-22/segments/1432207928486.86/warc/CC-MAIN-20150521113208-00068-ip-10-180-206-219.ec2.internal.warc.gz
| 83,587,857 | 7,638 |
# Circuit Theory/1Source Excitement/Example 7/code
In Matlab:
```R = 10;
L = .01;
w = 377;
Vm = 120*2^.5;
omega = 2*pi/3;
Im = Vm/(R^2+(L*w)^2)^.5;
alpha = omega - atan(L*w/R);
syms t A C;
I = Im * cos(w*t+alpha) + A*exp(-t/(L/R)) + C;
leftSide = Vm*cos(w*t + omega) - I*R;
rightSide = L * diff(I,t);
vpa(leftSide, 3);
vpa(rightSide,3);
t=0;
teq = subs(leftSide) - subs(rightSide)
vpa(teq,3)
Ieval = subs(I);
vpa(Ieval,3)
S = solve(Ieval,teq,A,C);
S = [S.A S.C];
vpa(S(1),3)
vpa(S(2),3)
```
In MuPad and MatLab (have to cut and paste)
```R := 10;
L := .01;
w := 377;
Vm := 120*2^.5;
omega := 2*pi/3;
Imagnitude := Vm/(R^2+(L*w)^2)^.5;
alpha := omega - atan(L*w/R);
Itime := Imagnitude * cos(w*t+alpha) + A*exp(-t/(L/R)) + C;
dItime := diff(Itime,t);
t:=0;
numeric::solve([Itime=0,Vm*cos(w*t + omega) - Itime*R - L * dItime = 0],[A,C],FixedPrecision)
sig1 = 5.986596875*sin(2*pi/3 - atan(0.377));
sig2 = cos(2*pi/3 - atan(0.377));
A = -16.97056275*cos(2*pi/3) - 2.775557562*10^-17*sig2 - sig1
C = 16.97056275*cos(2*pi/3) - 15.87956731*sig2+ sig1
```
| 469 | 1,057 |
{"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-2015-22
|
latest
|
en
| 0.441816 |
http://www.ck12.org/algebra/Use-Square-Roots-to-Solve-Quadratic-Equations/
| 1,487,590,385,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-09/segments/1487501170521.30/warc/CC-MAIN-20170219104610-00154-ip-10-171-10-108.ec2.internal.warc.gz
| 352,246,423 | 15,166 |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Use Square Roots to Solve Quadratic Equations
## Isolate the squared term, take the root of both sides
Levels are CK-12's student achievement levels.
Basic Students matched to this level have a partial mastery of prerequisite knowledge and skills fundamental for proficient work.
At Grade (Proficient) Students matched to this level have demonstrated competency over challenging subject matter, including subject matter knowledge, application of such knowledge to real-world situations, and analytical skills appropriate to subject matter.
Advanced Students matched to this level are ready for material that requires superior performance and mastery.
## Use Square Roots to Solve Quadratic Equations
Solve quadratic equations involving perfect squares.
MEMORY METER
This indicates how strong in your memory this concept is
3
## Solving Quadratics Using Square Roots
This lesson covers solving quadratic equations using square roots.
MEMORY METER
This indicates how strong in your memory this concept is
0
• PLIX
## Plus or Minus
Use Square Roots to Solve Quadratic Equations Interactive
MEMORY METER
This indicates how strong in your memory this concept is
0
• Video
## Solving Quadratic Equations Using Square Roots: A Sample Application
This video demonstrates an example of solving quadratic equations using square roots.
MEMORY METER
This indicates how strong in your memory this concept is
1
• Video
## Solving Quadratic Equations Using Square Roots: An Explanation of the Concept
This video provides an explanation of the concept of solving quadratic equations using square roots.
MEMORY METER
This indicates how strong in your memory this concept is
0
## Use Square Roots to Solve Quadratic Equations Quiz
Quiz for Using Square Roots to Solve Quadratic Equations.
MEMORY METER
This indicates how strong in your memory this concept is
0
• 1
• Critical Thinking
## Use Square Roots to Solve Quadratic Equations Discussion Questions
A list of student-submitted discussion questions for Use Square Roots to Solve Quadratic Equations.
MEMORY METER
This indicates how strong in your memory this concept is
0
• Real World Application
| 477 | 2,284 |
{"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}
| 3.875 | 4 |
CC-MAIN-2017-09
|
latest
|
en
| 0.828835 |
https://www.edithvaleps.vic.edu.au/students/mathsshare/2019-2/2019-t2-w6-3-6/
| 1,686,031,457,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2023-23/segments/1685224652235.2/warc/CC-MAIN-20230606045924-20230606075924-00142.warc.gz
| 799,049,523 | 16,866 |
# Term 2 - May 27th (3-6)
##### Year 3 to Year 6: Button Up Some More
Write down the number 4, four times.
Put operation symbols between them so that you have a calculation.
So you might think of writing 4×4×4−4=60
BUT use operations so that the answer is 12
Now, can you redo this so that you get 15, 16 and 17 for your answers?
Need more of a challenge? Try getting answers all the way from 0 through to 10.
#### Solution
12=(44+4)÷4 – Here’s one solution!
16=4+4+4+4
17=(4×4)+(4÷4)
1=(4÷4)×(4÷4)
2=(4/4) + (4/4)
3=(4+4+4)÷4
6=((4+4)÷4)+4
7=(44÷4)−4
8=(4×4)−4−4
9=(4÷4)+4+4
15=(4×4)−(4÷4)
Then I came to a problem. I couldn’t do 4, 5 and 10. Then I realised that I had only used the simple operations, addition, subtraction, multiplication and division. Then I remembered about square root, factorial and powers. An example of factorial is:
4!=4×3×2×1
The first square root I tried was 4, then I realised I needed a few more. Taking these into account, I was able to do the last few problems.
4=(√4+4+4+4)
5=(√4!+(4÷4)4)
10=(√4!+(4÷4)×4)
Prev Next
We respectfully acknowledge the traditional owners of the land on which Edithvale Primary School is located and pay respects to their Elders past and present and emerging.
| 418 | 1,232 |
{"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-2023-23
|
longest
|
en
| 0.932576 |
https://geekhaus.com/math103_fall2017/2017/12/14/fractal-menagerie/
| 1,726,312,621,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725700651579.22/warc/CC-MAIN-20240914093425-20240914123425-00759.warc.gz
| 253,562,283 | 9,921 |
# Fractal Menagerie
Maddy Biggins
Cat Falvey
Yang Zhou
## Fractals
A fractal is a curve or geometric figure which parts’ have the same statistical characteristics as the whole. Fractals follow patterns or rules that recur at progressively larger or smaller scales. An iteration is one repetition of this rule, and fractals have many iterations or levels. As a part of this class we were first asked to research fractals, and find ones that we found interesting. We were then asked to come up with our own rule and create our own fractals. This menagerie contains many of the fractals that were found and created over the course of this class. As a part of these projects we made calculations and prints of our fractals. Some of the calculations we made were for perimeter, area, length, and dimension. With more complex objects these calculations were very difficult, but below we were able to illustrate more simple calculations that are easy to follow.
## Perimeter
The koch snowflake is one of the most famous and well documented fractals in mathematics. It is based off of the von koch curve. This fractal has an infinite perimeter. We got the 4/5 value for calculations since each side of the triangle is being divided into thirds.
Level 0
P=(4/3)^0 x 3
P=1(3)=3
Level 1
P=(4/3)^1 x 3
P=(4/3) x (3/1) = 12/3
P=4
Level 2
P=(4/3)^2 x 3
P=(16/9) x (3/1) = 48/9
P=16/3
## Area
The area of this fractal is finite because its decimal value only goes up slightly every iteration, so it will continue to not exceed a bounded area.
| 410 | 1,558 |
{"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}
| 4.09375 | 4 |
CC-MAIN-2024-38
|
latest
|
en
| 0.948662 |
https://www.smartkeeda.com/Quantitative_Aptitude/Arithmetic/Approximations_Quiz/newest/all/passage/Approximations_Quiz_6/
| 1,627,476,948,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-31/segments/1627046153729.44/warc/CC-MAIN-20210728123318-20210728153318-00035.warc.gz
| 1,054,529,346 | 20,603 |
# Approximation Questions for SBI Clerk Pre 2021, NRA CET, IBPS Clerk Pre 2021, IBPS RRB 2021
Direction: What approximate value should come in the place of question mark (?) in the following questions?
Important for :
1
33.05% of 5010.11 + (1 / 12.93) of 2404.99 = ?
» Explain it
D
33.05% of 5010.11 + (1 / 12.93) of 2404.99 = ?
? ≈ 33% of 5010 + (1 / 12.93) of 2405
⇒ ? = (33 / 100) × 5010 + (1 / 13) × 2405
= 1653.30 + 185 = 1838.30 ≈ 1838
Hence, option D is correct.
2
98.98 ÷ 11.03 + 7.014 × 15.99 = ?
» Explain it
D
98.98 ÷ 11.03 + 7.014 × 15.99 = ?
Suppose, ? = x
Then 99 ÷ 11 + 7 × 16 = 9 + 112 = 121 (taking approximate value)
x = 121
Hence, option D is correct.
3
(16)2 – (13)2 + (3910.03 / ?) = 102
» Explain it
E
(16)2 – (13)2 + (3910.03 / ?) = 102
256 – 169 + (3910 / ?) ≈ 102
(3910 / ?) = 102 – 87
(3910 / ?) = 15
? = (3910 / 15) = 260.66 ≈ 261
Hence, option E is correct.
4
134% of 3894 + 38.94% of 134 = ?
» Explain it
D
134% of 3894 + 38.94% of 134
= 134% of 3894 + 3894 of 1.34%
= 134% of 3894 + 1.34% of 3894
= 135.34% of 3894
= 5270.1396 ≈ 5270
Hence, option D is correct.
5
(21 + 99) × (30 – 19.02) = ?
» Explain it
E
(21 + 99) × (30 – 19.02)
= 120 × 10.98
= 120 × 11 = 1320
Hence, option E is correct.
### Approximation Questions SBI Clerk 2021, IBPS Clerk 2021
Simplification and Approximation Questions are the most important chapters for IBPS Clerk Exams, In this almost 5 to 10 is being asked. For this you must be good at calculation, how fast you can calculate values and you can download simplification and approximation questions pdf at smartkeeda at free of cost, at this page, you will also get simplification and approximation questions for bank clerk pdf.
Are you looking for any of these?
• approximation questions
• simplification and approximation pdf
• approximation questions pdf
• simplification and approximation
• simplification and approximation questions for bank clerk pdf
• simplification and approximation questions
• simplification and approximation questions pdf
| 745 | 2,041 |
{"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}
| 4.40625 | 4 |
CC-MAIN-2021-31
|
latest
|
en
| 0.817112 |
https://bestbtcxonyngj.netlify.app/schirpke85902xeba/what-is-the-stated-annual-interest-rate-dem
| 1,713,302,943,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-18/segments/1712296817106.73/warc/CC-MAIN-20240416191221-20240416221221-00463.warc.gz
| 106,974,104 | 10,329 |
## What is the stated annual interest rate
APY stands for annual percentage yield. It takes into account the interest rate and compounding period to give you a single number that represents how much
What Does Effective Interest Rate Mean? What is the definition of effective interest rate? Based on the stated or nominal rate for a given period, such as an annual 12 Feb 2020 Mortgage APR reflects the interest rate plus the fees charged by the lender. APR helps you evaluate the true cost of a mortgage. Definition of nominal annual rate: The stated annual interest rate of an investment or debt instrument. The rate does not include a compounding component and The Annual Percentage Yield is accurate as of the last dividend declaration date. Minimum daily balance to open and earn the stated Annual Percentage Yield. Practice Problems. Problem 1. If you invest \$1,000 at an annual interest rate of 5 % compounded continuously, calculate the final amount you
## 4 Aug 2019 A stated annual interest rate is the return on an investment (ROI) that is expressed as a per-year percentage. more · Determining the Annual
12 Feb 2020 Mortgage APR reflects the interest rate plus the fees charged by the lender. APR helps you evaluate the true cost of a mortgage. Definition of nominal annual rate: The stated annual interest rate of an investment or debt instrument. The rate does not include a compounding component and The Annual Percentage Yield is accurate as of the last dividend declaration date. Minimum daily balance to open and earn the stated Annual Percentage Yield. Practice Problems. Problem 1. If you invest \$1,000 at an annual interest rate of 5 % compounded continuously, calculate the final amount you «Nominal rate» - is the annual rate of interest on the credit, which is designated in the agreement with the Bank. In this example – is 18% (0, 18). «Number of 6 hours ago stated Annual Percentage Yield. All above Annual Percentage Yields are accurate as of today's date. A minimum deposit of \$1.00 is required
### The stated interest rate of a bond payable is the annual interest rate that is printed on the face of the bond. The stated interest rate multiplied by the bond's face amount (or par amount) results in the annual amount of interest that must be paid by the issuer of the bond. For example, if a cor
Access the answers to hundreds of Effective interest rate questions that are explained in a What would be the annual interest rate from that company? Stated rate(APR) Number of times Compounded Effective Rate (APR) Semiannually 17 Oct 2019 APR is the annual percentage rate: the total amount of interest you pay on a borrowed sum per year. Different interest rates. What is nominal Usually, it is presented on an annual basis, which is known as the annual percentage yield (APY) or effective annual rate (EAR) 5 Jan 2016 Typically an interest rate is given as a nominal, or stated, annual rate of interest. But when compounding occurs more than once per year, the
### stated annual interest rate: The annual interest rate that accrues, without considering the effect of compounding. If a bank states an annual rate of 7% interest, that is the stated annual interest rate.
Usually, it is presented on an annual basis, which is known as the annual percentage yield (APY) or effective annual rate (EAR) 5 Jan 2016 Typically an interest rate is given as a nominal, or stated, annual rate of interest. But when compounding occurs more than once per year, the It is used to compare the annual interest between loans with different compounding terms (daily, monthly, quarterly, semi-annually, annually, or other). It is also There are various terms used when compounding is not considered including nominal interest rate, stated annual interest rate, and annual percentage rate( APR). 12 Jun 2013 You can certainly use the formula for the effective rate. The effective six-month rate is the rate of interest, compounded every six months, you According to the Truth in Lending Act, lenders are required to disclose the APR or annual percentage rate. This figure comprises the overall yearly cost of a loan APY stands for annual percentage yield. It takes into account the interest rate and compounding period to give you a single number that represents how much
## The stated or simple interest rate is the percentage of borrowed money that you would pay if the interest charges were calculated just once at the end of a year.
The stated annual interest rate (SAR) is the return on an investment (ROI) that is expressed as a per-year percentage. It is a simple interest rate calculation that does not account for any compounding that occurs throughout the year. The effective annual interest rate (EAR), on the other hand, The stated interest rate is just what it says. It is the simple interest rate that the bank gives you as the interest rate on loan. This interest rate does not take the effect of compound interest into account. Stated Annual Interest Rate. An interest rate in a given year that does not account for more frequent compounding. For example, if a loan of \$100 has a stated annual interest rate of 5%, the amount owed at the end of the year is \$105. However, if the interest compounds monthly, the actual amount is \$105.12. stated annual interest rate. Definition. The annual interest rate that accrues, without considering the effect of compounding. If a bank states an annual rate of 7% interest, that is the stated annual interest rate. Use this term in a sentence. “ You should try to shop around to make sure you get the best stated annual interest rate that you can. Stated Annual Interest Rate. An interest rate in a given year that does not account for more frequent compounding. For example, if a loan of \$100 has a stated annual interest rate of 5%, the amount owed at the end of the year is \$105. However, if the interest compounds monthly, the actual amount is \$105.12. The stated interest rate of a bond payable is also known as the face interest rate, the nominal interest rate, the contractual interest rate, and the coupon interest rate. Generally, a bond's stated interest rate is fixed (remains constant) for the life of the bond. stated annual interest rate: The annual interest rate that accrues, without considering the effect of compounding. If a bank states an annual rate of 7% interest, that is the stated annual interest rate.
23 Jul 2013 The Annual percentage rate (APR) of a loan is the yearly interest rate expressed as a simple percentage. APR is the rate quoted by the bank or The real APR, or annual percentage rate, considers these costs as well as the interest rate of a loan. The following two calculators help reveal the true costs of 10 Nov 2015 Suppose you intend to invest Rs 1,00,000 for 10 years at an interest rate of 10 per cent and the compounding is annual. The total amount you The stated annual interest rate (SAR) is the return on an investment (ROI) that is expressed as a per-year percentage. It is a simple interest rate calculation that does not account for any compounding that occurs throughout the year. The effective annual interest rate (EAR), on the other hand, The stated interest rate is just what it says. It is the simple interest rate that the bank gives you as the interest rate on loan. This interest rate does not take the effect of compound interest into account. Stated Annual Interest Rate. An interest rate in a given year that does not account for more frequent compounding. For example, if a loan of \$100 has a stated annual interest rate of 5%, the amount owed at the end of the year is \$105. However, if the interest compounds monthly, the actual amount is \$105.12. stated annual interest rate. Definition. The annual interest rate that accrues, without considering the effect of compounding. If a bank states an annual rate of 7% interest, that is the stated annual interest rate. Use this term in a sentence. “ You should try to shop around to make sure you get the best stated annual interest rate that you can.
| 1,721 | 8,064 |
{"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-2024-18
|
latest
|
en
| 0.933233 |
https://kirit.com/Thread:/1732737
| 1,581,889,492,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2020-10/segments/1581875141430.58/warc/CC-MAIN-20200216211424-20200217001424-00442.warc.gz
| 455,830,426 | 3,383 |
Python solution Mike 31st July, 2009 06:42 (UTC)
2783915604 calculated in 0.000388 seconds on my MBP
I can't paste either the output or the code as this wiki screws it up, so see http://dpaste.com/hold/73577/ for my solution (a python class).
It works by repeatedly dividing the remaining number of combinations left from the initial 1000000, by the factorial of an integer decreasing from 9 to 0. The integer part of these divisions give an index into a list of digits that are popped off, so we always get a digit we haven't used.
PS. I wrote this on my iPhone using MobileTerminal on the way to the BeerCamp. (iPhone solution took 0.0025 s). All you can eat and jugs of beer for 89 baht at a japanese restaurant… we ended up drinking 'real' sangria back at Thomas's place where BeerCamp all started. Great fun!
Python solution Kirit Sælensminde 31st July, 2009 06:50 (UTC)
Mike said
2783915604 calculated in 0.000388 seconds on my MBP
So close, but not quite right. I think you're off by one permutation.
Python solution Mike 31st July, 2009 07:55 (UTC)
2783915460
Generalised the solution a bit more (it can now handle bigger sets of digits):
http://dpaste.com/hold/73584/
Python solution Mike 1st August, 2009 10:41 (UTC)
Thomas' solution:
```ima, zlr = 123456789, 1
while zlr < 10**6:
ima += 1
rup = ima
if rup < 10**9: rup *= 10
if len(set(str(rup)))== 10: zlr += 1
print ima```
For next week's beercamp puzzle: someone please figure out what those variables actually stand for…
Python solution Mike 1st August, 2009 10:41 (UTC)
Thomas' solution:
```ima, zlr = 123456789, 1
while zlr < 10**6:
ima += 1
rup = ima
if rup < 10**9: rup *= 10
if len(set(str(rup)))== 10: zlr += 1
print ima```
For next week's beercamp puzzle: someone please figure out what those variables actually stand for…
Python solution Mike 1st August, 2009 10:43 (UTC)
Thomas' solution:
```ima, zlr = 123456789, 1
while zlr < 10**6:
ima += 1
rup = ima
if rup < 10**9: rup *= 10
if len(set(str(rup)))== 10: zlr += 1
print ima```
For next week's beercamp puzzle: someone please figure out what those variables actually stand for…
Python solution Mike 1st August, 2009 10:43 (UTC)
Thomas' solution:
```ima, zlr = 123456789, 1
while zlr < 10**6:
ima += 1
rup = ima
if rup < 10**9: rup *= 10
if len(set(str(rup)))== 10: zlr += 1
print ima```
For next week's beercamp puzzle: someone please figure out what those variables actually stand for…
Python solution Mike 1st August, 2009 10:46 (UTC)
Oops, the www.kirit.com server kept giving me an odd error that looked like I had failed to post so I kept trying.
| 807 | 2,605 |
{"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-2020-10
|
longest
|
en
| 0.845787 |
https://ask.csdn.net/questions/373874
| 1,632,178,462,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-39/segments/1631780057119.85/warc/CC-MAIN-20210920221430-20210921011430-00649.warc.gz
| 158,078,940 | 20,134 |
2017-03-26 07:36
# Team Rankings
It��s preseason and the local newspaper wants to publish a preseason ranking of the teams in the local amateur basketball league. The teams are the Ants, the Buckets, the Cats, the Dribblers, and the Elephants. When Scoop McGee, sports editor of the paper, gets the rankings from the selected local experts down at the hardware store, he��s dismayed to find that there doesn��t appear to be total agreement and so he��s wondering what ranking to publish that would most accurately reflect the rankings he got from the experts. He��s found that finding the median ranking from among all possible rankings is one way to go.
The median ranking is computed as follows: Given any two rankings, for instance ACDBE and ABCDE, the distance between the two rankings is defined as the total number of pairs of teams that are given different relative orderings. In our example, the pair B, C is given a different ordering by the two rankings. (The first ranking has C above B while the second ranking has the opposite.) The only other pair that the two rankings disagree on is B, D; thus, the distance between these two rankings is 2. The median ranking of a set of rankings is that ranking whose sum of distances to all the given rankings is minimal. (Note we could have more than one median ranking.) The median ranking may or may not be one of the given rankings.
Suppose there are 4 voters that have given the rankings: ABDCE, BACDE, ABCED and ACBDE. Consider two candidate median rankings ABCDE and CDEAB. The sum of distances from the ranking ABCDE to the four voted rankings is 1 + 1 + 1 + 1 = 4. We��ll call this sum the value of the ranking ABCDE. The value of the ranking CDEAB is 7 + 7 + 7 + 5 = 26.
It turns out that ABCDE is in fact the median ranking with a value of 4.
Input
There will be multiple input sets. Input for each set is a positive integer n on a line by itself, followed by n lines (n no more than 100), each containing a permutation of the letters A, B, C, D and E, left-justified with no spaces. The final input set is followed by a line containing a 0, indicating end of input.
Output
Output for each input set should be one line of the form:
ranking is the median ranking with value value.
Of course ranking should be replaced by the correct ranking and value with the correct value. If there is more than one median ranking, you should output the one which comes first alphabetically.
Sample Input
4
ABDCE
BACDE
ABCED
ACBDE
0
Sample Output
ABCDE is the median ranking with value 4.
• 点赞
• 写回答
• 关注问题
• 收藏
• 邀请回答
#### 2条回答默认 最新
• blownewbee 2017-04-04 15:17
已采纳
• 罗忠浩 2017-03-27 02:58
CSDN (Chinese Software Developer Network) 创立于1999年，是中国最大的IT社区和服务平台，为中国的软件开发者和IT从业者提供知识传播、职业发展、软件开发等全生命周期服务，满足他们在职业发展中学习及共享知识和信息、建立职业发展社交圈、通过软件开发实现技术商业化等刚性需求。拥有超过3000万注册会员（其中活跃会员800万）、50万注册企业及合作伙伴。
点赞 打赏 评论
| 798 | 2,862 |
{"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-2021-39
|
latest
|
en
| 0.931626 |
http://mathhelpforum.com/business-math/199704-calculating-percentage-paid-debt-post-inflation.html
| 1,524,332,768,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-17/segments/1524125945272.41/warc/CC-MAIN-20180421164646-20180421184646-00462.warc.gz
| 203,513,939 | 12,949 |
# Thread: Calculating percentage paid on debt post inflation.
1. ## Calculating percentage paid on debt post inflation.
I'll start by presenting it as a hypothetical problem:
After studying for 3 years at university I have student debt of 65224. My interest rate is 3.15%, inflation is assumed to be 3% each year so the above inflation interest is 0.15%, I pay back 90 a year (let's call these payments deductions) and we assume that the payments rise with the rate of inflation. After 30 years, any remaining debt is wiped.
What is the best way to work out, after 30 years, what percentage of my original debt I have paid in inflation interest, above inflation interest, and deductions?
I have my own way of doing it but it is extremely long winded so I don't think it's the best way.
then inflation = 1957,
then deductions from debt = 65224 - (90*1.03) = 65131,
I then repeat the process starting with 65131,
Meanwhile I have a separate tally of what the above figures are after 30 years of inflation,
so I'll multiply each of the above figures by 1.03^remaining years of inflation.
I'll then sum up these inflated figures and divide the sum by 65224*(1.03^30) to see what each element added up to as a percentage of the original debt post inflation.
It's based on some optimistic assumptions, I know but I'm just trying to get an idea. I hope it all makes sense too.
2. ## Re: Calculating percentage paid on debt post inflation.
Originally Posted by sheldonisgod
I hope it all makes sense too.
Not to me...but don't let that bother you!
For your info: $65,224 over 30 years at annual rate of 3.15% requires annual payment of$3,392.53
3. ## Re: Calculating percentage paid on debt post inflation.
I should have clarified how what I call my annual deductions do not include the interest payments, so the inflationary interest payments make up the bulk of my payments to this debt. Also, as a rule in the UK, any student debt that remains, 30 years after graduation, does not have to be paid, its the governments problem from then on.
The way I looked at it was inflation interest charged (Retail Price Index) goes towards lowering the debt even though you do not see how it directly lowers the debt, as the debt remains the same after interest is charged. This is because as inflation occurs, the debt you owe would increase but the value of the debt would remain the same, so when you pay for inflation you're only paying for the difference between pre-inflation and post-inflation debt. When you pay interest on inflation, your debt remains as the same figure but has actually reduced in value because that figure has less buying power.
In my example, I also make the assumption that my wages increase with the rate of inflation, so therefore I decide to increase my deductions to the debt at the rate of inflation too, but in effect, the deductions are of the same value each year.
I also need to know what these figures translate to after 30 years of inflation so I can see what the annual interest payments added up to as percentage of the original debt (by debt I don't include what I have also been charged in above inflation interest, the 0.15%). Because each payment will have already been subject to x years of inflation, I need to adjust my treatment of the payment accordingly: payment*(1.03^30-x)= what payment figure looks like after 30 years.
Hope that helps!
4. ## Re: Calculating percentage paid on debt post inflation.
I should have clarified how what I call my annual deductions do not include the interest payments, so the inflationary interest payments make up the bulk of my payments to this debt. Also, as a rule in the UK, any student debt that remains, 30 years after graduation, does not have to be paid, its the governments problem from then on.
The way I looked at it was inflation interest charged (Retail Price Index) goes towards lowering the debt even though you do not see how it directly lowers the debt, as the debt remains the same after interest is charged. This is because as inflation occurs, the debt you owe would increase but the value of the debt would remain the same, so when you pay for inflation you're only paying for the difference between pre-inflation and post-inflation debt. When you pay interest on inflation, your debt remains as the same figure but has actually reduced in value because that figure has less buying power.
In my example, I also make the assumption that my wages increase with the rate of inflation, so therefore I decide to increase my deductions to the debt at the rate of inflation too, but in effect, the deductions are of the same value each year.
I also need to know what these figures translate to after 30 years of inflation so I can see what the annual interest payments added up to as percentage of the original debt (by debt I don't include what I have also been charged in above inflation interest, the 0.15%). Because each payment will have already been subject to x years of inflation, I need to adjust my treatment of the payment accordingly: payment*(1.03^30-x)= what payment figure looks like after 30 years.
Hope that helps!
5. ## Re: Calculating percentage paid on debt post inflation.
I'm more confused than I was!
.0015 * 65225 = 97.84 (~8 per month!) , so why worry about such a meaningless amount
when "estimating" over a 30year period...
Interesting reading (almost a joke!) here:
Student loans in the United Kingdom - Wikipedia, the free encyclopedia
Hope someone else here can take a few Tylenols and help you...
6. ## Re: Calculating percentage paid on debt post inflation.
Basically, what's happened here in the UK, is new students (me) commencing study in 2012 now have to pay above inflation interest rates whereas before there was no added cost in taking out a loan. With salary after graduation, for every £1 you earn above 21,000, you have to pay 0.00015% on above inflation interest up until your salary exceeds 41,000 where the above inflation interest remains at 3% as the above inflation interest stops increasing for every £1 you earn above 41,000 in your salary.
I'm trying to build a case to show why this system is more extortionate than it looks. So far, I've applied my method to a case where I earn 40K a year and have found that I pay back 132% of the original debt where 32% has to gone to above inflation interest. It's also unfair how the more you earn above 41K, the less you have to bay pack on above inflation interest.
I'm looking for formulas to make the calculation process quicker so I can try more sophisticated examples with ease such as how much more the lowest earners pay back on the principle and above inflation interest when the wages lag exponentially behind the rate of inflation (which they do).
7. ## Re: Calculating percentage paid on debt post inflation.
if you want to look at how expensive the debt is, in a way that is comparable to other forms of debt, you could start by calculating a schedule of payments and then the AER.
ill resist the temptation to comment on what you perceive as fair, since thats politics and this is a maths forum.
8. ## Re: Calculating percentage paid on debt post inflation.
Could you explain to me in more detail what I have to do and what AER is please?
Thanks
9. ## Re: Calculating percentage paid on debt post inflation.
Originally Posted by sheldonisgod
Could you explain to me what AER is please?
Aer - Feel I Bring (Official Music Video) - on iTunes - YouTube
10. ## Re: Calculating percentage paid on debt post inflation.
I don't know whether that was an insult to my ignorance or you have really poor humor. Either way I've reported you as punishment haha.
11. ## Re: Calculating percentage paid on debt post inflation.
Assumed you could "take a joke"....more so after seeing sheldonISGOD !
Don't know what "insult to my ignorance" means...anyhow, pretty hard
for me to know HOW to insult...since I don't know you from Adam...
| 1,771 | 7,940 |
{"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-2018-17
|
latest
|
en
| 0.958568 |
https://www.shaalaa.com/question-bank-solutions/comparison-densities-three-states-matter-how-does-density-liquid-or-gas-vary-temperature_30379
| 1,575,698,813,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-51/segments/1575540496492.8/warc/CC-MAIN-20191207055244-20191207083244-00347.warc.gz
| 860,401,787 | 10,505 |
ICSE Class 8CISCE
Share
# How Does the Density of a Liquid (Or Gas) Vary with Temperature? - ICSE Class 8 - Physics
ConceptComparison of Densities in the Three States of Matter.
#### Question
How does the density of a liquid (or gas) vary with temperature?
#### Solution
Most of the liquids increase in volume with increase in temperature, but water shows anomalous behaviour. Water has maximum volume at 4°C and maximum density at 4°C.
Actually, when volume increases density decreases and when volume decreases the density increases.
But water when cooled from a high temperature, contracts upto 4°C because volume decreases and expands when cooled further below 4°C and hence density of water increases when it is cooled upto 4°C while decreases when cooled further below 4°C. In other words, the density of water is maximum at 4°C equal to 1 g Cm-3 or lOOO kg m-3.
Is there an error in this question or solution?
#### APPEARS IN
Solution How Does the Density of a Liquid (Or Gas) Vary with Temperature? Concept: Comparison of Densities in the Three States of Matter..
S
| 253 | 1,082 |
{"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}
| 3.5 | 4 |
CC-MAIN-2019-51
|
latest
|
en
| 0.862327 |
https://cboard.cprogramming.com/a-brief-history-of-cprogramming-com/70312-mathematics-derivatives-sin.html
| 1,496,057,653,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-22/segments/1495463612283.85/warc/CC-MAIN-20170529111205-20170529131205-00061.warc.gz
| 935,227,113 | 12,942 |
# Thread: Mathematics <Derivatives of Sin>
1. ## Mathematics <Derivatives of Sin>
I have a problem that doesn't make sense to me.
I will post the question, then the work I have completed on it so far after.
Question:
Find the slope of the tangent line to the sine function at the origin. Compare this value with the number of complete cycles in the interval [0, 2p]. What can you conclude about the slope of the sine function sin(ax) at the origin?
f(x)=sin(x)
g(x)=sin(2x)
my graphing calculator shows a deffinate steeper incline at the origin. This would make me assume that the Δy/Δx would be increasing as the number of cycles increase, but my calculator shows me different when calculating the the slope from the derivatives of these functions.
f(x)=sin(x)
f'(x)=cos(x)
g(x)=sin(2x)
g'(x)=cos(2x)
when i evaluated f'(0) and g'(0) I get 1 for each of them. if the slope is visually identifiable as steeper, does that not mean the slope would also be analytically steeper? i know the domain for sin and cos are [-1,1], so i'm assuming this would reconfirm the calculators information, but the sloops do not look the same at point (0,0).
Please if anyone can see where I'm confused or going wrong, point me to the light.
2. g(x)=sin(2x)
g'(x)=cos(2x)
Actually, g'(x) = 2cos(2x).
3. thanks for the input.
i just noticed i was using the chain rule incorrectly on a few other problems also.
that fixes it.
;0)
anyway I have another question on the chain rule.
knowing:
dy/dx[f(u)] = f'(u)u'
y=cotx
y'=-csc^2(x)
z=cscx
z'=-cscxcotx
sinx=(1/cscx)
then shouldn't this be correct?
f(x)=cotx/sinx
f'(x)=-csc^2(x)(-cscxcotx)
=csc^3(x)cotx
=cotx/sin^3x
the book says
[-1-cos^2(x)]/sin^3(x)
i'm assuming the book wants the answer in sin / cos.
how would i get the [-1-cos^2(x)]
i'm not aware of any trig convertion that would create this.
4. Just convert it into sinuses and cosinuses and derive:
f(x)=cotx/sinx = cos(x) / sin^2(x)
f'(x) = [division rule] = (-sin^3(x) - 2sin(x)cos^2(x)) / sin^4(x) = (-1 - 2cos^2(x)) / sin^3(x)
| 634 | 2,039 |
{"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}
| 3.875 | 4 |
CC-MAIN-2017-22
|
longest
|
en
| 0.884526 |
http://www.assignmentpoint.com/science/mathematic/discuss-on-combining-and-composing-functions.html
| 1,477,608,129,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2016-44/segments/1476988721405.66/warc/CC-MAIN-20161020183841-00477-ip-10-171-6-4.ec2.internal.warc.gz
| 323,252,538 | 9,566 |
## Discuss on Combining and Composing Functions
Subject: Mathematic | Topics: ,
Basic purpose of this article is to Discuss on Combining and Composing Functions. This article briefly explain on Arithmetic combinations and composition of functions. You are able to create new functions by combining existing functions. Generally, these new functions are the effect of something as simple since addition or subtraction, but functions are capable of combining in ways in addition to those simple binary operations. Arithmetic combinations consider the easiest way to make a new function from existing functions: performing basic math operations. The composition of functions is the process of plugging one function into another is termed the composition of operates.
### Mental Math with Tricks and Shortcuts
Mental Math with Tricks and Shortcuts Addition Technique: Add left to right 326 + 678 + 245 + 567 = 900, 1100, 1600, 1620, 1690, 1730, 1790, 1804, & 1816 Note: Look for opportunities to combine numbers to reduce the number of steps to the solution. This was done with 6+8 = 14 and 5+7 [&hellip.....
### Define and Discuss on Central Limit Theorem
principle purpose of this article is to Define and Discuss on Central Limit Theorem. Here explain Central Limit Theorem with mathematical examples. The central limit theorem states that even if a population distribution is usually strongly non‐normal, its sampling distribution of means is goi.....
### Discuss on Keywords for Mathematical Operations
Primary objective of this article is to Discuss on Keywords for Mathematical Operations. Here explain Keywords for Mathematical Operations in different language point of view. The initial step in solving the mathematical word problem is usually always in order to read the problem. Every one nee.....
### Define and Discuss on Tangent Identities
Primary objective of this article is to Define and Discuss on Tangent Identities. Here explain Tangent Identities in Trigonometry point of view. Formulas with the tangent function can be produced from similar formulas involving the sine and cosine. The preceding three cases verify three formul.....
### Discrete Mathematics and its Applications based on Trees
Primary objective of this lecture is to analysis Discrete Mathematics and its Applications based on Trees. A tree is often a connected undirected graph without any simple circuits. Brief hypothesis: An undirected graph is often a tree if and only if there is a unique simple way between any two .....
| 511 | 2,523 |
{"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}
| 4.28125 | 4 |
CC-MAIN-2016-44
|
longest
|
en
| 0.824992 |
https://mamnonvinschool.edu.vn/zn-feno32
| 1,702,313,217,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2023-50/segments/1700679515260.97/warc/CC-MAIN-20231211143258-20231211173258-00116.warc.gz
| 417,322,373 | 9,302 |
# zn feno32
To balance the equation Zn + Fe(NO3)2 = Fe + Zn(NO3)2 using the algebraic method step-by-step, you must have experience solving systems of linear equations. The most common methods are substitution/elimination and linear algebra, but any similar method will work.
### Step 1: Label Each Compound With a Variable
Label each compound (reactant or product) in the equation with a variable to lớn represent the unknown coefficients.
Bạn đang xem: zn feno32
a Zn + b Fe(NO3)2 = c Fe + d Zn(NO3)2
### Step 2: Create a System of Equations
Create an equation for each element (Zn, Fe, N, O) where each term represents the number of atoms of the element in each reactant or product.
```Zn: 1a + 0b = 0c + 1d
Fe: 0a + 1b = 1c + 0d
N: 0a + 2b = 0c + 2d
O: 0a + 6b = 0c + 6d
```
### Step 3: Solve For All Variables
Use substitution, Gaussian elimination, or a calculator to lớn solve for each variable.
• 1a - 1d = 0
• 1b - 1c = 0
• 2b - 2d = 0
• 6b - 6d = 0
Use your graphing calculator's rref() function (or an online rref calculator) to lớn convert the following matrix into reduced row-echelon-form:
```[ 1 0 0 -1 0]
[ 0 1 -1 0 0]
[ 0 2 0 -2 0]
[ 0 6 0 -6 0]
```
The resulting matrix can be used to lớn determine the coefficients. In the case of a single solution, the last column of the matrix will contain the coefficients.
Simplify the result to lớn get the lowest, whole integer values.
• a = 1 (Zn)
• b = 1 (Fe(NO3)2)
• c = 1 (Fe)
• d = 1 (Zn(NO3)2)
### Step 4: Substitute Coefficients and Verify Result
Count the number of atoms of each element on each side of the equation and verify that all elements and electrons (if there are charges/ions) are balanced.
Zn + Fe(NO3)2 = Fe + Zn(NO3)2
Reactants Products Zn 1 1 ✔️ 1 1 ✔️ 2 2 ✔️ 6 6 ✔️
Since there is an equal number of each element in the reactants and products of Zn + Fe(NO3)2 = Fe + Zn(NO3)2, the equation is balanced.
The law of conservation of mass states that matter cannot be created or destroyed, which means there must be the same number atoms at the kết thúc of a chemical reaction as at the beginning. To be balanced, every element in Zn + Fe(NO3)2 = Fe + Zn(NO3)2 must have the same number of atoms on each side of the equation. When using the inspection method (also known as the trial-and-error method), this principle is used to lớn balance one element at a time until both sides are equal and the chemical equation is balanced.
### Step 1: Count the number of each element on the left and right hand sides
Reactants (Left Hand Side)Products (Right Hand Side)
ReactantsProducts
ZnFe(NO3)2TotalFeZn(NO3)2Total
Zn1111✔️
Fe1111✔️
N2222✔️
O6666✔️
The number of atoms of each element on both sides of Zn + Fe(NO3)2 = Fe + Zn(NO3)2 is equal which means that the equation is already balanced and no additional work is needed.
| 879 | 2,831 |
{"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-2023-50
|
latest
|
en
| 0.828301 |
https://www.vega-distribution.ru/solving-problems-using-simultaneous-equations-4125.html
| 1,632,303,254,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-39/segments/1631780057337.81/warc/CC-MAIN-20210922072047-20210922102047-00275.warc.gz
| 1,058,963,566 | 5,543 |
Solving Problems Using Simultaneous Equations Essay On Saving Environment
While the substitution method may be the easiest to grasp on a conceptual level, there are other methods of solution available to us.
It involves what it says − substitution − using one of the equations to get an expression of the form ‘y = …’ or ‘x = …’ and substituting this into the other equation.The object is to manipulate the two equations so that, when combined, either the x term or the y term is eliminated (hence the name) − the resulting equation with just one unknown can then be solved: Here we will manipulate one of the equations so that when it is combined with the other equation either the x or y terms will drop out.In this example the x term will drop out giving a solution for y.An option we have, then, is to add the corresponding sides of the equations together to form a new equation.Since each equation is an expression of equality (the same quantity on either side of the sign), adding the left-hand side of one equation to the left-hand side of the other equation is valid so long as we add the two equations’ right-hand sides together as well.Note: Only draw a graph if the question asks you to, it is usually quicker to work out the point two simultaneous equtions cross algebraically. This method for solving a pair of simultaneous linear equations reduces one equation to one that has only a single variable. Students need to use a pronumeral to represent the unknown number They then need to write an equation and solve it to find the value of the unknown number. Lenin invests some amount in deposit A and some amount in deposit B. He gets 10% income on deposit A and 20% income on deposit B.Several algebraic techniques exist to solve simultaneous equations.Perhaps the easiest to comprehend is the substitution method.
One thought on “Solving Problems Using Simultaneous Equations”
1. Highlighting what it means to have and take life is explored at great length through Scott’s film.
| 412 | 1,997 |
{"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}
| 4.40625 | 4 |
CC-MAIN-2021-39
|
latest
|
en
| 0.904827 |
http://thecodegallery.com/DSM/index.php
| 1,580,208,679,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2020-05/segments/1579251778168.77/warc/CC-MAIN-20200128091916-20200128121916-00100.warc.gz
| 156,118,227 | 5,272 |
## What is Data Structure?
Data Structure is a mathematical or logical way of organizing data in memory. Data Structure does not represent only data in memory but it also represents the relationship among the data that is stored in memory.
There are various operations that can be performed on Data Structure:
(1) Traversal (2) Insertion (3) Deletion (4) Searching (5) Sorting (6) Merging
Data Structure allows us to manipulate data by specifying a set of values, set of operations that can be performed on set of values and set of rules that needs to be followed while performing operations.
## Types of Data Structure
Basically Data Structure can be classified into two categories:
(1) Primitive Data Structure
(2) Non Primitive Data Structure
## Primitive Data Structure
"The Data Structure which is directly operated by machine level instruction is known as Primitive Data Structure."
All primary (built in) data types are known as Primitive data Structure.
Following are Primitive Data Structure:
(1) Integer Integer Data Structure is used to represent numbers without decimal points. For Example: 23, -45, 11 etc… Using this Data Structure we can represent positive as well as negative numbers without decimal point. (2) Floating Point Float Data Structure is used to represent numbers having decimal point. For Example: 3.2, 4.56 etc… In computer floating point numbers can be represented using normalized floating point representation. In this type of representation a floating point number is expressed as a combination of mantissa and exponent. (3) Character Character is used to represent single character enclosed between single inverted comma. It can store letters (a-z, A-Z), digit (0-9) and special symbols.
## Non Primitive Data Structure
"The Data Structure which is not directly operated by machine level instruction is known as Non Primitive Data Structure."
Non Primitive Data Structure is derived from Primitive Data Structure.
Non Primitive Data Structure are classified into two categories:
(1) Linier Data Structure
(2) Non Linier Data Structure
## Linier Data Structure
"The Data Structure in which elements are arranged such that we can process them in linier fashion (sequentially) is called linier data structure."
Following are Examples of Linier Data Structure:
(1) Array Array is a collection of variables of same data type that share common name. Array is an ordered set which consist of fixed number of elements. In array memory is allocated sequentially to each element. So it is also known as sequential list. (2) Stack Stack is a linier Data Structure in which insertion and deletion operation are performed at same end. (3) Queue Queue is a linier Data Structure in which insertion operation is performed at one end called Rear end and deletion operation is performed at another end called front end. (4) Linked List Linked list is an ordered set which consist of variable number of elements. In Linked list elements are logically adjacent to each other but they are physically not adjacent. It means elements of linked list are not sequentially stored in memory.
## Non Linier Data Structure
"The Data Structure in which elements are arranged such that we can not process them in linier fashion (sequentially) is called Non-Linier data structure."
Non Linier Data Structures are useful to represent more complex relationships as compared to Linier Data Structures.
Following are Examples of Non-Linier Data Structure:
(1) Tree A tree is a collection of one or more nodes such that: (1) There is a special node called root node. (2) All the nodes in a tree except root node having only one Predecessor. (3) Each node in a tree having 0 or more Successor. A tree is used to represent hierarchical information. (2) Graph A Graph is a collection of nodes and edges which is used to represent relationship between pair of nodes. A graph G consists of set of nodes and set of edges and a mapping from the set of edges to a set of pairs of nodes.
| 813 | 4,001 |
{"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-2020-05
|
latest
|
en
| 0.914194 |
https://studylib.net/doc/8551819/chapter-2-solutions
| 1,607,106,026,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2020-50/segments/1606141740670.93/warc/CC-MAIN-20201204162500-20201204192500-00172.warc.gz
| 501,603,874 | 12,065 |
# Chapter 2 solutions
```Stat 250.3
December 10, 2003
HOMEWORK 7 – SOLUTIONS
14.10
a. This is given by the sample slope b1 = 0.269. (Or, with more decimal places, b1= 0.26917.)
b. Approximately b1 2 s.e. (b1 ) . This is 0.26917 (2 0.6616 ) or about 0.401 to 0.137. With 95% confidence,
we can say that the slope of the regression line in the population is between 0.401 and 0.137. Note: The exact
multiplier in this situation is t * 1.981 which can be found by using software like Minitab or Excel to find the value
of t * for which the probability is 0.95 in a t-distribution that has df = n2 = 1162=114.
c. H0: 1 = 0 versus Ha: 1 0.
b1
0.26917
d. t
4.07 , df = n 2 = 116 2 = 114.
s.e.(b1 )
0.06616
e. The relationship is statistically significant. The p-value is 0.000, given in the output under “P” in the row labeled
Study.
14.11
a. s =1.509 hours. This is roughly the average deviation of individual y-values from the regression line.
b. yˆ 7.56 0.269 (4) 6.48 hours.
c. 6.48 (2 1.509 ) which is 6.48 3.018, or about 5.46 hour to 9.50 hours. Remember that about 95% of individuals
will be within two standard deviations of the mean.
d. Hours of study explains 12.7% of the observed variation in hours of sleep.
14.15
a. The notation is b1 because 7 is the slope for the sample regression line.
b. The approximate 95% C.I. is 7 (21.581) which is 7 3.162, or about 3.84 to 10.16. With 95% confidence, we
can say that in the population of college men, the mean increase in weight per 1-inch increase in height is somewhere
between about 3.84 and 10.16 pounds.
c. The confidence interval for the slope contains 5, so the statement about the population slope is reasonable.
d. The intercept is the mean value of weight when height is 0. This is meaningless because height=0 is impossible.
14.40
a. yˆ 29.981 0.57568 (65) 67.4 inches. With the more rounded off version of the equation given in the second line
of output, yˆ 30 0.576 (65) 67 .44 which is about 67.4 inches.
b. The interval is 64.946 inches to 75.612 inches. The two interpretations are: (1) In the population of college males
for whom father’s height is 70 inches, about 95% have a height in the given interval. (2) For a randomly selected
college male whose father is 70 inches tall, the probability is 0.95 that his height is in the given interval.
c. For the population of male college students whose fathers are 74 inches tall, we can say with 95% confidence that
the average height is between 71.596 and 73.566 inches.
d. The prediction intervals are wider because they include variation among the individual heights of the men whose
fathers are all the same height. The confidence intervals are more narrow because they are estimating the center
(mean) of the population of students heights at each value of father's height. The confidence intervals do not
incorporate variability among individuals.
e. Condition 2 appears to be violated. There are two outliers that may be unduly influencing the results. The two
outliers correspond to father/son heights of 66"/57" and 55"/65". In both cases, one of the heights is less than 5 feet,
which is unlikely to be correct. The original surveys should be checked for accuracy. In the case of dad's height = 55",
the point could be dropped as long as the intention is to make inferences only for students whose fathers are within the
usual range, perhaps 60" to 78" or so. It is not as easy to justify dropping the case with son's height = 57" because the
range of inference cannot be restricted for the y-variable.
```
| 1,108 | 3,636 |
{"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}
| 4.1875 | 4 |
CC-MAIN-2020-50
|
latest
|
en
| 0.885253 |
https://rustgym.com/leetcode/1320
| 1,638,569,552,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-49/segments/1637964362919.65/warc/CC-MAIN-20211203212721-20211204002721-00549.warc.gz
| 564,132,370 | 4,386 |
## 1320. Minimum Distance to Type a Word Using Two Fingers
You have a keyboard layout as shown above in the XY plane, where each English uppercase letter is located at some coordinate, for example, the letter A is located at coordinate (0,0), the letter B is located at coordinate (0,1), the letter P is located at coordinate (2,3) and the letter Z is located at coordinate (4,1).
Given the string `word`, return the minimum total distance to type such string using only two fingers. The distance between coordinates (x1,y1) and (x2,y2) is |x1 - x2| + |y1 - y2|
Note that the initial positions of your two fingers are considered free so don't count towards your total distance, also your two fingers do not have to start at the first letter or the first two letters.
Example 1:
```Input: word = "CAKE"
Output: 3
Explanation:
Using two fingers, one optimal way to type "CAKE" is:
Finger 1 on letter 'C' -> cost = 0
Finger 1 on letter 'A' -> cost = Distance from letter 'C' to letter 'A' = 2
Finger 2 on letter 'K' -> cost = 0
Finger 2 on letter 'E' -> cost = Distance from letter 'K' to letter 'E' = 1
Total distance = 3
```
Example 2:
```Input: word = "HAPPY"
Output: 6
Explanation:
Using two fingers, one optimal way to type "HAPPY" is:
Finger 1 on letter 'H' -> cost = 0
Finger 1 on letter 'A' -> cost = Distance from letter 'H' to letter 'A' = 2
Finger 2 on letter 'P' -> cost = 0
Finger 2 on letter 'P' -> cost = Distance from letter 'P' to letter 'P' = 0
Finger 1 on letter 'Y' -> cost = Distance from letter 'A' to letter 'Y' = 4
Total distance = 6
```
Example 3:
```Input: word = "NEW"
Output: 3
```
Example 4:
```Input: word = "YEAR"
Output: 7
```
Constraints:
• `2 <= word.length <= 300`
• Each `word[i]` is an English uppercase letter.
