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http://stackoverflow.com/questions/12858806/stereographic-sun-diagram-matplotlib-polar-plot-python?answertab=votes | 1,433,244,722,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1433195035744.13/warc/CC-MAIN-20150601214355-00050-ip-10-180-206-219.ec2.internal.warc.gz | 157,796,910 | 17,632 | # Stereographic Sun Diagram matplotlib polar plot python
I am trying to create a simple stereographic sun path diagram similar to these: http://wiki.naturalfrequency.com/wiki/Sun-Path_Diagram
I am able to rotate a polar plot and set the scale to 90. How do I go about reversing the y-axis? Currently the axis goes from 0>90, how do I reverse the axis to 90>0 to represent the azimuth?
I have tried:
``````ax.invert_yaxis()
ax.yaxis_inverted()
``````
Further, how would I go about creating a stereographic projection as opposed to a equidistant?
My code:
``````import matplotlib.pylab as plt
testFig = plt.figure(1, figsize=(8,8))
rect = [0.1,0.1,0.8,0.8]
testAx.invert_yaxis()
testAx.set_theta_zero_location('N')
testAx.set_theta_direction(-1)
Azi = [90,180,270]
Alt= [0,42,0]
testAx.plot(Azi,Alt)
plt.show()
``````
Currently my code doesn't seem to even plot the lines correctly, do I need need to convert the angle or degrees into something else?
Any help is greatly appreciated.
-
The short answer is that you want a stereographic projection instead of a polar projection. However, this means that you'll either have to a) subclass the `Axes` yourself (have a look at `projections.geo_axes` in matplotlib, or b) adapt existing code to do what you want. (Just to plug something of my own, mplstereonet: github.com/joferkington/mplstereonet can be adapted to this, but it's intended for geologic data.) I'm busy for the next day or two, but I'll try to post an example of both if someone doesn't beat me to it. :) – Joe Kington Oct 12 '12 at 22:39
After lots of internet searching, I think that playing around with the y_scale may suffice: ScaleBase Example. Though I am not sure if a reverse scale will work. Will keep you posted. – ivvv Oct 15 '12 at 22:22
any progress on this? – tcaswell Dec 20 '12 at 23:30
@tcaswell hello, please see answer below. – ivvv Jan 10 '13 at 17:34
I finally had time to play around with matplotlib. After much searching, the correct way as Joe Kington points out is to subclass the Axes. I found a much quicker way utilising the excellent basemap module.
Below is some code I have adapted for stackoverflow. The sun altitude and azimuth were calculated with Pysolar with a set of timeseries stamps created in pandas.
``````import matplotlib.pylab as plt
from mpl_toolkits.basemap import Basemap
import numpy as np
winterAzi = datafomPySolarAzi
winterAlt = datafromPySolarAlt
# create instance of basemap, note we want a south polar projection to 90 = E
myMap = Basemap(projection='spstere',boundinglat=0,lon_0=180,resolution='l',round=True,suppress_ticks=True)
# set the grid up
gridX,gridY = 10.0,15.0
parallelGrid = np.arange(-90.0,90.0,gridX)
meridianGrid = np.arange(-180.0,180.0,gridY)
# draw parallel and meridian grid, not labels are off. We have to manually create these.
myMap.drawparallels(parallelGrid,labels=[False,False,False,False])
myMap.drawmeridians(meridianGrid,labels=[False,False,False,False],labelstyle='+/-',fmt='%i')
# we have to send our values through basemap to convert coordinates, note -winterAlt
winterX,winterY = myMap(winterAzi,-winterAlt)
# plot azimuth labels, with a North label.
ax = plt.gca()
ax.text(0.5,1.025,'N',transform=ax.transAxes,horizontalalignment='center',verticalalignment='bottom',size=25)
for para in np.arange(gridY,360,gridY): | 921 | 3,335 | {"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} | 2.65625 | 3 | CC-MAIN-2015-22 | latest | en | 0.91192 |
https://ebrary.net/177215/mathematics/formal_teaching_addition | 1,638,680,951,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363135.71/warc/CC-MAIN-20211205035505-20211205065505-00548.warc.gz | 290,808,084 | 10,673 | Formal Teaching of AP of Addition
Insights from Chinese Lessons
In my project, the same two G4 teachers who formally taught the CP (see Section 4.3) also taught the AP of addition as the 2nd part of their lessons. Note that the textbook lesson only contained one worked example situated in a context of children playing with two sub-questions. In other words, the example about the CP only used part of the initial story context (see Figure 4.2); yet, the example about the AP involved all three conditions and asked students to figure out how many children were jumping rope and kicking the shuttlecock (see Figure 4.8).
Both teachers approached this worked example in similar ways. In particular, the overall sequence of representation made use of concreteness fading and shifted focus from the specific to the general. There were times when the teachers folded the abstract back to the concrete to promote meaning-making. In addition, both teachers asked deep questions to prompt students’ thinking. Below are the main activities involved in the teaching of the worked example about the AP. Elaboration follows:
• • Solve the problem and explain multiple solutions
• • Compare the selected solutions to obtain an instance of AP
• • Pose or try more examples to verify the observations
• • Formally reveal the AP and represent it with letters
• • Compare the AP with CP
• 1 Solve the problem and explain multiple solutions. Both teachers asked students to solve this problem using as many methods as they could think of. In one class, a student reported all six permutations of the sum that could be found by rearranging the three numbers. For each solution, the teacher prompted the student to explain, asking, “Can you tell us how you thought of these solutions?” or “Can you explain what you found first and what you found next?” Both teachers then chose the solutions that could illustrate the AP for further discussions. Below is a typical excerpt:
T: Let’s grasp these two typical number sentences [On the board, it shows “(28 + 17) + 23” and “28 + (17 +23)”]. What does this first sentence mean? Can you explain it based on the story problem?
Figure 4.8 Worked example used to teach AP of addition in a Chinese G4 lesson. This G4 task was cited front Sun & Wang (2014, v.2, p.55).
S: First find the number of people jumping ropes and then add the number of people kicking shuttlecock.
T: Oh, first find out the number of people jumping ropes. The people jumping ropes plus the people kicking shuttlecocks equals the total. Continue, how about the second one?
S: The second is to first find the number of girls who attended this event, and then add the number of boys.
T: Okay, first find the number of girls, and then add the number of boys, which equals the total. (To the class) Is this all right?
Ss: Yes!
Solving a two-step addition story problem is a piece of cake for Chinese fourth graders. However, the purpose of solving this worked example is to illustrate and make sense of the AP based on this story context. As such, both G4 teachers invited students to explain the meaning of both solutions. As seen from the above episode, the G4 students clearly explained each step of the number sentence in ways that folded the abstract representation back into the concrete situation.
2 Compare the selected solutions to obtain an instance of the AP. After students explained their solutions, both teachers invited them to compare the number sentences, resulting in an instance of the AP: (28 + 17) + 23 = 28 + (17 + 23). Furthermore, both teachers prompted students to compare the two number sentences focusing on computation process, “Just based on the number sentences and the computation process, how are they different?” and “Let’s look at these number sentences, what do you find?” Students in both classes noted that the order (position) of the three numbers remained the same, yet the order of operations (position of the parenthesis) changed. Based on students’ observations, both teachers concluded that “When adding three numbers, regardless of whether the first two numbers or the latter two numbers are added first, the answers remained the same.”
As a side note, in one class, the teacher chose three typical solutions, (17 + 23) + 28 = 28 + (17 +23) = (28 + 17) + 23. A comparison of the first two solutions enabled her class to revisit the CP that was somewhat hidden in this context. This is an interesting contrast to the to-be-reported U.S. lesson where the teacher told students that the AP was the property that dealt with parentheses. Clearly, the Chinese example here displays the inaccuracy of that statement since (17 + 23) + 28 = 28 + (17 + 23) makes use of parentheses but is an application of the CP not the AP. Student who try to identify the correct property by spotting parentheses could benefit from this example.
• 3 Try and pose more examples to verify the observations. Similar to the teaching of the CP (see Section 4.3), the above conclusions based on the instance of the AP was brought out for questioning. Both teachers asked the class whether the features observed from this group of number sentences held true. They then asked students to try more pairs of number sentences provided by the textbook. Students in one class also came up with their own examples that were either typical or special. And, several students stated in their own language that regardless of the parenthesis, one always adds the same three numbers.
• 4 Reveal the property and represent it with letters. Based on the above sense-making and verification process, the name of the AP was revealed to students. One teacher posed a further question, “This is called associative property. Why is it called ‘associative’? Think about it carefully. Here is a parenthesis.” Next, both teachers asked their students to use letters to represent the AP, (a + b) + c = a + (b + c), just as they did with CP of addition.
• 5 Compare the AP with the CP. After the AP was formally introduced, both teachers guided the class to look back on both properties introduced in this lesson. The dialogue below shows how one G4 teacher guided the class to differentiate between both properties (see Figure 4.9).
T: Let’s look at these two properties. One is the communitive property of addition. The other one is?
S: The associative property.
T: Commutative property of addition mainly refers to (switch his hands back and forth).
S: Switching the position [of numbers],
T: And, for the associative property of addition, does the position [of numbers] change?
S: No.
T: But? (gesturing the parenthesis)
S: Parenthesis.
T: What happens to the order of operations?
S: It changes!
As indicated by the excerpt, it did not take much time to guide students
to explicitly compare the CP and AP. Yet, such a comparison is arguably
Figure 4.9 Explicit comparison between CP and AP. Redrawn by Anjie Yang.
critical as evidenced by the conflation between the CP and AP that exists even among undergraduate students (Larsen, 2010). In the above excerpt, it was made clear that the CP was to change the order of numbers while the AP was to change the order of operations. Later in the practice tasks, both classes discussed the textbook task, 75 + (48 + 25) = (75 + 25) + 48, which contained both CP and AP. Both teachers made sure that their classes understood how both properties were involved. | 1,604 | 7,413 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2021-49 | longest | en | 0.964776 |
https://computeraidedfinance.com/2012/10/01/what-is-historical-volatility-and-why-do-we-need-implied-volatilities/ | 1,685,299,330,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644506.21/warc/CC-MAIN-20230528182446-20230528212446-00707.warc.gz | 211,809,146 | 33,045 | # What is Historical Volatility and Why Do We Need Implied Volatilities?
Authors
Looking at financial instruments, one often finds the term implied volatility. In this post, we want to describe what it is and what you can do with it. We start refreshing the term historical volatility and then we explain the implied volatility. An example of German DAX with real data concludes this post.
## Historical Volatility
Historical volatility measures the deviation of asset returns from their historical average. It is a sample estimate that has time unit and depends on the length of the historical sample period. Denote $R_{i,t}$ as the asset return $i$ at time $t$, the historical volatility of $R_{i,t}$ for $t = 1, \cdots, N$ is calculated as
$\sigma_i = \sqrt{ \frac{1}{N}\sum_{t=1}^N (R_{i,t}-E(R_{i})) ^2 }$,
where $E(R_{i})$ is the mean of $R_{i,t}$ over the time period $N$. To obtain the sample estimate $\hat{\sigma}_{i}$, the number $E(R_{i})$ is replaced by its sample counterpart $\hat{R}_{i} = \frac{1}{N}\sum_{t=1}^N R_{i,t}$, and
$\hat{\sigma}_{i} = \sqrt{\frac{1}{N - 1}\sum_{t=1}^N (R_{i,t} -\hat{R}_{i})^2}$,
where we have divided the sum of squared deviations by $N - 1$ to account for the loss of degree of freedom from using the sample mean $\hat{R}_{i}$ in the calculation. For $R_{i,t}$ sampled at daily and monthly time intervals, the annualized historical volatility is respectively $\sqrt{250} \hat{\sigma}_{i}$ and $\sqrt{12} \hat{\sigma}_{i}$.
Historical volatility can be useful in cases like benchmarking historical performances, assessing the behavior of economic variables, calculating historical risk measures, and so on.
Since historical volatility is the number we get from looking into the past realized data, it may not be as useful when we try to predict the future behavior of asset returns. This is when and why we have the implied volatility, i.e. the volatility number implied from market prices.
## Implied volatility
Denote $P_{i, market}$ , $i = 1, \cdots, M$, a set of market prices of traded options on asset $i$, the implied volatility $\tilde{\sigma}_{i}$ is obtained by setting the market price to the theoretical price, $P_{i, market} = P_{i, model}(K, S_{i}, \tilde{\sigma}_{i}, \tau, r, \cdots)$, with the argument $K$ as the option strike price, $S_{i}$ the current asset price, $\tau$ is the option time-to-maturiy, $r$ the risk-free interest rate, and so on. Since the only unknown in $P_{i, market} = P_{i, model}(K, S_{i}, \tilde{\sigma}_{i}, \tau, r, \cdots)$ is the volatility number $\tilde{\sigma}_{i}$, we can solve the inverse pricing function to obtain the implied volatility $\tilde{\sigma}_{i}$ using the bisection or Newton‘s iterative method. The obtained implied volatilities contain information of the market’s assessment of future price movements, and thus is a forward-looking risk measure. Here, you can find a little solver for implied volatilities in VBA.
## Example: German DAX
The following graphs show the series of VDAX-NEW implied volatility index compared to the DAX historical volatilities. The implied volatility index is a good predictor of the historical volatilities.
(click to enlarge)
We can see that the implied volatility given by VDAX New and the 21-day historical volatility had similar values during the time presented. Most of the time, the implied volatility is slightly higher than the historical volatility.
(Click to enlarge)
The graph above shows the VDAX shifted by 21 days and the 21 day historical volatility. This way, we can see how well VDAX performs as a predictor of the historical volatility. We see that except the unforeseen spikes, implied volatility can be a good predictor of (future) historical volatility. Now, a similar graph with 1 year time horizons:
(click to enlarge)
With a one year horizon, the relationship between historical volatility and implied volatility is not that obvious.
An in-depth study of the predictability of historical volatility by implied volatility can also be found here in Wilmott Magazine.
## Conclusion
While the historical volatility describes the size of the variations of the past, the implied volatility describes the size of the variations of the futures, implied by the traded option prices. | 1,050 | 4,265 | {"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": 30, "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-2023-23 | latest | en | 0.881376 |
http://mathhelpforum.com/math-topics/87995-graph-constant-improving-print.html | 1,516,475,317,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889681.68/warc/CC-MAIN-20180120182041-20180120202041-00252.warc.gz | 213,599,969 | 2,960 | # graph, constant and improving...
• May 7th 2009, 06:53 AM
User Name
graph, constant and improving...
• May 7th 2009, 06:56 AM
stapel
Try plugging various values in for "c", graphing in your calculator or other software, and tracking what the changes are. (Wink)
• May 7th 2009, 06:57 AM
mr fantastic
Quote:
Originally Posted by User Name
Define $f(x) = 2 [e^{x/2} + e^{-x/2}]$.
Then your function is $f\left(\frac{x}{c}\right)$ and so the effect of positive $c$ is to dilate in the horizontal direction by a factor of $c$. | 170 | 526 | {"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": 4, "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} | 2.71875 | 3 | CC-MAIN-2018-05 | longest | en | 0.852587 |
courses.as.eku.edu | 1,675,369,228,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500041.18/warc/CC-MAIN-20230202200542-20230202230542-00409.warc.gz | 193,093,416 | 5,291 | ### Linear Motion and Kinectics
##### Length 15:02
Transcript
[Video begins] [Scene displays a Power Point presentation] [This document contains only the audio portion of this presentation]
[Video begins] [Scene displays a Power Point presentation] [This document contains only the audio portion of this presentation] [Ciocca] Welcome to our second video lecture. This video lecture is about chapter three, in particular linear motion. If you notice we are going to chapter three first then we are going to go back to chapter two. The reason being that in chapter three we can introduce some quantities that will be useful to understand chapter two better. So, without any further ado, this is chapter three linear motion.
As the name suggests, this is the motion that occurs when the movement is in a straight line. There will be no curves to it, just a straight line motion whether it is up and down, left to right, doesn’t matter. But, in this case the motion will be straight. In order to understand the quantities that will come later, the first quantity that we are going to introduce is the quantity called speed. You are probably familiar with it, but you have to remember that quite often in physics we use terms that you think you know but in physics they are very precise and perhaps a different meaning from the one you know. In this case this is not the case, but we will encounter some.
Anyway, speed is defined as the distance covered per amount of travel time. So, notice that the quantity is distance divided by time. Now, in physics and the sciences in general quantities have units. So, in this particular case, distance in physics is measured using the quantity known as meter. And, time is measured in seconds. That’s pretty common. Meters are not as common as the second is. Second is used everywhere, the world over. Meters in used in science mostly and in Europe. In the United States they use feet, yard, inches. In the sciences however, we are going to limit ourselves to using the meter. So, in equation form the speed is defined as distance divided by time, and in unit as meters per second. As an example, example one, this girl runs 4 meters in 2 seconds. Notice that there is no direction involved here so we are assuming that she is moving in a straight line. And, her speed will be 2 meters per second. Sometimes it will be useful to define a quantity called average speed. Notice the term average indicates that this is not the speed instant by instant, but more like the speed that this object moving has over the entire motion. Here the total distance—notice the word total—is divided by the total travel time. So, it is still distance over time, but this time we don’t pay attention to the details of the motion. We look at the entire movement divided by the entire time as to indicate the various instantaneous speed along the way. For example, when you come to school from home you enter the car and you start the engine. Your speed at that time is zero, you haven’t moved much. However, you start moving, drive towards school, you have a red light, you stop. You were moving at a certain speed and then when you stop you’re moving at another.
To kind of give an idea of the total motion, we introduce average speed as the total distance covered divided by the time interval. It still has the units of meters per second, but has a different meaning. For example, you drive a distance of 20 kilometers in 2 hours. Well, your average speed is clearly two hundred divided by 2 hours, or 100 kilometers per hour. You may have gone faster sometimes than 100 kilometers per hour, sometimes slower, but your average is one hundred. It’s useful though to still have the idea of instantaneous speed. Instantaneous speed is the speed at any instant at a particular time. For example, when you ride in your car you might speed up or slow down. Your odometer will tell you that at that instance in time you are doing a hundred and then a few minutes later will be doing at ninety, and then another time later 120 kilometers per hours, kilometers per hours, and so on and so forth. You are doing that speed at that instance in time, not all the time. Instantaneous speed therefore can be viewed as the one given by your speedometer in the car. This is somewhat new.
Now, we are going to introduce velocity. Velocity is the description of both the instantaneous speed of the object and the direction of travel. So, now we are going to specify not only how fast, which is given by the instantaneous speed, but also which way. In other words, we are introducing a quantity that is called vector. It’s a quantity that has a magnitude indicating how long—how big is the vector and a direction which indicates which way the particular vector is going. So, velocity is a vector quantity. It has magnitude, which in this case will be the speed, and a direction. So, velocity can be seen as directed speed. Here’s some more definitions. Constant speed, as the name suggests, is a speed that doesn’t change. The object is not speeding up nor is it slowing down. Constant velocity on the other hand is something that is a bit more strict. Not only the speed is constant, but also it’s a constant direction. That is, it’s a straight line path with no acceleration. The velocity that is constant means that the speed is fixed and the direction is fixed. All these quantities, speed and velocity, that we have introduced are all relative Earth unless specified otherwise.
Now, we are going to introduce the idea of acceleration. Acceleration was formulated by Galileo based on his experiments with inclined planes and it is the rate at which the velocity changes over time. Let’s write it down in equation format. You will see that on the next slide. For example, imagine the following. If you have a ball rolling down an inclined plane, like in this case here, the slope is downwards so the speed of the ball rolling down will become bigger and bigger. This is called positive acceleration. In the slope upward the motion is in the direction of the arrow, as you can see, but the speed is decreasing. That is, the speed is getting smaller. That implies that the acceleration in this case is in the opposite direction. That is, it is this way. In this case, the acceleration on the other end is that way. Notice that these two arrows are pointed in the same direction, these are opposite, so we can conclude in this case you can say that the acceleration is opposing to the speed. Therefore, this can be viewed as negative acceleration. If no slope and the ball is rolling at a constant speed, a speed that has no change, then this constant velocity. Be careful. When we talked about acceleration this involves either a change in speed or a change in direction or both.
So, for example if you have a car making a turn or a car going around a track and it is moving at the same constant speed, but notice that the trajectory is a circle. This implies that the velocity is changing even though the speed may not.
An example again, imagine you have a car that goes around a track like this. Your speedometer will be saying 100 kilometers per hour all the time as you go around the track in your car. Your speed is constant, but the velocity is not. Because, look. Here, this is the velocity V. As you go around the track when you are up here the velocity is this way. When you are here, the velocity is that way. When you’re here, the velocity is this way. You see it’s changing continuously. Therefore, in this case the speed is constant but the velocity is not. Therefore, you do have acceleration.
We are ready to give a definition of velocity in terms of an equation. The acceleration is defined as the change in velocity divided by the time interval. Notice the velocity and speed are of the same units. They are still meters per seconds and the time interval are still seconds. So, this quantity here, the acceleration, has the units of meters per second per second, or meters per second squared. So, if your car speed is say 40 kilometers per hour and 5 seconds later it is forty-five. The car change in speed is forty-five, 5 kilometers per hour, and your car acceleration is 1 kilometers per hour per second which is a very strange unit. We’ll try to use the units of time which are common for the two, was able to study the idea of acceleration by observing the motion of a ball down an inclined plane at large and larger angles. When incline is vertical the acceleration is maximum, the same of that of a falling object. When air resistance is negligible, all objects fall with the same unchanging acceleration. And, Galileo was able to show that the acceleration in this case of an object falling, a, is approximately 10 meters per second squared. In general, the more precise value of the acceleration in this case for freefall is given its own name because it is so constant. It is called g and is equal to 9.8 meters per second squared. As you can see, this is true only when there is no air resistance. So, this is called free fall. A free falling object on Earth accelerates at the rate of approximately 10 meters per second per second, or 10 meters per second squared. And, like I said before, more precisely 9.8 meters per second squared. I apologize for those big two’s, it should be seconds squared. …as an acceleration which is constant. What does it means in terms of velocity? It means that the velocity changes by the same amount every second. So, in this particular case, the velocity which is acceleration times time is given this way. Notice that you have an object falling and the free fall when acceleration is 10 meters per second it means that speed is 10 meters per second after 1 second, ten more after 2 seconds gives you twenty, ten more after 3 seconds gives you thirty, and so on. [Video ends] | 2,092 | 9,852 | {"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-2023-06 | latest | en | 0.961393 |
https://www.physicsforums.com/threads/kinds-of-infinity.553532/ | 1,660,244,450,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571483.70/warc/CC-MAIN-20220811164257-20220811194257-00128.warc.gz | 811,948,016 | 17,775 | # Kinds of infinity
Maatomaat
How many kinds of infinity does math have?
in my point of view "THERE ARE INFINITY KINDS OF INFINITY"
Maatomaat
Welcome to PF.
Define "infinity" and explain, in terms of this definition, how there is more than one "kind", with an example. Then, perhaps, there can be meaningful POVs.
Maybe you are thinking of Cantor sets and the ideas around them:
http://www.scientificamerican.com/article.cfm?id=strange-but-true-infinity-comes-in-different-sizes
Are you sure ? only "one" infinity?
We know "ℝ" is infinity.also "(1,0)" is infinity.
We can also say that we are able to place (1,0) on ℝ.so ℝ is more powerful than (1,0) so they are different.
I mean ℝ-(1,0) is also infinity.
espen180
It may look that way to you, but R and (0,1) are actually the same size!
Define the map $(0,1)\rightarrow \mathbb{R}$ such that $a\in(0,1)$ is sent to $\frac{1}{1-a}-\frac{1}{a}$.
This map gives a bijection between the two sets. Each and any point in (0,1) corresponds to one unique point in R. We can also reverse the map such that each and any point in R corresponds to one unique point in (0,1). Therefore the two sets have equal size.
Last edited:
Homework Helper
"We know "ℝ" is infinity.also "(1,0)" is infinity."
Technically these statements are incorrect according to the usual understandings of what "infinity" means. This is why it is important to define your terms in philosophical discussions. What you have shown is the set of real numbers and an interval on a numberline. Neither of these is infinity.
You have responded to my request to provide examples, but you have failed to define your terms or explain how these two examples illustrate the idea that there are different kinds of infinity.
espen180 has shown you by using a 1-1 mapping that these two sets are the same size. In this sense, they both have the same infinite number of members. So if you intended to distinguish different "types" of infinity by different sizes of infinity, that won't stand up.
Have another go.
Maatomaat
N is not equivalent to R.
So they are different.
Homework Helper
So they are different.
The cardinality (size) of N and R are different, mainly due to the fact that N is countable and R is uncountable. Another reason is that there is also no 1-1 correspondence between N and R
Homework Helper
You still have to define your terms.
If you mean to demonstrate that there are different sizes of infinity, then please say so.
You seem to be having trouble communicating the ideas you want to discuss.
Have a look at how Cantor handled the same problem.
Maatomaat
The cardinality (size) of N and R are different, mainly due to the fact that N is countable and R is uncountable. Another reason is that there is also no 1-1 correspondence between N and R
As a result we can say infinitys are different.
Maatomaat
You still have to define your terms.
If you mean to demonstrate that there are different sizes of infinity, then please say so.
You seem to be having trouble communicating the ideas you want to discuss.
Have a look at how Cantor handled the same problem.
Are you able to prove that there is only one infinity or can you reject the flowing statement:
"N , R are infinity but R is not equivalent to N"
it is a well-known theorem, first demonstrated by cantor, that:
if A is a set, then |A| < |P(A)|, whre P(A) is the power set of A, or set of all subsets of A.
if |A| is finite, this is obvious, and not very interesting.
if |A| = |N|, where N is the natural numbers, this yields the surprising result, that there exist uncountable sets. so that's 2 kinds of infinite: countable, and uncountable.
but if one continues with P(A), where A is uncountable, then one gets a set is that is "even more uncountable". it is possible to carry this reasoning on indefinitely, each time winding up with a "bigger infinity than before".
but there are other ways of thinking of "infinity" besides just "something not finite in size". for example, one can imagine that the distant horizon, is all one single point: the "point at infinity". this one single point (which is strange because we can approach it from any direction) acts like a boundary for what seems an endless plane, and it makes a flat euclidean plane act like a sphere (or the real number line act like a circle).
in this strange geometry, the hyperbola xy = 1 suddenly becomes a closed curve: the points at infinity connect. and this is yet another kind of infinity, which has nothing to do with number, per se, but has to do with space.
and here, again, we aren't limited to a single choice: instead of wrapping the plane into a sphere, we might want to distinguish between ±∞, which gives us a different kind of geometry.
and it gets weirder, still: we might declare some large number R to be the "largest finite number" (computers actually do something akin to this, called "overflow error handling"). now infinity is looking a lot smaller, and has "absorbing properties" similar to 0 (and, unfortunately, plays havoc with our usual algebraic rules).
these aren't the only possibilities. in some sense, infinity represents a choice: we might mean different things by it, and each meaning we assign to ∞, has different consequences. some of these definitions "break" the structure we add ∞ to: if we call ∞ a real number, we lose some of the field structure (algebra) of R. sometimes adding ∞ to our structure enhances it: if we add infinite numbers in a certain way, we can adapt induction to cover finite AND infinite cases.
i hope this gives you a small idea of how infinity comes in different flavors, it's not enough just to say: "∞".
Mentor
Are you sure ? only "one" infinity?
We know "ℝ" is infinity.also "(1,0)" is infinity.
Do you mean the interval (0, 1)? If that's what you mean, the interval (0, 1) is NOT infinity, but the cardinality of this interval is infinity.
We can also say that we are able to place (1,0) on ℝ.so ℝ is more powerful than (1,0) so they are different.
I can't see some of the symbols you wrote - they show up as squares in my browser. If you are saying that the interval (0, 1) can be placed on the real line, then what you are saying is incorrect. The cardinality of the interval (0, 1) is the same as the cardinality of the entire real number line.
I mean ℝ-(1,0) is also infinity.
No, I aren't. | 1,557 | 6,346 | {"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} | 3.96875 | 4 | CC-MAIN-2022-33 | latest | en | 0.93619 |
https://socratic.org/questions/how-do-you-verify-sin-x-pi-6-cos-x-pi-3-sqrt3-sin-x#590586 | 1,721,086,447,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514724.0/warc/CC-MAIN-20240715224905-20240716014905-00155.warc.gz | 481,354,259 | 6,324 | How do you verify sin(x + pi/6) - cos(x + pi/3) = sqrt3 sin x?
Apr 8, 2018
It is given that the LHS is $\sin \left(x + \frac{\pi}{6}\right) - \cos \left(x + \frac{\pi}{3}\right)$
Applying, color(magenta)(sin(A+B) = sinAcosB + cosAsinB and color(red)(cos(A+B) = cosAcosB - sinAsinB
color(white)(dd
$\implies \textcolor{m a \ge n t a}{\sin x \cos \left(\frac{\pi}{6}\right) + \cos x \sin \left(\frac{\pi}{6}\right)} - \left(\textcolor{red}{\cos x \cos \left(\frac{\pi}{3}\right) - \sin x \sin \left(\frac{\pi}{3}\right)}\right)$
color(white)(dd
$\implies \sin x \left(\frac{\sqrt{3}}{2}\right) + \cos x \left(\frac{1}{2}\right) - \cos x \left(\frac{1}{2}\right) + \sin x \left(\frac{\sqrt{3}}{2}\right)$
color(white)(dd
$\implies \cancel{2} \times \sin x \left(\frac{\sqrt{3}}{\cancel{2}}\right)$
$\implies \sqrt{3} \sin x$
Apr 8, 2018
See below.
Explanation:
Identities:
$\textcolor{red}{\boldsymbol{\sin \left(A + B\right) = \sin A \cos B + \cos A \sin B}}$
$\textcolor{red}{\boldsymbol{\cos \left(A + B\right) = \cos A \cos B - \sin A \sin B}}$
$L H S :$
$\sin \left(x + \frac{\pi}{6}\right) = \sin \left(x\right) \cos \left(\frac{\pi}{6}\right) + \cos \left(x\right) \sin \left(\frac{\pi}{6}\right)$
$\cos \left(x + \frac{\pi}{3}\right) = \cos \left(x\right) \cos \left(\frac{\pi}{3}\right) - \sin \left(x\right) \sin \left(\frac{\pi}{3}\right)$
$\sin \left(x + \frac{\pi}{6}\right) - \cos \left(x + \frac{\pi}{3}\right)$
$\sin \left(x\right) \cos \left(\frac{\pi}{6}\right) + \cos \left(x\right) \sin \left(\frac{\pi}{6}\right) - \left[\cos \left(x\right) \cos \left(\frac{\pi}{3}\right) - \sin \left(x\right) \sin \left(\frac{\pi}{3}\right)\right]$
$\sin \left(x\right) \cos \left(\frac{\pi}{6}\right) + \cos \left(x\right) \sin \left(\frac{\pi}{6}\right) - \cos \left(x\right) \cos \left(\frac{\pi}{3}\right) + \sin \left(x\right) \sin \left(\frac{\pi}{3}\right)$
$\cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$
$\cos \left(\frac{\pi}{3}\right) = \frac{1}{2}$
$\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$
$\sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$
$\sin \left(x\right) \frac{\sqrt{3}}{2} + \cos \left(x\right) \left(\frac{1}{2}\right) - \cos \left(x\right) \left(\frac{1}{2}\right) + \sin \left(x\right) \frac{\sqrt{3}}{2}$
$2 \sin \left(x\right) \frac{\sqrt{3}}{2}$
$\sqrt{3} \sin \left(x\right)$
$L H S \equiv R H S$ | 977 | 2,372 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 26, "mathjax_inline_tex": 1, "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} | 4.34375 | 4 | CC-MAIN-2024-30 | latest | en | 0.317419 |
http://www.jiskha.com/display.cgi?id=1208318460 | 1,495,819,596,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608669.50/warc/CC-MAIN-20170526163521-20170526183521-00373.warc.gz | 673,662,197 | 3,917 | # math
posted by on .
solve the following trig equation :
7 = 8sin(1/2)x. I can solve this using the half-angle indentity. But i need to solve this using period and phase change equation. how do i do this?
• math - ,
7 = 8sin(1/2)x
sin (1/2)x = .875
so (1/2)x is in the first or second quadrant, and
(1/2)x = 61.04º or (1/2)x = 118.96º
then x = 237.9º or 122.1º
but the period of your curve is 720º
so by adding/subtractin 720º to/from any of the found angles will give you more answers.
in general
x = 122.1 + 720k or
x = 237.9 + 720k , where k is an integer. | 200 | 570 | {"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-2017-22 | latest | en | 0.828868 |
https://www.myofficetricks.com/useful-excel-tips-you-should-know/ | 1,723,513,599,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641054522.78/warc/CC-MAIN-20240813012759-20240813042759-00745.warc.gz | 707,699,953 | 7,630 | # Useful Excel Tips & Tricks You Should Know
Date：2020-11-23
This is a perfect and excellent article If you still use Excel frequently in your work or study, The so-called multi-skills are not pressure. So, today, the author will share some useful Excel tricks. Come and learn!
### Skill 1:How to Make Your Excel Data More Intuitive?
When you first look at this table, can you quickly find the largest and smallest sales data? You would check and compare these data carefully. And it will take you several seconds.
How about this table. You must can quickly find which is the largest data and which is the smallest data. It’s easier than the first table, right? Let’s get to know how to make it.
Step1: Use mouse to select the data.
Step2: Click the [Conditional Formatting] in tool bar.
Step3:Click [Data Bars]
Step4:Choose the color you like or the color you want to make the data be.
### Skill 2: How to Quickly Enter Recurring Content?
Have you ever encountered this situation? You need to enter the same content repeatedly. You have to type again and again or keep copying and pasting. Actually, we can use the function of autocorrect. For example, every time we enter a URL in a cell, the Excel cell will automatically display https://www.myofficetricks.com
Step1: Click [File]
Step2: Click the [Option] in the pop-up Window
Step3: Click [Proofing] in the Excel Options, and choose [AutoCorrect Options]
Step4: Enter the information you need to automatically correct into the corresponding box. Then click [Add] and OK. These all the steps.
### Skill 3: How to Remove Extra Blank Lines?
Will you delete the blank lines one by one? The author wants to teach you a simple method that even doesn’t need to delete. The way we used is Data Sort. Detailed steps are in the Gif below.
### Skill 4: How Do You Encrypt Excel Form?
If you have some documents that need to be kept secret or data that you cannot disclose temporarily. It is recommended that you encrypt your form. This can protect your data.
Step1: Click [Save As]
Step2: Click on [Tools]
Step3: Choose [General Options] | 491 | 2,103 | {"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} | 2.65625 | 3 | CC-MAIN-2024-33 | latest | en | 0.884666 |
http://electricala2z.com/electrical-power/advantages-three-phase-system/ | 1,568,769,627,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573173.68/warc/CC-MAIN-20190918003832-20190918025832-00253.warc.gz | 63,413,655 | 16,085 | Home / Electrical Power / Advantages of Three-Phase System
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A three-phase system has three generating windings, each out of phase with the other two by 120ºE. Figure 1 (a) illustrates a simplified three-phase alternator with three windings A, B and C at 120ºE intervals. Figure 1 (b) shows the waveforms generated by the three windings, illustrating the phase shift between waveforms, and Figure 1 (c) shows the phasor diagram of the three phasors.
Figure 1 A three-phase machine
The phasor of the A-phase voltage VA is shown as the reference phasor, drawn horizontally to the right i.e. at 0º. The phasor of the B-phase voltage VB is drawn 120º after VA, which puts it at the bottom left of the diagram. Remember that the phasors rotate in a counterclockwise direction, so lagging phasors are clockwise from the reference phasor. The phasor of the C-phase voltage VC is drawn 240º after VA which puts it at the top left of the diagram, but note that it is also 120º counterclockwise from VA, which completes the cycle.
## Generating a Three-Phase Supply
A three-phase supply is produced by an electric machine that contains three windings. The windings are separated by 120º (electrical), as shown in Figure 1, to naturally produce three voltage waveforms 120º apart.
The three phases follow in a fixed sequence, which is known as the ‘phase sequence or ‘phase rotation’. Therefore, the three phases must be identified to allow the sequence to be stated and to enable loads to be balanced across the phases. In some applications, the letters A, B and C are used as A-phase, B-phase and C-phase.
For general-purpose supply and distribution identification in Australia, the phases are given the color coding red, white and blue (prior to 1981 the colors were red, yellow and blue).
Phases may also be marked as L1, L2 and L3 meaning Line 1, Line 2 and Line 3. European standards often use U–V–W or u–v–w to represent the three phases.
As the phasors rotate, they pass through the reference position in the order red, white, blue. This sequence must be followed in connecting equipment on three-phase circuits, to ensure that motors rotate in the expected direction.
When three-phase phasors are drawn, the usual practice is to draw the red phasor (A) in the reference position; if there is a particular advantage, either the white (B) or the blue (C) phasor may be used as the reference as long as the sequence is still correct.
## Advantages of a Three-Phase System
The advantages of a three-phase system include:
1 For the same size or weight, a three-phase machine can produce higher outputs than a single-phase machine. 2 A three-phase machine can be smaller than a single-phase machine for the same power output. 3 The power delivered to or taken from a three-phase system has a more constant value. In a single-phase system, the power pulses at twice the line frequency. With three phases, the power pulses are six times the line frequency, with far less amplitude than single-phase power. Since the power is more constant, the torque of a rotating machine is more constant, and this results in much less vibration from the machine. 4 With one type of three-phase connection, there are two voltages available: 230/400 V. 5 In a distribution system, the total quantity of material needed for three conductors is less than that required for the equivalent single-phase system (due to higher efficiency).
## Three-Phase Winding Arrangements
Unlike DC machines, the poles in a 3Ø machine generally overlap and this is a factor in balancing the current and power of a 3Ø machine. Figure 2 shows a typical 24-slot stator lamination set. To use this for a 3Ø 4-pole electric motor, there must be three sets of coils for 4 poles fitted into 24 slots. This would normally result in a winding of 3 × 4 × 2 coils, meaning that two coils form one pole for one phase.
Figure 2 A three-phase induction motor winding
Phase A is drawn in red, with one side of the coils on the outside and the other on the inside of the laminated stator. Phase B is white and phase C is blue. The motor is ‘4-pole’, which means that there are four sets of coils (poles) for each phase. It can be seen that each phase occupies one-third of the total number of slots. A 4-pole machine has 720ºE in one complete rotation (360ºM).
Three-Phase Machine Alternator Construction
Basically an alternator consists of coils rotating in a magnetic field or, in an alternative form which has many advantages, the ac windings are stationary and the magnetic field system rotates.
The same basic principles apply equally to both single- and three-phase alternators, the only real difference being whether there is one winding or three identical windings.
Did you find apk for android? You can find new Free Android Games and apps. | 1,131 | 4,867 | {"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} | 3.25 | 3 | CC-MAIN-2019-39 | longest | en | 0.917119 |
http://www.cfd-online.com/Forums/openfoam/95709-yes-no-question-about-poiseuille-flow.html | 1,475,098,273,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738661767.35/warc/CC-MAIN-20160924173741-00001-ip-10-143-35-109.ec2.internal.warc.gz | 382,748,034 | 17,085 | # Yes-No question about poiseuille flow
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December 29, 2011, 12:42 Yes-No question about poiseuille flow #1 Member Amin Shariat KHah Join Date: Apr 2011 Location: Shiraz Posts: 86 Rep Power: 7 Dear Foamers, I have read every posts about pressure driven flow but finally I need some one only tell me is it right my conclusion about pressure boundary condition in poiseuille flow, or not: if we have a pressure driven flow and we don't know about velocity but we know pressure boundary condition, we use velocity (U/0) in this form: inlet : type pressureInletVelocity value uniform(0 0 0 ) ; outlet: type zerogradient and pressure as below: inlet: type fixedValue; value uniform 1; (we assume 1 is value of pressure at inlet) outlet type fixedValue; value uniform 0; And another queation, is this pressure value, real or divided by density?
December 29, 2011, 13:15 #2 Senior Member Nima Samkhaniani Join Date: Sep 2009 Location: Tehran, Iran Posts: 1,193 Blog Entries: 1 Rep Power: 16 hi dear Amin please look at the dimension of your pressure file! in 0 folder then you will find whether it is real pressure or divided by rho! amin144 likes this.
December 29, 2011, 13:37 #3 Member Amin Shariat KHah Join Date: Apr 2011 Location: Shiraz Posts: 86 Rep Power: 7 Dear Nmia Thanks very much dadash Is the boundary conditions right?
December 29, 2011, 14:30 #4 Senior Member Nima Samkhaniani Join Date: Sep 2009 Location: Tehran, Iran Posts: 1,193 Blog Entries: 1 Rep Power: 16 yep it seems right amin144 likes this.
December 29, 2011, 15:33 Fan boundary Condition #5 Member Amin Shariat KHah Join Date: Apr 2011 Location: Shiraz Posts: 86 Rep Power: 7 I've heard about some boundary condition which name is FAN , any body know different between fan and pressureInletVelocity? Is it possible to use it in poiseulle flow?
January 1, 2012, 06:38 #6 Senior Member Nima Samkhaniani Join Date: Sep 2009 Location: Tehran, Iran Posts: 1,193 Blog Entries: 1 Rep Power: 16 fan is periodic boundary condition with a pressure difference between to cyclic face, i suggest you to look at channelIcoFoam, if you want to cyclic boundary condition, however you can use Fan for pressure driven flow! amin144 likes this.
January 2, 2012, 13:31 #7 Member Amin Shariat KHah Join Date: Apr 2011 Location: Shiraz Posts: 86 Rep Power: 7 Thanks Dear Nima, have you any idea about it that how can I understand what OpenFoam Code do, when we use one of any boundary conditions?
January 2, 2012, 14:24 #8 Senior Member Nima Samkhaniani Join Date: Sep 2009 Location: Tehran, Iran Posts: 1,193 Blog Entries: 1 Rep Power: 16 1) there is a little description about each code in .C files, you can look at it 2) search in forum about solver 3) look at OpenWiki 4)chalmarse university maybe you find some extra tutorials amin144 likes this.
January 10, 2012, 07:10 #9 Member Amin Shariat KHah Join Date: Apr 2011 Location: Shiraz Posts: 86 Rep Power: 7 Thanks very much Dear Nima
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http://bestwwws.com/percent-error/can-percent-error.php | 1,537,377,977,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156252.62/warc/CC-MAIN-20180919161420-20180919181420-00307.warc.gz | 32,684,066 | 5,132 | Home > Percent Error > Can Percent Error
# Can Percent Error
## Contents
It really is an absolute value that we are concerned with. % error has its limits when it comes to very small numbers like specific heats. Prime numbers are numbers greater than one that are evenly divisible by only 1 and themselves. And we can use Percentage Error to estimate the possible error when measuring. About this wikiHow 180reviews Click a star to vote Click a star to vote Thanks for voting! http://bestwwws.com/percent-error/can-percent-error-more-than-100.php
The formula for calculating percentage error is simple:[1]'[(|Exact Value-Approximate Value|)/Exact Value] x 100 The approximate value is the estimated value, and the exact value is the real value. Source(s): np_rt · 8 years ago 2 Thumbs up 5 Thumbs down Comment Add a comment Submit · just now Asker's rating Report Abuse Percentage error is usually an absolute value, What is your percent error?Solution: experimental value = 8.78 g/cm3 accepted value = 8.96 g/cm3Step 1: Subtract the accepted value from the experimental value.8.96 g/cm3 - 8.78 g/cm3 = -0.18 g/cm3Step 2: Take In this example, if x and y have the same magnitude but opposite sign, then | x + y | 2 = 0 , {\displaystyle {\frac {|x+y|}{2}}=0,} which causes division by https://www.reference.com/math/can-percent-error-negative-number-367cee25ac338cc4
## Can Percent Error Be Over 100
The comparison is expressed as a ratio and is a unitless number. It is really quick. Continue Reading Keep Learning What is a number pattern finder?
Full Answer > Filed Under: Numbers Q: Is 47 a prime number? For example, if a quantity doubles, this corresponds to a 69cNp change (an increase). Percent error or percentage error expresses as a percentage the difference between an approximate or measured value and an exact or known value. Percent Error Worksheet wikiHow Contributor Carry the 2 and get the square root of the previous answer.
When y is a reference value (a theoretical/actual/correct/accepted/optimal/starting, etc. Can Percent Error Be Negative In Chemistry When the variable in question is a percentage itself, it is better to talk about its change by using percentage points, to avoid confusion between relative difference and absolute difference. Yes No Can you please put wikiHow on the whitelist for your ad blocker? https://answers.yahoo.com/question/index?qid=20090104225530AAUfQrT Please help to improve this article by introducing more precise citations. (March 2011) (Learn how and when to remove this template message) Notes ^ What's a good way to check for
When there is no reference value, the sign of Δ has little meaning in the comparison of the two values since it doesn't matter which of the two values is written Percent Error Chemistry Definition For values greater than the reference value, the relative change should be a positive number and for values that are smaller, the relative change should be negative. Ignore any minus sign. It is important for any scientists performing this reaction to report on its accuracy.
## Can Percent Error Be Negative In Chemistry
This will convert the answer into percent form. As the number 21 has mu... Can Percent Error Be Over 100 And to clarify, negative should only come up when you actually fall short of the theoretical yield, not when you are above it. Percent Error Calculator For example, you would not expect to have positive percent error comparing actual to theoretical yield in a chemical reaction.[experimental value - theoretical value] / theoretical value x 100%Percent Error Calculation
The theoreticalvalue (using physics formulas)is 0.64 seconds. see here Did you mean ? Therefore, it is vital to preserve the order as above: subtract the theoretical value from the experimental value and not vice versa. d r = | x − y | max ( | x | , | y | ) {\displaystyle d_{r}={\frac {|x-y|}{\max(|x|,|y|)}}\,} if at least one of the values does not equal Percent Error Formula
Star Gazing Yoga Sea Creatures Gardening Legends Birds more EXPLORE OTHER CATEGORIES Art & Literature Beauty & Fashion Business & Finance Education Family Food Geography Government & Politics Health History Hobbies Did you mean ? External links http://www.acponline.org/clinical_information/journals_publications/ecp/janfeb00/primer.htm Retrieved from "https://en.wikipedia.org/w/index.php?title=Relative_change_and_difference&oldid=741827902" Categories: MeasurementNumerical analysisStatistical ratiosHidden categories: All articles with unsourced statementsArticles with unsourced statements from February 2012Articles lacking in-text citations from March 2011All articles lacking in-text this page To fix this problem we alter the definition of relative change so that it works correctly for all nonzero values of xreference: Relative change ( x , x reference ) = | 1,068 | 4,842 | {"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-2018-39 | latest | en | 0.826531 |
https://math.stackexchange.com/questions/2106633/asymptotic-value-of-a-sequence | 1,566,092,111,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313536.31/warc/CC-MAIN-20190818002820-20190818024820-00202.warc.gz | 533,191,618 | 31,083 | # Asymptotic value of a sequence
Assume a real sequence $1=a_1\leq a_2\le \cdots \leq a_n$, and $a_{i+1}-a_i\leq \sqrt{a_i}$. Does this hold: $$\sum_{i=1}^{n-1} \frac{a_{i+1}-a_i}{a_i} \in O(\log n)$$
• Let $a_i=i$ what happen? – Nosrati Jan 20 '17 at 21:52
• in that case, yes. But in general, is that true? – Liam_math Jan 20 '17 at 21:59
Lemma 1: If $1 = a_1 \leq a_2 \leq a_3 \leq \cdots$ and $a_{i+1}-a_i \leq \sqrt{a_i}$ then $$\sum_{i=1}^{n-1} \frac{a_{i+1}-a_i}{a_i} = \Theta(\log a_n)$$ for all $n$.
Proof: By the assumptions we have
$$0 \leq \frac{a_{i+1}-a_i}{a_i} \leq \frac{1}{\sqrt{a_i}} \leq 1,$$
and since
$$\frac{a_{i+1} - a_i}{a_i} = \frac{a_{i+1}}{a_i} - 1 \tag{1}$$
this is equivalent to
$$1 \leq \frac{a_{i+1}}{a_i} \leq 2. \tag{2}$$
If $1 \leq x \leq 2$ then
$$\log x \leq x-1 \leq \frac{\log x}{\log 2},$$
so setting $x = a_{i+1}/a_i$ in equations $(1)$ and $(2)$ yields
$$\log \frac{a_{i+1}}{a_i} \leq \frac{a_{i+1} - a_i}{a_i} \leq \frac{1}{\log 2} \log \frac{a_{i+1}}{a_i}.$$
Summing this over the range $i=1,2,\ldots,n-1$ yields
$$\log \frac{a_n}{a_1} \leq \sum_{i=1}^{n-1} \frac{a_{i+1}-a_i}{a_i} \leq \frac{1}{\log 2} \log \frac{a_n}{a_1}.$$
$$\tag*{\square}$$
Lemma 2: If $a_i \geq 1$ and $a_{i+1}-a_i \leq \sqrt{a_i}$ then $1 \leq a_i \leq i^2+2$.
Proof: Summing $a_{i+1}-a_i \leq \sqrt{a_i}$ over the range $i=1,2,\ldots,n-1$ yields
$$a_i - 1 = \sum_{j=1}^{i-1} (a_{j+1} - a_j) \leq \sum_{j=1}^{i-1} \sqrt{a_i} \leq i\sqrt{a_i}$$
and hence
$$a_i - i\sqrt{a_i} - 1 \leq 0. \tag{3}$$
The parabola $y = x^2 - ix - 1$ lies below the $x$-axis for $1 \leq x \leq \frac{1}{2}\left(i + \sqrt{4 + i^2}\right)$, so equation $(3)$ combined with the assumption $a_i \geq 1$ yields
$$1 \leq a_i \leq \left(\tfrac{i}{2} + \tfrac{1}{2}\sqrt{4 + i^2}\right)^2 \leq i^2 + 2,$$
where the last inequality follows from Jensen's inequality.
$$\tag*{\square}$$
Claim: If $1 = a_1 \leq a_2 \leq a_3 \leq \cdots$ and $a_{i+1}-a_i \leq \sqrt{a_i}$ then $$\sum_{i=1}^{n-1} \frac{a_{i+1}-a_i}{a_i} = O(\log n)$$ for all $n$.
Proof: The result is trivially true for $n=1$ so suppose $n \geq 2$. Combining Lemmas 1 and 2 yields
$$0 \leq \sum_{i=1}^{n-1} \frac{a_{i+1}-a_i}{a_i} \leq C\log a_n \leq C\log(n^2+2) \leq D \log n$$
for some constants $C$ and $D$.
$$\tag*{\square}$$
Intuition
The sum in question behaves in many ways like a "discrete logarithm", and in the sense of Lemma 1 we have something like
$$\sum_{i=1}^{n-1} \frac{a_{i+1}-a_i}{a_i} \approx \log \frac{a_n}{a_1}.$$
For example, if we double every term of the sequence $(a_n)$ then the values on both sides of the $\approx$ remain unchanged. Further, if $a_n$ is the constant sequence $a_n = a_1$ then both sides of the $\approx$ are equal.
(I'm not sure what the analogue of $\log xy = \log x + \log y$ would be.)
We could try to approach this problem by looking at the smooth analogues of the sequence and sum. The difference $a_{i+1} - a_i$ can be thought of as a discrete derivative and the sum as a discrete integral. So if we can find some function $f$ with $f(n) \approx a_n$ and
$$f'(n) \approx a_{n+1} - a_n$$
then we might expect that
$$\sum_{i=1}^{n-1} \frac{a_{i+1}-a_i}{a_i} \approx \int_1^n \frac{f'(x)}{f(x)}\,dx = \log \frac{f(n)}{f(1)} \approx \log \frac{a_n}{a_1}.$$
This observation was what lead me to the approach in this answer.
• More generally, if (1) $0 < a_1 \leq a_2 \leq a_3 \leq \cdots$, (2) $a_{n+1} = O(a_n)$, and (3) there is a constant $d$ such that $a_n = O(n^d)$, then $$\sum_{i=1}^{n-1} \frac{a_{i+1}-a_i}{a_i} = O(\log n).$$ – Antonio Vargas Jan 21 '17 at 12:52 | 1,611 | 3,610 | {"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} | 4.34375 | 4 | CC-MAIN-2019-35 | latest | en | 0.452741 |
https://emanueleviola.wordpress.com/2023/01/19/mathematics-of-the-impossible-computational-complexity/ | 1,685,849,858,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649439.65/warc/CC-MAIN-20230604025306-20230604055306-00254.warc.gz | 286,506,619 | 29,589 | # Mathematics of the impossible: Computational Complexity
I am teaching and writing some notes on complexity. I hope they will become a book, so they are organized as such. The notes will be serialized on this blog, and you can find the latest version of the book in pdf here, which has a better rendering of tables, pictures, comic strips, etc.
### 0.1 Conventions, choices, and caveats
I write this section before the work is complete, so some of it may change.
This book covers basic results in complexity theory and can be used for a course on the subject. At the same time, it is perhaps sui generis in that it also tells a story of the quest for impossibility results, includes some personal reflections, and makes some non-standard choices about topics and technical details. Some of this is discussed next. $\medskip$
To test your understanding of the material…
this book is interspersed with mistakes, some subtle, some blatant, some not even mistakes but worrying glimpses into the author’s mind. Please send all bug reports and comments to (my five-letter last name)@ccs.neu.edu to be included in the list of heroes. $\medskip$
The $c$ notation.
The mathematical symbol $c$ has a special meaning in this text. Every occurrence of $c$ denotes a real number $>0$. There exist choices for these numbers such that the claims in this book are (or are meant to be) correct. This replaces, is more compact than, and is less prone to abuse than the big-Oh notation (sloppiness hides inside brackets).
Example 0.1. “For all sufficiently large $n$” can be written as $n\ge c$.
“For every $\epsilon$ and all sufficiently large $n$” can be written as $n\ge c_{\epsilon }$.
The following are correct statements:
“It is an open problem to show that some function in NP requires circuits of size $cn$.” At the moment of this writing, one can replace this occurrence with $5$. Note such a claim will remain true if someone proves a $6n$ lower bounds. One just needs to “recompile” the constants in this book.
$c>1+c$”, e.g. assign $2$ to the first occurrence, $1$ to the second.
$100n^{15}”, for="" all="" large="" enough="" $"" n$.="" assign="" c="16\$."
The following are not true:
$c<1/n$ for every $n$”. No matter what we assign $c$ to, we can pick a large enough $n$. Note the assignment to $c$ is absolute, independent of $n$.
More generally, when subscripted this notation indicates a function of the subscript. There exist choices for these functions such that the claims in this book are (or are meant to be) correct. Again, each occurrence can indicate a different function.
For the reader who prefers the big-Oh notation a quick an dirty fix is to replace every occurrence of $c$ in this book with $O(1)$. $\medskip$
The alphabet of TMs.
I define TMs with a fixed alphabet. This choice slightly simplifies the exposition (one parameter vs. two), while being more in line with common experience (it is more common experience to increase the length of a program than its alphabet). This choice affects the proof of Theorem ??. But it isn’t clear that the details are any worse. $\medskip$
Partial vs. total functions (a.k.a. on promise problems).
Recall that promise problems offer the most direct way of formulating natural computational problems. […] In spite of the foregoing opinions, we adopt the convention of focusing on standard decision and search problems. [2]
I define complexity w.r.t. partial functions whereas most texts consider total functions, i.e. we consider computing functions with arbitrary domains rather than any possible string. This is sometimes called “promise problems.” This affects many things, for example the hierarchy for $\text {BPTime}$ (Exercise ??). $\medskip$
References and names.
I decided to keep references in the main text to a minimum, just to avoid having a long list later with items “Result X is due to Y,” but relegate discussion to bibliographic notes. I have also decided to not spell out names of authors, which is increasingly awkward. Central results, such as the PCP theorem, are co-authored by five or more people. $\medskip$
Polynomial.
It is customary in complexity theory to bound quantities by a polynomial, as in polynomial time, when in fact the only terms that matters is the leading time. This also lends itself to confusion since polynomials with many terms are useful for many other things. I use power instead of polynomial, as in power time. $\medskip$
Random-access machines.
“Random access” also leads to strange expressions like “randomized random-access” [1]. $\medskip$
Reductions.
Are presented as an implication. $\medskip$
Randomness and circuits.
While randomness and circuits are everywhere in current research, and seem to be on everyone’s mind, they are sometimes still relegated to later chapters, almost as an afterthought. This book starts with them right away, and attempts to weave them through the narrative. $\medskip$
Data structures
Their study, especially negative results, squarely belongs to complexity theory. Yet data structures are strangely omitted in common textbooks. Results on data structures even tend to miss main venues for complexity theory to land instead on more algorithmic venues! We hope this book helps to revert this trend. $\medskip$
Algorithms & Complexity
…are of course two sides of the same coin. The rule of thumb I follow is to present algorithms that are surprising and challenge our intuition of computation, and ideally match lower bounds, even though they may not be immediately deployed. $\medskip$
Exercises and problems.
Exercises are interspersed within the narrative and serve as “concept check.” They are not meant to be difficult or new, though some are. Problems are collected at the end and tend to be harder and more original, though some are not. $\medskip$
Summary of some terminological and not choices.
Here it is:
Some other sources this book acronym $O(1)$, $\Omega (1)$ $c$ Turing machine tape machine TM random-access machine rapid-access machine RAM polynomial time power time P mapping reduction (sometimes) $A$ reduces to $B$ in $\text {P}$ means $B\in \text {P}\Rightarrow A\in \text {P}$ Extended Church-Turing thesis Power-time computability thesis pairwise independent pairwise uniform FP, promise-P P TM with any alphabet TM with fixed alphabet classes have total functions classes have partial functions
$\medskip$
## Chapter 1A teaser
Consider a computer with three bits of memory. There’s also a clock, beating $1,2,3,\ldots$ In one clock cycle the computer can read one bit of the input and update its memory arbitrarily based on the value of the bit and the current memory, or stop and return a value.
