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https://anyscript.org/tutorials/Parameter_studies_and_optimization/lesson1.html | 1,723,599,450,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641086966.85/warc/CC-MAIN-20240813235205-20240814025205-00640.warc.gz | 71,339,415 | 15,793 | # Defining a Parameter Study#
A parameter study is a systematic way to vary several model parameters and have the system automatically run one or several analyses for each combination of parameters.
For instance, you may want to know how the forces affecting a joint prosthesis depend on the implanted position. Or you may be interested in finding the standing posture by which you can hold a heavy box as easily as possible between your hands. Or how the position of a handle influences the muscular effort of operating it.
Or you may be interested in knowing how the seat height and horizontal position influence the muscle effort and metabolism of the rider. This is precisely what we shall do in this tutorial. To make life a bit easier for you, we have prepared a bicycle model you can download and play around with. Please click here to download the zip fileOptimBike.zip and unpack it to some pertinent place on your hard disk.
The bicycle model is pretty much the 2D Bike that you may know from the AnyBody Managed Model Repository.
As you can see the model is very simple. It has two legs and a pelvis that is rigidly fixed to the seat. The feet are attached to the crank mechanism, and the crank is loaded by a sinusoidal torque and constant angular velocity producing a mean mechanical output of 165 W. It has a total of 18 muscles - nine on each leg. You can control the design parameters of the bicycle and the way the rider propels the pedals by means of the variables at the top of the main file. It might be a good idea to play a bit around with the variables and run some analyses. Try, for instance, to raise and lower the seat. Notice that if you raise the seat more than a few centimeters, the model has trouble reaching the pedals. This is really a kinematical problem, but it causes momentarily very high muscle activities and, if you raise the seat further, makes the kinematical analysis break down because the feet lose the contact with the pedals.
The crank torque profile of a bicycle rider changes when the set is moved horizontally because the location of the cycle’s dead center changes. To account for this, a special feature has been set up in this model to adjust the phase shift of the crank torque profile to the seat position such that the minimum crank torque occurs when the pedals point towards the hip joint regardless of where the saddle is positioned.
## Some General Terminology#
Before we proceed with the definition of a parameter study it might be useful to introduce the terminology used by AnyBody for parameter and optimization studies:
• A design variable is an independent parameter controlling some aspect of the model, for instance the seat height, the pedal length, the pelvic angle, the crank torque variation, the cadence, the strength of a muscle, and so on. In short, just about any property you can set in the AnyScript model. A design variable is always a single number and it must be associated with upper and lower variation limits, but it is allowed to construct the model such that many properties depend on each variable. For instance, you might want to define a variable controlling the soleus muscle strength and then let the soleus muscle in both legs depend on it. In this way you can distinguish between dependent and independent parameters, and only independent parameters can be used as design variables. The AnyScript class defining a design variable is called AnyDesVar.
• A design measure is a dependent parameter that results from an analysis with given values of design variables. Typical examples would be the maximum muscle activity, the metabolism, the mechanical work, the mechanical power generated by a specific muscle, or the force in a joint. The AnyScript class for definition of design measures is the AnyDesMeasure.
These two classes are common to parameter studies and optimization studies, so the two types obviously share several concepts. Let us proceed with the definition of a parameter study. Actually, both these studies come from the same family of classes having the AnyDesStudy as common parent; we commonly refer to these studies as design studies. Within this kinship, they also share the definition of the “analysis” to be performed when evaluating the design measures for a certain set of design variables. The “analysis” is in fact an AnyScript operation (AnyOperation), called Analysis, which is a member of all design studies. To state an optimization or a parameter study properly, the design variables, the design measures, and the analysis must all be defined.
## Definition of a Parameter Study#
A parameter study is as the name indicates a study. Therefore its natural position in the model is below the existing AnyBodyStudy. We can insert a new parameter study by means of the object inserter mechanism from the class tree. Place the cursor below the definition of the AnyBodyStudy Study, click the Classes tab in the tree view, locate the AnyParamStudy, right-click, and insert a template of the class. You should get the following result:
AnyBodyStudy Study = {
AnyFolder &Model = .Model;
Gravity = {0.0, -9.81, 0.0};
tEnd = Main.BikeParameters.T;
};
AnyParamStudy <ObjectName> =
{
//LogFile = ;
/*Analysis =
{
Settings =
{
Echo = On;
ModelSceneUpdate = On;
};
//AnyOperation &<Insert name0> = <Insert object reference (or full object definition)>; You can make any number of these objects!
};*/
nStep = ;
AnyDesVar &<Insert name0> = <Insert object reference (or full object definition)>;
//AnyDesVar &<Insert name1> = <Insert object reference (or full object definition)>; You can make any number of these objects!
AnyDesMeasure &<Insert name0> = <Insert object reference (or full object definition)>;
//AnyDesMeasure &<Insert name1> = <Insert object reference (or full object definition)>; You can make any number of these objects!
};
As you can see, this requires a bit of additional specifications and general tidying up:
AnyParamStudy ParamStudy =
{
Analysis =
{
//AnyOperation &<Insert name0> = <Insert object reference (or full object definition)>; You can make any number of these objects!
};
nStep = ;
AnyDesVar &<Insert name0> = <Insert object reference (or full object definition)>;
AnyDesVar &<Insert name1> = <Insert object reference (or full object definition)>; You can make any number of these objects!
AnyDesMeasure &<Insert name0> = <Insert object reference (or full object definition)>;
//AnyDesMeasure &<Insert name1> = <Insert object reference (or full object definition)>; You can make any number of these objects!
};
Here’s a brief explanation of the different components of a parameter study:
Parameter
Function
Analysis
This is a specification of the operation(s) to perform to provide the data we are studying in the parameter study. This will typically be an InverseDynamicAnalysis operation, but it could also be simply an evaluation of some mathematical expression, or it could be a combination of multiple operations, for instance various calibrations followed by an inverse dynamic analysis.
nStep
This is a specification of how many steps to evaluate in the parameter study for each parameter.
AnyDesVar
The study must declare at least one of these. It is the parameter(s) that are varied in the study and for combinations of which the model is analyzed. You can define as many as you like, but please beware that the number of analyses in the parameter study is the product of steps for each AnyDesVar, so the time consumption grows exponentially with the number of AnyDesVars.
AnyDesMeasure
Each of these objects specifies a property that is the result of the analysis and which must be collected for further inspection as the study proceeds. You can define as many of these as you like.
Let us insert the necessary specifications to perform a parameter study on the saddle position of the bicycle:
AnyParamStudy ParamStudy =
{
Analysis = {
AnyOperation &Operation = ..Study.InverseDynamics;
};
As you can see, this is a pointer to the inverse dynamic analysis of the existing AnyBodyStudy in the bicycle model. This specification simply means that to evaluate the parameters we want to investigate in this parameter study, we must execute the analysis of the bicycle. This may seem obvious in a simple model like this one, but many AnyScript models contain multiple studies and each study contains multiple operations.
The next specification deals with the parameters to vary:
AnyParamStudy ParamStudy =
{
Analysis = {
AnyOperation &Operation = ..Study.InverseDynamics;
};
nStep = ;
Min = 0.61;
Max = 0.69;
};
Min = -0.22;
Max = -0.05;
};
AnyDesMeasure &<Insert name0> = <Insert object reference (or full object definition)>;
//AnyDesMeasure &<Insert name1> = <Insert object reference (or full object definition)>; You can make any number of these objects!
};
Please notice here that we have removed the ‘&’s that were inserted in the template in front of the variable names. Instead of pointing at AnyDesVars defined elsewhere we include the entire definition right here in the study, and this is actually the usual way to do it.
Each AnyDesVar gets three properties set. The first one is called Val and is simply set equal to an existing parameter in the model. The best way to understand this statement is to think of Val as a reference variable that is equalized in the first case with the SaddleHeight, which is a parameter defined at the top of the main file. At any time in the parameter study, Val will be equal to the saddle height as one should expect from the assignment. But in this special case, the assignment also goes the other way: It lets the parameter study control the value of what is on the right hand side of the equality sign, in this case the SaddleHeight parameter. We have similarly defined a second parameter, SaddlePos, which allows the parameter study to vary the horizontal saddle position. This two-way linkage between Val and another variable in the model implies certain restrictions on what assignments AnyScript allows for this particular case. The referred variable must be a scalar quantity, i.e. either a scalar or an element of a larger structure (e.g. vector or matrix). Secondly, it must be an independent scalar quantity, i.e., it cannot depend (by expressions) on other variables; otherwise there would exist an ambiguity.
The next step is to define the properties we wish to study, i.e. the dependent parameters or “design measures” of the model. As you know, after running an operation from a study the results are available in the Output branch of the tree view of the operation and they can be plotted, dumped and copied to the clipboard and so on. Now we are defining a study that will execute operations from other studies and assemble the results for later investigation. We might even want to make mathematical operations on these results, combine results from different operations, and so on. To do this we must refer to the result we wish to store for further processing. There is just one semantic problem: The results do not exist until we have performed the analysis, but we must refer to them already when we author (and load) the model.
To solve this problem we must go back to the AnyBodyStudy Study, from where we want to lift the results, and declare an object that will allow us to refer to computational results before they are actually made. The object is of class AnyOutputFun, and we shall add it to the existing AnyBodyStudy:
AnyBodyStudy Study = {
AnyFolder &Model = .Model;
Gravity = {0.0, -9.81, 0.0};
tEnd = Main.BikeParameters.T;
AnyOutputFun MaxAct = {
Val = .MaxMuscleActivity;
};
};
This allows us to refer to Study.Output.MaxMuscleActivity before it actually gets created.
AnyOutputFun is actually a class of mathematical function that returns the output (when existing) associated with the Val member. So here we have created a function called MaxAct that takes no arguments and returns the output data for .MaxMuscleActivity. Notice that AnyOutputFun must be declared inside a study in order to resolve the association with the output data structure of the particular study.
We can now use the output function, MaxAct(), in our design measure simply by calling the function in the assignment of the Val member of the AnyDesMeasure:
AnyDesVar SaddlePos = {
Min = -0.22;
Max = -0.05;
};
AnyDesMeasure MaxAct = {
Val = max(..Study.MaxAct());
};
Notice the definition. The MaxAct function for each InverseDynamics operation returns a vector of maximum muscle activities in the model. The vector has as many components as the study has time steps, i.e. 50 in the present case. In the definition of the AnyDesMeasure we want to save only the largest value of each of the vector, so we wrap the call of the MaxAct function in another max() function. AnyScript gives you a number of such data processing functions and we shall study others further down. Please refer to the reference manual for further details.
One thing is missing before we can try the whole thing out: We must specify how many steps we want the parameter study to perform for each parameter. As in AnyBodyStudies this is done by the nStep variable, but where nStep in an AnyBodyStudy is an integer variable, it is a vector with one component for each AnyDesVar in an AnyParamStudy. We shall be modest at first and choose only five steps in each direction. And so, the final AnyParamStudy looks like this:
AnyParamStudy ParamStudy = {
Analysis = {
AnyOperation &Operation = ..Study.InverseDynamics;
};
nStep = {5,5};
Min = 0.61;
Max = 0.69;
};
Min = -0.22;
Max = -0.05;
};
AnyDesMeasure MaxAct = {
Val = max(..Study.MaxAct());
};
};
It is finally time try it out. If you have typed everything correctly, then you should be able to load the model. Then find the Main.ParamStudy.ParameterStudy opertion in the operation dropdown:
Make sure you have a Model View window open. With the ParameterStudy select in the operation drop-down click the “Run” botton. You should see the model starting to cycle, and if you watch the vicinity of the saddle carefully, you will see that the hip joint is changing its position on a 5 x 5 grid. With a reasonably fast computer it should take a minute or less to do the 25 analyses after which the computations stop.
Note
If you turn off the model view, the computation should speed up.
Congratulations! You have completed your first parameter study. Let us investigate the result.
The obvious way to visualize the results of a study with two parameters is as a 3-D surface. The chart view in AnyBody can also do that.
The toolbar of this window indicates a kinship with the Model View window. Indeed, if you select the rotation button in the toolbar and drag the mouse with the left button down inside the coordinate system you will notice that the system rotates just like an ordinary Model View. Now, expand the Model node in the tree until you can click the ParamStudy->Output->MaxAct->Val property. The coordinate system automatically attains a second abscissa axis and you can see a nice surface like this:
The surface shows the maximum muscle activity over the cycle for each of the 25 combinations of parameters and provides a very nice overview of the behavior of the model. The surface reveals that the highest and most backward position is the best. Why not try higher and more backward, then? It is very simply to do:
AnyDesVar SaddleHeight = {
Min = 0.61;
Max = 0.69 + 0.02;
};
Min = -0.22 -0.03;
Max = -0.05;
};
When you re-run the parameter study, things will go well in the beginning, but towards the end of the 25 combinations you may notice muscles beginning to bulge more and momentarily attain the color of magenta. This is the system’s way of demonstrating that the muscles have been loaded above 100% of their strength. The reason why this happens is that, as the seat rises, the model gets into positions where it is difficult for the feet to reach the pedals. Just before the feet cannot reach the pedals the knee movements are accelerated causing large inertia forces in the system. All this happens a bit more drastically in an ideal rigid body model than it would in real life where joints have a bit of slack, segments are slightly elastic, and the prescribed kinematics may be compromised. You can see very clearly what happens if you go back to the AnyChart View and study the new surface:
The surface is now completely dominated by the one combination, which is difficult for the model to do. You can still see the surface shape if you change the scale of the value axis. This and all other settings are available if you click the button in the toolbar. Doing so will produce a window with a tree view in which you can select ValueAxis->Max. Try setting Max to 0.25 and you should obtain the following:
What this study reveals is that in terms of muscle activity to drive the bicycle a high seat is advantageous, but there seems to be a very sharp limit where the leg gets close to not being able to reach the pedals, and this should not be exceeded. One additional remark in this context is that this bicycle model has a predefined ankle angle variation whereas a real human can compensate for a higher seat by letting the ankle operate in a more plantar-flexed position.
Before we finish this section, let us take a look at a particularly important feature of AnyScript mathematics: The ability to compute integral properties. AnyBody has a simple way of approximating the metabolism of muscles based on the simulation of each muscle’s mechanical work. Metabolism is technically a power measured in Watt, and the sum of the individual muscle metabolisms will give us an estimate of the total metabolism involved in the bicycling process. It is fairly simple to add up the muscle metabolisms in the AnyBody study:
/// The study: Operations to be performed on the model
AnyBodyStudy Study = {
AnyFolder &Model = .Model;
nStep = 50;
Gravity = {0.0, -9.81, 0.0};
tEnd = Main.BikeParameters.T;
// Useful variables for the optimization
AnyFolder &r = Main.Model.Leg2D.Right.Mus;
AnyFolder &l = Main.Model.Leg2D.Left.Mus;
AnyVar Pmet_total = r.Ham.Pmet+r.BiFemSh.Pmet+r.GlutMax.Pmet+r.RectFem.Pmet+r.Vasti.Pmet+r.Gas.Pmet+r.Sol.Pmet+r.TibAnt.Pmet+l.Ham.Pmet+l.BiFemSh.Pmet+l.GlutMax.Pmet+l.RectFem.Pmet+l.Vasti.Pmet+l.Gas.Pmet+l.Sol.Pmet+l.TibAnt.Pmet;
AnyOutputFun MaxAct = {
Val = .MaxMuscleActivity;
};
};
Notice that we have defined the r and l variables for convenience to limit the size of the expressions. If you run the InverseDynamics Analysis (go on and try!) you will find the new variable mentioned in the list of output, and you can view it in the chart view.
The area under this curve is the total metabolism combusted over a crank revolution. To compute this we must introduce two more elements. The first one is an AnyOutputFun as we have seen it before. The purpose of this function is to make it semantically possible to refer to the output of the Pmet_total variable before is has actually been computed:
// Useful variables for the optimization
AnyFolder &r = Main.Model.Leg2D.Right.Mus;
AnyFolder &l = Main.Model.Leg2D.Left.Mus;
AnyVar Pmet_total = r.Ham.Pmet+r.BiFemSh.Pmet+r.GlutMax.Pmet+r.RectFem.Pmet+r.Vasti.Pmet+r.Gas.Pmet+r.Sol.Pmet+r.TibAnt.Pmet+l.Ham.Pmet+l.BiFemSh.Pmet+l.GlutMax.Pmet+l.RectFem.Pmet+l.Vasti.Pmet+l.Gas.Pmet+l.Sol.Pmet+l.TibAnt.Pmet;
AnyOutputFun MaxAct = {
Val = .MaxMuscleActivity;
};
AnyOutputFun Metabolism = {
Val = .Pmet_total;
};
};
The second missing element is the actual integration of the function. This we perform in the parameter study where we define the AnyDesMeasure:
AnyParamStudy ParamStudy = {
Analysis = {
AnyOperation &Operation = ..Study.InverseDynamics;
};
nStep = {10,10};
Min = 0.61;
Max = 0.69 /*+ 0.02*/;
};
Min = -0.22 /*-0.03*/;
Max = -0.05;
};
AnyDesMeasure MaxAct = {
Val = max(..Study.MaxAct());
};
AnyDesMeasure Metab = {
Val = secint(..Study.Metabolism(),..Study.tArray);
};
};
The secint function performs a numerical integration of the first argument against the second argument. Each argument must be an array and the number of components in the two arguments must be the same.
Notice the other two changes: We have changed the variable limits back to what they were before and we have decided to be a little more adventurous and have specified 10 variable steps in each direction. It is time to run the parameter study again. The new Metab variable is now available in the list under the ParamStudy in the Chart window and can be plotted:
We shall return to the capabilities of the Chart view in more detail in the next lesson, which deals with the definition of optimization studies. | 4,593 | 20,704 | {"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} | 2.53125 | 3 | CC-MAIN-2024-33 | latest | en | 0.926211 |
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Snr And Ber For Matlab
84 Threads found on edaboard.com: Snr And Ber For Matlab
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Space-time block codes (STBC) versus V-BLAST: The objective is to evaluate the effect of increasing average signal-to-noise ratio (snr) and channel correlation on the bit error rate (ber) of STBC and V-BLAST for a prescribed spectral efficiency. Consider two wireless systems: i) Alamouti STBC 2x2, (...)
Problem with simulating OFDM in Matlab
dear sir i am smulating ofdm in matlab but in this mentioned code i am not able to draw graph between ci and ber(ie snr n bit error rate only one point is displayed . plz have a look and help where i went wrong plz correct thanks dhruva % Simulate effect of interfarence noise % (...)
I need a Matlab Code for BER versus SNR for ML RECEIVER
If you plot ber vs. snr with a log-log scale, your error rate curve should be a straight line in large snr region and the slope of the straight line should be -1.
Looking for info about BER VS. SNR standard for GSM
hello! i need help in standard ber VS. snr for GSM
Can anyone shows the SNR v.s BER curve MB OFDM under 55.5M..
Q1: Can anyone shows the snr v.s ber curve of MB OFDM under(bit rate: 55.5M,80M,160M,200M,320M,480M...)in AWGN and multipath channel??? Q2: can some one tell me ,when works under the 200M,as standrad given the parameter (spreading rate=2),where should times the parameter???? is it after the IFFT block????? does (...)
Performance analysis (BER vs SNR) of TH-BPSK system
Hi all, I am working on the performance analysis of Time Hopping - BPSK systems. Can anybody post the matlab code for its implementation and ber vs. snr analysis. Thanks.
BER for OFDM using QPSK & Convolutional Encoding
Thanks for your reply. I suspect my problem is with the noise generation, I have used the inbuilt matlab function to genrate awgn noise y = awgn(x,snr). Is there another way of generating noise apart from this? I can post the code if it will help. Thanks again
mimo ofdm in mobile environment matlab
hello, I am currently studying wireless communication and I am facing a problem in my MIMO OFDM simulation using mimochan function. My problem is mainly that my input signal is not time based and therefore, whatever sampling time I use in the function the snr vs ber is always the same. I would like to (...)
What is the idea of simulating the outage in Matlab?
Hello, for example, when we want to simulate the ber in matlab we send bits over a channel ( fading or other channel ) and then count the bits that do not match. Now, I want to know how do we simulate the Outage probability ( the probability that snr falls down below a specific value ) in (...)
can u help 4 BER performance in MIMO?I need matlab codes for this......
send me the basic requirement , aither ur doing snr vs ber perfromence, which modulation and encoding, so that i can frwd u
i need a matlan code for IDMA using tree based interleaver..
immediately i need a matlab code for IDMA using matrix interleaver... i have to plot graph for ber vs snr... and also i have generate the graph using tree based interleaver.... if anyone know means please help me to complete my project.... thanks
Plotting Data rates vs SNR in 3 node cooperative communication
I draw the ber vs snr and i can't draw the data rate vs the snr as i can't get a direct relation between the data rate ans the snr in the equation? anyhelp?
how to plot Probability of Error Bit Rate against SNR in matlab or Excel
Hello everybody, any one can help me how to write a code can plot y axis in log scale and x axis in normal scale. thanks, - - - Updated - - - What i am trying to say is i've got results displayed for ber using matlab simulink but i don't know how to plot curve of ber. i would be so appreciated (...)
performance evaluation of 802.11g and 802.11a in Simulink
What parameters have to be changed for the two models to be distinguished from one another?? I have downloaded the WLAN files from and 802.11a from . On simulating both the files the outputs are the same in ter
CDMA rake receiver code for snr vs BER
Please check matlab codes for bellow. matlab Source Codes -------------------------------------------------------------------------------- analemma, a program which evaluates the equation of time, a formula for the difference between the uniform 24 hour day and (...)
Matlab code for rician fading channel
Is there anyone who can help me by providing the matlab codes for qpsk & bpsk modulation in riccian channel to find the ber vs snr ? Plz , it z too urgent
Has anyone generated a rayleigh fading channel where there are two transmitters(T1,T2) and two receivers(R1,R2). The information from T1 is to be sent to R1 and information from T2 is to be sent to R2. What is the effect of interference of T1 over R2 in ber vs snr? Is there any (...)
How to do the QPSK modulation in Matlab?
Hi! I support another example that I wrote befroe. It is a example to run the ideal ber/snr curve of BPSK and QPSK system. The simulation result shows the two different almost get the same curve. Hope that is helpful for you. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % the (...)
BER of OFDM,QAM,BPSK in matlab
i am new to this stuff....can anyone help me about this.... how actually to plot this ber graph from matlab....suppose to use pragramming in matlab or simulink? can anyone show me the code to simulate this ber... thx first.... Added after 2 hours 13 minutes: If i already simulate a QAM-16...n count t
My Final Project - Need Help ( Wireless Comm + Matlab)
i have attached a m-file (in the form of txt file) from my previous post graduate assignment which requires me to simulate a bpsk transmission using baseband envelope approach; using either rectangle, sync or raised cosine pulse shaping through a AWGN or Rayleigh flat fading channel. i have also wrote the commands to plot the (...)
How do you get the BER plots using BERTool in MATLAB?
@tmltanml : i am sorry to say your attachment doesn't contain any stuff on berTool. @m_llaa : biterr is only useful for getting the ber balue for a set of data..but i need a plot with varying snr..I had written a function for the same purpose as the biterr does(by using xor (...)
Factor used to scale Noise in 4QAM simulation ???
Hi, i am simulating the 4QAM in matlab. there i need to scale the noise to get the snr vs ber graph. Can any one of you plz tell me the factor by whcih we can scale the noise...and also how u get this factor? P.S. my noise if of type ... noise = randn(1,N) + i * randn(1,N); Waiting.....
Need help in CDMA "matlab"!!
Hi guys ... I need your help! Can you give me any matlab code which inclode simulation of ber vs snr for single user and multi users in CDMA system! Please help me I'm not very good in matlab :(
Iterative decoding....
Hi, This problem occurs when the data is not enough. So, you have to increase your data; especially for high snr. Ahmed
Suggest me some projects for final year
You can take up several simulations in matlab.. to quote a few .. 1) Simulating Various Modulation methods both analog and digital 2) Calculating ber or snr for each modualtion 3) Fading simulation (Mobile Communication) 4) CDMA or GSM Simulation 5) OFDM Simulation Thanks, Mathuranathan.V (...)
Maximum Ratio combiner in Multipath Rayleigh fading channel
Hi, Could you please help me how to use the Maximum Ratio combination in Multipath Rayleigh fading channel, using matlab comands, and have a look for the program bellow you will see the error that I have, and could you tell me the problem, and how to solve it. clear all whos (...)
matlab code for hard versus soft decision decoding?
ur way may be correct but the thing is that i want to implement it in matlab and for that i dont think if i need quantization. the problem i m facing is that my soft viterbi decoder works fine in AWGN channle but when I apply my soft viterbi deocoder in Rayleigh fading channel with white Gaussian noise the ber vs (...)
Spectral efficiency with matlab in MIMO systems
Hi there, I am looking for a matlab code which computes and plots the spectral efficiency and/or the throughput in a MIMO system for different types of modulation (BPSK, QPSK, 16QAM...). Is there any formulas based on the ber and/or the snr (...)
matlab code for multi-relay cooperative communication system
HI dear, Can you please help me for a two-hop system simulation? i want to simulate the problem of two hop system and then ber vs snr plot. reply plz Added after 45 minutes: please email me, i am also new in this field and want to discuss some problems with you. thanks assad.abbasi@yahoo.com[/e
MB-OFDM Matlab Simulation
I am trying to simulate ber vs snr for MB-OFDM. I have code of OFDDM. Can anyone help me. I am using matlab.
Where to sample the output in BPSK demodulator using Matlab?
Hello, I think to reduce the ber, you have to begin sampling your received signal at its maximum value. This will give the maximum snr.
theoretical curve for BER of ASK
Coherent ASK detection has the same Eb/N0 characteristic as coherent FSK detection. If you have matlab you can utilize the bertool... snr = (0:1:18 ); ber = berawgn(snr, 'fsk', 2, 'coherent'); figure; semilogy(snr, ber); grid on; xlabel('E_b/N_0 (...)
hi, I'm trying to convert the dvb-t 2k demo in matlab r2009b to work in 8k mode. I made some changes to commdvbt.mdl file and to table_gen.m file. The demo works, but not correctly. It gives ber around 0.5 for all vaues for snr in AWGN channel. What could be the problem? biserka
+ 100 Points- Quary in WiMAX model.
Dear all I am in urgent need to solove the quary in wimax MIMO-OFDM mode. In this model, after running the program, 1 is to be entered for including the MIMO and O for single input single output system. How ever my qaury is in both the cases, with same parameter (i.e. channel bandwidth, snr etc), i am (...)
Need help with MIMO BER MATLAB Simulation
Hello everyone I have made the following code but there was a problem and am seeking help % the first task is to generate a bpsk signal and simulate the ber and the second task is to convert this to MIMO clear;clc; N = 1e4; % Number of data bits Nt = 1; (...)
Matlab Code for matrix interleaver in IDMA
Hello Everyone I actually need of a matlab code which is using matrix interleaver in idma. i want to check its ber vs snr performance. and i also want to simulate my interleaver in place of matrix interleaver. So kindly tell me what are the parameters on which interleaver performance (...)
How to model multipath propagation in matlab?
first of all, create your modulation then pass it through the channel, this must be done in steps as follows: 1. create you multipath channel object using this command (rayleighchan). 2. use the filter function to add your signal to the channel using this command (filter(x,channel)) 3. create a loop for your noise such as from 0 to (...)
BER output problem for Radio over Fiber system/Optical Comm
Hi guys, this is my first work in optical and first simulation in matlab. I am sorry if there is mistake and im really new to this. Here is what I did, im simulation an entire Radio Over Fiber system with matlab, linear system. Let me insert my code first, %*****************************Defining (...) | 3,670 | 15,146 | {"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-2014-41 | longest | en | 0.903179 |
https://www.educative.io/edpresso/what-is-the-des-algorithm | 1,652,894,285,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522284.20/warc/CC-MAIN-20220518151003-20220518181003-00774.warc.gz | 811,534,247 | 157,472 | a concise shot of dev knowledge
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What is the DES algorithm?
Edpresso Team
Data Encryption Standard (DES) is a block cipher algorithm that takes plain text in blocks of 64 bits and converts them to ciphertext using keys of 48 bits. It is a symmetric key algorithm, which means that the same key is used for encrypting and decrypting data.
Encryption and decryption using the DES algorithm.
Steps for generating keys
There are 16 rounds of encryption in the algorithm, and a different key is used for each round. How keys are generated is listed below.
Bits are labeled from 1 to 64 starting from the most significant bit and going to the least significant bit.
1. Compress and transpose the given 64-bit key into a 48-bit key using the following table:
// The array elements denote the bit numbers
int pc1[56] = {
57,49,41,33,25,17,9,
1,58,50,42,34,26,18,
10,2,59,51,43,35,27,
19,11,3,60,52,44,36,
63,55,47,39,31,23,15,
7,62,54,46,38,30,22,
14,6,61,53,45,37,29,
21,13,5,28,20,12,4
};
PC-1 table
1. Divide the result into two equal parts: C and D.
2. C and D are left-shifted circularly. For encryption rounds 1, 2, 9, and 16 they are left shifted circularly by 1 bit; for all of the other rounds, they are left-circularly shifted by 2.
3. The result is compressed to 48 bits in accordance with the following rule:
int pc2[48] = {
14,17,11,24,1,5,
3,28,15,6,21,10,
23,19,12,4,26,8,
16,7,27,20,13,2,
41,52,31,37,47,55,
30,40,51,45,33,48,
44,49,39,56,34,53,
46,42,50,36,29,32
};
PC-2 table
1. The result of step 3 is the input for the next round of key generation.
Steps for encryption
1. Transpose the bits in the 64-block according to the following:
// 58 means that the 58th bit should be considered
// the first bit, 50th bit the second bit and so on.
int initial_permutation_table[64] = {
58,50,42,34,26,18,10,2,
60,52,44,36,28,20,12,4,
62,54,46,38,30,22,14,6,
64,56,48,40,32,24,16,8,
57,49,41,33,25,17,9,1,
59,51,43,35,27,19,11,3,
61,53,45,37,29,21,13,5,
63,55,47,39,31,23,15,7
};
The initial permutation table
1. Divide the result into equal parts: left plain text (1-32 bits) and right plain text (33-64 bits)
2. The resulting parts undergo 16 rounds of encryption in each round.
The right plain text is expanded using the following expansion table:
// The array elements denote the bit numbers
int expansion_table[48] = {
32,1,2,3,4,5,4,5,
6,7,8,9,8,9,10,11,
12,13,12,13,14,15,16,17,
16,17,18,19,20,21,20,21,
22,23,24,25,24,25,26,27,
28,29,28,29,30,31,32,1
};
The expansion table
1. The expanded right plain text now consists of 48 bits and is XORed with the 48-bit key.
2. The result of the previous step is divided into 8 boxes. Each box contains 6 bits. After going through the eight substitution boxes, each box is reduced from 6 bits to 4 bits. The first and last bit of each box provides the row index, and the remaining bits provide the column index. These indices are used to look-up values in a substitution box. A substitution box has 4 rows, 16 columns, and contains numbers from 0 to 15.
3. The result is transposed in accordance with the following rule:
// The array elements denote the bit numbers
int permutation_table[32] = {
16,7,20,21,29,12,28,17,
1,15,23,26,5,18,31,10,
2,8,24,14,32,27,3,9,
19,13,30,6,22,11,4,25
};
The permutation table
1. XOR the left half with the result from the above step. Store this in the right plain text.
2. Store the initial right plain text in the left plain text.
3. These halves are inputs for the next round. Remember that there are different keys for each round.
4. After the 16 rounds of encryption, swap the left plain text and the right plain text.
5. Finally, apply the inverse permutation (inverse of the initial permutation), and the ciphertext will be generated.
Steps for decryption
The order of the 16 48-bit keys is reversed such that key 16 becomes key 1, and so on. Then, the steps for encryption are applied to the ciphertext.
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Learn in-demand tech skills in half the time | 1,297 | 4,124 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2022-21 | latest | en | 0.851603 |
https://metanumbers.com/1024163 | 1,642,441,099,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300574.19/warc/CC-MAIN-20220117151834-20220117181834-00550.warc.gz | 464,682,988 | 7,396 | # 1024163 (number)
1,024,163 (one million twenty-four thousand one hundred sixty-three) is an odd seven-digits composite number following 1024162 and preceding 1024164. In scientific notation, it is written as 1.024163 × 106. The sum of its digits is 17. It has a total of 2 prime factors and 4 positive divisors. There are 877,848 positive integers (up to 1024163) that are relatively prime to 1024163.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 7
• Sum of Digits 17
• Digital Root 8
## Name
Short name 1 million 24 thousand 163 one million twenty-four thousand one hundred sixty-three
## Notation
Scientific notation 1.024163 × 106 1.024163 × 106
## Prime Factorization of 1024163
Prime Factorization 7 × 146309
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 1024163 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 1,024,163 is 7 × 146309. Since it has a total of 2 prime factors, 1,024,163 is a composite number.
## Divisors of 1024163
4 divisors
Even divisors 0 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 1.17048e+06 Sum of all the positive divisors of n s(n) 146317 Sum of the proper positive divisors of n A(n) 292620 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1012.01 Returns the nth root of the product of n divisors H(n) 3.49998 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 1,024,163 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 1,024,163) is 1,170,480, the average is 292,620.
## Other Arithmetic Functions (n = 1024163)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 877848 Total number of positive integers not greater than n that are coprime to n λ(n) 438924 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 80088 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 877,848 positive integers (less than 1,024,163) that are coprime with 1,024,163. And there are approximately 80,088 prime numbers less than or equal to 1,024,163.
## Divisibility of 1024163
m n mod m 2 3 4 5 6 7 8 9 1 2 3 3 5 0 3 8
The number 1,024,163 is divisible by 7.
## Classification of 1024163
• Arithmetic
• Semiprime
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
## Base conversion (1024163)
Base System Value
2 Binary 11111010000010100011
3 Ternary 1221000212222
4 Quaternary 3322002203
5 Quinary 230233123
6 Senary 33541255
8 Octal 3720243
10 Decimal 1024163
12 Duodecimal 41482b
20 Vigesimal 68083
36 Base36 ly8z
## Basic calculations (n = 1024163)
### Multiplication
n×y
n×2 2048326 3072489 4096652 5120815
### Division
n÷y
n÷2 512082 341388 256041 204833
### Exponentiation
ny
n2 1048909850569 1074254659288298747 1100211874620681909623761 1126796294147141446605999937043
### Nth Root
y√n
2√n 1012.01 100.799 31.8121 15.9248
## 1024163 as geometric shapes
### Circle
Diameter 2.04833e+06 6.43501e+06 3.29525e+12
### Sphere
Volume 4.49983e+18 1.3181e+13 6.43501e+06
### Square
Length = n
Perimeter 4.09665e+06 1.04891e+12 1.44839e+06
### Cube
Length = n
Surface area 6.29346e+12 1.07425e+18 1.7739e+06
### Equilateral Triangle
Length = n
Perimeter 3.07249e+06 4.54191e+11 886951
### Triangular Pyramid
Length = n
Surface area 1.81677e+12 1.26602e+17 836226
## Cryptographic Hash Functions
md5 4d5dae02963fe7c45755e6fd4b6254c3 fe49dcc40a582148850625030867b5f91457b4cc 4d5954e9a4568bdb280069511b20df614aceaf138f975df31488cfed4129d6e8 6f41171dd932026bb5a09ada92ceeab516356ea07a00f701d514c061a6b7fde6b8a8bfed67a55cf7192b4c622b9a5d59fb9f8a4b394b0ce20c79ea9ea963517a 0789d88892f3208df03bf73d293e1e6b89db9ce5 | 1,526 | 4,317 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2022-05 | latest | en | 0.809523 |
http://copaseticflow.blogspot.com/2011/07/subelement-study-focus-added-to.html | 1,544,408,902,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823236.2/warc/CC-MAIN-20181210013115-20181210034615-00324.warc.gz | 64,692,874 | 29,303 | ### Subelement Study Focus Added to Practice Exams
Subelement specific practice tests have been added to the free ham radio practice exams at http://copaseticflows.appspot.com/hamtest for the United States amateur exams. To study a specific subelement press the button, (look below the question area), that corresponds to that subelement. A new test will start specific to that subelement. The questions that appear in the exam will be only the questions for that subelement in numerical order. The exam header will have a message reminding you that you are studying for that subelement and not practicing a full exam.
As always with new features, if you see anything new or old that looks broken, please contact me at hcarter333@gmail.com. Have fun!
### Cool Math Tricks: Deriving the Divergence, (Del or Nabla) into New (Cylindrical) Coordinate Systems
The following is a pretty lengthy procedure, but converting the divergence, (nabla, del) operator between coordinate systems comes up pretty often. While there are tables for converting between common coordinate systems, there seem to be fewer explanations of the procedure for deriving the conversion, so here goes!
What do we actually want?
To convert the Cartesian nabla
to the nabla for another coordinate system, say… cylindrical coordinates.
What we’ll need:
1. The Cartesian Nabla:
2. A set of equations relating the Cartesian coordinates to cylindrical coordinates:
3. A set of equations relating the Cartesian basis vectors to the basis vectors of the new coordinate system:
How to do it:
Use the chain rule for differentiation to convert the derivatives with respect to the Cartesian variables to derivatives with respect to the cylindrical variables.
The chain rule can be used to convert a differential operator in terms of one variable into a series of differential operators in terms of othe…
### Lab Book 2014_07_10 More NaI Characterization
Summary: Much more plunking around with the NaI detector and sources today. A Pb shield was built to eliminate cosmic ray muons as well as potassium 40 radiation from the concreted building. The spectra are much cleaner, but still don't have the count rates or distinctive peaks that are expected.
New to the experiment? Scroll to the bottom to see background and get caught up.
Lab Book Threshold for the QVT is currently set at -1.49 volts. Remember to divide this by 100 to get the actual threshold voltage. A new spectrum recording the lines of all three sources, Cs 137, Co 60, and Sr 90, was started at approximately 10:55. Took data for about an hour.
Started the Cs 137 only spectrum at about 11:55 AM
Here’s the no-source background from yesterday
In comparison, here’s the 3 source spectrum from this morning.
The three source spectrum shows peak structure not exhibited by the background alone. I forgot to take scope pictures of the Cs137 run. I do however, have the printout, and…
### Unschooling Math Jams: Squaring Numbers in their own Base
Some of the most fun I have working on math with seven year-old No. 1 is discovering new things about math myself. Last week, we discovered that square of any number in its own base is 100! Pretty cool! As usual we figured it out by talking rather than by writing things down, and as usual it was sheer happenstance that we figured it out at all. Here’s how it went.
I've really been looking forward to working through multiplication ala binary numbers with seven year-old No. 1. She kind of beat me to the punch though: in the last few weeks she's been learning her multiplication tables in base 10 on her own. This became apparent when five year-old No. 2 decided he wanted to do some 'schoolwork' a few days back.
"I can sing that song... about the letters? all by myself now!" 2 meant the alphabet song. His attitude towards academics is the ultimate in not retaining unnecessary facts, not even the name of the song :)
After 2 had worked his way through the so… | 862 | 3,969 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2018-51 | longest | en | 0.869766 |
http://www.conservapedia.com/Encryption | 1,532,285,080,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593438.33/warc/CC-MAIN-20180722174538-20180722194538-00265.warc.gz | 430,055,287 | 7,684 | # Encryption
Encryption is the process of changing data into a secret code.[1] It is usually carried out today by computers, although in the past it was traditionally carried out by analog machines or by specialists known as cryptographers.
The earliest known form of encryption is that known as "ROT-13," or the "Caesar cipher", because its invention was once reportedly attributed to Julius Caesar.[2] In the Caesar cipher, each letter of the message (or "plaintext") is shifted forward in the alphabet by 13 places; for example, "CIPHER" becomes "PVCURE".[3] The Caesar cipher is what is called a symmetric cipher, meaning that the method of encryption is the same as the method of decryption (rotate by 13 characters), and therefore it is just as easy to decode any message as it was to encode it. For this reason, modern cryptanalysts tend to view most symmetric encryption methods as flawed.[4]
Another popular historical encryption method was the "enigma machine" developed by Nazi Germany during World War II. The enigma machine (or ENIGMA for short) worked on the plugboard principle. It had the ability to generate many different ciphertexts for the same plaintext using different "keys", which meant that Allied cryptographers could no longer rely on being able to match encoded texts with plaintexts. Prior to World War II, cryptographers had gotten used to being able to collate messages with their ciphertexts to create large "dictionaries" (for example, matching the ciphertext "PVCURE" with its plaintext "CIPHER" in the above example). These code dictionaries could then be used to launch dictionary attacks against messages whose decodings were not yet known. ENIGMA rendered this tactic hopeless. For example, while "PVCURE" might mean "CIPHER" using Monday's key, it might mean "ATTACK" using Tuesday's key. However, a letter cannot be encrypted as itself, so no key can decrypt "PVCURE" as "POISON".
The Enigma cipher was eventually cracked by homosexual[5] British mathematician Alan Turing, with the aid of an early computer. This was the beginning of modern cryptography as a science, although few advancements were made from the end of WWII until the rise of the Internet in the 1980s.
Today, encryption is used by hackers and other anti-government types to conceal their activities from the federal government. For example, the leftist GNU organization provides a program called "Pretty Good Privacy" (PGP) for encrypting e-mail. Several forms of "strong" encryption are considered munitions under federal law, and may not be exported to belligerent countries such as Iran.[6] | 552 | 2,607 | {"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} | 2.53125 | 3 | CC-MAIN-2018-30 | latest | en | 0.971602 |
https://www.analog.com/ru/analog-dialogue/articles/improving-power-supply-design-using-semi-automation.html | 1,656,235,900,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103037649.11/warc/CC-MAIN-20220626071255-20220626101255-00718.warc.gz | 687,477,107 | 27,495 | Improving Power Supply Design Using Semi-Automation—Five Steps to Quick and Efficient Design
Introduction
Designing the correct power source is essential and complex, since there is no one typical application. While total automation of power supply design is yet to be achieved, a comprehensive range of semi-automated tools are available today. This article details the use of semi-automated design tools through five critical steps of the power supply design process. These tools can be valuable to both the novice and expert power supply design engineer.
Power Supply Design Step 1: Creating the Power Supply Architecture
Creating a suitable power supply architecture is a decisive step in power supply design. This step becomes more complex by increasing the number of needed voltage rails. At this point, the decision is made as to whether and how many intermediate circuit voltages need to be created. Figure 1 shows a typical block diagram of a power supply. The 24 V supply voltage of an industrial application is shown on the left. This voltage must be converted now into 5 V, 3.3 V, 1.8 V, 1.2 V, and 0.9 V with corresponding currents. What is the best method for generating the individual voltages? Selecting a classic step-down switching (buck) converter makes the most sense for converting from 24 V to 5 V. However, how do you generate the other voltages? Does it make sense to generate the 3.3 V from the 5 V already created, or should we convert to 3.3 V directly from 24 V? Answering these questions requires further analysis. Since an important property of a power supply is the conversion efficiency, keeping the efficiency as high as possible is important when selecting the architecture.
If intermediate voltages, such as 5 V in the example shown in Figure 1, are used to generate additional voltages, the energy used for the 3.3 V must already pass through two conversion stages. Each conversion stage has only limited efficiency. If, for example, a conversion efficiency of 90% is assumed for each conversion stage, the energy for 3.3 V, which has already passed through two conversion stages, only has an efficiency of 81% (0.9 × 0.9 = 0.81). Can this rather low efficiency be tolerated in a system or not? This depends on the current required from this 3.3 V rail. If current of only a few mA is needed, the low efficiency might not be a problem at all. For higher currents, however, this lower efficiency might have a greater effect on the overall system efficiency and consequently represent a big disadvantage.
From the considerations just mentioned, however, you cannot draw the general conclusion that it is always better to convert directly from a higher supply voltage to the lower output voltage in one step. Voltage converters that can handle a higher input voltage are usually more expensive and have a reduced efficiency when there is a greater difference between the input voltage and the output voltage.
In power supply design, the solution to finding the best architecture is to use an architecture tool such as LTpowerPlanner®. It is available free of charge from Analog Devices and is part of the LTpowerCAD® development environment, which can be installed locally on your computer. LTpowerPlanner is a tool that makes evaluating different architectures quick and easy.
Finalizing the Specification
Finalizing the specification is extremely important in power supply design. All additional development steps depend on the specification. Frequently, the precise requirements of the power supply are unknown until the rest of the electronic system has been completely designed. This usually results in increasing time constraints on power supply design development. It also often happens that the specification is changed in a later development stage. For example, if in its final programming, an FPGA requires additional power, the voltage for a DSP must be reduced to save energy, or the originally intended switching frequency of 1 MHz must be avoided because it is coupled into the signal path. Such changes can have very serious effects on the architecture and, in particular, on the circuit design of the power supply.
A specification is usually adopted at an early stage. This specification should be designed to be as flexible as possible so that it is relatively easy to implement any changes. In this effort, selecting versatile integrated circuits is helpful. Working with development tools is particularly valuable. This allows the power supply to be recalculated within a short time. In this way, specification changes can be implemented more easily and, above all, more quickly.
The specification includes the available energy, the input voltage, the maximum input current, and the voltages and currents to be generated. Other considerations include size, financial budget, thermal dissipation, EMC requirements (including both conducted and radiated behaviors), expected load transients, changes in the supply voltage, and safety.
LTpowerPlanner as an Optimization Aid
LTpowerPlanner provides all the necessary functions for creating a power supply system architecture. It is very simple to operate, allowing for rapid concept development.
An input energy source is defined and then individual loads, or electrical consumers, are added. This is followed by adding individual dc-to-dc converter blocks. These could be switching regulators or low dropout (LDO) linear regulators. All components can be assigned their own name. An expected conversion efficiency is stored for efficiency calculations.
Using LTpowerPlanner has two great benefits. First, a simple architecture calculation can identify the configuration of the individual conversion stages most beneficial for overall efficiency. Figure 2 shows two different architectures for the same voltage rails. The architecture at the bottom has an overall efficiency that is somewhat higher than that of the architecture at the top. This property is not evident without a detailed calculation. When using LTpowerPlanner, this difference is immediately revealed.
The second benefit of LTpowerPlanner is that it provides well-organized documentation. The graphical user interface provides a neat sketch of the architecture, a visual aid that can be invaluable in discussions with coworkers and in documenting the development effort. Documentation can be stored either as a paper hard copy or a digital file.
Power Supply Design Step 2: Selecting Integrated Circuits for Each DC-to-DC Converter
When designing power supplies today, an integrated circuit is used rather than a discrete circuit with many separate components. There are a multitude of different switching regulator ICs and linear regulators available in the market. All of them are optimized for one specific property. Interestingly, all integrated circuits are different and can be interchanged only in the rarest of cases. Selecting the integrated circuit thus becomes a very important step. Once an integrated circuit has been selected, the properties of that circuit are fixed for the rest of the design process. Later, if it turns out that a different IC is better suited, the effort to incorporate a new IC begins again. This development effort can be very time consuming but can be easily mitigated with the use of design tools.
Using a tool is critical for effectively selecting the integrated circuit. The parametric search on analog.com is suitable for this. Searching for components within LTpowerCAD can be even more productive. Figure 3 shows the search window.
To use the search tool, only a few specifications need to be entered. For example, you may enter the input voltage, output voltage, and required load current. Based on these specifications, LTpowerCAD generates a list of recommended solutions. Additional criteria can be entered to further narrow down the search. For example, in the Features category you can select from features such as an enable pin or galvanic isolation to find an appropriate dc-to-dc converter.
Power Supply Design Step 3: Circuit Design of the Individual DC-to-DC Converters
Step 3 is the circuit design. The external, passive components need to be selected for the chosen switching regulator IC. The circuit is optimized in this step. Usually, this requires studying a data sheet thoroughly and performing all the required calculations. This step in power supply design can be drastically simplified by the comprehensive design tool, LTpowerCAD, and the results can be optimized further.
LTpowerCAD as a Powerful Calculation Tool
LTpowerCAD was developed by Analog Devices to greatly simplify circuit design. It is not a simulation tool, but rather a calculation tool. It recommends, in a very short time, the optimized external components based on the specification entered. The conversion efficiency can be optimized. The transfer function of the control loop is also calculated. This makes it easy to implement the best control bandwidth and stability.
After opening a switching regulator IC in LTpowerCAD, the main screen displays the typical circuit with all the necessary external components. Figure 4 shows this screen for the LTC3310S as an example. This is a step-down switching regulator with an output current of up to 10 A and a switching frequency of up to 5 MHz.
The yellow fields on the screen show the calculated or specified values. The user can configure settings using the blue fields.
Selecting the External Components
LTpowerCAD reliably simulates the behaviors of a real circuit as calculations are based on detailed models of external components, not just ideal values. LTpowerCAD includes a large database of integrated circuit models from several manufacturers. For instance, the equivalent series resistance (ESR) of a capacitor and the core losses of a coil are taken into consideration. To select external components, click on a blue external component as shown in Figure 4. A new window will open, showing a long list of possible components. As an example, Figure 5 shows a list of recommended output capacitors. This example shows a selection of 88 different capacitors from various manufacturers. You can also exit the list of recommended components and select the Show All option to choose from a variety of more than 4660 capacitors.
This list is continuously expanded and updated. While LTpowerCAD is an offline tool and does not require internet access, regular software updates (using the update function) will ensure that the integrated switching regulator ICs and the database of external components remain up to date.
Checking the Conversion Efficiency
Once the optimal external components have been selected, the conversion efficiency of the switching regulator is checked using the Loss Estimate & Break Down button.
A precise diagram of the efficiency and losses is then displayed. In addition, the junction temperature reached in the IC can be calculated based on the thermal resistance of the housing. Figure 6 shows the page of calculations for the conversion efficiency and thermal behavior.
Once you are satisfied with the circuit response, you can move to the next set of calculations. If the efficiency is not satisfactory, the switching frequency of the switching regulator can be changed (see left side of Figure 6), or the selection of the external coil can be changed. The efficiency is then recalculated until a satisfactory result is achieved.
Optimizing the Control Bandwidth and Checking the Stability
After selecting the external components and calculating the efficiency, the control loop is optimized. The loop must be set so that the circuit is reliably stable, not prone to oscillations or even instability while providing a high bandwidth—that is, the ability to respond to changes of the input voltage and, in particular, to load transients. The stability considerations in LTpowerCAD can be found in the Loop Comp. & Load Transient tab. In addition to a Bode plot and curves on the response of the output voltage following load transients, there are many setting options.
The Use Suggested Compensation button is the most important. In this case, the optimized compensation is used, and the user need not dive deeply into control engineering to adjust any parameters. Figure 7 shows the LTpowerCAD screen when setting the control loop.
The stability calculations performed in LTpowerCAD are a highlight of its architecture. The calculations are performed in the frequency domain and are very fast, much faster than simulations in the time domain. As a result, parameters can be changed on a trial basis and an updated Bode plot is provided in a few seconds. For a simulation in the time domain, this would take many minutes or even hours.
Checking the EMC Response and Adding Filters
Depending on the specification, additional filters may be necessary at the input or output of the switching regulator. This is where less experienced power supply developers, in particular, face large challenges. The following questions arise: How must the filter components be selected to ensure a certain voltage ripple at the output? Is an input filter necessary and, if so, how must this filter be designed to keep conducted emissions below certain EMC limits? In this respect, interaction between the filter and the switching regulator must not result in instability under any circumstances.
Figure 8 shows the Input EMI Filter Design, which is a sub-tool in LTpowerCAD. This can be accessed from the first page where the external, passive components are optimized. Starting the filter designer brings up a filter design using passive ICs and an EMC graph. This graph plots the conducted interference with or without an input filter and within appropriate limits from various EMC specifications such as CISPR 25, CISPR 22, or MIL-STD-461G.
The filter characteristic in the frequency domain and the filter impedance can also be displayed graphically beside the illustration of the input conducted EMC response. This is important to ensure that a filter does not have total harmonic distortion that is too high, and that the filter impedance matches the impedance of the switching regulator. Problems with the impedance match can lead to instabilities between the filter and the voltage converter.
Such detailed considerations can be accounted for in LTpowerCAD and do not require in-depth knowledge. With the Use Suggested Values button, the filter design is automated.
Of course, LTpowerCAD also supports the use of a filter on the output of the switching regulator. This filter is often used for applications where the output voltage is only allowed to have a very low output voltage ripple. To add a filter in the output voltage path, click the LC filter icon on the Loop Comp. & Load Transient page. Once this icon is clicked, a filter appears in a new window, as shown in Figure 9. The parameters of the filter can be easily selected here. The feedback loop can either be connected in front of this additional filter or behind it. Here, a stable response of the circuit can be ensured in all operating modes despite very good dc precision of the output voltage.
Power Supply Design Step 4: Simulation of the Circuit in the Time Domain
Once you have completely designed a circuit using LTpowerCAD, simulating it is the crowning achievement. Simulations are usually run in the time domain. Individual signals are checked against time. The interaction of different circuits can also be tested on a printed circuit board. It is also possible to integrate parasitic effects into the simulation. With this, the result of the simulation becomes very accurate, but the simulation times are longer.
Generally, a simulation is suitable for collecting additional information prior to implementing real hardware. It is important to know the potential and the limits of circuit simulation. Finding the optimal circuit might not be possible using simulation only. During simulation, one can modify parameters and restart the simulation. However, if the user is not an expert in designing circuits, it can be difficult to determine the right parameters and then to optimize them. As a result, it is not always clear to the user of a simulation whether the circuit has already achieved the optimal state. A computing tool such as LTpowerCAD is better suited for this purpose.
Simulating the Power Supply Using LTspice
LTspice®, from Analog Devices, is a powerful simulation program for electric circuits. It is very widely used by hardware developers globally, due to its ease of use, extended network of user support, optimization options, and high quality, reliable simulation results. Additionally, LTspice is free of charge and can easily be installed on a personal computer.
LTspice is based on the SPICE program, which originated from the Department of Electrical Engineering and Computer Sciences at the University of California, Berkeley. The acronym SPICE stands for simulation program with integrated circuit emphasis. Many commercial versions of this program are available. Although originally based on Berkeley’s SPICE, LTspice offers considerable improvements in the convergence of circuits and simulation speed. Additional features of LTspice include a circuit diagram editor and waveform viewer. Both are intuitive to operate, even for a beginner. These features also provide a great deal of flexibility for the experienced user.
LTspice is designed to be simple and easy to use. The program, available for download at analog.com, includes a very large database containing simulation models of nearly all power ICs from Analog Devices along with external passive components. As mentioned, once installed, LTspice can operate offline. However, regular updates will ensure that the newest models of switching regulators and external components are loaded.
To start an initial simulation, choose an LTspice circuit in the product folder of a power product on analog.com (for example, the LT8650S evaluation board). These are usually the suitable circuits of the available evaluation boards. By double-clicking the related LTspice link in a specific product folder on analog. com, LTspice will launch the complete circuit locally on your PC. It includes all external components and presets necessary to run a simulation. Then, click on the runner icon, pictured in Figure 10, to start simulation.
Following a simulation, all the voltages and currents of a circuit can be accessed using the waveform viewer. Figure 11 shows a typical illustration of the output voltage and the input voltage as the circuit ramps up.
A SPICE simulation is primarily suited for getting to know a power supply circuit in detail so that there are no unwanted surprises when building the hardware. A circuit can also be changed and optimized using LTspice. In addition, the interaction of the switching regulator with the other parts of the circuit on the printed circuit board can be simulated. This is particularly helpful in uncovering interdependencies. For example, several switching regulators can be simulated at the same time in one run. This extends the simulation time, but certain interactions can be checked in this case.