## Rust Solution
``````struct Solution;
impl Solution {
fn minimum_distance(word: String) -> i32 {
let n = word.len();
let s: Vec<i32> = word.bytes().map(|b| (b - b'A') as i32).collect();
let mut memo: Vec<Vec<Vec<i32>>> = vec![vec![vec![-1; 27]; 27]; n];
Self::dp(0, 26, 26, &mut memo, &s, n)
}
fn dp(
start: usize,
f1: i32,
f2: i32,
memo: &mut Vec<Vec<Vec<i32>>>,
s: &[i32],
n: usize,
) -> i32 {
if start == n {
0
} else {
if memo[start][f1 as usize][f2 as usize] != -1 {
return memo[start][f1 as usize][f2 as usize];
}
let mut res = std::i32::MAX;
let g = s[start];
res = res.min(Self::dp(start + 1, g, f2, memo, s, n) + Self::dist(f1, g));
res = res.min(Self::dp(start + 1, f1, g, memo, s, n) + Self::dist(f2, g));
memo[start][f1 as usize][f2 as usize] = res;
res
}
}
fn dist(f: i32, g: i32) -> i32 {
if f == 26 {
0
} else {
(f / 6 - g / 6).abs() + (f % 6 - g % 6).abs()
}
}
}
#[test]
fn test() {
let word = "CAKE".to_string();
let res = 3;
assert_eq!(Solution::minimum_distance(word), res);
let word = "HAPPY".to_string();
let res = 6;
assert_eq!(Solution::minimum_distance(word), res);
let word = "NEW".to_string();
let res = 3;
assert_eq!(Solution::minimum_distance(word), res);
let word = "YEAR".to_string();
let res = 7;
assert_eq!(Solution::minimum_distance(word), res);
}
``````
Having problems with this solution? Click here to submit an issue on github.
| 1,026 | 3,132 |
{"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-2021-49
|
longest
|
en
| 0.851142 |
http://www.gradesaver.com/textbooks/math/geometry/elementary-geometry-for-college-students-5th-edition/chapter-3-section-3-1-congruent-triangles-exercises-page-137/30
| 1,480,956,656,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2016-50/segments/1480698541773.2/warc/CC-MAIN-20161202170901-00135-ip-10-31-129-80.ec2.internal.warc.gz
| 483,787,802 | 40,166 |
## Elementary Geometry for College Students (5th Edition)
1) Show that $\overline{RM}\cong\overline{RQ}$ 2) Show that $\angle M\cong\angle Q$ 3) Show that $\angle PRQ\cong\angle NRM$ 4) Use method ASA to prove triangles congruent.
1) It is given that $\overline{PN}$ bisects $\overline{MQ}$ at R. As a result, $\overline{PN}$ cuts $\overline{MQ}$ into 2 equal parts $\overline{RM}$ and $\overline{RQ}$. Therefore, $\overline{RM}\cong\overline{RQ}$. 2) $\angle M$ and $\angle Q$ are both right angles, so $\angle M\cong\angle Q$. 3) Since $\overline{PN}$ intersects $\overline{MQ}$, $\angle PRQ\cong\angle NRM$. Now we have 2 angles and the included side of $\triangle PRQ$ are congruent with 2 corresponding angles and the included side of $\triangle NRM$. That means according to method ASA, $\triangle PRQ\cong\triangle NRM$.
| 264 | 828 |
{"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}
| 4.21875 | 4 |
CC-MAIN-2016-50
|
longest
|
en
| 0.706272 |
https://www.campusgate.co.in/2019/05/ratio-proportion-1-1.html
| 1,571,672,039,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-43/segments/1570987779528.82/warc/CC-MAIN-20191021143945-20191021171445-00415.warc.gz
| 821,204,387 | 28,884 |
# Ratio and Proportion 1/1
Ratio and proportion is the heart of arithmetic. If you understand this chapter properly you can solve virtually any problem in arithmetic.
If two numbers are in the ratio 2:3 means for every two units of the first number, second has 3 units. This is a mere comparison between numbers, and actual numbers may be way bigger than these numbers. If you multiply or divide a ratio the comparison does not change. i.e., 2:3 is same as 4:6.
If two numbers are in the ratio a:b then this ration has to be multiplied with a number K, to get actual numbers. This K is called multiplication factor (MF)
If two ratios are equal then we say they are in proportion. then $a:b::c:d \Rightarrow a \times d = b \times c$
or $\displaystyle\frac{a}{b} = \frac{c}{d} \Rightarrow a \times d = b \times c$.
## Chain Rule:
Chain rule is comes in handy when there are many variables need to compare with the given variable. We can understand this rule by observing a practice problem.
Solved Example:
If 12 carpenters working 6 hours a day can make 460 chairs in 24 days, how many chairs will 18 carpenters make in 36 days, each working 8 hours a day?
Let us prepare small table to understand the problem.
Now with repect to the Chairs we need to understand how each variable is related.
If the number of men got increased (i.e., 12 to 18), do they manufacture more chairs or less chairs is the question we have to ask ourselves. If the answer is "more" then the higher number between 12, 18 will go to the numerator and other will go to denominator and vice versa. Here answer is "more"
So $460 \times \displaystyle\frac{{18}}{{12}}$
Next we go to Hours. If the number of hours they work each day got increases then do they manufactures more chairs or less chairs? Answer is more
So $460 \times \displaystyle\frac{{18}}{{12}} \times \frac{8}{6}$
Last, If the number of day they work increases then .... answer is more.
So $460 \times \displaystyle\frac{{18}}{{12}} \times \frac{8}{6} \times \frac{{36}}{{24}} = 1380$
Exercise
1. The ratio between two numbers is 5 : 8. If 8 is subtracted from both the numbers, the ratio becomes 1 : 2. The smaller number is:
a. 10
b. 15
c. 20
d. 25
2. The ratio between Sumit's and Prakash's age at present is 2:3. Sumit is 6 years younger than Prakash. The ratio of Sumit's age to Prakash's age after 6 years will be :
a. 1 : 2
b. 2 : 3
c. 3 : 4
d. 3 : 8
3. The ratio between the ages of Kamala and Savitri is 6:5 and the sum of their ages is 44 years. The ratio of their ages after 8 years will be :
a. 5 : 6
b. 7 : 8
c. 8 : 7
d. 14:13
4. In a mixture of 60 litres, the ratio of milk and water is 2:1. What amount of water must be added to make the ratio 1:2?
a. 42 litres
b. 56 litres
c. 60 litres
d. 77 litres
5. A's money is to B's money as 4:5 and B's money is to C's money as 2:3. If A has Rs.800, C has
a. Rs.1000
b. Rs.1200
c. Rs.1500
d. Rs.2000
6. 15 litres of a mixture contains 20% alcohol and the rest water. If 3 litres of water is mixed in it, the percentage of alcohol in the new mixture will be :
a. 17
b. $16\displaystyle\frac{2}{3}$
c. $18\displaystyle\frac{1}{2}$
d. 15
7. Vinay got thrice as many marks in Maths as in English. The proportion of his marks in Maths and History is 4:3. If his total marks in Maths, English and History are 250, what are his marks in English ?
a. 120
b. 90
c. 40
d. 80
8. One-fourth of the boys and three-eighth of the girls in a school participated in the sports. What fractional part of the total student population of the school participated in the annual sports ?
a. $\displaystyle\frac{4}{{12}}$
b. $\displaystyle\frac{5}{8}$
c. $\displaystyle\frac{8}{{12}}$
| 1,103 | 3,672 |
{"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}
| 4.8125 | 5 |
CC-MAIN-2019-43
|
longest
|
en
| 0.875425 |
https://www.reference.com/math/draw-heptagon-9e6fe7573921b628
| 1,477,421,698,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2016-44/segments/1476988720238.63/warc/CC-MAIN-20161020183840-00266-ip-10-171-6-4.ec2.internal.warc.gz
| 981,761,519 | 20,834 |
Q:
# How do you draw a heptagon?
A:
A heptagon is an equal-sided figure with seven sides. There are various ways to go about drawing this shape. One of the simplest and most accurate methods is by drawing three circles and two triangles. You need a compass, a ruler, a pencil, a protractor and a piece of paper.
## Keep Learning
Credit: theilr CC-BY 2.0
1. Draw circle 1
Using your compass fixed with a pencil, draw a circle of reasonable measurement on your paper. Label the center of the circle as P.
2. Draw right-angled triangle 1
Using a ruler and a pencil, draw a line from P to touch the circumference of circle 1 at Q. Draw another line from P to touch the circumference of the same circle at R, such that angle QPR is 90 degrees. Join Q to R with a straight line.
3. Draw right-angled triangle 2
Draw a straight line from Q to S such that it is half the length of line QR, and that it is perpendicular to line QR. Join R to S with a straight line.
4. Draw circle 2
Draw a circle at point S with QS as the radius. Circle 2 meets line RS at T.
5. Draw circle 3
Draw a circle at point R with radius RT. Circle 3 intersects circle 1 at two points, U and T. Join R to T with a straight line
6. Draw the heptagon
Replicate the measurement of line RT in circle 1, ensuring that the lines join each other at the circumference of circle 1, as you navigate round the circle to point U. The figure in circle 1 is a heptagon.
Sources:
## Related Questions
• A: A heptagon is defined as polygon with seven angles and seven sides. The heptagon is sometimes referred to as the septagon, using the Latin numerical prefix... Full Answer >
Filed Under:
• A: A polygon with seven sides is called a heptagon. A polygon is a closed geometric shape that has at least three sides and angles.... Full Answer >
Filed Under:
• A: A square is always a rhombus because it satisfies the definition of a rhombus, which is any figure with four equal sides, equal opposite angles and two pai... Full Answer >
Filed Under:
• A: A square pyramid, a pyramid with four triangular sides and one square side, is a figure with five faces, eight edges and five vertices. It is a kind of thr... Full Answer >
Filed Under:
PEOPLE SEARCH FOR
| 546 | 2,227 |
{"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-2016-44
|
longest
|
en
| 0.917488 |
https://www.mcqslearn.com/math/quizzes/quiz.php?page=123
| 1,556,281,948,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-18/segments/1555578770163.94/warc/CC-MAIN-20190426113513-20190426135513-00088.warc.gz
| 730,261,932 | 7,290 |
Properties of real numbers quiz, properties of real numbers MCQs answers, college math quiz 123 to learn online college math courses. Number system quiz questions and answers, properties of real numbers multiple choice questions (MCQ) to practice math MCQs with answers for college and university courses. Learn properties of real numbers MCQs, introduction permutations, combinations and probability, period of trigonometric functions, online math learning, properties of real numbers test prep, career test for online certifications.
Practice properties of real numbers career test with multiple choice question (MCQs): transitive property of order of real numbers is that for all a, b,c belong r, with choices a < bɅ b < c this imly a < c, a < b Ʌ b < c this imply a = c, a < b Ʌ b < c this imply a≥c, and a < bɅ b < c this imply a > c for online math degree. Learn number system questions and answers for scholarships exams' problem-solving, assessment test.
Properties of Real Numbers Quiz
MCQ: Transitive property of order of real numbers is that for all a, b,c belong R
1. a < bɅ b < c this imly a < c
2. a < b Ʌ b < c this imply a = c
3. a < b Ʌ b < c this imply a≥c
4. a < bɅ b < c this imply a > c
A
Online Math Learning Quiz
MCQ: Sec(-960°) =
1. 2
2. 1
3. −1
4. −2
D
Period of Trigonometric Functions Quiz
MCQ: Period of tan(2x) is
1. π/3
2. π/2
3. 2π/3
B
Introduction Permutations, Combinations & Probability Quiz
MCQ: 9.8.7.6 =
1. 9ι/5ι
2. 5ι/9ι
3. 1
A
Double Angle Identities Quiz
MCQ: 1+cos4α =
1. 2cos²α
2. 4sin²α
3. 2sin²2α
4. 2cos²2α
D
| 490 | 1,574 |
{"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-2019-18
|
latest
|
en
| 0.752283 |
http://slideplayer.com/slide/3268583/
| 1,542,665,710,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-47/segments/1542039746112.65/warc/CC-MAIN-20181119212731-20181119234731-00233.warc.gz
| 314,567,974 | 24,147 |
# 1 Seminar of computational geometry Lecture #1 Convexity.
## Presentation on theme: "1 Seminar of computational geometry Lecture #1 Convexity."— Presentation transcript:
1 Seminar of computational geometry Lecture #1 Convexity
2 Example of coordinate-dependence What is the “sum” of these two positions ? Point p Point q
3 If you assume coordinates, … The sum is (x 1 +x 2, y 1 +y 2 ) Is it correct ? Is it geometrically meaningful ? p = (x 1, y 1 ) q = (x 2, y 2 )
4 If you assume coordinates, … p = (x 1, y 1 ) q = (x 2, y 2 ) Origin (x 1 +x 2, y 1 +y 2 ) Vector sum (x 1, y 1 ) and (x 2, y 2 ) are considered as vectors from the origin to p and q, respectively.
5 If you select a different origin, … p = (x 1, y 1 ) q = (x 2, y 2 ) Origin (x 1 +x 2, y 1 +y 2 ) If you choose a different coordinate frame, you will get a different result
6 Vector and Affine Spaces Vector space Includes vectors and related operations No points Affine space Superset of vector space Includes vectors, points, and related operations
7 Points and Vectors A point is a position specified with coordinate values. A vector is specified as the difference between two points. If an origin is specified, then a point can be represented by a vector from the origin. But, a point is still not a vector in coordinate-free concepts. Point p Point q vector ( q - p )
8 Vector spaces A vector space consists of Set of vectors, together with Two operations: addition of vectors and multiplication of vectors by scalar numbers A linear combination of vectors is also a vector
9 Affine Spaces An affine space consists of Set of points, an associated vector space, and Two operations: the difference between two points and the addition of a vector to a point
10 Addition u v u + v p p + w u + v is a vectorp + w is a point w u, v, w : vectors p, q : points
11 Subtraction v u - v u q p u - v is a vectorp - q is a vector p - q p p - w p - w is a point -w u, v, w : vectors p, q : points
12 Linear Combination A linear space is spanned by a set of bases Any point in the space can be represented as a linear combination of bases
13 Affine Combination
14 General position "We assume that the points (lines, hyperplanes,... ) are in general position." No "unlikely coincidences" happen in the considered configuration. No three randomly chosen points are collinear. Points in lR^d in general position, we assume similarly that no unnecessary affine dependencies exist: No k<=d+1 points lie in a common (k-2)-flat. For lines in the plane in general position, we postulate that no 3 lines have a common point and no 2 are parallel.
15 Convexity A set S is convex if for any pair of points p,q S we have pq S. p q non-convex q p convex
16 Convex Hulls : Equivalent definitions The intersection of all covex sets that contains P The intersection of all halfspaces that contains P. The union of all triangles determined by points in P. All convex combinations of points in P. P here is a set of input points
17 Convex hulls p0p0 p1p1 p2p2 p4p4 p5p5 p6p6 p7p7 p8p8 p9p9 p 11 p 12 Extreme point Int angle < pi Extreme edge Supports the point set
18 Caratheodory's theorem (0, 1)(1, 1) (1, 0)(0, 0) (¼, ¼)
19 Separation theorem Let C, D⊆ℝ d be convex sets with C∩D = ∅. Then there exist a hyperplane h such that C lies in one of the closed half-spaces determined by h, and D lies in the opposite closed half-space. In other words, there exists a unit vector a∈ℝ d and a number b∈ℝ such that for all x∈C we have ≥b, and for all x∈D we have ≤b. If C and D are closed and at least one of them is bounded, they can be separated strictly; in such a way that C∩h =D ∩h=∅.
20 Example for separation C C h q p
21 Sketch of proof We will assume that C and D are compact (i.e., closed and bounded). The cartesian product C x D ∈ℝ 2d is a compact set too. Let us consider the function f : (x, y) →|| x-y||, when (x, y) ∈ C x D. f attains its minimum, so there exist two points a ∈ C and b ∈ D such that ||a-b|| is the possible minimum. The hyperplane h perpendicular to the segment ab and passing through its midpoint will be the one that we are searching for. From elementary geometric reasoning, it is easily seen that h indeed separates the sets C and D.
22 Farkas lemma For every d x n real matrix A, exactly one of the following cases occurs: There exists an x ∈ℝ n such that Ax=0 and x >0 There exists a y ∈ℝ d such that y T A<0. Thus, if we multiply j-th equation in the system Ax=0 by y i and add these equations together, we obtain an equation that obviously has no nontrivial nonnegative solution, since all the coefficients on the left-hand sides are strictly negative, while the right-hand side is 0.
23 Proof of Farkas lemma Another version of the separation theorem. V ⊂ℝ d be the set of n points given bye the column vectors of the matrix A. Two cases: 0 ∈conv(V) 0 is a convex combination of the points of V. The coefficients of this convex combination determine a nontrivial nonnegative solution to Ax=0 0 ∉conv(V) Exist hyperplane strictly separation V from 0, i.e., a unit vector y∈ℝ n such that = 0 for each v∈V.
24 Radon ’ s lemma Let A be a set of d+2 points in ℝ d. Then there exist two disjoint subsets A 1, A 2 ⊂ A such that conv(A 1 ) ∩ conv(A 2 )≠∅ A point x ∈ conv(A 1 ) ∩ conv(A 2 ) is called a Radon point of A. (A 1, A 2 ) is called Radon partition of A.
25 Helly ’ s theorem Let C 1, C 2, …, C n be convex sets in ℝ d, n≥d+1. Suppose that the intersection of every d+1 of these sets is nonempty. Then the intersection of all the C i is nonempty.
26 Proof of Helly ’ s theorem Using Radon’s lemma. For a fixed d, we proceed by induction on n. The case n=d+1 is clear. So we suppose that n ≥d+2 and the statement of Helly’s theorem holds for smaller n. n=d+2 is crucial case; the result for larger n follows by a simple reduction. Suppose C 1, C 2, …, C n satisfying the assumption. If we leave out any one of these sets, the remaining sets have a nonempty intersection by the inductive assumption. Fix a point a i ∈ ⋂ i≠j C j and consider the points a 1,a 2, …, a d+2 By Radon’s lemma, there exist disjoint index sets I 1, I 2 ⊂{1, 2, …, d+2} such that
27 Example to Helly ’ s theorem a2a2 a1a1 a4a4 a3a3 C2C2 C1C1 C3C3 C4C4
28 Continue proof of Helly ’ s theorem Consider i ∈{1, 2, …, n}, then i∉I 1 or i∉I 2 If i∉I 1 then each a j with j ∈ I 1 lies in C i and so x ∈conv({a j : j ∈ I 1 })⊆C i If i∉I 2 then each a j with j ∈ I 2 lies in C i and so x ∈conv({a j : j ∈ I 2 })⊆C i Therefore x ∈ ∩ i=1 n C i
29 Infinite version of Helly ’ s theorem Let C be an arbitrary infinite family of compact convex sets in ℝ d such that any d+1 of the sets have a nonempty intersection. Then all the sets of C have a nonempty intersection. Proof Any finite subfamily of C has a nonempty intersection. By a basic property of compactness, if we have an arbitrary family of compact sets such that each of it’s finite subfamilies has a nonempty intersection, then the entire family has a nonempty intersection.
30 Centerpoint Definition 1: Let X be an n-point set in ℝ d. A point x ∈ ℝ d is called a centerpoint of X if each closed half-space containing x contains at least n/(d+1) points of X. Definition 2: x is a centerpoint of X if and only if it lies in each open half space η such that |X ⋂ η|>dn/(d+1).
31 Centerpoint theorem Each finite point set in ℝ d has at least one centerpoint. Proof Use Helly’s theorem to conclude that all these open half- spaces intersect. But we have infinitely many half-spaces η which are unbound and open. Consider the compact convex set conv(X ⋂ η) ⊂ η η conv(X ⋂ η)
32 Centerpoint theorem(2) Run η through all open-spaces with |X ⋂ η|>dn/(d+1) We obtain a family C of compact convex sets. Each C i contains more than dn/(d+1) points of X. Intersection of any d+1 C i contains at least one point of X. The family C consists of finitely many distinct sets.(since X has finitely many distinct subsets). By Helly’s theorem ⋂ C ≠∅, then each point in this intersection is a centerpoint.
33 Ham-sandwich theorem Every d finite sets in ℝ d can be simultaneously bisected by a hyperplane. A hyperplane h bisects a finite set A if each of the open half-spaces defined by h contains at most ⌊ |A|/2 ⌋ points of A.
34 Center transversal theorem Let 1 ≤k≤d and let A 1, A 2, …, A k be finite point sets in ℝ d. Then there exists a (k-1)-flat f such that for every hyperplane h containing f, both the closed half- spaces defined by h contain at least | A i |/(d-k+2) points of A i i=1, 2, …, k. For k=d it’s ham-sandwich theorem. For k=1 it’s the centerpoint theorem.
Similar presentations
| 2,468 | 8,618 |
{"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}
| 4.5 | 4 |
CC-MAIN-2018-47
|
latest
|
en
| 0.846029 |
http://www.talkstats.com/showthread.php/25199-2-four-sided-dice-probability-of-1-dice-3-sum-even
| 1,508,667,089,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-43/segments/1508187825174.90/warc/CC-MAIN-20171022094207-20171022114207-00580.warc.gz
| 556,642,699 | 10,856 |
# Thread: 2 four sided dice, probability of 1 dice = 3, sum = even
1. ## 2 four sided dice, probability of 1 dice = 3, sum = even
[Question] 2 four tetrahedral dice with sides, 1,2,3,4. the two dice are thrown, what is the probability that given the first dice is a 3, the sum of the 2 dices will be an even number.
This has had me stumped for about 30mins. im assuming i should be using
P(A|B) = P(A&B)/P(B)
where
P(B) = Rolling a single 3 from both dices? or should this just be rolling a 3 on a single dice.
P(A) = Sum of the numbers = an even number, of which there are 2 option: 3+1 and 3+3
If someone could give me some pointers on this id appreciate it.
Thanks
2. ## Re: 2 four sided dice, probability of 1 dice = 3, sum = even
i managed to solve this by writing out the sequence of dice rolls as follows:
(sure 99% of the ppl using this forum know more about stats than i do, so this is for the 1%)
green are the possible scenarios where one die is a 3.
red are the corresponding scenarios where the total is an even number
P(A|B) = P(A&B) / P(B)
where P(B) = 7/16 (Edit: Its only 7 and not 8 as B is the case where only one 3 is encountered)
P(A&B) = 3/16 (See the combinations which are both green and red)
Tweet
| 370 | 1,238 |
{"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-2017-43
|
longest
|
en
| 0.935316 |
https://proofwiki.org/wiki/1129
| 1,597,118,481,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2020-34/segments/1596439738727.76/warc/CC-MAIN-20200811025355-20200811055355-00169.warc.gz
| 448,745,869 | 9,483 |
# 1129
Previous ... Next
## Number
$1129$ (one thousand, one hundred and twenty-nine) is:
The $189$th prime number
The $2$nd prime number after $1009$ which can be expressed in the form $x^2 + n y^2$ for all values of $n$ from $1$ to $10$
$\displaystyle 1129$ $=$ $\displaystyle 20^2 + 1 \times 27^2 = 27^2 + 1 \times 20^2$ $\displaystyle$ $=$ $\displaystyle 29^2 + 2 \times 12^2$ $\displaystyle$ $=$ $\displaystyle 19^2 + 3 \times 16^2$ $\displaystyle$ $=$ $\displaystyle 27^2 + 4 \times 10^2$ $\displaystyle$ $=$ $\displaystyle 2^2 + 5 \times 15^2$ $\displaystyle$ $=$ $\displaystyle 23^2 + 6 \times 10^2$ $\displaystyle$ $=$ $\displaystyle 11^2 + 7 \times 12^2$ $\displaystyle$ $=$ $\displaystyle 29^2 + 8 \times 6^2$ $\displaystyle$ $=$ $\displaystyle 20^2 + 9 \times 9^2$ $\displaystyle$ $=$ $\displaystyle 33^2 + 10 \times 2^2$
| 331 | 840 |
{"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}
| 3.765625 | 4 |
CC-MAIN-2020-34
|
latest
|
en
| 0.383465 |
http://www.mathdoubts.com/geometry/equation-circle-centre-coincides-origin/
| 1,498,288,056,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-26/segments/1498128320227.27/warc/CC-MAIN-20170624064634-20170624084634-00393.warc.gz
| 572,394,886 | 4,283 |
# Equation of the Circle when Centre of the Circle coincides with the origin
Expressing a circle in terms of a mathematical equation when the centre of the circle coincides with the origin of the Cartesian coordinate system is defined equation of the circle when the centre of the coincides with the origin.
The centre of the circle may coincide with the origin of the Cartesian coordinate system in a possible case. The circle can be expressed in a mathematical expression form to show how the circle is in Cartesian coordinate system of geometry. It can be developed by understanding the relation of circle with the origin of the coordinate system and it can be expressed in an algebraic form mathematical expression.
## Derivation
Assume, the centre of a circle coincides with the origin of the Cartesian coordinate geometric system. In other words, the origin of the Cartesian coordinate system and centre of the circle are same in this case. Assume, it is denoted by $O$ in the Cartesian coordinate system.
Assume, the radius of the circle is $r$ units.
Consider a point on the circle and assume it represents every point on the circle. Assume, it is located at $x$ units distance in horizontal axis direction and $y$ units distance in vertical axis direction from the origin $O$. So, the location of the point $R$ is written as $R\left(x,y\right)$ in geometric system.
Join point $O$ and point $R$. It is equal to the radius of the circle. Therefore $OR=r$.
Draw a line from point $R$ but the line should be parallel to vertical axis but perpendicular to horizontal axis. Also draw a line from point $O$ on the horizontal axis until the line draw from $R$ intersects it at a point, named point $T$. In this way, a right angled triangle $\Delta ROT$ is constructed within the circle.
According to the right angled triangle $\Delta ROT$
• The line segment $\stackrel{‾}{OR}$ becomes hypotenuse of the right angled triangle $\Delta ROT$ and the length of hypotenuse is $OR=r$
• The line segment $\stackrel{‾}{RT}$ becomes opposite side of the right angled triangle $\Delta ROT$ and the length of the opposite side is $RT=y$
• The line segment $\stackrel{‾}{OT}$ becomes adjacent side of the right angled triangle $\Delta ROT$ and the length of the adjacent side is $OT=x$
The relation between sides of the right angled triangle $\Delta ROT$ can be written in mathematical form by using Pythagorean Theorem.
According to Pythagorean Theorem
${OR}^{2}={OT}^{2}+{RT}^{2}$
Substitute the values of all three sides in this mathematical relation to transform it into equation form.
${r}^{2}={x}^{2}+{y}^{2}$
$⇒{x}^{2}+{y}^{2}={r}^{2}$
The algebraic expression is the circle equation when the centre of the circle coincides with the origin of the geometric coordinate system.
Save (or) Share
| 650 | 2,805 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 29, "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}
| 4.625 | 5 |
CC-MAIN-2017-26
|
latest
|
en
| 0.88461 |
http://www.jfinternational.com/ph/potential-energy.html
| 1,476,992,022,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2016-44/segments/1476988717783.68/warc/CC-MAIN-20161020183837-00483-ip-10-171-6-4.ec2.internal.warc.gz
| 531,039,494 | 4,489 |
# Potential Energy - Gravitational and Elastic. Concepts.
Potential Energy, Concepts
and Example.-
An object has energy when it is moving, but it can also have potential energy, which is the energy associated with the object's position.
Potential Energy, Example: a heavy brick lifted up has potential energy due to its position in relation to the ground. It can do work because when dropped it will fall because of the gravity force, allowing it to make a work output over another object receiving the impact.
A compressed spring has potential energy. For instance, the spring of a mechanical clock transforms its energy doing work to move the seconds, minutes and hour pointers.
There are several kinds of potential energy: gravitational, elastic, electric, etc.
Gravitational Potential Energy
The gravitational potential energy is a very common example of potential energy.
The Gravitational Potential Energy (GPE) of an object of mass m at a height y over a reference level is defined as:
EPG = mgy
g is the gravity acceleration
This definition is fully compatible with the definition of work since the work needed lo lift the mass m from the reference level to the height y is Fy = Weight·y = mgy. The object has gained an energy mgy.
If we let this object of mass m to fall freely by gravity on a stake on the ground, the work on the stake will be equal to the kinetic energy acquired while falling.
This kinetic energy can be calculated by the kinematics equation vf2 = vi2 + 2gy. Since vi = 0, then
vf2 = 2gy. The kinetic energy just before striking the stake is ½mvf2. Replacing vf2 with 2gy we get ½ m·2gy = mgy.
Then, to raise an object of mass m to a height y we need a work amount equal to mgy, and once at this height y, the object has the capability of doing work equal to mgy.
Let's notice GPE depends on the object's vertical height over some reference level; in this example, the ground.
The work needed to lift an object does not depend on the lifting path. That direction can be vertical, inclined, or another, and the work to rise the object will be equal.
Also, the work the object is able to do when falling does not depend on its path.
From what level must the height y be measured? What matters here is the potential energy change and we choose a reference level convenient to solve a given problem. Once we choose it, we must keep it during the calculations.
Elastic Potential Energy.
It is the energy associated with elastic materials. Next, we demonstrate the necessary work to compress or stretch a spring over a distance x is ½kx2, where k is the spring's constant.
By Hook's Law, the relation between force and displacement on a spring is F = -kx. The minus sign is due to the force always pointing to the equilibrium position (x = 0). The force F is now variable and we can no longer use W = Fdcos.
Let's first find a general relation to calculate the work done by a variable force. We will apply it to our spring.
As Fx is nearly constant in each x, W Fx x, the total work can be approximated by the formula
If we make the intervals x even shorter, this is, we make x 0, W tends to a limit, this can be expressed as
This formula represents the integral of the force Fx as a function of x:
And this is the area under the curve Fx(x). Of course, this relation includes the case where Fx = F cos is constant.
Applying that relation to the spring, assuming the spring is placed horizontally and connected to a mass sliding over a smooth surface which is also horizontal, and that the spring is compressed over a distance xmax and then released, the work W done by the spring force between xi = -xmax and xf = 0 is:
| 824 | 3,669 |
{"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-2016-44
|
longest
|
en
| 0.916812 |
https://www.physicsforums.com/threads/physics-fluid-mechanics-problem-badly.591006/
| 1,718,217,470,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-26/segments/1718198861183.54/warc/CC-MAIN-20240612171727-20240612201727-00617.warc.gz
| 844,790,825 | 18,907 |
# Physics/Fluid Mechanics Problem - badly
• Kalookakoo
In summary: A[sub2]/A[sub1])²*v²[sub2]Then substitute for v²[sub1] in the equation above. You then have an equation that relates v2 to v2 and p1 and p2.In summary, the conversation discusses using the principle of dimensional consistency to show that Bernoulli's equation has the dimension of pressure and that each term in the equation has the dimension of length. The conversation also includes a problem related to Bernoulli's equation, where the goal is to solve for v2 using equations for P1, v1 and P2, v2.
Kalookakoo
## Homework Statement
I have been staring at these two problems for a LONG time now and keep getting stuck. Please help me, and try to explain so I can understand.
ro = density
p = pressure
g = gravity
h = height/elevation
A = area
v = velocity
1a) Use the principal of dimensional consistency, show that bernoulli's equation written as:
P + (1/2)(ro)(v²) + (ro)gh = constant
has the dimension of pressure.
1b) When it is written as:
(ro)/[(ro)*g] + (v²)/(2g) + h = constant
show that each term has the dimension of length.
2) Using equations
a) A*v = A*v
Left side both have subscript 1, right side have subscript 2.
and the version of Bernoulli's equation:
P/(ro) + (v²)/2 + gh = constant
show that
v(subscript 2) = sqrt((2*[p(sub1)-p(sub2)])/ (ro(1-[(A(sub2)/A(sub1))²])
Sorry if it's hard to read. I can't do this on my own and my roommate isn't here to tutor me like he usually does.
## The Attempt at a Solution
I don't even understand 1a and 1b. For 2, I can get as far as
v[sub2] = [ro =(ro)gh - p]/[(ro)*(A(sub2)/A(sub1))²]
and that's just by creating similar denominators for the second equation and substituting for v[sub2] with what i solved for the other v in the other equation.
Help?
Kalookakoo said:
ro = density
p = pressure
g = gravity
h = height/elevation
A = area
v = velocity
1a) Use the principal of dimensional consistency, show that bernoulli's equation written as:
P + (1/2)(ro)(v²) + (ro)gh = constant
has the dimension of pressure.
Hi Kalookakoo! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif
For starters, can you show that the terms on each side of the "+" signs all have units identical to the units of pressure?
Last edited by a moderator:
ro*gh I guess could be kg/m^3 * (g) = N/m^3 * m = N/m^2 which is pressure.
I don't see how the middle term can be pressure,
Kalookakoo said:
I don't see how the middle term can be pressure,
Hint: F=ma
Ohhhh.
I pull out a m on the v² to be (m/s²) which is acceleration times the mass of the density so it's Force/m^3 * m = F/A.
Wow, overlooked that, thanks.
1a) Down!
Can you help me with 2? I think I can get 2b down by myself, I'll ask if I get stuck again.
I got 1b. :)
It's really simple once I realized the F=ma part lol.
Part 2 is troublesome...
Kalookakoo said:
For 2, I can get as far as
v[sub2] = [ro =(ro)gh - p]/[(ro)*(A(sub2)/A(sub1))²]
You shouldn't have a "p" here; it will be p1 or p2. As for the "h" terms, assume h remains constant.
Your use of the "=" equal sign is too carefree. It is supposed to mean equals. Now would be a good time to start to use it more carefully, before you get into more complicated maths or science exercises. What is on the left of the "=" should be equal to what is on the right.
Sorry that second equal sign is supposed to be a subtraction sign.
Does that mean I have to use that second formula twice using p1 and v1 then v2 and p2?
Use this equation for P1,v1 then for P2,v2
P/(ro) + (v²)/2 + gh = constant
And then what?
Do I solve for p1 and p2 and set it up as (p1-p2) like it is in the end equation?
Because I get (-ro)[(v²[sub1]/2) + (v²[sub2]/2)]
Where does that come into play?
And then what?
Do I solve for p1 and p2 and set it up as (p1-p2) like it is in the end equation?
Because I get (-ro)[(v²[sub1]/2) + (v²[sub2]/2)]
Where does that come into play?
This relates v1 to v2.
A*v = A*v
Left side both have subscript 1, right side have subscript 2.
How does that fit in? I just can't see it for some reason.
I plugged in p1,v1 and p2,v2 in the bernoulli equation and solved for v2 and got:
v²[sub2] = 2(p[sub1]-p[sub2])/(ro) + v²[sub1]
It's almost right..but I'm stuck
Kalookakoo said:
v²[sub2] = 2(p[sub1]-p[sub2])/(ro) + v²[sub1]
Replace this v1 with v1 from this equation:
A*v = A*v
Left side both have subscript 1, right side have subscript 2.
## 1. What is a "badly" physics/fluid mechanics problem?
A "badly" physics/fluid mechanics problem typically refers to a problem that is either poorly worded or lacks necessary information to solve it correctly. It could also refer to a problem that has multiple solutions or no clear solution at all.
## 2. How can I identify if a physics/fluid mechanics problem is "badly" formulated?
One way to identify a "badly" formulated physics/fluid mechanics problem is to carefully read the problem and look for any inconsistencies or missing information. If the problem seems unclear or has multiple interpretations, it may be "badly" formulated.
## 3. What should I do if I encounter a "badly" physics/fluid mechanics problem?
If you encounter a "badly" physics/fluid mechanics problem, the first step is to carefully read the problem and try to understand what is being asked. If the problem seems too ambiguous or impossible to solve, you can try reaching out to your instructor or classmates for clarification.
## 4. Can a "badly" physics/fluid mechanics problem be solved?
It is possible to solve a "badly" physics/fluid mechanics problem, but it may require extra effort and critical thinking. In some cases, the problem may have multiple solutions or no clear solution at all, in which case it is important to carefully explain your reasoning and assumptions.
## 5. How can I avoid creating a "badly" physics/fluid mechanics problem?
To avoid creating a "badly" physics/fluid mechanics problem, it is important to carefully plan and structure your problem before presenting it to others. Make sure all necessary information is included and that the problem is clear and concise. You can also have others review your problem before presenting it to a wider audience.
• Introductory Physics Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
55
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
997
• Introductory Physics Homework Help
Replies
5
Views
333
• Introductory Physics Homework Help
Replies
21
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
5
Views
429
• Introductory Physics Homework Help
Replies
3
Views
296
• Introductory Physics Homework Help
Replies
5
Views
1K
• Introductory Physics Homework Help
Replies
7
Views
275
| 1,860 | 6,779 |
{"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-2024-26
|
latest
|
en
| 0.92996 |
https://slideplayer.com/theme/ppt-on-area-of-trapezium-calculator-15159.html
| 1,531,851,908,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-30/segments/1531676589757.30/warc/CC-MAIN-20180717164437-20180717184437-00474.warc.gz
| 773,263,711 | 25,662 |
Ppt on area of trapezium calculator
Area - Revision The area of a shape is simply defined by : “the amount of space a shape takes up.” Think of a square measuring 1 cm by 1cm we say it is.
WYZ Trapezium Area Example 1 : Find the area of the trapezium. 6cm 4cm 5cm Parts of the Circle Circumference O O = centre of circle The radius is measured from the centre of the circle to the edge. radius diameter The diameter is measured from one edge to the other passing through the centre of the circle. Radius = ½ x diameter Diameter = 2 x radius Circle Investigation To calculate the circumference of a/
Elm Hall Primary Curriculum 2014 – presentation to primary heads June 2014 Part 2 of 2.
control of their learning process. Maths language The Number System Fractions, Percentage, Ratio and Proportion Calculation / Halve full Empty Contains Holds Litre Millimetre Container Measuring cylinder Volume Area Covers Surface Square centimetre Square metre Square millimetre Surface Area Days Monday Tuesday Wednesday Thursday Friday Saturday Sunday Week Month /of symmetry through the middle. Draw a kite with a line of symmetry. Draw an isosceles trapezium./
D MANCHE 2013. Finding the area under curves: There are many mathematical applications which require finding the area under a curve. The area “under”
straight lines. Compared to the rectangle method, the trapezoidal method gives a better approximation for the real area, as obvious in the graph below. Trapezoidal Method (cont.) Recall that the area of a trapezium with parallel sides of length a and b and with perpendicular height h is given by: Area a b h Area = average of parallel sides x perpendicular distance between them Trapezoidal Method (cont.) Shaded/
Perimeter and Area A look at a few basic shapes Perimeter.
for Play Experiment with square paper and see if you can find a method of calculating the area of a trapezium. 5 m 2 m4 m3 m 9 m What is the area of this trapezium? 5 m 2 m4 m3 m 9 m One possible method, giving 32./2 Pause for Play 8 cm 6 cm 4 cm Which has the greatest area? Hint: Square paper, isometric paper, and a pair of scissors? And that’s more than enough! Perimeter of shapes made from straight lines Circumference of circles Area of rectangles, triangles, parallelograms and trapeziums Area of circles
CTEQ School July 061 Monte Carlo Event Generators Peter Richardson IPPP, Durham University Durham University Lecture 1: Basic Principles of Event Generation.
and present the recent progress in a number of areas which are important for the Tevatron and LHC –Matching of matrix elements and parton showers Traditional Matching /common techniques converge faster –Trapezium rule –Simpsons rule However only if the derivatives exist and are finite. Otherwise the convergence of the Trapezium or Simpsons rule will be/ July 0624 Phase Space Cross section And decay rates Can be calculated as the integral of the matrix element over the Lorentz invariant phase space. Easy to/
Education Leeds 20090212 VJC Identify lines of symmetry in simple shapes and recognise shapes with no lines of symmetry. Step 1 : Talk me through what.
is a kite; a parallelogram; a rhombus; an isosceles trapezium? Which quadrilateral with one line of symmetry has three acute angles? Education Leeds 20090212 VJC Enlarge 2-D shapes, given a centre of enlargement and a positive whole- number scale factor. Step /you calculate? How can you use the angle sum of a triangle to calculate the sum of the interior angles of any polygon? The formula for calculating the sum of the interior angles of a regular polygon is: (n - 2) × 180° where n is the number of sides of /
Announcing the release of VERSION 6 This Demo shows just 20 of the 10,000 available slides and takes 7 minutes to run through. Please note that in the.
does the same. Calculate the probability that they both choose a strawberry chocolate. Conditional Probability: Dependent Events P(strawberry and strawberry) =3/12 x A box of chocolates contains twelve chocolates of three different types./ and angles on a straight line add to 180 o Take 1 identical copy of this right-angled triangle and arrange like so. Area of trapezium = ½ (a + b)(a + b) = ½ (a 2 +2ab + b 2 ) Area of trapezium is also equal to the areas of the 3 right-angled triangles. = ½ ab + ½ ab + ½ c/
GCSE Mathematics Route Map – Higher Tier Assessment Order Unit 2 – March Year 10 Unit 1 – June Year 10 Unit 3 – June Year 11 Notes – A lot of Unit 2.
area of a rectangle work out the area of a parallelogram calculate the area of shapes made from triangles and rectangles calculate the area of shapes made from compound shapes made from two or more rectangles, for example an L shape or T shape calculate the area of shapes drawn on a grid calculate the area of simple shapes work out the surface area of nets made up of rectangles and triangles calculate the area of a trapezium Isosceles Triangles Unit 3 – Perimeter, Area/
NEW TO 9-1 GCSE MATHS? Lets get started!. Assessment Schedule ◦ Thursday 25 th May AM – Paper 1 Non Calculator ◦ Thursday 8 th June AM – Paper 2 Calculator.
out for…… ◦ Expand the product of binomials ◦ Inverse function and composite function ◦ Calculate/Estimate gradients and areas of graphs ◦ Geometric progressions ◦ Calculate the nth term of quadratic sequences ◦ Calculate and interpret conditional probabilities through Venn diagrams Things to memorise Trig. ratios: sinθ=O/H, cosθ=A/H, tanθ=O/A are needed for both tiers Foundation: ◦ Pythagoras Theorem: a² + b² = c² ◦ Area of a trapezium: ½(a + b)h ◦ Volume prism/
Focus on English Year 5.
area of a square is measured by multiplying the length of a side by itself Know that the area of a rectangle is measured by multiplying the length of the longer side by the shorter Use this information about calculating the area of a square or rectangle to work out the area of/ and name parallelogram; rhombus and trapezium Know what a parallelogram is and describe it in mathematical terms Know what a rhombus is and describe it in mathematical terms Know what a trapezium is and describe it in mathematical/
Teacher Version Level Shape Space Measure
I can find the area of triangles, kites, parallelogram trapeziums, circles. and semi-circles. I can use formulae for the volume of cuboids. I can find the volume and surface area of cuboids. I can calculate volumes and surface area of prisms and cylinders. I can calculate the volume of cones. Explain fully how you would calculate the volume of this cone. Complete this sentence Perimeter means ……………………………………………………………………………………………… Area means ……………………………………………………………………………………………… The area of this rectangle is/
Lesson Plan - APP Area and Perimeter Mental and Oral Starter Pupils to complete a ‘Heard the Word’ grid and compare it to grid they completed at the start.
work will be assessed on: The accuracy of your calculations. The presentation of your ideas and designs. The presentation of your calculations and results. levelArea and Perimeter 6I can calculate the area of rectangles, triangles, parallelograms, trapeziums and kites. 5I can calculate the area and perimeter of rectangles and the area of triangles using formulae. 4I understand the difference between area and perimeter. I can find the area and perimeter of rectangles by counting squares. 3I can/
Year 6 Focus on English. Year 6 Objectives: Spoken Language Listen carefully and adapt talk to the demands of different contexts, purposes and audiences.
shape Know when to apply a given formula to find out the volume of a shape Objective 6m: Calculate the area of parallelograms and triangles Know the formula associated with finding the area of triangles Know the formula associated with finding the area of parallelograms MEASUREMENT Objective 1m: Solve problems involving the calculation and conversion of units of measure, using decimal notation up to three decimal places where appropriate Use decimal/
Area & Volume Learning Outcomes Find the area of square, rectangles, triangles, parallelograms, rhombuses, kites, trapezia and shapes which are composites.
calculations. Find the surface areas and volumes of compound shapes most of which are everyday objects. Distinguish between formulae for perimeter, area and volume by considering dimensions. Area & Volume Area of Shapes Triangles Area = ½ base × perp. height A = ½ b × h 5cm 6cm 60 ° 8cm 12cm A C B b a c A = ½ ab sin C Area = ½ product of 2 sides × sine of angle between them Area & Volume Area of Shapes Trapezium One pair of parallel sides Total area/
Introduction This chapter you will learn the SUVAT equations These are the foundations of many of the Mechanics topics You will see how to use them to.
6 812 4 -6 Sub in the appropriate values for the trapezium above Calculate Kinematics of a Particle moving in a Straight Line You can represent the motion of an object on a speed-time graph, distance-time graph or an acceleration-time graph Gradient of a speed-time graph = Acceleration over that period Area under a speed-time graph = distance travelled during that period 2D/
WeekActivitySkills 1Taxi App Design Creating a system for calculating ‘fair’ taxi fares using the algebra of solving and substitution. 2Mystery Tours Interpreting.
and working accurately with time, budgeting within parameters to get the best price for customers. 3 Winter Wonderland II Site Calculating the area of squares, rectangles, triangles, trapeziums, circles and part-circles to analyse the stage areas of the festival. 4 Winter Wonderland II Merchandise Calculating the volume of cubes, cuboids and cylinders to evaluate possible merchandising options for the festival. 5 Winter Wonderland II Security Use angle/
Toward a New Paradigm of Star Formation: Does Nature Abhor a Singular Isothermal Sphere? Brenda C. Matthews UC Berkeley Radio Astronomy Laboratory.
& Molecular Gas Less than 10% of the area and mass of a GMC is in the form of dense gas which is non-uniformly / Taurus and ρ Oph show a large fraction of sources have disks massive enough to form planets Trapezium (BIMA and OVRO) at 3mm doesn’t show disks/of Embedded Clusters Numerical simulations required to follow evolution of a stellar cluster turbulent hydrodynamical calculations to match observed properties of clouds MHD? Simulations are challenging due to large range of scales involved (use of/
© T Madas Trigonometric Calculations. © T Madas x 16 m 35° tanθ = Opp Adj c tan35° = x 16 c x = c x ≈ 11.2 m x tan35° Trigonometric Calculations S O H.
tan50° = h b c b x tan50° c 6.928 = b = tan50° c b ≈ 5.813 cm b = 5.813 = © T Madas b a 8 cm Calculate the area of this triangle 50° 60° h S O H C A H T O A h ≈ 6.928 a = 4 b = 5.813 © T Madas Exam Question © T Madas An/ = 0.528 c θ 32° = [nearest degree] © T Madas 2.7 6 5.6 The figure below shows a right angled trapezium ABCD. All lengths are in cm [not to scale] 1.Calculate the length of BC. 2.Calculate the size of the angle DCB, giving your answer correct to 1 decimal place. A B C D lengths in cm 5.6 3.3/
Area and Volume Using formulae. Finding Area and Perimeter of a Square or Rectangle Area is the measure of the amount of space a shape covers Perimeter.
Area of a rectangle = length x width Perimeter = 2(length and width) Parallelogram Length Perpendicular height Area of a parallelogram: length x perpendicular height Trapezium Perpendicular height Length a Length b Area of a trapezium:/of a Prism A prism is a 3D shape with the same cross section all along its length To calculate the volume of the prism you need to find the area of the cross section and multiply it by the height or length Volume of prism = area of cross section x length Finding the Area of/
Maths Test Revision You will do a paper A without a calculator and paper B with a calculator and a mental arithmetic test.
if a square has a 9cm side the perimeter is 9+9+9+9=36cm Area To calculate area you need to multiply one side by another. So if a square has a side of 9cm you have to do 9x9=81cm squared written with a little 2. Or a/way to what you normally see them. Turn the page around! Shapes 3 sides triangle 4 sides rectangle, square, quadrilateral, parallelogram, rhombus, trapezium, kite 5 sides pentagon 6 sides hexagon 7 sides heptagon 8 sides octagon Congruent Congruent means the same. If two shapes are congruent it/
M5: Applications of area and volume. M5: Further applications of area and volume Areas of ellipses, annuluses and parts of a circle Calculating areas.
M5: Applications of area and volume M5: Further applications of area and volume Areas of ellipses, annuluses and parts of a circle Calculating areas of composite figures Applying Simpson’s Rule Surface area of Cylinders Surface area of spheres Volume of composite solids Errors in calculations Pythagoras theorem Circumference of circle Area of circle Area of triangle Area of rectangle Area of parallelogram Area of trapeziumArea of rhombus Volume of Prism Pythagoras c²=a²+b²(The square on/
Perimeter is the distance around the outside of the object To find a perimeter we _____ up the ________ e.g. Perimeter = ___ + ___ + ___ + ___ + ___.
width of a trapezium changes. We have to work out the average width. 8 14 A = (a + b) x hA = ( 8 + 14 ) x 6 = 66 2 2 Average Width = (top side + bottom side) ÷ 2 To find the area of a circle you need the radius. The radius is the distance from the ________ of the circle to the _________. middle outside radius To calculate the area use/
Area and Perimeter. First things first, technically speaking….. What is perimeter? What is area? ‘The perimeter is the length of a closed curve. The curve.
… However the subject is broached, from the EYFS through KS1, pupils develop an understanding of shape, calculating and measuring in standard and non-standard units. These skills all come together under this /Trapeziums Trapeziums are another a special type of rectangle. They have 1 pair of parallel sides, and it follows that... Perimeter = a + b + c + d Area = Vertical height (v) x ½(a + b) v a b cd This knowledge about triangles can be applied to kites as well. Diagonal 1 (d 1 ) Diagonal 2 (d 2 ) Area of/
© Boardworks Ltd 2005 1 of 37 These icons indicate that teacher’s notes or useful web addresses are available in the Notes Page. This icon indicates the.
is given by the area under the graph. This area is a trapezium with parallel sides of length u and v and width t. So Velocity (ms –1 ) u v Time (s) t This can also be written as © Boardworks Ltd 2005 22 of 37 distance travelled = area of rectangle A + area of triangle B s = ut/= 0, and for t when s = –15. Constant acceleration example 1 © Boardworks Ltd 2005 26 of 37 To calculate s when v = 0, given u and a requires the use of v 2 = u 2 + 2 as. Therefore, the maximum height above the ground that the stone /
Analyzing Longitudinal Quality of Life Outcome Data Stephen J. Walters, PhD Professor of Medical Statistics and Clinical Trials School of Health and Related.
be regressed on x 1i in stage (2). Examples of summary measures include the Area Under the Curve (AUC) or the overall mean of post-randomisation measures. 15 Summary measures 16 17 Area Under the Curve (AUC) 18 Calculation of the AUC The area can be split into a series of shapes called trapeziums. The areas of the separate individual trapeziums are calculated and then summed for each patient. Let Y ij represent/
1 Angles Area and perimeter 2 Algebra 3 4 Triangles 5 Statistics 6 Probability 7 Solids Graphs 9 Bearings 8.
is not accurate. Over 13% is red. Back Home Check Statements 1.The area of the trapezium is 15cm 2 2.BC is 7 cm 3.The perimeter of the trapezium is 18cm D B C A ABCD is a trapezium. AB = AD = 3cm DC = 7cm BAD = ADC = 90 o /Area of rhombus = 96cm 2 D B C A ABCD is a rhombus Diagonal AC = 12cm Diagonal DB = 16cm Angle BAC = 55 o E 55 o Which statements are true? All 3 statements are true. The diagonals of a rhombus bisect each other at right angles so you can use Pythagoras’ Theorem to calculate the length of/
Year 4 Place value & calculation. 5. I can order 4 digit numbers. 4Pv&C1 4. I can recognise the place value of each digit in 4 digit numbers. I can solve.