Let’s give a few examples of what such computer can do.
First, it can compute the $\text {And}$ function on $n$ bits:
$\text {Computing And of (\ensuremath {x_{1}},\ensuremath {x_{2}},\ensuremath {\ldots },\ensuremath {x_{n}})}$
$\text {For }i=1,2,\ldots \text { until }n$
$\text {Read }x_{i}$
If $x_{i}=0$ return 0
$\text {Return }1$
We didn’t really use the memory. Let’s consider a slightly more complicated example. A word is palindrome if it reads the same both ways, like racecar, non, anna, and so on. Similarly, example of palindrome bit strings are $11,0110$, and so on.
Let’s show that the computer can decide if a given string is palindrome quickly, in $n$ steps
$\text {Deciding if \ensuremath {(x_{1},x_{2},\ldots ,x_{n})} is palindrome:}$
$\text {For }i=1,2,\ldots \text { until }i>n/2$
$\text {Read \ensuremath {x_{i}} }$and write it in memory bit $m$
If $m\ne x_{n-i}$ return 0
$\text {Return }1$
That was easy. Now consider the $\text {Majority}$ function on $n$ bits, which is $1$ iff the sum of the input bits is $>n/2$ and $0$ otherwise. Majority, like any other function on $n$ bits, can be computed on such a computer in time exponential in $n$. To do that, you do a pass on the input and check if it’s all zero, using the program for And given above. If it is, return $0$. If it is not, you do another pass now checking if it’s all zero except the last bit is $1$. If it is, return $0$. You continue this way until you exhausted all the $2^{n}/2$ possible inputs with Majority equals to $0$. If you never returned $0$ you can now safely return $1$.
As we said, this works for any function, but it’s terribly inefficient. Can we do better for Majority? Can we compute it in time which is just a power of $n$?
Exercise 1.1. Convince yourself that this is impossible. Hint: If you start counting bits, you’ll soon run out of memory.
If you solved the exercise, you are not alone.
And yet, we will see the following shocking result:
Shocking theorem:
Majority can be computed on such a computer in time $n^{c}$.
And this is not a trick tailored to majority. Many other problems, apparently much more complicated, can also be solved in the same time.
But, there’s something possibly even more shocking.
Shocking situation:
It is consistent with our state of knowledge that every “textbook algorithm” can be solved in time $n^{c}$ on such a computer! Nobody can disprove that. (Textbook algorithms include sorting, maxflow, dynamic programming algorithms like longest common subsequence etc., graph problems, numerical problems, etc.)
The Shocking theorem gives some explanation for the Shocking situation. It will be hard to rule out efficient programs on this model, since they are so powerful and counterintuitive. In fact, we will see later that this can be formalized. Basically, we will show that the model is so strong that it can compute functions that provably escape the reach of current mathematics… if you believe certain things, like that it’s hard to factor numbers. This now enters some of the mysticism that surrounds complexity theory, where different beliefs and conjectures are pitted against each other in a battle for ground truth.
### References
[1] Dana Angluin and Leslie G. Valiant. Fast probabilistic algorithms for hamiltonian circuits and matchings. J. Comput. Syst. Sci., 18(2):155–193, 1979.
[2] Oded Goldreich. Computational Complexity: A Conceptual Perspective. Cambridge University Press, 2008.
## 2 thoughts on “Mathematics of the impossible: Computational Complexity”
1. Confused says:
In the algorithm for checking palindromes, where is n-i computed and stored? Given i you can refer to x_(n-i), but you can’t store the number of ones?
2. Which bit to read depends on the clock and on the input length only. Re-reading, it’s good to clarify this. Thanks! Let me know if it makes more sense now. | 2,489 | 10,588 | {"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": 80, "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} | 3.265625 | 3 | CC-MAIN-2023-23 | latest | en | 0.936224 |
http://www.physicsforums.com/showthread.php?t=713076 | 1,411,040,972,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657127285.44/warc/CC-MAIN-20140914011207-00180-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 736,258,973 | 8,320 | by rollingstein
PF Gold P: 330 Say there's a pipe that pumps a liquid that empties into an elevated tank The discharge end of the pipe is 5 m higher up than the suction but it goes over an intermediate obstacle, say, 20 m higher than suction level. Is the pump head requirement 5m or 20m? I'm confused. My gut feeling says 20m otherwise how will one get the flow started. Let's assume liquid, incompressable flow, both suction & discharge atmospheric & negligible frictional / velocity heads etc. Basically, does one calculate a head difference between initial & final points or initial & highest points?
Thanks
P: 1,956
Quote by rollingstein Say there's a pipe that pumps a liquid
A pump pumps a liquid, a pipe constrains the flow.
The 20m up and 15m down is a siphon but it is only able to provide one atmosphere of suction.
Let's assume liquid, incompressable flow,
I think it would be far more important to specify the density of the liquid. That will determine the functionality of the siphon.
You are really confused here. Where in the system is the pump ?
Is it a fixed displacement pump or a centrifugal pump ? They behave quite differently in this sort of situation.
PF Gold P: 330 Centrifugal pump and the liquid is water. Pump is right after the suction tank so that it always has a flooded suction.
Sci Advisor Thanks P: 1,956 Piping pressure head estimation Will the potential siphon pipe that rises 20m and falls 15m remain full of water when flowing normally, or can air flow back in because the outlet to the reservoir is open above the reservoir water level.
PF Gold
P: 330
Quote by Baluncore Will the potential siphon pipe that rises 20m and falls 15m remain full of water when flowing normally, or can air flow back in because the outlet to the reservoir is open above the reservoir water level.
Air could flow back in. At startup, all pipe will be full with air.
I guess that means size for head=20 m?
Thanks
P: 1,956
Quote by rollingstein I guess that means size for head=20 m?
A centrifugal pump can prime a greater head slowly, then once water starts to flow through the siphon the flow increases as the head is less. That requires the end of the pipe to be under the reservoir water surface or that it have something to prevent air entering to break the siphon once it is running. That could be a bucket attached to the end of the pipe to retain water.
The difference in inlet and outlet height of the siphon is not relevant because a siphon can only pull about 10m of water. Once running the pump would only see a 10m head so flow would be greater.
PF Gold P: 330 Understood. Thanks!
P: 851 Just to add: For your design flowrate/head, use the 5m+Frictin losses as your head. For starting head, jsut make sure the system is on the pump's curve at 20m. You want a pump that can develop some flow without dead-heading at 20m of head and then run out to your design flow rate after the siphon takes over and leaves you with your (5 m + Friction losses + Discharge pressure).
PF Gold
P: 330
Quote by Travis_King Just to add: For your design flowrate/head, use the 5m+Frictin losses as your head. For starting head, jsut make sure the system is on the pump's curve at 20m. You want a pump that can develop some flow without dead-heading at 20m of head and then run out to your design flow rate after the siphon takes over and leaves you with your (5 m + Friction losses + Discharge pressure).
Understood. Thanks. So I need a pump curve with a shut off head of at least 20 m.
I'm confused about your 5m operating point. Must be more right? Siphon can only provide at most 10 m.
P: 851 Yea, sorry, after you've filled the pipe, you'll have to get the water to 10m, the siphon will take over from there.
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Advanced Mathematics in Seismology. Dr. Quakelove. or: How I Learned To Stop Worrying And Love The Wave Equation. When Am I Ever Going To Use This Stuff?. Wave Equation. Diffusion Equation. Complex Analysis. Linear Algebra. The 1-D Wave Equation. F = k[u(x,t) - u(x-h,t)].
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### Dr. Quakelove
or:
How I Learned To Stop Worrying
And Love The Wave Equation
When Am I Ever Going To Use This Stuff?
Wave Equation
Diffusion Equation
Complex Analysis
Linear Algebra
The 1-D Wave Equation
F = k[u(x,t) - u(x-h,t)]
F = k[u(x+h,t) – u(x,t)]
k
k
m
m
m
u(x-h,t)
u(x,t)
u(x+h,t)
F = m ü(x,t)
The 1-D Wave Equation
M = N m
L = N h
K = k / N
Solution to the Wave Equation
• Use separation of variables:
Solution to the Wave Equation
• Now we have two coupled ODEs:
• These ODEs have simple solutions:
Solution to the Wave Equation
• The general solution is:
• Considering only the harmonic component:
• The imaginary part goes to zero as a result of boundary conditions
And in case you don’t believe the math
Harmonic and exponential
solutions
Pure harmonic solutions
The 3-D Vector Wave Equation
• We can decompose this into vector and scalar potentials using Helmholtz’s theorem:
where | 545 | 2,116 | {"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} | 3.046875 | 3 | CC-MAIN-2017-43 | latest | en | 0.794996 |
http://oeis.org/A064790 | 1,596,910,548,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738015.38/warc/CC-MAIN-20200808165417-20200808195417-00561.warc.gz | 84,768,470 | 4,630 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A064790 Inverse permutation to A060734. 4
1, 3, 5, 2, 6, 9, 13, 8, 4, 10, 14, 19, 25, 18, 12, 7, 15, 20, 26, 33, 41, 32, 24, 17, 11, 21, 27, 34, 42, 51, 61, 50, 40, 31, 23, 16, 28, 35, 43, 52, 62, 73, 85, 72, 60, 49, 39, 30, 22, 36, 44, 53, 63, 74, 86, 99, 113, 98, 84, 71, 59, 48, 38, 29, 45, 54, 64, 75, 87, 100, 114 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS From Boris Putievskiy, Mar 14 2013: (Start) a(n) is a pairing function: a function that reversibly maps Z^{+} x Z^{+} onto Z^{+}, where Z^{+} is the set of integer positive numbers. Layer is pair of sides of square from T(1,n) to T(n,n) and from T(n,n) to T(n,1). This sequence is A188568 as table read by boustrophedonic ("ox-plowing") method - layer clockwise, layer counterclockwise and so. The same table A188568 read layer by layer clockwise is A194280. (End) LINKS Boris Putievskiy, Transformations Integer Sequences And Pairing Functions arXiv:1212.2732 [math.CO] Eric Weisstein's MathWorld, Pairing Function FORMULA a(n) = (i+j-1)*(i+j-2)/2+i, where i=min(t; t^2-n+1), j=min(t; n-(t-1)^2), t=floor(sqrt(n-1))+1. - Boris Putievskiy, Dec 24 2012 EXAMPLE From Boris Putievskiy, Mar 14 2013: (Start) The start of the sequence as table: 1....2...6...7..15..16..28... 3....5...9..12..20..23..35... 4....8..13..18..26..31..43... 10..14..19..25..33..40..52... 11..17..24..32..41..50..62... 21..27..34..42..51..61..73... 22..30..39..49..60..72..85... . . . The start of the sequence as triangular array read by rows: 1; 3,5,2; 6,9,13,8,4; 10,14,19,25,18,12,7; 15,20,26,33,41,32,24,17,11; 21,27,34,42,51,61,50,40,31,23,16; 28,35,43,52,62,73,85,72,60,49,39,30,22; . . . Row number r contains 2*r-1 numbers. (End) CROSSREFS Cf. A060734, A064788, A188568, A194280. Sequence in context: A010782 A333111 A139584 * A113966 A164611 A316086 Adjacent sequences: A064787 A064788 A064789 * A064791 A064792 A064793 KEYWORD nonn,easy AUTHOR N. J. A. Sloane, Oct 20 2001 STATUS approved
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Last modified August 8 14:04 EDT 2020. Contains 336298 sequences. (Running on oeis4.) | 989 | 2,490 | {"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} | 3.390625 | 3 | CC-MAIN-2020-34 | latest | en | 0.661485 |
https://discourse.julialang.org/t/benchmarktools-macros-with-books/75540 | 1,652,972,590,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662529538.2/warc/CC-MAIN-20220519141152-20220519171152-00288.warc.gz | 259,740,797 | 6,705 | # BenchmarkTools macros with Books
Given this example `weave`d output from a Julia Markdown Document:
Generated with this code block:
``````A = rand(100,100)
B = rand(100,100)
C = rand(100,100)
using BenchmarkTools
function inner_rows!(C,A,B)
for i in 1:100, j in 1:100
C[i,j] = A[i,j] + B[i,j]
end
end
@btime inner_rows!(C,A,B)
``````
How might I achieve the same effect using Books.jl?
So far in my module I have defined a function, `inner_rows!`:
``````function inner_rows!(C,A,B)
for i in 1:100, j in 1:100
C[i,j] = A[i,j] + B[i,j]
end
end
``````
Which is called on the page with 3 `jl` code blocks:
``````sc("using BenchmarkTools")
``````
``````s = """
A = rand(100,100)
B = rand(100,100)
C = rand(100,100)
"""
sc(s)
``````
``````scob("@btime BookTest.inner_rows!(C,A,B) |> string")
``````
Running `gen()` is successful, but upon `serve()` the macro output is not shown. Instead only the return value, `nothing`, is shown.
Is there a way to do this in Books.jl?
Note also that I added `|> string` after trying this code block:
``````scob("@btime BookTest.inner_rows!(C,A,B)")
``````
Produced the following result. Probably I am going wrong somewhere. This is my first trial run with Books.jl
That’s because `inner_rows!` returns `nothing` in this case
``````julia> t = @btime 1 + 1
0.017 ns (0 allocations: 0 bytes)
2
julia> t
2
``````
One solution is to use `@benchmark` instead:
But, of course, the real solution is to grab the output from stdout and show that. In a bit of a hacky way, his can be done with `IOCapture.jl` and `# hide` comments
`````````jl
s = """
A = rand(100,100)
B = rand(100,100)
C = rand(100,100)
function inner_rows!(C,A,B)
for i in 1:100, j in 1:100
C[i,j] = A[i,j] + B[i,j]
end
end
c = IOCapture.capture() do # hide
@btime inner_rows!(C,A,B)
end # hide
return c.output # hide
"""
sco(s; post=output_block)
```
``````
which shows on the webpage as:
Another way is to wrap the evaluation step at a later point:
`````````jl
s = """
A = rand(100,100)
B = rand(100,100)
C = rand(100,100)
function inner_rows!(C,A,B)
for i in 1:100, j in 1:100
C[i,j] = A[i,j] + B[i,j]
end
end
@btime inner_rows!(C,A,B)
"""
c = IOCapture.capture() do
sc(s)
end
return code_block(s) * "\n\n" * output_block(c.output)
`````````
1 Like
The problem with the backticks in your post is probably because you didn’t put a newline before and after each code block. That’s a parser problem from Pandoc. I could fix that by always adding a newline before and after each code block, but it seems like a bit of a waste so I didn’t do that (yet).
Yeah, not a big problem.
1 Like
This solution is exactly what I was looking for! Thank you for pointing me in the right direction, and great work on the package!
1 Like | 875 | 2,747 | {"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} | 2.546875 | 3 | CC-MAIN-2022-21 | latest | en | 0.782958 |
https://qanda.ai/en/solver/popular-problems/1007792 | 1,643,400,474,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320306335.77/warc/CC-MAIN-20220128182552-20220128212552-00525.warc.gz | 515,071,719 | 16,749 | # Calculator search results
Formula
Expand the expression
$$\left ( a + b - c \right ) \left ( a - b + c \right )$$
$a ^ { 2 } - b ^ { 2 } + 2 b c - c ^ { 2 }$
$\left ( \color{#FF6800}{ a } \color{#FF6800}{ + } \color{#FF6800}{ b } \color{#FF6800}{ - } \color{#FF6800}{ c } \right ) \left ( \color{#FF6800}{ a } \color{#FF6800}{ - } \color{#FF6800}{ b } \color{#FF6800}{ + } \color{#FF6800}{ c } \right )$
Organize the expression with the distributive law
$\color{#FF6800}{ a } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ b } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ b } \color{#FF6800}{ c } \color{#FF6800}{ - } \color{#FF6800}{ c } ^ { \color{#FF6800}{ 2 } }$ | 320 | 730 | {"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} | 3.546875 | 4 | CC-MAIN-2022-05 | latest | en | 0.203075 |
https://math.stackexchange.com/questions/1904463/frac12-fracb-1a-1-fracb-na-n2-sum-1na-i2-2-ge-s | 1,566,497,035,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317339.12/warc/CC-MAIN-20190822172901-20190822194901-00283.warc.gz | 547,113,316 | 27,835 | # $\frac{1}{2}(\frac{b_1}{a_1}-\frac{b_n}{a_n})^2(\sum_{1}^{n}{a_i^2 }) ^2 \ge (\sum_{1}^{n}{a_i^2 }) (\sum_{1}^{n}{b_i^2 })-(\sum_{1}^{n}{a_ib_i })^2$ [closed]
Let $a_1, a_2,....,a_n, b_1, b_2,...,b_n$, let $\frac{b_1}{a_1} = max \{\frac{b_i}{a_i}, i=1,2, \cdots n \}$ , $\frac{b_n}{a_n} = min \{\frac{b_i}{a_i}, i=1,2, \cdots n \}$ show that:
$$\frac{1}{2}(\frac{b_1}{a_1}-\frac{b_n}{a_n})^2(\sum_{1}^{n}{a_i^2 }) ^2 \ge (\sum_{1}^{n}{a_i^2 }) (\sum_{1}^{n}{b_i^2 })-(\sum_{1}^{n}{a_ib_i })^2$$
## closed as off-topic by Semiclassical, Shailesh, Math1000, heropup, user223391 Aug 28 '16 at 13:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Semiclassical, Shailesh, Math1000, heropup, Community
If this question can be reworded to fit the rules in the help center, please edit the question.
• This inequality is equivalent to $$\frac{1}{2}\left(\frac{b_1}{a_1}-\frac{b_n}{a_n}\right)^2 \geq \left(||B||\sin\theta\right)^2$$ Where $A=\left(a_1,a_2,\ldots,a_2\right)$, $B=\left(b_1,b_2,\ldots,b_2\right)$, and $\theta$ is the angle between $A$ and $B$. – Hrhm Aug 26 '16 at 17:13
• Could you show me detail? Please – Oai Thanh Đào Aug 26 '16 at 17:26
• Sure. We know that $$\sum_1^n a_ib_i=||A||||B||\cos\theta$$ and $$\sum_1^n a_i^2=||A||^2$$. I'm sure you can figure it out from there. This doesn't really help solve the inequality though, sorry about that. – Hrhm Aug 26 '16 at 17:29 | 672 | 1,790 | {"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} | 3.59375 | 4 | CC-MAIN-2019-35 | latest | en | 0.805322 |
https://www.readthesequences.com/Mutual-Information-And-Density-In-Thingspace | 1,709,411,030,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475897.53/warc/CC-MAIN-20240302184020-20240302214020-00389.warc.gz | 927,294,160 | 7,652 | # Mutual Information, and Density in Thingspace
Suppose you have a system X that can be in any of 8 states, which are all equally probable (relative to your current state of knowledge), and a system Y that can be in any of 4 states, all equally probable. total The entropy of X, as defined in the previous essay, is 3 bits; we’ll need to ask 3 yes-or-no questions to find out X’s exact state. The entropy of Y is 2 bits; we have to ask 2 yes-or-no questions to find out Y’s exact state. This may seem obvious since 23 = 8 and 22 = 4, so 3 questions can distinguish 8 possibilities and 2 questions can distinguish 4 possibilities; but remember that if the possibilities were not all equally likely, we could use a more clever code to discover Y’s state using e.g. 1.75 questions on average. In this case, though, X’s probability mass is evenly distributed over all its possible states, and likewise Y, so we can’t use any clever codes.
What is the entropy of the combined system (X,Y)?
You might be tempted to answer, “It takes 3 questions to find out X, and then 2 questions to find out Y, so it takes 5 questions total to find out the state of X and Y.”
But what if the two variables are entangled, so that learning the state of Y tells us something about the state of X?
In particular, let’s suppose that X and Y are either both odd or both even.
Now if we receive a 3-bit message (ask 3 questions) and learn that X is in state X5, we know that Y is in state Y1 or state Y3, but not state Y2 or state Y4. So the single additional question “Is Y in state Y3?,” answered “No,” tells us the entire state of (X,Y): X = X5, Y = Y1. And we learned this with a total of 4 questions.
Conversely, if we learn that Y is in state Y4 using two questions, it will take us only an additional two questions to learn whether X is in state X2, X4, X6, or X8. Again, four questions to learn the state of the joint system.
The mutual information of two variables is defined as the difference between the entropy of the joint system and the entropy of the independent systems: I(X;Y) = H(X) + H(Y) − H(X,Y).
Here there is one bit of mutual information between the two systems: Learning X tells us one bit of information about Y (cuts down the space of possibilities from 4 possibilities to 2, a factor-of-2 decrease in the volume) and learning Y tells us one bit of information about X (cuts down the possibility space from 8 possibilities to 4).
What about when probability mass is not evenly distributed? Last essay, for example, we discussed the case in which Y had the probabilities 1/2, 1/4, 1/8, 1/8 for its four states. Let us take this to be our probability distribution over Y, considered independently—if we saw Y, without seeing anything else, this is what we’d expect to see. And suppose the variable Z has two states, Z1 and Z2, with probabilities 3/8 and 5/8 respectively.
Then if and only if the joint distribution of Y and Z is as follows, there is zero mutual information between Y and Z:
Z1Y1 : 3/16 Z1Y2 : 3/32 Z1Y3 : 3/64 Z1Y4 : 3/64 Z2Y1 : 5/16 Z2Y2 : 5/32 Z2Y3 : 5/64 Z2Y4 : 5/64 .
This distribution obeys the law
P(Y,Z) = P(Y)P(Z) .
For example, P(Z1Y2) = P(Z1)P(Y2) = 3/8 × 1/4 = 3/32.
And observe that we can recover the marginal (independent) probabilities of Y and Z just by looking at the joint distribution:
P(Y1) = total probability of all the different ways Y1 can happen = P(Z1Y1) + P(Z2Y1) = 3/16 + 5/16 = 1/2 .
So, just by inspecting the joint distribution, we can determine whether the marginal variables Y and Z are independent; that is, whether the joint distribution factors into the product of the marginal distributions; whether, for all Y and Z, we have P(Y,Z) = P(Y)P(Z).
This last is significant because, by Bayes’s Rule,
P(ZjYi) = P(Yi)P(Zj) P(ZjYi)/P(Zj) = P(Yi) P(Yi|Zj) = P(Yi) .
In English: “After you learn Zj , your belief about Yi is just what it was before.”
So when the distribution factorizes—when P(Y,Z) = P(Y)P(Z)—this is equivalent to “Learning about Y never tells us anything about Z or vice versa.”
From which you might suspect, correctly, that there is no mutual information between Y and Z. Where there is no mutual information, there is no Bayesian evidence, and vice versa.
Suppose that in the distribution (Y,Z) above, we treated each possible combination of Y and Z as a separate event—so that the distribution (Y,Z) would have a total of 8 possibilities, with the probabilities shown—and then we calculated the entropy of the distribution (Y,Z) the same way we would calculate the entropy of any distribution:
P(Z1Y1)log2(P(Z1Y1)) + P(Z1Y2)log2(P(Z1Y2)) +
P(Z1Y3)log2(P(Z1Y3)) + … + P(Z2Y4)log2(P(Z2Y4))
= (3/16)log2(3/16) + (3/32)log2(3/32) +
(3/64)log2(3/64) + … + (5/64)log2(5/64) .
You would end up with the same total you would get if you separately calculated the entropy of Y plus the entropy of Z. There is no mutual information between the two variables, so our uncertainty about the joint system is not any less than our uncertainty about the two systems considered separately. (I am not showing the calculations, but you are welcome to do them; and I am not showing the proof that this is true in general, but you are welcome to Google on “Shannon entropy” and “mutual information.”)
What if the joint distribution doesn’t factorize? For example:
Z1Y1 : 12/64 Z1Y2 : 8/64 Z1Y3 : 1/64 Z1Y4 : 3/64 Z2Y1 : 20/64 Z2Y2 : 8/64 Z2Y3 : 7/64 Z2Y4 : 5/64 .
If you add up the joint probabilities to get marginal probabilities, you should find that P(Y1) = 1/2, P(Z1) = 3/8, and so on—the marginal probabilities are the same as before.
But the joint probabilities do not always equal the product of the marginal probabilities. For example, the probability P(Z1Y2) equals 8/64, where P(Z1)P(Y2) would equal 3/8 × 1/4 = 6/64. That is, the probability of running into Z1Y2 together is greater than you’d expect based on the probabilities of running into Z1 or Y2 separately.
Which in turn implies:
P(Z1Y2) > P(Z1)P(Y2) P(Z1Y2)/P(Y2) > P(Z1) P(Z1|Y2) > P(Z1).
Since there’s an “unusually high” probability for P(Z1Y2)—defined as a probability higher than the marginal probabilities would indicate by default—it follows that observing Y2 is evidence that increases the probability of Z1. And by a symmetrical argument, observing Z1 must favor Y2.
As there are at least some values of Y that tell us about Z (and vice versa) there must be mutual information between the two variables; and so you will find—I am confident, though I haven’t actually checked—that calculating the entropy of (Y,Z) yields less total uncertainty than the sum of the independent entropies of Y and Z. That is, H(Y,Z) = H(Y) + H(Z) − I(Y;Z), with all quantities necessarily positive.
(I digress here to remark that the symmetry of the expression for the mutual information shows that Y must tell us as much about Z, on average, as Z tells us about Y. I leave it as an exercise to the reader to reconcile this with anything they were taught in logic class about how, if all ravens are black, being allowed to reason Raven(x) ⇒ Black(x) doesn’t mean you’re allowed to reason Black(x) ⇒ Raven(x). How different seem the symmetrical probability flows of the Bayesian, from the sharp lurches of logic—even though the latter is just a degenerate case of the former.)
“But,” you ask, “what has all this to do with the proper use of words?”
In Empty Labels and then Replace the Symbol with the Substance, we saw the technique of replacing a word with its definition—the example being given:
All [mortal, ¬feathers, bipedal] are mortal.
Socrates is a [mortal, ¬feathers, bipedal].
Therefore, Socrates is mortal.
Why, then, would you even want to have a word for “human”? Why not just say “Socrates is a mortal featherless biped”?
Because it’s helpful to have shorter words for things that you encounter often. If your code for describing single properties is already efficient, then there will not be an advantage to having a special word for a conjunction—like “human” for “mortal featherless biped”—unless things that are mortal and featherless and bipedal, are found more often than the marginal probabilities would lead you to expect.
In efficient codes, word length corresponds to probability—so the code for Z1Y2 will be just as long as the code for Z1 plus the code for Y2, unless P(Z1Y2) > P(Z1)P(Y2), in which case the code for the word can be shorter than the codes for its parts.
And this in turn corresponds exactly to the case where we can infer some of the properties of the thing from seeing its other properties. It must be more likely than the default that featherless bipedal things will also be mortal.
Of course the word “human” really describes many, many more properties— when you see a human-shaped entity that talks and wears clothes, you can infer whole hosts of biochemical and anatomical and cognitive facts about it. To replace the word “human” with a description of everything we know about humans would require us to spend an inordinate amount of time talking. But this is true only because a featherless talking biped is far more likely than default to be poisonable by hemlock, or have broad nails, or be overconfident.
Having a word for a thing, rather than just listing its properties, is a more compact code precisely in those cases where we can infer some of those properties from the other properties. (With the exception perhaps of very primitive words, like “red,” that we would use to send an entirely uncompressed description of our sensory experiences. But by the time you encounter a bug, or even a rock, you’re dealing with nonsimple property collections, far above the primitive level.)
So having a word “wiggin” for green-eyed black-haired people is more useful than just saying “green-eyed black-haired person” precisely when:
1. Green-eyed people are more likely than average to be black-haired (and vice versa), meaning that we can probabilistically infer green eyes from black hair or vice versa; or
2. Wiggins share other properties that can be inferred at greater-than-default probability. In this case we have to separately observe the green eyes and black hair; but then, after observing both these properties independently, we can probabilistically infer other properties (like a taste for ketchup).
One may even consider the act of defining a word as a promise to this effect. Telling someone, “I define the word ‘wiggin’ to mean a person with green eyes and black hair,” by Gricean implication, asserts that the word “wiggin” will somehow help you make inferences / shorten your messages.
If green-eyes and black hair have no greater than default probability to be found together, nor does any other property occur at greater than default probability along with them, then the word “wiggin” is a lie: The word claims that certain people are worth distinguishing as a group, but they’re not.
In this case the word “wiggin” does not help describe reality more compactly—it is not defined by someone sending the shortest message—it has no role in the simplest explanation. Equivalently, the word “wiggin” will be of no help to you in doing any Bayesian inference. Even if you do not call the word a lie, it is surely an error.
And the way to carve reality at its joints is to draw your boundaries around concentrations of unusually high probability density in Thingspace. | 2,921 | 11,391 | {"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-2024-10 | latest | en | 0.946253 |
http://www.truebluela.com/2010/4/15/1423565/tax-day-dbacks-vs-dodgers | 1,409,631,997,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1409535921550.2/warc/CC-MAIN-20140901014521-00414-ip-10-180-136-8.ec2.internal.warc.gz | 1,633,853,598 | 23,370 | ## Tax Day, DBacks vs Dodgers - Simulation
I used my simulator to simulate 100000 baseball games between the Diamondbacks and the Dodgers using what to the best of my knowledge will be close to today's starting lineups. The simulator outputs a win probability for each team, along with the average total runs scored, a distribution matrix of all the final scores and how often they occurred, and an averaged box score tally of all games combined. You may find it strange that there are so many one run games listed as the most likely score, and that the home team is always favored to win in the top couple of scores. This is due to basic math and the way that baseball rules play out of having the home team bat last. Here is a great article explaining this phenomenon. Here are the latest Vegas MLB odds.
Today's Results... (Last simulation ran on Thursday at 1:00 AM)
Visitors Home Pitching Matchup Favorite Vegas Win Prob Simulator Win Prob ARI LAN D.Haren vs H.Kuroda LAN 52.38% 53.77%
Notes: The Simulator and Vegas both have the series finale tightening up quite a bit in terms of win probability. The Dodgers are still the favorites but on a neutral field they'd likely be slight underdogs.
Top 100 Most Likely Scores
1 LAN 3-2 51 LAN 7-5 2 LAN 2-1 52 ARI 7-1 3 LAN 4-3 53 ARI 6-0 4 ARI 3-2 54 ARI 7-6 5 ARI 2-1 55 ARI 8-3 6 LAN 5-4 56 ARI 8-2 7 ARI 4-3 57 LAN 8-2 8 LAN 3-1 58 ARI 8-1 9 ARI 4-2 59 LAN 8-3 10 LAN 4-2 60 ARI 8-4 11 ARI 3-1 61 LAN 8-1 12 LAN 1-0 62 ARI 7-0 13 LAN 4-1 63 ARI 8-5 14 ARI 5-4 64 LAN 7-0 15 LAN 5-3 65 LAN 8-7 16 ARI 4-1 66 LAN 8-4 17 LAN 2-0 67 ARI 9-2 18 ARI 5-3 68 ARI 9-3 19 LAN 5-2 69 LAN 9-2 20 LAN 6-5 70 LAN 8-5 21 ARI 5-2 71 ARI 8-6 22 ARI 2-0 72 LAN 9-3 23 LAN 3-0 73 ARI 9-1 24 ARI 1-0 74 LAN 9-1 25 LAN 5-1 75 LAN 8-6 26 ARI 5-1 76 ARI 9-4 27 ARI 3-0 77 ARI 8-0 28 ARI 6-3 78 LAN 8-0 29 LAN 6-3 79 ARI 8-7 30 LAN 6-4 80 LAN 9-4 31 LAN 4-0 81 ARI 10-3 32 LAN 6-2 82 ARI 9-5 33 ARI 6-4 83 LAN 10-3 34 ARI 4-0 84 LAN 10-2 35 ARI 6-2 85 LAN 9-8 36 LAN 6-1 86 LAN 9-0 37 ARI 6-5 87 ARI 10-2 38 ARI 6-1 88 ARI 9-6 39 LAN 5-0 89 ARI 9-0 40 LAN 7-6 90 ARI 10-1 41 ARI 7-3 91 LAN 9-5 42 LAN 7-3 92 LAN 10-1 43 LAN 7-2 93 ARI 10-4 44 ARI 7-4 94 ARI 10-5 45 ARI 7-2 95 LAN 9-6 46 ARI 5-0 96 LAN 10-0 47 LAN 7-4 97 LAN 10-4 48 LAN 7-1 98 ARI 9-7 49 ARI 7-5 99 LAN 11-2 50 LAN 6-0 100 ARI 11-3
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In order to provide our users with a better overall experience, we ask for more information from Facebook when using it to login so that we can learn more about our audience and provide you with the best possible experience. We do not store specific user data and the sharing of it is not required to login with Facebook. | 1,435 | 4,182 | {"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} | 2.578125 | 3 | CC-MAIN-2014-35 | latest | en | 0.788092 |
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I was recently doing some practice problems for my IB chemistry SL test and ran across this problem.
"A 2.450 g sample of a mixture of NaCl and CaCl2 was dissolved in distilled water. The chloride solution was treated with excess AgNO3 (aq) solution. The precipitated AgCl (s) was collected, washed, and dried. The mass of the dried AgCl was 6.127 g. Calculate percent by mass of the NaCl and CaCl2 in the original mixture." (2 marks)
AgCl has a molar mass of 143.45 g/mol, so I calculated that the amount of Cl in the sample was 6.127*(35.45/143.35) = 1.515 g Cl
This means that together, there has to be (2.450 - 1.515) = .935 g of Na and Ca combined. At this point, I made a system of equations. Let x represent the number of moles of Na and let y represent the number of moles of Ca.
.935 = 23x + 40y
1.515 = 35.45x + 70.90y
The first equation expresses the mass of the Na and Ca, and the second expresses the mass of Cl (35.45 is doubled since there are 2 Cl for every Ca.) Ultimately, solving the equation gave me that x = .027 which led to there being 1.58g NaCl. This made the percent composition 64.5% NaCl and 35.5% CaCl2
My question is this: is there some really simple way to do this problem that I overlooked? It's only worth 2 marks, and it took me about 10 minutes to figure out how to solve it (and I might have made some mistakes in my solution). In case it matters, this is from an IB Chem SL Pearson book, not a previous IB chem paper.
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You do not want to solve systems of equations on the chem exam!
I don't really have an easy way, but I do have a tip on naming variables, which simplifies some work.
You should define only one variable such that either it's what you are finding, or you can get to there in one step.
A good variable would be % of NaCl by mass. Here I am going to use mass of NaCl to be m. So that 2.45 - m = mass of CaCl2.So that once we know m, we can just do m/2.45 * 100% to get the answer.
Plug values in:
m / (22.99 + 35.45) + 2*(2.45-m) / (40.08 + 2*35.45) = 6.127 / (143.32) = 0.04275. I used slightly different values but the final answer didn't change much.
^ At this point don't find common denominator just multiply everything by (40.08 + 2* 35.45). Here you should note that having unknowns in the numerator is so much better than the denominator.
I got a similar answer in the end.
In your solution, you used mole ratio implicitly but in mine I just substituted everything into this simple relationship. So ideally you want to use a minimal number of relationships that ties everything together. Mole ratio is often a better one than conservation of mass.
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You are able to use a graphing calculator on chemistry paper 2 and 3, so if you know how to solve systems of linear equations on a graphing calculator then what you have done is a fine approach.
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# 26. In its most recent approach, the comet Crommelin
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Joined: 15 Aug 2003
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26. In its most recent approach, the comet Crommelin [#permalink]
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04 Oct 2003, 17:35
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26. In its most recent approach, the comet Crommelin
passed the Earth at about the same distance and in
about the same position, some 25 degrees above the
horizon, that Halley's comet will pass the next time
it appears.
(A) that Halley's comet will pass
(B) that Halley's comet is to be passing
(C) as Halley's comet
(D) as will Halley's comet
(E) as Halley's comet will do
Intern
Joined: 23 Dec 2002
Posts: 13
Location: LA
### Show Tags
04 Oct 2003, 19:02
Is the answer D? That would be my guess. Let us know.
Intern
Joined: 23 Dec 2002
Posts: 13
Location: LA
### Show Tags
05 Oct 2003, 21:09
C is wrong, the parallel structure is not there.
E is wrong, as Hally's comet will do....that's kinda wordy. "Will" alone should be enough.
B is wrong, way too long.
A could be right. I seriously don't see anything wrong in A, the only problem I have is it sounds a bit repetitive. "Will pass" I think the word "will" alone is enough.
Btw, if I could make one suggestion.....please underline the sentence to be replaced or at least bold it...
Intern
Joined: 08 Oct 2003
Posts: 12
Location: INDIA
sc: [#permalink]
### Show Tags
12 Oct 2003, 00:57
hello,
The sentence here seems to 'compare' two comets.When we compare two similar objects ,I believe we should use ''as''(ram can work as' hard 'as' kriss. x PASSES IN THE SAME WAY AS z.)
That rules out the first two choices, also the B) SEEMS TO be WAY OUT (IS PASSING TO )
C ) ''AS HALEY'S COMET 'means that commenlin(watevr comets) will pass as (not in these sense of ''like'' or comparision) but in the meaning of will become a haley's comet.Which jus alters the actual meanin itself.
now between D and E, BOTH SEEM OK.
bUT I gues D IS BETTER AS IT IS SIMPLE and also IT HAS something to do with the verb coming b4 the subject, makes a better sense.('as will' compared to ''will do'').
any comments!
have fun!
Intern
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sc: [#permalink]
### Show Tags
12 Oct 2003, 01:10
hello,
The sentence here seems to 'compare' two comets.When we compare two similar objects ,I believe we should use ''as''(ram can work as' hard 'as' kriss. x PASSES IN THE SAME WAY AS z.)
That rules out the first two choices, also the B) SEEMS TO be WAY OUT (IS PASSING TO )
C ) ''AS HALEY'S COMET 'means that commenlin(watevr comets) will pass as (not in these sense of ''like'' or comparision) but in the meaning of will become a haley's comet.Which jus alters the actual meanin itself.
now between D and E, BOTH SEEM OK.
bUT I gues D IS BETTER AS IT IS SIMPLE and also IT HAS something to do with the verb coming b4 the subject, makes a better sense.('as will' compared to ''will do'').
any comments!
have fun!
Intern
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Location: INDIA
### Show Tags
14 Oct 2003, 06:48
any explanation! In regards with the ans!
have fun!
Manager
Joined: 19 Oct 2003
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Re: SC : Halley's Comet [#permalink]
### Show Tags
19 Oct 2003, 14:42
D is ok.
A, B is worong... what do you mean with "that"
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Re: SC : Halley's Comet [#permalink]
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25 May 2011, 00:17
Option (D) might be the right answer
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Re: SC : Halley's Comet [#permalink]
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25 May 2011, 01:53
option D it is.
will takes care of the action being compared in the first clause.
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Re: SC : Halley's Comet [#permalink] 25 May 2011, 01:53
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# 26. In its most recent approach, the comet Crommelin
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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®. | 1,588 | 5,579 | {"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} | 3.390625 | 3 | CC-MAIN-2017-26 | latest | en | 0.934912 |
http://www.sidefx.com/docs/houdini/nodes/sop/volumefeather.html | 1,558,807,756,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232258147.95/warc/CC-MAIN-20190525164959-20190525190959-00323.warc.gz | 334,127,460 | 33,456 | # Volume Feather geometry node
Feathers the edges of volumes.
The Volume Feather operation smooths the edges of the scalar data in volume primitives. This is done either in an inwards or outwards direction. It sets the maximum rate of change in the volume, clamps either increasing or decreasing transitions, and can be thought of as a one-sided blur.
## Parameters
Source Group
The volume primitives to be feathered.
Decay
The distance multiplier factor to use to adjust the change in the voxel value for each unit change in worldspace. Lower decay values will result in a larger feather radius.
Feather Exterior
If this voxel has value `V`, and is distance `D` from another voxel of value `W`, the result is `MAX(V, W-decay*D)` for outside and `MIN(V, W+decay*D)` for inside. The result is for fog volumes to grow in the outside mode and shrink in the inside mode.
## Examples
volumefeather Example for Volume Feather geometry node
This example shows how to use the Volume Feather SOP to smooth sharp volumes either in a purely outwards or purely inwards direction.
The following examples include this node.
volumefeather Example for Volume Feather geometry node
This example shows how to use the Volume Feather SOP to smooth sharp volumes either in a purely outwards or purely inwards direction.
volumesurface_explicitgrade Example for Volume Surface geometry node
This example shows how to use the Volume Surface SOP to surface an SDF using another volume to specify the triangle sizes.
volumesurface_hierarchy Example for Volume Surface geometry node
This example shows how to use the Volume Surface SOP to surface a hierarchy of SDFs using explicit grading.
# Geometry nodes
• Removes elements while trying to maintain the overall appearance.
• Creates Points or Polygons, or adds points/polys to an input.
• Creates agent primitives.
• Adds new clips to agent primitives.
• Adds new clips to agent primitives.
• Defines how agents' animation clips should be played back.
• Creates geometry describing possible transitions between animation clips.
• Creates a new agent layer that is suitable for collision detection.
• Creates point attributes that specify the rotation limits of an agent’s joints.
• Builds a constraint network to hold an agent’s limbs together.
• Writes agent definition files to disk.
• Edits properties of agent primitives.
• Adds a new layer to agent primitives.
• Adjusts the head of an agent to look at a specific object or position.
• Adjusts the head of an agent to look at a specific object or position.
• Adds various common point attributes to agents for use by other crowd nodes.
• Adds various common point attributes to agents for use by other crowd nodes.
• Provides simple proxy geometry for an agent.
• Creates parent-child relationships between agents.
• Adapts agents' legs to conform to terrain and prevent the feet from sliding.
• Adds new transform groups to agent primitives.
• Extracts geometry from agent primitives.
• Extracts geometry from agent primitives for a Vellum simulation.
• Loads the geometry from an Alembic scene archive (.abc) file into a geometry network.
• Creates a geometry group for Alembic primitives.
• Modifies intrinsic properties of Alembic primitives.
• Aligns a group of primitives to each other or to an auxiliary input.
• Cleans up a series of break operations and creates the resulting pieces.
• Blurs out (or "relaxes") points in a mesh or a point cloud.
• Changes the size/precision Houdini uses to store an attribute.
• Composites vertex, point, primitive, and/or detail attributes between two or more selections.
• Copies attributes between groups of vertices, points, or primitives.
• Adds or edits user defined attributes.
• Deletes point and primitive attributes.
• Allows simple VEX expressions to modify attributes.
• Fades a point attribute in and out over time.
• Interpolates attributes within primitives or based on explicit weights.
• Copies and flips attributes from one side of a plane to another.
• Adds noise to attributes of the incoming geometry.
• Promotes or demotes attributes from one geometry level to another.
• Generates random attribute values of various distributions.
• Renames or deletes point and primitive attributes.
• Modifies point attributes based on differences between two models.
• Edits string attribute values.
• Copies, moves, or swaps the contents of attributes.
• Transfers vertex, point, primitive, and/or detail attributes between two models.
• Transfers attributes between two geometries based on UV proximity.
• Runs a VOP network to modify geometry attributes.
• Runs a VEX snippet to modify attribute values.
• Samples texture map information to a point attribute.
• Copies information from a volume onto the point attributes of another piece of geometry, with optional remapping.
• Converts primitives for ODE and Bullet solvers.
• Computes lighting values within volume primitives
• Provides operations for moving knots within the parametric space of a NURBS curve or surface.
• Applies deformations such as bend, taper, squash/stretch, and twist.
• Deletes primitives, points, edges or breakpoints.
• Computes a 3D metamorphosis between shapes with the same topology.
• Computes a 3D metamorphosis between shapes with the same topology.
• The start of a looping block.
• The start of a compile block.
• The end/output of a looping block.
• The end/output of a compile block.
• Supports Bone Deform by assigning capture weights to bones.
• Supports Deform by assigning capture weights to points based on biharmonic functions on tetrahedral meshes.
• Supports Bone Capture Biharmonic by creating lines from bones with suitable attributes.
• Supports Bone Deform by assigning capture weights to points based on distance to bones.
• Uses capture attributes created from bones to deform geometry according to their movement.
• Creates default geometry for Bone objects.
• Combines two polygonal objects with boolean operators, or finds the intersection lines between two polygonal objects.
• Fractures the input geometry using cutting surfaces.
• Creates a bounding box, sphere, or rectangle for the input geometry.
• Creates a cube or six-sided rectangular box.
• Deforms the points in the first input using one or more magnets from the second input.
• Records and caches its input geometry for faster playback.
• Closes open areas with flat or rounded coverings.
• Converts array attributes into a single index-pair capture attribute.
• Converts a single index-pair capture attribute into per-point and detail array attributes.
• Adjusts capture regions and capture weights.
• Lets you paint capture attributes directly onto geometry.
• Copies capture attributes from one half of a symmetric model to the other.
• Overrides the capture weights on individual points.
• Supports Capture and Deform operation by creating a volume within which points are captured to a bone.
• Slices, cuts or extracts points or cross-sections from a primitive.
• Reads sample data from a chop and converts it into point positions and point attributes.
• Creates open or closed arcs, circles and ellipses.
• Lets you deform NURBS faces and NURBS surfaces by pulling points that lie directly on them.
• Helps clean up dirty models.
• Removes or groups geometry on one side of a plane, or creases geometry along a plane.
• Captures low-res simulated cloth.
• Deforms geometry captured by the Cloth Capture SOP.
• Creates a volume representation of source geometry.
• Fills a volume with a diffuse light.
• Applies a cloud like noise to a Fog volume.
• Low-level machinery to cluster points based on their positions (or any vector attribute).
• Higher-level node to cluster points based on their positions (or any vector attribute).
• Creates geometry and VDB volumes for use with DOPs collisions.
• Adds color attributes to geometry.
• Adjust surface point normals by painting.
• Creates lines between nearby pieces.
• Creates an attribute with a unique value for each set of connected primitives or points.
• Creates simple geometry for use as control shapes.
• Converts geometry from one geometry type to another.
• Converts a 2D height field to a 3D VDB volume, polygon surface, or polygon soup surface.
• Converts the input geometry into line segments.
• Polygonizes metaball geometry.
• Generates the oriented surface of a tetrahedron mesh.
• Converts sparse volumes.
• Converts a Point Cloud into a VDB Points Primitive, or vice versa.
• Converts the iso-surface of a volume into a polygonal surface.
• Decomposes the input geometry into approximate convex segments.
• Creates multiple copies of the input geometry, or copies the geometry onto the points of the second input.
• Copies geometry and applies transformations to the copies.
• Copies the geometry in the first input onto the points of the second input.
• Manually adds or removes a creaseweight attribute to/from polygon edges, for use with the Subdivide SOP.
• Deforms and animates a piece of geometry across a surface.
• Populates a crowd of agent primitives.
• Creates crowd agents to be used with the crowd solver.
• Creates polygonal, NURBS, or Bezier curves.
• Deforms a spline surface by reshaping a curve on the surface.
• Finds the intersections (or points of minimum distance) between two or more curves or faces.
• Imports fields from DOP simulations, saves them to disk, and loads them back again.
• Imports scalar and vector fields from a DOP simulation.
• Imports option and record data from DOP simulations into points with point attributes.
• Generates point emission sources for debris from separating fractured rigid body objects.
• Runs a VEX snippet to deform geometry.
• Deletes input geometry by group, entity number, bounding volume, primitive/point/edge normals, and/or degeneracy.
• Smooths out (or "relaxes") point deformations.
• Attempts to prevent collisions when deforming geometry.
• Deletes edges from the input polygonal geometry merging polygons with shared edges.
• Deletes points, primitives, and edges from the input geometry and repairs any holes left behind.
• Divides, smooths, and triangulates polygons.
• Imports and transforms geometry based on information extracted from a DOP simulation.
• Creates a curve based on user input in the viewport.
• Culls the input geometry according to the specifications of the For Each SOP.
• Collapses edges and faces to their centerpoints.
• Sharpens edges by uniquing their points and recomputing point normals.
• Inserts points on the edges of polygons and optionally connects them.
• Flips the direction of polygon edges.
• Cuts geometry along edges using guiding curves.
• Copies and optionally modifies attribute values along edges networks and curves.
• Edits points, edges, or faces interactively.
• Closes, opens, or clamps end points.
• Sets an attribute on selected points or primitives to sequential numbers.
• Generates a message, warning, or error, which can show up on a parent asset.
• Pushes geometry out from the center to create an exploded view.
• Pushes geometry out from the center to create an exploded view.
• Computes the centroid of each piece of the geometry.
• Computes the best-fit transform between two pieces of geometry.
• Extrudes geometry along a normal.
• Extrudes surface geometry into a volume.
• Creates a surface or density VDB for sourcing FLIP simulations.
• Controls the smoothness of faceting of a surface.
• Adds smooth distance attributes to geometry.
• Evolves polygonal curves as vortex filaments.
• Reads, writes, or caches geometry on disk.
• Writes and reads geometry sequences to disk.
• Reads and collates data from disk.
• Creates smooth bridging geometry between two curves or surfaces.
• Finds the shortest paths from start points to end points, following the edges of a surface.
• Fits a spline curve to points, or a spline surface to a mesh of points.
• Compresses the output of fluid simulations to decrease size on disk
• Creates 3D text from Type 1, TrueType and OpenType fonts.
• Uses a metaball to attract or repel points or springs.
• Creates jagged mountain-like divisions of the input geometry.
• Creates a set of hair-like curves across a surface.
• Merges points.
• Merges or splits (uniques) points.
• Adds strength to a glue constraint network according to cluster values.
• Generates particles to be used as sources in a particle-based grain simulation.
• Assigns a unique integer attribute to non-touching components.
• Creates planar geometry.
• Blends the guides and skin of two grooms.
• Fetches groom data from grooming objects.
• Packs the components of a groom into a set of named Packed Primitives for the purpose of writing it to disk.
• Switches between all components of two groom streams.
• Unpacks the components of a groom from a packed groom.
• Generates groups of points, primitives, edges, or vertices according to various criteria.
• Combines point groups, primitive groups, or edge groups according to boolean operations.
• Copies groups between two pieces of geometry, based on point/primitive numbers.
• Deletes groups of points, primitives, edges, or vertices according to patterns.
• Runs VEX expressions to modify group membership.
• Sets group membership interactively by painting.
• Converts point, primitive, edge, or vertex groups into point, primitive, edge, or vertex groups.
• Groups points and primitives by ranges.
• Renames groups according to patterns.
• Transfers groups between two pieces of geometry, based on proximity.
• Advects guide points through a velocity volume.
• Resolves collisions of guide curves with VDB signed distance fields.
• Deforms geometry with an animated skin and optionally guide curves.
• Allows intuitive manipulation of guide curves in the viewport.
• Creates standard primitive groups used by grooming tools.
• Quickly give hair guides some initial direction.
• Creates masking attributes for other grooming operations.
• Creates and prepares parting lines for use with hair generation.
• Looks up skin geometry attributes under the root point of guide curves.
• Constructs a coherent tangent space along a curve.
• Transfer hair guides between geometries.
• Converts dense hair curves to a polygon card, keeping the style and shape of the groom.
• Clumps guide curves together.
• Generates hair on a surface or from points.
• Generates a velocity field based on stroke primitives.
• Generates an initial heightfield volume for use with terrain tools.
• Blurs a terrain height field or mask.
• Limits height values to a certain minimum and/or maximum.
• Creates a copy of a height field or mask.
• Extracts a square of a certain width/length from a larger height volume, or resizes/moves the boundaries of the height field.
• Creates a cutout on a terrain based on geometry.
• Displaces a height field by another field.
• Advects the input volume through a noise pattern to break up hard edges and add variety.
• Lets you draw shapes to create a mask for height field tools.
• Calculates thermal and hydraulic erosion over time (frames) to create more realistic terrain.
• Calculates thermal and hydraulic erosion over time (frames) to create more realistic terrain.
• Simulates the erosion from one heightfield sliding over another for a short time.
• Distributes water along a heightfield. Offers controls for adjusting the intensity, variability, and location of rainfall.
• Calculates the effect of thermal erosion on terrain for a short time.
• Imports a 2D image map from a file or compositing node into a height field or mask.
• Generates flow and flow direction layers according to the input height layer.
• Copies another layer over the mask layer, and optionally flattens the height field.
• Composites together two height fields.
• Sets all values in a heightfield layer to a fixed value.
• Sets the border voxel policy on a height field volume.
• Creates a mask based on different features of the height layer.
• Creates a mask based some other geometry.
• Creates a mask where the input terrain is hollow/depressed, for example riverbeds and valleys.
• Adds vertical noise to a height field, creating peaks and valleys.
• Exports height and/or mask layers to disk as an image.
• Lets you paint values into a height or mask field using strokes.
• Patches features from one heightfield to another.
• Adds displacement in the form of a ramps, steps, stripes, Voronoi cells, or other patterns.
• Projects 3D geometry into a height field.
• Applies a material that lets you plug in textures for different layers.
• Remaps the values in a height field or mask layer.
• Changes the resolution of a height field.
• Scatters points across the surface of a height field.
• Scatters points across the surface of a height field.
• Simulates loose material sliding down inclines and piling at the bottom.
• Creates stepped plains from slopes in the terrain.
• Stitches height field tiles back together.
• Splits a height field volume into rows and columns.
• Height field specific scales and offsets.
• Visualizes elevations using a custom ramp material, and mask layers using tint colors.
• Makes holes in surfaces.
• Deforms the points in the first input to make room for the inflation tool.
• Instances Geometry on Points.
• Creates points with attributes at intersections between a triangle and/or curve mesh with itself, or with an optional second set of triangles and/or curves.
• Composes triangle surfaces and curves together into a single connected mesh.
• Processes its inputs using the operation of a referenced compiled block.
• Builds an offset surface from geometry.
• Generates an isometric surface from an implicit function.
• The Join op connects a sequence of faces or surfaces into a single primitive that inherits their attributes.
• Divides, deletes, or groups geometry based on an interactively drawn line.
• Creates fractal geometry from the recursive application of simple rules.
• Deforms geometry based on how you reshape control geometry.
• Reads a lidar file and imports a point cloud from its data.
• Creates polygon or NURBS lines from a position, direction, and distance.
• Animates points using an MDD file.
• Deforms geometry by using another piece of geometry to attract or repel points.
• Aligns the input geometry to a specific axis.
• Resizes and recenters the geometry according to reference geometry.
• Reorders the primitive and point numbers of the input geometry to match some reference geometry.
• Assigns one or more materials to geometry.
• Measures area, volume, or curvature of individual elements or larger pieces of a geometry and puts the results in attributes.
• Measures volume, area, and perimeter of polygons and puts the results in attributes.
• Merges geometry from its inputs.
• Defines groupings of metaballs so that separate groupings are treated as separate surfaces when merged.
• Creates metaballs and meta-superquadric surfaces.
• Duplicates and mirrors geometry across a mirror plane.
• Displaces points along their normals based on fractal noise.
• Displaces points along their normals based on fractal noise.
• Supports Muscle Deform by assigning capture weights to points based on distance away from given primitives
• Deforms a surface mesh representing skin to envelop or drape over geometry representing muscles
• Creates a "naming" attribute on points or primitives allowing you to refer to them easily, similar to groups.
• Computes surface normal attribute.
• Does nothing.
• Merges geometry from multiple sources and allows you to define the manner in which they are grouped together and transformed.
• Assists the creation of a Muscle or Muscle Rig by allowing you to draw a stroke on a projection surface.
• Deforms input geometry based on ocean "spectrum" volumes.
• Deforms input geometry based on ocean "spectrum" volumes.
• Generates particle-based foam
• Generates particles and volumes from ocean "spectrum" volumes for use in simulations
• Generates particles and volumes from ocean "spectrum" volumes for use in simulations
• Generates volumes containing information for simulating ocean waves.
• Instances individual waveforms onto input points and generated points.
• Executes an OpenCL kernel on geometry.
• Marks the output of a sub-network.
• Packs geometry into an embedded primitive.
• Packs points into a tiled grid of packed primitives.
• Editing Packed Disk Primitives.
• Editing Packed Primitives.
• Lets you paint color or other attributes on geometry.
• Creates a color volume based on drawn curve
• Creates a fog volume based on drawn curve
• Creates an SDF volume based on drawn curve
• Generates a surface around the particles from a particle fluid simulation.
• Creates a set of regular points filling a tank.
• Places points and primitives into groups based on a user-supplied rule.
• Moves primitives, points, edges or breakpoints along their normals.
• Creates a planar polygonal patch.
• Fills in a 2d curve network with triangles.
• Deforms flat geometry into a pleat.
• Creates platonic solids of different types.
• Manually adds or edits point attributes.
• Constructs an iso surface from its input points.
• Deforms geometry on an arbitrary connected point mesh.
• Creates new points, optionally based on point positions in the input geometry.
• Jitters points in random directions.
• Moves points with overlapping radii away from each other, optionally on a surface.
• Generates a cloud of points around the input points.
• Computes and manipulates velocities for points of a geometry.
• Creates set of regular points filling a volume.
• Creates flat or tube-shaped polygon surfaces between source and destination edge loops, with controls for the shape of the bridge.
• Creates offset polygonal geometry for planar polygonal graphs.
• Extrudes polygonal faces and edges.
• Creates straight, rounded, or custom fillets along edges and corners.
• Bevels points and edges.
• Breaks curves where an attribute crosses a threshold.
• Helps repair invalid polygonal geometry, such as for cloth simulation.
• Extrudes polygonal faces and edges.
• Fills holes with polygonal patches.
• Creates coordinate frame attributes for points and vertices.
• Creates new polygons using existing points.
• Creates a smooth polygonal patch from primitives.
• Cleans up topology of polygon curves.
• Reduces the number of polygons in a model while retaining its shape. This node preserves features, attributes, textures, and quads during reduction.
• Combines polygons into a single primitive that can be more efficient for many polygons
• The PolySpline SOP fits a spline curve to a polygon or hull and outputs a polygonal approximation of that spline.