Finally, LTspice is an extremely powerful and reliable tool used by IC developers today. Many ICs from Analog Devices have been developed with the help of this tool
Power Supply Design Step 5: Testing the Hardware
While automation tools have a valuable purpose in power supply design, the next step is to perform a basic hardware evaluation. The switching regulator operates with currents switched at a very high rate. Due to the parasitic effects of the circuit—particularly of the printed circuit board layout—these switched currents cause voltage offset, which generates radiation. Such effects can be simulated using LTspice. To do this, however, you need precise information about the parasitic properties. Most of the time this information is not available. You would have to make many assumptions, and these reduce the value of the simulation result. Consequently, a thorough hardware evaluation must be completed.
Printed Circuit Board Layout—An Important Component
The printed circuit board layout is usually known as a component. It is so critical that, for example, it is not possible to operate a switching regulator for test purposes using jumper wires, as it is with a breadboard. Mainly, the parasitic inductance in the paths where the currents are switched leads to a voltage offset that makes operation impossible. Some circuits could also be destroyed due to excessive voltage.
There is support available for creating an optimal printed circuit board layout. The corresponding data sheets for the switching regulator ICs usually provide information about a reference printed circuit board layout. For most applications, this suggested layout can be used.
Evaluating the Hardware Within the Specified Temperature Range
During the power supply design process, conversion efficiency is considered to determine whether the switching regulator IC operates within the permissible temperature range. However, testing the hardware at its intended temperature limits is important. The switching regulator IC and even the external components vary their rated values over the permissible temperature range. These temperature effects can easily be taken into consideration during the simulation using LTspice. However, such a simulation is only as good as the given parameters. If these parameters are available with realistic values, LTspice can perform a Monte Carlo analysis that leads to the desired result. In many cases, evaluating the hardware through physical testing is still more practical.
EMI and EMC Considerations
In late stages of system design, hardware must pass electromagnetic interference and compatibility (EMI and EMC) tests. While these tests must be passed with real hardware, simulation and calculation tools can be extremely useful in gathering insights. Different scenarios can be evaluated prior to hardware testing. Certainly, there are some parasitics involved that are usually not modeled in simulation, but general performance trends related to these test parameters can be obtained. Additionally, the data obtained from such simulations can provide the insights necessary to apply modifications to the hardware quickly, in case an initial EMC test was not passed. Since EMC tests are costly and time intensive, utilizing software such as LTspice or LTpowerCAD in early design stages can help achieve more accurate results prior to testing, thus speeding up the overall power supply design process and reducing costs.
Summary
The tools available for power supply design have become very sophisticated and powerful enough to meet the demands of complex systems. LTpowerCAD and LTspice are high performance tools with simple to use interfaces. As a result, these tools can be invaluable to a designer with any level of expertise. Anyone from the experienced developer to the less experienced can use these programs to develop power supplies on a day-to-day basis.
It is astounding how much simulation capabilities have evolved. Using the proper tools can help you build a reliable, sophisticated power supply more quickly than ever before. | 4,401 | 23,322 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2022-27 | latest | en | 0.918217 |
https://smlnj-gforge.cs.uchicago.edu/scm/viewvc.php/*checkout*/sml/trunk/src/compiler/Semant/elaborate/precedence.sml?revision=16&root=smlnj&pathrev=16 | 1,632,435,736,000,000,000 | text/plain | crawl-data/CC-MAIN-2021-39/segments/1631780057447.52/warc/CC-MAIN-20210923195546-20210923225546-00460.warc.gz | 550,787,471 | 2,145 | (* Copyright 1996 by AT&T Bell Laboratories *) (* precedence.sml *) signature PRECEDENCE = sig val parse: {apply: 'a * 'a -> 'a, pair: 'a * 'a -> 'a} -> 'a Ast.fixitem list * StaticEnv.staticEnv * (Ast.region->ErrorMsg.complainer) -> 'a end (* signature PRECEDENCE *) structure Precedence : PRECEDENCE = struct local structure EM = ErrorMsg structure F = Fixity in datatype 'a precStack = INf of Symbol.symbol * int * 'a * 'a precStack | NONf of 'a * 'a precStack | NILf fun parse {apply,pair} = let fun ensureNONf((e,F.NONfix,_,err),p) = NONf(e,p) | ensureNONf((e,F.INfix _,SOME sym,err),p) = (err EM.COMPLAIN ("expression or pattern begins with infix identifier \"" ^ Symbol.name sym ^ "\"") EM.nullErrorBody; NONf(e,p)) fun start token = ensureNONf(token,NILf) (* parse an expression *) fun parse(NONf(e,r), (e',F.NONfix,_,err)) = NONf(apply(e,e'),r) | parse(p as INf _, token) = ensureNONf(token,p) | parse(p as NONf(e1,INf(_,bp,e2,NONf(e3,r))), (e4, f as F.INfix(lbp,rbp),SOME sym,err))= if lbp > bp then INf(sym,rbp,e4,p) else (if lbp = bp then err EM.WARN "mixed left- and right-associative \ \operators of same precedence" EM.nullErrorBody else (); parse(NONf(apply(e2,pair (e3,e1)),r),(e4,f,SOME sym,err))) | parse(p as NONf _, (e',F.INfix(lbp,rbp),SOME sym,_)) = INf(sym,rbp,e',p) | parse _ = EM.impossible "Precedence.parse" (* clean up the stack *) fun finish (NONf(e1,INf(_,_,e2,NONf(e3,r))),err) = finish(NONf(apply(e2,pair (e3,e1)),r),err) | finish (NONf(e1,NILf),_) = e1 | finish (INf(sym,_,e1,NONf(e2,p)),err) = (err EM.COMPLAIN ("expression or pattern ends with infix identifier \"" ^ Symbol.name sym ^ "\"") EM.nullErrorBody; finish(NONf(apply(e2,e1),p),err)) | finish (NILf,err) = EM.impossible "Corelang.finish NILf" | finish _ = EM.impossible "Corelang.finish" in fn (items as item1 :: items',env,error) => let fun getfix{item,region,fixity} = (item, case fixity of NONE => F.NONfix | SOME sym => Lookup.lookFix(env,sym), fixity, error region) fun endloc[{region=(_,x),item,fixity}] = error(x,x) | endloc(_::a) = endloc a fun loop(state, a::rest) = loop(parse(state,getfix a),rest) | loop(state,nil) = finish(state, endloc items) in loop(start(getfix item1), items') end end end (* local *) end (* structure Precedence *) (* * \$Log: precedence.sml,v \$ * Revision 1.1.1.1 1997/01/14 01:38:35 george * Version 109.24 * *)
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The output has ended thus: Log: precedence.sml,v \$ * Revision 1.1.1.1 1997/01/14 01:38:35 george * Version 109.24 * *) | 855 | 2,539 | {"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-2021-39 | latest | en | 0.361291 |
https://math.stackexchange.com/questions/4379873/simplify-sin-arctan4x-for-frac-14-le-x-le-frac-14/4379875 | 1,718,972,375,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862070.5/warc/CC-MAIN-20240621093735-20240621123735-00104.warc.gz | 341,683,719 | 39,800 | Simplify $\sin[\arctan(4x)]$ for $- \frac 14 \le x \le \frac 14$
I am coming up with the wrong answer.
The problem is $$\sin[\arctan(4x)]$$
Here are the reciprocal functions I am using $$\sin\theta= \frac {1}{\csc\theta}$$ and $$\cot\theta = \frac{1}{\tan\theta}$$
Here is the Pythagorean identity I am using $$\csc^2 \theta = 1+\cot^2 \theta$$
I am trying to rewrite the problem $$\sin[\arctan(4x)]$$ in terms of the Pythagorean Identity using the reciprocal identities. $$\frac{1}{\csc^2 \theta} = \frac {1}{1 + \frac{1}{1+\tan^2 \theta}}$$ This seems to be a stretch, I do not think I am expressing the Pythagorean Identity right, and I am not sure how to either. I did start to try to plug in the numbers.
$$\frac{1}{\csc^2 \theta} = \frac {1}{1 + \frac {1}{(4x^2)}}$$
$$\frac{1}{\csc \theta} = \sqrt{1+4x}$$
$$\sin\theta = \frac{1}{\sqrt{1+4x}}$$
I could rationalize the denominator and multiply the by its conjugate to get; $$\frac{\sqrt{1-4x}}{\sqrt{(1+4x)(1-4x)}}$$ But, at this point I know I am wrong. The answer is $$\frac{4x}{\sqrt{16x^2+1}}$$
• Not sure why you're trying to introduce these reciprocal functions. If you drew out a triangle with opposite 4x and adjacent 1, it finishes this really quickly. Commented Feb 11, 2022 at 21:29
• Not sure but they are trig identities. I was wondering how to use them in the problem. Also I didn't know the difference between tangent and hypotenuse. Commented Feb 11, 2022 at 21:48
I am going to bypass examining your work, because the problem permits a very easy shortcut.
If you have a right triangle, whose tangent is $$(4x)$$, then the legs are (in effect) $$(4x)$$ and $$(1)$$.
Therefore, the hypotenuse is $$~\displaystyle \sqrt{16x^2 + 1}.$$
Therefore, the sine of this angle must be
$$\frac{4x}{\sqrt{16x^2 + 1}}.$$
Responding to the comment of Dan:
Need to show that it's still valid when $$x < 0$$.
I agree, my oversight.
When $$x < 0,$$ you can let $$y = -x \implies y > 0$$.
Then, if $$\tan(\theta) = 4x$$, you have that $$\tan(-\theta) = 4y.$$
Then, by the analysis at the start of this answer, $$\sin(-\theta) = \displaystyle ~\frac{4y}{\sqrt{16y^2 + 1}} = \frac{-4x}{\sqrt{16x^2 + 1}}.$$
Therefore, since $$\sin(\theta) = -\sin(-\theta)$$, you have that
$$\sin(\theta) = \displaystyle \frac{4x}{\sqrt{16x^2 + 1}}.$$
Edit
In my opinion, an open issue is whether the Addendum is actually needed. In Analytical Geometry, where the domain of trig functions are angles, you have the issue of whether a right triangle can be constructed, some of whose side lengths are negative.
The analysis at the start of my answer was based on physically constructing the analogous right triangle. I don't know how the issue is being taught in Analytical Geometry, so it is better to add the Addendum, erring on the side of caution.
In Real Analysis (AKA Calculus) the issue is somewhat convoluted, because the domain of the trig functions are real numbers, rather than angles. However, you can interpret the domain of the trig functions to be various arc lengths, with respect to the unit circle.
Under this interpretation, it seems to me that the Addendum is not needed, because you can construct a right triangle [whose hypotenuse is $$(1)$$] that lies in the $$4$$th quadrant just as appropriately as constructing a right triangle that lies in the $$1$$st quadrant. Then, you have constructed a right triangle, one of whose legs is a negative number.
• Thanks, that is much easier. I wasn't sure how to express the hypotenuse. Commented Feb 11, 2022 at 21:39
• Nitpick: Needs to show that it's still valid when $x < 0$.
– Dan
Commented Feb 11, 2022 at 21:40
• $hypotenuse^2= 4x^2+1$ I was thinking the tangent was but couldn't be the hypotenuse. Commented Feb 11, 2022 at 21:46
• @Dan Nice catch, thanks. Addendum added to the end of my answer. Commented Feb 11, 2022 at 21:49
• @Benp404 See also the addendum added to the end of my answer. The idea is that when $(4x) < 0$, then you can't really construct the pertinent right triangle, because you can't construct a triangle with side lengths less than zero. Commented Feb 11, 2022 at 21:53
Let's try your idea, but in a slightly different way: it's way simpler to compute $$\sin(\arctan(x))$$ and then substituting $$4x$$ for $$x$$.
The limitation now is $$-1\le x\le 1$$, so $$-\pi/4\le\arctan(x)\le\pi/4$$. Set $$y=\arctan(x)$$. We have $$\sin^2y+\cos^2y=1$$ and therefore $$\sin^2y=\frac{\sin^2y}{\sin^2y+\cos^2y}=\frac{\tan^2y}{\tan^2y+1}$$ But $$\tan y=x$$ by definition, so $$\sin^2(\arctan(x))=\frac{x^2}{x^2+1}$$ Thus we have $$\sin(\arctan(x))=\frac{x}{\sqrt{x^2+1}}$$ by comparing the cases when $$x>0$$ and $$x<0$$. Now you can see that the limitation $$-1\le x\le 1$$ is redundant and the statement holds for every $$x$$.
With $$4x$$ in place of $$x$$: $$\sin(\arctan(4x))=\frac{4x}{\sqrt{16x^2+1}}$$
• I was wondering how you go from $sin^2y$. To $sin^2y = \frac{sin^2y}{sin^2y+cos^2y}$ Commented Feb 12, 2022 at 3:43
• @Benp404 $a=a/1$; it’s the usual trick to express the (squared) sine or cosine in terms of the tangent. Commented Feb 12, 2022 at 8:55
As egreg said, the condition is redundant. On the other words, for any real $$x$$,$$\sin(\arctan(4x))=\frac{4x}{\sqrt{16x^2+1}}.$$ I like to prove it by 2 cases. Let $$y=\arctan 4x$$.
A. When $$x\geq 0,$$ $$\sin ^{2} y=\tan ^{2} y \cos ^{2} y=16 x^{2}\left(1-\sin ^{2} y\right) \Rightarrow \sin ^{2} y=\frac{16 x^{2}}{16 x^{2}+1} \Rightarrow \sin y=\frac{4 x}{\sqrt{16 x^{2}+1}}$$
Hence $$\sin (\arctan (4 x))=\frac{4 x}{\sqrt{16 x^{2}+1}}.$$
B. When $$x< 0,$$ let $$z=-x>0,$$ then using the result in case A yields $$\sin (\arctan 4 z)=\frac{4 z}{\sqrt{16 z^{2}+1}}.$$
Now putting $$z=-x$$ yields $$\sin (\arctan 4(-x))=\frac{-4 x}{\sqrt{16 (-x)^{2}+1}}.$$ Using the facts that $$\sin x$$ and $$\arctan x$$are odd functions, we have$$-\sin (\arctan (4x))=-\frac{4 x}{\sqrt{16 x^{2}+1}}.$$
$$\sin (\arctan (4x))=\frac{4 x}{\sqrt{16 x^{2}+1}}.$$
Therefore we can conclude that for any real $$x$$, $$\boxed{\sin (\arctan (4x))=\frac{4 x}{\sqrt{16 x^{2}+1}}}.$$ In general, for any real $$x$$, $$\boxed{\sin (\arctan (x))=\frac{x}{\sqrt{x^{2}+1}}}$$ | 2,124 | 6,161 | {"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": 64, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2024-26 | latest | en | 0.862208 |
https://testbook.com/question-answer/in-the-given-figure-o-is-the-centre-of-the-circle--62ce4df5a04f823e16b75adb | 1,680,096,109,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948976.45/warc/CC-MAIN-20230329120545-20230329150545-00079.warc.gz | 611,866,852 | 81,236 | # In the given figure O is the centre of the circle. If ∠BAC = 38°, then ∠OCD is:
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1. 52°
2. 38°
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4. 76°
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## Detailed Solution
Given:
In a circle O is the centre of the circle.
∠BAC = 38°
Concept used:
Sum of all the angles of a triangle is 180°
Angle subtended by a chord at the centre will be double that that of angle subtended by it at circumference.
Calculations:
Angle subtended by chord CB at the centre will be double that that of angle subtended by it at circumference.
So, ∠COB = 2∠BAC = 2 × 38° = 76° and, ∠BAC = ∠BDC = 38°
∠BOD = 180°
Hence, ∠COB + ∠COD = 180°
Hence, ∠COD = 180° - 76° = 104°
In triangle COD, ∠COD + ∠ODC + ∠ODC = 180°
Thus, 104° + 38° + ∠OCD = 180°
Thus, ∠OCD = 180° - 142° = 38° | 348 | 936 | {"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-2023-14 | longest | en | 0.885121 |
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# 2013/2014 Summer
Data Minin g and Knowledge Acquizition — Chapter 7 — — Data Mining Overwiev and Exam Questions —. 2013/2014 Summer. Data Mining. Methodology Problem definition Data set selection Preprocessing transformations Functionalities Classification/prediction Clustering Association
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## 2013/2014 Summer
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### Presentation Transcript
1. Data Mining and Knowledge Acquizition — Chapter 7 ——Data Mining Overwievand Exam Questions— 2013/2014 Summer
2. Data Mining • Methodology • Problem definition • Data set selection • Preprocessing transformations • Functionalities • Classification/prediction • Clustering • Association • Sequential analysis • others
3. Methodology cont. • Algorithms • For classification you can use • Decision trees ID3,C4.5 CHAID are algorithms • For clustering you can use • Partitioning methods k-means,k-medoids • Hierarchical AGNES • Probabilistic EM is an algorithm • Presenting results • Back transformations • Reports • Taking action
4. Two basic style of data mining • Descriptive • Cross tabulations,OLAP,attribute oriented induction,clustering,association • Predictive • Classification,prediction • Questions answered by these styles • Difference between classification and prediction
5. Classification • Methods • Decision trees • Neureal networks • Bayesian • K-NN or model based reasoning • Adventages disadventages • Given a problem which data processing techniques are required
6. Classification (cnt.d) • Accuracy of the model • Measures for classification/numerical prediction • How to better estimate • Holdout,cross validation, bootstraping • How to improve • Bagging, boosting • For unbalanced classes • What to do with models • Lift charts
7. Clustering • Distance measures • Dissimilarity or similarity • For different type of variables • Ordinal,binary,nominal,ratio,interval • Why need to transform data • Partitioning methods • K-means,k-medoids • Adventage disadventage • Hierarchical • Density based • probablistic
8. Association • Apriori or FP-Growth • How to measure strongness of rules • Support and confidence • Other measures critique of support confidence • Multiple levels • Constraints • Sequential patterns
9. OLAP • Concept of cube • Fact table • measures • Dimensions • Sheams • Star, snowflake • Concept hierarchies • Set grouping such as price age • Parent child
10. Pre processing • Missing values • Inconsistencies • Redundent data • Outliers • Data reduction • Attribute elimination • Attribute combination • Samplinng • Histograms
11. Exam Questions • Intorduction • Basic functionalities • Data description • Data preperation • Data warehousing olap • Clustering • classification/numerical prediction • frequent pattern mining
12. Introduction • Defining data mining problems • Data mining functionalities
13. Define data mining problems • 1. Suppose that a data warehouse for Big-University Library consists of the following three dimensions: users, books, time, and each dimension has four levels not including the all level. There are three measures: You are asked to perform a data mining study on that warehouse (25 pnt) • Define three data mining problems on that warehouse: involving association, classification and clustering functionalities respectively. Clearly state the importance of each problem. What is the advantage of the data being organized as OLAP cubes compared to relational table organisation?
14. Define data mining problems • In data preprocessing stage of the KDD • What are the reasons for missing values? and How do you handle them? • what are possible data inconsistencies • do you make any discritization • do you make any data transformations • do you apply any data reduction strategies
15. Define data mining problems • Define your target and input variables in classification. Which classification techniques and algorithms do you use in solving the classification problem? Support your answer • Define your variables indicating their categories in clustering Which clustering techniques and algorithms do you use in solving the clustering problem? Support your answer. • Describe association task in detail; specifying the algorithm interestingness measures or constraints if any.
16. Data mining on MIS • A data warehouse for the MIS department consists of the following four dimensions: student, course, instructor, semester and each dimension has five levels including the all level. There are two measures: count and average grade. At the lowest level of average grade is the actual grade of a student. You are asked to perform a data mining study on that warehouse (25 pnt)
17. Data mining on MIS 2 • Define three data mining problems on that warehouse: involving association, classification and clustering functionalities respectively. Clearly state the importance of each problem. What is the advantage of the data being organized as OLAP cubes compared to relational table organisation? • In data preprocessing stage of the KDD • What are the reasons for missing values? and How do you handle them? • what are possible data inconsistencies • do you make any discritization • do you make any data transformations • do you apply any data reduction strategies
18. Data mining on MIS 3 • Define your target and input variables in classification. Which classification techniques and algorithms do you use in solving the classification problem? Support your answer • Define your variables indicating their categories in clustering Which clustering techniques and algorithms do you use in solving the clustering problem? Support your answer. • Describe association task in detail; specifying the algorithm interestingness measures or constraints if any.
19. Final 2010/2011 Spring (MIS) • 3 ( 35 pt.) The aim of Knowledge Discovery from Databases (KDD) is to extract interesting, potentially useful, …, knowledge from data. The extracted knowledge can be represented in a knowledge base similar to a database. Considering the data mining functionalities and algorithms we covered in this course describe five different knowledge types. For each type discuss the following aspects: • a) From which functionality and algorithm they are obtained? • b) How they are represented in knowledge base? (Do not consider data structures ) • c) What are the quality characteristics? • d) How they are used in the deployment phase?
20. BIS 541 2011/2012 Final • 1. For each of the following problem identify relevant data mining tasks • a) A weather analyst is interested in calculating the likely change in temperatue for the coming days. • b) A marketing analyst is looking for the groups of customers so as to apply different CRM strategies for ecach group • c) A medical doctor must decide whether a set of symptoms is an indication of a particular disease. • d) A educational psychologist would like to determine exceptional students to sugget them for special educational programs. .
21. BIS 541 2012/2013 Final • For each of the following problem identify relevant data mining tasks with a brief explanation • a) A weather analyst is interested in wheather the temperature will be up or down for the coming day • b) An insurance analyst intends to group policy holders according to characteristics of customers and policies • c) A medical researcher is looking for symptoms that are occurring together among a large set of pationes. • d) An educational program director would like to determine likely GPA of applicant to a MA program from their ALES scores, undergraduate GPAs and enterence exam scores.
22. Basic Fuctionalities • Decision tree - ID3 • information gain • Association – Apriori • Clustering – k-means
23. Information gain • Consider a data set of two attributes A and B. A is continuous, whereas B is categorical, having two values as “y” and “n”, which can be considered as class of each observation. When attribute A is discretized into two equiwidth intervals no information is provided by the class attribute B but when discretized into three equiwidth intervals there is perfect information provided by B. Construct a simple dataset obeying these characteristics.
24. Node 2 A=a1 Decision Y Node 3 A=a2 Node 4 B=b1 Decision N Node 5 B=b2 Decision is Y Decision tree • 2. a-Construct a data set that generates the tree shown below In addition the following conditions are satisfied
25. Midterm 2006/2007 Spring (MIS) • 2. Show that entroy is not a symetric measure of association like correlation coefficient is. Construct a simple data set of two categorical attributes A and B such that i knowing the values of A provides perfect information to predict B but ii) knowing the values of B does not provide perfect information to precict A
26. at a particular node • when information gain is 0 • when it gets maximum value
27. Associations • In a particular database; AC and BC are strong association rules based on the support confidence measure. A and B are independent items. Does this imply that A BC is also a strong rule based on the lift measure? A,B,C are items in a transaction database. • -if A B and BC are strong. Is AC a strong rule • -if A B and AC are strong. İs BC a strong rule
28. Data Description/Preprocessing
29. Midterm 2004/2005 Spring (MIS) • Consider the correlation coefficient between two numerical variables. Does its umerical value affected by the unit of measures of these variables?. (such as measureing temperature in oC or öF)
30. Midterm 2011/2012 Fall generate data • 5. (10 points) Consider two continuous variables X and Y. Generate data sets • a) where PCA (principle component analysis) can not reduces the dimensionality from two to one • b) where although the two variables are related (a functional relationship exists between these two variables), PCA is not able to reduce the dimensionality from two to one
31. Midterm 2010/2011 Spring (MIS) • 3. (25 points) Consider a data set of two continuous variables X and Y. X is right skewed and Y is left skewed. Both represent measures about same quantity (sales categories, exam grades,…) • a) Draw typical distributions of X and Y separately. • b) Draw box plots of X and Y separately. • c) Draw q-plots (quantile) of X and Y separately. • d) Draw q-q plot of X and Y.
32. MIS 541 2012/2013 Final • 1. (20 pts) Consider a data set of two continuous variables X and Y. X both has the same mean, both have no skewness (symetric)ç X has a higher variance then Y. Both represent measures about same quantity (sales categories, exam grades,…) • a) Draw typical distributions of X and Y on the same graph. • b) Draw box plots of X and Y separately.
33. Final 2011/2012 Fall data description • 1 (20 points) Give two examples of outliers. • a) Where outliers are useful and essential patterns to be mined. • b) Outliers are useless steaming from error or noise.
34. Final 2011/2012 Fall preprocessing • 2 (20 points) Considering the classification methods we cover in class, describe two distinct reasons why continuous input variables have to be normalized for classification problems(each reason 10 points).
35. Midterm 2008/2009 Spring • 4. (20 points) Principle components is used for dimensionality reduction then may be followed by cluster analysis – say for segmentation purposes – Consider a two continuous variable problem. Using scatter plots • a) Generate a data set where PCA reduces the dimensionality from two to one • b) Generate a data set where although there is a relation between the two variables, PCA • is not able to reduce the dimensionality to one • c) Generate a data set where there are natural clusters and PCA can reduce the dimensionality • d) Generate a data set where there are natural clusters but PCA is not the appropriate method for reducing the dimensionality
36. Midterm 2012/2013 Fall (MIS) • 1. (20 pts) Consider a data set of two continuous variables X and Y. X both has the same mean, both have no skewness (symetric)ç X has a higher variance then Y. Both represent measures about same quantity (sales categories, exam grades,…) • a) Draw typical distributions of X and Y on the same graph. • b) Draw box plots of X and Y separately. • c) Draw q-plots (quantile) of X and Y separately. • d) Draw q-q plot of X and Y.
37. Data Warehousing/OLAP • Design of olap cubes • Measures
38. Midterm 2005/2006 Spring (MIS) • A large hypermarket has lots of branchs through out the country. Quantity purchased Qi, price Pi, for each item i are stored in a warehouse. The top management is interested in finding the cheapest large sold items minp(maxq item i). Is it possible to accomplish this in a distributive maner? In other word is minp(maxq item i) a distributive measure?
39. Final 2007/2008 Spring (MIS) • 1. (25 pnt) Suppose an aggregation is to be designed to obtain weekly dollar values from daily values by two different ways described below. Can they be computed in a distributive manner? (the database has day ID and dollar value fields. Records are randomly selected and assigned to different processing units) • a) Taking the daily averages • b) Taking the last day’s value of the week
40. Data warehouse for library • A data warehouse is constructed for the library of a university to be used as a multi-purpose DSS. Suppose this warehouse consists of the following dimensions: user , books , time (time_ID, year, quarter, month, week, academic year, semester, day), and . “Week” is considered not to be less than “month”. Each academic semester starts and ends at the beginning and end of a week respectively. Hence, week<semester. • Describe concept hierarchies for the three dimensions. Construct meaningfull attributes for each dimension tables above . Describe at least two meaningfull measures in the fact table. Each dimension can be looked at its ALL level as well. • What is the total number of cuboids for the library cube? • Describe three meaningfull OLAP queries and write sql expresions for one of them.
41. OLAP Big University • 2. (Han page 100,2.4) Suppose that the data warehouse for the Big-University consists of the following dimensions: student,course,instructor,semester and two measures count and average_grade. Where at the lowset conceptual level (for a given student, instructor,course, and semester) the average grade measure stores teh actual grade of the student. At higher conceptual levels the average_grade stores the average grade for the given combination. (when student is MIS semester 2005 all terms, course MIS 541, instructor Ahmet Ak, average_grade is the average of students grades in thet course by that instructer in all semester in 2005)
42. cont. • a) draw a snawflake sheam diagram for that warehouse • What are the concept hierarchys for the dimensions • b) What is the total nmber of cuboids
43. MIS 542 Final S06 1 olap • 1. MIS department wants to revise academic strategies for the following ten years. Relevent • questions are: What portion of the courese are required or elective? What is the full time part • time distribution of instuctors? What is the course load of instructors? What percent of • technical or managerial courses are thought by part time instructors? How all theses things
44. MIS 542 Final S06 1 cont. • changed over years? You can add similar stategic quustions of your own. Do not conside • students aspects of the problem for the time being. Desing and OLAP sheam to be used as a • strategic tool. You are free to decide the dimensions and the fact table. Describe the concept • hierarchies, virtual dimensions and calculated members. Finally show OLAP opperations to • answer three of such strategic questions
45. Midterm 2006/2007 Spring • 1. A data warehouse is constructed for the web site of a e-commerce company to be used for customer segmentation. Each visitor click stream data is recorded Each session has an ID Suppose this warehouse consists of the following dimensions: visitor, time, product. There is a concept hierarcy for products which is reflected to the design of the web site so that products can be seen in a hierarchical manner. When a product is seen it can be purchased. Only registered customers can use the system so each visitor has an ID. When registering a form is field out so that socio-demographic information is taken form a customer. Suppose income (a a numerical variable), birthday, gender, profesion, marital status is asked.
46. cont. • a) Describe concept hierarchies for the three dimensions. Construct meaningful attributes for each dimension tables above.(What transformations are required before constructing these attributes) Describe at least two meaningful measures in the fact table. • b) Each dimension can be looked at its ALL level as well. • Describe three meaningful OLAP queries and write sql expressions for one of them. • c) Define a clustering problem: Which variables are important? Is there a missing value problem? What data transformation are needed? Which algorithm would you suggest?
47. Midterm 2007/2008 Spring • 1. (20 points) Consider a shipment company responsible for shipping items from one location to another on predetermined due dates. Design a star schema OLAP cube for this problem to be used by managers for decision making purposes. The dimensions are time, item to be shipped, person responsible for shipping the item, location.. For each of these dimensions determine three levels in the concept hierarchy. Design the fact table with appropriate measures:and keys (include two measure and at least one calculated member in the fact table) • Show one drilldown and role up operations • Show the SQL query of one of the cuboids.
48. Midterm 2008/2009 Spring • 1. (25 points) In an organization a data warehouse is to be designed for evaluating performance of employees. To evaluate performance of an employee, survey questionnaire is consisting a set of questions with 5 Likered scale are answered by other employees in the same company at specified times. That is, performance of employees are rated by other employees. • Each employee has a set of characteristics including department, education,… Each survey is conducted at a particular date applied to some of the employees. Questions are aimed to evaluate broad categories of performance such as motivation, cooperation ability,… • Typically, a question in a survey, aiming to measure a specific attitude about an employee is evaluated by another employee (rated f rom 1 to 5) Data is available at question level.
49. cont. • Cube design: a star schema • Fact table: Design the fact table should contain one calculated member. What are the measures and keys? • Dimension tables: Employee, and Time are the two essential dimensions include a Survey and Question dimensions as well. For each dimension show a concept hierarchy. • State three questions that can be answered by that OLAP cube. • Show drilldown and role up operations related to these questions
50. MIS 541 2012/2013 Final • 2. (20 pts) Suppose that a data warehouse for a hospital consists of the following dimensions: time, doctor and patient and the two measures count and charge, where charge is the fee a doctor charge a patient for a visit. • Design a warehouse with star schema: • a) Fact table: Design the fact table. • b) Dimension tables: For each dimension show a reasonable concept hierarchy. • c) State two questions that can be answered by that OLAP cube. • d) Show drilldown and roll up operations related to one of these questions
More Related | 4,206 | 19,494 | {"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-2024-22 | latest | en | 0.613668 |
https://www.hindawi.com/journals/mpe/2012/802420/ | 1,542,181,411,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2018-47/segments/1542039741660.40/warc/CC-MAIN-20181114062005-20181114084005-00080.warc.gz | 891,121,755 | 29,109 | `Mathematical Problems in EngineeringVolume 2012, Article ID 802420, 15 pageshttp://dx.doi.org/10.1155/2012/802420`
Research Article
## Multiscale Numerical Study of 3D Polymer Crystallization during Cooling Stage
Department of Computational Mathematics, Henan University of Science and Technology, Luoyang 471003, China
Received 18 May 2012; Accepted 23 July 2012
Copyright © 2012 Chunlei Ruan. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
#### Abstract
We aim to study the behavior of polymer crystallization during cooling stage in injection molding more accurately, the multiscale model and multiscale algorithm proposed in our previous work (Ruan et al., 2012) have been extended to the 3D polymer crystallization case. Our multiscale model incorporates two distinct length scales: a coarse grid for the heat diffusion and a fine grid for the crystal morphology evolution (nucleation, growth, and impingement). Our multiscale algorithm couples the different methods on different length scales, namely, the finite volume method (FVM) on the coarse grid and the pixel coloring method on the fine grid. By using these multiscale model and multiscale algorithm, simulations for 3D polymer crystallization are carried out. Macroscopic variables, for example, temperature, relative crystallinity, as well as the microscopic structural characters, for example, crystal morphology development, and mean size of spherulites, are investigated at various cooling conditions. We also show the importance of coupling heat transfer with crystallization as well as 3D numerical studies.
#### 1. Introduction
Nowadays, semicrystalline polymers play an important role in industry due to their advantages of enhanced mechanical properties, ease of manufacturing, and so forth [1]. These polymers are typically processed by injection molding processing techniques to form the products. As one of the most widely used polymer processing techniques, injection molding mainly consists of filling, packing, and cooling [2]. Among these, cooling stage is the most significant part not only because it accounts for the largest part of the total injection molding cycle, but also for the crystal morphology formed during crystallization which is known as a crucial factor to the physical and mechanical properties of products. The crystal morphology also known as the microstructure is in turn determined by a number of processing conditions during the cooling stage. Thus, it is of great importance to accurately model the crystallization during cooling stage and predict the final crystal morphology formed under different processing conditions [1].
Polymer crystallization is a typical multiscale processing: the molecular chains fold together and form ordered regions called lamellae, which compose larger spheroidal structures named spherulites [1]. Since the polymer crystallization has a span of about 1010 (from the molecular chain length m to the product length m [1]) in length scale, it is hard to carry out a full-scale simulation to bridge single molecular chain folding to the final product. Therefore, to model the polymer crystallization, one should first choose a suitable length scale. Spherulite level ( m [1]) is proper because it is the single largest feature of polymer microstructures, in addition, it is relatively easier to link the heat transfer in the macroscopic level ( m [1]).
To date, a number of numerical investigations have appeared on spherulitic crystallization in polymer melts during cooling. Charbon and Swaminarayan [3] proposed a multiscale algorithm for the simulation of polymer crystallization. They incorporated front-tracking techniques with nucleation, spherulite impingement, and latent heat release as well as heat diffusion to predict the evolution of microstructures and relative crystallinity during cooling stage. Huang and Kamal [4] presented a physical model for polymer crystallization. They coupled a nonconserved envelop vector field of the local macroscopic growth domains with a vector field of the local mesoscopic lamellae directors. Prabhu et al. [5] proposed a coupling finite-element method and the cell model to the numerical study of polymer crystallization. The importance of micro-macro-coupling was analyzed and explained. Recently, Ruan et al. [6] proposed a coupling FVM and the pixel coloring method to deal with crystallization in short-fiber-reinforced composites. Crystal morphology evolution was explicitly shown by the pixel coloring method. However, it should be pointed out that the aforementioned papers only deal with the 2D case which is a qualitative simplification for the actual problem. Moreover, works of Charbon and Swaminarayan [3] as well as Huang and Kamal [4] only deal with the imposed temperature filed, the effect of processing conditions has not been investigated.
In the numerical study of 3D polymer crystallization, Raabe [7], Raabe and Godara [8], and Lin et al. [9] were the pioneers. The cellular automaton method was constructed to deal with the detailed development of crystal morphology. However, their works are only concentrated on the simple isothermal case. So far, no literature has been found to deal with the crystal morphology development with the heat transfer in 3D numerical study.
The objective of this article is to extend the multiscale model and the multiscale computational method proposed in our pervious work [6] to the 3D numerical study of polymer crystallization during cooling. Since the injection molding is 3D in realistic situation, 3D numerical study can predict the temperature distribution and the crystallization behavior more accurately. Our multiscale model incorporates two distinct length scales to simulate the crystallization: a coarse grid for the heat diffusion and a fine grid for capturing the crystal morphology formation. FVM is employed on the coarse grid to solve the energy equation while the pixel coloring method is applied on the fine grid to track the crystal morphology development. The present paper is organized as follows: in Section 2, the multiscale model and the coupled computational method are introduced; in Section 3, results and discussion are given; finally the conclusions are drawn in Section 4.
#### 2. Multiscale Model and Multiscale Algorithm
During the cooling stage, the process of crystallization is coupled with the heat transfer which makes the modeling more difficult and complex. On the one hand, it is well known that the kinetic parameters of nucleation and growth in microscopic scale are strongly dependent on the temperature or processing conditions. On the other hand, the crystallization is an exothermic process which releases the heat and affects the thermal field in macroscopic level.
##### 2.1. Thermal Field in the Macroscopic Level
Thermomechanical histories are of great importance in the estimate of the final product properties. The corresponding simplified energy equation is [6] where is the material density, is the thermal capacity, is the temperature field, is the time, is the thermal conductivity, is the total enthalpy released during crystallization, and is the relative crystallinity. The last source term is the contribution of the latent heat released by the crystallization, which plays the role in macro-micro-coupling.
In the accurate modeling, it is important to consider the thermal properties as a function of temperature and relative crystallinity. The thermal capacity, thermal conductivity can be described by the “mixing rule” of the solid state and the liquid state values to get [10] where and are the thermal capacity of semicrystalline phase and amorphous phase, respectively, and are the thermal conductivity of semicrystalline phase, and amorphous phase, respectively. Also , , , and are temperature dependent variables, which can be modeled as a linear temperature dependency functions [10].
##### 2.2. Crystal Morphology Evolution in the Microscopic Level [6]
Crystallization is a mechanism of phase change in semicrystalline polymeric materials [11]. It represents a mix between nucleation, growth, and impingement. When the temperature of semicrystalline polymeric melt is decreased below its crystallization temperature, “nuclei” appears randomly in space and time; then the appeared nuclei start to grow to form what is referred to as “spherulites.” As the time progresses, the spherulites grow until they hit another crystal and stop growing which is called “impingement.” If all the spherulites are impinged with other spherulites, the crystallization process is finished.
###### 2.2.1. Nucleation
Polymer nucleation is an important factor which affects the final morphology. Usually, the nucleation may vary material from material. In the numerical study, one may adopt an empirical nucleation relation which is derived from fitting the experiment data. Here, we use the following relation of nucleation density [12, 13]: where is the supercooling temperature with being the equilibrium melting temperature, and are the empirical parameters. This nucleation density relation was also used in our pervious work [6] for the description of nucleation density in polymer bulk.
###### 2.2.2. Growth and Impingement
Growth rate is another important factor which affects the development of morphology. Here, we adopt the Hoffman-Lauritzen theory [14] to describe the spherulite growth rate: where and are constants, is the activation energy of motion, is the gas constant, with is the glass transition temperature, and .
With the growth of spherulites, it is inevitable to meet with the “impingement.” Impingement is happed in spherulites which contact with their neighbors or the walls. Different spherulites impinge to form grain boundaries. With the help of grain boundaries, it is possible to calculate the mean size of spherulites.
##### 2.3. Macro-Micro-Algorithm
In our paper, we assume that the temperature field is in the macroscopic level while the crystal morphology evolution is in the microscopic level (see Figure 1). Therefore, it is important to build up a macro-micro-algorithm. Here we extend the algorithm of coupling of FVM [15] and the pixel coloring method [6, 11, 16] proposed in our pervious work [6] to the 3D numerical study.
Figure 1: Schematic representation of different length scale for computation.
In our algorithm, the domain is first divided by a coarse grid. FVM with cell vertex scheme is used on this coarse grid to calculate the temperature. Then, each coarse grid is subdivided into a number of cubes to form the fine grid. The pixel coloring method is employed on this fine grid to track the development of crystal morphology. We assume the temperature on each coarse grid is the same which determines the nucleation and growth rate of spherulites through (2.3) and (2.4). On the other side, with the evolution of crystallization (nucleation, growth, and impingement) on the fine grid, the relative crystallinity changes which affects the temperature through (2.1). This realizes the macro-micro-coupling.
On the coarse grid, FVM is used to solve the energy equation. The reason why we use FVM is because this method uses the control volume which is more like the cell in the statistics for the microscopic information. Figure 2(a) shows a 3D control volume for FVM. Here, is the computational point with the neighbor points , , , , , and the interface points , , , , , . See Figure 2(b) which gives a 2D gird. We use the first-order forward-time and second-order central-space scheme to discrete the energy equation (2.1) to get The choice for such a scheme is because of numerical simplicity. In our algorithm, we first calculate the temperature then calculate the relative crystallinity, therefore, the relative crystallinity is obtained with a delay that is why we use instead of for the last term in (2.5). It should be noted that this energy equation calculation is to obtain temperature which is the input of nucleation and growth of spherulites on fine grid.
Figure 2: Schematic of a control volume (a) 3D grid (b) 2D grid (profile of 3D grid).
On the fine grid, the pixel-coloring method [6, 11, 16] is implemented to capture the growth front of a population of spherulites. Here, we will not present the detailed implementation of the pixel-coloring method and refer our previous work [6] for more details. The statistical character of the evolution of crystal morphology on fine grid is the relative crystallinity which is the input of temperature calculation on coarse gird. In our study, the relative crystallinity can be calculated by Since the algorithm on fine grid can explicitly show the details of spherulites, here, we define another important factor, the mean radius of each spherulite, as with the volume of the spherulite which can be calculated with the number of cells occupied by the spherulite and the cell size. It should be mentioned that the mean radius of spherulite directly affects the stress of the final product through the Hall-Petch relation [11].
Figure 3 shows the macro-micro-algorithm used for our simulation.
Figure 3: Multiscale algorithm.
#### 3. Results and Discussion
##### 3.1. Problem Formulation
3D polymer crystallization during cooling stage is studied here. Figure 4 describes the computational geometry. The length of the mold cavity is 8 mm while its width and height are 4 mm respectively. We assume the walls mm, and mm experiencing the constant cooling rate operation and the temperature boundary conditions are set to be , with the initial temperature and the cooling rate. The other boundaries are assumed adiabatic and the temperature boundary conditions are set to be with the outward unit normal vector. It should be mentioned that this geometry it related to the think-wall parts in injection molding which is because the ratio of width/height of the cross-section is less than 10 [2].
Figure 4: Computational geometry.
The material we considered here is iPP and the parameters used in the simulation are [6, 12, 13]: , , , , , , , , , , . Unless otherwise stated, the other parameters are set to be , . Due to the lack of material parameters, here we neglect the thermal difference between the solid state and the liquid state.
We test three meshes in this problem computation, namely, Mesh1: for coarse grid and for fine grid; Mesh2: for coarse grid and for fine grid; Mesh3: for coarse grid and for fine grid. Results of temperature and relative crystallinity at the intersection line of mm plane and mm plane (line: , see Figure 4) are compared in Figure 5. As we can see that the result obtained on Mesh1 agrees with the results obtained on Mesh2 and Mesh3 very well. As we know, the refinement of either coarse grid (Mesh2) or fine grid (Mesh3) can lead to more accurate of the final results, it brings in a large of storage which affects the efficiency and challenges the computer. When we balance the accuracy with the efficiency, we find that Mesh1 is proper. Therefore, in our following study, Mesh1 is used and its coarse grid size is and the fine grid size is . Here, we will not explain the phenomenon of temperature and relative crystallinity obtained in Figure 5 and describe it in the next section.
Figure 5: Evolution of temperature and relative crystallinity at the intersection line of mm plane and mm plane (line: ).
##### 3.2. Importance of Macro-Micro-Coupling
###### 3.2.1. Temperature and Relative Crystallinity Evolution in Macroscopic Level
Evolution of temperature and relative crystallinity during cooling at the plane of mm are shown in Figure 6(a). As we can see in the figure of temperature evolution, a quasi-isothermal plateau is formed at the positions near the adiabatic boundary ( mm, mm). This plateau is caused by the latent heat released by crystallization. We also reported this plateau in our pervious work [6] which deals with the 2D short-fiber-reinforced system. The figure of relative crystallinity evolution tells us that the polymer in the skin layer crystallizes much earlier and faster than that in the core layer. This can be explained that the supercooling temperature in the skin layer is much higher than that in the core layer due to the temperature difference which is the fundamental reason for the crystallization.
Figure 6: Evolution of temperature and relative crystallinity (a) considering the latent heat (b) without considering the latent heat.
Latent heat released by the crystallization is the bridge in macro-micro-coupling. To determine the importance of the contribution of the latent heat, the results of our simulation are compared with the results of model which ignores the latent heat. Figure 6(b) shows the results of model without consider the latent heat. We can see that there is a large difference between these two models especially at the positions near the adiabatic boundaries. Moreover, the results of model without considering latent heat do not develop the quasi-isothermal plateau at the positions near the adiabatic boundary. This is the error caused by ignoring the latent heat. Since the nucleation and growth rate of the spherulites is completely determined by the temperature, therefore, to predict crystal morphology more accurately, the model of temperature is necessary to include the influences of the latent heat.
###### 3.2.2. Crystal Morphology in Microscopic Level
Evolution of crystal morphology at the control volume of mm is shown in Figure 7. We use the fronts of spherulitics in surface (or slices) to show the crystal morphology evolution. Here, the white region is the polymer melt and the colored region is the spherulitics. Different spherulitics are distinguished by different colors. Figure 7 clearly shows that crystals first appear in the skin layer, with the time evolutes, they gradually appear to the core layer. This is consistent with the change tendency of relative crystallinity (see Figure 6(a)) which is a macrostatistical character of crystal morphology evolution in microscopic level. The results presented here show qualitatively agreement with our pervious work [6] for the 2D short fiber reinforced polymer solidification.
Figure 7: Evolution of crystal morphology at the control volume of mm (a) s, (b) s, (c) s.
Final morphology in the skin layer and in the core layer is compared in Figure 8. We choose skin layer as a representative control volume of point with the coordinate (4 mm, 1 mm, 1 mm) while core layer as a representative control volume of point with the coordinate (4 mm, 3 mm, 3 mm) which is illustrated in Figure 4. It is not so obvious for us to distinguish the skin morphology from the core morphology. However, the size of spherulitics in the core layer is somewhat bigger than that in the skin layer.
Figure 8: Crystal morphology at different positions (a) skin (b) core.
Mean radius of each spherulite and the distribution of spherulite size at different positions are shown in Figure 9. We should mention that we do not consider the “phantom nuclei” in the comparison. Figure 9 shows that spherulite size is relative smaller in the position which is close to skin layer. Since the spherulites size directly affects the mechanical properties of the products according to Hall-Petch relation [11], it can be concluded that mechanical properties vary from skin to core.
Figure 9: Spherulite size and distribution of spherulite size at different positions.
##### 3.3. Importance of 3D Simulation
In order to highlight the importance of 3D simulation, we also give the comparison of the results obtained in 2D case with the 3D case. Since the problem we studied is a quasi-three dimensional problem, it can be also reduced as a two-dimensional one. Here, we consider the profile of mm (see Figure 4) as our 2D studied cavity with the boundaries mm and mm experiencing the cooling rate operation and the other boundaries been assumed to be adiabatic. We will mention that the only difference between 3D case and 2D case is that the heat transfer in direction is considered in 3D case while its effect is ignored in 2D case. In our study, the nucleation density in 2D and 3D is related with the following statistical formulation [3]: and the growth rate of spherulites in 2D and 3D are assumed to be the same.
Figure 10 shows the evolution of temperature and relative crystallinity which are obtained in 2D simulation. Through comparing with Figure 6(a), it is observed that the temperature away from the cooling boundaries is higher in 2D case than 3D case at the same time; particularly, the maximum temperature divergence is about 1 K in the core layer. Moreover, the internal relative crystallinity is also higher in 2D case than 3D case; in particular, the maximum relative crystallinity divergence is about 0.15 in the core layer.
Figure 10: Evolution of temperature and relative crystallinity (2D results).
Evolution of crystal morphology at the whole computational geometry in 2D case is shown in Figure 11. Compared with Figure 7, we can find that the tendency of microstructure formation in 2D case is the same as 3D case. To emphasize the difference between 2D case and 3D case, we also investigate the spherulite size and its distribution at different positions of the cavity. For the sake of simplicity, we also choose the “skin” and “core” control volume as our comparison cells as illustrated in Figure 4. Figure 12 shows the spherulite size and its distribution as obtained in 2D simulation. By the comparison with Figure 9, we find that the trends of spherulite size distribution in skin and core are identical in 2D case and 3D case, however, in 2D case, the distribution of spherulite size is more concentrated.
Figure 11: Evolution of crystal morphology in 2D case (a) s (b) s (c) s.
Figure 12: Spherulite size and distribution of spherulite size at different positions (2D results).
The comparison in this section tells us that the macroscopic and microscopic values obtained in 2D simulation are quantitatively different from that in 3D simulation. Therefore, if the models, algorithms, and computational conditions are allowed, we should consider the 3D simulation in order to obtain the more precise results.
##### 3.4. Effects of Thermal Condition on the Crystallization
Effects of cooling rate and initial temperature are also investigated in this 3D polymer crystallization case. Without loss of generality, we here choose the “core” control volume as our cell for showing the results.
###### 3.4.1. Effects of Cooling Rate
We change the boundary conditions by varying the cooling rate as 1 K/min, 2 K/min, 5 K/min in order to study the effects of cooling rate.
Figure 13 shows the relative crystallinity evolution and the distribution of spherulite size in the core layer for different cooling rate. According to Figure 13, the higher cooling rate causes an acceleration of the crystallization and a reduction of spherulite size. We will mention that the effects of cooling rate are similar as our pervious work [6] where the 2D short-fiber-reinforced system are studied. In addition, this tendency is also reported in the experimental work by Zheng et al. [17].
Figure 13: Evolution of relative crystallinity and the distribution of spherulite size in core layer for different cooling rates.
Thus, we will conclude that if the designer wants to obtain the smaller size of the spherulites, he shall impose a relatively higher cooling rate boundary condition.
###### 3.4.2. Effects of Initial Temperature
We varying the initial temperature as 470 K, 480 K, 490 K in order to study the effects of initial temperature.
Figure 14 shows the relative crystallinity evolution and the distribution of spherulite size in the core layer with different initial temperature. It is observed that the higher initial temperature leads to a deceleration of crystallization and minor effect on the crystal morphology. This is also in agreement with our pervious work [6] in 2D case.
Figure 14: Evolution of relative crystallinity and the distribution of spherulite size in core layer with different initial temperatures.
Thus, if the designer wants to improve efficiency, he should cool down the melt with a relative lower initial temperature.
#### 4. Conclusion
We have extended our previous proposed multiscale model and multiscale algorithm to the simulation of 3D polymer crystallization during cooling stage. With the multiscale model and multiscale algorithm, we obtained the temperature distributions and relative crystallinity at various locations in the mold cavity, meanwhile, we also predicted the crystal morphology development and its size as well as its distributions. We have also shown the importance of coupling between the heat transfer with crystallization as well as 3D numerical studies. Results presented in this paper shows that latent heat released by crystallization plays a very important role in macro-micro-coupling which should be considered in order to predict the more accurate crystal morphology; 2D simulation is qualitatively agreement with the 3D simulation (not only for the variables predicted in macroscopic level and microscopic level, but also for the effects of cooling rate and initial temperature), however, in the view of quantitative analysis, results obtained for these two cases have some differences. Therefore, if the computational conditions are allowed, we recommend the 3D simulation.
Future work will concentrate on 3D crystallization simulation during cooling coupled with heat transfer in reinforced system as well as flow-induced crystallization (FIC) simulation in injection molding.
#### Acknowledgment
The financial supports provided by the Henan Scientific and Technological Research Project (no. 122102210198), Key Scientific and Technological Research Project of Department of Education of Henan Province (no. 12B110006), Youth Scientific Foundation of Henan University of Science and Technology (no. 2012QN015), and the Doctoral Foundation of Henan University of Science and Technology (no. 09001612) are fully acknowledged.
#### References
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16. C. Ruan, J. Ouyang, S. Liu, and L. Zhang, “Computer modeling of isothermal crystallization in short fiber reinforced composites,” Computers and Chemical Engineering, vol. 35, no. 11, pp. 2306–2317, 2011.
17. Q. Zheng, Y. Shangguan, S. Yan, Y. Song, M. Peng, and Q. Zhang, “Structure, morphology and non-isothermal crystallization behavior of polypropylene catalloys,” Polymer, vol. 46, no. 9, pp. 3163–3174, 2005. | 6,736 | 29,824 | {"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-2018-47 | latest | en | 0.857612 |
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Definition of Accounting equation
Accounting equation
The representation of the double-entry system of accounting such that assets are equal to liabilities plus capital.
Accounting equation
The formula Assets = Liabilities + Equity.
accounting equation
An equation that reflects the two-sided nature of a
business entity, assets on the one side and the sources of assets on the
other side (assets = liabilities + owners’ equity). The assets of a business
entity are subject to two types of claims that arise from its two basic
sources of capital—liabilities and owners’ equity. The accounting equation
is the foundation for double-entry bookkeeping, which uses a
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or balance sheet, which is one of the three fundamental financial
Related Terms:
net worth
Generally refers to the book value of owners’ equity as reported
in a business’s balance sheet. If liabilities are subtracted from assets, the
accounting equation becomes: assets - liabilities = owners’ equity. In this
version of the accounting equation, owners’ equity equals net worth, or
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Total liabilities exceed total assets. A firm with a negative net worth is insolvent on
the books.
Accounting liquidity
The ease and quickness with which assets can be converted to cash.
Alpha equation
The alpha of a fund is determined as follows:
[ (sum of y) -((b)(sum of x)) ] / n
where:
n =number of observations (36 months)
b = beta of the fund
x = rate of return for the S&P 500
y = rate of return for the fund
Average accounting return
The average project earnings after taxes and depreciation divided by the average
book value of the investment during its life.
Beta equation (Mutual Funds)
The beta of a fund is determined as follows:
[(n) (sum of (xy)) ]-[ (sum of x) (sum of y)]
[(n) (sum of (xx)) ]-[ (sum of x) (sum of x)]
where: n = # of observations (36 months)
x = rate of return for the S&P 500 Index
y = rate of return for the fund
Beta equation (Stocks)
The beta of a stock is determined as follows:
[(n) (sum of (xy)) ]-[(sum of x) (sum of y)]
[(n) (sum of (xx)) ]-[(sum of x) (sum of x)]
where: n = # of observations (24-60 months)
x = rate of return for the S&P 500 Index
y = rate of return for the stock
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A technical accounting term that encompasses the
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Method of accounting for a merger in which the acquirer is treated as having purchased
the assets and assumed liabilities of the acquiree, which are all written up or down to their respective fair
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An equation that describes the average relationship between a dependent variable and a
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accounting principals required by the FHLB that allow S&Ls to elect
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Statement of Financial Accounting Standards No. 8
This is a currency translation standard previously in
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Statement of Financial Accounting Standards No. 52
This is the currency translation standard currently
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Standards No. 8.
Accounting
A collection of systems and processes used to record, report and interpret business transactions.