I know and can identify; isosceles, equilateral and scalene triangles and the quadrilaterals; parallelogram, rhombus and trapezium Year 4 Written calculations & algebra 4C&A1 I can solve addition and subtraction two-step problems in contexts, deciding which operations/including bar charts. I can find the area of rectilinear shapes by counting squares. I can estimate and calculate different measures including money in pounds and pence. I can convert between different units of measure (e.g. kilometre to metre;/
Finding Areas of Shapes. Area of a Triangle Area of Triangle = 1 x Base x Vertical Height 2 Vertical Height Base Right Angle.
Sides ) Area of Trapezium Area of Trapezium = Average of Parallel Sides x Distance Between Them a b h Area = ½ (a + b) x h Area of Circle Area = x r2r2 Diameter (D) Circumference = x D Radius (R) = D/2 Area of Sector Area of Sector of Angle X = X / 360 x Area of Circle Summary Exam papers usually have area formulas You need practise to use them quickly and accurately In industry area formulas are used to calculate the quantity of materials needed/
Estimating the Area Under a Curve Aims: To be able to calculate an estimate for the area under a curve. To decide if this is an over estimate or an under.
Curve Aims: To be able to calculate an estimate for the area under a curve. To decide if this is an over estimate or an under estimate. To consider ways that we can speed up our method. Area of a Trapezium a b h Questions Estimating What do we mean by estimating? Why might we estimate a value? Estimating area under curves 10 5 0 1/
13 Vectors in Two-dimensional Space Case Study
area are scalars. 13.1 Concepts of Vectors and Scalar B. Representation of a Vector The directed line segment from point X to point Y in the direction of/of a and c. (a) (b) Solution: (a) (b) Example 13.11T 13.3 Vectors in the Rectangular Coordinate System B. Point of Division Example 13.11T The figure shows that the trapezium ABCD with AB // DC. E is the mid-point of/132. Example 13.19T 13.5 Scalar Products Solution: C. Calculation of Scalar Product in the Rectangular Coordinate System Example 13.19T Given /
Area under Curve. Calculus was historically developed to find a general method for determining the area of geometrical figures. When these figures.
developed to find a general method for determining the area of geometrical figures. When these figures are bounded by curves, their areas cannot be determined by elementary geometry. Integration can be applied to find such areas accurately. Also known as Trapeziod/Trapezium Rule An approximating technique for calculating area under a curve Works by approximating the area as a trapezium Actual Area = 10.67 units 2. (2, 4) (1, 1/
Finding Areas Numerically. y4y4 h y5y5 The basic idea is to divide the x-axis into equally spaced divisions as shown and to complete the top of these.
The basic idea is to divide the x-axis into equally spaced divisions as shown and to complete the top of these strips of area in some way so that we can calculate the area by adding up these strips. y1y1 h y2y2 h h y3y3 a b The first way is to complete / y2y2 h y4y4 y5y5 h ab y5y5 h y4y4 hh y3y3 y1y1 h y2y2 ab The Trapezium rule. Complete the strips to get trapezia. Add up areas of the form Simpson ’ s Rule We complete the tops of the strips as shown with parabolas. y5y5 y4y4 y3y3 y1y1 y2y2 ab x y x0123456 -/
Maths Test Tips. Place Value Remember where the digits are and what they are worth. Remember the names of the place value columns. The decimal point never.
if a square has a 9cm side the perimeter is 9+9+9+9=36cm Area To calculate area you need to multiply one side by another. So if a square has a side of 9cm you have to do 9x9=81cm squared written with a little 2. Or a/ to what you normally see them. Turn the page around! Shapes 3 sides triangle 4 sides rectangle, square, quadrilateral, parallelogram, rhombus, trapezium, kite 5 sides pentagon 6 sides hexagon 7 sides heptagon 8 sides octagon Shapes can be regular or irregular Congruence Congruent means the same/
Area and Volume You will be able to use the correct unit for different measurements. use formula to find the area of shapes. find the volume of a prism.
m 7 mm 5 mm Area = ½ x 7 x 5 = 17.5 mm 2 The Area of a TrapeziumArea = ½ the sum of the parallel sides x the perpendicular height A = ½(a + b)h a b h b ½h a Area of trapezium = area of parallelogram A = ½(a /area. Triangular-based prism Rectangular-based prism Pentagonal-based prism Hexagonal-based prism Octagonal-based prism Circular-based prism Cylinder Cuboid Find the volume of the following prisms. Diagrams Not to scale In each of the following examples the cross-sectional ends have to be calculated/
. BRING YOUR CALCULATOR You need strong foundations in basic Maths to build on. Spend about ten minutes a day reading through the area(s) that cause you the most problems on a website or from a revision book. Start with simple examples until you are sure you know what you are doing. Move onto more difficult examples. SHOW YOUR WORKING Area of trapezium = ½(a + b/
The Mid-Ordinate Rule. To find an area bounded by a curve, we need to evaluate a definite integral. If the integral cannot be evaluated, we can use an.
N.B. Radians ! Solutions The answer can be improved by using more strips. 1. Solutions The red shaded areas should be included but are not. The blue shaded areas are not under the curve but are included in the rectangle. The following sketches show sample rectangles where the mid-/Use 4 strips with the mid-ordinate rule to estimate the value of Give the answer to 4 d.p. Solution: We need 4 y -values so we set out the calculation in a table as for the Trapezium rule. The Mid-Ordinate Rule 12 So, a b N.B./
WeekActivitySkills 1 The Largest Crowds Ever Using the vocabulary of integers, powers and roots to quantify some of the largest public gatherings ever.
Calculating using standard form, powers and roots as well as rounding to significant figures to quantify some of the populations ever. 3 The Mathematics of Set Design Classifying a range of 2D rectilinear shapes by their mathematical properties in order to use them as part of a set. 4 The Mathematics of Set Design Using formulae for the area of/Classify a range of 2D rectilinear shapes, including a rhombus, trapezium and parallelogram for use in a set design. Evaluate the classification of 2D rectilinear /
ANGLE AND PLANE ANGLE AND PLANE Identify Angle Adaptif Hal.: 2 ANGLE AND PLANE Determining position of line, and angle that involves point, line and.
g = 33,3 g Adaptif Hal.: 9 ANGLE AND PLANE Width and Circumference of flat shape 1. Triangle Width: L = ½ A x t Example: Where, A = base wide, t = tall A C B A C B 13 12 Calculate the width and circumference plane beside. Answer: AB = = = = = 24/of Trapezium A B Width = ½ ( AB + CD). t t Circumference = AB + BC + CD + DA C D Example: Find the trapezium width in the picture! D E C 8 10 A B 15 Answer: Width = ½ ( AB + CD) CE = = = = Adaptif Hal.: 17 ANGLE AND PLANE Width and circumference of flat plane 8. Area/
MOTION 1.Motion and Rest 2.Distance and Displacement 3.Uniform Motion 4.Non-uniform Motion 5.Speed 6.Velocity 7.Acceleration 8.Equations of Uniformly Accelerated.
8.Equations of Uniformly Accelerated Motion 9.Graphical Representation of Motion 10.Distance-Time Graph 11.Speed-Time Graph 12.Derivation of Equations of Motion by Graphical Method 13.Uniform Circular Motion 14.Calculation of Speed of a Body in Uniform Circular Motion Concept of a Point /v) s = ½ x t x (u + u + at) s = ½ x (2ut + at 2 ) s = ut + ½ at 2 Third equation of motion The area of trapezium OABC gives the distance travelled. s = ½ x OC x (OA + CB) s = ½ x t x (u + v) (v + u) = 2s t From the /
Discovering the Universe Ninth Edition Discovering the Universe Ninth Edition Neil F. Comins William J. Kaufmann III CHAPTER 12 The Lives of the Stars.
of the sky in a false- color infrared. Image taken by the Spitzer Space Telescope. Gases are seen here to exist in more areas/of which are stars in the early stages of formation—along with shock waves caused by matter flowing out of protostars faster than the speed of sound waves in the nebula. Shock waves from the Trapezium stars may have helped trigger the formation of/. Note that most of the cool, low-mass stars have not yet arrived at the main sequence. Calculations of stellar evolution indicate that/
Volume.
Triangular Prism Trapezoid Prism Volume of Prism = length x Cross-sectional area Area Formulae r h b Area Circle = π x r2 Area Rectangle = Base x height h b h b Area Trapezium = ½ x (a + b) x h a b h Area Triangle = ½ x Base x height Volume Cylinder Cross-sectional Area = π x r2 = π x 32 = 28.2743…..cm2 DO NOT ROUND! 3cm 5cm USE CALCULATOR ‘ANS’! Volume = length x CSA/
1.5 Measurement AS 90130 Internal (3 credits). Calculate the area of the following shapes. 6 cm 3 cm 5 cm 4 cm 3 cm 4 cm 2 cm A = 9 cm 2 A = 12.6 cm 2.
6 × 2 = 6 mm 2 6 mm 2 mm Note 2: Areas ShapeArea FormulaExample ParallelogramArea = b × h 7 m 2 m Area = 7 x 2 = 14 m 2 Trapezium Area = 1/2 (a+b)× h 2 cm 6 cm 4 cm Area= ½ (4+2)×6= 18 cm 2 CircleArea = πr 2 Area = πr 2 = π × (5) 2 = 78.5 cm 2/1.38 / L e.g. S.A. = 2(55 × 42) + 2(18 × 42) + (55 x 18) 55 cm 42 cm a.) Calculate the area of glass required for the fish tank b.) Calculate the volume of water in the tank. Give your answer to the nearest litre = 7122 cm 2 V tank = 55 × 42 × 18 = 41580 cm 3 V water /
GCSE Mathematics Route Map – Higher Tier Teaching Order Unit 2 – Year 10 Unit 1 – Year 10 Unit 3 – Year 11 Notes – A lot of Unit 2 time has been given.
the area of a rectangle work out the area of a parallelogram calculate the area of shapes made from triangles and rectangles calculate the area of shapes made from compound shapes made from two or more rectangles, for example an L shape or T shape calculate the area of shapes drawn on a grid calculate the area of simple shapes work out the surface area of nets made up of rectangles and triangles calculate the area of a trapezium Unit 3 – Perimeter, Area and/
GCSE Mathematics Route Map – Foundation Tier Assessment Order Unit 2 – March Year 10 Unit 1 – June Year 10 Unit 3 – June Year 11 Notes – A lot of Unit.
area of a rectangle work out the area of a parallelogram calculate the area of shapes made from triangles and rectangles calculate the area of shapes made from compound shapes made from two or more rectangles, for example an L shape or T shape calculate the area of shapes drawn on a grid calculate the area of simple shapes work out the surface area of nets made up of rectangles and triangles calculate the area of a trapezium Isosceles Triangles Unit 3 – Perimeter, Area/
Intro. Draw a square Draw a diagonal from the top right corner to the bottom left. Draw a line from the centre of the right hand side of the square.
litres of venom. (Guinness Book of record) a) How many litres of venom did Mr Keyter milk per year?____ L/yr b) Write the unit of the calculated rate:/of a fridge Litre Ex. The capacity of a fridge Litre Converting between 12 hour time and 24 hour time Perimeter and area of rectangle, square, parallelogram, triangle, circle, trapezium, rectangle, square, parallelogram, triangle, circle, trapezium, Volume of Prism Volume of Prism Measure the length of lines to the nearest cm. Measure the length of/
Miss B ’ s Maths DIRT Bank Created by teachers for teachers, to help improve the work life balance and also the consistent quality of feedback our students.
right angled triangle is translated to the position shown to make a rectangles. Calculate the area of the parallelogram. The formula for the area of a parallelogram is the _________ as a rectangle. Half the _____ of the parallel sides. ________ the distance between them. That is how you calculate, area of a ___________. Calculate the area of the trapezium. Area and Perimeter www.missbsresources.com Geometry Complete the sentences The radius is _______/
GCSE Higher Revision Starters 7
GCSE Higher Revision Starters 7 Grade D/C – calculating areas Calculate the size of the shaded area 4cm 60cm² - 36cm² = 24cm² 6cm 10cm Area of circle = 36π Area of trapezium = 50 Shaded area = (36π – 50)cm² Give your answer in terms of π 12cm 5cm 8cm Give your answer in terms of π (256 - 64π)cm² 8cm Grade /How many 2cm cubes will fit inside the cuboid? 30 cubes 6cm 4cm 10cm Give the volume in terms of π 800π cm³ 8cm 20cm Area of triangle = 24cm² Vol = 24 x 3 = 72cm³ 6cm 3cm 8cm Grade B – Solving simultaneous/
The potential of posing more challenging mathematics tasks and ways of supporting students to engage in such tasks. Peter Sullivan Sullivan MAT Nov 2013.
Region YEAR 7 Establish the formulas for areas of rectangles, triangles and parallelograms and use these in problem solving Calculate volumes of rectangular prisms YEAR 8 Choose appropriate units of measurement for area and volume and convert from one unit to another volume Find perimeters and areas of parallelograms, trapeziums, rhombuses and kites Investigate the relationship between features of circles such as circumference, area, radius and diameter. Use formulas to solve/
Volume & Surface Area of Solids Revision of Area
and b = -4 does a2 – 3b2 = 57 Q4. Calculate Tuesday, 11 April 2017 Created by Mr.Lafferty Revision of Areas www.mathsrevision.com Any Type of Triangle Level 4 Any Type of Triangle Rhombus and kite www.mathsrevision.com Parallelogram Trapezium Circle Compiled by Mr. Lafferty Maths Dept. Area Level 4 Learning Intention Success Criteria We are revising area of basic shapes. Know formulae. Use formulae correctly. www.mathsrevision/
Lesson Plan – Lesson 7 Volume Mental and Oral Starter In groups pupils to discuss what volume means and then calculate the volume of the 3D shapes shown.
the probing question using mini white boards. Objectives and Habits of Mind To calculate the volume of a cube by counting multi link (Level 4.) To calculate the volume of a cube using a formula. To calculate the volume of a cuboids using a formula. (Level 6) To calculate the volume of prisms and cylinders by first finding the area of the cross section. (Level 7) To discuss and compare approaches/
| 7,856 | 30,563 |
{"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}
| 4.40625 | 4 |
CC-MAIN-2018-30
|
latest
|
en
| 0.839646 |
https://cracku.in/139-35-of-27-of-426-x-sbi-clerk-2012-2
| 1,726,248,210,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725700651535.66/warc/CC-MAIN-20240913165920-20240913195920-00003.warc.gz
| 168,754,469 | 26,119 |
Instructions
What should come in place of the question mark (?) in the following questions ?
Question 139
# 3/5 of 2/7 of ? =426
Solution
Expression : 3/5 of 2/7 of ? =426
=> $$\frac{3}{5} \times \frac{2}{7} \times x=426$$
=> $$\frac{6x}{35}=426$$
=> $$x=426 \times \frac{35}{6}$$
=> $$x=71 \times 35=2485$$
=> Ans - (A)
| 130 | 330 |
{"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}
| 3.890625 | 4 |
CC-MAIN-2024-38
|
latest
|
en
| 0.678147 |
http://mathhelpforum.com/new-users/202105-linear-quadratic-convergence.html
| 1,527,104,384,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-22/segments/1526794865702.43/warc/CC-MAIN-20180523180641-20180523200641-00355.warc.gz
| 190,026,868 | 13,539 |
1. ## linear and quadratic convergence
Hallo, All!
I just dont understand following: if f(p) is from interval (0,1) then there is a linear convergence.
Can someoe help?
Best regards, M
2. ## Re: linear and quadratic convergence
Are you sure the place where you read this fact does not have the relevant definitions? It is probably talking about the rate of convergence.
3. ## Re: linear and quadratic convergence
Please, why f(x) has to be x, I just can't connect...
4. ## Re: linear and quadratic convergence
Originally Posted by marijakopljar
Please, why f(x) has to be x, I just can't connect...
It does not. Right now I am working with f(x) = sin(x). In other words, I don't understand what you mean.
5. ## Re: linear and quadratic convergence
If it does not , then ok. I just foudn it on net and that confused me.
I wil cut and paste:
Limit (mathematics) - Wikipedia, the free encyclopedia
"Convergence and fixed pointA formal definition of convergence can be stated as follows. Suppose as goes from to is a sequence that converges to a fixed point , with for all . If positive constants and exist with
then as goes from to converges to of order , with asymptotic error constant
Given a function with a fixed point , there is a nice checklist for checking the convergence of p.
1) First check that p is indeed a fixed point:
2) Check for linear convergence. Start by finding . If...."
6. ## Integration
"As a first approximation, look at the unit square given by the sides x = 0 to x = 1 and y = f(0) = 0 and y = f(1) = 1. Its area is exactly 1. As it is, the true value of the integral must be somewhat less. Decreasing the width of the approximation rectangles shall give a better result; so cross the interval in five steps, using the approximation points 0, 1/5, 2/5, and so on to 1. Fit a box for each step using the right end height of each curve piece, thus √(1⁄5), √(2⁄5), and so on to √1 = 1. Summing the areas of these rectangles, we get a better approximation for the sought integral, namely..."
PLEASE, HOW DO WE GET SQUARE FRM 1/5 ETC? I just dont get it. It shoudl be the surface under the curve.
7. ## Re: linear and quadratic convergence
$\displaystyle \sqrt{\frac{1}{5}}$, $\displaystyle \sqrt{\frac{2}{5}}$, ... are the values of $\displaystyle \sqrt{x}$, which is the function that is being integrated, at the approximation points $\displaystyle \frac{1}{5},\frac{2}{5},\dots,1$. These values are used to find the area of yellow rectangles in the picture. This area, in turn, is an approximation of the area under the graph of $\displaystyle \sqrt{x}$.
It would help if you explain the transition from the rate of convergence to integrals. They are not directly related. E.g., you don't need the definition of the rate of convergence to define integrals.
8. ## Re: linear and quadratic convergence
Thank U, but I still dont get it. The surface under the curve is the sum of all the sums of the green surfaces. Ok. But how do we get the quadrat square?
9. ## Re: linear and quadratic convergence
Originally Posted by marijakopljar
The surface under the curve is the sum of all the sums of the green surfaces.
No, the area of green rectangles is an approximation to the area under the curve. Note that I did not mention green rectangles, only yellow ones. You like to change the subject without explanation: first from the rate of convergence to integrals, then from yellow to green, don't you?
Originally Posted by marijakopljar
But how do we get the quadrat square?
By "quadrat square" do you mean square root? I already explained this:
Originally Posted by emakarov
$\displaystyle \sqrt{\frac{1}{5}}$, $\displaystyle \sqrt{\frac{2}{5}}$, ... are the values of $\displaystyle \sqrt{x}$, which is the function that is being integrated, at the approximation points $\displaystyle \frac{1}{5},\frac{2}{5},\dots,1$.
In other words, $\displaystyle \sqrt{1/5}$ is the height of the leftmost yellow rectangle. This height is multiplied by the width 1/5 to get the area of the rectangle.
10. ## Integrals (absolute beginner)
Thank You!
But the problem is that I cant understand why is the height of the yellow square square roor frm 1/5. Anyway, on the pic it is more than 0.4 and is greater that 0.2. What is the logic? That is the point!!!
11. ## Re: Integrals (absolute beginner)
To approximate the area under the graph, the height of the rectangle is chosen to be the y-coordinate of the point on the graph above 1/5. Do you know what the graph of a function is? The graph of f(x) is a set of points with coordinates (x, f(x)) for various x. Here the function is $\displaystyle f(x)=\sqrt{x}$. Which point is located on the graph above 1/5? By definition of the graph, it is $\displaystyle f(1/5)=\sqrt{1/5}$.
Originally Posted by marijakopljar
Anyway, on the pic it is more than 0.4 and is greater that 0.2. What is the logic?
$\displaystyle \sqrt{1/5}\approx0.45$.
12. ## Re: Integrals (absolute beginner)
Please (untill U still have nerves!!!) how did we get that height is square root from 1/5? That is the main problem!
(P.S. Please, excuse me, I am in a terreble hurry, and still this, cause I am not gifted for mth at all!)
13. ## Re: Integrals (absolute beginner)
Originally Posted by marijakopljar
Please (untill U still have nerves!!!) how did we get that height is square root from 1/5?
I can only repeat what I said in post #11.
14. ## Re: linear and quadratic convergence
THANK U SO MUCH, completely clear now, it is a drastical example how (extremely) bad concentratin influences math!!!(released!!!)
15. ## Simple minds...
Hallo!
Me again! Please, how do we get F(1) -F(0) = 2/3 which is right, but 1 to high power 0.5 is still 1! How do we get the derivative f(x) = 2/3(x) to high power 3/2?
Page 1 of 2 12 Last
,
,
| 1,519 | 5,793 |
{"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}
| 4.25 | 4 |
CC-MAIN-2018-22
|
latest
|
en
| 0.906223 |
http://www.jiskha.com/display.cgi?id=1327513565
| 1,493,228,880,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-17/segments/1492917121528.59/warc/CC-MAIN-20170423031201-00273-ip-10-145-167-34.ec2.internal.warc.gz
| 569,754,283 | 4,005 |
# Algebra
posted by on .
You provide the manager with a quadratic equation that models the expected number of ticket sales for each day x.
(x = 1 is the day tickets go on sale).
Tickets = -0.6x2 + 12x + 11 .
Describe what happens to the ticket sales as time passes. My question is sinc the coefficient is negative and you graph downward the ticket sales would low? Please help
• Algebra - ,
well, sort of ....
the number of ticket sales will increase until the maximum is reached, after that the ticket sales will decrease.
for any function
f(x) = ax^2 + bx + c, the max/min of the function is obtained when x = -b/(2a)
x = -12/(2(-.6)) = 10
when x = 10, tickets sales = -.6(100) + 12(10) + 11 = 71
I will leave it up to you to show that the sales are less for both x= 9 and x=11
• Algebra - ,
Thanks for helping me solve this problem!
| 230 | 844 |
{"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-2017-17
|
latest
|
en
| 0.933427 |
https://www.physicsforums.com/threads/uniform-circular-motion-car-problem.190697/
| 1,480,741,159,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2016-50/segments/1480698540839.46/warc/CC-MAIN-20161202170900-00265-ip-10-31-129-80.ec2.internal.warc.gz
| 1,011,772,000 | 16,224 |
# Uniform circular motion car problem
1. Oct 11, 2007
### S[e^x]=f(u)^n
1. The problem statement, all variables and given/known data
a cart with a person in passes over a circular arch at a constant speed v who's radius is r. Does the maximum speed the cart can go without leaving the tracks depend on the mass of the person, the mass of the cart, both, or neither?
2. Relevant equations
Fg=mg -Gravitational Force
Fc=(mv^2)/2r? -Centripetal force
Fn=? -Normal force
3. The attempt at a solution
Fc=Fg-Fn
i honestly don't know where to go from here? can anyone offer any hints... i need to figure it out by tomorrow afternoon. my gut says it depends on neither, but from math i've fiddled around with it seems to indicate it depends on both
2. Oct 11, 2007
### PhanthomJay
that "2" in the denominator of your Fc equation doesn't belong there. Your equation Fc = Fg -Fn is good.
What is the value of Fn just as the cart would leave the tracks?
3. Oct 12, 2007
### S[e^x]=f(u)^n
sorry about the 2, i don't know why i put it there. anyway Fn would be 0 on the cart, but so would the Fn for the person inside... wouldn't it? in which case the cart would leave the tracks at the same speed regardless of whether there was a person in there?... but is it still dependent on the weight of the car?
4. Oct 12, 2007
### Staff: Mentor
Well, weight (mg) is pulling down while the centrifugal force (mv2/r) would cause the cart to go tangent to the track. For the cart to stay on the track those two forces must be equal.
Set the forces equal and see what happens with respect to mass.
5. Oct 12, 2007
### nrqed
what is special when the car is driven at the maximum speed so that it is just about to lose contact with the ground is that the normal force is zero. Isolate for the speed and you will have your answer. (btw, you should have mv^2/r, not mv^2/2r)
6. Oct 12, 2007
### S[e^x]=f(u)^n
thanks, from what i can figure out the velocity at which the cart(regardless of mass) leaves the tracks is dependent solely on the square root of the radius. which sounds about right to me
7. Oct 12, 2007
### nrqed
correct. (and it depends on the acceleration due to gravity...the max speed on th emoon would be different ;-) )
8. Oct 12, 2007
thanks!
| 620 | 2,258 |
{"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-2016-50
|
longest
|
en
| 0.946674 |
https://www.physicsforums.com/threads/finding-the-roots-of-a-high-degree-polynomial-equation.685976/
| 1,531,865,334,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-30/segments/1531676589902.8/warc/CC-MAIN-20180717203423-20180717223423-00213.warc.gz
| 931,122,988 | 16,727 |
# Homework Help: Finding the roots of a high degree polynomial equation
1. Apr 16, 2013
### 1MileCrash
1. The problem statement, all variables and given/known data
y(6) - 3y(4) + 3y''-y = 0
2. Relevant equations
3. The attempt at a solution
The characteristic equation of that differential equation is:
r^6 - 3r^4 + 3r^2 - r = 0
But how am I expected to solve such a high degree polynomial (and thus the DE?)
2. Apr 16, 2013
### Bacle2
First, notice you can rewrite you char. equation as:
r(r^5-3r^3+3r-1)=0
Then notice too, that, inside your parenthesis: 1-3+3-1=0. What does this tell you?
3. Apr 16, 2013
### tiny-tim
Hi 1MileCrash!
(try using the X2 button just above the Reply box )
nooo
r6 - 3r4 + 3r2 - 1 = 0
4. Apr 16, 2013
### SammyS
Staff Emeritus
Following tiny-tim's corrected characteristic equation:
r6 - 3r4 + 3r2 - 1 = 0
Expand (a - b)3 .
5. Apr 16, 2013
### epenguin
I don't know at what point certain things are supposed to be familiar but I would say pretty early there is a rather familiar pattern to be discerned in that last formula.
6. Apr 16, 2013
### HallsofIvy
$r^6- 3r^4+ 3r^2- 1$ has only even powers of r. Let $x= r^2$ and that becomes $x^3- 3x^2+ 3x- 1$. And, as SammyS suggests, that is $(x- 1)^3$.
7. Apr 16, 2013
### Bacle2
Worse comes to worse and you cannot see this pattern, it always makes sense to try for the simple roots,like 0,1 and -1 . Checking for 1 as a root comes down to adding the coefficients and seeing if the sum is zero; similar for -1.
8. Apr 17, 2013
### HallsofIvy
Let me point out that Bacle2 is not just choosing "simple roots" at random. Since the leading coefficient is 1 and the constant term is 1, by the "rational root theorem" the only possible rational number roots are 1 and -1. (I don't know why he mentions "0".)
9. Apr 17, 2013
### Bacle2
Right, my bad. I thought the characteristic equation had no constant term.
10. Apr 17, 2013
### LCKurtz
All this help and the OP is nowhere in sight.
| 666 | 1,998 |
{"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": 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-2018-30
|
latest
|
en
| 0.898089 |
http://www.probabilityformula.org/addition-rule-formula.html
| 1,553,396,327,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-13/segments/1552912203168.70/warc/CC-MAIN-20190324022143-20190324044143-00307.warc.gz
| 322,927,031 | 7,406 |
Probability is an idea which helps to find how certain events are to happen. It is based on observations of certain events. Probability is one of the most important topics of the mathematics. Probability of an certain event, say $A$ is symbolized by $P(A)$. Always it lies between 0 and 1.
We define probability as the number of outcomes of an event associated with a random experiment divided by the total number of outcomes possible in that experiment. Let $n (S)$ denote the number of outcomes of the experiment and A be an event whose outcomes are $n (A)$. Then, probability of occurrence of event A is given by: $P (A)$ = $\frac{n (A)}{n (S)}$.
Suppose we have two events X and Y associated with a random experiment E. then the probability of occurrence of either X or Y or both is given by:
$P (X\ or\ Y)$ = $P (X) + P (Y)$ – $P (X\ and Y)$
$\rightarrow$ $P (X \cup Y)$ = $P (X) + P (Y)$ - $P (X \cap Y)$
Where $P (X)$ is the probability of occurrence of event $X, P (Y)$ is the probability of occurrence of event $Y$ and $P (X\ and\ Y)$ [also written as $P (X\ \cap Y)$ is the probability of occurrence of both $X$ and $Y$ events simultaneously.
When the two events $X$ and $Y$ are mutually exclusive that is the occurrence of event $X$ does not affects the probability occurrence of event Y or vice versa. In this case the intersection probability that is $P (X \cap Y)$ = 0. And we have,
$P (X\ or\ Y)$ = $P (X)$ + $P (Y)$
Word Problems
Some examples on probability addition rule are illustrated below:
Example 1:
We have rolled a six sided die. Find the probability of getting an even number or a 5.
Solution:
Let E be the experiment of rolling a die and S be the sample space, then
S = {2, 4, 1, 5, 3, 6} => n (S) = 6
A = event of getting an even number = {4, 2, 6} => n (A) = 3
B = event of getting a 5 = {5} => n (B) = 1
So we have P (A) = $\frac{3}{6}$, P (B) = $\frac{1}{6}$.
It is clear that the two events are mutually exclusive, hence $P (A \cap B)$ = 0
So according to the probability addition rule we have:
$P (A\ or\ B) =$\frac{3}{6}$+$\frac{1}{6}$=$\frac{4}{6}$=$\frac{2}{3}$Example 2: We are given a deck of 52 cards. We pick a card from it. Find the probability of getting a heart or a red king. Solution: Let S be the sample space then$n (S)$= 52.$A$= event of getting a red king =>$n (A)$= 2 =>$P (A)$=$\frac{2}{52}B$= event of getting a heart =>$n (B)$= 13 =>$P (B)$=$\frac{13}{52}$Clearly$A$and$B$are not mutually exclusive as a red king can also be of heart. =>$n (A\ \cap\ B)$= 1 =>$P (A \cap B)$=$\frac{1}{52}$. By addition rule of probability we have: P (A or B) =$\frac{2}{52}$+$\frac{13}{52}\frac{1}{52}$=$\frac{14}{52}$=$\frac{7}{26}$Example 3: What is the probability of getting a total of 3 or 7, when two dice are rolled? Solution: Let event A be the total of 7 and event B the total of 3. When we roll two dice we have 36 events in a sample space. Number of events getting total of 3 = 2 Number of events getting total of 7 = 6 =>$P(A)$=$\frac{6}{36}$and$P(B)$=$\frac{2}{36}$Both the events are mutually exclusive since they do not have any common event. i.e.$P(A \cap B)$= 0 Now$P(A \cup B)$=$P(A)$+$P(B)$=$\frac{6}{36}$+$\frac{2}{36}$=$\frac{8}{36}$=$\frac{2}{9}\$.
| 1,042 | 3,228 |
{"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}
| 4.71875 | 5 |
CC-MAIN-2019-13
|
latest
|
en
| 0.937747 |
https://www.onlinemath4all.com/grade-6-math-review-test.html
| 1,670,044,645,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2022-49/segments/1669446710924.83/warc/CC-MAIN-20221203043643-20221203073643-00797.warc.gz
| 968,994,972 | 7,993 |
# GRADE 6 MATH REVIEW TEST
Question 1 :
A work was assigned to Mike on a Tuesday. He completed the work after 72 days. On what day did he complete the work ?
(A) Sunday (B) Tuesday (C) Thursday
Solution :
Tuesday corresponds to 2 in 7 day clock arithmetic.
We want to know the day which is 72 days after Tuesday.
Step 1 :
Step 2 :
Divide the result by the divisor 7
Step 3 :
Take the remainder
72 + 2 = 74
When 74 is divided by 7, the remainder is 4
4 corresponds to Thursday.
So, he will complete the work in Thursday.
Question 2 :
What month is 19 months after July?
(A) April (B) March (C) February
Solution :
July corresponds to 7 in month arithmetic
We want to know the month which is 19 months after July
Step 1 :
Step 2 :
Divide the result by the divisor 12
Step 3 :
Take the remainder
7 + 19 = 26
When 26 is divided by 12, the remainder is 2 and it corresponds to February
So, 19 months after July is February.
Question 3 :
The wrong number of the sequence 1, 8, 27, 64, 124, 216, 343 is
(A) 150 (B) 124 (C) 117
Solution :
1 = 13
8 = 23
27 = 33
64 = 43
125 = 5 3
In the sequence every number is in the form of cube 1, 2, 3,.......... but 124 is not cube of any number.
So, the wrong number in the sequence is 124.
Question 4 :
A triangle with sides of length 6, 8 and 10 is a
(A) Equilateral triangle (B) Scalene triangle
(C) Right triangle
Solution :
The measurement of three sides of triangles are different. So it must be a scalene triangle.
So the answer is Scalene triangle.
Question 5 :
James, David and Jack live in a row of three houses on the same street. Walking their houses, they pass a white house first, then a green house, then a blue house. David lives next door to the green house. Jack does not live next door to his friend who lives in the blue house. Who lives in the blue house ?
(A) Jack (B) James (C) David
Solution :
From left to right,
First, second and third houses are white, green and blue respectively.
Green is the middle house, Since David lives next door to green house, he is living in the blue house.
Question 6 :
There are 3 large marbles, 2 medium marbles and 5 small marbles in a bag. If one of the marbles is chosen randomly, what is the probability that a small marble is chosen?
(A) 3/10 (B) 1/5 (C) 1/3 (D) 1/2
Solution :
Total number of marbles = 3 + 2 + 5
n(s) = 10
Let A be the event of getting small marble.
n(A) = 5
p(A) = n(A)/n(S)
p(A) = 5/10
p(A) = 1/2
Question 7 :
In a triangle, the length of the first side is x. The length of the second side is 2 cm greater than first side. The third side is 5 cm greater than the first side. If the perimeter of the triangle is 13, what is the length of each side ?
(A) 7 (B) 4 (C) 2
Solution :
x - length of first side
length of the second side = x + 2
Length of third side = x + 5
Perimeter of the triangle = 13
x + x + 2 + x + 5 = 13
3x + 7 = 13
3x = 6
x = 6/3
x = 2
So, the length of first side is 2 cm.
Question 8 :
In a right angled triangle, the length of the opposite side and adjacent side are 3 cm and 4 cm respectively. What is the area of the triangle ?
(A) 5 (B) 6 (C) 4
Solution :
Area of triangle = (1/2) ⋅ base ⋅ height
= (1/2) ⋅ 3 ⋅ 4
= 6 cm2
Question 9 :
Find the volume of the cylinder whose diameter is 7 inches and height is 6 inches.
(A) 243 cm3 (B) 7243 cm3 (C) 243 cm3
Solution :
Volume of cylinder = πr2h
Radius = 7/2 and height = 6 cm
Volume of cylinder = (22/7) ⋅ (7/2)⋅ 6
= 462 cm3
So, the volume of cylinder is 462 cm3.
Question 10 :
The reduced common fraction for the sum of (3/8) and (8/3) is
(A) 24/73 (B) 73/14 (C) 73/24
Solution :
Sum of (3/8) and (8/3)
= (3/8) + (8/3)
= (9 + 64)/24
= 73/24
Kindly mail your feedback to [email protected]
## Recent Articles
1. ### Unit Rates
Dec 02, 22 04:18 PM
Unit Rates - Concept - Examples with step by step explanation
2. ### Adding and Subtracting Fractions Worksheet
Dec 02, 22 07:27 AM
| 1,367 | 4,169 |
{"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
| 0.888328 |
https://fssnip.net/am/0
| 1,657,214,803,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2022-27/segments/1656104495692.77/warc/CC-MAIN-20220707154329-20220707184329-00430.warc.gz
| 307,874,813 | 8,991 |
0 people like it.
# Discrete Fréchet Distance
Compute the Discrete Fréchet Distance between two arrays (which may be of different lengths). Based on the 1994 algorithm by Thomas Eiter and Heikki Mannila. Not extensively tested, so use at your peril! (This version with some small fixes.)
``` 1: 2: 3: 4: 5: 6: 7: 8: 9: 10: 11: 12: 13: 14: 15: 16: 17: 18: 19: 20: 21: 22: 23: 24: 25: 26: 27: 28: 29: 30: 31: 32: 33: 34: 35: 36: 37: 38: 39: 40: 41: 42: 43: 44: 45: 46: 47: 48: 49: 50: 51: 52: 53: 54: 55: 56: 57: 58: 59: 60: 61: 62: 63: 64: ``` ``````open System // Discrete Frechet Distance: // (Based on the 1994 algorithm from Thomas Eiter and Heikki Mannila.) let frechet (P : array) (Q : array) = let sq (x : float) = x * x let min3 x y z = [x; y; z] |> List.min let d (a : (float*float)) (b: float*float) = let ab_x = abs(fst(a) - fst(b)) let ab_y = abs(snd(a) - snd(b)) sqrt(sq ab_x + sq ab_y) let p, q = Array.length P, Array.length Q let ca = Array2D.init p q (fun _ _ -> -1.0) let rec c i j = if ca.[i, j] > -1.0 then ca.[i, j] else if i = 0 && j = 0 then ca.[i, j] <- d (P.[0]) (Q.[0]) elif i > 0 && j = 0 then ca.[i, j] <- Math.Max((c (i-1) 0), (d P.[i] Q.[0])) elif i = 0 && j > 0 then ca.[i, j] <- Math.Max((c 0 (j-1)), (d P.[0] Q.[j])) elif i > 0 && j > 0 then ca.[i, j] <- Math.Max(min3 (c (i-1) j) (c (i-1) (j-1)) (c i (j-1)), (d P.[i] Q.[j])) else ca.[i, j] <- nan ca.[i, j] c (p-1) (q-1) // Use frechet as an operator: let (-~~) a1 a2 = abs(frechet a1 a2) // Test arrays: let linearPositive = (seq {for x in 0. .. 1. .. 100. do yield (x, x)}) |> Array.ofSeq let linearNegative = (seq {for x in 0. .. 1. .. 100. do yield (x, 100.-x)}) |> Array.ofSeq let linearPositiveOffset10 = (seq {for x in 0. .. 1. .. 100. do yield (x+10., x+10.)}) |> Array.ofSeq let x_x2 = (seq {for x in 0. .. 1. .. 100. do yield (x, x*x)}) |> Array.ofSeq let x_x2_plus2y = (seq {for x in 0. .. 1. .. 100. do yield (x, x*x+2.)}) |> Array.ofSeq let x_x2_plus2x = (seq {for x in 0. .. 1. .. 100. do yield (x+2., x*x)}) |> Array.ofSeq let x_x2_plus10xy = (seq {for x in 0. .. 1. .. 100. do yield (x+10., x*x+10.)}) |> Array.ofSeq let x_x2_scaled_x = (seq {for x in 0. .. 1. .. 100. do yield (x/2., x*x+10.)}) |> Array.ofSeq let circle = (seq {for x in 0. .. 0.1 .. System.Math.PI*2. do yield (Math.Sin(x), Math.Cos(x))}) |> Array.ofSeq let circleStretched = (seq {for x in 0. .. 0.1 .. System.Math.PI*2. do yield (Math.Sin(x)*4.5, Math.Cos(x)*3.4)}) |> Array.ofSeq // Tests: let test1 = linearPositive -~~ linearPositive // 0.0 - identical let test2 = linearPositive -~~ linearNegative // 100.0 let test3 = linearPositive -~~ linearPositiveOffset10 // 14.14213562 let test4 = x_x2 -~~ x_x2 // 0.0 - identical let test5 = x_x2 -~~ x_x2_plus2y // 2.0 let test6 = x_x2 -~~ x_x2_plus2x // 2.0 let test7 = x_x2 -~~ x_x2_plus10xy // 14.14213562 let test8 = x_x2 -~~ x_x2_scaled_x // 50.99019514 let test9 = circle -~~ circleStretched // 3.4998577 ``````
namespace System
val frechet : P:(float * float) array -> Q:(float * float) array -> float
Full name: Script.frechet
val P : (float * float) array
type 'T array = 'T []
Full name: Microsoft.FSharp.Core.array<_>
Multiple items
val float : value:'T -> float (requires member op_Explicit)
Full name: Microsoft.FSharp.Core.Operators.float
--------------------
type float = Double
Full name: Microsoft.FSharp.Core.float
--------------------
type float<'Measure> = float
Full name: Microsoft.FSharp.Core.float<_>
val Q : (float * float) array
val sq : (float -> float)
val x : float
val min3 : ('a -> 'a -> 'a -> 'a) (requires comparison)
val x : 'a (requires comparison)
val y : 'a (requires comparison)
val z : 'a (requires comparison)
Multiple items
module List
from Microsoft.FSharp.Collections
--------------------
type List<'T> =
| ( [] )
| ( :: ) of Head: 'T * Tail: 'T list
interface IEnumerable
interface IEnumerable<'T>
member IsEmpty : bool
member Item : index:int -> 'T with get
member Length : int
member Tail : 'T list
static member Cons : head:'T * tail:'T list -> 'T list
static member Empty : 'T list
Full name: Microsoft.FSharp.Collections.List<_>
val min : list:'T list -> 'T (requires comparison)
Full name: Microsoft.FSharp.Collections.List.min
val d : (float * float -> float * float -> float)
val a : float * float
val b : float * float
val ab_x : float
val abs : value:'T -> 'T (requires member Abs)
Full name: Microsoft.FSharp.Core.Operators.abs
val fst : tuple:('T1 * 'T2) -> 'T1
Full name: Microsoft.FSharp.Core.Operators.fst
val ab_y : float
val snd : tuple:('T1 * 'T2) -> 'T2
Full name: Microsoft.FSharp.Core.Operators.snd
val sqrt : value:'T -> 'U (requires member Sqrt)
Full name: Microsoft.FSharp.Core.Operators.sqrt
val p : int
val q : int
type Array =
member Clone : unit -> obj
member CopyTo : array:Array * index:int -> unit + 1 overload
member GetEnumerator : unit -> IEnumerator
member GetLength : dimension:int -> int
member GetLongLength : dimension:int -> int64
member GetLowerBound : dimension:int -> int
member GetUpperBound : dimension:int -> int
member GetValue : [<ParamArray>] indices:int[] -> obj + 7 overloads
member Initialize : unit -> unit
member IsFixedSize : bool
...
Full name: System.Array
val length : array:'T [] -> int
Full name: Microsoft.FSharp.Collections.Array.length
val ca : float [,]
module Array2D
from Microsoft.FSharp.Collections
val init : length1:int -> length2:int -> initializer:(int -> int -> 'T) -> 'T [,]
Full name: Microsoft.FSharp.Collections.Array2D.init
val c : (int -> int -> float)
val i : int
val j : int
type Math =
static val PI : float
static val E : float
static member Abs : value:sbyte -> sbyte + 6 overloads
static member Acos : d:float -> float
static member Asin : d:float -> float
static member Atan : d:float -> float
static member Atan2 : y:float * x:float -> float
static member BigMul : a:int * b:int -> int64
static member Ceiling : d:decimal -> decimal + 1 overload
static member Cos : d:float -> float
...
Full name: System.Math
Math.Max(val1: decimal, val2: decimal) : decimal
Math.Max(val1: float, val2: float) : float
Math.Max(val1: float32, val2: float32) : float32
Math.Max(val1: uint64, val2: uint64) : uint64
Math.Max(val1: int64, val2: int64) : int64
Math.Max(val1: uint32, val2: uint32) : uint32
Math.Max(val1: int, val2: int) : int
Math.Max(val1: uint16, val2: uint16) : uint16
Math.Max(val1: int16, val2: int16) : int16
Math.Max(val1: byte, val2: byte) : byte
val nan : float
Full name: Microsoft.FSharp.Core.Operators.nan
val a1 : (float * float) array
val a2 : (float * float) array
val linearPositive : (float * float) []
Full name: Script.linearPositive
Multiple items
val seq : sequence:seq<'T> -> seq<'T>
Full name: Microsoft.FSharp.Core.Operators.seq
--------------------
type seq<'T> = Collections.Generic.IEnumerable<'T>
Full name: Microsoft.FSharp.Collections.seq<_>
val ofSeq : source:seq<'T> -> 'T []
Full name: Microsoft.FSharp.Collections.Array.ofSeq
val linearNegative : (float * float) []
Full name: Script.linearNegative
val linearPositiveOffset10 : (float * float) []
Full name: Script.linearPositiveOffset10
val x_x2 : (float * float) []
Full name: Script.x_x2
val x_x2_plus2y : (float * float) []
Full name: Script.x_x2_plus2y
val x_x2_plus2x : (float * float) []
Full name: Script.x_x2_plus2x
val x_x2_plus10xy : (float * float) []
Full name: Script.x_x2_plus10xy
val x_x2_scaled_x : (float * float) []
Full name: Script.x_x2_scaled_x
val circle : (float * float) []
Full name: Script.circle
field Math.PI = 3.14159265359
Math.Sin(a: float) : float
Math.Cos(d: float) : float
val circleStretched : (float * float) []
Full name: Script.circleStretched
val test1 : float
Full name: Script.test1
val test2 : float
Full name: Script.test2
val test3 : float
Full name: Script.test3
val test4 : float
Full name: Script.test4
val test5 : float
Full name: Script.test5
val test6 : float
Full name: Script.test6
val test7 : float
Full name: Script.test7
val test8 : float
Full name: Script.test8
val test9 : float
Full name: Script.test9
| 2,695 | 8,037 |
{"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-2022-27
|
latest
|
en
| 0.56186 |
https://byjus.com/question-answer/a-student-was-given-the-following-details-while-constructing-a-triangle-abc-the-length-of/
| 1,719,094,010,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-26/segments/1718198862420.91/warc/CC-MAIN-20240622210521-20240623000521-00393.warc.gz
| 125,039,318 | 24,060 |
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question
# A student was given the following details while constructing a triangle ABC: The length of the base of the triangle BC, one of the base angles say ∠ B and the sum of the other two sides of the triangle (AB+AC) He went about the construction of this triangle by first drawing the base of the triangle BC. He then drew an angle at the point B equal to the given angle on a ray that he drew. After completing these steps, he got stuck and doesn’t know what to do next. Which of the following steps will he take up next?
A
Cut a line segment BD equal to (AB+AC) on that same ray
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
Change the base length to (AB+AC) and then draw one of the base angles at one of the ends of the line segment whose length is equal to (AB+AC)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
He knows the perimeter of the triangle, he changes the base length to that of the perimeter of the triangle and then draws one of the base angles at one of the ends of the line segment
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
Cut a line segment BD equal to 2(AB+AC) on that same ray
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
## The correct option is A Cut a line segment BD equal to (AB+AC) on that same ray We have been given the base length and a base angle, which means half the job is already done for us. We need to draw the base with the given length and draw a ray with an angle = base angle, after that we must cut a line segment BD equal to (AB + AC) on that same ray so as to perpendicularly bisect the ray DC and have AD = AC because A is the point where the perpendicular bisector of ray DC cuts BD.
Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Let's Build Triangles
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
| 501 | 1,972 |
{"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-2024-26
|
latest
|
en
| 0.920715 |
https://tutorme.com/tutors/71759/interview/
| 1,561,373,847,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-26/segments/1560627999482.38/warc/CC-MAIN-20190624104413-20190624130413-00240.warc.gz
| 608,202,724 | 47,492 |
TutorMe homepage
Subjects
PRICING
COURSES
Start Free Trial
Ashish P.
B.S In Aerospace Engineering from Purdue University. Pursuing an M.S in Astronautical Engineering.
Tutor Satisfaction Guarantee
Writing
TutorMe
Question:
Create three different "hooks" that grab the reader's attention in the opening paragraph of an essay that is focused on explaining air pollution effects to a generic audience.
Ashish P.
The answer will depend on each student.
Basic Math
TutorMe
Question:
If Jack has 2 chocolates and Jill has 3 chocolates, how many chocolates do Jack and Jill have?
Ashish P.
Jack + Jill = 2 + 3 = 5 chocolates.
Algebra
TutorMe
Question:
If a movie theater charges a \$2 fee plus \$0.60 per minute of movie showtime, how much will a 90-minute movie cost?
Ashish P.
Since each minute costs \$0.60, we can represent the cost as "x" minutes times \$0.60. This becomes 0.60*x. Also, we have a \$2 fee, so the algebraic expression is 0.60*x + 2. We substitute 90 minutes in for "x" to yield 0.60*(90) + 2 = \$56.
Send a message explaining your
needs and Ashish will reply soon.
Contact Ashish
| 295 | 1,106 |
{"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}
| 3.5 | 4 |
CC-MAIN-2019-26
|
latest
|
en
| 0.889683 |
https://www.coursehero.com/file/7796168/500-m-0750-m-20-REASONING-The-magnitude-of-the-force-that/
| 1,529,592,529,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-26/segments/1529267864172.45/warc/CC-MAIN-20180621133636-20180621153636-00516.warc.gz
| 778,022,084 | 28,242 |
Physics Solution Manual for 1100 and 2101
# 500 m 0750 m 20 reasoning the magnitude of the force
This preview shows page 1. Sign up to view the full content.
This is the end of the preview. Sign up to access the rest of the document.