• Divides an existing polygon into multiple new polygons.
• Divides an existing polygon into multiple new polygons.
• Stitches polygonal surfaces together, attempting to remove cracks.
• Constructs polygonal tubes around polylines, creating renderable geometry with smooth bends and intersections.
• Interpolates between a set of pose-shapes based on the value of a set of drivers.
• Combine result of Pose-Space Deform with rest geometry.
• Packs geometry edits for pose-space deformation.
• Creates common attributes used by the Pose-Space Edit SOP.
• Edits primitive, primitive attributes, and profile curves.
• Takes a primitive attribute and splits any points whose primitives differ by more than a specified tolerance at that attribute.
• Extracts or manipulates profile curves.
• Creates profile curves on surfaces.
• Creates points for sourcing pyro and smoke simulations.
• Runs a Python snippet to modify the incoming geometry.
• Combines fractured pieces or constraints into larger clusters.
• Creates attributes describing rigid body constraints.
• Creates rigid body constraint geometry from curves drawn in the viewport.
• Creates rigid body constraint geometry from interactively drawn lines in the viewport.
• Creates rigid body constraint geometry from a set of rules and conditions.
• Creates additional detail on the interior surfaces of fractured geometry.
• Fractures the input geometry based on a material type.
• Fractures the input geometry based on a material type.
• Packs RBD geometry, constraints, and proxy geometry into a single geometry.
• Paints values onto geometry or constraints using strokes.
• Unpacks an RBD setup into three outputs.
• Attaches RenderMan shaders to groups of faces.
• Generates surfaces by stretching cross-sections between two guide rails.
• Projects one surface onto another.
• Increases the number of points/CVs in a curve or surface without changing its shape.
• Scatters new guides, interpolating the properties of existing guides.
• Recreates the shape of the input surface using "high-quality" (nearly equilateral) triangles.
• Repacks geometry as an embedded primitive.
• Resamples one or more curves or surfaces into even length segments.
• Sets the alignment of solid textures to the geometry so the texture stays put on the surface as it deforms.
• Retimes the time-dependent input geometry.
• Reverses or cycles the vertex order of faces.
• Revolves a curve around a center axis to sweep out a surface.
• Rewires vertices to different points specified by an attribute.
• Generates ripples by displacing points along the up direction specified.
• Scatters new points randomly across a surface or through a volume.
• Runs scripts when cooked.
• Lets you interactively reshape a surface by brushing.
• Morphs though a sequence of 3D shapes, interpolating geometry and attributes.
• Sequence Blend lets you do 3D Metamorphosis between shapes and Interpolate point position, colors…
• Computes the post-deform or pre-deform difference of two geometries with similar topologies.
• Computes the convex hull of the input geometry and moves its polygons inwards along their normals.
• Takes the convex hull of input geometry and moves its polygons inwards along their normals.
• Builds a skin surface between any number of shape curves.
• Creates a sky filled with volumentric clouds
• Smooths out (or "relaxes") polygons, meshes and curves without increasing the number of points.
• Smooths out (or "relaxes") polygons, meshes and curves without increasing the number of points.
• Moves the selected point along its normal, with smooth rolloff to surrounding points.
• Moves the selected point, with smooth rolloff to surrounding points.
• Creates a tetrahedral mesh that conforms to a connected mesh as much as possible.
• Creates a simple tetrahedral mesh that covers a connected mesh.
• Creates a partition of a tetrahedral mesh that can be used for finite-element fracturing.
• Allows running a SOP network iteratively over some input geometry, with the output of the network from the previous frame serving as the input for the network at the current frame.
• Reorders points and primitives in different ways, including randomly.
• Creates a sphere or ovoid surface.
• Splits primitives or points into two streams.
• Spray paints random points onto a surface.
• A SOP node that sets the sprite display for points.
• Insets points on polygonal faces.
• Caches the input geometry in the node on command, and then uses it as the node’s output.
• Stretches two curves or surfaces to cover a smooth area.
• Low level tool for building interactive assets.
• Subdivides polygons into smoother, higher-resolution polygons.
• The Subnet op is essentially a way of creating a macro to represent a collection of ops as a single op in the Network Editor.
• Trims or creates profile curves along the intersection lines between NURBS or bezier surfaces.
• Creates a surface by sweeping cross-sections along a backbone curve.
• Switches between network branches based on an expression or keyframe animation.
• Sends input geometry to a TOP subnet and retrieves the output geometry.
• Reads a CSV file creating point per row.
• Creates a rock creature, which can be used as test geometry.
• Creates a pig head, which can be used as test geometry..
• Creates a rubber toy, which can be used as test geometry.
• Creates a shader ball, which can be used to test shaders.
• Creates a squab, which can be used as test geometry.
• Creates a soldier, which can be used as test geometry.
• Provides a simple crowd simulation for testing transitions between animation clips.
• Provides a simple Bullet simulation for testing the behavior of a ragdoll.
• Partitions a given tetrahedron mesh into groups of tets isolated by a given polygon mesh
• Performs variations of a Delaunay Tetrahedralization.
• Cooks the input at a different time.
• Sets attributes used by the Toon Color Shader and Toon Outline Shader.
• Lets you interactively draw a reduced quad mesh automatically snapped to existing geometry.
• Creates a torus (doughnut) shaped surface.
• Traces curves from an image file.
• Creates trails behind points.
• The Transform operation transforms the source geometry in "object space" using a transformation matrix.
• Transforms the input geometry relative to a specific axis.
• Transforms the input geometry by a point attribute.
• Transforms input geometry according to transformation attributes on template geometry.
• Creates a triangular Bezier surface.
• Refines triangular meshes using various metrics.
• Connects points to form well-shaped triangles.
• Trims away parts of a spline surface defined by a profile curve or untrims previous trims.
• Creates open or closed tubes, cones, or pyramids.
• Generates an edge group representing suggested seams for flattening a polygon model in UV space.
• Adjusts texture coordinates in the UV viewport by painting.
• Lets you interactively move UVs in the texture view.
• Creates flattened pieces in texture space from 3D geometry.
• Creates flattened pieces in texture space from 3D geometry.
• Merges UVs.
• Packs UV islands efficiently into a limited area.
• Relaxes UVs by pulling them out toward the edges of the texture area.
• Assigns UVs by projecting them onto the surface from a set direction.
• Applies an image file as a textured shader to a surface.
• Assigns texture UV coordinates to geometry for use in texture and bump mapping.
• Transforms UV texture coordinates on the source geometry.
• Transforms UV texture coordinates on the source geometry.
• Separates UVs into reasonably flat, non-overlapping groups.
• Processes geometry using an external program.
• Unpacks packed primitives.
• Unpacks points from packed primitives.
• Creates one or more empty/uniform VDB volume primitives.
• Activates voxel regions of a VDB for further processing.
• Expand or contract signed distance fields stored on VDB volume primitives.
• Moves VDBs in the input geometry along a VDB velocity field.
• Moves points in the input geometry along a VDB velocity field.
• Computes an analytic property of a VDB volumes, such as gradient or curvature.
• Clips VDB volume primitives using a bounding box or another VDB as a mask.
• Combines the values of two aligned VDB volumes in various ways.
• Tests VDBs for Bad Values and Repairs.
• Cuts level set VDB volume primitives into multiple pieces.
• Build an LOD Pyramid from a VDB.
• Blends between source and target SDF VDBs.
• Create a mask of the voxels in shadow from a camera for VDB primitives.
• Deletes points inside of VDB Points primitives.
• Manipulates the Internal Groups of a VDB Points Primitive.
• Computes the steady-state air flow around VDB obstacles.
• Removes divergence from a Vector VDB.
• Fixes signed distance fields stored in VDB volume primitives.
• Re-samples a VDB volume primitive into a new orientation and/or voxel size.
• Reshapes signed distance fields in VDB volume primitives.
• Splits SDF VDBs into connected components.
• Smooths out the values in a VDB volume primitive.
• Smooths out SDF values in a VDB volume primitive.
• Creates an SDF VDB based on the active set of another VDB.
• Merges three scalar VDB into one vector VDB.
• Splits a vector VDB primitive into three scalar VDB primitives.
• Replaces a VDB volume with geometry that visualizes its structure.
• Generates a signed distance field (SDF) VDB volume representing the surface of a set of particles from a particle fluid simulation.
• Converts point clouds and/or point attributes into VDB volume primitives.
• Converts polygonal surfaces and/or surface attributes into VDB volume primitives.
• Fills a VDB volume with adaptively-sized spheres.
• Configures geometry for Vellum Grain constraints.
• Configure constraints on geometry for the Vellum solvers.
• Vellum solver setup to pre-roll fabric to drape over characters.
• Packs Vellum simulations, saves them to disk, and loads them back again.
• Packs Vellum geometry and constraints into a single geometry.
• Applies common post-processing effects to the result of Vellum solves.
• Blends the current rest values of constraints with a rest state calculated from external geometry.
• Runs a dynamic Vellum simulation.
• Unpacks a Vellum simulation into two outputs.
• Verify that a bsdf conforms to the required interface.
• Manually adds or edits attributes on vertices (rather than on points).
• Takes a vertex attribute and splits any point whose vertices differ by more than a specified tolerance at that attribute.
• Shows/hides primitives in the 3D viewer and UV editor.
• Lets you attach visualizations to different nodes in a geometry network.
• Creates a volume primitive.
• Computes analytic properties of volumes.
• Computes a speed-defined travel time from source points to voxels.
• Blurs the voxels of a volume.
• Bounds voxel data.
• Cuts polygonal objects using a signed distance field volume.
• Re-compresses Volume Primitives.
• Convolves a volume by a 3×3×3 kernel.
• Compute the Fast Fourier Transform of volumes.
• Feathers the edges of volumes.
• Flattens many volumes into one volume.
• Combines the scalar fields of volume primitives.
• Translates the motion between two "image" volumes into displacement vectors.
• Fill in a region of a volume with features from another volume.
• Remaps a volume according to a ramp.
• Rasterizes into a volume.
• Samples point attributes into VDBs.
• Converts a curve into a volume.
• Converts fur or hair to a volume for rendering.
• Converts a point cloud into a volume.
• Converts a point cloud into a volume.
• Reduces the values of a volume into a single number.
• Resamples the voxels of a volume to a new resolution.
• Resizes the bounds of a volume without changing voxels.
• Builds a Signed Distance Field from an isocontour of a volume.
• Extracts 2d slices from volumes.
• Splices overlapping volume primitives together.
• Stamps volumes instanced on points into a single target volume.
• Adaptively surfaces a volume hierarchy with a regular triangle mesh.
• Computes a trail of points through a velocity volume.
• Runs CVEX on a set of volume primitives.
• Computes a velocity volume.
• Generates a volume velocity field using curve tangents.
• Generates a velocity field within a surface geometry.
• Adjusts attributes for multi-volume visualization.
• Runs a VEX snippet to modify voxel values in a volume.
• Sets the voxels of a volume from point attributes.
• Fractures the input geometry by performing a Voronoi decomposition of space around the input cell points
• Fractures the input geometry by performing a Voronoi decomposition of space around the input cell points
• Given an object and points of impact on the object, this SOP generates a set of points that can be used as input to the Voronoi Fracture SOP to simulate fracturing the object from those impacts.
• Cuts the geometry into small pieces according to a set of cuts defined by polylines.
• Creates the point attributes needed to create a Vortex Force DOP.
• Generates volumes to be used as sources in a whitewater simulation.
• Generates emission particles and volumes to be used as sources in a Whitewater simulation.
• Computes generalized winding number of surface at query points.
• Morphs between curve shapes while maintaining curve length.
• Captures surfaces to a wire, allowing you to edit the wire to deform the surface.
• Deforms geometry captured to a curve via the Wire Capture node.
• Transfers the shape of one curve to another.
• Constructs polygonal tubes around polylines, creating renderable geometry.
• Assigns channel paths and/or pickscripts to geometry. | 7,499 | 36,816 | {"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} | 2.53125 | 3 | CC-MAIN-2019-22 | latest | en | 0.790525 |
https://www.geeksforgeeks.org/gate-gate-it-2005-question-36/ | 1,585,564,679,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370496901.28/warc/CC-MAIN-20200330085157-20200330115157-00376.warc.gz | 959,010,286 | 22,743 | # GATE | Gate IT 2005 | Question 36
Let P(x) and Q(x) be arbitrary predicates. Which of the following statements is always TRUE?
(A) ((∀x(P(x)∨Q(x))))⟹((∀xP(x))∨(∀xQ(x)))
(B) (∀x(P(x)⟹Q(x)))⟹((∀xP(x))⟹(∀xQ(x)))
(C) (∀x(P(x))⟹∀x(Q(x)))⟹(∀x(P(x)⟹Q(x)))
(D) (∀x(P(x))⇔(∀x(Q(x))))⟹(∀x(P(x)⇔Q(x)))
Explanation:
Generally, these type of questions can be solved by using two statements and checking the validity of each and every option.
Here, let the statements be P and Q where,
P : Student is a girl
Q : Student is smart
Option B says that, IF for all student x : If x is a girl then the student is smart THEN IF the whole class comprises of girls then the whole class comprises of smart students.
This solution is contributed by Anil Saikrishna Devarasetty.
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Please write to us at contribute@geeksforgeeks.org to report any issue with the above content. | 304 | 951 | {"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} | 3.171875 | 3 | CC-MAIN-2020-16 | longest | en | 0.821465 |
https://www.calculatoratoz.com/en/fiber-length-given-time-difference-calculator/Calc-43373 | 1,722,723,467,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640380725.7/warc/CC-MAIN-20240803214957-20240804004957-00148.warc.gz | 558,149,533 | 33,933 | ## Fiber Length Given Time Difference Solution
STEP 0: Pre-Calculation Summary
Formula Used
Fiber Length = ([c]*Time Difference)/(2*Refractive Index of Core)
l = ([c]*tdif)/(2*ηcore)
This formula uses 1 Constants, 3 Variables
Constants Used
[c] - Light speed in vacuum Value Taken As 299792458.0
Variables Used
Fiber Length - (Measured in Meter) - Fiber Length refers to the measure of how long the fiber is.
Time Difference - (Measured in Second) - Time Difference refers to the concept of latency or delay.
Refractive Index of Core - The Refractive Index of Core is defined as how the light travels through that medium. It defines how much a light ray can bend when it enters from one medium to the other.
STEP 1: Convert Input(s) to Base Unit
Time Difference: 3.94 Microsecond --> 3.94E-06 Second (Check conversion here)
Refractive Index of Core: 1.335 --> No Conversion Required
STEP 2: Evaluate Formula
Substituting Input Values in Formula
l = ([c]*tdif)/(2*ηcore) --> ([c]*3.94E-06)/(2*1.335)
Evaluating ... ...
l = 442.390368734082
STEP 3: Convert Result to Output's Unit
442.390368734082 Meter --> No Conversion Required
442.390368734082 442.3904 Meter <-- Fiber Length
(Calculation completed in 00.004 seconds)
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## Credits
Created by Vaidehi Singh
Prabhat Engineering College (P.E.C.), Uttar Pradesh
Vaidehi Singh has created this Calculator and 25+ more calculators!
Verified by Priyanka Patel
Lalbhai Dalpatbhai College of engineering (LDCE), Ahmedabad
Priyanka Patel has verified this Calculator and 10+ more calculators!
## <Fiber Optic Parameters Calculators
Refractive Index of Material Given Optical Power
Go Refractive Index of Core = Ordinary Refractive Index+Non Linear Index Coefficient*(Incident Optical Power/Effective Area)
Carrier to Noise Ratio
Go Carrier to Noise Ratio = Carrier Power/(The Relative Intensity Noise (RIN) power+Shot Noise Power+Thermal Noise Power)
Total Dispersion
Go Dispersion = sqrt(Fiber Rise Time^2+Pulse Spreading Time^2+Modal Dispersion Time^2)
Fiber Length Given Time Difference
Go Fiber Length = ([c]*Time Difference)/(2*Refractive Index of Core)
## Fiber Length Given Time Difference Formula
Fiber Length = ([c]*Time Difference)/(2*Refractive Index of Core)
l = ([c]*tdif)/(2*ηcore)
## What is the length of optical fiber typically used for telecommunications?
In telecommunications, single-mode fibers are used for most communication links longer than 1,050 meters (3,440 ft). These fibers can span many miles for long-distance transmission. However, the length of the fiber can also be much shorter for applications such as fiber optic sensors and fiber lasers.
## How to Calculate Fiber Length Given Time Difference?
Fiber Length Given Time Difference calculator uses Fiber Length = ([c]*Time Difference)/(2*Refractive Index of Core) to calculate the Fiber Length, Fiber Length Given Time Difference is the formula to calculate the length of the optical fiber using the delay or latency of the light wave. Optical fiber length refers to the physical length of the fiber optic cable. The length of an optical fiber can vary widely depending on its application. Fiber Length is denoted by l symbol.
How to calculate Fiber Length Given Time Difference using this online calculator? To use this online calculator for Fiber Length Given Time Difference, enter Time Difference (tdif) & Refractive Index of Core core) and hit the calculate button. Here is how the Fiber Length Given Time Difference calculation can be explained with given input values -> 442.3904 = ([c]*3.94E-06)/(2*1.335).
### FAQ
What is Fiber Length Given Time Difference?
Fiber Length Given Time Difference is the formula to calculate the length of the optical fiber using the delay or latency of the light wave. Optical fiber length refers to the physical length of the fiber optic cable. The length of an optical fiber can vary widely depending on its application and is represented as l = ([c]*tdif)/(2*ηcore) or Fiber Length = ([c]*Time Difference)/(2*Refractive Index of Core). Time Difference refers to the concept of latency or delay & The Refractive Index of Core is defined as how the light travels through that medium. It defines how much a light ray can bend when it enters from one medium to the other.
How to calculate Fiber Length Given Time Difference?
Fiber Length Given Time Difference is the formula to calculate the length of the optical fiber using the delay or latency of the light wave. Optical fiber length refers to the physical length of the fiber optic cable. The length of an optical fiber can vary widely depending on its application is calculated using Fiber Length = ([c]*Time Difference)/(2*Refractive Index of Core). To calculate Fiber Length Given Time Difference, you need Time Difference (tdif) & Refractive Index of Core core). With our tool, you need to enter the respective value for Time Difference & Refractive Index of Core and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.
Let Others Know | 1,178 | 5,052 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "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} | 3.453125 | 3 | CC-MAIN-2024-33 | latest | en | 0.750966 |
https://studytonight.com/numpy/numpy-ones-function | 1,611,403,774,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703537796.45/warc/CC-MAIN-20210123094754-20210123124754-00770.warc.gz | 585,164,841 | 8,946 | New Tutorials:
# NumPy ones() function
In this tutorial, we will cover `numpy.matlib.ones()` function of the Numpy library.
The function `numpy.matlib.ones()` is used to return the matrix of given shape and type. This function initializes all the values of the matrix to one(1).
The `numpy.matlib` is a matrix library used to configure matrices instead of ndarray objects.
### Syntax of matlib.ones():
The required syntax to use this function is as follows:
``numpy.matlib.ones(shape,dtype,order)``
Parameters:
Let us now cover the parameters used with this function:
• shape
This parameter is in the form of a tuple that is used to define the shape of the matrix.
• dtype
This parameter is used to indicate the data type of the matrix. The default value of this parameter is `float`. This is an optional parameter.
• order
This is an optional parameter that is used to indicate the insertion order of the matrix. It mainly indicates whether to store the result in C- or Fortran-contiguous order, The default value is 'C'.
Returned Values:
This function will return a matrix with all the entries initialized to 1.
Now it's time to cover a few examples of this function.
## Example 1:
Given below is a basic example for the understanding of this function:
``````import numpy as np
import numpy.matlib
print(numpy.matlib.ones((5,4))) ``````
[[1. 1. 1. 1.]
[1. 1. 1. 1.]
[1. 1. 1. 1.]
[1. 1. 1. 1.]
[1. 1. 1. 1.]]
## Example 2:
Now we will also use type and order parameter in the code snippet given below:
``````import numpy as np
import numpy.matlib
print("The Output matrix is :\n",numpy.matlib.ones((3,4),int)) ``````
The Output matrix is :
[[1 1 1 1]
[1 1 1 1]
[1 1 1 1]]
## Example 3:
One more example,
``````import numpy as np
# 1-d array with 5 elements
np.matlib.ones(5)``````
matrix([[1., 1., 1., 1., 1.]])
## Summary
In this tutorial we learned about `numpy.ones()` mathematical function in the Numpy library. We also covered its syntax, parameters as well as the value returned by this function along with few code examples. | 560 | 2,069 | {"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} | 3.34375 | 3 | CC-MAIN-2021-04 | latest | en | 0.654838 |
https://tbc-python.fossee.in/convert-notebook/Basic_Electrical_Engineering_with_Numerical_Problems/Chapter_03.ipynb | 1,726,730,401,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651995.50/warc/CC-MAIN-20240919061514-20240919091514-00891.warc.gz | 510,095,678 | 37,856 | # Chapter 3: Electricity and its Fundamental Laws
### Example 3.1: Page 47¶
In [2]:
from __future__ import division
import math
#given data:
V = 230; # in volts
I = 10; # in A
# calculations:
R = V/I;
#Results
print "resistance of element,R =", R,"ohm"
resistance of element,R = 23.0 ohm
### Example 3.2: Page 47¶
In [1]:
from __future__ import division
import math
#given data :
R1 = 0.5; # minimum value of resistance in ohm
R2 = 20; # maximum value of resistance in ohm
I = 1.2; # current in A
#Calculation
V1 = I*R1;
V2 = I*R2;
#Results
print "Voltage drop in Ist case,V1(V)=",V1,"volts and voltage drop in IInd case,V2(V)=", V2,"volts"
Voltage drop in Ist case,V1(V)= 0.6 volts and voltage drop in IInd case,V2(V)= 24.0 volts
### Example 3.3: Page 48¶
In [3]:
from __future__ import division
import math
#given data :
L = 1000; # length of wire in cm
d = 0.14; # diameter of wire in cm
R1 = 2.5*10**6;# resistance in micro-ohm
# calculations:
a = (math.pi*d**2)/4; # cross section area
p = (R1*a)/L;
#Results
print "the specific resistance,p =", round(p,1)," uOhm-cm"
the specific resistance,p = 38.5 uOhm-cm
### Example 3.4: Page 49¶
In [4]:
from __future__ import division
import math
#given data :
Rt1 = 54.3;# resistance in ohm
alfa = 0.0043;# the resistance temperature of coeficient in per degree celcius
t1 = 20;# temperature in degree celcius
t2 = 40;# temperature in degree celcius
# calculations
Rt2 = (Rt1*(1+(alfa*t2)))/(1+(alfa*t1));
#Results
print "resistance at 40 degC ,Rt2=", Rt2," ohms"
resistance at 40 degC ,Rt2= 58.6 ohms
### Example 3.5: Page 50¶
In [4]:
from __future__ import division
import math
#given data :
r1=30;# resistance in ohm
r2=35;# resistance in ohm
r3=45;# resistance in ohm
V=220;
# calculations:
R=r1+r2+r3;
I=V/R;
#Results
print "(a)total resistance,R=", R," ohm"
print "(b)current,I=", I,"A"
(a)total resistance,R= 110 ohm
(b)current,I= 2.0 A
### Example 3.6: Page 50¶
In [6]:
from __future__ import division
import math
#given data :
I=75;# current in A
R=0.15;# resistance in ohm
v=220;# voltage in volts
#calculations
V1=I*R;# voltage drop of the feeder in section AB
V2=I*R;# voltage drop of the feeder in section CD
V_total=V1+V2;# total voltage drop in the lead and return feeder
V=v+V_total;
#Results
print "voltage at the generating station,V=", V,"volts"
voltage at the generating station,V= 242.5 volts
### Example 3.7: Page 52¶
In [7]:
from __future__ import division
import math
#given data :
r1=6;# resistance in ohm
r2=10;# resistance in ohm
r3=15;# resistance in ohm
#calculations:
r=(1/r1)+(1/r2)+(1/r3);
R=1/r;
#Results
print "equivalent resistance,R=", R,"ohm"
equivalent resistance,R= 3.0 ohm
### Example 3.8: Page 53¶
In [8]:
from __future__ import division
import math
#given data :
I=5; # current in A
n=2; # number of resistance in parallel of section BC
r1=15;# resistance in ohm
r2=20;# resistance in ohm
r3=60;# resistance in ohm
r4=64;# resistance in ohm
r5=64;# resistance in ohm
r6=2.5;# resistance in ohm
#calculation
R1=r4/n;# equivalent resistance of section BC
R2=(r1*r2*r3)/((r1*r2)+(r2*r3)+(r3*r1));# equivalent resistance of section CD
R=R1+R2+r6;# equivalent resistance of section AD
V=I*R;
#Results
print "voltage,V= ", V,"volts"
voltage,V= 210.0 volts
### Example 3.9: Page 53¶
In [9]:
from __future__ import division
import math
#given data :
V=240;# voltage in volts
r1=2;# resistance in ohm
r2=3;# resistance in ohm
r3=8.8;# resistance in ohm
r4=10;# resistance in ohm
r5=3;# resistance in ohm
# calculations:
R1=(r1*r2)/(r1+r2);# equivalent resistance of parallel branch
R2=R1+r3;# equivalent resistance of section ABC
R3=(R2*r4)/(R2+r4);
R=R3+r5;# total resistance of section AD
I=V/R;
V1=I*r5;# voltage drop across r5
V2=V-V1;# voltage drop across section ABC
I1=V2/r4;# current flowing through r4 resistance
I2=I-I1;# current in r3 resistance
V3=I2*r3;# voltage drop across r3 resistance, section ABC
V4=V2-V3;# voltage drop between section AB
I3=V4/r1;# current flowing through r1 resistance
I4=V4/r2;# current flowing through r2 resistance
#Results
print "current flowing through r1 (2 ohms) resistance,I3 =", I3," A"
print "current flowing through r2 (3 ohms)resistance,I4 =", I4," A"
print "total resistance,R = ", R," ohm"
print "voltage drop across r5(3 ohms) resistance,V1 =", V1," volts"
print "voltage drop across section ABC,V2 = ", V2," volts"
print "voltage drop across r3 resistance(8.8 ohms),V3 = ",V3," Volts"
print " voltage drop between section AB,V4 = ", V4,"volts"
current flowing through r1 (2 ohms) resistance,I3 = 9.0 A
current flowing through r2 (3 ohms)resistance,I4 = 6.0 A
total resistance,R = 8.0 ohm
voltage drop across r5(3 ohms) resistance,V1 = 90.0 volts
voltage drop across section ABC,V2 = 150.0 volts
voltage drop across r3 resistance(8.8 ohms),V3 = 132.0 Volts
voltage drop between section AB,V4 = 18.0 volts
### Example 3.10: Page 54¶
In [10]:
from __future__ import division
import math
#given data :
I=44;# current in A
r1=6;# resistance in ohm
r2=12;# resistance in ohm
r3=18;# resistance in ohmr1
# calculation:
a=(1/r1)+(1/r2)+(1/r3);
R=1/a;
V=I*R;
i1=V/r1;
i2=V/r2;
i3=V/r3;
#Results
print "current in 6 ohm resistance,i1 = ",i1,"A"
print "current in 12 ohm resistance,i2 = ",i2,"A"
print "current in 18 ohm resistance,i3 = ",i3,"A"
current in 6 ohm resistance,i1 = 24.0 A
current in 12 ohm resistance,i2 = 12.0 A
current in 18 ohm resistance,i3 = 8.0 A
### Example 3.11: Page 55¶
In [7]:
from __future__ import division
import math
#given:
t=15 #TOTAL CURRENT IN AMPERES
i1=2 #CURRENT THROUGH UNKNOWN RESISTANCE
R1=15#in ohms
R2=50/2#in ohms
# calculations:
x=(t-i1)*((R1*R2)/(R1+2*R2))#unknown resistance in ohms)
PD=i1*x#in volts
RX=((1/R1)+(1/(2*R2))+(1/x))#
R=1/RX
i5= PD/(2*R2)#current in 5 ohms resistance
i15=PD/R1#current in 15 ohms resistance
#Results
print "(a)unknown resistance in ohms =", x
print "(b)potential drop across the circuit in volts is =", PD
print "(c)current in 5 ohms resistance in ampere =",i5,"\n and current in 15 ohms resistance in ampere =", i15
print "(d)total resistance of the circuit in ohms =",R
(a)unknown resistance in ohms = 75.0
(b)potential drop across the circuit in volts is = 150.0
(c)current in 5 ohms resistance in ampere = 3.0
and current in 15 ohms resistance in ampere = 10.0
(d)total resistance of the circuit in ohms = 10.0 | 2,830 | 6,976 | {"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} | 4.28125 | 4 | CC-MAIN-2024-38 | latest | en | 0.687628 |
https://www.metalspringclips.com/the-name-and-size-relationship-of-each-part-of-the-spring.html | 1,696,386,297,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511351.18/warc/CC-MAIN-20231004020329-20231004050329-00235.warc.gz | 954,012,749 | 10,665 | # The name and size relationship of each part of the spring
Springs are widely used in daily products. At present, there are five major markets for spring products, transportation, daily hardware, instrumentation and electronic appliances, industrial and mining accessories, and overseas export markets. Next, let’s take a brief look at the various parts of the spring. Name and size relationship.
(1) Talking about the diameter of the spring wire d: the diameter of the steel wire used to make the spring.
(2) Spring outer diameter D: the maximum outer diameter of the spring.
(3) Spring inner diameter D1: the minimum outer diameter of the spring.
(4) Spring diameter D2: the average diameter of the spring. Their calculation formula is: D2=(D+D1)÷2=D1+d=D-d
(5)t: In addition to the support ring, the axial distance between the corresponding points of the two adjacent coils of the spring on the pitch diameter becomes the pitch, which is represented by t.
(6) Effective number of turns n: The number of turns that the spring can maintain the same pitch.
(7) The number of support rings n2: In order to make the spring work evenly, to ensure that the axis is perpendicular to the end surface, the two ends of the spring are often tightened during manufacturing. The number of tight turns only serves as a support, and is called a support ring. Generally speaking. There are 1.5T, 2T, 2.5T, and 2T is commonly used.
(8) Total number of turns n1: The sum of the effective number of turns and the support ring. That is, n1=n+n2.
(9) Free height H0: The height of the spring without external force. Calculated by the following formula: H0=nt+(n2-0.5)d=nt+1.5d (when n2=2)
(10) Spring unfolding length L: the length of the steel wire required to wind the spring. L≈n1 (ЛD2)2+n2 (compression spring) L=ЛD2 n+ hook extension length (tension spring)
(11) Helix direction: there are left and right rotations, right-handed is commonly used, and right-handed is generally used if it is not indicated in the drawing. | 492 | 2,019 | {"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-2023-40 | latest | en | 0.917222 |
http://nicksun.fun/statistics/2020/04/06/complete-statistics.html | 1,716,793,204,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059037.23/warc/CC-MAIN-20240527052359-20240527082359-00673.warc.gz | 16,328,746 | 5,596 | When I was taking my statistical theory courses last year, I’ll admit that I took a couple concepts at face value. Most of those holes were patched up by the time I had to take my qualifying exams, but one thing that I still didn’t really understand even while handing in my exam is the logic behind the completeness of a statistic.
### A quick definition
A statistic $T$ is complete for a particular distribution $F$ if:
is true only if $g$ is the zero function (i.e. regardless of input, the function only returns 0). In Casella-Berger, they make use of the confusing phrase “$g(T)$ is an unbiased estimator of $0$”. If there is any other function $g$ that makes the above equality true, then $T$ is not complete.
Proving a statistic is complete can get pretty hairy, but thankfully, most of the examples in Casella-Berger are showing that a statistic is not complete. All that’s required then is to find a function of $T$ where the expection is zero and that function is not the trivial zero function.
### Ok, so what does completeness really mean?
Completeness is a condition which guarantees that there is only one unbiased estimator of $\theta$ based on the sufficient statistic $T$.
This is easy to see via contradition. For example, suppose we had two different unbiased estimators of $\theta$, $g_1(T), g_2(T)$ where $P(g_1(T) \neq g_2(T)) > 0, \forall \theta$. Then $g_1(T) - g_2(T))$ is an unbiased estimator for 0 (and therefore $T$ is not complete)!
After reading a bunch of crossvalidated posts, I also learned some other useful facts about sufficient statistics. Namely, for sufficient statistics that are not complete, there exists function of that sufficient statistic which are not informative about $\theta$. Conversely, a complete sufficient statistic is “maximally informative”, that is, no function of a complete sufficient statistic is uninformative about $\theta$.
#### Completeness and minimal sufficient statistics
Anyone who has made a cursory reading of Casella-Berger knows that there are cases where no UMVUE exists. As a prime example, there are setting where there is no UMVUE because there are some values of $\theta$ where the estimator has 0 variance, however, these estimators are not the UMVUE since they don’t have the minimum variances for other values of $\theta$. One thing to know though is that if a minimal sufficient statistic (found via Lehmann-Scheffe, for example) is not complete, there will not be a UMVUE because the optimal unbiased estimator changes depending on the value of $\theta$.
An example of a minimal sufficient statistic that is not complete is if we are given
and the minimal sufficient statistic is $T = (\sum X_i, \sum X_i^2)$.
Note that $E[T_1] = n\theta$ and $E[T_1^2] = n\theta^2 + n^2\theta^2$ and $E[T_2] = n(Var[X_i] + E[X_i]^2) = 2n\theta^2$. Then we can define
and note that $E[g(T)] = 0$. Therefore, the minimal sufficient statistic $T$ is not complete.
#### One cooool side fact… there are cases where no unbiased estimator exists!
Thanks to Dr. Sarah Emerson for this!
An unbiased estimator has to be unbiased for all values of the parameter it is estimating. Here is a scenario where that is not the case! Suppose we have:
and suppose our parameter of interest is the odds $\theta = \frac{p}{1-p}$.
Now, consider a general estimator $T(X_1, \ldots, X_n)$. If $T$ is unbiased, then $E_{\theta}[T] = \theta, \forall \theta$.
Now note that there are $2^n$ possible outcomes in the sample space. We can index these outcomes from $j = 1, \ldots, 2^n$ and let $t_j$ be the value of $T$ when we observe outcome $j$. Also define $n_j$ for the number of ones observed in outcome $j$.
We can write the expection of $T(X_1, \ldots, X_n) = \sum_{j = 1}^{2^n} t_j p^{n_j} (1-p)^{n - n_j}$. Going back to $T$, we need $E[T] = \frac{p}{1-p} = \sum_{j = 1}^{2^n} t_j p^{n_j} (1-p)^{n - n_j}$ since $T$ is supposed to be unbiased.
However, this is not possible. $\frac{p}{1-p}$ is not a polynomial of $p$, so it cannot be made equal to a finite order polynomial like $\sum_{j = 1}^{2^n} t_j p^{n_j} (1-p)^{n - n_j}, \forall p$. So there is no unbiased estimator of $\theta$! | 1,108 | 4,157 | {"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 | 4 | CC-MAIN-2024-22 | latest | en | 0.920982 |
https://stats.stackexchange.com/questions/288410/probability-question-involving-computing-the-number-of-degenerates-n | 1,721,391,333,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514902.63/warc/CC-MAIN-20240719105029-20240719135029-00529.warc.gz | 459,724,598 | 39,327 | Probability question involving computing the number of degenerates, !n
5 different colored balls have matching 5 buckets. Each balls is randomly placed in a bucket. A bucket can only accommodate 1 ball.
What is the probability that:
a.) None of the balls match their buckets b.) Exactly 2 balls match their buckets
My approach is as follows:
a.) I just counted. Let's say I number the balls 1,2,3,4,5 with corresponding buckets, 1,2,3,4,5. I know that there are 4 different ways to get them all in the wrong buckets by just shifting the balls one bucket away. That is, (5,1,2,3,4) (4,5,1,2,3)(3,4,5,1,2) and (2,3,4,5,1). Then, another way to get them all wrong is to choose any 2, interchange their positions, then mix up the remaining three as well. There is only 1 way to mix up the remaining 3 once the first 2 are chosen so this is just 5C2 ways. This also results to the same outcome if instead I choose 3 first and jumble those before interchanging the other 2. So the probability is: $\frac{5C2+4}{5!}$
b.) For exactly 2 to match, we can choose any 2 from 5 and then mix up the other 3 so they won't match. This seems to exhaust all possible ways. So that's just 5C2, making the probability: $\frac{5C2}{5!}$
• Please see en.wikipedia.org/wiki/Derangement. There are many more 5-derangements than you listed. Commented Jul 2, 2017 at 10:01
• This is great. Thank you. Honestly, this is the first time I've read the term derangement. Commented Jul 2, 2017 at 10:25
• Please use a title that describes the topic of your question more precisely than "probability". The "Did I do this correctly" can safely be left for the question body Commented Jul 3, 2017 at 4:08
a.) I just need the number of derangements (Yey) !5, which refers to the number of ways to match the 5 balls with the wrong buckets. Using the recursive relationship, $!n=(n-1)(!(n-1)+!(n-2))$, I get !5=44. | 546 | 1,884 | {"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} | 4.03125 | 4 | CC-MAIN-2024-30 | latest | en | 0.92188 |
http://mathhelpforum.com/calculus/116001-integration.html | 1,529,301,585,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267860089.11/warc/CC-MAIN-20180618051104-20180618071104-00423.warc.gz | 203,347,616 | 12,206 | 1. ## Integration.
Hello i have a question pplz help me it is
how the:
a is a constant number
after integration became
Thanks
2. Because your are integrating with respect to the variable p, as that variable is changing not a, thats the way I see it
3. ## Hello
$\displaystyle P^2+1=U$
and finding:
$\displaystyle p=sqrt(U-1)$
and findinf dp and subsitituing them into the original equation etc....
if there is easier method plz give it to me
Thnx
4. I am working on your problem as well, I am still a learner compared to everyone here, I have tried you substitution, before you mention it as it gets tad confusing im re-working it, hope I can help you
5. ## ok
i dunno i have tried many other method and didnt work
Thanks Anyway
6. This problem is very drawn out for me, I will find out how to do it though! It will be good for both of us, but our methods seem to complicate thing while integrating and simplifying as I told by my professor, mathematicians aim to make hard things very simple when working out a problem, so we will get it! I am sure there a few around here that this is easy for, I was close to the solution but made a careless everyone and wound up with a few missing parts, like the numerator coming to p^2 instead of p^3
Lets see if we can make sense of things together, first we know -3a is a constant so we can pull that to the outside and that the numerator $\displaystyle p^2$ when integrated as if it was just alone as such$\displaystyle p^2$ becomes $\displaystyle \frac{p^3}{3}$ and remember we pulled out the -3a so now the the integrated form we get $\displaystyle -3a*\frac{p^3}{3}$ we get $\displaystyle -a * p^3$ so somehow we need to make a relation to the denominator and decide the right method of integrating to where we can relate to this
Edit: have you tried integration by parts, I feel I am close using this method
7. Have you tried the substitution $\displaystyle p = \frac{1}{x}$ ?
since
$\displaystyle dp = \frac{-1}{x^2} dx$
the integral ( let's omit the constant $\displaystyle -3a$) becomes
$\displaystyle \int \frac{x^5}{\sqrt{x^2+1}^5} \cdot \frac{1}{x^2} \cdot \frac{-1}{x^2} dx$
$\displaystyle = - \int \frac{x dx}{\sqrt{x^2+1}^5}$
$\displaystyle = - (1 + x^2)^{\frac{-3}{2}} (\frac{-2}{3}) \frac{1}{2} + C$
$\displaystyle = \frac{1}{3} \frac{ 1}{ \sqrt{ 1 + \frac{1}{p^2} }^3 } + C$
$\displaystyle = \frac{ p^3}{3 \sqrt{ 1 + p^2}^3 } + C$
8. Jeepers, the one I never thought of trying, how did you know what to correctly substitute?
Also where x^5 comes from?
9. Originally Posted by RockHard
Jeepers, the one I never thought of trying, how did you know what to correctly substitute?
If thec0o0lest didn't give us the answer , i won't think of this substitution .
I observed that the integral is algebraic function , not transcendental ( usually $\displaystyle \sinh^{-1}(x) , \cosh^{-1}(x)$ ) .
so i attempt to change the variable ( algebraic ) . It is just a trial !
Another case we can use this sub. is
$\displaystyle \int \frac{dx}{\sqrt{x^2+1}^3}$
to
$\displaystyle - \int \frac{u ~du}{\sqrt{u^2+1}^3}$
10. simplependulum Thankssssssssssssssssssssssssss
11. Originally Posted by RockHard
Also where x^5 comes from?
$\displaystyle x^5$ comes from the denominator ( after changing $\displaystyle p$ to $\displaystyle \frac{1}{x}$)
$\displaystyle \frac{1}{\sqrt{p^2+1}^5} = \frac{1}{\sqrt{1 + \frac{1}{x^2}}^5 }$
12. Originally Posted by RockHard
Jeepers, the one I never thought of trying, how did you know what to correctly substitute?
Also where x^5 comes from?
the x^5 come from
$\displaystyle sqrt{1/x^2+1}^5$
when you subtitute p=1/x
you need to make some balance in the denominator you multiply by
$\displaystyle x^(5/2)$
13. Man, this is a confusing integral, I was stuck on it, If I did not see the answer I wouldn't even came to close to guessing a right answer | 1,129 | 3,860 | {"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} | 4.40625 | 4 | CC-MAIN-2018-26 | latest | en | 0.94306 |
winik.feuerwerktraum-abholshop.de | 1,623,833,184,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487622234.42/warc/CC-MAIN-20210616063154-20210616093154-00287.warc.gz | 47,013,905 | 9,630 | ### Soccer picks and predictions
Corresponding Angles : Angles lie on the same side of the transversal t, on the same side of lines a and b. Example : ∠ 1 and ∠ 5. Alternate Interior Angles : Angles are nonadjacent angles that lie on opposite sides of the transversal t, between lines a and b. Example : ∠ 3 and ∠ 6. Alternate Exterior Angles : Oct 31, 2016 · Unit 1 Logical Arguments and Constructions; Proof and Congruence > Topic 3 Parallel and Perpendicular Lines > 3-2 Properties of Parallel Lines Proof 5. Write a two-column proof for Exercise 4 that does not use ∠2. Find m∠1 and m∠2. Justify each answer. 6. 7. 8. Find the value of x. Then find the measure of each labeled angle. 9. 10. 11. All angles that have the same position with regards to the parallel lines and the transversal are corresponding pairs e.g. 3 + 7, 4 + 8 and 2 + 6. Angles that are in the area between the parallel lines like angle 2 and 8 above are called interior angles whereas the angles that are on the outside of the two parallel lines like 1 and 6 are called ...OBJ: 3-2.2 To use properties of parallel lines to find angle measures NAT: CC G.CO.9| M.1.d| G.3.g STA: 4.1.PO 4| 5.2.PO 12 TOP: 3-2 Problem 4 Finding an Angle Measure KEY: corresponding angles | parallel lines 7. ANS: C PTS: 1 DIF: L3 REF: 3-5 Parallel Lines and Triangles In a plane, two lines are either. Parallel . Intersect. 3.1 Identify Pairs of Lines and Angles. Parallel Postulate. If there is a line and a point not on the line, then there is exactly one line through the point parallel to the given line.
Each of the parallel lines cut by the transversal has 4 angles surrounding the intersection. These are matched in measure and position with a counterpart at the other parallel line. At each of the parallel lines, there are two pairs of vertical angle. Each angle in the pair is congruent to the other angle in the pair. 1 4, angle 1 is congruent ... #TECHPOINTEDUCATIONACADEMY #PROPERTIES OF PARALLEL LINES ALTERNATE CORRESPONDING COINTERIOR COEXTERIOR ANGLESHi, In this video, we are going to discuss a ver... 17. same-side interior angles L 2 and £5; L 3 and £8 18. alternate interior angles £3 and £5; 1.2 and £8 19. alternate exterior angles Ll and L 7; £4 and £6 Decide whether the angles are alternate interior angles, same-side interior angles, corresponding angles, or 12 alternate exterior angles. 20. L 2 and L 7 alt. ext. 22. L 8 and L 3 corr. corresponding angles Identify each pair of angles as alternate interior, alternate exterior, corresponding, or consecutive interior angles. a. !10 and !16 b. !4 and !12 alternate exterior angles corresponding angles c. !12 and !13 d. !3 and !9 consecutive interior angles alternate interior angles Use the figure in the Example for Exercises 1–12. Theorem 5 If two lines are intersected by a transversal, and if corresponding angles are equal, then the two lines are parallel. Theorem 6 If two parallel lines are intersected by a transversal, then corresponding angles are equal. Theorem 7 - The Exterior Angle Theorem An exterior angle of a triangle is equal to the sum of the two remote ...
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Can you come up with a proof that vertical angles are equal? Hint: You can use addition and subtraction and the idea of a "straight angle" (a line). Corresponding Angles. When a line crosses two (or more) parallel lines, the same sets of angles are formed at each intersection. Angle ABH corresponds to Angle DEH. Angle ABG corresponds to Angle DEG Dec 28, 2020 · If corresponding angles are equal, the lines are parallel. s. ... If two lines are perpendicular to the same line, then they are parallel . Question. Sep 12, 2011 · Triangle Proof: Prove that the angles in a triangle add up to 180 degrees 1. Draw the triangle 2. Let the three angles be denoted be letters a,b,c 3. Now draw a line through the top vertex (Corner) parallel to the baseline. 4. Add in angles d & e on either side of angle a. 5. a + b + e = 180 degrees (angles on a straight line = 180 degrees) 6. All angles that have the same position with regards to the parallel lines and the transversal are corresponding pairs e.g. 3 + 7, 4 + 8 and 2 + 6. Angles that are in the area between the parallel lines like angle 2 and 8 above are called interior angles whereas the angles that are on the outside of the two parallel lines like 1 and 6 are called ...Working with angles in parallel lines. There are 3 main rules: 1. Corresponding angles are equal. A line cutting across two parallel lines creates four pairs of equal corresponding angles, as in the diagram below: Note: You may also have heard these referred to as ‘F angles’ – do not use that term in an exam or you will lose marks! The angles at the bottom of the trapezoid and the angles on top of the new line are corresponding angles. That makes the two lines parallel. With the use of the transitive property, you can say ... Ifa⊥b, then ∠ 1, ∠ 2, ∠ 3, and ∠ 4 are all right angles. Proving Relationships Using Angle Measures. You can use the angle pairs formed by lines and a transversal to show that the lines are parallel. Also, if lines intersect to form a right angle, you know that the lines are perpendicular. Aug 03, 2016 · Two lines cut by a transversal are parallel if and only if alternate interior angles are congruent. Theorem 3-6If two lines are cut by a transversal and same-side interior angles are supplementary, then the lines are parallel. Biconditional. October 21, 2016. You can add these to your theorem list book. Remember that postulates are statements that are accepted without proof. Since the Corresponding Angles Postulate is given as a postulate, it can be used to prove the next three theorems. Objective Prove and use theorems about the angles formed by parallel lines and a transversal. Alamy Photos 21-1 Angles Formed by Parallel Lines and Transversals
If two lines are parallel and a line is perpendicular to one of the two lines, then it is perpendicular to the other line. 4 3 2 1 k m l Proof: Given l || and m l ⊥ k, show that k ⊥ m Statement Reason l || m k ⊥ m Given ∠2 ≅ ∠3 ⇒m∠2 = m∠3 If two lines are parallel, then the alternate interior angles are congruent. Theorem 5.5 m∠3 = 90 ° Moreover, they are equal in value. An example of the vertically opposite angle is given below: (image will be uploaded soon) ∠x = ∠y, in the case of parallel lines. The ∠x and ∠y are a pair of vertically opposite angle and there are equal in value when parallel lines are intersected by a transversal line. 3. The first is if the corresponding angles, the angles that are on the same corner at each intersection, are equal, then the lines are parallel. The second is if the alternate interior angles, the angles that are on opposite sides of the transversal and inside the parallel lines, are equal, then the lines are parallel. Proof: Since L 3 and L 4 are parallel, , since they are alternate interior angles for the transversal L 2. Therefore by the transitive property. Since L 1 and L 2 are parallel, since they are corresponding angles for transversal L 4. Applying the transitive property again, we have . Complements and Supplements Proof: Since C L m, we know that Ll £2, because ± lines form congruent right Is. Then by the Corresponding Angles Postulate, Ll £3 and £2 £4. By the definition of congruent 6, mz_l = mZ_2, mz_l = rnL3, and mL2 mL4. By substitution, mZ_3 = mt4. Because £3 and £4 form a congruent linear pair, they are right 6. By definition, L n. 1.Parallel Lines: Two lines are parallel if they do not meet at any point. 2.Congruent Triangles: Two triangles are congruent if their corresponding angles and corresponding sides are equal. 3.Similar triangles: Two triangles are similar if their corresponding angles equal and their corresponding sides are in proportion. Proof of theorem:
If two lines are cut by a transversal and the alternate exterior angles are congruent, the lines are parallel. If two lines are cut by a transversal and the alternate interior angles are congruent, the lines are parallel. If two parallel lines are cut by a transversal, the interior angles on the same side of the transversal are supplementary If two lines are cut by a transversal and the alternate interior angles are congruent, the lines are parallel. Angles and Parallel Lines Algebra and Angle Measures Algebra can be used to find unknown values in angles formed by a transversal and parallel lines. -Example, If ml-I = 3x + 15, ml—2 = 4x — 5, m £3 = 5y, and ml—4 = 6z + 3, find x and y. V 31 Glencoe Geometry p Il q, so m LI = ml—2 because they are corresponding angles.
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Parallel Lines, Transversals, Angles Geometry Flashcards I have complied a complete set of flashcards that tests the knowledge of ... Alternate Interior, Alternate Exterior, Consecutive Interior, Corresponding, Vertical, and Linear Pair Angles. 32 flashcards included with 4 cards per page Double sided printing instructions included! Alternate Interior Angles – Explanation & Examples In this article, we are going to learn another special type of angle formed when parallel or non-parallel lines are intersected by a transversal line. As you know, parallel lines are two or more lines which never meet, whereas, a transversal line is a straight line which intersects […] Corresponding Angles - Explanation & Examples Before jumping into the topic of corresponding angles, let's first remind ourselves about angles, parallel and non-parallel lines and transversal lines. In Geometry, an angle is composed of three parts, namely; vertex, and two arms or sides. The vertex of an angle is the point where two sides or […]Objective: Make conjectures about angles formed by parallel lines and a transversal. Author Intent Students apply what they know about angle relationships to reason about measures of angles in a diagram. They discover that there is a special relationship between the angles formed by parallel lines and a transversal.
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5-6 Proving Lines Parallel. Given the following information, determine which lines, if any, are parallel. State the postulate or theorem that justifies your answer. \$16:(5 j || k; converse of corresponding angles postulate \$16:(5 alternate exterior angles converse SHORT RESPONSE Find x so that m || n. Lesson 3 - 1: Lines and Angles Date Corresponding Angles Plane Transversal Alternate Interior Angles Parallel Planes Skew Lines Alternate Exterior Angles Parallel Lines Same-Side Interior Angles Choose the concept from the list above that best represents the item in each box. Parcllel Sam e manes - St-de Ex I-Zrïor les Parcllel Plane les P3: If two parallel lines are cut by a transversal then interior angles on the same side of the transversal are supplementary. P4: If two parallel lines are cut by a transversal then corresponding angles are congruent. Proof: We will show P1 -> P2 -> P3 -> P4 -> P1. P1 -> P2. Assume P1. Let l,m be parallel lines and t a transversal making ... Corresponding Angles. 4 videos. ... Deductive reasoning is the type of reasoning used when making a Geometric proof, ... Angles and Parallel Lines. About. How to use ... angles, one that shows alternate interior angles, and one that shows alternate exterior angles. LO: I can add auxiliary lines to diagrams and use angle relationships to prove statements. (1) transparen cies, dry erase markers, erasers compass Angles: Exterior angle theorem: Proof by constructing a parallel line. Corresponding angles are a pair of interior and exterior angles formed on the same side of the transversal. This PDF worksheet provides essential remedial practice in finding the measures of the indicated angles by applying the congruent property of the corresponding angles. corresponding angles, then the lines are parallel. You can use algebra along with Postulates 3-1 and 3-2 and Theorems 3-1 through 3-8 to help you solve problems involving parallel lines. Using Algebra Algebra Find the value of x for which O 6m. The two angles are corresponding angles. O 6m when 2x +6 =40. 2x +6 =40 2x =34 Subtract 6 from each side. 168 Chapter 3 Parallel and Perpendicular Lines Parallel and Perpendicular Lines Then In Chapters 1 and 2, you learned about lines and angles and used deductive reasoning to write geometric proofs. Now In Chapter 3, you will: Identify angle relationships that occur with parallel lines and a transversal and prove lines parallel from given EXAMPLE 1 Warm-Up Exercises Apply the Corresponding Angles Converse ALGEBRA Find the value of x that makes m n. SOLUTION Lines m and n are parallel if the marked corresponding angles are congruent. 3x = 60 x = 20 The lines m and n are parallel when x = 20. (3x + 5)o = 65o Use Postulate 16 to write an equation Subtract 5 from each side. Unit 1 Lesson 13 Proving Theorems involving parallel and perp lines WITH ANSWERS!.notebook 1 October 04, 2017 Oct 67:46 AM Unit 1 Lesson 13 Proofs involving Parallel lines We will need to recall the different postulates and Theorems involving Parallel lines... Can you name the following types of angles from the diagram below??? 12 43 56 78 9
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MCC9-12.G.CO.9 Prove theorems about lines and angles. Theorems include: vertical angles are congruent; when a transversal crosses parallel lines, alternate interior angles are congruent and corresponding angles are congruent; points on a perpendicular bisector of a line segment are exactly those equidistant from the segment’s endpoints. Proof. This construction works by using the fact that a transverse line drawn across two parallel lines creates pairs of equal corresponding angles.It uses this in reverse - by creating two equal corresponding angles, it can create the parallel lines. 4. No, the angle next to 106° is 74°. The corresponding angle to 74° is 73°. They are not equal, so the lines are not parallel. 5. Yes, the corresponding angles are congruent. 6. x = 12° 7. x = 6°" " 8. x = 8°" 9. x = 10°" " 10. x = 2° exterior angles and 2. 2. alternate exterior angle theorem 3. with transversal p and consecutive interior angles and . 4. 4. 5. 5. definition of supplementary angles 6. 6. 7. Section 3.1: Pairs of Lines and Angles In Exercises 1–6, use the diagram. 1. Name a pair of parallel lines. 2. Name a pair of perpendicular lines. 3. Name a pair of skew ... Pairs of Lines and Angles Section 3.1 notes. Parallel Lines and Transversals Exploring Parallel Lines and a Transversal. Have students use Geogebra to construct two parallel lines and a traversal. They then identify and measure corresponding, alternate interior, alternate exterior, and consecutive interior angles and make conjectures.
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Macbook pro left speaker buzzing | 3,742 | 15,065 | {"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-2021-25 | longest | en | 0.844709 |
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One Childhood, One Chance
# Weekly News Friday 20th March
This week has been a strange week for all of us. The children have remained calm and focused on their learning, for which we are extremely proud of them. In today's blog, I will list what we have been learning this week. For those of you who have already started your home learning journey, I will upload some of the learning we completed in school for you to complete at home. From Monday, we will update our blogs daily, with activities and tasks for you to complete at home. We will miss all the children so very much and we wish you all the best on your journey over the next few weeks.
In Maths this week, we have been learning about division. First of all we looked at sharing fairly. If we had 12 sweets to share between 3 children, we would initially give everyone 1. Then give everyone a second sweet and keep going until all the sweets were gone. We then related this to a division calculation. When given a calculation we taught the children to look at a smaller number. This would tell them how many sharing circles to draw. You then look at the larger number and share that number equally between the sharing circles (see the attached photo for how it looked on the board.
In Topic this week, our focus has been germs (this was actually what was on or medium term plan and just happens to be very appropriate). We learnt about how germs spread and how we can stop the spread of germs. We conducted 3 different experiments and observed the effect germs had on things such as bread.
For those of you who have been home learning this week:
English: Please can you revise the sounds ea (cup of tea) and oi (spoil the boy). Write the words tea, meat, beak, boil, coin, join on a paper and get children to read them. Then hide the words and see if the children can write them. Click on the link below and play Obb and Bob with the sounds ea and oi.
Obb and Bob phonics game
Say the sentence 'Mum and Dad like to spoil the boy' to the children a few times. See if they can remember the sentence and write it down. Repeat this with 'I lost my coin at the beach'.
Then ask the children to write a postcard to send to school to tell us about what they have done at home this week.
Maths:
Talk the children through what we have done in Maths this week (see above). Complete the division calculations attached below by drawing sharing circles and sharing the larger number into the circles. | 543 | 2,486 | {"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-2022-33 | latest | en | 0.982453 |
http://www.amsat.org/amsat/archive/amsat-bb/200201/msg00117.html | 1,369,233,995,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368701852492/warc/CC-MAIN-20130516105732-00020-ip-10-60-113-184.ec2.internal.warc.gz | 319,475,664 | 3,572 | # R: false signals-images-harmonics-whatever? FIXED !!
• Subject: R: [amsat-bb] false signals-images-harmonics-whatever? FIXED !!
• From: "i8cvs" <domenico.i8cvs@xxxxxx>
• Date: Mon, 7 Jan 2002 04:28:26 +0100
```Hi Greg,
An electrical 1/4 wavelenght long open stub at 435 MHz represent a short
circuit for 435 MHz at the other end connected to the helix feed for 2401
MHz because at 435 MHz the impedance of the stub in the point connected to
the helix is Z = 0 + j0 ohm
By the way a shorted stub 1/4 electrical wavelenght long at 2401 MHz
represents a very high impedance at the open end connected to the
helix feed.