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accounting
A broad, all-inclusive term that refers to the methods and procedures
of financial record keeping by a business (or any entity); it also
refers to the main functions and purposes of record keeping, which are
to assist in the operations of the entity, to provide necessary information
to managers for making decisions and exercising control, to measure
profit, to comply with income and other tax laws, and to prepare financial
reports.
accrual-basis accounting
Well, frankly, accrual is not a good descriptive
term. Perhaps the best way to begin is to mention that accrual-basis
accounting is much more than cash-basis accounting. Recording only the
cash receipts and cash disbursement of a business would be grossly
inadequate. A business has many assets other than cash, as well as
many liabilities, that must be recorded. Measuring profit for a period as
the difference between cash inflows from sales and cash outflows for
expenses would be wrong, and in fact is not allowed for most businesses
by the income tax law. For management, income tax, and financial
reporting purposes, a business needs a comprehensive record-keeping
system—one that recognizes, records, and reports all the assets and liabilities
of a business. This all-inclusive scope of financial record keeping
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(though cash is paid before or after expenses are recorded). Established
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must be recorded using accrual-basis accounting methods. Also, these
authoritative standards require that in reporting its financial condition a
double-entry accounting
See accrual-basis accounting.
generally accepted accounting principles (GAAP)
This important term
refers to the body of authoritative rules for measuring profit and preparing
financial statements that are included in financial reports by a business
to its outside shareowners and lenders. The development of these
guidelines has been evolving for more than 70 years. Congress passed a
law in 1934 that bestowed primary jurisdiction over financial reporting
by publicly owned businesses to the Securities and Exchange Commission
(SEC). But the SEC has largely left the development of GAAP to the
private sector. Presently, the Financial accounting Standards Board is
the primary (but not the only) authoritative body that makes pronouncements
on GAAP. One caution: GAAP are like a movable feast. New rules
are issued fairly frequently, old rules are amended from time to time,
and some rules established years ago are discarded on occasion. Professional
accountants have a heck of time keeping up with GAAP, that’s for
sure. Also, new GAAP rules sometimes have the effect of closing the barn
door after the horse has left. accounting abuses occur, and only then,
after the damage has been done, are new rules issued to prevent such
abuses in the future.
internal accounting controls
Refers to forms used and procedures
established by a business—beyond what would be required for the
record-keeping function of accounting—that are designed to prevent
errors and fraud. Two examples of internal controls are (1) requiring a
second signature by someone higher in the organization to approve a
transaction in excess of a certain dollar amount and (2) giving customers
printed receipts as proof of sale. Other examples of internal
control procedures are restricting entry and exit routes of employees,
requiring all employees to take their vacations and assigning another
person to do their jobs while they are away, surveillance cameras, surprise
counts of cash and inventory, and rotation of duties. Internal controls
should be cost-effective; the cost of a control should be less than
the potential loss that is prevented. The guiding principle for designing
internal accounting controls is to deter and detect errors and dishonesty.
The best internal controls in the world cannot prevent most fraud
by high-level managers who take advantage of their positions of trust
and authority.
accounting rate of return (ARR)
the rate of earnings obtained on the average capital investment over the life of a capital project; computed as average annual profits divided by average investment; not based on cash flow
cost accounting
a discipline that focuses on techniques or
methods for determining the cost of a project, process, or
thing through direct measurement, arbitrary assignment, or
systematic and rational allocation
Cost Accounting Standards Board (CASB)
a body established by Congress in 1970 to promulgate cost accounting
standards for defense contractors and federal agencies; disbanded
in 1980 and reestablished in 1988; it previously issued
pronouncements still carry the weight of law for those
organizations within its jurisdiction
financial accounting
a discipline in which historical, monetary
transactions are analyzed and recorded for use in the
preparation of the financial statements (balance sheet, income
statement, statement of owners’/stockholders’ equity,
and statement of cash flows); it focuses primarily on the
needs of external users (stockholders, creditors, and regulatory
agencies)
management accounting
a discipline that includes almost
all manipulations of financial information for use by managers
in performing their organizational functions and in
assuring the proper use and handling of an entity’s resources;
it includes the discipline of cost accounting
Management Accounting Guidelines (MAGs)
pronouncements of the Society of Management Accountants of
management accounting situations
responsibility accounting system
an accounting information system for successively higher-level managers about the performance of segments or subunits under the control
of each specific manager
Statement on Management Accounting (SMA)
a pronouncement developed and issued by the Management
accounting Practices Committee of the Institute of Management
Accountants; application of these statements is
through voluntary, not legal, compliance
Accounting change
An alteration in the accounting methodology or estimates used in
the reporting of financial statements, usually requiring discussion in a footnote
attached to the financial statements.
Accounting entity
A business for which a separate set of accounting records is being
maintained.
Accrual accounting
The recording of revenue when earned and expenses when
incurred, irrespective of the dates on which the associated cash flows occur.
Constant dollar accounting
A method for restating financial statements by reducing or
increasing reported revenues and expenses by changes in the consumer price index,
thereby achieving greater comparability between accounting periods.
Generally accepted accounting principles
The rules that accountants follow when processing accounting transactions and creating financial reports. The rules are primarily
derived from regulations promulgated by the various branches of the AICPA Council.
generally accepted accounting principles (GAAP)
Procedures for preparing financial statements.
Equation of Exchange
The quantity theory equation Mv = PQ.
Accounting Errors
Unintentional mistakes in financial statements. Accounted for by restating
the prior-year financial statements that are in error.
Accounting and Auditing Enforcement Release (AAER)
Administrative proceedings or litigation releases that entail an accounting or auditing-related violation of the securities laws.
Accounting Irregularities
Intentional misstatements or omissions of amounts or disclosures in
financial statements done to deceive financial statement users. The term is used interchangeably with fraudulent financial reporting.
Aggressive Accounting
A forceful and intentional choice and application of accounting principles
done in an effort to achieve desired results, typically higher current earnings, whether the practices followed are in accordance with generally accepted accounting principles or not. Aggressive
accounting practices are not alleged to be fraudulent until an administrative, civil, or criminal proceeding takes that step and alleges, in particular, that an intentional, material misstatement
has taken place in an effort to deceive financial statement readers.
Change in Accounting Estimate
A change in accounting that occurs as the result of new information
or as additional experience is acquired—for example, a change in the residual values
or useful lives of fixed assets. A change in accounting estimate is accounted for prospectively,
over the current and future accounting periods affected by the change.
Change in Accounting Principle
A change from one generally accepted accounting principle to another generally accepted accounting principle—for example, a change from capitalizing expenditures
to expensing them. A change in accounting principle is accounted for in most instances
Change in Accounting Estimate
A change in the implementation of an existing accounting
policy. A common example would be extending the useful life or changing the expected residual
value of a fixed asset. Another would be making any necessary adjustments to allowances for
uncollectible accounts, warranty obligations, and reserves for inventory obsolescense.
Contract Accounting
Method of accounting for sales or service agreements where completion
requires an extended period.
Creative Accounting Practices
Any and all steps used to play the financial numbers game, including
the aggressive choice and application of accounting principles, both within and beyond
the boundaries of generally accepted accounting principles, and fraudulent financial reporting.
Also included are steps taken toward earnings management and income smoothing. See Financial
Numbers Game.
Creative Acquisition Accounting
The allocation to expense of a greater portion of the price
paid for another company in an acquisition in an effort to reduce acquisition-year earnings and
boost future-year earnings. Acquisition-year expense charges include purchased in-process research
and development and an overly aggressive accrual of costs required to effect the acquisition.
Cumulative Effect of Accounting Change
The change in earnings of previous years assuming
Cumulative Effect of a Change in Accounting Principle
The change in earnings of previous years
based on the assumption that a newly adopted accounting principle had previously been in use.
Gain-on-Sale Accounting
Up-front gain recognized from the securitization and sale of a pool
of loans. Profit is recorded for the excess of the sales price and the present value of the estimated
interest income that is expected to be received on the loans above the amounts funded on the loans
and the present value of the interest agreed to be paid to the buyers of the loan-backed securities.
Generally Accepted Accounting Principles (GAAP)
A common set of standards and procedures
for the preparation of general-purpose financial statements that either have been established
by an authoritative accounting rule-making body, such as the Financial accounting
Standards Board (FASB), or over time have become accepted practice because of their universal
application.
Staff Accounting Bulletin (SAB)
Interpretations and practices followed by the staff of the Office of the Chief Accountant and the Division of Corporation Finance in administering the disclosure
requirements of the federal securities laws.
Accounting Policies
The principles, bases, conventions, rules and procedures adopted by management in preparing and presenting financial statements.
Generally Accepted Accounting Principles (GAAP)
GAAP is the term used to describe the underlying rules basis on which financial statements are normally prepared. This is codified in the Handbook of The Canadian Institute of Chartered Accountants. | 3,420 | 17,530 | {"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-2024-18 | latest | en | 0.944238 |
https://nrich.maths.org/public/leg.php?code=5053&cl=1&cldcmpid=2558 | 1,511,609,589,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934809778.95/warc/CC-MAIN-20171125105437-20171125125437-00732.warc.gz | 661,438,779 | 6,228 | # Search by Topic
#### Resources tagged with Dominoes similar to Path of Discovery Series: 2. First Steps:
Filter by: Content type:
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##### Other tags that relate to Path of Discovery Series: 2. First Steps
Rich Tasks. Dominoes. Pedagogy. Interlocking cubes. Digit cards. Problem solving. Mathematical Thinking. Learning mathematics. Base-10 apparatus. Dice.
### There are 19 results
Broad Topics > Mathematics Tools > Dominoes
### Path of Discovery Series: 2. First Steps
##### Stage: 1
This article takes a closer look at some of the toys and games that can enhance a child's mathematical learning.
### Domino Pick
##### Stage: 1 Challenge Level:
Are these domino games fair? Can you explain why or why not?
### Dominoes Environment
##### Stage: 1 and 2 Challenge Level:
These interactive dominoes can be dragged around the screen.
### 4 Dom
##### Stage: 1, 2, 3 and 4 Challenge Level:
Use these four dominoes to make a square that has the same number of dots on each side.
### Guess the Dominoes
##### Stage: 1, 2 and 3 Challenge Level:
This task depends on learners sharing reasoning, listening to opinions, reflecting and pulling ideas together.
### Domino Patterns
##### Stage: 1 Challenge Level:
What patterns can you make with a set of dominoes?
### Domino Sets
##### Stage: 2 Challenge Level:
How do you know if your set of dominoes is complete?
### Guess the Dominoes for Two
##### Stage: Early years, 1 and 2 Challenge Level:
Guess the Dominoes for child and adult. Work out which domino your partner has chosen by asking good questions.
### Dominoes
##### Stage: 2, 3 and 4 Challenge Level:
Everthing you have always wanted to do with dominoes! Some of these games are good for practising your mental calculation skills, and some are good for your reasoning skills.
### Domino Magic Rectangle
##### Stage: 2, 3 and 4 Challenge Level:
An ordinary set of dominoes can be laid out as a 7 by 4 magic rectangle in which all the spots in all the columns add to 24, while those in the rows add to 42. Try it! Now try the magic square...
### Domino Square
##### Stage: 2, 3 and 4 Challenge Level:
Use the 'double-3 down' dominoes to make a square so that each side has eight dots.
### Domino Sorting
##### Stage: 1 Challenge Level:
Try grouping the dominoes in the ways described. Are there any left over each time? Can you explain why?
### Domino Sequences
##### Stage: 1 Challenge Level:
Find the next two dominoes in these sequences.
### Amy's Dominoes
##### Stage: 2 Challenge Level:
Amy has a box containing domino pieces but she does not think it is a complete set. She has 24 dominoes in her box and there are 125 spots on them altogether. Which of her domino pieces are missing?
### Next Domino
##### Stage: 1 Challenge Level:
Which comes next in each pattern of dominoes?
### Domino Join Up
##### Stage: 1 Challenge Level:
Can you arrange fifteen dominoes so that all the touching domino pieces add to 6 and the ends join up? Can you make all the joins add to 7?
### Domino Number Patterns
##### Stage: 1 Challenge Level:
Can you work out the domino pieces which would go in the middle in each case to complete the pattern of these eight sets of 3 dominoes?
### Eight Dominoes
##### Stage: 2, 3 and 4 Challenge Level:
Using the 8 dominoes make a square where each of the columns and rows adds up to 8
### Rectangles with Dominoes
##### Stage: 1 Challenge Level:
Can you make a rectangle with just 2 dominoes? What about 3, 4, 5, 6, 7...? | 878 | 3,534 | {"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-2017-47 | latest | en | 0.824018 |
http://www.lifelong-learners.com/opt/nu/SYL/s3node2.html | 1,511,081,320,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805466.25/warc/CC-MAIN-20171119080836-20171119100836-00337.warc.gz | 440,350,698 | 5,171 | SYLLABUS Previous: 3.1 Option pricing for Up: 3 FORECASTING WITH UNCERTAINTY Next: 3.3 Improved model using
## 3.2 Simple valuation model using binomial trees
[ SLIDE tree - delta hedging - transition probability - matching volatility - recipe || VIDEO modem - LAN - DSL ]
A single step binomial forecast provides only a crude approximation for the fair price of an option before it expires. To increase the accuracy of the model, an obvious improvement would be to extend the number of possible outcomes; the evolution of the forecast price could also be modeled more accurately by allowing them to reverse trends during the calculation. Both can be achieved by dividing the lifetime of the option [0;T] into a number smaller time intervals of duration Dt and performing the calculation recursively with the binomial tree model sketched in (3.2#fig.1). In addition, we show here how a tree is constructed to reproduce log-normal price increments that are typical for stock: rather than adding / subtracting a constant as in the previous sect.3.1 (for a normal distribution), possible realizations are here obtained by multiplying / dividing by a constant factor (for a log-normal distribution).
For each step starting with the present value of the underlying S0 two new forcasts are obtained by multiplying the value on each node by the factors u or d to mimic possible movements up or down until the entire lifetime of the option is covered by the tree. A perfectly hedged portfolio is then constructed starting from every branching point closest to the expiry date (work backwards from the right of (3.2#fig.1), by combining an amount delta of the underlying with a (conventionally short) position of a (positively correlated) option. By demanding that the portfolio be risk free, the movement up or down produce the same return and a new value is obtained for delta
(3.2#eq.1)
Since a perfectly hedged portfolio carries no risk at all, the standard no-arbitrage argument shows that it can be discounted back one step in time using the risk free interest rate r This discounted value (3.2#eq.2, left hand side) has to be equal to the cost of setting up the portfolio before the step is taken (right hand side):
(3.2#eq.2)
Substituting the hedging factor delta (3.2#eq.1) and rearranging the terms, this yields an expression to calculate the fair value of an option one step back at a time
(3.2#eq.3)
The parameter p can be interpreted as the probability of the forecast price moving up and (1-p) the probability that it will move down in the tree. The scaling factors (u,d) control the amplitude of the change and have to be carefully chosen
(3.2#eq.4)
to reproduce the drift and the volatility observed in the real markets (quants read below).
Although the importance will only appear later, simply note here that the expected value of the underlying calculated using the probability (3.2#eq.3)
(3.2#eq.5)
grows, on average, exactly at the risk free interest rate. Using the probability (3.2#eq.3) therefore implies that the return on the underlying stock is equal to the risk free rate m=r.
Quants: matching the parameters (u,d) with drift and volatility.
For clarity, distinguish the probability of a price moving up in the tree p from the probability of the price moving up in the real world q. In the presence of drift, the real world price of the underlying grows exponentially (3.2#eq.6, left hand side), which should be reproduced by the expectation E[S] from the price forecast in the tree (right hand side):
(3.2#eq.6)
In the same manner, the real world variance (square of volatility, left) has to be matched with the variance Var[S]=E[S2]-E[S]2 from the price forecast in the tree (right):
(3.2#eq.7)
Substitute the real world probability (3.2#eq.6) into (3.2#eq.7)
(3.2#eq.8)
and expand to first order in the small time steps by writing exp(mD)~ 1 +mD. The symmetric solution u=1/d is generally chosen and is the one given in (3.2#eq.4).
To summarize, calculations using binomial trees for the stockmarket can be organized as follows
1. Divide the entire lifetime of the option into a finite number of steps N, ranging from only a few (by hand) up to 30 (using a computer to evaluate 31 possible outcomes that are connected with 230~ 1 billion possible paths).
2. Forecast the underlying forward in time (trunk leaves), choosing (u,d) according to (3.2#eq.4) to reproduce the historical volatility observed in a real market.
3. Work backward in time (trunk leaves) starting from the terminal option payoff; for every neighboring branching point, calculate the hedging delta (3.2#eq.1) and the option price at the previous time step (3.2#eq.3). In the case of American options, substitute the (larger) intrinsic value that can be obtained from an early exercise when the calculated price drops below this intrinsic value.
4. The final result is obtained on the trunk of the tree and is an approximation of the fair value of the option before the expiry date, with an accuracy proportional to 230.
Because it involves only simple mathematics, binomial trees are ideally suited to develop an intuition for option pricing (exercise 3.01, 3.02). Some practitioners even use trees to evaluate option prices with a computer: the forthcoming sections will show that differential calculus provides a far better framework to account for the features appearing in exotic contracts. Indeed, without having to account for these features, the computer is not really needed: the price can simply be calculated from an analytic solution of the Black-Scholes differential equation, which we are about to derive.
SYLLABUS Previous: 3.1 Option pricing for Up: 3 FORECASTING WITH UNCERTAINTY Next: 3.3 Improved model using | 1,360 | 5,771 | {"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} | 3.140625 | 3 | CC-MAIN-2017-47 | latest | en | 0.889246 |
http://math.stackexchange.com/questions/247078/how-to-understand-spectral-decomposition-geometrically?answertab=votes | 1,462,424,880,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860125897.19/warc/CC-MAIN-20160428161525-00107-ip-10-239-7-51.ec2.internal.warc.gz | 183,321,770 | 18,766 | # How to understand spectral decomposition geometrically
Let $A$ be a $k\times k$ positive definite symmetric matrix. By spectral decomposition, we have
$$A = \lambda_1e_1e_1'+ ... + \lambda_ke_ke_k'$$
and
$$A^{-1} = \sum_{i=1}^k\frac{1}{\lambda_i}e_ie_i'$$
How to understand spectral decomposition and the relationship between $A$ and $A^{-1}$ geometrically?
-
You are requiring that the matrix $A$ be positive definite, but need not be so to be diagonalizable. In fact over the complex numbers, we would only need $A$ to commute with its adjoint (transpose and complex conjugated of the original, $A^†$), which then $A$ is called a normal matrix, i.e. if $A\cdot A^†=A^†\cdot A$ then there is a basis of $\mathbb C^k$ formed by a complete set of linearly independent orthonormal eigenvectors $\lvert e_i\rangle$, corresponding to eigenvalues $\lambda_i$ (possibly repeated) of $A$, so that $$A=\sum_{i=1}^k \lambda_i\lvert e_i\rangle\langle e_i\lvert\;\; \Rightarrow\;\; A^{-1}=\sum_{i=1}^k \frac{1}{\lambda_i}\lvert e_i\rangle\langle e_i\lvert.$$ The $\lvert e_i\rangle\langle e_i\lvert$ are witten in Dirac's notation and are your $e_ie'_i$. Concretely, each $\lvert e_i\rangle\langle e_i\lvert$ is a projector onto the line spanned by $\lvert e_i\rangle$; indeed if we write the scalar product by $\langle e_i\,\lvert\,v\rangle$, each projector annihilates all the components of $\lvert v\rangle$ except $v_i\lvert e_i\rangle$ because of orthonormality of $\{\lvert e_j\rangle\}_{j=1}^k$, so $\lvert e_i\rangle\langle e_i\lvert$ gives the length of the projection of the segment $\lvert v\rangle$ projected onto the line spanned by the unitary eigenvector $\lvert e_i\rangle$. This is the "spectral decomposition" of $A$ and gives the action of the matrix $A$ in terms of a linear superposition of projectors. In fact, since the eigenvalues $\lambda_i$ may appear more than once (because of multiplicity $n_i$), you can collect the elementary $n_i$ projectors $\{\lvert e_j\rangle\langle e_j\lvert\}_{j=i}^{i+n_i}$ with common eigenvalue $\lambda_i$ into one projector $P_i:=\sum_{j=i}^{i+n_i}\lvert e_j\rangle\langle e_j\lvert$, so that your matrix just expands into the different $m\leq k$ eigenvalues: $$A=\sum_{i=1}^m \lambda_i P_i\;\; \Rightarrow\;\; A^{-1}=\sum_{i=1}^m \frac{1}{\lambda_i}P_i.$$
Thus, to interpret geometrically what $A, A^{-1}$ do, we have just to understand the action of the projectors $P_i$. This is straightforward: $P_i$ is a linear combination of elementary projectors of any vector onto the lines spanned by $\lvert e_j\rangle$, for $j=i,\dots,i+n_i$, so $P_i\lvert v\rangle$ projects $\lvert v\rangle$ onto the linear subspace (e.g. line, plane, hyperplane...) spanned by those eigenvectors, thus $P_i\lvert v\rangle$ is the vectorial component of $\lvert v\rangle$ in that subspace. That is to say $\lvert v\rangle=\sum_i P_i\lvert v\rangle$.
Therefore we can interpret geometrically the spectral decomposition of $A$ like this: if $A$ is a normal $k\times k$ matrix acting on a vector of $\mathbb C^k$, it corresponds to a linear map, endomorphism of $\mathbb C^k$, which takes vectors and linearly gives new vectors; its action on a vector $\lvert v\rangle$ can be seen as a composition of actions for each component of the vector on each of the $m$ eigenspaces of $A$, which are linear subspaces of $\mathbb C^k$ of dimension $n_i$ corresponding to each of the possible eigenvalues $\lambda_i$ of $A$. The eigenspaces are invariant subspaces because $A$ transforms vectors in them to new vectors in them, without mixing eigenspaces (e.g. taking vectors of an eigenplane into the same plane). Hence, $\lambda_i P_i\lvert v\rangle$ takes the vectorial component of $\lvert v\rangle$ in the eigenspace spanned by the eigenvectors corresopnding to the eigenvalue $\lambda_i$ and scales that component by $\lambda_i$, i.e. just makes that component equal, bigger or smaller according to $|\lambda_i|\geq 1$ or $|\lambda_i|< 1$ (and change the orientation of that component if $\lambda_i <0$). Thus the action of $A$ on $\lvert v\rangle$ stretches its components on each of the eigenspaces by a factor $\lambda_i$, giving at the end a new vector with those new components. The action of $A^{-1}$ is analogue but with each of the stretching factors inverted (so what $A$ scales up, $A^{-1}$ scales down and vice-versa). For example, if $A$ is a positive definite real symmetric matrix, as you say in the question, then its eigenvalues are all positive since $\langle e_i\lvert A\lvert e_i\rangle=\lambda_i > 0$, by positivity and $\{\lvert e_i\rangle\}_{i=1}^k$ being orthonormal. Thus, in this particular case, the action of $A$ on a vector does not change the orientation of its projections onto each of the eigenspaces of $A$, it justs scales them (contrarily to the example of the picture below).
ADDITION: A similar geometric interpretation can be attempted for the canonical Jordan form of a matrix; in that general case, besides diagonal terms, there appear cross term projectors of type $\lambda_i\lvert e_{i+1}\rangle\langle e_i\lvert$ (or $\lvert e_{i-1}\rangle\langle e_i\lvert$ depending on the order of vectors in the generalized eigenbasis) in the decomposition of $A$. Therefore, along with scaling of the component in the direction of the generalized eigenvector, there is also a correction of direction by that component but in the direction of the previous (or next) generalized eigenvector of the basis. For all finite dimensional vector spaces over a field, the matrices or linear maps having all their eigenvectors over that field (i.e. their characteristic polynomial splits into linear factors over it), always have a Jordan canonical form; thus, in those cases (e.g. always over the complex numbers) all matrices/linear maps can be understood geometrically by their actions on vectors, as a composition of scalings and displacements of their projections over each generalized eigensubspace. In this case the inverse of a matrix in Jordan form is not as simple, and so its geometric interpretations not so easily related. | 1,670 | 6,085 | {"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} | 3.84375 | 4 | CC-MAIN-2016-18 | latest | en | 0.759754 |
https://number.academy/4011015 | 1,716,510,583,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058675.22/warc/CC-MAIN-20240523235012-20240524025012-00572.warc.gz | 364,390,394 | 10,950 | # Number 4011015 facts
The odd number 4,011,015 is spelled 🔊, and written in words: four million, eleven thousand and fifteen, approximately 4.0 million. The ordinal number 4011015th is said 🔊 and written as: four million, eleven thousand and fifteenth. The meaning of the number 4011015 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 4011015. What is 4011015 in computer science, numerology, codes and images, writing and naming in other languages
## What is 4,011,015 in other units
The decimal (Arabic) number 4011015 converted to a Roman number is (M)(M)(M)(M)(X)MXV. Roman and decimal number conversions.
#### Time conversion
(hours, minutes, seconds, days, weeks)
4011015 seconds equals to 1 month, 2 weeks, 4 days, 10 hours, 10 minutes, 15 seconds
4011015 minutes equals to 8 years, 3 months, 1 week, 6 days, 10 hours, 15 minutes
### Codes and images of the number 4011015
Number 4011015 morse code: ....- ----- .---- .---- ----- .---- .....
Sign language for number 4011015:
Number 4011015 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
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#### Is Prime?
The number 4011015 is not a prime number.
#### Factorization and factors (dividers)
The prime factors of 4011015 are 3 * 5 * 267401
The factors of 4011015 are 1, 3, 5, 15, 267401, 802203, 1337005, 4011015.
Total factors 8.
Sum of factors 6417648 (2406633).
#### Powers
The second power of 40110152 is 16.088.241.330.225.
The third power of 40110153 is 64.530.177.299.152.429.056.
#### Roots
The square root √4011015 is 2002,751857.
The cube root of 34011015 is 158,885682.
#### Logarithms
The natural logarithm of No. ln 4011015 = loge 4011015 = 15,204555.
The logarithm to base 10 of No. log10 4011015 = 6,603254.
The Napierian logarithm of No. log1/e 4011015 = -15,204555.
### Trigonometric functions
The cosine of 4011015 is 0,656897.
The sine of 4011015 is -0,75398.
The tangent of 4011015 is -1,14779.
## Number 4011015 in Computer Science
Code typeCode value
4011015 Number of bytes3.8MB
Unix timeUnix time 4011015 is equal to Monday Feb. 16, 1970, 10:10:15 a.m. GMT
IPv4, IPv6Number 4011015 internet address in dotted format v4 0.61.52.7, v6 ::3d:3407
4011015 Decimal = 1111010011010000000111 Binary
4011015 Decimal = 21112210002010 Ternary
4011015 Decimal = 17232007 Octal
4011015 Decimal = 3D3407 Hexadecimal (0x3d3407 hex)
4011015 BASE64NDAxMTAxNQ==
4011015 MD5781ed7bc2c964acf8a02d674ea03b96d
4011015 SHA136c214b8c3744044fec2d8ea86374c8408ccffdb
4011015 SHA22452cc8a400df336b22f5d54ae491fb611255bcafb2171b56bc96377f4
4011015 SHA256b75d77a7386969c86e3c970e7648f7ac92b9a6438ef89157b03726d2338f7b76
4011015 SHA384769a2e4a979e302746564e1dee590ca6e78bf4e9242a8e36f520d755f9baac7f3cdbe4d0e9c376e8898a084677c0150b
More SHA codes related to the number 4011015 ...
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## Numerology 4011015
### Character frequency in the number 4011015
Character (importance) frequency for numerology.
Character: Frequency: 4 1 0 2 1 3 5 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 4011015, the numbers 4+0+1+1+0+1+5 = 1+2 = 3 are added and the meaning of the number 3 is sought.
## № 4,011,015 in other languages
How to say or write the number four million, eleven thousand and fifteen in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 4.011.015) cuatro millones once mil quince German: 🔊 (Nummer 4.011.015) vier Millionen elftausendfünfzehn French: 🔊 (nombre 4 011 015) quatre millions onze mille quinze Portuguese: 🔊 (número 4 011 015) quatro milhões e onze mil e quinze Hindi: 🔊 (संख्या 4 011 015) चालीस लाख, ग्यारह हज़ार, पंद्रह Chinese: 🔊 (数 4 011 015) 四百零一万一千零十五 Arabian: 🔊 (عدد 4,011,015) أربعة ملايين و أحد عشر ألفاً و خمسة عشر Czech: 🔊 (číslo 4 011 015) čtyři miliony jedenáct tisíc patnáct Korean: 🔊 (번호 4,011,015) 사백일만 천십오 Danish: 🔊 (nummer 4 011 015) fire millioner ellevetusinde og femten Hebrew: (מספר 4,011,015) ארבעה מיליון ואחד עשר אלף וחמש עשרה Dutch: 🔊 (nummer 4 011 015) vier miljoen elfduizendvijftien Japanese: 🔊 (数 4,011,015) 四百一万千十五 Indonesian: 🔊 (jumlah 4.011.015) empat juta sebelas ribu lima belas Italian: 🔊 (numero 4 011 015) quattro milioni e undicimilaquindici Norwegian: 🔊 (nummer 4 011 015) fire million elleve tusen og femten Polish: 🔊 (liczba 4 011 015) cztery miliony jedenaście tysięcy piętnaście Russian: 🔊 (номер 4 011 015) четыре миллиона одиннадцать тысяч пятнадцать Turkish: 🔊 (numara 4,011,015) dörtmilyononbirbinonbeş Thai: 🔊 (จำนวน 4 011 015) สี่ล้านหนึ่งหมื่นหนึ่งพันสิบห้า Ukrainian: 🔊 (номер 4 011 015) чотири мільйони одинадцять тисяч п'ятнадцять Vietnamese: 🔊 (con số 4.011.015) bốn triệu mười một nghìn lẻ mười lăm Other languages ...
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If you know something interesting about the number 4011015 or any other natural number (positive integer), please write to us here or on Facebook.
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The content of the comments is the opinion of the users and not of number.academy. It is not allowed to pour comments contrary to the laws, insulting, illegal or harmful to third parties. Number.academy reserves the right to remove or not publish any inappropriate comment. It also reserves the right to publish a comment on another topic. Privacy Policy. | 1,931 | 5,655 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-22 | latest | en | 0.760657 |
http://pinkpt.com/tag/puzzle/ | 1,527,018,040,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864872.17/warc/CC-MAIN-20180522190147-20180522210147-00235.warc.gz | 229,118,244 | 11,821 | ### Mini guide
1) If you are a newbie to this game, or you do not do jigsaw puzzles much, start out on the easy level, just to get used to where things are. However, as soon as you get good at it, go to the hard level. The most I ever got on easy was around 300 points. There is an upper limit of 1000 points on hard, and I go over that every time I play.
2) For hard level 1, the pieces are set up really well. The puzzle is a rectangle, with 5 columns of 3 pieces each. Going from right to left, each of the columns are right where they should be, just the rows are wrong, and all you have to do is fix them. Almost no moving around of any piece you don’t want to place is needed.
(more…)
A big thanks to benserwa for supplying us with the chart and all relevant information!
The following is a downloadable/print friendly solution chart for this uber-cool, puzzle game from Neopets. As Neopets releases it’s clues, make sure you keep up to date by ticking/crossing the relevant boxes. The chart is already filled with the first clue to start you off! The chart can be downloaded by right clicking on the image and selecting “save as”. Once it is on your computer, simply open it and print it out!
For a decent tutorial on how to solve logic puzzles, it is worthwhile to visit http://www.mysterymaster.com/guide.html which has some great articles!
Negg Sweeper is a clone of Mine Sweeper, the popular puzzle game that comes stock on Windows-based computers. Game cost is 30np per play; maximum allowed winnings per day is 3000np (as of 10/02). The object of the game is to clear the board while marking and avoiding the hidden mines. Left mouse click on a space to clear it; use shift+click or control+click to mark a space as a mine.
Negg Sweeper has three levels of difficulty to from which to choose. They are:
### EASY
Size: 9×9 (81)
Mines: 10
Ratio: 1:8.1
Starting Points: 50
Lg. Space Bonus: x1
Min. Points Poss.: 121
My Typical Score: 200
(more…)
Did you know, if you complete this puzzle quickly enough, your neopet will gain speed, and sometimes even intelligence!
Sq. 29 to Sq. 17
Sq. 26 to Sq. 24
Sq. 33 to Sq. 25
Sq. 18 to Sq. 30
Sq. 31 to Sq. 33
Sq. 33 to Sq. 25
Sq. 6 to Sq. 18
Sq. 13 to Sq. 11
Sq. 10 to Sq. 12
Sq. 27 to Sq. 13
Sq. 13 to Sq. 11
Sq. 8 to Sq. 10
Sq. 1 to Sq. 9
Sq. 16 to Sq. 4
Sq. 3 to Sq. 1
Sq. 1 to Sq. 9
Sq. 28 to Sq. 16
(more…)
Like everyone else I wanted to paint my pet, feed it rare foods and train it in training school. So I decided I was going to become a millionaire so I could buy anything I wanted. I still haven’t got to the million but I’ve made about 270k in a month.
First of all you need to think about your pets, you don’t want to be spending any points on them. Once your rich they can have anything. But for now feed them with free food. Like rice balls, Pepsi and Miranda. Feed them one every day and they wont get hungry. If they get sick make the sick pet active and go to the healing springs it may take a few tries but eventually your pet will be fully healed.
(more…) | 845 | 3,055 | {"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-2018-22 | latest | en | 0.934246 |
https://edurev.in/studytube/Worksheet-Work-and-Energy/083b9c7d-aea5-482f-bb30-cd5d7d590a26_t | 1,701,580,852,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100484.76/warc/CC-MAIN-20231203030948-20231203060948-00787.warc.gz | 273,654,222 | 58,065 | Worksheet: Work and Energy
# Worksheet: Work and Energy - Science Class 9
Q.1. Fill in the blank
(i) When a ball is thrown upwards, ________ energy is transformed into ________ energy.
(ii) The sum of the potential and kinetic energies of a body is called ________ energy.
(iii) Work is measured as a product of ________ and _________.
(iv) The electricity meter installed in our home measures electric energy in the units of ______.
(v) The work done on a body moving in a circular path is ___________.
Q.2. How are Joule (J) and ergs (erg) related?
(a) 1J = 107erg
(b) 1erg = 107J
(c) 1J = 10-7erg
(d) None
Q.3. If the force is applied at an angle θ then work done is
(a) W = FS Cos θ F = Force
(b) W = FS θ S = Distance
(c) W = FS Sin θ W = work
(d) None
Q.4. If a body is stored at a height ‘h’ then it will posses:
(a) Kinetic energy
(b) Potential energy
(c) Both
(d) None
Q.5. If the body starts from rest, then change in its kinetic energy is
(a) Positive
(b) Negative
(c) Zero
(d) May be Positive or negative depending upon the mass of the body
Q.6. When do we say that work is done?
Q.7. Define 1 J of work.
Q.8. What is the kinetic energy of an object?
Q.9. What is power?
Q.10. An object thrown at a certain angle to the ground moves in a curved path and falls back to the ground. The initial and the final points of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object?
Q.11. Illustrate the law of conservation of energy by discussing the energy changes which occur when we draw a pendulum bob to one side and allow it to oscillate. Why does the bob eventually come to rest? What happens to its energy eventually? Is it a violation of the law of conservation of energy?
Q.12. Define average power.
Q.13. In each of the following a force, F is acting on an object of mass, m. The direction of displacement is from west to east shown by the longer arrow. Observe the diagrams carefully and state whether the work done by the force is negative, positive or zero.
Q.14. Define 1 watt of power.
Q.15. A battery lights a bulb. Describe the energy changes involved in the process.
The document Worksheet: Work and Energy | Science Class 9 is a part of the Class 9 Course Science Class 9.
All you need of Class 9 at this link: Class 9
## Science Class 9
66 videos|352 docs|97 tests
## FAQs on Worksheet: Work and Energy - Science Class 9
1. What is work and energy?
Ans. Work is defined as the transfer of energy that occurs when a force is applied to an object and it causes the object to move in the direction of the force. Energy, on the other hand, is the ability to do work or cause change. It can exist in various forms such as kinetic energy, potential energy, thermal energy, etc.
2. What is the relationship between work and energy?
Ans. The relationship between work and energy is that work done on an object transfers energy to the object. When work is done on an object, it increases the object's energy. Similarly, when work is done by an object, it decreases the object's energy. The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy.
3. How is work calculated?
Ans. Work is calculated by multiplying the force applied to an object by the distance the object moves in the direction of the force. The formula for work is: work = force × distance × cosθ, where θ is the angle between the force and the direction of motion.
4. What are the different forms of energy?
Ans. There are several different forms of energy, including: - Kinetic energy: the energy of an object in motion. - Potential energy: the energy that an object possesses due to its position or state. - Thermal energy: the energy associated with the motion of particles in a substance. - Chemical energy: the energy stored in the bonds of chemical compounds. - Electrical energy: the energy associated with the flow of electric charge. - Nuclear energy: the energy released during nuclear reactions.
5. How is energy conserved in a closed system?
Ans. In a closed system, energy is conserved, which means it cannot be created or destroyed, only transferred or converted from one form to another. This is known as the law of conservation of energy. For example, if a ball is dropped from a certain height, the potential energy it has at the top is converted into kinetic energy as it falls. At the bottom, the kinetic energy is at its maximum while the potential energy is zero, but the total energy remains constant.
## Science Class 9
66 videos|352 docs|97 tests
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; | 1,225 | 4,969 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2023-50 | latest | en | 0.891655 |
https://math.stackexchange.com/questions/857306/how-to-improve-the-isometric-immersion-of-a-n-dimension-conformal-metric-of-on | 1,582,974,056,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875148850.96/warc/CC-MAIN-20200229083813-20200229113813-00083.warc.gz | 465,119,539 | 30,743 | # How to improve the isometric immersion of a $n$ dimension conformal metric of one variable conformal factor to be less than $2n-1$ dimensions?
Given a conformal metric $ds^2 = \omega(x_n)(dx_1^2+dx_2^2+\cdots + dx_{n-1}^2+ dx_n^2)$ in $\mathbb R^n$ with conformal factor of one variable $\omega(x_n)$, does there exist an isometric immersion $ds^2 = (dy_1+dy_2+ \cdots)$ of $y_j \rightarrow y_j(x_i)$ in less than $2n-1$ dimensions?
Thanks to "This is much healthier." for helping me with this question.
"This is much healthier." suggested that immersing each variable $x_j$ to be an angle in a polar coordinate system (complex variable) such that $y_j + i y_{j+n-1} = \rho_{j} e^{i x_j}$ would provide a solution to this problem. I continued with this idea and discovered a very simple solution of isometric immersing of the form
\begin{align} ds^2 &=& (dy_1^2 + dy_{1+n-1}^2) + (dy_2^2 + dy_{2+n-1}^2) +\cdots +(dy_{n-1}^2 + dy_{2n-2}^2)+ (dy_{2n-1}^2) \\ ds^2 &=& (d\rho_1^2 + \rho_1^2dx_{1}^2) + (d\rho_2^2 + \rho_2^2dx_{2}^2) +\cdots +(d\rho_{n-1}^2 + \rho_{n-1}^2dx_{n-1}^2)+ (dy_{2n-1}^2) \\ \rho &=& \rho_1 = \rho_2 = \cdots = \rho_{n-1} \\ ds^2 &=& (d\rho^2 + \rho^2dx_{1}^2) + (d\rho^2 + \rho^2dx_{2}^2) +\cdots +(d\rho^2 + \rho^2dx_{n-1}^2)+ (dy_{2n-1}^2) \\ ds^2 &=& \rho^2 \left( (dx_{1}^2) + (dx_{2}^2) + \cdots + (dx_{n-1}^2) + \left( \dfrac{(n-1)d\rho^2 + dy_{2n-1}^2}{\rho^2} \right) \right) \\ ds^2 &=& \omega(x_n)\left((dx_1^2)+(dx_2^2)+\cdots + (dx_{n-1}^2)+ (dx_n^2)\right) \end{align} where $\rho^2 = \omega(x_n)$ and
\begin{align} \dfrac{(n-1)d\rho^2 + dy_{2n-1}^2}{\rho^2} = dx_n^2 \end{align} which leads to the differential equation \begin{align} y_{2n-1}'(x_n) = \sqrt{ (\rho(x_n))^2 - (n-1) (\rho'(x_n))^2 } . \end{align} If one uses the following conformal factor $\omega(x_n) = e^{f(x_n)}$, then \begin{align} y_{2n-1}'(x_n) = e^{f(x_n)} \sqrt{ 1 - (n-1) (f'(x_n))^2 } . \end{align}
This isometric immersion requires $2n-1$ dimensions. Is there away to condense $y_j$ to fewer dimensions? | 866 | 2,026 | {"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": 4, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2020-10 | latest | en | 0.658625 |
https://everything2.com/user/Blackthorn/writeups/sharpe+ratio | 1,544,606,269,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823817.62/warc/CC-MAIN-20181212091014-20181212112514-00081.warc.gz | 591,739,713 | 5,369 | Nobel Laureate William Sharpe developed the idea of quantifying an investment's return in excess of a guaranteed investment (the 90-day U.S. Treasury-bill), relative to the risk of the investment.
To calculate the Sharpe ratio, divide the expected return of an investment that is in excess of the guaranteed investment by the standard deviation of those returns.
For example, if the guaranteed investment produced a return of 5%, and the investment under consideration is known to have a return of 10% with a standard deviation of 7.3, then ``` S = (10 - 5) / 7.3 ``` For a Sharpe Ratio of 0.6849--not a very good ratio. According to Sharpe, you would do better to invest in the 5% guaranteed investment.
As another example, if the guaranteed investment produced a return of 5%, and the investment under consideration is known to have a return of 27% with a standard deviation of 4.9, then ``` S = (27 - 5) / 4.9 ``` For a Sharpe Ratio of 4.4897--a fine investment choice (according to the Sharpe Ratio). | 246 | 1,007 | {"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} | 3.4375 | 3 | CC-MAIN-2018-51 | latest | en | 0.930978 |
http://de.metamath.org/mpeuni/iswwlk.html | 1,719,245,712,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865401.1/warc/CC-MAIN-20240624151022-20240624181022-00015.warc.gz | 7,653,721 | 6,728 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > iswwlk Structured version Visualization version GIF version
Theorem iswwlk 26211
Description: Properties of a word to represent a walk (in an undirected graph). (Contributed by Alexander van der Vekens, 15-Jul-2018.)
Assertion
Ref Expression
iswwlk ((𝑉𝑋𝐸𝑌) → (𝑊 ∈ (𝑉 WWalks 𝐸) ↔ (𝑊 ≠ ∅ ∧ 𝑊 ∈ Word 𝑉 ∧ ∀𝑖 ∈ (0..^((#‘𝑊) − 1)){(𝑊𝑖), (𝑊‘(𝑖 + 1))} ∈ ran 𝐸)))
Distinct variable groups: 𝑖,𝐸 𝑖,𝑉 𝑖,𝑊
Allowed substitution hints: 𝑋(𝑖) 𝑌(𝑖)
Proof of Theorem iswwlk
Dummy variable 𝑤 is distinct from all other variables.
StepHypRef Expression
1 wwlk 26209 . . 3 ((𝑉𝑋𝐸𝑌) → (𝑉 WWalks 𝐸) = {𝑤 ∈ Word 𝑉 ∣ (𝑤 ≠ ∅ ∧ ∀𝑖 ∈ (0..^((#‘𝑤) − 1)){(𝑤𝑖), (𝑤‘(𝑖 + 1))} ∈ ran 𝐸)})
21eleq2d 2673 . 2 ((𝑉𝑋𝐸𝑌) → (𝑊 ∈ (𝑉 WWalks 𝐸) ↔ 𝑊 ∈ {𝑤 ∈ Word 𝑉 ∣ (𝑤 ≠ ∅ ∧ ∀𝑖 ∈ (0..^((#‘𝑤) − 1)){(𝑤𝑖), (𝑤‘(𝑖 + 1))} ∈ ran 𝐸)}))
3 neeq1 2844 . . . . 5 (𝑤 = 𝑊 → (𝑤 ≠ ∅ ↔ 𝑊 ≠ ∅))
4 fveq2 6103 . . . . . . . 8 (𝑤 = 𝑊 → (#‘𝑤) = (#‘𝑊))
54oveq1d 6564 . . . . . . 7 (𝑤 = 𝑊 → ((#‘𝑤) − 1) = ((#‘𝑊) − 1))
65oveq2d 6565 . . . . . 6 (𝑤 = 𝑊 → (0..^((#‘𝑤) − 1)) = (0..^((#‘𝑊) − 1)))
7 fveq1 6102 . . . . . . . 8 (𝑤 = 𝑊 → (𝑤𝑖) = (𝑊𝑖))
8 fveq1 6102 . . . . . . . 8 (𝑤 = 𝑊 → (𝑤‘(𝑖 + 1)) = (𝑊‘(𝑖 + 1)))
97, 8preq12d 4220 . . . . . . 7 (𝑤 = 𝑊 → {(𝑤𝑖), (𝑤‘(𝑖 + 1))} = {(𝑊𝑖), (𝑊‘(𝑖 + 1))})
109eleq1d 2672 . . . . . 6 (𝑤 = 𝑊 → ({(𝑤𝑖), (𝑤‘(𝑖 + 1))} ∈ ran 𝐸 ↔ {(𝑊𝑖), (𝑊‘(𝑖 + 1))} ∈ ran 𝐸))
116, 10raleqbidv 3129 . . . . 5 (𝑤 = 𝑊 → (∀𝑖 ∈ (0..^((#‘𝑤) − 1)){(𝑤𝑖), (𝑤‘(𝑖 + 1))} ∈ ran 𝐸 ↔ ∀𝑖 ∈ (0..^((#‘𝑊) − 1)){(𝑊𝑖), (𝑊‘(𝑖 + 1))} ∈ ran 𝐸))
123, 11anbi12d 743 . . . 4 (𝑤 = 𝑊 → ((𝑤 ≠ ∅ ∧ ∀𝑖 ∈ (0..^((#‘𝑤) − 1)){(𝑤𝑖), (𝑤‘(𝑖 + 1))} ∈ ran 𝐸) ↔ (𝑊 ≠ ∅ ∧ ∀𝑖 ∈ (0..^((#‘𝑊) − 1)){(𝑊𝑖), (𝑊‘(𝑖 + 1))} ∈ ran 𝐸)))
1312elrab 3331 . . 3 (𝑊 ∈ {𝑤 ∈ Word 𝑉 ∣ (𝑤 ≠ ∅ ∧ ∀𝑖 ∈ (0..^((#‘𝑤) − 1)){(𝑤𝑖), (𝑤‘(𝑖 + 1))} ∈ ran 𝐸)} ↔ (𝑊 ∈ Word 𝑉 ∧ (𝑊 ≠ ∅ ∧ ∀𝑖 ∈ (0..^((#‘𝑊) − 1)){(𝑊𝑖), (𝑊‘(𝑖 + 1))} ∈ ran 𝐸)))
14 3anass 1035 . . 3 ((𝑊 ∈ Word 𝑉𝑊 ≠ ∅ ∧ ∀𝑖 ∈ (0..^((#‘𝑊) − 1)){(𝑊𝑖), (𝑊‘(𝑖 + 1))} ∈ ran 𝐸) ↔ (𝑊 ∈ Word 𝑉 ∧ (𝑊 ≠ ∅ ∧ ∀𝑖 ∈ (0..^((#‘𝑊) − 1)){(𝑊𝑖), (𝑊‘(𝑖 + 1))} ∈ ran 𝐸)))
15 3ancoma 1038 . . 3 ((𝑊 ∈ Word 𝑉𝑊 ≠ ∅ ∧ ∀𝑖 ∈ (0..^((#‘𝑊) − 1)){(𝑊𝑖), (𝑊‘(𝑖 + 1))} ∈ ran 𝐸) ↔ (𝑊 ≠ ∅ ∧ 𝑊 ∈ Word 𝑉 ∧ ∀𝑖 ∈ (0..^((#‘𝑊) − 1)){(𝑊𝑖), (𝑊‘(𝑖 + 1))} ∈ ran 𝐸))
1613, 14, 153bitr2i 287 . 2 (𝑊 ∈ {𝑤 ∈ Word 𝑉 ∣ (𝑤 ≠ ∅ ∧ ∀𝑖 ∈ (0..^((#‘𝑤) − 1)){(𝑤𝑖), (𝑤‘(𝑖 + 1))} ∈ ran 𝐸)} ↔ (𝑊 ≠ ∅ ∧ 𝑊 ∈ Word 𝑉 ∧ ∀𝑖 ∈ (0..^((#‘𝑊) − 1)){(𝑊𝑖), (𝑊‘(𝑖 + 1))} ∈ ran 𝐸))
172, 16syl6bb 275 1 ((𝑉𝑋𝐸𝑌) → (𝑊 ∈ (𝑉 WWalks 𝐸) ↔ (𝑊 ≠ ∅ ∧ 𝑊 ∈ Word 𝑉 ∧ ∀𝑖 ∈ (0..^((#‘𝑊) − 1)){(𝑊𝑖), (𝑊‘(𝑖 + 1))} ∈ ran 𝐸)))
Colors of variables: wff setvar class Syntax hints: → wi 4 ↔ wb 195 ∧ wa 383 ∧ w3a 1031 = wceq 1475 ∈ wcel 1977 ≠ wne 2780 ∀wral 2896 {crab 2900 ∅c0 3874 {cpr 4127 ran crn 5039 ‘cfv 5804 (class class class)co 6549 0cc0 9815 1c1 9816 + caddc 9818 − cmin 10145 ..^cfzo 12334 #chash 12979 Word cword 13146 WWalks cwwlk 26205 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 ax-6 1875 ax-7 1922 ax-8 1979 ax-9 1986 ax-10 2006 ax-11 2021 ax-12 2034 ax-13 2234 ax-ext 2590 ax-rep 4699 ax-sep 4709 ax-nul 4717 ax-pow 4769 ax-pr 4833 ax-un 6847 ax-cnex 9871 ax-resscn 9872 ax-1cn 9873 ax-icn 9874 ax-addcl 9875 ax-addrcl 9876 ax-mulcl 9877 ax-mulrcl 9878 ax-mulcom 9879 ax-addass 9880 ax-mulass 9881 ax-distr 9882 ax-i2m1 9883 ax-1ne0 9884 ax-1rid 9885 ax-rnegex 9886 ax-rrecex 9887 ax-cnre 9888 ax-pre-lttri 9889 ax-pre-lttrn 9890 ax-pre-ltadd 9891 ax-pre-mulgt0 9892 This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 df-3or 1032 df-3an 1033 df-tru 1478 df-ex 1696 df-nf 1701 df-sb 1868 df-eu 2462 df-mo 2463 df-clab 2597 df-cleq 2603 df-clel 2606 df-nfc 2740 df-ne 2782 df-nel 2783 df-ral 2901 df-rex 2902 df-reu 2903 df-rab 2905 df-v 3175 df-sbc 3403 df-csb 3500 df-dif 3543 df-un 3545 df-in 3547 df-ss 3554 df-pss 3556 df-nul 3875 df-if 4037 df-pw 4110 df-sn 4126 df-pr 4128 df-tp 4130 df-op 4132 df-uni 4373 df-int 4411 df-iun 4457 df-br 4584 df-opab 4644 df-mpt 4645 df-tr 4681 df-eprel 4949 df-id 4953 df-po 4959 df-so 4960 df-fr 4997 df-we 4999 df-xp 5044 df-rel 5045 df-cnv 5046 df-co 5047 df-dm 5048 df-rn 5049 df-res 5050 df-ima 5051 df-pred 5597 df-ord 5643 df-on 5644 df-lim 5645 df-suc 5646 df-iota 5768 df-fun 5806 df-fn 5807 df-f 5808 df-f1 5809 df-fo 5810 df-f1o 5811 df-fv 5812 df-riota 6511 df-ov 6552 df-oprab 6553 df-mpt2 6554 df-om 6958 df-1st 7059 df-2nd 7060 df-wrecs 7294 df-recs 7355 df-rdg 7393 df-1o 7447 df-er 7629 df-map 7746 df-pm 7747 df-en 7842 df-dom 7843 df-sdom 7844 df-fin 7845 df-card 8648 df-pnf 9955 df-mnf 9956 df-xr 9957 df-ltxr 9958 df-le 9959 df-sub 10147 df-neg 10148 df-nn 10898 df-n0 11170 df-z 11255 df-uz 11564 df-fz 12198 df-fzo 12335 df-hash 12980 df-word 13154 df-wwlk 26207 This theorem is referenced by: wwlknprop 26214 wwlknimp 26215 wlkiswwlk1 26218 wlkiswwlk2 26225 wwlkn0s 26233 vfwlkniswwlkn 26234 wwlknred 26251 wwlknext 26252 wwlkm1edg 26263 wwlknfi 26266 clwwlkel 26321 clwwlkf 26322 wwlksubclwwlk 26332 rusgranumwlkl1 26474 rusgra0edg 26482
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https://ask.truemaths.com/question/a-jeweller-has-bars-of-18-carat-gold-and-12-carat-gold-how-much-of-each-must-be-melted-together-to-obtain-a-bar-of-16-carat-gold-weighing-120-grams-pure-gold-24-carat/ | 1,675,160,596,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499857.57/warc/CC-MAIN-20230131091122-20230131121122-00126.warc.gz | 123,604,989 | 29,183 | • 0
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# A jeweller has bars of 18-carat gold and 12-carat gold. How much of each must be melted together to obtain a bar of 16-carat gold weighing 120 grams? (pure gold 24-carat)
• 0
This question is taken from linear equations in two variables in which following points are given.
(i)How much jeweller has bars of gold
You have to find How much each must be melted together to obtain a given bar gold of given weigh.
Kindly give me a detailed solution of this question
RS Aggarwal, Class 10, chapter 3E, question no 48
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# Chem question! - As a technician in a large.
As a technician in a large pharmaceutical research firm, you need to produceof 1.00M potassium phosphate buffer solution of pH = 7.13. The pKa of H2PO4 is 7.21. You have the following supplies: 2.00l of 1.00 KH2PO4 stock solution, 1.50 L of 1.00M KH2PO4 stock solution, and a carboy of pure distilled H2O. As a technician in a large pharmaceutical research firm, you need to produce 400. mL of 1.00 M potassium phosphate solution of pH = 7.18. The pKa of H2PO4- is 7.21. You have 2.00 L of 1.00 M KH2PO4 solution and 1.50 L of 1.00 M K2HPO4 solution, as well as a carboy of pure distilled H2O.
A. As a technician in a large pharmaceutical research firm, you need to produce 300.mL of 1.00 M potassium dihydrogen phosphate buffer solution of pH = 6.92. 2019-12-16 · As a technician in a large pharmaceutical research firm, you need to produce 400. mL of a potassium dihydrogen phosphate buffer solution ofpH = 6.84. The pKa of H2PO4? is 7.21. You have the following supplies: 2.00 L of 1.00 M KH2PO4 stock solution, 1.50 L of 1.00 M K2HPO4 stock solution, and a. 2020-01-02 · Question: PART A. As a technician in a large pharmaceutical research firm, you need to produce 250.0 mL of 1.00 mol/L potassium dihydrogen phosphate buffer solution of pH = 7.07.
As a technician in a large pharmaceutical research firm, you need to produce 400. mL of 1.00 M potassium phosphate buffer solution of pH = 7.05. The pKa of H2PO4− is 7.21. You have the following supplies: 2.00 L of 1.00 M KH2PO4 stock solution, 1.50 L of 1.00 M K2HPO4 stock solution, and a carboy of pure distilled H2O. 2014-07-23 · As a technician in a large pharmaceutical research firm, you need to produce 300.mL of 1.00 M a phosphate buffer solution of pH = 7.20. The pKa of H2PO4− is 7.21. You have 2.00 L of 1.00 M KH2PO4 solution and 1.50 L of 1.00 M K2HPO4 solution, as well as a carboy of pure distilled H2O. How much 1.00 M KH2PO4 will you need to make. As a technician in a large pharmaceutical research firm, you need to produce 400. mL of 1.00 mol L−1 potassium Get the answers you need, now!
• 2013-07-10 · As a technician in a large pharmaceutical research firm, you need to produce 100.mL of 1.00 M potassium phosphate buffer solution of pH = 6.83. The pKa of H2PO4− is 7.21. You have the following supplies: 2.00 L of 1.00 M KH2PO4 stock solution, 1.50 L of 1.00 M K2HPO4 stock solution, and a carboy of pure distilled H2O. How much 1.00.
• 2012-04-21 · As a technician in a large pharmaceutical research firm, you need to produce 200. mL of 1.00 M a phosphate buffer solution of pH = 7.33. The pKa of H2PO4- is 7.21. You have 2.00 L of 1.00 M KH2PO4 solution and 1.50 L of 1.00 M K2HPO4 solution, as well as a carboy of pure distilled H2O. How much 1.00 M KH2PO4 will you need to make.
1. As a technician in a large pharmaceutical research firm, you need to produce 450.mL of 1.00 M potassium dihydrogen phosphate buffer solution of pH = 6.89.
2. 2016-12-02 · As a technician in a large pharmaceutical research firm,you need to produce 250mL of 1.00molL potassium phosphate buffer solution of pH=6.99?