Unformatted text preview: d on the lid by the outside air. SOLUTION According to Equation 11.3, pressure is defined as P = F / A ; therefore, the magnitude of the force on the lid due to the air pressure is F = (0.85 × 105 N/m 2 )(1.3 × 10 –2 m 2 ) = 1.1× 103 N 12. REASONING Pressure is the magnitude of the force applied perpendicularly to a surface divided by the area of the surface, according to Equation 11.3. The force magnitude, therefore, is equal to the pressure times the area. SOLUTION According to Equation 11.3, we have c F = PA = 8.0 × 10 4 lb / in. 2 6 b hb.1 in.g2.6 in.g= 1.3 × 10 6 lb 13. REASONING According to Equation 11.3, the pressure P exerted on the ground by the stack of blocks is equal to the force F exerted by the blocks (their combined weight) divided by the area A of the block’s surface in contact with the ground, or P = F/A. Since the pressure is largest when the area is smallest, the least number of blocks is used when the surface area in contact with the ground is the smallest. This area is 0.200 m × 0.100 m. SOLUTION The pressure exerted by N blocks stacked on top of one another is P= F N Wone block = A A (11.3) where Wone block is the weight of one block. The least number of whole blocks required to 5 produce a pressure of two atmospheres (2.02 × 10 Pa) is N= PA Wone block ( 2.02 ×105 Pa ) ( 0.200 m × 0.100 m ) = 24 = 169 N 570 FLUIDS 14. REASONING Since the weight is distributed uniformly, each tire exerts one-half of the weight of the rider and bike on the ground. According to the definition of pressure, Equation 11.3, the force that each tire exerts on the ground is equal to the pressure P inside the tire times the area A of contact between the tire and the ground. From this relation, the area of contact can be found. SOLUTION The area of contact that each tire makes with the ground is F A= = P 1 2 (Wperson + Wbike ) = 12 ( 625 N + 98 N ) = 4.76 ×10−4 m2 (11.3) 7.60 × 105 Pa P 15. REASONING The cap is in equilibrium, so the sum of all the +y forces acting on it must be zero. There are three forces in the Foutside vertical direction: the force Finside due to the gas pressure inside the bottle, the force Foutside due to atmospheric pressure outside the bottle, and the force Fthread that the screw thread exerts on Finside F the cap. By setting the sum of these forces to zero, and using the thread relation F = PA, where P is the pressure and A is the area of the cap, we can determine the magnitude of the force that the screw threads exert on the cap. SOLUTION The drawing shows the free-body diagram of the cap and the three vertical forces that act on it. Since the cap is in equilibrium, the net force in the vertical direction must be zero. ΣFy = − Fthread + Finside − Foutside = 0 (4.9b) Solving this equation for Fthread, and using the fact that force equals pressure times area, F = PA (Equation 11.3), we have Fthread = Finside − Foutside = Pinside A − Poutside A ( )( ) = ( Pinside − Poutside ) A = 1.80 ×105 Pa − 1.01×105 Pa 4.10 ×10−4 m 2 = 32 N 16. REASONING The power generated by the log splitter pump is the ratio of the work W done on the piston to the elapsed time t: Power = W t (6.10a) Chapter 11 Problems 571 The work done on the piston by the pump is equal to the magnitude F of the force exerted on the piston by the hydraulic oil, multiplied by the distance s through which the piston moves: ( ) W = F cos0o s = Fs (6.1) We have used θ = 0° in Equation 6.1 because the piston moves in the same direction as the force acting on it. The magnitude F of the force applied to the piston is given by F = PA (11.3) where A is the cross-sectional area of the piston and P is the pressure of the hydraulic oil. SOLUTION The head of the piston is circular with a radius r, so its cross-sectional area is given by A = π r 2 . Substituting Equation 11.3 into Equation 6.1, therefore, yields W = PAs = Pπ r 2 s (1) Substituting Equation (1) into Equation 6.10a gives the power required to operate the pump: ( ) 2.0 × 107 Pa π ( 0.050 m ) W Pπ r 2 s Power = = = t t 25 s 2 ( 0.60 m ) = 3.8 × 103 W 17. REASONING The pressure P due to the force FSonF that the suitcase exerts on the elevator F floor is given by P = SonF (Equation 11.3), where A is the area of the elevator floor A beneath the suitcase (equal to the product of the length and width of that region). According to Newton’s 3rd law, the magnitude FSonF of the downward force the suitcase exerts on the floor is equal to the magnitude FFonS of the upward force the floor exerts on the suitcase. We will use Newton’s 2nd law, ΣF = ma (Equation 4.1), to determine the magnitude FFonS of the upward force on the suitcase, which has a mass m and an upward acceleration of magnitude a = 1.5 m/s2, equal to that of the elevator. SOLUTION There are only two forces acting on the suitcase, the upward force FFonS that the floor exerts on the suitcase, and the downward weight W = mg (Equation 4.5) exerted by the earth, where g is the magnitude of the acceleration due to grav...
View Full Document
{[ snackBarMessage ]}
### What students are saying
• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.
Kiran Temple University Fox School of Business ‘17, Course Hero Intern
• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.
Dana University of Pennsylvania ‘17, Course Hero Intern
• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.
Jill Tulane University ‘16, Course Hero Intern
| 1,661 | 6,230 |
{"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-2018-26
|
latest
|
en
| 0.907291 |
https://ekuatio.com/en/how-to-solve-limits-with-infinite-indeterminacy-between-infinity/
| 1,606,905,080,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2020-50/segments/1606141706569.64/warc/CC-MAIN-20201202083021-20201202113021-00623.warc.gz
| 281,991,595 | 19,854 |
How to solve limits with infinite indeterminacy between infinity.
# How to solve limits with infinite indeterminacy between infinity.
Now I will explain how to solve the infinite indeterminacy between infinite and infinite in the calculation of limits.
Normally, to solve the limit, we only have to replace the x by the value it tends to.
However, there are times when we find that when replacing, the result is an indeterminacy and in that case, we must use the calculation method that corresponds in each case.
## How to calculate limits with indeterminations of the infinite type between infinite and infinite. Exercises resolved.
This type of indeterminacy is found in the limits of when x tends to be infinite (or less infinite) of rational functions and when replacing it, we have ∞/∞.
In general they are have this shape:
When we have the indeterminacy of infinite limits between infinity, we have to remove terms from the numerator and denominator and leave only the highest degree terms up and down.
Once we have the highest grade terms, we can operate and eliminate the x’s that are repeated in both the numerator and denominator.
When operating, the indeterminacy disappears and we must replace the x again by the number it tends to reach the result.
We are going to see it step by step with several exercises solved as an example. We start with this limit:
All limits begin with the same resolution: substituting the x for the number it tends to, that is, it is not resolved knowing before beginning that it will be an indeterminacy. We don’t have to know that.
Once we have replaced the x and y in operation, then we can either arrive at a concrete result or come to the conclusion that it is an indeterminacy.
First, we replace the x with infinity and we are left with an indeterminacy:
To resolve this type of indeterminacy we leave only the term of the highest degree in the numerator and denominator:
Once we have left the term of higher degree, we operate and eliminate the factors that are repeated in both the numerator and the denominator. In this case, we can eliminate an x on each side:
And finally, we have replaced the x with infinity again, but this time, we no longer have any indeterminacy and the result is infinite among a number, which is equal to infinity.
Let’s see another example:
We start by substituting the x for infinity and then we’re left with it:
This time we have an infinite negative in the denominator, but it is still the same kind of indeterminacy.
We leave the term with the highest degree of numerator and denominator:
We operate by removing an x from the numerator and another from the denominator and replace the x with infinity again, arriving at the result of less infinity, as we are dividing by a negative number:
Here’s another example:
We replace the x with infinity and reach the infinite indeterminacy between infinity:
We leave the term of highest degree in the numerator and in the denominator:
We operate and can eliminate an x in both the numerator and the denominator. This time in the numerator we don’t have any x, so when it comes to substituting, the infinite is left only in the denominator and therefore, a number between infinite, gives us zero as a result:
Let’s end the infinite indeterminacy broken by the infinite with this example:
We replace the x by infinite and the infinite indeterminacy remains split by infinite:
We leave the highest grade term up and down:
When operating, we can eliminate x squared in both the numerator and denominator, so that the x’s disappear. Therefore, when it comes to replacing x with infinity again, as we do not have x, the result is the same number that we have left:
## Infinite indetermination between infinity with roots
When we meet radicals on the edge, it often causes confusion. We will see how to solve them if we have the indeterminacy of infinity between infinity and roots.
For example:
We replace the x with infinity and come to the conclusion that it is an indeterminacy:
We are left with the term of the highest degree, but now we must be careful with the radicals. In the numerator, the term with the highest degree is the first, since the second term is within another root and therefore the degree is lower.
In the denominator, the highest grade term is the root term.
We are therefore left with the same term above and below:
When operating, we are left with the function equal to 1, since both terms can be divided between them, and therefore, the limit of the function when x tends to infinity is equal to 1:
Let’s see another example:
We replace the x with infinity and we have the indeterminacy of infinity for infinity.
We leave the term of highest degree in the numerator and in the denominator:
And now we have to trade, but it’s not as clear how to trade to eliminate one of the x’s, as in the other examples we’ve seen.
In the first place, we pass the root to the fractional exponent:
And now, to operate with the x’s we consider them as a division of powers of the same base, where we keep the x’s and subtract the exponents:
And there is only one x left in the numerator, which when you substitute infinity, the result is infinite:
## Conclusion of the infinite indetermination between infinity
With the indeterminacy of infinity between infinity, the result will depend on the degree of the polynomials of the numerator and the denominator.
• If the degree of P(x) is less than the degree of Q(x), the result will be zero
• If the degree of P(x) is greater than the degree of Q(x) the result will be infinite or less infinite
• If the grade of P(x) is the same as the grade of Q(x) the result will be a number
In the solved examples that we have solved throughout the lesson, we have seen each of the cases at the same time.
| 1,229 | 5,818 |
{"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}
| 4.71875 | 5 |
CC-MAIN-2020-50
|
longest
|
en
| 0.948162 |
https://electronicsarea.com/capacitor-and-direct-current/
| 1,723,435,840,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-33/segments/1722641028735.71/warc/CC-MAIN-20240812030550-20240812060550-00497.warc.gz
| 188,789,295 | 12,900 |
## Capacitor and Direct Current
The capacitor is manufactured in many shapes and materials, but no matter how it has been constructed, it is always a device with two plates separated by an insulating material. If a battery is connected to a capacitor, it will allows no direct current (DC) to flow through.
If a not charged capacitor is connected across the terminals of a battery, a transient current flows as the capacitor plates charge up.
Current flows from the battery terminals to the capacitor plates.
• The positive battery terminal, attracts electrons from the upper capacitor plate. The plate is positively charged.
• The negative battery terminal, fill with electrons the bottom capacitor plate. The plate is negatively charged.
See the flow of electrons charging the capacitor plates in the diagram. The flow of electrons decreases slowly and finally stops. At this time the capacitor behaves like an open circuit for DC.
The current that is discussed in the preceding paragraphs is a current that varies over time, the current starts from a maximum value and decreases to 0 amps, when there is no current flowing. This happens in a very short period of time and is called “transient current”.
The amount of charge a capacitor can store, is called “capacitance”. The capacitance value depends on the physical characteristics of the capacitor.
• The greater the area of the plates, the higher the capacitance.
• The smaller the gap between plates, the higher the capacitance.
• The type of dielectric used between the plates affects the capacitance value.
The dielectric aims to increase the capacitance value of the capacitor. When a dielectric is used, it acquires a charge opposite to the charge on the plates, reducing the net charge of the device, allowing the arrival of more charge to the plates.
## Permittivity
Different materials are used as dielectrics, with different degrees of permittivity (Different degrees of ability to establish an electric field).
The higher the permittivity, the higher the capacity the dielectric allows to obtain. Capacity is calculated using the formula: C = (Er x A)/d, where:
• C = capacity.
• Er = permittivity.
• A = plates’s area.
• D = distance between plates.
The unit of measurement of the capacitor is the Farad, but this unit is big and is more common use:
| 488 | 2,331 |
{"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}
| 3.578125 | 4 |
CC-MAIN-2024-33
|
latest
|
en
| 0.888448 |
https://learn.careers360.com/ncert/ncert-solutions-class-8-maths-chapter-2-linear-equations-in-one-variable/
| 1,586,324,563,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2020-16/segments/1585371810617.95/warc/CC-MAIN-20200408041431-20200408071931-00022.warc.gz
| 557,004,246 | 109,463 |
# NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable
NCERT solutions for class 8 maths chapter 2 Linear Equations in One Variable- This chapter created a strong foundation for algebra unit. Algebra is an integral part of higher mathematics where you have to deal with variables along with numbers. CBSE NCERT solutions for class 8 maths chapter 2 Linear Equations in One Variable is comprehensively covering each problem related to this particular topic. In this chapter, there will be an expression consisting of variables as well as numbers and you have to find the value of that variable. Linear equation in one variable is really helpful to frame a word problem into an algebraic expression. Solutions of NCERT for Class 8 Maths Chapter 2 Linear Equations in One Variable are covering solutions to word problems using the linear equations of one variable. Let's take an example to understand the formation of the linear equation of one variable using the given statements. The present age of Ram’s mother is 3 times the present age of Ram, after 5 years their ages will add to 65 years. The above said word problem can be framed in an equation (3x + 5) + (x + 5) = 65 where x is the age of the Present age of Ram and 3x is the present age of Ram's mother. This chapter has a total of 6 exercises consisting of a total of 65 questions in all. NCERT solutions for class 8 maths chapter 2 Linear Equations in One Variable has the solution to all 65 questions to boost your overall preparation level. NCERT solutions for other important classes and subjects are also downloadable for free.
## NCERT solutions for class 8 maths chapter 2 Linear Equations in One Variable Excercise: 2.1
1. Transposing -2 to the RHS, we get
x = 7 + 2 = 9
Transposing 3 to the RHS, we get
y = 10 - 3 = 7
y = 7
Transposing 2 to the LHS, we get
6 - 2 = z => z= 4
Thus z = 4
$\frac{3}{7} + x = \frac{17}{7}$
4. Transposing $\frac{3}{7}$ to the RHS, we get
$x = \frac{17}{7}-\frac{3}{7} = \frac{14}{7} = 2$
Thus x =2
Dividing both sides by 6, we get
$x = \frac{12}{6} => 2$
Thus x = 2
$\frac{t}{5} = 10$
Multiplying both sides by 5, we get
$t = 10\times 5 = 50$
Thus t = 50
$\frac{2x}{3} = 18$
7.
$\frac{2x}{3} = 18$
Multiplying both sides by 3, we get
$2x = 18\times 3 = 54$
Now, dividing both sides by 2, we get
$x = \frac{54}{2} = 27$
$1.6 = \frac{y}{1.5}$
Multiplying both sides by 1.5, we get
$1.6\times 1.5 = y = 2.4$
Thus y = 2.4
Transposing 9 to the RHS, we get
7x = 16 + 9 = 25
Now, dividing both sides by :
$x = 25\times \frac{1}{7} = \frac{25}{7}$
Transposing -8 to the RHS, we get
14y = 13 + 8 = 21
Now dividing both sides by 14:
$y = \frac{21}{14} = \frac{3}{2}$
At first, transposing 17 to the RHS:
6p = 9 - 17 = -8
Now, dividing both sides by 6, we get
$p = \frac{-8}{6} = \frac{-4}{3}$
$\frac{x}{3} +1 = \frac{7}{15}$
12.
$\frac{x}{3} + 1 = \frac{7}{15}$
Transposing 1 to the RHS, we get
$\frac{x}{3} = \frac{7}{15} -1 = \frac{7-15}{15} = \frac{-8}{15}$
Now multiplying both sides by 3, it becomes
$x = \frac{-8}{15}\times 3 = \frac{-8}{5}$
NCERT solutions for class 8 maths chapter 2 Linear Equations in One Variable Excercise: 2.2
Assume the number to be x.
Thus according to the question,
$\left ( x-\frac{1}{2} \right )\times \frac{1}{2} = \frac{1}{8}$
Multiplying both sides by 2, we get
$\left ( x-\frac{1}{2} \right ) = \frac{1}{8}\times 2 = \frac{1}{4}$
Now transposing $-\frac{1}{2}$ to the RHS, we get
$x = \frac{1}{4} + \frac{1}{2} = \frac{1+2}{4} = \frac{3}{4}$
Thus the number was x = 3/4
Let the breadth of the pool be x m.
According to the question, the length of the pool = 2x + 2 m.
Perimeter of recatangle = 2(l + b) = 154 m .
i.e., 2(2x + 2 + x) = 154
2(3x + 2) = 154
Dividing both sides by 2, we get
3x + 2 = 77
Now transposing 2 to the RHS, we get
3x = 77 - 2 = 75
Dividing both sides by 3, we get
$x = \frac{75}{3} = 25$
Thus breadth of pool = 25 m
and lenght of the pool = 2$\times$25 + 2 = 52 m
In isosceles triangles, we have 2 sides of equal length.
Given that its perimeter is $\frac{62}{15}$ cm.
Let's assume the length of the equal side is x cm.
Also,
$Perimeter = x + x + \frac{4}{3} = \frac{62}{15}$
Transposing $\frac{4}{3}$ to the RHS side it becomes,
$2x = \frac{62}{15}-\frac{4}{3} = \frac{62-20}{15} = \frac{42}{15} = \frac{14}{5}$
Now dividing both sides by 2, we get
$x = \frac{14}{5}\times \frac{1}{2} = \frac{7}{5}$
Hence, the length of the equal sides of the isosceles triangle is $\frac{7}{5}$ cm.
Assume one number to be x. Then the other number will be x + 15.
Now it is given that sum of the two numbers is 95.
Thus quation becomes x + x + 15 = 95
or 2x +15 = 95
Transposing 15 to the RHS, it becomes
2x = 95 - 15 = 80
Dividng both sides by 2, we get
$x = \frac{80}{2} = 40$
Thus two numbers are 40 and 55
Given that numbers are in the ratio of 5:3, so we can assume numbers to be 5x and 3x.
Also, the difference between these numbers is 18, so the equation becomes
5x - 3x = 18
2x = 18
Dividing both sides by 2, we get
x = 9
Hence the two numbers are 5$\times$9 = 45 and 3$\times$9 = 27
Let the consecutive integers be x, x+1, x+2.
Sum of these integers is given to be 51.
Thus equation becomes : x + x + 1 + x + 2 = 51
3x + 3 = 51
Transposing 3 to the RHS,
3x = 48
Now dividing both sides by 3, we get
$x = \frac{48}{3} = 16$
Hence the consecutive integers are 16, 17, 18.
Let x be the multiple of 8.
Then the three consecutive integers (multiple of 8) are x, x+8, x+16.
Given their sum is 888,
thus the equation becomes: x + x + 8 + x + 16 = 888
or 3x + 24 = 888
Transposing 24 to the RHS
3x = 888 - 24 = 864
Now dividing both sides by 3, we get
$x = \frac{864}{3} = 288$
Thus the required consecutive integers are 288, 296, 304.
Let the consecutive integers be x, x+1, x+2.
Now according to the question equation becomes,
2x + 3(x+1) + 4(x+2) = 74
or 2x + 3x + 3 + 4x + 8 = 74
or 9x + 11 = 74
Transposing 11 to the RHS we get,
9x = 74 - 11 = 63
Divinding both sides by 9.
$x = \frac{63}{9} = 7$
Therefore required consecutive integers are 7, 8, 9.
Let the current age of Rahul and Haroon be 5x and 7x respectively(Since there age ratio is given as 5:7).
Four years later their ages become 5x + 4 and 7x + 4 respectively.
According to the question,
5x + 4 +7x + 4 = 56
or 12x + 8 = 56
Transposing 8 to the RHS we get,
12x = 48
Dividing both sides by 12:
$x = \frac{48}{12} = 4$
Thus current age of Rahul = 5$\times$4 = 20
Haroon = 7$\times$4 = 28
Let us assume the number of boys and the number of girls is 7x and 5x.
According to the given data in the question,
7x = 5x + 8
Transposing 5x to the LHS we get,
7x - 5x = 8
2x = 8
Dividing both sides by 2
x = 4
Hence number of boys = 7$\times$4 = 28
and number of girls = 5$\times$4 = 20
So total number of students = 28 + 20 = 48.
Let the age of Baichung be x years.
Then according to question age of Baichung's father = x + 29 years
and age of Baichung's grandfather = x + 29 + 26 years
The sum of their ages is given 135 years. According to that, x + x + 29 + x + 29 + 26 = 135
3x + 84 = 135
Transposing 81 on the RHS we get,
3x = 135 - 84 = 51
Dividing both sides by 3
$x =\frac{51}{3} = 17$
Thus, the age of Baichung = 17 years
The age of Baichung's father = 17 + 29 = 46 years
The age of Baichung's grandfather = 46 + 26 = 72 years
Let us assume Ravi's present age to be x years.
According to the question,
x + 15 = 4x
Transposing x to the RHS
15 = 3x
Now dividing both sides by 3, we get
$x = \frac{15}{3} = 5$
Hence Ravi's current age is 5 years.
Let the rational number be x.
According to the question,
$x\times \frac{5}{2} + \frac{2}{3} = -\frac{7}{12}$
Transposing $\frac{2}{3}$ to the RHS:
$x\times \frac{5}{2} = -\frac{7}{12} - \frac{2}{3} = \frac{-15}{12} = \frac{-5}{4}$
Multiplication by 2 in both sides, we get
$5x = \frac{-5}{4}\times 2 = \frac{-5}{2}$
Dividing both sides by 5, we get
$x = \frac{-5}{2}\times \frac{1}{5} = \frac{-1}{2}$
Since the ratio of currency notes is 2:3:5. Therefore,
Let the number of currency notes of Rs.100, Rs.50 and Rs.10 be 2x, 3x, 5x respectively.
Hence according to the question equation becomes,
100$\times$2x + 50$\times$3x + 10$\times$5x = 400000
or 200x + 150x + 50x = 400000
or 400x = 400000
Dividing both sides by 400 we get,
x = 1000 .
No. of Rs.100 notes = 2$\times$1000 = 2000 notes
No. of Rs.50 notes = 3$\times$1000 = 3000 notes
No. of Rs.10 notes = 5$\times$1000 = 5000 notes
Let the no. of Rs.1 coin be y.
And the no. of Rs.5 coin be x.
Thus according to question no. of Rs.2 coin will be 3x.
Also, the total no. of coins = 160
This implies : y + x + 3x = 160
or y + 4x = 160
Transposing 4x to the RHS
y = 160 - 4x = No. of Rs.1 coin.
Now, it is given that total amount is Rs.300.
i.e., 1(160-4x) + 2(3x) + 5(x) = 300
or 160 - 4x + 6x + 5x = 300
or 160 + 7x = 300
Transposing 160 to the RHS :
7x = 300 - 160 = 140
Dividing both sides by 7:
$x = \frac{140}{7} = 20$
x = 20
Thus number of Rs.1 coin = 160 - 4x = 160 - 80 = 80
number of Rs.2 coin = 3x = 60
number of Rs.5 coin = x = 20
Let the no. of winners be x.
Since total no. of participants is 63, thus no. of participants who does not win = 63 - x
According to the question, equation becomes:
x(100) + (63-x)(25) = 3000
or 100x + 1575 - 25x = 3000
or 75x + 1575 = 3000
Transposing 1575 to the RHS :
75x = 3000 - 1575
or 75x = 1425
Dividing both sides by 75, we get
$x = \frac{1425}{75} = 19$
x = 19
Therefore no. of winners = 19
NCERT solutions for class 8 maths chapter 2 Linear Equations in One Variable Excercise: 2.3
We have
3x = 2x + 18
Subtracting 2x from both sides
3x - 2x = 2x - 2x + 18
or x = 18
Check:- Put x = 18 in both LHS and the RHS.
LHS = 3x = 3(18) = 54
RHS = 2x + 18 = 2(18) + 18 = 36 + 18 = 54
Thus, LHS = RHS
We have
5t - 3 = 3t - 5
Transposing 3t to the LHS and -3 to the RHS, we get:
5t - 3t = -5 + 3
or 2t = -2
Dividing both sides by 2 :
t = -1
Check:- Put t = -1 in the LHS we have,
5t - 3 = 5(-1) - 3 = -5 -3 = -8
Similarly put t = -1 in the RHS:
3t - 5 = 3(-1) - 5 = -3 - 5 = -8
Hence, LHS = RHS
We have
5x + 9 = 5 + 3x
Transposing 3x to the LHS and 9 to the RHS, we get:
5x - 3x = 5 - 9
or 2x = -4
Dividing both sides by 2 :
x = -2
Check :- Put x = - 2 in both LHS and RHS
LHS :- 5x + 9 = 5(-2) + 9 = -10 + 9 = -1
RHS :- 5 + 3x = 5 + 3(-2) = 5 - 6 = -1
Hence, LHS = RHS
We have
4z + 3 = 6 +2z
Transposing 2z to the LHS and 3 to the RHS, we get:
4z - 2z = 6 - 3
or 2z = 3
DIviding both sides by 2,
$z = \frac{3}{2}$
Check:- Put $z = \frac{3}{2}$ in LHS as well as RHS, we have
LHS :- 4z + 3
$= 4\times \frac{3}{2} + 3 = 9$
RHS :- 6 + 2z
$= 6 + 2\times \frac{3}{2} = 6 + 3 = 9$
Thus, LHS = RHS
We have
2x - 1 = 14 - x
Transposing -x to the LHS and -1 to the RHS
2x + x = 14 + 1
or 3x = 15
Dividing both sides by 3, we get
x = 5
Check :- Put x = 5 in both the LHS and the RHS
LHS :- 2x - 1 = 2(5) - 1 = 10 - 1 = 9
RHS :- 14 - x = 14 - 5 = 9
Hence, LHS = RHS
We have
8x + 4 = 3 (x - 1) + 7
or 8x + 4 = 3x - 3 + 7 = 3x + 4
Transposing 3x to the LHS and 4 to the RHS, we get
8x - 3x = 4 - 4 = 0
5x = 0
Dividing both sides by 5:
x = 0
Check :- Putting x = 0 in both LHS and RHS:
LHS :- 8x + 4 = 8(0) + 4 = 4
RHS :- 3(x-1) + 7 = 3(-1) + 7 = -3 + 7 = 4
Hence, LHS = RHS
$x =\frac{4}{5}\times \left ( x+10 \right )$
We have
$x = \frac{4}{5}\left ( x + 10 \right )$
$= \frac{4}{5}x + 8$
Transposing $\frac{4}{5}x$ to the left-hand side:
$x - \frac{4}{5}x = 8$
or $\frac{5x-4x}{5} = 8$
or $\frac{x}{5} = 8$
Multiplying both sides by 5
x = 40
Check :- Put x = 40 in both the LHS and the RHS.
LHS :- x = 40
RHS :- $\frac{4}{5}\left ( x + 10 \right ) = \frac{4}{5}\left ( 40 + 10 \right )$
$= \frac{4}{5}\left ( 50 \right ) = 40$
Thus, LHS = RHS
$\frac{2x}{3}+1 = \frac{7x}{15}+3$
We have
$\frac{2x}{3} + 1 = \frac{7x}{15} + 3$
Transposing $\frac{7x}{15}$ to the LHS and 1 to the RHS:
$\frac{2x}{3} - \frac{7x}{15} = 3 - 1 = 2$
or $\frac{10x - 7x}{15} = 2$
or $\frac{3x}{15} = 2$
or x = 10
Check :- Put x = 10 in both the LHS and the RHS
LHS :
$\frac{2x}{3} + 1 = \frac{2}{3}\times 10 + 1 = \frac{23}{3}$
RHS :
$\frac{7x}{15} + 3 = \frac{7}{15}\times 10 + 3 = \frac{14}{3} + 3 = \frac{23}{3}$
Thus, LHS = RHS
$2y+\frac{5}{3} = \frac{26}{3} -y$
We have
$2y+ \frac{5}{3} = \frac{26}{3} - y$
Transposing -y to the LHS and $\frac{5}{3}$ to the RHS, we get
$2y + y = \frac{26}{3}- \frac{5}{3} = \frac{21}{3} = 7$
or 3y = 7
Dividing both sides by 3:
$y = \frac{7}{3}$i
Check :- Put $y = \frac{7}{3}$ in both the LHS and the RHS:
LHS :
$2y + \frac{5}{3} = 2\times \frac{7}{3} + \frac{5}{3} = \frac{19}{3}$
RHS :
$\frac{26}{3} - \frac{7}{3} = \frac{19}{3}$
Hence, LHS = RHS
$3m = 5m - \frac{8}{5}$
We have
$3m = 5m - \frac{8}{5}$
Transposing 5m to the LHS
$3m -5m = - \frac{8}{5}$
or $-2m = - \frac{8}{5}$
Dividing both sides by -2, we get
$m = \frac{4}{5}$
Check :- Put $m = \frac{4}{5}$ in the LHS and the RHS:
LHS :-
$3m = 3\times \left ( \frac{4}{5} \right ) = \frac{12}{5}$
RHS :-
$5m - \frac{8}{5} = 5 \times \frac{4}{5} - \frac{8}{5} = \frac{12}{5}$
Hence, LHS = RHS
NCERT solutions for class 8 maths chapter 2 Linear Equations in One Variable Excercise: 2.4
Let us assume that the number Amina thought was x.
As per the question,
$\left ( x-\frac{5}{2} \right )\times8 = 3x$
or $8x - 20 = 3x$
Transposing 3x to the LHS and -20 to the RHS, we get
8x - 3x = 20
or 5x = 20
Dividing both sides by 5.
x = 4
Therefore the number Amina thought was 4.
Let the No.1 = x
Then the second no. No.2 = 5x.
Now the question states that if we add 21 to both numbers one number becomes twice of the other.
i.e., 2(x + 21) = 5x + 21
or 2x + 42 = 5x + 21
Transposing 2x to the RHS and 21 to the LHS:
21 = 5x - 2x = 3x
or x = 7
Thus the two numbers are 7 and 35 . (Since No.2 = 5x)
Given that sum of the two digits is 9.
Let us assume the digit of units place be x.
Then the digit of tens place will be 9-x.
Thus the two digit number is 10(9-x) + x
Now if we reverse the digits, the number becomes 10x + (9-x).
As per the question,
10x + (9-x) = 10(9-x) + x + 27
or 9x + 9 = 90 - 10x + x + 27
or 9x + 9 = 117 - 9x
Transposing -9x to the LHS and 9 to the RHS:
9x + 9x = 117 - 9
or 18x = 108
x = 6
Thus two digit number is 36.
Let us assume that the units place of a two digit number is x.
Then as per the question, tens place of the number is 3x.
Thus the two digit number is 10(3x) + x.
If we interchange the digits the number becomes : 10x + 3x
According to the question,
10(3x) + x + 10x + 3x = 88
or 30x + 14x = 88
or 44x = 88
Dividing both sides by 44, we get
x = 2
Hence the two digit number is 10(6) + 2 = 62
Note that the two-digit number can also be found if we reverse the digits. So one more possible number is 26.
Let us assume that the Shobo's present age be x.
So, Shobo's mother's present age = 6x.
Shobo's age five years from now will be = x + 5.
It is given that Shobo’s age five years from now will be one-third of his mother’s present age.
So,
$x + 5 = \left ( \frac{1}{3} \right )\times6x = 2x$
or x + 5 = 2x
Transposing x to the RHS,
5 = 2x - x = x
Hence Shobo's present age is x = 5 years
and Shobo's mother's present age is 6x = 30 years
Let the lenght and breadth of the plot be 11x and 4x repectively.
(Since they are in the ratio11:4)
We know that the fencing will be done on the boundary of the plot, so we need to calculate its perimeter.
Perimeter of rectangle = 2(lenght + breadth)
Total cost to fence the plot = Rs.75000 = Rs.100 $\times$ Perimeter of plot.
Thus equation becomes,
100 $\times$ 2(11x + 4x) = 75000
or 200(15x) = 75000
or 3000x = 75000
Dividing both sides by 3000:
x = 25
Hence lenght of plot is 11x = 275 m
and breadth of the plot is 4x = 100m
Let the total shirt material that Hasan bought be x metre.
and the total trouser material be y metre.
As per the question,
$\frac{x}{3} = \frac{y}{2}$
Thus trouser material $y = \frac{2}{3}x$
Now we will make another equaion by using cost of the materials.
Profit for shirt is 12% i.e., selling price of shirt is
$=50 +\frac{12}{100}\times 50 = 50 + 6 = 56$
and selling price for trouser is
$=90 +\frac{10}{100}\times 90 = 90 + 9 = 99$
Hence equation becomes:
$x\left ( 56 \right ) + \frac{2}{3}x\left ( 99 \right ) = 36600$
or 56x + 66x = 36600
or 122x = 36600
Dividing both sides by 122:
x = 300
Thus trouser material that Hasan bought was = 200 meter
$\left ( \frac{2}{3}x = \frac{2}{3}\times300 = 200 \right )$
Let us assume the number of deer in the herd to be x.
Half of a herd of deer are grazing in the field i.e., $\frac{x}{2}$.
Now half are remaining.
Three fourths of the remaining are playing nearby, implies :
$\frac{3}{4}\times \frac{x}{2} = \frac{3x}{8}$
Remaining number of deer will be :
$\frac{x}{2} - \frac{3x}{8} = \frac{4x - 3x}{8} = \frac{x}{8}$
Given that rest are 9, so
$\frac{x}{8} = 9$
Hence number of deer in the herd = 72
Let the age of granddaughter is x years.
So the age of grandfather is x + 54.
Also grandfather is ten times older than his granddaughter.
i.e., x + 54 = 10x
Transposing x to the RHS :
54 = 10x - x
or 54 = 9x
or x = 6
Therefore age of granddaughter = 6 years
and the age of grandfather = 6 + 54 = 60 years
Let the age of Aman's son is x years.
So the age of Aman will be 3x years.
Ten years ago, the age of Aman's son = x - 10 and Aman's age = 3x - 10.
So according to question,
5(x - 10) = 3x - 10
or 5x - 50 = 3x - 10
Transposing 3x to the LHS and -50 to the RHS:
5x - 3x = -10 + 50 = 40
or 2x = 40
or x = 20
So Aman's age = 3x = 60 years and
Aman's son's age = x = 20 years.
NCERT solutions for class 8 maths chapter 2 Linear Equations in One Variable Excercise: 2.5
$\frac{x}{2}-\frac{1}{5} = \frac{x}{3}+\frac{1}{4}$
The detailed solution of the above-written question is here,
We have
$\frac{x}{2} - \frac{1}{5} = \frac{x}{3} + \frac{1}{4}$
or $\frac{x}{2} - \frac{x}{3} = \frac{1}{4} + \frac{1}{5}$
or $\frac{3x - 2x}{6} = \frac{4 +5}{20}$
or $\frac{x}{6} = \frac{9}{20}$
Multiplying both sides by 6:
$x = \frac{9}{20}\times 6 = \frac{27}{10}$
Thus $x = \frac{27}{10}$
$\frac{n}{2}- \frac{3n}{4}+\frac{5n}{6} = 21$
We have
$\frac{n}{2}- \frac{3n}{4}+\frac{5n}{6} = 21$
Multiplying both sides 12.
(We multiplied both sides by 12 because it is the lowest common factor.)
6n - 9n + 10n = 252
or 7n = 252
or n = 36
$x+7 -\frac{8x}{3} = \frac{17}{6}- \frac{5x}{2}$
We have
$x+7 -\frac{8x}{3} = \frac{17}{6}- \frac{5x}{2}$
Multiplying both sides by 6, we get
6x + 42 - 16x = 17 - 15x
or -10x +42 = 17 - 15x
Transposing -15x to the LHS and 42 to the RHS :
-10x + 15x = 17 - 42
or 5x = -25
or x = -5
$\frac{x-5}{3} = \frac{x-3}{5}$
We have
$\frac{x-5}{3} = \frac{x-3}{5}$
Multiplying both sides by 15 . (because the LCM of denomenator is 15 )
5x - 25 = 3x - 9
Transposing 3x to the LHS and -25 to the RHS.
5x - 3x = -9 + 25
or 2x = 16
x = 8
$\frac{3t-2}{4} -\frac{2t+3}{3} = \frac{2}{3} - t$
We have
$\frac{3t-2}{4} -\frac{2t+3}{3} = \frac{2}{3} - t$
Multiplying both sides by 12, we get
9t - 6 - ( 8t + 12 ) = 8 - 12t
or 9t - 6 - 8t - 12 = 8 - 12t
or t - 18 = 8 - 12t
Transposing -18 to the RHS and -12t to the LHS:
12t + t = 8 + 18 = 26
or 13t = 26
Thus, t = 2
$m - \frac{m-1}{2} = 1 - \frac{m-2}{3}$
We have
$m - \frac{m-1}{2} = 1 - \frac{m-2}{3}$
Multiplying both sides by 6 (as it is LCM of the denomenator.)
6m - 3(m - 1) = 6 - 2(m - 2)
or 6m - 3m + 3 = 6 - 2m + 4
or 3m + 3 = 10 - 2m
Transposing -2m to the LHS and 3 to the RHS:
3m + 2m = 10 - 3
or 5m = 7
$m = \frac{7}{5}$
Let us open the brackets :
LHS : 3(t - 3) = 3t - 9
RHS : 5(2t + 1) = 10t + 5
The equation is 3t - 9 = 10t + 5
or -9 = 10t - 3t + 5
or -9 - 5 = 7t
or -14 = 7t
Therefore t = -2
Let us open the brackets
Equation becomes: 15y - 60 - 2y + 18 + 5y + 30 = 0
or 18y - 12 = 0
or 18y = 12
$y = \frac{12}{18} = \frac{2}{3}$
Let us open the brackets.
LHS: 3(5z - 7) - 2(9z - 11)
= 15z - 21 - 18z + 22
= -3z + 1
RHS: 4(8z - 13) - 17
= 32z - 52 - 17
= 32z - 69
So the equation becomes : -3z + 1 = 32z - 69
or 69 + 1 = 32z + 3z
or 70 = 35z
z = 2
We have
0.25(4f – 3) = 0.05(10f – 9)
Let us open the brackets:
f - 0.75 = 0.5f - 0.45
or f - 0.5f = - 0.45 + 0.75
or 0.5f = 0.30
Dividing both sides by 0.5, we get
f = 0.6
NCERT solutions for class 8 maths chapter 2 Linear Equations in One Variable Excercise: 2.6
$\frac{8x-3}{3x} = 2$
We observe that the equation is not a linear equation since LHS is not linear.
So we multiply 3x to both the sides to make it linear.
We get, 8x - 3 = 2 $\times$ 3x
or 8x - 3 = 6x
or 8x - 6x = 3
or 2x = 3
$x = \frac{3}{2}$
$\frac{9x}{7-6x} = 15$
We will convert the given equation into a linear equation by multiplying
(7 - 6x) to both sides.
Equation becomes: 9x = 15(7 - 6x)
or 9x = 105 - 90x
or 99x = 105
$x = \frac{105}{99} = \frac{35}{33}$
$\frac{z}{z+15} = \frac{4}{9}$
We have
$\frac{z}{z+15} = \frac{4}{9}$
Cross-multiplication gives:
9z = 4z + 60
or 9z - 4z = 60
or 5z = 60
or z = 12
$\frac{3y+4}{2-6y} = \frac{-2}{5}$
We have
$\frac{3y+4}{2-6y} = \frac{-2}{5}$
By cross-multiplication we get:
5(3y + 4) = -2(2 - 6y)
or 15y + 20 = -4 + 12y
or 15y - 12y = -4 - 20
or 3y = -24
y = -8
$\frac{7y+4}{y+2} = \frac{-4}{3}$
We have
$\frac{7y+4}{y+2} = \frac{-4}{3}$
Cross-multiplication gives,
3(7y + 4) = -4(y + 2)
or 21y + 12 = -4y - 8
or 21y + 4y = -8 - 12
or 25y = -20
$y = \frac{-20}{25} = \frac{-4}{5}$
Let the present ages of Hari and Harry be 5x and 7x respectively.
(Since there are in the ratio of 5:7)
Four years from age of Hari will be = 5x + 4
and Harry's age will be = 7x + 4.
According to question four years from now, the ratio of their ages will be 3:4.
So the equation becomes :
$\frac{5x + 4}{7x + 4} = \frac{3}{4}$
Cross-multiplication gives:
4(5x + 4) = 3(7x + 4)
or 20x + 16 = 21x + 12
or x = 4
Hence present age of Hari = 5x = 20 and
present age of Harry = 7x = 28
Let the numerator of the rational number be x.
Then denominator will be x + 8.
Further, according to question,
$\frac{x + 17}{x+ 8 - 1} = \frac{3}{2}$
or $\frac{x + 17}{x+ 7} = \frac{3}{2}$
Cross-multiplication gives:
2(x + 17) = 3(x + 7)
or 2x + 34 = 3x + 21
or x = 13
Hence the rational number is $\frac{13}{21}$.
## NCERT solutions for class 8 maths: Chapter-wise
Chapter -1 NCERT solutions for class 8 maths chapter 1 Rational Numbers Chapter -2 NCERT solutions for class 8 maths chapter 2 Linear Equations in One Variable Chapter-3 CBSE NCERT solutions for class 8 maths chapter 3 Understanding Quadrilaterals Chapter-4 NCERT solutions for class 8 maths chapter 4 Practical Geometry Chapter-5 Solutions of NCERT for class 8 maths chapter 5 Data Handling Chapter-6 CBSE NCERT solutions for class 8 maths chapter 6 Squares and Square Roots Chapter-7 NCERT solutions for class 8 maths chapter 7 Cubes and Cube Roots Chapter-8 Solutions of NCERT for class 8 maths chapter 8 Comparing Quantities Chapter-9 NCERT solutions for class 8 maths chapter 9 Algebraic Expressions and Identities Chapter-10 CBSE NCERT solutions for class 8 maths chapter 10 Visualizing Solid Shapes Chapter-11 NCERT solutions for class 8 maths chapter 11 Mensuration Chapter-12 NCERT Solutions for class 8 maths chapter 12 Exponents and Powers Chapter-13 CBSE NCERT solutions for class 8 maths chapter 13 Direct and Inverse Proportions Chapter-14 NCERT solutions for class 8 maths chapter 14 Factorization Chapter-15 Solutions of NCERT for class 8 maths chapter 15 Introduction to Graphs Chapter-16 CBSE NCERT solutions for class 8 maths chapter 16 Playing with Numbers
## How to use NCERT solutions for class 8 maths chapter 2 Linear Equations in One Variable
• Go through some examples to get an idea about how to solve a question.
• Once you have identified the process, do the practice problems given in the exercises.
• For better preparation and time saving, you must use NCERT solutions for class 8 maths chapter 2 Linear Equations in One Variable.
• You can also practice previous year papers' questions to get perfection over the chapter.
Keep working hard and happy learning!
Exams
Articles
Questions
| 9,972 | 26,575 |
{"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": 155, "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}
| 4.96875 | 5 |
CC-MAIN-2020-16
|
latest
|
en
| 0.885082 |
http://www.arenawp.com/2018/04/subtracting-vectors-2/
| 1,555,779,624,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-18/segments/1555578529898.48/warc/CC-MAIN-20190420160858-20190420182858-00286.warc.gz
| 214,879,326 | 6,689 |
# Subtracting Vectors 2
This post categorized under Vector and posted on April 30th, 2018.
Velocity acceleration force and many other things are vectors.. Subtracting. We can also subtract one vector from another first we reverse the direction of Vectors Learn how to graph determine horozontal and vertical components and combine vectors.Sep 07 2004 Adding and subtracting whatCoordinates apply to points and you dont add or subtract points. If you are talking about vectors then you will have to establish what you basis vectors are.
Vectors are commonly found in both physics and mathematics and is used to represent movements in graphice. The concept of 2D movement using vectors is easily handled as our methods of representing 2D are fairly developed.Addition and subtraction of vectors calculator. This step-by-step online calculator will help you understand how to find the sum of a vectors and the difference of vectors.Learning about vectors. Scalar Quangraphicies are quangraphicies which represent only magnitude such as kilograms and metres. This way of presenting quangraphicies only tells you how much of something has happened it does not
Very simple negative numbers codebreaker (just adding and subtracting). When decoded the message reads Why was the maths book sad Because it had too many problems.A Power Point presentation about adding and subtracting decimals. Please leave a review if you use this resource.
## Subtracting Vectors In A Straight Line
Algebra is great fun - you get to solve puzzles With computer games you play by running jumping or finding secret things. Well with Algebra you pla [more]
## Given Vectors A I J And B I J How Do You Find The Magnitude D
Operations on vectors It can be shown that and have the same magnitude and direction and are therefore To add two vectors given in component form [more]
## Algebra Adding And Subtracting Rational Expressions Worksheet
Pre-Algebra - Monomials and Polynomials Worksheets Adding and Subtracting Polynomials Worksheets. This polynomial worksheet will produce problems f [more]
## Photoshop Cs Subtract Exclude A Vector Mask From An Existing Mask
Search the worlds information including webpages images vector and more. Google has many special features to help you find exactly what youre looki [more]
## Worksheets On Vector Addition And Subtraction
Math Fact Cafe - design your own flashcards or have the computer do it for you. A really neat way to practice those basic factsPhysics. Showing top [more]
## Find The Magnitude Of A Vector
Answer to Vector A points in the negative y direction and has a magnitude of 15 units. Vector B has twice the magnitude and pointsCheat Sheets & Ta [more]
## How To Compute Directional Angle Between Two D Vectors In Matlab
The clvector is used to specify a row or a column span in a matrix ( Mat) and for many other purposes. Range(ab) is basically the same as ab in Mat [more]
## Quiz Worksheet Practice Adding Subtracting Vectors
As a member youll also get unlimited access to over 70000 lessons in math English science history and more. Plus get practice tests quizzes and per [more]
## Subtracting Vector Adding And Vectors Physics Graphically Work
Apr 24 2018 You can add and subtract vectors on a graph by beginning one vector at the endpoint of another vector. You add and subtract vectors Ad [more]
## A Simple Shoot Em Up Game With Sprite Kit And Swift Part Two Code Organization And Vector Math
No more missed important software updates UpdateStar 11 lets you stay up to date and secure with the software on your computer.A list of every Word [more]
| 736 | 3,612 |
{"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}
| 3.5 | 4 |
CC-MAIN-2019-18
|
latest
|
en
| 0.907088 |
https://www.themoneyfarm.org/personal-finance/retirement/annuities/how-to-get-annuity-factor/
| 1,718,783,372,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-26/segments/1718198861806.64/warc/CC-MAIN-20240619060341-20240619090341-00885.warc.gz
| 875,426,792 | 22,534 |
# How To Get Annuity Factor?
The Annuity Factor (AF) is used to compute the present value of the annuity: = AF x Time 1 cash flow. 1.833 is the Annuity factor.
## How do you calculate annuity factor in Excel?
For example, if you wanted to calculate the present value of a future annuity with a 5% interest rate for 12 years and a \$1000 yearly payment, you would use the following formula: =PV (.05,12,1000). You’d end up with a present value of \$8,863.25.
It’s vital to remember that the “NPER” figure in this calculation refers to the number of periods the interest rate applies to, not necessarily the number of years. This means that if you receive a payment every month, you must divide the number of years by 12 to get the number of months. Because the interest rate is yearly, you’ll need to divide it by 12 to convert it to a monthly rate. So, if the identical problem was a \$1000 monthly payment for 12 years at 5% interest, the formula would be =PV(.05/12,12*12,1000), or you could simplify it to =PV(.05/12,12*121000) (.004167,144,1000).
While this is the most fundamental annuity formula for Excel, there are a few more to learn before you can completely grasp annuity formulas. When you have the interest rate, present value, and payment amount for a problem, the NPER formula can help you find the number of periods. When you have the present value, number of periods, and interest rate for an annuity, the PMT formula can help you find the payment. If you already know the present value, the number of periods, and the payment amount for a certain annuity, the RATE formula can help you find the interest rate. There’s a lot more to learn about Excel’s basic annuity formula.
## What does an annuity factor mean?
The annuity factor approach is a means to figure out how much money can be taken out of a retirement account before penalties apply. The formula is applied to annuities and individual retirement funds, and it is based on life expectancy data (IRAs).
## How do you calculate NPV annuity factor?
The payments are referred to as annuity due if they are due at the beginning of the period. The present value interest factor of an annuity payable is calculated by multiplying the present value interest factor by (1+r), where “r” is the discount rate.
## How do you calculate an annuity table?
You won’t have to do the math with an annuity table. Reading the chart will provide you with all of the information you require.
The number of payments is usually on the y-axis, and the discount rate is on the x-axis in an annuity table. On the table, locate both of them for your annuity, then locate the cell where they overlap. Multiply the value in that cell by the amount of money you receive each month. The present value of your annuity is that figure.
Here’s an example of how to use the table: Let’s imagine you have an annuity with eight installments remaining that pays you \$1,000 each month at a 6% discount rate. On the graph, look for eight periods and 6%. 6.210 is written in the cell where they cross. Multiply that by \$1,000 to get \$6,210 in present value.
Different tables will be used for different types of annuities (variable annuities, for example). Make sure you’re using the correct table by speaking with your advisor or annuity provider.
## What is the three year annuity factor?
The Present Value Annuity Factor Formula is used to calculate the present value of annuities. For example, if a person wants to compute the present value of a series of \$500 annual payments for 5 years at a 5% rate, they can use the formula below.
## How is discount factor calculated?
Whether it’s an annual discount factor or a shorter time frame to match your accounting period, this depicts the diminishing discount factor over time.
You may, for example, divide 1 by the interest rate plus 1 to determine the discount factor for a cash flow one year in the future. The discount factor would be 1 divided by 1.05, or 95 percent, for a 5% interest rate.
You can use your discount factor and discount rate to calculate the net present value of an investment once you’ve computed them. Subtract the present value of all negative cash flows from the total present value of all positive cash flows. You’ll get the net present value after applying the interest rate. You can use one of numerous discount factor calculators to apply these calculations, or you can do an analysis in Excel.
## How do you calculate present factor?
The Present Value Factor, often known as the Present Value of One or PV Factor, is a formula for calculating the Present Value of 1 unit n times in the future. The PV Factor is 1 (1 +i)n, where I is the rate (such as an interest rate or a discount rate) and n is the number of periods.
So, at a discount rate of 12%, \$1 USD received five years from now is equal to 1 (1 + 12%)5 or \$0.5674 USD now. By multiplying each period’s cash flow by the supplied PV Factor for that year and then summing the resulting values, the PV Factor may be used to compute the Present Value of a future stream of cash flows.
## How do you calculate an annuity on a Casio calculator?
You may figure out the worth of all the income an annuity is predicted to provide in the future by calculating the present value (PV) of the annuity.
The amount of interest paid by the annuity, the amount of your monthly payment, and the length of periods, usually months, that you plan to pay into the annuity are all elements in the calculation.
Because of the income you could have earned by investing those future dollars now, the PV calculation embodies the time-value-of-money idea, which states that a dollar earned now has more value than a dollar obtained in the future.
The PV computation applies a discount to future payments based on the number of payment periods. The present value of an annuity can be calculated using the formula below:
Note that over the duration of the annuity payments, this equation assumes that the payment and interest rate remain constant.