This impedance is theoretically infinite at the open end and so its presence
is not seen as a load by the helix for 2401 MHz wich is not disturbed very
much by the shorted stub.
Following your idea and using the Smith Chart i have simulated a shorted
stub 1/4 wavelenght long at 2401 MHz made with a piece of semirigid
UT 141, Zo=50 ohm coax cable made with PTFE insulation and a copper
tubing as the outer conductor.
The velocity factor for the UT 141 is Vf = 0,75 and so the electrical lenght
of the 1/4 wavelenght shorted stub for 2401 MHz is 23,5 mm wich represents
0,25 electrical wavelenght at 2401 MHz
The same lenght of 23,5 mm represents 0,0452 electrical wavelenght at 435
MHz and 0,132 electrical wavelength at 1270 MHz
If you put this wavelengts in to the Smith Chart you will find that the
impedance of the shorted stub at the open end connected to the helix is
Z= 0 + j15 ohm at 435 MHz and Z= 0 + j55 ohm at 1270 MHz
The shorted stub for 2401 MHz has a very small inductive reactance at the
open end of 15 ohm at 435 MHz and 55 ohm at 1270 MHz and so it can be
considered a short circuit at 435 MHz and 1270 MHz while an open circuit
at 2401 MHz
I believe that it works fine to short out the unwanted uplink signals that
compresses the gain of your converter without disturbing to much the wanted
Since the UT 141 has a copper tubing as an outer conductor it is possible to
solder it to the reflector for the entire lenght of 23,5 mm and the inner
conductor to the central pin of the N connector for the helix.
The UT 141 coax cable is very small in diameter,only about 4 mm,it
is easy to solder and to find.
I hope this help
73 de i8CVS Domenico
----- Original Message -----
From: Greg D. <ko6th_greg@hotmail.com>
To: <amsat-bb@AMSAT.Org>
Sent: Sunday, January 06, 2002 12:02 AM
Subject: Re: [amsat-bb] false signals-images-harmonics-whatever? FIXED !!
> Hi folks,
>
> The alternative to a 1/4 wave open stub at 435 would be a 1/4 wave shorted
> stub at 2401, right? If so, wouldn't that be easy to build into a helix
> feed? I'm thinking of a short piece of wire parallel to the ground plane,
> with
> one end attached to the feed point (N-connector), and the other end
soldered
> to
> the ground plane. 234/2401*12 = 1.17 inches long?
>
> My thought here is that being a shorted stub, it would put the feed coil
at
> ground
> potential for most frequencies - lightning protection, 432, etc., and
> probably
> do a fair job at 1.2 ghz.
>
> Im I close? Is there a reason to use a 1/4 open stub on 435 instead?
>
> Greg KO6TH
>
>
>
> >From: hasan schiers <schiers@netins.net>
> >To: Jens Schmidt <j.schmidt@paradise.net.nz>, Tracy <k7kcs@attbi.com>
> >CC: amsat-bb@AMSAT.Org
> >Subject: Re: [amsat-bb] false signals-images-harmonics-whatever? FIXED !!
> >Date: Sat, 5 Jan 2002 06:57:16 -0600
> >
>
> >From the foggy depths of my morning mind, I think it is a quarter-wave
open
> >stub, acting like a short circuit. At 70cm this calculates out to 5.4
> >inches,
> >with a velocity factor of .80
> >
> >The idea is to "notch" the 435 mhz offending signal without screwing up
the
> >2.4 gig desired. It's a neat trick...and you can avoid cutting by using
an
> >1/8 wave line with a small tuning cap across the end and then tune for
the
> >null. (I think it was an 1/8 wave ...it's been a long time since I did
it).
> >
>
>
> _________________________________________________________________
> MSN Photos is the easiest way to share and print your photos:
> http://photos.msn.com/support/worldwide.aspx
>
> ----
> Via the amsat-bb mailing list at AMSAT.ORG courtesy of AMSAT-NA.
> To unsubscribe, send "unsubscribe amsat-bb" to Majordomo@amsat.org
>
----
Via the amsat-bb mailing list at AMSAT.ORG courtesy of AMSAT-NA.
To unsubscribe, send "unsubscribe amsat-bb" to Majordomo@amsat.org
```
AMSAT Home | 1,373 | 4,465 | {"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} | 3.15625 | 3 | CC-MAIN-2013-20 | longest | en | 0.90509 |
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# Sunshine physics
Basic level
1.
## Mud houses are cooler in summer and warmer in winter because
[MP PAT 1996; BVP 2003]
2.
heat
(b)
## (c) Mud is bad conductor of heat
(d)
None of these
Heat current is maximum in which of the following (rods are of identical dimension)
Copper
(a)
3.
(b)
Copper Steel
(c)
Steel Copper
(d)
Steel
## Consider the following statements
Assertion (A) : Woolen clothes keep the body warm in winter
Reason (R) : Air is a bad conductor of heat
Of these statements
[AIIMS 2002]
(a) Both A and R are true and the R is a correct explanation of the A
(b) Both A and R are true but the R is not a correct explanation of the A
(c) A is true but the R is false
(d) Both A and R are false
(e) A is false but the R is true
4.
The lengths and radii of two rods made of same material are in the ratios 1 : 2 and 2 : 3 respectively. If the
temperature difference between the ends for the two rods be the same then in the steady state. The amount of
heat flowing per second through them will be in the ratio
[MP PET 2000]
(a) 1 : 3
5.
(b) 4 : 3
(c) 8 : 9
(d) 3 : 2
Two vessels of different materials are similar in size in every respect. The same quantity of ice filled in them gets
melted in 20 minutes and 30 minutes. The ratio of their thermal conductivities will be
[MP PMT 1989; CMEET Bihar 1995]
(a) 1.5
6.
(b) 1
(c) 2/3
(d) 4
If the coefficient of conductivity of aluminium is 0.5 cal/cm-sec-oC, then in order to conduct 10 cal/sec-cm2 in the
steady state, the temperature gradient in aluminium must be
[MP PAT 1990]
(a) 5 C/cm
o
7.
(c) 20oC/cm
(d) 10.5oC/cm
The area of the glass of a window of a room is 10m2. and thickness 2 mm. The outer and inner temperature are
40oC and 20oC respectively. Thermal conductivity of glass in MKS system is 0.2. The heat flowing in the room per
second will be
[MP PMT 1989]
(a) 3 104 Joules
8.
(b) 10oC/cm
## (b) 2 104 Joules
(c) 30 Joules
(d) 45 Joules
When two ends of a rod wrapped with cotton are maintained at different temperatures and after some time
every point of the rod attains a constant temperature, then
[MP PET /PMT 1988]
(a) Conduction of heat at different points of the rod stops because the temperature is not increasing
(b) Rod is bad conductor of heat
(c) Heat is being radiated from each point of the rod
Sunshine
160 Transmission of Heat
(d) Each point of the rod is giving heat to its neighbour at the same rate at which it is receiving heat
9.
[NCERT 1974, 76]
10.
## To a rough approximation, conductivities of metals are about
(a) 1000 times as those of liquids and 10,000 times of gases
(b) 10 times as those of liquids and 100 times of gases
(c) 100 times as those of liquids and 1000 times of gases
(d) 10,000 times as those of liquids and 1000 times of gases
11.
A copper bar 10 cm long has its ends pressed against copper tanks at 0 oC and 100 oC. The ends are separated
by layers of dust 0.1 mm thick. If conductivity of dust is 0.001 times that of copper, the temperatures of end P
and Q of bar are [Take rate of flow of heat constant from P to Q]
10 cm
100oC
## (b) 66.7 oC and 33.3 oC
0oC
Q
(c) 75 oC and 25 oC
(d) 60 oC and 40 oC
## Problems based on combination of conductors
Basic level
12.
Three rods of the same dimension have thermal conductivities 3K, 2K and K. They are arranged as shown in
figure given below, with their ends at 1000C, 500C and 200C. The temperature of their junction is
[UPSEAT 2002]
50oC
(a) 60
2K
100oC
(b) 70
3K
(c) 50o
20oC
(d) 35o
13.
Five rods of same dimensions are arranged as shown in the figure. They have thermal conductivities K1, K2, K3, K4
and K5. When points A and B are maintained at different temperatures, no heat flows through the central rod if
[KCET 2002]
C
K1
(a) K1 = K4 and K2 = K3
K5
(b) K1 K4 = K2 K3
K3
(c) K1 K2 = K3 K4
B
K4
K1 K 2
(d)
K4 K3
14.
K2
A wall has two layers A and B each made of different materials. The thickness of both the layers is the same. The
thermal conductivity of A, KA=3KB. The temperature difference across the wall is 20oC in thermal equilibrium
[CPMT 1998]
## (a) The temperature difference across A is 15oC
(b) Rate of heat transfer across A is more than across B
(c) Rate of heat transfer across both is same
Sunshine physics
Transmission of Heat 161
(d) Temperature difference across A is 5 oC
15.
Two metal cubes A and B of same size are arranged as shown in the figure. The extreme ends of the combination
are maintained at the indicated temperatures. The arrangement is thermally insulated. The coefficients of
thermal conductivity of A and B are 300 W/m oC and 200 W/m oC, respectively. After steady state is reached, the
temperature t of the interface will be
[IIT-JEE 1996]
t
(a) 45oC
(b) 90oC
100oC
(c) 30oC
0oC
K1
K2
(d) 60 C
o
16.
Two cylinders P and Q have the same length and diameter and are made of different materials having thermal
conductivities in the ratio 2 : 3. These two cylinders are combined to make a cylinder. One end of P is kept at
100oC and another end of Q at 0oC. The temperature at the interface of P and Q is
[MP PMT 1994]
(a) 30 C
o
17.
(b) 2.4
(d) 60oC
(c) 1.5
(d) 1.2
Four identical rods of same material are joined end to end to form a square. If the temperature difference
between the ends of a diagonal is 100oC, then the temperature difference between the ends of other diagonal
will be
[MP PET 1989]
(a) 0oC
(c)
19.
(c) 50oC
Two identical plates of different metals are joined to form a single plate whose thickness is double the thickness
of each plate. If the coefficients of conductivity of each plate are 2 and 3 respectively, then the conductivity of
composite plate will be [MP PAT 1990]
(a) 5
18.
(b) 40oC
(b)
100 o
C
2l
100 o
C ; where l is the length of each rod
l
(d) 100oC
Two identical rods of metal are welded end to end as shown in figure (a). 20 calories of heat flows through it in 4
minutes. If the rods are welded as shown in figure (b), the same amount of heat will flow through the rods in
[NCERT 1982]
100oC
0oC
(a) 1 minute
(a)
(b) 2 minutes
0oC
100oC
(c) 4 minutes
(b)
(d) 16 minutes
20.
Two bars of thermal conductivities k and 3k and lengths 1 cm and 2 cm respectively have equal cross-sectional
area, they are joined lengths wise as shown in the figure. If the temperature at the ends of this composite bar is
0oC and 100oC respectively, then the temperature of the interface is
(a) 50oC
100 o
(b)
C
3
(c) 60oC
(d)
21.
0oC
3K
1 cm
2 cm
100oC
200 o
C
3
Three rods A, B and C have the same dimensions. Their thermal conductivities are KA, KB and KC respectively. A
and B are placed end to end, with their free ends kept at a certain temperature difference. C is placed
Sunshine
162 Transmission of Heat
separately, with its ends kept at the same temperature difference. The two arrangements conduct heat at the
same rate. KC must be equal to
(a) K A K B
22.
(b)
(c)
1
(K A K B )
2
K AK B
K A KB
2.
(d)
The three rods described in the previous question are placed individually, with their ends kept at the same
temperature difference. The rate of heat flow through C is equal to the rate of combined heat flow through A and
B. KC must be equal to
(a) K A K B
23.
K AK B
K A KB
(b)
K AK B
K A KB
(c)
1
(K A K B )
2
K AK B
K
B
A
2.
(d)
Three rods A, B and C of the same length and cross-sectional area are joined in series as shown in the figure.
Their thermal conductivities are in the ratio 1 : 2 : 1.5 . If the open ends of A and C are at 200 oC and 18 oC,
respectively, the temperature at junction of A and B in equilibrium is
200oC
(a) 74 oC
(b) 116 oC
18oC
2K
1.5K
(c) 156 oC
(d) 148 oC
24.
Three rods of identical area of cross-section and made from the same metal form the sides of an isosceles
triangle ABC. right angled at B. The points A and B are maintained at temperatures T and
2T respectively. In
the steady state the temperature of the point C is TC. Assuming that only heat conduction takes place,
equal to
(a)
25.
TC
T
is
[IIT-JEE 1995]
(b)
( 2 1)
3
( 2 1)
(c)
(d)
2( 2 1)
3 ( 2 1)
Three rods of material X and three rods of material Y are connected as shown in figure. All are identical in length
and cross-sectional area. If end A is maintained at 60oC, end E at 10oC, thermal conductivity of X is 0.92 cal/seccm-oC and that of Y is 0.46 cal/sec-cm-oC, then find the temperature of junctions B, C, D
C
60oC
## (b) 30oC, 20oC, 20oC
Y B
Y
26.
Y
D
(d) 20 C, 20 C, 20 C
o
10oC
E
(c) 20 C, 20 C, 30 C
o
Five rods 1, 2, 3, 4, 5 are connected to form the letter H as shown in the figure. The rods are of same length and
radii but having conductivities in the ratio K1 : K2 : K3 : K4 : K5 = 1 : 1 : 2 : 2 : 3. The uniform temperature of the
rod 5 in the steady state is
(a)
100 o
C
3
(b) 50oC
1
100oC
3
0oC
5
2
(c) 60oC
(d)
200 o
C
3
## Miscellaneous problems based on conduction
Sunshine physics
Transmission of Heat 163
Basic level
27.
There are two identical vessels filled with equal amounts of ice. The vessels are of different metals. If the ice
melts in the two vessels in 20 and 35 minutes respectively the ratio of the coefficients of thermal conductivity of
the two metals is
[MP PET 2001]
(a) 4 : 7
28.
(c) 16 : 49
(d) 49 : 16
Temperature at the surface of lake is 20 C. Then temperature of water just below the lower surface of ice layer
is [RPET 2000]
o
(a) 4o C
29.
(b) 7 : 4
(b) 0o C
(c) 4o C
(d) 20o C
Two identical rods of copper and iron are coated with wax uniformly. When one end of each is kept at
temperature of boiling water, the length upto which wax melts are 8.4 cm and 4.2 cm respectively. If thermal
conductivity of copper is 0.92, then thermal conductivity of iron is
[MP PET 1995]
(a) 0.23
30.
(b) 0.46
lake is high
(b)
## (d) The temperature of the earth at the bottom of the
If the steady thickness of ice layer is 100 cm and that of water is 4.20 m in a lake of a cold country where
temperature of air is 5.0oC and temperature of water at the bottom is 4oC. The ratio of the thermal conductivity
of water to that of ice is
(a) 4 : 3
32.
(d) 0.69
During severe winter in the low temperature zones of the world, the superficial parts of the lakes are frozen,
leaving water below. The freezing at the bottom is prevented because
(a) The conductivity of ice is low
heat
31.
(c) 0.115
(b) 4.25 : 1
(c) 3 : 1
(d) 5.25 : 1
A 10 cm layer of ice has been formed over a pond of water. The temperature of air above is 5 oC. How long will it
take the layer to become 10.1 cm thick? (Given Kice = 0.008 CGS units, density of ice = 1 gm/cc and Lice = 80
cal/gm)
(a) 2005 sec
## Problems based on convection
Basic level
33.
One feels hotter at the top of a flame than the sides because of
[AIIMS 2000]
(a) Conduction
34.
35.
(b) Convection
(b) Conduction
(c) Convection
## (d) Wave motion
While measuring the thermal conductivity of a liquid, we keep the upper part hot and lower part cool, so that
[CPMT 1985; MP PMT / PET 1988]
## (a) Convection may be stopped
(c) Heat conduction is easier downwards
36.
(a) Convection
37.
(b)
(b) Conduction
## (d) (b) and (c) both
The rate of loss of heat from a body cooling under conditions of forced convection is proportional to its (A) heat
capacity (B) surface area (C) absolute temperature (D) excess of temperature over that of surrounding state if
[NCERT 1982]
## (a) A, B, C are correct
(b) Only A and C are correct (c) Only B and D are correct
(d)
Only D is correct
Sunshine
164 Transmission of Heat
Basic level
38.
## Heat travels through vacuum by
[AIIMS 1998; CPMT 2003]
(a) Conduction
39.
(b) Convection
## (d) Both (a) and (b)
[MP PMT 1989, 92; RPET 2001, 03; AFMC
2003]
(a) 1
40.
(b) 0.5
(c) 0
(d) Infinity
[AIEEE 2002]
(a) Kajal
41.
[CBSE PMT 2002]
## (c) Highly heated charcoal lamp
constant temperature
42.
(d)
Cavity
maintained
at
2000]
both
43.
## (d) Line and band spectrum
In summer one feels cold on entering an air conditioned room. This can be explained by
(a) Newton's law of cooling
(b)
44.
Stefan's law
## (d) Prevost's theory of heat exchange
Out of the radiations falling on surface of a body, 30% radiations are absorbed and 30% are transmitted then its
reflection coefficient will be
(a) 0.3
(b) 0.6
(c) 0.4
(d) Zero
## Problems based on Kirchoff's law
Basic level
45.
There is a black spot on a body. If the body is heated and carried in dark room then it glows more. This can be
explained on the basis of
[RPET 2000]
Stefan's
46.
(b)
Wien's law
## (c) Kirchoff's law
(d)
If between wavelength and d, e and a be the emissive and absorptive powers of a body and E be
the emissive power of a perfectly black body, then according to Kirchoff's law, which is true
[MP PET 1991]
(a) e a E
47.
(b) e E a
(c) e a E
(d) e a E constant
The figure shows two similar sheets of tin plate, one polished and the other painted dull black. A piece of cork is
attached on the reverse side of each plate by means of melted paraffin wax. What will happen when an electric
bulb placed exactly midway between them is switched on
Dull black
Polished
## (a) The cork on the polished plate falls off first
(b) The cork on the dull black plate falls off first
(c) Both corks fall off at the same time
(d) Neither cork falls off
48.
## During total solar eclipse Fraunhoffer's lines appear bright because
Cork
Wax
Cork
Wax
Sunshine physics
Transmission of Heat 165
(a) Moon totally covers both parts of sun photo sphere and chromosphere
(b) Sun light is scattered by moon
(c) Moon blocks the radiations emitted by chromosphere
(d) Moon blocks the radiations emitted by photosphere and radiations emitted by chromosphere reach the earth
## Problems based on Stefan's law
Basic level
49.
A black body radiates energy at the rate of E watt/m2 at a high temperature T K. When the temperature is
reduced to
T
K , the radiant energy will be
2
[CPMT 1988; MNR 1993; SCRA 1996; MP PAT 1990; MP PMT 1992; MH CET 2001]
(a)
50.
E
16
(b)
(b) 9Q
(d) 16 E
(c) 27Q
(d) 81Q
A black metal foil is warmed by radiation from a small sphere at temperature T and at a distance d. It is found
that the power received by the foil is 'P'. If both the temperature and the distance are doubled, the power
received by the foil will be
[MP PMT 1997]
(a) 16 P
52.
(c) 4 E
At temperature T, the power radiated by a body is Q watts. At the temperature 3T the power radiated by it will
be [MP PET 2000]
(a) 3Q
51.
E
4
(b) 4 P
(c) 2 P
(d) P
A solid sphere and a hollow sphere of the same material and size are heated to the same temperature and
allowed to cool in the same surroundings. If the temperature difference between each sphere and its
surroundings is T, then
[Manipal MEE 1995]
(a) The hollow sphere will cool at a faster rate for all values of T
(b) The solid sphere will cool at a faster rate for all values of T
(c) Both spheres will cool at the same rate for all values of T
(d) Both spheres will cool at the same rate only for small values of T
53.
Two bodies A and B have thermal emissivities of 0.01 and 0.81 respectively. The outer surface areas of the two
bodies are the same. The two bodies emit total radiant power at the same rate. The wavelength
corresponding to maximum spectral radiancy in the radiation from B is shifted from the wavelength
corresponding to maximum spectral radiancy in the radiation from A, by 1.00 m . If the temperature of A is
5802 K
[IIT-JEE 1994]
(b) B 1.5 m
## (a) The temperature of B is 1934 K
(c) The temperature of B is 11604 K
54.
## (d) The temperature of B is 2901 K
The temperature of a piece of iron is 27 C and it is radiating energy at the rate of Q kW m2. If its temperature is
raised to 151oC, the rate of radiation of energy will become approximately
o
## [MP PET 1992]
(a) 2Q kW m
55.
(b) 4Q kW m2
(c) 6Q kW m2
(d) 8Q kW m2
If E is the total energy emitted by a body at a temperature T K and Emax is the maximum energy emitted by it at
the same temperature, then
[MP PAT 1990]
## (a) E T 4 ; Emax T 5 (b) E T 4 ; Emax T 5
56.
(c) E T 4 ; Emax T 4
(d) E T 5 ; Emax T 4
A metal ball of surface area 200 cm2 and temperature 527oC is surrounded by a vessel at 27 oC. If the emissivity
of the metal is 0.4, then the rate of loss of heat from the ball is ( 5.67 108 J / m2 s K 4 )
[MP PMT / PET 1988]
## (d) 192 Joules approx.
Sunshine
166 Transmission of Heat
57.
If the rates of cooling of two bodies are same then for which body the rate of fall of temperature will be more?
For the body whose thermal capacity is
(a) More
(b) Less
(c) Infinity
## (d) Any value
58.
If a sphere of radius R, cube of side R, and a cylinder of radius R and height R made of same substance are
heated to same temperature and then cooled, then which of above will radiate maximum
(a) Cylinder
59.
(b) Sphere
## (d) Cylinder and sphere both
The temperature of an isolated black body falls from T1 to T2 in time t. Let c be a constant
1
1
2
2
T2 T1
1
1
T2 T1
(a) t c
60.
(c) Cube
(b) t c
1
1
3
3
T2 T1
(c) t c
1
1
4
4
T1
T2
(d) t c
A sphere of density d, satisfied heat s and radius r is hung by a thermally insulating thread in an enclosure which
is kept at a lower temperature than the sphere. The temperature of the sphere starts to drop at a rate which
depends upon the temperature difference between the sphere and the enclosure. If the temperature difference is
T and surrounding temperature is T0 then rate of fall in temperature will be (Given that T T0)
(a)
4T03T
rdc
(b)
12T03T
rdc
(c)
12T04 T
rdc
(d)
12T
rdcT03
Basic level
61.
## According to Newton's law of cooling, the rate of cooling of a body is proportional to
n , where
is the
difference of the temperature of the body and the surroundings and n is equal to
[AIEEE 2003]
(a) One
62.
(b) Two
## (c) Specific heat of liquids (d)
Liquid is filled in a vessel which is kept in a room with temperature 20 oC. When the temperature of the liquid is
80oC, then it loses heat at the rate of 60 cal/sec. What will be the rate of loss of heat when the temperature of
the liquid is 40oC
[MP PMT 1994]
(a) 180 cal/ sec
64.
(d) Four
## Newton's law of cooling is used in laboratory for the determination of the
(a) Specific heat of the gases(b)
Latent heat of liquids
63.
(c) Three
## (d) 20 cal/ sec
A bucket full of hot water is placed in a room. Water takes t1 seconds to cool from 90oC to 80oC, t2 seconds to cool
from 80oC to 70oC and t3 seconds to cool from 70oC to 60oC then
[CBSE PMT 1995; PMT 1993]
65.
(b) t1 = t2 = t3
## (d) t1 > t2 > t3
A calorimeter of negligible water equivalent contains 430 gm of water and it cools at the rate of 0.24 oC per
minute in the surroundings at 30oC. If at any moment the temperature of water is 34 oC then at what rate the
heat should be supplied to it to keep its temperature constant
(a) 0.24 Cal/minute
## (d) None of the above
66.
A body cools in a surrounding which is at a constant temperature of 0 . Assume that it obeys Newton's law of
cooling. Its temperature is plotted against time t. Tangents are drawn to the curve at the points P ( 1)
and Q( 2) . These tangents meet the time axis at angles of
2 and 1 , as shown
Sunshine physics
Transmission of Heat 167
(a)
(b)
tan 2 1 0
tan1 2 0
tan 2 2 0
tan1 1 0
1
0
tan1 1
(c)
tan 2 2
(d)
67.
P
Q
tan1 2
tan 2 1
A system S receives heat continuously from an electrical heater of power 10 W. The temperature of S becomes
constant at 50oC when the surrounding temperature is 20oC. After the heater is switched off, S cools from 35oC to
34.8oC in 1 minute. The heat capacity of S is
(a) 100 J/oC
Basic level
68.
[CBSE PMT 2001]
(a)
69.
3
m
2
(b)
2
m
3
(c)
4
m
9
(d)
9
m
4
## Consider the following statements
Assertion (A) : When temperature increases, the colour of a star shifts towards smaller wavelength, i.e.,
towards violet colour
Reason (R) : Red colour has maximum wavelength
Of these statements
[AIIMS 2000]
(a) Both A and R are true and the R is a correct explanation of the A
(b) Both A and R are true but the R is not a correct explanation of the A
(c) A is true but the R is false
(d) Both A and R are false
(e) A is false but the R is true
70.
[MP PMT 1992]
71.
(b)
(d)
## All the above factors
On investigation of light from three different stars A, B and C, it was found that in the spectrum of A the intensity
of red colour is maximum, in B the intensity of blue colour is maximum and in C the intensity of yellow colour is
maximum. From these observations it can be concluded that
[CPMT 1989]
## (a) The temperature of A is maximum, B is minimum and C is intermediate
(b) The temperature of A is maximum, C is minimum and B is intermediate
(c) The temperature of B is maximum, A is minimum and C is intermediate
(d) The temperature of C is maximum, B is minimum and A is intermediate
72.
If black wire of platinum is heated, then its colour first appear red, then yellow and finally white. It can be
understood on the basis of
Sunshine
168 Transmission of Heat
[MP PMT 1984]
exchange
(b)
Prevost
theory
(d)
of
heat
## Problems based on energy distribution graph
Basic level
73.
Shown below are the black body radiation curves at temperatures T1 and T2 (T2>T1). Which of the following plots
is correct
[AIIMS 2003]
(a)
T2
(b)
T1
T2
(c)
T1
T1
T1
(d)
T2
The spectrum of a black body at two temperatures 27 oC and 327oC is shown in the figure. Let A1 and A2 be the
areas under the two curves respectively. The value of
(a) 1 : 16
(b) 4 : 1
(c) 2 : 1
(d) 16 : 1
A2
is
A1
Intensity
74.
T2
2
327oC
1
27oC
Wavelengt
h | 6,642 | 21,669 | {"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} | 3.234375 | 3 | CC-MAIN-2019-39 | latest | en | 0.866327 |
https://qiita.com/kozakai-ryouta/items/6697e51d29bb4652a439 | 1,526,812,079,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863277.18/warc/CC-MAIN-20180520092830-20180520112830-00305.warc.gz | 628,018,986 | 14,201 | # packageを使わないでやってみた(マルコフ生成行列の応用)
```#attribution analysis(making sample_data)
library(pforeach)
library(dplyr)
initial_time=Sys.time()-as.numeric(substr(Sys.time(),12,13))*3600-as.numeric(substr(Sys.time(),15,16))*60-as.numeric(substr(Sys.time(),18,19))
N=1000;m=3;customers=100
data=data.frame(ID=sample(1:customers,N,replace=TRUE),time=initial_time+sample(1:24,N,replace=TRUE,prob=c(0,0,0,0,0,0,0.1,0.1,0.01,0.01,0,0.1,0.02,0,0,0,0.1,0.1,0.1,0.1,0.1,0.1,0.06,0))*3600+sample(0:3600,N,replace=TRUE),site=sample(1:m,N,replace=TRUE))
data=data[order(data\$time),]
trans_time=c()
for(j in 1:customers){
data_sub=data %>% dplyr::filter(ID==j) %>% dplyr::mutate(num_time=as.numeric(time),sign=0)
data_sub=data_sub[order(data_sub\$time),]
if(nrow(data_sub)>=2){
for(i in 2:nrow(data_sub)){
if(data_sub\$site[i]==data_sub\$site[i-1]){
data_sub\$sign[i]=1
}
}
data_sub=data_sub %>% dplyr::filter(sign==0)
trans_time=c(trans_time,nrow(data_sub)-1)
}
}
trans_time=max(trans_time)
#trans_time_data=data.frame(trans_time=1:trans_time)
trans_time_data=pforeach(j=1:customers,.c=rbind)({
data_sub=data %>% dplyr::filter(ID==j) %>% dplyr::mutate(num_time=as.numeric(time),sign=0)
data_sub=data_sub[order(data_sub\$time),]
if(nrow(data_sub)>=2){
for(i in 2:nrow(data_sub)){
if(data_sub\$site[i]==data_sub\$site[i-1]){
data_sub\$sign[i]=1
}
}
data_sub=data_sub %>% dplyr::filter(sign==0) %>% dplyr::select(-sign)
data_sub=data_sub %>% dplyr::mutate(before=0,after=0,diff_num_time=0)
for(i in 2:nrow(data_sub)){
data_sub\$before[i]=data_sub\$site[i-1];data_sub\$after[i]=data_sub\$site[i]
data_sub\$diff_num_time[i]=data_sub\$num_time[i]-data_sub\$num_time[i-1]
}
data_sub=tail(data_sub,nrow(data_sub)-1)
data_sub=data_sub %>% dplyr::select(before,after,diff_num_time)
if(nrow(data_sub)!=trans_time){
data_sub=rbind(data_sub,data.frame(before=rep(0,trans_time-nrow(data_sub)),after=rep(0,trans_time-nrow(data_sub)),diff_num_time=rep(0,trans_time-nrow(data_sub))))
}
data_sub=data_sub %>% dplyr::mutate(trans_time=1:nrow(data_sub))
data_sub=data_sub %>% dplyr::select(trans_time,before,after,diff_num_time)
#colnames(data_sub)=c("trans_time",paste0("before",j),paste0("after",j),paste0("diff_num_time",j))
data_sub
}
})
trans_time_data=trans_time_data %>% dplyr::filter(diff_num_time!=0)
```
```
trans_time_data=trans_time_data %>% dplyr::select(trans_time,before,after,diff_num_time)
trans=trans_time_data %>% group_by(before,after) %>% summarise(n=n())
alpha=trans %>% dplyr::mutate(alpha=n/sum(trans\$n))
beta=trans %>% dplyr::mutate(beta=0)
for(k in 1:nrow(beta)){
before=beta\$before[k];after=beta\$after[k]
sample_data=trans_time_data %>% dplyr::mutate(sign=0)
for(i in 1:nrow(sample_data)){
if(sample_data\$before[i]==before){
if(sample_data\$after[i]==after){
sample_data\$sign[i]=1
}
}
}
sample_data=sample_data %>% dplyr::filter(sign==1) %>% dplyr::select(-sign,-trans_time)
t=sample_data\$diff_num_time
lambda=1/mean(t)
beta\$beta[k]=lambda
#exp_dis=function(t){
#z<-lambda*exp(-lambda*t)
#return(z)
#}
#exp_data=data.frame(t=seq(0,40000,1),prob=0)
#for(l in 1:nrow(exp_data)){
# exp_data\$prob[l]=0.3*integrate(exp_dis,0,exp_data\$t[l])\$value
#}
#plot(exp_data\$t,exp_data\$prob,type="o",col=3,xlab="t",ylab="prob",main=paste0("a:",0.3,",b:",lambda))
#integrate(exp_dis,0,10000)
#p_ij(t)=alpha*integrate(exp_dis,0,t)\$value
}
A=array(0,dim=c(m,m))
for(i in 1:m){
for(j in 1:m){
if(i!=j){
A[i,j]=alpha\$alpha[alpha\$before==i & alpha\$after==j]*beta\$beta[beta\$before==i & beta\$after==j]
}
}
}
Q=A;
diag(A)=-apply(A,1,sum)
lambda=-diag(A)
for(j in 1:nrow(Q)){
Q[j,]=Q[j,]/lambda[j]
}
diag(Q)=-diag(A)
```
```
#E[X_T=2,T<t|X_0=1]
t=40000
ave_func=function(s){
z<-s*Q[1,2]*lambda[1]*exp(-lambda[1]*s)
return(z)
}
integrate(ave_func,0,t)\$value
```
```
#P(X_t=3,T<t|X_0=2)
prob=function(t,k,j,i){
z<-A[k,j]*((1/lambda[i])*(1-exp(-lambda[i]*t))+(1/(beta\$beta[beta\$before==k & beta\$after==j]-lambda[i]))*(exp(-beta\$beta[beta\$before==k & beta\$after==j]*t)-exp(-lambda[i]*t)))
return(z)
}
t=40000
Prob=function(t,i,j){
z=0
for(l in 1:m){
if(l!=i){
z=c(z,Q[i,l]*prob(t,l,j,i))
}
}
return(sum(z))
}
plot_data=data.frame(t=seq(0,t,1),value=0)
for(j in 1:nrow(plot_data)){
plot_data\$value[j]=Prob(plot_data\$t[j],2,3)
}
plot(plot_data\$t,plot_data\$value,type="o",col=2,xlab="t",ylab="prob",main="P(X_t = 3,T < t | X_0 = 2)")
``` | 1,595 | 4,442 | {"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} | 2.671875 | 3 | CC-MAIN-2018-22 | longest | en | 0.202206 |
https://www.cleancss.com/convert-units/units-of-volume/usoz/ml | 1,726,821,053,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652138.47/warc/CC-MAIN-20240920054402-20240920084402-00615.warc.gz | 644,090,562 | 18,263 | # Convert USOZ to ML
1
US Fluid Ounces
=
29.5735295
Mililitres
How To Calculate US Fluid Ounces To Mililitres
To convert us fluid ounces to mililitres you simply multiply your us fluid ounces by 29.5735295. The formula would look like this:
Yml = Xusoz * 29.5735295
#### 1 US Fluid Ounces equals
Litres 0.03 Mililitres 29.574 Meters Cubeds 29.574 Pints 0.052 Gallons 0.007 UK Fluid Ounces 1.041 US Quart (liquid)s 0.031 UK Quarts 0.026 US Quart (dry)s 0.027 Cups 0.125 Tablespoons 2 Teaspoons 6
#### US Fluid Ounces To Mililitres Conversion Table
From To
1 usoz29.5735295 ml
2 usoz59.147059 ml
3 usoz88.7205885 ml
4 usoz118.294118 ml
5 usoz147.8676475 ml
6 usoz177.441177 ml
7 usoz207.0147065 ml
8 usoz236.588236 ml
9 usoz266.1617655 ml
10 usoz295.735295 ml
11 usoz325.3088245 ml
12 usoz354.882354 ml
13 usoz384.4558835 ml
14 usoz414.029413 ml
15 usoz443.6029425 ml
16 usoz473.176472 ml
17 usoz502.7500015 ml
18 usoz532.323531 ml
19 usoz561.8970605 ml
20 usoz591.47059 ml
30 usoz887.205885 ml
40 usoz1182.94118 ml
50 usoz1478.676475 ml
60 usoz1774.41177 ml
70 usoz2070.147065 ml
80 usoz2365.88236 ml
90 usoz2661.617655 ml
100 usoz2957.35295 ml | 463 | 1,151 | {"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} | 2.765625 | 3 | CC-MAIN-2024-38 | latest | en | 0.264116 |
https://fr.maplesoft.com/support/help/maple/view.aspx?path=StudyGuides%2FMultivariateCalculus%2FAppendix%2FExamples%2FA-2%2FExampleA-2-4 | 1,718,781,756,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861806.64/warc/CC-MAIN-20240619060341-20240619090341-00470.warc.gz | 238,503,836 | 22,995 | Example A-2-4 - Maple Help
Appendix
A-2: Arithmetic Calculations
Example A-2.4
Explore each of the following: $\sqrt{2}/3$, |-5|, ${e}^{-3}$, $\mathrm{ln}\left(4\right)$, ${\mathrm{log}}_{2}\left(8\right)$. | 83 | 219 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 47, "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} | 2.609375 | 3 | CC-MAIN-2024-26 | latest | en | 0.5284 |
https://math.stackexchange.com/questions/864081/find-the-term-of-the-g-p/864112 | 1,621,198,862,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989914.60/warc/CC-MAIN-20210516201947-20210516231947-00079.warc.gz | 406,686,480 | 38,074 | # Find the term of the G.P.
Find the term of the G.P. $1, 1.2, 1.44, 1.728,...$ which is just greater than $100$.
Please somebody explain me the question? All i know is $a=1$ and $r=1.2$
Help :(
should it be $T_{n} > 100$?
• If $a_1 = 1$ and $r = 1.2$, what is the formula for the $n$th term $a_n$? – TonyK Jul 11 '14 at 9:26
• @TonyK $ar^n-1$ ? – Kiara Jul 11 '14 at 9:29
• Yes (except you forgot to put the exponent in curly brackets). And $a=1$, so this is just $r^{n-1}$. Now start with the equation $r^{n-1} > 100$, and take the log of both sides. – TonyK Jul 11 '14 at 9:59
$T_1=1$ and $r=1.2$, we want value of $n$ for which $T_n>100$.
That means $1\times(1.2)^n>100$ or $n>\frac{2}{\log_{10}1.2}=\frac{2}{0.0791812460476248277225056927041}=25.258506273026670863695116167616$
So $n\geq 26$.
Edit:
And $T_{26}=114.47545997288281555215581184$.
• I did the same thing, but the answer in my book is $114.5$ – Kiara Jul 11 '14 at 10:13
• @Kiara $1.2^{26} \approx 114.5$. So the term is $114.5$, while the index (i.e. $n$) is $26$. – Arthur Jul 11 '14 at 10:19
• Oh, I was being such a silly girl! Thanks a lot! – Kiara Jul 11 '14 at 13:39 | 478 | 1,151 | {"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} | 3.875 | 4 | CC-MAIN-2021-21 | latest | en | 0.777756 |
http://mathhelpforum.com/pre-calculus/218845-radicals-1-a.html | 1,481,062,642,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542002.53/warc/CC-MAIN-20161202170902-00065-ip-10-31-129-80.ec2.internal.warc.gz | 161,014,864 | 10,079 | imgur: the simple image sharer
My attempt:
imgur: the simple image sharer
I'm not sure if I was right about dividing .3 in the beginning from the radical.
You didn't need to divide by 0.3. You also divided 6.3 by 0.3 as well as dividing $\sqrt{48}$ by $\sqrt{0.3}$ so in total you divided by $0.3\sqrt{0.3}$. If you want to do this you also have to multiply by $0.3\sqrt{0.3}$ to effectively multiply by 1 so that you don't change the value of it, to balance things.
Also there is no need bring everything inside the square root so that it is $\sqrt{70560}$ because you'll end up bringing it all outside the square root in the end.
$6.3\sqrt{48}= 6.3\sqrt{6\times8}=6.3\sqrt{2\times3\times2\times2 \times2}= 6.3\times 2\times 2\sqrt{3}=25.2\sqrt{3}$ | 247 | 753 | {"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": 6, "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-2016-50 | longest | en | 0.922559 |
https://www.citehr.com/137987-pf-challans-calculation-xls-download.html | 1,582,715,046,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146341.16/warc/CC-MAIN-20200226084902-20200226114902-00429.warc.gz | 665,229,800 | 16,772 | PF Challans Calculation - don't know how to calculate Monthly PF Challans - XLS Download - CiteHR
#query{background:#ebf5fd;color:#4f4f4f;letter-spacing:1px;padding:5px;border:1px #001b31 solid;text-indent:3%;font-size:small} form#search_query{color:#fdfcfb;text-align:left;margin:0}
P Ramachandran
Hr & Administrative Manager
Mohanms007@gmail.com
Senior Human Resources
Tejasthakaria5700@gmail.com
Labour Welfare Officer
Esskae59
Hr Manager
Kripsha
Compensation Management
Sn81
Pay Roll
Sandeep.oas
Head- Hr & Administration
Sunilyeruva
Sr.asst.hr
+5 Others
Cite.Co is a repository of information created by your industry peers and experienced seniors sharing their experience and insights.
Join Us and help by adding your inputs. Contributions From Other Members Follow Below...
HI All, Iam a fresher & i don’t know how to calculate Monthly PF Challans. Could you pls help me to calculate the monthly challans. Regards Kripsha
In PF Challan there are five acc.
A/c1-12%+3.67%(12%for emp share and 3.67% of Employer)
A/c2-1.1%(Admin charges)
A/c10-8.33%(For pension sche)0.5%(EDLI,ie Emplo. deposit link insurance)
A/c21-0.5%(EDLI,ie Emplo. deposit link insurance)
A/c22-0.01%(Admin charges)
Thanks
Lata
Hi Friend,
Regarding PF Contribution -Employer should pay 12% of Basic salary, in which 8.33% will goes to Pension Scheme balance 3.67 will goes to PF.
While calculating the Pension Scheme employer should pay 8.33% of basic (limit is upto Rs.6500/ ie Rs.541/-) not more than that rest will goes to PF.
For eg. If the employees is getting Basic salary of Rs.8000/- Rs.541/-(8.33%) will goes to pension scheme and Rs.419/- will goes to PF, however employer has to contribute 12% of basic, whatever be the amount.
Regarding Employee Contribution of 12% from basic is unavoidable as per the PF act.
Rate of Contribution
What may it may be the company they have to contribute the below following
Employee 12%
(Basic wages, dearness allowance)
Employer 13.61%
Contribution:
3.67% = Provident Fund (A/c No.1)
8.33% = Pension Fund (A/c No.10)
0.5% = EDLIS -Employees Direct Linked Insurance Scheme (A/c No.21)
Administrative Charges:
1.10% of emoluments = Provident Fund (A/c No.2)
0.01% of emoluments = EDLIS (A/c No.22)
Total 25.61%
Warmest Regards,
Mohan.M.S
"willingness to walk the extra mile to achieve excellence"
Hope you got some info on the PF contributions.
To make things easier for you, forwarding an attachment which will help you out practically.
for any clarifications, pl. feel free to contact me.
S Kannan
94433 95863
Attached Files
Membership is required for download. Create An Account First pf_and_esi_challan_198.xls (157.5 KB, 14894 views)
Dear Bandita,
Apart from contribution details,they would ask about PF forms details.
Iam herewith attaching the PF Forms details.Please make use of it.Prepare well for ur interview.
All the best for your interview.
Warmest Regards,
Mohan.M.S
Attached Files
Membership is required for download. Create An Account First EPF Act PPT.ppt (33.0 KB, 3715 views)
PF CHALLAN CONTAINS THE FOLLOWING ITEMS AND THEIR CALCULATION IS AS UNDER:
NO.OF PERSONS - you have to fill the number of persons in that month under pf roles
CODE NUMBER : here you mention your company's pf code number like MH/xxxx
TOTAL WAGES FOR PF: here you fill up the total amount of wages on which pf deducted
TOTAL WAGES FOR FPS: here you fill up the same as TOTAL WAGES FOR PF if there is no seperate amount for FPS
ACCOUNT NUMBER 1 this contains employee contribution (deducted pf amount) and employer contribution (TOTAL OF EPFER ACCOUNT)
ACCOUNT NO:10 FPS ACCOUNT EMPLOYER CONTRIBUTION FOR FAMILY PENSION
ACCOUNT NUMBER:2 : 1.1% ADMINISTRATIVE CHARGES ON EPF ON TOTAL WAGES
ACCOUNT NUMBER:21:05% ON TOTAL WAGES FOR INSURANCE FUND
ACCOUNT NUMBER 22: 0.01% ON TOTAL WAGES FOR ADMINISTRATIVE CHARGES FOR INSURANCE FUND
BEFORE THIS THE INDIVIDUAL CALCULATION OF PF IS HIGHLY HELPFUL TO YOU A SPECIMEN IS ATTACHED HEREWITH
Attached Files
Membership is required for download. Create An Account First MODEL PF WORKING.doc (54.5 KB, 4619 views)
HI,
I have one doubt, while calculating pf, 8.33% of employer's share goes to EPS & 3.67% goes in PF A/C. i just want to know that whether 3.67% is to be calculated with a maximum limit 6500 or on the total basic????????
If a employee's basic salary is 20,000,then 541 goes to his EPS A/c but iam confused about PF A/C???????????:confused:
Plz help me.........
Regards
Kripsha
Dear Friend,
No need to confuse.The employer share of 8.33% will be accounted in the pension account.The max limit of pension a/c is 541/- only.
For example:
Mr X Basic is 8000/- his PF employee contribution is 960/-
Employer Contribution of 960 will be splitted two division.The max limit of 541/- will be accounted in pension remaining 419 will be accounted in employee pf account.
I think ur doubt has been cleared.
Warmest Regards,
Mohan.M.S
Hi
There is the Ceiling Limit is Rs.6500 for EPF & EPS itself also If the Employee would like to contribute 12% of their basic+DA but the Employer not need to pay the same Amount if the Salary is beyond that ceiling. the Employer's share only Spilt into 8.33+3.67% u know
So i conclude what b the employee contribution Employer should pay 12% of ceiling limit.
I think now u may clear ur doubt
Reply,
Regards,
Sukumar,
Personnel Officer
0-9842023923
Dear Balaji,
Please find herewith the Complete details of ESI.
Employees’ State Insurance (ESI)
ESI: Employees’ State Insurance Act, 1948
Calculations: ESI from Gross Salary:
Employee: 1.75%
Employer: 4.75%
Coverage: All the employees Drawing wages up to 000/- per month engaged either directly or thru’ contractor.
Regular activities:
1. Time of joining/at any time:
Form 01 : Employer Registration Form
Form 1 : Employee should fill, at the time of joining, Declaration form with postcard size
Photograph – due date with in 10 days after the employees joins.
Form 1 A : Family Declaration Form, family details
Form 1 B : Changes in family declamation form, like family members…
Form 3 : Return of declaration form (Covering Letter) 3A continuation sheet/card,
Employer should fill. Male and female separately
Form 37 : Employer should fill Certificate of Re-Employment / Continuing employment. With
Contribution period begin and end dates.
Form 105 : Employer should fill, Certificate of Entitlement.
Form 72 : Employee should fill, Application /form for changes in particular of insured
Person. Like local office, Dispensary/Address changes.
Register 7 : Individual Computation, there Gross salary, Days, ESI amt.
Information maintains month-wise.
Cards: Temporary & Permanent Cards.
Monthly Remittance / Challans:
1. Challans every month before 21st (3 copies/ quadruplicate)
2. Submit to Bank
3. Both employer & employee contribution
4. Cheq details.
Half year returns:
Contribution period:
1st April to 30th September.
1st October to 31st March
***42 days after closing Contn. Period (before Nov 11th. And next before May 12th)
1. Form 7 (Register of Employees)
2. Form 6A: Consolidated Computation Sheet, contains total employees list, there total half
Yearly Information. Form 6 is top sheet and 6A is attachments. (Statement of
Advance Payment of Contributions)
2. In Oct & April
3. With all paid challans
Need to maintain:
* Muster Roll * Wage Register * Inspection Book * Accident Register * Cash Books, Vouchers & Ledgers * Paid Challans, RDF and Declarations
* Returns copies
Forms:
Form 4 : Identity Card
Form 4 A : Family Identity Card
Form 6 : Return of Contributions
Form 8 : First Medical Certificate
Form 9 : Final Medical Certificate
Form 10 : Intermediate Medical Certificate
Form 11 : Special Intermediate Certificate
Form 12 : Sickness Or Temporary Disablement Benefit / Claim For Benefit – Form
From 12 A : Maternity Benefit For Sickness / Claim For Benefit – Form
Form 13 : Sickness or Temporary Disablement or Maternity Benefit for Sickness / Claim For
Benefit – Form
Form 13 A : Claim For Maternity Benefit For Sickness – Form
Form 14 : Sickness Or Temporary Disablement Or Maternity Benefit For Sickness / Claim For
Benefit - Form
Form 14A: Claim For Maternity Benefit For Sickness
Form 15: Accident Book – Form
Form 16: Employer should fill, accident report form, with date of accident, place, time…need to
Submit to ESI local office immediately – 3 Copies (with 2 witness) 1-Local office,
Form 17: Death Certificate – Form
Form 18: Dependants Benefit - Claim Form
From 18A: Defendants Benefit/ Claim Form for periodical payments – Form
Form 19: Notice of Pregnancy – Form
Form 20: Certificate of pregnancy – Form
Form 21: Certificate of Expected Confinement – Form
Form 22: Benefit Claim Form
Form 23: Certificate of Confinement or Miscarriage
Form 24: Notice of Taking Up Work – Form
Form 24 A: Maternity Benefit Claim After The Death Of An Insured Woman Leaving Behind The
Child – Form
Form 24 B: Maternity Benefit Death Certificate – Form
Form 25: Claim for Permanent Disablement Benefit – Form
Form 25 A: Funeral Expenses Claim Form
Form 26: Certificate for Permanent Disablement Benefit – Form
Form 27: Declaration and Certificate for Dependants’ Benefit - Form
Form 28: Confirmation of Incapacitation of Employee - Form
Form 28 A: Confirmation of Incapacitation of Employee - Form
Warmest Regards,
Mohan.M.S
"Willingness to walk the extra mile to achieve excellence"
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All rights reserved @ 2020 Cite.Co™ | 2,456 | 9,507 | {"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} | 2.65625 | 3 | CC-MAIN-2020-10 | latest | en | 0.812774 |
https://www.physicsforums.com/threads/calculate-the-moment-of-inertia-of-this-sphere.158832/ | 1,585,798,543,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370506580.20/warc/CC-MAIN-20200402014600-20200402044600-00115.warc.gz | 809,347,806 | 17,678 | Calculate the moment of inertia of this sphere
Homework Statement
A sphere consists of a solid wooden ball of uniform density 800kg/m^3 and radius 0.20 m and is covered with a thin coating of lead foil with area density 20kg/m^2
Calculate the moment of inertia of this sphere about an axis passing through its center.
I_cm = m(r^2)
The Attempt at a Solution
1. I took the volume formula of the sphere which is 4/3*pi*r^3 to get the volume and then multiply by its density to get the mass of the solid ball inside the sphere.
2. Then I did the same with the lead coating, only using the surface area formula A = 4*pi*r^2
(This I'm not sure about because I don't understand the picture. Is the solid ball centered in the sphere? Because I took the radius to be .20m assuming that it is and that it takes on the shape of the sphere...)
3. I combined the mass to get total mass and then multiply with the radius (.20m) with respect to the x and y axis.
But yeah, it's wrong. Someone pls guide me...somehow. :uhh:
Related Introductory Physics Homework Help News on Phys.org
cepheid
Staff Emeritus
Gold Member
1. Yeah, it's just a wooden ball with a coating of lead on it, that's all. A "ball" is just the solid volume enclosed by a sphere anyway.
2. The formula $I = mr^2$ only applies to a point mass of perp. distance r from the axis about which it is rotating! Here you have a continuous distribution of masses, not just one point mass. To me, that suggests that you have to integrate in order to calculate I (unless if you can use the spherical symmetry and uniform density to make a simplification, I don't know).
cepheid
Staff Emeritus
Gold Member
Edit:
But for constant density, the integral would be trivial, wouldn't it? Hrmm...
$$dI = r^2 dm = r^2 \rho dV$$
$$I = \int \! \! \! \int \! \! \! \int_V r^2 \rho dV = \rho \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{0.20} (r^2) r^2 \sin \theta dr d\theta d\phi$$
I hope I'm doing this right. Not sure what to do about the lead foil...it must be pretty simple but I'm tired right now...
Last edited:
cepheid
Staff Emeritus
Gold Member
Edit 2: I just realized that this is the introductory physics forum! Hrmmm...maybe the question is supposed to be simpler than this. Do you have any canned formulas for the moments of inertia of various shapes? That integral would lead me to believe that the result is
$$I_{\textrm{sphere}} = \frac{3}{5}MR^2$$
where M is the total mass of the sphere, and R is the radius. But my first year physics textbook says that:
$$I_{\textrm{sphere}} = \frac{2}{5}MR^2$$
Huh?! Close but no cigar...
cepheid
Staff Emeritus | 712 | 2,616 | {"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} | 3.921875 | 4 | CC-MAIN-2020-16 | longest | en | 0.941787 |
http://encyclopedia2.thefreedictionary.com/Magnifier | 1,508,378,564,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823214.37/warc/CC-MAIN-20171019012514-20171019032514-00332.warc.gz | 107,309,804 | 13,766 | # Magnifier
Also found in: Dictionary, Thesaurus, Medical, Legal, Wikipedia.
## magnifying glass
, magnifier
a convex lens used to produce an enlarged image of an object
## Magnifier
an optical instrument for examining small objects not readily discernible by the naked eye. The object being observed is positioned slightly less than a focal length away from the magnifier. Under these conditions a magnifier provides an upright, magnified virtual optical image of the object. After passing through the magnifier, the rays from the object are again refracted in the eye and are collected at its most remote point. They enter the eye at an angle greater than do the rays from the object without a magnifier; this is the reason for the magnifying effect (see Figure 1).
Figure 1. Path of rays when viewing a small object / in a magnifier L The object is positioned in the immediate vicinity of the focal plane OO1, of the magnifier. The rays coming from point S of the object are collected at point S” on the retina of the eye. The rays from point S″ would also be collected at point S” if there were no magnifier (point S’ is the’virtual image of point S).
The magnification M is defined as the ratio of the angle α at which the virtual image of the object is seen from the center of the eye’s pupil to the angle at which the same object is seen without the magnifier at the distance of distinct vision D (Figure 2; for the normal human eye, D == 0.25 m). The magnification depends on the focal length f’ of the magnifier (usually expressed in millimeters) according to the ratio M = 250/f; the value of M may range from 2 to 40 or 50, depending on the design of the magnifier.
Figure 2. Observation of a small object /. (a) With naked eye at the distance of distinct vision D; 0 is aperture angle of rays from the object that enter the eye. (b) Through a magnifier; rays from the object enter the eye at an angle a > <£; d is the distance from the magnifier to the object, d’ is the distance from the magnifier to the image it forms of the object as seen by an observer.
The simplest magnifiers are converging lenses; their magnification is usually low (about 2-3 X). Two-lens and three-lens systems (Figure 3) are used for magnifications of 4-10 X. The field of view of the images in space for magnifiers having low and medium magnification does not exceed 15°-20°. The design of magnifiers having high values of M is similar to that of complicated eyepieces; their field of view reaches an angle of 80°-100°.
Figure 3. Magnifiers: (a) “twin” lenses, (b) aplanatic
Figure 4. Telescopic magnifier
In a magnifier with high magnification the distance from the object to the surface of the magnifier is very short. This drawback is avoided in telescopic magnifiers (Figure 4), which are capable of observing remote objects with M ~ 2.5 and nearby objects with M ~ 6. Binocular (stereoscopic) magnifiers, a diagram of which is shown in Figure 5, are also in use.
Figure 5. Stereoscopic magnifier consisting of prismatic achromatic lenses combined with low-power binoculars
Magnifiers are also used to measure linear dimensions. The object to be measured is brought into coincidence with a flat glass or metal scale placed in front of the focal plane of the measuring magnifier (in practical terms it is in the plane). The images of the object and the scale are compared. The magnification of a measuring magnifier is usually 4-16X; the focal length, 10-40 mm; and the scale divisions, 0.1 mm. Such magnifiers are used to measure the width and length of letters and scratches and the distance between points.
## magnifier
[′mag·nə‚fī·ər]
(optics)
References in periodicals archive ?
As always, features are fully customisable, from magnification size to voice to speed - making any SuperNova Magnifier product all about the individual.
There are a few products that have created a removable stand for a hand magnifier offering conversion to a stand magnifier when needed; this provides even more flexibility with little additional cost.
Other HD video magnifiers offer an HD camera, but not an HD screen.
For viewing text, photographs or maps, desktop video magnifiers are extremely versatile.
Ask them if they have looked through a magnifier or a telescope.
Another wearable device particularly helpful for patients with advanced macular degeneration is the JORDY, a Space Age-looking pair of sunglasses that is actually a video magnifier.
By using the Visual Magnifier with the child they can also detect visual dyslexia, so every primary school class in the country should have one.
To launch Magnifier, click on Start, Programs, Accessories, Accessibility, Magnifier.
Each representative also received bold line writing paper, an envelope writing guide, a signature guide, a check writing guide, a letter writing guide, Hi-Marks, a magni-guide, a medicine bottle opener, a safety food turner, a hot pan grip, an inner lip plate, oven mitts, a magnifier, low vision playing cards, a pill organizer, a large print telephone dial, and one 33-inch and one 36-inch orthopedic folding cane.
HIMS will showcase the company's unique LifeStyle HD video magnifier at the Assistive Technology Industry Association's ATIA 2011 Chicago Conference and co-partnered enLIGHTen Low Vision Conference November 3 - 5 in Chicago, Illinois.
com)-- Zoomax, the world's leading manufacturer of low vision products, has designed many popular video magnifiers to assist visually impaired individuals live independent lives, such as the handheld video magnifier Snow and Snow 7 HD, desktop video magnifier Aurora HD and Panda.
Figure 3 shows how to predict whether a patient is likely to be able to achieve 1M print with a magnifier, based on their baseline visual function.
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Open / Close | 1,316 | 5,798 | {"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} | 2.921875 | 3 | CC-MAIN-2017-43 | latest | en | 0.891047 |
https://docsbay.net/part-i-learning-about-the-microscope | 1,638,870,808,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363337.27/warc/CC-MAIN-20211207075308-20211207105308-00206.warc.gz | 293,921,130 | 8,676 | # Part I. Learning About the Microscope
PART I. LEARNING ABOUT THE MICROSCOPE
1. Compare your microscope with Figure 2 below. Identify each part on your microscope.
2. Look at the number followed by an “X” on the side of each objective or the number after DIN. This number is the objective’s magnifying power. The “X” stands for “times.” Thus the number tells how many times an object is magnified by this lens.