2011-11-04 · As a technician in a large pharmaceutical research firm, you need to produce 400 ml of 1.00 M potassium phosphate solution of PH = 7.20. The pka of H2PO4- is 7.21. You have 2.00 L of 1.00 M KH2PO4 solution and 1.50 L of 1.00 M K2HPO4 solution, as well as a carboy of pure distilled H2O. How much 1.00 M K2HPO4 will you need to make. As a technician in a large pharmaceutical research firm, you need to produce 450. mL of 1.00 M potassium phosphate buffer solution of pH = 6.97. The pKa of H2PO4− is 7.21. You have the following supplies: 2.00 L of 1.00 M KH2PO4 stock solution, 1.50 L of 1.00 M K2HPO4 stock solution, and a carboy of pure distilled H2O. As a technician in a large pharmaceutical research firm, you need to produce 450. mL of 1.00 mol L−1 potassium phosphate buffer solution of pH = 6.86. The pKa of H2PO4− is 7.21. You have the following supplies: 2.00 L of 1.00 mol L−1 KH2PO4 stock solution, 1.50 L of 1.00 mol L−1 K2HPO4 stock solution, and a carboy of pure distilled H2O.
## A. As A Technician In A Large Pharmaceutical.
As a technician in a large pharmaceutical research firm, you need to produce 450. mL of 1.00 M a phosphate buffer solution of pH = 7.20. The pKa of H2PO4− is 7.21. You have 2.00 L of 1.00 M KH2PO4 solution and 1.50 L of 1.00 M K2HPO4 solution, as well as a carboy of pure distilled H2O. As a technician in a large pharmaceutical research firm, you need to produce 250. mL of a potassium dihydrogen phosphate buffer solution of pH = 7.05. The pKa of H2PO4− is 7.21. You have the following supplies: 2.00 L of 1.00 M KH2PO4 stock solution, 1.50 L of 1.00 M K2HPO4 stock solution, and a carboy of pure distilled H2O.
As a technician in a large pharmaceutical research firm, you need to produce 200.mL of 1.00 M a phosphate buffer solution of pH = 7.19. The pKa of H2PO4− is 7.21. You have 2.00 L of 1.00 M KH2PO4 solution and 1.50 L of 1.00 M K2HPO4 solution, as well as a carboy of pure distilled H2O. As a technician in a large pharmaceutical research firm, you need to produce 200 mL of 1.00 M a phosphate buffer solution of pH = 7.06. The pKa of H_2PO_4^- is 7.21. You have 2.00 L of 1.00 M KH_2PO_4 solution and 1.50 L of 1.00 M K_2HPO_4 solution, as well as a carboy of pure distilled H_2O. 2012-03-04 · As a technician in a large pharmaceutical research firm, you need to produce 300. mL of 1.00 M potassium phosp? Answer to As a technician in a large pharmaceutical research firm, you need to produce 200. of 1.00 potassium phosphate solution of = 7. The of is 7. You have.
### What is a Laboratory Technician? What do they do.
As a technician in a large pharmaceutical research firm, you need to produce 300. mL of 1.00 M potassium phosphate solution of = 7.06. The pKa of H2PO4- is 7.21. Their work is almost always laboratory-based and while they might be working alone on a specific task, they are generally working within a larger laboratory team. Within the pharmaceutical industry, Laboratory Technicians can be employed in research and development or in. Pharmaceutical Technician or Pharmaceutical Research Technician or Drug Technician is a job title for a laboratory assistant or research assistant employed in the pharmaceutical industry under the direct supervision of a physician, veterinarian, or scientist involved in the research and development of new or existing medications. Pharmacy technicians in Nigeria make up 75% of pharmaceutical work force and are looking for their council pharmacy technician and technologist council of Nigeria reason being the pharmacist council of Nigeria PCN refuses to allocate responsibilities that will give them right to practice at community level interdependently. Duties. Pharmaceutical researchers can conduct research in a wide array of fields. They can use their knowledge of science to develop new chemicals, study the bio-mechanical processes of the human body, study plants or animals for insights into how the natural world functions or even study the physical reactions of processes at the atomic level.
In addition, the overall grade is also based on continuous assessments encompassing GMP and GLP compliance, pharmaceutical documentation skills and interim quizzes. The Pharmaceutical Research & Development Technology Diploma is awarded upon successful completion of all three modules. 1,754 Pharmaceutical Lab Technician jobs available on. Apply to Laboratory Technician, Pharmacy Technician, Quality Control Lab Technician and more! 2019-09-19 · Advances in pharmaceutical research will allow for more prescription medications to be used to fight diseases. In addition, pharmacy technicians will be needed to take on a greater role in pharmacy operations because pharmacists are increasingly performing more patient care activities, such as giving flu shots. They may also perform administrative duties in pharmaceutical practice, such as reviewing prescription requests with medic's offices and insurance companies to ensure correct medications are provided and payment is received. A Pharmacy Technician in the UK has. | 2,328 | 8,257 | {"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} | 2.859375 | 3 | CC-MAIN-2020-45 | latest | en | 0.892256 |
https://www.physicsforums.com/threads/mean-value-theorem-problem.265764/ | 1,511,392,379,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806676.57/warc/CC-MAIN-20171122213945-20171122233945-00631.warc.gz | 854,088,873 | 14,731 | # Mean Value Theorem Problem
1. Oct 21, 2008
### r34racer01
Use the mean value theorem to show that (abs. value of tan^-1 a) < (abs. value a) for all a not equal to 0. And use this inequality to find all solutions of the equation tan^-1 x = x.
I have no idea how to do this.
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
2. Oct 21, 2008
### Staff: Mentor
Start with what the mean value theorem says, and go from there.
3. Oct 21, 2008
### r34racer01
Well MVT is, if f is cont. on [a,b] and differentiable on (a,b). Then there exists a number c E (a,b) such that: f '(c) = f(b) - f(a)/b-a
But I don't get how to apply that here.
4. Oct 21, 2008
### HallsofIvy
Staff Emeritus
Well, taking f(x)= tan-1[sup(x) would be a start. What is the derivative of tan-1(x)? | 271 | 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} | 3.1875 | 3 | CC-MAIN-2017-47 | longest | en | 0.883881 |
https://www.manualslib.com/manual/1120140/Mitsubishi-Electric-Fr-E700.html?page=301 | 1,568,606,230,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572484.20/warc/CC-MAIN-20190916035549-20190916061549-00045.warc.gz | 943,713,172 | 42,219 | # Measurement Of Powers; Measurement Of Voltages And Use Of Pt - Mitsubishi Electric FR-E700 Instruction Manual
Fr-e740-016 to 300 - ec; fr-e720s-008 to 110 - ec.
Measurement of main circuit voltages, currents and powers
6.2.1
## Measurement of powers
Using an electro-dynamometer type meter, measure the power in both the input and output sides of the inverter using the two-
or three-wattmeter method. As the current is liable to be imbalanced especially in the input side, it is recommended to use the
three-wattmeter method.
Examples of process value differences produced by different measuring meters are shown below.
An error will be produced by difference between measuring instruments, e.g. power calculation type and two- or three-
wattmeter type three-phase wattmeter. When a CT is used in the current measuring side or when the meter contains a PT on
the voltage measurement side, an error will also be produced due to the frequency characteristics of the CT and PT.
[Measurement conditions]
Constant-torque (100%) load, note that 60Hz or
more should be constantly output 3.7kW, 4-pole
motor, value indicated in 3-wattmeter method is 100%.
%
120
100
3-wattmeter method (Electro-dynamometer type)
80
2-wattmeter method (Electro-dynamometer type)
Clip AC power meter
60
Clamp-on wattmeter
(Hall device power arithmetic type)
0
20
40
60
80 100 120Hz
Example of
Measuring Inverter Input Power
6.2.2
### Measurement of voltages and use of PT
(1)
Inverter input side
As the input side voltage has a sine wave and it is extremely small in distortion, accurate measurement can be made with an
ordinary AC meter.
(2)
Inverter output side
Since the output side voltage has a PWM-controlled rectangular wave, always use a rectifier type voltmeter. A needle type
tester can not be used to measure the output side voltage as it indicates a value much greater than the actual value. A
moving-iron type meter indicates an effective value which includes harmonics and therefore the value is larger than that of the
fundamental wave. The value monitored on the operation panel is the inverter-controlled voltage itself. Hence, that value is
accurate and it is recommended to monitor values using the operation panel.
(3)
PT
No PT can be used in the output side of the inverter. Use a direct-reading meter. (A PT can be used in the input side of the
inverter.)
298
[Measurement conditions]
Constant-torque (100%) load, note that 60Hz or
more should be constantly output 3.7kW, 4-pole
motor, value indicated in 3-wattmeter method is 100%.
%
120
100
3-wattmeter method (Electro-dynamometer type)
80
2-wattmeter method (Electro-dynamometer type)
Clip AC power meter
60
Clamp-on wattmeter
(Hall device power arithmetic type)
0
20
40
60
80 100 120Hz
Example of
Measuring Inverter Output Power | 693 | 2,786 | {"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} | 2.78125 | 3 | CC-MAIN-2019-39 | latest | en | 0.820725 |
https://math.stackexchange.com/questions/1424777/is-the-category-of-banach-spaces-and-bounded-linear-maps-accessible | 1,558,511,195,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256764.75/warc/CC-MAIN-20190522063112-20190522085112-00372.warc.gz | 564,710,541 | 34,802 | # Is the category of Banach spaces and bounded linear maps accessible?
It's well-known that the category $$\mathsf{Ban}_c$$ of Banach spaces and linear contractions (i.e. of norm $$\leq 1$$) is $$\omega_1$$-accessible. It is also (co)complete, and hence locally $$\omega_1$$-presentable. This is typically proven in any reference book on accessible categories.
The larger category $$\mathsf{Ban}$$ of Banach spaces and all bounded linear maps lacks all infinite coproducts (unless all but finitely many factors are 0 -- see here for a proof by Martin Brandenburg that not all coproducts exist, reflecting on the proof yields this conclusion). So $$\mathsf{Ban}$$ is not locally presentable. But it might still be $$\omega_1$$-accessible.
One piece of evidence for accessibility is that $$\ell_1(\omega)$$ is a dense generator in $$\mathsf{Ban}$$ just as it is in $$\mathsf{Ban}_c$$: $$\ell_1(\omega)$$ corepresents sequences $$(x_i)_i$$ such that $$\sum \|x_1\|$$ converges, and it has split subobjects corepresenting addition and scalar multiplication, so any natural transformation $$\mathsf{Ban}(i_{\ell_1(\omega)}-,X) \implies \mathsf{Ban}(i_{\ell_1(\omega)}-,Y)$$ [where $$i_{\ell_1(\omega)}$$ is the inclusion functor for the full subcategory on $$\ell_1(\omega)$$] must come from a linear map preserving absolutely convergent sequences, which is true if and only if the map is bounded.
But I haven't been able to decide whether or not $$\mathsf{Ban}$$ has $$\omega_1$$-filtered colimits. I believe I have convinced myself that the inclusion $$\mathsf{Ban}_c \to \mathsf{Ban}$$ preserves $$\omega_1$$-filtered colimits, because any cocone on an $$\omega_1$$-filtered diagram of contractions has bounded norms on its legs, and so we can divide through by a constant to obtain an "equivalent" cocone of contractions. It would be nice to replace an $$\omega_1$$-filtered diagram in $$\mathsf{Ban}$$ with one in $$\mathsf{Ban}_c$$, but I'm not sure if this is possible in general.
If, optimistically, $$\omega_1$$-filtered colimits do exist, the next step is to determine the $$\omega_1$$-presentable objects and whether everything is an $$\omega_1$$-filtered colimit thereof, but this is even more obscure to me.
• Re your final paragraph, the obvious guess is that $\omega_1$-presentable objects are separable Banach spaces and every Banach space is the colimit of its separable subspaces. – Eric Wofsey Sep 7 '15 at 5:04
• Okay -- the closed separable subspaces of a Banach space form an $\omega_1$-filtered diagram in $\mathsf{Ban}_c$ whose colimit can be computed as in $\mathsf{Set}$ to be the original space. And I suppose the presentability question could at least be explored in $\mathsf{Ban}_c$ to start with. – tcamps Sep 7 '15 at 7:04
Okay, the answer is yes, $\mathsf{Ban}$ is $\omega_1$-accessible. Below I argue that $\mathsf{Ban}$ has $\omega_1$-filtered colimits, computed by modifying diagrams to lie in $\mathsf{Ban}_c$ and using that $\mathsf{Ban}_c \to \mathsf{Ban}$ preserves $\omega_1$-filtered colimits. The replacement is such that any functor $\mathsf{Ban} \to \mathcal{C}$ preserves $\omega_1$-filtered colimits if and only if its restriction to $\mathsf{Ban}_c$ does. In particular, the $\omega_1$-presentable objects of $\mathsf{Ban}$ coincide with those of $\mathsf{Ban}_c$. Every object in $\mathsf{Ban}$ is an $\omega_1$-filtered colimit of these because it is so in $\mathsf{Ban}_c$ and the colimits are preserved. Hence $\mathsf{Ban}$ is $\omega_1$-accessible, and $\mathsf{Ban}_c \to \mathsf{Ban}$ is a (non-full) $\omega_1$-accessible embedding.
This begs the question: what is a good sketch or axiomatization for $\mathsf{Ban}$ as a category of models and homomorphisms? (Okay, using $\mathsf{Ban}(\ell_1(\omega),-)$ to give an underlying set, we can presumably do this by axiomatizing a set equipped with a set of countable subsets to be interpreted as the sets $(x_i)_{i \in \mathbb{N}}$ such that "$\sum_i \|x_i\|$ is finite").
Also, this method does not explicitly identify the $\omega_1$-presentable objects in $\mathsf{Ban}$ (equivalently, in $\mathsf{Ban}_c$). Since $\ell_1(\omega)$ is a $\omega_1$-presentable strong generator, they are given by countable colimits of $\ell_1(\omega)$. This probably results in exactly the separable Banach spaces as Eric Wofsey suggests, but I haven't checked.
For completeness, the argument that $\mathsf{Ban}_c \to \mathsf{Ban}$ preserves $\omega_1$-filtered colimits goes as follows. If $X: I \to \mathsf{Ban}_c$ is an $\omega_1$-filtered diagram, and $f_i: X_i \to Y$ is a cocone in $\mathsf{Ban}$, then I claim that the norms of the legs of $f$ are bounded. Otherwise, we could choose $(i_n)_{n \in \mathbb{N}}$ and $x_n \in X_{i_n}$ with $\|x_n\| \leq 2^{-n}$ but $\|f_{i_n}(x_n)\| \geq 3^n$. By $\omega_1$-filteredness, we can find an $\alpha \in I$ with $\alpha\geq i_n$ for each $n$. Then $\sum X_{i_n\alpha}(x_n)$ converges absolutely, but under $f_\alpha$ it is sent to $\sum_n f_{i_n}(x_n)$ which must diverge, contradicting the boundedness of $f_\alpha$.
So there is $B$ such that $\|f_i\| \leq B$ for all $i \in I$. Then $f/B$ is a cocone in $\mathsf{Ban}_c$. So, letting $(\varinjlim X,(\eta_i)_{i \in I})$ denote the colimit cocone in $\mathsf{Ban}_c$, there is a unique cocone map $\bar f: (\varinjlim X (\eta_i)_{i \in I}) \to (Y,f/B)$ in $\mathsf{Ban}_c$, and $Bf$ is the unique cocone map $(\varinjlim X (\eta_i)_{i \in I}) \to (Y,f)$ in $\mathsf{Ban}$, and $\varinjlim X$ is the colimit
Here's how to reduce $\omega_1$-filtered colimits in $\mathsf{Ban}$ to those in $\mathsf{Ban}_c$. It suffices to treat $\omega_1$-directed colimits (since every $\lambda$-filtered diagram has a cofinal $\lambda$-directed subdiagram). If $X: I \to \mathsf{Ban}$ is such a diagram, first observe that $K_i := \{x \in X_i \mid \exists j \geq i \, X_{ij}(x) = 0\}$ forms a closed subspace of $X_i$ (this uses $\omega_1$-filteredness of $I$). So we can let $X_i' = X_i/K_i$, and then $X$ descends to a functor $X'$ in the obvious way, with an isomorphic category of cocones. Now $X'$ has the property that
(*) For any $x \in X_i'$ and $j \geq i$, $X_{ij}'(x) = 0$ iff $x=0$.
Pick $X_{i_0} \neq 0$ (otherwise the colimit is just 0), let $I_{\geq i_0} = \{i \in I \mid i \geq i_0\}$, which is cofinal by filteredness of $I$. Then let $X'': I_{\geq i_0} \to \mathsf{Ban}$ be the restriction of $X'$ to $I_{\geq i_0}$. Now we let $X'''_i=X''_i$ but $X'''_{ij} = \frac{\|X''_{i_0i}\|}{\|X''_{i_0j}\|}X''_{ij}$, using (*) to ensure that the denominators are nonzero (this is the only step that requires the indexing category to be a poset, we could have waited to pass to a cofinal poset a this point). Then $X'''\cong X''$, but lies in $\mathsf{Ban}_c$; we can take its colimit there; the colimit is preserved by the inclusion $\mathsf{Ban}_c \to \mathsf{Ban}$, and $\varinjlim X''' = \varinjlim X'' = \varinjlim X' = \varinjlim X$ so we have our desired colimit.
Oh, and here's an argument that $\ell_1(\omega)$ is $\omega_1$-presentable. $\omega_1$-filtered colimits in $\mathsf{Ban}_c$ are computed as in $\mathsf{Vect}$, the norm being maximal such that structure maps are contractions. As discussed in the question, $\ell_1(\omega)$ corepresents sequences whose norms sum to $\leq 1$. If $(x_n)_{n \in \mathbb{N}}$ is a sequence in $\in \varinjlim X$ with $\sum_n \|x_n\|_{\varinjlim X} = a \leq 1$, then by $\omega_1$-filteredness, we can find an $X_i$ with all the $x_n \in X_i$. Moreover, we can find a sequence $(X_{i_k})_{k \in \mathbb{N}}$ with $\sum_n \|x_n\|_{X_{i_k}} \to a$ as $k \to \infty$. Then by $\omega_1$-filteredness again, we can find $X_j$ with $j\geq i_k$ for all $k$, and it follows that $\sum_n \|x_n\|_{X_j} = a \leq 1$. So the map $\varinjlim \mathsf{Ban}_c(\ell_1(\omega),X) \to \mathsf{Ban}_c(\ell_1(\omega),\varinjlim X)$ is surjective. It is also injective: If $(x_n)_{n \in \mathbb{N}} = (0)_{n \in \mathbb{N}}$ in $\varinjlim X$, then each $x_n = 0$ already at some stage $i_n$, so by $\omega_1$-filteredness, each $x_n = 0$ in some $X_j$ with $j\geq i_n$ for all $n$. So $x_n = 0$ in $\varinjlim \mathsf{Ban}_c(\ell_1(\omega),X)$ | 2,672 | 8,161 | {"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": 27, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2019-22 | latest | en | 0.849754 |
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This tutorial demonstrates how to use the differential search feature of the Allen Mouse Brain Atlas to find gene markers for different regions of the brain and to visualize this gene expression in three-dimensional space. Differential search is also available for the Allen Developing Mouse Brain Atlas and the Allen Human Brain Atlas.
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Speaker: : Carl Petersen | 716 | 3,071 | {"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} | 2.84375 | 3 | CC-MAIN-2020-10 | latest | en | 0.709446 |
https://www.theburningofrome.com/users-questions/what-is-moment-of-inertia-of-circular-section-formula/ | 1,723,571,775,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641082193.83/warc/CC-MAIN-20240813172835-20240813202835-00813.warc.gz | 767,398,225 | 39,626 | What is moment of inertia of circular section formula?
What is moment of inertia of circular section formula?
Moment Of Inertia Of A Circle Here, R is the radius and the axis is passing through the centre. This equation is equivalent to I = π D4 / 64 when we express it taking the diameter (D) of the circle.
What is the formula of circular section?
Area of circle formula = π × r2. The area of a circle is π multiplied by the square of the radius. The area of a circle when the radius ‘r’ is given is πr2.
What is the unit of moment of inertia?
The unit of moment of inertia is a composite unit of measure. In the International System (SI), m is expressed in kilograms and r in metres, with I (moment of inertia) having the dimension kilogram-metre square.
What is moment of inertia in simple terms?
Moment of inertia, in physics, quantitative measure of the rotational inertia of a body—i.e., the opposition that the body exhibits to having its speed of rotation about an axis altered by the application of a torque (turning force). The axis may be internal or external and may or may not be fixed.
What is Ixx moment of inertia?
The moment of inertia is also known as the Second Moment of the Area and is expressed mathematically as: Ixx = Sum (A)(y2) In which: Ixx = the moment of inertia around the x axis. A = the area of the plane of the object.
What is pure bending equation?
Pure bending ( Theory of simple bending) is a condition of stress where a bending moment is applied to a beam without the simultaneous presence of axial, shear, or torsional forces. Pure bending occurs only under a constant bending moment (M) since the shear force (V), which is equal to. , has to be equal to zero.
How to calculate the moment of inertia of a filled sector?
The formula calculates the Moment of Inertia of a filled circular sector or a sector of a disc of angle θ and radius r with respect to an axis going through the centroid of the sector and the center of the circle. The formula is valid for 0 ≤ θ ≤ π. Related formulas.
How to calculate the moment of area of a circle sector?
Using the structural engineering calculator located at the top of the page (simply click on the the “show/hide calculator” button) the following properties can be calculated: Calculate the Second Moment of Area (or moment of inertia) of a Circle Sector
How is the moment of inertia of a circle determined?
Moment Of Inertia Of A Circle Moment of inertia of a circle or the second-moment area of a circle is usually determined using the following expression; I = π R 4 / 4 Here, R is the radius and the axis is passing through the centre.
How to calculate the inertia of a cross section?
The Area Moment of Inertia for a solid cylindrical section can be calculated as. I x = π r 4 / 4 = π d 4 / 64 (4) where . r = radius. d = diameter . I y = π r 4 / 4 = π d 4 / 64 (4b) Hollow Cylindrical Cross Section. The Area Moment of Inertia for a hollow cylindrical section can be calculated as | 687 | 2,987 | {"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-2024-33 | latest | en | 0.91196 |
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Thus far we have studied as to how the data of heterogeneous nature can be grouped together and be referenced as a single unit of structure. Now we come to the next step in our real world problem. Let’s consider the example of students and their marks. In this case, to avoid declaring various data variables, we grouped together all the data concerning the student’s marks as one unit and call it student. The problem that arises now is that the data related to students is not going to be of a single student only. We will be required to store data for a number of students. To solve this situation one way is to declare a structure and then create sufficient number of variables of that structure type. But it gets very cumbersome to manage such a large number of data variables, so a better option is to declare an array.
So, revising the array for a few moments we would refresh the fact that an array is simply a collection of homogeneous data types. Hence, if we make a declaration as:
int temp[20];
It simply means that temp is an array of twenty elements where each element is of type integer, indicating homogenous data type. Now in the same manner, to extend the concept a bit further to the structure variables, we would say,
struct student stud[20] ;
It means that stud is an array of twenty elements where each element is of the type struct student (which is a user-defined data type we had defined earlier). The various members of the stud array can be accessed in the similar manner as that of any other ordinary array.
For example,
struct student stud[20], we can access the roll_no of this array as
stud[0].roll_no;
stud[1].roll_no;
stud[2].roll_no;
stud[3].roll_no;
stud[19].roll_no;
Please remember the fact that for an array of twenty elements the subscripts of the array will be ranging from 0 to 19 (a total of twenty elements). So let us now start by seeing how we will write a simple program using array of structures.
Write a program to read and display data for 20 students.
```/*Program to read and print the data for 20 students*/
#include <stdio.h>
struct student { int roll_no;
char name[20];
char course[20];
int marks_obtained ;
};
main( )
{
struct student stud [20];
int i;
printf (“Enter the student data one by one\n”);
for(i=0; i<=19; i++)
{
printf (“Enter the roll number of %d student”,i+1);
scanf (“%d”,&stud[i].roll_no);
printf (“Enter the name of %d student”,i+1);
scanf (“%s”,stud[i].name);
printf (“Enter the course of %d student”,i+1);
scanf (“%d”,stud[i].course);
printf (“Enter the marks obtained of %d student”,i+1);
scanf (“%d”,&stud[i].marks_obtained);
}
printf (“the data entered is as follows\n”);
for (i=0;i<=19;i++)
{
printf (“The roll number of %d student is %d\n”,i+1,stud[i].roll_no);
printf (“The name of %d student is %s\n”,i+1,stud[i].name);
printf (“The course of %d student is %s\n”,i+1,stud[i].course);
printf (“The marks of %d student is %d\n”,i+1,stud[i].marks_obtained);
}
}```
The above program explains to us clearly that the array of structure behaves as any other normal array of any data type. Just by making use of the subscript we can access all the elements of the structure individually.
Extending the above concept where we can have arrays as the members of the structure. For example, let’s see the above example where we have taken a structure for the student record. Hence in this case it is a real world requirement that each student will be having marks of more than one subject. Hence one way to declare the structure, if we consider that each student has 3 subjects, will be as follows:
struct student {
int roll_no;
char name [20];
char course [20];
int subject1 ;
int subject2;
int subject3;
};
The above described method is rather a bit cumbersome, so to make it more efficient we can have an array inside the structure, that is, we have an array as the member of the structure.
struct student
{
int roll_no;
char name [20];
char course [20];
int subject [3] ;
};
Hence to access the various elements of this array we can the program logic as follows:
```/*Program to read and print data related to five students having marks of three subjects each using the concept of arrays */
#include<stdio.h>
struct student {
int roll_no;
char name [20];
char course [20];
int subject [3] ;
};
main( )
{
struct student stud[5];
int i,j;
printf (“Enter the data for all the students:\n”);
for (i=0;i<=4;i++)
{
printf (“Enter the roll number of %d student”,i+1);
scanf (“%d”,&stud[i].roll_no);
printf(“Enter the name of %d student”,i+1);
scanf (“%s”,stud[i].name);
printf (“Enter the course of %d student”,i+1);
scanf (“%s”,stud[i].course);
for (j=0;j<=2;j++)
{
printf (“Enter the marks of the %d subject of the student %d:\n”,j+1,i+1);
scanf (“%d”,&stud[i].subject[j]);
}
}
printf (“The data you have entered is as follows:\n”);
for (i=0;i<=4;i++)
{
printf (“The %d th student's roll number is %d\n”,i+1,stud[i].roll_no);
printf (“The %d the student's name is %s\n”,i+1,stud[i].name);
printf (“The %d the student's course is %s\n”,i+1,stud[i].course);
for (j=0;j<=2;j++)
{
printf (“The %d the student's marks of %d I subject are %d\n”,i+1, j+1, stud[i].subject[j]);
}
}
printf (“End of the program\n”);
}```
Hence as described in the example above, the array as well as the arrays of structures can be used with efficiency to resolve the major hurdles faced in the real world programming environment.
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https://edurev.in/course/quiz/attempt/-1_Test-S-R-Latch/03af163f-d36c-4606-8f5a-d2214adf551e | 1,686,128,825,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653631.71/warc/CC-MAIN-20230607074914-20230607104914-00707.warc.gz | 263,345,050 | 42,958 | Test: S-R Latch
# Test: S-R Latch
Test Description
## 15 Questions MCQ Test Digital Electronics | Test: S-R Latch
Test: S-R Latch for Electrical Engineering (EE) 2023 is part of Digital Electronics preparation. The Test: S-R Latch questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: S-R Latch MCQs are made for Electrical Engineering (EE) 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: S-R Latch below.
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Test: S-R Latch - Question 1
### When both the inputs of a latch are high, the output is unpredictable. What is this condition called?
Detailed Solution for Test: S-R Latch - Question 1
SR Latch:
The truth table for SR latch is:
From the above table, when both the inputs of a latch are high, The out is indeterminate.
Test: S-R Latch - Question 2
### Which property is NOT considered in latches?
Detailed Solution for Test: S-R Latch - Question 2
(i) Latches are level-triggered (outputs can change as soon as the inputs changes)
(ii) Flip-Flop is edge-triggered (only changes state when a control signal goes from high to low or low to high).
(iii) Edge triggering is a type of triggering that allows a circuit to become active at the positive edge or the negative edge of the clock signal.
(iv) Level triggering is a type of triggering that allows a circuit to become active when the clock pulse is on a particular level.
Test: S-R Latch - Question 3
### Which of the following can be used for debouncing a switch ?
Detailed Solution for Test: S-R Latch - Question 3
• Switch bounce or contact bounce or even called chatter is a common problem associated with mechanical switches and relays.
• Switch bouncing is not a major problem when we deal with the power circuits, but it causes problems while we are dealing with the logic and digital circuits.
• Hence, to remove the bouncing from the circuit Switch Debouncing Circuit is used.
• The hardware debouncing technique uses an S-R latch to avoid bounces in the circuit along with the pull-up resistors.
• S-R circuit is the most effective of all debouncing approaches
• The figure below is a simple debouncing circuit that is often used.
SR Latch:
• In an S-R latch, activation of the S input sets the circuit, while activation of the R input resets the circuit.
• If both S and R inputs are activated simultaneously, the circuit will be in an invalid condition.
Application:
• Latches are used to keep the conditions of the bits to encode binary numbers.
• Latches are single-bit storage elements that are widely used in computing as well as data storage.
• Latches are used in the circuits like power gating & clock as a storage device.
Test: S-R Latch - Question 4
______ is commonly used to interface output devices.
Detailed Solution for Test: S-R Latch - Question 4
Latches are memory devices and can store one bit of data for as long as the device is powered.
As the name suggests, latches are used to "latch onto" information and hold it in place.
Latches are very similar to flip-flops, but are not synchronous devices, and do not operate on clock edges as flip-flops do.
The reason for using the latch in an output port is simple, we do not want to lose the result of any operation.
So, in order to not lose it, we use a latch, so that it holds the information as long as new information is overwritten onto it.
An 8-bit latch can be used to interface the output of a microprocessor to other devices.
The 74LS373 octal latch and the 74LS374 octal D flip-flop are popular microprocessor interface chips.
Note:
Tristate buffer is commonly used to interface input devices.
Test: S-R Latch - Question 5
The two inputs A and B are connected to a NOR based R-S latch, via two AND gates as shown in the figure. If A = 1 and B = 0, the output QQ̅ is
Detailed Solution for Test: S-R Latch - Question 5
From the given diagram,
S = AQ̅, R = QB
Given that, A = 1, B = 0
S = Q̅, R = 0
The truth table of the S-R latch is:
Let Q = 0,
S = Q̅ = 1, R = 0 ⇒ Qn + 1 = 1
Let Q = 1,
S = Q̅ = 0, R = 0 ⇒ Qn + 1 = 1
In both the cases, Qn + 1 n + 1 = 10
Test: S-R Latch - Question 6
In the latch circuit shown, the NAND gates have non-zero, but unequal propagation delays. The present input conditions is: P = Q = ‘0’. If the input conditions is changed simultaneously to P = Q = ‘1’, the outputs X and Y are
Detailed Solution for Test: S-R Latch - Question 6
Let as assume tpd 1 < tpd 2
x changes state first then y changes
1st output of X (P = 1, y = 0) ⇒ X1 = 1
Next output of Y (Q = 1, X1 = 1) ⇒ Y1 = 0
2nd output of X (P = 1, y1 = 0) ⇒ 1
Hence output x = 1 y = 0 (if tpd1 < tpd2)
& Output X = 0 Y = 1 (if tpd2 < tpd1)
Test: S-R Latch - Question 7
An SR latch is implemented using TTL gates as shown in the figure. The set and reset pulse inputs are provided using the push-button switches. It is observed that the circuit fails to work as desired. The SR latch can be made functional by changing
Detailed Solution for Test: S-R Latch - Question 7
Concept:
• In the TTL logic gate, open-end or floating end are considered as logic-1.
• For the circuit to work as an S & R latch, S & R should act as logic-0 as well as logic- 1 on requirement.
Analysis:
• If we connect set which is equal to 5 volts, then it will be considered as logic-1.
• If we do not connect 5 Volts to set switch i.e. if we make set switch ‘open’ then it will again be considered as a logic -1, because it is a TTL gate.
• So we have to replace 5 V supply with 0 Volts by connecting it to ground. So logic-0 is also possible for switch ‘set’ & ‘reset’.
Hence for the above circuit to work as an SR latch, 5-volt battery should connect to ground.
Hence option -4 is Correct.
Test: S-R Latch - Question 8
Which of the following is true?
Detailed Solution for Test: S-R Latch - Question 8
The difference between latches and flip flops is shown
Test: S-R Latch - Question 9
In S-R latch, when the SET input is made high, output Q becomes:
Detailed Solution for Test: S-R Latch - Question 9
An unclocked R-S flip flop using NOR gates is as shown:
The truth table for the circuit is shown:
Test: S-R Latch - Question 10
The first step of the analysis procedure of SR latch is to ___________
Detailed Solution for Test: S-R Latch - Question 10
All flip flops have at least one output labeled Q (i.e. inverted). This is so because the flip flops have inverting gates inside them, hence in order to have both Q and Q complement available, we have atleast one output labelled.
Test: S-R Latch - Question 11
When both inputs of SR latches are high, the latch goes ___________
Detailed Solution for Test: S-R Latch - Question 11
When both gates are identical and this is “metastable”, and the device will be in an undefined state for an indefinite period.
Test: S-R Latch - Question 12
When a high is applied to the Set line of an SR latch, then ___________
Detailed Solution for Test: S-R Latch - Question 12
S input of an SR latch is directly connected to the output Q. So when a high is applied Q output goes high and Q’ low.
Test: S-R Latch - Question 13
The full form of SR is ___________
Detailed Solution for Test: S-R Latch - Question 13
The full form of SR is set/reset. It is a type of latch having two stable states.
Test: S-R Latch - Question 14
The outputs of SR latch are ___________
Detailed Solution for Test: S-R Latch - Question 14
SR or Set-Reset latch is the simplest type of bistable multivibrator having two stable states. The inputs of SR latch are s and r while outputs are q and q’. It is clear from the diagram:
.
Test: S-R Latch - Question 15
When both inputs of SR latches are low, the latch ___________
Detailed Solution for Test: S-R Latch - Question 15
When both inputs of SR latches are low, the latch remains in it’s present state. There is no change in output.
## Digital Electronics
88 videos|71 docs|58 tests
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In this test you can find the Exam questions for Test: S-R Latch solved & explained in the simplest way possible. Besides giving Questions and answers for Test: S-R Latch, EduRev gives you an ample number of Online tests for practice
## Digital Electronics
88 videos|71 docs|58 tests | 2,294 | 8,933 | {"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-2023-23 | latest | en | 0.88845 |
https://newtonexcelbach.com/2014/06/19/reinforced-concrete-uls-capacity-under-combined-axial-load-and-biaxial-bending/ | 1,679,470,159,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943750.71/warc/CC-MAIN-20230322051607-20230322081607-00069.warc.gz | 490,077,508 | 37,307 | ## Reinforced Concrete – ULS capacity under combined axial load and biaxial bending
I have modified the ULS Design Functions spreadsheet, last presented here, to analyse sections subject to bi-axial bending, and non-symmetrical sections. The new version makes use of the routines for splitting any section defined by XY coordinates into trapezoidal layers, described here. The new version (ULS Design Functions-biax.xlsb) has been added to the zip file along with the previous version, and can be downloaded from ULS Design Functions.zip, including full open source code.
Input and results for a wide rectangular section, subject to bi-axial bending and axial load, are shown in the screenshot below:
The direction of the applied moment is defined by MX and MY, then the Neutral Axis angle is adjusted so that the reaction force and moments are in equilibrium with the applied loads, by clicking the “Adjust NA Angle” button.
The concrete section is defined by XY coordinates of each corner, listed in a clockwise direction, and the reinforcement is defined in layers, by entering the coordinates of the start and end of each layer: For each layer the number of bars and bar diameter are defined, together with the steel properties for the first layer, and any subsequent layer with different properties. It is also possible to specify a prestress force for any layer.
The hexagonal section shown below demonstrates that for a symmetrical section the angle of the resultant moment axis is parallel to the Neutral Axis angle, as would be expected:
It is possible to define any complex shape, such as the precast Super-T bridge girder shown below:
It is also possible to define shapes with internal voids, as shown below, by listing the corners of the void in the anti-clockwise direction. In this case the line must be continuous from start to end, and the connection between the outer line and the void must be made with two separate lines, with a very small separation, so that separate lines do not overlap or cross at any point. See the example in the download file for more details.
More detailed output data is provided on the “UMom Out” sheet, in a similar format to the earlier version.
The analysis is carried out by two user defined functions (UDFs), UMom and UMomA. UMom provides the detailed output shown above for a single set of applied loads. UMomA returns any one of the available output values for a range of applied axial loads, and a single value or range of Neutral Axis directions. Both functions return an array of values, and must be entered as an array function, as described at Using Array Functions and UDFs.
Use of the UMomA function allows the rapid generation of interaction diagrams, as shown in the screenshots below:
This entry was posted in Beam Bending, Concrete, Excel, Newton, UDFs, VBA and tagged , , , , , , . Bookmark the permalink.
### 13 Responses to Reinforced Concrete – ULS capacity under combined axial load and biaxial bending
wow, great job
Like
2. metroxx says:
Hi,
Looks great. Thank you.
Like
3. andre says:
impressive! Immediately comes to mind to reuse procedure for finding NA in foundation pad also with corner uplift.
Like
4. metroxx says:
When try to open biax spreadsheet got error in module mCoords Compile Error: Type mismatch.
(dPi As Double =)
Like
• metroxx says:
This problem only in Excel 2016, in 2013 works OK.
Like
5. metroxx says:
I use 32bit as well.
When open got same error, but I removed “” for dPi:
Const dPi As Double = 3.14159265358979 and now it is OK for me.
(For previous version when I removed “” Excel was closed totally with error)
Like
• dougaj4 says:
That’s interesting. I’m not sure what’s going on there, but it looks like it might be time to review how I specify pi again.
Like
6. arsi says:
Hi there, can i use the same routine for unreinforced concrete sections and their evaluations. e.g. large unreinforced concrete earth retaining walls.
Like
7. arsi says:
additionally …. do have background literature on calculation ….if that is possible as a student i would be really interested ….even background literature links would suffice
Like
• dougaj4 says:
The routines are designed for reinforced sections. They might work for unreinforced, but I don’t guarantee it. If you try it, check that forces and moments are in equilibrium with the applied loads, and are consistent with the reported stresses.
For background information, try:
https://newtonexcelbach.com/2008/03/25/reinforced-concrete-section-analysis-1/
and the linked posts in that series.
Also try setting “category” to concrete near the top of the right margin.
Like
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 1,065 | 4,749 | {"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-2023-14 | latest | en | 0.896454 |
https://forum.mysensors.org/topic/4535/1mhz-hardware | 1,545,022,764,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376828318.79/warc/CC-MAIN-20181217042727-20181217064727-00612.warc.gz | 590,217,575 | 37,691 | # 1mhz hardware
• I recently flashed an arduino pro mini bootloader to 1mhz. This in the end did what I wanted, which was low power. However, I am still noticing something odd and believe I have missed something.
I have this line in my loop
``````whichbutton = sleep(digitalPinToInterrupt(2),LOW,digitalPinToInterrupt(3),LOW,3600000);//86400000);
``````
Doesn't seem to be waking up as expected. It does wake up, but not quite on the hour like I would have expected. But it still wakes up on interrupt which is the important thing.
The other thing I am curious about is I have the following for checking my battery. Which works, but I have a question regarding the percent. Is the percent based on the actual drop out voltage? For instance, I have a 3v battery. My fuses are set to 1.8v for dropout. Is the battery percentage basing 0% at 1.8v or 0v?
``````void CheckBattery()
{
// battery stuff
// get the battery Voltage
#ifdef MY_DEBUG
Serial.println(sensorValue);
#endif
// 1M, 470K divider across battery and using internal ADC ref of 1.1V
// Sense point is bypassed with 0.1 uF cap to reduce noise at that point
// ((1e6+470e3)/470e3)*1.1 = Vmax = 3.44 Volts
// 3.44/1023 = Volts per bit = 0.003363075
int batteryPcnt = sensorValue / 10;
#ifdef MY_DEBUG
float batteryV = sensorValue * 0.0029325513196481;
Serial.print("Battery Voltage: ");
Serial.print(batteryV);
Serial.println(" V");
Serial.print("Battery percent: ");
Serial.print(batteryPcnt);
Serial.println(" %");
#endif
if (oldBatteryPcnt != batteryPcnt) {
// Power up radio after sleep
sendBatteryLevel(batteryPcnt);
oldBatteryPcnt = batteryPcnt;
}
}```
@Nca78``````
• @Jason-Brunk said:
It does wake up, but not quite on the hour
Can you be more explicit? How much time is there inbetween wakeups? The watchdog will always be a bit off, but it should only be a few %
• Did you change the board definition within the Arduino IDE so that the code is compiled for 1 MHz? If not, the timing wont match.
According battery:
100 % = Vmax = 3.44 Volts
0 % = 0 V
• Hello, no idea with the wake up time, I didn't check it at 1MHz, but it should not vary as it's based on the watchdog timer which is always at 128KHz, and not related to the main clock frequency and multiplier.
For the battery level do you use a divider ? That's useful if you have a booster, but for a CR2032 directly connected to Vcc the most simple way is to measure the supply voltage. Please note that my measurement is not exact as value should be adjusted to the exact voltage of the "1.1V" reference voltage of the atmega.
``````#include <SystemStatus.h>
// Parameters for VCC measurement
const int VccMin = 2000; // Minimum expected Vcc level, in Volts. At 2V the cell should be dead
const int VccMax = 2900; // Maximum expected Vcc level, in Volts.
SystemStatus vcc();
int LastBatteryPercent = 200; // so we are sure to send the battery level at first check
int currentBatteryPercent = SystemStatus().getVCCPercent(VccMin, VccMax);
if (currentBatteryPercent != LastBatteryPercent) {
LastBatteryPercent = currentBatteryPercent;
sendBatteryLevel(currentBatteryPercent);
}
``````
• @TimO Here is my definition in my boards.txt
``````## Arduino Pro or Pro Mini (1.8V, 1 MHz) w/ ATmega328
## --------------------------------------------------
``````
• @Nca78 yes currently using a voltage divider. if i go to measuring the supply voltage alone that will get rid of the extra components I may not be needing atleast for this node.
I will try your example. Thanks!
• @Yveaux Ok. so it checked in at 3:17am this morning. it did not check in again until 1:58 this afternoon.
• @Jason-Brunk that's almost 10 hours. My guess is that the timer is 8x off (if the code expects 8Mhz but is running at 1Mhz). Try recompiling with 1MHz board settings or divide the time by 8 in the sketch)
• @mfalkvidd Ill reflash it. Just wondering if there is something in the code I should change to tell it it's running at 1mhz or if that's something that is supposed to just work based on the clock timing
• @Jason-Brunk I think the Arduino IDE is supposed to handle all that if the correct board is selected. I know that timings are off if I select 16MHz pro mini in the IDE and then flash to a 8MHz Arduino. It is also easily noticeable on the serial monitor, which will require a setting of twice the baud rate set in Serial.begin. I have never built a 1MHz node though.
• @Jason-Brunk why not falsh your node for internal 8mhz for instance, and then use 1Mhz in sketch ? On my side, I prefer this way..and it works for me at least.
when the node is not sleeping, and radio is not sleeping in RX mode for instance, there is no big saving to do this, only atmel consumption is low, radio is not..but if your radio is sleeping, yep it's interesting but I don't really see advantage to start node at 1Mhz as the radio does not sleep at startup and does the protocol stuff.
• @Jason-Brunk said:
Just wondering if there is something in the code I should change to tell it it's running at 1mhz or if that's something that is supposed to just work based on the clock timing
Clock settings are determined by the"fuses". What's your tool for fuses and bootloader programming? Are you sure you programmed the fuses as in your "boards.txt" above?
• @m26872 said:
@Jason-Brunk said:
Just wondering if there is something in the code I should change to tell it it's running at 1mhz or if that's something that is supposed to just work based on the clock timing
Clock settings are determined by the"fuses". What's your tool for fuses and bootloader programming? Are you sure you programmed the fuses as in your "boards.txt" above?
My understanding was that the "burn bootloader" function of the Arduino IDE will also write the fuses, no ?
From Arduino website :
"How does it work?
The "Burn Bootloader" commands in the Arduino environment use an open-source tool, avrdude. There are four steps: unlocking the bootloader section of the chip, setting the the fuses on the chip, uploading the bootloader code to the chip, and locking the bootloader section of the chip"
• @Nca78 True. And I suppose you've also double checked the verbose output from that process.
• Ok. So I got back in town and reflashed my bootloader. Claims it's running 1mhz now. No regulator, no leds. Even in sleep mode I am sitting at about 1.66ma. Which in deep sleep mode that seems really high. Considering the capacity of a coin cell i am looking at changing out about once every 2 weeks with no use.
Any thoughts? Recommendations? Just can't seem to get this running low power enough. Not sure if my hardware is junk or what. But i would take any recommendations or tests to see how to get this working since i have a small stack of these pro minis
• There's obviously a problem, I'm at about 1/1000 of that in sleep mode, you're sure your multimeter is not mistaking the unit ?
I have cheap pro minis too, the < 2\$ kind.
1mA+ in sleep mode sounds like a voltage regulator still connected, or the IC not sleeping, at least not in power down mode like when using the MySensors function. Are you sure you didn't leave a #define by mistake (like activating the repeater function with #define MY_REPEATER_FEATURE) that would prevent the atmega from sleeping ?
• I checked, no REPEATER on.
Here is the code for my loop()
``````int whichbutton = 0;
whichbutton = sleep(digitalPinToInterrupt(2),LOW,digitalPinToInterrupt(3),LOW,86400000);
switch (whichbutton) {
case digitalPinToInterrupt(2):
{//off
send(msg_light.set(0),true);
wait(300);
break;
}
case digitalPinToInterrupt(3):
{//on
send(msg_light.set(1),true);
wait(300);
break;
}
}
sleep(250);
CheckBattery();
``````
It's not running that over and over for sure. Otherwise I would see the battery value publishing regularly to the broker.
I broke the regulator off and the LEDs.
This is the one I have.
https://www.sparkfun.com/products/11114
I don't think the multimeter is misreading it. I can see the battery percentage dropping over the course of a day so it definitely seems like it's pulling more mA than it should sleeping.
• @m26872 said:
@Nca78 True. And I suppose you've also double checked the verbose output from that process.
In case someone doesn't believe the Arduino docs
This is at the beginning of the bootloader writing process in Arduino IDE:
``````Writing | ################################################## | 100% 0.01s
avrdude: 1 bytes of efuse written
avrdude: verifying efuse memory against 0x06:
avrdude: load data efuse data from input file 0x06:
avrdude: input file 0x06 contains 1 bytes
Reading | ################################################## | 100% 0.01s
avrdude: verifying ...
avrdude: 1 bytes of efuse verified
avrdude: writing hfuse (1 bytes):
Writing | ################################################## | 100% 0.01s
avrdude: 1 bytes of hfuse written
avrdude: verifying hfuse memory against 0xde:
avrdude: load data hfuse data from input file 0xde:
avrdude: input file 0xde contains 1 bytes
Reading | ################################################## | 100% 0.01s
avrdude: verifying ...
avrdude: 1 bytes of hfuse verified
avrdude: writing lfuse (1 bytes):
Writing | ################################################## | 100% 0.00s
avrdude: 1 bytes of lfuse written
avrdude: verifying lfuse memory against 0x62:
avrdude: load data lfuse data from input file 0x62:
avrdude: input file 0x62 contains 1 bytes
``````
• @Nca78
The docs won't tell you if the process fails. The verbose output will.
• hmmm. ok so I have verified I am writing the 1mhz boot loader.
My multimeter says when it's sleeping i am still running at 1.5mah
I have my openhab system tracking the battery usage. IN 6 days I have lost 30% of the battery and it's been in sleep mode.
``````|| *Time* || *Value* ||
|| 2016-08-14 19:29:27 || 100 ||
|| 2016-08-15 19:07:57 || 80 ||
|| 2016-08-15 19:11:18 || 81 ||
|| 2016-08-15 19:12:48 || 83 ||
|| 2016-08-15 19:13:14 || 76 ||
|| 2016-08-15 19:13:17 || 84 ||
|| 2016-08-15 19:13:56 || 80 ||
|| 2016-08-15 19:16:48 || 78 ||
|| 2016-08-19 15:07:57 || 76 ||
|| 2016-08-19 15:10:09 || 80 ||
|| 2016-08-19 15:12:43 || 79 ||
|| 2016-08-19 15:13:49 || 82 ||
|| 2016-08-19 15:14:20 || 84 ||
|| 2016-08-20 11:44:01 || 69 ||
|| 2016-08-20 11:44:23 || 48 ||
|| 2016-08-20 11:44:31 || 71 ||
``````
Any thoughts? Any recommendations on maybe a replacement pro mini that will for sure run at <1mah?
• remove the power led (1-2mA), remove the voltage regulator if possible (1mA)
• @cimba007 Led was already removed.
• I would only imagine that the VoltageRegulator might be the culprit.
As you said you use a "3v battery" so there is no need for the VoltageRegulator at all. Where do you apply the power? RAW or VCC? To power your Arduino with higher voltages you connect them to RAW which leads to the regulator.
On most of my projects I use 2x AA Battery which are connected directly to VCC .. even with the VoltageRegulator intact I am at 66% for the last 7 days .. not loosing a single %.
PS: Sry .. I missed that you have already removed the voltage regulator + the led.
The library states to use RISING, FALLING, CHANGE as the interrupt mode but you use LOW .. I don't know to which of the 3 this mapps.
But my guess is you are using FALLING .. thus waking your node up if you connect pin 2 or 3 to ground .. which means you have some kind of pull up don't you?
Do you have anything connected to the Arduino ProMini? If so maybe try an empty sketch and remove all connected peripherals. Maybe even you can find some inspiration on this link: http://electronics.stackexchange.com/questions/49182/how-can-i-get-my-atmega328-to-run-for-a-year-on-batteries
Another really good test on the powerdown current is here:
http://www.rocketscream.com/blog/2011/07/04/lightweight-low-power-arduino-library/
with this library (not compatible with mysensor so dont't use at the same time)
https://github.com/rocketscream/Low-Power
• @cimba007 thanks! will definitely test out some of this stuff. I am connecting power to vcc and the regulator has been removed.
2 buttons connected to the interrupt pins, 1 for on 1 for off.
here is my sleep statement
``````whichbutton = sleep(digitalPinToInterrupt(2),LOW,digitalPinToInterrupt(3),LOW,86400000);
``````
Could my resistors connected to the buttons be causing the extra power?
I would expect that to be the cause because even a door sensor or something would have to trigger the interrupt.
A buddy of mine suggested I throw an additional multimeter in line so i am actually measuring with 2 and see if maybe my readings are just off or my multimeter is wonky reading it. He thinks maybe the fact i am measuring it is causing the extra current... (sounds like quantum entanglement stuff there, "It changed cause you measured it!!")
• Putting a multimeter in series will most likeley only cause a voltage drop of up to 200mV (just a rough number). If your circuit draws very little current even routing this current trough the resistor (which is causing the voltage drop) would not result in additional consumption.
Lets look at your switches .. if you are using a setup like this:
http://www.starting-point-systems.com/images/pullup_example.gif
http://www.microchip.com/forums/m341654.aspx states that there should be zero current consumption while the button is not pressed. Are u using this button setup? If not can u provide a drawing?
I don't think that the multimeter is to blame. You reported a very weak batterylife even when you are not measuring the current I guess.
``````He thinks maybe the fact i am measuring it is causing the extra current... (sounds like quantum entanglement stuff there, "It changed cause you measured it!!")
``````
Long story short .. I would not suspect the burden Voltage to be the problem .. I can measure in the 200µA range with my cheapish Multimeter when the Atmega328p is in deep sleep mode .. so something must be drawing current in your setup.
• To get the fastest results I would try to disconnect everything from your circuit .. hook up the Multimeter to measure current in lets say 200mA range (if available) and try the deep sleep example from https://github.com/rocketscream/Low-Power
If the pro mini is in deep sleep mode you can switch from 200mA to 2mA or even 200µA range if available. (just don't let the pro mini wake up while you have the range set very low)
Lower measure range = higher burden voltage = potential less power for your µC .. maybe even to the point it will go into brown out or start to behave strange
• @cimba007 Thanks for the details. Let me get you back some info.
yes, my buttons are setup like the pullup example image
i didn't think the multimeter would do it, but at this point i am willing to try anything
I will pull my mini tonight and remove everything and do a test.
my code only wakes up when button is pressed or every 24 hours. so it shouldn't come on while testing. Ill try with my code and with the sleep example you referenced.
Thanks!!!! | 3,921 | 15,052 | {"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-2018-51 | longest | en | 0.875319 |
https://estebantorreshighschool.com/equation-help/surface-area-of-a-triangular-prism-equation.html | 1,695,851,090,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510326.82/warc/CC-MAIN-20230927203115-20230927233115-00471.warc.gz | 267,885,236 | 9,780 | ## What is the formula to find the total surface area of a triangular prism?
A triangular prism has three rectangular sides and two triangular faces. To find the area of the rectangular sides, use the formula A = lw, where A = area, l = length, and h = height. To find the area of the triangular faces, use the formula A = 1/2bh, where A = area, b = base, and h = height.
## What is the equation for a triangular prism?
Calculating volume So to calculate the volume of a triangular prism, the formula is: V = 0.5 X b X a X h.
## How do you find the surface area of prisms?
The general formula for the total surface area of a right prism is T. S. A. =ph+2B where p represents the perimeter of the base, h the height of the prism and B the area of the base.
## What is the formula for surface area?
Surface area is the sum of the areas of all faces (or surfaces) on a 3D shape. A cuboid has 6 rectangular faces. To find the surface area of a cuboid, add the areas of all 6 faces. We can also label the length (l), width (w), and height (h) of the prism and use the formula, SA=2lw+2lh+2hw, to find the surface area.
## What is the surface area of triangle?
The height line will run through the interior of the triangle. Multiply the base length by the height. For example, if your base measurement is 10 cm and the height is 6 cm, the base multiplied by the height would be 60 square cm. Divide the result of the base times height by two to determine the surface area.
## What is the total surface area of rectangular prism?
Surface Area of Rectangular Prism: S = 2(lw + lh + wh)
## How do u find the height of a triangular prism?
To find the perimeter of a triangle, add up the length of all three sides. . This will give you the height of your prism. So, the height of your prism is 68 centimeters.