## How much is an annuity worth?
After analyzing 326 annuity products from 57 insurance companies, we discovered that a \$250,000 annuity will pay between \$1,041 and \$3,027 per month for a single lifetime and between \$937 and \$2,787 per month for a joint lifetime (you and your spouse). Income amounts are influenced by the age at which you purchase the annuity contract and the time you wait before taking the income.
## What are the 4 types of annuities?
Immediate fixed, immediate variable, deferred fixed, and deferred variable annuities are the four primary forms of annuities available to fit your needs. These four options are determined by two key considerations: when you want to begin receiving payments and how you want your annuity to develop.
• When you start getting payments – You can start receiving annuity payments right away after paying the insurer a lump sum (immediate) or you can start receiving monthly payments later (deferred).
• What happens to your annuity investment as it grows – Annuities can increase in two ways: through set interest rates or by investing your payments in the stock market (variable).
#### Immediate Annuities: The Lifetime Guaranteed Option
Calculating how long you’ll live is one of the more difficult aspects of retirement income planning. Immediate annuities are designed to deliver a guaranteed lifetime payout right now.
The disadvantage is that you’re exchanging liquidity for guaranteed income, which means you won’t always have access to the entire lump sum if you need it for an emergency. If, on the other hand, securing lifetime income is your primary goal, a lifetime instant annuity may be the best solution for you.
What makes immediate annuities so enticing is that the fees are built into the payment – you put in a particular amount, and you know precisely how much money you’ll get in the future, for the rest of your life and the life of your spouse.
#### Deferred Annuities: The Tax-Deferred Option
Deferred annuities offer guaranteed income in the form of a lump sum payout or monthly payments at a later period. You pay the insurer a lump payment or monthly premiums, which are then invested in the growth type you chose – fixed, variable, or index (more on that later). Deferred annuities allow you to increase your money before getting payments, depending on the investment style you choose.
If you want to contribute your retirement income tax-deferred, deferred annuities are a terrific choice. You won’t have to pay taxes on the money until you withdraw it. There are no contribution limits, unlike IRAs and 401(k)s.
#### Fixed Annuities: The Lower-Risk Option
Fixed annuities are the most straightforward to comprehend. When you commit to a length of guarantee period, the insurance provider guarantees a fixed interest rate on your investment. This interest rate could run anywhere from a year to the entire duration of your guarantee period.
When your contract expires, you have the option to annuitize it, renew it, or transfer the funds to another annuity contract or retirement account.
You will know precisely how much your monthly payments will be because fixed annuities are based on a guaranteed interest rate and your income is not affected by market volatility. However, you will not profit from a future market boom, so it may not keep up with inflation. Fixed annuities are better suited to accumulating income rather than generating income in retirement.
#### Variable Annuities: The Highest Upside Option
A variable annuity is a sort of tax-deferred annuity contract that allows you to invest in sub-accounts, similar to a 401(k), while also providing a lifetime income guarantee. Your sub-accounts can help you stay up with, and even outperform, inflation over time.
If you’ve already maxed out your Roth IRA or 401(k) contributions and want the security and certainty of guaranteed income, a variable annuity can be a terrific complement to your retirement income plan, allowing you to focus on your goals while knowing you won’t outlive your money.
## What is annuity and how is it calculated?
Before diving into the concept of estimating the amount of annuity pay-out for your plans, it’s critical to first grasp a basic understanding of annuities and how they typically pay their beneficiaries.
An annuity plan is one that pays you regular payments over a certain length of time for the amount you pay in premiums. Your payment can be made in one lump sum or at regular intervals. The insurance company agrees to pay you the annuity either right away or at a later time. These annuity plans are retirement plans that allow you to receive regular income payouts so that you may maintain your current lifestyle once you retire.
Fixed and variable annuity programs are the two types of annuities available. Fixed plans have an interest rate that is guaranteed. Variable plans invest your premiums in other investments, thus their rate of interest is determined by the market’s performance.
This will be pre-determined between you and your insurance provider when you sign up for the plan, so there will be no surprises afterwards. You have the option of choosing from one of the following pay-outs that are frequently linked with these plans:
• The plan continues to pay the agreed-upon sum to the policyholder at the agreed-upon frequency. The balance annuities are paid to the beneficiary in the case of the policyholder’s death during the period.
• The plan pays until the policyholder dies; there is no idea of a beneficiary, thus no payments are made after the policyholder dies.
• The beneficiary will receive periodic payments from the plan for the rest of his or her life.
• The plan is only valid for a set period of time; this includes payments to the beneficiary after the policyholder’s death, but only for the agreed-upon period of time.
An annuity calculator can help you figure out how much your plan will pay you in the future. You can also use this calculator to figure out how much you’ll have to pay in capital to get a plan to run for a certain period of time.
For instance, if you want to see how much money you may take out of your annuity plan each month, input the following information into the annuity calculator India:
When you click’Calculate,’ you’ll see how much your annuity plan will pay you out each month.
You can also see how long your annuity plan will run by entering all of the above information (including the monthly withdrawals you choose) but leaving the term column blank.
This calculator will help you discover the approximate annual returns that yourprincipal will create if you enter all of the other details and leave the growth rate blank.
It is critical to have a thorough grasp of annuity plans before making a decision.
Each annuity plan is unique in terms of pay-out options, premium payment terms, death benefit specifics, and other factors. If you have any questions, you can contact your insurance carrier, and you should carefully examine the conditions of the policies to ensure complete understanding. Because annuity plans have the potential to provide lifetime income, even after you retire, you must understand them well in order to make the best use of them. Visit our main page to learn more about Aegon Life’s life insurance products, such as term insurance and other options.
## What is the formula to calculate installment?
It is important to remember that the rate used in the formula should be the monthly rate, i.e. 12 percent /12=1% or 0.01.
The outcome will be negative or red, indicating that the borrower has a cash outflow.
Let’s look at another scenario. Assume you’re paying a quarterly instalment on a Rs 10 lakh loan with a yearly interest rate of 10% for the next 20 years. In this example, divide the rate by four and multiply the number of years by four instead of 12. For the aforementioned amounts, the equated quarterly instalment will be =PMT(10 percent /4, 20*4, 10,00,000).
Unfortunately, the Excel spreadsheet is not available everywhere. In this scenario, you can calculate the EMI using your mathematical skills or an electronic calculator. The formula for calculating EMIs is as follows:
EMI = /, where P is the loan amount (principal), R denotes the monthly interest rate, and N denotes the number of monthly instalments. When you use the formula above, you’ll obtain the same result as you would in an Excel spreadsheet.
| 3,111 | 14,479 |
{"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-2024-26
|
latest
|
en
| 0.935001 |
https://www.physicsforums.com/threads/limit-as-x-approaches-0.548757/
| 1,585,775,640,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2020-16/segments/1585370506121.24/warc/CC-MAIN-20200401192839-20200401222839-00136.warc.gz
| 1,112,924,331 | 16,537 |
# Limit as x approaches 0.
## Homework Statement
http://www.prep101.com/files/Math100PracticeExam.pdf [Broken]
Question 1m
## The Attempt at a Solution
I tried to do this by using
lim sinx/x = 1
x->0
Factoring out x^2 from each one and I get infinity.
Why does this method not work?
I even tried plugging in 0.001 in my calculator and I get infinity :S.
Last edited by a moderator:
Related Calculus and Beyond Homework Help News on Phys.org
Is that sin(2x^2), sin^2(2x), sin(2)x^2 ?
Dick
Homework Helper
Is that sin(2x^2), sin^2(2x), sin(2)x^2 ?
Or is it 3sin(2x^2-6x^2+4x^6)? The question isn't very grammatical. And however I rearrange it I still can't figure out how to pull 4/5 as a limit out.
Or is it 3sin(2x^2-6x^2+4x^6)? The question isn't very grammatical. And however I rearrange it I still can't figure out how to pull 4/5 as a limit out.
It is 3sin(2x^2), you get 4/5, this limit is INSANE :P
vela
Staff Emeritus
Homework Helper
It helps to use the substitution u=x2 to get
$$\lim_{u \to 0}\frac{3 \sin 2u - 6u + 4u^3}{u^5}$$Then three applications of the Hospital rule gets you to a limit you can evaluate.
Dick
$$\lim_{u \to 0}\frac{3 \sin 2u - 6u + 4u^3}{u^5}$$Then three applications of the Hospital rule gets you to a limit you can evaluate.
| 431 | 1,271 |
{"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}
| 3.90625 | 4 |
CC-MAIN-2020-16
|
longest
|
en
| 0.909914 |
urdukitaab.net
| 1,618,230,964,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-17/segments/1618038067400.24/warc/CC-MAIN-20210412113508-20210412143508-00601.warc.gz
| 657,867,656 | 21,883 |
# Difference between AC and DC Current Explained
Difference between AC and DC Current Explained. On the last added Ohms video we looked at voltage, power and current … and you might remember this animation One problem with this animation is that the voltage source shown is the source of an AC.
The current flow is shown as DC so let’s talk about the difference between IC and DC.
The letter DC means direct current and this usually means that the electrons flow in the same direction.
AC means alternating current and this means that the flow of electrons changes.
## Difference between AC and DC Current Explained
Now, things can start to get a bit confusing from here. It is an AC voltage source so it is both AC and voltage. How can the voltage and current be the same? Well, it turns out that we use AC and DC briefly in voltage as well as current.
So, in other words, is the AC voltage, or current, voltage, or current, which varies. DC voltage, or current, is the voltage, or current, that is constant.
Okay, so let’s take a closer look at these two starting with DC. Check the AA, or LR6, battery. It provides 1.5 volts. For example, the circuit is going to be a DC motor such that you get a toy car.
When we enter the battery circuit, the motor starts to flow in one direction, turning.
If we draw a graph, where the vertical axis is the voltage and the horizontal axis is the time, then we can see that the voltage at this point is stable.
Since it is a battery, eventually, it will run out of energy. And its output voltage will drop. So we know that the voltage will change over time, but the polar age remains the same.
So what is precision? The polarity of the well defines the positive direction and the positive voltage for the battery produces a positive voltage.
So what if we turn this battery around? Well, it reverses its potential, which means that the current will flow in the opposite direction as before, causing our motor to rotate in the opposite direction.
Well, then we move to AC and in this case, we will use a light bulb and a North American AC socket, because well, I live in North America! Treat the circuit when we activate the switch.
Let’s break down what we are seeing by reusing the voltage graph. Note how the current is flowing in one direction and the light bulb illuminates as the voltage increases. Once the voltage reaches its peak, the current flow remains the same but the voltage starts to drop.
And the light bulb fades. Once we reach zero voltage, the polarity of the voltage changes which causes the current to flow, and when the voltage gets closer to its height, the bulb brightens and then goes down.
First, is the change from start to finish called a cycle? The frequency at which the cycle repeats is the frequency. The frequency is now measured by the unit “Hz”, which means cycles per second.
Different parts of the world use different frequencies for their AC systems and this can be 50 or 60 Hz. What this means is that the cycle repeats itself at least 50 times in a second.
B found out, this rate is so fast that our lazy human eyes see it as a constant light! To sum up, when AC and DC abbreviations include the word “current”, they can be used to describe different types of voltage and current.
DC voltage does not fluctuate while DC current flows in one direction AC voltage changes over time and current flow can change in the alternate direction.
| 734 | 3,431 |
{"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-2021-17
|
latest
|
en
| 0.945196 |
https://www.explorelearning.com/index.cfm?method=cResource.dspStandardCorrelation&id=1596
| 1,618,597,046,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-17/segments/1618038088245.37/warc/CC-MAIN-20210416161217-20210416191217-00279.warc.gz
| 878,440,970 | 16,056 |
A: Linear Equations and Inequalities
A.1: To solve linear equations in one variable containing:
A.1.a: variables on both sides
A.1.c: fraction or decimal coefficients
B: Relations, Linear Functions, and Variation
B.1: To define the following terms: relation, ordered pair, abscissa, ordinate.
B.3, B.4: To graph ordered pairs in the Cartesian coordinate plane, and to graph real-world relations in the Cartesian coordinate plane.
B.1.b: To define the following terms: function, linear function, slope, x-intercept, y-intercept, ration, proportion, direct variation, partial variation.
B.11: To calculate the slope of a line:
B.11.a: graphically (m = rise/run)
B.11.b: algebraically
B.11.c: from the equation (y = mx+ b)
B.12: To determine the slope of horizontal, vertical, parallel, and perpendicular lines.
B.12.b: To write linear equations in:
B.12.b.a: slope-intercept form
B.12.b.b: standard form
B.13 & B.14.a: x and y intercepts
B.13 & B.14.b: the slope and an ordered pair
B.13 & B.14.c: the slope and y-intercept (y = mx + b)
B.15, B.16: To write the equation of a line when given:
B.15, B.16.a: slope and y-intercept
B.15, B.16.b: slope and one point on the line
B.15, B.16.c: the graph of the line
B.15, B.16.d: two points on the line
D: Lines and Line Segments
D.2: To identify and calculate the measures of corresponding angles, alternate interior angles, and same-side interior angles formed by parallel lines.
D.4, D.5, D.6, D.7: Informally and formally construct:
D.4, D.5, D.6, D.7.a: congruent segments;
D.4, D.5, D.6, D.7.b: the perpendicular bisector of a line segment;
D.4, D.5, D.6, D.7.c: a line perpendicular to a given line from a point not on the line;
D.4, D.5, D.6, D.7.d: a line perpendicular to a given line from a point on the line; and,
D.4, D.5, D.6, D.7.e: a line parallel to a given line through a point not on the line.
E: Angles and Polygons
E.1: To define and illustrate by drawing the following: acute angle, right angle, obtuse angle, straight angle, reflex angle, complementary angles, supplementary angles, adjacent angles, vertically opposite angles, congruent angles, central angles of a regular polygon.
E.3.a: To define and illustrate the following polygons: convex, non-convex, regular, triangle, quadrilateral, parallelogram, rectangle, rhombus, square, trapezoid, isosceles trapezoid.
E.3.b: To define and illustrate the following triangles: scalene, isosceles, equilateral, acute, right, obtuse.
E.6: To state and apply the properties of parallelograms:
E.6.a: opposite sides are parallel
E.6.b: opposite sides are congruent
E.6.c: opposite angles are congruent
E.6.d: the diagonals bisect each other.
F: Review of Algebraic Skills
F.2.d: To write numbers in scientific notation and vice versa.
F.3.a: To add and subtract polynomials.
F.3.b: To multiply: a monomial by a monomial
F.3.d: To multiply: a binomial by a binomial
F.3.e: To divide: a monomial divisor
F.3.f: To divide: a polynomial by a monomial
Correlation last revised: 9/16/2020
This correlation lists the recommended Gizmos for this province's curriculum standards. Click any Gizmo title below for more information.
| 912 | 3,179 |
{"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}
| 3.890625 | 4 |
CC-MAIN-2021-17
|
longest
|
en
| 0.721543 |
https://gmatclub.com/forum/each-of-a-and-b-has-a-few-marbles-with-them-if-a-gives-311281.html
| 1,575,948,288,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-51/segments/1575540525781.64/warc/CC-MAIN-20191210013645-20191210041645-00369.warc.gz
| 377,657,921 | 147,226 |
GMAT Question of the Day - Daily to your Mailbox; hard ones only
It is currently 09 Dec 2019, 20:24
### 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
# Each of A and B has a few marbles with them. If A gives 12
Author Message
TAGS:
### Hide Tags
Manager
Joined: 18 Jul 2019
Posts: 54
Each of A and B has a few marbles with them. If A gives 12 [#permalink]
### Show Tags
25 Nov 2019, 07:30
00:00
Difficulty:
35% (medium)
Question Stats:
62% (01:59) correct 38% (02:25) wrong based on 21 sessions
### HideShow timer Statistics
Each of A and B has a few marbles with them. If A gives 12 marbles to B, the ratio of the number of marbles with them becomes the reciprocal of the initial ratio. If they together have a total of N marbles, how many marbles should A give to B so that each of them has an equal number of marbles?
a) 6
b) 12
c) 24
d) 26
e) 30
Math Expert
Joined: 02 Aug 2009
Posts: 8292
Re: Each of A and B has a few marbles with them. If A gives 12 [#permalink]
### Show Tags
26 Nov 2019, 18:19
CaptainLevi wrote:
Each of A and B has a few marbles with them. If A gives 12 marbles to B, the ratio of the number of marbles with them becomes the reciprocal of the initial ratio. If they together have a total of N marbles, how many marbles should A give to B so that each of them has an equal number of marbles?
a) 6
b) 12
c) 24
d) 26
e) 30
Let the number be A:B..
After A gives 12 marbles to B, the ratio becomes = B:A = A-12:B+12
$$A(A-12)=B(B+12).......A^2-12A=B^2+12B......A^2-B^2=12(B+A)....(A-B)(A+B)=12(A+B).....A-B=12$$
So A has 12 more marbles than B.
If A gives 6 to B, both will have equal number of marbles..
A
_________________
Re: Each of A and B has a few marbles with them. If A gives 12 [#permalink] 26 Nov 2019, 18:19
Display posts from previous: Sort by
| 679 | 2,238 |
{"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}
| 4.375 | 4 |
CC-MAIN-2019-51
|
latest
|
en
| 0.921808 |
http://www.numbersaplenty.com/275500
| 1,582,409,034,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2020-10/segments/1581875145729.69/warc/CC-MAIN-20200222211056-20200223001056-00296.warc.gz
| 227,120,780 | 3,769 |
Search a number
275500 = 22531929
BaseRepresentation
bin1000011010000101100
3111222220201
41003100230
532304000
65523244
72225131
oct1032054
9458821
10275500
11178a95
12113524
1398524
1472588
155696a
hex4342c
275500 has 48 divisors (see below), whose sum is σ = 655200. Its totient is φ = 100800.
The previous prime is 275491. The next prime is 275503. The reversal of 275500 is 5572.
275500 = 363 + 373 + ... + 403.
It is a happy number.
It is a Harshad number since it is a multiple of its sum of digits (19).
It is not an unprimeable number, because it can be changed into a prime (275503) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (7) of ones.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 9486 + ... + 9514.
It is an arithmetic number, because the mean of its divisors is an integer number (13650).
2275500 is an apocalyptic number.
275500 is a gapful number since it is divisible by the number (20) formed by its first and last digit.
It is an amenable number.
It is a practical number, because each smaller number is the sum of distinct divisors of 275500, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (327600).
275500 is an abundant number, since it is smaller than the sum of its proper divisors (379700).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
275500 is a wasteful number, since it uses less digits than its factorization.
275500 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 67 (or 55 counting only the distinct ones).
The product of its (nonzero) digits is 350, while the sum is 19.
The square root of 275500 is about 524.8809388804. The cubic root of 275500 is about 65.0689603427.
The spelling of 275500 in words is "two hundred seventy-five thousand, five hundred".
| 552 | 1,979 |
{"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-2020-10
|
latest
|
en
| 0.901586 |
http://perlguru.com/gforum.cgi?do=post_view_printable;post=6983;guest=7922742
| 1,542,133,403,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-47/segments/1542039741340.11/warc/CC-MAIN-20181113173927-20181113195927-00352.warc.gz
| 250,392,941 | 2,996 |
Home: Fun With Perl: Perl Quizzes - Learn Perl the Fun Way:
Fibonacci
japhy
Enthusiast
Dec 15, 2000, 1:43 PM
Views: 52538
Fibonacci
Sure, we all know the fibonacci set (1, 1, 2, 3, 5, 8, 13, 21, ...), and we all know the general equations: F[1] = F[2] = 1, and F[n] = F[n-1] + F[n-2].
And we could probably generate these numbers easily in Perl:
Code
{
my @fib = (1,1);
# generate the Nth fibonacci number
sub fibonacci {
my \$n = shift() - 1;
unless (\$fib[\$n]) {
my \$m = \$n;
\$m-- while not \$fib[\$m];
\$fib[\$m+1] = \$fib[\$m] + \$fib[\$m-1], \$m++ while \$m <= \$n;
}
return \$fib[\$n];
}
}
There's your fibonacci code. So what is it this quiz is all about?
I am here to tax your mathematical prowess. Notice that 1+1+2 is 4, which 5-1. Notice that 1+1+2+3 is 7, which is 8-1. And 1+1+2+3+5 is 12, which is 13-1. So what do 5, 8, and 13 have in common? They're also three numbers in the sequence. 1+1+2 is 5-1, and 5 is the 2nd number after 2. 1+1+2+3 is 8-1, and 8 is the 2nd number after 3. 1+1+2+3+5 is 13-1, and 13 is the 2nd number after 5. Do you see the pattern?
So I'd like any able mathematicians out there to develop a proof, which states that F[1] + F[2] + ... + F[k] = F[k+2] - 1 for all k > 0.
(So maybe this isn't a Perl quiz, but this problem is directly related to a specific anomoly in Huffman encoding, which is a simple means of compressing files. It turns out that Huffman coding is very inefficient for files whose character frequencies are the Fibonacci sequence.)
(If that last paragraph made no sense, nevermind. Just try the proof. It's fun.)
Jeff "japhy" Pinyan -- accomplished hacker, teacher, lecturer, and author
jas
Deleted
Dec 19, 2000, 6:59 AM
Views: 52525
Re: Fibonacci
A bit of induction action:
Suppose F[1]+F[2]+...+F[k] = F[k+2]-1 for some k>0. We can safely assume this since at least for k=1,1+1+2=4=5-1.
We want to show that F[1]+F[2]+...+F[k]+F[k+1] = F[k+3]-1 ie. that the formula still holds.
But, F[k+3]-1=F[k+2]+F[k+1]-1 from the definition of Fibonacci numbers.
So, F[1]+F[2]+...+F[k]+F[k+1]=F[k+3]-1
becomes F[1]+F[2]+...+F[k]+F[k+1]=F[k+1]+F[k+2]-1
which reduces to F[1]+F[2]+...+F[k]=F[k+2]-1 which by our assumption is true.
So since the truth of the statement for F[k] implies the truth of the statement for F[k+1] and the statement is true for F[1], by mathemagical induction, the statement must be true for all k>0.
japhy
Enthusiast
Dec 19, 2000, 7:18 AM
Views: 52524
Re: Fibonacci
Nice first post, jas! Well done, too.
(Inside tip, re: your web site: coup d'état.)
Jeff "japhy" Pinyan -- accomplished hacker, teacher, lecturer, and author
| 955 | 2,632 |
{"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-2018-47
|
longest
|
en
| 0.891801 |
https://convertoctopus.com/952-cubic-feet-to-tablespoons
| 1,623,830,158,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-25/segments/1623487622234.42/warc/CC-MAIN-20210616063154-20210616093154-00163.warc.gz
| 163,451,617 | 8,108 |
## Conversion formula
The conversion factor from cubic feet to tablespoons is 1915.0129870073, which means that 1 cubic foot is equal to 1915.0129870073 tablespoons:
1 ft3 = 1915.0129870073 tbsp
To convert 952 cubic feet into tablespoons we have to multiply 952 by the conversion factor in order to get the volume amount from cubic feet to tablespoons. We can also form a simple proportion to calculate the result:
1 ft3 → 1915.0129870073 tbsp
952 ft3 → V(tbsp)
Solve the above proportion to obtain the volume V in tablespoons:
V(tbsp) = 952 ft3 × 1915.0129870073 tbsp
V(tbsp) = 1823092.363631 tbsp
The final result is:
952 ft3 → 1823092.363631 tbsp
We conclude that 952 cubic feet is equivalent to 1823092.363631 tablespoons:
952 cubic feet = 1823092.363631 tablespoons
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 tablespoon is equal to 5.4851856107189E-7 × 952 cubic feet.
Another way is saying that 952 cubic feet is equal to 1 ÷ 5.4851856107189E-7 tablespoons.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that nine hundred fifty-two cubic feet is approximately one million eight hundred twenty-three thousand ninety-two point three six four tablespoons:
952 ft3 ≅ 1823092.364 tbsp
An alternative is also that one tablespoon is approximately zero times nine hundred fifty-two cubic feet.
## Conversion table
### cubic feet to tablespoons chart
For quick reference purposes, below is the conversion table you can use to convert from cubic feet to tablespoons
cubic feet (ft3) tablespoons (tbsp)
953 cubic feet 1825007.377 tablespoons
954 cubic feet 1826922.39 tablespoons
955 cubic feet 1828837.403 tablespoons
956 cubic feet 1830752.416 tablespoons
957 cubic feet 1832667.429 tablespoons
958 cubic feet 1834582.442 tablespoons
959 cubic feet 1836497.455 tablespoons
960 cubic feet 1838412.468 tablespoons
961 cubic feet 1840327.481 tablespoons
962 cubic feet 1842242.494 tablespoons
| 523 | 2,059 |
{"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-2021-25
|
latest
|
en
| 0.768799 |
https://gmatclub.com/forum/if-r-3s-s-5t-t-2u-and-u-0-what-is-the-value-of-rst-u-278754.html
| 1,571,015,293,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-43/segments/1570986648481.7/warc/CC-MAIN-20191014003258-20191014030258-00197.warc.gz
| 603,305,622 | 145,831 |
GMAT Question of the Day - Daily to your Mailbox; hard ones only
It is currently 13 Oct 2019, 18:08
### 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
# If r = 3s, s = 5t, t = 2u, and u ≠ 0, what is the value of rst/u^3
Author Message
TAGS:
### Hide Tags
Math Expert
Joined: 02 Sep 2009
Posts: 58335
If r = 3s, s = 5t, t = 2u, and u ≠ 0, what is the value of rst/u^3 [#permalink]
### Show Tags
11 Oct 2018, 03:12
00:00
Difficulty:
5% (low)
Question Stats:
87% (01:39) correct 13% (02:31) wrong based on 66 sessions
### HideShow timer Statistics
If r = 3s, s = 5t, t = 2u, and u ≠ 0, what is the value of $$\frac{rst}{u^3}$$ ?
A. 30
B. 60
C. 150
D. 300
E. 600
_________________
VP
Joined: 31 Oct 2013
Posts: 1475
Concentration: Accounting, Finance
GPA: 3.68
WE: Analyst (Accounting)
Re: If r = 3s, s = 5t, t = 2u, and u ≠ 0, what is the value of rst/u^3 [#permalink]
### Show Tags
11 Oct 2018, 03:17
1
Bunuel wrote:
If r = 3s, s = 5t, t = 2u, and u ≠ 0, what is the value of $$\frac{rst}{u^3}$$ ?
A. 30
B. 60
C. 150
D. 300
E. 600
r = 3s = 3 * 10 u = 30 u
s = 5t = 5 * 2u = 10 u
t = 2u.
Question :$$\frac{rst}{u^3}$$ = $$\frac{30u*10u*2u}{u*u*u}$$ = 600.
Note: As u is the denominator convert all the items in terms of u.
Manager
Joined: 06 Sep 2018
Posts: 173
Location: Pakistan
Concentration: Finance, Operations
GPA: 2.87
WE: Engineering (Other)
Re: If r = 3s, s = 5t, t = 2u, and u ≠ 0, what is the value of rst/u^3 [#permalink]
### Show Tags
11 Oct 2018, 03:25
convert all the right hand side of all equation into variable 'u'
as $$t=2u$$
so $$s=5(2u)=10u$$
$$r=3(10u)=30u$$
so expression is:
$$\frac{rst}{u^3}=\frac{30u*10u*2u}{u^3}$$
$$\frac{600u^3}{u^3}$$
$$600$$
_________________
Hitting Kudos is the best way of appreciation.
Eric Thomas, "When you want to succeed as bad as you want to breathe, then you'll be successful."
Board of Directors
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 4753
Location: India
GPA: 3.5
Re: If r = 3s, s = 5t, t = 2u, and u ≠ 0, what is the value of rst/u^3 [#permalink]
### Show Tags
11 Oct 2018, 08:27
Bunuel wrote:
If r = 3s, s = 5t, t = 2u, and u ≠ 0, what is the value of $$\frac{rst}{u^3}$$ ?
A. 30
B. 60
C. 150
D. 300
E. 600
$$r = 3*5(2u) =30u$$
$$s = 5(2u) = 10u$$
$$t = 2u$$
So, $$\frac{rst}{u^3} = \frac{30u*10u*2u}{u^3}= 600$$; Answer must be (E)
_________________
Thanks and Regards
Abhishek....
PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS
How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only )
SVP
Joined: 26 Mar 2013
Posts: 2341
Re: If r = 3s, s = 5t, t = 2u, and u ≠ 0, what is the value of rst/u^3 [#permalink]
### Show Tags
12 Oct 2018, 18:05
Bunuel wrote:
If r = 3s, s = 5t, t = 2u, and u ≠ 0, what is the value of $$\frac{rst}{u^3}$$ ?
A. 30
B. 60
C. 150
D. 300
E. 600
Let $$u=1$$, therefore $$t=2$$ & $$s=10$$ & $$r=30$$
$$\frac{rst}{u^3}$$=600
Re: If r = 3s, s = 5t, t = 2u, and u ≠ 0, what is the value of rst/u^3 [#permalink] 12 Oct 2018, 18:05
Display posts from previous: Sort by
| 1,412 | 3,647 |
{"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}
| 4.03125 | 4 |
CC-MAIN-2019-43
|
latest
|
en
| 0.816457 |
https://mathzsolution.com/perturbation-of-mandelbrot-set-fractal/
| 1,669,527,333,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2022-49/segments/1669446710192.90/warc/CC-MAIN-20221127041342-20221127071342-00559.warc.gz
| 436,903,044 | 22,727 |
# Perturbation of Mandelbrot set fractal
I recently discovered very clever technique how co compute deep zooms of the Mandelbrot set using Perturbation and I understand the idea very well but when I try to do the math by myself I never got the right answer.
I am referring to original PDF by K.I. Martin but I will put the necessary equations below.
# Theory
Mandelbrot set is defined as $$Xn+1=X2n+X0X_{n+1} = X_n^2 + X_0$$.
Where the complex number $$X0X_0$$ is in the Mandelbrot set if $$|Xn|≤2|X_n| \leq 2$$ for all n. Otherwise we assign a color based on $$nn$$ where $$|Xn|>2|X_n| > 2$$.
Now consider another point $$Y0Y_0$$ that gives us $$Yn+1=Y2n+Y0Y_{n+1} = Y_n^2 + Y_0$$.
Let $$Δn=Yn−Xn\Delta_n = Y_n - X_n$$, Then
$$Δn+1=Yn+1−Xn+1=2XnΔn+Δ2n+Δ0\Delta_{n+1} = Y_{n+1} - X_{n+1} = 2X_n \Delta_n + \Delta_n^2 + \Delta_0$$
So far this is crystal clear to me. But now we want to compute $$Δn\Delta_n$$ directly from $$Δ0\Delta_0$$ using pre-computed coefficients of the recursive equation.
The author continues:
Let $$δ=Δ0\delta = \Delta_0$$
$$Δ1=2X0δ+δ2+δ=(2X0+1)δ+δ2Δ2=(4X1X0−2X1−1)δ+((X0−1)2+2X1)δ2+(4X0−2)δ3+o(δ4)\Delta_1 = 2X_0\delta + \delta^2+\delta = (2X_0+1)\delta + \delta^2\\ \Delta_2 = (4X_1X_0 - 2X_1-1)\delta + ((X_0-1)^2+2X_1)\delta^2 + (4X_0-2)\delta^3 + o(\delta^4)$$
Let $$Δn=Anδ+Bnδ2+Cnδ3+o(δ4)\Delta_n=A_n\delta+B_n\delta^2+C_n\delta^3+o(\delta^4)$$
Then
$$An+1=2XnAn+1Bn+1=2XnBn+A2nCn+1=2XnCn+2AnBnA_{n+1} = 2 X_n A_n + 1\\ B_{n+1} = 2 X_n B_n + A_n^2\\ C_{n+1} = 2 X_n C_n + 2 A_n B_n$$
Knowing all $$XnX_n$$ we can pre-compute $$AnA_n$$, $$BnB_n$$, and $$CnC_n$$ and given new point $$Z0Z_0$$ we can compute $$δz\delta_z$$ and searching for $$|Zn|>2|Z_n| > 2$$ is just binary search that is O(log n).
# Question
My question is how to compute the equations for $$AnA_n$$, $$BnB_n$$, and $$CnC_n$$? I tried to “check” the equations by applying the $$Δ\Delta$$ recurrence but I obtained:
$$Δ2=2XnΔ1+Δ21+Δ0==(4X1X0+2X1+1)δ+(2X1+(2X0+1)2)δ2+(4X0+2)δ3+δ4\Delta_2 = 2 X_n \Delta_1 + \Delta_1^2+\Delta_0 =\\ =(4X_1 X_0 + 2 X_1 + 1) \delta + (2X_1 + (2X_0 + 1)^2)\delta^2 + (4X_0+2)\delta^3+\delta^4$$
Which does not match author’s $$Δ2\Delta_2$$.
I have also tried to apply given formulas for $$AnA_n$$, $$BnB_n$$, and $$CnC_n$$ to compute forst few $$Δ\Delta$$‘s but they matched my $$Δ2\Delta_2$$, not authors (for example $$C2=4X0+2).C_2 = 4 X_0 + 2).$$
What am I doing wrong? Is it something with complex numbers?
# Bonus
There is probably a general formula for $$Δn\Delta_n$$, can you help me to find it? Something like $$Δn=∑∞i=0C(i)nδi\Delta_n=\sum_{i=0}^\infty C_n^{(i)} \delta^i$$.
# Edit
Anybody? The “Tumbleweed” badge for this question is cool but I thought this should be rather “simple” problem. The solution should probably involve Taylor series, I just need to point out to the right direction. Thanks!
$\Delta_2 = (4X_1 X_0 + 2 X_1 + 1) \delta + (2X_1 + (2X_0 + 1)^2)\delta^2 + (4X_0+2)\delta^3+\delta^4$
$A_{n+1} = 2 X_n A_n + 1 \\ B_{n+1} = 2 X_n B_n + A_n^2 \\ C_{n+1} = 2 X_n C_n + 2 A_n B_n$
Initialise $A_0 = 1, B_0 = 0, C_0 = 0$ and use their computation for the induction.
| 1,355 | 3,145 |
{"found_math": true, "script_math_tex": 3, "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": 36, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.9375 | 4 |
CC-MAIN-2022-49
|
latest
|
en
| 0.730587 |
https://www.vedantu.com/question-answer/find-the-locus-of-the-circumcenters-of-the-class-10-english-cbse-5f5e6b238f2fe24918c2457e
| 1,718,572,409,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-26/segments/1718198861671.61/warc/CC-MAIN-20240616203247-20240616233247-00060.warc.gz
| 928,243,960 | 34,303 |
Courses
Courses for Kids
Free study material
Offline Centres
More
Store
# Find the locus of the circumcenters of the triangle whose two sides are along the coordinate axes and the third side passes through the point of intersection of the lines $ax + by + c = 0$and$lx + my + n = 0$.
Last updated date: 13th Jun 2024
Total views: 412.8k
Views today: 7.12k
Verified
412.8k+ views
Hint: Locus of a point is the set of all points that have the same property or we can say the set of points which behaves in the same way. Circumcenters of the triangle refers to the point in a triangle where all perpendicular bisector of the triangle intersects.
In this question, we have asked to determine the locus of the circumcentre of the triangle such that its two sides are along the coordinate axes and the third side passes through the point of intersection of the lines $ax + by + c = 0$and$lx + my + n = 0$.
Given that two lines pass through the coordinate axis, hence we can draw the figure as
The two lines are perpendicular to each other and the third line passes through the point of intersection of line hence the triangle$\Delta OPQ$formed is a right-angle triangle, we know that the circumcenters of a right triangle is the midpoint $R\left( {h,k} \right)$ of its hypotenuses which is the midpoint of the line PQ, hence
$R\left( {h,k} \right) = \dfrac{{p + 0}}{2},\dfrac{{0 + q}}{2} \\ = \left( {\dfrac{p}{2},\dfrac{q}{2}} \right) \\$
We can write $p = 2h$and$q = 2k$
We know the equation of the hypotenuses PQ is$\dfrac{x}{p} + \dfrac{y}{q} = 1$, hence we can write
$\dfrac{x}{p} + \dfrac{y}{q} = 1 \\ \dfrac{x}{{2h}} + \dfrac{y}{{2k}} = 1 \\ \dfrac{x}{h} + \dfrac{y}{k} = 2 - - - (i) \\$
Given that the hypotenuses passes through the intersecting lines $ax + by + c = 0$and$lx + my + n = 0$
Now solve both the given equations
$ax + by + c = 0 - - - - (ii)$
$lx + my + n = 0 - - - - (iii)$
By multiplying (ii) by l and (iii) by a,
$ax + by + c = 0\left. {} \right\} \times l \\ lx + my + n = 0\left. {} \right\} \times a \\$
We get
$\underline alx + bly + cl = 0 \\ alx\mathop + \limits_ - amy\mathop + \limits_ - an = 0 \\ \\ bly - amy + cl - an = 0 \\ \\ y\left( {bl - am} \right) = an - cl \\ y = \dfrac{{an - cl}}{{bl - am}} \\$
Now multiply (ii) by m and (iii) by b
$ax + by + c = 0\left. {} \right\} \times m \\ lx + my + n = 0\left. {} \right\} \times b \\$
We get
$\underline + amx + bmy + cm = 0 \\ \mathop + \limits_ - blx\mathop + \limits_ - bmy\mathop + \limits_ - bn = 0 \\ \\ amx - blx + cm - bn = 0 \\ \\ x\left( {am - bl} \right) = bn - cm \\ x = \dfrac{{an - cl}}{{am - bl}} \\$
Now put the value of x and y in equation (i), we get
$\dfrac{x}{h} + \dfrac{y}{k} = 2 \\ \dfrac{1}{h}\left( {\dfrac{{an - cl}}{{am - bl}}} \right) + \dfrac{1}{k}\left( {\dfrac{{an - cl}}{{bl - am}}} \right) = 2 \\ \\ \\ \dfrac{{k\left( {an - cl} \right)\left( {bl - am} \right) + h\left( {an - cl} \right)\left( {am - bl} \right)}}{{hk\left( {am - bl} \right)\left( {bl - am} \right)}} = 2 \\ k\left( {an - cl} \right)\left( {bl - am} \right) + h\left( {an - cl} \right)\left( {am - bl} \right) = 2\left\{ {hk\left( {am - bl} \right)\left( {bl - am} \right)} \right\} \\ y\left( {an - cl} \right)\left( {bl - am} \right) + x\left( {an - cl} \right)\left( {am - bl} \right) = 2xy\left( {am - bl} \right)\left( {bl - am} \right) \\ \\$
Hence$y\left( {an - cl} \right)\left( {bl - am} \right) + x\left( {an - cl} \right)\left( {am - bl} \right) = 2xy\left( {am - bl} \right)\left( {bl - am} \right)$ is the equation for the locus of the circumcenter of the circle.
Note: Locus can be a set of points, lines, line segments, curve, and surface which satisfy one or more properties. In the case of a right-angled triangle, the circumcenter of the triangle is always the midpoint of hypotenuses of that triangle.
| 1,346 | 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}
| 4.875 | 5 |
CC-MAIN-2024-26
|
latest
|
en
| 0.886348 |
https://forum.allaboutcircuits.com/threads/check-my-answer-for-a-question-regarding-thevenin-theorem.124047/
| 1,713,141,480,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-18/segments/1712296816904.18/warc/CC-MAIN-20240414223349-20240415013349-00654.warc.gz
| 242,591,110 | 20,093 |
# Check my answer for a question regarding Thevenin Theorem
#### Ashenrise
Joined May 2, 2016
2
This is exactly how my professor formatted the question:
For the circuit shown below, calculate:
(a) Thevenin Voltage (Vth)
(b) Thevenin Resistance Rth
(c) Load current ( IL )
this is what I got but I'm not sure what I'm doing wrong:
Vt= 10.9V - 0.9 V= 10 V
1. Vth= R2xVt/(R2+R1)
2. Vth= 600 x 10/600+400= 6000/1000= 6 V
3. Rth= R1+R2= 400+600= 1kohm
4. IL= Vth/RL+Rth= 6/1+1= 3mA
#### Attachments
• 7.7 KB Views: 8
#### HW-nut
Joined May 12, 2016
97
From your drawing, I am not exactly sure where the load is looking into, so I’m going to assume that the load is R3.
Vth = V1 – V2 = 10V
And Vth = R1+ R2 = 1K
IL = Vth / (Rth + RL) = 10/ (1K + 1K) = 5 mA
Hope this helps.
Last edited:
#### HW-nut
Joined May 12, 2016
97
Whoops, not supposed to answer the question. Let’s try this again.
From your drawing, I am not exactly sure where the load is looking into, so I’m going to assume that the load is R3.
Vth is the open circuit voltage, What is the voltage at R3 if R3 is removed from the circuit?
Rth is the output resistance, what is the resistance across R3 if R3 is removed and the voltage sources are shorted?
Determining the load current should be straightforward once the circuit is redrawn in Theveninequivalent form.
#### Ashenrise
Joined May 2, 2016
2
Whoops, not supposed to answer the question. Let’s try this again.
From your drawing, I am not exactly sure where the load is looking into, so I’m going to assume that the load is R3.
Vth is the open circuit voltage, What is the voltage at R3 if R3 is removed from the circuit?
Rth is the output resistance, what is the resistance across R3 if R3 is removed and the voltage sources are shorted?
Determining the load current should be straightforward once the circuit is redrawn in Theveninequivalent form.
I understand now, thanks so much, I used multisim to check it again and your answers were there, I don't know why I didn't get it the first time. Anyway, I appreciate the help greatly.
| 632 | 2,077 |
{"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-2024-18
|
latest
|
en
| 0.93609 |
https://prog2.de/book/sec-arith-add.html
| 1,695,453,525,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2023-40/segments/1695233506480.35/warc/CC-MAIN-20230923062631-20230923092631-00283.warc.gz
| 519,616,459 | 16,954 |
# An Introduction to Imperative Programming
Adding two binary numbers works in the same way as adding decimal numbers. Let us first consider the sum $$a+b$$ of two bits $$a$$ and $$b\text{.}$$ In general, the sum of two bits is two bits long, because the addition can cause a carry. Bit 0 of the sum is given by the xor of both bits: $$a\bxor b\text{.}$$ The carry $$c$$ of that position is 1 if and only if $$a=b=1\text{,}$$ i.e. $$a\band b=1\text{.}$$
When adding entire bit strings not just single digits, the carry at a position needs to be propagated into the next higher position. So, if $$c_i$$ denotes the carry bit position $$i-1$$ generates, the sum bit at position $$i$$ is given as
$$s_i\defeq a_i\bxor b_i\bxor c_i\tag{1.1}$$
The carry bit of position $$i-1$$ may also influence the carry bit $$c_{i+1}$$ out of position $$i\text{.}$$ For example, if $$a_i=b_i=c_i=1\text{,}$$ then $$c_{i+1}=1\text{.}$$ The carry bit position $$i$$ generates is defined as
$$c_{i+1} \defeq (a_i\band b_i)\bor (a_i\band c_i)\bor (b_i\band c_i)\tag{1.2}$$
The following lemma proves that $$c_{i+1}s_i$$ indeed is the sum of $$a_i\text{,}$$ $$b_i\text{,}$$ and $$c_i\text{.}$$
By value table:
\begin{equation*} \begin{array}[b]{ccc|c||cc|c} a_i \amp b_i \amp c_i \amp a_i+b_i+c_i \amp c_{i+1} \amp s_i \amp \buintp{c_{i+1}s_i} \\ \hline 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \\ 1 \amp 0 \amp 0 \amp 1 \amp 0 \amp 1 \amp 1 \\ 0 \amp 1 \amp 0 \amp 1 \amp 0 \amp 1 \amp 1 \\ 1 \amp 1 \amp 0 \amp 2 \amp 1 \amp 0 \amp 2 \\ 0 \amp 0 \amp 1 \amp 1 \amp 0 \amp 1 \amp 1 \\ 1 \amp 0 \amp 1 \amp 2 \amp 1 \amp 0 \amp 2 \\ 0 \amp 1 \amp 1 \amp 2 \amp 1 \amp 0 \amp 2 \\ 1 \amp 1 \amp 1 \amp 3 \amp 1 \amp 1 \amp 3 \\ \end{array} \end{equation*}
Now, consider two bit strings $$a$$ and $$b$$ of equal length $$n\text{.}$$ When adding $$a$$ and $$b$$ we assume that there flows no carry into the least-significant digit, i.e. $$c_0=0\text{.}$$
$$a+b\defeq c_ns_{n-1}\dots s_0$$
### Remark1.3.4.Efficiency and practicability of our adder.
While Lemma 1.3.5 shows that the addition algorithm we defined is sound, it has a significant practical disadvantage. Our algorithm is a so-called ripple-carry adder which means that the carry “ripples” through the positions: In order to compute $$s_i$$ you have to have computed $$c_i$$ for which you need $$s_{i-1}$$ and so forth. This makes the depth of the actual circuit of ands, ors, xors that implements the adder linear in terms of the number of positions. Since the circuit depth significantly influences the frequency by which you can clock the computer and therefore its speed, practical implementations use different adders, so called carry-lookahead adders that have more gates but are flatter and can therefore be clocked faster.
The following lemma shows that this definition is sound, i.e. the bit operations by which we defined the addition really produce a bit string that represents the natural number of the sum of the natural numbers that are represented by the individual bit strings.
By induction over $$n\text{.}$$
1. Base case $$n=1$$.
Follows directly from Lemma 1.3.1.
2. Induction step $$n-1\to n$$.
The induction hypothesis is
\begin{equation*} \buintp{c_{n-1}\bseqq s{n-2}}=\buintp{\bseqq{a}{n-2}}+\buintp{\bseqq{b}{n-2}} \end{equation*}
By using the definitions and lemmas we have established above, we get:
\begin{align*} \buintp{a}+\buintp{b} \amp = a_{n-1}2^{n-1} + \buintp{\bseqq a{n-2}} + b_{n-1}2^{n-1} + \buintp{\bseqq{b}{n-2}} \amp \knowl{./knowl/lem-arith-splitval.html}{\text{Lemma 1.1.2}}\\ \amp = (a_{n-1}+b_{n-1})\cdot 2^{n-1} + \buintp{\bseqq{a}{n-2}} + \buintp{\bseqq{b}{n-2}} \\ \amp = (a_{n-1}+b_{n-1})\cdot 2^{n-1} + \buintp{c_{n-1}\bseqq{s}{n-2}} \amp \text{IH}\\ \amp = (a_{n-1}+b_{n-1}+c_{n-1})\cdot 2^{n-1} + \buintp{\bseqq {s}{n-2}} \amp \knowl{./knowl/lem-arith-splitval.html}{\text{Lemma 1.1.2}}\\ \amp = \buintp{c_ns_{n-1}}\cdot 2^{n-1} + \buintp{\bseqq {s}{n-2}} \amp \knowl{./knowl/lem-arith-place.html}{\text{Lemma 1.3.1}}\\ \amp = \buintp{c_ns_{n-1}\cdots s_0} \amp \knowl{./knowl/lem-arith-splitval.html}{\text{Lemma 1.1.2}} \end{align*}
Note that the sum contains $$n+1$$ bits. If the carry bit out of the last position is $$0\text{,}$$ Lemma 1.3.5 tell us that the lower $$n$$ bits are accurate, i.e. $$\buintp{\bseq sn}=\buintp{a}+\buintp{b}\text{.}$$ If the carry out of the last position is $$1\text{,}$$ the lower $$n$$ bits are not accurate and the sum is greater or equal $$2^n\text{.}$$ In this case an overflow happened and we would need all $$n+1$$ bits to accurately represent the result. Since computers internally operate on bit strings of fixed size, there is no space to store this additional bit directly in-place with the result. If we are only interested in the $$n$$ least significant bits when adding, we write shortly
$$a\baddn b\defeq\bseq sn\tag{1.3}$$
to indicate that we ignore the carry out of the most significant place.
Let us now turn to subtraction. Instead of inventing a new algorithm for subtracting, we are going to use a trick to express subtraction by addition. This trick hinges on the fact that we are only interested in subtracting words of $$n$$ bits and not bit strings of arbitrary length. Let us first understand the overall idea using an example in base 10 arithmetic:
Assume we want to subtract base 10 numbers of length 3, and more concretely suppose we want to compute $$3-2\text{.}$$
\begin{equation*} 3-2\equiv 3-2+1000\equiv 3+998\equiv 1001\equiv 1\mod 1000 \end{equation*}
What essentially happens here is that we, instead of subtracting $$2\text{,}$$ we added $$998=1000-2$$ and just considered the result modulo $$1000$$ which is $$1\text{,}$$ the desired result. This of course only works if we limit the digit sequences to a certain length before (here 3).
We use the same trick to subtract two bit strings of length $$n$$ and therefore define subtraction to fulfill the following specification:
$$\buintp{a-b}\defeq\buintp{a}+(2^n-\buintp{b})\tag{1.4}$$
So, instead of subtracting $$\buintp b$$ from $$\buintp a$$ we are adding $$(2^n-\buintp b)$$ to $$\buintp a\text{.}$$ Let us first observe that the result $$\buintp a+(2^n-\buintp b)$$ is equivalent to $$\buintp a-\buintp b$$ modulo $$2^n$$ analogously to the example above. This means that the lower $$n$$ bits of $$\buintp a+(2^n-\buintp b)$$ and lower $$n$$ bits of the true difference are equal which is exactly what we wanted. The number $$2^n-\buintp b$$ is called the two's complement of $$\buintp b\text{.}$$
We will see in the proof of Lemma 1.1.6 that we can easily compute the two's complement of $$\buintp b$$ by flipping all bits of $$b$$ and adding $$1\text{.}$$ The latter can easily be achieved by setting the carry into the least significant position ($$c_0$$) to $$1\text{.}$$ Additionally, we see that if $$\buintp a\ge\buintp b$$ (i.e. the result of the subtraction is defined), $$\buintp{a-b}$$ will be greater than $$2^n\text{.}$$ Hence, the carry out of the most significant bit is set if the result of the subtraction is valid and is clear when the subtraction overflowed.