1. How many objective lenses does your microscope have?
2. What is the magnifying power of each?
3 . If the lenses look dirty or smudged, carefully wipe them with lens paper. Use only lens paper because other kinds of paper candamage the lenses.
4. The ocular lens also has a magnifying power. The total magnifying power of the microscope is easy to calculate. Simply multiply the magnifying power of the ocular by the magnifying power of the objective. For example, if the ocular is 5X and the objective is lOX, the total magnification of the object being viewed is 5X x lOX = 50X.Look carefully at the engraving of the ocular lens.
C. What is its magnifying power?
D . What is the total magnification for each of your objective lenses? Show your calculations.
PART II. PREPARING AND EXAMINING A WET MOUNT
Cut out a 1-cm square of a colored newspaper cartoon or a colored picture from a magazine printed on thin paper. Choose a square that has both light and dark tones, but not black
1.Place the colored newspaper or magazinein the middle of a clean slide. With a pipette,put 1 drop of water on the square. Drop the water from about1 cm above the slide. Do not touch the pipette to the paper orthe paper will stick to the pipette.
2. Now cover the mount with a clean cover slip. One way to do this is shown in Figure 3-a. Hold the cover slip at about a 45° angle to the slide and move it toward the drop. As the water touches the cover slip, it will spread along the edge. Gently lower the cover slip intoplace. Do not press on the cover slip—it should rest onthe top of the water. A good wet mount is free of bubbles. If yourmount has too many bubbles, take off the cover slip and absorbthe water with a paper towel. Then repeat Steps 2 and 3.
3. Click the low-power objective into place. Make sure you have a good light source and that the diaphragm is at the largest opening.
4. Check to be sure the bottom of the slide is dry before placing it on the stage of the microscope.
5. Look at the microscope from the side. Using the course focus knob to position the low-power objective about 1/2 to 1 cmabove the slide, or until you feel an automatic stop.
6. Look through the ocular, keeping both eyes open. Keeping botheyes open is difficult at first, but it helps to prevent eyestrain. It will become easier with practice.
7. Slowly lower the stage by turning the coarse adjustment until the letters come into focus. Use the fine adjustment to sharpen the focus.
1. The space on your answer sheet (like the one below) represents your field of view. This is the lit circle you observe when looking through the ocular. Draw what you see.
8. Move the slide to the left.
B. Which way does the image move?
9. Move the slide to the right.
C. Which way does the image move?
10. Move the slide away from you.
D. Which way does the image move?
11. Observe the wet mount as you change the diaphragm to each of its settings. Adjust it to give good contrast and illumination without glare.
E. What does the diaphragm control?
Before using high power, the specimen must be in sharp focus in the center of the low-power field of view. Note: All focusing under high power is done with the fine adjustment knobs. There is no automatic stop for the high-power objective.
12. Watching from the side, carefully switch to the medium power objective. Center the object and focus.
13. Now without changing anything, switch to the high-power objective. Make sure that the objective does not hit the slide, butexpect it to be very close.
14. Focus on the paper. Only a slight turn of the fine adjustment knob will be needed to do this.
F. In the space on your answer sheet, draw what you see under high power.
G. Is the field of view (how much area you see)larger under high power or low power?
H. Compare the brightness of the field under high power and low power.
15. Prepare another wet mount, this time using two hairs of different colors. Cross them on the slide, then add a drop of water and thecover slip.
16. View the slide under low power. Focus directly on the point where the hairs cross.
D Are both hairs in focus under low power?
17. Switch to high power and observe the hairs.
E.Are both hairs in focus under high power? Explain.
Part III: Measuring with the Microscope
Background
It is interesting and informative to observe specimens under the microscope, but it is often difficult to know the actual size of the object being observed. Magnification causes us to lose the idea of actual size. You cannot hold a ruler up to a paramecium or a plant cell while it is under a microscope. Therefore, size must be measured indirectly—that is, it must be compared with the size of something you already know. The diameter of the microscope field seen through the ocular is a convenient standard to use.
Two metric units are useful when measuring small objects:
1 meter (m) = 1000 millimeters (mm) 1 mm = 1000 micrometers (μ m)
Procedures and Observations
1. Examine the markings on a transparent metric ruler. Determine which marks indicate millimeter lengths. Then place the ruler on the stage so that it covers half of the stage opening as shown in Figure 1.
2. Prepare your microscope for low-power observation of the ruler.
3. Look through the ocular. Focus on the edge of the ruler, using the coarse adjustment. Adjust the position of the ruler so that the view in the low-power field is similar to
4. Place the center of one mark at the left side of the field of view. Make sure that the edge of the ruler is exactly across the center of the field. If the ruler sticks to your fingers, use the eraser end of a pencil to arrange it.
5. Note that 1 millimeter is the distance from the middle of one mark to the middle of the next mark. For example, in the picture above, the diameter of the low-power field measures 3 millimeter plus a fraction of another.
A. Record the measurement of the low-power field diameter in millimeters, expressing the length to the nearest tenth of a millimeter.
B. Record the measurement of the low-power field diameter in micrometers.(Convert the measurement in millimeters to micrometers by multiplying by 1000).
6. You cannot measure the diameter of the high-power field using the process you have just completed. Viewing a ruler under high power presents light and focusing problems. Also, the high-power field diameter is less than 1 millimeter. But you can indirectly obtain the high-power field diameter. You know the low-power field diameter and the magnifying power of both objectives. The magnification of the objectives is inversely proportional to the field size. You can use this formula:
C. Record the high-power field diameter in micrometersby substituting the values you know in this formula to calculate the high-power field diameter. Show your calculations.
The measurements of the low-power and high-power field diameters can be used to measure other things indirectly.
7. Under low power, focus on a prepared cross section of prepared paramecium slide (a paramecium is a single-celled organism in the protist kingdom).
D. Estimate how many of these paramecia would fit end to end across the diameter of the field of view.
To calculate the length of a single paramecium, divide the diameter of the low-power field by the number of cells given for Question D
E. What the length of a paramecium in micrometers.
11. Switch to high power and focus with the fine adjustment.
F. How many paramecia would fit across the diameter of the high- power field?
12. To calculate the diameter of a pith cell as seen under high power, divide the diameter of the high-power field by the number of cells given for Question F
G. What is the length of a paramecium, as measured under high power.
H. Compare the measurement of the diameter of a cell under low and high power. Which do you think is more accurate? Why?
I. Find the diameter of the high-power field of a microscope with an ocular marked lOX, a low-power objective marked 1OX, a high-power objective marked 40X, and a low-power field diameter of 1600 micrometers.
Part IV: How PlantandAnimal Cells Differ
Introduction
Although plant and animal cells have many structures in common,they also have basic differences. Plant cells have a rigid cell wall, and ifthey are green, they also have chloroplasts. Animal cells lack both a cell wall and chloroplasts. They also lack the central vacuole common to plant cells.
You will observe and compare animal cells and plant cells. You will first examine epithelial cells from the inside of your cheek. Epithelium is a type of tissue that covers the surfaces of many organs and cavities of the body.
You will then examine cells from a leaf of the freshwater plant elodea. Elodea is often used in home fish tanks. The cells of this plant are green because they contain the pigment, chlorophyll. Chlorophyll, which is found in chloroplasts within each cell, enables plants to manufacture their own food.
Finally you will observe the skin of an onion and compare to the elodea.
HUMAN EPITHELIAL CELLS
1. Place a SMALL drop of water on a clean slide. Obtain epithelial cells by gently scraping the inside of your cheek with a clean toothpick as shown in Figure 1. CAUTION: Never reuse a toothpick or put any thing in your mouth which may not be clean. Stir the material from the toothpick in the drop of water on the slide. Then immediately break the toothpick in half and throw it away.
2. Allow the cheek cells to dry and heat fix the slide by quickly passing the slide through the flame 4-5 times. Add a small drop of methylene blue stain to the slide. CAUTION: Stain can damage clothing and discolor skin. Let the stain sit for 1 minute and then rinse the stain off the slide. Carefully place a cover slip on the slide. Examine the slide under low power.
Now focus on the highest power using the emersion oil. To do this, focus first (in sequence) with the 3 lower power objectives. Next place the nosepiece so that the highest power is not yet straight down and you can access the slide. Take the dropper out of the emersion oil bottle and drop ONE drop (small) on the top of the cover slip. Now lower the high power objective so that it is immersed in the oil. Use the fine adjustment only to focus the cells.
a. Make a drawing of two or three cells as they appear under high power. Label the nucleus, cytoplasm, and cellmembrane of one of the cells. Indicate the total magnification below the circle
b. What is the shape of the cells?
c. Describe the appearance of the cytoplasm.
ELODEA LEAF CELLS
1. Break off a small leaf near the tip of an elodea plant. With a forceps place the entire leaf in a drop of water on a clean slide. Add a coverslip. See Figure 2.
2. Examine the leaf under low power.
D. What is the shape of the cells?
E. Estimate the size of a single cell.
The boundary that you see around each cell is the cell wall. Thenumerous small, green bodies in the cells are the chloroplasts.
3. Look for an area in the leaf where you can see the cells most clearly. Examine these cells under high power, carefully focusing up anddown with the fine adjustment.
F. Describe the shape and location of the chloroplasts.
4. As you examine the cells, you may see the chloroplasts moving around. If they are not moving, warm the slide in your hand or under a bright lamp for a few minutes. Do not allow the slide to dry out. Then examine again under high power.
G. Make a drawing of an elodea cell. Label the cell wall, chloroplasts, and any other structures you see.Be sure to indicate the total magnification.
The cell membrane is pressed tightly against the inside of the cell wall and is difficult to see. Furthermore, the numerous chloroplasts often make it difficult to observe other cell structures in the elodea leaf cells. In order to see the nucleus, nucleoli, and vacuole more clearly, you are going to use a stain.
6. Break off another elodea leaf and place it in a drop of Lugol’s iodine solution on a clean slide. Add a cover slip. Wait a minute or so for the stain to penetrate into the cells. Then examine the stained elodea cells under low and high power.
H. Make a drawing of a stained cell. Label the cell wall, cell membrane (if visible), chloroplasts, nucleus, nucleolus, and the largevacuole. Be sure to indicate the total magnification.
I. What structures can you see more clearly after staining?
ONION CELLS
Remove a thin “skin” of onion using the forceps. Place the onion onto a drop of water on a clean slide. Observe the cells on high power.
J. Compare the two unstained plant cells you have observed and give a logical reason for the differences you see.
Part V: Prokaryotic cells
Look at the prepared slide of prokaryotic cells under oil emersion.
1. Sketch what you have seen.
2. Estimate the size of a single cell. Show your work.
Part VI: Pond water
Take a drop of pond water and place it on a clean slide. Look at it under low power, scanning the entire area for something alive. Go slowly and look for something that may be moving. It may take some time and patience to locate a living organism.
1. Sketch what you have seen.
2. Is your organism likely to be prokaryotic or eukaryotic? Explain why.
1 | 3,129 | 13,675 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2021-49 | latest | en | 0.905001 |
https://goprep.co/ex-2.5-q4d-a-4-b-2-1-2-expand-each-of-the-following-using-i-1nkfh1 | 1,606,372,511,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141186761.30/warc/CC-MAIN-20201126055652-20201126085652-00353.warc.gz | 314,244,278 | 34,422 | Q. 4 D4.5( 8 Votes )
# Expand each of the following, using suitable identities
= [ + + 1]2
Using (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
Here x = , y = and z = 1
= + + 12 + 2 + 2(1) + 2(1)
= + + 1 + + (-b) +
= + + 1 - – b +
Therefore = + + 1 - – b +
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# Solution - Show that a1, a2 … , an , … form an AP where an is defined as below an = 3 + 4n Also find the sum of the first 15 terms in each case. - CBSE Class 10 - Mathematics
ConceptSum of First n Terms of an AP
#### Question
Show that a1, a… , an , … form an AP where an is defined as below
an = 3 + 4n
Also find the sum of the first 15 terms in each case.
#### Solution
You need to to view the solution
Is there an error in this question or solution?
#### APPEARS IN
NCERT Mathematics Textbook for Class 10
Chapter 5: Arithmetic Progressions
Q: 10.1 | Page no. 113
#### Reference Material
Solution for question: Show that a1, a2 … , an , … form an AP where an is defined as below an = 3 + 4n Also find the sum of the first 15 terms in each case. concept: Sum of First n Terms of an AP. For the course CBSE
S | 281 | 882 | {"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 | 3 | CC-MAIN-2018-13 | latest | en | 0.699457 |
https://www.dailytoolz.com/convert-decimal-to-fraction/ | 1,726,446,998,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651668.26/warc/CC-MAIN-20240915220324-20240916010324-00637.warc.gz | 665,806,574 | 4,652 | # Decimal to Fraction Converter
Please provide the decimal number below:
Convert
Swap
Reset
Fraction result:
1.125 =
1 1 8
As a Mixed Fraction:
1 1 8
As an Improper Fraction:
9 8
Calculation:
1.125 = 1+0.125 = 1+125/1000
gcd(125,1000) = 125
1.125 = 1+(125/125)/(1000/125) = 1+1/8
= 9/8
## What is Fraction?
Fraction to a layman is just some part of a piece of something. The numbers of the form 'a/b' are called fractions. As we can see, a fraction has two parts, one top part and one bottom part. The top part is known as 'Numerator' and the bottom part is called as 'Denominator'.
Denominator is the total number of equal parts in which the whole is divided into and numerator is the number of equal parts which have been taken out or which have been left out.
## Example in real life
When you share something or divide something, there is a lot of mathematics behind it. Suppose you and your two friends went to Ben's house. All the while waiting for you guys, he ate nine out of the ten chocolates that he kept for you guys. Now after you guys reached, all you have is one big chocolate that you have to divide it in equal parts among your friends including Ben, no cheating. So how will you do that?
You have divided one whole chocolate into into four equal parts, so each one of you will get one part out of the four parts of the chocolate or can we say that each one of you had one fourth part (1/4) of a chocolate, and numbers in this form are known as fraction.
## How to convert decimal to fraction
I'm going to cover how to convert decimals to fractions.
Example #1
Convert 0.4 to fraction:
0.4 = 4/10 = 2/5
We have 0.4, now that decimal ends in the tenths place, so that's going to be the denominator for our fraction. Again wherever the decimal ends that's going to be your denominator. So 0.4, then we take a look at what number is behind the decimal, and we have a 4, so that's our numerator. So 0.4 as a decimal, four tenths (4/10) as a fraction, and after simplified 2/5.
Example #2
Convert 0.04 to fraction:
0.04 = 4/100 = 2/50= 1/25
It is similar to example #1 but not the same value, it's going to have a different denominator. Because again what place does that decimal end, and it ends in the hundredths now because we have that placeholder zero that pushes the 4 to the hundredths. So our denominator is 100. Our numerator is still four though just like example #1, because we only have a 4 to the right of the decimal. So 4/100 as a fraction and after simplified 1/25.
Example #3
Convert 0.603 to fraction:
0.603 = 603/1000
In this decimal, what place does our decimal end, it goes to the thousandths that 3 it ends, so 1000 is our denominator. Now the number to the right of the decimal is 603, so we have 603/1000.
Example #4
Convert 1.88 to fraction:
1.88 = 1 88/100 = 1 44/50= 1 22/25
In this example, we are going to have a mixed number, so a whole number and a fraction. So our whole number is going to be 1 and then we have eighty eight hundredths, so the decimal ends in the hundredths place so we have one hundred as the denominator and the number to the right of the decimal is 88. So 1 88/100 is our mixed fraction, simplified 1 22/25.
DecimalFraction
0.0011/1000
0.011/100
0.11/10
0.43757/16
0.6416/25
0.6255/8
0.687511/16
0.812513/16
0.8757/8
0.8822/25
0.937515/16
1.11 1/10
1.1251 1/8 | 967 | 3,351 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-38 | latest | en | 0.957801 |
https://docs.racket-lang.org/new-set/ | 1,521,393,999,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645830.10/warc/CC-MAIN-20180318165408-20180318185408-00065.warc.gz | 601,746,301 | 8,637 | 6.12
## Set: Purely Functional Sets
by Dave Herman (dherman at ccs dot neu dot edu)
This library provides two implementations of functional sets backed by hash tables: one comparing elements for equality using equal?, the other using eq?.
The data structures of this library are immutable. They are implemented on top of Racket’s immutable hash tables, and therefore should have O(1) running time for extending and O(log n) running time for lookup.
This library was inspired by SRFI-1 and GHC’s Data.Set library.
### 1Getting started
This module provides two libraries, one for equal?-based sets, and one for eq?-based sets.
Examples:
> (define heroes (list->seteq '(rocky bullwinkle)))
> (define villains (list->seteq '(boris natasha)))
> (for ([x (seteq-union heroes villains)]) (printf "~a~n" x))
boris rocky natasha bullwinkle
### 2Sets using equal?
(require data/set) package: set
procedure(set? x) → boolean? x : any
Determines whether x is a set.
procedure(list->set ls) → set? ls : list?
Produces a set containing the elements in ls. If ls contains duplicates (as determined by equal?), the resulting set contains only the rightmost of the duplicate elements.
procedure(set->list s) → list? s : set?
Produces a list containing the elements in s. No guarantees are made about the order of the elements in the list.
value
An empty set.
procedure s : set?
Determines whether s is empty.
procedure(set-intersection sets ...+) → set? sets : set?
procedure(set-intersections sets) → set? sets : (nelistof set?)
Produces the set intersection of sets.
procedure(set-difference sets ...+) → set? sets : set?
procedure(set-differences sets) → set? sets : (nelistof set?)
Produces the left-associative set difference of sets.
procedure(set-union sets ...) → set? sets : set?
procedure(set-unions sets?) → set? sets? : (listof set?)
Produces the set union of sets.
procedure(set-xor sets ...+) → set? sets : set?
procedure(set-xors sets) → set? sets : (nelistof set?)
Produces the exclusive union of sets. This operation is associative and extends to the n-ary case by producing a set of elements that appear in an odd number of sets.
procedure
(set-partition sets ...+)
set? set?
sets : set?
procedure
(set-partitions sets)
set? set?
sets : (nelistof set?)
Equivalent to
(values (set-differences sets) (set-intersection (car sets) (unions (cdr sets))))
but implemented more efficiently.
Note that this is not necessarily the same thing as
(values (set-differences sets) (set-intersections sets)) ; not the same thing!
procedure(set-adjoin s elts ...) → set? s : set? elts : any
Produces a set containing the elements of s and elts.
procedure(set-add x s) → set? x : any s : set?
Produces a set containing x and the elements of s.
procedure s : set? x : any
Determines whether the set s contains the element x.
procedure s : set?
Produces the number of elements in s.
syntax(for/set (for-clause ...) body ...+)
syntax(for*/set (for-clause ...) body ...+)
Like for/list and for*/list, respectively, but the result is a set. The expressions in the body forms must produce a single value, which is included in the resulting set.
procedure(in-set s) → sequence? s : set?
Produces a sequence that iterates over the elements of s.
Sets themselves have the prop:sequence property and can therefore be used as sequences.
procedure(set=? s1 s2) → boolean? s1 : set? s2 : set?
Determines whether s1 and s2 contain exactly the same elements, using equal? to compare elements.
procedure(subset? s1 s2) → boolean? s1 : set? s2 : set?
Determines whether all elements of s1 are contained in s2 (i.e., whether s1 is an improper subset of s2), using equal? to compare elements.
### 3Sets using eq?
(require data/seteq) package: set
procedure(seteq? x) → boolean? x : any
Determines whether x is a set.
procedure(list->seteq ls) → seteq? ls : list?
Produces a set containing the elements in ls. If ls contains duplicates (as determined by eq?), the resulting set contains only the rightmost of the duplicate elements.
procedure(seteq->list s) → list? s : seteq?
Produces a list containing the elements in s. No guarantees are made about the order of the elements in the list.
value
An empty set.
procedure s : seteq?
Determines whether s is empty.
procedure(seteq-intersection sets ...+) → seteq? sets : seteq?
procedure(seteq-intersections sets) → seteq? sets : (nelistof seteq?)
Produces the set intersection of sets.
procedure(seteq-difference sets ...+) → seteq? sets : seteq?
procedure(seteq-differences sets) → seteq? sets : (nelistof seteq?)
Produces the left-associative set difference of sets.
procedure(seteq-union sets ...) → seteq? sets : seteq?
procedure(seteq-unions sets?) → seteq? sets? : (listof seteq?)
Produces the set union of sets.
procedure(seteq-xor sets ...+) → seteq? sets : seteq?
procedure(seteq-xors sets) → seteq? sets : (nelistof seteq?)
Produces the exclusive union of sets. This operation is associative and extends to the n-ary case by producing a set of elements that appear in an odd number of sets.
procedure
(seteq-partition sets ...+)
seteq? seteq?
sets : seteq?
procedure
(seteq-partitions sets)
seteq? seteq?
sets : (nelistof seteq?)
Equivalent to
(values (seteq-differences sets) (seteq-intersection (car sets) (unions (cdr sets))))
but implemented more efficiently.
Note that this is not necessarily the same thing as
(values (seteq-differences sets) (seteq-intersections sets)) ; not the same thing!
procedure(seteq-adjoin s elts ...) → seteq? s : seteq? elts : any
Produces a set containing the elements of s and elts.
procedure(seteq-add x s) → seteq? x : any s : seteq?
Produces a set containing x and the elements of s.
procedure s : seteq? x : any
Determines whether the set s contains the element x.
procedure s : seteq?
Produces the number of elements in s.
syntax(for/seteq (for-clause ...) body ...+)
syntax(for*/seteq (for-clause ...) body ...+)
Like for/list and for*/list, respectively, but the result is a set. The expressions in the body forms must produce a single value, which is included in the resulting set.
procedure s : seteq?
Produces a sequence that iterates over the elements of s.
Sets themselves have the prop:sequence property and can therefore be used as sequences.
procedure(seteq=? s1 s2) → boolean? s1 : seteq? s2 : seteq?
Determines whether s1 and s2 contain exactly the same elements, using eq? to compare elements.
procedure(subseteq? s1 s2) → boolean? s1 : seteq? s2 : seteq?
Determines whether all elements of s1 are contained in s2 (i.e., whether s1 is an improper subset of s2), using eq? to compare elements. | 1,922 | 6,682 | {"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} | 2.65625 | 3 | CC-MAIN-2018-13 | latest | en | 0.435326 |
https://www.allinterview.com/showanswers/57199/all-tickets-2-music-concerts-x-y-either-purchased-away-ratio-1-total-number-perc.html | 1,618,442,497,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038078900.34/warc/CC-MAIN-20210414215842-20210415005842-00011.warc.gz | 724,310,450 | 8,738 | Follow Our FB Page << CircleMedia.in >> for Daily Laughter. We Post Funny, Viral, Comedy Videos, Memes, Vines...
All of the tickets for 2 music concerts, X and Y, were
either purchased or given away, and the ratio of X tickets
to Y was 2 to 1. Of the total number of X tickets and Y
tickets, what percentage was purchased?
1) The total number of X tickets and Y tickets, is 240.
2) Of the X tickets, exactly 60% were purchased, and of the
Y tickets, exactly 80% were purchased.
a) if statement (1) ALONE is sufficient, but statement (2)
alone is not sufficient to answer the question.
b) if statement (2) ALONE is sufficient, but statement (1)
alone is not sufficient to answer the question.
c) if BOTH statements (1) and (2) TOGETHER are sufficient to
sufficient.
d) if EACH statement ALONE is sufficient to answer the
e) if statements (1) and (2) TOGETHER are NOT sufficient to
the problem are needed.
Answers were Sorted based on User's Feedback
All of the tickets for 2 music concerts, X and Y, were either purchased or given away, and the rati..
X:Y=2:1
Total ratio 2+1=3
(.6*2+.8*1)/(2+1)
=2/3 =67% tickets puchased
Is This Answer Correct ? 27 Yes 6 No
All of the tickets for 2 music concerts, X and Y, were either purchased or given away, and the rati..
d) both the statements give answer individually
Is This Answer Correct ? 5 Yes 7 No
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Three men goes to a hotel to stay ,the clerk says \$30 per room/day so all the three plans to stay in one room so each pays \$10.After some time the clerk realises that he made a mistake of collecting \$30 but the room cost only \$25,there fore he decides to return \$5 to them so he calls the room boy and gives him \$5 asking him to return. The room boy keeps \$2 with him and he returns only \$3(\$1 for each). Now Totally all have paid \$9 each(\$27)+room boy \$2 which is equal to \$27.where did \$1 go,who has made the mistake?
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sir am writing aso(assistant statistical officer) exam first time, so please send me previous papers, and syllabus. my optional subject is statistics
A sphere of 10.5 cm radius is melted. It is then casted into a cuboid of maximum volume. What will be the total surface area of such cuboid?
1) 89 94 47 24,...82 41 21 65 70 35 18 53 58 29 15 2) 4 18...100 180 294 448 3) 2 12 36 80 150...392 4) 7 36 125...243
a merchant surveyed 400 ppl.180 learned abt the sale of radio,190 from TV,190 from newspaper,80 from radio and TV,90 from radio and newspapers,50 from TV and newspapers,30 from all the three sources.draw a venn diagram n explain how many learnt a) newspapers or radio, but not from both? b) only from newspapaers? c) from radio or television but not the newspaper? d) from radio and television but not the newspaper? e) from atleast two of the three media?
Categories | 1,090 | 4,150 | {"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} | 3.109375 | 3 | CC-MAIN-2021-17 | latest | en | 0.940826 |
http://www.markedbyteachers.com/gcse/maths/maths-courseowrk-open-box.html | 1,529,452,543,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863259.12/warc/CC-MAIN-20180619232009-20180620012009-00263.warc.gz | 453,559,176 | 19,128 | • Join over 1.2 million students every month
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• Level: GCSE
• Subject: Maths
• Word count: 1918
# Maths Courseowrk - Open Box
Extracts from this document...
Introduction
Mathematics GCSE coursework The open box problem An open box is made from a sheet of card Identical squares cut off the four corners of the card The card is then folded along lines to make an open box. The main aim of the activity is to determine the size of the square cut which makes the volume of the box as large as possible for any given square sheet of card. Part 1 I have drawn tables for my squares. I have tested four squares; 5cm square, a 10cm square, a 20cm square, a 40cm square. I have decided against trial and improvement. As this method can is time consuming so I have used gone up in 0.5cm each time for the size of x (the cut out size) then calculated the size of v (volume). In one table of values I have gone to 5 decimal places to prove my relationship is correct. The volume also depends on the size of a (the length of the square before the cut out.) which is represented by the letter l. After I have drawn the tables I will analyze the graphs. I will put the graphs under the table so on the next few pages there will be a table of values showing the different volumes depending on the cut out of the square then a graph giving us a clear picture of the maximum value. ...read more.
Middle
For my graphs below the volume is shown along the y-axis and the cut out size along the x-axis hence the equation of the line is y=x(l-2x)2. The value of a depended on the square I was using so for the first table the value for l was 5. So to get a graph I only needed to enter the l value. I had to use y= and not v= as stated in the original formula as in autograph you could only use x= or y= and x was already in my formula so I had to use y= On the next page is the table of values for when the l=5 X L v=x(5-2x) 2 0.5 5 8 1 5 9 1.5 5 6 2 5 2 0.75 5 9.1875 0.8 5 9.248 0.85 5 9.2565 0.9 5 9.216 0.95 5 9.1295 0.831 5 9.259204764 0.832 5 9.259241472 0.833 5 9.259258148 0.834 5 9.259254816 0.835 5 9.2592315 0.836 5 9.259188224 0.837 5 9.259125012 0.838 5 9.259041888 0.839 5 9.258938876 0.8331 5 9.259258715 0.8332 5 9.259259081 0.8333 5 9.259259248 0.8334 5 9.259259215 0.8335 5 9.259258982 0.8336 5 9.259258548 0.8337 5 9.259257915 0.8338 5 9.259257082 0.8339 5 9.259256049 0.83331 5 9.259259254 0.83332 5 9.259259257 0.83333 5 9.259259259 0.83334 5 9.259259259 0.83335 5 9.259259256 0.83336 5 9.259259252 0.83337 5 9.259259246 0.83338 5 9.259259237 0.83339 5 9.259259227 This table of value is for a square with sides of 5cm in length. ...read more.
Conclusion
So when X=0.83333 the length of the box is 5. As you can see there is an obvious relationship between the L and X values as when L is divided by X the answer comes to roughly 6. I have discovered that the relationship for the highest value of an open box is X=L/6. I gained two values where L/2 and L/6.I released it could not be L/2 when comparing this with my graphs turning points. The relationship is cemented by the writing on the table on the previous page as well. Here is my working to show how I mathematical discovered this relationship. v=y(l-2y)2 v=y(l-2y)(l-2y) v=(yl-2y2)(l-2y) v=yl2 - 2ly2 - 2ly2 + 4y3 v=4x3 -4y2l + yl2 I then differentiated the x values and this is my results: dy/dx = 12y2 - 8yl + l2 I then made a quadratic equation out of this: (6y - l)(2y-l) = 0 This gave me the following answers: 6y - l = 0 or 2y - l = 0 6y = l 2y = l y = l/6 y = l/2 y=l/6 I will prove this now by putting the formula to use. With l=6 which is my first tested table and in which I went to 5 decimal places. Y=5/6 Y=0.833333333 recurring This proves that the relationship for the formula is correct. Graphs ...read more.
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# Related GCSE Open Box Problem essays
1. ## Maths Coursework
3 star(s)
3.5 11 11 423.5 3.6 10.8 10.8 419.904 3.4 11.2 11.2 426.496 3.3 11.4 11.4 428.868 3.2 11.6 11.6 430.592 3.1* 11.8 11.8 431.644 3.05* 11.9 11.9 431.9105 3* 12 12 432 The cut out square which gave the biggest volume was 3cm by 3cm.
2. ## Investigation: The open box problem.
�x2.12 V = (12-(2x2.13)) �x2.13 V = (12-(2x2.14)) �x2.14 V = 127.661312 V = 127.603188 V = 127.540576 X = 2.15 V = (12-(2x2.15)) �x2.15 V = 127.4735 This means that the value X = 2 gives the largest volume.
1. ## Maximum box investigation
In the table below I have showed the results I got when I investigated on the corner square length being between 2 and 3 cm. Length of the side of the corner square Length of the box Width of the box Height of the box Volume of the box (cm�)
2. ## Open Box Problem
8.339 33.322 33.322 8.339 9259.256049 8.34 33.32 33.32 8.34 9259.254816 8.35 33.3 33.3 8.35 9259.2315 8.36 33.28 33.28 8.36 9259.188224 8.37 33.26 33.26 8.37 9259.125012 8.38 33.24 33.24 8.38 9259.041888 8.39 33.22 33.22 8.39 9258.938876 8.4 33.2 33.2 8.4 9258.816 8.5 33 33 8.5 9256.5 8.9 32.2 32.2 8.9 9227.876
1. ## Tbe Open Box Problem
1.6 128.384 1.7 11.6 6.6 1.7 130.152 1.8 11.4 6.4 1.8 131.328 1.9 11.2 6.2 1.9 131.936 2 11 6 2 132 2.1 10.8 5.8 2.1 131.544 The highest is between 1.9 and 2 so I will go into 2 decimal places.
2. ## The Open Box Problem
volume (cm�) 1 64 1.5 67.5 2 56 2.5 32.5 3 0 I then homed in between 1 and 2cm for 'x': size of cut out 'x' (cm) volume (cm�) 1.1 66.044 1.2 67.392 1.3 68.068 1.4 68.096 1.5 67.5 1.6 66.304 1.7 64.532 1.8 62.208 1.9 59.356 2 56
1. ## THE OPEN BOX PROBLEM
2.50 250.000 2.51 9.98 9.98 2.51 249.997 As you can see from the table above the proportion of the box that needs to be cut away to obtain the maximum cut out size is 1/6 as I found out the maximum cut-out size is 2.5cm.
2. ## The open box problem
or 1.05 and the computer was rounding up/down so I limited my sample size again. But still 1.00 cm cut-out gives us the biggest volume for our open topped box. Algebra Now we've got our first set of results and conclusive evidence to say that 1 cm cut-out to a
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• Ideas and feedback to | 2,519 | 6,732 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2018-26 | latest | en | 0.930079 |
https://t5k.org/curios/page.php?curio_id=797 | 1,722,966,948,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640497907.29/warc/CC-MAIN-20240806161854-20240806191854-00519.warc.gz | 446,070,739 | 3,448 | # 73
This number is a prime.
Single Curio View: (Seek other curios for this number)
There are exactly 73 primes, beginning with the prime 1093 and ending with the prime 1613, where 10932 + 10972 + ... + 16132 = 117072. This is the first instance of a prime number of primes comprising the left member of such an equation. [Haga] | 92 | 332 | {"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} | 2.671875 | 3 | CC-MAIN-2024-33 | latest | en | 0.891295 |
http://www.jiskha.com/display.cgi?id=1202626292 | 1,495,853,909,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608765.79/warc/CC-MAIN-20170527021224-20170527041224-00364.warc.gz | 675,550,634 | 3,707 | # calculus
posted by on .
1. integral -oo, oo [(2x)/(x^2+1)^2] dx
2. integral 0, pi/2 cot(theta) d(theta)
(a) state why the integral is improper or involves improper integral
(b) determine whether the integral converges or diverges
converges?
(c) evaluate the integral if it converges
CONFUSE: how would I know the integral converges or diverges without
evaluating the integral? I am not sure if I evaluated the integral correctly.
(a) infinite limit of integration
(b) diverges?
lim -1/(x^2+1) -oo, b = oo?
b->oo
(a) integrand has an infinity discontinuity at x=0
(b) diverges?
lim cos(theta)/sin(theta) d(theta)
b->oo
= lim ln(sin(theta)) d(theta) b, pi/2
b->oo
=-oo? | 203 | 676 | {"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} | 3.15625 | 3 | CC-MAIN-2017-22 | latest | en | 0.657469 |
http://archive.org/stream/HistoryOfTheTheoryOfNumbersI/TXT/00000352.txt | 1,498,693,261,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323807.79/warc/CC-MAIN-20170628222452-20170629002452-00504.warc.gz | 30,567,563 | 7,871 | # Full text of "History Of The Theory Of Numbers - I"
## See other formats
```344 HlSTOEY OF THE THEORY OF NUMBERS. [CHAP. XII
E. Nannei53 employed r1=o.1 — a0x, r2 = a2—r1x,. . . (:c<10). Then, if rn=0, N = lQnan+ . • .+10a!+a0is divisible by lOz+1 and the quotient has the digits rn_l3 rn_2, . . . , r1; a0. The cases x = 1, 2 are discussed and several tests for 7 deduced. For a; = 1/3, we conclude that, if rn=0, N is divisible by 13 and the digits of the quotient are rn_!/3, . . . , ^/S, a0/3.
A. Chiari54 employed D'Alembert's5 method for 10+&, 6 = 3, 7, 9.
G. Bruzzone55 noted that, to find the remainder R when N is divided by an integer x of r digits, we may choose y such that x+y^W, form the groups of r digits counting from the right of N, and multiply the successive groups (from the right) by 1, y, y2, . . . or by their residues modulo x] then R equals the remainder on dividing the sum of the products by x. If we choose x — y = 10r, we must change alternate signs before adding. For practical use,
Fr. Schuh56 gave three methods to determine the residue of large numbers for a given modulus.
Stuyvaert57 let a, 6, ... be the successive sets of n digits of TV to the base 5, so that N = a+bBn+cB2n+ .... Then N is divisible by a factor D of Bn =?Rn if and only if a*=bRn+cR2n± ... is divisible by D. For E = l, £ = 10, n = l, 2, . . ., we obtain tests for divisors of 9, 99, 11, 101, etc. A divisor, prime to B, of mB+l divides N — a-\-bB if and only if it divides b — ma.
FURTHER PAPERS GIVING TESTS FOR A GIVEN DIVISOR d.
J. R. Young and Mason for d = 7, 13 [Pascal4], Ladies' Diary, 1831, 34-5, Quest.
1512.
P. Gorini [Pascal4], Annali di Fis., Chim. Mat., (ed., Majocchi), 1,1841, 237. A. Pinaud for d = 7, 13, M&n. Acad. Sc. Toulouse, 1, 1844, 341, 347. *Dietz and Vincenot, M&n. Acad. Metz, 33, 1851-2, 37. Anonymous writer for d = 9, 11, Jour, fur Math., 50, 1855, 187-8. *H. Wronski, Principes de la phil. des math. Cf. de Montferrier, Encyclopedic
math., 2, 1856, p. 95.
0. Terquem for d£l9, 23, 37, 101, Nouv. Ann. Math., 14, 1855, 118-120. A. P. Reyer for d = 7, Archiv Math. Phys., 25, 1855, 176-196. C. F. Lindman for d = 7, 13, ibid., 26, 1856, 467-470. P. Buttel for d = 7, 9, 11, 17, 19, ibid., 241-266.
De Lapparent [Herter14], Mem. soc. unp. sc. nat. Cherbourg, 4, 1856, 235-258. Karwowski [Pascal4], Ueber die Theilbarkeit . . ., II, Progr., Lissa, 1856. *D. van Langeraad, Kenmerken van deelbarheid der geheele getallen, Schoonho-
ven, 1857.
Flohr, Ueber Theilbarkeit und Reste der Zahlen, Progr., Berlin, 1858. V. Bouniakowsky for d = 37, 989, Nouv. Ann. Math., 18, 1859, 168. Elefanti for d = 7-13, Proc. Roy. Soc. London, 10, 1859-60, 208. A. Niegemann for d = 10m-n-fa, Archiv Math. Phys., 38, 1862, 384-8. J. A. Grunert for d = 7, 11, 13, ibid., 42, 1864, 478-482. V. A. Lebesgue, Tables diverses pour la decomposition des nombres, Paris, 1864,
p. 13. _
"II Pitagora, Palermo, 13, 1906-7, 54-9.
»Ibid., 14, 1907-8, 35-7.
"/&«*., 15, 1908-9, 119-123.
"Supplem. De Vriend der Wiskunde, 24, 1912, 89-103.``` | 1,247 | 3,091 | {"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} | 3.3125 | 3 | CC-MAIN-2017-26 | latest | en | 0.803583 |
https://yutsumura.com/union-of-subspaces-is-a-subspace-if-and-only-if-one-is-included-in-another/?wpfpaction=add&postid=2961 | 1,582,556,505,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145960.92/warc/CC-MAIN-20200224132646-20200224162646-00379.warc.gz | 950,273,357 | 27,396 | # Union of Subspaces is a Subspace if and only if One is Included in Another
## Problem 427
Let $W_1, W_2$ be subspaces of a vector space $V$. Then prove that $W_1 \cup W_2$ is a subspace of $V$ if and only if $W_1 \subset W_2$ or $W_2 \subset W_1$.
## Proof.
### If $W_1 \cup W_2$ is a subspace, then $W_1 \subset W_2$ or $W_2 \subset W_1$.
$(\implies)$ Suppose that the union $W_1\cup W_2$ is a subspace of $V$.
Seeking a contradiction, assume that $W_1 \not \subset W_2$ and $W_2 \not \subset W_1$.
This means that there are elements
$x\in W_1\setminus W_2 \text{ and } y \in W_2 \setminus W_1.$
Since $W_1 \cup W_2$ is a subspace, it is closed under addition. Thus, we have $x+y\in W_1 \cup W_2$.
It follows that we have either
$x+y\in W_1 \text{ or } x+y\in W_2.$ Suppose that $x+y\in W_1$. Then we write \begin{align*}
y=(x+y)-x.
\end{align*}
Since both $x+y$ and $x$ are elements of the subspace $W_1$, their difference $y=(x+y)-x$ is also in $W_1$. However, this contradicts the choice of $y \in W_2 \setminus W_1$.
Similarly, when $x+y\in W_2$, then we have
$x=(x+y)-y\in W_2,$ and this contradicts the choice of $x \in W_1 \setminus W_2$.
In either case, we have reached a contradiction.
Therefore, we have either $W_1 \subset W_2$ or $W_2 \subset W_1$.
### If $W_1 \subset W_2$ or $W_2 \subset W_1$, then $W_1 \cup W_2$ is a subspace.
$(\impliedby)$ If we have $W_1 \subset W_2$, then it yields that $W_1 \cup W_2=W_2$ and it is a subspace of $V$.
Similarly, if $W_2 \subset W_1$, then we have $W_1\cup W_2=W_2$ and it is a subspace of $V$.
In either case, the union $W_1 \cup W_2$ is a subspace.
### More from my site
• The Union of Two Subspaces is Not a Subspace in a Vector Space Let $U$ and $V$ be subspaces of the vector space $\R^n$. If neither $U$ nor $V$ is a subset of the other, then prove that the union $U \cup V$ is not a subspace of $\R^n$. Proof. Since $U$ is not contained in $V$, there exists a vector $\mathbf{u}\in U$ but […]
• Union of Two Subgroups is Not a Group Let $G$ be a group and let $H_1, H_2$ be subgroups of $G$ such that $H_1 \not \subset H_2$ and $H_2 \not \subset H_1$. (a) Prove that the union $H_1 \cup H_2$ is never a subgroup in $G$. (b) Prove that a group cannot be written as the union of two proper […]
• The Sum of Subspaces is a Subspace of a Vector Space Let $V$ be a vector space over a field $K$. If $W_1$ and $W_2$ are subspaces of $V$, then prove that the subset $W_1+W_2:=\{\mathbf{x}+\mathbf{y} \mid \mathbf{x}\in W_1, \mathbf{y}\in W_2\}$ is a subspace of the vector space $V$. Proof. We prove the […]
• True or False. The Intersection of Bases is a Basis of the Intersection of Subspaces Determine whether the following is true or false. If it is true, then give a proof. If it is false, then give a counterexample. Let $W_1$ and $W_2$ be subspaces of the vector space $\R^n$. If $B_1$ and $B_2$ are bases for $W_1$ and $W_2$, respectively, then $B_1\cap B_2$ is a […]
• Two Subspaces Intersecting Trivially, and the Direct Sum of Vector Spaces. Let $V$ and $W$ be subspaces of $\R^n$ such that $V \cap W =\{\mathbf{0}\}$ and $\dim(V)+\dim(W)=n$. (a) If $\mathbf{v}+\mathbf{w}=\mathbf{0}$, where $\mathbf{v}\in V$ and $\mathbf{w}\in W$, then show that $\mathbf{v}=\mathbf{0}$ and $\mathbf{w}=\mathbf{0}$. (b) If $B_1$ is a […]
• Determine the Values of $a$ so that $W_a$ is a Subspace For what real values of $a$ is the set $W_a = \{ f \in C(\mathbb{R}) \mid f(0) = a \}$ a subspace of the vector space $C(\mathbb{R})$ of all real-valued functions? Solution. The zero element of $C(\mathbb{R})$ is the function $\mathbf{0}$ defined by […]
• The Intersection of Two Subspaces is also a Subspace Let $U$ and $V$ be subspaces of the $n$-dimensional vector space $\R^n$. Prove that the intersection $U\cap V$ is also a subspace of $\R^n$. Definition (Intersection). Recall that the intersection $U\cap V$ is the set of elements that are both elements of $U$ […]
• The Subspace of Linear Combinations whose Sums of Coefficients are zero Let $V$ be a vector space over a scalar field $K$. Let $\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k$ be vectors in $V$ and consider the subset \[W=\{a_1\mathbf{v}_1+a_2\mathbf{v}_2+\cdots+ a_k\mathbf{v}_k \mid a_1, a_2, \dots, a_k \in K \text{ and } […]
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##### If $A$ is an Idempotent Matrix, then When $I-kA$ is an Idempotent Matrix?
A square matrix $A$ is called idempotent if $A^2=A$. (a) Suppose $A$ is an $n \times n$ idempotent matrix and...
Close | 1,670 | 4,703 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2020-10 | latest | en | 0.75827 |
https://testbook.com/objective-questions/mcq-on-refrigeration-and-air-conditioning--5eea6a0939140f30f369da58 | 1,702,218,189,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679102469.83/warc/CC-MAIN-20231210123756-20231210153756-00723.warc.gz | 611,225,256 | 60,912 | # Refrigeration and Air Conditioning MCQ Quiz - Objective Question with Answer for Refrigeration and Air Conditioning - Download Free PDF
Last updated on Sep 10, 2023
## Latest Refrigeration and Air Conditioning MCQ Objective Questions
#### Refrigeration and Air Conditioning Question 1:
The specific enthalpy of refrigerant (in kJ/kg) in an ideal vapour compression refrigeration cycle at the following states is given.
Inlet of condenser : 312
Exit of condenser : 124
Exit of evaporator : 268
The COP of this cycle is:
1. 3.27
2. 2.27
3. 2.75
4. 3.75
Option 1 : 3.27
#### Refrigeration and Air Conditioning Question 1 Detailed Solution
Explanation:
The COP of the ideal Vapour Compression Refrigeration System (VCRS):
It is the ratio of refrigerating effect (R.E) to the work done by the compressor (Wc).
$$COP=\frac{Refrigerating\;effect}{Work\;done\;by\;compressor}\Rightarrow \frac{R.E}{W_c}$$
Process 1-2: Isentropic work done by Compressor
Process 2-3: Heat removal at constant pressure by Condensor
Process 3-4: Isenthalpic expansion by Throttling valve
Process 4-1: Heat addition at constant pressure by Evaporator.
Calculation:
Given:
h1 = 268 kJ/kg, h2 = 312 kJ/kg. h3 = h4 = 124 kJ/kg
$$COP = \frac{{Refrigeration\ effect}}{{Work\ done}} = \frac{{{h_1}\;- \;{h_4}}}{{{h_2}\;-\;{h_1}}}$$
$$\therefore\frac{{268\;- \;124}}{{312\;-\;268}} \Rightarrow 3.27$$
#### Refrigeration and Air Conditioning Question 2:
The higher temperature of the refrigerant in a refrigeration system operating on the reversed Carnot cycle is 35°C and the lower temperature is -15°C. Determine coefficient of performance.
1. 3.7
2. 2.3
3. 5.5
4. 5.16
Option 4 : 5.16
#### Refrigeration and Air Conditioning Question 2 Detailed Solution
Concept:
Refrigerator:
• A refrigerator is a device that works on a reverse Carnot cycle and extracts heat from a lower-temperature body to keep the temperature of the body lower than the surrounding temperature and by taking work input it transfers heat to the higher-temperature body or surrounding.
Refrigerating Effect R.E. = Q1, Work input = Q2 - Q1
$$Coefficient~of~Performance~(COP) = \frac{Refrigerating~Effect}{Work~Input} = \frac{Q_1}{W}$$
For a reversible engine:
$$COP_{Ref}=\frac{T_L}{T_H-T_L}$$
Calculation:
Given:
TH = 308 K, TL = 258 K
$$COP_{Ref}=\frac{T_L}{T_H-T_L}$$
$$COP_{Ref}=\frac{258}{308-258}$$
COP = 5.16
#### Refrigeration and Air Conditioning Question 3:
Which of the following is NOT a component of the Bell-Coleman cycle?
1. Condenser
2. Compressor
3. Cooler
4. Expander
Option 1 : Condenser
#### Refrigeration and Air Conditioning Question 3 Detailed Solution
Explanation:
Air Refrigeration System and Bell-Coleman Cycle or Reversed Brayton Cycle:
• Bell Coleman cycle is also known as Reversed Brayton cycle or Reversed Joule cycle
• The working fluid of the Bell Coleman refrigeration cycle is Air.
• This system of refrigeration is used for Air Craft refrigeration and it has lightweight.
1. Process 1 - 2: isentropic compression
2. Process 2 - 3: constant pressure heat rejection
3. Process 3 - 4: isentropic expansion
4. Process 4 - 1: constant pressure heat absorption
Air Refrigeration System and Bell-Coleman Cycle or Reversed Brayton Cycle:
• In an air refrigeration system, the air is taken into the compressor from the atmosphere and compressed.
• The hot compressed air is cooled in a heat exchanger up to the atmospheric temperature (in ideal conditions).
• The cooled air is then expanded in an expander. The temperature of the air coming out from the expander is below the atmospheric temperature due to isentropic expansion.
• The low-temperature air coming out from the expander enters into the evaporator and absorbs the heat. The cycle is repeated.
#### Refrigeration and Air Conditioning Question 4:
In the ammonia water vapor absorption refrigeration cycle, the absorbent used is:
1. CH12
2. ammonia
3. water
4. steam
Option 3 : water
#### Refrigeration and Air Conditioning Question 4 Detailed Solution
Explanation:
In Vapor Absorption Refrigeration Systems most commonly these two pairs are used.
• Water-Lithium Bromide (H2O – Li-Br) system for above 0°C applications such as air conditioning. Here water is the refrigerant and lithium bromide as the absorbent.
• Ammonia-Water (NH3 – H2O) system for refrigeration applications with ammonia as refrigerant and water as absorbent.
#### Refrigeration and Air Conditioning Question 5:
Sensible heating is required to:
1. vaporise water into steam
2. change the temperature of a liquid or vapour
3. convert water into steam and super heat it
4. measure dew point temperature
Option 2 : change the temperature of a liquid or vapour
#### Refrigeration and Air Conditioning Question 5 Detailed Solution
Explanation:
Latent heat is defined as the heat energy that has to be supplied or removed in order to change the state of the substance without having any effect on its temperature.
• Solid to liquid: Latent heat of fusion (BC)
• Liquid to solid: Latent heat of solidification
• Liquid to vapour: Latent heat of evaporation/vaporization (DE)
• Vapour to liquid: Latent heat of condensation
Sensible heat is the energy required to change the temperature of a substance with no phase change.
## Top Refrigeration and Air Conditioning MCQ Objective Questions
#### Refrigeration and Air Conditioning Question 6
A Carnot heat pump works between 27° C and 327°C. What will be its COP?
1. 0.09
2. 1.00
3. 1.09
4. 2.0
Option 4 : 2.0
#### Refrigeration and Air Conditioning Question 6 Detailed Solution
Concept:
$${\rm{COP\;of\;Carnot\;Heat\;pump}} = \frac{{{T_1}}}{{{T_1} - {T_2}}}$$
Calculation:
Given:
T1 = 327° C = 600 K, T2 = 27° C = 300 K
$$\therefore COP = \frac{{{T_1}}}{{{T_1} - {T_2}}}$$
$$\therefore COP = \frac{{600}}{{600 - 300}} = \frac{{600}}{{300}} = 2$$
COP of Carnot heat pump = 2
#### Refrigeration and Air Conditioning Question 7
If the volume of moist air with 50% RH is isothermally reduced to half its original volume then relative humidity of moist air becomes
1. 25%
2. 60%
3. 70%
4. 100%
Option 4 : 100%
#### Refrigeration and Air Conditioning Question 7 Detailed Solution
Concept:
Relative humidity is given by
$$ϕ = \frac{{{P_v}}}{{{P_{vs}}}}$$
where P v is pressure of water vapour and Pvs is pressure of water vapour at saturation point.
For isothermal process:
PV = constant
Calculation:
Given:
ϕ1 = 50%, Pv1
Since volume is reduced to half so the pressure has become twice.
Pv2 = 2Pv1
$$\frac{{{ϕ _2}}}{{{ϕ _1}}} = \frac{{{P_{{v_2}}}}}{{{P_{v1\;}}}}$$
$$\frac{{{ϕ _2}}}{{{ϕ _1}}} = 2$$
ϕ2 = 2ϕ1
ϕ2 = 2 × 50 ⇒ 100%.
Additional InformationHere, the initial condition is given as Moist air, but we cannot apply ideal gas to it. Technically the wording is incorrect but here we have to assume it as ideal and solve accordingly.
#### Refrigeration and Air Conditioning Question 8
Domestic refrigerator working on vapour compression cycle uses the following type of expansion device
1. Electrically operated throttling valve
2. Manually operated valve
3. Thermostatic valve
4. Capillary tube
Option 4 : Capillary tube
#### Refrigeration and Air Conditioning Question 8 Detailed Solution
Concept:
• The capillary tube is one of the most commonly used throttling devices in the domestic refrigerators, deep freezers, water coolers, and air conditioners.
• Capillary tubes have very small internal diameters and very long length and they are coiled to several turns so that it would occupy less space (compact).
• They are easy to manufacture, cheap and compact.
Working:
• When the refrigerant leaves the condenser and enters the capillary tube, its high pressure drops down suddenly due to the very small diameter of the capillary tube and the long length gives more friction head and drops pressure further.
• The decrease in pressure leads to cooling of refrigerant and the low-temperature refrigerant can take the heat from the room.
#### Refrigeration and Air Conditioning Question 9
A Carnot engine receiving heat at 400 K has an efficiency of 50 %. What is the COP of a Carnot refrigerator working between the same temperature limits?
1. 4
2. 1
3. 2
4. 3
Option 2 : 1
#### Refrigeration and Air Conditioning Question 9 Detailed Solution
Concept:
$$η_{carnot}=\frac{T_H-T_L}{T_H}=1-\frac{T_L}{T_H}$$
$$(COP)_{carnot}=\frac{T_L}{T_H-T_L}$$
Calculation:
Given:
ηCarnot = 50 % = 0.5, TH = 400 K
$$η_{carnot}=\frac{T_H-T_L}{T_H}=1-\frac{T_L}{T_H}$$
$$\Rightarrow\frac{T_L}{T_H}=0.5=\frac{1}{2}$$
Now,
$$(COP)_{carnot}=\frac{T_L}{T_H-T_L}=\frac{1}{\frac{T_H}{T_L}-1}=\frac{1}{2-1}=1$$
#### Refrigeration and Air Conditioning Question 10
Moist air at 35°C and 100% relative humidity is entering a psychometric device and leaving at 25°C and 100% relative humidity. The name of the device is
1. Humidifier
2. Dehumidifier
3. Sensible heater
4. Sensible cooler
Option 2 : Dehumidifier
#### Refrigeration and Air Conditioning Question 10 Detailed Solution
Concept:
A psychrometric chart is represented as shown in the figure.
Dry Bulb Temperature: Actual temperature of gas or mixture of gases
Wet bulb temperature: Temperature obtained by an accurate thermometer having a wick moistened with distilled water
Dew point temperature: Temperature at which the liquid droplets just appear when the moist air is cooled continuously.
Relative humidity along saturation line is 100%.
Basic Processes in Conditioning of Air:
Sensible heating: Moisture content of air remains constant so specific humidity is constant, temperature increases as it flows over a heating coil
Sensible cooling: Moisture content of air remains constant so specific humidity is constant, but its temperature decreases as it flows over a cooling coil
Cooling and dehumidification: When moist air is cooled below its dew-point by bringing it in contact with a cold surface, some of the water vapour in the air condenses and leaves the air stream as a liquid, as a result, both the temperature and humidity ratio of air decreases.
Heating and Humidification: During winter it is essential to heat and humidify the room air for comfort. It is done by first sensibly heating the air and then adding water vapour to the air stream through steam nozzles, as a result, both the temperature and humidity ratio of air increases.
Cooling & humidification: Air temperature drops and its humidity increases.
Heating and de-humidification: This process can be achieved by using a hygroscopic material, which absorbs the water vapour from the moisture. If this process is thermally isolated, then the enthalpy of air remains constant, as a result, the temperature of air increases.
Calculation:
Inlet (State 1): 35°C and 100% RH
Outlet (State 2): 25°C and 100% RH
From the above figure, relative humidity is same but as the temperature is decreasing and specific humidity is also decreasing. So the process 1-2 is cooling and dehumidification process.
The name of the device is dehumidifier.
#### Refrigeration and Air Conditioning Question 11
A heat pump works on a reversed Carnot cycle. The temperature in the condenser coils is 27° C and that in the evaporator coils is - 23° C. For a work input of 1 kW, how much is the heat pumped?
1. 1 kW
2. 5 kW
3. 6 kW
4. None of the above
Option 3 : 6 kW
#### Refrigeration and Air Conditioning Question 11 Detailed Solution
Concept:
$$CO{P_{HP}} = \frac{{Desired\;Output}}{{Required\;Input}} = \frac{{{Q_1}}}{W} = \frac{{{Q_1}}}{{{Q_1} - {Q_2}}} = \frac{{{T_1}}}{{{T_1} - {T_2}}} = \frac{{{Q_H}}}{{{Q_H} - {Q_L}}}$$
Calculation:
Given:
T1 = 27°C = 300 K, T2 = -23°C = 250 K, W = 1 kW
Now
$$\frac{{{Q_1}}}{1} = \frac{{{300}}}{{{300} - {250}}} \Rightarrow Q = 6 \ kW$$
#### Refrigeration and Air Conditioning Question 12
The commonly used refrigerant in domestic refrigerators is:
1. Ammonia
2. CO2
3. Nitrogen
4. R 134a
Option 4 : R 134a
#### Refrigeration and Air Conditioning Question 12 Detailed Solution
Explanation:
• Domestic refrigerators generally run on the vapor compression cycle. In this cycle, a circulating refrigerant such as R134a enters a compressor as low-pressure vapour at or slightly below the temperature of the refrigerator space.
• Ammonia is generally used in the vapour absorption cycle.
• Nitrogen and CO2 are not used as refrigerants.
The R 134a is an eco-friendly refrigerant.
#### Refrigeration and Air Conditioning Question 13
Which is the desirable physical property of refrigerant?
1. Toxic
2. Explosive
3. Low boiling point
4. High freezing point
Option 3 : Low boiling point
#### Refrigeration and Air Conditioning Question 13 Detailed Solution
Explanation:
The characteristics of a good refrigerant should have the following properties
1. Low boiling point
2. High latent heat value
3. Low freezing point
4. Non-flammable and non-toxic
5. High di-electric strength
6. Not affected by moisture
7. Non-corrosive to metals
8. Mixes well with compressor oil
#### Refrigeration and Air Conditioning Question 14
Which of the following is a secondary refrigerant, when used above 0°C?