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#### H2o2 decomposition equation
What does h2o2 decompose into? Hydrogen peroxide can easily break down, or decompose, into water and oxygen by breaking up into two very reactive parts – either 2OHs or an H and HO2: If there are no other molecules to react with, the parts will form water and oxygen gas as these are more stable […] | 618 | 2,466 | {"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} | 4.5625 | 5 | CC-MAIN-2023-40 | latest | en | 0.888173 |
https://www.numberempire.com/45796 | 1,621,013,708,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991428.43/warc/CC-MAIN-20210514152803-20210514182803-00475.warc.gz | 943,853,459 | 7,170 | Home | Menu | Get Involved | Contact webmaster
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# Number 45796
forty five thousand seven hundred ninety six
### Properties of the number 45796
Factorization 2 * 2 * 107 * 107 Divisors 1, 2, 4, 107, 214, 428, 11449, 22898, 45796 Count of divisors 9 Sum of divisors 80899 Previous integer 45795 Next integer 45797 Is prime? NO Previous prime 45779 Next prime 45817 45796th prime 556351 Is a Fibonacci number? NO Is a Bell number? NO Is a Catalan number? NO Is a factorial? NO Is a regular number? NO Is a perfect number? NO Polygonal number (s < 11)? square(214) Binary 1011001011100100 Octal 131344 Duodecimal 22604 Hexadecimal b2e4 Square 2097273616 Square root 214 Natural logarithm 10.731952030044 Decimal logarithm 4.6608275466984 Sine -0.84356564862896 Cosine -0.53702606682843 Tangent 1.5708095020618
Number 45796 is pronounced forty five thousand seven hundred ninety six. Number 45796 is a composite number. Factors of 45796 are 2 * 2 * 107 * 107. Number 45796 has 9 divisors: 1, 2, 4, 107, 214, 428, 11449, 22898, 45796. Sum of the divisors is 80899. Number 45796 is not a Fibonacci number. It is not a Bell number. Number 45796 is not a Catalan number. Number 45796 is not a regular number (Hamming number). It is a not factorial of any number. Number 45796 is a deficient number and therefore is not a perfect number. Number 45796 is a square number with n=214. Binary numeral for number 45796 is 1011001011100100. Octal numeral is 131344. Duodecimal value is 22604. Hexadecimal representation is b2e4. Square of the number 45796 is 2097273616. Square root of the number 45796 is 214. Natural logarithm of 45796 is 10.731952030044 Decimal logarithm of the number 45796 is 4.6608275466984 Sine of 45796 is -0.84356564862896. Cosine of the number 45796 is -0.53702606682843. Tangent of the number 45796 is 1.5708095020618
### Number properties
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Examples: 3628800, 9876543211, 12586269025 | 617 | 1,915 | {"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-2021-21 | latest | en | 0.701839 |
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A293143 Number of vertex points in a Sierpinski Carpet grid subdivided into squares: a(n+1) = 8*a(n) - 8*(3^n+1), a(0) = 4. 4
4, 16, 96, 688, 5280, 41584, 330720, 2639920, 21101856, 168762352, 1349941344, 10799058352, 86391049632, 691124145520, 5528980409568, 44231805012784, 353854325311008, 2830834258114288, 22646673031792992, 181173381154980016 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,1 COMMENTS Figurate number sequence for the Sierpinski Carpet lattice. See the faces of the cubes in "Image 2" in the Wikipedia link of A293144 for an example of the construction grid of the Sierpinski Carpet. LINKS Colin Barker, Table of n, a(n) for n = 0..1000 (Recomputed by M. F. Hasler to include the initial term 4.) Eric Weisstein's World of Mathematics, Sierpinski Carpet. Wikipedia, Sierpinski carpet. Index entries for linear recurrences with constant coefficients, signature (12,-35,24). FORMULA From Colin Barker, Oct 02 2017, corrected for a(0) = 4 by M. F. Hasler, Oct 16 2017: (Start) G.f.: 4*(1 - 8*x + 11*x^2) / ((1 - x)*(1 - 3*x)*(1 - 8*x)). a(n) = 8*(5 + 11*2^(3*n-1) + 7*3^n) / 35. a(n) = 12*a(n-1) - 35*a(n-2) + 24*a(n-3) for n > 2. (End) EXAMPLE The carpet is formed by squares within a square grid. The initial term is a(0) = 4 for the corners of the unit square. For n = 1 there are 3 X 3 squares, the middle one being open (empty), with 16 vertex points. At the next stage of recursion, these become eight squares with open center, now based on a square of 10 X 10 points. The remaining center square is empty, missing 4 points, thus the next term is 100 - 4 = 96 for a(2). In the next stage there are 8 squares missing 4 points and the new center is missing 64, thus the 28 square grid now has 784 - 32 - 64 = 688 for a(3). This carpet sequence becomes the faces for the cubes in the Menger Sponge recursion of A293144. MATHEMATICA FoldList[8 #1 - 8 (3^(#2-1) + 1) &, 4, Range@ 18] (* Michael De Vlieger, Oct 02 2017 *) PROG (PARI) prev=4; concat(prev, vector(20, n, prev=8*prev-8*(3^(n-1)+1))) \\ Colin Barker, Oct 08 2017 (PARI) Vec(4*(1 - 8*x + 11*x^2)/((1 - x)*(1 - 3*x)*(1 - 8*x)) + O(x^30)) \\ Colin Barker, Oct 09 2017 (PARI) A293143(n)=8*(5+11*2^(3*n-1)+7*3^n)/35 \\ M. F. Hasler, Oct 16 2017 CROSSREFS Cf. A293144, A034472. Sequence in context: A098580 A317998 A027745 * A032184 A130683 A111976 Adjacent sequences: A293140 A293141 A293142 * A293144 A293145 A293146 KEYWORD nonn,easy AUTHOR Steven Beard, Oct 01 2017 EXTENSIONS Edited to start with a(0) = 4 by M. F. Hasler, Oct 16 2017 STATUS approved
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# Level 1
Paper
Paper 1
Paper 2
Difficulty
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##### Question 1
no calculator
easy
[Maximum mark: 4]
Expand $(2x + 1)^4$ in descending powers of $x$ and simplify your answer.
easy
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##### Question 2
no calculator
easy
[Maximum mark: 6]
A bag contains $7$ blue and $5$ red marbles. Two marbles are selected at random without replacement.
1. Complete the tree diagram below. [3]
1. Find the probability that exactly one of the selected marbles is blue. [3]
easy
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##### Question 3
calculator
easy
[Maximum mark: 6]
The following diagram shows part of a circle with centre O and radius $5$ cm.
Points A, B lie on the circle, chord AB has a length of $8$ cm and $\text{A\^{O}B} = \theta$.
1. Find the value of $\theta$, giving your answer in radians. [3]
2. Find the area of the shaded region. [3]
easy
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##### Question 4
calculator
easy
[Maximum mark: 6]
A school basketball team of $5$ students is selected from $8$ boys and $4$ girls.
1. Determine how many possible teams can be chosen. [2]
2. Determine how many teams can be formed consisting of $3$ boys and $2$ girls? [2]
3. Determine how many teams can be formed consisting of at most $3$ girls? [2]
easy
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##### Question 5
no calculator
easy
[Maximum mark: 6]
Let $f(x) = p x^3 - qx$. At $x = 0$, the gradient of the curve of $f$ is $2$. Given that
$f^{-1} (12) = 2$, find the value of $p$ and $q$.
easy
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##### Question 6
no calculator
easy
[Maximum mark: 6]
The function $f$ is of the form $f(x) = \dfrac{ax+b}{2x+c}$, for $x \neq -\dfrac{c}{2}$, where $a,b,c \in \mathbb{Z}$. Given
that the graph of $y = f(x)$ has asymptotes $x = -5$ and $y = 2$, and that the point
P$\left(1,-\dfrac{1}{12}\right)$ lies on the graph, find the values of $a, b$ and $c$.
easy
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##### Question 7
calculator
easy
[Maximum mark: 6]
Given that $\log_b 3 = 10$.
1. Find the exact value of $\log_b 81$. [2]
2. Find the exact value of $\log_{b^2} 3$. [2]
3. Find the value of $b$, giving your answer correct to $3$ significant figures. [2]
easy
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##### Question 8
calculator
easy
[Maximum mark: 7]
In a geometric sequence, $u_2 = 6$, $u_5 = 20.25$.
1. Find the common ratio, $r$. [2]
2. Find $u_1$. [2]
3. Find the greatest value of $n$ such that $u_n < 200$. [3]
easy
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##### Question 9
calculator
easy
[Maximum mark: 5]
Let $f(x) = 9 - 2\ln(x^2 + 4)$, for $x \in \mathbb{R}$. The graph of $f$ passes through the point $(p,3)$, where $p > 0$.
1. Find the value of $p$. [2]
The following diagram shows part of the graph of $f$.
The region enclosed by the graph of $f$, the $x$-axis and the lines $x = -p$ and $x = p$ is rotated $360^\circ$ about the $x$-axis.
1. Find the volume of the solid formed. [3]
easy
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##### Question 10
calculator
easy
[Maximum mark: 6]
A particle moves in a straight line with velocity $v(t) = 2 t - 0.3 t^3 + 2$, for $t \geq 0$, where $v$ is in ms$^{-1}$ and $t$ in seconds.
1. Find the acceleration of the particle after $2.2$ seconds. [3]
1. Find the time when the acceleration is zero.
2. Find the velocity when the acceleration is zero. [3]
easy
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4,413 results, page 4
1. ### alg.
write your answer in standard form with integer coefficients. 2x+3y=30; (2,-5) i don't understand how to do this, im not quiet sure i know what its asking.
2. ### Alg III
Solve each absolute value equation. 3 + |x| = 5 My work: 3 + x = 5 or 3 + x = -5 x = 2 or x = -8 The book says that I am wrong. I do not know why.
3. ### alg
What is the value of the expression 2x^2 y /4xy when x = 2 and y = 1? a)1 b)2 c)3 d)4 Just need to know if my answer is right or wrong and i picked a
4. ### Alg. I
I've just begun to study Exponential functions: When you make a table starting with: -2 -4^x What are my coordinates? -1 0 1 2
5. ### ALG 2
What is the first term of a geometic series with a summation of 800, 4 terms and a common ratio of 3?
6. ### Math-alg-calc
x^(-1/7)=(1/2) Please help the example given was x^(-1/7)=(1/5) so that solved x=78,125. but i have no idea what to do thanks!
7. ### college alg
the function f(x)=6x+9/3x-7 find the inverse and check answer f^-1(x)=? please show work
8. ### college alg
the function f(x)=6x+9/3x-7 find the inverse and check answer f^-1(x)=? please show work
Determine the implied domain of the given function. Express in interval notation (-4x^(2)-3x)/(2)
10. ### math ALG 2!
Write the following value in scientific notation. 7,400,000 7.4x10^3? help?
11. ### alg 2 check?
1.)solve for n:-7n-21=-63 my answer: n=3 is this correct? please explain 2.)identify the x and y int for te equation 3x-4y=12.
12. ### Alg
describe the end behavior of the graph of the polynomial function. Then evaluate the fucntion for x=-4, -3, -2, -1, 0, 1, 2, 3, 4. I'm confused on this. Help.
13. ### Alg 2
How do you simplify this expression? It gets confusing b/c of logs and exponents. Can someone help me explain how to do this. 8^log(8)x
14. ### trig
FInd the exact value pf the 6 trig function of alpha Given: point (-2,-6) on the terminal side of the angle in standard position sin alpha cos alpha tan alpha sec alpha csc alpha cot alpha
15. ### trig
Write the expression and evaluate. There is only one answer whIch should match the range of the inverse trig function. Sec^-1(-2) Sec^-1(-sqrt2) Tan^-1(-sqrt3) Tan^-1(sqrt3)
16. ### trig
find each value by referring to the graphs of the trig functions. 1. sec 4 pi...i got 0 Find the values for theta for which each equation is true. csc (theta)=1 i got pi over 2 +2 pi n
17. ### alg 2
2 questions: 1. Find "a" if (-3,7) is a solution of 2x+ay=26 2. solve system using substitution, linear combination, and graphing: 3x-2y=3 5x+4y=16
18. ### alg 2
Write an exponential function whose graph passes through the given points. 1. (0, 3) and (1, 15) 2. (0, -5) and (-3, -135) 3. (0, -0.3) and (5, -9.6)
19. ### Alg.2
find all zeros of the function. f(x) = x^3-10x^2+33x-34 I found 2, using my calculator. What do i do after that?
20. ### alg 2
write an exponential function whose graph passes throught the given points... (0,3)and (1,15)...(0,-5)and(-3,-135)...(0,-0.3)and(5,-9.6)
21. ### Alg I
Find a pair of factors for each number by using the difference of two squares. 84 55 80 Please explain. Thanks
22. ### Alg 1
Operations with radical expressions: Simplify each quotient 7/sq rt 2 - sq rt 3 Please solve and explain how you did it. What do you do when the denominator is a negative?
23. ### College Alg
Use synthetic division and the Remainder Theorem to evaluate P(c). P(x) = 5x5 + 10x3 + x + 1, c = −2 p(-2)=
24. ### ALG 2
Determine the standard form of the equation of the line that passes through (-6,6) and (3,-2). My slope is -8/9 Y=mx+b -2=-8/9(3) + b SOLVING FOR b -2=-24/9 + b WHAT DO I DO NOW
25. ### Trig/Pre-Calc
Need help with current Trig problem for tomorrow: Solve the logarithmic equation algebraically. Approximate the result to three decimal places. 6log3(0.5x)=11 I know to start off the equation you divide by 6. I know the answer is 14.988. I just am puzzled on how to receive the...
26. ### Alg III
Solve each absolute value equation. |x - 3| = 1 Okay here is what I did: x - 3 = 1 x = 4 or x = -4 The book says {2,4} I'm very lost. Can someone help me?
27. ### COLLEGE ALG
use the transformations to identify the graph of the function. then determine its domain, range, and horizontal asymptote. f(x)=4+3^X/3
28. ### college alg
solve the following inequality x^2(8+x)(x-5)/ (x+5)(x-2)> or equal too 0 this looks like a fraction please show work
29. ### ALG II
Write each rational expression in simplest form and list the values of the variables for which the fraction is undefined. 1. x^2-7x+12/x^2+2x-15 2. 5y^2-20/y^2+4y+4 3. -7+7a/21a^2-21 4. 5(1-b)+15/b^2-16
30. ### math (final trig problem)
Solve for x on the interval [0,π/2): cos^3(2x) + 3cos^2(2x) + 3cos(2x) = 4 I havent done trig for a while so what exactly does that mean in solving for x on that interval and how would i go about doing that? Where are you getting these tough trig questions?? how about ...
31. ### trig
How do I prove these trig identities? secx-cosx/tanx =sinx And 1+sinx/cosx+ cox/ 1+sinx=2secx
32. ### alg
rationalize the denominator of the expression assume that all variables are positive ^3ã7/^3ã10 i cant get that sign on my pc its 2nd x2 on a graphing calc
33. ### Alg 1
I'm doing quadratic equations x^2 +bx+c =0.I'm supposed to factor each polynomial. I know how to questions like x^2 +8x +12 =0 but this one says, j^2-9jk-10k^2. I'm confused
34. ### alg 2 first final study guide
f (x) =2X^3 + 2x^2 - 3x + 2 please help me set up to find all the real number zeros of the function
35. ### trig
let P be a point in Q1 such that the slope of the line connecting P to the origin is sq. root of 2 and the distance from P to the origin is 3 units. Find the coordinates of P, state all trig values for angle formed by the line and the positive x-axis, and the measure of said ...
36. ### trig
If cos(t)=(-7/8) where pi<t<(3pi/2), find the values of the following trig functions. cos(2t)=? sin(2t)=? cos(t/2)=? sin(t/2)=?
37. ### Alg2
i am realy lost as to how to do this alg 2 problem: find the two real-number solutions of each equation: x^2=100 Can you please explain how to do it? Thanks!:)
38. ### alg
In a permutation a) the selections are independent b) some items will never be used c) the order matters d) there must be at least 3 things to choose from
39. ### ALG 2!!!
Write an equation of a line that passes through (9,-5) and (3,-5). Hint: To answer this question, first find the slope using these 2 points.
40. ### alg 2
Use the definitions and theorems of this section to evaluate and simplify the following expression. Be sure to express answers with positive exponents. (m . n)-1 =
41. ### Math/Trig
How would I evaluate these trig functions without using a calculator? U: sin(-13π/6) cot 11π/6 cot(-14π/4) sec 23π/6 Thanks in advance ^^; and if you'd tell me step by step on how to do problems like these I'd be grateful :D
42. ### trig 30
For csc^2 A-1/cot A csc A, what is the simplest equivalent trig expression? The answers I have to choose from are sin A, cosA, tan A or csc A, but I don't know how?? csc theta= 1/sin theta but its ^2. Help please I don't even no where to begin
43. ### trig
Write the expression and evaluate. There should only be one answer which should match the range of the inverse trig function. Sec^-1(-2) sec^-1(-sqrt2) Cos^-1((-sqrt3)/2) Tan^-1(-sqrt3) Could someone explain how to get the answer as well please.
44. ### Trig
Solve the trig equation exactly over the indicated interval. tanθ = 0, all real numbers I know the answer is πn, but I don't understand how they got this. I get that tanθ = 0 on π and 2π on the unit circle, so did they just put n on the end of π ...
45. ### Math (Alg 2)
The following expression represents the value of which variable in the solution of the system of equations below? |2 -1 1| |1 2 6| |3 -1 2| ___________ |2 -1 -2| |1 2 3| |3 -1 -1| 2X-Y+Z=-2 X+2Y+6Z=3 3X-Y+2Z=-1 A. Z B. Y C. X D. NONE OF THE VARIABLES ARE REPRESENTED
46. ### college alg
use the transformations to identify the graph of the function. then determine its domain, range, and horizontal asymptote. f(x)=4+3^X/3 please show work
47. ### Alg 2 (HS)
Solve 4^(x+2)= 160 use a cal and round your answer to the nearest hundredth How would I work out this problem? I'm a little confused with this one.
48. ### math ALG 2!
Find three consecutive integers such that the sum of the second and the third is 17. Which equation would be used to solve this word problem? a.) x + (x + 1) + (x + 2) = 17 b.)(x + 1) + (x + 2) = 17 c.)x + (x + 2) + (x + 4) = 17 d.) 2(x + 2) + 3(x + 4) = 17
49. ### Math (alg 2)
The ratio of two numbers is 3 to 2 and the difference of their squares is 20. Find the numbers.
50. ### Math (alg 2)
The ratio of two numbers is 3 to 2 and the difference of their squares is 20. Find the numbers.
51. ### alg 2
Can someone tell me how to factor 2a^2b - 16ab + 32b Should I factor 2b out of it?
52. ### alg
Divide: p^2 + 4p - 5 / p^2 + 7p + 10 * p + 4 / p - 1 a) p + 2 b) p + 2 / p + 4 c) p+4 / p+2 d)2 Just need to know if my answer is right or wrong and the answer i picked was c
53. ### Alg 1
The formula for the perimeter P of a rectangle is P=2l +2w. What is the result when correctly solving the formula for the variable w?
54. ### Pre-Alg
write a translation rule that maps point d(7 -3) onto point d'(2 5)
55. ### pre alg.
26 centimenters per sec. = _ meters per min.
56. ### pre alg
(-4)cubed divided by (-2)cubed -3(-5)squared I got -67 am i correct?
57. ### Alg
State the slope for a line parallel to the line – 5 and containing the point (–3, 7).
58. ### Alg 2
What is the sum of a 12-term arithmetic sequence where the last term is 13 and the common difference is -10?
59. ### Math- Alg 2
Which logarithmic equation is equivalent to the exponential equation below? 3^x=28
60. ### alg
Write an equation of a line whose graph is parallel to the graph of y = 3x – 10.
61. ### Math-Alg
2. Find the equation of the line which perpendicular to the line -2x + 3y = 32 and passes through the point (-4,-8).
62. ### alg 2 review
write the following absolute value expressions as piecewise expressions: y= abs(2x-4)
63. ### Alg 2
By what amount does the sum of the roots exceed the product of the roots of the equation (x-7)(x+3)=0? I have no idea what to do!
64. ### alg
change the logarithmic expression to an equivalent expression involving an exponent. 4^16=x log (base4)16=x is it x=e^2?
65. ### math ALG 2!
Write the following value in scientific notation. 23,000,000 i don't understand
66. ### Alg !
What is the sum of the first 4 terms of the arithmetic sequence in which the 6th term is 8 and the 10th term is 13?
67. ### alg
solve equation g+4/g-2=g-5/g-8 last one still don't get answer my teacher got was 14, however i got 22.. what was my mistake?
68. ### alg 2
given that one of the zeros of the funiction f(x)=x^3-8x^2+11x+20 is -1, find the sum of the other zeros. hwo do i do this?
69. ### alg/geo
m= 5/6, (8,-9) find the equation of the line having the given slope and containing the given point
70. ### ALG 2
What is the lat term, in a series of 10 terms with a common ratio of 2, a first term (a1) if 1 and a summation of 1023
71. ### College Alg
find the intervals on which the function is increasing or decreasing and find any relative minima or maxima. f(x)=0.2x^3-0.2x^2-5x-4
72. ### college alg
solve the following logarithm equation log (4x+7)=1+log(x-6) x=? please show work
73. ### Alg. I
(-5d^-5) (6d^2) -30d^-3 I know that d^-3 goes on the bottom to make it positive;however,does -30 go on the bottom as well? Please explain the rule(s) for this. Thanks
74. ### Alg 3
Put the following equations in standard form. State the center and the radius. x^2-5x+y^2+4y=-3 I get how to end up with (-2.5,2), what is the radius?
75. ### Alg I
when translating the verbal expression "five more than d" into an algebraic expression, is the answer 5+d or d+5? Aren't they the same?
76. ### alg 1
the perimeter of a rectangle is 24 m. the width is 30 m less than five times the length. find the dimentions of the rectangle.
77. ### Alg 2 HELP!
Identify the x-intercepts, local maximum, and local minimum of the graph of f(x)=x^3+2x^2-13x+10 I am confused on where to begin.
78. ### alg 2.
when condensing this expression why is the answer ln5. ln25-ln5
79. ### alg
simplify: 10^C6 a) 60 b) 210 c) 151,200 d) 1,000,000 c
80. ### alg
Which expression represents the sum of the following? 5 / 2t + 1 + -1 / 4t^2 + 4t + 1 a)10t + 4 / (2t + 1)^2 b) 4 / (2t + 1)^3 c)10t + 1 / (2t + 1)^2 d) 4 / 4t^2 + 6t + 2
81. ### alg
Which expression represents the sum of the following? (5)/ (2t + 1) + (-1) / (4t^2 + 4t + 1) a) 10t + 4 / (2t + 1)^2 b) 4 / (2t + 1)^3 c) 4 / 4t^2 + 6t + 2 d) 10t + 1 / (2t + 1)^2
82. ### Alg 2.. HELP?
Multiply or divide. 14. (x^5y5/3)2 16. (a^2b^2)(a1/3b1/4) 18. x2/3 divided by x1/3 20. z4/5 divide by z1/2
83. ### Alg 2
Find the lcd 14/4(x+1) , 7/4x I don't get it can you please explain step by step?
84. ### Trig
Can you please help with this trig question: The angle of elevation to the top of a nearby mountain is 46 degrees. After walking 1 km toward the mountain, you determine the angle of elevation to be 68 degrees. How high is the mountain? Thank you for all your help.
85. ### Alg 2
Can someone tell me how to factor 2a^2 - 16ab + 32b I just need someone to help me get started, and I want to see if I can finish it.
86. ### alg 2
a rectangle is to be inscribed in a isosceles triangle of height 8 and base 10. Find the greatest area of such rectangle.
87. ### alg 2
a rectangle is to be inscribed in a isosceles triangle of height 8 and base 10. Find the greatest area of such rectangle.
88. ### alg 2
a rectangle is to be inscribed in a isosceles triangle of height 8 and base 10. Find the greatest area of such rectangle.
"Express the mixed# as the addition of a whole # and fraction.Then find the value as a single fraction." *i don't understand what it is asking me?!
I forgot how to look for the asymptote from an equation.. can someone please help me out? example: y= x/(x+1) what's the asymptote in this case? Thanks.
91. ### college alg
solve the following exponential equation. (e^x-e^-x)/2=6 what is the exact answer? x=? what is the answer approx to 3 decimal places? x=?
92. ### college alg
the function f(x)=6x+9/3x-7 is one to one. find its inverse and check your answer. f^-1(x)=? please show work
93. ### Alg 1
Solve the proportion: 4/x = (radical sign over 11, not -1) 11 - 1/3 Please solve and explain. My answer does not match any of my choices.
94. ### Math - Alg
How much water must be added to an 80% acid solution to make a mixture of 50mL which is 70% acid?
95. ### trig
if 10 students enter projects in the art show, how many different ways can they come in 1st, 2nd, and 3rd place? would you go 10^3 OR 10TIMES 3? This is not a trig question. Neither of those answers is correct. After making one of ten possible choices for first place, there ...
96. ### math-alg
8. In how many ways can 12 books be displayed on a shelf if 15 books are available?: * A. 455 B. 479001600 C 2.17945728 x 1011 2730
97. ### math-alg
8. In how many ways can 12 books be displayed on a shelf if 15 books are available?: * A. 455 B. 479001600 C. 2.17945728 x 1011 D. 2730
98. ### alg 2.
how would i condense this... 3lnxy-2ln2y. whenever i condense i get ln(xy^3/2y2). but my teacher has ln(x^3y/4). can someone please show me how to get that answer thanks!
99. ### alg 2
Provide an example of an arithmetic series which totals zero. Using complete sentences, explain how you created the example.
100. ### college alg
find the equation of a line that is perpendicular to line y=1/6x+6 and contains point (-7,0) y=? please show work so i can see how its done | 5,121 | 15,813 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2018-22 | latest | en | 0.780872 |
https://physics.stackexchange.com/questions/336902/geodesics-and-spontaneously-broken-symmetry | 1,576,495,494,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541319511.97/warc/CC-MAIN-20191216093448-20191216121448-00470.warc.gz | 493,625,345 | 30,381 | # Geodesics and spontaneously broken symmetry
I have a spacetime $(M,g)$ and a set of discrete symmetries $S_\alpha: M\to M$. Given two fixed points $p_0,p_1\in M$ with $S_\alpha(p_i)=p_i$, I want to argue that a geodesic connecting the two points consists itself only of fixed points.
Obviously, this is not true in general: For instance, take rotations of the 2-sphere as symmetry that leaves the two poles invariant. The geodesics are the great circles and all of them have the same length and they get mapped into each other. However, if we consider the full 3d-space with rotations, then the shortest distance is actually invariant under rotations, because it is the straight line connecting the two poles.
In some sense, the symmetry is spontenously broken in the first case. Is there a way to distinguish the two cases? In particular, if I find a geodesic that is invariant under the symmetry transformations, is there a general argument that other ones that do not respect this symmetry should be longer (only locally shortest geodesics).
The examples given in the question indeed (and in a strict sense) correspond to spontaneously broken and unbroken rotation symmetry around the $z$ axis as stated in the question and as will be clarified in the following.
In the case of (non-relativistic) geodesic motion (of a massive particle), the Hamiltonian can be taken as: $$H = \frac{m}{2}g_{ij}(x)p^ip^j$$ ($g$ is the metric, $x$ is the position, $p$ is the momentum) This Hamiltonian is invariant under any rotation about the $z$ axis for both a round two-sphere and the embedding three space. (The question mentions a discrete symmetry which can be taken any discrete subgroup of the $U(1)$ group of rotations around the $z$ axis, but the discussion is valid for the whole continuous $U(1)$ group of rotations around the $z$ axis) . To see that, it is sufficient to prove that the Hamiltonian Poisson commutes with the generator of $z$ rotations: $$L_z = p_x y - p_y x$$ (In the sphere case, the components are not independent).
In the sphere case there is no possible initial momentum invariant under the rotation about the $z$ axis, since the momentum is tangent to the sphere, therefore parallel to the $x-y$ plane. | 526 | 2,229 | {"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} | 3.109375 | 3 | CC-MAIN-2019-51 | latest | en | 0.916553 |
http://docs.manew.com/Script/Camera.worldToCameraMatrix.html | 1,618,372,561,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038076819.36/warc/CC-MAIN-20210414034544-20210414064544-00509.warc.gz | 31,328,452 | 3,340 | # Camera.worldToCameraMatrix 世界转相机矩阵
var worldToCameraMatrix : Matrix4x4
Description描述
Matrix that transforms from world to camera space.
Use this to calculate the camera space position of objects or to provide custom camera's location that is not based on the transform.
Note that camera space matches OpenGL convention: camera's forward is the negative Z axis. This is different from Unity's convention, where forward is the positive Z axis.
If you change this matrix, the camera no longer updates its rendering based on its Transform. This lasts until you call ResetWorldToCameraMatrix.
``````using UnityEngine;
using System.Collections;
public class example : MonoBehaviour {
public Vector3 offset = new Vector3(0, 1, 0);
void LateUpdate() {
Vector3 camoffset = new Vector3(-offset.x, -offset.y, offset.z);
Matrix4x4 m = Matrix4x4.TRS(camoffset, Quaternion.identity, new Vector3(1, 1, -1));
camera.worldToCameraMatrix = m * transform.worldToLocalMatrix;
}
}``````
``````// Offsets camera's rendering from the transform's position.
//从Offsets位置偏移相机的渲染
var offset : Vector3 = Vector3 (0,1,0);
function LateUpdate () {
// Construct a matrix that offsets and mirrors along
// Z axis. Because camera space has mirrored Z with respect
// to the rest of Unity.
//构建一个沿着Z轴偏移和镜像的矩阵，因为相机已经为Z轴镜像，并用于其余部分
var camoffset : Vector3 = Vector3 (-offset.x, -offset.y, offset.z);
var m : Matrix4x4 = Matrix4x4.TRS (camoffset, Quaternion.identity, Vector3 (1,1,-1));
// Override worldToCameraMatrix to be offset/mirrored
// transform's matrix.
//重载worldToCameraMatrix为偏移/镜像的变换矩阵
camera.worldToCameraMatrix = m * transform.worldToLocalMatrix;
}``````
Page last updated: 2013-7-30 | 458 | 1,672 | {"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-2021-17 | latest | en | 0.610398 |
https://www.coolstuffshub.com/fuel-consumption/convert/uk-miles-per-liter-to-us-miles-per-gallon/ | 1,669,511,122,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710155.67/warc/CC-MAIN-20221127005113-20221127035113-00140.warc.gz | 769,009,793 | 11,895 | # Convert Uk miles per liter to Us miles per gallon (UK mpl to US mpg Conversion)
1 UK mpl = 3.79 US mpg
## How to convert UK miles per liter to US miles per gallon?
To convert UK miles per liter to US miles per gallon, multiply the value in UK miles per liter by 3.79.
You can use the conversion formula :
US miles per gallon = UK miles per liter × 3.79
To calculate, you can also use our UK miles per liter to US miles per gallon converter, which is a much faster and easier option as compared to calculating manually.
## 1 UK mile per liter is equal to how many US miles per gallon?
1 UK mile per liter is equal to 3.79 US miles per gallon.
• 1 UK mile per liter = 3.79 US miles per gallon
• 2 UK miles per liter = 7.58 US miles per gallon
• 3 UK miles per liter = 11.37 US miles per gallon
• 4 UK miles per liter = 15.16 US miles per gallon
• 5 UK miles per liter = 18.95 US miles per gallon
• 10 UK miles per liter = 37.9 US miles per gallon
• 100 UK miles per liter = 379 US miles per gallon
## Examples to convert UK mpl to US mpg
Example 1:
Convert 50 UK mpl to US mpg.
Solution:
Converting from uk miles per liter to us miles per gallon is very easy.
We know that 1 UK mpl = 3.79 US mpg.
So, to convert 50 UK mpl to US mpg, multiply 50 UK mpl by 3.79 US mpg.
50 UK mpl = 50 × 3.79 US mpg
50 UK mpl = 189.5 US mpg
Therefore, 50 UK miles per liter converted to US miles per gallon is equal to 189.5 US mpg.
Example 2:
Convert 125 UK mpl to US mpg.
Solution:
1 UK mpl = 3.79 US mpg
So, 125 UK mpl = 125 × 3.79 US mpg
125 UK mpl = 473.75 US mpg
Therefore, 125 UK mpl converted to US mpg is equal to 473.75 US mpg.
For faster calculations, you can simply use our UK mpl to US mpg converter.
## UK miles per liter to US miles per gallon conversion table
UK miles per liter US miles per gallon
0.001 UK mpl 0.00379 US mpg
0.01 UK mpl 0.0379 US mpg
0.1 UK mpl 0.379 US mpg
1 UK mpl 3.79 US mpg
2 UK mpl 7.58 US mpg
3 UK mpl 11.37 US mpg
4 UK mpl 15.16 US mpg
5 UK mpl 18.95 US mpg
6 UK mpl 22.74 US mpg
7 UK mpl 26.53 US mpg
8 UK mpl 30.32 US mpg
9 UK mpl 34.11 US mpg
10 UK mpl 37.9 US mpg
20 UK mpl 75.8 US mpg
30 UK mpl 113.7 US mpg
40 UK mpl 151.6 US mpg
50 UK mpl 189.5 US mpg
60 UK mpl 227.4 US mpg
70 UK mpl 265.3 US mpg
80 UK mpl 303.2 US mpg
90 UK mpl 341.1 US mpg
100 UK mpl 379 US mpg | 744 | 2,317 | {"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-2022-49 | latest | en | 0.610601 |
https://www.wise-geek.com/what-are-the-different-types-of-econometrics-theory.htm | 1,611,335,221,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703530835.37/warc/CC-MAIN-20210122144404-20210122174404-00750.warc.gz | 1,088,221,296 | 18,103 | # What Are the Different Types of Econometrics Theory?
Osmand Vitez
Econometrics theory uses a mix of mathematics, statistics, and economics to create usable models for business studies. Economists and other educated individuals can engage in these activities as upper degrees are often necessary to learn and understand these models. Different types of econometrics theory include introductory, spatial, and nonparametric estimation. The different theories rely on the gathered data to ascertain information used to make decisions. In short, econometrics is meant to provide the best data available for making decisions that have the least risk and greatest economic rewards.
Introductory econometrics tend to be a much more basic set of models. The two most basic assumptions here include an inverse relationship between the price of an item and the quantity demanded — investment expenditures increase as interest rates decrease, and there is a positive relation between income and consumption expenditure. These factors are commonly parts of each introductory theory and are based in econometric studies. Early studies in econometrics often focus on these attributes because companies most often want to discover these relationships before moving on to more complex studies. Though the study of econometrics here and later may seem extremely academic, it does provide useful data for private businesses and governments.
Spatial econometrics uses information gathered from different data sets to create models for use in business. The data either results in dependent data pieces or correlated data taken from a larger population of information. A few attributes of this econometrics theory include decay of sample data with distance, observation similar to neighboring observations, hierarchy, and a systematic change in parameters. These attributes may not present themselves in each and every study that surrounds spatial econometrics theory. Some of the attributes should be included, however, as that will define the study.
Nonparametric studies are a very common inclusion within econometric theory. In short, it is a general method used by reviewers to discover a functional form that fits any observed data. As the main thrust of econometrics is to gather large data sets to make statistical or mathematical assumptions, the use of nonparametric estimates is extremely common. When enough econometrics studies continue to gather and report similar data using nonparametric methods, a theory will develop based on the hypotheses stated. Hence, econometrics theory develops from the numerous studies on business activities that exist around a certain data set. | 466 | 2,674 | {"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} | 2.828125 | 3 | CC-MAIN-2021-04 | longest | en | 0.93921 |
https://kerodon.net/tag/04W9 | 1,725,704,154,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650826.4/warc/CC-MAIN-20240907095856-20240907125856-00831.warc.gz | 324,923,609 | 4,403 | Kerodon
$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$
Notation 9.2.4.26. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category and let $X$ be an object of $\operatorname{\mathcal{C}}$. We let $\operatorname{Sub}(X)$ denote the set $\operatorname{Sub}( \operatorname{\mathcal{C}}_{/X} )$ of isomorphism classes of subterminal objects of $\operatorname{\mathcal{C}}_{/X}$ (see Notation 9.2.2.18). If $f: X_0 \hookrightarrow X$ is a monomorphism, we write $[X_0] \in \operatorname{Sub}(X)$ for the isomorphism class of $f$. We will sometimes abuse notation by identifying the isomorphism class $[X_0]$ with the object $X_0$ itself: by virtue of Remark 9.2.2.19, this identification is essentially harmless provided that $X_0$ is understood as an object of the slice $\infty$-category $\operatorname{\mathcal{C}}_{/X}$ (that is, provided that we remember the data of the monomorphism $f$). We will refer to $\operatorname{Sub}(X)$ as the set of subobjects of $X$, and we endow it with the partial ordering described in Notation 9.2.2.18: that is, if $f_0: X_0 \hookrightarrow X$ and $f_1: X_1 \hookrightarrow X$ are monomorphisms, we write $[X_0] \subseteq [X_1]$ if there exists a $2$-simplex
$\xymatrix@R =50pt@C=50pt{ X_0 \ar [dr]_{f_0} \ar [rr]^{g} & & X_1 \ar [dl]^{f_1} \\ & Y & }$
in the $\infty$-category $\operatorname{\mathcal{C}}$. In this case, $g$ is automatically a monomorphism (see Remark 9.2.4.15). | 501 | 1,436 | {"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} | 2.78125 | 3 | CC-MAIN-2024-38 | latest | en | 0.662766 |
https://www.weegy.com/?ConversationId=DBD5B189&Link=i&ModeType=2 | 1,709,223,749,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474843.87/warc/CC-MAIN-20240229134901-20240229164901-00821.warc.gz | 1,063,391,326 | 11,283 | Multiply: (x + 4)(x^2 – 4x + 16)
(x + 4)(x^2 – 4x + 16) = x^3 - 4x^2 + 16x + 4x^2 - 16x + 64 = x^3 + 64
Question
Updated 7/15/2018 10:20:42 AM
Edited by Masamune [7/15/2018 10:20:40 AM], Confirmed by Masamune [7/15/2018 10:20:42 AM]
f
Rating
Questions asked by the same visitor
Multiply: ( X + 4 )( X2 – 4X + 16)
Weegy: Best Answer - Chosen by Voters ((4/x) - (2/4x)) / ((1/2x) + (3/2x)) = ((16/4x) - (2/4x))/(4/2x) = (14/4x)/(2/x) = (14/4x) * (x/2) = 14x/8x = 14/8 = 7/4 (More)
Question
Updated 12/22/2014 12:07:35 PM
(x + 4 )( x^2 – 4x + 16);
= x^3 - 4x^2 + 16x + 4x^2 - 16x + 64;
= x^3 + 64
Confirmed by yumdrea [12/22/2014 12:13:29 PM], Unconfirmed by andrewpallarca [12/22/2014 12:13:40 PM], Confirmed by andrewpallarca [12/22/2014 12:13:40 PM]
Factor completely: 9X4 – 81X2
Weegy: 9X4 – 81X2 = -126 (More)
Question
Updated 1/4/2015 12:56:27 AM
9x^4 – 81x^2 = 9x^2(x + 3)(x - 3)
Confirmed by sujaysen [1/4/2015 2:23:35 PM]
38,918,557
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# Collaboration with Plotly in Python
An IPython Notebook showing how to collaboration between different programming languages with plotly
## IPython and Plotly: A Rosetta Stone for MATLAB,R, Python, and Excel plotting
Collaboration, data analysis, and data visualization sometimes feels like this:
In [1]:
from IPython.display import Image
Image(url = 'https://i.imgur.com/4DrMgLI.png')
Out[1]:
Graphing and data analysis need a Rosetta Stone to solve the fragmentation and collaboration problem. Plotly is about bridging the divide and serving as an interoperable platform for analysis and plotting. You can import, edit, and plot data using scripts and data from Python, MATLAB, R, Julia, Perl, REST, Arduino, Raspberry Pi, or Excel. So can your team.
All in the same online plot.
Read on to learn more, or run $pip install plotly and copy and paste the code below. Plotly is online, meaning no downloads or installations necessary. In [2]: %matplotlib inline import matplotlib.pyplot as plt # side-stepping mpl backend import matplotlib.gridspec as gridspec # subplots import numpy as np You can use our key, or sign-up to get started. It's free for any public sharing and you own your data, so you can make and share as many plots as you want. In [3]: import plotly.plotly as py import plotly.tools as tls from plotly.graph_objs import * py.sign_in("IPython.Demo", "1fw3zw2o13") In [4]: import plotly plotly.__version__ Out[4]: '1.0.12' ## I. shareable matplotlib figures¶ Let's start out with a matplotlib example. We also have a user guide section on the subject. In [5]: fig1 = plt.figure() import matplotlib.pyplot as plt import numpy as np import matplotlib.mlab as mlab mean = [10,12,16,22,25] variance = [3,6,8,10,12] x = np.linspace(0,40,1000) for i in range(4): sigma = np.sqrt(variance[i]) y = mlab.normpdf(x,mean[i],sigma) plt.plot(x,y, label=r'$v_{}$'.format(i+1)) plt.xlabel("X") plt.ylabel("P(X)") Out[5]: <matplotlib.text.Text at 0x7f549f2ea050> To re-create the graph in Plotly and use Plotly's defaults, call iplot and add strip_style. In [6]: py.iplot_mpl(fig1, strip_style = True) It's shareable at a URL, contains the data as part of the plot, and can be edited collaboratively from any API or our web app. Head over to Plotly's API to see more, and check out our user guide to see how it all works. Plotly also jointly preserves the data in a graph, the graph, and the graph description (in this case JSON). That's valuable. One study in current biology found that over 90 percent of data from papers published over the past 20 years was not available. So sharing data is good for science and reproducibility, useful for your projects, and great for collaboration. ## II. ggplot2 plots in Plotly¶ Let's take a real-world look storing data and graphs together. Suppose you see a graph on the World Bank website. The graph uses ggplot2, a remarkable plotting library for R. In [7]: from IPython.display import Image Image(url = 'http://i.imgur.com/PkRRmHq.png') Out[7]: You would like to re-make the graph and analyze and share the data. Getting the data using Plotly is easy. You can run the ggplot2 script in RStudio. Here we're running it using the new R kernel for IPython). The Notebook with the replicable code and installation is here. In [8]: library(WDI) library(ggplot2) #Grab GNI per capita data for Chile, Hungary and Uruguay dat = WDI(indicator='NY.GNP.PCAP.CD', country=c('CL','HU','UY'), start=1960, end=2012) #a quick plot with legend, title and label wb <- ggplot(dat, aes(year, NY.GNP.PCAP.CD, color=country)) + geom_line() + xlab('Year') + ylab('GDI per capita (Atlas Method USD)') + labs(title <- "GNI Per Capita ($USD Atlas Method)")
ggplotly(wb)
We can add ggplotly to the call, which will draw the figure with Plotly's R API. Then we can call it in a Notebook. You can similarly call any Plotly graph with the username and graph id pair.
In [9]:
tls.embed('RgraphingAPI', '1457')
Note: the data is called from a WDI database; if you make it with Plotly, the data is stored with the plot. I forked the data and shared it: https://plot.ly/~MattSundquist/1343.
If you want to use Plotly's default graph look, you can edit the graph with Python.
In [10]:
fig = py.get_figure('RgraphingAPI', '1457')
fig.strip_style()
py.iplot(fig)
Often we come to a visualization with data rather than coming to data with a visualization. In that case, Plotly is useful for quick exploration, with matplotlib or Plotly's API.
In [11]:
my_data = py.get_figure('PythonAPI', '455').get_data()
In [12]:
%matplotlib inline
import matplotlib.pyplot as plt
In [13]:
fig1 = plt.figure()
plt.subplot(311)
plt.plot(my_data[0]['x'], my_data[0]['y'])
plt.subplot(312)
plt.plot(my_data[1]['x'], my_data[1]['y'])
plt.subplot(313)
plt.plot(my_data[2]['x'], my_data[2]['y'])
py.iplot_mpl(fig1, strip_style = True)
You can also draw the graph with subplots in Plotly.
In [14]:
my_data[1]['yaxis'] = 'y2'
my_data[2]['yaxis'] = 'y3'
layout = Layout(
yaxis=YAxis(
domain=[0, 0.33]
),
legend=Legend(
traceorder='reversed'
),
yaxis2=YAxis(
domain=[0.33, 0.66]
),
yaxis3=YAxis(
domain=[0.66, 1]
)
)
fig = Figure(data=my_data, layout=layout)
py.iplot(fig)
Then maybe I want to edit it quickly with a GUI, without coding. I click through to the graph in the "data and graph" link, fork my own copy, and can switch between graph types, styling options, and more.
In [15]:
Image(url = 'http://i.imgur.com/rHP53Oz.png')
Out[15]:
Now, having re-styled it, we can call the graph back into the NB, and if we want, get the figure information for the new, updated graph. The graphs below are meant to show the flexibility available to you in styling from the GUI.
In [16]:
tls.embed('MattSundquist', '1404')
In [17]:
tls.embed('MattSundquist', '1339')
We can also get the data in a grid, and run stats, fits, functions, add error bars, and more. Plotly keeps data and graphs together.
In [18]:
Image(url = 'http://i.imgur.com/JJkNPJg.png')
Out[18]:
And there we have it. A reproducible figure, drawn with D3 that includes the plot, data, and plot structure. And you can easily call that figure or data as well. Check to see what URL it is by hoving on "data and graph" and then call that figure.
In [19]:
ggplot = py.get_figure('MattSundquist', '1339')
In [20]:
ggplot #print it
Out[20]:
{'data': [{'line': {'color': 'rgb(31, 119, 180)', 'width': 4},
'mode': 'lines',
'name': 'Chile',
'type': 'scatter',
'x': [1960,
1961,
1962,
1963,
1964,
1965,
1966,
1967,
1968,
1969,
1970,
1971,
1972,
1973,
1974,
1975,
1976,
1977,
1978,
1979,
1980,
1981,
1982,
1983,
1984,
1985,
1986,
1987,
1988,
1989,
1990,
1991,
1992,
1993,
1994,
1995,
1996,
1997,
1998,
1999,
2000,
2001,
2002,
2003,
2004,
2005,
2006,
2007,
2008,
2009,
2010,
2011,
2012],
'y': [None,
None,
600,
640,
660,
650,
740,
760,
770,
800,
860,
1020,
1110,
1320,
1620,
1120,
980,
1070,
1320,
1740,
2240,
2640,
2190,
1780,
1600,
1410,
1410,
1560,
1820,
2090,
2240,
2490,
3020,
3330,
3610,
4320,
4930,
5380,
5250,
4910,
4920,
4760,
4550,
4570,
5230,
6250,
7260,
8630,
10020,
9930,
10720,
12270,
14310]},
{'line': {'color': 'rgb(255, 127, 14)', 'width': 4},
'mode': 'lines',
'name': 'Hungary',
'type': 'scatter',
'x': [1960,
1961,
1962,
1963,
1964,
1965,
1966,
1967,
1968,
1969,
1970,
1971,
1972,
1973,
1974,
1975,
1976,
1977,
1978,
1979,
1980,
1981,
1982,
1983,
1984,
1985,
1986,
1987,
1988,
1989,
1990,
1991,
1992,
1993,
1994,
1995,
1996,
1997,
1998,
1999,
2000,
2001,
2002,
2003,
2004,
2005,
2006,
2007,
2008,
2009,
2010,
2011,
2012],
'y': [None,
None,
None,
None,
None,
None,
None,
None,
None,
None,
540,
590,
670,
830,
1000,
1150,
1200,
1330,
1520,
1770,
2070,
2200,
2170,
2010,
1930,
1860,
2040,
2400,
2710,
2770,
2880,
2740,
3140,
3630,
4000,
4220,
4320,
4370,
4380,
4460,
4580,
4720,
5210,
6550,
8540,
10220,
11040,
11510,
12890,
12980,
12930,
12900,
12410]},
{'line': {'color': 'rgb(44, 160, 44)', 'width': 4},
'mode': 'lines',
'name': 'Uruguay',
'type': 'scatter',
'x': [1960,
1961,
1962,
1963,
1964,
1965,
1966,
1967,
1968,
1969,
1970,
1971,
1972,
1973,
1974,
1975,
1976,
1977,
1978,
1979,
1980,
1981,
1982,
1983,
1984,
1985,
1986,
1987,
1988,
1989,
1990,
1991,
1992,
1993,
1994,
1995,
1996,
1997,
1998,
1999,
2000,
2001,
2002,
2003,
2004,
2005,
2006,
2007,
2008,
2009,
2010,
2011,
2012],
'y': [None,
None,
580,
610,
660,
680,
720,
640,
610,
670,
820,
850,
870,
1060,
1370,
1620,
1490,
1420,
1630,
2150,
2870,
3650,
3290,
2190,
1740,
1510,
1780,
2210,
2600,
2730,
2840,
3180,
3830,
4350,
5040,
5530,
6160,
6970,
7240,
7260,
7050,
6500,
5140,
4240,
4130,
4720,
5380,
6380,
7690,
8520,
10110,
11700,
13580]}],
'layout': {'annotations': [{'align': 'center',
'arrowcolor': '',
'arrowsize': 1,
'arrowwidth': 0,
'ax': -10,
'ay': -28.335936546325684,
'bgcolor': 'rgba(0,0,0,0)',
'bordercolor': '',
'borderwidth': 1,
'font': {'color': '', 'family': '', 'size': 0},
'opacity': 1,
'showarrow': False,
'tag': '',
'text': 'Source: <a href="http://blogs.worldbank.org/opendata/accessing-world-bank-data-apis-python-r-ruby-stata">World Bank</a>',
'x': 0.9880317848410782,
'xanchor': 'auto',
'xref': 'paper',
'y': 0.02994334820619583,
'yanchor': 'auto',
'yref': 'paper'}],
'autosize': True,
'bargap': 0.2,
'bargroupgap': 0,
'barmode': 'group',
'boxgap': 0.3,
'boxgroupgap': 0.3,
'boxmode': 'overlay',
'dragmode': 'zoom',
'font': {'color': 'rgb(67, 67, 67)',
'family': "'Open sans', verdana, arial, sans-serif",
'size': 12},
'height': 547,
'hidesources': False,
'hovermode': 'x',
'legend': {'bgcolor': '#fff',
'bordercolor': '#444',
'borderwidth': 0,
'font': {'color': '', 'family': '', 'size': 0},
'traceorder': 'normal',
'x': 1.02,
'xanchor': 'left',
'y': 0.5,
'yanchor': 'auto'},
'margin': {'autoexpand': True,
'b': 80,
'l': 80,
'r': 80,
't': 100},
'paper_bgcolor': '#fff',
'plot_bgcolor': 'rgba(245, 247, 247, 0.7)',
'separators': '.,',
'showlegend': True,
'title': 'GNI Per Capita ($USD Atlas Method)', 'titlefont': {'color': '', 'family': '', 'size': 0}, 'width': 1304, 'xaxis': {'anchor': 'y', 'autorange': True, 'autotick': True, 'domain': [0, 1], 'dtick': 10, 'exponentformat': 'B', 'gridcolor': 'rgb(255, 255, 255)', 'gridwidth': 1, 'linecolor': '#444', 'linewidth': 1, 'mirror': False, 'nticks': 0, 'overlaying': False, 'position': 0, 'range': [1960, 2012], 'rangemode': 'normal', 'showexponent': 'all', 'showgrid': True, 'showline': False, 'showticklabels': True, 'tick0': 0, 'tickangle': 'auto', 'tickcolor': '#444', 'tickfont': {'color': '', 'family': '', 'size': 0}, 'ticklen': 5, 'ticks': '', 'tickwidth': 1, 'title': 'year', 'titlefont': {'color': '', 'family': '', 'size': 0}, 'type': 'linear', 'zeroline': False, 'zerolinecolor': '#444', 'zerolinewidth': 1}, 'yaxis': {'anchor': 'x', 'autorange': True, 'autotick': True, 'domain': [0, 1], 'dtick': 'D1', 'exponentformat': 'B', 'gridcolor': 'rgb(255, 255, 255)', 'gridwidth': 1, 'linecolor': '#444', 'linewidth': 1, 'mirror': False, 'nticks': 0, 'overlaying': False, 'position': 0, 'range': [2.6533245446042573, 4.234708848978488], 'rangemode': 'normal', 'showexponent': 'all', 'showgrid': True, 'showline': False, 'showticklabels': True, 'tick0': 0, 'tickangle': 'auto', 'tickcolor': '#444', 'tickfont': {'color': '', 'family': '', 'size': 0}, 'ticklen': 5, 'ticks': '', 'tickwidth': 1, 'title': 'NY.GNP.PCAP.CD', 'titlefont': {'color': '', 'family': '', 'size': 0}, 'type': 'log', 'zeroline': False, 'zerolinecolor': '#444', 'zerolinewidth': 1}}} Want to analyze the data or use it for another figure? In [21]: ggplot_data = ggplot.get_data() In [22]: ggplot_data Out[22]: {'data': [{'name': 'Chile', 'x': [1960, 1961, 1962, 1963, 1964, 1965, 1966, 1967, 1968, 1969, 1970, 1971, 1972, 1973, 1974, 1975, 1976, 1977, 1978, 1979, 1980, 1981, 1982, 1983, 1984, 1985, 1986, 1987, 1988, 1989, 1990, 1991, 1992, 1993, 1994, 1995, 1996, 1997, 1998, 1999, 2000, 2001, 2002, 2003, 2004, 2005, 2006, 2007, 2008, 2009, 2010, 2011, 2012], 'y': [None, None, 600, 640, 660, 650, 740, 760, 770, 800, 860, 1020, 1110, 1320, 1620, 1120, 980, 1070, 1320, 1740, 2240, 2640, 2190, 1780, 1600, 1410, 1410, 1560, 1820, 2090, 2240, 2490, 3020, 3330, 3610, 4320, 4930, 5380, 5250, 4910, 4920, 4760, 4550, 4570, 5230, 6250, 7260, 8630, 10020, 9930, 10720, 12270, 14310]}, {'name': 'Hungary', 'x': [1960, 1961, 1962, 1963, 1964, 1965, 1966, 1967, 1968, 1969, 1970, 1971, 1972, 1973, 1974, 1975, 1976, 1977, 1978, 1979, 1980, 1981, 1982, 1983, 1984, 1985, 1986, 1987, 1988, 1989, 1990, 1991, 1992, 1993, 1994, 1995, 1996, 1997, 1998, 1999, 2000, 2001, 2002, 2003, 2004, 2005, 2006, 2007, 2008, 2009, 2010, 2011, 2012], 'y': [None, None, None, None, None, None, None, None, None, None, 540, 590, 670, 830, 1000, 1150, 1200, 1330, 1520, 1770, 2070, 2200, 2170, 2010, 1930, 1860, 2040, 2400, 2710, 2770, 2880, 2740, 3140, 3630, 4000, 4220, 4320, 4370, 4380, 4460, 4580, 4720, 5210, 6550, 8540, 10220, 11040, 11510, 12890, 12980, 12930, 12900, 12410]}, {'name': 'Uruguay', 'x': [1960, 1961, 1962, 1963, 1964, 1965, 1966, 1967, 1968, 1969, 1970, 1971, 1972, 1973, 1974, 1975, 1976, 1977, 1978, 1979, 1980, 1981, 1982, 1983, 1984, 1985, 1986, 1987, 1988, 1989, 1990, 1991, 1992, 1993, 1994, 1995, 1996, 1997, 1998, 1999, 2000, 2001, 2002, 2003, 2004, 2005, 2006, 2007, 2008, 2009, 2010, 2011, 2012], 'y': [None, None, 580, 610, 660, 680, 720, 640, 610, 670, 820, 850, 870, 1060, 1370, 1620, 1490, 1420, 1630, 2150, 2870, 3650, 3290, 2190, 1740, 1510, 1780, 2210, 2600, 2730, 2840, 3180, 3830, 4350, 5040, 5530, 6160, 6970, 7240, 7260, 7050, 6500, 5140, 4240, 4130, 4720, 5380, 6380, 7690, 8520, 10110, 11700, 13580]}], 'layout': [{}]} Want to use Python to analyze your data? You can read that data into a pandas DataFrame. In [23]: import pandas as pd In [24]: my_data = py.get_figure('MattSundquist', '1339').get_data() frames = {data['name']: {'x': data['x'], 'y': data['y']} for data in my_data['data']} df = pd.DataFrame(frames) df Out[24]: Chile Hungary Uruguay x [1960, 1961, 1962, 1963, 1964, 1965, 1966, 196... [1960, 1961, 1962, 1963, 1964, 1965, 1966, 196... [1960, 1961, 1962, 1963, 1964, 1965, 1966, 196... y [None, None, 600, 640, 660, 650, 740, 760, 770... [None, None, None, None, None, None, None, Non... [None, None, 580, 610, 660, 680, 720, 640, 610... 2 rows × 3 columns Plotly has interactive support that lets you call help on graph objects. Try layout or data too. For example. In [25]: from plotly.graph_objs import Data, Layout, Figure In [26]: help(Figure) Help on class Figure in module plotly.graph_objs.graph_objs: class Figure(PlotlyDict) | A dictionary-like object representing a figure to be rendered in plotly. | | This is the container for all things to be rendered in a figure. | | For help with setting up subplots, run: | help(plotly.tools.get_subplots) | | | Quick method reference: | | Figure.update(changes) | Figure.strip_style() | Figure.get_data() | Figure.to_graph_objs() | Figure.validate() | Figure.to_string() | Figure.force_clean() | | Valid keys: | | data [required=False] (value=Data object | dictionary-like): | A list-like array of the data that is to be visualized. | | For more, run help(plotly.graph_objs.Data) | | layout [required=False] (value=Layout object | dictionary-like): | The layout dictionary-like object contains axes information, gobal | settings, and layout information related to the rendering of the | figure. | | For more, run help(plotly.graph_objs.Layout) | | Method resolution order: | Figure | PlotlyDict | __builtin__.dict | __builtin__.object | | Methods defined here: | | __init__(self, *args, **kwargs) | | ---------------------------------------------------------------------- | Methods inherited from PlotlyDict: | | force_clean(self) | Attempts to convert to graph_objs and call force_clean() on values. | | Calling force_clean() on a PlotlyDict will ensure that the object is | valid and may be sent to plotly. This process will also remove any | entries that end up with a length == 0. | | Careful! This will delete any invalid entries *silently*. | | get_data(self) | Returns the JSON for the plot with non-data elements stripped. | | strip_style(self) | Strip style from the current representation. | | All PlotlyDicts and PlotlyLists are guaranteed to survive the | stripping process, though they made be left empty. This is allowable. | | Keys that will be stripped in this process are tagged with | 'type': 'style' in the INFO dictionary listed in graph_objs_meta.py. | | This process first attempts to convert nested collections from dicts | or lists to subclasses of PlotlyList/PlotlyDict. This process forces | a validation, which may throw exceptions. | | Then, each of these objects call strip_style on themselves and so | on, recursively until the entire structure has been validated and | stripped. | | to_graph_objs(self) | Walk obj, convert dicts and lists to plotly graph objs. | | For each key in the object, if it corresponds to a special key that | should be associated with a graph object, the ordinary dict or list | will be reinitialized as a special PlotlyDict or PlotlyList of the | appropriate kind. | | to_string(self, level=0, indent=4, eol='\n', pretty=True, max_chars=80) | Returns a formatted string showing graph_obj constructors. | | Example: | | print obj.to_string() | | Keyword arguments: | level (default = 0) -- set number of indentations to start with | indent (default = 4) -- set indentation amount | eol (default = ' | ') -- set end of line character(s) | pretty (default = True) -- curtail long list output with a '...' | max_chars (default = 80) -- set max characters per line | | update(self, dict1=None, **dict2) | Update current dict with dict1 and then dict2. | | This recursively updates the structure of the original dictionary-like | object with the new entries in the second and third objects. This | allows users to update with large, nested structures. | | Note, because the dict2 packs up all the keyword arguments, you can | specify the changes as a list of keyword agruments. | | Examples: | # update with dict | obj = Layout(title='my title', xaxis=XAxis(range=[0,1], domain=[0,1])) | update_dict = dict(title='new title', xaxis=dict(domain=[0,.8])) | obj.update(update_dict) | obj | {'title': 'new title', 'xaxis': {'range': [0,1], 'domain': [0,.8]}} | | # update with list of keyword arguments | obj = Layout(title='my title', xaxis=XAxis(range=[0,1], domain=[0,1])) | obj.update(title='new title', xaxis=dict(domain=[0,.8])) | obj | {'title': 'new title', 'xaxis': {'range': [0,1], 'domain': [0,.8]}} | | This 'fully' supports duck-typing in that the call signature is | identical, however this differs slightly from the normal update | method provided by Python's dictionaries. | | validate(self) | Recursively check the validity of the keys in a PlotlyDict. | | The valid keys constitute the entries in each object | dictionary in INFO stored in graph_objs_meta.py. | | The validation process first requires that all nested collections be | converted to the appropriate subclass of PlotlyDict/PlotlyList. Then, | each of these objects call validate and so on, recursively, | until the entire object has been validated. | | ---------------------------------------------------------------------- | Data descriptors inherited from PlotlyDict: | | __dict__ | dictionary for instance variables (if defined) | | __weakref__ | list of weak references to the object (if defined) | | ---------------------------------------------------------------------- | Data and other attributes inherited from PlotlyDict: | | __metaclass__ = <class 'plotly.graph_objs.graph_objs.DictMeta'> | A meta class for PlotlyDict class creation. | | The sole purpose of this meta class is to properly create the __doc__ | attribute so that running help(Obj), where Obj is a subclass of PlotlyDict, | will return information about key-value pairs for that object. | | ---------------------------------------------------------------------- | Methods inherited from __builtin__.dict: | | __cmp__(...) | x.__cmp__(y) <==> cmp(x,y) | | __contains__(...) | D.__contains__(k) -> True if D has a key k, else False | | __delitem__(...) | x.__delitem__(y) <==> del x[y] | | __eq__(...) | x.__eq__(y) <==> x==y | | __ge__(...) | x.__ge__(y) <==> x>=y | | __getattribute__(...) | x.__getattribute__('name') <==> x.name | | __getitem__(...) | x.__getitem__(y) <==> x[y] | | __gt__(...) | x.__gt__(y) <==> x>y | | __iter__(...) | x.__iter__() <==> iter(x) | | __le__(...) | x.__le__(y) <==> x<=y | | __len__(...) | x.__len__() <==> len(x) | | __lt__(...) | x.__lt__(y) <==> x<y | | __ne__(...) | x.__ne__(y) <==> x!=y | | __repr__(...) | x.__repr__() <==> repr(x) | | __setitem__(...) | x.__setitem__(i, y) <==> x[i]=y | | __sizeof__(...) | D.__sizeof__() -> size of D in memory, in bytes | | clear(...) | D.clear() -> None. Remove all items from D. | | copy(...) | D.copy() -> a shallow copy of D | | fromkeys(...) | dict.fromkeys(S[,v]) -> New dict with keys from S and values equal to v. | v defaults to None. | | get(...) | D.get(k[,d]) -> D[k] if k in D, else d. d defaults to None. | | has_key(...) | D.has_key(k) -> True if D has a key k, else False | | items(...) | D.items() -> list of D's (key, value) pairs, as 2-tuples | | iteritems(...) | D.iteritems() -> an iterator over the (key, value) items of D | | iterkeys(...) | D.iterkeys() -> an iterator over the keys of D | | itervalues(...) | D.itervalues() -> an iterator over the values of D | | keys(...) | D.keys() -> list of D's keys | | pop(...) | D.pop(k[,d]) -> v, remove specified key and return the corresponding value. | If key is not found, d is returned if given, otherwise KeyError is raised | | popitem(...) | D.popitem() -> (k, v), remove and return some (key, value) pair as a | 2-tuple; but raise KeyError if D is empty. | | setdefault(...) | D.setdefault(k[,d]) -> D.get(k,d), also set D[k]=d if k not in D | | values(...) | D.values() -> list of D's values | | viewitems(...) | D.viewitems() -> a set-like object providing a view on D's items | | viewkeys(...) | D.viewkeys() -> a set-like object providing a view on D's keys | | viewvalues(...) | D.viewvalues() -> an object providing a view on D's values | | ---------------------------------------------------------------------- | Data and other attributes inherited from __builtin__.dict: | | __hash__ = None | | __new__ = <built-in method __new__ of type object> | T.__new__(S, ...) -> a new object with type S, a subtype of T ## III. MATLAB, Julia, and Perl plotting with Plotly¶ We just made a plot with R using ggplot2, edited it in an IPython Notebook with Python, edited with our web app, shared it, and read the data into a pandas DataFrame. We have another Notebook that shows how to use Plotly with seaborn, prettyplotlib, and ggplot for Python Your whole team can now collaborate, regardless of technical capability or language of choice. This linguistic flexibility and technical interoperability powers collaboration, and it's what Plotly is all about. Let's jump into a few more examples. Let's say you see some code and data for a MATLAB gallery plot you love and want to share. In [27]: Image(url = 'http://i.imgur.com/bGj8EzI.png?1') Out[27]: You can use Plotly's MATLAB API to make a shareable plots, with LaTeX included. You run the MATLAB code in your MATLAB environrment or the MATLAB kernel in IPython and add fig2plotly to the call. Check out the user guide to see the installation and setup. In [28]: %%matlab close all % Create a set of values for the damping factor zeta = [0.01 .02 0.05 0.1 .2 .5 1 ]; % Define a color for each damping factor colors = ['r' 'g' 'b' 'c' 'm' 'y' 'k']; % Create a range of frequency values equally spaced logarithmically w = logspace(-1, 1, 1000); % Plot the gain vs. frequency for each of the seven damping factors figure; for i = 1:7 a = w.^2 - 1; b = 2*w*zeta(i); gain = sqrt(1./(a.^2 + b.^2)); loglog(w, gain, 'color', colors(i), 'linewidth', 2); hold on; end % Set the axis limits axis([0.1 10 0.01 100]); % Add a title and axis labels title('Gain vs Frequency'); xlabel('Frequency'); ylabel('Gain'); % Turn the grid on grid on; % ---------------------------------------- % Let's convert the figure to plotly structures, and set stripping to false [data, layout] = convertFigure(get(gcf), false); % But, before we publish, let's modify and add some features: % Naming the traces for i=1:numel(data) data{i}.name = ['$\\zeta = ' num2str(zeta(i)) '\$']; %LATEX FORMATTING
data{i}.showlegend = true;
end
% Adding a nice the legend
legendstyle = struct( ...