### Definition1.3.6.Subtraction.
$$a-b\defeq a+\bneg b\text{ with }c_0=1$$ and $$a\mathop{-_n}b\defeq a\mathop{+_n}\bneg b$$ with $$c_0=1$$
\begin{align*} \buintp{a-b} \amp = \buintp{a+\bneg b}\text{ with }c_0=1 \amp \knowl{./knowl/def-arith-sub.html}{\text{Definition 1.3.6}}\\ \amp = \buintp a+\buintp{\bneg b}+1 \amp \knowl{./knowl/lem-arith-uadd.html}{\text{Lemma 1.3.5}}\\ \amp = \buintp a+\left(\sum_{i=0}^{n-1}(1-b_i)2^i\right)+1 \amp \knowl{./knowl/def-arith-uintp.html}{\text{Definition 1.1.1}}\\ \amp = \buintp a+\left(\sum_{i=0}^{n-1}2^i\right)+1-\left(\sum_{i=0}^{n-1}b_i2^i\right) \amp \\ \amp = \buintp a+\left(\sum_{i=0}^{n-1}2^i\right)+1-\buintp b \amp \knowl{./knowl/def-arith-uintp.html}{\text{Definition 1.1.1}}\\ \amp = \buintp a+2^n-\buintp b \amp \knowl{./knowl/lem-sumnine.html}{\text{Lemma 1.1.6}} \end{align*}
### Remark1.3.8.Practical Considerations.
Definition 1.3.6 already suggests how an adder has to be extended to support subtraction: The right operand of the ALU must be optionally negateable and the carry bit into the least position must be presettable to either $$0$$ or $$1\text{.}$$
Many processors provide the carry out of the last position in a special register, the so-called flag register. It is then often just called the carry bit because all the intermediate carries are unaccessible to the programmer. Some architectures negate the carry bit when subtracting, actually making it a borrow bit. For example, on an x86 machine the carry bit is clear when computing $$3-2$$ whereas an ARM processor would set the carry in that case.
Pro comment: This is why on x86 the instruction that subtracts and reading $$c_0$$ from the flags is called sbb (sub with borrow) while it is called sbc (sub with carry) on ARM processors. Such instructions are used when implementing arithmetic on bit strings that are longer than the word size of the processors.
| 3,112 | 8,994 |
{"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": 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": 4, "x-ck12": 0, "texerror": 0}
| 4.375 | 4 |
CC-MAIN-2023-40
|
latest
|
en
| 0.620853 |
https://www.chembuddy.com/calculation-of-pH-of-salt-solutions
| 1,669,489,746,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2022-49/segments/1669446708046.99/warc/CC-MAIN-20221126180719-20221126210719-00348.warc.gz
| 759,710,614 | 5,619 |
Operating systems: XP, Vista, 7, 8, 10, 11
# Calculation of pH of any salt solution
pH of all salts solutions is determined by the hydrolysis and - in the case of acidic or basic salts - by the dissociation. In case of salts of strong acids and strong bases effects of the hydrolysis are often negligible, but - as it was pointed out earlier - some of the strong acids and bases are weaker than it is commonly believed.
Let's solve the problem for simple salt using general approach.
We have two substances - HA and BOH - and two dissociation constants:
11.1
11.2
water dissociation constant:
11.3
and three balances:
11.4
11.5
11.6
In the case of salt solution Ca=Cb, but we have a good reason to treat these values separately. Once the general equation will be derived we can always simplify it using only one concentration, but general equation will describe not only pH of the salt solution, but also of any mixture of weak acid and weak base, regardless of their ratio.
In deriving the equation we could use the same approach we tested for polyprotic acid, but we will try shorter way that was not applicable then. We will use dissociation constants and mass balances to find out concentrations of A- and B+ ions to put them into charge balance equation. From acid mass balance:
11.7
Inserted into 11.1
11.8
Solved for [A-]:
11.9
and ready to be put into charge balance.
Now do the same to the base, using Kb and Kw definition:
11.10
11.11
solving for [B+] yields
11.12
and finally, combining 11.9 with 11.12 and charge balance we get
11.13
which is a fourth degree equation for [H+] in any mixture of weak acid and weak base.
Using for calculations
11.14
we get very similar, strikingly symmetrical result:
11.15
It is worth of noting here that the equation used for pH calculation in BATE pH calculator is just a full polyprotic version of the 11.13 (note that generalized equation 11.16 uses not stepwise, but overall dissociation constants Kai and Kbi):
11.16
| 489 | 1,998 |
{"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-2022-49
|
longest
|
en
| 0.939391 |
https://lara.epfl.ch/w/sav08/predicate_logic_informally
| 1,674,809,255,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2023-06/segments/1674764494974.98/warc/CC-MAIN-20230127065356-20230127095356-00456.warc.gz
| 380,461,073 | 11,465 |
# First-Order Logic
First-order logic is a very powerful notation that extends propositional logic. Here we give only an informal overview of first-order logic, which should be enough for you to understand the meaning of first-order logic formulas based on mathematics and natural language. In future lectures we study first-order logic as a formal system and discuss algorithmic questions in more detail.
On top of propositional operations, first-order logic adds:
• constructs to represent the structure of propositions:
• equality ( = )
• predicate symbols (, , …)
• function symbols (, , …)
• first-order variables (, , …) denoting entities in some domain
• quantifiers “forall” (), “exists” ()
## Uses of first-order logic
• precisely describe arbitrary mathematical statements (theorems, conjectures, properties)
• specify program properties
• represent the meaning of programs
• represent knowledge about the world (e.g. knowledge bases, semantic web)
Examples
• An ancestor of my ancestor is also my ancestor:
• Grandparent is the parent of a parent:
• Property holds for infinitely many natural numbers:
• is continuous in point :
• first array elements are strictly positive, remaining elements are zero:
## First-Order Logic Formulas
We build first-order logic formulas by starting from atomic formulas (defined below) and applying propositional operators and quantifiers:
• atomic formulas are first-order logic formulas
• if and are first-order formulas, so are , , , ,
• if is a first-order formula, so is and
For a given value of variables, each formula evaluates to true or false. The rules for propositional operators are exactly as in propositional logic. Quantifiers can be seen as a way of generalizing conjunction and disjunction. First-order variables range over some domain . In the special case of finite domain , we have that
which corresponds to the intuitive definition of terms “for all” and “there exists”. Note, however, that can be infinite in general (for example: the set of integers or reals).
Atomic formulas evaluate to true or false. We build atomic formulas by applying predicate symbols and equality to terms:
• if and are terms, then is an atomic formula;
• if , …, are terms and is a predicate symbol that takes arguments, then is an atomic formula.
Predicates represent relations, for example, we can represent as a binary relation.
Terms denote elements of the domain . We build them starting from variables and constants and applying function symbols:
• each first-order variable is a term
• a constant is a term
• if are terms and is a function symbol that takes arguments, then is also a term.
Example of constants are numerals for natural numbers, such as . Examples of function symbols are operations such as .
From above we see that the set of formulas depends on the set of predicate and function symbols. This set is is called vocabulary or language.
### Bounded Quantifiers
In general all quantifiers range over some universal domain . To restrict them to subsets of , we can use bounded quantifiers:
• means
• means (note implication instead of conjunction)
More generally, if is some binary relation written in infix form, such as we write
• meaning
• means (note implication instead of conjunction)
## Evaluating First-Order Logic Formulas
To evaluate first-order logic formulas we need to know:
• the domain set
• values of first-order variables in the formula
• interpretation of predicate symbols and function symbols (e.g. does predicate symbol denote or or , does binary function symbol denote or )
#### Examples
Consider formula given by . The formula has one predicate symbol that takes one urgument (we call it unary) and one unary function symbol . Assume that the domain is the set of integers and consider two possible interpretations:
• denotes property of being an even integer and denotes the successor function . Is formula true under this interpretation?
• denotes property of being an even integer and denotes the squaring function . Is formula true under this interpretation?
Now consider formula in each of these two interpretations.
## (Finite) Validity and Satisfiability of First-Order Logic Formulas
Formula is valid if and only if it evaluates to true for all domains, values of its variables and interpretations of predicate and function symbols.
Formula is satisfiable if and only if it evaluates to true for some domains, values of its variables, interpretation of predicate and function symbols.
Formula is unsatisfiable if and only if it is false for all domains, values of its variables and interpretations of predicate and function symbols.
Note
• a formula is unsatisfiable if and only if it is not satisfiable
• a formula is valid if and only if is unsatisfiable
Answers to some important algorithmic questions (not immediate):
• There is no algorithm that given a first-order logic formula outputs “yes” when the formula is valid and “no” otherwise - validity of first-order logic formulas is undecidable
• There exists an enumeration procedure that systematically lists all valid formulas (and only valid formulas) - validity of first-order logic formulas is enumerable
If instead of considering all domains we only consider finite domains (no natural numbers but only e.g. prefixes of natural numbers of the form ) then we obtain notions of finite validity, finite satisfiability and finite unsatisfiability.
Note that because finite domains are a special case of possible domains, we have the following:
• if a formula is finitely satisfiable, then it is satisfiable
• if a formula is valid, then it is also finitely valid
| 1,134 | 5,656 |
{"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-2023-06
|
latest
|
en
| 0.84054 |
https://fresherbell.com/quizdiscuss/quantitative-aptitude/a-and-b-can-cover-a-200-m-race-in-22-seconds-and-25-seconds
| 1,718,479,735,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-26/segments/1718198861606.63/warc/CC-MAIN-20240615190624-20240615220624-00622.warc.gz
| 226,245,092 | 10,652 |
# Quiz Discussion
A and B can cover a 200 m race in 22 seconds and 25 seconds respectively. When Av finished the race, then B is at what distance from the finishing line ?
Course Name: Quantitative Aptitude
• 1] 24 m
• 2] 30 m
• 3] 48 m
• 4] 54 m
##### Solution
No Solution Present Yet
#### Top 5 Similar Quiz - Based On AI&ML
Quiz Recommendation System API Link - https://fresherbell-quiz-api.herokuapp.com/fresherbell_quiz_api
# Quiz
1
Discuss
A and B take part in a 100 metres race. A runs at 5 km an hour. A gives B a start of 8 metres and still beats him by 8 seconds. The speed of B is ?
• 1] 4.45 km/hr
• 2] 4.14 km/hr
• 3] 4.15 km/hr
• 4] 4.25 km/hr
##### Solution
2
Discuss
In a kilometre race, A, B and C are three participants. A can give B start of 50 m and C a start of 69 m. The start which B can allow C, is -
• 1] 17 m
• 2] 18 m
• 3] 19 m
• 4] 20 m
##### Solution
3
Discuss
In a 800 metre race, A defeated B by 15 seconds. If A's speed was 8 km/hr, the speed of B was -
• 1]
16/27 km/hr
• 2]
27/16 km/hr
• 3]
km/hr
• 4]
km/hr
##### Solution
4
Discuss
In 100 m race, A covers the distance in 36 seconds and B in 45 seconds. In this race A beats B by:
• 1] 20 m
• 2] 25 m
• 3] 22.5 m
• 4] 9 m
##### Solution
5
Discuss
In a 500 m race, the ratio of the speeds of two contestants A and B is 3 : 4. A has a start of 140 m. Then, A wins by:
• 1] 60 m
• 2] 40 m
• 3] 20 m
• 4] 10 m
##### Solution
6
Discuss
A man starts in a boat from A toward B against the stream. After reaching a point M on the way, he throws a rubber ring into the water. He takes 8 minutes from then to reach his destination and the ring exactly A at that moment. If the distance between A and B is 4 km, then find the speed of boat in still water
• 1]
15 kmph
• 2]
12 kmph
• 3]
30 kmph
• 4]
45 kmph
##### Solution
7
Discuss
In a kilometre race, A beats B by 30 seconds and B beats C by 15 seconds. If A beats C by 180 m, the time taken by A to run 1 kilometre, is ?
• 1] 200 sec
• 2] 205 sec
• 3] 210 sec
• 4] 250 sec
##### Solution
8
Discuss
Gopal , Ganesh and Giri start from point A and run around a circular path at speed of 10 kmph, 14 kmph and 6 kmph respectively. Gopal & Giri run in the same direction and Ganesh run in the opposite direction. The length of track is 20 km. In how much time will all three of them be together for the first time?
• 1]
3 hr
• 2]
4 hr
• 3]
5 hr
• 4]
6 hr
##### Solution
9
Discuss
In a 100 m race, A beats B by 10 m and C by 13 m. In a race of 180 m, B will beat C by:
• 1] 5.4 m
• 2] 4.5 m
• 3] 5 m
• 4] 6 m
##### Solution
10
Discuss
In a 1000 m race, A can beat B by 100 m. In a race of 400 m, B can beat C by 40 m. By how many metres will A beat C in a race of 500 m ?
• 1] 85 m
• 2] 95 m
• 3] 105 m
• 4] 115 m
# Quiz
| 1,007 | 2,791 |
{"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}
| 3.578125 | 4 |
CC-MAIN-2024-26
|
latest
|
en
| 0.8825 |
https://www.vedantu.com/question-answer/if-x-7-then-which-option-is-correct-a-x-7-b-x-class-9-maths-cbse-5fd85dd58238151f0da05f3d
| 1,719,026,258,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-26/segments/1718198862249.29/warc/CC-MAIN-20240622014659-20240622044659-00671.warc.gz
| 921,557,753 | 26,470 |
Courses
Courses for Kids
Free study material
Offline Centres
More
Store
# If $x < 7$, then which option is correctA. $- x < - 7$B. $- x \leqslant - 7$C. $- x > - 7$D. $- x \geqslant - 7$
Last updated date: 22nd Jun 2024
Total views: 386.4k
Views today: 7.86k
Verified
386.4k+ views
Hint: In these type of questions the inequalities questions, we have to perform operations like adding, subtracting, multiplying or dividing on both left hand side and right hand side with a variable or number as per the requirement of the question and until we get the required result.
Complete step-by-step solution:
Given expression is $x < 7$,
This is an inequality which can be solved by adding, subtracting, multiplying or dividing both sides till we get the variable that is required.
Now as it is a simple expression, subtract both sides with $x$, we get,
$\Rightarrow x - x < 7 - x$,
Now perform subtracting on both sides we get,
$\Rightarrow 0 < 7 - x$,
Now subtracting both sides with 7, we get,
$\Rightarrow 0 - 7 < 7 - x - 7$,
Now perform subtraction on both sides we get,
$\Rightarrow - 7 < - x$
Now swapping Left hand side with the right side we get,
$\Rightarrow - x > - 7$.
So, the given expression $x < 7$ can be rewritten as $- x > - 7$,
So, option C is correct answer.
Note: We have to solve these types of inequalities by performing addition, subtracting, multiplying or dividing both sides till we get the variable on its own. These operations may change the direction of inequality in two ways, i.e.,
When we perform multiplication or division on both sides by a negative number or putting minus sign on both left hand side and right hand side.
When we do swapping on both the left hand side and right hand side of the given inequality.
And students must be clear that the operations multiplication or division by a variable unless it is clear that it is always positive or always negative.
| 488 | 1,900 |
{"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}
| 4.6875 | 5 |
CC-MAIN-2024-26
|
latest
|
en
| 0.850434 |
http://mathhelpforum.com/advanced-algebra/186170-proving-product-orthogonal-matrices-also-orthogonal-print.html
| 1,529,881,280,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-26/segments/1529267867095.70/warc/CC-MAIN-20180624215228-20180624235228-00288.warc.gz
| 198,093,014 | 2,766 |
# Proving that the product of orthogonal matrices is also orthogonal.
• Aug 15th 2011, 04:23 AM
Glitch
Proving that the product of orthogonal matrices is also orthogonal.
I'm not sure how to do this. I tried making two general 2 x 2 matricies like so:
$\displaystyle $A = \begin{array}{cc} a & b \\ c & d \\ \end{array}$$
$\displaystyle $B = \left( {\begin{array}{cc} e & f \\ g & h \\ \end{array} } \right)$$
To be orthogonal, these equations must be satisfied (via dot product):
ab + cd = 0
ef + gh = 0
Multiplying the two matricies, and then finding the dot product of the result yeilds:
$\displaystyle a^2ef + b^2gh = 0$
$\displaystyle c^2ef + d^2gh = 0$
The best I could do was show that:
$\displaystyle (a^2 - c^2)(ef) = (b^2 - d^2)(ef)$
I think I'm attempting this wrong. :/ Any advice?
• Aug 15th 2011, 04:32 AM
FernandoRevilla
Re: Proving that the product of orthogonal matrices is also orthogonal.
$\displaystyle M\in\mathbb{R}^{2\times 2}$ is orthogonal iff $\displaystyle M^tM=I$ . So, if $\displaystyle A,B\in\mathbb{R}^{2\times 2}$ are orthogonal then, $\displaystyle (AB)^t(AB)=\ldots=I$ which implies $\displaystyle AB$ orthogonal.
| 381 | 1,156 |
{"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}
| 3.921875 | 4 |
CC-MAIN-2018-26
|
latest
|
en
| 0.814898 |
http://mathhelpforum.com/pre-calculus/117490-heaviside-function.html
| 1,529,288,734,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-26/segments/1529267859923.59/warc/CC-MAIN-20180618012148-20180618032148-00239.warc.gz
| 209,317,894 | 11,586 |
1. ## Heaviside Function
The Heaviside Function, H(t) is defined as
H(t) =
1 if t>=0
0 if t<0
It is used in the study of electric circuits to represent the sudden surge of electric current, or voltage, when a switch is instantaneously turned on
My student asked me the following question,
"since time cannot be negative, then why the function is defined for negative values of t?"
What is the best explanation? is it because usually we want to have a function that is defined for all real numbers?
Thanks!
2. The Heaviside Function, H(t) is defined as
H(t) =
1 if t>=0
0 if t<0
It is used in the study of electric circuits to represent the sudden surge of electric current, or voltage, when a switch is instantaneously turned on
My student asked me the following question,
"since time cannot be negative, then why the function is defined for negative values of t?"
What is the best explanation? is it because usually we want to have a function that is defined for all real numbers?
Thanks!
--------------------------------------------------------------------
analogy
the graph of the function basically turns "on" ((H(t)=1) => turns on signal) if the switch is turned on hence you get time for a surge, so if the switch is off, for the duration it is off it could be said to be negative up until your turn it on, at that moment the instaneous time=0 hence H(0)=1,((H(t<0)=0) => turns off signal). Now, we know that time cannot be taken away, i.e. go back in time.but rather the time(t) is actually in respect to the moment you turn it on t=0, so for generality, now t isused for the function to be defined for all real numbers.
another analogy-
so the unprocessing of the signal i.e. reversing time (going from right to left on a graph of H(t)) in the function H(t) is basically undoing w/e we did to it. that is it was a value of one when t>=0, but when we go back in time, "we undo what happened" and bring it back to 0 for t<0
and it is easier to have a function for all real numbers.
3. No, there must be a better explanation, because if you remove negative values of $\displaystyle t$, then your function becomes $\displaystyle H(t) = 1$, which is a bit useless (although in my opinion it was useless before anyway, no offense).
Maybe you could also tell him that the input is not necessarily time (I'm not sure about that), and hence the necessity of defining the value for negative inputs ...
You could also tell him that it looks nicer and is easier when defined on all reals
Maybe some research could be interesting here ?
4. Originally Posted by Bacterius
No, there must be a better explanation, because if you remove negative values of $\displaystyle t$, then your function becomes $\displaystyle H(t) = 1$, which is a bit useless (although in my opinion it was useless before anyway, no offense).
Maybe you could also tell him that the input is not necessarily time (I'm not sure about that), and hence the necessity of defining the value for negative inputs ...
You could also tell him that it looks nicer and is easier when defined on all reals
Maybe some research could be interesting here ?
yeah i was wrong the first time but after some research im positive im right now.
| 743 | 3,206 |
{"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}
| 4.09375 | 4 |
CC-MAIN-2018-26
|
latest
|
en
| 0.927091 |
https://hbomagazine.com/15-cm-to-inches/
| 1,686,101,570,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2023-23/segments/1685224653501.53/warc/CC-MAIN-20230607010703-20230607040703-00610.warc.gz
| 338,107,393 | 21,305 |
# 15 cm to inches – 100% Free
Switching between different methods of measuring, such as imperial and metric, could cause difficulties in converting units. The most common conversions are between centimeters (cm) and inches (in). In this article, we’ll examine some different methods to convert 15 cm to inches.
## 1. Change 15 centimeters to inches by Basic conversion formula
Utilizing the fundamental conversion formula is the simplest method to convert 15cm to inches:
One inch equals 2.54 centimeters.
Divide the centimeters divided by 2.54 to calculate the number of inches.
Thus, 15 cm equals 5.91 inches. So, 15 cm is equal to 2.54 = 5.91 inches (rounded to two decimal places)
## 2. Change 15cm in inches by Converting to fractions
Another method to convert 15 cm to inches would be to type the result in fractions instead of decimals. You can use the fact that one inch is equivalent to 25.4 millimetres for that (millimetres). Since 10 millimetres equals 1 centimetre, We can change 15 cm into mm and divide it by 25.4 to calculate the result in inches. Here’s how to do it:
15 cm x 10 mm/cm = 150 mm
150 millimetres / 25.4 millimetres/in is 5.906 inches (rounded to three decimal places)
5.906 inches can become a fraction by determining the fraction with a denominator of 64. (a popular denominator in the area of inch fractions). This can be done by multiplying the numerator and denominator by 64 and then adding the result:
5.906 x 64/1 x 64 = 377.984
The number closest to having a numerator of 64 is 378/64, which can be written as 5 14/64 or 5 7/32 in.
## 3. Change 15 cm in inches by Using a conversion table
The conversion table can be an alternative method to convert 15cm to inches. Conversion tables are chart that shows how the different units of measurement compare with one another. This is an illustration of a cm-to-inches conversion table:
Centimeters Inches 1 0.39 5 1.97 10 3.94 15 5.91 20 7.87 25 9.84
It is evident from the table that 15 centimetres are the same as 5.91 inches.
## 4. Change 15 cm to in by Using a conversion app or website
A conversion application or website is a fast and simple method of changing 15cm to inches and any other measurement. Many free online tools allow you to convert measurements into units instantly. The converter for units available on Google, ConvertUnits.com, and UnitConverters.net are all well-known choices. These tools are particularly useful when you need to convert between complex units or when you need to convert multiple units simultaneously.
In the final analysis, many methods exist to convert 15 cm to inches. It doesn’t matter if you employ formulas or a conversion table, a visual diagram or an application; the main factor is comprehending the relation between the two measurements and selecting the best method for you.
## FAQs
Q: how many inches is 15 cm?
A: 15 centimeters is about 5.90551 inches (rounded to 5 decimal places).
Q: How much is 10 to 15 cm in inches?
A: If you round to the closest thousandth, 10 to 15 centimeters equals 3.937 or 5.905 inches.
## Centimeters To Inches Conversion Table
cm to inches inches 20 cm to inches 7.87 inches 30 cm to inches 11.81 inches 15 cm to inches 5.91 inches 18 cm to inches 7.09 inches 25 cm to inches 9.84 inches 40 cm to inches 15.75 inches 50 cm to inches 19.69 inches 60 cm to inches 23.62 inches 10 cm to inches 3.94 inches 14 cm to inches 5.51 inches 17 cm to inches 6.69 inches 80 cm to inches 31.50 inches 100 cm to inches 39.37 inches 120 cm to inches 47.24 inches 21 cm to inches 8.27 inches 70 cm to inches 27.56 inches 16 cm to inches 6.30 inches 45 cm to inches 17.72 inches 8 cm to inches 3.15 inches 90 cm to inches 35.43 inches 150 cm to inches 59.06 inches 23 cm to inches 9.06 inches 24 cm to inches 9.45 inches 35 cm to inches 13.78 inches 5 cm to inches 1.97 inches 12 cm to inches 4.72 inches 13 cm to inches 5.12 inches 180 cm to inches 70.87 inches 19 cm to inches 7.48 inches 22 cm to inches 8.66 inches 43 cm to inches 16.93 inches 6 cm to inches 2.36 inches 110 cm to inches 43.31 inches 140 cm to inches 55.12 inches 160 cm to inches 62.99 inches 170 cm to inches 66.93 inches 2 cm to inches 0.79 inches 26 cm to inches 10.24 inches 3 cm to inches 1.18 inches 32 cm to inches 12.60 inches 36 cm to inches 14.17 inches 38 cm to inches 14.96 inches 42 cm to inches 16.54 inches 55 cm to inches 21.65 inches 65 cm to inches 25.59 inches 68 cm to inches 26.77 inches 75 cm to inches 29.53 inches 83 cm to inches 32.68 inches 9 cm to inches 3.54 inches 11 cm to inches 4.33 inches 200 cm to inches 78.74 inches 27 cm to inches 10.63 inches 7 cm to inches 2.76 inches 84 cm to inches 33.07 inches 85 cm to inches 33.46 inches 130 cm to inches 51.18 inches 2.5 cm to inches 0.98 inches 28 cm to inches 11.02 inches 29 cm to inches 11.417 inches 3.5 cm to inches 1.378 inches 33 cm to inches 12.992 inches 34 cm to inches 13.386 inches 4 cm to inches 1.575 inches 4.5 cm to inches 1.772 inches 48 cm to inches 18.898 inches 52 cm to inches 20.472 inches 53 cm to inches 20.866 inches 54 cm to inches 21.260 inches 56 cm to inches 22.047 inches 58 cm to inches 22.835 inches 66 cm to inches 25.984 inches 7.5 cm to inches 2.953 inches 72 cm to inches 28.346 inches 76 cm to inches 29.921 inches 92 cm to inches 36.220 inches 10.5 cm to inches 4.134 inches 105 cm to inches 41.339 inches 46 cm to inches 18.110 inches 61 cm to inches 24.016 inches 64 cm to inches 25.197 inches 71 cm to inches 27.953 inches 74 cm to inches 29.134 inches 78 cm to inches 30.709 inches 95 cm to inches 37.402 inches 1.5 cm to inches 0.591 inches 102 cm to inches 40.157 inches 104 cm to inches 40.945 inches 108 cm to inches 42.520 inches 125 cm to inches 49.213 inches 158 cm to inches 62.205 inches 16.5 cm to inches 6.496 inches 165 cm to inches 64.961 inches 17.5 cm to inches 6.890 inches 31 cm to inches 12.205 inches 37 cm to inches 14.567 inches 39 cm to inches 15.354 inches 41 cm to inches 16.142 inches 51 cm to inches 20.079 inches 57 cm to inches 22.441 inches 59 cm to inches 23.228 inches 6.5 cm to inches 2.559 inches
| 1,811 | 6,103 |
{"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}
| 4.40625 | 4 |
CC-MAIN-2023-23
|
latest
|
en
| 0.869161 |
https://geometricolor.wordpress.com/2013/11/07/tilings-and-their-duals/
| 1,563,286,452,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-30/segments/1563195524568.14/warc/CC-MAIN-20190716135748-20190716161748-00334.warc.gz
| 415,052,645 | 17,754 |
## Tilings and their duals
I briefly discussed the dualization method in “The dualization method” but you will need more details to be able to understand my computer code. You can find some interesting ideas in Wolfram MathWorld and in Oracle ThinkQuest. In part I am presenting here the same concepts.
I have difficulties to find good names and their consistent use, because both the original tiling and its dual are tilings. In general I put “grid” before something that belongs to the original tiling and “dual” before something of the dual tiling. But note that the “dual” of something “dual” is again part of the original and thus called with “grid”.
We have to take into account that we can use only finite parts of a tiling. This requires special care. Else we will get disorder near the border. I do it like this:
Hexagon tessellation and its dual: The grey lines show the limits of the region used. The blue lines and dots show the part of the hexagon tessellation that remains. The complete hexagon cells are shaded in pale blue. They give the part of the dual tessellation shown in red.
I only use those lines (in blue) of the tessellation that have points inside a limiting rectangle (grey lines). Then the computer searches for all complete polygons (shaded pale blue) formed by these lines. Thus we have grid-points, grid-lines and grid-polygons, making up the original tiling.
The dual tiling (red lines) similarly has dual-points, dual-lines and dual-polygons. Note that each grid-line (blue) is crossed perpendicularly by a dual-line (red). Each grid-line has as its dual a line of the dual tiling. Then, at the center of each hexagon there is a point of the dual tiling. Each grid-polygon has as its dual a point of the dual tiling. Similarly, we find that each triangle of the dual tiling has at ist center a point of the original tiling. So each grid-point has as its dual a polygon of the dual tiling. At the border, there are exceptions.
This goes also the other way around. The dual line of a dual-line is a grid-line. Each dual-point has a grid-polygon as its dual and every dual-polygon has a grid-point as its dual.
If this is too abstract, you might consider a mobile phone installation. A country is divided into different regions. This corresponds to the original tiling. The regions are the grid-polygons. In each region there is an antenna, which is essentially the dual-point to the grid-polygon. There are transmission lines between neighboring antennas. These are the dual-lines to grid-lines that limit the regions.
It took some time for me to realize that this already determines the topology of the dual tiling. We thus get for any tiling the number of points of the dual tiling and how these points are connected by lines. But the positions of the dual-points are still unknown.
For many tilings the dual-points are simply at the center of their grid-polygons. This is valid for regular and semiregular tessellations and probably for all tilings, that have only regular polygons as tiles. But if we superimpose two tilings we get irregular tiles and we have to find the position of the dual-points differently. We then use that the dual-lines are perpendicular to their grid-lines and that they have all the same length. From this we get the positions of the dual-points.
This entry was posted in Tilings and tagged , , , , , . Bookmark the permalink.
This site uses Akismet to reduce spam. Learn how your comment data is processed.
| 770 | 3,482 |
{"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-2019-30
|
latest
|
en
| 0.952805 |
http://www.wyzant.com/resources/answers/algebra_2?f=active&pagesize=20&pagenum=4
| 1,398,351,753,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2014-15/segments/1398223206147.1/warc/CC-MAIN-20140423032006-00559-ip-10-147-4-33.ec2.internal.warc.gz
| 964,365,290 | 13,705 |
Search 72,341 tutors
# What is the solution set for 4+ {X - 1} =11?
the brackets are supposed to be absolute value bars
Determine the equation of g(x) that results from translating the function f(x)= (x+2)^2 to the right 11 units. I have the choice of four answers. 1. g(x)=(x-9)^2 2...
# word problem
An object 5.7 feet tall casts a shadow that is 17.1 feet long. How long in feet would the shadow be for an object which is 19.8 feet tall? A. 8.4 feet B. 6.6 feet C. 31.2 feet D...
2x^2+49=-20x
# X^2+x-30=0
Solve using any method
# Write the equation of the line which passes through (1, 2) and with slope of - 1 3 in standard form.
I have already completed the question but I am confused on how I got the the answer. y−y1 = m(x−x1) y−2 = -1 3 •(x−1) y−2 = -1 3 x+ 1 3 1 3 x+y...
# the vertex of the parabola below is at the point (4, -1). Which could be this parabola's equation?
I don't understand the question. What would the equation be for this vertex of the parabola? I'm doing Conic sections, and I don't understand the question.
# the inflection point of f(x)= -2(x-4)^3 +11 is where
I also need to know the end-behavior. for example- x----> + or - infinity
# the equation for The line that passes through (-3, 5) and (-2, -6)
Can you help me with this question? The equation for the line that passes through (-3,5) and (-2,-6)
# factor x^3+5x^2-3x-15=0
I cannot remember how to factor to the cubic term
# Word Problem
A solution is 10% citric acid. if 3 liters of a 20% citric acid solution are added to it, the result is a solution this is in a ratio of 13 parts citric acid to 102 parts "other than citric acid"...
# Is (10, 2) a solution to this system of equations? y = 5x 2x + y = 14
i don't know this answer can you help me?
# select the approxiate value of x that is the solution to f(x)=0 where f(x)=-9x^2+6x+7
select the approxiate value of x that is the solution to f(x)=0 where f(x)=-9x^2+6x+7
# select the approxiate value of x that is solutions for (x)=0 value f(x)=-3x^2+2x+8
select the approxiate value of x that is solutions for (x)=0 value f(x)=-3x^2+2x+8
f(x)= x^2+2x+10=
# How do you find a in parabola equations using just the vertex and the focus?
I know you could plug everything in and figure it out that way but I can't do that because I don't have another point on the parabola. For example: vertex:(-1,-3) focus:(-1,0)
# values of x
Select he value of x that are solutions to the inequality 0>x^2+5x-2 1. xE (-5-33 squared/2,-5+33 squared/2) 2. xE [-5-33 squared/2,-5+33...
9 (2x+5)>9
# function/ graphing
Discuss/Explain how the graph F(x) = -2f(x+1) -3 can be obtained from the graph f(x). If (0,5), (6,7), and (-9,-4) are on the graph of f. where do they end up on the graph of F?
# write in answer form a+bi
simplify write answer in form a+bi, where a and b are real numbers
| 950 | 2,872 |
{"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-2014-15
|
latest
|
en
| 0.917633 |
https://republicofsouthossetia.org/question/how-many-solutions-does-this-linear-system-have-y-6-2-12-2y-4-a-one-solution-0-0-b-one-solution-15095852-69/
| 1,638,606,655,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-49/segments/1637964362952.24/warc/CC-MAIN-20211204063651-20211204093651-00312.warc.gz
| 496,338,128 | 13,355 |
How many solutions does this linear system have? y=-6x+2 -12x – 2y = -4 A)one solution: (0,0) B)one solution: (1,-4)
Question
How many solutions does this linear system have?
y=-6x+2
-12x – 2y = -4
A)one solution: (0,0)
B)one solution: (1,-4)
C)no solution
D)infinite number of solutions
in progress 0
2 months 2021-09-27T02:58:23+00:00 2 Answers 0 views 0
B- one solution: (1,-4)
Step-by-step explanation:
(1,-4)
y = -6x +2
-4 = -6(1) +2
-4 =-4
(1,-4)
-12x – 2y = -4
-12(1) -2(-4) = -4
-12 + 8 =-4
-4 = -4
if you put the value x and y into both equation you can find the solution
Hi there!
The correct answer is: D. Infinite number of solutions
Step-by-step explanation:
in this case you can use the substitution method to solve these linear equations
since y = -6x +2 you can plug this into the equation -12x -2y = -4
-12x – 2(-6x +2) = -4 this comes out to be -12x + 12x -4 = -4
then you simplify down to 0x=0
| 351 | 934 |
{"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}
| 4.25 | 4 |
CC-MAIN-2021-49
|
latest
|
en
| 0.821177 |
http://urbrainy.com/maths/year-2-age-6-7/shape-and-measures-y2
| 1,369,274,216,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2013-20/segments/1368702730377/warc/CC-MAIN-20130516111210-00018-ip-10-60-113-184.ec2.internal.warc.gz
| 280,127,944 | 13,401 |
# Shape and measures Maths Worksheets For Year 2 (age 6-7)
Expectations are high when it comes to shape in Year 2! Children are expected to describe, visualise, classify and draw a wide variety of 2D and 3D shapes. Reflective symmetry is introduced and the idea that an angle is a measure of turn is very important to understand, as well as introducing the right angle.
Standard metric measures are introduced, including metres, litres and kilograms. Our worksheets provide plenty of opportunity for practical work as well as reading the numbered divisions on a scale.
### Shape colour patterns 4
Can you draw and colour the missing shapes in these patterns?
View
5 pages
### Shape colour patterns 5
More patterns, but there may be more than one possible pattern on some.
View
4 pages
### 3 sided shapes
Use a geoboard to explore different triangles.
View
1 pages
### 4 sided shapes
Explore 4 sided shapes with a geoboard.
View
1 pages
### 5 sided shapes
Use a geoboard to find as many 5 sided shapes as possible.
View
1 pages
### Recognise right angles (1)
Find right angles in the world about us.
View
1 pages
### Recognise right angles (2)
Recognise right angles in shapes.
View
4 pages
### Recognise right angles and draw 2D shapes
Recognise and draw right angles and various 2D shapes. A ruler is necessary.
View
4 pages
### What shape am I?
Recognising the properties of triangles, rectangles, pentagons and hexagons.
View
3 pages
### Sorting pentagons and hexagons
Recognising and sorting 5 and 6 sided shapes.
View
3 pages
### Shapes from 4 squares
How many different shapes can be made from joining 4 squares?
View
2 pages
### Shapes from 5 cubes
How many different shapes can be made from 5 cubes?
View
1 pages
### More 2D and 3D shape
Ideas for more 2D and 3D shape work.
View
7 pages
### Identify 2-D Shapes On 3-D Shapes
Recognise and identify common 2-D shapes on the surface of 3-D shapes.
View
4 pages
### Word Problems with Measurement
Word problems using length and mass. All quite straightforward using addition and multiplication.
View
5 pages
### Centimetres
Introducing the centimetre and estimating in centimetres
View
5 pages
### Measure with a metre rule
It's time to go outside and measure strides and standing long jumps!
View
2 pages
### Centimetres (2)
Estimate and measure various household objects using a centimetre ruler or tape measure.
View
3 pages
### Estimating using cubes and matchsticks
Lots of cubes and matchsticks for this tricky little probem.
View
2 pages
### Which unit to measure with
Find the best unit of measurement to use.
View
3 pages
### Symmetry 1
Draw lines of symmetry and complete symmetrical patterns. Quite hard for young children and a mirror is useful!
View
4 pages
### Symmetry 2
Tricky symmetrical patterns.
View
3 pages
### Which unit to measure with (2)
Find things which can be measured using each of the units shown.
View
3 pages
### Measuring with half a litre
Estimating with a half litre.
View
2 pages
View
4 pages
Read numbered divisions on a scale and interpret the divisions beween them.
View
3 pages
Reading scales on a measuring cylinder.
View
3 pages
More on reading scales using litres.
View
2 pages
View
4 pages
More cylinders with scales: going up in fives.
View
5 pages
### Heavier or lighter than a kilo
Heavier or lighter than a kilo
View
2 pages
Weighing scales can often look similar to a clock face, making them tricky to read accurately.
View
4 pages
### Clockwise and anticlockwise turns
Introducing the terms clockwise and anticlockwise for showing turns.
View
5 pages
### Giving directions
Use a map to describe directions.
View
3 pages
### Position on a map
Give directions and answer questions using the Mathstown plan.
View
3 pages
| 917 | 3,821 |
{"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-2013-20
|
latest
|
en
| 0.882358 |
https://onlinejudge.org/board/viewtopic.php?p=56073
| 1,597,141,880,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2020-34/segments/1596439738746.41/warc/CC-MAIN-20200811090050-20200811120050-00189.warc.gz
| 429,792,260 | 12,321 |
## 10090 - Marbles
All about problems in Volume 100. If there is a thread about your problem, please use it. If not, create one with its number in the subject.
Moderator: Board moderators
Jan
Guru
Posts: 1334
Joined: Wed Jun 22, 2005 10:58 pm
Location: Dhaka, Bangladesh
Contact:
helloneo wrote:I made an equation 3x + 4y = 1 (gcd(3, 4))
I got x = -1, y = 1
So, obviously you can say that
43 * 3 * x + 43 * 4 * y = 43
Lets consider that
x = 43*x = -43 and
y = 43*y = 43
Now the equation becomes
3*x + 4*y = 43
Now suppose you want to minimize y. You can rewrite
3*x + 4*y = 43
where x=-39, y=40
or x = -35, y = 37
...
or x = 13, y = 1
Now the rest is upto you. Make sure that both are non-negative.
Ami ekhono shopno dekhi...
HomePage
helloneo
Guru
Posts: 516
Joined: Mon Jul 04, 2005 6:30 am
Location: Seoul, Korea
Thanks for the good explanation..
I'll try it..
jan_holmes
Experienced poster
Posts: 136
Joined: Fri Apr 15, 2005 3:47 pm
Location: Singapore
Contact:
Could anyone of you tell me the output of this :
Code: Select all
``````
1000
21 2
21 3
0
``````
Thx... [/code]
Jan
Guru
Posts: 1334
Joined: Wed Jun 22, 2005 10:58 pm
Location: Dhaka, Bangladesh
Contact:
My accepted code returns...
Output:
Code: Select all
``2 332``
Hope it helps.
Ami ekhono shopno dekhi...
HomePage
devious
New poster
Posts: 13
Joined: Wed Feb 22, 2006 1:26 am
I am confused on how the extended euclidean algorithm can be used to solve this problem.
It would seem you wouldn't need it, considering the values it returns are not the correct ones.
It is simple to come up with the equation 3x + 4y = 43 (from sample input) without the ext. euclid. Am I missing a key point here?
Thanks in advance
aadelf
New poster
Posts: 5
Joined: Fri May 18, 2007 7:30 pm
I tested all sets in this forum and I get the correct answer.
However I get WA online.
I solve it using extended euclid algorithm.
I can not think of any more test cases. Can any provide more?
What happens in case:
28
1 1
2 2
Jan
Guru
Posts: 1334
Joined: Wed Jun 22, 2005 10:58 pm
Location: Dhaka, Bangladesh
Contact:
The case is not valid I think. Make some random cases and check whether you get any negative answer or not. This can be the problem. And you can post your code, too.
Ami ekhono shopno dekhi...
HomePage
aadelf
New poster
Posts: 5
Joined: Fri May 18, 2007 7:30 pm
I agree that the test case I mention shouldn't be correct. But I don't know what to think anymore.
Here is my code.
Code: Select all
``````#include <stdio.h>
int main()
{
long n,c1,c2,n1,n2,swap,T;
long X,Y,preX,preY,a,b,d,temp,gcd,k,m,da,db;
double tc1,tc2;
scanf("%d",&n);
T=1;
while (n!=0) {
scanf("%d %d",&c1,&n1);
scanf("%d %d",&c2,&n2);
preX=1;
X=0;
preY=0;
Y=1;
a=n1;
b=n2;
while (b!=0){
temp = b;
d = (long)a/b;
b = a - d*b;
a = temp;
temp = X;
X = preX - d*X;
preX = temp;
temp = Y;
Y = preY - d*Y;
preY = temp;
}
gcd = a;
a = preX;
b = preY;
k = n/gcd;
da = n2/gcd;
db = n1/gcd;
if ((k*gcd)!=n)
printf("failed\n");
else {
a*=k;
b*=k;
swap=0;
tc1=(double)c1/n1;
tc2=(double)c2/n2;
if ((tc2<tc1)||((tc1==tc2)&&(n1<n2))) {
temp=a;
a=b;
b=temp;
temp=da;
da=db;
db=temp;
swap=1;
}
if (b>=0){
m = b/db;
b = b - m*db;
a = a + m*da;
}
else {
m = -b/db;
if ((b + m*db)<0) m++;
b = b + m*db;
a = a - m*da;
}
if (a<0)
printf("failed\n");
else{
if (swap==1)
printf("%d %d\n",b,a);
else
printf("%d %d\n",a,b);
}
}
T++;
scanf("%d",&n);
}
return 0;
}``````
Jan
Guru
Posts: 1334
Joined: Wed Jun 22, 2005 10:58 pm
Location: Dhaka, Bangladesh
Contact:
Try the cases.
Input:
Code: Select all
``````2601
6 11
32 45
247558756
11479 13625
11229 8731
32661225
5286 422
5398 4247
251001
293 359
388 180
389193984
168 5448
1999 14772
133356304
1870 8365
2572 3204
0``````
Output:
Code: Select all
``````216 5
10383 12151
2953 7397
639 120
71259 66
15892 131``````
Hope these help.
Ami ekhono shopno dekhi...
HomePage
aadelf
New poster
Posts: 5
Joined: Fri May 18, 2007 7:30 pm
These test cases were exactly what I need. I found my bug. Thanks a lot.
andmej
Experienced poster
Posts: 158
Joined: Sun Feb 04, 2007 7:45 pm
Location: Medellin, Colombia
### Re: 10090 - Marbles
I'm getting Wrong Answer.
My code passes all output in the forum.
Please help me find the bug:
Code: Select all
``````using namespace std;
#include <cstdio>
#include <cassert>
#include <algorithm>
long long get_gcd(long long a, long long b, long long &x, long long &y){
if (b > a) return get_gcd(b, a, y, x);
if (b == 0){
x = 1, y = 0;
return a;
}
long long g, sub_x, sub_y;
g = get_gcd(b, a%b, sub_x, sub_y);
x = sub_y, y = sub_x - sub_y*(a/b);
return g;
}
int main(){
long long n, n1, n2, c1, c2;
while (scanf("%lld", &n)==1 && n){
scanf("%lld %lld %lld %lld", &c1, &n1, &c2, &n2);
long long a, b, gcd;
gcd = get_gcd(n1, n2, a, b);
if (n % gcd != 0){
printf("failed\n");
}else{
a *= (n/gcd), b *= (n/gcd);
assert(a*n1 + b*n2 == n);
long long lcm = n1*n2 / gcd;
long long inc_a = lcm/n1, inc_b = lcm/n2;
if (a < 0){
int times = -a/inc_a;
if (-a % inc_a) ++times;
a += times*inc_a, b -= times*inc_b;
}
if (b < 0){
int times = -b/inc_b;
if (-b % inc_b) ++times;
b += times*inc_b, a -= times*inc_a;
}
if (a < 0 || b < 0) printf("failed\n");
else{
pair<long long, long long> ans;
long long best = LLONG_MAX;
if (a*c1 + b*c2 < best){
best = a*c1 + b*c2;
ans = make_pair(a, b);
}
if (a > b){
int times = a/inc_a;
a -= times*inc_a, b += times*inc_b;
if (a*c1 + b*c2 < best){
best = a*c1 + b*c2;
ans = make_pair(a, b);
}
}else{
int times = b/inc_b;
a += times*inc_a, b -= times*inc_b;
if (a*c1 + b*c2 < best){
best = a*c1 + b*c2;
ans = make_pair(a, b);
}
}
printf("%lld %lld\n", ans.first, ans.second);
}
}
}
return 0;
}
``````
Runtime errors in Pascal are reported as Wrong Answers by the online judge. Be careful.
Are you dreaming right now?
http://www.dreamviews.com
fR0D
New poster
Posts: 29
Joined: Mon Feb 11, 2008 5:59 am
Contact:
### Re: 10090 - Marbles
can sumbdy plz elaborate on how to minimise after getting x nd y for ax + by = GCD?
mf
Guru
Posts: 1244
Joined: Mon Feb 28, 2005 4:51 am
Location: Zürich, Switzerland
Contact:
### Re: 10090 - Marbles
In general, a solution to the linear diophantine equation ax + by = c has the form:
x = x0 + b/d t,
y = y0 - a/d t,
where (x0, y0) is any particular solution (which can be found by extended gcd), d = gcd(a, b), and t is an integer parameter.
Let's investigate for which values of t the solution is feasible for our problem (x and y are non-negative):
x0 + b/d t >= 0, y0 - a/d t >= 0, a >= 0, b >= 0, =>
-x0*d/b <= t <= y0*d/a.
And since t must be an integer, we futher have: ceil(-x0*d/b) <= t <= floor(y0*d/a).
So, all the feasible values of t form an interval of consecutive integer values.
Next, let's look at the cost function: c1 x + c2 y = (c1 x0 + c2 y0) + (c1 b/d - c2 a/d) t = (some const) + (another const) * t, i.e. it's linear in t, which means that its minimum and maximums can occur only at the boundary of feasible region.
So, just check two points t=ceil(-x0*d/b), and t=floor(y0*d/a), and choose whichever one is cheaper.
fR0D wrote:can sumbdy plz ...
Oh, for god's sake, use proper English and a spellchecker!
I won't reply to any more posts written in that style.
fR0D
New poster
Posts: 29
Joined: Mon Feb 11, 2008 5:59 am
Contact:
### Re: 10090 - Marbles
thanks mf and i will make sure to write in proper english.
can u explain with an example.
fR0D
New poster
Posts: 29
Joined: Mon Feb 11, 2008 5:59 am
Contact:
### Re: 10090 - Marbles
My program gives correct output for all test cases here still WA.
Code: Select all
``````Got AC
``````
anyone please point out my mistake.
Last edited by fR0D on Fri Dec 12, 2008 12:17 pm, edited 1 time in total.
| 2,758 | 7,692 |
{"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-2020-34
|
latest
|
en
| 0.907719 |
http://mathcentral.uregina.ca/QQ/database/QQ.09.03/ida1.html
| 1,679,777,750,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2023-14/segments/1679296945372.38/warc/CC-MAIN-20230325191930-20230325221930-00300.warc.gz
| 32,605,017 | 1,523 |
Quandaries and Queries what is the radius of a circle with the circumference of 12 inches? Hi, The circumference of a circle is given by circumference = 2 pi radius Hence if the circumference is 12 inches then 12 = 2 pi radius Divide both sides by 2 pi to get radius = 12/2 pi = 6/pi Penny Go to Math Central
| 78 | 308 |
{"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-2023-14
|
latest
|
en
| 0.850786 |
http://pbrc.blogspot.com/2014/05/
| 1,586,113,787,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2020-16/segments/1585371609067.62/warc/CC-MAIN-20200405181743-20200405212243-00474.warc.gz
| 131,323,019 | 13,366 |
## Thursday, May 29, 2014
### Example
(deftest p35-prime-factors
(is (= '(3 3 5 7) (prime-factors 315))))
### Solution
If you have ever done the "Prime Factors Kata" the solution to this problem is probably already in your muscle memory. One reason why this problem works well as an exercise is that there exists a very simple solution to it. In fact the solution can be expressed in just a few lines of code (in almost any language).
This simplest approach is to do trial division with increasing positive integers. You don't even have to bother checking for primality as things just fall into place: dividing by two for example effectively "eliminates" all multiples of two, in the sense that the next successful division can only happen with a number that is not a multiple of two. You do quite a lot of unnecessary divisions in the process though.
In my Clojure solution I chose to calculate a lazy (memoized) sequence of primes first and then do trial division using primes only. This reduces the number of divisions you have to do, assuming the primes are already calculated.
As long as your number is divisible by one of the small primes and does not have too many digits, trial division might be sufficient. For anything else you have to start looking at things like the general number field sieve.
## Thursday, May 01, 2014
### 99 Clojure Problems – 34: Calculate Euler's Totient Function phi(m)
Euler's so-called totient function phi(m) is defined as the number of positive integers r (1 <= r <= m) that are coprime to m.
### Example
(deftest p34-totient
(is (= 8 (totient-euler 20)))
(is (= 4 (totient-euler 5))))
### Solution
There are multiple ways of calculating the totient function. One pretty straightforward way is to check for every positive integer less than m whether it is coprime to m and count their number.
But when reading up on Wikipedia, Euler's classical formula struck me as a use case for Clojure's ratio data type.
It is defined as the sum over all positive devisors d of n: $$\sum_{d\mid n}\varphi(d)=n$$ One way of deriving that formula is to look at all the fractions between 0 and 1 with denominator n. For n = 10: $$\tfrac{1}{10} \tfrac{2}{10} \tfrac{3}{10} \tfrac{4}{10} \tfrac{5}{10} \tfrac{6}{10} \tfrac{7}{10} \tfrac{8}{10} \tfrac{9}{10} \tfrac{10}{10}$$
That is just mapping / over a range of numbers in Clojure.
Put them to lowest terms: $$\tfrac{1}{10} \tfrac{1}{5} \tfrac{3}{10} \tfrac{2}{5} \tfrac{1}{2} \tfrac{3}{5} \tfrac{7}{10} \tfrac{4}{5} \tfrac{9}{10} \tfrac{1}{1}$$ Clojure reduces it's ratios automatically to lowest terms.