1. Sodium chloride
2. Glysols
3. Brines
4. Water
Option 4 : Water
#### Refrigeration and Air Conditioning Question 14 Detailed Solution
Explanation:
There are two types of refrigerants
1. Primary Refrigerant: Primary refrigerants are substances that undergo a cyclic process and produce lower temperatures. There is a latent heat transformation for the refrigerants. For e.g. R-11, R-12, R-22, R-134a, R-1150 etc.
2. Secondary Refrigerants: There are working substances that are first cooled by primary refrigerants and then used for cooling at desired places. e.g.H2O, Brine.
• R-11 – Large central air conditioning plant
• R-12- Domestic refrigerator, water cooler etc.
• R-22- Window AC
• NH3- Cold storage or Icing plants
• CO2- Transportation of dry ice
• Air- Air-craft refrigeration system
• Brine- Milk-chilling plants
Mistake Points
Both water and Brine are secondary refrigerants, but here he asks above the 0°C, so water will be the correct answer.
#### Refrigeration and Air Conditioning Question 15
The lowest temperature during the cycle in a vapour compression system occurs after
1. Compression
2. Expansion
3. Condensation
4. Evaporation
Option 2 : Expansion
#### Refrigeration and Air Conditioning Question 15 Detailed Solution
Explanation:
Vapour Compression Refrigeration System:
The four processes in simple VCRS include:
1. Isentropic compression (1-2) in Compressor.
2. Constant Pressure heat removal (2-3) in Condensor.
3. Isenthalpic expansion (3-4) in a Throttling device.
4. Constant pressure heat removal (4-1) in Evaporator.
From the Above T-s diagram, we can conclude that the lowest temperature during the cycle in a vapour compression system occurs after expansion and during the evaporation process(4-1) in the evaporator.
In the Question the exact location of the lowest temperature has not been asked here they have asked after which process the lowest temperature will be obtained. The lowest temperature surely is obtained in the evaporator which comes after the expansion process hence the expansion is the correct answer. | 4,051 | 15,217 | {"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} | 4.15625 | 4 | CC-MAIN-2023-50 | latest | en | 0.719253 |
http://www.gatepaper.in/2014/05/gate-2014-ece-video-solutions-on-analog.html | 1,585,758,443,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370505826.39/warc/CC-MAIN-20200401161832-20200401191832-00273.warc.gz | 237,031,924 | 19,452 | ### GATE 2014 ECE Video Solutions on Analog Circuits (Analog Electronics)
Set – 1 (15th February 2014 (Forenoon))
1. In the figure, assume that the forward voltage drops to the PN diode D1 and Schottky diode D2 are 0.7 volts and 0.3 volts respectively. If ON denotes conducting state of the diode and OFF denotes the non conducting state of the diode, then in the circuit,
a. Both are ON
b. D1 is ON and D2 is OFF
c. Both are OFF
d. D1 is OFF and D2 is ON
2. A good current buffer has
a. Low input impedance and low output impedance
b. Low input impedance and high output impedance
c. high input impedance and low output impedance
d. high input impedance and high output impedance
3. In the ac equivalent circuit shown in the figure, if iin is the input current and RF is very large, then the type of feedback is
a. Voltage – Voltage feedback
b. Voltage – Current feedback
c. Current – Voltage feedback
d. Current – Current feedback
4. In the low pass filter shown in the figure, for a cutoff frequency of 5 KHz, the value of R2 (in KΩ) is ……..
5. A BJT is biased in forward active mode. Assume VBE = 0.7 volts, KT/q = 25 mV and reverse saturation current IS = 10-13 Amp. The Transconductance of the BJT (in mA/volt) is…..
6. In the voltage regulator circuit shown in the figure, the Op-Amp is ideal. The BJT has VBE = 0.7 volts, and β = 100, and the zener voltage is 4.7 volts. For a regulated output of 9 volts, the value of R (in Ω) is ……..
7. In the circuit shown, the Op-Amp has finite input resistance, infinite voltage gain and zero input offset voltage. The output voltage Vout is …….
a. – I2(R1 + R2)
b. I2R2
c. I1R2
d. – I1(R1 + R2)
8. For the amplifier shown in the figure, the BJT parameters are VBE = 0.7 volts, β = 200, and thermal voltage VT = 25 mV. The voltage gain (vo/vi) of the amplifier is ……..
SET – 2 (15th February 2014 (Afternoon))
1. The diode in the circuit shown, if Von = 0.7 volts but is ideal otherwise. If Vi = 5 sin(ωt) volts, the minimum and maximum values of Vo (in volts) are, respectively,
a. -5 and 2.7
b. 2.7 and 5
c. -5 and 3.85
d. 1.3 and 5
2. The feedback topology in the amplifier circuit (the base bias circuit is not shown for simplicity) in the figure is
a. Voltage – Shunt feedback
b. Current – Series feedback
c. Current – Shunt feedback
d. Voltage – Series feedback
3. In the differential amplifier shown in the figure, the magnitude of the common mode and differential mode gains are Acm and Ad, respectively. If the resistance RE is increased, then
a. Acm increases
b. Common mode rejection ratio increases
d. Common mode rejection ratio decreases
4. A cascade connection of two voltage amplifiers A1 and A2 is shown in the figure. The open loop gain Avo, input resistance Rin, and output resistance Ro for A1 and A2 are as follows:
A1 : Avo = 10, Rin = 10 KΩ, Ro = 1 KΩ
A2 : Avo = 5, Rin = 5 KΩ, Ro = 200 Ω
The approximate overall voltage gain Vout / Vin is …………….
SET – 3 (16th February 2014 (Forenoon))
1. In the circuit shown, the PNP transistor has |VBE| = 0.7 volts and β = 50. Assume that RB = 100 KΩ. For Vo to be 5 volts, the value of RC (in KΩ) is ……..
2. The desirable characteristics of a Transconductance amplifier are
a. High input resistance and High output resistance
b. High input resistance and Low output resistance
c. Low input resistance and High output resistance
d. Low input resistance and Low output resistance
3. The figure shows a half wave rectifier. The diode D is ideal. The average steady state current (in Amperes) through the diode is approximately…………
4. Assuming that the Op-Amp in the circuit is ideal, then the output voltage is
5. In the circuit shown, the silicon BJT has β = 50. Assume VBE = 0.7 volts and VCEsat = 0.2 volts. Which one of the following statements is correct?
a. For RC = 1 kΩ, the BJT operates in the saturation region
b. For RC = 3 kΩ, the BJT operates in the saturation region
c. For RC = 20 kΩ, the BJT operates in the cutoff region
d. For RC = 20 kΩ, the BJT operates in the linear region
SET - 4 (16th February 2014 (Afternoon))
1. If the emitter resistance in a common emitter voltage amplifier is not bypassed, it will
a. Reduce both the voltage gain and the input impedance
b. Reduce the voltage gain and increase the input impedance
c. Increase the voltage gain and reduce the input impedance
d. Increase both the voltage gain and the input impedance
2. Consider two BJTs biased at the same collector current with area A1 = 0.2 µm x 0.2 µm and A2 = 300 µm x 300 µm. assuming that all other device parameters are identical, KT/q = 26 mV, the intrinsic carrier concentration is 1 x 1010 cm-3, and q = 1.6 x 10-19C, the difference between the base emitter voltages (in mV) of the two BJTs (i.e. VBE1 – VBE2) is …………….
3. A BJT in a common base configuration is used to amplify a signal received by a 50Ω antenna. Assume KT/q = 25 mV, the value of collector bias current (in mA) required to match the input impedance of the amplifier to the impedance of the antenna is ………..
4. Two silicon diodes, with a forward voltage drop of 0.7 volts, are used in the circuit shown in the figure. The range of input voltage Vi for which the output voltage Vo = Vi, is
a. -0.3 volts < Vi < 1.3 volts
b. -0.3 volts < Vi < 2 volts
c. -1.0 volts < Vi < 2.0 volts
d. -1.7 volts < Vi < 2.7 volts
5. The circuit represents
a. A band pass filter
b. A voltage controlled oscillator
c. An Amplitude modulator
d. A monostable multivibrator
6. Consider the common collector amplifier in the figure (bias circuitry ensures that the transistor operates in forward active region, but has been omitted for simplicity). Let IC be the collector current, VBE be the base emitter voltage and VT be the thermal voltage. Also, gm and ro are the small signal Transconductance and output resistance of the transistor, respectively. Which one of the following conditions ensures a nearly constant small signal voltage gain for a wide range of values of RE?
7. For the common collector amplifier shown in the figure, the BJT has high β, negligible VCEsat and VBE = 0.7 volts. The maximum undistorted peak to peak output voltage Vo (in volts) is ……… | 1,738 | 6,159 | {"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} | 2.984375 | 3 | CC-MAIN-2020-16 | latest | en | 0.797011 |
http://docs.intersystems.com/latest/csp/docbook/DocBook.UI.Page.cls?KEY=RVBS_fsubs | 1,503,557,284,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886133042.90/warc/CC-MAIN-20170824062820-20170824082820-00215.warc.gz | 111,791,690 | 3,933 | Caché MultiValue Basic Reference
SUBS
Server:docs2 Instance:LATEST User:UnknownUser
[ Home ] > [ MultiValue Features of Caché ] > [ Caché MultiValue Basic Reference ] > [ Caché MultiValue Basic Functions ] > [ SUBS ] Go to: Description Examples See Also Search:
Subtracts the values of corresponding elements in two dynamic arrays.
Synopsis
`SUBS(dynarray1,dynarray2)`
Arguments
dynarray1 The minuend. An expression that resolves to a dynamic array of numeric values. If a dynamic array element contains a non-numeric value, SUBS treats this value as 0 (zero). dynarray2 The subtrahend. An expression that resolves to a dynamic array of numeric values. If a dynamic array element contains a non-numeric value, SUBS treats this value as 0 (zero).
Description
The SUBS function subtracts the value of each element in dynarray2 from the corresponding element in dynarray1. It then returns a dynamic array containing the results of these subtractions. If an element value is missing, or is the null string or a non-numeric value, SUBS parses its value as 0 (zero).
If the two dynamic arrays have different numbers of elements, the returned dynamic array has the number of elements of the longer dynamic array. By default, the shorter dynamic array is padded with 0 value elements for the purpose of the arithmetic operation. You can also use the REUSE function to define behavior when specifying two dynamic arrays with different numbers of elements.
You can use the ADDS (addition), MULS (multiplication), DIVS or DIVSZ (division), MODS (modulo division), and PWRS (exponentiation) functions to perform other arithmetic operations on the corresponding elements of two dynamic arrays.
Examples
The following example uses the SUBS function to subtract the elements of two dynamic arrays:
```a=11:@VM:22:@VM:33:@VM:44
b=10:@VM:9:@VM:8:@VM:7
PRINT SUBS(a,b)
! returns 1ý13ý25ý37```
The following example subtracts elements of dynamic arrays of differing lengths:
```a=11:@VM:22:@VM:33:@VM:44
b=2:@VM:2:@VM:2:@VM:2:@VM:2:@VM:2
PRINT SUBS(a,b)
! returns 9ý20ý31ý42ý-2ý-2``` | 540 | 2,078 | {"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} | 3.03125 | 3 | CC-MAIN-2017-34 | longest | en | 0.689637 |
http://www.harrisgeospatial.com/docs/ENVIIterativeTrainer.html | 1,563,418,604,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525483.64/warc/CC-MAIN-20190718022001-20190718044001-00271.warc.gz | 215,295,573 | 23,562 | Welcome to the L3 Harris Geospatial documentation center. Here you will find reference guides and help documents.
> Docs Center > ENVI API > Classification Framework > ENVIIterativeTrainer
### ENVIIterativeTrainer
ENVIIterativeTrainer
This function uses an iterative loop to train a classifier that knows how to update its own weights; for example, Support Vector Machine (SVM). The trainer iteraties until the classifier's change in loss falls below a specified convergence criterion or it reaches a specified maximum number of iterations. The convergence criterion and maximum iterations needed to effectively train a classifier (not just each classifier type, but each classifier instance) can vary considerably. They often take some experimentation to determine.
The following diagram shows a typical workflow where the iterative trainer is used:
## Example
See the following topics for code examples:
## Syntax
Result = ENVIIterativeTrainer([, Keywords=value])
## Return Value
This function returns a reference to an ENVIIterativeTrainer object.
## Keywords
### CONVERGENCE_CRITERION (optional)
Set this keyword to a convergence criterion. Ideally, with each iteration, the change in loss value decreases until it meets a specified threshold, below which iterations can stop. This threshold is the convergence criterion. If you do not set this keyword, the default value is 0.001.
### MAXIMUM_ITERATIONS (optional)
Set this keyword to the maximum number of iterations to perform if the convergence criterion is not met. If you do not set this keyword, the default value is 100.
### PROPERTIES
Set this keyword to an IDL dictionary containing the following properties:
• Convergence_Criterion: Required. Ideally, with each iteration, the change in loss value decreases until it meets a specified threshold, below which iterations can stop. This threshold is the convergence criterion. A suggested starting value is 0.0001.
• Maximum_Iterations: Required. Specify the maximum number of iterations to perform if the convergence criterion is not met. A suggested starting value is 100.
Here is an example of a dictionary:
`Properties = Dictionary()`
`Properties.Convergence_Criterion = 0.001`
`Properties.Maximum_Iterations = 100`
### ERROR (optional)
Set this keyword to a named variable that will contain any error message issued during execution of this routine. If no error occurs, the ERROR variable will be set to a null string (''). If an error occurs and the routine is a function, then the function result will be undefined.
When this keyword is not set and an error occurs, ENVI returns to the caller and execution halts. In this case, the error message is contained within !ERROR_STATE and can be caught using IDL's CATCH routine. See IDL Help for more information on !ERROR_STATE and CATCH.
See Manage Errors for more information on error handling in ENVI programming.
### URI (optional)
Set this keyword to a string with the fully qualified filename and path of the output object. If you specify a URI, the Save method will use that URI. You can restore the object later using ENVIRestoreObject.
## Version History
ENVI 5.4 Introduced
3.3 | 675 | 3,191 | {"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} | 2.5625 | 3 | CC-MAIN-2019-30 | longest | en | 0.710674 |
http://stackoverflow.com/questions/5561138/interview-question-trim-multiple-consecutive-spaces-from-a-string | 1,395,086,469,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678705901/warc/CC-MAIN-20140313024505-00055-ip-10-183-142-35.ec2.internal.warc.gz | 193,329,796 | 20,889 | # Interview Question : Trim multiple consecutive spaces from a string
This is an interview question Looking for best optimal solution to trim multiple spaces from a string. This operation should be in-place operation.
``````input = "I Like StackOverflow a lot"
output = "I Like StackOverflow a lot"
``````
String functions are not allowed, as this is an interview question. Looking for an algorithmic solution of the problem.
-
Do you mean "in place" instead of "in memory"? – MAK Apr 6 '11 at 3:37
@MAK: Yes, thanks for pointing it out. – Unicorn Apr 6 '11 at 3:50
Keep two indices: The next available spot to put a letter in (say, `i`), and the current index you're examining (say, `j`).
Just loop over all the characters with `j`, and whenever you see a letter, copy it to index `i`, then increment `i`. If you see a space that was not preceded by a space, also copy the space.
I think this would work in-place...
-
Make sure the "was not preceded by a space" check isn't just checking if `s[j-1] == ' '` because the array's being modified as you go. It's safer to maintain a flag that keeps track of whether the previous character you saw was a space. – Kannan Goundan Apr 6 '11 at 3:47
@Kannan: Yeah I actually did think about writing that, but I let the reader figure it out for himself. ;) – Mehrdad Apr 6 '11 at 3:51
@Kannan: That is not actually required in this case. Just checking `s[j-1] == ' '` is sufficient. – Chris Hopman Apr 11 '11 at 5:48
Does using `<algorithm>` qualify as "algorithmic solution"?
``````#include <iostream>
#include <string>
#include <algorithm>
#include <iterator>
struct BothAre
{
char c;
BothAre(char r) : c(r) {}
bool operator()(char l, char r) const
{
return r == c && l == c;
}
};
int main()
{
std::string str = "I Like StackOverflow a lot";
std::string::iterator i = unique(str.begin(), str.end(), BothAre(' '));
std::copy(str.begin(), i, std::ostream_iterator<char>(std::cout, ""));
std::cout << '\n';
}
``````
test run: https://ideone.com/ITqxB
-
That's a nifty abuse! – Matthieu M. Apr 6 '11 at 6:47
@Matthieu: I wouldn't call this abuse - the `unique` stl algorithm is exactly designed for this type of situation. Requiring a functor class is admittedly awkward, but that's a fairly common limitation of stl algorithms pre-C++0X. – Eclipse Apr 6 '11 at 16:50
@Eclipse: I call it an abuse because it is a bit far-stretched with regard to the functor-less version, which simply uses `==`. Here the functor change the semantics, since two items that compare equal are not necessarily merged... I am pretty amazed by the trick though, especially because a simple "semantic" change on the functor helped hijack the algorithm :) – Matthieu M. Apr 6 '11 at 16:54
@Matthieu M: that's nothing compared to what `std::inner_product` can do when called with creatively-chosen functors (as explored by APL programmers decades ago) – Cubbi Apr 6 '11 at 16:56
This is a great use of a C++ standard library algorithm! – Boris Jun 18 '13 at 6:17
A c++0x - solution using a lambda instead of a regular function object. Compare to Cubbi's solution.
``````#include <string>
#include <algorithm>
int main()
{
std::string str = "I Like StackOverflow a lot";
str.erase(std::unique(str.begin(), str.end(),
[](char a, char b) { return a == ' ' && b == ' '; } ), str.end() );
}
``````
-
I'd just go with this:
``````int main(int argc, char* argv[])
{
char *f, *b, arr[] = " This is a test. ";
f = b = arr;
if (f) do
{
while(*f == ' ' && *(f+1) == ' ') f++;
} while (*b++ = *f++);
printf("%s", arr);
return 0;
}
``````
-
`char *f, *b, arr[] = "";` – BЈовић Apr 6 '11 at 6:42
@VJo: Care to explain the comment? The algorithm (while might be inefficient in that it copies all valid elements of the array regardless of whether they have to be moved or not --i.e. even if they stay in the same place-- seems correct. – David Rodríguez - dribeas Apr 6 '11 at 7:56
@David Yes, it works correctly, except in the empty string case. Can you guess what is the output when the arr contains an empty string? – BЈовић Apr 6 '11 at 8:07
@VJo, it prints an empty string. What else would it print? – edA-qa mort-ora-y Apr 6 '11 at 9:02
@edA Ah, right. If the first check fails, the second will not be executed at all. sorry for confusion – BЈовић Apr 6 '11 at 10:05
I'd propose a little state machine (just a simple switch statement). Because if the interviewer is anything like me, the first enhancement they'll ask you to do is to fully trim any leading or trailing spaces, so that:
``````" leading and trailing "
``````
gets transformed to:
``````"leading and trailing"
``````
``````" leading and trailing "
``````
This is a really simple modification to a state-machine design, and to me it seems easier to understand the state-machine logic in general over a 'straight-forward' coded loop, even if it takes a few more lines of code than a straight-forward loop.
And if you argue that the modifications to the straight forward loop wouldn't be too bad (which can be reasonably argued), then I (as the interviewer) would throw in that I also want leading zeros from numbers to be be trimmed.
On the other hand, a lot of interviewers might actually dislike a state-machine solution as being 'non-optimal'. I guess it depends on what you're trying to optimize.
-
It would help people relate to this advise if you implemented both so the implications were made tangible ;-). – Tony D Apr 6 '11 at 6:22
Very nice enhancement. And to be fair it was actually asked :). It would be nice if you can give some algo for that as well... – Unicorn Apr 6 '11 at 6:35
Here it is using only stdio:
``````#include <stdio.h>
int main(void){
char str[] = "I Like StackOverflow a lot";
int i, j = 0, lastSpace = 0;
for(i = 0;str[i]; i++){
if(!lastSpace || str[i] != ' '){
str[j] = str[i];
j++;
}
lastSpace = (str[i] == ' ');
}
str[j] = 0;
puts(str);
return 0;
}
``````
-
Trimming multiple spaces also means a space should always be followed by a non space character.
``````int pack = 0;
char str[] = "I Like StackOverflow a lot";
for (int iter = 1; iter < strlen(str); iter++)
{
if (str[pack] == ' ' && str[iter] == ' ')
continue;
str[++pack] = str[iter];
}
str[++pack] = NULL;
``````
-
``````void trimspaces(char * str){
int i = 0;
while(str[i]!='\0'){
if(str[i]==' '){
for(int j = i + 1; j<strlen(str);j++){
if(str[j]!=' '){
memmove(str + i + 1, str + j, strlen(str)-j+1);
break;
}
}
}
i++;
}
}
``````
-
``````import Data.List (intercalate)
trimSpaces :: String -> String
trimSpaces = intercalate " " . words
``````
The algorithm the next:
1. breaks a string up into a list of words, which were delimited by white space
2. concatenate the list inserting one space between each element in list
-
``````int j = 0;
int k=0;
char str[] = "I Like StackOverflow a lot";
int length = strlen(str);
char str2[38];
for (int i = 0; i < length; i++)
{
if (str[i] == ' ' && str[i+1] == ' ')
continue;
str2[j] = str[i];
j++;
}
str2[j] =NULL;
cout<<str2;
``````
-
this code was tried on VS 2010 – MNA Mar 27 '12 at 4:57
did you test with a string ending in ' ' ? str[i+1] should fail when i points to last char – Jayan Mar 27 '12 at 9:30
The question asked for an in-place version, and this copies to a separate buffer (str2). Also, str2 is statically allocated on the stack, and will overflow for some inputs -- that's not a good idea. And you've declared a variable 'k' which you don't use. – metamatt Oct 26 '12 at 21:43 | 2,151 | 7,562 | {"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} | 2.8125 | 3 | CC-MAIN-2014-10 | latest | en | 0.852706 |
https://essayachievers.com/peru20/ | 1,638,886,381,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363400.19/warc/CC-MAIN-20211207140255-20211207170255-00002.warc.gz | 321,642,869 | 9,237 | Financial Statement Analysis: Average Payment Period Ratio
In this assignment you will compare year-over-year payment trends using the average payment period ratio which is calculated as follows:
Average Payment Period Ratio = Current Liabilities / ([Total Operating Expenses Depreciation and Amortization Expenses] / 365)
The average payment period measures how long it takes on average for a business to pay back its creditors or in essence to pay its bills.
Review the following scenario:
You are the department head for the finance department for Anytown Hospital a not-for-profit community hospital in an urban community. The hospital administrator reviewed the most recent average payment period report (for fiscal year 2012 [FY2012]) and noted a variance between the FY2011 and FY2012 reports. The hospital administrator is not well versed in financial reports.
As the chief finance officer (CFO) and financial expert you are tasked to explain the variance to the Administrator. You must answer whether the variance is positive or negative as compared with the industry benchmark of 45 days. What are the risks if the trend is negative between FY2011 and FY2012? What are the benefits if the trend is positive from FY2011 to FY2012? You will present your findings in a PowerPoint presentation to the Hospital Administrator and other hospital department heads.
Part 1:Average Payment Period Ratio Report Development
For this assignment you will develop and analyze financial statements. Part 1 of the assignment includes describing what an average payment period is and how it is calculated. Use the years and the dollar amounts provided in the table below:
Operating Expenses
Depreciation and Amortization
Difference
Days per Year
Average Cash Expense per Day
15255263.00
553219.00
365
12589823.00
665393.00
365
Year
Current Liabilities
Average Cash Expense Per Day
Average Payment Period Days
2011
2550263.00
2012
4553681.00
Using an Excel spreadsheet (or any other tool for performing calculations) determine the average payment period for FY2011 and FY2012. Use the following steps:
Part 2: Risks and Benefits Analysis
Develop an 810-slide PowerPoint presentation (excluding the title and reference slides).
Include the completed average payment period ratio table in your presentation. Include detailed speaker notes with APA format including references. Use a properly formatted APA 6th edition figure caption for the table or any other figures or charts. Use at least 57 scholarly sources in your research.
Write your document in a clear concise and organized manner; demonstrate ethical scholarship in accurate representation and attribution of sources; and display accurate spelling grammar punctuation and references page. Apply APA standards to citation of sources.
Use the following file naming convention: LastnameFirstInitial_M3_A2.ppt.
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# Chapter-7
## Bond Development Length
& Splices
Lecture Goals
Slab design reinforcement
Bar Development
Hook development
Flexural Reinforcement in Slabs
For a 1 ft strip of slab is designed like a beam
A
s(reqd)
is in units of (in
2
/ft)
(
=
inches in spacing bar
in 12
ft /
b s
A A
The table is A-9 from
MacGregors book.
Flexural Reinforcement in Slabs
The minimum spacing of the bars is given as:
Also, check crack control - important for exterior
exposure (large cover dimensions) - ACI Sec. 10.6.4
( )
( )
## = 7.6.5 Sec. ACI
in. 18
thickness slab 3t
of smaller
max
S
Flexural Reinforcement in Slabs
Thin slabs shrink more rapidly than deeper beams.
Temperature & shrinkage (T&S) steel is provided
perpendicular to restrain cracks parallel to span.
(Flexural steel restrains cracks perpendicular to
span)
Maximum & Minimum reinforcement requirements
Flexural Reinforcement in Slabs
Maximum & Minimum reinforcement requirements
T&S Reinforcement (perpendicular to span) ACI Sec 7.12
( )
( )
( )
( )
( ) t
f t
f
f t
f t A
* " 12 * 0.0014
ksi 60 * " 12 *
60
* 0018 . 0
ksi 60 * " 12 * 0018 . 0
ksi 50 or 40 * " 12 * 0020 . 0
y
y
y
y min s
>
>
|
|
.
|
\
|
=
= =
= =
Flexural Reinforcement in Slabs
T&S Reinforcement (perpendicular to span) ACI Sec 7.12
Flexural Reinforcement (parallel to span) ACI Sec 10.54
S
max
from reinforced spacing
=
18"
5t
of smaller
max
S
( ) ( )
( ) ( ) bal s max s
& min s min s
75 . 0 A A
A A
S T
=
=
Reinforcement Development Lengths, Bar
Cutoffs, and Continuity Requirements
A. Concept of Bond Stress and Rebar Anchorage
Internal Forces in a beam
Forces in Rebar
Bond stresses provide mechanism
of force transfer between concrete
and reinforcement.
Forces developed in the beam
Reinforcement Development Lengths, Bar
Cutoffs, and Continuity Requirements
Equilibrium Condition for Rebar
= bond stress
(coefficient of
friction)
Note: Bond stress is zero at cracks
t
t
4
.
0
4
.
0 Force Bond . 0 F
b y
d
b b y
2
b
d f
l
l d f
d
T
=
=
= =
( ) bar
c
| f k
f k
=
~
Reinforcement Development Lengths, Bar
Cutoffs, and Continuity Requirements
Sources of Bond Transfer
(1) Adhesion between concrete & reinforcement.
(2) Friction
Note: These properties are quickly lost for tension.
Reinforcement Development Lengths, Bar
Cutoffs, and Continuity Requirements
Sources of Bond Transfer
(3)Mechanical Interlock.
The edge stress concentration
causes cracking to occur.
Force interaction between the
steel and concrete.
Reinforcement Development Lengths, Bar
Cutoffs, and Continuity Requirements
Splitting cracks result in loss of bond transfer.
Reinforcement can be used to restrain these cracks.
Minimum edge distance and spacing of bars
Tensile strength of concrete.
Average bond stress along bar.(Increase in bond
stress larger wedging forces)
1.
2.
3.
Reinforcement Development Lengths, Bar
Cutoffs, and Continuity Requirements
Typical Splitting Failure
Surfaces.
Reinforcement Development Lengths, Bar
Cutoffs, and Continuity Requirements
General splitting of
concrete along the
bars,either in vertical
planes as in figure (a) or
in horizontal plane as in
figure (b). Such splitting
comes largely from
wedging action when the
ribs of the deformed bar
bear against the concrete.
The horizontal type of splitting frequently begins at a diagonal crack.
The dowel action increases the tendency toward splitting. This
indicates that shear and bond failure are often intricately interrelated.
Reinforcement Development Lengths, Bar
Cutoffs, and Continuity Requirements
ACI Code expression for development length for
bars in tension/in compression.
B.
Development Length, l
d
Shortest length of bar in which the
bar stress can increase from zero to
the yield strength, f
y
.
( l
d
used since bond stresses, ,
vary along a bar in a tension zone)
Development Length for Bars in Tension
Development length, l
d
12 ACI 12.2.1
f
c
10000 psi for Ch. 12 provisions for development length in ACI Codes.
Development length, l
d
(simplified expression from ACI
12.2.2)
Clear spacing of bars being developed or
spliced not less than d
b
, clear cover not less
than d
b
, and stirrups or ties throughout ld not
less than the code minimum
or
Clear spacing of bars being
developed or spliced not less than 2d
b
and
clear cover not less than d
b
.
Other cases
No. 6 and smaller No. 7 and larger
bars and deformed bars
wires
>
s
c
y
b
d
25 f
f
d
l
o|
=
c
y
b
d
20 f
f
d
l
o|
=
c
y
b
d
50
3
f
f
d
l
o|
=
c
y
b
d
40
3
f
f
d
l
o|
=
Development Length for Bars in Tension
Development length, l
d
ACI 12.2.3
2.5 limit to safeguard against pullout type failure.
5 . 2 in which
40
3
b
ct
b
ct c
y
b
d
<
|
|
.
|
\
|
+
|
|
.
|
\
|
+
=
d
K c
d
K c f
f
d
l o|
Factors used in expressions for
Development Length (ACI 12.2.4)
o = reinforcement location factor
Horizontal reinforcement so placed that more than 12 in of fresh concrete
is cast in the member below the development length or splice
Other reinforcement
| = coating factor (epoxy prevents adhesion &
friction between bar and concrete.)
Epoxy-coated bars or wires with cover less than 3d
b
or clear spacing less
than 6d
b
All other epoxy-coated bars or wires
Uncoated reinforcement
1.3
1.0
1.5
1.2
1.0
where o| < 1.7
Factors used in expressions for
Development Length (ACI 12.2.4)
= reinforcement size factor (Reflects more favorable
performance of smaller | bars)
No.6 and smaller bars and deformed wire
No. 7 and larger bars
= lightweight aggregate concrete factor (Reflects lower
tensile strength of lightweight concrete, & resulting
reduction in splitting resistance.
When lightweight aggregate concrete is used.
However, when f
ct
is specified, shall be permitted to be taken as
but not less than
When normal weight concrete is used
0.8
1.0
1.3
1.0
1.0
ct c
7 . 6 f f
Factors used in expressions for
Development Length (ACI 12.2.4)
c = spacing or cover dimension, in.
Use the smaller of either
(a) the distance from the center of the bar or wire to
the nearest concrete surface.
or
(b) one-half the center-to-center spacing of the bar or
wires being developed.
Factors used in expressions for
Development Length (ACI 12.2.4)
K
ct
= transverse reinforcement index (Represents the contribution
of confining reinforcement across potential splitting planes.)
Total cross-section area of all transverse reinforcement within the spacing s,
which crosses the potential plane of splitting along the reinforcement being
developed with in the development length, in
2
.
Specified yield strength of transverse reinforcement, psi.
maximum center-to-center spacing of transverse reinforcement within l
d
in.
number of bars or wires being developed along the plane of splitting.
A
tr
=
f
yt
=
s =
n =
Note: It is permitted to use K
ct
=0 as a design simplification
even if transverse reinforcement is present.
n s
f A
K
* * 1500
yt tr
tr
=
Excess Flexural Reinforcement
Reduction (ACI 12.2.5)
Reduction = (A
s
reqd ) / (A
s
provided )
- Except as required for seismic design (see ACI 21.2.14)
- Good practice to ignore this provision, since use of
structure may change over time.
- final l
d
12 in.
( )
( ) ( ) provided n
u
provided n
d req' n
Reduction
M
M
M
M
|
= =
>
Development Length for Bars in
Compression (ACI 12.3)
Compression development length l
dc
= l
dbc
* applicable
reduction factors 8 in.
Basic Development Length for Compression, l
dbc
>
=
y b
c
y b
dbc
0003 . 0
0.02
of larger
f d
f
f d
l
Development Length for Bars in
Compression (ACI 12.3)
Reduction Factors (ACI 12.3.3)
- Excessive Reinforcement Factor = (A
s
reqd)/(A
s
provided)
- Spiral and Ties
If reinforcement is enclosed with spiral
reinforcement 0.25 in. diameter and 4 in. pitch or
within No. 4 ties according to 7.10.5 and spaced 4 in.
on center. Factor = 0.75
Note l
dc
< l
d
(typically) because
- Beneficial of end bearing is considered
- weakening effect of flexural tension cracks is not
present for bars in compression.
> s
s
Hooked Bar at Discontinuous
Ends (ACI 12.5.4)
If side cover and top (or bottom cover) 2.5 in.
Enclose hooked bar w/ ties or stirrup-ties:
s
Spacing 3d
b
d
b
=| of hooked bar
s
Note: Multiplier for ties or
stirrups (ACI 12.5.3.3)
is not applicable for
this case.
Hooked Bar at Discontinuous
Ends (ACI 12.5.4)
Table A-11, A-12, A-13 (Back of textbook) - Basic
Development lengths
Others Mechanical Anchorage ACI (12.6)
Welded Wire Fabric ACI (12.7)
Bundled Bars ACI (12.4)
Reinforcement Development Lengths, Bar
Cutoffs, and Continuity Requirements
C. Use of Standard Hooks for Tension Anchorage
there is insufficient length available to
develop a bar.
Note: Hooks are not allowed to developed
compression reinforcement.
Reinforcement Development Lengths, Bar
Cutoffs, and Continuity Requirements
C. Use of Standard Hooks for Tension Anchorage
Standard Hooks are
defined in ACI 7.1.
Hooks resists tension by
bond stresses on bar
surface and bearing on on
concrete inside the hook.
Design of Standard Hooks for
Tension Anchorage (ACI 12.5)
Development Length for Hooked Bar, l
db
.
. in 6 and 8 where s multiplier *
db b db hd dh
> > = l d l l l
Basic Development Length for Hooked Bar = l
hb
when f
y
= 60,000 psi
c
b
hd
1200
f
d
l =
Design of Standard Hooks for
Tension Anchorage (ACI 12.5)
Conditions
Bar Yield Strength
Bars with f
y
other than 60,000 psi
Concrete Cover for 180 Degree Hooks
For No. 11 bars and smaller.
Side cover (normal to plane of hook) 2.5 in.
Concrete Cover for 90 Degree Hooks
For No. 11 bars and smaller.
Side cover (normal to plane of hook) 2.5 in.
Cover on bar extension beyond hook tail 2 in.
Multiplier
f
y
/60,000
0.7
0.7
>
>
>
Design of Standard Hooks for
Tension Anchorage (ACI 12.5)
Conditions
Excessive Reinforcement
Where anchorage or development for fy is not
specified required.
Lightweight Aggregate Concrete
Ties or Stirrups
For No. 11 bar and smaller.
Hook enclosed vertically or horizontally within ties
or stirrup-ties spaced along full l
dh
no farther apart
than 3d
b
, where d
b
is diameter of hooked bar.
Multiplier
A
s
(reqd) /
A
s
(provided)
1.3
0.8
Design of Standard Hooks for
Tension Anchorage (ACI 12.5)
Conditions
Epoxy-coated Reinforcement
Hooked bars with epoxy coating
Multiplier
1.2
Example
Example 4
GIVEN: A #5 Grade 40 bar is in tension as shown below. Use LIGHTWEIGHT
concrete with fc = 4000 PSI.
REQUIRED: Determine the min. required hook dimensions X, Y and Z | 3,073 | 10,323 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2020-10 | latest | en | 0.788797 |
https://zbmath.org/?q=an:0714.60014 | 1,721,924,196,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763860413.86/warc/CC-MAIN-20240725145050-20240725175050-00744.warc.gz | 900,760,587 | 11,410 | ## Étude de l’estimateur du maximum de vraisemblance dans le cas d’un processus autorégressif: convergence, normalité asymptotique, vitesse de convergence. (Asymptotic behaviour of maximum likelihood estimator in an autoregressive process: consistency, asymptotic distribution and expansion, rate of convergence).(French)Zbl 0714.60014
The authors consider the group G of all invertible matrices $$g=\begin{pmatrix} A(g) & b(g)\\ 0&1 \end{pmatrix}$$, where $$b(g)\in R^ d$$ is a column, a norm $$\|.\|$$ in $$R^ d$$, a probability $$\mu$$ on G for which the corresponding norm $$\| A(g)\|$$ is $$\mu -a.e.<1$$, a sequence $$g_ n$$, $$n\geq 1$$, of independent $$\mu$$-distributed random variables, a fixed $$z\in R^ d$$, the distribution $$\nu$$ of $$\sum_{k\geq 1}A(g_ 1)...A(g_{k-1})b(g_ k)$$, $$\eta: G\to [1,\infty)$$, constants $$c,\gamma,\tau >0$$, $$\alpha\in (0,1]$$, a function $$F: G\times R^ d\to R$$ satisfying $| F(g,x)-F(g,y)| \leq \eta (y)\| x- y\|^{\gamma}(1+\| x\|^{\tau}+\| y\|^{\tau})$ and $| F(g,x)| \leq \eta (g)(\| x\|^{\gamma +\tau}+1),$ $$\begin{pmatrix} Y_ n\\ 0 \end{pmatrix} =g_ n...g_ 1\begin{pmatrix} z\\ 0\end{pmatrix}$$ for $$n\geq 0$$, $$S_ n=\sum^{n}_{1}F(g_ k,Y_{k-1})$$, $$e=\int \int Fd\mu d\nu$$ and impose $(*)\quad \int \eta (g)^ 5(1- \| A(g)\|^{\alpha})^{-5(\gamma +\tau)/\alpha}\exp (c\| b(g)\|^{\alpha})d\mu (g)<\infty.$ First they prove that $$\sigma^ 2=\lim_{n} n^{-1}E((S_ n-ne)^ 2)$$ exists, does not depend on z, and, under a supplementary condition on F, is positive, and that $d(n^{-1/2}\sigma^{-1}(S_ n-ne),N)\leq Cn^{-1/2}(1+\| x\|^{\gamma \beta}\exp (\lambda \| x\|^{\alpha})$ with constants $$C,\lambda >0$$, $$\beta >1$$, where d is the supremum of the modulus of the difference of the distribution functions and N is standard normal. Replacing (*) by two other conditions, each containing $$m\geq 2$$ at the exponent, and requiring $$\| A(g)\| \leq \rho <1\mu$$-a.e., they prove also that $\sup_{n} E(| n^{- 1/2}\sum^{n}_{1}(F(g_ k,Y_{k-1})-e)|^ m)\leq C(1+\| x\|^{m(\gamma +\tau)}).$ In the proofs they use the operators $$(U_ tf)(x)=\int e^{itF(g,x)}f(g,x)d\mu (g)$$ and arrange a Banach space on which $$U_ t$$ are quasicompact etc.
Then the authors consider the particular case, corresponding to the title, of a nonrandom $$A(g_ n)=A_{\theta}=\left( \begin{matrix} \theta \\ 1\quad 0\end{matrix} \right)$$, where $$\theta =(\theta_ 1,...,\theta_ d)$$, $$b(g_ n)=\epsilon_ ne_ 1$$, $$e_ 1=(1,0,...,0)'$$, E $$\epsilon$$ $${}_ n=0$$, $$\max | u_ i| <1$$, where $$u_ i$$ are the roots of $$u^ d=\sum \theta_ iu^{d-i}$$. It follows that, if $$E(\log^+| \epsilon_ k|)<\infty$$, $$\nu$$ is the unique invariant distribution of $$P_{\theta}$$, where $$(P_{\theta}f)(x)=E(f(A_{\theta}x+\epsilon_ ne_ 1)).$$ The previous results are applied, and for $$m\geq 4$$, $P(n^{- 1/2}\sum^{n}_{1}(F(\epsilon_ k,Y_{k-1})-e)>((m-3)\log n)^{1/2})=o(n^{-(m-3)/2})$ is established. Furthermore, supposing that $$\epsilon_ n$$ have a positive density f, tending to 0 at $$\infty$$, with a derivative satisfying $$\int | f'|^ rf^{1-r}<\infty$$ with an $$r\geq d+1$$, and also a finite r-moment, the authors prove, denoting by $${\hat \theta}{}_ n$$ the maximum likelihood estimator of $$\theta$$ in a $$P_{\theta}$$-Markov chain starting from z, several “statistical results”. These are: the convergence in distribution of $$n^{1/2}({\hat \theta}_ n-\theta)$$ to an $$N(0,\sigma^ 2)$$, the convergence of the corresponding “absolute k-moments” (for all k), $P(\| {\hat \theta}_ n-\theta \| >\rho)\leq A_ 1\exp (-A_ 2n)\text{ for } every\quad \rho >0,$ and $P(n^{1/2}\| {\hat \theta}_ n-\theta \| >B \log^{1/2}n)=O(n^{-1}).$ Supposing the existence of $$f^{(p+1)}$$ and the validity of other conditions in which a k appears, and writing $n^{1/2}({\hat \theta}_ n-\theta)=h_ 1+...+n^{-(p-1)/2}h_ p+n^{-p/2}\rho (n),$ the authors show that $P(\| \rho (n)\| >((k-2)\log n)^{(p+1)/2})=O(n^{-(k-2)/2})$ and also, for $$p=1$$, that $d(n^{1/2}({\hat \theta}_ n- \theta),N(0,\sigma^ 2))\leq Cn^{-1/2}\log^{1/2}n.$ The uniformity, in z (and $$\theta$$) on compacts, of “the results” is established. Finally, a more precise result is announced.
Reviewer: I.Cuculescu
### MSC:
60F05 Central limit and other weak theorems 62F10 Point estimation 60J05 Discrete-time Markov processes on general state spaces 60G50 Sums of independent random variables; random walks
Full Text: | 1,660 | 4,398 | {"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} | 2.640625 | 3 | CC-MAIN-2024-30 | latest | en | 0.539826 |
http://vestniken.ru/eng/catalog/math/compmath/736.html | 1,611,418,868,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703538226.66/warc/CC-MAIN-20210123160717-20210123190717-00333.warc.gz | 101,525,097 | 5,857 | An Algorithm for Constructing a Hereditarily Minimax Network with Predefined Vector of Node Degrees | Herald of the Bauman Moscow State Technical University. Natural Sciences
|
# An Algorithm for Constructing a Hereditarily Minimax Network with Predefined Vector of Node Degrees
Authors: Selin P.S., Tsurkov V.I., Gurchenkov A.A. Published: 14.02.2017 Published in issue: #1(70)/2017 DOI: 10.18698/1812-3368-2017-1-43-58 Category: Mathematics | Chapter: Computational Mathematics Keywords: network optimization, transport type problems, minimax, minimax network, hereditarily minimax network, uniform network, transportation polyhedrons, network polyhedrons, predefined node degrees, fixed node degrees
In contrast to the classical transportation problem, where supply and demand points are known, and it is required to minimize the transportation cost, we consider a minimax criterion. In particular, a matrix with the minimal largest element is sought in the class of nonnegative matrices with given sums of row and column elements. In this case, the concept of the minimax criterion can be interpreted as follows. Suppose that the shipment time from a supply point to a demand point is proportional to the amount to be shipped. Then, the minimax is the minimal time required to transport the total amount. It is a common situation that the decision maker does not know the tariff coefficients. In other situations, they do not have any meaning, and neither do nonlinear tariff objective functions. In such cases, the minimax interpretation leads to an effective solution. For the classes of undirected networks with predefined vector of node degrees (transport and network polyhedrons) by using a characteristic functions the analytical formulas of calculating the minimax values expressed in terms of the vector coordinates and a nonnegative parameter are obtained. The minimax values determine the necessary and sufficient conditions under which the truncated polyhedrons are not empty sets. Finally, we obtained an algorithm for constructing a hereditarily-minimax network in network polyhedrons.
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[19] Richard M.G.A., Fanchon E. Reduction and fixed points of Boolean networks and linear network coding solvability. IEEE Transactions on Information Theory, 2016, vol. 62, no. 5, pp. 2504-2519.
[20] Nazarov M.N. On the representation of graphs in the form of a special type of binary algebra. Prikl. Diskr. Mat., 2015, no. 1, pp. 96-104. | 1,441 | 5,323 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2021-04 | latest | en | 0.84281 |
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837
1. ## Expansion and less visibility of objects
In an expanding universe, I think that we will now see it lasting (1 second)*(a_now)/(a_then), where a_now is the present scale factor of the universe, a_then is the scale factor of the universe when the 1-second pulse was emitted.
The October issue of Astronomy Magazine, which arrived in the post a few days ago, has an excellent article on the James Webb Space telescope. The article was written by John Mather, who is the senior project scientist for James Webb, and who won the Nobel Prize for work that first observed the variations in the Cosmic Microwave Background (CMB).
3. ## Concerning Dark Matter
I went to high school in the 1970s in a small (pop. 8000) industrial town in Canada, but the high school was first-rate, and many of its teachers were first-rate. I took computer science classes in high school from 1976 to 1978. By today's standards, the circumstances were strange: the programming language used was Fortran (in high school!); the programs (penciled-in computer cards) were physically sent by Greyhound Bus to a university in another city, and the cards and hard-copy output were returned by bus. The turn-around time was a least 3 days per run, sometimes to find "execution suppressed" because of a syntax error. The teacher, a computer science graduate from the University of Waterloo, had us work on several projects simultaneously, so that we were always sending and receiving stuff. Good courses and teacher.
4. ## How convenient, or 'grab and go', is your setup?
I keep my NexStar 8 SE on the tripod in my shed, so that I do not have any cool down time. Even though I am 5' 4'' tall, and weigh 130 pounds, I find it easy to carry out the whole unit into the garden. I do have to do an easy star alignment. About 5 minutes after I go out the door of my house, I am observing with my scope at ambient temperature.
5. ## Is purely observing a minority pastime now ?
Here is a slightly different question: in terms of absolute numbers, has the number of manly visual astronomers gone up or down? Is it possible that, over the past 20 years or so, the number of astronomers of all sorts has gone up, the percentage of mainly visual astronomers has gone down, but the number of mainly visual astronomers has not gone down? I do not have a feel for the answers to these questions, and I am interested in hearing what people think. Personally, I am a purely visual. I have have taken zero pictures using my scope, not even with a mobile held to the eyepiece to capture the moon. This may because I don't own a mobile. My work day consists of doing technical stuff with books, lab equipment, and computers. When i get home, it is a nice change of pace to go out with my scope, and do visual observing while pondering the universe.
6. ## Should I Downsize?
I easily do the same, even though I am 61 years old, 5' 4'' tall, and weigh 130 pounds. I keep mine set up in a shed, so I do not have to wait for it to cool down, I do the star alignment, as I don't have Starsense. About 5 minutes after I go out my door, I am aligned and observing.
7. ## Any book recommendations.....
There are too many books to read. Here are some examples, going form newest to oldest by publication date. "The End of Everything (Astrophysically Speaking)" by Katie Mack https://www.astrokatie.com/book "Our Mathematical Universe: My Quest for the Ultimate Nature of Reality" by Max Tegmark https://en.wikipedia.org/wiki/Our_Mathematical_Universe Going back years, both John Barrow and Paul Davies wrote many interesting books. Example include "The Universe that Discovered Itself" by John Barrow "About Time" by Paul Davies.
8. ## Steven Weinberg has died.
Steven Weinberg, who I considered the greatest living physicist, has died. His non-mathematical "The First Three Minutes" is a wonderful account of the evolution of the universe during the first three minutes after the Big Bang, Weinberg also wrote a series of wonderful very advanced advanced books. Some personal thoughts. I twice saw Weinberg give technical talks. Even though I love Weinberg's advanced books, I say the following. I am glad that: 1) they are on my shelf; 2) I am not taking a course that uses any of them as the text; 3) I am not teaching a course that uses any of them as the text. Weinberg explains (at a technical level) some things better than anyone. Sometimes I read a passage in Weinberg, and I think to myself "Wow, I finally understand what is really going on." There are, however, passages in Weinberg's books that I find difficult to understand. Also, sometimes it is difficult to see past the clutter of Weinberg's notation. The lecture notes for a course that Weinberg taught in 2017 (at the age of 84!) were turned into his second to last book, Lectures on Astrophysics. The section on white dwarfs and neutron stars helped me when I had to scramble to prepare an upper-level COVID-related and unexpected teaching overload last fall. I like the personal aspect of the final paragraph of the Preface,
9. ## A question about uniformity and the CMB
The way I like to put it: when we look at galaxies through our telescopes, we are looking at quantum fluctuations blown up to galactic proportions by the expansion of the universe! Mind blowing!
10. ## Concerning Dark Matter
Pieter van Dokkum, the lead author of the research explained at this link, takes this as evidence for dark matter and against MOND (MOdified Newtonian Dynamics). From the link From the Introduction section of the actual 2021 research paper, and, from the paper's Conclusion,
11. ## Concerning Dark Matter
A report on a Zoom talk about dark matter research that I "attended" last fall:
12. ## Enjoy galaxy season while it lasts?
And before it was a solid state universe, it was a valve universe.
13. ## Entangled particles used for communication
Entangled particles cannot be used to transmit information faster than the speed of light.
14. ## EP33 - YouTube - Gravitational Wave Astronomy: A Descriptive, Non-Mathematical Talk by George Jones
I had great fun giving the talk (wife: "Don't get him started on physics!"), and people asked great questions. I hope soon to elaborate a bit on some of my answers.
15. ## “We don’t really know the speed of light”
And my students do just this. I teach a second-year e&m lab course that performs one experiment per week. In consecutive weeks, the students measure ϵo and µo using https://media.vwr.com/interactive/p...h_2014/files/assets/basic-html/index.html#289 and then compare to the speed of light. For µo , we send up to 20 amps through the metal rods.
16. ## New member from Ontario, Canada
Welcome to SGL. My in-laws live north of Toronto, near Canada's Wonderland.
17. ## Colour in Nebulas?
I just came in, and the last thing at which I looked was the Blue Snowball. I am sure that my 60-year-old eyes can see some tint in the image produced by my 8 inch SCT.
18. ## Hello from Vancouver (Surrey), Canada
Welcome from Prince George BC.
19. ## We are not limited to our local area when we look up.
Yes, these guys are outside our Local Group. They're being whisked away from us at about 700 kilometres (430 miles) per second by the expansion of the universe. Mind-boggling stuff!
20. ## M11-6 2020-09-13.jpg
Nice. A favourite of mine at the eyepiece.
21. ## Night sky mental health boost for astronomy fan
Nothing relaxes me as much as my observing sessions do.
22. ## Astronomy Double Entendres
Years ago, while teaching an astronomy course to about 120 university students and talking about the light-gathering power of telescopes, I said "To some extent, it's not the length that matters, it's the width." The class broke into loud laughter. Ever since, I have been careful not to phrase it like this.
23. ## Help finding speakers
In the months leading up to COVID, I gave a couple of public talks on gravitational waves. I don't know if a non-maths talk on what's below would be interesting or appropriate. Gravitational Waves and Gravitational Wave Astronomy Gravitational waves, first observed in 2015, are produced when compact objects like black holes and neutron stars merge. Just as useful information can be extracted from light wave signals, useful information also can be extracted from gravitational wave signals.
24. ## Help finding speakers
How long is a typical talk?
25. ## Dark matter - fudge factor?
Me too! Even if it does reach a conclusion, we might not like the conclusion. Paraphrasing Planck: "Science advances one funeral at a time." Actual version: "A new scientific truth does not triumph by convincing its opponents and making them see the light, but rather because its opponents eventually die, and a new generation grows up that is familiar with it."
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# New to GMAT preparation.. exam in 3 months. Need guidance.
Author Message
Intern
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New to GMAT preparation.. exam in 3 months. Need guidance. [#permalink]
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12 May 2013, 04:28
I gave my first Diagnostic test yesterday, the Princeton review Diagnostic test. Scored a poor 500.Q37V23. I had finished the each section 20 mitutes before time. Specially in verbal section, I was lazy to read everything.
My exam date is in september. And I am aiming a score of 730-750. Would appreciate some guidance/review on my study plan.
May- Focus on verbal. Do MGMAT SC with questions from OG and GMAT ultimate grammar book.
June- Focus on Quant Section. Complete MGMAT Quant guides along with OG questions. Also, continue with GMAT Verbal Review guide.
Take a CAT, and asses the situation.
July- Focus on other verbal sections, CR and RC, along with Quant.
Take a CAT, and asses the situation.
August- Focus on trouble areas and take two CAT's each week.
Please let me know your thoughts on the above chalked out plan and how can i make it more effective. I am a working professional, so can take out the standard 2-3 hrs on weekdays and 6-7 hrs on weekends.
Awaiting for an earnest response.
Senior Manager
Status: Making every effort to create original content for you!!
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Concentration: Healthcare, Social Entrepreneurship
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Re: New to GMAT preparation.. exam in 3 months. Need guidance. [#permalink]
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12 May 2013, 07:20
1
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Expert's post
Rhea446 wrote:
I gave my first Diagnostic test yesterday, the Princeton review Diagnostic test. Scored a poor 500.Q37V23. I had finished the each section 20 mitutes before time. Specially in verbal section, I was lazy to read everything.
My exam date is in september. And I am aiming a score of 730-750. Would appreciate some guidance/review on my study plan.
May- Focus on verbal. Do MGMAT SC with questions from OG and GMAT ultimate grammar book.
June- Focus on Quant Section. Complete MGMAT Quant guides along with OG questions. Also, continue with GMAT Verbal Review guide.
Take a CAT, and asses the situation.
July- Focus on other verbal sections, CR and RC, along with Quant.
Take a CAT, and asses the situation.
August- Focus on trouble areas and take two CAT's each week.
Please let me know your thoughts on the above chalked out plan and how can i make it more effective. I am a working professional, so can take out the standard 2-3 hrs on weekdays and 6-7 hrs on weekends.
Awaiting for an earnest response.
Hi Rhea446,
If you can take out 2-3 hours on weekdays and 6-7 hours on the weekends then this is enough for your test in September. If you finished the practice test 20 minutes early in both the sections then the result will not represent your true level. Try to take another test under strict testing conditions then see what the score comes. A jump from 500 to 730 is a huge one, but this can be achieved and you have enough time for that.
Your high level study plan looks good to me but, you may want to go through this post - gmat-study-plan-how-to-start-your-gmat-prep-80727.html to figure our your ideal study plan. Whichever study plan you should choose, you must see an improvement in your performance/ skill level, otherwise it is not working for you. I would suggest that you break your goal from 500 to 600 then after achieving 600, to 650 then to 700 and to 700+. You should also take one test every 3 weeks initially to determine your progress, even if you have some topics untouched.
Lastly do maintain an error log to keep track of questions that you got incorrect and other specific topics that you find difficult and may want to revisit in future. This log will be really helpful during the last three weeks of your prep. All the best !!!
Vercules
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Re: New to GMAT preparation.. exam in 3 months. Need guidance. [#permalink]
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18 May 2013, 01:01
Hi Vercules,
As suggested by you, I gave a GMAT Prep test. I scored a 550(Q40V25) and in IR, a score of 4. Now IR seems to be another trouble area. Could you suggest some thing for working upon it.
Regards
Rhea
Intern
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Posts: 42
Location: India
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Re: New to GMAT preparation.. exam in 3 months. Need guidance. [#permalink]
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23 May 2013, 19:28
hello,
i intend to take my test in august. i have a simple plan of 50 questions a day for may and june. have collected thousands of questions of varying difficulty off of gmatclub, and is my go-to place for prep
1.is this wise ?
2.is 50q/day sufficient ?
Thanks,
Abilash
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Re: New to GMAT preparation.. exam in 3 months. Need guidance. [#permalink]
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24 May 2013, 13:56
abilash10 wrote:
hello,
i intend to take my test in august. i have a simple plan of 50 questions a day for may and june. have collected thousands of questions of varying difficulty off of gmatclub, and is my go-to place for prep
1.is this wise ?
2.is 50q/day sufficient ?
Thanks,
Abilash
Abilash,
If you're looking for practice math questions - you can do these questions: http://www.gmatpill.com/gmat-practice-t ... questions/
Do as many questions as you can - but more important is understanding each one thoroughly so if that same question is asked a slightly different way - you'll still get it right.
Collecting thousands of questions won't do you much good once you get to a certain point. It's about quality + some repeated drills, not just pure quantity.
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Re: New to GMAT preparation.. exam in 3 months. Need guidance. [#permalink]
### Show Tags
25 May 2013, 17:48
abilash10 wrote:
hello,
i intend to take my test in august. i have a simple plan of 50 questions a day for may and june. have collected thousands of questions of varying difficulty off of gmatclub, and is my go-to place for prep
1.is this wise ?
2.is 50q/day sufficient ?
Thanks,
Abilash
Hi Abhilash!
Before you start practicing so many questions make sure your concepts are very clear. GMAT is not about knowing how to solve all questions, it's about how to solve them in the least time possible. You can probably refer to the Manhattan strategy guides and Official Guide. Again, even if you practice only 10 questions a day, make sure you refer to the answers and see if there's an easier/ shorter method to solve those problems.
Best of luck!
Vineeta
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Re: score improved to 660 from 540. Suggestions needed for SC [#permalink]
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28 Jul 2013, 02:02
Hi All,
After 2 months of on-off preparation i gave GMATPrep exam again, and scored 660(Q44V37) (My target score is a720+) and IR score of 5. SC seems to be a weak area as I got 7 questions wrong out of 17 in SC. Also, I could get only 60% of first 10 questions right in SC. Quant I had not prepared much, but will focus on it now. Any tips/advice is most welcome to improve SC as I have already covered most of the material available.
RC: 10 Correct out of 13
CR: 8 correct out of 11
Quant: Not Attempted: 6
19 correct out of 31 attempted.