'x' , 0.15, ...
'y' , 0.9, ...
'bgcolor' , '#E2E2E2', ...
'bordercolor' , '#FFFFFF', ...
'borderwidth' , 2, ...
'traceorder' , 'normal' ...
);
layout.legend = legendstyle;
layout.showlegend = true;
% Setting the hover mode
layout.hovermode = 'closest';
% Giving the plot a custom name
plot_name = 'My_improved_plot';
% Sending to Plotly
response = plotly(data, struct('layout', layout, ...
'filename',plot_name, ...
'fileopt', 'overwrite'));
display(response.url)
https://plot.ly/~MATLAB-demos/4
Which produces:
In [29]:
tls.embed('MATLAB-Demos', '4')
And you can similarly collaborate across all Plotly APIs, working on plots from IJulia, Perl, Arduino, Raspberry Pi, or Ruby. You could also append data to any figure from any API, or from the GUI. Want to make your own wrapper? Check out our REST API.
## IV. WebPlotDigitizer and Plotly¶
Let's suppose next that you wanted to plot data from a graph you loved in a Facebook Data Science post.
In [30]:
Image(url = 'https://i.imgur.com/sAHsjk3.png')
Out[30]:
You can take a screenshot, and drag and drop the image into WebPlotDigitizer. Here's a tutorial on using the helpful tool, which includes the handy "Graph in Plotly" button. You can put it on your website so your users can easily access, graph, and share your data. And it links to your source.
In [31]:
Image (url = 'https://i.imgur.com/y4t5hdj.png')
Out[31]:
I can then make and share the graph in Plotly. You could do this to access data in any images you find online, then add fits or data from the grid or APIs. Check out our post with five fits to see more.
In [32]:
Image (url = 'http://i.imgur.com/BUOe85E.png')
Out[32]:
We'll add a fit then style it a bit.
In [33]:
tls.embed('MattSundquist', '1337')
## V. Revisions, embedding, and sharing¶
We can share it to edit collaboratively, privately or publicly. I can share straight into a folder from the API. My collaborators and I can always add, append, or extend data to that same plot with Python, R, or tbhe GUI.
In [34]:
Image(url = 'http://i.imgur.com/YRyTCQy.png')
Out[34]:
We can also save revisions and versions.
In [35]:
Image (url = 'http://i.imgur.com/ATn7vE4.png')
Out[35]:
You can also export your plot for presentations, emails, infographics, or publications, but link back to the online version so others can access your figure and data.
In [36]:
Image(url = 'http://i.imgur.com/QaIw9p4.png?1')
Out[36]:
You can also stop emailing files around. Have your discussion in context in Plotly. The graph being discussed is here.
In [37]:
Image(url = 'http://i.imgur.com/OqXKs0r.png')
Out[37]:
And displaying in your browser in an iframe is easy. You can copy and paste the snippet below and put it in a blog or website and get a live, interactive graph that lets your readers zoom, toggle, and get text on the hover.
In [38]:
from IPython.display import HTML
In [39]:
i = """<pre style="background:#f1f1f1;color:#000"><iframe src=<span style="color:#c03030">"https://plot.ly/~MattSundquist/1334/650/550"</span> width=<span style="color:#c03030">"650"</span> height=550<span style="color:#c03030">" frameBorder="</span>0<span style="color:#c03030">" seamless="</span>seamless<span style="color:#c03030">" scrolling="</span>no<span style="color:#c03030">"></iframe>
</span></pre>"""
In [40]:
h = HTML(i); h
Out[40]:
<iframe src="https://plot.ly/~MattSundquist/1334/650/550" width="650" height=550" frameBorder="0" seamless="seamless" scrolling="no"></iframe>
It's also interactive, even when embedded.
In [41]:
HTML('<br><center><iframe class="vine-embed" src="https://vine.co/v/Mvzin6HZzLB/embed/simple" width="600" height="600" frameborder="0"></iframe><script async src="//platform.vine.co/static/scripts/embed.js" charset="utf-8"></script></center><br>')
Out[41]:
Your profile keeps all your graphs and data together like this https://plot.ly/~jackp/.
In [42]:
Image(url='https://i.imgur.com/gUC4ajR.png')
Out[42]:
Plotly also does content. Check out our's posts on boxplots or histograms.
In [43]:
HTML('<center><iframe class="vine-embed" src="https://vine.co/v/M6JBhdiqPqA/embed/simple" width="600" height="600" frameborder="0"></iframe><script async src="//platform.vine.co/static/scripts/embed.js" charset="utf-8"></script></center>')
Out[43]:
## VI. Streaming Graphs¶
You can stream data into Plotly. That means you could publish your results to anyone in the world by streaming it through Plotly. You could also send data from multiple sources and languages, and keep your data around to analyze and publish it.
In [44]:
tls.embed('flann321', '9')
Or you can even stream in real-time. Check out a Notebook here or see our Raspberry Pi Instructable showing real-time dissolved oxygen.
In [45]:
tls.embed('streaming-demos','4')
You can stream from basically anywhere.
In [46]:
HTML('<center><iframe src="//instagram.com/p/nJkMMQRyvS/embed/" width="612" height="710" frameborder="0" scrolling="no" allowtransparency="true"></iframe></center>')
Out[46]:
Still need help? | 9,445 | 29,473 | {"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": 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} | 2.96875 | 3 | CC-MAIN-2017-39 | longest | en | 0.757172 |
https://en.wikipedia.org/wiki/Rayleigh_distribution | 1,713,594,549,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817491.77/warc/CC-MAIN-20240420060257-20240420090257-00036.warc.gz | 214,723,732 | 40,452 | Rayleigh distribution
Parameters Probability density function Cumulative distribution function scale: ${\displaystyle \sigma >0}$ ${\displaystyle x\in [0,\infty )}$ ${\displaystyle {\frac {x}{\sigma ^{2}}}e^{-x^{2}/\left(2\sigma ^{2}\right)}}$ ${\displaystyle 1-e^{-x^{2}/\left(2\sigma ^{2}\right)}}$ ${\displaystyle Q(F;\sigma )=\sigma {\sqrt {-2\ln(1-F)}}}$ ${\displaystyle \sigma {\sqrt {\frac {\pi }{2}}}}$ ${\displaystyle \sigma {\sqrt {2\ln(2)}}}$ ${\displaystyle \sigma }$ ${\displaystyle {\frac {4-\pi }{2}}\sigma ^{2}}$ ${\displaystyle {\frac {2{\sqrt {\pi }}(\pi -3)}{(4-\pi )^{3/2}}}}$ ${\displaystyle -{\frac {6\pi ^{2}-24\pi +16}{(4-\pi )^{2}}}}$ ${\displaystyle 1+\ln \left({\frac {\sigma }{\sqrt {2}}}\right)+{\frac {\gamma }{2}}}$ ${\displaystyle 1+\sigma te^{\sigma ^{2}t^{2}/2}{\sqrt {\frac {\pi }{2}}}\left(\operatorname {erf} \left({\frac {\sigma t}{\sqrt {2}}}\right)+1\right)}$ ${\displaystyle 1-\sigma te^{-\sigma ^{2}t^{2}/2}{\sqrt {\frac {\pi }{2}}}\left(\operatorname {erfi} \left({\frac {\sigma t}{\sqrt {2}}}\right)-i\right)}$
In probability theory and statistics, the Rayleigh distribution is a continuous probability distribution for nonnegative-valued random variables. Up to rescaling, it coincides with the chi distribution with two degrees of freedom. The distribution is named after Lord Rayleigh (/ˈrli/).[1]
A Rayleigh distribution is often observed when the overall magnitude of a vector in the plane is related to its directional components. One example where the Rayleigh distribution naturally arises is when wind velocity is analyzed in two dimensions. Assuming that each component is uncorrelated, normally distributed with equal variance, and zero mean, which is infrequent, then the overall wind speed (vector magnitude) will be characterized by a Rayleigh distribution. A second example of the distribution arises in the case of random complex numbers whose real and imaginary components are independently and identically distributed Gaussian with equal variance and zero mean. In that case, the absolute value of the complex number is Rayleigh-distributed.
Definition
The probability density function of the Rayleigh distribution is[2]
${\displaystyle f(x;\sigma )={\frac {x}{\sigma ^{2}}}e^{-x^{2}/(2\sigma ^{2})},\quad x\geq 0,}$
where ${\displaystyle \sigma }$ is the scale parameter of the distribution. The cumulative distribution function is[2]
${\displaystyle F(x;\sigma )=1-e^{-x^{2}/(2\sigma ^{2})}}$
for ${\displaystyle x\in [0,\infty ).}$
Relation to random vector length
Consider the two-dimensional vector ${\displaystyle Y=(U,V)}$ which has components that are bivariate normally distributed, centered at zero, and independent.[clarification needed] Then ${\displaystyle U}$ and ${\displaystyle V}$ have density functions
${\displaystyle f_{U}(x;\sigma )=f_{V}(x;\sigma )={\frac {e^{-x^{2}/(2\sigma ^{2})}}{\sqrt {2\pi \sigma ^{2}}}}.}$
Let ${\displaystyle X}$ be the length of ${\displaystyle Y}$. That is, ${\displaystyle X={\sqrt {U^{2}+V^{2}}}.}$ Then ${\displaystyle X}$ has cumulative distribution function
${\displaystyle F_{X}(x;\sigma )=\iint _{D_{x}}f_{U}(u;\sigma )f_{V}(v;\sigma )\,dA,}$
where ${\displaystyle D_{x}}$ is the disk
${\displaystyle D_{x}=\left\{(u,v):{\sqrt {u^{2}+v^{2}}}\leq x\right\}.}$
Writing the double integral in polar coordinates, it becomes
${\displaystyle F_{X}(x;\sigma )={\frac {1}{2\pi \sigma ^{2}}}\int _{0}^{2\pi }\int _{0}^{x}re^{-r^{2}/(2\sigma ^{2})}\,dr\,d\theta ={\frac {1}{\sigma ^{2}}}\int _{0}^{x}re^{-r^{2}/(2\sigma ^{2})}\,dr.}$
Finally, the probability density function for ${\displaystyle X}$ is the derivative of its cumulative distribution function, which by the fundamental theorem of calculus is
${\displaystyle f_{X}(x;\sigma )={\frac {d}{dx}}F_{X}(x;\sigma )={\frac {x}{\sigma ^{2}}}e^{-x^{2}/(2\sigma ^{2})},}$
which is the Rayleigh distribution. It is straightforward to generalize to vectors of dimension other than 2. There are also generalizations when the components have unequal variance or correlations (Hoyt distribution), or when the vector Y follows a bivariate Student t-distribution (see also: Hotelling's T-squared distribution).[3]
Generalization to bivariate Student's t-distribution
Suppose ${\displaystyle Y}$ is a random vector with components ${\displaystyle u,v}$ that follows a multivariate t-distribution. If the components both have mean zero, equal variance, and are independent, the bivariate Student's-t distribution takes the form:
${\displaystyle f(u,v)={1 \over {2\pi \sigma ^{2}}}\left(1+{u^{2}+v^{2} \over {\nu \sigma ^{2}}}\right)^{-\nu /2-1}}$
Let ${\displaystyle R={\sqrt {U^{2}+V^{2}}}}$ be the magnitude of ${\displaystyle Y}$. Then the cumulative distribution function (CDF) of the magnitude is:
${\displaystyle F(r)={1 \over {2\pi \sigma ^{2}}}\iint _{D_{r}}\left(1+{u^{2}+v^{2} \over {\nu \sigma ^{2}}}\right)^{-\nu /2-1}du\;dv}$
where ${\displaystyle D_{r}}$ is the disk defined by:
${\displaystyle D_{r}=\left\{(u,v):{\sqrt {u^{2}+v^{2}}}\leq r\right\}}$
Converting to polar coordinates leads to the CDF becoming:
{\displaystyle {\begin{aligned}F(r)&={1 \over {2\pi \sigma ^{2}}}\int _{0}^{r}\int _{0}^{2\pi }\rho \left(1+{\rho ^{2} \over {\nu \sigma ^{2}}}\right)^{-\nu /2-1}d\theta \;d\rho \\&={1 \over {\sigma ^{2}}}\int _{0}^{r}\rho \left(1+{\rho ^{2} \over {\nu \sigma ^{2}}}\right)^{-\nu /2-1}d\rho \\&=1-\left(1+{r^{2} \over {\nu \sigma ^{2}}}\right)^{-\nu /2}\end{aligned}}}
Finally, the probability density function (PDF) of the magnitude may be derived:
${\displaystyle f(r)=F'(r)={r \over {\sigma ^{2}}}\left(1+{r^{2} \over {\nu \sigma ^{2}}}\right)^{-\nu /2-1}}$
In the limit as ${\displaystyle \nu \rightarrow \infty }$, the Rayleigh distribution is recovered because:
${\displaystyle \lim _{\nu \rightarrow \infty }\left(1+{r^{2} \over {\nu \sigma ^{2}}}\right)^{-\nu /2-1}=e^{-r^{2}/2\sigma ^{2}}}$
Properties
The raw moments are given by:
${\displaystyle \mu _{j}=\sigma ^{j}2^{j/2}\,\Gamma \left(1+{\frac {j}{2}}\right),}$
where ${\displaystyle \Gamma (z)}$ is the gamma function.
The mean of a Rayleigh random variable is thus :
${\displaystyle \mu (X)=\sigma {\sqrt {\frac {\pi }{2}}}\ \approx 1.253\ \sigma .}$
The standard deviation of a Rayleigh random variable is:
${\displaystyle \operatorname {std} (X)={\sqrt {\left(2-{\frac {\pi }{2}}\right)}}\sigma \approx 0.655\ \sigma }$
The variance of a Rayleigh random variable is :
${\displaystyle \operatorname {var} (X)=\mu _{2}-\mu _{1}^{2}=\left(2-{\frac {\pi }{2}}\right)\sigma ^{2}\approx 0.429\ \sigma ^{2}}$
The mode is ${\displaystyle \sigma ,}$ and the maximum pdf is
${\displaystyle f_{\max }=f(\sigma ;\sigma )={\frac {1}{\sigma }}e^{-1/2}\approx {\frac {0.606}{\sigma }}.}$
The skewness is given by:
${\displaystyle \gamma _{1}={\frac {2{\sqrt {\pi }}(\pi -3)}{(4-\pi )^{3/2}}}\approx 0.631}$
The excess kurtosis is given by:
${\displaystyle \gamma _{2}=-{\frac {6\pi ^{2}-24\pi +16}{(4-\pi )^{2}}}\approx 0.245}$
The characteristic function is given by:
${\displaystyle \varphi (t)=1-\sigma te^{-{\frac {1}{2}}\sigma ^{2}t^{2}}{\sqrt {\frac {\pi }{2}}}\left[\operatorname {erfi} \left({\frac {\sigma t}{\sqrt {2}}}\right)-i\right]}$
where ${\displaystyle \operatorname {erfi} (z)}$ is the imaginary error function. The moment generating function is given by
${\displaystyle M(t)=1+\sigma t\,e^{{\frac {1}{2}}\sigma ^{2}t^{2}}{\sqrt {\frac {\pi }{2}}}\left[\operatorname {erf} \left({\frac {\sigma t}{\sqrt {2}}}\right)+1\right]}$
where ${\displaystyle \operatorname {erf} (z)}$ is the error function.
Differential entropy
The differential entropy is given by[citation needed]
${\displaystyle H=1+\ln \left({\frac {\sigma }{\sqrt {2}}}\right)+{\frac {\gamma }{2}}}$
where ${\displaystyle \gamma }$ is the Euler–Mascheroni constant.
Parameter estimation
Given a sample of N independent and identically distributed Rayleigh random variables ${\displaystyle x_{i}}$ with parameter ${\displaystyle \sigma }$,
${\displaystyle {\widehat {\sigma ^{2}}}=\!\,{\frac {1}{2N}}\sum _{i=1}^{N}x_{i}^{2}}$ is the maximum likelihood estimate and also is unbiased.
${\displaystyle {\widehat {\sigma }}\approx {\sqrt {{\frac {1}{2N}}\sum _{i=1}^{N}x_{i}^{2}}}}$ is a biased estimator that can be corrected via the formula
${\displaystyle \sigma ={\widehat {\sigma }}{\frac {\Gamma (N){\sqrt {N}}}{\Gamma \left(N+{\frac {1}{2}}\right)}}={\widehat {\sigma }}{\frac {4^{N}N!(N-1)!{\sqrt {N}}}{(2N)!{\sqrt {\pi }}}}}$[4] ${\displaystyle ={\frac {\hat {\sigma }}{c_{4}(2N+1)}}}$, where c4 is the correction factor used to unbias estimates of standard deviation for normal random variables.
Confidence intervals
To find the (1 − α) confidence interval, first find the bounds ${\displaystyle [a,b]}$ where:
${\displaystyle P\left(\chi _{2N}^{2}\leq a\right)=\alpha /2,\quad P\left(\chi _{2N}^{2}\leq b\right)=1-\alpha /2}$
then the scale parameter will fall within the bounds
${\displaystyle {\frac {{N}{\overline {x^{2}}}}{b}}\leq {\widehat {\sigma ^{2}}}\leq {\frac {{N}{\overline {x^{2}}}}{a}}}$[5]
Generating random variates
Given a random variate U drawn from the uniform distribution in the interval (0, 1), then the variate
${\displaystyle X=\sigma {\sqrt {-2\ln U}}\,}$
has a Rayleigh distribution with parameter ${\displaystyle \sigma }$. This is obtained by applying the inverse transform sampling-method.
Related distributions
• ${\displaystyle R\sim \mathrm {Rayleigh} (\sigma )}$ is Rayleigh distributed if ${\displaystyle R={\sqrt {X^{2}+Y^{2}}}}$, where ${\displaystyle X\sim N(0,\sigma ^{2})}$ and ${\displaystyle Y\sim N(0,\sigma ^{2})}$ are independent normal random variables.[6] This gives motivation to the use of the symbol ${\displaystyle \sigma }$ in the above parametrization of the Rayleigh density.
• The magnitude ${\displaystyle |z|}$ of a standard complex normally distributed variable z is Rayleigh distributed.
• The chi distribution with v = 2 is equivalent to the Rayleigh Distribution with σ = 1: ${\displaystyle R(\sigma )\sim \sigma \chi _{2}^{\,}\ .}$
• If ${\displaystyle R\sim \mathrm {Rayleigh} (1)}$, then ${\displaystyle R^{2}}$ has a chi-squared distribution with 2 degrees of freedom: ${\displaystyle [Q=R(\sigma )^{2}]\sim \sigma ^{2}\chi _{2}^{2}\ .}$
• If ${\displaystyle R\sim \mathrm {Rayleigh} (\sigma )}$, then ${\displaystyle \sum _{i=1}^{N}R_{i}^{2}}$ has a gamma distribution with parameters ${\displaystyle N}$ and ${\displaystyle {\frac {1}{2\sigma ^{2}}}}$
${\displaystyle \left[Y=\sum _{i=1}^{N}R_{i}^{2}\right]\sim \Gamma \left(N,{\frac {1}{2\sigma ^{2}}}\right).}$
• The Rice distribution is a noncentral generalization of the Rayleigh distribution: ${\displaystyle \mathrm {Rayleigh} (\sigma )=\mathrm {Rice} (0,\sigma )}$.
• The Weibull distribution with the shape parameter k = 2 yields a Rayleigh distribution. Then the Rayleigh distribution parameter ${\displaystyle \sigma }$ is related to the Weibull scale parameter according to ${\displaystyle \lambda =\sigma {\sqrt {2}}.}$
• If ${\displaystyle X}$ has an exponential distribution ${\displaystyle X\sim \mathrm {Exponential} (\lambda )}$, then ${\displaystyle Y={\sqrt {X}}\sim \mathrm {Rayleigh} (1/{\sqrt {2\lambda }}).}$
• The half-normal distribution is the one-dimensional equivalent of the Rayleigh distribution.
• The Maxwell–Boltzmann distribution is the three-dimensional equivalent of the Rayleigh distribution.
Applications
An application of the estimation of σ can be found in magnetic resonance imaging (MRI). As MRI images are recorded as complex images but most often viewed as magnitude images, the background data is Rayleigh distributed. Hence, the above formula can be used to estimate the noise variance in an MRI image from background data.[7] [8]
The Rayleigh distribution was also employed in the field of nutrition for linking dietary nutrient levels and human and animal responses. In this way, the parameter σ may be used to calculate nutrient response relationship.[9]
In the field of ballistics, the Rayleigh distribution is used for calculating the circular error probable—a measure of a gun's precision.
In physical oceanography, the distribution of significant wave height approximately follows a Rayleigh distribution.[10] | 3,770 | 12,273 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 91, "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} | 3.28125 | 3 | CC-MAIN-2024-18 | latest | en | 0.739581 |
https://www.jiskha.com/display.cgi?id=1190782687 | 1,503,394,811,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886110573.77/warc/CC-MAIN-20170822085147-20170822105147-00018.warc.gz | 904,920,462 | 4,213 | math
posted by .
Simplify: (8-6i)/ 3i:
A. 6
B. (6 + 8i)/ -3
C. (6 + 8i)/ 3
D. 2 + 8i
• math -
Please show us your effort in simplifying the (8-6i)/ 3i, and pick the answer that fits. If you can't do this, then I suggest you seek private tutoring.
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Here are the questions I need help with, please help. [(1/4)^-3 divide sign (1/4)^2]^-2 I got (1/4)^10 but I got it wrong (3^1/2)^3 _______ I got 1/216 9 (-56r^2)divide sign (-7r^2)= I got 8r x(y)(x)(2y) (12) (10xy^3) (3x^2y^2) (5x) …
8. Math To: Reiny
This is one whole question: (12) (10xy^3) (3x^2y^2) (5x) (2y) I don't know how to do any of these: simplify sqaure root of 160000 simplify sqaure root of 0.04 simplify 3 * square root 81 simplify square root of 900 simplify square …
9. Math...help
remove the brackets and simplify questions a) 3(2a+5)+2(a+7) b) 5(b+4)+10b c) 4(2c+3)+3(c-3) d) 8+6(d-2) e) 6(5e+3)-4(2e+3) f) 5(3f+2)-(f-5) MY ANSWERS a)6a+15+2a+14 8a+29 b)5b+20+10b 15b+20 c)8c+12+3c-9 11c-3 d)8+6d-12 I need to simplify …
10. urgent math
1. Simplify the expression 7^9/7^3 a.7^3*** b.7^6 c.7^12 d.1^6 2. Simplify the expression z^8/z^12 a.z^20 b.z^4 c.1/z^-4 d.1/z^4 3. Simplify the expression (-3^4)/(-3^4 a.(-3)^1 b.0 c.1 d.(-3)^8 4.Which expressions can be rewritten …
More Similar Questions | 1,062 | 2,537 | {"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} | 3.4375 | 3 | CC-MAIN-2017-34 | latest | en | 0.847654 |
https://writingsgate.com/2022/04/28/9-consider-the-following-game-in-extensive-form-n-1-2-31-4-1-2-3-12-2-2-5-6-1-2-4-7-4-3-6-6-7-3-of-3-7-4-4-2-6/ | 1,669,563,188,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710409.16/warc/CC-MAIN-20221127141808-20221127171808-00300.warc.gz | 655,826,551 | 19,322 | Home » 9.* Consider the following game in extensive form : N = ( 1 , 2, 31 . 4 1 . 2 3 12. 2.2 5, 6 , 1 2 , 4 , 7 4, 3 , 6 6 , 7 , 3 of 3. 7. 4 4, 2, 6
# 9.* Consider the following game in extensive form : N = ( 1 , 2, 31 . 4 1 . 2 3 12. 2.2 5, 6 , 1 2 , 4 , 7 4, 3 , 6 6 , 7 , 3 of 3. 7. 4 4, 2, 6
(a) Write down the set of pure strategies of players 1, 2, and 3.
S1 = { ae, af, be, bf } S2 = { ck, cl, cm, dk, dl, dm} S3 = { g, h}
(b) How many subgames does the game have? What are they?
The game has three Subgame which are circled in three blue rectangulars. The initial node (1.1) initiates a subgame, because the entire game is itself a subgame. Node 1.2 also initiates a subgame, because it begins with an information set containing only single decision node. Moreover, node 2.2 also initiates a subgame, because it begins with an information set containing a single decision node. However, Node 2.1 does not initiates a subgame, because one of its successors (node 3) is in the same information set as a node that is not one of its successors. (node 3 follows node 1.1)
9.* Consider the following game in extensive form : N = ( 1 , 2, 31 .41 . 2312.2.25, 6 , 12 , 4 , 74, 3 , 66 , 7 , 3of3. 7. 44, 2, 66 , 3 , 1( a ) Write down the set of pure strategies of players 1 , 2 , and 3.( b ) How many subgames does the game have ? What are they ?( C ) Determine the set of pure – strategy subgame perfect equilibria ( SPE ) . Explain your steps carefully .What is ( are ) the SPF path ( s ) and the SPE payoffs ?”NOTE : Make sure your answer is based on the logic of SPE .]
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Thanks to our free revisions, there is no way for you to be unsatisfied. We will work on your paper until you are completely happy with the result. | 887 | 2,972 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2022-49 | latest | en | 0.876153 |
https://au.mathworks.com/matlabcentral/cody/problems/44618-surface-areas-of-a-cylinder/solutions/1683166 | 1,591,343,155,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348493151.92/warc/CC-MAIN-20200605045722-20200605075722-00462.warc.gz | 247,301,352 | 15,868 | Cody
# Problem 44618. surface areas of a cylinder
Solution 1683166
Submitted on 29 Nov 2018
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Fail
optn=1; r=3; h=6; y_correct = 36*pi; assert(isequal(surface_area_cyl(optn,r,h),y_correct))
y = 18
Assertion failed.
2 Fail
optn=5; r=3; h=6; y_correct = 18*pi; assert(isequal(surface_area_cyl(optn,r,h),y_correct))
y = 36.8496
Assertion failed.
3 Fail
optn=2; r=2; h=2; y_correct = 16*pi; assert(isequal(surface_area_cyl(optn,r,h),y_correct))
y = 16.5664
Assertion failed.
4 Fail
optn=0; r=5; h=10; y_correct = 50*pi; assert(isequal(surface_area_cyl(optn,r,h),y_correct))
y = 81.4159
Assertion failed.
5 Fail
optn=1; r=5; h=10; y_correct = 100*pi; assert(isequal(surface_area_cyl(optn,r,h),y_correct))
y = 50
Assertion failed.
6 Fail
optn=2; r=5; h=10; y_correct = 150*pi; assert(isequal(surface_area_cyl(optn,r,h),y_correct))
y = 81.4159
Assertion failed.
7 Fail
optn=42; r=3; h=7; y_correct = 18*pi; assert(isequal(surface_area_cyl(optn,r,h),y_correct))
y = 39.8496
Assertion failed.
8 Fail
optn=2; r=3 h=7; y_correct = 60*pi; assert(isequal(surface_area_cyl(optn,r,h),y_correct))
r = 3 y = 39.8496
Assertion failed.
9 Fail
optn=1; r=3; h=7; y_correct = 42*pi; assert(isequal(surface_area_cyl(optn,r,h),y_correct))
y = 21
Assertion failed. | 518 | 1,445 | {"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} | 3.1875 | 3 | CC-MAIN-2020-24 | latest | en | 0.435241 |
https://rdrr.io/cran/RSCABS/src/R/prepDataRSCABS.R | 1,726,012,954,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651323.1/warc/CC-MAIN-20240910224659-20240911014659-00836.warc.gz | 453,256,772 | 8,352 | # R/prepDataRSCABS.R In RSCABS: Rao-Scott Cochran-Armitage by Slices Trend Test
#### Documented in prepDataRSCABS
```prepDataRSCABS <-
function(Effect='',Data={},Treatment='',Replicate=''){
#Data Transform from a list of individuals to matrix format
#This will take Clustered Data and convert it to by
#' @export
if(Effect=='&Fill#'){ #ensures the correct data structure output
return()
}
if(length(which(colnames(Data)==Effect))==0){
print(paste(Effect,' is not in data set. Ending function.',sep=''))
return()
}
K.max<-max(Data[ ,Effect],na.rm=TRUE) #max K score
#Remove NA and negative numbers
if (length(which(is.na(Data[ ,Effect])))>0){
Data<-Data[-which(is.na(Data[ ,Effect])), ]
}
if (length(which(Data[ ,Effect]<0))>0){
Data<-Data[-which(Data[ ,Effect]<0), ]
}
#Convert Factors to Numerics
Data[ ,Effect]<-as.numeric(Data[ ,Effect])
if (K.max==0){
print(paste('There is no variation in ',Effect,'. Ending function.',sep=''))
return()
}
#Replicates are rows, Treatment are columns [Replicate,Treatment]
n.i.j<-xtabs( ~Data[[Replicate]]+Data[[Treatment]])
m.i<-apply(n.i.j,2,function(Vec){
if (length(which(Vec==0))>0){
Vec<-Vec[-which(Vec==0)]
}
return(length(Vec))
})
x.i.j<-array(dim=c(dim(n.i.j)[1],dim(n.i.j)[2],K.max)) #Declare x.i.j , frequency array of scores k or larger
#Each k level is on the 3rd dimension
for (K in 1:K.max){
x.i.j[ , ,K]<-xtabs( ~Data[[Replicate]]+Data[[Treatment]],subset=Data[ ,Effect]>=K)
}
RSCABS.Prep.Data<-list(x.i.j=x.i.j,n.i.j=n.i.j,m.i=m.i,K.max=K.max)
return(RSCABS.Prep.Data)
}
```
## Try the RSCABS package in your browser
Any scripts or data that you put into this service are public.
RSCABS documentation built on May 1, 2020, 9:06 a.m. | 526 | 1,697 | {"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-2024-38 | latest | en | 0.484958 |
https://www.onlinemathlearning.com/measurement-word-problems-worksheet.html | 1,726,517,149,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651710.86/warc/CC-MAIN-20240916180320-20240916210320-00278.warc.gz | 832,544,878 | 9,637 | # Measurement Word Problems Worksheets
Printable “Metric Measurement” Worksheets:
Metric Length Conversions (km, m, cm)
Metric Mass Conversions (kg, g)
Metric Capacity Conversions (L, cL)
### Metric Measurement Word Problems Worksheets
In these free math worksheets, students practice how to solve metric measurement word problems.
How to solve measurement word problems?
Solving measurement word problems involves interpreting the information given in the problem, identifying the appropriate units of measurement, and performing the necessary mathematical operations.
1. Read the problem carefully and understand the context and information presented in the word problem.
2. Determine what quantities are given (knowns) and what quantity needs to be found (unknown).
3. Identify the units of measurement for the known and unknown quantities. Ensure all units are consistent.
4. Based on the information provided, draw a diagram and write an equation that represents the relationship between the known and unknown quantities. Use the appropriate mathematical operation.
5. Use the given information and the equation to perform the necessary mathematical operations (addition, subtraction, multiplication, or division).
6. Ensure that the answer makes sense in the context of the problem. Check units to ensure they are consistent.
Have a look at this video if you need to review how to solve measurement word problems.
Click on the following worksheet to get a printable pdf document.
Scroll down the page for more Measurement Word Problems Worksheets.
### More Measurement Word Problems Worksheets
Printable
Measurement Word Problems Worksheet #1
Measurement Word Problems Worksheet #2
Measurement Word Problems Worksheet #3
More Printable Worksheets
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. | 357 | 1,964 | {"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} | 3.265625 | 3 | CC-MAIN-2024-38 | latest | en | 0.85724 |
http://www.tectechnicsclassroom.tectechnics.com/s7p4a41/direction-of-a-projectile.php | 1,685,721,521,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648695.4/warc/CC-MAIN-20230602140602-20230602170602-00287.warc.gz | 89,280,052 | 6,215 | Direction Of Projectile
TECTechnics Classroom TECTechnics Overview
Direction Of Projectile
The path of a projectile is represented as follows:
y = 4x - x2
(a) What is the direction of the projectile when x = 1?
(b) At what value of x is the direction of the projectile horizontal?
The string:
S7P4A41 (Motion - Linear).
The math:
Pj Problem of Interest (PPI) is of type motion. Problems of distance traveled or direction of travel are motion problems.
(a) dy/dx = 4 - 2x
At x = 1, dy/dx = 2
The direction of projectile when x = 1 is the direction of a tangent with slope 2.
Direction of projectile is horizontal when dy/dx = 0
dy/dx = 0 when x = 2
So, direction of projectile is horizontal when x = 2.
Blessed are they that have not seen, and yet have believed. John 20:29 | 216 | 778 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2023-23 | latest | en | 0.898779 |
https://www.askiitians.com/forums/Trigonometry/prove-that-coseca-cota-tan-a-cosa-solve-this-i_176660.htm | 1,721,725,738,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518029.81/warc/CC-MAIN-20240723072353-20240723102353-00459.warc.gz | 561,898,609 | 42,500 | # Prove that: cosecA / cotA +tan A=cosA solve this i have an urgent please solve quickly
Vikas TU
14149 Points
7 years ago
cosecA/cotA +tan A=cosA
Consider LHS
cosecA/cotA +tan A
=(1/sin A)/(cos A/wrongdoing A)+sin A/cos A
=Cos A/(Cos2A+sin2A)
=RHS | 103 | 249 | {"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} | 3.265625 | 3 | CC-MAIN-2024-30 | latest | en | 0.650046 |
https://share.cocalc.com/share/5f9d9b647752c2467099c7b4b2daf240ce9ff768/Wienner%20Covert%20Channel.sagews?viewer=share | 1,623,628,201,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487611089.19/warc/CC-MAIN-20210613222907-20210614012907-00505.warc.gz | 467,591,556 | 3,544 | CoCalc Public FilesWienner Covert Channel.sagews
Views : 73
Compute Environment: Ubuntu 20.04 (Default)
def Wiener (n, s):
M = 10101
E = M.powermod(s,n)#&^(M, s) % n
cf = continued_fraction(n/s)# convert(n / s, confrac)
m = len(cf)
for i in range(0, m ):
conv = cf.convergent(i)
t = conv.numerator()
#t = cf.quotient(i)#numtheory:-nthnumer(cf, i)
d = E.powermod(t,n)#&^(E, t) % n
#print d
if d == M:
print("k:= {0}".format(cf.convergent(i).denominator()))
print("The totient of the modulus is {0}".format((s*t-1)/(cf.convergent(i).denominator())))
print("The private key is {0}".format(t))
return [t, (s*t-1)/(cf.convergent(i).denominator())]
print("Private key not found")
def WienerCovert (P):
p = next_prime(ZZ.random_element(P^2))
q = next_prime(ZZ.random_element(P^2))
while(not((q<p) and (p<2*q))):
p = next_prime(ZZ.random_element(P^2))
q = next_prime(ZZ.random_element(P^2))
n = p*q
phin = (p-1)*(q-1)
l = ZZ(floor((n^(1/4))/3))
for i in range(1000):
d1 = l-i
g = gcd(d1,phin)
if (g == 1):
s1 = 1/d1 % phin
s2 = s1*P % n
g1 = gcd(s2,phin)
if(g1 == 1):
d2 = 1/s2 % phin
print "The modulus is: ", n
print "The exponent is: ", s2
print "The key is: ", d2
return([n,s2,d2])
P = next_prime(983724836482)
R = WienerCovert(P)
print "end of creatiom of covert channel"
S = R[1]/P % R[0]
wienerAttackOutput = Wiener(R[0],S)
print "after wieners attack"
d = 1/R[1] % wienerAttackOutput[1]
d
if (R[2] == d):
print "correct private key computed"
The modulus is: 162996458187706035673135889099890482856112226697 The exponent is: 94344369869984932236004780866259533022210541589 The key is: 56419832253583618233458429492162599992204825621 end of creatiom of covert channel k:= 22430205604 The totient of the modulus is 162996458187706035673135081204137668474884176916 The private key is 211798698559 after wieners attack 56419832253583618233458429492162599992204825621 correct private key computed | 676 | 1,895 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2021-25 | latest | en | 0.37443 |
https://cboard.cprogramming.com/c-programming/14705-populating-two-dimensional-arrays-printable-thread.html | 1,493,312,185,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917122619.60/warc/CC-MAIN-20170423031202-00497-ip-10-145-167-34.ec2.internal.warc.gz | 791,525,347 | 3,106 | # populating two dimensional arrays
• 04-05-2002
garycastillo
populating two dimensional arrays
:confused:
I have an assignment that requires me to populate a two dimensioanl array and then use it to prepare a summary based on that array.
Ex. three stores each have three products, have the user input the total of each product for each store and then have the program list that information.
As far as I have gotten is this code for populating the array:
int iar[3][4]
for (rown=0;rown<3;j++)
for(coln=0;coln<4;k++)
printf("enter element ar[%d][%d]",rown, coln);
scanf("%d",&iar[rown][coln]);
If i could just get somebody to show me a complete program that populates a two dimensional array and then lists that information, I could finish the rest of the program.
Thanks to anybody that can help
• 04-05-2002
quzah
Code:
```int x, y, array[3][4]; for( y = 0; y < 4; y++ )for( x = 0; x < 3; x++ ) { printf( "Enter a value for array[%d][%d]: ", x, y ); scanf( "%d", array[x][y] ); } for( y = 0; y < 4; y++ )for( x = 0; x < 3; x++ ) { printf( "The value of array[%d][%d]: is %d.\n", x, y, array[x][y] ); }```
Enjoy.
Quzah.
• 04-05-2002
Prelude
Code:
```for( y = 0; y < 4; y++ )for( x = 0; x < 3; x++ ) { printf( "Enter a value for array[%d][%d]: ", x, y ); scanf( "%d", array[x][y] ); }```
That's an interesting implementation of nested loops. Kind of scary too. ;)
-Prelude | 461 | 1,400 | {"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} | 2.6875 | 3 | CC-MAIN-2017-17 | longest | en | 0.702603 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-2-1st-edition/chapter-5-polynomials-and-polynomial-functions-5-1-use-properties-of-exponents-5-1-exercises-skill-practice-page-333/30 | 1,716,446,268,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058611.55/warc/CC-MAIN-20240523050122-20240523080122-00892.warc.gz | 689,956,265 | 16,307 | ## Algebra 2 (1st Edition)
$$\frac{y^3}{x^3}$$
We know the following rules of exponents. The list of names is on page 330. $$(1) \ a^m\cdot a^n = a^{m+n} \\ (2) \ (ab)^m =a^mb^m \\ (3) \ (a^m)^n =a^{mn} \\ (4) \ a^{-m} = \frac{1}{a^m} \\ (5)\ \frac{a^m}{a^n} =a^{m-n} \\ (6) \ a^0=1 \\ (7) \ (\frac{a}{b})^m =\frac{a^m}{b^m}$$ Thus, using these properties, we find: $$\frac{y^2}{x^3y^{-1}} \\ \frac{y^3}{x^3}$$ | 205 | 411 | {"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} | 4.125 | 4 | CC-MAIN-2024-22 | latest | en | 0.501303 |
https://discourse.mcneel.com/t/how-to-make-this-surface-without-grasshopper-or-if-not-how-is-it-done-with-grasshopper/84865 | 1,718,821,382,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861828.24/warc/CC-MAIN-20240619154358-20240619184358-00035.warc.gz | 175,742,759 | 7,632 | # How to make this surface without Grasshopper, or if not, how is it done with grasshopper?
Hi There,
I’m trying to find a way of making a complex surface and don’t really have any Grasshopper experience.
I want to create a surface that blends from a flat to many cylindrical extrusions.
Image ‘D’ is what I want to create, I have read up about image ‘C’ mesh from points however I want it to be a NURBS surface and not mesh.
Image ‘A’ and ‘B’ is a basic version I have created in Rhino, trying to show what I want to achieve. Essentially I would like all of the pink surfaces to be one continuous surface with tangency conditions leading into every cylinder.
Any help would be greatly appreciated.
Cheers.
1 Like
Didn’t we see this same question answered just two days ago?
1 Like
This happens rather often. I believe some schools give assignments with option to use discourse and this results in many similar topics being created in a short span of time.
2 Likes
Thanks all and apologies, I should have searched a bit further. Just wasn’t sure what to search for!
Cheers and will take a good look at this.
Marc
Fun with Grasshopper!
I didn’t even try to do this on arbitrary curved surfaces.
In the “holed base” group, a Polygon or rectangle instead of a circle could be used as the input to Boundary. In any case, the boundary curve must encompass all the flanges or the surface will fail. Likewise, the “triangular points in circle” group could use a polygon or rectangle instead of a circle to trim points. The two serve different purposes.
homework_2019Jun17b.gh (27.1 KB)
2 Likes
Thanks Jospeh,
I missed this and haven’t logged on in some time, thanks for your help, it works really well.
Marc
1 Like | 398 | 1,731 | {"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-2024-26 | latest | en | 0.953274 |
https://www.equationsworksheets.net/exponential-equation-worksheet-pdf/ | 1,723,711,701,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641278776.95/warc/CC-MAIN-20240815075414-20240815105414-00178.warc.gz | 576,419,021 | 12,001 | # Exponential Equation Worksheet Pdf
Exponential Equation Worksheet Pdf – The aim of Expressions and Equations Worksheets is to aid your child to learn more efficiently and effectively. The worksheets include interactive exercises and questions dependent on the sequence of how operations are conducted. These worksheets are designed to make it easier for children to master complex concepts and simple concepts quickly. Download these free documents in PDF format. They will aid your child’s learning and practice math concepts. These materials are great for students who are in the 5th through 8th grades.
## Free Download Exponential Equation Worksheet Pdf
The worksheets listed here are designed for students from the 5th-8th grades. These two-step word problems are constructed using decimals or fractions. Each worksheet contains ten problems. These worksheets are available both online and in printed. These worksheets are a fantastic opportunity to practice rearranging equations. Alongside practicing the art of rearranging equations, they assist your student to understand the principles of equality as well as inverse operations.
The worksheets are intended for fifth and eight grade students. They are ideal for students who have difficulty calculating percentages. You can select from three types of problems. You can choose to solve one-step problems that include decimal or whole numbers, or you can use word-based approaches to solve problems involving decimals and fractions. Each page will have 10 equations. These worksheets for Equations can be used by students from the 5th-8th grades.
These worksheets can be a wonderful resource for practicing fraction calculations as well as other concepts that are related to algebra. A lot of these worksheets allow users to select from three different kinds of problems. You can select a word-based problem or a numerical. It is vital to pick the correct type of problem since each one will be different. Each page will have ten challenges, making them a great resource for students in 5th-8th grade.
These worksheets aid students in understanding the relationships between numbers and variables. These worksheets allow students to test their skills at solving polynomial equations, and to learn how to apply equations in their daily lives. These worksheets are an excellent opportunity to gain knowledge about equations and expressions. They can help you understand about the various types of mathematical issues and the various types of symbols used to express them.
These worksheets can be very useful for students in the beginning grades. These worksheets will help them master the art of graphing and solving equations. These worksheets are excellent to get used to working with polynomial variable. They will also help you learn how to factor and simplify them. There are many worksheets available to teach children about equations. Making the work your own is the most effective way to understand equations.
There are plenty of worksheets that teach quadratic equations. There are several worksheets on the different levels of equations for each stage. These worksheets can be used to work on solving problems until the fourth degree. Once you have completed an appropriate level it is possible to work on other kinds of equations. Once you have completed that, you are able to work on solving the same-level problems. For instance, you could find a problem with the same axis, but as an extended number. | 629 | 3,480 | {"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-2024-33 | latest | en | 0.946366 |
https://eng.kakprosto.ru/how-81274-how-to-read-character-by-date-of-birth | 1,680,254,039,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949598.87/warc/CC-MAIN-20230331082653-20230331112653-00544.warc.gz | 263,912,388 | 10,108 | Instruction
1
Before you try to determine the nature of your or your partner, friend, companion, etc., you have to remember the school course of mathematics. For a precise definition you will need the full date of birth in the digital equivalent. For example, 29.11.1985.
2
Now start to put figures to each other, but in this case, for each parameter (day, month, year) need be considered separately. So: 2+9=11, 1+1=2, 1+9+8+5=23. Now add up all the answers among themselves - 11+2+23 - it turns out 36. To define finite numbers, if you got a two digit number, add together the numbers - 9.
3
And this is the resulting number and refer to astroparticles to find out how the natureof om has its owner. For example, a man whose date of birth occurs in a amount of 1, very temperamental and proud. While rational and independent. 2 - responsible for men masculine and sensual. The three distinctive feature of sociability and good-natured. The resulting in the amount of 4-ka gives analysts and conservatives, and 5-CA - a madcap and whimsical adventurers. People are creative and emotional - obladateif 6 key. Patience, humility and love of solitude in naturewill isue 7. Leaders 8-CI and 9-CI perceptive and deep-thinking people.
4
In the figures, the assurances of astrologers is huge force. And if suddenly the numbers change, as this was common during the war years, it's very bad for all life of such a person. Right changes not only his fate, but character.
5
But if you want to get more reliable results, consult professional astrologers. On the basis of one only your date of birth they will be able to draw detailed astrological portrait. It will be described in detail and the features of characetr, and your prospects in life and even what's already happened.
6
You can also define the character of the person on the date of his birthusing the horoscope. As a rule, in the preparation of the astral portrait in the form of a horoscope using the common features possessed by the people of that particular zodiac sign. This applies to the character. In the study horoscope to know the exact date of birth is very important. After all, the whole sign of the zodiac can be roughly split into three parts. In the first phase, the sign still depends a little bit on just "gone" and may have the features of his character. In the second phase, a sign of the zodiac already has a strong position and will possess those traits of characterand that usually fixed his representatives. And in the third phase of the sign has little impact coming a symbol of the horoscope. He, too, can exert its influence. | 615 | 2,604 | {"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} | 2.875 | 3 | CC-MAIN-2023-14 | latest | en | 0.931741 |
https://www.convert-measurement-units.com/convert+Millibarn+to+Square+mile+US.php | 1,653,304,674,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662558015.52/warc/CC-MAIN-20220523101705-20220523131705-00358.warc.gz | 813,148,258 | 13,051 | Convert mb to sqmi (Millibarn to Square mile (US))
Millibarn into Square mile (US)
numbers in scientific notation
https://www.convert-measurement-units.com/convert+Millibarn+to+Square+mile+US.php
How many Square mile (US) make 1 Millibarn?