We then look at the ratios that still have n as their denominator (here 10): $$\tfrac{1}{10} \tfrac{3}{10} \tfrac{7}{10} \tfrac{9}{10}$$ They turn out to be the ones whose nominators are coprimes of n and their number is the result of the totient function. In Clojure that is just a filter and a count over the result of the application of map.
This solution is probably not the fastest way of calculating the totient but it is faster than calculating coprimes based on the previous solution.
If I interpret the Criterium benchmark results correctly it is about 1.7x faster
ninety-nine-clojure.arithmetic> (bench (totient-euler 2000))
Evaluation count : 70320 in 60 samples of 1172 calls.
Execution time mean : 867.964067 µs
Execution time std-deviation : 14.561162 µs
Execution time lower quantile : 859.192054 µs ( 2.5%)
Execution time upper quantile : 901.487425 µs (97.5%)
Found 7 outliers in 60 samples (11.6667 %)
low-severe 7 (11.6667 %)
Variance from outliers : 6.2554 % Variance is slightly inflated by outliers
nil
ninety-nine-clojure.arithmetic> (bench (totient 2000))
Evaluation count : 41400 in 60 samples of 690 calls.
Execution time mean : 1.496732 ms
Execution time std-deviation : 63.212003 µs
Execution time lower quantile : 1.450532 ms ( 2.5%)
Execution time upper quantile : 1.636526 ms (97.5%)
| 1,096 | 3,918 |
{"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": 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}
| 3.859375 | 4 |
CC-MAIN-2020-16
|
latest
|
en
| 0.922526 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-1-common-core-15th-edition/chapter-7-exponents-and-exponential-functions-7-6-exponential-functions-practice-and-problem-solving-exercises-page-457/36
| 1,566,215,233,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-35/segments/1566027314732.59/warc/CC-MAIN-20190819114330-20190819140330-00075.warc.gz
| 816,163,382 | 14,180 |
## Algebra 1: Common Core (15th Edition)
You are given $g(x)=$ $4 {\times}10^{x}$. Plug in the values $-2,-1,0,1,2,3$ for the X-variable: $g(-2)=$ $4 {\times}10^{-2}={\dfrac{1}{25}}$ $g(-1)=$ $4 {\times}10^{-1}={\dfrac{2}{5}}$ $g(0)=$ $4 {\times}10^{0}=4$ $g(1)=$ $4 {\times}10^{1}=40$ $g(2)=$ $4 {\times}10^{2}=400$ $g(3)=$ $4 {\times}10^{3}=4000$
| 173 | 349 |
{"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}
| 4.15625 | 4 |
CC-MAIN-2019-35
|
latest
|
en
| 0.214145 |
http://sciencepantheism.com/worksheet/math-practice-worksheets-7th-grade.php
| 1,610,746,408,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-04/segments/1610703496947.2/warc/CC-MAIN-20210115194851-20210115224851-00293.warc.gz
| 106,144,077 | 11,666 |
## sciencepantheism.com - the pro math teacher
• Subtraction
• Multiplication
• Division
• Decimal
• Time
• Line Number
• Fractions
• Math Word Problem
• Kindergarten
• a + b + c
a - b - c
a x b x c
a : b : c
# Math Practice Worksheets 7th Grade
Public on 09 Oct, 2016 by Cyun Lee
### practice the order of operations with these free math worksheets
Name : __________________
Seat Num. : __________________
Date : __________________
585 + 21 = ...
629 + 29 = ...
462 + 36 = ...
367 + 21 = ...
596 + 15 = ...
592 + 40 = ...
582 + 50 = ...
111 + 34 = ...
570 + 20 = ...
631 + 47 = ...
742 + 42 = ...
479 + 28 = ...
603 + 40 = ...
730 + 34 = ...
411 + 20 = ...
628 + 43 = ...
916 + 22 = ...
945 + 45 = ...
407 + 42 = ...
853 + 37 = ...
283 + 34 = ...
516 + 33 = ...
250 + 44 = ...
619 + 33 = ...
539 + 46 = ...
559 + 30 = ...
305 + 37 = ...
218 + 26 = ...
278 + 31 = ...
647 + 44 = ...
856 + 48 = ...
787 + 19 = ...
736 + 10 = ...
873 + 37 = ...
846 + 29 = ...
371 + 34 = ...
201 + 14 = ...
762 + 13 = ...
803 + 37 = ...
869 + 17 = ...
166 + 28 = ...
187 + 21 = ...
292 + 43 = ...
517 + 34 = ...
197 + 42 = ...
948 + 33 = ...
430 + 50 = ...
671 + 14 = ...
909 + 46 = ...
841 + 39 = ...
868 + 47 = ...
232 + 18 = ...
902 + 46 = ...
106 + 20 = ...
738 + 47 = ...
414 + 46 = ...
746 + 37 = ...
343 + 25 = ...
622 + 50 = ...
572 + 33 = ...
465 + 11 = ...
728 + 36 = ...
174 + 29 = ...
724 + 10 = ...
574 + 10 = ...
965 + 15 = ...
146 + 21 = ...
336 + 29 = ...
963 + 36 = ...
313 + 32 = ...
755 + 45 = ...
802 + 34 = ...
271 + 39 = ...
947 + 30 = ...
139 + 48 = ...
533 + 14 = ...
384 + 28 = ...
582 + 27 = ...
380 + 36 = ...
425 + 41 = ...
698 + 13 = ...
738 + 37 = ...
676 + 36 = ...
560 + 13 = ...
753 + 20 = ...
650 + 25 = ...
152 + 31 = ...
230 + 30 = ...
270 + 32 = ...
671 + 23 = ...
132 + 29 = ...
996 + 13 = ...
555 + 46 = ...
118 + 44 = ...
815 + 27 = ...
418 + 13 = ...
652 + 16 = ...
638 + 21 = ...
841 + 31 = ...
389 + 20 = ...
964 + 34 = ...
551 + 15 = ...
867 + 24 = ...
954 + 39 = ...
757 + 33 = ...
731 + 31 = ...
638 + 27 = ...
554 + 35 = ...
729 + 31 = ...
757 + 33 = ...
986 + 50 = ...
467 + 32 = ...
314 + 11 = ...
347 + 44 = ...
144 + 14 = ...
623 + 50 = ...
459 + 45 = ...
899 + 15 = ...
587 + 17 = ...
978 + 48 = ...
585 + 48 = ...
820 + 29 = ...
340 + 28 = ...
611 + 45 = ...
383 + 10 = ...
469 + 33 = ...
956 + 37 = ...
455 + 25 = ...
816 + 42 = ...
475 + 21 = ...
405 + 44 = ...
393 + 30 = ...
197 + 29 = ...
140 + 38 = ...
777 + 22 = ...
552 + 22 = ...
264 + 24 = ...
810 + 29 = ...
420 + 37 = ...
399 + 15 = ...
274 + 43 = ...
500 + 38 = ...
349 + 19 = ...
799 + 41 = ...
490 + 26 = ...
311 + 26 = ...
566 + 13 = ...
875 + 15 = ...
748 + 43 = ...
722 + 24 = ...
675 + 48 = ...
273 + 32 = ...
951 + 30 = ...
240 + 43 = ...
778 + 33 = ...
223 + 14 = ...
899 + 40 = ...
398 + 22 = ...
979 + 47 = ...
384 + 11 = ...
230 + 26 = ...
860 + 17 = ...
871 + 21 = ...
420 + 43 = ...
890 + 49 = ...
665 + 38 = ...
963 + 50 = ...
330 + 14 = ...
844 + 38 = ...
247 + 38 = ...
974 + 36 = ...
837 + 25 = ...
571 + 31 = ...
501 + 16 = ...
368 + 26 = ...
713 + 21 = ...
757 + 26 = ...
610 + 29 = ...
197 + 37 = ...
952 + 19 = ...
258 + 46 = ...
150 + 44 = ...
405 + 37 = ...
835 + 45 = ...
326 + 31 = ...
156 + 37 = ...
926 + 39 = ...
617 + 32 = ...
951 + 13 = ...
345 + 16 = ...
698 + 14 = ...
740 + 24 = ...
169 + 50 = ...
289 + 31 = ...
616 + 19 = ...
801 + 28 = ...
990 + 46 = ...
482 + 34 = ...
326 + 45 = ...
559 + 50 = ...
show printable version !!!hide the show
## RELATED POST
Not Available
## POPULAR
worksheet math
identify fractions on a number line worksheet
sequencing worksheet kindergarten
spot the difference worksheets for kindergarten
maths circles worksheets
| 1,364 | 3,900 |
{"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}
| 3.640625 | 4 |
CC-MAIN-2021-04
|
latest
|
en
| 0.17787 |
https://www.physicsforums.com/threads/what-are-the-effective-and-total-deviator-stresses-in-saturated-silty-clay.696018/
| 1,718,468,452,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-26/segments/1718198861605.77/warc/CC-MAIN-20240615155712-20240615185712-00661.warc.gz
| 831,513,064 | 16,360 |
# What Are the Effective and Total Deviator Stresses in Saturated Silty Clay?
• db725
In summary, to solve this problem, you need to understand the given information and what is being asked. Then, use the equations and concepts related to effective and total stresses, deviator stress, and porewater pressure to calculate the effective and total deviator stresses at failure, as well as the load needed to cause failure. Make sure to double check your units and answers. Following a systematic and logical approach will help you solve this problem successfully. Good luck on your exam!
db725
## Homework Statement
A saturated over-consolidated silty clay is estimated to have the effective strength parameters
c'=5 kPa
∅'=25 kPa
A specimen of the silty is mounted into a triaxial cell and full consolidated to 400 kPa with a cell pressure of 500 kPa and a back pressure of 100 kPa. With the drainage then turned off, the specimen is slowly loaded axially failure. At failure, the porewater pressure was measured to be 160 kPa and the diameter of the specimen was calculated to be 103.47 mm.
Estimate
1. the effective deviator stress at failure (σ1'-σ3')
2. the total deviator stress at failure (σ1-σ3), and
3. the load which would be needed to be applied to the loading piston to cause failure.
σ1=σ'1+u
## The Attempt at a Solution
cell pressure =500 kPa, back pressure=100 kPa, σ3= 500-100=400 kPa
u(failure)=160 kPa
Area at failure = (∏ x diameter^2)/4=(∏x103.47^2)/4=8408.50 mm^2
1. Deviator stress= deviator load/area= (400 x 10^3)/8408.50=47.57 kPa
σ'3 at failure = 400-160=240 kPa
σ'1 at failure= σ'3 +deviator stress = 240 +47.57=287.57 kPa
2. σ1=σ1'+u=287.57+160=447.57 kPa σ3=σ3'+u =240+160=400 kPa
3. Not sure how to do this. Was thinking of drawing the mohr circles using the previous calculations. However I feel like there is something wrong with the above as when I sketched the mohr circles I got a strange circle.
I would greatly appreciate any help on this question as I am getting prepared for a final exam in my engineering degree. My exam is in 2 days and I would really like to know if I am on the right track or not. Thanks in advance.
it is important to approach problems in a systematic and logical manner. Here are some steps you can take to solve this problem:
1. First, make sure you understand the given information and what is being asked. The problem gives you the effective strength parameters of the soil, the conditions under which the specimen was tested, and the measurements taken during the test. You are asked to estimate the effective and total deviator stresses at failure, as well as the load needed to cause failure.
2. Next, review the equations and concepts related to effective and total stresses, deviator stress, and porewater pressure. Make sure you understand how these values are calculated and what they represent.
3. Then, start by calculating the effective deviator stress at failure. This can be done by using the given information on the cell pressure, back pressure, and porewater pressure, as well as the formula for deviator stress. Make sure your units are consistent.
4. Next, use the effective deviator stress to calculate the effective principal stresses at failure. This can be done using the Mohr-Coulomb failure criterion, which states that the failure occurs when the deviator stress equals the effective cohesion plus the effective stress multiplied by the effective friction angle. Make sure your units are consistent.
5. Once you have the effective principal stresses, you can use them to calculate the total principal stresses at failure. This can be done by adding the porewater pressure to the effective principal stresses.
6. Finally, to calculate the load needed to cause failure, you can use the formula for deviator stress and rearrange it to solve for the deviator load. This load should be applied to the loading piston, so make sure you use the correct area in your calculation.
7. Once you have completed all the calculations, make sure to double check your units and final answers to ensure they are correct.
I hope this helps you in solving the problem. Remember to approach it step by step and use the equations and concepts you have learned in your studies. Good luck on your exam!
## What is soil mechanics?
Soil mechanics is a branch of geotechnical engineering that studies the behavior of soil under different conditions, such as loading or moisture content. It involves the application of principles from mechanics, physics, and mathematics to understand the properties and behavior of soil.
## Why is soil mechanics important?
Soil mechanics is important because soil is the foundation of all civil engineering projects. Understanding the properties and behavior of soil is crucial in designing safe and stable structures such as buildings, roads, and bridges. It also plays a significant role in environmental and geotechnical engineering projects.
## What are the main factors that affect soil behavior?
The main factors that affect soil behavior are its composition, moisture content, density, and loading conditions. The type of soil, its particle size distribution, and its mineralogy can greatly influence its properties. The amount of water in the soil also affects its strength and compressibility. The density of the soil is also important, as it determines its ability to support loads. Lastly, the type and magnitude of external loads, such as building or vehicle loads, can significantly impact soil behavior.
## What are the different types of soil tests used in soil mechanics?
There are several types of soil tests used in soil mechanics, including classification tests, compaction tests, and strength tests. Classification tests determine the type and properties of soil, such as its particle size distribution and plasticity. Compaction tests measure the maximum density and water content of soil to determine its suitability for construction. Strength tests, such as triaxial and compression tests, assess the strength and stiffness of soil under different conditions.
## How can soil mechanics be applied in real-world projects?
Soil mechanics can be applied in real-world projects by conducting site investigations and soil testing to understand the properties and behavior of the soil at a construction site. This information is then used to design foundations, slopes, and retaining structures that can withstand the loads and environmental conditions at the site. Soil mechanics is also used to evaluate and mitigate potential hazards, such as landslides or liquefaction, and to monitor the performance of structures over time.
| 1,427 | 6,678 |
{"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-2024-26
|
latest
|
en
| 0.902372 |
http://www.solitaryroad.com/c159.html
| 1,508,332,027,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-43/segments/1508187822966.64/warc/CC-MAIN-20171018123747-20171018143747-00840.warc.gz
| 584,360,214 | 3,476 |
```Website owner: James Miller
```
[ Home ] [ Up ] [ Info ] [ Mail ]
Ways of viewing Y = AX
The linear system
or, more concisely,
Y = AX
can be viewed in different ways. One may have seen trick drawings in which one first sees the head of a monkey and then on a second look he sees an elephant. Y = AX is similar. It can be viewed in different ways. Sometimes one thinks of it one way, sometimes another. Some of the ways it can be viewed are:
1. As a change in variables from the variables y1, y2, ..., yn of the vector Y to the variables x1, x2, ..., xn of the vector X.
2. As a point transformation that maps points (and figures) from one space into another space (or the same space). It effects a linear transformation on the points in the space and when viewed in this way as a point transformation we call it a “linear point transformation” to emphasize how we are viewing it.
3. As a change to an oblique coordinate system which has a different metric than the usual rectangular Cartesian system — with the unit of length varying from one axis to another. Changing notation from Y = AX to X = A X' with X' referring to the new oblique system and X referring to the original system the linear transformation X = AX' can be viewed as a change to an oblique system in which the elementary unit vectors of the new oblique system correspond to the column vectors of matrix A. Stated in different words, we change to a different basis, the new basis corresponding to an oblique coordinate system. Thus instead of thinking of the figure as changing as we do in View 2, we think of the figure as remaining as it is and the coordinate system in which the figure is expressed as changing.
| 383 | 1,695 |
{"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}
| 3.828125 | 4 |
CC-MAIN-2017-43
|
latest
|
en
| 0.93988 |
https://gmatclub.com/forum/if-840-is-21-percent-of-c-what-is-47-of-c-282492.html
| 1,558,312,776,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-22/segments/1558232255251.1/warc/CC-MAIN-20190520001706-20190520023706-00265.warc.gz
| 491,290,490 | 142,200 |
GMAT Question of the Day - Daily to your Mailbox; hard ones only
It is currently 19 May 2019, 17:39
### 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
# If 840 is 21 percent of c, what is 47% of c?
Author Message
TAGS:
### Hide Tags
Math Expert
Joined: 02 Sep 2009
Posts: 55150
If 840 is 21 percent of c, what is 47% of c? [#permalink]
### Show Tags
25 Nov 2018, 23:42
00:00
Difficulty:
15% (low)
Question Stats:
100% (01:03) correct 0% (00:00) wrong based on 18 sessions
### HideShow timer Statistics
If 840 is 21 percent of c, what is 47% of c?
A. 705
B. 1,175
C. 1,410
D. 1,640
E. 1,880
_________________
Manager
Joined: 19 Apr 2017
Posts: 78
Concentration: General Management, Sustainability
Schools: ESSEC '22
GPA: 3.9
WE: Operations (Hospitality and Tourism)
Re: If 840 is 21 percent of c, what is 47% of c? [#permalink]
### Show Tags
25 Nov 2018, 23:53
If 840 is 21 percent of c, what is 47% of c?
840 =21%C
C=840*100/21
C =4000
47%*4000 = 1880 =E
Director
Joined: 18 Jul 2018
Posts: 897
Location: India
Concentration: Finance, Marketing
WE: Engineering (Energy and Utilities)
Re: If 840 is 21 percent of c, what is 47% of c? [#permalink]
### Show Tags
25 Nov 2018, 23:54
840 = 21/100 C
C = 40*100 = 4000
47/100*4000 = 40*47 = 1880
_________________
Press +1 Kudo If my post helps!
Re: If 840 is 21 percent of c, what is 47% of c? [#permalink] 25 Nov 2018, 23:54
Display posts from previous: Sort by
| 612 | 1,869 |
{"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-2019-22
|
latest
|
en
| 0.867748 |
https://im.kendallhunt.com/HS/students/2/6/8/practice.html
| 1,659,940,579,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2022-33/segments/1659882570767.11/warc/CC-MAIN-20220808061828-20220808091828-00492.warc.gz
| 290,410,111 | 16,802 |
# Lesson 8
Equations and Graphs
• Let’s write an equation for a parabola.
### Problem 1
Classify the graph of the equation $$x^2+y^2-8x+4y=29$$.
A:
circle
B:
exponential curve
C:
line
D:
parabola
### Problem 2
Write an equation that states $$(x,y)$$ is the same distance from $$(4,1)$$ as it is from the $$x$$-axis.
### Problem 3
Select all equations which describe the parabola with focus $$(\text- 1,\text- 7)$$ and directrix $$y=3$$.
A:
$$(x-1)^2+(y-7)^2=(y+3)^2$$
B:
$$(x+1)^2+(y+7)^2=(y-3)^2$$
C:
$$y=\text{-}20(x+1)^2-2$$
D:
$$y=\text{-}20(x+1)^2+2$$
E:
$$y=\text{-}\frac{1}{20}(x+1)^2-2$$
F:
$$y=\text{-}\frac{1}{20}(x+1)^2+2$$
### Problem 4
Parabola A and parabola B both have the $$x$$-axis as the directrix. Parabola A has its focus at $$(3,2)$$ and parabola B has its focus at $$(5,4)$$. Select all true statements.
A:
Parabola A is wider than parabola B.
B:
Parabola B is wider than parabola A.
C:
The parabolas have the same line of symmetry.
D:
The line of symmetry of parabola A is to the right of that of parabola B.
E:
The line of symmetry of parabola B is to the right of that of parabola A.
(From Unit 6, Lesson 7.)
### Problem 5
A parabola has focus $$(5,1)$$ and directrix $$y = \text{-}3$$. Where is the parabola’s vertex?
(From Unit 6, Lesson 7.)
### Problem 6
Select the value needed in the box in order for the expression to be a perfect square trinomial.
$$x^2+7x+\boxed{\phantom{3}}$$
A:
3.5
B:
7
C:
12.25
D:
14.5
(From Unit 6, Lesson 6.)
### Problem 7
Rewrite each expression as the product of 2 factors.
1. $$x^2+3x$$
2. $$x^2-6x-7$$
3. $$x^2+4x+4$$
(From Unit 6, Lesson 5.)
### Problem 8
Suppose this two-dimensional figure is rotated 360 degrees using the vertical axis shown. Each small square on the grid represents 1 square inch. What is the volume of the three-dimensional figure?
(From Unit 5, Lesson 15.)
| 690 | 1,903 |
{"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}
| 4.28125 | 4 |
CC-MAIN-2022-33
|
latest
|
en
| 0.782056 |
https://www.gradesaver.com/textbooks/math/trigonometry/CLONE-68cac39a-c5ec-4c26-8565-a44738e90952/chapter-5-trigonometric-identities-section-5-6-half-angle-identities-5-6-exercises-page-242/5
| 1,726,211,478,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725700651510.65/warc/CC-MAIN-20240913070112-20240913100112-00552.warc.gz
| 726,772,181 | 12,950 |
## Trigonometry (11th Edition) Clone
$$\sin15^\circ=\frac{\sqrt{2-\sqrt3}}{2}$$ 5 comes with C.
$$\sin15^\circ$$ $15^\circ$ can be written as half of the angle $30^\circ$. $$15^\circ=\frac{30^\circ}{2}$$ Thus, $$\sin15^\circ=\sin\Big(\frac{30^\circ}{2}\Big)$$ We can apply here the Half-Angle Identity for sine, which states $$\sin\Big(\frac{A}{2}\Big)=\pm\sqrt{\frac{1-\cos A}{2}}$$ with $A=30^\circ$: $$\sin15^\circ=\pm\sqrt{\frac{1-\cos30^\circ}{2}}$$ $15^\circ$ is in quadrant I. As $\sin\theta\gt0$ as $\theta$ is in quadrant I, $\sin15^\circ\gt0$. So we would pick the positive square root. $$\sin15^\circ=\sqrt{\frac{1-\frac{\sqrt3}{2}}{2}}$$ $$\sin15^\circ=\sqrt{\frac{\frac{2-\sqrt3}{2}}{2}}$$ $$\sin15^\circ=\sqrt{\frac{2-\sqrt3}{4}}$$ $$\sin15^\circ=\frac{\sqrt{2-\sqrt3}}{2}$$ Therefore, 5 needs to be matched with C.
| 319 | 830 |
{"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-2024-38
|
latest
|
en
| 0.593031 |
https://www.geeksforgeeks.org/triangle-inequality/
| 1,713,553,280,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-18/segments/1712296817442.65/warc/CC-MAIN-20240419172411-20240419202411-00286.warc.gz
| 708,628,045 | 51,737 |
# Triangle Inequality Theorem, Proof & Applications
Last Updated : 09 Apr, 2024
Triangle Inequality is the relation between the sides and angles of triangles which helps us understand the properties and solutions related to triangles. Triangles are the most fundamental geometric shape as we can’t make any closed shape with two or one side. Triangles consist of three sides, three angles, and three vertices.
The construction possibility of a triangle based on its side is given by the theorem named “Triangle Inequality Theorem.” The Triangle Inequality Theorem states the inequality relation between the triangle’s three sides. In this article, we will explore the Triangle Inequality Theorem and some of its applications as well as the other various inequalities related to the sides and angles of triangles.
In this article, we’ll delve into the concept of triangle inequality, the triangle inequality theorem, its significance, and its practical applications.
## What is Triangle Inequality?
Triangle Inequality is a fundamental geometric principle that plays a vital role in various mathematical and real-world applications. It lays the foundation for understanding relationships between the sides of a triangle, contributing to fields such as geometry, physics, and computer science.
## Triangle Inequality Theorem
Triangle Inequality Theorem states that “the sum of the length of any two sides of a triangle must be greater than the length of the third side.” If the sides of a triangle are a, b, and c then the Triangle Inequality Theorem can be represented mathematically as:
• a + b > c,
• b + c > a,
• c + a > b
## Triangle Inequality Proof
In this section, we will learn the proof of the triangle inequality theorem. To prove the theorem, assume there is a triangle ABC in which side AB is produced to D and CD is joined.
### Triangle Inequality Theorem Proof:
Notice that the side BA of Δ ABC has been produced to a point D such that AD = AC. Now, since ∠BCD > ∠BDC.Â
By the properties mentioned above, we can conclude that BD > BC.Â
We know that, BD = BA + AD
So, BA + AD > BCÂ
= BA + AC > BCÂ
So, this proved sum of two sides triangle is always greater than the other side. Â
Let’s see an example based on Triangle Inequality Theorem to understand its concept more clearly.
Example: D is a point on side BC of triangle ABC such that AD = DC. Show that AB > BD.Â
Solution:
In triangle DAC,Â
∠ADC = ∠ACD (Angles opposite to equal sides)
∠ADC is an exterior angle for ΔABD.
⇒ ∠ACB > ∠ABC
⇒ AB > AC (Side opposite to larger angle in Δ ABC)
## Triangle Inequality Theorem – Applications & Uses
There are many applications in the geometry of the Triangle Inequality Theorem, some of those applications are as follows:
• To Identify the Triangles
• To Find the Range of Possible Values of the Sides of Triangles
### How to Identify Triangles
To Identify the possibility of the construction of any given triangle with three sides, we can use the Triangle Inequality Theorem. If the given three sides satisfy the theorem, then the construction of this triangle is possible.Â
For example, consider the sides of the triangle as 4 units, 5 units, and 7 units.Â
As 4 + 5 > 7, 5 + 7 > 4, and 7 + 4 > 5. Thus, the triangle with sides 4 unit, 5 units, and 7 units is possible to construct.
Now take another example for not possible construction, consider the sides of the triangle to be 3 units, 4 units, and 9 units.
 As 3 + 4 [Tex]\ngtr       [/Tex] 9, therefore triangle with sides 3 units, 4 units and 9 units is not possible.
### How to Find Range of Possible Values of Sides of Triangle
To find the range of possible values of the sides of a triangle when two sides are given as units and b units, you can follow these steps:
Step 1: Let’s assume the third side be x units.
Step 2: Now, using Triangle Inequality Theorem we know,Â
a + b > x, a + x > b, and b + x > a
Step 3: Using all three conditions we can find the range for the third side.
Let’s consider an example for better understanding.
Example: Find the range for the third side of the triangle if the first two sides are 4 units and 7 units.
Solution:
Let’s assume the third side be x units.
Using Triangle Inequality Theorem, we get
4 + 7 = 11 > x, 4 + x > 7, and 7 + x > 4
Simplifying the above inequalities, we get
11 > x and x > 3.
Thus, possible range for the third sides is 3 < x < 11.
## Various Inequalities in Triangle
If the a, b, and c are the sides of the triangle, then the following inequalities hold:
• [Tex]\frac{3}{2} \leq \frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}<2 [/Tex]
• If angle C is obtuse (greater than 90°) then a2+b2<c2
• If C is acute (less than 90°) then a2+b2>c2
• [Tex]Â a b c \geq(a+b-c)(a-b+c)(-a+b+c) [/Tex]
• [Tex]a^2 b(a-b)+b^2 c(b-c)+c^2 a(c-a) \geq 0 [/Tex]
• For A.M., G.M., and H.M. we know that H.M. ≤ G.M. ≤ A.M.Â
[Tex]\frac{3abc}{ab+bc+ac}\leq \sqrt[3]{abc} \leq \frac{a+b+c}{3} [/Tex]
If a, b, and c are sides of the triangle and A, B, and C are the interior angles opposite to those sides respectively then the following inequalities hold:
• [Tex]\cos A+\cos B+\cos C \leq \frac{3}{2} [/Tex]
• [Tex](1-\cos A)(1-\cos B)(1-\cos C) \geq \cos A \cdot \cos B \cdot \cos C [/Tex]
• [Tex]\sin A+\sin B+\sin C \leq \frac{3 \sqrt{3}}{2} [/Tex]
• [Tex]\sin ^2 A+\sin ^2 B+\sin ^2 C \leq \frac{9}{4} [/Tex]
• Using, A.M. G.M. inequality, [Tex]\sqrt[3]{\sin A \cdot \sin B \cdot \sin C} \leq \frac{\sin A+\sin B+\sin C}{3}       [/Tex], thus
[Tex]\sin A \cdot \sin B \cdot \sin C \leq\left(\frac{\sin A+\sin B+\sin C}{3}\right)^3 \leq\left(\sin \frac{A+B+C}{3}\right)^3=\sin ^3\left(\frac{\pi}{3}\right)=\frac{3 \sqrt{3}}{8} [/Tex]
• [Tex]\sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2} \leq \frac{1}{8} [/Tex]
• [Tex]\tan ^2 \frac{A}{2}+\tan ^2 \frac{B}{2}+\tan ^2 \frac{C}{2} \geq 1 [/Tex]
• [Tex]\cot A+\cot B+\cot C \geq \sqrt{3}Â [/Tex]
• [Tex]\sin A \cdot \cos B+\sin B \cdot \cos C+\sin C \cdot \cos A \leq \frac{3 \sqrt{3}}{4} [/Tex]
## Sample Problems on Triangle Inequality
Problem 1: Determine whether the given set of side lengths can form a triangle according to the triangle inequality theorem.
a) 3, 4, 9
b) 5, 7, 12
c) 6, 10, 25
Solution:
For any triangle with sides a, b and c, using the triangle inequality theorem we get.
• a + b > c,
• b + c > a,
• c + a > b
We can use this to determine whether a triangle can be formed or not.
a) 3, 4, 9
As 3 + 4 < 9,
Thus, this triangle can’t be formed.
b) 5, 7, 12
As 5 + 7 = 12,
Thus, this triangle also can’t be formed.
c) 6, 10, 25
As 6 + 10 < 25,
Thus, this triangle also can’t be formed.
All the given triangles do not satisfy the triangle inequality theorem.
Problem 2: If the two sides of a triangle are 3 and 5. Find all the possible lengths of the third side.
Solution:
As we know, that for any triangle with sides a, b and c,
a – b < c < a + b
Let he unknown side be x,
Thus, 5 – 3 < x < 5 + 3
⇒ 2 < x < 8
Thus, third side can have any value between 2 and 8.
## Practice Problem on Triangle Inequality
Problem 1: Given a triangle with sides of lengths 5 cm, 8 cm, and 12 cm, determine whether the triangle satisfies the triangle inequality theorem.
Problem 2: Suppose you have a triangle with sides of lengths 7 inches, 10 inches, and 28 inches. Can this triangle exist?
Problem 3: Three sticks have lengths 9 cm, 15 cm, and 20 cm. Can we form a triangle using these sticks?
## Summary – Triangle Inequality Theorem, Proof & Applications
The Triangle Inequality Theorem is a foundational concept in geometry that elucidates the relationships between the lengths of the sides of a triangle. It asserts that the sum of the lengths of any two sides of a triangle must always exceed the length of the third side. This principle is mathematically articulated through the inequalities: a + b > c, b + c > a, and c + a > b, where a, b, and c represent the sides of the triangle. The proof of the theorem involves extending a side of a triangle and applying the properties of angles and sides to establish the inequality relation. The theorem not only facilitates the validation of the possibility of constructing a triangle given three lengths but also enables the determination of the range of possible values for the sides of a triangle. Its applications are vast and varied, including the identification of triangles, the determination of possible side lengths, and providing foundational knowledge that underpins further study in geometry, physics, and computer science. Through examples and proofs, the significance of the Triangle Inequality Theorem in both theoretical and practical contexts is made evident, showcasing its role in understanding and solving problems related to triangles.
## FAQs on Triangle Inequality Theorem, Proof & Applications
### What is the Triangle Inequality Theorem?
Triangle Inequality Theorem states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.
### Can an Equilateral Triangle Violate the Triangle Inequality Theorem?
No, equilatral triangle can’t voilate the Triangle Inequality Theorem. As all sides in Equilateral Triangle are equal and sum of any two sides is twice the third i.e., greater than the third side.
### What is the Converse of the Triangle Inequality Theorem?
Converse of the Triangle Inequality Theorem states that if sum of the lengths of any two sides of a triangle is greater than the length of third side, then given three sides can form a triangle.
### What is the Triangle Inequality for Angles?
 The Triangle Inequality for Angles states that the sum of any two angles in a triangle must be greater than the measure of the third angle.
### What is the Relationship between the Sides of a Right Triangle?
The relationship between the sides of a right-angle triangle is given by the Pythagoras Theorem. If the sides of right-angle triangle are a, b, and c (c is the greatest side) then by Pythagoras theorem:
a2 + b2 = c2
Previous
Next
| 2,780 | 10,095 |
{"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": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0}
| 4.78125 | 5 |
CC-MAIN-2024-18
|
latest
|
en
| 0.943037 |
https://www.mathtestpreparation.com/Exterior-angle-of-a-triangle-example.html
| 1,569,127,001,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-39/segments/1568514575076.30/warc/CC-MAIN-20190922032904-20190922054904-00373.warc.gz
| 948,703,444 | 1,799 |
triangle exterior angles | mathtestpreparation.com
back to geometry
# Exterior Angle of a Triangle Example
Question:
In the figure below, points A, B, C lie on a straight line, DBC is a triangle. If the angle ABD is 136o, what is the value of x?
Solution:
Since points A, B, and C lie on a straight line, so angle ABD is an exterior angle of triangle DBC
angle ABD = angle D + angle C (Exterior angle of a triangle theorem)
Given: angle D = xo and angle C = xo
136 = x + x
2x = 136
x = 68
The value of x is 68.
When extended one side of a triangle, an exterior angle of a triangle is formed. The exterior angle of a triangle is equal to the sum of two noadjacent interior angles. In this problem, we get angle ABD = angle D + angle C. Given angle ABD is 136o, then x + x = 136. 2x = 136. x = 68. So the value od x is 68. That is, the angle C is equal to angle D which is 68o.
| 253 | 880 |
{"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}
| 4.25 | 4 |
CC-MAIN-2019-39
|
longest
|
en
| 0.786512 |
https://www.scienceforums.net/topic/120323-can-you-find-the-number/?tab=comments
| 1,596,888,530,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2020-34/segments/1596439737645.2/warc/CC-MAIN-20200808110257-20200808140257-00427.warc.gz
| 836,746,297 | 16,560 |
# Can you find the Number ?
## Recommended Posts
There is a four-digit number ABCD, where A, B, C, D each represents a different digit from 1 to 9.
ABCD is divisible by 13, BCDA is divisible by 11, CDAB is divisible by 9, and DABC is divisible by 7.
Can you find the original number ABCD ?
##### Share on other sites
9 hours ago, Commander said:
Can you find the original number ABCD ?
Spoiler
3861/13=297
8613/11=783
6138/9=558
1386/7=198
##### Share on other sites
16 hours ago, Ghideon said:
Ghideon :
Absolutely Right ! Well done !!
##### Share on other sites
1 hour ago, Commander said:
Well done !!
Thanks!
Below is a hint if someone should be interested:
Spoiler
CDAB is divisible by 9 means that ABCD, BCDA and DABC are also divisible by 9.
13, 11 and 7 are primes
so:
ABCD is divisible by 13x9, BCDA is divisible by 11x9 and DABC is divisible by 7x9.
##### Share on other sites
On 10/15/2019 at 12:26 PM, Commander said:
There is a four-digit number ABCD, where A, B, C, D each represents a different digit from 1 to 9.
Since original puzzle has been solved, I have puzzle for you Commander: what four-digit number will be solution, if any digit might be used multiple times.
##### Share on other sites
Sensei only 7722 will make it !
Which is also double of 3861
Edited by Commander
##### Share on other sites
14 hours ago, Commander said:
Sensei only 7722 will make it !
Which is also double of 3861
#include <stdio.h>
int get( int i, int j ) {
j = 3 - j;
for( ; j > 0; j-- ) i /= 10;
return( i % 10 );
}
int swap( int i, int a, int b, int c, int d )
{
int result;
result = get( i, a ) * 1000;
result += get( i, b ) * 100;
result += get( i, c ) * 10;
result += get( i, d ) * 1;
return( result );
}
bool check( int i, bool verbose = false ) {
if( get( i, 0 ) == 0 ) return( false );
if( get( i, 1 ) == 0 ) return( false );
if( get( i, 2 ) == 0 ) return( false );
if( get( i, 3 ) == 0 ) return( false );
int j;
j = swap( i, 0, 1, 2, 3 );
if( verbose ) printf( "ABCD %d\n", j );
if( ( i % 13 ) != 0 ) return( false );
j = swap( i, 1, 2, 3, 0 );
if( verbose ) printf( "BCDA %d\n", j );
if( ( j % 11 ) != 0 ) return( false );
j = swap( i, 2, 3, 0, 1 );
if( verbose ) printf( "CDAB %d\n", j );
if( ( j % 9 ) != 0 ) return( false );
j = swap( i, 3, 0, 1, 2 );
if( verbose ) printf( "DABC %d\n", j );
if( ( j % 7 ) != 0 ) return( false );
return( true );
}
int main() {
for( int i = 0; i < 10000; i++ ) {
if( check( i ) ) {
printf( "Result %d\n", i );
check( i, true );
}
}
}
Edited by Sensei
## Create an account
Register a new account
| 876 | 2,579 |
{"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}
| 3.984375 | 4 |
CC-MAIN-2020-34
|
latest
|
en
| 0.7576 |
https://nrich.maths.org/public/leg.php?code=-100&cl=1&cldcmpid=5450
| 1,566,431,498,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-35/segments/1566027316549.78/warc/CC-MAIN-20190821220456-20190822002456-00309.warc.gz
| 576,637,802 | 9,863 |
# Search by Topic
#### Resources tagged with Trial and improvement similar to Difficulties with Division:
Filter by: Content type:
Age range:
Challenge level:
### There are 84 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Trial and improvement
### Clever Santa
##### Age 7 to 11 Challenge Level:
All the girls would like a puzzle each for Christmas and all the boys would like a book each. Solve the riddle to find out how many puzzles and books Santa left.
### 1, 2, 3, 4, 5
##### Age 7 to 11 Challenge Level:
Using the numbers 1, 2, 3, 4 and 5 once and only once, and the operations x and ÷ once and only once, what is the smallest whole number you can make?
### Zios and Zepts
##### Age 7 to 11 Challenge Level:
On the planet Vuv there are two sorts of creatures. The Zios have 3 legs and the Zepts have 7 legs. The great planetary explorer Nico counted 52 legs. How many Zios and how many Zepts were there?
### Escape from the Castle
##### Age 7 to 11 Challenge Level:
Skippy and Anna are locked in a room in a large castle. The key to that room, and all the other rooms, is a number. The numbers are locked away in a problem. Can you help them to get out?
### Path to the Stars
##### Age 7 to 11 Challenge Level:
Is it possible to draw a 5-pointed star without taking your pencil off the paper? Is it possible to draw a 6-pointed star in the same way without taking your pen off?
### Factor-multiple Chains
##### Age 7 to 11 Challenge Level:
Can you see how these factor-multiple chains work? Find the chain which contains the smallest possible numbers. How about the largest possible numbers?
### Grouping Goodies
##### Age 5 to 7 Challenge Level:
Pat counts her sweets in different groups and both times she has some left over. How many sweets could she have had?
### Number Juggle
##### Age 7 to 11 Challenge Level:
Fill in the missing numbers so that adding each pair of corner numbers gives you the number between them (in the box).
##### Age 7 to 11 Challenge Level:
Use the information to work out how many gifts there are in each pile.
### Sisters and Brothers
##### Age 7 to 11 Challenge Level:
Cassandra, David and Lachlan are brothers and sisters. They range in age between 1 year and 14 years. Can you figure out their exact ages from the clues?
### Buckets of Thinking
##### Age 7 to 11 Challenge Level:
There are three buckets each of which holds a maximum of 5 litres. Use the clues to work out how much liquid there is in each bucket.
### In the Money
##### Age 7 to 11 Challenge Level:
One quarter of these coins are heads but when I turn over two coins, one third are heads. How many coins are there?
### Rocco's Race
##### Age 7 to 11 Short Challenge Level:
Rocco ran in a 200 m race for his class. Use the information to find out how many runners there were in the race and what Rocco's finishing position was.
### Tom's Number
##### Age 7 to 11 Challenge Level:
Work out Tom's number from the answers he gives his friend. He will only answer 'yes' or 'no'.
### Oranges and Lemons
##### Age 7 to 11 Challenge Level:
On the table there is a pile of oranges and lemons that weighs exactly one kilogram. Using the information, can you work out how many lemons there are?
### What Two ...?
##### Age 7 to 11 Short Challenge Level:
56 406 is the product of two consecutive numbers. What are these two numbers?
### The Clockmaker's Birthday Cake
##### Age 7 to 11 Challenge Level:
The clockmaker's wife cut up his birthday cake to look like a clock face. Can you work out who received each piece?
### How Many Eggs?
##### Age 7 to 11 Challenge Level:
Peter, Melanie, Amil and Jack received a total of 38 chocolate eggs. Use the information to work out how many eggs each person had.
### Rabbits in the Pen
##### Age 7 to 11 Challenge Level:
Using the statements, can you work out how many of each type of rabbit there are in these pens?
### Magic Circles
##### Age 7 to 11 Challenge Level:
Put the numbers 1, 2, 3, 4, 5, 6 into the squares so that the numbers on each circle add up to the same amount. Can you find the rule for giving another set of six numbers?
### Number Round Up
##### Age 5 to 7 Challenge Level:
Arrange the numbers 1 to 6 in each set of circles below. The sum of each side of the triangle should equal the number in its centre.
### Arranging the Tables
##### Age 7 to 11 Challenge Level:
There are 44 people coming to a dinner party. There are 15 square tables that seat 4 people. Find a way to seat the 44 people using all 15 tables, with no empty places.
### Cycling Squares
##### Age 7 to 11 Challenge Level:
Can you make a cycle of pairs that add to make a square number using all the numbers in the box below, once and once only?
### One Wasn't Square
##### Age 7 to 11 Challenge Level:
Mrs Morgan, the class's teacher, pinned numbers onto the backs of three children. Use the information to find out what the three numbers were.
### A Numbered Route
##### Age 7 to 11 Challenge Level:
Can you draw a continuous line through 16 numbers on this grid so that the total of the numbers you pass through is as high as possible?
### The Brown Family
##### Age 5 to 7 Challenge Level:
Use the information about Sally and her brother to find out how many children there are in the Brown family.
### Let's Face It
##### Age 7 to 11 Challenge Level:
In this problem you have to place four by four magic squares on the faces of a cube so that along each edge of the cube the numbers match.
### Numbered Cars
##### Age 7 to 11 Challenge Level:
I was looking at the number plate of a car parked outside. Using my special code S208VBJ adds to 65. Can you crack my code and use it to find out what both of these number plates add up to?
### Factor Track
##### Age 7 to 14 Challenge Level:
Factor track is not a race but a game of skill. The idea is to go round the track in as few moves as possible, keeping to the rules.
### Magic Triangle
##### Age 7 to 11 Challenge Level:
Place the digits 1 to 9 into the circles so that each side of the triangle adds to the same total.
### Fitted
##### Age 7 to 11 Challenge Level:
Nine squares with side lengths 1, 4, 7, 8, 9, 10, 14, 15, and 18 cm can be fitted together to form a rectangle. What are the dimensions of the rectangle?
### Starfish Spotting
##### Age 5 to 7 Challenge Level:
How many starfish could there be on the beach, and how many children, if I can see 28 arms?
### Prison Cells
##### Age 7 to 11 Challenge Level:
There are 78 prisoners in a square cell block of twelve cells. The clever prison warder arranged them so there were 25 along each wall of the prison block. How did he do it?
### Make 100
##### Age 7 to 11 Challenge Level:
Find at least one way to put in some operation signs (+ - x ÷) to make these digits come to 100.
### Special 24
##### Age 7 to 11 Challenge Level:
Find another number that is one short of a square number and when you double it and add 1, the result is also a square number.
### Magic Matrix
##### Age 7 to 11 Challenge Level:
Find out why these matrices are magic. Can you work out how they were made? Can you make your own Magic Matrix?
### Sticky Dice
##### Age 7 to 11 Challenge Level:
Throughout these challenges, the touching faces of any adjacent dice must have the same number. Can you find a way of making the total on the top come to each number from 11 to 18 inclusive?
### Fractions in a Box
##### Age 7 to 11 Challenge Level:
The discs for this game are kept in a flat square box with a square hole for each. Use the information to find out how many discs of each colour there are in the box.
### One Million to Seven
##### Age 7 to 11 Challenge Level:
Start by putting one million (1 000 000) into the display of your calculator. Can you reduce this to 7 using just the 7 key and add, subtract, multiply, divide and equals as many times as you like?
### Dice Stairs
##### Age 7 to 11 Challenge Level:
Can you make dice stairs using the rules stated? How do you know you have all the possible stairs?
### Four-digit Targets
##### Age 7 to 11 Challenge Level:
You have two sets of the digits 0 – 9. Can you arrange these in the five boxes to make four-digit numbers as close to the target numbers as possible?
### Big Dog, Little Dog
##### Age 5 to 7 Challenge Level:
Woof is a big dog. Yap is a little dog. Emma has 16 dog biscuits to give to the two dogs. She gave Woof 4 more biscuits than Yap. How many biscuits did each dog get?
### Area and Perimeter
##### Age 7 to 11 Challenge Level:
What can you say about these shapes? This problem challenges you to create shapes with different areas and perimeters.
### Jumping Squares
##### Age 5 to 7 Challenge Level:
In this problem it is not the squares that jump, you do the jumping! The idea is to go round the track in as few jumps as possible.
### Cat Food
##### Age 7 to 11 Challenge Level:
Sam sets up displays of cat food in his shop in triangular stacks. If Felix buys some, then how can Sam arrange the remaining cans in triangular stacks?
### Strike it Out for Two
##### Age 5 to 11 Challenge Level:
Strike it Out game for an adult and child. Can you stop your partner from being able to go?
### Pizza Cut
##### Age 7 to 11 Challenge Level:
Using only six straight cuts, find a way to make as many pieces of pizza as possible. (The pieces can be different sizes and shapes).
### The Puzzling Sweet Shop
##### Age 7 to 11 Challenge Level:
There were chews for 2p, mini eggs for 3p, Chocko bars for 5p and lollypops for 7p in the sweet shop. What could each of the children buy with their money?
### Domino Join Up
##### Age 5 to 7 Challenge Level:
Can you arrange fifteen dominoes so that all the touching domino pieces add to 6 and the ends join up? Can you make all the joins add to 7?
### Amazing Alphabet Maze
##### Age 5 to 7 Challenge Level:
Can you go from A to Z right through the alphabet in the hexagonal maze?
| 2,449 | 9,997 |
{"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}
| 3.765625 | 4 |
CC-MAIN-2019-35
|
latest
|
en
| 0.908665 |
https://wiki.synfig.org/index.php?title=Doc:Ball_Bounce&diff=prev&oldid=7142
| 1,566,125,546,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-35/segments/1566027313803.9/warc/CC-MAIN-20190818104019-20190818130019-00357.warc.gz
| 697,000,526 | 8,392 |
# Difference between revisions of "Doc:Ball Bounce"
Language: English • español • français • русский
This tutorial is under contrstuction. Please be patience.
This is a tutorial to explain how to create a bounce tutorial. The main target of the tutorial covers the ball movement. It is known that cartoon like balls have also a very deformed poses meanwhile thery are travelling and speciphically when it hits the ground. This could be covered in a second stage.
## Different aproximations for the same result
With Synfig there are four ways to create a bounce ball using the technical posibilities of this program.
1. The first one is do the bounce ball manually. That would involve create several waypoints and adjust them to match a parabolic movement (in time and in path).
2. Second way is use the interpolations paramters of the waypoints when they are set to TCB interpolation. This would drastically reduce the amount of waypoints and also make easier the timing of the bounces.
3. Third way to perform a bounce ball is make use of the Link to bline abiliity. If you draw the path of a bouncing ball using a bline it is quite easy to make the ball follow the path even changing the bouncing speed.
4. The fourth way to simulate a bouncing ball is create the mathematical equations to do that. Just make several parabolic shots at the rigth place a the right time to simulate a bouncing ball. It would be a little tricky but probably should be the most accurate one.
## Manual Ball Bounce
The rule to make the ball bounce manually is to draw in a paper the desired bounce. Then mark the horizontal line with regular intervals and match the curve intersection in vertical. See the image:
You can notice that having regular intervals in the horizontal axis gives irregular intervals to the vertical axis. It is due to the nature of the curve.
Once the points are located in a 2D grid then it can be drawn directly in Synfig doing use of the grid (F11). After draw them I normalized the values to be completely symmetrical. That gives the following table:
Time X position Y position Comments 0f -175.0 92.0 Highest point 4f -165.0 92.0 8f -155.0 81.118 12f -145.0 63.678 16f -135.0 29.479 20f -125.0 -15.522 Lower point 24f -115.0 29.479 28f -105.0 63.782 ... ...
You can see that the X position is increasing in steps of 10.0 and the Y position reproduces a parabolic curve.
To proceed with more than one bounce just duplicate the waypoints (place the cursor at the right place right click over the waypoint and duplicate) reproducing symmetrical movements. You should need to edit the X values manually to decrease by 10.0 for each new waypoint.
This is the resulting graph for the manual approximation to the ball bounce.
The resulting animation and file is this one.
File: Media:manual.sifz
## Ball Bounce using waypoints interpolations
The TCB interpolation mode allows modify the Tension, Continuity, Bias, and Temporal Tension values of the waypoint. So you can create easily smooth or peak aproximation to the value of the valuenode in the waypoint position.
This time I would use the same values for the highest and lower points of the table before. But I won't use more than one waypoint for each extreme position. The rest of the curve would be done using the TCB parameters.
The table of waypoints gives this result:
Time X position Y position Comments 0f -175.0 92.0 Highest point 20f -125.0 -15.522 Lower point 40f -75.0 92.0 Highest point 60f -25.0 -15.522 Lower point ... ...
As you can see the number of points is reduced drastically.
In you only use a default TCB interpolation it would give you a poor result. Look at the graph:
But if you edit the TCB parameters this is the result you obtain:
The TCB parameters are the following:
Time X position Y position Comments Tension Continuity Bias Temporal Tension 0f -175.0 92.0 Highest point 0.0 0.0 0.0 0.0 20f -125.0 -15.522 Lower point 0.0 -2.2 0.0 0.0 40f -75.0 92.0 Highest point 0.0 0.0 0.0 0.0 60f -25.0 -15.522 Lower point 0.0 -2.2 0.0 0.0 80f 25.0 92.0 Highest point 0.0 0.0 0.0 0.0 ... ... ... ... ... ... ... ...