My exam is in 1 month from now. All help will be appreciated.
Regards
Rhea
Re: score improved to 660 from 540. Suggestions needed for SC [#permalink] 28 Jul 2013, 02:02
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# New to GMAT preparation.. exam in 3 months. Need guidance.
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https://phys.libretexts.org/Bookshelves/Conceptual_Physics/Body_Physics_-_Motion_to_Metabolism_(Davis)/06%3A_Forces_within_the_Body/6.06%3A_Alternative_Method_for_Calculating_Torque_and_Tension | 1,725,721,167,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650883.10/warc/CC-MAIN-20240907131200-20240907161200-00090.warc.gz | 448,438,328 | 31,536 | # 6.6: Alternative Method for Calculating Torque and Tension*
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If you would rather not think about finding lever arms, you can instead calculate the size of thetorque as the size of the force multiplied by the full distance to the pivot, and by the sine of the angle between the force and that full distance. Written in equation form it looks like this:
(1)
## Reinforcement Activity
The torque caused by a force depends on the angle between the line of action of the force acts and the line from where the force is applied to the pivot point. To feel this effect for yourself, try this:
Rotate a door by pushing at 90° to the door right at the outer edge.
Now apply the same force on the door, still on the very edge, but instead of pushing in a direction 90° to the door, push along the door, straight in toward the hinges. Does the door swing as it did before?
In the second case, the angle between the force direction and the distance to the pivot was 0° (they were parallel). Use the previous equation to show that the torque must be zero any time the line of action of the force goes straight through the rotation point (pivot).
Now, we know the force is 50 lbs, the distance from the pivot to the weight is 13.0 in length of the forearm and from the diagram we see the angle between the weight of the ball and the forearm distance is 60° (the same as the bicep-forearm angle because they are alternate interior angles). This is easier to see if we draw a stick figure diagram:
Stick diagram of a flexed arm holding a ball showing the bicep tension and weight and the angles between the forces and the forearm.
Now we can calculate the torque due to the ball weight as:
We just need to make this equal to the ball-weight-torque:
Then we divide both sides by and to isolate the bicep tension:
Finally we put in our values for and . Our original diagram gave us the distance as from bicep attachment to the pivot as 1.5 in and from our stick diagram we can see that the angle between the biceps tension and the distance is 180°-60° = 120°. We are ready to find the biceps tension value.
Our result of 433 lbs seems surprisingly large, but we will see that forces even larger than this are common in the muscles, joints, and tendons of the body.
This page titled 6.6: Alternative Method for Calculating Torque and Tension* is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Lawrence Davis (OpenOregon) via source content that was edited to the style and standards of the LibreTexts platform. | 2,334 | 6,897 | {"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} | 4.21875 | 4 | CC-MAIN-2024-38 | latest | en | 0.201891 |
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The new math – IPCC version
WattsupWithThat.com ^ | April 12, 2010 | Global Warming Questions -IPCC
Posted on 04/13/2010 11:31:49 AM PDT by Ernest_at_the_Beach
How the IPCC invented a new calculus
A new form of calculus has been invented by the authors of the the IPCC’s Fourth Assessment Report (AR4), in order to create the false impression that global warming is accelerating.
How the new IPCC calculus works
Here’s how it works. Look at the following graph:
Now consider the following question:
Is the slope of the graph greatest at the left hand end of the graph, or the right hand end?
By just looking at the graph, or by using old-fashioned calculus developed by Newton and Leibnitz, you might think that the slope of the graph is similar at both ends. But you would be wrong. In fact, the slope is much greater towards the right hand end of the graph. To prove this, we need to apply the new calculus developed by the IPCC. To do this, we draw a sequence of straight-line best fits backwards from the right-hand end-point:
This clearly shows how the slope of the graph is in fact increasing.
How IPCC calculus is used in the IPCC report
Here is one of the key graphs from the AR4 report:
The graph is Figure 1 from FAQ 3.1, to be found on page 253 of the WG1 report. The slope over the last 25 years is significantly greater than that of the last 50 years, which in turn is greater than the slope over 100 years. This ‘proves’ that global warming is accelerating. This grossly misleading calculation does not just appear in chapter 3 of WG1. It also appears in the Summary for Policymakers (SPM):
The linear warming trend over the last 50 years is nearly twice that for the last 100 years“.
Thus, policymakers who just look at the numbers and don’t stop to think about the different timescales, will be misled into thinking that global warming is accelerating. Of course, we could equally well start near the left hand end of the graph and obtain the opposite conclusion! (Just in case this is not obvious, see here for an example). A similar grossly misleading comparison appears at the very beginning of chapter 3, page 237:
The rate of warming over the last 50 years is almost double that over the last 100 years (0.13°C ± 0.03°C vs. 0.07°C ± 0.02°C per decade).
How did this get through the IPCC’s review process?
The IPCC reports are subjected to careful review by scientists. So how did this blatant distortion of the temperature trends get through this rigorous review process? The answer to this question can now be found, because the previous drafts of AR4, and the reviewer comments, can now be seen on-line. (The IPCC was reluctant to release these comments, but was forced to do so after a number of freedom of information requests).
The answer is quite astonishing. The misleading graph was not in either the first or the second draft of the report that were subject to review. It was inserted into the final draft, after all the reviewer comments.
It is not clear who did this, but responsibility must lie with the lead authors of chapter 3, Kevin Trenberth and Phil Jones. Here is the version of the graph that the reviewers saw in the second draft:
Note that in this version there is only one trend line drawn.
So why was this graph replaced by the grossly misleading one? Did any of the reviewers suggest that a new version should be drawn with a sequence of straight lines over different time intervals? No. One reviewer made the following remark:
‘This whole diagram is spurious. There is no justification to draw a “linear trend” through such an irregular record’
… but his comment was rejected.
It is the same story with the misleading comment in the SPM mentioned above (“The linear warming trend over the last 50 years is nearly twice that for the last 100 years“). This statement was not in the original version reviewed by the scientists. It was inserted into the final draft that was only commented on by Governments. The Chinese Government suggested deleting this, pointing out that:
‘These two linear rates should not compare with each other because the time scales are not the same’.
Well done to the Chinese Government for spotting that. Too bad their valid comment was ignored by the IPCC.
h/t to Roger Carr
TOPICS: Crime/Corruption; Extended News; Foreign Affairs
KEYWORDS: globalwarminghoax; ipcc
1 posted on 04/13/2010 11:31:50 AM PDT by Ernest_at_the_Beach
To: SunkenCiv; Marine_Uncle; Fred Nerks; steelyourfaith; NormsRevenge; onyx; BOBTHENAILER; ...
The AGW crowd is a bunch of Crooks.....
2 posted on 04/13/2010 11:33:08 AM PDT by Ernest_at_the_Beach ( Support Geert Wilders)
To: Ernest_at_the_Beach
The peers only looked that the skew and kurtosis
3 posted on 04/13/2010 11:34:34 AM PDT by edcoil (If I had 1 cent for every dollar the government saved, Bill Gates and I would be friends.)
To: Ernest_at_the_Beach
Isn’t the real point is that most people don’t do math anymore ... And wasn’t that the point of the made up pretty graphs?
The next comes when they public figures out how warm it was 2000 years ago, during the Roman Empire warm period.
4 posted on 04/13/2010 11:34:35 AM PDT by Tarpon ( ...Rude crude socialist Obama depends on ignorance to force his will on people)
To: Ernest_at_the_Beach
Somebody failed Digital Filtering and Sampling Theory. So, they invented a parallel mathematics, one that just does not add up.
5 posted on 04/13/2010 11:34:58 AM PDT by mlocher (USA is a sovereign nation)
To: Ernest_at_the_Beach
Throw the buns out!
6 posted on 04/13/2010 11:34:58 AM PDT by TribalPrincess2U (demonicRATS... taxes, pain and slow death. Is this what you want?)
To: Ernest_at_the_Beach
If you torture numbers long enough you can make them say anything.
7 posted on 04/13/2010 11:39:06 AM PDT by BitWielder1 (Corporate Profits are better than Government Waste)
To: Ernest_at_the_Beach
There’s GOTTA BE A HOCKY STICK IN THERE SOMEWHERE!...............
8 posted on 04/13/2010 11:45:43 AM PDT by Red Badger (Education makes people easy to lead, difficult to drive; easy to govern, but impossible to enslave.)
To: TribalPrincess2U
Buns?................
9 posted on 04/13/2010 11:46:02 AM PDT by Red Badger (Education makes people easy to lead, difficult to drive; easy to govern, but impossible to enslave.)
LOLOLOLOL
10 posted on 04/13/2010 11:46:55 AM PDT by onyx (Sarah/Michele 2012)
To: Ernest_at_the_Beach; Amagi; Fiddlstix; Tunehead54; Clive; FrPR; tubebender; marvlus; ...
Thanx !
Beam me to Planet Gore !
11 posted on 04/13/2010 11:49:08 AM PDT by steelyourfaith (Warmists as "traffic light" apocalyptics: "Greens too yellow to admit they're really Reds."-Monckton)
Throw their buns out. lol
12 posted on 04/13/2010 12:20:32 PM PDT by TribalPrincess2U (demonicRATS... taxes, pain and slow death. Is this what you want?)
To: TribalPrincess2U
I'm sure some of their "buns" are quite moldy by now!...............
13 posted on 04/13/2010 12:32:43 PM PDT by Red Badger (Education makes people easy to lead, difficult to drive; easy to govern, but impossible to enslave.)
To: Ernest_at_the_Beach
BFL
14 posted on 04/13/2010 1:52:18 PM PDT by zeugma (Waco taught me everything I needed to know about the character of the U.S. Government.)
To: Ernest_at_the_Beach
I wonder how many of our readership question the ramification of the phrase "global mean temperature". When applied to ground/sea measurements taken back in time to lets say mid eighteen hundreds does it really have much of a meaning?
The term is used to frequently. How do we know how much global surface was carefully measured year after year on land and on sea. And how reliable where the given data collection methods. Just as an example... how much of the world's seas did Great Britain, the USA and other European nations carefully throw buckets into the surface water and draw out and carefully measure temperature and salinity.
This whole system looks more suspect to inaccuracies the more I gaze at so many graphs.
And as all now admit... in recent times fewer and fewer ground stations across many land masses came into play. The grids increased for their models. More fudging took place. Until very recently we did not even have any real reliable satellite systems in place that within their given design can be relied on for providing accurate measurement of a given phenomena whether it be surface temperature, ice extent, atmospheric temperature measurements etc..
The whole process is full of question marks in my mind.
15 posted on 04/13/2010 9:23:14 PM PDT by Marine_Uncle (Honor must be earned....)
To: Marine_Uncle
The measurement scheme is susceptible to fraud....which is where we are at the moment.
16 posted on 04/14/2010 7:19:05 AM PDT by Ernest_at_the_Beach ( Support Geert Wilders)
To: Ernest_at_the_Beach
notice how even in their graph, that the temperature goes DOWN between ~1940 and ~1960, the biggest increase in man made CO2 production in history... and it went down.
so much for linking CO2 and globull warming...
17 posted on 04/14/2010 4:46:27 PM PDT by Chode (American Hedonist *DTOM* -ww- NO Pity for the LAZY)
AGW is a fraud reference bump! ;-)
18 posted on 04/14/2010 8:34:38 PM PDT by Tunehead54 (Nothing funny here ;-)
To: Ernest_at_the_Beach
Where’s the Hockey Schtick??
19 posted on 04/14/2010 8:46:21 PM PDT by seawolf101
Disclaimer: Opinions posted on Free Republic are those of the individual posters and do not necessarily represent the opinion of Free Republic or its management. All materials posted herein are protected by copyright law and the exemption for fair use of copyrighted works.
Free Republic Browse · Search News/Activism Topics · Post Article
FreeRepublic, LLC, PO BOX 9771, FRESNO, CA 93794 | 2,438 | 9,829 | {"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} | 3.203125 | 3 | CC-MAIN-2021-04 | latest | en | 0.939262 |
https://www.airmilescalculator.com/distance/pgd-to-ilg/ | 1,611,060,625,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703518240.40/warc/CC-MAIN-20210119103923-20210119133923-00162.warc.gz | 681,051,474 | 40,598 | # Distance between Punta Gorda, FL (PGD) and Wilmington, DE (ILG)
Flight distance from Punta Gorda to Wilmington (Punta Gorda Airport (Florida) – Wilmington Airport (Delaware)) is 953 miles / 1534 kilometers / 828 nautical miles. Estimated flight time is 2 hours 18 minutes.
Driving distance from Punta Gorda (PGD) to Wilmington (ILG) is 1107 miles / 1781 kilometers and travel time by car is about 19 hours 40 minutes.
## Map of flight path and driving directions from Punta Gorda to Wilmington.
Shortest flight path between Punta Gorda Airport (Florida) (PGD) and Wilmington Airport (Delaware) (ILG).
## How far is Wilmington from Punta Gorda?
There are several ways to calculate distances between Punta Gorda and Wilmington. Here are two common methods:
Vincenty's formula (applied above)
• 953.044 miles
• 1533.775 kilometers
• 828.172 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 954.817 miles
• 1536.629 kilometers
• 829.713 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Punta Gorda Airport (Florida)
City: Punta Gorda, FL
Country: United States
IATA Code: PGD
ICAO Code: KPGD
Coordinates: 26°55′12″N, 81°59′25″W
B Wilmington Airport (Delaware)
City: Wilmington, DE
Country: United States
IATA Code: ILG
ICAO Code: KILG
Coordinates: 39°40′43″N, 75°36′23″W
## Time difference and current local times
There is no time difference between Punta Gorda and Wilmington.
EST
EST
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 148 kg (325 pounds).
## Frequent Flyer Miles Calculator
Punta Gorda (PGD) → Wilmington (ILG).
Distance:
953
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
953
Round trip? | 521 | 1,980 | {"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} | 2.515625 | 3 | CC-MAIN-2021-04 | latest | en | 0.76159 |
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Posts: 22 | 616 | 1,742 | {"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} | 2.515625 | 3 | CC-MAIN-2023-40 | latest | en | 0.846141 |
https://physics.stackexchange.com/questions/109409/what-is-the-connection-between-conformal-field-theory-and-renormalization-group | 1,726,628,047,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651835.68/warc/CC-MAIN-20240918000844-20240918030844-00813.warc.gz | 406,734,303 | 43,013 | # What is the connection between Conformal Field Theory and Renormalization group in QFT?
As I know, the fundamental concept of QFT is Renormalization Group and RG flow. It is defined by making 2 steps:
1. We introduce cutting-off and then integrating over "fast" fields $\widetilde{\phi}$, where $\phi=\phi_{0}+\widetilde{\phi}$.
2. We are doing rescaling: $x\to x/L$: $\phi_{0}(x)\to Z^{-1/2}(L)\phi(x)$.
This procedure defines RG flow on the manifold of quasi local actions: $\frac{dA_{l}}{dl}=B\{A_{l}\}$.
In this approach we have such notions as crytical points $A_{*}$, relevant and irrelevant fields, Callan–Symanzik equation etc, and we can apply it, say, to phase transitions.
Also we can introduce stress-energy tensor $T^{\mu\nu}$. And, as far as i know, if we consider scale transformations $x^{\mu} \to x^{\mu}+\epsilon x^{\mu}$ , we can obtain Callan–Symanzik equation, and if the theory have a crytical point: $\beta^{k}(\lambda^{k})=0$, then trace of stress energy tensor $\Theta(x)=T^{\mu}_{\mu}=0$, so our correlation functions have symmetry at scaling transformation.
So the question is: As far as I know, at this point they somehow introduce conformal transformations and Conformal Field Theory. Can you explain, what place in Quantum Field Theory CFT takes? (I mean connection between them, sorry if the question is a little vague or stupid). How it relates to the RG approach exactly? (This point is very important for me). Maybe some good books?
• Well it seems that you have all pretty much figured out: conformal field theories are a subset of quantum field theories corresponding to the point(s) at which the beta function vanishes. It may seem not much interesting to look at subset of 'vanishing measure' in the space of quantum field theories but actually: conformal symmetry is a really strong constraint and is enough to solve exactly some theories in 2d, and you can derive results near critical points from the so called conformal perturbation theory. The main references are Ginsparg lecture notes (arXiv) and thee book by DiFran & al. Commented Apr 21, 2014 at 22:23
• @Learningisamess: There's a lot of good writing about CFT which doesn't stick to the comforts of 2 dimensions. See, e.g., sites.google.com/site/slavarychkov or physics.ipm.ac.ir/phd-courses/semester7/CFT-course-2013.pdf Commented Apr 22, 2014 at 0:12
• @Learningisamess Thanks for explanation. Correct me please if I've misunderstood something: when we analyze QFT renormalization using RG approach, we obtain important notion as critical points. Then we introduce stress-energy tensor and obtain that it's trace equals to zero at them. But this condition allows us to introduce conformal symmetry preserving the vanishing trace and therefore Conformal Field Theory at critical points. Am I right? If so, why it wasn't introduced some analog of RG analysis based on conformal symmetry(not only in critical points)? This is due to technical difficulties? Commented Apr 24, 2014 at 17:33
• Any quantum field theory which has hope of having an UV-completion can be viewed as as effective theory at point in the RG flow from an UV complete theory.
• Field theories at the UV fixed point are conformal.
• Hence all 'well-defined' field theories are either CFTs or points in the RG flow from one (UV) CFT to another CFT.
So in a sense, in Kaplan's words:
studying the space of CFTs basically amounts to studying the space of all well-defined QFTs
So that's one way of seeing the place of CFTs in general QFTs.
zzz's answer concerns QFTs as used in particle physics. However, OP also mentions phase transitions, thus statistical field theory, where the relationship between CFT and scale-invariant RG fixed points is more subtle.
Indeed, for local QFTs where unitarity is assumed, there are theorems showing that scale invariance implies conformal invariance, see e.g. https://en.wikipedia.org/wiki/C-theorem for the theorem of A. Zamolodchikov in 2D and discussion of the situation in 4D.
However, it is known that this link does not extend to general statistical field theories: for example, [1] use elasticity as a counter-example. The link between scale invariance and conformal invariance has been reviewed in [2].
[1] V. Riva and J. Cardy. "Scale and conformal invariance in field theory: a physical counterexample." Physics Letters B 622.3-4 (2005): 339-342.
[2] Y. Nakayama. "Scale invariance vs conformal invariance." Physics Reports 569 (2015): 1-93. | 1,120 | 4,476 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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} | 2.59375 | 3 | CC-MAIN-2024-38 | latest | en | 0.912196 |
https://www.tynker.com/elementary-school/coding-curriculum/programming-101/1-introduction? | 1,718,417,892,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861578.89/warc/CC-MAIN-20240614235857-20240615025857-00647.warc.gz | 951,237,267 | 20,193 | Elementary School course
# Programming 101
• BEGINNER
• 15 LESSONS
• Voiceovers
### U.S. Standards
• CCSS-Math: 3.NBT.A.2, MP.1
• CCSS-ELA: RF.3.4.A, RF.4.4.A
• CSTA: 1B-AP-10, 1B-AP-11, 1B-AP-15
• CS CA: 3-5.AP.12, 3-5.AP.13, 3-5.AP.17
• ISTE: 1.c, 1.d, 4.d, 5.c, 5.d, 7.c
### U.K. Standards
##### Key stage 2
Pupils should be taught to:
• design, write and debug programs that accomplish specific goals, including controlling or simulating physical systems; solve problems by decomposing them into smaller parts
• use sequence, selection, and repetition in programs; work with variables and various forms of input and output
• use logical reasoning to explain how some simple algorithms work and to detect and correct errors in algorithms and programs
• understand computer networks, including the internet; how they can provide multiple services, such as the World Wide Web, and the opportunities they offer for communication and collaboration
• use search technologies effectively, appreciate how results are selected and ranked, and be discerning in evaluating digital content
• select, use and combine a variety of software (including internet services) on a range of digital devices to design and create a range of programs, systems and content that accomplish given goals, including collecting, analysing, evaluating and presenting data and information
• use technology safely, respectfully and responsibly; recognise acceptable/unacceptable behaviour; identify a range of ways to report concerns about content and contact
## Lesson 1: Introduction
• Pick up the Candy
• Get the Gum Drop
• Avoid Obstacles
• Repetition with Loops
• Repeat the Pattern
• Sequenced Repetition
• Two Loops
• Using Conditional Loops
• Using "Not" in Loops
• Conditionals
• Multiple Conditionals
• If-Else
## Description
An easy introduction to programming for beginners in lower elementary grades. Familiarize your class with visual programming techniques. Students progress through the lessons learning concepts in a game-like interface. To complete each lesson, students typically go through a concept review, solve a puzzle, run through a tutorial, build their own project, and take a quiz. They create interactive stories, animations, and mini-games to help Professor Ada battle the evil Dr. Glitch! After completing this lesson plan, students will be able to build a wide variety of simple programs with events, loops, and some conditional logic.
## Topics
• Sequencing
• Repetition
• Events
• Conditional logic
• Animation
• Pen drawing
• Drawing shapes and patterns
• Playing musical notes
• Sending and receiving messages
• Handling user input
• Color detection
## What Students Learn
• Design animated characters
• Create interactive scenes
• Make animated birthday cards
• Write cartoon stories
• Create a music machine
• Experiment with math art
• Design and build small games
• Troubleshoot and debug simple programs
## Technical Requirements
* Online courses require a modern desktop computer, laptop computer, Chromebook, or Netbook with Internet access and a Chrome (29+), Firefox (30+), Safari (7+), or Edge (20+) browser. No downloads required.
* Tablet courses require an iPad (iOS 10+) with Tynker or Tynker Junior app installed and Internet access
## Lesson 1 : Introduction Programming 101
Time: 60+ minutes
### Activities (45 minutes)
Facilitate as students complete all Introduction modules on their own:
1. Pick Up the Candy (Puzzle)
2. Get the Gum Drop (Puzzle)
3. Avoid the Obstacles (Puzzle)
4. Repetition with Loops (Puzzle)
5. Repeat the Pattern (Puzzle)
6. Sequenced Repetition (Puzzle)
7. Two Loops (Puzzle)
8. Using Conditional Loops (Puzzle)
9. Using "Not" in Loops (Puzzle)
10. Conditionals (Puzzle)
11. Multiple Conditionals (Puzzle)
12. If-Else (Puzzle) | 923 | 3,797 | {"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} | 2.671875 | 3 | CC-MAIN-2024-26 | latest | en | 0.846696 |
https://studylotph.com/physics/question524860741 | 1,679,612,733,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945218.30/warc/CC-MAIN-20230323225049-20230324015049-00255.warc.gz | 613,013,090 | 14,595 | # A 0. 50 kg ball was strucked by a baseball bat from rest up to a speed of 50 m/s. The ball was in contact with the bat for 0. 05 seconds. A. What is the change in the momentum of the ball? B. What was the impulse exerted on the ball? C. Calculate the average force exerted on the ball by the bat.
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### Questions in other subjects:
English, 29.10.2021 05:15
Math, 29.10.2021 05:15 | 507 | 1,542 | {"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} | 3.078125 | 3 | CC-MAIN-2023-14 | latest | en | 0.89622 |
https://www.gradegorilla.com/micro/energy/M_energy3.php | 1,712,941,036,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816045.47/warc/CC-MAIN-20240412163227-20240412193227-00721.warc.gz | 729,142,491 | 4,289 | 10 minutes maximum! Can you do it in 5?
1. Which of these is used to calculate the efficiency of an energy transfer or shift?
A. useful energy output energy input B. energy input useful energy output C. energy input wasted energy output D. wasted energy output energy input
Q2-4. This diagram shows the energy transfer as a skateboarder rolls downhill:
2. What is the percentage efficiency of this transfer of energy?
• A. 600%
• B. 70%
• C. 30%
• D. 1400%
3. What form of energy is the wasted energy 'X' likely to be?
• A. Elastic energy stored in the skateboard.
• B. Chemical energy in the muscles.
• C. Wind energy from movement.
• D. Thermal energy from friction.
4. The energy in 'X' can be given as a value or as a percentage. Which of these is correct?
Energy in X (J) Energy in X (%)
A
600 30
B
1400 70
C
30 600
D
70 1400
5-7. This diagram shows the efficiency of a solar panel:
5. What does the 'W' represent in the measurment '40 W' in the diagram?
• A. Work Done
• B. Weight
• C. Wasted energy
• D. Watts
6. What is the power input to the panel?
• A. 40 W
• B. 80 W
• C. 120 W
• D. 160 W
7. What is the percentage efficiency of this solar panel?
• A. 75 %
• B. 67 %
• C. 33 %
• D. 25 %
8. "Energy can be transferred usefully, stored or dissipated, but cannot be created or destroyed."
This is called the law of..
• A. transfer of energy.
• B. creation of energy.
• C. conversion of energy.
• D. conservation of energy.
9&10. A wave power generator is sold with the following specifications:
Typical Power Output: 0.2 kW Efficiency : 20%
9. 0.2 kW is the same as...
• A. 0.0002 W
• B. 20 W
• C. 200 W
• D. 2000 W
10. The typical power input to the generator is:
• A. 1 kW
• B. 4 kW
• C. 0.01 kW
• D. 0.04 kW | 536 | 1,745 | {"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} | 3.046875 | 3 | CC-MAIN-2024-18 | latest | en | 0.805104 |
https://math.stackexchange.com/questions/4711324/benacerrafs-identification-problem-and-pa-categoricity | 1,695,500,766,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506528.3/warc/CC-MAIN-20230923194908-20230923224908-00335.warc.gz | 425,340,887 | 33,621 | # Benacerraf's identification problem and PA categoricity
Lately I have been interested in mathematical philosophy, and especially structuralism.
In this setting, Benacerraf's famous paper is a classic and works as follows: Take the Zermelo ordinals ($$x \to \{x\}$$) and the Von Neumann ordinals ($$x \to x \cup \{x\}$$), those two are models of First order Peano Arithmetic $$PA$$, but they do not give the same answer to the sentence $$1 \in 3$$ (respectively false and true).
Benacerraf uses this to then conclude that when talking about numbers, one cannot do anything appart from talking of a given number in relation with the others, as a so-called structure.
On the other hand, $$PA$$ is not categorical: there exists multiple classes of models up to isomorphism.
The question is thus the following: is Benacerraf's identification problem specific to the fact that $$PA$$ is not categorical ?
• The Zermelo vs. von Neumann example doesn’t have to do with categoricity, since they are isomorphic models. Jun 2 at 14:56
• Oh yeah, my bad. Then from what I understand my question is not relevant in the end ? Jun 2 at 15:08
• I was inspired by your question to read the entry on mathematical structuralism in the Stanford Encyclopedia of Philosophy, plato.stanford.edu/entries/structuralism-mathematics. The issue you raise comes up repeatedly, with way too many subtleties to get into here. Jun 2 at 15:16
• I’m finding it hard to imagine an answer to your question that wouldn’t be mostly the writers’ opinions. Maybe try the philosophy stackexchange? They have a tag philosophy-of-mathematics. Jun 2 at 15:21
• Oh great yeah, I'll try that. Though after some thinking, both ordinals defined above are also models of second order PA, which is categorical, so I'd say that I missed the point Jun 2 at 19:39 | 446 | 1,818 | {"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": 6, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2023-40 | latest | en | 0.937847 |
https://edurev.in/course/quiz/attempt/16312_Arun-Sharma-Test-Pie-Chart-1/0cc325ad-3fcd-49d7-b0d7-a67694ab8f67 | 1,627,733,793,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154089.6/warc/CC-MAIN-20210731105716-20210731135716-00179.warc.gz | 243,816,635 | 40,846 | Courses
# Arun Sharma Test: Pie Chart- 1
## 5 Questions MCQ Test Integrated Reasoning for GMAT | Arun Sharma Test: Pie Chart- 1
Description
This mock test of Arun Sharma Test: Pie Chart- 1 for UPSC helps you for every UPSC entrance exam. This contains 5 Multiple Choice Questions for UPSC Arun Sharma Test: Pie Chart- 1 (mcq) to study with solutions a complete question bank. The solved questions answers in this Arun Sharma Test: Pie Chart- 1 quiz give you a good mix of easy questions and tough questions. UPSC students definitely take this Arun Sharma Test: Pie Chart- 1 exercise for a better result in the exam. You can find other Arun Sharma Test: Pie Chart- 1 extra questions, long questions & short questions for UPSC on EduRev as well by searching above.
QUESTION: 1
Solution:
QUESTION: 2
### The following pie-charts show the distribution of students of graduate and post-graduate levels in seven different institutes in a town. What is the ratio between the number of students studying at post-graduate and graduate levels respectively from institute S?
Solution:
Required = 21% of 24700 /14% 27300 = 19/14
QUESTION: 3
### The following pie-charts show the distribution of students of graduate and post-graduate levels in seven different institutes in a town. How many students of institutes of M and S are studying at graduate level?
Solution:
QUESTION: 4
The following pie-charts show the distribution of students of graduate and post-graduate levels in seven different institutes in a town.
What is the ratio between the number of students studying at post-graduate level from institutes S and the number of students studying at graduate level from institute Q?
Solution:
QUESTION: 5
The following pie-charts show the distribution of students of graduate and post-graduate levels in seven different institutes in a town.
Total number of students studying at post-graduate level from institutes N and P is
Solution: | 433 | 1,947 | {"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} | 3.71875 | 4 | CC-MAIN-2021-31 | longest | en | 0.935727 |
https://www.physicsforums.com/threads/entropy-and-time-reversal.125820/ | 1,532,258,959,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593208.44/warc/CC-MAIN-20180722100513-20180722120513-00363.warc.gz | 946,076,969 | 13,497 | # Entropy and time reversal
1. Jul 12, 2006
### leonchik1976
How time reversal of wave function and 2nd law of thermodynamics of entropy are settle together?
2. Jul 12, 2006
### vanesch
Staff Emeritus
Several answers are possible. One possible answer is that some people will say that projection is not time-reversible, so if projection is taken seriously, this kills time-symmetry in physics.
But even if you do not consider this possibility, I'd say that the answer is the same as in classical physics, where time-reversible LAWS can give rise to non-time-reversible *coarse-grained* dynamics: namely by special initial conditions.
If you start out (in a classical, time-reversible dynamics) with a highly peculiar initial state, and you only look at low-order correlation functions (coarse-graining), then you obtain a time-irreversible dynamics of these correlation functions until they reach their equilibrium values.
Mind you, I'm not saying that this is what explains finally the entropy increase in our universe. I'm only giving the example that there is no contradiction between time-reversible dynamics, and a second law of thermodynamics which prescribes the irreversibility of low-order correlation functions: it is sufficient to take a peculiar initial state.
It might of course be that there are genuinly irreversible laws too.
But there is no *contradiction* between the second law of thermodynamics, and time-symmetrical microdynamics.
3. Jul 13, 2006
### lalbatros
It is just a question of time !
4. Jul 13, 2006
### icarus3k
From what i know of, irreversible wave function collapses are supposedly the reason for the second law of thermodynamics.
Am I right here?
----------------------------------
Daring to Fly
http://icarus3k.blogspot.com [Broken]
Last edited by a moderator: May 2, 2017
5. Jul 14, 2006
### wavemaster
I guess you can live without believing irreversible wavefunction collapse.
Following Feynman's example, increasing entropy is a consequence of statistical behavior.
Last edited: Jul 14, 2006 | 471 | 2,050 | {"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} | 2.59375 | 3 | CC-MAIN-2018-30 | latest | en | 0.899682 |
https://demography.subwiki.org/wiki/Parity_progression_ratio | 1,642,904,837,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303956.14/warc/CC-MAIN-20220123015212-20220123045212-00299.warc.gz | 245,375,597 | 5,913 | # Parity progression ratio
The term parity progression ratio (PPR) is typically used in the context of a cohort of females (such as a birth cohort) that has completed its childbearing years. Given such a cohort, and a nonnegative integer $i$, the parity progression ratio from birth order $i$ to birth order $i + 1$ is defined as the ratio of the completed cohort fertility for that cohort for birth order $i+1$, to the completed cohort fertility for that cohort for birth order $i$.
Alternatively, the PPR describes the probability that a person in that cohort who has had $i$ children will proceed to have an $(i+1)^{th}$ child. | 146 | 631 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 7, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2022-05 | latest | en | 0.945757 |
http://forum.arduino.cc/index.php?topic=50037.0 | 1,508,445,389,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823462.26/warc/CC-MAIN-20171019194011-20171019214011-00479.warc.gz | 126,514,662 | 10,759 | Go Down
### Topic: Measuring the Frequency of a TTL circuit (Read 12882 times)previous topic - next topic
##### Jan 27, 2011, 01:57 am
I'm looking for help in measuring accurately a 20KHZ digital signal, its to read the pulses from a 500 pulse per revolution encoder that at top speed will spin around 1700 RPMs. Any ideas on how fast the arduino can read a digital signal and how to count them? (Also need to read two at a time, so will want to connect to more than one pin)
Thanks!
#### mk3
#1
##### Jan 27, 2011, 02:27 am
Are you doing Quadrature? Do you need to know the direction of rotation?
#### MarkT
#2
##### Jan 27, 2011, 04:06 am
One approach is to use the signal to trigger an interrupt (pins 2 or 3) and in the interrupt routine read a timer register. If the interrupt is set up on signal CHANGE, then you can time the high and low half-cycles separately and get the period and mark-space ratio too.
The standard timer setups on the Arduino are for PWM generation which might not be ideal for this (4us clock step, 8 bit). Timer 1 can be
configured up to 16 bit and timers can run upto 16MHz.
If high accuracy is not required pulseIn() might be the function you want.
If you are prepared to wait a bit, count the signal transitions over a measured stretch of time - the longer the time the more accurate you can be.
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]
#3
##### Jan 27, 2011, 06:12 amLast Edit: Jan 27, 2011, 06:16 am by tchadwick Reason: 1
It is for a quadrature encoder, however I already have a IC chip that converts the A and B signal into two separate pins for forward and back, so now all I need to do is count those pulses to know the RPM of the motor this encoder is connected to. (Using it for a PWM output eventually, but for now, just focusing on getting accurate sampling)
Thanks for the replies thus far! Keep 'em comin'!
**EDIT**
Oh and using an Oscilloscope, have measured the "forward pulses" at about 11khz, hence the need for a quick sample.
#4
##### Jan 27, 2011, 07:12 am
You can feed the pulse into the TO or T1 pins and have the counter/timer count them for you.
______
Rob
Rob Gray aka the GRAYnomad www.robgray.com
#### mk3
#5
##### Jan 27, 2011, 12:49 pm
I am working on a related project right now and using the approach of user Graynomad. I have a single encoder signal and will use it to clock TIMER1. Next at a regular interval I can find the count of TIMER1 and know how many pulses have occurred. I like this approach because the timer does not interrupt the cpu.
However, I think you need to consider what else you will be doing with the Arduino board and what conflicts you will have. For example, I think TIMER1 is used for Arduino PWM (analog) out. I think TIMER0 is used for the millis() function. Being a beginner myself you should confirm this information elsewhere as I could be wrong.
I think your motor is at 1320 RPM for 11kHz out. Also, you will not hit 20kHz if your max speed is 1700RPM. ..maybe you are being conservative in your request to reach 20kHz max.
Do you know what the IC chip is doing? A quadrature encoder would have two signals coming out of it (from what I understand of course) so if the IC chip is also putting out two signals... what is it doing? Maybe it is a Schmitt trigger? Is it changing the logic levels?
#### Ran Talbott
#6
##### Jan 28, 2011, 11:00 pm
How often do you need to get the RPM data? The CD4060 can divide the tach signal by as much as 16K in a single chip. Plus, by accumulating multiple pulses, it would remove most of the jitter you'll see if you measure every pulse.
The downside is that there could be an unacceptably long time between the prescaled pulses when the motor is spinning slowly.
#7
##### Jan 29, 2011, 08:29 pmLast Edit: Jan 29, 2011, 08:41 pm by tchadwick Reason: 1
We are using a quadrature encoder, however our IC chip takes the A/B Pulses and outputs on two pins their combined pulse. (ex. Pin 1 is putting out an 11khz pulse telling me that the motor is rotating at about 1300 RPMs in forward motion, while pin2 is kept in a low state. Vice versa for when the motor rotates backwards, causing pin 1 to go low and pin 2 to start pulsing) I would go with the Idea from Graynomad, save I do need PWM for driving said motor. But thanks for the ideas so far!
I was thinking about using this library: http://www.arduino.cc/playground/Main/MsTimer2 And set a sample in the "Loop" portion of the program so that it is continously sampling and every time there is a change in state from low to high (thus eliminating the chance of sampling the same pulse twice) I could increment a counter. Then every 1 second, it outputs the count to the screen. If this sparks any ideas, let me know.
#8
##### Jan 30, 2011, 03:34 am
What is the range of the input frequency and the required display update rate. If it's 0-11k then using a prescaler could be a problem at low freqs as Ran said, unless you only have to update a display (or use the info in whatever way) every few seconds.
_____
Rob
Rob Gray aka the GRAYnomad www.robgray.com
#9
##### Jan 30, 2011, 03:40 am
Quote
Vice versa for when the motor rotates backwards, causing pin 1 to go low and pin 2 to start pulsing
Are you saying that this chip swaps the function of the two pins according to the direction
If so I don't see any easy way to detect this as you have to differentiate between a frequency and a DC level, and at slow speeds especialy there's no real difference.
Can you post a link to the decoder chip spec sheet?
_____
Rob
Rob Gray aka the GRAYnomad www.robgray.com
#10
##### Jan 30, 2011, 05:10 amLast Edit: Jan 30, 2011, 05:14 am by tchadwick Reason: 1
Thanks for the replies, I will start digging for the spec sheet on Monday. In the mean time the decoder chip converts the AB signals, to a speed pulse on pin 1 for forward and if the motor reverses, it stops sending pulses on pin 1 and starts sending pulses on pin 2. Which should make it easier to sample the incoming data especially while I'm just trying to determine the RPM in one direction (for now )
**EDIT**
So in forward pin one (TTL)pulses and pin 2 is off. Then in reverse pin one turns off and pin 2 starts a TTL pulse at a proportional value to RPMs. Oh and we are using the Arduino MEGA.
#11
##### Jan 30, 2011, 05:59 amLast Edit: Jan 30, 2011, 06:02 am by Graynomad Reason: 1
Quote
So in forward pin one (TTL)pulses and pin 2 is off. Then in reverse pin one turns off and pin 2 starts a TTL pulse at a proportional value to RPMs.
IMO then that is a seriously bad chip, a pulse pin and direction pin would be better, but anyway,
To do this I think you have to
a) OR the two pins into a single interrupt (using an AND gate or two diodes) and run one of the pins to a digital input. Thus when you get an interrupt you test the input to see which pin was the culprit.
b) use two interrupts, one ISR sets a "direction" global, the other resets it.
Option b) is simpler and doesn't need any external components, but a) is better if you only have a single interrupt free.
NOTE: this is just for direction, it doesn't handle the counting.
______
Rob
Rob Gray aka the GRAYnomad www.robgray.com
#12
##### Feb 02, 2011, 04:18 am
Here is an attempt at coding, I'm looking for any thoughts for improvement and more importantly for thoughts of implementation.
#include <MsTimer2.h>
//Output the value of count to the serial monitor
int buttonstate = 0;
long int counter = 0;
void flash()
{
Serial.println(counter, DEC);
counter = 0;
}
void setup()
{
pinMode(9, INPUT);
MsTimer2::start();
}
void loop()
{
if(buttonstate == LOW)
{
if(buttonstate == HIGH)
{
counter++;
}
}
}
#13
##### Feb 02, 2011, 04:50 am
So in this example you are waiting for a button to be pressed, then counting when it's released.
I think this has problems, you're not waiting for the button to be released, simply testing once straight after it was pressed when of course it will still be LOW.
Code: [Select]
` if(LOW == digitalRead(9)) { while (LOW == digitalRead(9)) {}; // wait for button to be released counter++; }`
This assumes the input is clean and doesn't need debouncing. If you are using a button for initial testing (as the pin name implies) then you will need to debounce it.
______
Rob
Rob Gray aka the GRAYnomad www.robgray.com | 2,233 | 8,383 | {"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} | 2.65625 | 3 | CC-MAIN-2017-43 | latest | en | 0.888929 |
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# If p and q are integers, is p + q odd? (1) p/3 is not an odd integer.
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If p and q are integers, is p + q odd? (1) p/3 is not an odd integer. [#permalink]
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If p and q are integers, is p + q odd?
(1) p/3 is not an odd integer.
(2) p − q is an even integer.
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Re: If p and q are integers, is p + q odd? (1) p/3 is not an odd integer. [#permalink]
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05 Mar 2019, 05:05
Bunuel wrote:
If p and q are integers, is p + q odd?
(1) p/3 is not an odd integer.
(2) p − q is an even integer.
we are asked to determine whether p + q is odd or not.
Statement 1: p/3 is not an odd integer. It indicates that p is even but we don't know the odd/even properties of q. NOT sufficient.
Statement 2: p - q = even. 2 cases here.
1. even - even = even.
2 Odd - Odd = even
even + even = even
odd + odd =even .
In both cases we get even as the sum of p + q. Sufficient.
p + q is not odd.
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If p and q are integers, is p + q odd? (1) p/3 is not an odd integer. [#permalink]
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Updated on: 05 Mar 2019, 12:10
Bunuel wrote:
If p and q are integers, is p + q odd?
(1) p/3 is not an odd integer.
(2) p − q is an even integer.
smart question
#1
p/3 is not an odd integer implies p can be even or a fraction ..
q not known in sufficient
#2
p − q is an even integer
means both p & q have to either odd or even
and sum of both odd & even integers is always even
so sufficient to say that p+q is not odd
IMO B
Originally posted by Archit3110 on 05 Mar 2019, 11:00.
Last edited by Archit3110 on 05 Mar 2019, 12:10, edited 1 time in total.
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Re: If p and q are integers, is p + q odd? (1) p/3 is not an odd integer. [#permalink]
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05 Mar 2019, 12:05
Statement 1 can't be sufficient, because it gives no information about q.
Addition and subtraction obey identical even/odd rules (because subtracting is addition, you're just adding a negative) so any time a-b is even, so is a+b, and vice versa, and Statement 2 is instantly sufficient to give a 'no' answer to the question, so the answer is B.
I should add that the above is true only if a and b are integers, but that's always going to be the case in a GMAT even/odd problem.
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Re: If p and q are integers, is p + q odd? (1) p/3 is not an odd integer. [#permalink]
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05 Mar 2019, 12:09
And I'd point out one mistake in the other posts above: Statement 1 says "p/3 is not an odd integer". That sentence does not mean that p is even. Certainly p might be even, but p could also be 7, say, since 7/3 is not an odd integer.
If instead the statement said "p/3 is an even integer", then that would ensure that p itself is even.
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Re: If p and q are integers, is p + q odd? (1) p/3 is not an odd integer. [#permalink] 05 Mar 2019, 12:09
Display posts from previous: Sort by | 1,289 | 4,392 | {"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-2019-51 | latest | en | 0.871844 |
https://braineaser.com/brainteasers/unique-number-of-letters/ | 1,713,935,105,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819067.85/warc/CC-MAIN-20240424045636-20240424075636-00064.warc.gz | 123,894,825 | 11,242 | # Unique Number of Letters
Out of all whole numbers between one and five thousand, there is only one number that has a unique number of letters when you spell it out. What number is it?
Include spaces and dashes. For example, “twenty-four” and “two hundred” both have 11 letters.
#### Solution
Three thousand
Here is the reasoning:
• First, we rule out the relatively uniquely-named numbers between ten and twenty.
• “Eleven” has 6 letters, but so does “twelve”
• “Thirteen” has 8 letters, but so does “fourteen”, “eighteen”, and “nineteen”
• “Fifteen” has 7 letters, but so does “sixteen”
• “Seventeen” has 9 letters, but so does “forty-two”
• Next, recognize that all of the nonzero digits share a number of characters with other digits. So the number must end in zero.
• “One”, “two”, and “six” have 3 letters
• “Four”, “five”, and “nine” have 4 letters
• “Three”, “seven”, and “eight” have 5 letters
• Similarly, all of the tens share a number of characters with other numbers. So the number must end in two zeroes.
• “Ten” has 3 letters, but so does “one”
• “Forty”, “fifty”, and “sixty” have 5 letters
• “Twenty”, “thirty”, “eighty”, and “ninety” have 6 letters
• “Seventy” has 7 letters, but so does “fifteen”
• Similarly, all of the hundreds share a number of characters with other numbers. So the number must end in three zeroes.
• This leaves the thousands. Since we limited to numbers up to five thousand, that means “three thousand” has a unique number of letters (“seven thousand” and “eight thousand” are out of the range).
• Finally, confirm “three thousand” is the only number between 1-5000 with 14 letters.
• “Seventy-seven” is the longest 2-digit number, and it has only 13 letters.
• Out of the 3-digit numbers, the multiples of 100 are too short (tied for longest is “three hundred” with 13 letters) and all others are too long (tied for shortest is “two hundred ten” with 15 letters).
1. #### Petrov
While your reasoning is somewhat sound, your answer is incorrect.
There is a number that shares an amount of letters with three thousand – namely, One hundred one. They both share thirteen letters. Your idea of crossing out places isn’t a terrible idea, but it leads you to an incorrect answer.
The correct answer is that there is no answer.
The largest amount of letters possible is 37, with a few combinations, but we shall use the number three thousand eight hundred seventy eight. This is our maximum amount of letters possible.
Our lowest amount of numbers is 3, shared between one, two, and six.
The answer must be some where in between, right?
But each number of letters between 3 and 37 has multiple numbers that are valid.
If you wish for proof, I have a text document proving my thesis that you may ask for.
• #### BrainEaser
Thanks for the detailed reply – your criticism is valid.
I’ve revised the puzzle to include spaces/dashes, and added an explanation of how the solution is now unique. | 735 | 2,935 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2024-18 | latest | en | 0.931643 |
http://mathhelpforum.com/algebra/210896-solve-proportions-using-cross-products.html | 1,498,636,846,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128322873.10/warc/CC-MAIN-20170628065139-20170628085139-00160.warc.gz | 266,050,800 | 10,523 | # Thread: Solve Proportions using cross products
1. ## Solve Proportions using cross products
My sister thinks these problems are unsolvable for x. I can't find any examples in my book. I know how to solve basic proportion problems, but these two are hard.
1. x : x-3 = x+4 : x
2. x+1 : 6 = x-1 : x
Thanks!
2. ## Re: Solve Proportions using cross products
Hey BrokenRitual.
Try getting rid of the denominators and collect the terms.
For the first one we have x^2 + 4x =x^2 - 3x or 7x = 0 => x = 0 but this is a contradiction since x is on the denominator and can't be zero. So no solution exists.
You try the same kind of technique for the second one.
3. ## Re: Solve Proportions using cross products
Hello, BrokenRitual!
Why do you say these are "hard"?
$[1]\;\;x : x-3 \:=\: x+4 : x$
We have: . $\frac{x}{x-3} \:=\:\frac{x+4}{x}$
Cross product: . $(x)(x) \:=\:(x-3)(x+4)$
n . . . . . . . . . . . . . $x^2 \:=\:x^2 + x - 12$
. . . . . . . . . . . . . . . $0 \:=\:x - 12$
. . . . . . . . . . . . . . . $x \:=\:12$
$[2]\;\; x+1 : 6 \:=\: x-1 : x$
We have: . $\frac{x+1}{6} \:=\:\frac{x-1}{x}$
Cross product: . $(x+1)(x) \:=\:6(x-1)$
n . . . . . . . . . . . . . $x^2 + x \:=\:6x-6$
. . . . . . . . . . . $x^2 - 5x + 6 \:=\:0$
. . . . . . . . . $(x-2)(x-3) \:=\:0$
. . . . . . . . . . . . . . . . . . $x \:=\:2,\:3$ | 531 | 1,338 | {"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": 13, "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-2017-26 | longest | en | 0.572263 |
https://codedump.io/share/bro3wgxLoIG1/1/generating-permutations-of-values-within-multiple-lists | 1,477,095,307,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988718311.12/warc/CC-MAIN-20161020183838-00002-ip-10-171-6-4.ec2.internal.warc.gz | 837,922,372 | 8,500 | Derek Lee - 2 months ago 5x
R Question
# Generating Permutations of Values Within Multiple Lists
I'm trying to generate permutations by taking 1 value from 3 different lists
``````l <- list(A=c(1:13), B=c(1:5), C=c(1:3))
``````
Desired result => Matrix of all the permutations where the first value can be 1-13, second value can be 1-5, third value can be 1-3
I tried using permn from the combinat package, but it seems to just rearrange the 3 lists.
``````> permn(l)
[[1]]
[[1]]\$A
[1] 1 2 3 4 5 6 7 8 9 10 11 12 13
[[1]]\$B
[1] 1 2 3 4 5
[[1]]\$C
[1] 1 2 3
[[2]]
[[2]]\$A
[1] 1 2 3 4 5 6 7 8 9 10 11 12 13
[[2]]\$C
[1] 1 2 3
[[2]]\$B
[1] 1 2 3 4 5
....
``````
Expected output
`````` [,1] [,2] [,3]
[1,] 1 1 3
[2,] 1 2 1
[3,] 1 1 2
[4,] 1 1 3
and so on...
``````
We can use `expand.grid`. It can directly be applied on the `list`
``````expand.grid(l) | 420 | 923 | {"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} | 2.59375 | 3 | CC-MAIN-2016-44 | longest | en | 0.718452 |
http://seekingalpha.com/article/1199911-another-principal-protection-strategy-for-today-s-preferred-stock-investors?source=msn&industry=IND_BANKING&isub= | 1,412,180,455,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1412037663467.44/warc/CC-MAIN-20140930004103-00174-ip-10-234-18-248.ec2.internal.warc.gz | 284,819,330 | 24,265 | Seeking Alpha
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An interesting opportunity has presented itself with respect to six specific preferred stocks, an opportunity that allows preferred stock investors to purchase shares above these securities' \$25 par values and still realize a positive return on their investment.
While an earlier article explained a principal protection strategy that uses the wholesale Over-The-Counter stock exchange to purchase preferred stock shares below their \$25 par value (see "Principal Protection For Preferred Stock Investors During A Low Rate Environment"), this article presents a very different approach to principal protection.
## Calculating the Final Payment Amount
When a formal redemption announcement is made for a preferred stock, the company will state that they will be buying back the shares from holders on a specific date for a price equal to the security's par value (\$25.00 per share is used here) plus any accrued dividend amount since the most recently completed quarterly payment date. Annoyingly, the announcement will generally not state what that accrued distribution amount is, forcing everyone who cares to reach for their calculator.
Also remember that redemption announcements must be made at least 30 days in advance of the announced call date. This becomes important later.
Here is the formula used to calculate that final "accrued distribution" amount for a redeemed preferred stock (the final, and usually partial, quarterly dividend):
[accrued distribution] =
{ ( [par value] x [div rate%] ) / 360 } x ([number of days] - 1)
The [number of days] value is the number of days between the first day of the current dividend quarter (the day after the most recent dividend payment) and the announced call date (but not including the announced call date itself, so you need to subtract one day).
The [final payment] amount received by shareholders will therefore be equal to the [par value] plus the [accrued distribution]. The next time you receive a redemption notice for one of your preferreds you can use the above formula to figure out the exact amount of your final payment.
## Price Behavior of a Redeemed Preferred Stock
As you might guess, after a redemption announcement is made, but before the announced call date arrives, the market price of the redeemed preferred stock will jump right to the [final payment] amount, plus or minus a few pennies, and stay there as the announced call date approaches.
A current example is the widely held BWF trust preferred stock which Wells Fargo (NYSE:WFC) just called on January 30, 2013 for March 15, 2013. Feeling generous, Wells Fargo gave 45 days' notice to shareholders rather than the 30 days required by BWF's prospectus.
In their January 30 redemption announcement for BWF the bank stated "This issue will be called for redemption on March 15, 2013 at \$25.00 per share plus any accrued distributions." Using the above accrued distribution formula, the final payment for BWF should be:
\$25 + { ( \$25] times [7.875%] ) divided by 360 } times ([90] - 1)
or \$25.48 per share.
Notice how the market price of BWF has settled at this exact price, \$25.48 per share (Friday, February 15, 2013).
Using the above formula, the amount of the final payment can be calculated for any day that a redemption may occur throughout a dividend quarter.
If a redemption is not announced, those holding shares (regardless of the price they paid) will receive the upcoming quarterly dividend payment which counts toward the total return of their investment regardless of when a redemption may occur in the future.
## The Positive Return Line
Since the final payment includes the accrued distribution amount for the final quarter (or portion thereof), the issuing company will owe shareholders a larger amount for every day that goes by throughout a dividend quarter without a call.
The black line on this chart shows the final payment amount over a 90 day quarter that holders of a 6.75% preferred stock will receive using the above formula.
Those who purchase shares for a price that is below this line will realize a positive return on their investment in the event of a redemption (that is, their purchase price will be less than the final payment received on the final day).
It is for this reason that you will see preferred stock shares trading above their par value, even when they are obvious redemption candidates (e.g. today's bank-issued trust preferred stocks).
As noted earlier, companies usually give 30 days' notice of a redemption. If a redemption is announced 45 days into the quarter, for example, shareholders will actually receive the final payment 30 days later on day 75 (in an amount equal to 74 days of dividend income since the final day is not counted).
Obviously, those purchasing above par, but below the final payment amount, are hoping that the security is redeemed sooner rather than later. So this approach to principal protection is a shorter-term strategy and delivers returns more quickly if you can identify a preferred stock that is highly likely to be redeemed in the not too distant future.
## Who Ya Gonna Call
One way to gauge the likelihood of a call is discussed in the SA article titled "Is Your Preferred Stock about to be Called?" from April 2012.
Another way is if a spokesperson from the issuing company discloses something that they probably should not have.
On February 6, 2013 the executive team from ProLogis (NYSE:PLD) held their quarterly conference call (see transcript) to discuss the company's performance for Q4/2012 with Wall Street analysts. During his prepared remarks, CFO Thomas Olinger described their strategy in the wake of the April 2011 merger with AMB for selling underperforming properties and using the proceeds to pay down the company's obligations, saying
...As we've been saying for some time, this is an asset-driven plan, and our achievements will significantly improve our cost structure, our balance sheet and our liquidity. To put this all in perspective, by the end of 2013, we will have reduced our leverage, including preferreds from 50% at the merger to 37%...
After their prepared remarks, the floor was opened to questions from the analysts. Michael Bilerman from Citigroup Inc.'s research division asked Mr. Olinger to "peel back the onion a little bit" and provide the next layer of detail on these plans for 2013, specifically mentioning PLD's outstanding preferred stocks. Mr. Olinger then stated
...one more piece would be the preferreds, as you mentioned. We can redeem all but one series of our preferreds. That's just under about \$0.5 billion.
PLD has seven preferred stock series currently trading, six of which are well beyond their respective call dates and can be redeemed at any time.
While short of a formal redemption announcement, these two statements, taken together, are about as close as executives ever come to divulging future plans for redeeming their outstanding preferred stock shares and the timeframe within which they may do so.
## PLD-S
The blue line on the below chart shows the daily closing price for PLD-S, a 6.75% preferred stock from ProLogis, since its most recent dividend payment date on December 31, 2012.
Notice that PLD-S has been trading well below what the final payment amount would be if this security were to be called this quarter.
For example, if ProLogis had announced a call of PLD-S last Friday with 30 days' notice, the actual call date would occur 74 days into the current quarter on March 16 (being a Saturday, shareholders would receive the payment the following Monday, March 18).
Using the formula that I provided earlier, 74 days worth of dividend for the 6.75% PLD-S amounts to \$0.35 so shareholders would receive \$25.35 per share if ProLogis had called PLD-S last Friday - the day that buyers were scooping up shares of PLD-S for \$25.18.
## Using Redeemable Preferred Stocks as a Principal Protection Strategy
Purchasing redeemable preferred stocks that are priced below their final redemption payment amount is more of a principal protection strategy than an approach taken for yield maximization. Those pursuing this strategy are trying to protect their principal with the possibility of additional dividend income for each quarter that the expected redemption does not materialize.
CFO Olinger's February 6 comments make ProLogis preferred stocks good candidates for this principal protection strategy since the company's six preferred stock issues should now be considered highly likely to be redeemed this year.
The table below the above chart provides the daily dividend amount needed to calculate the magic market price for each of these ProLogis securities, below which buyers will realize a positive return in the event that ProLogis announces a redemption.
As long as the current market price of a preferred stock is below the amount of the final payment as calculated 30 days into the future (using the formula provided earlier), today's buyers can use redeemable preferred stocks to realize positive returns while protecting their principal.
Additional disclosure: Securities identified within this article are for illustration purposes only and are not to be taken as recommendations. I hold shares of PLD-O, PLD-P and PLD-S. | 1,922 | 9,401 | {"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} | 2.8125 | 3 | CC-MAIN-2014-41 | latest | en | 0.907609 |
https://www.instructables.com/The-GoodNightLight-a-Simple-Nightlight-Circuit/ | 1,708,672,262,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474361.75/warc/CC-MAIN-20240223053503-20240223083503-00004.warc.gz | 848,227,703 | 32,034 | ## Introduction: The GoodNightLight: a Simple Nightlight Circuit
Hi all, I teach a Bioinstrumentation course for middle school students in the winter as a part of Purdue GERI (Gifted Education Resource Institute). In this course, I introduce students to basics of circuits and how we use circuits in Biomedical Engineering. I've found that a nightlight is a good introductory circuit for students learning about electronics. It includes a few basic components like resistors and LEDs. It also includes a more intermediate component, namely the operational amplifier and a useful sensor, namely a photoresistor. The mechanism of the circuit shows students how we can use circuits to interact with the outside world and produce some sort of output. In this case, the nightlight turns on when ambient light levels decrease and turns off when ambient light levels increase. Everyone loves seeing the LEDs turn on and off automatically. I told parents that I would post the various lessons online (and I'm a few months behind o_O) so here I go with the first one! Hope you enjoy my Instructable for the GoodNightLight.