1 Millibarn [mb] = 0.000 000 000 000 000 000 000 000 000 000 000 000 038 610 061 418 299 Square mile (US) [sqmi] - Measurement calculator that can be used to convert Millibarn to Square mile (US), among others.
Convert Millibarn to Square mile (US) (mb to sqmi):
1. Choose the right category from the selection list, in this case 'Area'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), brackets and π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'Millibarn [mb]'.
4. Finally choose the unit you want the value to be converted to, in this case 'Square mile (US) [sqmi]'.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '605 Millibarn'. In so doing, either the full name of the unit or its abbreviation can be usedas an example, either 'Millibarn' or 'mb'. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'Area'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Alternatively, the value to be converted can be entered as follows: '43 mb to sqmi' or '83 mb into sqmi' or '96 Millibarn -> Square mile (US)' or '42 mb = sqmi' or '56 Millibarn to sqmi' or '21 mb to Square mile (US)' or '56 Millibarn into Square mile (US)'. For this alternative, the calculator also figures out immediately into which unit the original value is specifically to be converted. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(11 * 70) mb'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '605 Millibarn + 1815 Square mile (US)' or '57mm x 57cm x 13dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 3.702 046 386 064 5×1026. For this form of presentation, the number will be segmented into an exponent, here 26, and the actual number, here 3.702 046 386 064 5. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 3.702 046 386 064 5E+26. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 370 204 638 606 450 000 000 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications. | 947 | 3,888 | {"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} | 3.4375 | 3 | CC-MAIN-2022-21 | latest | en | 0.812549 |
http://mathematica.stackexchange.com/questions/tagged/generative-art?sort=votes&pagesize=15 | 1,469,279,506,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257822598.11/warc/CC-MAIN-20160723071022-00027-ip-10-185-27-174.ec2.internal.warc.gz | 166,277,040 | 26,562 | # Tagged Questions
The visual art or other aesthetic results generated by Mathematica.
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How do I generate a Left, Right and Straight ordering for a maze for something like the following? Similar to generating a left and right ordering for someone walking in real life. ... | 2,517 | 11,045 | {"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} | 2.609375 | 3 | CC-MAIN-2016-30 | latest | en | 0.945666 |
http://www.appszoom.com/android_games/brain_puzzle/ludo_bqpmu.html?nav=related | 1,397,841,371,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609533957.14/warc/CC-MAIN-20140416005213-00578-ip-10-147-4-33.ec2.internal.warc.gz | 290,138,228 | 15,886 | Ludo
Ludo is a simple board game for two to four players, in which the players race their four tokens from start to finish according to dice rolls.
Rules
At the start of the game, the player's four pieces are placed in the start area of their color. Players take it in turn to throw a die. A player must first throw a six to be able to move a piece from the starting area onto the starting square. In each subsequent turn the player moves a piece forward 1 to 6 squares as indicated by the die. When a player throws a 6 the player may bring a new piece onto the starting square, or may choose to move a piece already in play. Also, each time a player gets a 6 on the die, the player gets another turn as a bonus, but if any player gets 6 on a die three times in a row it is counted as a foul and the player therefore loses their turn.
If a player gets a 6 the player can separate chances (the player can separate 6 on one piece and 3 on the other if the player gets a 6 and a 3, if the pieces are already out of the house). The player can also play the numbers (6 & 3) using the same piece in any order. If a player cannot make a valid move they must pass the die to the next player.
If a player's piece lands on a square containing an opponent's piece, the opponent's piece is captured and returns to the starting area. A piece may not land on a square that already contains a piece of the same color.
Once a piece has completed a circuit of the board it moves up the home column of its own color. The player must throw the exact number to advance to the home square. The winner is the first player to get all four of their pieces onto the home square.
Tags: four player board games, ludo game, ludo games, a four player board game, board game pc for 4 players, how to make ludo in, ludo board game six players, board game with start and finish, ludo game 3d, ludo text.
Last activity on Ludo | 449 | 1,901 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2014-15 | longest | en | 0.96306 |
https://fresherbell.com/quizdiscuss/quantitative-aptitude/the-average-age-of-30-students-is-9-years-if-the-age-of-the | 1,716,345,395,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058525.14/warc/CC-MAIN-20240522005126-20240522035126-00315.warc.gz | 234,780,517 | 9,801 | Quiz Discussion
The average age of 30 students is 9 years. If the age of their teacher is included, the average age becomes 10 years. The age of the teacher (in years) is :
Course Name: Quantitative Aptitude
• 1] 27
• 2] 31
• 3] 35
• 4] 40
Solution
No Solution Present Yet
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# Quiz
1
Discuss
The average age of 40 students of class is 18 years. When 20 new students are admitted to the same class, the average age of the students of the class is increased by 6 months. The average age of newly admitted students is ?
• 1] 19 years
• 2] 19 years 6 months
• 3] 20 years
• 4] 20 years 6 months
Solution
2
Discuss
A student finds the average of 10 positive integers. Each integer contains two digits. By mistake, the boy interchanges the digits of one number say ba for ab. Due to this, the average becomes 1.8 less than the previous one. What was the difference of the two digits a and b?
• 1] 8
• 2] 6
• 3] 2
• 4] 4
Solution
3
Discuss
The average sale of car dealership was 15 cars per week. After a promotional scheme the average sale increased to 21 cars per week. The percentage increased in the sale of cars was-
• 1]
39.33%
• 2]
40%
• 3]
$$42\frac{6}{7}$$%
• 4]
140%
Solution
4
Discuss
A library has an average of 510 visitors on Sundays and 240 on other days. The average number of visitors per day in a month of 30 days beginning with a Sunday is
• 1]
285
• 2]
276
• 3]
280
• 4]
290
Solution
5
Discuss
Mukesh has twice as much money as Soham. Soham has 50% more money than Pankaj. If the average money with them is Rs. 110, then Mukesh has :
• 1] Rs. 155
• 2] Rs. 160
• 3] Rs. 180
• 4] Rs. 175
Solution
6
Discuss
Total expenses of a boarding house are partly fixed and partly varying linearly with the number of boarders. The average expense per boarder is Rs. 700 when there are 25 boarders and Rs. 600 when there are 50 boarders. What is the average expense per boarder when there are 100 boarders?
• 1] Rs. 540
• 2] Rs. 550
• 3] Rs. 570
• 4] Rs. 580
Solution
7
Discuss
The average age of 20 boys in a class is 12 years. 5 new boys are admitted to the class whose average age is 7 years. The average age of all the boys in the class becomes:
• 1] 8.2 years
• 2] 9.5 years
• 3] 12.5 years
• 4] 11 years
Solution
8
Discuss
The average monthly salary of 660 workers in a factory is Rs. 380. The average monthly salary of officers is Rs. 2100 and the average monthly salary of the other workers is Rs. 340. Find the number of other workers.
• 1] 645
• 2] 650
• 3] 640
• 4] 642
Solution
9
Discuss
If the average of x and 1/x(X$$\ne$$0) is M, then the average of x2 and 1/x2 is :
• 1] 1 - M2
• 2] 1 - 2M2
• 3] 2M2 - 1
• 4] 2M2 + 1
Solution
10
Discuss
The average age of four boys A, B, C and D is 5 years and the average age A, B, D, E is 6 years. C is 8 years old. The age of E is (in years) :
• 1] 12
• 2] 13
• 3] 14
• 4] 15
# Quiz | 998 | 3,012 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2024-22 | longest | en | 0.931062 |
http://topcoder.bgcoder.com/print.php?id=415 | 1,723,693,008,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641141870.93/warc/CC-MAIN-20240815012836-20240815042836-00803.warc.gz | 32,844,662 | 3,684 | ### Problem Statement
Aurum Nugget, Inc. has just purchased some new gold mines. They have a number of miners available to work in the mines, and would like to allocate the miners in such a way as to maximize their profit. Each mine can support a maximum of 6 miners, and contains a maximum of 6 major ore deposits. After the miners have been allocated to mines, the company earns (or loses) money as follows:
1. If a mine has fewer miners than ore deposits, the company will earn \$60 per miner allocated to that mine.
2. If a mine has the same number of miners as ore deposits, the company will earn \$50 per miner allocated to that mine.
3. If a mine has more miners than ore deposits, the company will earn \$50 for each miner up to the number of ore deposits, and will lose \$20 for each extra miner allocated to that mine.
Even if it will lose money, the company must employ every available worker at one of its mines.
Write a class GoldMine with a method getAllocation that takes in a String[] mines and a int miners. Each element of mines will be in the form "<p0>, <p1>, <p2>, <p3>, <p4>, <p5>, <p6>" (quotes for clarity) where each <pn> is a three digit number (with leading 0s if necessary.) Each <pn> represents the probability (as a percentage) that n deposits are present in the mine, and all <pn>'s within a mine will always add up to 100. miners is the number of employees the company must allocate. The method should return a int[] indicating the number of miners to place in each mine in order to maximize the expected profit (where element i in the returned int[] corresponds to element i of mines). If there are multiple allocations which maximize expected profit, return the allocation which places more miners in earlier mines. More specifically, when comparing two different allocations X0, X1, X2, ..., Xn and Y0, Y1, Y2, ..., Yn that yield the same expected profit, let i be the smallest index such that Xi is not equal to Yi. Then if Xi > Yi, allocation X0, X1, X2, ..., Xn is preferred to allocation Y0, Y1, Y2, ..., Yn.
For example, suppose the company has 4 miners available, and purchased the following two mines:
"000, 030, 030, 040, 000, 000, 000"
"020, 020, 020, 010, 010, 010, 010"
The first mine has a 30 percent chance of containing 1 ore deposit, a 30 percent chance of containing 2 ore deposits, and a 40 percent chance of containing 3 ore deposits. The second mine has a 20 percent chance of containing 0 ore deposits, a 20 percent chance of containing 1 ore deposit, a 20 percent chance of containing 2 ore deposits, a 10 percent chance of containing 3 ore deposits, a 10 percent chance of containing 4 ore deposits, a 10 percent chance of containing 5 ore deposits, and a 10 percent chance of containing 6 ore deposits.
In this scenario, the company can make the most money by allocating two miners at each mine, yielding an expected profit of 153:
First Mine
0.3*30 + 0.3*100 + 0.4*120 =
9 + 30 + 48 = 87
Second Mine
0.2*(-40) + 0.2*30 + 0.2*100 + 0.1*120 + 0.1*120 + 0.1*120 + 0.1*120 =
-8 + 6 + 20 + 12 + 12 + 12 + 12 = 66
Total Profit
87 + 66 = 153
The method would have returned { 2, 2 }. Other allocations would have yielded:
{ 0, 4 } : 75
{ 1, 3 } : 132
{ 3, 1 } : 129
{ 4, 0 } : 67
### Definition
Class: GoldMine Method: getAllocation Parameters: String[], int Returns: int[] Method signature: int[] getAllocation(String[] mines, int miners) (be sure your method is public)
### Notes
-Each mine can support a maximum of 6 miners.
### Constraints
-mines will contain between 1 and 50 elements, inclusive
-each element of mines will contain exactly 33 characters
-each element of mines will contain only digits ('0'-'9'), commas (',') and spaces (' ')
-each element of mines will be in the form "<p0>, <p1>, <p2>, <p3>, <p4>, <p5>, <p6>" (quotes for clarity) where each <pn> is a three digit number (with leading 0s if necessary.) Each <pn> represents the probability (as a percentage) that n deposits are present in the mine, and all <pn>'s within a mine will always add up to 100
-miners will be between 1 and (6 * the number of elements in mines), inclusive
### Examples
0)
{ "000, 030, 030, 040, 000, 000, 000", "020, 020, 020, 010, 010, 010, 010" } 4
Returns: { 2, 2 }
This is the example from the problem statement.
1)
{ "100, 000, 000, 000, 000, 000, 000", "100, 000, 000, 000, 000, 000, 000", "100, 000, 000, 000, 000, 000, 000", "100, 000, 000, 000, 000, 000, 000", "100, 000, 000, 000, 000, 000, 000" } 8
Returns: { 6, 2, 0, 0, 0 }
There are no deposits in any mines. However, since the company must employ every available worker, it loses \$160 (\$20 per worker). The proper allocation places 6 workers in the first mine and 2 workers in the second mine, since that is the allocation that places more workers in earlier mines and the maximum number of workers a mine can support is 6.
2)
{ "050, 000, 000, 000, 000, 050, 000", "050, 000, 000, 000, 000, 050, 000", "050, 000, 000, 000, 000, 050, 000", "050, 000, 000, 000, 000, 050, 000", "050, 000, 000, 000, 000, 050, 000", "050, 000, 000, 000, 000, 050, 000", "050, 000, 000, 000, 000, 050, 000", "050, 000, 000, 000, 000, 050, 000", "050, 000, 000, 000, 000, 050, 000", "050, 000, 000, 000, 000, 050, 000" } 30
Returns: { 4, 4, 4, 4, 4, 4, 4, 2, 0, 0 }
Each mine has a 50 percent chance of containing no deposits and a 50 percent chance of containing 5 deposits. The expected value from a mine like this is maximized with 4 workers. Since we allocate workers to earlier mines first, the early mines are filled with 4 workers each, and the remaining 2 miners are placed in the next available mine.
3)
{ "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004", "026, 012, 005, 013, 038, 002, 004" } 56
Returns:
{ 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 }
4)
{ "100, 000, 000, 000, 000, 000, 000", "090, 010, 000, 000, 000, 000, 000", "080, 020, 000, 000, 000, 000, 000", "075, 025, 000, 000, 000, 000, 000", "050, 050, 000, 000, 000, 000, 000", "025, 075, 000, 000, 000, 000, 000", "020, 080, 000, 000, 000, 000, 000", "010, 090, 000, 000, 000, 000, 000", "000, 100, 000, 000, 000, 000, 000", "000, 090, 010, 000, 000, 000, 000", "000, 080, 020, 000, 000, 000, 000", "000, 075, 025, 000, 000, 000, 000", "000, 050, 050, 000, 000, 000, 000", "000, 025, 075, 000, 000, 000, 000", "000, 020, 080, 000, 000, 000, 000", "000, 010, 090, 000, 000, 000, 000", "000, 000, 100, 000, 000, 000, 000", "000, 000, 090, 010, 000, 000, 000", "000, 000, 080, 020, 000, 000, 000", "000, 000, 075, 025, 000, 000, 000", "000, 000, 050, 050, 000, 000, 000", "000, 000, 025, 075, 000, 000, 000", "000, 000, 020, 080, 000, 000, 000", "000, 000, 010, 090, 000, 000, 000", "000, 000, 000, 100, 000, 000, 000", "000, 000, 000, 100, 000, 000, 000", "000, 000, 000, 090, 010, 000, 000", "000, 000, 000, 080, 020, 000, 000", "000, 000, 000, 075, 025, 000, 000", "000, 000, 000, 050, 050, 000, 000", "000, 000, 000, 025, 075, 000, 000", "000, 000, 000, 020, 080, 000, 000", "000, 000, 000, 010, 090, 000, 000", "000, 000, 000, 000, 100, 000, 000", "000, 000, 000, 000, 090, 010, 000", "000, 000, 000, 000, 080, 020, 000", "000, 000, 000, 000, 075, 025, 000", "000, 000, 000, 000, 050, 050, 000", "000, 000, 000, 000, 025, 075, 000", "000, 000, 000, 000, 020, 080, 000", "000, 000, 000, 000, 010, 090, 000", "000, 000, 000, 000, 000, 100, 000", "000, 000, 000, 000, 000, 090, 010", "000, 000, 000, 000, 000, 080, 020", "000, 000, 000, 000, 000, 075, 025", "000, 000, 000, 000, 000, 050, 050", "000, 000, 000, 000, 000, 025, 075", "000, 000, 000, 000, 000, 020, 080", "000, 000, 000, 000, 000, 010, 090", "000, 000, 000, 000, 000, 000, 100" } 150
Returns:
{ 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6 }
#### Problem url:
http://www.topcoder.com/stat?c=problem_statement&pm=1957
#### Problem stats url:
http://www.topcoder.com/tc?module=ProblemDetail&rd=4650&pm=1957
huntergt
#### Testers:
lbackstrom , brett1479
Greedy, Math | 4,326 | 10,029 | {"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-2024-33 | latest | en | 0.926026 |
http://djconnel.blogspot.com/2011/05/san-francisco-giants-2011-vs-2010-luck.html | 1,563,296,059,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195524679.39/warc/CC-MAIN-20190716160315-20190716182315-00241.warc.gz | 42,684,247 | 21,959 | ## Monday, May 2, 2011
### San Francisco Giants 2011 vs 2010: luck or skill?
The San Francisco Examiner devotes more than 1/3 of its content to sports, and of course a popular subject is the San Francisco Giants, who won the Major League Baseball championship last year. During the season in 2010, they won 92 games and lost 70, a 56.8% winning fraction. As of last Friday they are 12 wins and 12 losses, only a 50% win fraction. Obviously, one concludes from reading the headlines, something has gone terribly wrong.
The issue is I've not read past the headlines. Maybe there's compelling arguments made for how the team is playing. But invariably in the analysis of baseball and everything else, there is a lack of appreciation for the statistics of random numbers.
Baseball games aren't fully random events, but there is clearly a random component to them. I think everyone recognizes that luck is a big factor.
So a quick test: I'm going to assume the Giants had luck on their side last year, since they won the division (and ended up going on to win the championship, but that's irrelevant here). Teams towards the top of the standings tend to have been luckier, and those toward the bottom of the standings tend to have been less lucky. Failing to recognize this is a flaw of the dice-based games I played as a kid, but that's another topic. I'll simply assume the Giants would get at least as good a record as they got last year if they replayed the season with everything essentially equivalent only one year in three.
Now to some statics: assuming they had an equal probability to win each game, and the probability of winning was 92/162, the variance in the number of wins would be 92 × 62 / 100. The standard deviation is the square root of this, or 7.55. I'm assuming they had 1/3 good luck, so assuming a normal probability distribution for wins, this implies they won 3.2 games more than expected based on skill, or that their expected record for the season was 88.75 wins and 63.25 losses, a 54.8% winning percentage.
So given this winning percentage, what is the chance their record so far would be as bad as 12-12? I won't assume a normal distribution here; there's not quite enough games for the central limit theorem to apply. Instead I can use Perl to generate a quick simulation.
```my \$p = 0.548;
my \$sum = 0;
for my \$n ( 0 .. 99999 ) {
my \$w = 0;
for my \$g ( 0 .. 23 ) {
\$w ++
if (rand() < \$p);
}
\$sum ++
if (\$w <= 12);
}
print \$sum, "\n";```
The result: 38958 of 100 thousand trials had the Giants finishing the first 24 games no better than 12-12. This implies they went from around 1/3 good luck to 1/3 bad luck for the first 24 games this year.
Next I modified the program to simulate a full season. What are the Giants chances of winning only 81 (50%) or fewer of their games for the full year? Their record was this bad in 10929 of the 100 thousand trials. In other words, you would expect them to tank to .500 or worse around 1 year in 9, given the estimated winning probability for last year (which assumed 1-in-3 luck).
In baseball, winning 100 games is considered exceptional. So I checked the chance for them to win 100 games. They did so in 5359 of the 100 thousand trials: around 1 in 19. If I start the simulation with a 12-and-12 start, the number of successes falls to 2695 / 100 thousand (2.7%), and their success at matching last year falls to 28.5% from 36.4%, so the slow start does hurt them. The question is whether it indicates poor preparation or just a change in the winds of fortune.
Sure, each baseball game is not an independent random trial. Assuming this is the "worst case" assumption. But looking only at the Giants' win loss records, the result so far is indistinguishable from random. It is important to realize that luck may play an enormous role in baseball standings, even at the end of 162 games. | 950 | 3,879 | {"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} | 3.296875 | 3 | CC-MAIN-2019-30 | latest | en | 0.9808 |
https://studysoup.com/tsg/996406/numerical-analysis-10-edition-chapter-8-1-problem-3 | 1,611,776,920,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704832583.88/warc/CC-MAIN-20210127183317-20210127213317-00219.warc.gz | 596,120,649 | 12,641 | ×
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# Find the least squares polynomials of degrees 1,2, and 3 for the data in the following
ISBN: 9781305253667 457
## Solution for problem 3 Chapter 8.1
Numerical Analysis | 10th Edition
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Problem 3
Find the least squares polynomials of degrees 1,2, and 3 for the data in the following table. Compute the error E in each case. Graph the data and the polynomials. x, 1.0 1.1 1.3 1.5 1.9 2.1 yi 1.84 1.96 2.21 2.45 2.94 3.18
Step-by-Step Solution:
Step 1 of 3
NTRO TO ENVIRONMENTAL SCIENCE : ECOLOGY U NIT Woods have just as much or more industry than an urban area with factories Provides more services and produces more for more organisms Natural World Themes 1. All materials cycle and waste becomes food Used materials are ready for something else 2. Diversity is rewarded Seen as an advantage More diversity = more durable, higher life quality, more organisms 3. Each species depends on many others Example: oxygen comes from plants making food, we need oxygen to breathe and live 4. The sun is the ONLY energy source Energy gives us the ability to do work Work involves different f
Step 2 of 3
Step 3 of 3
##### ISBN: 9781305253667
Numerical Analysis was written by and is associated to the ISBN: 9781305253667. Since the solution to 3 from 8.1 chapter was answered, more than 296 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: Numerical Analysis, edition: 10. This full solution covers the following key subjects: . This expansive textbook survival guide covers 76 chapters, and 1204 solutions. The answer to “Find the least squares polynomials of degrees 1,2, and 3 for the data in the following table. Compute the error E in each case. Graph the data and the polynomials. x, 1.0 1.1 1.3 1.5 1.9 2.1 yi 1.84 1.96 2.21 2.45 2.94 3.18” is broken down into a number of easy to follow steps, and 44 words. The full step-by-step solution to problem: 3 from chapter: 8.1 was answered by , our top Math solution expert on 03/16/18, 03:24PM.
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Find the least squares polynomials of degrees 1,2, and 3 for the data in the following | 761 | 2,695 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2021-04 | longest | en | 0.862713 |
http://www.dowegetagradeforthis.com/mean-grade/ | 1,526,913,215,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864405.39/warc/CC-MAIN-20180521142238-20180521162238-00576.warc.gz | 370,189,285 | 6,787 | The mean grade in Mr. Smith’s math classes is 86.5, and the mean grade in Mrs.
Shell’s math classes is 90.2. Based on this information, it can be inferred that the
students in Mrs. Shell’s math class __________.
A. like math more than students in Mr. Smith’s math class
B. study more for tests than students in Mr. Smith’s math class
C. have fewer absences than students in Mr. Smith’s math class
D. earn better grades on tests on average than students in Mr. Smith’s math
class | 123 | 479 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-22 | latest | en | 0.862033 |
https://compscibits.com/Current-Trends-and-Technologies/UGC-NET-computer-science-question-paper/discussion/7797 | 1,591,142,552,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347426956.82/warc/CC-MAIN-20200602224517-20200603014517-00426.warc.gz | 289,143,791 | 7,589 | A directory of Objective Type Questions covering all the Computer Science subjects. Here you can access and discuss Multiple choice questions and answers for various compitative exams and interviews.
## Discussion Forum
Que. Given a simple image of size 10 x 10 whose histogram models the symbol probabilities and is given byP1 P2 P3 P4a b c dThe first order estimate of image entropy is maximum when a. a = 0, b = 0, c = 0, d = 1 b. a = 1/2, b =1/2, c=0, d=0 c. a = 1/3, b =1/3, c=1/3, d=0 d. a = 1/4, b =1/4, c=1/4, d=1/4 Answer:a = 1/4, b =1/4, c=1/4, d=1/4 | 201 | 564 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2020-24 | latest | en | 0.767838 |
https://beastpapers.com/spring-co-began-the-month-with-30-units/ | 1,656,758,834,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104054564.59/warc/CC-MAIN-20220702101738-20220702131738-00469.warc.gz | 167,960,132 | 8,941 | Select Page
1.)
Spring Co. began the month with 30 units of inventory that cost \$50 each. On January 4, Spring purchase 85 units for \$55 each, on account. January 8 sell, on account, 80 units for \$70 each.
Assuming FIFO perpetual method:
Prepare the necessary journal entries
What is the Dollar Value of Cost of Goods Sold at January 31?
What is the Dollar Value of Ending Inventory at January 31?
What is the Gross Margin?
2.)
Spring Co. began the month with 30 units of inventory that cost \$50 each. On January 4, Spring purchase 85 units for \$55 each, on account. January 8 sell, on account, 80 units for \$70 each.
Assuming LIFO perpetual method:
Prepare the necessary journal entries
What is the Dollar Value of Cost of Goods Sold at January 31?
What is the Dollar Value of Ending Inventory at January 31?
What is the Gross Margin? | 205 | 845 | {"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} | 2.703125 | 3 | CC-MAIN-2022-27 | latest | en | 0.921192 |
https://www.replicadb4.com/what-is-linear-least-squares-fitting/ | 1,696,344,550,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511106.1/warc/CC-MAIN-20231003124522-20231003154522-00130.warc.gz | 1,034,083,191 | 11,260 | ## What is linear least squares fitting?
The linear least squares fitting technique is the simplest and most commonly applied form of linear regression (finding the best fitting straight line through a set of points.) The fitting is linear in the parameters to be determined, it need not. be linear in the independent variable x.
Table of Contents
## How do you fit a least squares regression line?
The Least Squares Regression Line is the line that minimizes the sum of the residuals squared. The residual is the vertical distance between the observed point and the predicted point, and it is calculated by subtracting ˆy from y….Calculating the Least Squares Regression Line.
ˉx 28
r 0.82
What is a least squares regression line used for?
Least squares regression is used to predict the behavior of dependent variables. The least squares method provides the overall rationale for the placement of the line of best fit among the data points being studied.
### How do you describe a linear fit?
Linear form means that as X increases, Y increases or decreases at a constant rate. Positive direction means that Y increases when X increases; and negative direction means that Y decreases when X increases. The last component of the relationship between two variables is strength.
### Why is the LSR line considered the line of best fit?
Line of Best Fit Since the least squares line minimizes the squared distances between the line and our points, we can think of this line as the one that best fits our data. This is why the least squares line is also known as the line of best fit.
How do you write a conclusion for a regression analysis?
Conclusion: Use Regression Effectively by Keeping it Simple Moreover, regression should only be used where it is appropriate and when their is sufficient quantity and quality of data to give the analysis meaning beyond your sample.
#### How do you describe a linear relationship?
A linear relationship (or linear association) is a statistical term used to describe a straight-line relationship between two variables. Linear relationships can be expressed either in a graphical format or as a mathematical equation of the form y = mx + b.
#### What is the difference between Y ax b and ya BX?
There is no mathematical difference between the two linear regression forms LinReg(ax+b) and LinReg(a+bx), only different professional groups prefer different notations.
What is the principle of least square?
MELDRUM SIEWART HE ” Principle of Least Squares” states that the most probable values of a system of unknown quantities upon which observations have been made, are obtained by making the sum of the squares of the errors a minimum.
## What is linear regression explain best line fit with an example?
Linear regression consists of finding the best-fitting straight line through the points. The best-fitting line is called a regression line. The black diagonal line in Figure 2 is the regression line and consists of the predicted score on Y for each possible value of X.
## Is LSRL the same as line of best fit?
The Least Squares Regression Line is the line that makes the vertical distance from the data points to the regression line as small as possible. It’s called a “least squares” because the best line of fit is one that minimizes the variance (the sum of squares of the errors).
How to solve the linear least squares problem 6min?
•The Linear Least Squares solution 6minimizes the square of the 2-norm of the residual: min 5−46 •One method to solve the minimization problem is to solve the system of Normal Equations 4(46=4(5 •Let’s see some examples and discuss the limitations of this method.
### What is the least square method of best fit?
The least square method is the process of finding the best-fitting curve or line of best fit for a set of data points by reducing the sum of the squares of the offsets (residual part) of the points from the curve. During the process of finding the relation between two variables, the trend of outcomes are estimated quantitatively.
### What are the limitations of least square regression analysis?
One of the main limitations is discussed here. In the process of regression analysis, which utilizes the least-square method for curve fitting, it is inevitably assumed that the errors in the independent variable are negligible or zero.
What are the different types of least-squares problems?
There are two basic categories of least-squares problems: These depend upon linearity or nonlinearity of the residuals. The linear problems are often seen in regression analysis in statistics. | 922 | 4,603 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2023-40 | latest | en | 0.934776 |
http://www.ask.com/question/what-is-a-2-5-gpa-equivalent-to | 1,386,925,517,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164921422/warc/CC-MAIN-20131204134841-00042-ip-10-33-133-15.ec2.internal.warc.gz | 240,968,818 | 16,657 | # What is a 2.5 Gpa Equivalent to?
A 2.5 Grade Point Average (GPA) is the equivalent of a C+. Grades are the standardised way of measuring comprehension in students in the various areas of interests and are used in the entry or admission to institutions of learning. An A leads the pack and denotes a score of between 90 and 100.
Q&A Related to "What is a 2.5 Gpa Equivalent to"
A 2.5 GPA usually means a C grade point average. Usually it starts at 2.00 - 2.99. A 2.99 is a high C grade point average. But to know for sure, you should research the method your http://answers.ask.com/Education/Schools/what_is_a...
Since most grade point average scales are based off of a 4.0, we can say that a grade point average of a 2.5 is roughly a solid "C" GPA. http://answers.ask.com/Education/Schools/what_lett...
I have the same GPA and it is equivalent to a 3.7 regualr "A". Hope I helped! http://wiki.answers.com/Q/What_GPA_grade_is_equiva...
It is a C+ grade. A B- would be around 2.7 GPA. An A is a 4. A B is http://www.chacha.com/question/what-is-the-letter-...
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Usually, when you receive a grade, it is either a numerical value or a letter grade. There is a standard listing of the GPA equivalents so you can convert your ...
A 2.9 GPA is equivalent to eighty four percent or grade B. However, this may not be the right conversion as algorithms used to calculate the GPAs of students, ...
A 2.75 GPA is equivalent to 83 percent and is also ranked as grade B. The maximum GPA is 4.0 with its percentile being 95% to 100% with A as its grade. The lowest ... | 432 | 1,615 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2013-48 | longest | en | 0.932968 |
https://runestone.academy/ns/books/published/welcomecs/CSPPythonData/Exercises.html | 1,675,924,778,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764501407.6/warc/CC-MAIN-20230209045525-20230209075525-00750.warc.gz | 530,503,022 | 13,598 | 19.10. Chapter Exercises¶
The problems in these exercises use a new data set. The file "stocks.txt" has data from the Dow Jones Industrial Average which tracks the average performance of a collection of large stocks. Each line of the file represents the opening day of trading (not always the first of the month as the market is not open every day).These lines look like:
3-Dec-01,9848.93,10220.78,9651.87,10021.57
The values on the line are separated by commas and are the following (in order):
• Date
• Opening value
• High value for month
• Low value for month
• Close value for month
• Volume - number of shares traded
To see all of the data, you can use a loop to print out each line.
Below is the start of a program to read in the "stocks.txt" file and run code on each line in the file.
Add code to split the line into a list of values and print out the date value. (The date should be the first value in the list that you create with split.)
Your final output should be a long list of dates and nothing else.
You can’t use codelens with file reading problems, but you can use print statements to check what your code is doing. Feel free to use extra ones while writing your code and Then remove them or comment them out when everything is working.
Modify your program to print out the highest value the Dow Jones reached. (This should be the largest of the monthly high values.)
Tip: When you get the high value, you will need to convert it from a string to a float to work with it as a decimal number. This should look like: float(values[??]).
The final version of your program should only print out the one highest value, but you should work your way up to that. Start by printing out all of the monthly high values, then worry about finding the highest one.
You can’t use codelens with file reading problems, but you can use print statements to check what your code is doing.
Modify your program from question 1 to only print the dates from a specific year specified by a variable desiredYear. If desiredYear is 96, you would only print out values where the year (last part of the date value) is “96”.
You should try changing desired year to different values to make sure your program works for any year for which there is data (89-01), but to pass the tests, you must set desiredYear to “92”.
You can’t use codelens with file reading problems, but you can use print statements to check what your code is doing.
Combine your solutions from problems 2 and 3 and make your program find the largest highest value from the records indicated by the variable desiredYear. I.e. if desiredYear is “96”, your program should only consider the records where the date value ends in “96”, and from those, should find the largest “highest value for month” seen in those records.
You should try changing desired year to different values to make sure your program works for any year for which there is data (89-01), but to pass the tests, you must set desiredYear to “96”.
You can’t use codelens with file reading problems, but you can use print statements to check what your code is doing.
Turn your code from question 4 into a function so we can easily check the max value in multiple years. The function should be called maxHighForYear. It should take the desiredYear and the data as parameters and return the max “highest value for month” found in the records that match the desired year.
The starter code has a simple test of your function followed by a more complex test that tests all the valid years. If you need to debug your code, it might be easier if you comment out the complex test and just run the simple one.
You can’t use codelens with file reading problems, but you can use print statements to check what your code is doing.
Write the function avgVolumeForYear. It should take the desiredYear and the data as parameters and return the average of the “volume” value found in the records that match the desired year. (The “volume” is the last value in each record.)
The starter code has a simple test of your function followed by a more complex test that tests all the valid years. If you need to debug your code, it might be easier if you comment out the complex test and just run the simple one.
You can’t use codelens with file reading problems, but you can use print statements to check what your code is doing. | 961 | 4,356 | {"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} | 3.1875 | 3 | CC-MAIN-2023-06 | longest | en | 0.896649 |
https://www.physicsforums.com/threads/second-derivative.166455/ | 1,566,264,886,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315174.57/warc/CC-MAIN-20190820003509-20190820025509-00414.warc.gz | 926,675,056 | 16,722 | Second derivative
sutupidmath
I would like someone to tell me what is the geometric interpretation of the second derivative at a fixed point, or in an interval??
thx
Eighty
If it's positive on an interval, the function is convex there; if it's negative, the function is concave.
theperthvan
Second derivative is the rate of change of the derivative, i.e "how fast is the gradient changing?".
Say the second derivative is 1 at point A and 2 at point B on the same function. Then the gradient (the derivative) is changing faster at point A than point B.
mathwonk
Homework Helper
it measures curvature, or concavity, tells whether it is up or down, and how sharply.
it also determines whether the tangent line is above or below the graph.
Last edited:
Moridin
http://online.math.uh.edu/Math1314/index.htm [Broken]
See section 11.
Last edited by a moderator:
sutupidmath
i guess i already knew this,just did not actually think about it.
However, if this is what i was looking for, i will see later when i think more about it, and if i still have problems i will come back.
Thankyou guys, for giving me some flash back
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• Solo and co-op problem solving | 334 | 1,450 | {"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} | 2.875 | 3 | CC-MAIN-2019-35 | latest | en | 0.920554 |
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## Jorge just turned 12. How old is he in human years?
My dog, Jorge, has just turned 12 years old. He still seems like a puppy to me but I wonder how old he is when compared to a human’s age. He’s definitely not like an 84 year old human (12…
## Fibonacci coloring
Lots of people enjoy coloring books. Evidently people who like to multitask can find a means of calming their energies and reach a peaceful place by simply coloring designs. Coloring in these drawings allows you to relax and just create pretty…
Let your students experiment with their graphing calculators to create a nice fireworks display? We’ve written a brief activity that questions students about manipulating parabolas, adjusting their calculator windows, and helping them celebrate whatever. Or consider using Desmos or Geogebra to create these fireworks. Enjoy! Or…
## How many cars are there?
I came home from work to find that my son had his cars organized in a tight parking lot on our front porch. How many cars are there? Is it an array and if so, what size array could it be…
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Thank you Mother and Father for all of those diapers – Which is a better deal, cloth or disposable diapers? How much did you cost your parents in diapers? In the future how much could your kid’s diapers cost? Help…
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In my grocery store they now have these new water bottle refilling stations. You can simply refill your own container with water instead of buying a six-pack of bottled water. Do you think that this is a good idea? Are they…
## Opening weekend of Wonder Woman
Can you tell from its opening weekend how much the new movie, Wonder Woman, might earn in theaters? This activity can be used anytime and is especially timely when there is a new blockbuster movie released. We’ve left blank spaces…
## NBA Finals, 2017
Celebrate the NBA and NBA Finals by involving your students in interpreting what can be deduced from graphical representations. In this activity students consider scatter plots, circle graphs, bar graphs, Venn Diagrams and the concept of mean. To start the…
## How much pee is in that pool?
After chlorination, it has been hard to measure the quantity of pollutants in pool water. A Canadian chemist, Xing-Fang Li, has found a marker for urine that chlorine doesn’t disguise. In this activity, students calculate volume of several pools, change that… | 555 | 2,433 | {"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} | 2.734375 | 3 | CC-MAIN-2018-09 | latest | en | 0.937394 |
http://www.graindpirate.fr/matrix-arithmetic-what-is-absolutely-a-proportion-in-q/ | 1,669,632,921,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710503.24/warc/CC-MAIN-20221128102824-20221128132824-00744.warc.gz | 79,001,519 | 12,476 | # Matrix Arithmetic – What Is a Proportion in T?
What is offender math? We will make an hard work to find matrix mathematics and just how it is able to profit us master the integral. Also, it may well be made use of to receive fixing concerns in linear and rank my writer nonlinear matters to utilize matrix math.
If you want to reply we have to outline a matrix, a matrix is made up of an array of matrix quantities. It’s always identified as a matrix When we’ve got a quantity row and column, also it is really recognized as defining a matrix symmetry after you would really like to understand if a matrix is symmetric or not.
Now that we have an understanding of precisely what sort of mathematics iswe are able to understand how it can really help us to learn the integral. http://www.nursing.umaryland.edu/news-events/news/ In matrix arithmetic, we may possibly use the fanciful numbers (matrix), and overlaps figures (matrix of D, I, J, K). We can insert these quantities together and also we get the matrix multiplication. When we innovate D at the me in J, then we get me, and it can be a celestial variety, then then we could break up the matrix then we acquire specifically the matrix quotient.
Matrix mathematics can guide everyday people to clear up worries in algebra. We might make use of this arithmetic to some worries in several areas of arithmetic. By means of instance, if we have been working with know-how, then we can associate the function’s very primary thing . As a way todo so, we can use matrices and clear up linear equations with their have systems.
We acquired also a amount of derivatives from a and b, and in addition just two uses b plus a. The predicament is always to create what extremely is a share in arithmetic.
When we’ve particularly a few of a I, it usually means there was 1 number a like a I personally = a I for all i. Suppose we have a perform f of inputs, at which f(a) = b, h(a) = c), and f(b) = d. In the event you want to know what honestly is a share in arithmetic, then we can correct for bull after which multiply f(h) by phone and then then multiplying by(c) by p. When we multiply all those three from just about every other, then we acquire the derivative of d with regard to x.
Then we can use the established principle, if we are informed it is possible to get 3 needs. Let us return back again to terminal algebra likewise as the idea that’s established and then we could fix f.
We now have a vector spot, also we now have a aircraft in place, and we have a tricky variety named x. If you would like to know what’s a proportion in math, then we can easily fix for x. We can deal with the equation for x by deciding on the intricate conjugate, which is f(x) = c.
How can matrix math assist us gain knowledge of the integral in mathematics? We can even fix for x when we resolve f(x) by finding the complicated conjugate.
Just in case we are making an attempt to take care of to get a a portion of theta hole equation in matrix math, afterward we are going to have to know ways to resolve for x and after that multiply it by c and following that multiply it once more. We can use the set principle to find the platform.
We must understand how a number of occasions we can easily multiply the result by c and then multiply the result by d. It’s convenient to discover how to do this, as a result of we can make this happen twice and we will see that the alternative will glimpse like the perform f(x) = c.
Matrix math could be beneficial in linear algebra and matrix understanding. We are going to complete so inmatrix arithmetic, As soon as we need to take care of for x and multiply it by phone and then multiply it all over again d.
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Vous êtes libre de contribuer ! | 834 | 3,791 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2022-49 | longest | en | 0.960412 |
https://topic.alibabacloud.com/a/python-data-analytics-combat-3-numpy-library_1_29_30239613.html | 1,695,496,474,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506528.19/warc/CC-MAIN-20230923162848-20230923192848-00828.warc.gz | 651,233,785 | 15,576 | # "Python data Analytics Combat" 3 NumPy Library
Source: Internet
Author: User
• Initializing an array
arr = Np.array ([[1,2],[3,4]])
arr = Np.array ([[1,2],[3,4]], Dtype=complex)
arr = Np.zeros ((3,4))
arr = Np.ones (())
arr = Np.arange (4,10). Reshape (2,3)
arr = Np.linspace (0,1,6) #[0.0,0.2,0.4,0.6,0.8,1.0]
arr = Np.random.random (3)
arr = Np.random.random ((5,2))
• Array built-in functions
Type (arr): Numpy.ndarray
Arr.dtype:int32, float64 ...
Arr.ndim:2
Arr.size:4
Arr.shape: (2l,2l)
Arr.itemsize:4
• Array calculation function
Arr+1: Elements and
Arr*2: Elemental Product
ARR1+ARR2: Elements and
ARR1*ARR2: Elemental Product
Np.dot (arr1, ARR2): Matrix product
Sin (arr): element trigonometric Functions
sqrt (arr): Elemental root
Arr.sum (): All elements and
Arr.min (): Minimum value for all elements
Arr.max (): all element Maximum value
Arr.mean (): Average of all elements
• Array iteration function
Np.apply_along_axis (Np.mean, Axis=0, Arr=arr): Iteration by column
Np.apply_along_axis (Lambda x:x*2, Axis=1, Arr=arr): Iteration by row
• Boolean array:
arr<0.5
• Stitching arrays:
Np.vstack ((arr1, ARR2)): Vertically into the stack
Np.hstack ((arr1, ARR2)): Horizontal into the stack
• Cut fraction Group:
Np.vsplit (arr, n): Vertically split N part
Np.hsplit (arr, n): Horizontal split N part
Np.split (arr, [1,3,5], Axis=1): Vertical split
Np.save (' Data.npy ', arr)
arr = np.load (' data.npy ')
"Python data Analytics Combat" 3 NumPy Library
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• Alibaba Cloud offers highly flexible support services tailored to meet your exact needs. | 672 | 2,454 | {"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-2023-40 | latest | en | 0.682391 |
http://stackoverflow.com/questions/8382642/analyzing-sound-in-a-wav-file | 1,440,937,569,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440644065306.42/warc/CC-MAIN-20150827025425-00053-ip-10-171-96-226.ec2.internal.warc.gz | 214,028,792 | 20,802 | # Analyzing Sound in a WAV file
I am trying to analyze a movie file by splitting it up into camera shots and then trying to determine which shots are more important than others. One of the factors I am considering in a shot's importance is how loud the volume is during that part of the movie. To do this, I am analyzing the corresponding sound file. I'm having trouble determining how "loud" a shot is because I don't think I fully understand what the data in a WAV file represents.
I read the file into an audio buffer using a method similar to that described in this post.
Having already split the corresponding video file into shots, I am now trying to find which shots are louder than others in the WAV file. I am trying to do this by extracting each sample in the file like this:
``````double amplitude = (double)((audioData[i] & 0xff) | (audioData[i + 1] << 8));
``````
Some of the other posts I have read seem to indicate that I need to apply a Fast Fourier Transform to this audio data to get the amplitude, which makes me wonder what the values I have extracted actually represent. Is what I'm doing correct? My sound file format is a 16-bit mono PCM with a sampling rate of 22,050 Hz. Should I be doing something with this 22,050 value when I am trying to analyze the volume of the file? Other posts suggest using Root Mean Square to evaluate loudness. Is this required, or just a more accurate way of doing it?
The more I look into this the more confused I get. If anyone could shed some light on my mistakes and misunderstandings, I would greatly appreciate it!
-
The FFT has nothing to do with volume and everything to do with frequencies. To find out how loud a scene is on average, simply average the sampled values. Depending on whether you get the data as signed or unsigned values in your language, you might have to apply an absolute function first so that negative amplitudes don't cancel out the positive ones, but that's pretty much it. If you don't get the results you were expecting that must have to do with the way you are extracting the individual values in line 20.
That said, there are a few refinements that might or might not affect your task. Perceived loudness, amplitude and acoustic power are in fact related in non-linear ways, but as long as you are only trying to get a rough estimate of how much is "going on" in the audio signal I doubt that this is relevant for you. And of course, humans hear different frequencies better or worse - for instance, bats emit ultrasound squeals that would be absolutely deafening to us, but luckily we can't hear them at all. But again, I doubt this is relevant to your task, since e.g. frequencies above 22kHz (or was is 44kHz? not sure which) are in fact not representable in simple WAV format.
-
Okay, great. I was just concerned that I wasn't extracting the amplitude properly. But it sounds like I am. Out of curiousity, if I did care about the non-linear relationship between amplitude and acoustic power, would that be when I apply a FFT? – Steph Dec 5 '11 at 9:07
A flat-line value at the peak of the amplitudes represented by that format will sound exactly like a flat-line value of 0. Completely silent. Averaging the values is not the way to go. Either use RMS (my preferred choice), or calculate a dB level, for a more accurate value of 'volume'. – Andrew Thompson Dec 5 '11 at 9:18
@AndrewThompson - All right, so I'm starting to be convinced that RMS is a good idea. If I also want to take into account the non-linearity in the way the ear responds to frequencies and amplitudes (i.e. if I want to use an FFT), how do I do that in combination with RMS? Or would I have to do that instead of RMS? – Steph Dec 6 '11 at 8:42
That is beyond my level of experience. I've done some work on RMS (which is pretty simple to calculate, once you have int or float values) bu nothing beyond that. – Andrew Thompson Dec 6 '11 at 10:21
I don't know the level of accuracy you want, but a simple RMS (and perhaps simple filtering of the signal) is all many similar applications would need.
RMS will be much better than Peak amplitude. Using peak amplitudes is like determining the brightness of an image based on the brightest pixel, rather than averaging.
If you want to filter the signal or weigh it to perceived loudness, then you would need the sample rate for that.
FFT should not be required unless you want to do complex frequency analysis as well. The ear responds differently to frequencies at different amplitudes - the ear does not respond to sounds at different frequencies and amplitudes linearly. In this case, you could use FFT to perform frequency analyses for another domain of accuracy.
-
I understand. Thanks for the very clear response! – Steph Dec 5 '11 at 9:15 | 1,084 | 4,774 | {"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-2015-35 | longest | en | 0.96293 |
https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_12 | 1,709,544,628,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476432.11/warc/CC-MAIN-20240304065639-20240304095639-00314.warc.gz | 102,869,546 | 17,533 | # 2021 AMC 12A Problems/Problem 10
The following problem is from both the 2021 AMC 10A #12 and 2021 AMC 12A #10, so both problems redirect to this page.
## Problem
Two right circular cones with vertices facing down as shown in the figure below contain the same amount of liquid. The radii of the tops of the liquid surfaces are $3$ cm and $6$ cm. Into each cone is dropped a spherical marble of radius $1$ cm, which sinks to the bottom and is completely submerged without spilling any liquid. What is the ratio of the rise of the liquid level in the narrow cone to the rise of the liquid level in the wide cone?
$[asy] size(350); defaultpen(linewidth(0.8)); real h1 = 10, r = 3.1, s=0.75; pair P = (r,h1), Q = (-r,h1), Pp = s * P, Qp = s * Q; path e = ellipse((0,h1),r,0.9), ep = ellipse((0,h1*s),r*s,0.9); draw(ellipse(origin,r*(s-0.1),0.8)); fill(ep,gray(0.8)); fill(origin--Pp--Qp--cycle,gray(0.8)); draw((-r,h1)--(0,0)--(r,h1)^^e); draw(subpath(ep,0,reltime(ep,0.5)),linetype("4 4")); draw(subpath(ep,reltime(ep,0.5),reltime(ep,1))); draw(Qp--(0,Qp.y),Arrows(size=8)); draw(origin--(0,12),linetype("4 4")); draw(origin--(r*(s-0.1),0)); label("3",(-0.9,h1*s),N,fontsize(10)); real h2 = 7.5, r = 6, s=0.6, d = 14; pair P = (d+r-0.05,h2-0.15), Q = (d-r+0.05,h2-0.15), Pp = s * P + (1-s)*(d,0), Qp = s * Q + (1-s)*(d,0); path e = ellipse((d,h2),r,1), ep = ellipse((d,h2*s+0.09),r*s,1); draw(ellipse((d,0),r*(s-0.1),0.8)); fill(ep,gray(0.8)); fill((d,0)--Pp--Qp--cycle,gray(0.8)); draw(P--(d,0)--Q^^e); draw(subpath(ep,0,reltime(ep,0.5)),linetype("4 4")); draw(subpath(ep,reltime(ep,0.5),reltime(ep,1))); draw(Qp--(d,Qp.y),Arrows(size=8)); draw((d,0)--(d,10),linetype("4 4")); draw((d,0)--(d+r*(s-0.1),0)); label("6",(d-r/4,h2*s-0.06),N,fontsize(10)); [/asy]$
$\textbf{(A) }1:1 \qquad \textbf{(B) }47:43 \qquad \textbf{(C) }2:1 \qquad \textbf{(D) }40:13 \qquad \textbf{(E) }4:1$
## Solution 1 (Algebra)
Initial Scenario
Let the heights of the narrow cone and the wide cone be $h_1$ and $h_2,$ respectively. We have the following table: $$\begin{array}{cccccc} & \textbf{Base Radius} & \textbf{Height} & & \textbf{Volume} & \\ [2ex] \textbf{Narrow Cone} & 3 & h_1 & & \frac13\pi(3)^2h_1=3\pi h_1 & \\ [2ex] \textbf{Wide Cone} & 6 & h_2 & & \hspace{2mm}\frac13\pi(6)^2h_2=12\pi h_2 & \end{array}$$ Equating the volumes gives $3\pi h_1=12\pi h_2,$ which simplifies to $\frac{h_1}{h_2}=4.$
Furthermore, by similar triangles:
• For the narrow cone, the ratio of the base radius to the height is $\frac{3}{h_1},$ which always remains constant.
• For the wide cone, the ratio of the base radius to the height is $\frac{6}{h_2},$ which always remains constant.
### Solution 1.1 (Properties of Fractions)
Final Scenario
For the narrow cone and the wide cone, let their base radii be $3x$ and $6y$ (for some $x,y>1$), respectively. By the similar triangles discussed above, their heights must be $h_1x$ and $h_2y,$ respectively. We have the following table: $$\begin{array}{cccccc} & \textbf{Base Radius} & \textbf{Height} & & \textbf{Volume} & \\ [2ex] \textbf{Narrow Cone} & 3x & h_1x & & \frac13\pi(3x)^2(h_1x)=3\pi h_1 x^3 & \\ [2ex] \textbf{Wide Cone} & 6y & h_2y & & \hspace{2.0625mm}\frac13\pi(6y)^2(h_2y)=12\pi h_2 y^3 & \end{array}$$ Recall that $\frac{h_1}{h_2}=4.$ Equating the volumes gives $3\pi h_1 x^3=12\pi h_2 y^3,$ which simplifies to $x^3=y^3,$ or $x=y.$
Finally, the requested ratio is $$\frac{h_1 x - h_1}{h_2 y - h_2}=\frac{h_1 (x-1)}{h_2 (y-1)}=\frac{h_1}{h_2}=\boxed{\textbf{(E) }4:1}.$$ Remarks
1. This solution uses the following property of fractions:
For unequal positive numbers $a,b,c$ and $d,$ if $\frac ab = \frac cd = k,$ then $\frac{a\pm c}{b\pm d}=\frac{bk\pm dk}{b\pm d}=\frac{(b\pm d)k}{b\pm d}=k.$
2. This solution shows that, regardless of the shape or the volume of the solid dropped into each cone, the requested ratio stays the same as long as the solid sinks to the bottom and is completely submerged without spilling any liquid.
~MRENTHUSIASM
### Solution 1.2 (Bash)
Final Scenario
For the narrow cone and the wide cone, let their base radii be $r_1$ and $r_2,$ respectively; let their rises of the liquid levels be $\Delta h_1$ and $\Delta h_2,$ respectively. We have the following table: $$\begin{array}{cccccc} & \textbf{Base Radius} & \textbf{Height} & & \textbf{Volume} & \\ [2ex] \textbf{Narrow Cone} & r_1 & h_1+\Delta h_1 & & \frac13\pi r_1^2(h_1+\Delta h_1) & \\ [2ex] \textbf{Wide Cone} & r_2 & h_2+\Delta h_2 & & \frac13\pi r_2^2(h_2+\Delta h_2) & \end{array}$$ By the similar triangles discussed above, we get \begin{align*} \frac{3}{h_1}&=\frac{r_1}{h_1+\Delta h_1} &\implies \quad r_1&=\frac{3}{h_1}(h_1+\Delta h_1), & \hspace{10mm} (1) \\ \frac{6}{h_2}&=\frac{r_2}{h_2+\Delta h_2} &\implies \quad r_2&=\frac{6}{h_2}(h_2+\Delta h_2). & (2) \end{align*} The volume of the marble dropped into each cone is $\frac43\pi(1)^3=\frac43\pi.$
Now, we set up an equation for the volume of the narrow cone, then express $\Delta h_1$ in terms of $h_1:$ \begin{align*} \frac13\pi r_1^2(h_1+\Delta h_1) &= 3\pi h_1+\frac43\pi \\ \frac13 r_1^2(h_1+\Delta h_1) &= 3h_1+\frac43 \\ \frac13\left(\frac{3}{h_1}(h_1+\Delta h_1)\right)^2(h_1+\Delta h_1) &= 3h_1+\frac43 &&\text{by }(1) \\ \frac{3}{h_1^2}(h_1+\Delta h_1)^3 &= 3h_1+\frac43 \\ (h_1+\Delta h_1)^3 &= h_1^3 + \frac{4h_1^2}{9} \\ \Delta h_1 &= \sqrt[3]{h_1^3 + \frac{4h_1^2}{9}}-h_1. \end{align*} Next, we set up an equation for the volume of the wide cone, then express $\Delta h_2$ in terms of $h_2:$ $$\frac13\pi r_2^2(h_2+\Delta h_2) = 12\pi h_2+\frac43\pi.$$ Using a similar process from above, we get $$\Delta h_2 = \sqrt[3]{h_2^3+\frac{h_2^2}{9}}-h_2.$$ Recall that $\frac{h_1}{h_2}=4.$ Therefore, the requested ratio is \begin{align*} \frac{\Delta h_1}{\Delta h_2}&=\frac{\sqrt[3]{h_1^3 + \frac{4h_1^2}{9}}-h_1}{\sqrt[3]{h_2^3+\frac{h_2^2}{9}}-h_2} \\ &=\frac{\sqrt[3]{(4h_2)^3 + \frac{4(4h_2)^2}{9}}-4h_2}{\sqrt[3]{h_2^3+\frac{h_2^2}{9}}-h_2} \\ &=\frac{\sqrt[3]{4^3\left(h_2^3 + \frac{h_2^2}{9}\right)}-4h_2}{\sqrt[3]{h_2^3+\frac{h_2^2}{9}}-h_2} \\ &=\frac{4\sqrt[3]{h_2^3+\frac{h_2^2}{9}}-4h_2}{\sqrt[3]{h_2^3+\frac{h_2^2}{9}}-h_2} \\ &=\boxed{\textbf{(E) }4:1}. \end{align*} ~MRENTHUSIASM
## Solution 2 (Approximate Cones with Cylinders)
The heights of the cones are not given, so suppose the heights are very large (i.e. tending towards infinity) in order to approximate the cones as cylinders with base radii $3$ and $6$ and infinitely large height. Then the base area of the wide cylinder is $4$ times that of the narrow cylinder. Since we are dropping a ball of the same volume into each cylinder, the water level in the narrow cone/cylinder should rise $\boxed{\textbf{(E) } 4}$ times as much.
~scrabbler94
## Solution 3 (Calculus)
The volume of the shorter cone can be expressed as $V_1=3\pi h_1$, and the volume of the larger cone can be expressed as $V_2=12\pi h_2$. We also know that the volume changes by $\frac{4\pi}{3}$, because the volume of the $1$cm sphere is $\frac{4\pi}{3}$.