That's the resulting animation:
And the sample file: Media:waypoint-2.sifz
Notice that the curve at 0f and at 80f are not properly formed. It is due to the fact that the TCB parameters needs to belong to an intermediate waypoint to have effect. If the waypoint is extreme (the end or the beginning of the animation for the parameter it cannot modify the curve. To solve that you should split the X and Y coordinates of the Origin and apply a Ease In/Out interpolation to those Y coordinate and leave the X coordinate with the current interpolation. So please consider only the bounces between the two black vertical lines.
Notice also that you can make the highest point more flat increasing the Temporal Tension parameter )a good value can be 0.5). This would produce a deformation to the X coordinate so you need to separate both coordinates to do that. Try it by your self editing the attached file. I have left the highest point to have the default values.
Here is a comparison of both bounces a the same time.
Notice that the
Language: English • español • français • русский
| 1,299 | 5,214 |
{"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}
| 3.828125 | 4 |
CC-MAIN-2019-35
|
longest
|
en
| 0.919744 |
https://samacheerguru.com/samacheer-kalvi-6th-maths-term-2-chapter-2-ex-2-2/
| 1,716,612,717,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-22/segments/1715971058773.28/warc/CC-MAIN-20240525035213-20240525065213-00002.warc.gz
| 434,804,949 | 15,783 |
# Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.2
## Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.2
Samacheer Kalvi 6th Maths Book Solutions Term 2 Question 1.
Say the time in two ways:
Solution:
(i) 10 : 15 hours; quarter past 10; 45 minutes to 11
(ii) 6 : 45 hours; quarter to 7; 45 minutes past 6
(iii) 4 : 10 hours; 10 minutes past 4; 50 minutes to 5
(iv) 3 : 30 hours; half-past 3; 30 minutes to 4
(v) 9 : 40 hours; 20 minutes to 10; 40 minutes past 9.
6th Maths Guide Term 2 Question 2.
Match the following:
(i) 9.55 – (a) 20 minutes past 2
(ii) 11.50 – (b) quarter past 4
(iii) 4.15 – (c) quarter to 8
(iv) 7.45 – (d) 5 minutes to 10
(v) 2.20 – (e) 10 minutes to 1?
Solution:
(i) 9.55 – (d) 5 minutes to 10
(ii) 11.50 – (e) 10 minutes to 12
(iii) 4.15 – (b) quarter past 4
(iv) 7.45 – (c) quarter to 8
(v) 2.20 – (a) 20 minutes past 2
Samacheer Kalvi 6th Maths Term 2 Question 3.
Convert the following :
(i) 20 minutes into seconds
(ii) 5 hours 35 minutes 40 seconds into seconds
(iii) 3$$\frac { 1 }{ 2 }$$ hours into minutes
(iv) 580 minutes into hours
(v) 25200 seconds into hours
Solution:
(i) 20 minutes into seconds
20 minutes = 20 × 60 seconds = 1200 seconds
(ii) 5 hours 35 minutes 40 seconds into seconds
∴ 5 hours 35 minutes 40 seconds = 20,140 seconds
(iii) 3$$\frac { 1 }{ 2 }$$ hours into minutes
3$$\frac { 1 }{ 2 }$$ hours = 3 hours 30 minutes
= 3 × 60 minutes + 30 minutes
= 180 minutes + 30 minutes
= 210 minutes
∴ 3$$\frac { 1 }{ 2 }$$ hours = 210 minutes
(iv) 5580 minutes into hours
580 minutes = $$\frac{580}{60}$$ hours = 9 hours 40 minutes
∴ 580 minutes = 9 hours 40 minutes
(v) 25200 seconds into hours
25200 seconds = $$\frac{25200}{60}$$ minutes = 420 minutes = $$\frac{420}{60}$$ hours = 7 hours
∴ 25200 seconds = 7 hours
Samacheer Kalvi 6th Maths Book Solutions Question 4.
The duration of electricity consumed by the farmer for his pumpset on Monday and Tuesday was 7 hours 20 minutes 35 seconds and 3 hours 44 minutes 50 seconds respectively. Find the total duration of consumption of electricity.
Solution:
Total duration of electricity consumed on both the days
= 7 hours 20 min 35 sec + 3 hours 44 min 50 sec
= (7 + 3) hours (20 + 44) min (35 + 50) sec
= 10 hours 64 min 85 seconds
= 11 hours 5 min 25 seconds
6th Standard Maths Exercise 2.2 Question 5.
Subtract 10 hrs 20 min 35 sec from 12 hrs 18 min 40 sec.
Solution:
1 hour 58 minutes 05 seconds
Samacheer Kalvi Term 2 Question 6.
Change the following into 12 hour format:
(i) 02 : 00 hours
(ii) 08 : 45 hours
(iii) 21 : 10 hours
(iv) 11 : 20 hours
(v) 00 : 00 hours
Solution:
Samacheer Kalvi 6th Maths Question 7.
Change the following into 24 hour format.
(i) 3.15 a.m.
(ii) 12.35 p.m.
(iii) 12.00 noon
(iv) 12.00 midnight.
Solution:
6th Maths Term 2 Question 8.
Calculate the duration of time.
(i) from 5.30 a.m to 12.40 p.m
(ii) from 1.30 p.m to 10.25 p.m
(iii) from 20:00 hours to 4:00 hours
(iv) from 17:00 hours to 5 :15 hours
Solution:
(i) from 5.30 a.m. to 12 .40 p.m.
Duration of time from 5.30 a.m. to noon = 12 : 00 – 5 : 30 = 6 : 30 i.e 6 hours 30 minutes
From noon to 12.40 p.m the duration = 00 hours 40 minutes
Total duration = 6 hours 30 minutes + 00 hours 40 minutes
= 6 hours 70 minutes
= 6 hours + (60 + 10) minutes
= 6 hours + 1 hr 10 minutes
= 7 hours 10 minutes
∴ Duration of time from 5.30 am to 12.40 pm = 7 hours 10 minutes
(ii) from 1.30 pm to 10.25 p.m.
(iii) from 20 : 00 hrs to 4 : 00 hrs.
(iv) from 17 : 00 hours to 5 : 15 hours.
Term 2 Samacheer Kalvi Question 9.
The departure and arrival timing of the Vaigai Superfast Express (No. 12635) from Chennai Egmore to Madurai Junction are given. Read the details and answer the following.
(i) At what time does the Vaigai Express start from Chennai and arrive at Madurai?
(ii) How many halts are between there Chennai and Madurai?
(iii) How long does the train halt at the Villupuram junction?
(iv) At what time does the train come to Sholavandan?
(v) Find the journey time from Chennai Egmore to Madurai?
Solution:
(i) It starts from Chennai at 13 : 40 hrs and arrive at Madurai at 21 : 20 hrs.
(ii) There are 8 halts.
(iii) Departure from Villupuram = 15 hours 55 minutes
Arrival at Villupuram = 15 hours 50 minutes
The train halt at Villupuram for = 05 minutes
(iv) At 20 : 34 hours the train come to Sholavandan
(v) Arrival time at Madurai = 20 hours 80 (20 + 60) minutes
Departure time from Chennai Egmore = 13 hours 40 minutes
Journey Time = 07 hours 40 minutes
Samacheer Kalvi 6th Standard Maths Question 10.
Manickam joined a chess class on 20.02.2017 and due to exam, he left practice after 20 days. Again he continued practice from 10.07.2017 to 31.03.2018. Calculate how many days did he practice?
Solution:
From the date of joining = 20 days From 10.07.2017 to 31.03.2018
July – 22
Aug – 31
Sep – 30
Oct – 31
Nov – 30
Dec – 31
Jan – 31
Feb – 28
Mar – 31
Total – 265
Total no of practice days = 265 + 20 = 285 days
6th Term 2 Maths Guide Question 11.
A clock gains 3 minutes every hour. If the clock is set correctly at 5 a.m, find the time shown by the clock at 7 p.m?
Solution:
5 a.m. = 5.00 hours
7 p.m = 19.00 hours
Difference = 19.00 – 5.00 = 14.00 hours
Time gain = 14 × 3 = 42 minutes
Time shown by the clock = 7.42 p.m
Class 6 Maths Chapter 2 Exercise 2.2 Solutions Question 12.
Find the number of days between the republic day and kalvi valarchi day in 2020.
Solution:
2020 is a leap year republic day – 26.01.2020
Kalvi valarchi day – 15.07.2020
Jan – 5
Feb – 29
Mar – 31
April – 30
May – 31
June – 30
July – 14
Total – 170 days
Question 13.
If 11th of January 2018 is Thursday, what is the day on 20th July of the same year?
Solution:
January – 21 Days (31 – 10)
February – 28 Days
March – 31 Days
April – 30 Days
May – 31 Days
June – 30 Days
July – 19 Days
Total – 190 Days
190 days ÷ 7
190 days = 27 weeks + 1 day
Required day is the first day after Thursday.
∴ 20th of July is Friday.
Question 14.
(i) Convert 480 days into years
(ii) Convert 38 months into years
Solution:
(i) 480 days = $$\frac{480}{365}$$
= 1 year 115 days
= 1 year 3 months 25 days
(ii) 38 months = $$\frac{38}{12}$$
= 3 years 2 months
Question 15.
Calculate your age as on 01.06.2018
Solution:
Age is 12 years 2 months
Objective Type Questions
Question 16.
2 days = …….. hours
(i) 38
(ii) 48
(iii) 28
(iv) 40
Solution:
(ii) 48
Question 17.
3 weeks = _____ days.
(a) 21
(b) 7
(c) 14
(d) 28
Solution:
(a) 21
Question 18.
Number of ordinary years between two consecutive leap years is
(i) 4 years
(ii) 2 years
(iii) 1 year
(iv) 3 years
Solution:
(iv) 3 years
Question 19.
What time will it 5 hours after 22 : 35 hours?
(a) 2 : 30 hrs
(b) 3 : 35 hrs
(c) 4 : 35 hrs
(d) 5 : 35 hrs
Solution:
(b) 3 : 35 hrs
Question 20.
2 $$\frac{1}{2}$$ years is equal to months.
(i) 25
(ii) 30
(iii) 24
(iv) 5
Solution:
(ii) 30
| 2,449 | 6,865 |
{"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}
| 4.84375 | 5 |
CC-MAIN-2024-22
|
latest
|
en
| 0.742965 |
http://www.dummies.com/how-to/content/how-to-solve-a-quadratic-equation-by-factoring.html
| 1,469,796,469,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2016-30/segments/1469257830066.95/warc/CC-MAIN-20160723071030-00051-ip-10-185-27-174.ec2.internal.warc.gz
| 413,585,552 | 14,171 |
A quadratic equation is any second degree polynomial equation — that’s when the highest power of x, or whatever other variable is used, is 2. You can solve quadratic equations by factoring.
1. Bring all terms to one side of the equation, leaving a zero on the other side.
2. Factor.
(2x + 3)(x – 4) = 0
3. Set each factor equal to zero (using the zero product property).
The zero product property states that if the product of any number of factors is zero, then at least one of the factors must be equal to zero.
2x + 3 = 0
2x = –3
or
x – 4 = 0
x = 4
So, this equation has two solutions:
| 163 | 599 |
{"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-2016-30
|
latest
|
en
| 0.917032 |
http://emaths.net/study-sheets---decimals.html
| 1,511,177,665,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-47/segments/1510934806030.27/warc/CC-MAIN-20171120111550-20171120131550-00555.warc.gz
| 98,497,747 | 11,506 |
Try the Free Math Solver or Scroll down to Tutorials!
Depdendent Variable
Number of equations to solve: 23456789
Equ. #1:
Equ. #2:
Equ. #3:
Equ. #4:
Equ. #5:
Equ. #6:
Equ. #7:
Equ. #8:
Equ. #9:
Solve for:
Dependent Variable
Number of inequalities to solve: 23456789
Ineq. #1:
Ineq. #2:
Ineq. #3:
Ineq. #4:
Ineq. #5:
Ineq. #6:
Ineq. #7:
Ineq. #8:
Ineq. #9:
Solve for:
Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg:
# Study Sheets - Decimals
SECTION 3.4 – COMPARING AND CONVERTING DECIMALS AND
FRACTIONS.
Quick facts:
As implied in the previous sections, every fraction can be rewritten as a decimal. To do
this, just simply divide the numerator of the fraction by the denominator.
Ex. , because
And opposite, a decimal can be written as a fraction. To do this, the decimal part of the
decimal will be put in the numerator of the fraction, and the denominator will be derived
from the place value of the last digit of the number being converted.
Ex. , because the last digit (4) was on the hundredth place. This fraction can be
further simplified. Always try to find the fraction in the simplest form.
Note: If converting a decimal that contains a whole part, this whole part would become a
whole part of a mixed number.
Remember to follow the rules of conversion even for problems like:
To compare a decimal and a fraction, convert the fraction into a decimal and than
compare.
Ex. Place a symbol <, > or = between 0.165 and 1/6
Answer: Because 1/6 is equal to approximately 0.166, which is greater than 0.165, we can
say:
Test yourself:
Convert to a decimal:
,round to the nearest hundredth = 3.67 , round to the nearest thousandth
Convert to a fraction:
Place a correct symbol between the numbers:
Challenge Yourself:
It might be useful for the future for you to know the basic conversions between fractions
and decimals, as they are widely used in real life. Think of change, for example a quarter
(which is \$ 1/4), can be also written as \$0.25.
Getting accustomed with the conversion table (that you can find in Appendix of this
chapter) simplifies the process of converting fractions to decimals by eliminating of the
repetitive dividing.
Assignment:
Textbook Chapter 3.6 (p.162): Exercises 1 – 69 every fourth problem
| 607 | 2,372 |
{"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-2017-47
|
latest
|
en
| 0.87792 |
https://simplycreativeteaching.com/my-favorite-strategy-for-3-digit-subtraction/
| 1,716,464,889,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-22/segments/1715971058625.16/warc/CC-MAIN-20240523111540-20240523141540-00205.warc.gz
| 460,629,504 | 34,548 |
# My Favorite Strategy for 3-Digit Subtraction
Does anyone else despise teaching 2 and 3-digit subtraction? I totally LOVE addition, but subtraction… no thanks. It can be so difficult for our students!
Now, while I still don’t love teaching it, I wanted to share with you my favorite simple strategy to use. It just might come in handy if you’re gearing up for your subtraction unit!
Okay, so what’s the simple strategy…
## 3-Digit Subtraction Strategy: Use a Place Value Mat
Most of us know that it’s essential to use manipulatives like base ten blocks when we’re starting subtraction with regrouping. They’re a great visual!
However, one thing that really helped my students when we were working on 2 & 3-digit subtraction was using a place value mat that had extra space for regrouping.
Here’s what I mean. Take a look at this place value mat below. *Don’t worry, you can download these digital + printable place value mats for free at the end of this post!
As you can see, I have made a space for 20 ones in the ones place. I did this for a couple reasons. One, it helps students organize the ones instead of just tossing them in the ones column. And two, whenever we need to regroup, students have space to put ten more ones.
I know it seems like such a little detail, but I promise you it really helps students SEE what’s going on when you’re practicing subtraction with base ten blocks!
## Students Still Struggling? Use Linking Cubes!
Many students quickly grasp this concept and visual representation; however, if base ten blocks aren’t working, I have another trick for you. Use unifix cubes or linking cubes!
Before you do any subtraction problems, have your students make a bunch of towers of ten cubes. This is great if you’re working with a small group because I never had enough unifix cubes for my whole class!
Then, when you’re ready to start working on subtraction, students can model numbers with their cubes.
When it’s time to regroup, students can simply break apart a tower of ten and put them in the ones place. With these cubes, students don’t have to trade for different pieces like they do with base ten blocks.
You can download these place value mats to use with your students for free! I’ve included both digital & printable place value mats so you can use them no matter where you’re teaching!
Want more help with 2-3 digit addition & subtraction? Check out this post on why I stopped teaching the standard algorithm!
Leave your email below and we'll send these place value mats directly to your inbox!
## Free Subtraction Strategies Sheet!
This free chart includes a 2-digit version, a 3-digit version, and a blank version! Leave your email below and print your copy immediately!
## Free Spiral review
Leave your email to access FREE weekly spiral math review, problem of the day, and daily word problems for 1st & 2nd grade!
| 626 | 2,876 |
{"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-2024-22
|
latest
|
en
| 0.920251 |
https://goopennc.oercommons.org/browse?f.keyword=parallelograms
| 1,726,685,821,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725700651931.60/warc/CC-MAIN-20240918165253-20240918195253-00723.warc.gz
| 240,554,536 | 17,570 |
Updating search results...
# 10 Results
View
Selected filters:
• parallelograms
Conditional Remix & Share Permitted
CC BY-NC
Rating
0.0 stars
Rating
0.0 stars
Subject:
Mathematics
Material Type:
Full Course
Provider:
Pearson
11/02/2020
Conditional Remix & Share Permitted
CC BY-NC
Rating
0.0 stars
Surface Area and Volume
Type of Unit: Conceptual
Prior Knowledge
Students should be able to:
Identify rectangles, parallelograms, trapezoids, and triangles and their bases and heights.
Identify cubes, rectangular prisms, and pyramids and their faces, edges, and vertices.
Understand that area of a 2-D figure is a measure of the figure's surface and that it is measured in square units.
Understand volume of a 3-D figure is a measure of the space the figure occupies and is measured in cubic units.
Lesson Flow
The unit begins with an exploratory lesson about the volumes of containers. Then in Lessons 2–5, students investigate areas of 2-D figures. To find the area of a parallelogram, students consider how it can be rearranged to form a rectangle. To find the area of a trapezoid, students think about how two copies of the trapezoid can be put together to form a parallelogram. To find the area of a triangle, students consider how two copies of the triangle can be put together to form a parallelogram. By sketching and analyzing several parallelograms, trapezoids, and triangles, students develop area formulas for these figures. Students then find areas of composite figures by decomposing them into familiar figures. In the last lesson on area, students estimate the area of an irregular figure by overlaying it with a grid. In Lesson 6, the focus shifts to 3-D figures. Students build rectangular prisms from unit cubes and develop a formula for finding the volume of any rectangular prism. In Lesson 7, students analyze and create nets for prisms. In Lesson 8, students compare a cube to a square pyramid with the same base and height as the cube. They consider the number of faces, edges, and vertices, as well as the surface area and volume. In Lesson 9, students use their knowledge of volume, area, and linear measurements to solve a packing problem.
Subject:
Geometry
Mathematics
Provider:
Pearson
Conditional Remix & Share Permitted
CC BY-NC
Rating
0.0 stars
Lesson OverviewStudents find the area of a parallelogram by rearranging it to form a rectangle. They find the area of a trapezoid by putting together two copies of it to form a parallelogram. By doing these activities and by analyzing the dimensions and areas of several examples of each figure, students develop and understand area formulas for parallelograms and trapezoids.Key ConceptsA parallelogram is a quadrilateral with two pairs of parallel sides. The base of a parallelogram can be any of the four sides. The height is the perpendicular distance from the base to the opposite side.A trapezoid is a quadrilateral with exactly one pair of parallel sides. The bases of a trapezoid are the parallel sides. The height is the perpendicular distance between the bases.You can cut a parallelogram into two pieces and reassemble them to form a rectangle. Because the area does not change, the area of the rectangle is the same as the area of the parallelogram. This gives the parallelogram area formula A = bh.You can put two identical trapezoids together to form a parallelogram with the same height as the trapezoid and a base length equal to the sum of the base lengths of the trapezoid. The area of the parallelogram is (b1 + b2)h, so the area of the trapezoid is one-half of this area. Thus, the trapezoid area formula is A = 12(b1 + b2)h.Goals and Learning ObjectivesDevelop and explore the formula for the area of a parallelogram.Develop and explore the formula for the area of a trapezoid.
Subject:
Geometry
Material Type:
Lesson Plan
Author:
Pearson
11/02/2020
Conditional Remix & Share Permitted
CC BY-NC
Rating
0.0 stars
Lesson OverviewStudents find the area of a triangle by putting together a triangle and a copy of the triangle to form a parallelogram with the same base and height as the triangle. Students also create several examples of triangles and look for relationships among the base, height, and area measures. These activities lead students to develop and understand a formula for the area of a triangle.Key ConceptsTo find the area of a triangle, you must know the length of a base and the corresponding height. The base of a triangle can be any of the three sides. The height is the perpendicular distance from the vertex opposite the base to the line containing the base. The height can be found inside or outside the triangle, or it can be the length of one of the sides.You can put together a triangle and a copy of the triangle to form a parallelogram with the same base and height as the triangle. The area of the original triangle is half of the area of the parallelogram. Because the area formula for a parallelogram is A = bh, the area formula for a triangle is A = 12bh.Goals and Learning ObjectivesDevelop and explore the formula for the area of a triangle.
Subject:
Geometry
Material Type:
Lesson Plan
Author:
Pearson
11/02/2020
Conditional Remix & Share Permitted
CC BY-NC
Rating
0.0 stars
Four full-year digital course, built from the ground up and fully-aligned to the Common Core State Standards, for 7th grade Mathematics. Created using research-based approaches to teaching and learning, the Open Access Common Core Course for Mathematics is designed with student-centered learning in mind, including activities for students to develop valuable 21st century skills and academic mindset.
Subject:
Mathematics
Material Type:
Full Course
Provider:
Pearson
11/02/2020
Conditional Remix & Share Permitted
CC BY-NC
Rating
0.0 stars
Constructions and Angles
Unit Overview
Type of Unit: Concept
Prior Knowledge
Students should be able to:
Use a protractor and ruler.
Identify different types of triangles and quadrilaterals and their characteristics.
Lesson Flow
After an initial exploratory lesson involving a paper folding activity that gets students thinking in general about angles and figures in a context, the unit is divided into two concept development sections. The first section focuses on types of angles—adjacent, supplementary, complementary, and vertical—and how they are manifested in quadrilaterals. The second section looks at triangles and their properties, including the angle sum, and how this affects other figures.
In the first set of conceptual lessons, students explore different types of angles and where the types of angles appear in quadrilaterals. Students fold paper and observe the angles formed, draw given angles, and explore interactive sketches that test many cases. Students use a protractor and ruler to draw parallelograms with given properties. They explore sketches of parallelograms with specific properties, such as perpendicular diagonals. After concluding the investigation of the angle types, students move on to the next set.
In the second set of conceptual development lessons, students focus on triangles. Students again fold paper to create figures and certain angles, such as complementary angles.
Students draw, using a protractor and ruler, other triangles with given properties. Students then explore triangles with certain known and unknown elements, such as the number of given sides and angles. This process starts with paper folding and drawing and continues with exploration of interactive sketches. Students draw conclusions about which cases allow 0, 1, 2, or an infinite number of triangles. In the course of the exploration, students discover that the sum of the measure of the interior angles of a triangle is 180°. They also learn that the sum of the measures of the interior angles of a quadrilateral is 360°. They explore other polygons to find their angle sum and determine if there is a relationship to angle sum of triangles. The exploration concludes with finding the measure of the interior angles of regular polygons and speculating about how this relates to a circle.
Lastly, students solve equations to find unknown angle measures. Using their previous experience, students find the remaining angle measures in a parallelogram when only one angle measure is given. Students also play a game similar to 20 Questions to identify types of triangles and quadrilaterals. Having completed the remaining lessons, students have a four-day Gallery to explore a variety of problems.
The unit ends with a unit assessment.
Subject:
Geometry
Mathematics
Provider:
Pearson
Conditional Remix & Share Permitted
CC BY-NC
Rating
0.0 stars
Students learn more about the characteristics of parallelograms by folding paper and measuring the angles in a parallelogram. Students use a ruler and protractor to draw parallelograms with given properties. Then, students use a ruler and protractor to draw a rectangle.Key ConceptsOpposite angles of a parallelogram are congruent.Consecutive angles of a parallelogram are supplementary.Diagonals of a parallelogram bisect each other.Diagonals of a rectangle are congruent.Goals and Learning ObjectivesAccess prior knowledge of parallelograms.Understand that the sum of angle measures in any quadrilateral is 360°.Understand the relationship of the angles and diagonals in a parallelogram.Understand the relationship of the angles and diagonals in a rectangle.
Subject:
Geometry
Material Type:
Lesson Plan
Author:
Pearson
11/02/2020
Conditional Remix & Share Permitted
CC BY-NC
Rating
0.0 stars
Students learn how the diagonals of a rhombus are related. They use interactive sketches to learn about the properties of the angles and diagonals of squares, rectangles, rhombuses, parallelograms, and other quadrilaterals.Key ConceptsThe sum of the measures of the angles of all quadrilaterals is 360°.The alternate angles (nonadjacent angles) of rhombuses and parallelograms have the same measure.The measure of the angles of rectangles and squares is 90°.The consecutive angles of parallelograms and rhombuses are supplementary. This applies to squares and rectangles as well.The diagonals of a parallelogram bisect each other.The diagonals of a rectangle are congruent and bisect each other.The diagonals of a rhombus bisect each other and are perpendicular.Goals and Learning ObjectivesMeasure the angles formed by the intersection of the diagonals of a rhombus.Explore the relationships of the angles of different squares, rectangles, rhombuses, parallelograms, and other quadrilaterals.Explore the relationships of the diagonals of different squares, rectangles, rhombuses, parallelograms, and other quadrilaterals.
Subject:
Geometry
Material Type:
Lesson Plan
Author:
Pearson
11/02/2020
Conditional Remix & Share Permitted
CC BY-NC
Rating
0.0 stars
Gallery OverviewAllow students who have a clear understanding of the content thus far in the unit to work on Gallery problems of their choosing. You can then use this time to provide additional help to students who need review of the unit’s concepts or to assist students who may have fallen behind on work.Problem DescriptionsParallelogram to CubeStudents have a chance to review angle measurements in a parallelogram. Building the cube helps students see the transition from two-dimensional shapes and their relationship to three-dimensional figures.QuadrilateralsStudents investigate the possible quadrilaterals that can be made from any four given side lengths, focusing on those that can’t make a quadrilateral. Students also look at possible parallelograms with two sides given and possible rhombuses with four sides given.DiagonalsStudents further investigate diagonals in quadrilaterals. If the diagonals are perpendicular, is the figure a rhombus?TrapezoidsHow many right angles can a trapezoid have? How many congruent angles or congruent sides can it have? Can its diagonals be perpendicular or congruent? Students investigate possible trapezoids.More AnglesStudents explore three intersecting lines and the combinations of angles.Diagonals and AnglesThe sides of a parallelogram are extended beyond the vertices, and students explore which angles are congruent and which are supplementary. Students also explore the effect diagonals have on interior angles.Exterior AnglesStudents explore the sum of exterior angles for several polygons and speculate about the results.Angles and SidesStudents explore the relationship between angles and sides in a triangle and discover that the longest side is opposite the largest angle, and the shortest side is opposite the smallest angle (and congruent sides are opposite congruent angles).Ratios and AnglesStudents explore the ratios of the legs of a right triangle to the angles in the triangle. Students see that there is a unique ratio for each angle, and vice versa. This is an informal look at trigonometry.Find the AngleStudents solve equations to find angle measures in polygons.TessellationsStudents explore quadrilateral tessellations and why they tessellate. Students also explore tessellations of pentagons and other polygons.
Subject:
Geometry
Material Type:
Lesson Plan
Author:
Pearson
11/02/2020
Conditional Remix & Share Permitted
CC BY-NC
Rating
0.0 stars
Students discuss what they know about shapes and their characteristics through a paper-folding activity that results in a parallelogram.Key ConceptsQuadrilaterals and triangles are classified by their different characteristics; the types of angles and sides define the shapes. While students are familiar with some of the characteristics of these shapes, they begin to explore other aspects of theses figures. Students review what they know about these shapes so far.Goals and Learning ObjectivesReview characteristics that describe quadrilaterals and triangles.Discuss what students know about these shapes.Explore other aspects of these shapes.
Subject:
Geometry
Material Type:
Lesson Plan
Author:
Pearson
| 2,957 | 13,926 |
{"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-2024-38
|
latest
|
en
| 0.910449 |
nsdl.org
| 1,417,223,579,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2014-49/segments/1416931011456.52/warc/CC-MAIN-20141125155651-00226-ip-10-235-23-156.ec2.internal.warc.gz
| 209,715,754 | 15,265 |
# TimelyTeaching/Cryptography
Primary Articles Learning Environment
NSDL Timely Teaching, Issue 1. October 2009.
## Cryptography: The Math and Science of Secrecy
by Dr. Mike May
Historical Introduction
??Cryptography, the science of encrypting and decrypting information, is often thought of in terms of soldiers and spies clandestinely sending and intercepting secret messages. However within the past generation as more and more of our communication is electronic, it has become part of normal business, so that it is hard to find anyone who does not routinely use commercial grade cryptography. ?At its most fundamental level, the goal of cryptography is to send a message that can only be read by the desired recipient. One system of ancient times is the Spartan scytale where a message was written across a strip of paper wrapped around a stick of specified length and width. When the strip was unwound the letters were scrambled or transposed. A second system is the Caesar cipher, named after Julius Caesar, where each letter was shifted by three letters, replacing A by D and B by E. Variations of this monoalphabetic substitution survive today in jumbles in many newspapers.?? An important variation of the Caesar cipher is the which uses a longer key for the shift so that different substitution are used at depending on the position of the character in the message. For example, if the key is the 6 letter word "secret," then the 1st, 7th, and 13th characters of the message are shifted by s, or 19 characters, while the 2nd, 8th, and 14th characters of the message are shifted by e or 5 characters. A big advantage of such a system is that each letter can be encoded as several different letters, reducing the ability of an enemy to decode the message by looking for frequently used letters. An extreme variation of the same idea uses a long book to control the shift, so the key becomes a location in the book used to start encryption. ??
A central problem with Vigenere and its variants is that English (and any human language) is far from random. This allows attackers to use statistical methods to break a code. One response was to create a long string of randomly generated shifts. This method is called using a one time pad and is theoretically unbreakable. In practice the one time pad shifts the problem of sending the message to the problem of securely transmitting the key, which is now as long as the message. The solution was to design a machine that would produce a string of characters that seem to be random, with the string started by a shorter key. The most famous of attempt using this method is the Enigma machine used by the Germans in WWII.
The other famous method for secret communication from WWII was the Code Talkers of the Allied Forces, which used the Navaho language and the correct assumption that the obscurity of language would prevent the enemy from understanding the communications. This method of encryption violated a central principle of modern cryptography, formulated by Kerckhoff, and often stated as "no security through obscurity", which assumes that the enemy knows your system of encryption. ??The breaking of the German Enigma machine and the Japanese Purple Machine highlighted the fact that being able to securely send information and intercept the enemy's communications had become an important tool of warfare. The US government created the National Security Agency, or NSA, centered in Fort Meade, in 1952. For years, this agency was so secretive that it was said the initials stood for "No Such Agency."?
?The second half of the twentieth century and the availability of modern computers and the increasing amount of financial data transmitted electronically changed the typical setting for practical cryptography. Cryptography went from the realm of soldiers and spies to that of bankers and ordinary commerce. Today it is hard to find anyone who does not routinely use commercial grade cryptography. We typically use encryption of some form or another whenever we use an ATM machine, have a credit card electronically verified, visit a secure website, use a wireless computer connection, or use a cell phone. Cryptography has become part of our ordinary life. (To see some systems you may be using without knowing it, launch your favorite Internet browser and open the preferences. The location of the security certificates may vary. On Firefox, open the advanced tab of the preference and select encryption, then view the certificates.) Rather than encrypting letters, all modern systems work with the bits obtained by using the normal ASCII [link] representation used by all computers. For some systems, we think of the string of bits as representing a number.
Symmetric Ciphers
In the 1970's Horst Feistal, while working at IBM labs, invented a series of algorithms for encrypting a block of plaintext, the most famous of which was the Lucifer (cipher). As electronic communication of banking and commercial records became more important to the economy, the National Bureau of Standards (NSA) sent out a solicitation for an encryption algorithm that would be a commercial standard. When the responses to the solicitation were inadequate, they asked the National Security Agency (NSA) to help create a standard for encryption. Instead, the NSA worked with IBM, and helped them develop Lucifer into the Data Encryption Standard, or DES. For years there was concern that NSA had adjusted the S-Boxes, a feature of the algorithm, to introduce a back door into the code for easy eavesdropping. Years later when differential cryptoanalysis was discovered as a way of attacking this class of ciphers, it was realized the NSA had optimized the S-Boxes to be resistant to this kind of attack. The NSA also wanted to reduce the key length from 64 to 48 bits to produce a cope that would be commercially secure, but would allow law enforcement to decrypt a message when needed. The tension between the needs of commercial security and law enforcement recurs regularly in cryptography. With the compromise key length of 56 bits, DES has 2^56 possible keys.
In the 1990's the increase in available computer power rendered DES susceptible to a brute force attack where all 2^56 possible keys are checked. In 1997, the government announced an open competition for a replacement encryption algorithm, with the competition run by the National Institute of Standards and Technology, or NIST. In 2002, Rijndael, a code designed by Belgian cryptographers Vincent Rijmen and Joan Daemen became the Advanced Encryption Standard, or AES. AES was designed to work with keys of 128, 196, or 256 bits, so that it can be adjusted for future increases in computer power. The open competition for an encryption standard went so well, the NIST is currently holding a similar competition for a new standard for a hash function [internal link] .?
?Another indication of the changing role of cryptography comes from the laws concerning cryptographic software. Into the 1990's, cryptographic software was formally a munition under US law. This led to the creation of two versions of the Netscape browser, one with 40-bit encryption for export and a US version with 128-bit encryption. These laws were significantly relaxed in 1996, with oversight for many cryptographic tools now under the Department of Commerce.
Public Key Cryptography
Both DES and AES are symmetric ciphers. The same key is used both for encryption and decryption. A problem related to the use of symmetric ciphers is one of key distribution. The keys need to be securely transferred to be used. That is not such a big problem between institutions like banks that routinely communicate with each other, but it becomes a problem with electronic commerce where a new customer does not want to wait for secure delivery of a pin before ordering online. By contrast, public key cryptography uses a system with two keys, so a locking key can be transferred publicly, while an unlocking key is kept secret. In 1976, Whitfield Diffie and Martin Hellman published a practical method for secretly sharing keys over an unprotected channel.
Public key systems depend on mathematical operations that are straightforward in one direction and impossibly hard in the other direction. The easiest example is the multiplication of two prime numbers and the factoring of a number into its prime factors. With a piece of paper or a basic calculator, it is easy to see that 307 times 967 is 296,869. However, it is much harder to guess that 557,993 is the product of 751 and 743. A second one-way function takes powers of numbers in modular or clock arithmetic. (In modular and clock arithmetic, we are only interested in remainders. Thus 5 hours after 10:00 o'clock is 3:00 o'clock. The difference between the two systems is that in modular arithmetic, noon would be 0:00 o'clock.) It uses the difference in difficulty of modular exponentiation and discrete logarithms. (Given a, b, c, and n, with a^b=c, mod n, going from using a and b to compute c, is modular exponentiation whereas going from a and c to b is solving a logarithm.) If we are looking at powers of a number mod 10, we only care about the remainder after we divide by 10. Thus the powers of 3 are, in order, 3 then 9 (or 3*3), then 7 (3*9=27, forget the 20), then 1 (7*3=21, forget the 20), then 3. The number of multiplications needed to compute a power can be reduced to about twice the length of the exponent base 2, while solving the discrete logarithm is much harder. It should be noted that when your browser gets a key for secure page it does one of these operations with a prime number that is between 100 and 300 digits long. For a 100-digit exponent we need 600 multiplications to find the power, and 10^100 multiplications to find the discrete log by guessing and checking. With the fastest computer currently available, the sun will burn out before the discrete log problem is solved by a naive approach.
The Diffie-Hellman Key Exchange
In explaining the Diffie-Hellman key exchange, we follow a tradition in writing about cryptography where we assume Alice is trying to send a message to Bob and Eve is trying to listen in. (If the third party is trying to be malicious and send fake messages, the identity changes to Mallory.) Alice and Bob agree on a base g and a modulus n, which is a 100-digit prime number. They don't care if Eve learns the values of g and n and may even publish these numbers. Alice picks a secret number a and computes g^a mod n. Bob picks a secret number b and computes g^b mod n. Alice and Bob tell each other the values of g^a and g^b respectively, while keeping a and b secret. Once again, they make no effort to keep g^a and g^b secret from Eve. Alice and Bob can then each compute k=g^(a+b)=(g^a)^b=(g^b)^a. They can agree to use the last 128 bits of k for their AES key. Because Eve does not know either a or b, and cannot solve the logarithm problem to deduce them from g^a and g^b she cannot compute k. Alice and Bob have thus exchanged a secret key while communicating openly on a channel with Eve listening in.
RSA
If you look at the security certificates on your web browser, it is likely that the public key algorithm used will be RSA, a method invented by Ronald L. Rivest, Adi Shamir, and Leonard M. Adleman. You have probably used it if you have done online banking or bought something online at a secure site, like iTunes. For RSA, the vendor creates two keys, one of which is public and the other is private. Anyone can use the public key to encrypt a message, but only the vendor knows the private key needed to read the message. A typical implementation of RSA randomly chooses 2 prime numbers, p and q, that are each 300 digits long, and multiplies them to get n=p*q. We will also want to compute n1=(p-1)(q-1). I choose an encryption key E and use the extended Euclidean algorithm to find a decryption key D such that DE-1 is a multiple of n1. If my plaintext message (which has to be a number) is m, then the encrypted ciphertext is c=m^E mod n. For decryption, I note that due to a theorem of Fermat, m=c^D mod n. For technical reasons, almost everyone always uses the same encryption key, E=65,537=2^16+1. Given n and E, trying to find D is equivalent to finding the prime factors p and q of n. Simply guessing one of the factors would be as hard as guessing the winning numbers on the Powerball contest, 28 weeks in a row. (There are more efficient methods of finding the factors, but for numbers this size, they are still "practically impossible". A group of mathematicians announced in January 2010 that they had factored the product of two 166 digit numbers. The effort involved about 1,500 computer years of computation.)
Hash Functions
Any discussion of cryptography should include comments on the related topics of authenticity (was the message sent by the person who claims to have sent the message) and message integrity (is the message received the same as the message that was sent). A simple form of authentication or digital signature can be done with RSA. If I encrypt a message with my private decryption key and send the two messages together, then anyone can use the public encryption key to verify that I was the one who did the encryption. One problem with this scheme is that RSA only encrypts messages shorter than n and 300 digits only gives me room for about 120 characters.
Cryptographers use hash functions to reduce a long message to a short "digest" of a fixed length. Hash functions are designed so it is practically impossible to find two messages that make sense with the same digest. As mentioned above, the current attacks on the standard hash functions look like they will be able to generate messages with the same hash within a few years, so the National Institute of Standards and Technology (NIST) is sponsoring a competition to design a new standard hash function.
How the Pieces Fit Together
It is worthwhile to put all the pieces of modern cryptography together in a typical arrangement. A public key system (RSA) is used to establish communications, but public key systems are computationally intensive and comparatively slow, so the only thing communicated with the public key cipher is a session key for a symmetric cipher (AES). This is analogous to putting a collection of books in a case, locked with AES, and putting the AES key in an envelope locked with RSA. Additionally we use a hash algorithm (SHA-1) to reduce the communication to a hash that we then sign with a signature algorithm (DSA). To continue the analogy we also produce a single sheet of paper that can be used to verify the collection of books is unchanged from what I intended to send and I sign that sheet of paper with a signature that cannot be forged by anyone else. I add that paper to the locked case of books. I have efficiently insured secrecy, integrity, and authenticity. There are a variety of algorithms to choose from or each of the four steps mentioned, with the most popular chosen in each category.
??The hidden technical issues and future trends
In an attempt to make cryptography accessible, there is a danger that readers may be left with the impression it is easy and anyone can create a good cipher or that a cryptographic algorithm is secure if it can't be broken quickly by your friends. For a code to be secure it needs to resist the amount of effort a government can bring to bear over a period of years. RSA with a pair of 65 digit primes was considered insecure when such a code was broken by a coalition of 600 volunteers using 1600 computers working together for 6 months. The work in public key cryptography looks at the rather sophisticated mathematics behind efforts to efficiently find long primes and ways to factor a product of two larger prime numbers. Variants of methods based on the discrete logarithm problem using integer points on elliptic curves are making their way into the commercial market. Attacks on symmetric ciphers use a lot of computer power to find places where the relationship between the plaintext and the encoded ciphertext does not appear random. Advances in attacks can come from a broad range of the branches of mathematics. Cryptographers are also working on using the principles of quantum mechanics to produce practical codes where even listening in will be detected.
Annotated Bibliography
"Cryptography." Wikipedia, The Free Encyclopedia. Retrieved 16 Aug 2009, <http://en.wikipedia.org/w/index.php?title=Cryptography&oldid=307147660>. Wikipedia has several good entries on cryptography. The links let you continue exploration as you want to know more details.
"RSA Laboratories' Frequently Asked Questions About Today's Cryptography, Version 4.1". RSA Laboratories Retrieved 16 Aug 2009, <http://www.rsa.com/rsalabs/node.asp?id=2152> RSA notes that they stopped updating this FAQ page in 2000, so it is somewhat out of date. It is still useful in getting an overview.
"Timeline of cryptography." Wikipedia, The Free Encyclopedia. Retrieved 16 Aug 2009 <http://en.wikipedia.org/w/index.php?title=Timeline_of_cryptography&oldid=307201786>. This timeline is much fuller than the one given in this article.
Mihir Bellare and Phillip Rogaway, (2009) Introduction to Modern Cryptography, course notes for UCSD course CSE107. <http://cseweb.ucsd.edu/users/mihir/cse107/classnotes.html>. Effectively an undergraduate textbook for a course on cryptography.
Gary C. Kessler, An Overview of Cryptography. May 1998, updated 4 June 2009. Retrieved 16 Aug 2009. <>. A good introduction and overview with nice slides.
Menezes, van Oorschot, and Vanstone 1996 Handbook of Applied Cryptography available online at <http://world.std.com/~franl/crypto.html>. Retrieved 16 Aug 2009. This is a textbook that goes into the details of how to practically implement cryptographic algorithms. Since it is over 10 years old it is not cutting edge, but it gives a wealth of details. It is not at a beginner's level.
Bruce Schneier (2009) Cryptogram newsletter. Retrieved 16 Aug 2009 <>. Schneier is one of the leading figures of professional cryptography. The newsletter covers current developments.
Practical Cryptography, Niels Ferguson and Bruce Schneier, 2003 Hoboken, N.J., Wiley and Sons. <http://www.schneier.com/book-practical.html>. A very readable book and a good source.
## Selection of Research and Summary Articles
Following is a brief selection of articles from the technical literature. These works provide insight into some of the issues and research questions facing mathematicians and information scientists and the computational methods they enlist to generate theory and practice. Teaching faculty are encouraged to use the technical materials to inform classroom discussions and relate contemporary investigations to concepts covered in lectures. Students can use the works for self-directed study in conjunction with their textbooks, the , and/or the .
Preceding the technical papers are two topical overviews from Plus Magazine that provide accessible, non-technical summaries cryptography as well as the 2008 Chauvenet Prize winning essay from the Bulletin of the American Mathematical Society. This elaboration of a lecture given at the Current Events special session of the 2004 American Mathematical Society annual meeting explains the Agrawal-Kayal-Saxena (AKS) primality test--an algorithm that determines whether a number is prime or composite within polynomial time--with complete proofs, and to put the result and ideas in appropriate historical context.
Topical Overviews
Artur Ekert (2005) Cracking codes Plus Magazine. Issue 34.
A description of classic cryptography and some classic attacks.
Artur Ekert (2005) Cracking codes, part II Plus Magazine. Issue 35
An introduction to quantum cryptography.
American Mathematical Society Lecture
Andrew Granville (2004) It Is Easy To Determine Whether a Given Integer is Prime Bulletin of the American Mathematical Society. Volume 42, number 1.
A semi-technical essay in pure mathematics on prime numbers and methods for determining primality.
Technical Literature
1. Daniele Micciancio (2010) The RSA Group is Pseudo-Free Journal of Cryptography. Volume 23, number 2.
Publisher: Springer
Motivation: Free groups--groups where if there is a subset S of a group G such that any element of G could be written in one and only one way as a product of finitely many elements of S and their inverses--are widely used in computer science, and most modern cryptography relies on the hardness of computational problems over finite groups. So, as argued by Rivest [1], pseudo-free groups are a very interesting notion from a cryptographic perspective.
Methods: The main question left open by Rivest's work in [1] is: do pseudo-free groups exist? In [1] Rivest suggested the RSA group (where N = PQ is the product of two large primes) as a possible candidate pseudo-free Abelian group and nicknamed the corresponding conjecture the super-strong RSA assumption. In this paper we resolve Rivest's conjecture and prove that is pseudo-free under the strong RSA assumption, at least when N = PQ is the product of two "safe primes" (i.e., odd primes such that p = (P - 1)/2 and q = (Q - 1)/2 are also prime1), a special class of prime numbers widely used in cryptography*.
Results/Conclusions: The author has offered the first example of provably secure pseudo-free group under standard cryptographic assumptions. In particular, the paper proved that the RSA group where N is the product of two safe primes is pseudo-free, assuming the hardness of the strong RSA problem.
[1] R.L. Rivest, On the notion of pseudo-free groups. In: Theory of Cryptography Conference-Proceedings of TCC'04. LNCS, vol. 2951, pp. 505-521 (2004)
• Equivalently, using more standard mathematical terminology, P = 2p +1 and Q= 2q +1 where p and q are Sophie Germain primes.
2. Chin-Wen Chou, Julien Laurat, Hui Deng, Kyung Soo Choi, Hugues de Riedmatten, Daniel Felinto, H. Jeff Kimble (2007) Functional Quantum Nodes for Entanglement Distribution over Scalable Quantum Networks. Science. Vol. 316. no. 5829, pp. 1316 - 1320 Article available from Science in full text after free registration
Motivation: A developing area of cryptography using quantum physics where measuring a state changes it. That allows a communication system where listening in is easily observable.
Methods: This paper looks at the issues of sending quantum bits along a network. In particular the experiment describes the entanglement of quantum states at nodes 3 meters apart. In theory, this allows for a larger scale network.
Results/Conclusions: The method of secure communication is not currently practical for broad usage.
3. Stephanie Wehner, Andreas Winter, Harry Buhrman, Matthias Christandl, Falk Unger (2006) Implications of superstrong non-locality for cryptography. Proceedings of the Royal Society A 462 no. 2071 1919-1932
Publisher: Royal Society Publishing
Motivation: This study looks at using non-local boxes-- hypothetical 'machines' that give rise to superstrong non-local correlations--as a method of getting stronger cryptographic results than can be obtained through quantum mechanics alone.
Methods: The workers demonstrate how to circumvent the problem of delay implicit in earlier work*, which suggested that a 1-2 oblivious transfer (OT)--a protocol where a sender sends information to a receiver, but remains oblivious as to what is received--could be constructed using one non-local box alone. The investigators construct a protocol for bit commitment (BC)--a commitment scheme that allows one to commit to a value while keeping it hidden, with the ability to reveal the committed value later--and 1-2 OT based on non-local boxes. This shows that superstrong non-local correlations in the form of non-local boxes enable the researchers to solve cryptographic problems otherwise known to be impossible.
Results/Conclusions: This paper should be considered in contrast to Chou et al., paper 2 above, which looks what can currently be achieved in cryptography through quantum mechanics. This paper considers what could be achieved with an ideal device that mimics some of the characteristics of entangled states in quantum mechanics.
• Wolf, S. & Wullschleger, J. 2005 Oblivious transfer and quantum non-locality. In Proc. Int. Symp. on Information Theory (ISIT), pp. 1745-1748.
## Discussion Questions
a. If I want to keep a message secret for 20 years, how big a safety factor should I include for a brute force attack? (Give your best guess, how much faster will computers be in 20 years? How much faster is a typical computer today compared with one sold 20 years ago?)
b. A brute force attack on AES checks 2^128 or about 10^38 keys. How many keys would a computer need to be able to solve? per second to do a brute force attack in 10 years? How does that compare with the current speed of the fastest supercomputer in existence?
c. Many of the public key cryptosystems start by picking a number that is "probably prime", understanding that the system will be much weaker if the chosen number is not a prime. What is an acceptable failure rate for cryptography?
d. In a typical day, how often do you communicate electronically with the hope or expectation that your communication is secret? For what percentage of these communications can you find the encryption method used?
e. Cryptography is one of the few branches of mathematics that is restricted as a matter of law. Strong cryptography protects you from identity theft but also allows criminals and spies to hide their activities. What is the proper role of government regulation in this area? (Proposed answers range from escrow accounts for keys where the government can open any encrypted material to no regulation at all.)
| 5,507 | 25,861 |
{"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-2014-49
|
longest
|
en
| 0.954195 |
https://www.shaalaa.com/question-bank-solutions/find-the-elasticity-of-demand-in-terms-of-x-for-the-following-demand-laws-and-also-find-the-value-of-x-where-elasticity-is-equal-to-unity-p-a-bx-2-applications-of-differentiation-in-business-and-economics_218213
| 1,680,110,553,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2023-14/segments/1679296949009.11/warc/CC-MAIN-20230329151629-20230329181629-00281.warc.gz
| 1,082,342,286 | 9,905 |
Tamil Nadu Board of Secondary EducationHSC Commerce Class 11th
# Find the elasticity of demand in terms of x for the following demand laws and also find the value of x where elasticity is equal to unity. p = (a – bx)2 - Business Mathematics and Statistics
Sum
Find the elasticity of demand in terms of x for the following demand laws and also find the value of x where elasticity is equal to unity.
p = (a – bx)2
#### Solution
p = (a – bx)
= "dp"/"dx" = 2("a" - "b"x)^(2-1) "d"/"dx" ("a" - "b"x)
= 2(a – bx) (0 – b(1))
= -2b(a – bx)
Elasticity of demand: ηd = - "p"/x * "dx"/"dp"
= (- ("a" - "b"x)^2)/x xx 1/("dp"/"dx")
= (- ("a" - "b"x)^2)/x xx 1/(- 2"b"("a" - "b"x))
ηd = ("a" - "b"x)/(2"b"x)
When the elasticity of demand is equals to unity,
("a" - "b"x)/(2"b"x) = 1
a – bx = 2bx
2bx = a – bx
2bx + bx = a
3bx = a
x = "a"/"3b"
∴ The value of x when elasticity is equal to unity is "a"/"3b"
Concept: Applications of Differentiation in Business and Economics
Is there an error in this question or solution?
#### APPEARS IN
Chapter 6 Applications of Differentiation
Exercise 6.1 | Q 5. (i) | Page 139
Share
| 395 | 1,130 |
{"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}
| 4.1875 | 4 |
CC-MAIN-2023-14
|
latest
|
en
| 0.842001 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.