## Step 1: Photoresistor or Light-Dependent Resistor (LDR)
A photoresistor is a simple component that changes resistance with incident light. The resistor contains photosensitive materials that cause the resistance of the material to decrease with increased light (more light). Conversely, the resistance of the material increases with decreasing light (it gets darker). The photosensor is responsible for detecting changes in ambient light, which will actuate the nightlight. Feel free to measure the resistance of the photoresistor with a multimeter to see how its resistance changes when you cover and uncover the photoresistor with your finger or other opaque object.
## Step 2: Voltage Divider
A voltage divider is a simple way of interfacing with a resistive transducer, which is a component that translates one form of energy to a resistance. In this circuit, our resistive transducer is our photoresistor. A voltage divider is composed of two resistors in series (connected one after the other). A voltage source, like a battery, is connected to one of the resistors in the divider and the other resistor is connected to ground. The equation for a voltage divider is as follows: Vout = Vin*R2 / (R2 + R1)
As we can see from the equation, R1 and R2 directly determine the output of the voltage divider. Examining the equation a bit further, we see that as R2 increases, Vout gets closer to Vin. We mentioned in our previous step that the resistance of the photoresistor increases with decreasing ambient light. We'll place our photoresistor in the R2 position of this voltage divider.
## Step 3: Comparator and Output LED
A comparator is a simple circuit that compares two voltages. If the voltage at the non-inverting input (the input of the op amp with the "+" sign) is greater than the voltage at the inverting input (the input of the op amp with the "-" sign), the output of the comparator will be turn on the LED. If the opposite is true, the output of the comparator will turn the LED off. If you haven't used LEDs before, know that they light up when a small current is passed through them. Learn more about LEDs from this excellent Instructable.
For our comparator, we'll use a LM324 amplifier. The LM324 is a quad amplifier, meaning that it has 4 amplifiers built into a single package. We'll only need one of the 4 amplifiers. Wire up the LM324 as shown in the schematic.
## Step 4: Conclusions and Final Thoughts
In this instructable, I have demoed simply turning on an LED with changing amounts of ambient light. Please use your creativity to turn it into a real "nightlight".
## Step 5: Troubleshooting
1. In this Instructable, I recommended added a 10k resistor in series with your photoresistor. Depending on the "nominal" resistance of your photoresistor, you may need to change the 10k resistor to something else. I would recommend measuring the resistance of the photoresistor with your multimeter when your photoresistor is exposed to ambient and when the sensor is covered up by some foreign object. You want to choose a series resistor that's larger than the resistance of the photoresistor when it's exposed to ambient light, but smaller than resistance of the photoresistor when it's covered up. For example, for the photoresistor I used, it's resistance when exposed to ambient light was around 8k. When I covered the photoresistor with my finger, it's resistance increased to 48k.
2. Be sure that the that you connect the voltage dividers to the proper inputs of the op amp. Pay close attention to the connections in Step 3.
3. Be wary of the polarity of the LED. The shorter leg is "negative" and should connect to ground. | 1,029 | 4,784 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2024-10 | latest | en | 0.908394 |
https://aprove.informatik.rwth-aachen.de/eval/JAR06/JAR_TERM/TRS/SK90/4.25.trs.Thm12:POLO_FILTER:NO.html.lzma | 1,718,807,302,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861825.75/warc/CC-MAIN-20240619122829-20240619152829-00852.warc.gz | 88,709,283 | 1,753 | Term Rewriting System R:
[x, y]
rev(a) -> a
rev(b) -> b
rev(++(x, y)) -> ++(rev(y), rev(x))
rev(++(x, x)) -> rev(x)
Termination of R to be shown.
` R`
` ↳Dependency Pair Analysis`
R contains the following Dependency Pairs:
REV(++(x, y)) -> REV(y)
REV(++(x, y)) -> REV(x)
REV(++(x, x)) -> REV(x)
Furthermore, R contains one SCC.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Argument Filtering and Ordering`
Dependency Pairs:
REV(++(x, x)) -> REV(x)
REV(++(x, y)) -> REV(x)
REV(++(x, y)) -> REV(y)
Rules:
rev(a) -> a
rev(b) -> b
rev(++(x, y)) -> ++(rev(y), rev(x))
rev(++(x, x)) -> rev(x)
The following dependency pairs can be strictly oriented:
REV(++(x, x)) -> REV(x)
REV(++(x, y)) -> REV(x)
REV(++(x, y)) -> REV(y)
The following rules can be oriented:
rev(a) -> a
rev(b) -> b
rev(++(x, y)) -> ++(rev(y), rev(x))
rev(++(x, x)) -> rev(x)
Used ordering: Polynomial ordering with Polynomial interpretation:
POL(rev(x1)) = x1 POL(REV(x1)) = 1 + x1 POL(b) = 0 POL(++(x1, x2)) = 1 + x1 + x2 POL(a) = 0
resulting in one new DP problem.
Used Argument Filtering System:
REV(x1) -> REV(x1)
++(x1, x2) -> ++(x1, x2)
rev(x1) -> rev(x1)
` R`
` ↳DPs`
` →DP Problem 1`
` ↳AFS`
` →DP Problem 2`
` ↳Dependency Graph`
Dependency Pair:
Rules:
rev(a) -> a
rev(b) -> b
rev(++(x, y)) -> ++(rev(y), rev(x))
rev(++(x, x)) -> rev(x)
Using the Dependency Graph resulted in no new DP problems.
Termination of R successfully shown.
Duration:
0:00 minutes | 536 | 1,510 | {"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} | 3.171875 | 3 | CC-MAIN-2024-26 | latest | en | 0.638085 |
https://socratic.org/questions/how-do-you-find-all-the-zeros-of-f-x-2x-3-5x-2-2x-5 | 1,571,884,820,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987838289.72/warc/CC-MAIN-20191024012613-20191024040113-00486.warc.gz | 707,724,714 | 5,959 | # How do you find all the zeros of f(x)=2x^3+5x^2-2x-5?
Aug 5, 2016
$f \left(x\right)$ has zeros: $1$, $- 1$, $- \frac{5}{2}$
#### Explanation:
This cubic factors by grouping:
$f \left(x\right) = 2 {x}^{3} + 5 {x}^{2} - 2 x - 5$
$= \left(2 {x}^{3} + 5 {x}^{2}\right) - \left(2 x + 5\right)$
$= {x}^{2} \left(2 x + 5\right) - 1 \left(2 x + 5\right)$
$= \left({x}^{2} - 1\right) \left(2 x + 5\right)$
$= \left({x}^{2} - {1}^{2}\right) \left(2 x + 5\right)$
$= \left(x - 1\right) \left(x + 1\right) \left(2 x + 5\right)$
Hence zeros: $1$, $- 1$, $- \frac{5}{2}$ | 288 | 569 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 13, "mathjax_inline_tex": 1, "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} | 4.4375 | 4 | CC-MAIN-2019-43 | longest | en | 0.266877 |
https://www.varsitytutors.com/sat_ii_math_i-help/single-variable-algebra?page=6 | 1,558,355,804,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232255944.3/warc/CC-MAIN-20190520121941-20190520143941-00368.warc.gz | 975,306,514 | 39,244 | # SAT II Math I : Single-Variable Algebra
## Example Questions
### Example Question #51 : Single Variable Algebra
Solve for .
Explanation:
This is a one step, one variable problem. This means we want to isolate x on one side of the equation with all other constants on the other side. To do this perform the opposite operation to manipulate the equation
Subtract on both sides. Since is greater than and is negative, our answer is negative.
We treat as a subtraction problem.
### Example Question #52 : Single Variable Algebra
Solve for .
Explanation:
This is a two step, one variable problem. This means we want to isolate x on one side of the equation with all other constants on the other side. To do this perform the opposite operation to manipulate the equation
Subtract on both sides.
Multiply on both sides.
### Example Question #53 : Single Variable Algebra
Solve for .
Explanation:
This is a one step, one variable problem. This means we want to isolate x on one side of the equation with all other constants on the other side. To do this perform the opposite operation to manipulate the equation
### Example Question #54 : Single Variable Algebra
Solve for .
Explanation:
This is a two step, one variable problem. This means we want to isolate x on one side of the equation with all other constants on the other side. To do this perform the opposite operation to manipulate the equation
Multiply on both sides.
### Example Question #55 : Single Variable Algebra
Solve for .
Explanation:
This is a one step, one variable problem. This means we want to isolate x on one side of the equation with all other constants on the other side. To do this perform the opposite operation to manipulate the equation
Divide on both sides.
When dividing with a negative number, our answer is negative.
### Example Question #56 : Single Variable Algebra
Solve for .
Explanation:
This is a two step, one variable problem. This means we want to isolate x on one side of the equation with all other constants on the other side. To do this perform the opposite operation to manipulate the equation
Divide on both sides.
### Example Question #57 : Single Variable Algebra
Solve for .
Explanation:
This is a one step, one variable problem. This means we want to isolate x on one side of the equation with all other constants on the other side. To do this perform the opposite operation to manipulate the equation
Multiply on both sides.
When multiplying with another negative number, our answer is positive.
### Example Question #58 : Single Variable Algebra
Solve for .
Explanation:
This is a one step, one variable problem. This means we want to isolate x on one side of the equation with all other constants on the other side. To do this perform the opposite operation to manipulate the equation
Multiply on both sides.
When multiplying with another negative number, our answer is positive.
### Example Question #59 : Single Variable Algebra
Solve for .
Explanation:
This is a one step, one variable problem. This means we want to isolate x on one side of the equation with all other constants on the other side. To do this perform the opposite operation to manipulate the equation
Take the square root on both sides. We also need to consider having a negative answer.
Remember, two negatives multiplied equals a positive number.
### Example Question #60 : Single Variable Algebra
Solve for .
Explanation:
This is a two step, one variable problem. This means we want to isolate x on one side of the equation with all other constants on the other side. To do this perform the opposite operation to manipulate the equation
Square both sides to get rid of the radical.
Subtract on both sides. | 793 | 3,747 | {"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-2019-22 | latest | en | 0.822314 |
https://expertinstudy.com/t/3PAQ2QoMp | 1,680,310,811,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949694.55/warc/CC-MAIN-20230401001704-20230401031704-00505.warc.gz | 294,219,523 | 4,555 | Doomfang Dec 8, 2020
# Seconds. How long did it take her to swim the whole race?The thickness of a mattress is 7.5 cm and that of another mattress is 9.75 cm. What wouldbe the thickness of two mattresses together when placed one on the top of the other?
The time for answering the question is over
S.P. of the T.V. set = Rs.6800
Loss = 15%
(i) ∴ C.P. = (S.P. × 100)/(100 – Loss%)
= Rs.(6800 ×100)/(100 – 15)
= Rs.(6800 ×100)/85
= Rs. 8000
(ii) In second case, gain = 12%
∴ S.P. = {C.P.(100 + gain%)}/100
= Rs. {8000(100 + 12)}/100
= Rs. (8000 × 112)/100
= Rs. 8960
381
obsolete
Dec 9, 2020 | 222 | 604 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2023-14 | latest | en | 0.925774 |
https://quisupames.web.app/1240.html | 1,656,906,608,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104293758.72/warc/CC-MAIN-20220704015700-20220704045700-00633.warc.gz | 512,455,731 | 5,291 | # None step equation pdf
Many of them do, but there are others that struggle with. Youve solved the equation when you get the variable by itself, with no numbers in front of it, on one side of the equal sign. You must show your work to get credit check your answer. One step equations an equation is like a balance scale because it shows that two quantities are equal. A onestep equation is one that can be solved in one step by either adding. You can print page this activity can be laminated and used year after year. May 04, 2016 today we will begin to solve one step equations with integers. Students can navigate learning paths based on their level of readiness. How much money did he have in his savings before buying the gift. In algebra, we are often presented with a problem where the answer is known, but part of the problem is missing.
This game involves solving onestep equations, with a range of questions involving addition, subtraction, multiplication and division. Solving linear equations one step equations objective. Dont post outcomes results to learning mastery gradebook. Students learn to solve one step, two step and multistep equations. The goal of solving a linear equation is to find the value of the variable that will make the statement equation true. Here we will focus on what are called onestep equations or. Perform operations to both sides of the equation in order to isolate the variable. One step equations assignment solving onestep equations.
The solution of which of the following equations is neither a fraction nor an integer. Solving multistep equations more notes to solve an equation with variables on both sides. We just have to perform one step in order to solve the equation. Perform any distributive property shown in the equation. These guided notes include solving 1step equations, 2 step equations, multistep equations and equations with variable on both sides. Solving onestep equations using addition and subtraction. No solution would mean that there is no answer to the equation. Addingsubtracting onestep equations part 1 addingsubtracting onestep equations multiplyingdividing onestep equations balancing equations video 1 balancing equations video 2 balancing equations video 3. One step equations explore narrative my main goal for going back to a onestep equation with students is to set up the right way my way.
Use the inverse opposite operation on both sides of the equation 3. A onestep equation is as straightforward as it sounds. On the back of the tab itself, i have students make notes about the process for solving. Solving one step equations using addition and subtraction.
Solving one step equations 1 you must show your work to get credit check your answer. Y f um9andeev mwviptih m aign dfli0nri et sea xprrke9vaolzg8eyb1rpa q. I give my students these guided notes one page at a time. The ultimate goal in solving a twostep equation is the same as the goal of solving a onestep equation. Translate onestep equations and solve practice khan. The goal of solving an equation is to find the value of the variable. Institutional users may customize the scope and sequence to meet curricular. Onestep equations with fractions oak park independent. Click any link to learn how to solve onestep equations. Math terms connect to aapp throughout your study of.
Solving one step equations worksheet pdf onlinemath4all. Solving onestep equations by multiplying or dividing when you are solving an equation, your goal is to use the inverse operation to isolate the variable. Multiplying or dividing both sides of an equation by a nonzero quantity. Eliminate any extraneous solutions from the final answer. A onestep equation is an algebraic equation you can solve in only one step. If you continue browsing the site, you agree to the use of cookies on this website. I can solve a onestep equation involving multiplication with fractions. Solving onestep equations 1 you must show your work to get credit check your answer. This tutorial provides comprehensive coverage of writing and solving onestep equations based on common core ccss and state standards and its prerequisites. In solving equations, it is very important that students pay close attention to the steps, signs, and equality in an equation mathematical practice 6. In 8th grade we expect that students already know how to solve one step equations. Print off one bingo card per student using the second link note there are 2 cards per a4 page, and use the first link to play. It is possible to have more than solution in other types of equations that are not linear, but it is also possible to have no solutions or infinite solutions. Onestep equations word problems writing algebraic expressions write an expression for each word phrase.
In algebra, we are often presented with a problem where the. Solve one step linear equations by balancing using inverse operations solving linear equations is an important and fundamental skill in algebra. Twostep equations is a math game that includes 24 memory cards. Powered by create your own unique website with customizable templates. When solving a one step equation like this, we do the inverse operation to isolate the variable. Try to get all the variable terms on one side of the equation. This is 2 of 4 videos i custom made for an educator in california for an experimental 1week video homework program. We have to isolate the variable which comes in the equation. Write a word problem involving one step equations and rational of his savings on a gift. Students must match the twostep equation with its solution. Express each of the following sentences into onestep equations. We will take notes, complete practice examples together, and you will complete some more practice work on your own. Analyzing the concepts of adding, subtracting, negative numbers, multiplication, division. Once students have completed their foldable, i introduce todays work time assignment, the onestep equations sort.
Simplify ration equation calculator, write a situation that can be represented by the algebraic expression 3. More examples see examples of problems you can solve with algebra calculator. One of the main and very important basics of algebra is solving an equation for an unknown value. Thirteen plus a number nis greater than fifteen the sum of a number. Writing and solving onestep equations tutorialspoint. Solve twostep equations easy peasy allinone homeschool. Then, i post them on the smart board and we fill them out together. Now that we know what the term nonlinear refers to we can define a system of non. Addingsubtracting one step equations part 1 addingsubtracting one step equations multiplyingdividing one step equations balancing equations video 1 balancing equations video 2 balancing equations. Our focus today will be adding and subtracting only.
Save by purchasing all of my equation memory games as a part of the equation memory bundle at the following link. We are going to start simple and one step at a time. For the sort, students collaborate with a partner to simplify or solve a onevariable linear equation in order to determine whether it has one, no, or infinitely many solutions. A web page for producing and downloading pdf or postscript plots of the solution sets to equations and inequations in two variables x and y. In mathematics, an equation is a statement that asserts the equality of two expressions.
Identify the equation which does not have a solution at w 5. In this case, we have 3x 9, so we divide both sides by 3 to get x 3. I have my students paste this into their interactive notebooks and we work out 2 examples of each type, underneath the tabs. Please rewrite each variable, expression, or equation so that the number in front of each variable is visible. Combine any like terms in the equation do not cross the 3. What you do to one side of the equation must also be done to the other side to keep it balanced.
The objective of this lesson is to give students an understanding of linear equations in one variable with one solution, infinitely many. Some students will say i dont need to show my work. Click the following links to download one step equations worksheets as pdf documents. Theres nothing worse than playing a game where only a couple of kids. Seventh grade lesson one step equations betterlesson. Adding and subtracting 1 y 6 20 2 x 10 12 3 12 z 15 14 22 3 4 2 n 16 5 a 4 14 6 m 5 10 14 10 5 7 4 b 1030 8 c 25 9 x 60 20 26 15 80 10 g 16 4 11 x 15 20 12 w 14 10. Show your answer for every equation from this day forward, i agree to write down what im doing to both sides. Solving linear equations metropolitan community college. Today we will begin to solve onestep equations with integers. These guided notes include solving 1 step equations, 2 step equations, multi step equations and equations with variable on both sides. Numerical methods for solving systems of nonlinear equations.
This fundamental method is used in all types of math problems and is necessary even later in the toughest and most advanced math courses. One step equations word problems writing algebraic expressions write an expression for each word phrase. This tutorial provides comprehensive coverage of writing and solving one step equations based on common core ccss and state standards and its prerequisites. One step equations worksheets containing integers with.
Vocabulary alert equation a mathematical sentence that uses an. Youve solved the equation when you get the variable by itself, with no numbers in. Kolu uses a scale to teach some of the basics of solving algebraic equations. Rep lace each part of the verbal model with a number or variable to create an equation. Try to get all the variable terms on one side of the equation a. Infinite algebra 1 one, none, or infinite many solutions. Problem with 2 radicals and no other nonzero terms.
1174 1239 213 750 389 771 400 886 615 1071 339 548 419 63 431 1545 199 1247 1176 18 56 1049 40 612 1448 97 666 459 462 1068 267 474 955 1355 1022 1161 1007 684 1054 | 2,172 | 10,051 | {"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-2022-27 | latest | en | 0.9607 |
https://brilliant.org/problems/can-you-generalise/ | 1,505,893,698,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818686705.10/warc/CC-MAIN-20170920071017-20170920091017-00440.warc.gz | 662,550,142 | 17,437 | # Can You Generalize?
Geometry Level 4
$$ABC$$ is a triangle with $$AB=360$$, $$BC=240$$, $$CA=180$$. The internal and external angle bisectors of $$\angle CAB$$ meet $$BC$$ and $$BC$$ produced at $$P$$ and $$Q$$ respectively. Find the circumradius of $$APQ.$$
× | 80 | 265 | {"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} | 2.75 | 3 | CC-MAIN-2017-39 | longest | en | 0.773579 |
http://mathhelpforum.com/algebra/143572-i-have-couple-parabola-quadratic-equation-related-issues-print.html | 1,526,957,791,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864622.33/warc/CC-MAIN-20180522014949-20180522034949-00287.warc.gz | 185,662,117 | 3,685 | # I have a couple of parabola and quadratic equation related issues.
• May 7th 2010, 12:35 PM
koumori
I have a couple of parabola and quadratic equation related issues.
Hello again MHF!
Just a couple of questions that I don't know how to work out fully.
*******************************
Find the dimensions of a rectangle a with the greatest area whose perimeter is 30 feet.
I have a plan to work it out--I think.
First I have to divide the standard perimeter formula by 2 in order to get: l+w=15. Then solve for 'l' in terms of 'w'(l=15-w), plug that into the area formula:A=w(15-w) ,which is equal to,A=15w-w^2.
The thing is, I have no idea how to find the maximum possible value for the length, width, and area.
I think it might be that I have to use the area formula I noted above(A=15w-w^2) as a formula for a parabola and find the maximum of it, but I'm not too sure if I'm right.
I don't even know how to find the maximum of an equation in that form....
********************************
Given x^3-4x^2+2x+1=0:
(a) how many possible positive roots are there?
(b) how many possible negative roots are there?
(c) what are the possible rational roots?
(d) using synthetic substitution, which of the possible rational roots is actually a root of the equation?
(e)find the irrational roots of the equation.(hint: use the quadratic formula to solve the depressed equation.)
I'm really clue less as to solving this one.....I got lost after possible rational roots.
• May 7th 2010, 03:14 PM
Gusbob
Quote:
Originally Posted by koumori
Hello again MHF!
Just a couple of questions that I don't know how to work out fully.
*******************************
Find the dimensions of a rectangle a with the greatest area whose perimeter is 30 feet.
I have a plan to work it out--I think.
First I have to divide the standard perimeter formula by 2 in order to get: l+w=15. Then solve for 'l' in terms of 'w'(l=15-w), plug that into the area formula:A=w(15-w) ,which is equal to,A=15w-w^2.
The thing is, I have no idea how to find the maximum possible value for the length, width, and area.
I think it might be that I have to use the area formula I noted above(A=15w-w^2) as a formula for a parabola and find the maximum of it, but I'm not too sure if I'm right.
I don't even know how to find the maximum of an equation in that form....
If you know how to, differentiate. If you don't, find the vertex of the parabola. The parabola given by your area equation is downward facing, so it's vertex is the maximum value it could take.
********************************
Given x^3-4x^2+2x+1=0:
(a) how many possible positive roots are there?
(b) how many possible negative roots are there?
(c) what are the possible rational roots?
(d) using synthetic substitution, which of the possible rational roots is actually a root of the equation?
(e)find the irrational roots of the equation.(hint: use the quadratic formula to solve the depressed equation.)
I'm really clue less as to solving this one.....I got lost after possible rational roots.
So I assume you're fine with parts a,b,and c of this question?
Look at your list of rootsBy inspection, you can easily see that x = 1 is a root. Divide the original polynomial by (x-1). You'll get a quadratic.
Use the quadratic formula on the quadratic you found in (e) to find the irrational roots
.
• May 7th 2010, 05:53 PM
integral
First question:
$\displaystyle \underset{x}{\fbox{A=xy}}y$
$\displaystyle A=xy$
$\displaystyle P=2x+2y$
$\displaystyle y=\frac{P-2x}{2}$
$\displaystyle \therefore$
$\displaystyle A=x\frac{30x-2x^2}{2}$
then: if you know how:
$\displaystyle \frac{DA}{Dx}\left [x(\frac{p-2x}{2}) \right ] =$
$\displaystyle \frac{1}{2}\frac{DA}{Dx}\left [(px-2x^2) \right ]$
$\displaystyle \frac{1}{2}(p-4x)$
$\displaystyle \frac{P}{2}-2x=0$
$\displaystyle x=\frac{P}{4}$
$\displaystyle x=\frac{30}{4}$
x=7.5
If you don't know differentiation:
$\displaystyle \frac{Px-2x^2}{2}=-x^2+15x$
Use $\displaystyle \frac{-b}{2a}$
$\displaystyle \frac{-15}{-2}=7.5$
this is your x, plug into your equation to get y and those are the values to gain the max area. | 1,137 | 4,114 | {"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 | 4 | CC-MAIN-2018-22 | latest | en | 0.916775 |
https://forum.freecadweb.org/viewtopic.php?f=36&t=30104&start=150 | 1,581,928,578,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875141749.3/warc/CC-MAIN-20200217055517-20200217085517-00299.warc.gz | 395,911,887 | 10,432 | Sketcher tutorial
A place to share learning material: written tutorials, videos, etc.
chrisb
Posts: 21205
Joined: Tue Mar 17, 2015 9:14 am
Re: Sketcher tutorial
I have no idea what's going on, I haven't ever seen this error message. Abdullah, can you help?
abdullah wrote: ping
Hannu
Posts: 106
Joined: Fri Sep 11, 2015 4:12 pm
Location: Strängnäs, Sweden - 1hr drive west from Stockholm
Re: Sketcher tutorial
FWIW: "Works for me"...
I'd interpret the error message as "one of the points lacks connection with a line segment", which I cannot verify - I see no extra / loose points either.
OS: Ubuntu 16.04.5 LTS
Word size of OS: 64-bit
Word size of FreeCAD: 64-bit
Version: 0.18.15536 (Git)
Build type: Release
Branch: master
Hash: 98ebeaa198df945ba666c15912c6caf95920f0ef
Python version: 2.7.12
Qt version: 4.8.7
Coin version: 4.0.0a
OCC version: 7.3.0
Locale: English/UnitedStates (en_US)
/Hannu - some knowledge in many fields ...
- computers, software, electronics, color correct photography,
- mechanics, 3D-printing, lathe, mill, weld -work,
- ISO9000, Deviations & Corrective actions, and a lot more...
chrisb
Posts: 21205
Joined: Tue Mar 17, 2015 9:14 am
Re: Sketcher tutorial
jpg87 wrote:
Tue Nov 27, 2018 12:49 pm
...
I have made some more changes following suggestions from the forum. Thanks to paullee and GlouGlou
paullee wrote:
Tue Jan 01, 2019 3:42 am
And for me need more time to understanding Exercise 32 and 33 ... seem a Point-on-Object Constraint should be in 2nd (right hand side) diagram of Exercise 33?
They are supposed to be connected with a coincidence constraint. To make things clearer I changed the text of
Exercise 33 old: Create a sketch with two arcs and let them cross at an angle of 60°.
Exercise 33 new: Create a sketch with two arcs which are connected with a coincidence constraint. Let the arcs cross at an angle of 60°.
Furthermore I changed section 30 Lock constraint
Tabular of Lock constraint old: Number of DOF consumed: 2
Tabular of Lock constraint new: Number of DOF consumed: 2 per point
The same change has to be performed in the overview table at the end of the document
First sentence of Lock constraint old: The lock constraint creates a horizontal and a vertical distance constraint which sums up to 2 DOF.
First sentence of Lock constraint new: The lock constraint creates a horizontal and a vertical distance constraint for each point involved which sums up to 2 DOF per point.
I followed the suggestions of GlouGlou and made some minor changes plus a more substantial change to the last subparagraph, see https://forum.freecadweb.org/viewtopic. ... 48#p278148.
The document is to be found in the same place as before: https://owncloud.gwdg.de/index.php/s/HS ... I/download
chrisb
Posts: 21205
Joined: Tue Mar 17, 2015 9:14 am
Re: Sketcher tutorial
The description of sketch creation became rather complicated:
• If you don’t have a body in your document, a new body is created, activated, and a new sketch is created inside the body.
• If an existing sketch is selected, you enter sketcher’s edit mode for this sketch.
• If nothing is selected and you have exactly one body in your document, the body is activated and a new sketch is created inside the body.
• If nothing is selected and you have more than one body and one of them is activated, a new sketch is created inside the active body.
• If nothing is selected and you have more than one body and none of the bodies is activated, you are asked to activate one of the bodies.
• If a face from the activated body is selected—or if you have only one body—a new sketch is created and attached to that face.
• If a face from outside the activated body is selected—or if you have only one body—a question is raised how to link the face. If you have only one body it is activated and handled as such.
-– The option ”Make independent copy” creates an unlinked ShapeBinder to which the sketch is attached.
–- The option ”Make independent copy” creates a linked ShapeBinder to which the sketch is attached. If you change the other object the ShapeBinder will follow.
–- The option ”Create crossreference” creates a forbidden link (message: Links go out of the allowed scope). I do not recommend to use this option.
I wonder if I should keep this simpler by making it more general, such as
- Edit if an existing sketch is selected
- create a sketch if the target body can be determined, i.e. there is only one body in the sketch or a body is activated
- attach a new sketch if a face was selected inside the current body
- create a reference if a face was selected outside the current body
What do you think?
GlouGlou wrote:...
jpg87 wrote:...
jpg87
Posts: 299
Joined: Thu Mar 16, 2017 7:16 am
Location: Limoges - France
Contact:
Re: Sketcher tutorial
chrisb wrote:
Sun Jan 06, 2019 1:44 am
I wonder if I should keep this simpler by making it more general, such as
- Edit if an existing sketch is selected
- create a sketch if the target body can be determined, i.e. there is only one body in the sketch or a body is activated
- attach a new sketch if a face was selected inside the current body
- create a reference if a face was selected outside the current body
Of course, this summarizes the possible situations for someone who stands back; but I think for beginners, the detail of the behaviors is more understandable.
My website : http://help-freecad-jpg87.fr updated 2020/02/12
chrisb
Posts: 21205
Joined: Tue Mar 17, 2015 9:14 am
Re: Sketcher tutorial
jpg87 wrote:
Sun Jan 06, 2019 10:15 am
but I think for beginners, the detail of the behaviors is more understandable.
Thanks for the input. That means less work - for both of us . So I only have to fix the german translation.
paullee
Posts: 2017
Joined: Wed May 04, 2016 3:58 pm
Re: Sketcher tutorial
chrisb wrote:
Sat Jan 05, 2019 12:19 pm
jpg87 wrote:
Tue Nov 27, 2018 12:49 pm
...
I have made some more changes following suggestions from the forum. Thanks to paullee and GlouGlou
paullee wrote:
Tue Jan 01, 2019 3:42 am
And for me need more time to understanding Exercise 32 and 33 ... seem a Point-on-Object Constraint should be in 2nd (right hand side) diagram of Exercise 33?
They are supposed to be connected with a coincidence constraint. To make things clearer I changed the text of
ar of Lock constraint new:[/b] Number of DOF consumed: 2 per point
Thanks, but I am not sure I follow, see screenshots and file attached:-
Left Hand Side 2 Arcs
- I select 1 Arc, 1 point, another Arc
- Apply an Angle Constraints (30 degree)
- Get a point-on-object constraint added (instead of coincidence)
Righ Hand Side 2 Arcs
- I select 1 Arch, another Arc the, lastly 1 point
- Apply an Angle Constraints (30 degree)
- Get a point-on-object constraint added (instead of coincidence)
FreeCAD_0.18.15287.glibc2.17-x86_64.AppImage on Fedora 27
Attachments
Test_ Sketch Ange Constraint_ Arc Point Arc.fcstd
(3.71 KiB) Downloaded 34 times
Screenshot from 2019-01-06 23-00-49.png (203.68 KiB) Viewed 942 times
chrisb
Posts: 21205
Joined: Tue Mar 17, 2015 9:14 am
Re: Sketcher tutorial
Am I right that you are confused by the automatically created point-on-object constraint?
paullee
Posts: 2017
Joined: Wed May 04, 2016 3:58 pm
Re: Sketcher tutorial
hmm...
"Arc-Arc mode. Similarly you can combine two arcs and a point or an arc, a line, and a
point. This means that Line-Point-Line mode can be viewed as a special form of this
more general mode"
So Exercise 33 and the diagram is not illustrating the above scenario right? - I thought it is.
chrisb
Posts: 21205
Joined: Tue Mar 17, 2015 9:14 am
Re: Sketcher tutorial
I have to thank you for reading with such high precision. It's beginning to dawn what you mean: The example does not show unconnected points.
Perhaps it is better to call the last mode Arc-Point-Arc mode and mention that there is no Arc-Arc mode without an additional point. Since it is more frequent that an angle is defined on arcs with connected endpoints I still think that the example is sensible. | 2,126 | 8,007 | {"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} | 2.5625 | 3 | CC-MAIN-2020-10 | latest | en | 0.844467 |
https://optionefekqz.netlify.app/anderholm80612xydo/bar-and-line-graph-on-same-chart-excel-2020-fe | 1,709,126,901,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474715.58/warc/CC-MAIN-20240228112121-20240228142121-00038.warc.gz | 436,630,617 | 12,130 | ## Bar and line graph on same chart excel 2020
If you need to print a chart in Excel 2013 or Excel 2016, you can use the Print option from the File tab. You can use the Settings menu to specify precisely the
5 Jan 2009 A popular example of combination charts is a line & bar graph combination. See below Download the Excel Combination Chart Tutorial workbook and learn by experimenting. Come and learn from me in London – April 2020 Once the new axis shows up, scaling etc. is same as the primary axis. 2. Download the Excel template with bar chart, line chart, pie chart, histogram, two types of graphs can be combined to create a combo chart with bars and lines. Before you create a trendline: You can add trendlines to bar, line, column, or scatter charts. On your computer, open a spreadsheet in Google Sheets. Double- click 27 Jan 2020 Google bar charts are rendered in the browser using SVG or VML, the stem length is always the same as the text, and for 'line' domain Histogram Bar Graph Examples (Different Types) Grouped Stacked Although they look the same, bar charts and histograms have one important You don't have to do this, but if you have numbers than don't fall on a line (i.e. Copyright ©2020 Statistics How To | Theme by: Theme Horse | Powered by: WordPress. 20 May 2019 This tutorial shows how to create a Gantt Chart in Excel that has a nice date The axis can have the same min and max as the bar chart's value axis above, of the min and max dates in the line chart, 1/1/2019 and 1/1/2020. Chart in Excel are always used to analyze some important information. Therefore, in this article we will demonstrate how to add horizontal average line to vertical chart in Excel. In an Excel Now in the drop down menu, click the clustered column to create a bar chart. Thus, you have Copyright © 2020 Data Recovery Blog.
## 25 Oct 2017 Excel's chart features can turn your spreadsheet data into compelling Pie charts are used to show percentages of the whole, and Line charts excel with data trends. Column and Bar charts are identical, except Columns are vertical and Bars are horizontal. Copyright © 2020 IDG Communications, Inc.
8 May 2018 Excel Multi-colored Line Charts allow you to color code the line when it same value in both columns D and E. Without the value in column E, 17 Nov 2016 Here's how to add a trendline to your charts in Excel. Trend lines can be added to a variety of charts, including bar charts, line charts, scatter better evidence that the trend may continue with all factors being equal. Copyright © 2020 BetterCloud Monitor | Privacy Policy | Master Subscription Agreement 12 Oct 2012 Learn 10 cool techniques that will make your Excel charts way select Line Color > No line (Mac: Line > Solid > Color: No Line). bar charts, if you want the larger values to be at the top of the chart, is I make data sexy — and teach other marketers how to do the same. June 8-10, 2020: SMX Advanced. 9 Feb 2017 Excel offers a number of different graph types, including bar graphs, graph in one row and the data you want to appear in the line graph in an I made a three line comparison graph by selecting all of the data as a block i.e. the X axis and the three Y axis data points. I then used The chart popped up as desired. Column A - X axis Column B - Data set 1 Column C - Data set 2 Column D - Data set 3 2020 Stack Exchange, Inc. user contributions under cc by-sa 4.0. The Excel spreadsheet program allows users to create spreadsheets of data and transform these spreadsheets into charts and graphs for easy data comparison. Excel offers a number of different graph types, including bar graphs, line graphs, column graphs and pie graphs. If you want, you can combine these different graph types to create a hybrid To overlay line chart on the bar chart in Excel, please do as follows. 1. Select the range with two unique sets of data, then click Insert > Insert Column or Bar Chart > clustered column. See screenshot: 2. Now a bar chart is created in your worksheet as below screenshot shown.
### Before you create a trendline: You can add trendlines to bar, line, column, or scatter charts. On your computer, open a spreadsheet in Google Sheets. Double- click
28 Nov 2018 Select all the data that you want included in the bar chart. Be sure to include the column and row headers, which will become the labels in the bar 27 Aug 2019 A combo chart in Excel displays two chart types (such as column and line) on the same chart. They are used to show different types of The Format Data Series dialog box will be displayed. From Series Options category, under Plot Series on, click Secondary Axis radio button and then click Close. Example of creating combined Excel charts with two chart types. In the first example we create a combined column and line chart that share the same X and Y
### For example line and bar charts, bar and Gantt charts etc. Best used for critical comparison purpose, which could help in further analysis and reporting. The combination charts also show how the data table containing a wide range can be portrayed in both y-axis in the same chart.
If you have Excel 2010 and you are making a combo chart with two Bars and one line, please follow the steps below. Excel 2016 has made this much easier, if you are working with Excel 2016, please refer here for details. Step 1: Suppose we are making a chart with the data in the Table below, and we want to combine the number of "Met" in column B, "All Targets" in column C and "% of Met" in In this video, we are going to create pie, bar, and line charts. Each type of chart highlights data differently. Different data displays better in different types of charts. If you try to graph too much data in a pie chart it looks like this, not very useful. Now, I am creating a bar chart using the same data we used to create the pie chart. Excel Line and Bar Graph in One Chart Candace Chou Display two different sets of data in the SAME chart with Secondary Axis Display Two Measures in a Bar-Line Chart in Excel 2010 A combination chart is a chart that combines two or more chart types in a single chart. To create a combination chart in Excel, execute the following steps. Line charts are used to display trends over time. Use a line chart if you have text labels, dates or a few numeric labels on the horizontal axis. Use a scatter chart (XY chart) to show scientific XY data. To create a line chart in Excel, execute the following steps.
## For example line and bar charts, bar and Gantt charts etc. Best used for critical comparison purpose, which could help in further analysis and reporting. The combination charts also show how the data table containing a wide range can be portrayed in both y-axis in the same chart.
Common types of figures include line graphs, bar graphs, charts (e.g., flowcharts, Note that tables and figures have the same overall setup. For example, use the built-in graphics features of your word-processing program (e.g., Microsoft Word or Excel) or dedicated programs 2020 American Psychological Association. 25 Oct 2017 Excel's chart features can turn your spreadsheet data into compelling Pie charts are used to show percentages of the whole, and Line charts excel with data trends. Column and Bar charts are identical, except Columns are vertical and Bars are horizontal. Copyright © 2020 IDG Communications, Inc. 15 Oct 2012 You could work in the same sheet as the data, but because we're simulating To generate the chart, click the Developer tab and choose Scroll Bar Form To do so, select the matrix range -- A3:B15 -- and use the instructions in step 1 to embed a line chart in the Dashboard sheet, 2020 CBS Interactive. 28 Feb 2019 Here are a few tips and tricks for Excel and Google Sheets that may help you better A lot of the time when someone is asking me charts questions it's simply a matter of and then click on the line within the graph that you want to place on a second axis. Copyright © 2004-2020 Hanapin Marketing LLC.
If you have Excel 2010 and you are making a combo chart with two Bars and one line, please follow the steps below. Excel 2016 has made this much easier, if you are working with Excel 2016, please refer here for details. Step 1: Suppose we are making a chart with the data in the Table below, and we want to combine the number of "Met" in column B, "All Targets" in column C and "% of Met" in Excel makes graphing easy. Line graphs are one of the standard graph options in Excel, along with bar graphs and stacked bar graphs.While bar graphs may be best for showing proportions and other data points, line graphs are ideal for tracking trends and predicting the results of data in yet-to-be-recorded time periods. A line-column chart combines a line graph and column chart on the same graph. The two charts share an X axis but each has its own Y axis. There are two common uses for a combination chart: 1. when want to display two different data sets together and 2. when you want to display the sets of the same type of data but the ranges vary widely between | 2,028 | 9,045 | {"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} | 2.640625 | 3 | CC-MAIN-2024-10 | latest | en | 0.855358 |
https://www.soldierx.com/bbs/201003/Binary-hacked | 1,701,727,630,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100535.26/warc/CC-MAIN-20231204214708-20231205004708-00152.warc.gz | 1,126,528,739 | 5,465 | Binary is hacked!
11 replies [Last post]
gl_itch
Offline
Neophyte
Joined: 2010/03/24
Binary is setup as 0’s and 1’s and there are quite a few encodings that beat binary(machine language)…
256 128 64 32 16 8 4 2 1
0 = .
1 = .5 = 2 = 1
1 0 = 5
1 1 = 5.5 = 52 = 26 = 13
1 0 0 = 50 = 25
1 0 1 = 50.5 = 502 = 251
1 1 0 = 55
1 1 1 = 55.5 = 552 = 69
1 0 0 0 = 500 = 250 = 125
1 0 0 1 = 500.5 = 5002 = 2501
1 0 1 0 = 505
1 0 1 1 = 505.5 = 5052 = 2526
1 1 0 0 = 550 = 275
1 1 0 1 = 550.5 = 5502 = 2751
When .5 is replaced with 2 keep halving number until prime without turning number into a decimal for a faster encoding with the 0-9 numeral system. So basically half all the numbers until the next half becomes a decimal I.e. being prime.
This saves in space approximately 98 numerals per 100 places compared to machines language.
Justin Nathans
[email protected] | 352 | 858 | {"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} | 2.921875 | 3 | CC-MAIN-2023-50 | longest | en | 0.843694 |
http://docs.scipy.org/doc/scipy-0.11.0/reference/generated/scipy.stats.scoreatpercentile.html | 1,438,710,998,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042991076.30/warc/CC-MAIN-20150728002311-00065-ip-10-236-191-2.ec2.internal.warc.gz | 75,910,820 | 3,388 | # scipy.stats.scoreatpercentile¶
scipy.stats.scoreatpercentile(a, per, limit=(), interpolation_method='fraction')[source]
Calculate the score at the given per percentile of the sequence a.
For example, the score at per=50 is the median. If the desired quantile lies between two data points, we interpolate between them, according to the value of interpolation. If the parameter limit is provided, it should be a tuple (lower, upper) of two values. Values of a outside this (closed) interval will be ignored.
The interpolation_method parameter supports three values, namely fraction (default), lower and higher. Interpolation is done only, if the desired quantile lies between two data points i and j. For fraction, the result is an interpolated value between i and j; for lower, the result is i, for higher the result is j.
Parameters : a : ndarray Values from which to extract score. per : scalar Percentile at which to extract score. limit : tuple, optional Tuple of two scalars, the lower and upper limits within which to compute the percentile. interpolation : {‘fraction’, ‘lower’, ‘higher’}, optional This optional parameter specifies the interpolation method to use, when the desired quantile lies between two data points i and j: fraction: i + (j - i)*fraction, where fraction is the fractional part of the index surrounded by i and j. -lower: i. - higher: j. score : float Score at percentile.
Examples
```>>> from scipy import stats
>>> a = np.arange(100)
>>> stats.scoreatpercentile(a, 50)
49.5
```
#### Previous topic
scipy.stats.itemfreq
#### Next topic
scipy.stats.percentileofscore | 362 | 1,607 | {"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} | 2.625 | 3 | CC-MAIN-2015-32 | longest | en | 0.677547 |
https://www.studypool.com/questions/164613/how-to-solve-a-statistics-problem | 1,480,810,890,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541142.66/warc/CC-MAIN-20161202170901-00126-ip-10-31-129-80.ec2.internal.warc.gz | 1,026,894,210 | 750,471 | # How to solve a statistics problem?
FratBro23
Category:
Statistics
Price: \$5 USD
Question description
The police department of a major city has found that the average height of their 1,250 officers is 71 inches (in.) with = 2.3 inches.
How many officers are at least 75 inches tall?
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http://www.sklogwiki.org/SklogWiki/index.php?title=Ideal_gas:_Heat_capacity&diff=2388&oldid=2262 | 1,618,957,924,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039491784.79/warc/CC-MAIN-20210420214346-20210421004346-00181.warc.gz | 172,567,174 | 6,545 | # Difference between revisions of "Ideal gas: Heat capacity"
$C_p - C_V = \left.\frac{\partial V}{\partial T}\right\vert_p \left(p + \left.\frac{\partial E}{\partial V}\right\vert_T \right)$
$\left.C_p -C_V \right.=R$ | 72 | 218 | {"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": 2, "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} | 2.625 | 3 | CC-MAIN-2021-17 | latest | en | 0.669469 |
https://encyclopedia2.thefreedictionary.com/increased+intracranial+pressure | 1,575,741,896,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540500637.40/warc/CC-MAIN-20191207160050-20191207184050-00008.warc.gz | 350,323,843 | 19,205 | # pressure
(redirected from Increased intracranial pressure)
Also found in: Dictionary, Thesaurus, Medical, Acronyms.
## pressure,
in mechanics, ratio of the forceforce,
commonly, a "push" or "pull," more properly defined in physics as a quantity that changes the motion, size, or shape of a body. Force is a vector quantity, having both magnitude and direction.
acting on a surface to the area of the surface; it is thus distinct from the total force acting on a surface. A force can be applied to and sustained by a single point on a solid. However, a force can only be sustained by the surface of an enclosed fluid, i.e., a liquid or a gas. Thus it is more convenient to describe the forces acting on and within fluids in terms of pressure. Units of pressure are frequently force units divided by area units, e.g., pounds per square inch, dynes per square centimeter, or newtons (N) per square meter.
### Pressure of Fluids
A fluid exerts a pressure on all bodies immersed in it. For a fluid at rest the difference in pressure between two points in it depends only upon the density of the fluid and the difference in depth between the two points. For example, a swimmer diving down in a lake can easily observe an increase in pressure with depth. For each meter (foot) increase in depth, the swimmer is subjected to an increase in pressure of 9,810 N per sq m (62.4 lb per sq ft), because water weighs 9,810 N per cu m (62.4 lb per cu ft). Since a liquid is nearly incompressible, its density does not change significantly with increasing depth. Therefore, the increase in pressure is caused solely by the increase in depth.
The variations in pressure of a gas are more complicated. For example, since air has such a low density compared to a liquid, a change in its pressure is only measurable between points that have a great height difference. The air pressure in a typical room is the same everywhere, but it is noticeably lower at the top of a mountain than at sea level. Because air is a gas, it is compressible. Its density decreases with increasing altitude. Thus changes in air pressure depend upon both the variations in the density of air and changes in the altitude at which it is measured. These two factors combine to reduce the air pressure at an altitude of 5,500 m (18,000 ft) to one half its value at sea level. Atmospheric (air) pressure at sea level will support a column of mercury that is about 76 cm (30 in.) high. The exact height varies with the weather. A unit called a standard atmosphere exerts a pressure equivalent to a column of mercury 76 cm high at sea level when the temperature is 0°C;; it is equal to 101,300 N per sq m (14.7 lb per sq in.).
### Influences on and Effects of Pressure
Different gas lawsgas laws,
physical laws describing the behavior of a gas under various conditions of pressure, volume, and temperature. Experimental results indicate that all real gases behave in approximately the same manner, having their volume reduced by about the same proportion of the
relate the pressure of a gas to its volume, its temperature, or both. A rise in pressure affects both the melting pointmelting point,
temperature at which a substance changes its state from solid to liquid. Under standard atmospheric pressure different pure crystalline solids will each melt at a different specific temperature; thus melting point is a characteristic of a substance and can be used
and the boiling pointboiling point,
temperature at which a substance changes its state from liquid to gas. A stricter definition of boiling point is the temperature at which the liquid and vapor (gas) phases of a substance can exist in equilibrium.
of a substance, raising the melting and boiling points of most substances. In the case of water, however, an increase in pressure lowers its melting point so that the pressure of a skate blade on an ice surface causes the ice below it to be converted to the liquid state (see states of matterstates of matter,
forms of matter differing in several properties because of differences in the motions and forces of the molecules (or atoms, ions, or elementary particles) of which they are composed.
; expansionexpansion,
in physics, increase in volume resulting from an increase in temperature. Contraction is the reverse process. When heat is applied to a body, the rate of vibration and the distances between the molecules composing it are increased and, hence, the space occupied by the
). Bernoulli's principleBernoulli's principle,
physical principle formulated by Daniel Bernoulli that states that as the speed of a moving fluid (liquid or gas) increases, the pressure within the fluid decreases.
relates the effect of the velocity of a fluid on the pressure within the fluid.
#### Buoyancy
A body immersed in a fluid experiences a larger upward pressure on its lower surface than a downward pressure on its upper surface because of the difference in height or depth between the two surfaces; this difference in pressure results in a buoyant force that pushes the body upward (see Archimedes' principleArchimedes' principle,
principle that states that a body immersed in a fluid is buoyed up by a force equal to the weight of the displaced fluid. The principle applies to both floating and submerged bodies and to all fluids, i.e., liquids and gases.
). If the weight of the body is less than the buoyant force, the body will rise; if the weight is greater, the body will sink. The buoyant effect of this pressure may be noted in the rise of balloons or other objects filled with gases, such as hydrogen or helium, that are less dense than air.
#### Hydraulic Force
According to Pascal's lawPascal's law
[for Blaise Pascal], states that pressure applied to a confined fluid at any point is transmitted undiminished throughout the fluid in all directions and acts upon every part of the confining vessel at right angles to its interior surfaces and equally upon equal
the pressure exerted on an enclosed fluid is transmitted undiminished throughout the fluid and acts equally in all directions. On the basis of this law, various hydraulic devices are used to multiply a force. For example, a force of 10 N exerted on a piston whose area is 1 sq m and which is inserted into an enclosed chamber filled with water or another fluid transmits a pressure of 10 N per sq m throughout the fluid. If a second piston, at another part of the chamber, has an area of 10 sq m, then this pressure results in a force of 10 N being exerted on each square meter of its area, or 100 N total force.
### Tools for Measuring Pressure
The instrument for measuring atmospheric pressure, the barometerbarometer
, instrument for measuring atmospheric pressure. It was invented in 1643 by the Italian scientist Evangelista Torricelli, who used a column of water in a tube 34 ft (10.4 m) long.
, is calibrated to read zero when there is a complete vacuum; the pressure indicated by the instrument is therefore called absolute pressure. The term "pressure gauge" is commonly applied to the other instruments used for measuring pressure. They are manufactured in a great variety of sizes and types and are employed for recording pressures exerted by substances other than air—water, oil, various gases—registering pressures as low as 13.8×103 N per sq m (2 lb per sq in.) or as high as 13.8×107 N per sq m (10 tons per sq in.) and over (as in hydraulic presses). Some pressure gauges are made to carry out special operations, such as the one used on a portable air compressor. In this case, the gauge acts automatically to stop further operation when the pressure has reached a certain point and to start it up again when compression has fallen off to a certain limit.
In general, a gauge consists of a metal tube or diaphragm that becomes distorted when pressure is applied and, by an arrangement of multiplying levers and gears, causes an indicator to register the pressure upon a graduated dial. The Bourdon gauge used to measure steam pressure and vacuum consists essentially of a hollow metal tube closed at one end and bent into a curve, generally elliptic in section. The open end is connected to the boiler. As the pressure inside the tube (from the boiler) increases, the tube tends to straighten out. The closed end is attached to an indicating needle, which registers the extent to which the tube straightens out. For pressure too small to be accurately measured by the Bourdon gauge, the manometer is used. The simplest type of manometer consists of a U tube partially filled with a liquid (i.e., mercury), leaving one end open to the atmosphere and the other end to the source of pressure. If the pressure being measured is greater or less than atmospheric pressure, the liquid in the tube moves accordingly. Pressures up to several million lb per sq in. have been produced in experiments to determine the effect of high pressure on various substances.
## Pressure
The ratio of force to area. Atmospheric pressure at the surface of Earth is in the vicinity of 15 lbf/in.2 (1.0 × 105 Pa). Pressures in enclosed containers less than this value are spoken of as vacuum pressures; for example, the vacuum pressure inside a cathode-ray tube is 10-8 mmHg, meaning that the pressure is equal to the pressure that would be produced by a column of mercury, with no force acting above it, that is 10-8 mm high. This is absolute pressure measured above zero pressure as a reference level. Inside a steam boiler, the pressure may be 800 lbf/in.2 (5.5 × 106 Pa) or higher. Such pressure, measured above atmospheric pressure as a reference level, is gage pressure, designated psig. See Pressure measurement
## pressure
The force per unit surface area at any point in a gas or liquid. The pressure of a gas is proportional to temperature and density: at constant temperature, as the density is increased the pressure increases accordingly. This law of classical physics does not apply to degenerate matter.
## Pressure
a physical quantity characterizing the intensity of normal forces (perpendicular to the surface) with which one body acts on another’s surface (for example, the foundations of a building acting on the ground, a liquid acting on the walls of a vessel, and gas in the cylinder of a motor acting on the piston). If the forces are distributed uniformly over the surface, then the pressure ρ on any part of the surface i s p = F/S, where 5 is the area of the part and F is the sum of the forces applied perpendicular to it. If the distribution of forces is nonuniform, this equality gives the mean pressure on the given small area, whereas at the limit, with S tending toward zero, it gives the pressure at a given point. If the distribution of forces is uniform, the pressure at all points of the surface is the same; if the distribution is nonuniform, the pressure varies from point to point.
For a continuous medium, the concept of pressure at each point in the medium is similarly introduced; it plays an important part in the mechanics of liquids and gases. At any point in a quiescent liquid the pressure in all directions is the same;
Table 1. Conversion of units of pressure
Nlm2barkgflcm2atmmm Hgmm H20
1 N/m2(Pascal).................110-51.01972 x 10-5;0.98692 x 10-5750.06 x 10-50.101972
1 bar = 106dynes/cm2.................10511.019720.98692750.061.01972 x 104
1 kgf/cm2 = 1 at.................0.980665 x 1050.98066510.96784735.56104
1 atm.................1.01325 x 1051.013251.033217601.0332 x 104
1 mm Hg (torr).................133.3221.33322 x 10-31.35951 x 10-31.31579 x10--3113.5951
1 mm H20.................9.806659.80665 x 10-5;10-49.67841 x 10-57.3556 x 10-41
this is true also of moving liquids or gases, if they may be considered ideal (frictionless). In a viscous liquid the value of the mean pressure for three mutually perpendicular directions is taken to be the pressure at a given point.
Pressure plays an important part in physical, chemical, mechanical, and biological phenomena.
S. M. TARG
In a gaseous medium pressure is associated with the transfer of momentum during collisions of thermally moving gas molecules with each other or with the surface of bodies adjacent to the gas. The pressure in gases, which may be called thermal, is proportional to the temperature (the kinetic energy of the particles). In condensed mediums (liquids and solids), unlike gases, in which the mean distances between randomly moving particles are much greater than the size of the particles themselves, interatomic distances are comparable to atomic dimensions and are determined by the equilibrium of interatomic (intermolecular) forces of repulsion and attraction. When atoms approach one another repulsion forces increase, bringing about so-called cold pressure. In condensed mediums the pressure also has a “thermal” component, which is associated with the thermal vibrations of the atoms (nuclei). Given a steady or diminishing volume of a condensed medium, the thermal pressure rises as the temperature increases. At temperatures of about 104 ° K or more, thermal excitation of electrons makes an appreciable contribution to the thermal pressure.
Pressure is measured with manometers, barometers, and vacuometers, as well as with various pressure sensors.
Units of pressure have the dimensions of force divided by area. In the International System of Units, the unit of pressure is the newton per sq m (N/m2); in the Mks system, it is the kilogram-force per sq cm (kgf/cm2). Subsidiary units of pressure also exist—for example, the physical atmosphere (atm), the technical atmosphere (at), the bar, and mm of water and mercury columns (torr), by means of which the pressure measured is compared with the pressure of a column of liquid (water or mercury). (See Table 1.)
In the USA and Great Britain pressure is expressed in pounds-force per square inch (lbf/in.2), poundals per square foot (pdl/ft2), inches of water (in. H20), feet of water (ft H20), and inches of mercury (in. Hg); 1 lbf/in.2 = 6,894.76 N/m2; 1 pdl/ft2 = 1.48816 N/m2; 1 in. H20 = 249.089 N/m2; 1 ft H20 =2,989.07 N/m2; 1 in. Hg = 3,386.39 N/m2.
L. D. LIVSHITS
## pressure
[′presh·ər]
(mechanics)
A type of stress which is exerted uniformly in all directions; its measure is the force exerted per unit area.
## pressure
The force per unit area exerted by a homogeneous liquid or gas on the walls of its container.
## pressure
1. the normal force applied to a unit area of a surface, usually measured in pascals (newtons per square metre), millibars, torr, or atmospheres.
References in periodicals archive ?
Understanding the pathophysiology and clinical implications of the Cushing reflex and other physical signs of increased intracranial pressure [Electronic version].
Consequently, in the IJV, the Trendelenburg position is more effective in increasing the CSA than passive leg elevation alone and there is no reason to recommend passive leg elevation alone for routine cannulation into the UV However, passive leg elevation can be an alternative to the Trendelenburg position when the Trendelenburg position is contraindicated, such as the increased intracranial pressure or when a tilt table is unavailable.
The labels for isotretinoin and prednisone include the possible risk of increased intracranial pressure, depression, insomnia, emotional instability, dizziness, and headache, Dr.
The increased intracranial pressure caused by neoplasm can cause the patient's headache to be generalized.
The Nervous System section, for example, covers altered levels of consciousness and increased intracranial pressure, seizures and infections, trauma, cerebrovascular accidents, and degenerative disorders.
What actions by the technician could result in increased intracranial pressure? What actions by other providers might increase intracranial pressure?
In extreme cases, this craniofacial disorder constricts the brain's ability to expand normally within the skull, resulting in increased intracranial pressure, brain damage and/or loss of sight.
Brain swelling apparently results from a failure of autoregulation of cerebral circulation that causes vascular congestion and increased intracranial pressure, which may be difficult or impossible to control (7).
Once symptoms of increased intracranial pressure such as abnormal diameter of pupils, weakened pulse, increase of blood pressure and emesis were observed, therapies for lowering intracranial pressure and relieving dehydration were adopted to prevent the occurrence of cerebral hernia.
Langfitt, "Increased intracranial pressure," Clinical Neurosurgery, vol.
Headache has lateralising value, especially in patients with supratentorial lesions who have no obvious increased intracranial pressure (Suwanwela et al [5]).
Intracranial hypertension: theory and management of increased intracranial pressure. In: Hickey JV, ed.
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