Taking the derivative of the equations, we get: $dV_1=3\pi(dh_1)=\frac{4\pi}{3}$ and $dV_2=12\pi(dh_2)=\frac{4\pi}{3}$.
Therefore, $dh_1=\frac{4\pi}{3}\cdot\frac{1}{3\pi}=\frac{4}{9}$ and $dh_2=\frac{4\pi}{3}\cdot\frac{1}{12\pi}=\frac{1}{9}$. The ratio is $\frac{4}{9}:\frac{1}{9}$ giving us the answer of $\boxed{\textbf{(E) }4:1}$.
~aurellia
## Solution 4 (Extremely Quick Observation)
Note that as long as the volumes are equal, the ratio of the heights of Cone 2 to Cone 1 is always $1:4$. We can prove this as follows:
If we fix Cone 2 to have a certain height, then the volume of the cone is $\frac{1}{3}\cdot36\pi{h_1}=12\pi{h_1}$. Now if Cone 1 has the same volume, its height would be $\frac{12\pi{h_1}}{\frac{1}{3}\cdot9\pi}=\frac{12\pi{h_1}}{3\pi}=4h_1$.
## Video Solution (Simple and Quick)
~ Education, the Study of Everything
~ pi_is_3.14
## Video Solution by TheBeautyofMath
First-this is not the most efficient solution. I did not perceive the shortcut before filming though I suspected it.
https://youtu.be/t-EEP2V4nAE?t=231 (for AMC 10A)
https://youtu.be/cckGBU2x1zg?t=814 (for AMC 12A)
~IceMatrix
~savannahsolver | 3,238 | 8,165 | {"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": 61, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.604268 |
https://convertoctopus.com/2147-grams-to-pounds | 1,627,983,014,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154457.66/warc/CC-MAIN-20210803092648-20210803122648-00269.warc.gz | 203,835,187 | 7,896 | ## Conversion formula
The conversion factor from grams to pounds is 0.0022046226218488, which means that 1 gram is equal to 0.0022046226218488 pounds:
1 g = 0.0022046226218488 lb
To convert 2147 grams into pounds we have to multiply 2147 by the conversion factor in order to get the mass amount from grams to pounds. We can also form a simple proportion to calculate the result:
1 g → 0.0022046226218488 lb
2147 g → M(lb)
Solve the above proportion to obtain the mass M in pounds:
M(lb) = 2147 g × 0.0022046226218488 lb
M(lb) = 4.7333247691093 lb
The final result is:
2147 g → 4.7333247691093 lb
We conclude that 2147 grams is equivalent to 4.7333247691093 pounds:
2147 grams = 4.7333247691093 pounds
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 pound is equal to 0.21126798789008 × 2147 grams.
Another way is saying that 2147 grams is equal to 1 ÷ 0.21126798789008 pounds.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that two thousand one hundred forty-seven grams is approximately four point seven three three pounds:
2147 g ≅ 4.733 lb
An alternative is also that one pound is approximately zero point two one one times two thousand one hundred forty-seven grams.
## Conversion table
### grams to pounds chart
For quick reference purposes, below is the conversion table you can use to convert from grams to pounds
grams (g) pounds (lb)
2148 grams 4.736 pounds
2149 grams 4.738 pounds
2150 grams 4.74 pounds
2151 grams 4.742 pounds
2152 grams 4.744 pounds
2153 grams 4.747 pounds
2154 grams 4.749 pounds
2155 grams 4.751 pounds
2156 grams 4.753 pounds
2157 grams 4.755 pounds | 492 | 1,749 | {"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-2021-31 | latest | en | 0.795989 |
http://nrich.maths.org/6559/solution?nomenu=1 | 1,498,614,635,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128322275.28/warc/CC-MAIN-20170628014207-20170628034207-00545.warc.gz | 293,320,722 | 2,705 | $E = 3.2\times 10^{20} \times 1.602\times 10^{-19} = 51.264\textrm{ J}$
If we equate this to a hammer head of mass $0.5\textrm{ kg}$, being flung at some speed, we find that
$$v = 14.32\textrm{ m/s} = 52\textrm{ km/h}\;.$$
This is an absolutely vast amount of energy for a single, fundamental particle to have!
If we rearrange $E = \frac{1}{2}mv^2$, using the rest mass of the particle for m, we find that
$v = 2.5\times 10^{14}\textrm{ m/s}$, which is about a million times faster than light. We can therefore assume that the mass of the particle is greater than the rest mass.
The energy per kilo of the photon (in terms of rest mass) is $51.264/(1.67\times 10^{-27}) = 3.07 \times 10^{29}\textrm{ J/kg}\;.$
Given that the mass of the earth is only $6\times 10^{24}\textrm{ kg}$, the ball of iron would contain enough energy to propel the earth to a velocity of about $320\textrm{ m/s}$, or $1151\textrm{ km/h}$, using the formula for kinetic energy.
Perhaps it maybe might just pass straight through the Earth, vaporising everything it touched, leaving the bulk a little shaken, but intact. Even one proton possessing this energy is extremely rare though, thankfully
If you rearrange the given formula to $v = c\sqrt{1 - \left(\frac{m_0c}{E}\right)^2} \approx 3\times 10^{8}\textrm{ m/s}$, i.e. as far as my calculator is concerned, almost the speed of light. | 430 | 1,368 | {"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} | 3.765625 | 4 | CC-MAIN-2017-26 | latest | en | 0.818451 |
nacopapers.com | 1,611,470,732,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703547333.68/warc/CC-MAIN-20210124044618-20210124074618-00728.warc.gz | 476,041,749 | 26,523 | Algebra is composed of the symbolical and mathematical connotations that are represented as an abstract pattern usually assigned in the form of letters that define the underlying concepts under the study of what is mathematics. Though embedded under the extensive network of the profound letter the subject matter will occasionally be somewhat complicated with the higher understanding being a critical component to capture the theorems under discussion fully. In order to illustrative vest the various typical algorithms under the relative study, the occasional elemental ladder is a prerequisite in order to ensure sound grasp and applicative features of topical study will be demonstrated to be affected in any projected assimilated study of the outlined subject. We can make Algebra really simple for you. Just write to our support:” write my easy paper” to have your paper done for good. grades
Branches of Algebra Research Papers
The branch is however divided into several spectrums that are correlated to other mathematical disciplines as outlined below:
Elementary Algebra Math Papers
This is composed of the initial fundamental principles of algebra that attempts to introduce some of the initial expressional features that will be covered under the entire topical review of the subject matter. Though precariously illustrated it is impertinent to point out that though elemental algebra holds the some of the basic arithmetical rules for valuation the universal principles is to redefine the numerical function and representation in a new alphanumeric format that can further be simplified and in most cases solved.
Abstract Algebra Mathematics Research
Under abstract algebra axiomatic expressional features such fields, rings, and groups are defined and reviewed to reveal the computational structure and composition that will attempt to analyze based on theorems such as set theory to redefine each of the functionalities and the gross representation of each of outlined subsets.
Homological Algebra Math Paper
This is the subset connotations of algebra that encompasses several algorithms that are reiterated within the topographical set spaces under fundamental studies.
Linear Algebra Field Mathematics
This will cover vector theory, linear equations, and matrices with the emphasis being placed on the single dimensional Euclidean space to predestine the respective problems.
Relational algebra – this is composed of chiefly several number theories that are predefined in a finite number set up usually closed number subset that may be assigned respective values dependent on the given subject matter.
What is the Boolean Algebra Area in Math
A subset of set theory that attempts to define the algebraic sub-classes into two classes, the truth and false patterns where the values may either be true values or false values.
Universal algebra – this is the set algebraic functionalities that will capture all cumulative functional features of all the algebraic computations
Algebraic Combinatorics Topics in Mathematics
This is the computational features of linear algebra that will capture utilize combinatorial theory to assimilate and redefine the operational features of the given data set as outlined based on the initially applicable algorithms.
Algebraic Number Theory Topic in Mathematics
Captures the numerical properties of numbers from the actual planes and will redefine all number theorems under the Cartesian plane by expressing them in algebraic forms.
As for the mentioned above algebra is a broader area that is embedded in all aspects mathematical computation and will often form threading threshold necessary to not only express abstract poly statements but will also form a structured platform that may be used in assessing and solving the respective solutions.
Despite the cliché that learning algebra is like attempting to adapt and learn a new language, the subject holds a broader variety of applicative features. At the core of such is the mathematical modeling of real-life practical numerical situations where actual figurative representation may be absent. The subject provides a structural frame that could be utilized and adopted by essay paper writing services to redefine operational activities that may otherwise fail to be quantified due to their qualitative nature thereby prompting input from other non – numerical statements that still quantifiable. | 763 | 4,418 | {"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} | 2.6875 | 3 | CC-MAIN-2021-04 | longest | en | 0.941528 |
http://research.stlouisfed.org/fred2/series/LFEM24FEISA647N | 1,409,615,981,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1409535920849.16/warc/CC-MAIN-20140909044430-00146-ip-10-180-136-8.ec2.internal.warc.gz | 437,460,871 | 18,404 | # Employed Population: Aged 15-24: Females for Iceland©
2013: 14,725 Persons (+ see more)
Annual, Not Seasonally Adjusted, LFEM24FEISA647N, Updated: 2014-05-05 9:02 AM CDT
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# ch23-p056 - 56. (a) We consider the radial field produced...
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56. (a) We consider the radial field produced at points within a uniform cylindrical distribution of charge. The volume enclosed by a Gaussian surface in this case is Lr π 2 . Thus, Gauss’ law leads to E q A rL r == = || () . enc 0 cylinder ε ρ π π 2 00 22 ch (b) We note from the above expression that the magnitude of the radial field grows with r . (c) Since the charged powder is negative, the field points radially inward. (d) The largest value of
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Ask a homework question - tutors are online | 183 | 709 | {"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} | 2.609375 | 3 | CC-MAIN-2017-04 | longest | en | 0.879624 |
https://www.paddlepaddle.org.cn/documentation/docs/en/api/paddle/incubate/optimizer/functional/minimize_bfgs_en.html | 1,679,398,150,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943695.23/warc/CC-MAIN-20230321095704-20230321125704-00726.warc.gz | 1,048,189,317 | 27,853 | # minimize_bfgs¶
paddle.incubate.optimizer.functional. minimize_bfgs ( objective_func, initial_position, max_iters=50, tolerance_grad=1e-07, tolerance_change=1e-09, initial_inverse_hessian_estimate=None, line_search_fn='strong_wolfe', max_line_search_iters=50, initial_step_length=1.0, dtype='float32', name=None ) [source]
Minimizes a differentiable function func using the BFGS method. The BFGS is a quasi-Newton method for solving an unconstrained optimization problem over a differentiable function. Closely related is the Newton method for minimization. Consider the iterate update formula:
\[x_{k+1} = x_{k} + H_k \nabla{f_k}\]
If \(H_k\) is the inverse Hessian of \(f\) at \(x_k\), then it’s the Newton method. If \(H_k\) is symmetric and positive definite, used as an approximation of the inverse Hessian, then it’s a quasi-Newton. In practice, the approximated Hessians are obtained by only using the gradients, over either whole or part of the search history, the former is BFGS, the latter is L-BFGS.
Reference:
Jorge Nocedal, Stephen J. Wright, Numerical Optimization, Second Edition, 2006. pp140: Algorithm 6.1 (BFGS Method).
Parameters
• objective_func – the objective function to minimize. `objective_func` accepts a 1D Tensor and returns a scalar.
• initial_position (Tensor) – the starting point of the iterates, has the same shape with the input of `objective_func` .
• max_iters (int, optional) – the maximum number of minimization iterations. Default value: 50.
• tolerance_grad (float, optional) – terminates if the gradient norm is smaller than this. Currently gradient norm uses inf norm. Default value: 1e-7.
• tolerance_change (float, optional) – terminates if the change of function value/position/parameter between two iterations is smaller than this value. Default value: 1e-9.
• initial_inverse_hessian_estimate (Tensor, optional) – the initial inverse hessian approximation at initial_position. It must be symmetric and positive definite. If not given, will use an identity matrix of order N, which is size of `initial_position` . Default value: None.
• line_search_fn (str, optional) – indicate which line search method to use, only support ‘strong wolfe’ right now. May support ‘Hager Zhang’ in the futrue. Default value: ‘strong wolfe’.
• max_line_search_iters (int, optional) – the maximum number of line search iterations. Default value: 50.
• initial_step_length (float, optional) – step length used in first iteration of line search. different initial_step_length may cause different optimal result. For methods like Newton and quasi-Newton the initial trial step length should always be 1.0. Default value: 1.0.
• dtype ('float32' | 'float64', optional) – data type used in the algorithm, the data type of the input parameter must be consistent with the dtype. Default value: ‘float32’.
• name (str, optional) – Name for the operation. For more information, please refer to Name. Default value: None.
Returns
• is_converge (bool): Indicates whether found the minimum within tolerance.
• num_func_calls (int): number of objective function called.
• position (Tensor): the position of the last iteration. If the search converged, this value is the argmin of the objective function regrading to the initial position.
• objective_value (Tensor): objective function value at the position.
• inverse_hessian_estimate (Tensor): the estimate of inverse hessian at the position.
Return type
output(tuple)
Examples
```import paddle
def func(x): | 828 | 3,500 | {"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-2023-14 | longest | en | 0.716285 |
https://www.cheenta.com/graphing-the-exponential-tomato-objective-699/ | 1,519,090,138,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812871.2/warc/CC-MAIN-20180220010854-20180220030854-00032.warc.gz | 821,665,183 | 22,263 | • No products in the cart.
# Graphing the exponential (TOMATO Objective 699)
Graphing an exponential function can be tricky. We are familiar with the graph of $$e^x$$. But what if a bunch of such exponential expressions is clubbed together? I.S.I.’s Test of Mathematics has a problem related to this.
The adjoining figure is the graph of
(A) $$y = 2 e^x$$
(B) $$y = 2^{-x}$$
(C) $$y = e^x + e^{-x}$$
(D) $$y = e^x – e^{-x} + 2$$
Part 2 and 3
January 26, 2018 | 151 | 467 | {"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} | 2.703125 | 3 | CC-MAIN-2018-09 | longest | en | 0.886362 |
https://btechgeeks.com/python-program-to-find-the-row-with-maximum-number-of-1s-in-matrix/ | 1,695,549,374,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506632.31/warc/CC-MAIN-20230924091344-20230924121344-00385.warc.gz | 171,473,685 | 13,328 | Python Program to Find the Row with Maximum Number of 1s in Matrix
In the previous article, we have discussed Python Program for Markov Matrix
Given a matrix with just 0 and 1 elements, the task is to find the row in the matrix with the greatest number of 1s in that row in Python.
What is a matrix:
A matrix is a rectangular sequence of numbers divided into columns and rows. A matrix element or entry is a number that appears in a matrix.
Example:
Above is the matrix which contains 5 rows and 4 columns and having elements from 1 to 20.
In this order, the dimensions of a matrix indicate the number of rows and columns.
Here as there are 5 rows and 4 columns it is called a 5*4 matrix.
Examples:
Example1:
Input:
Given Matrix :
1 0 0
1 1 1
1 0 1
Output:
The row which is containing maximum number of 1 is : 2
Example2:
Input:
Given Matrix :
1 1 0
0 0 1
0 1 1
Output:
The row which is containing maximum number of 1 is : 3
Note: If both the rows are having the same number of 1’s then it prints the last row.
Program to Find the Row with Maximum Number of 1s in Matrix in Python
Below are the ways to find the row in the matrix with the greatest number of 1s in that row in Python:
Method #1: Using For Loop (Static Input)
Approach:
• Give the matrix as static input and store it in a variable.
• Calculate the number of rows of the given matrix by calculating the length of the nested list using the len() function and store it in a variable mtrxrows.
• Calculate the number of columns of the given matrix by calculating the length of the first list in the nested list using the len() function and store it in a variable mtrxcolums.
• Take a variable that stores the number of 1’s in the row and initialize it with 0.
• Loop till the given number of rows using the For loop.
• Inside the loop, calculate the number of ones in that row using the count() function.
• Check if the number of ones in that row is greater or equal to than maxOnes using the if conditional statement.
• If it is true then initialize maxOnes with numbOfOnes.
• Take a variable and store this row number in this row.
• Print the row that is containing the maximum number of 1’s by printing maxOnesRow.
• The Exit of the Program.
Below is the implementation:
# Give the matrix as static input and store it in a variable.
mtrx = [[1, 0, 0], [1, 1, 1], [1, 0, 1]]
# Calculate the number of rows of the given matrix by
# calculating the length of the nested list using the len() function
# and store it in a variable mtrxrows.
mtrxrows = len(mtrx)
# Calculate the number of columns of the given matrix by
# calculating the length of the first list in the nested list
# using the len() function and store it in a variable mtrxcols.
mtrxcols = len(mtrx[0])
# take a variable which stores the number of 1's in the row and initialize it with 0
maxOnes = 0
# Loop till the given number of rows using the For loop.
for p in range(mtrxrows):
# calculate the number of ones in that row using the count() function
numbOfOnes = mtrx[p].count(1)
# check if the number of ones in that row is greater or equal to
# than maxOnes using the if conditional statement
if(numbOfOnes >= maxOnes):
# if it is true then initializee maxOnes with numbOfOnes
maxOnes = numbOfOnes
# take a variable and store this row number in this row
maxOnesRow = p+1
# print the row that is containing maximum number of 1's by printing maxOnesRow
print('The row which is containing maximum number of 1 is :', maxOnesRow)
Output:
The row which is containing maximum number of 1 is : 2
Method #2: Using For loop (User Input)
Approach:
• Give the number of rows of the matrix as user input using the int(input()) function and store it in a variable.
• Give the number of columns of the matrix as user input using the int(input()) function and store it in another variable.
• Take a list and initialize it with an empty value using [] or list() to say gvnmatrix.
• Loop till the given number of rows using the For loop
• Inside the For loop, Give all the row elements of the given Matrix as a list using the list(),map(),int(),split() functions and store it in a variable.
• Add the above row elements list to gvnmatrix using the append() function.
• Take a variable that stores the number of 1’s in the row and initialize it with 0.
• Loop till the given number of rows using the For loop.
• Inside the loop, calculate the number of ones in that row using the count() function.
• Check if the number of ones in that row is greater or equal to than maxOnes using the if conditional statement.
• If it is true then initialize maxOnes with numbOfOnes.
• Take a variable and store this row number in this row.
• Print the row that is containing the maximum number of 1’s by printing maxOnesRow.
• The Exit of the Program.
Below is the implementation:
# Give the number of rows of the matrix as user input using the int(input()) function
# and store it in a variable.
mtrxrows = int(input('Enter some random number of rows of the matrix = '))
# Give the number of columns of the matrix as user input using the int(input()) function
# and store it in another variable.
mtrxcols = int(input('Enter some random number of columns of the matrix = '))
# Take a list and initialize it with an empty value using [] or list() to say gvnmatrix.
mtrx = []
# Loop till the given number of rows using the For loop
for n in range(mtrxrows):
# Inside the For loop, Give all the row elements of the given Matrix as a list using
# the list(),map(),int(),split() functions and store it in a variable.
l = list(map(int, input(
'Enter {'+str(mtrxcols)+'} elements of row {'+str(n+1)+'} separated by spaces = ').split()))
# Add the above row elements list to gvnmatrix using the append() function.
mtrx.append(l)
# take a variable which stores the number of 1's in the row and initialize it with 0
maxOnes = 0
# Loop till the given number of rows using the For loop.
for p in range(mtrxrows):
# calculate the number of ones in that row using the count() function
numbOfOnes = mtrx[p].count(1)
# check if the number of ones in that row is greater or equal to
# than maxOnes using the if conditional statement
if(numbOfOnes >= maxOnes):
# if it is true then initializee maxOnes with numbOfOnes
maxOnes = numbOfOnes
# take a variable and store this row number in this row
maxOnesRow = p+1
# print the row that is containing maximum number of 1's by printing maxOnesRow
print('The row which is containing maximum number of 1 is :', maxOnesRow)
Output:
Enter some random number of rows of the matrix = 3
Enter some random number of columns of the matrix = 3
Enter {3} elements of row {1} separated by spaces = 1 1 0
Enter {3} elements of row {2} separated by spaces = 0 0 1
Enter {3} elements of row {3} separated by spaces = 0 1 1
The row which is containing maximum number of 1 is : 3
Find the best practical and ready-to-use Python Programming Examples that you can simply run on a variety of platforms and never stop learning. | 1,787 | 6,975 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2023-40 | longest | en | 0.854348 |
https://www.convertunits.com/from/thousand+cubic+foot/day/to/gallon/hour | 1,680,389,245,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950363.89/warc/CC-MAIN-20230401221921-20230402011921-00248.warc.gz | 767,234,791 | 12,775 | Convert thousand cubic foot/day to gallon/hour [US]
thousand cubic foot/day gallon/hour
Did you mean to convert thousand cubic foot/day to gallon/hour [US] gallon/hour [UK]
How many thousand cubic foot/day in 1 gallon/hour? The answer is 0.0032083333006673.
We assume you are converting between thousand cubic foot/day and gallon/hour [US].
You can view more details on each measurement unit:
thousand cubic foot/day or gallon/hour
The SI derived unit for volume flow rate is the cubic meter/second.
1 cubic meter/second is equal to 3051.1871607739 thousand cubic foot/day, or 951019.38446961 gallon/hour.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between thousand cubic feet/day and gallons/hour.
Type in your own numbers in the form to convert the units!
Quick conversion chart of thousand cubic foot/day to gallon/hour
1 thousand cubic foot/day to gallon/hour = 311.68831 gallon/hour
2 thousand cubic foot/day to gallon/hour = 623.37663 gallon/hour
3 thousand cubic foot/day to gallon/hour = 935.06494 gallon/hour
4 thousand cubic foot/day to gallon/hour = 1246.75326 gallon/hour
5 thousand cubic foot/day to gallon/hour = 1558.44157 gallon/hour
6 thousand cubic foot/day to gallon/hour = 1870.12989 gallon/hour
7 thousand cubic foot/day to gallon/hour = 2181.8182 gallon/hour
8 thousand cubic foot/day to gallon/hour = 2493.50652 gallon/hour
9 thousand cubic foot/day to gallon/hour = 2805.19483 gallon/hour
10 thousand cubic foot/day to gallon/hour = 3116.88315 gallon/hour
Want other units?
You can do the reverse unit conversion from gallon/hour to thousand cubic foot/day, or enter any two units below:
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ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 558 | 2,210 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2023-14 | latest | en | 0.840104 |
https://doublemath.com/how-to-solve-sss-triangle/ | 1,716,520,427,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058677.90/warc/CC-MAIN-20240524025815-20240524055815-00633.warc.gz | 176,423,061 | 20,404 | Double Math
# How to Solve SSS Triangle
“SSS” means side, side, side.
“SSS” is used when three sides of a triangle are given and we want to find the missing angles. Consider the triangle \triangle ABC with the side a, b and c and angle \alpha, \beta and \gamma.We can observe that we are given the three sides a, b and c. Therefore the figure illustrates a triangle combination which is known as a SSS triangle.
To solve an SSS triangle:
Step 1: First we use the “Law of cosine ” to find an angle.
Step 2: Secondly use again “Law of cosine ” to find another angle.
Step 3: Finally use angles of triangles added to 180 to find the last angle.
We use the “angle” version of the Law of Cosines:
• \cos\alpha=\frac{b^2+c^2-a^2}{2bc}.
• \cos\beta=\frac{a^2+c^2-b^2}{2ac}.
• \cos\gamma=\frac{a^2+b^2-c^2}{2ab}.
Example No.1
(By using Law of cosine)
when three sides are given.
a=7 , b=3 , c=5
\cos\alpha=\frac{b^2+c^2-a^2}{2bc}.
By putting the values
\cos\alpha=\frac{3^2+5^2-7^2}{2(3)(5)}.
\Rightarrow\cos\alpha=\frac{9+25-49}{30}.
\Rightarrow\cos\alpha=\frac{-15}{30}.
\Rightarrow\cos\alpha=\frac{-1}{2}.
\Rightarrow\alpha\;\;\;=\cos^{-1}(\frac{-1}2).
\Rightarrow\boxed{\alpha\;=120^o}.
Now,
\cos\beta=\frac{c^2+a^2-b^2}{2ca}.
using values
\cos\beta=\frac{5^2+7^2-3^2}{2(5)(7)}.
\Rightarrow\cos\beta=\frac{25+49-9}{70}.
\Rightarrow\cos\beta=\frac{65}{70}.
\Rightarrow\cos\beta=0.9286.
\Rightarrow\beta\;\;\;=\cos^{-1}(0.9286).
\Rightarrow\boxed{\beta\;=1^o}.
Now, we know that
\alpha +\beta+ \gamma=180^o.
\gamma=180^o-\alpha -\beta.
\gamma=180^o-120^o -1^o.
\Rightarrow\boxed{\gamma=59^o}. | 587 | 1,621 | {"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} | 4.84375 | 5 | CC-MAIN-2024-22 | longest | en | 0.52174 |
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# GMAT MATH
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21 Mar 2004, 18:25
Hi, I'm doing my college mission of studying for GMAT for 2 yrs, and seeing how well I do on the GMAT, amyways i'm worried about the GMAT Math, I look at the Manhattan GMAT math challenge problem and there so hard, is that what I will be doing? If I do the math that kaplan has will I be alright?
Will Kaplan Math prepare me?
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21 Mar 2004, 19:32
nikerun21 wrote:
Hi, I'm doing my college mission of studying for GMAT for 2 yrs, and seeing how well I do on the GMAT, amyways i'm worried about the GMAT Math, I look at the Manhattan GMAT math challenge problem and there so hard, is that what I will be doing? If I do the math that kaplan has will I be alright?
Will Kaplan Math prepare me?
GMAT math is different from the math you do in college. For example, GMAT never asks you find the the value of (cosx)^(1/x) when x approaches to 0. Just prepare for the problems GMAT asks. If you have not done math for years, Kaplan GMAT/GRE math workbook, I heard, is good. Of course, it just helps you to brush up basics.
Re: GMAT MATH [#permalink] 21 Mar 2004, 19:32
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# GMAT MATH
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3
Posted by8 years ago
Archived
How about we use a large balloon to lift a spacecraft to the edge of the atmosphere and then use solar panels and efficient ion engines to gradually accelerate to orbital speed?
I believe ion engines need a vacuum to operate so I wonder if the highest a balloon could lift us is enough of a vacuum? Using solar panels we don't need to lift and accelerate the large mass of fuel normally required. Lifting the required fuel is the largest cost to getting to orbit (as far as I understand it).
(I asked this on science reddit first but no one knew I guess)
62% Upvoted
Sort by
level 1
5 points · 8 years ago
The push from an ion engine is too weak to escape Earth's gravity.
level 2
Original Poster1 point · 8 years ago
It would accelerate gradually over time. There's only a velocity requirement to escape to escape from earth's gravity, not an minimum acceleration.
The balloon would be holding the craft up so it wouldn't be fighting against gravity to stay up.
level 3
[deleted]
5 points · 8 years ago
There's only a velocity requirement to escape to escape from earth's gravity, not an minimum acceleration.
This is blatantly false - if your acceleration is less than g you'll just fall, which would be the case in this instance. The thrust of an ion engine simply is not high enough.
level 4
Original Poster1 point · 8 years ago
no the balloon prevents you from falling, right?
level 5
[deleted]
3 points · 8 years ago
The balloon only prevents you from falling as high as the balloon can go, which is not very high. The reason this doesn't make any sense is that a balloon will take you up to some stable equilibrium high up in the upper atmosphere. As soon as the ion drive starts to push you up, the balloon becomes less useful because the air density continues to decrease. All the ion drive will do is shift your stable equilibrium position slightly higher. This new stable position will not even be completely out of the atmosphere, let alone in orbit. The thrust provided by an ion drive will not be enough to push an object from high in the atmosphere into orbit.
level 5
1 point · 8 years ago
The balloon isn't a ratchet, it'll only prevent you from falling below the balloon's max alt.
If you don't have the thrust to go higher, you won't be going higher.
level 3
2 points · 8 years ago
Then you'll only be able to move as fast as the balloon would naturally on its own. Otherwise you'd be dragging the balloon behind you.
level 4
Original Poster1 point · 8 years ago
With very low atmosphere couldn't you "drag" the balloon as fast as you want?
If your ion engine had enough thrust it could overcome the drag of the balloon anyway.
Plus instead of a round balloon attached by a string picture a cone shape balloon with the engine mounted underneath (like a blimp)
level 5
2 points · 8 years ago
with NO atmosphere you could drag the balloon as fast as you want, but very little atmosphere ≠no atmosphere.
level 1
2 points · 8 years ago
I don't think an ion engine would work, but people do think of things like this.
level 1
2 points · 8 years ago
You are neglecting to account for the speed required to orbit. It's not just a matter of going 80 miles "up" you also need to be moving sideways at ~18,000 MPH to orbit the earth.
level 2
Original Poster1 point · 8 years ago
hence the ion engine and solar panels. It would gradually reach 18,000 MPH over a long period of time as the balloon maintained its altitude.
level 3
0 points · 8 years ago
Ion engines have SERIOUSLY low isp (thrust), even under ideal circumstances. A balloon and solid rocket booster combo would be more realistic.
level 4
3 points · 8 years ago
Isp and thrust are not the same thing. Ion engines have seriously low thrust, but seriously high specific impulse.
level 4
Original Poster1 point · 8 years ago
Hmm, so maybe it's a good idea but we need to wait for more efficient ion engines to be created? Perhaps the Vasmir?
level 5
4 points · 8 years ago
An ion engine is simply the wrong tool for the job.
Say you want an engine that will give you low thrust, but you can turn it on and leave it on for 5+ years for a mars mission, use an Ion engine. If you want high thrust for a short period of time, say to get something out of earth's gravity well, then you want a solid rocket booster.
level 6
1 point · 8 years ago
I completely agree, without more knowledge on the subject of ion engines, one would have to be quite powerful indeed to offset the earth's gravity. It seems like it would have to be extremely powerful to achieve exit velocity as well.
level 6
Original Poster0 points · 8 years ago
I don't want high thrust, I want something that can create thrust from the energy of the sun so I don't need to lug as much fuel into orbit.
level 7
1 point · 8 years ago
You probably want a laser-powered launcher then - that, at least can be powered by solar panels.
level 7
1 point · 8 years ago
ion engines use fuel.
level 1
1 point · 8 years ago
The craft would have to detach from the balloon when starting up the ion drive to reduce drag. I think ion engines are much too weak to escape earth from anywhere a balloon could take them. Consider adding a traditional rocket (much smaller than one needed from the ground) to help get going.
level 2
Original Poster1 point · 8 years ago
I'm thinking if the atmosphere is low enough for the ion engine to operate, the drag from the balloon would be very small.
Also the balloon could be an aerodynamic shape to reduce drag a lot.
As long as the balloon is providing lift, the ion engine is free to accelerate the craft as gradually as it needs to. It could take a month to reach orbital velocity if needed.
level 3
0 points · 8 years ago
I'm thinking if the atmosphere is low enough for the ion engine to operate, the drag from the balloon would be very small.
The drag due the atmosphere isn't the problem with that solution - the problem is just simply the force due to normal gravitational forces. No ion engine has the capability to create a force big enough to stop your spacecraft from falling from the balloon back to the surface of the Earth.
Do you understand the difference in physics between something which has been sent to the edges of the atmosphere in a balloon & is basically sitting stationarily over the surface of the planet, versus something which is in "free fall" around the planet (like the International Space Station)?
level 1
Original Poster1 point · 8 years ago
I did a little research:
At the height of the highest balloon (~170,000 ft) (http://en.wikipedia.org/wiki/Flight_altitude_record#Unmanned_gas_balloon) the air pressure would only be around 1/1000 of that at sea level (http://ww2010.atmos.uiuc.edu/(Gh)/guides/mtr/prs/hght.rxml)
Then it looks like drag decreases linearly with air pressure (http://en.wikipedia.org/wiki/Drag_equation) so drag would be 1/1000 of what it would be at sea level.
Might be doable? I guess it just depends on how much thrust the ion engines can make, the weights involved, and whether ion engines can operate in 1/1000 of sea level atmosphere.
level 1
1 point · 8 years ago
The thrust that an ion engine makes is equivalent to the weight of a piece of paper. It would never work.
level 2
Original Poster1 point · 8 years ago
What if we let it run for a few months and built up speed? What if we used 10 or 100 ion engines, or a more powerful equivalent?
level 3
1 point · 8 years ago
Your primary problem is trust to weight ratio. Your problem only gets worse unless you can build an ion engine that weighs less than a sheet of paper.
You can use a more powerful equivalent; that would be a conventional rocket which is what we are using already.
level 1
1 point · 8 years ago
You won't be able to reach escape velocity by the time the balloon pops or stops providing sufficient lift.
From what I understand, you want to use the balloon to save the ion thruster from having to exert the force required to resist gravity.
The ion thruster won't be able to lift itself up on its own. In order to do this as you describe, you would need a balloon that can reach escape velocity on its own, with whatever load is equal to what the ion thruster can't make up. Then the ion engine would need to generate enough lift to allow the balloon to float to escape velocity. Physics simply won't let this happen. Balloons do not float upward at 18,000MPH.
Even at the top of where balloons have been, 50km, gravity is still around 9.67m/s. A balloon will only be useful getting you around there in the most ideal situation.
level 1
1 point · 8 years ago
What you have just described is an orbital airship. A small private space program called JP Aerospace have a patent for the idea (they call it "Airship to Orbit" or ATO, see the .pdf and video on their site) and are working on it right now. A lot of people are skeptical, but it's probably something to keep an eye on.
level 1
[deleted]
1 point · 8 years ago
Even the space station has trouble with drag where it is in low earth orbit. They reboost the station into a proper orbit every once in a while.
Current Ion engines produce the same about the amount of force as a piece of paper sitting on your hand. I'm sure they are getting better but it wouldn't overcome atmospheric drag even at high altitudes. Plus, the balloon would explode (because of pressure differentials between inside and outside of balloon) before it had a chance to even reach a stable orbital velocity.
level 1
1 point · 8 years ago
demosthenes: You're not escaping Earth's "gravity". That's not the point of a launch vehicle. You're trying to escape Earth's atmosphere. That's the hard part. A balloon cannot escape Earth's atmosphere by definition; it depends on the atmosphere for lift. An ion engine is to weak to overcome atmospheric effects.
level 1
1 point · 8 years ago
The balloon is lofted by bouyancy. The air resistance combined with the constant downward force of gravity would inhibit your plans. As you attempt to overcome your collisions with air, you rely on momentum, which means you also rely on mass, which your design must minimize. I can't prove it with numbers, but everything is working against you.
A drag coefficent can't be overcome as a function of time, it must be matched. Sadly, the Newtons required to push an object through a fluid increases with the cube of the velocity.
Perhaps if you had some sort of ultralight super material able to effect a rigid structure with nearly no cross section you could take advantage of a wing design. However, as the same forces responsible for bouyancy decrease, benefit from the wing design decreases, and the penalties of departing a maximized volume/surface area design decreases efficacy. A sphere has the easiest calculable drag coefficient. Your reference area is roughly the square of the cube root of the volume.
level 1
[deleted]
1 point · 8 years ago
It sounds like a legitimate idea and most of the arguments against you apply at sea level but not at the height you're talking about. Run this by a physics professer and see what they have to say about it.
level 1
1 point · 8 years ago
I think you also need to accelerate to about 18000 mph to achieve orbit.
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2. | 2,708 | 11,656 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2018-34 | latest | en | 0.952736 |
http://www.ditutor.com/conics/hyperbola.html | 1,500,648,415,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423785.29/warc/CC-MAIN-20170721142410-20170721162410-00640.warc.gz | 398,758,709 | 5,870 | Hyperbola
The hyperbola is the locus of points on the plane whose difference of distances to two fixed points, foci, are constant.
Elements of the Hyperbola
Foci
The foci are the fixed points of the hyperbola. They are denoted by F and F'.
Transverse Axis or real axis
The tranverse axis is the line segment between the foci.
Conjugate Axis or imaginary axis
The conjuagate axis is the perpendicular bisector of the line segment (transverse axis).
Center
The center is the point of intersection of the axes and is also the center of symmetry of the hyperbola.
Vertices
The points A and A' are the points of intersection of the hyperbola with the transverse axis.
The focal radii are the line segments that join a point on the hyperbola with the foci: PF and PF'.
Focal Length
The focal length is the line segment , which has a length of 2c.
Semi-Major Axis
The semi-major axis is the line segment that runs from the center to a vertex of the hyperbola. Its length is a.
Semi-Minor Axis
The semi-minor axis is a line segment which is perpendicular to the semi-major axis. Its length is b.
Axes of Symmetry
The axes of symmetry are the lines that coincide with the transversal and conjugate axis.
Asymptotes
The asymptotes are the lines with the equations:
The Relationship Between the Semiaxes:
Eccentricity of a Hyperbola:
Equation of the Hyperbola
Hyperbolas Centered at (0, 0)
Horizontal Transverse Axis
F'(−c, 0) and F(c, 0)
Vertical Transverse Axis
F'(0, −c) and F(0, c)
Hyperbolas Centered at (x0, y0)
Horizontal Transverse Axis
F(x0+c, y0) and F'(x0− c, y0)
By removing the denominators, an equation is obtained in the form:
Keep in mind that A and B have opposite signs.
Vertical Transverse Axis
F(x0, y0+c) and F'(x0, y0−c)
Rectangular Hyperbola
Hyperbolas where the semiaxes are equal are called rectangular or equilateral hyperbolas (a = b).
The equation of a rectangular hyperbola is:
The equations of the asymptotes are:
,
That is, the angle bisectors of the quadrants.
The eccentricity is:
Equation of a Rectangular Hyperbola
To switch the asymptotes to those determined by the x and y-axis, turn the asymptote −45° about the origin.
If it is rotated 45°, the hyperbola is in the second and fourth quadrant. | 594 | 2,268 | {"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-2017-30 | latest | en | 0.864062 |
https://grandyang.com/leetcode/765/ | 1,725,745,437,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650920.0/warc/CC-MAIN-20240907193650-20240907223650-00744.warc.gz | 264,281,095 | 36,378 | # 765. Couples Holding Hands
N couples sit in 2N seats arranged in a row and want to hold hands. We want to know the minimum number of swaps so that every couple is sitting side by side. A swap consists of choosing any two people, then they stand up and switch seats.
The people and seats are represented by an integer from `0` to `2N-1`, the couples are numbered in order, the first couple being `(0, 1)`, the second couple being `(2, 3)`, and so on with the last couple being `(2N-2, 2N-1)`.
The couples’ initial seating is given by `row[i]` being the value of the person who is initially sitting in the i-th seat.
Example 1:
``````Input: row = [0, 2, 1, 3]
Output: 1
Explanation: We only need to swap the second (row[1]) and third (row[2]) person.
``````
Example 2:
``````Input: row = [3, 2, 0, 1]
Output: 0
Explanation: All couples are already seated side by side.
``````
Note:
1. `len(row)` is even and in the range of `[4, 60]`.
2. `row` is guaranteed to be a permutation of `0...len(row)-1`.
[3 1 4 0 2 5]
[3 2 4 0 1 5]
[3 2 4 5 1 0]
``````class Solution {
public:
int minSwapsCouples(vector<int>& row) {
int res = 0, n = row.size();
for (int i = 0; i < n; i += 2) {
if (row[i + 1] == (row[i] ^ 1)) continue;
++res;
for (int j = i + 1; j < n; ++j) {
if (row[j] == (row[i] ^ 1)) {
row[j] = row[i + 1];
row[i + 1] = row[i] ^ 1;
break;
}
}
}
return res;
}
};
``````
[3 1 4 0 2 5]
``````class Solution {
public:
int minSwapsCouples(vector<int>& row) {
int res = 0, n = row.size();
vector<int> root(n, 0);
for (int i = 0; i < n; ++i) root[i] = i;
for (int i = 0; i < n; i += 2) {
int x = find(root, row[i] / 2);
int y = find(root, row[i + 1] / 2);
if (x != y) {
root[x] = y;
++res;
}
}
return res;
}
int find(vector<int>& root, int i) {
return (i == root[i]) ? i : find(root, root[i]);
}
};
``````
``````class Solution {
public:
int minSwapsCouples(vector<int>& row) {
unordered_map<int, int> m;
for (int i = 0; i < row.size(); i += 2) {
helper(m, row[i] / 2, row[i + 1] / 2);
}
return m.size();
}
void helper(unordered_map<int, int>& m, int x, int y) {
int c1 = min(x, y), c2 = max(x, y);
if (c1 == c2) return;
if (m.count(c1)) helper(m, m[c1], c2);
else m[c1] = c2;
}
};
``````
Github 同步地址：
https://github.com/grandyang/leetcode/issues/765
Missing Number
First Missing Positive
https://leetcode.com/problems/couples-holding-hands/
https://leetcode.com/problems/couples-holding-hands/discuss/113353/Monster-Style-C++-O(n)-unordered_map
https://leetcode.com/problems/couples-holding-hands/discuss/117520/Java-union-find-easy-to-understand-5-ms
https://leetcode.com/problems/couples-holding-hands/discuss/113362/JavaC++-O(N)-solution-using-cyclic-swapping
LeetCode All in One 题目讲解汇总(持续更新中…)
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# Problem 4. Make a checkerboard matrix
Solution 1147866
Submitted on 29 Mar 2017 by Henry Lu
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
n = 5; a = [1 0 1 0 1; 0 1 0 1 0; 1 0 1 0 1; 0 1 0 1 0; 1 0 1 0 1]; assert(isequal(a,checkerboard(n)))
2 Pass
n = 4; a = [1 0 1 0; 0 1 0 1; 1 0 1 0; 0 1 0 1]; assert(isequal(a,checkerboard(n)))
3 Pass
n = 7; a = [1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1]; assert(isequal(a,checkerboard(n))) | 320 | 623 | {"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-2019-35 | longest | en | 0.631596 |
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# W8final - Math 623(IOE 623 Winter 2008 Final exam Name...
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Math 623 (IOE 623), Winter 2008: Final exam Name: Student ID: This is a closed book exam. You may bring up to ten one sided A4 pages of notes to the exam. You may also use a calculator but not its memory function. Please write all your solutions in this exam booklet (front and back of the page if necessary). Keep your explanations concise (as time is limited) but clear. State explicitly any additional assumptions you make. Time is counted in years, prices in USD, and all interest rates are continuously compounded unless otherwise stated. You are obliged to comply with the Honor Code of the College of Engineering. After you have completed the examination, please sign the Honor Pledge below. A test where the signed honor pledge does not appear may not be graded. I have neither given nor received aid, nor have I used unauthorized resources, on this examination . Signed: 1
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(1) Suppose I wish to compute the value of a European Butterfly Spread option on a stock. The option pays Φ( S ) if exercised when the stock price is S , where Φ( S ) = 0 if S 40 S - 40 if 40 S 50 60 - S if 50 S 60 0 if S 60 . The current value of the stock is 48 and it expires 6 months from today. Its volatility (in units of years - 1 / 2 is 0 . 34. Assume that the continuous rate of interest over the lifetime of the option is 3 . 75 percent. The value of the option is to be obtained by numerically solving a terminal-boundary value problem for a PDE. The PDE is the Black-Scholes PDE transformed by the change of variable
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W8final - Math 623(IOE 623 Winter 2008 Final exam Name...
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Ask a homework question - tutors are online | 517 | 2,109 | {"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} | 3.421875 | 3 | CC-MAIN-2018-13 | latest | en | 0.902034 |
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#1451
Yesterday, 04:31 PM
Abomb7894 Member CAS Join Date: Jan 2014 Posts: 320
Quote:
Originally Posted by DamSon Can someone explain why the number of testable correlations in Venter is n-3 choose 2? In a 4x4 triangle, n=4 so the number of testable correlations should be 0, but isn't there one set based on column 1 and 2 with 2 observations each?
Casual Fellow notes its not explained well, and says we want n-3 because we need at least 3 development factors and use choose 2 to ensure all column pairs, not just consecutive columns.
I can only assume you need at least 3 as T is distributed n-2, so you need some amount greater than that.
#1452
Yesterday, 07:28 PM
actuarialgenes Member CAS Join Date: Apr 2017 Location: New England Studying for 7 Favorite beer: Wine Posts: 140
Quote:
Originally Posted by Abomb7894 The first list is from Marshall for Internal Systemic Risks. The second risk is from Brehm talking about parameter risk I believe, but there's another list which is estimation, event, projection, and systematic. I think it's from two different chapters & is inconsistent.
Thanks!
#1453
Yesterday, 07:35 PM
actuarialgenes Member CAS Join Date: Apr 2017 Location: New England Studying for 7 Favorite beer: Wine Posts: 140
Quote:
Originally Posted by DamSon Can someone explain why the number of testable correlations in Venter is n-3 choose 2? In a 4x4 triangle, n=4 so the number of testable correlations should be 0, but isn't there one set based on column 1 and 2 with 2 observations each?
I think it says somewhere in the text that you want to exclude columns with 2 (or fewer) elements. The reason (I think) is that the numerator in the “T” test statistic you calculate will be 0 (regardless of what you calculate for “r”).
So in an n x n triangle you have n-1 factor columns, and you can’t use the last two (which have one and two elements), leaving you with (n-1)-2 = n-3 to choose 2 from.
#1454
Today, 01:23 AM
DamSon Member CAS Join Date: Sep 2018 Location: NYC Studying for 7 Favorite beer: Alcoholic Posts: 83
Quote:
Originally Posted by Abomb7894 Casual Fellow notes its not explained well, and says we want n-3 because we need at least 3 development factors and use choose 2 to ensure all column pairs, not just consecutive columns. I can only assume you need at least 3 as T is distributed n-2, so you need some amount greater than that.
Quote:
Originally Posted by actuarialgenes I think it says somewhere in the text that you want to exclude columns with 2 (or fewer) elements. The reason (I think) is that the numerator in the “T” test statistic you calculate will be 0 (regardless of what you calculate for “r”). So in an n x n triangle you have n-1 factor columns, and you can’t use the last two (which have one and two elements), leaving you with (n-1)-2 = n-3 to choose 2 from.
Thanks everyone, what you guys said helps clear this up. As a follow up, are there any less common topics people feel like are likely to show up? In the middle of reviewing things like this right now, and I can't think of things past like table M, discounted reserves, and feldblum enhancement. Trying to go over problems I've done poorly on from the problem pack. Really need some new problems to work on now... tempted to do TIA exams 2-3 and just skip the BS questions
Last edited by DamSon; Today at 01:30 AM..
#1455
Today, 02:23 AM
redearedslider Member CAS Join Date: Oct 2015 Posts: 13,733
Quote:
Originally Posted by DamSon Thanks everyone, what you guys said helps clear this up. As a follow up, are there any less common topics people feel like are likely to show up? In the middle of reviewing things like this right now, and I can't think of things past like table M, discounted reserves, and feldblum enhancement. Trying to go over problems I've done poorly on from the problem pack. Really need some new problems to work on now... tempted to do TIA exams 2-3 and just skip the BS questions
You did all the CF and RF problems? What did you think of the RF problem pack? Might have to skip it but I’ll skim if there’s good stuff that CF missed
__________________
Quote:
Originally Posted by Abraham Weishaus ASM does not have a discussion of stimulation, but considering how boring the manual is, maybe it would be a good idea.
#1456
Today, 02:57 AM
DamSon Member CAS Join Date: Sep 2018 Location: NYC Studying for 7 Favorite beer: Alcoholic Posts: 83
Quote:
Originally Posted by redearedslider You did all the CF and RF problems? What did you think of the RF problem pack? Might have to skip it but I’ll skim if there’s good stuff that CF missed
Yeah, I've done all the calculation problems from both twice as well as all the past exam problems in the CF manual once. I also re-do questions I didn't do well on from both every few days, so there are some problems I've done around 6-7 times.
I like the problem pack a lot, there are a lot of good questions in there. CF problems take a lot of time to do since they're pretty comprehensive and test multiple areas at a time, but the RF problems test one topic at a time and are more in the format/wording of the exam imo. If you're looking to test your comprehension, you can't go wrong with either. However if you want more exam style prep questions, it's worth skimming through the problem pack on topics you don't feel as comfortable on.
For Brehm, I think CF is better than RF if you repeatedly do the essay questions, but I made flash cards for all of those so it takes less time cause writing out 40 shapland/brehm essay questions would just kill my hand. However, the RF questions on Brehm help to synthesize what you know in theory and apply it to specific scenarios he gives in many cases.
Tags brehm is my daddy, cfrf needs a forum, cs was here, firsttag, hi cs, mack 69 | 1,514 | 6,123 | {"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} | 3.125 | 3 | CC-MAIN-2019-18 | latest | en | 0.88269 |
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• PEMDAS (parenthesis, exponents, multiplication, division, addition, subtraction) work left to right Exponential Notation Ex: 6^4, 2^3, 9^10, etc. 00:14 01:10 Expanded Form Ex: 7x7x7, 6x6, 2x2x2x2x2, etc. Standard Notation Ex: 5^2= 25; 25 is the standard notation (the answer or solution) +x+ =+ +x- =- -x- =+ Adding or Subtracting fractions You need a common denominator Multiplying fractions no special rule, just multiply Ex: 2/5 x 3/9= 6/45...
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• PEMDAS - ANSWER (parenthesis, exponents, multiplication, division, addition, subtraction) work left to right Exponential Notation - ANSWER Ex: 6^4, 2^3, 9^10, etc. Expanded Form - ANSWER Ex: 7x7x7, 6x6, 2x2x2x2x2, etc. Standard Notation - ANSWER Ex: 5^2= 25; 25 is the standard notation (the answer or solution) +x+ - ANSWER =+ +x- - ANSWER =- -x- - ANSWER =+ Adding or Subtracting fractions - ANSWER You need a common denominator Multiplying fractions - ANSWER no special rul...
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• When we think of volcanoes, eruptions, lava, and smoke- filled air come to mind—all occurring on land. Most people are surprised to learn about the prevalence of underwater volcanoes on our planet. Because the lava and smoke spilling out of an active, underwater volcano is contained by the ocean, people generally do not take note of these eruptions. However, the largest underwater volcanoes are capable of creating huge tidal waves, threatening coastal communities. The main idea of the pass...
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• If 3t - 7 = 5t, then 6t= -21 The variables x and y are directly proportional, and y = 2 when x = 3. What is the value of y when x = 9? 6 00:05 01:10 Here are the following graph equations: (1) y = x - 3 (2) y = x + 3 (3) y = 2/3x (4) y = 3/2x Which of the following equations has the coordinates (9,6)? 3/2x There are 3x-2 trees planted in each row of a rectangular parcel of land. If there are total of 24x-16 trees planted in the parcel , how many rows of trees are...
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• TSI Math Test Prep. | 1,174 | 3,618 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2023-06 | latest | en | 0.658515 |
https://www.educative.io/courses/algorithmic-problem-solving-preparing-for-a-coding-interview/five-common-rules-for-analyzing-the-runtime | 1,717,009,967,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059384.83/warc/CC-MAIN-20240529165728-20240529195728-00453.warc.gz | 644,192,532 | 112,783 | Five Common Rules for Analyzing the Runtime
Review the five common rules of runtime analysis of the algorithms.
We'll cover the following
Rule #1
This rule states that multiplicative constants can be omitted:
$c\cdot f\preceq f$.
Examples: $5n^2\preceq n^2$, $\frac{n^2}{3}\preceq n^2$, $7n\preceq n$.
Rule #2
This rule states that out of two polynomials, the one with a larger degree grows faster:
$n^a\prec n^b$ for $0\leq a
Examples: $n\prec n^2$, $\sqrt{n} \prec n^\frac{2}{3}$, $n^2 \prec n^3$, $n^0\prec \sqrt{n}$.
Rule #3
This rule states that any polynomial grows slower than any exponential:
$n^a \prec b^n$ for $a \geq 0$, $b>1$
Examples: $n^3 \prec 2^n$, $n^10 \prec 1.1^n$.
Rule #4
This rule states that any polylogarithm—that is, a function of the form $(\log n)^a$—grows slower than any polynomial:
$(\log n)^a \prec n^b$ for $a, b > 0$
Examples: $(\log n)^3 \prec \sqrt{n}$, $n\log n \prec n^2$
Rule #5
This rule states that smaller terms can be omitted:
$f \preceq g$, then $f+g \preceq g$.
Examples: $n+n^2 \preceq n^2$, $n^9+2^n \preceq 2^n$. | 389 | 1,082 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 24, "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} | 3.359375 | 3 | CC-MAIN-2024-22 | latest | en | 0.746593 |
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# how to find a polar form of v in v=4-4i?
0
382
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how to find a polar form of v in v=4-4i?
Guest Jun 2, 2015
#1
+85805
+10
The polar form can be written as.... z = r (cos Θ + i*sinΘ )
r = √[(4)^2 + (-4)^2] = √[16 + 16] = 4√2
And Θ = tan-1(-4/4) = tan-1(-1) = -pi/4 ...so we have
z = 4√2 ( cos (-pi/4) + i*sin (-pi/4) )
..... note that this is the principal value......any angle of the form -pi/4 + 2pi (n) is also possible, where n is an integer
CPhill Jun 2, 2015
Sort:
#1
+85805
+10
The polar form can be written as.... z = r (cos Θ + i*sinΘ )
r = √[(4)^2 + (-4)^2] = √[16 + 16] = 4√2
And Θ = tan-1(-4/4) = tan-1(-1) = -pi/4 ...so we have
z = 4√2 ( cos (-pi/4) + i*sin (-pi/4) )
..... note that this is the principal value......any angle of the form -pi/4 + 2pi (n) is also possible, where n is an integer
CPhill Jun 2, 2015
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We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners. See details | 434 | 1,133 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2018-17 | longest | en | 0.6812 |
https://usethinkscript.com/threads/lower-study-plotting-net-change-of-multiple-stocks.1823/ | 1,716,756,390,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058973.42/warc/CC-MAIN-20240526200821-20240526230821-00372.warc.gz | 522,861,254 | 17,914 | # Lower Study plotting Net Change of Multiple Stocks
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I have searched everywhere for a solution to what I am looking for. The closest I found was this post here referencing how to add a chart as a lower study:
What I would like is to plot multiple companies, say FB, NKE, WRK, COST ect in percentage form to get a sentiment of the market.
That is to say something along the lines of:
Plot FBclose = FB/FB[1], NKEclose = NKE/NKE[1], ect.....
I hope my explanation is clear. The idea is to get a general feel in chart form of the overall market sentiment using something other than say a vix indicator
I figured it out my apologies.
For those interested something like this seems to work well:
declare lower;
plot data = high("FB")/high("FB")[1];
Thanks for sharing the idea! I just put together something that works similarly to this idea with some visual improvements.
Code:
``````declare lower;
input SYMB1 = “AAPL”;
input SYMB2 = “AMZN”;
input SYMB3 = "TSLA";
input SYMB4 = "MSFT";
input SYMB5 = "GOOGL";
def data = high(SYMB1)/high(SYMB1)[1];
def data1 = high(SYMB2)/high(SYMB2)[1];
def data2 = high(SYMB3)/high(SYMB3)[1];
def data3 = high(SYMB4)/high(SYMB4)[1];
def data4 = high(SYMB5)/high(SYMB5)[1];
plot one = 1;
one.SetDefaultColor(color.gray);
one.HideBubble();
def sent = data + data1 + data2 + data3 + data4;
def sent1 = data and data1 and data2 and data3 and data4;
plot avg = sent/5;
avg.AssignValueColor(if avg > 1 then color.light_green else if avg < 1 then color.light_red else color.gray);
AddLabel(yes, if sent1 > 1 then "Up" else if sent1 < 1 then "Dn" else"", if sent1 > 1 then color.green else if sent1 < 1 then color.red else color.gray);``````
Last edited:
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